Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why is the volume of a sphere $\frac{4}{3}\pi r^3$? I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?
| I am no where near as proficient in math as any of the people who answered this before me, but nonetheless I would like to add a simplified version;
A cylinder's volume is:
$$\pi r^2h$$
A cone's volume is $\frac{1}{3}$ that of a cylinder of equal height and radius:
$$\frac{1}{3}\pi r^2h$$
A sphere's volume is two cone... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "108",
"answer_count": 19,
"answer_id": 10
} |
Proof that the sum of two Gaussian variables is another Gaussian The sum of two Gaussian variables is another Gaussian.
It seems natural, but I could not find a proof using Google.
What's a short way to prove this?
Thanks!
Edit: Provided the two variables are independent.
| I posted the following in response to a question that got closed as a duplicate of this one:
It looks from your comment as if the meaning of your question is different from what I thought at first. My first answer assumed you knew that the sum of independent normals is itself normal.
You have
$$
\exp\left(-\frac12 \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
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Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Can someone give a simple explanation as to why the harmonic series
$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$
doesn't converge, on the other hand it grows very slowly?
I'd prefer an easily comprehensible explanation rather tha... | Let's group the terms as follows:$$A=\frac11+\frac12+\frac13+\frac14+\cdots\\ $$
$$
A=\underbrace{(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9})}_{\color{red} {9- terms}}
+\underbrace{(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{99})}_{\color{red} {90- terms}}\\+\underbrace{(\frac{1}{101}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "159",
"answer_count": 25,
"answer_id": 9
} |
Simple numerical methods for calculating the digits of $\pi$ Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few ... | The first method that I applied successfully with function calculator was approximation of circle by $2^k$-polygon with approximating sides with one point on the circle and corners outside the circle. I started with unit circle that was approximated by square and the equation $\tan(2^{-k} \pi/4) \approx 2^{-k} \pi/4$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 0
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,... | it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).
If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, that series (which is not the or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 12,
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisi... | Given the $n$th case, you want to consider the $(n+1)$th case, which involves the number $(n+1)^3 + 2(n+1)$. If you know that $n^3+2n$ is divisible by $3$, you can prove $(n+1)^3 + 2(n+1)$ is divisible by $3$ if you can show the difference between the two is divisible by $3$. So find the difference, and then simplify... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 13,
"answer_id": 7
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Perfect numbers, the pattern continues The well known formula for perfect numbers is
$$
P_n=2^{n-1}(2^{n}-1).
$$
This formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$:
$$
496=1+2+4+8+16+31+62+124+248
$$
one can see that the first pattern is a sequence o... | You've observed that $P_5 = 2^4(2^5-1) = 496$ can also be written as the sum of the first 9 powers of two minus the sum of the first four powers of two.
Sums of powers of two
Powers of two written in binary look like $1, 10, 100, 1000, \cdots$ but you can also write them like this $1, 1+1, 11+1, 111+1, \cdots$. This ex... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | There is a simple way of proving that $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ using the following well-known series identity: $$\left(\sin^{-1}(x)\right)^{2} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2 \binom{2n}{n}}.$$ From the above equality, we have that $$x^2 = \frac{1}{2}\sum_{n=1}^{\infty}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 39
} |
Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I have some trouble to do that and I'd glad with any help I may get.
| The algebraic $\frac{1}{12}(17 + \sqrt{97})$ is not a root of the equation
\begin{eqnarray}
-2 x^3 + 10 x^2 - 17 x + 8 = (2 x^2) (5 x - x^3)^{1/3}
\end{eqnarray}
Plugging it in, you find that the left hand side is real and equal to
\begin{eqnarray}
\tfrac{1}{216}(-149 - 37 \sqrt{97}) = -2.37689 \dots
\end{eqnarray}
Th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Probability of cumulative dice rolls hitting a number Is there a general formula to determine the probability of unbounded, cumulative dice rolls hitting a specified number?
For Example, with a D6 and 14:
5 + 2 + 3 + 4 = 14 : success
1 + 1 + 1 + 6 + 5 + 4 = 17 : failure
| Assuming the order matters (i,e 1+2 is a different outcome from 2+1)
The probability of getting the sum $n$ with dice numbered $1,2,\dots,6$ is the coefficient of $x^n$ in
$$\sum_{j=0}^{\infty}(\frac{x+x^2+x^3+x^4+x^5+x^6}{6})^j = \frac{6}{6-x-x^2-x^3-x^4-x^5-x^6}$$
Writing it as partial fractions (using roots of $6-x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/9210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Can't Solve an Integral According to the solution manual:
$\int \frac{x}{\sqrt{1-x^{4}}}dx = \frac{1}{2}\arcsin x^{2}+C$
My solution doesn't seem to be working. I know another way of solving it (setting $u=x^{2}$) but the fact that this way of solving it doesn't work bothers me.
$$\text{set }u=1-x^{4}\text{ so } dx=\fr... | Your solution is an antiderivative of the original function. You can always check whether your solution is correct by taking its derivative. This also implies that the book solution and your solution differ by a constant.
For this specific problem, imagine the right triangle with sides $x^2$ and $\sqrt{1-x^4}$ and hypo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?
| $$\sin^2A+\sin^2B+\sin^2C$$
$$=1-(\cos^2A-\sin^2B)+1-\cos^2C$$
$$=2-\cos(A+B)\cos(A-B)-\cos C\cdot\cos C$$
$$=2-\cos(\pi-C)\cos(A-B)-\cos\{\pi-(A+B)\}\cdot\cos C$$
$$=2+\cos C\cos(A-B)+\cos(A+B)\cdot\cos C\text{ as }\cos(\pi-x)=-\cos x$$
$$=2+\cos C\{\cos(A-B)+\cos(A+B)\}$$
$$=2+2\cos A\cos B\cos C$$
$(1)$ If $2+2\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Rational Numbers and Uniqueness Let $x$ be a positive rational number of the form $\displaystyle x = \sum\limits_{k=1}^{n} \frac{a_k}{k!}$ where each $a_k$ is a nonnegative integer with $a_k \leq k-1$ for $k \geq 2$ and $a_n >0$. Prove that $a_1 = [x]$, $a_k = [k!x]-k[(k-1)!x]$ for $k = 2, \dots, n$, and that $n$ is t... | Since $a_k \le k-1$ for $k \ge 2$ we have
$$ \lfloor x \rfloor = \left\lfloor \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor \le a_1
+ \left\lfloor \sum_{k=2}^n \frac{k-1}{k!} \right\rfloor = a_1 $$
as the latter term is $0$ since $ \sum_{k=2}^n \frac{k-1}{k!} = \sum_{k=2}^n \lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \rbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem:
$$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$
| Every question of the form: prove by induction that
$$\sum_{k=1}^n f(k)=g(n)$$
can be done by verifying two facts about the functions
$f$ and $g$:
*
*$f(1)=g(1)$
and
*$g(n+1)-g(n)=f(n+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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If $4^x + 4^{-x} = 34$, then $2^x + 2^{-x}$ is equal to...? I am having trouble with this:
If $4^x + 4^{-x} = 34$, then what is $2^x + 2^{-x}$ equal to?
I managed to find $4^x$ and it is:
$$4^x = 17 \pm 12\sqrt{2}$$
so that means that $2^x$ is:
$$2^x = \pm \sqrt{17 \pm 12\sqrt{2}}.$$
Correct answer is 6 and I am not... | You haven't done anything wrong! To complete your answer, one way you can see the answer is $6$ is to guess that
$$17 + 12 \sqrt{2} = (a + b\sqrt{2})^2$$
Giving us
$$17 = a^2 + 2b^2, \ \ ab = 6$$
Giving us
$$a = 3, \ \ b = 2$$
Thus $$ \sqrt{17 + 12 \sqrt{2}} = 3 + 2\sqrt{2}$$
which gives $$2^x + 2^{-x} = 6$$
(And sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/17291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrate Form $du / (a^2 + u^2)^{3/2}$ How does one integrate
$$\int \dfrac{du}{(a^2 + u^2)^{3/2}}\ ?$$
The table of integrals here: http://teachers.sduhsd.k12.ca.us/abrown/classes/CalculusC/IntegralTablesStewart.pdf
Gives it as: $$\frac{u}{a^2 ( a^2 + u^2)^{1/2}}\ .$$
I'm getting back into calculus and very rusty. ... | A trigonometric substitution does indeed work.
We want to express $(a^2 + u^2)^{3/2}$ as something without square roots. We want to use some form of the Pythagorean trigonometric identity $\sin^2 x + \cos^2 x = 1$. Multiplying each side by $\frac{a^2}{\cos^2 x}$, we get $a^2 \tan^2 x + a^2 = a^2 \sec^2 x$, which is in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/17666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
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How does teacher get first step? Below are the steps the teacher took to solve:
$y = \sqrt{3}\sin x + \cos x$ find min and max on $[0, 2\pi)$
Step 1: = $2\sin(x + \pi/6))$
How does the teacher get this first step?
Note: No calculus please.
| picakhu's answer is the simplest way to see how it works having already arrived at $y=2\sin(x+\frac{\pi}{6})$ (use the identity there to expand this form). In general, given $a\sin x+b\cos x$ (let's say for $a,b>0$), it is possible to arrive at a similar equivalent form:
$$\begin{align}
a\sin x+b\cos x
&=a\left(\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/17716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$n$th derivative of $e^{1/x}$ I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula
$$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$$
I ... | We may obtain a recursive formula, as follows:
\begin{align}
f\left( t \right) &= e^{1/t} \\
f'\left( t \right) &= - \frac{1}{{t^2 }}f\left( t \right) \\
f''\left( t \right) &= - \frac{1}{{t^2 }}f'\left( t \right) + f\left( t \right)\frac{2}{{t^3 }} \\
&= - \frac{1}{{t^2 }}\left\{ {\frac{1}{{t^2 }}f\left( t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/18284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 6,
"answer_id": 0
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decompose some polynomials [ In first, I say "I'm sorry!", because I am not a Englishman and I don't know your language terms very well. ]
OK, I have some polynomials (like $a^2 +2ab +b^2$ ). And I can't decompress these (for example $a^2 +2ab +b^2 = (a+b)^2$).
Can you help me? (if you can, please write the name or for... | For
1) $(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2$
think about rearranging
$$(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2=a^2x^2+2adx+d^2-(b^2x^2+2bcx+c^2)$$
The same idea can be applied to all your questions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the sum of the following series How can I find the sum of the following series?
$$
\sum_{n=0}^{+\infty}\frac{n^2}{2^n}
$$
I know that it converges, and Wolfram Alpha tells me that its sum is 6 .
Which technique should I use to prove that the sum is 6?
| It is equal to $f(x)=\sum_{n \geq 0} n^2 x^n$ evaluated at $x=1/2$.
To compute this function of $x$, write $n^2 = (n+1)(n+2)-3(n+1)+1$, so that $f(x)=a(x)+b(x)+c(x)$ with:
$a(x)= \sum_{n \geq 0} (n+1)(n+2) x^n = \frac{d^2}{dx^2} \left( \sum_{n \geq 0} x^n\right) = \frac{2}{(1-x)^3}$
$b(x)=\sum_{n \geq 0} 3 (n+1) x^n = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Showing $\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=e^{ax}\cos(bx).$ I've got:
$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}.$$
Could someone show me how it simplifies to:
$e^{ax} \cos(bx)$?
It looks like the denominator is canceled by the terms that are being added, but then how do I get rid of one of the cos... | You use the distributive law, which says that $(X+Y)\cdot Z=(X\cdot Z)+(Y\cdot Z)$ for any $X$, $Y$, and $Z$. In your case, we have $X=b^2$, $Y=a^2$, and $Z=\cos(bx)$, and so
$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=\frac{e^{ax}((b^2+a^2)\cos(bx))}{(a^2+b^2)}=\frac{e^{ax}((a^2+b^2)\cos(bx))}{(a^2+b^2)}=e^{ax}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is the simplification of $\frac{\sin^2 x}{(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)}$? What is the simplification of $$\frac{\sin^2 x}{(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)} \space \text{?}$$
| What does $1 + \sin^2 x + \sin^4 x + \sin^6 x + ....$ simplify to?
Or better, what does $1 + x^2 + x^4 + x^6 + ....$ simplify to?
Or better, what does $1 + x + x^2 + x^3 + ....$ simplify to?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/21182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ Using congruences, show that the following is always an integer for every integer
value of $n$:
$$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$
| Lets show that $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$. To do this, we will show that it is divisible by $3$ and $5$ for every $n$.
Recall that for a prime $p$, $x^p\equiv x \pmod{p}$. (Fermat's Little Theorem) Then, looking modulo 5 we see that
$$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$
Now looking... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Chain rule for multi-variable functions So I have been studying the multi-variable chain rule. Most importantly, and this is what I must have overlooked, is it's not always clear to me how to see which variables are functions of other variables, so that you know when to use the chain rule. For example, if you have:
$... | If we have an explicit function $z = f(x_1,x_2,\ldots,x_n)$, then
$$\displaystyle \frac{dz}{dt} = \frac{\partial z}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial z}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial z}{\partial x_n} \frac{dx_n}{dt}$$
If we have an implicit function $f(z,x_1,x_2,\ldots,x_n) = 0$, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*}
\lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}
&=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\... | Others have already pointed out a sign error. One way to avoid such is to first simplify the problem by changing variables. Let $\rm\ z = \sqrt{4+x}\ $ so $\rm\ x = z^2 - 4\:.\:$ Then
$$\rm \frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}\ =\ \frac{\frac{1}z - \frac{1}2}{z^2-4}\ =\ \frac{-(z-2)}{2\:z\:(z^2-4)}\ =\ \frac{-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22704",
"timestamp": "2023-03-29T00:00:00",
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Question regarding Hensel's Lemma Hensel's Lemma
Suppose that f(x) is a polynomial with integer coefficients, $k$ is an integer with $k \geq 2$, and $p$ a prime.
Suppose further that $r$ is a solution of the congruence $f(x) \equiv 0 \pmod{p^{k-1}}$. Then,
If $f'(r) \not\equiv 0 \pmod{p}$, then there is a unique inte... | You go off track at the word "Hence". If $f'(3)\equiv 0\pmod 5$ and $f(3)\equiv 0 \pmod{25}$ (I assume you've done this correctly; I didn't check), that means that $x\equiv 3,8,13,18,23 \pmod{25}$ are all solutions modulo 25. You only verified that there are no solutions modulo 125 which are 3 modulo 25. There may stil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/26685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus:
$|z-1| = 2|z+1|$
that is, where the modulus of $z -1$ is twice the modulus of $z+1$
I've solved this problem algebraically (by letting $z=x+iy$) as follows:
$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2... | Just to add on to Aryabhata's comment above. The map $f(z) = \frac{1}{z}$ for $ z \in \mathbb{C} -\{0\}$, $f(0) = \infty$ and $f(\infty) = 0$ is a circle preserving homeomorphism of $\bar{\mathbb{C}}$. To see this, one needs to prove that it is continuous on $\bar{\mathbb{C}}$, and since $f(z)$ is an involution proving... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/27199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
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Understanding a proof by descent [Fibonacci's Lost Theorem] I am trying to understand the proof in Carmichaels book Diophantine Analysis but I have got stuck at one point in the proof where $w_1$ and $w_2$ are introduced.
The theorem it is proving is that the system of diophantine equations:
*
*$$x^2 + y^2 = z^2$$
... | $u^2$ and $v^2$ are $m$ and $n$, respectively, which are coprime. Then since $(u^2+v^2)+(u^2-v^2)=2u^2$ and $(u^2+v^2)-(u^2-v^2)=2v^2$, the only factor that $u^2+v^2$ and $u^2-v^2$ can have in common is a single factor of $2$. Since their product is the square $w^2$, that leaves the two possibilities given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/27309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now
| You don't have to square the equation in the first place.
Let $y = \sqrt{(5+2\sqrt{6})^x}$, then $\frac{1}{y} = \sqrt{(5-2\sqrt{6})^x}$. Hence you have $y + \frac{1}{y} = 10$ i.e. $y^2 + 1 = 10y$ i.e. $y^2-10y+1 = 0$.
Hence, $(y-5)^2 =24 \Rightarrow y = 5 \pm 2 \sqrt{6}$.
Hence, $$\sqrt{(5+2\sqrt{6})^x} = 5 \pm 2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/28157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
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Tips for understanding the unit circle I am having trouble grasping some of the concepts regarding the unit circle. I think I have the basics down but I do not have an intuitive sense of what is going on. Is memorizing the radian measurements and their corresponding points the only way to master this? What are some way... | It is probably useful to memorize a table like this:
\begin{align}
\theta & & \sin\theta & & \cos \theta
\\
0 & & \frac{\sqrt{0}}{2} = 0 & & \frac{\sqrt{4}}{2} = 1
\\
\frac{\pi}{6} & & \frac{\sqrt{1}}{2} = \frac{1}{2} & & \frac{\sqrt{3}}{2}
\\
\frac{\pi}{4} & & \frac{\sqrt{2}}{2} & & \frac{\sqrt{2}}{2}
\\
\frac{\pi}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/31163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Representation of integers How can I prove that every integer $n>=170$ can be written as a sum of five positive squares? (i.e. none of the squares are allowed to be zero).
I know that $169=13^2=12^2+5^2=12^2+4^2+3^2=10^2+8^2+2^2+1^2$, and $n-169=a^2+b^2+c^2+d^2$ for some integers $a$, $b$, $c$, $d$, but do I show it?
T... | Hint: let $n-169 = a^2+b^2+c^2+d^2$; if $a,b,c,d \neq 0$ then ... if $d = 0$ and $a,b,c \neq 0$ then ... if $c = d = 0$ and $a,b \neq 0$ then ... if $b = c = d = 0$ and $a \neq 0$ then ... if $a = b = c = d = 0$ then - wait, that can't happen!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the best way to show that no positive powers of this matrix will be the identity matrix?
Show that no positive power of the matrix $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ equals $I_2$.
I claim that given $A^{n}, a_{11} = 1$ and $a_{12} >0, \forall n \in \mathbb{N}$. This is the case for ... | Your solution seems OK to me. You can also find $A^n$ explicitly:
let $E=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$.
Then $A=I+E$ and $E^2=0$. So $(I+nE)(I+E)=I+(n+1)E$ and so, by induction, $A^n=I+nE\ne I$ for $n\ge1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/32059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solving differential equation Below is my work for a particular problem that is mixing me up, since no matter how many times, I can't get my answer to match the book solution.
Given ${f}''(x)= x^{-\frac{3}{2}}$ where $f'(4)= 2$ and $f(0)= 0$, solve the differential equation.
$$f'(x)= \int x^{-\frac{3}{2}} \Rightarrow \... | Nothing to worry about! There is a minor slip, $-2(4)^{-1/2}=-2/2=-1$. You got $-4$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/33033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An inequality on a convex function An exercise in my textbook asked to prove the following inequality, valid for all $a,b,c,d \in R $
$$\left(\frac{a}{2} + \frac{b}{3} + \frac{c}{12} + \frac{d}{12}\right)^4 \leq \frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}$$
There is a straightforward proof using Con... | By Holder
$$\frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}=\left(\frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}\right)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{12} + \frac{1}{12}\right)^3\geq$$
$$\geq\left(\frac{|a|}{2} + \frac{|b|}{3} + \frac{|c|}{12} + \frac{|d|}{12}\right)^4\geq\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/33225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
The locus of the intersection point of two perpendicular tangents to a given ellipse
For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.
I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof.
| If all you want is a proof that the locus is a circle, we may assume that the ellipse is given by
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
Ignoring vertical tangents for now,
if a line $y=mx+k$ is tangent to the ellipse, then plugging in this value of $y$ into the equation for the ellipse gives
$$\frac{x^2}{a^2} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/33520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Using binomial expansion to derive Pascal's rule $\displaystyle \binom{n}{k}=\binom{n-1}{k} + \binom{n-1}{k-1}$
$\displaystyle \left(1+x\right)^{n} = \left(1+x\right)\left(1+x\right)^{n-1}$
How do I use binomial expansion on the second equations for the right hand side and convert it to the first equation? The left han... | Binomial expansion of both sides of $$\left(1+x\right)^{n} = \left(1+x\right)\left(1+x\right)^{n-1}$$ gives $$\sum_{k=0}^n \binom{n}{k} x^k = \left(1+x\right)\sum_{k=0}^{n-1} \binom{n-1}{k} x^k$$ by distributivity on the right hand side we find $$\left(\sum_{k=0}^{n-1} \binom{n-1}{k} x^k \right)+\left(\sum_{k=0}^{n-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/38900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do get combination sequence formula? What would be a closed-form formula that would determine the ith value of the sequence
1, 3, 11, 43, 171...
where each value is one minus the product of the previous value and 4?
Thanks!
| The sequence can be written using the recurrence formula: $y_n = a+by_{n-1}$ . Then using the first 3 terms, one gets $y_1=1, a=-1, b=4 $
Sometimes it is easy to convert a recurrence formula to a closed-form, by studding the structure of the math relations.
$y_1 = 1$
$y_2 = a+by_1$
$y_3 = a+by_2 = a+b(a+by_1) = a+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/40036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Primitive polynomials of finite fields there are two primitive polynomials which I can use to construct $GF(2^3)=GF(8)$:
$p_1(x) = x^3+x+1$
$p_2(x) = x^3+x^2+1$
$GF(8)$ created with $p_1(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3 = \alpha + 1$
$\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$
$\alp... | The generator $\alpha$ for your field with the first description cannot be equal to the generator $\beta$ for your field with the second description. An isomorphism between $\mathbb{F}_2(\alpha)$ and $\mathbb{F}_2(\beta)$ is given by taking $\alpha \mapsto \beta + 1$; you can check that $\beta + 1$ satisfies $p_1$ iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/40326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers Given a polynomial such as $X^4 + 4X^3 + 6X^2 + 4X + 1,$ where the coefficients are symmetrical, I know there's a trick to quickly find the zeros. Could someone please refresh my memory?
| Hint: This particular polynomial is very nice, and factors as $(X+1)^4$.
Take a look at Pascal's Triangle and the Binomial Theorem for more details.
Added: Overly complicated formula
The particular quartic you asked about had a nice solution, but lets find all the roots of the more general $$ax^{4}+bx^{3}+cx^{2}+bx... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially
$$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$
For the first one, I was able to express $k^2$ in terms of the binomial coefficients by considering a set $X$ of cardinality $2k$ and par... | For the first one, $\displaystyle \sum_{k=1}^{n} k^2$, you can probably try this way.
$$k^2 = \binom{k}{1} + 2 \binom{k}{2}$$
This can be proved using combinatorial argument by looking at drawing $2$ balls from $k$ balls with replacement.
The total number of ways to do this is $k^2$.
The other way to count it is as fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/43317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ How do I show that:
$$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$
This is actually problem B $4371$ given at this link. Looks like a... | Use $\sin(x) = \cos(\frac{\pi}2 - x)$, we can rewrite this as:
$$\frac{1}{\cos^2 \frac{3\pi}{7}} + \frac{1}{\cos^2 \frac{2\pi}{7}} + \frac{1}{\cos^2 \frac{\pi}{7}}$$
Let $a_k = \frac{1}{\cos \frac{k\pi}7}$.
Let $f(x) = (x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)$.
Now, using that $a_k = - a_{7-k}$, this can be written a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 0
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain ... | Notice that $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ and hence
$$(n+1)^3 = \sum_{k=0}^n \left[ (k+1)^3 - k^3\right] = 3\sum_{k=0}^n k^2 + 3\sum_{k=0}^n k + \sum_{k=0}^n 1$$
which gives you
$$\begin{align}
\sum_{k=1}^n k^2
& = \frac{1}{3}(n+1)^3 - \frac{1}{2}n(n+1) - \frac{1}{3}(n+1) \\
& = \frac{1}{6}(n+1) \left[ 2(n+1)^2 - 3n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "145",
"answer_count": 32,
"answer_id": 6
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Problem finding zeros of complex polynomial I'm trying to solve this problem
$$ z^2 + (\sqrt{3} + i)|z| \bar{z}^2 = 0 $$
So, I know $ |z^2| = |z|^2 = a^2 + b ^2 $ and $ \operatorname{Arg}(z^2) = 2 \operatorname{Arg} (z) - 2k \pi = 2 \arctan (\frac{b}{a} ) - 2 k\pi $ for a $ k \in \mathbb{Z} $. Regarding the other term... | Here is an alternative to solving it using polar form.
Let $z=a+bi$, so that $\bar{z}=a-bi$ and $|z|=\sqrt{a^2+b^2}$. Then you want to solve
$$(a+bi)^2+(\sqrt{3}+i)\sqrt{a^2+b^2}(a-bi)^2=0,$$
which expands to
$$(a^2-b^2)+2abi+(\sqrt{3}+i)\sqrt{a^2+b^2}\left((a^2-b^2)-2abi\right)=0$$
Thus, we need both the real part and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and
$6 \mid ab(a^2+b^2)(a^2-b^2)$?
Indeed this method implies studying nu... | You need to show $ab(a^2 - b^2)(a^2 + b^2)$ is a multiple of 2,3, and 5 for all $a$ and $b$.
For 2: If neither $a$ nor $b$ are even, they are both odd and $a^2 \equiv b^2 \equiv 1 \pmod 2$, so that 2 divides $a^2 - b^2$.
For 3: If neither $a$ nor $b$ are a multiple of 3, then $a^2 \equiv b^2 \equiv 1 \pmod 3$, so 3 di... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the perimeter of any triangle given the three altitude lengths The altitude lengths are 12, 15 and 20. I would like a process rather than just a single solution.
| First,
$\begin{aligned}\operatorname{Area}(\triangle ABC)=\frac{ah_a}2=\frac{bh_b}2=\frac{ch_c}2\implies \frac1{h_a}&=\frac{a}{2\operatorname{Area}(\triangle ABC)}\\\frac1{h_b}&=\frac{b}{2\operatorname{Area}(\triangle ABC)}\\\frac1{h_c}&=\frac{c}{2\operatorname{Area}(\triangle ABC)} \end{aligned}$
By the already mentio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/55440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Determining which values to use in place of x in functions When solving partial fractions for integrations, solving x for two terms usually isn't all that difficult, but I've been running into problems with three term integration.
For example, given
$$\int\frac{x^2+3x-4}{x^3-4x^2+4x}$$
The denominator factored out to ... | To find the constants in the rational fraction $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$
you may use any set of 3 values of $x$, provided that the denominator $x^3-4x^2+4x\ne 0$.
The "standard" method is to compare the coefficients of $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Nested Square Roots $5^0+\sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt\dots}}}$ How would one go about computing the value of $X$, where
$X=5^0+ \sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+\sqrt{5^{16}+\sqrt{5^{32}+\dots}}}}}}$
I have tried the standard way of squaring then trying some trick, but nothing is working. I have also look... | The trick is to pull out a $\sqrt{5}$ factor from the second term:
$$
\frac{\sqrt{5^1+ \sqrt{5^2 + \sqrt{5^4 + \sqrt{5^8 + \cdots}}}}}{\sqrt{5}}
= \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}},
$$
which I call $Y$ for convenience. To see why this is true, observe that
$$ \begin{align*} \frac{\sqrt{5^1+x}}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/61012",
"timestamp": "2023-03-29T00:00:00",
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Canonical to Parametric, Ellipse Equation I've done some algebra tricks in this derivation and I'm not sure if it's okay to do those things.
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2\theta + \sin^2\theta$$
Can I really do this next step?
$$\frac{x^2}{a^2} = \cos^2\theta\qua... | The idea behind your argument is absolutely fine. Any two non-negative numbers $u$ and $v$ such that $u+v=1$ can be expressed as $u=\cos^2\theta$, $v=\sin^2\theta$ for some $\theta$. This is so obvious that it probably does not require proof. Set $u=\cos^2\theta$. Then $v=1-\cos^2\theta=\sin^2\theta$.
The second di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/61071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Does the Schur complement preserve the partial order? Let
$$\begin{bmatrix}
A_{1} &B_1 \\ B_1' &C_1
\end{bmatrix} \quad \text{and} \quad \begin{bmatrix}
A_2 &B_2 \\ B_2' &C_2
\end{bmatrix}$$
be symmetric positive definite and conformably partitioned matrices. If
$$\begin{bmatrix}
A_{1} &B_1 \\ B_1' &C_1
\end{bma... | For a general block matrix $X=\begin{pmatrix}A&B\\C&D\end{pmatrix}$, the Schur complement $S$ to the block $D$ satisfies
$$
\begin{pmatrix}A&B\\C&D\end{pmatrix}
=\begin{pmatrix}I&BD^{-1}\\&I\end{pmatrix}
\begin{pmatrix}S\\&D\end{pmatrix}
\begin{pmatrix}I\\D^{-1}C&I\end{pmatrix}.
$$
So, when $X$ is Hermitian,
$$
\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| Let the induction hypothesis be
$$ (1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$
Now consider:
$$ (1+2+3+\cdots+n + (n+1))^2 $$
$$\begin{align}
& = \color{red}{(1+2+3+\cdots+n)^2} + (n+1)^2 + 2(n+1)\color{blue}{(1+2+3+\cdots+n)}\\
& = \color{red}{(1^3+2^3+3^3+\cdots+n^3)} + (n+1)^2 + 2(n+1)\color{blue}{(n(n+1)/2)}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
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Solve to find the unknown I have been doing questions from the past year and I come across this question which stumped me:
The constant term in the expansion of $\left(\frac1{x^2}+ax\right)^6$
is $1215$; find the value of $a$. (The given answer is: $\pm 3$ )
Should be an easy one, but I don't know how to begin. Som... | Sofia,Sorry for the delayed response.I was busy with other posts.
you have two choices.One is to use pascals triangle and the other one is to expand using the binimial theorem.
You can compare the expression $$\left ( \frac{1}{x^2} + ax \right )^6$$ with $$(a+x)^6$$ where a = 1/x^2 and x = ax,n=6.Here'e the pascal tria... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On factorizing and solving the polynomial: $x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$ The actual problem is to find the product of all the real roots of this equation,I am stuck with his factorization:
$$x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x... | In regard to the first part of your question ("wild guessing"), the point was to note that the polynomial can be expressed as the sum of three polynomials, grouping same coefficients:
$$ P(x)= x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2
= A(x)+B(x)+C(x)$$
with
$$\begin{eqnarra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving for Inequality $\frac{12}{2x-3}<1+2x$ I am trying to solve for the following inequality:
$$\frac{12}{2x-3}<1+2x$$
In the given answer,
$$\frac{12}{2x-3}-(1+2x)<0$$
$$\frac{-(2x+3)(2x-5)}{2x-3}<0 \rightarrow \textrm{ How do I get to this step?}$$
$$\frac{(2x+3)(2x-5)}{2x-3}>0$$
$$(2x+3)(2x-5)(2x-3)>0 \textrm{ v... | $$ \frac{12}{2x-3} - (1-2x) = \frac{12 - (1+2x)(2x-3) }{2x-3} = \frac{ 12 - (2x-3+4x^2-6x)}{2x-3} $$
$$= - \frac{4x^2-4x-15}{2x-3} = - \frac{(2x+3)(2x-5)}{2x-3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/63250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can the following be calculated? How can the following series be calculated?
$$S=1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots+(1+2+3+4+\cdots+2011)$$
| Let $S$ be our sum. Then
$$S=\binom{2}{2}+\binom{3}{2}+\binom{4}{2} + \cdots + \binom{2012}{2}=\binom{2013}{3}=\frac{2013\cdot 2012\cdot 2011}{3 \cdot 2 \cdot 1}.$$
Justification: We count, in two different ways, the number of ways to choose $3$ numbers from the set
$$\{1,2,3,4,\dots, n,n+1\}.$$
(For our particular ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/65465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Combinatorics-number of permutations of $m$ A's and at most $n$ B's Prove that the number of permutations of $m$ A's and at most $n$ B's equals $\dbinom{m+n+1}{m+1}$.
I'm not sure how to even start this problem.
| By summing all possibilities of $n$, we get that the number of permutations $P_n$ satisfies
$$P_n = \binom{m+n}{n} + \binom{m+(n-1)}{(n-1)} + \ldots + \binom{m+0}{0} = \sum_{i=0}^n \binom{m + i}{i}$$
Note that
$$\binom{a}{b} = \binom{a-1}{b} + \binom{a-1}{b-1}$$
Repeatedly applying this to the last term, we get
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/65947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$ Given the geometric series:
$1 + x^2 + x^4 + x^6 + x^8 + \cdots$
We can recast it as:
$S = 1 + x^2 \, (1 + x^2 + x^4 + x^6 + x^8 + \cdots)$, where $S = 1 + x^2 + x^4 + x^6 + x^8 + \cdots$.
This recasting is possible only because there is an infin... | The $n$th partial sum of your series is
$$
\begin{align*}
S_n &= 1+x^2+x^4+\cdots +x^{2n}= 1+x^2(1+x^2+x^4+\cdots +x^{2n-2})\\
&= 1+x^2S_{n-1}
\end{align*}
$$
Assuming your series converges you get that
$$
\lim_{n\to\infty}S_n=\lim_{n\to\infty}S_{n-1}=S.
$$
Thus $S=1+x^2S$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lesser-known integration tricks I am currently studying for the GRE math subject test, which heavily tests calculus. I've reviewed most of the basic calculus techniques (integration by parts, trig substitutions, etc.) I am now looking for a list or reference for some lesser-known tricks or clever substitutions that are... | When integrating rational functions by partial fractions decomposition, the trickiest type of antiderivative that one might need to compute is
$$I_n = \int \frac{dx}{(1+x^2)^n}.$$
(Integrals involving more general quadratic factors can be reduced to such integrals, plus integrals of the much easier type $\int \frac{x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/70974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "169",
"answer_count": 8,
"answer_id": 6
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Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$
| Suppose $\frac{a}{b}$ and $\frac{c}{d}$ are fractions. That is, $a$, $b$, $c$, $d$ are whole numbers and $b\ne0$, $d\ne0$. In addition we require $c\ne0$.
Let $\frac{a}{b}\div\frac{c}{d}=A$. Then by definition of division of fractions , $A$ is a unique fraction so that $A\times\frac{c}{d}=\frac{a}{b}$.
However, $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/71157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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On the meaning of being algebraically closed The definition of algebraic number is that $\alpha$ is an algebraic number if there is a nonzero polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $p(\alpha)=0$.
By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will... | Let $p(x) = a_0+a_1x+\cdots +a_{n-1}x^{n-1} + x^n$ be a polynomial with coefficients in $\overline{\mathbb{Q}}$. For each $i$, $0\leq i\leq n-1$, let $a_i=b_{i1}, b_{i2},\ldots,b_{im_i}$ be the $m_i$ conjugates of $a_i$ (that is, the "other" roots of the monic irreducible polynomial with coefficients in $\mathbb{Q}$ th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove for which $n \in \mathbb{N}$: $9n^3 - 3 ≤ 8^n$ A homework assignment requires me to find out and prove using induction for which $n ≥ 0$ $9n^3 - 3 ≤ 8^n$ and I have conducted multiple approaches and consulted multiple people and other resources with limited success. I appreciate any hint you can give me.
Thanks i... | Let $f(n)=n^3-3$, and let $g(n)=8^n$. We compute a little, to see what is going on.
We have $f(0) \le g(0)$; $f(1)\le g(1)$; $f(2) > g(2)$; $f(3) \le g(3)$; $f(4) \le g(4)$. Indeed $f(4)=573$ and $g(4)=4096$, so it's not even close.
The exponential function $8^x$ ultimately grows incomparably faster than the polynomia... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series.
| I will try to give an answer using basic complex variables here.
This calculation is very simple in spite of some more complicated intermediate expressions that appear.
Suppose we are trying to show that
$$\sum_{q=0}^\infty
{p+q\choose q} {2p+m\choose m-2q}
= 2^{m-1} \frac{2p+m}{m} {m+p-1\choose p}.$$
Introduce the i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
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How do I prove equality of x and y? If $0\leq x,y\leq\frac{\pi}{2}$ and $\cos x +\cos y -\cos(x+y)=\frac{3}{2}$, then how can I prove that $x=y=\frac{\pi}{3}$?
Your help is appreciated.I tried various formulas but nothing is working.
| You could also attempt an geometric proof. First, without loss of generality you can assume
$0 <x,y < \frac{\pi}{2}$.
Construct a triangle with angles $x,y, \pi-x-y$.
Let $a,b,c$ be the edges. Then by cos law, you know that
$$\frac{a^2+b^2-c^2}{2ab}+ \frac{a^2+c^2-b^2}{2ac}+ \frac{b^2+c^2-a^2}{2bc}=\frac{3}{2}$$
and y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Counting words with parity restrictions on the letters Let $a_n$ be the number of words of length $n$ from the alphabet $\{A,B,C,D,E,F\}$ in which $A$ appears an even number of times and $B$ appears an odd number of times.
Using generating functions I was able to prove that $$a_n=\frac{6^n-2^n}{4}\;.$$
I was wondering ... | I don’t know whether you’d call them combinatorial, but here are two completely elementary arguments of a kind that I’ve presented in a sophomore-level discrete math course. Added: Neither, of course, is as nice as Didier’s, which I’d not seen when I posted this.
Let $b_n$ be the number of words of length $n$ with an o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prime reciprocals sum Let $a_i$ be a sequence of $1$'s and $2$'s and $p_i$ the prime numbers.
And let $r=\displaystyle\sum_{i=1}^\infty p_i^{-a_i}$
Can $r$ be rational, and can r be any rational $> 1/2$ or any real?
ver.2:
Let $k$ be a positive real number and let $a_i$ be $1 +$ (the $i$'th digit in the binary decimal... | The question with primes in the denominator:
The minimum that $r$ could possibly be is $C=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}$. However, a sequence of $1$s and $2$s can be chosen so that $r$ can be any real number not less than $C$. Since $\sum\limits_{i=1}^\infty\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)$ diverges... | {
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How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$? How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?
| Since $\log(x)$ is concave,
$$
\log\left(\frac{ax+by}{a+b}\right)\ge\frac{a\log(x)+b\log(y)}{a+b}\tag{1}
$$
Rearranging $(1)$ and exponentiating yields
$$
\left(\frac{ax+by}{a+b}\right)^{a+b}\ge x^ay^b\tag{2}
$$
Plugging $x=a^3$ and $y=b^3$ into $(2)$ gives
$$
\left(\frac{a^4+b^4}{a+b}\right)^{a+b}\ge a^{3a}b^{3b}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the Laurent expansion of $\frac{1}{\sin^3(z)}$ on $0<|z|<\pi$? How do you find the Laurent expansion of $\frac{1}{\sin^3(z)}$ on $0<|z|<\pi$? I would really appreciate someone carefully explaining this, as I'm very confused by this general concept! Thanks
| use this formula
$$\sum _{k=1}^{\infty } (-1)^{3 k} \left(-\frac{x^3}{\pi ^3 k^3 (\pi k-x)^3}-\frac{x^3}{\pi ^3 k^3 (\pi k+x)^3}+\frac{3 x^2}{\pi ^2 k^2 (\pi k-x)^3}-\frac{3 x^2}{\pi ^2 k^2 (\pi k+x)^3}-\frac{x^3}{2 \pi k (\pi k-x)^3}-\frac{x^3}{2 \pi k (\pi k+x)^3}+\frac{x^2}{(\pi k-x)^3}-\frac{x^2}{(\pi k+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Is there an easy way to determine when this fractional expression is an integer? For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer?
$$z=\frac{(1-x)-(1+x)y}{(1+x)+(1-x)y}$$
The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several fact... | Since $$ \frac{(1-x)-(1+x)y}{(1+x)+(1-x)y} = \frac{ xy+x+y-1}{xy-x-y-1} = 1 + \frac{2(x+y) }{xy-x-y-1} $$
and $ 2x+2y < xy - x -y - 1 $ if $ 3(x+y) < xy - 1 .$ Suppose $ x\leq y$, then $ 3(x+y) \leq 6y \leq xy-1 $ if $ x\geq 7. $ So all solutions must have $0\leq x< 7 $ so it is reduced to solving $7$ simpler Diophant... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Sketch the graph of $y = \frac{4x^2 + 1}{x^2 - 1}$ I need help sketching the graph of $y = \frac{4x^2 + 1}{x^2 - 1}$.
I see that the domain is all real numbers except $1$ and $-1$ as $x^2 - 1 = (x + 1)(x - 1)$. I can also determine that between $-1$ and $1$, the graph lies below the x-axis.
What is the next step? I... | You can simplify right away with
$$
y = \frac{4x^2 + 1}{x^2 - 1} = 4+ \frac{5}{x^2 - 1} =4+ \frac{5}{(x - 1)(x+1)}
$$
Now when $x\to\infty$ or $x\to -\infty$, adding or subtracting 1 doesn't really matter hence that term goes to zero. When $x$ is quite large, say 1000, the second term is very small but positive henc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lambert series expansion identity I have a question which goes like this:
How can I show that $$\sum_{n=1}^{\infty} \frac{z^n}{\left(1-z^n\right)^2} =\sum_{n=1}^\infty \frac{nz^n}{1-z^n}$$ for $|z|<1$?
| Hint: Try using the expansions
$$
\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\dots
$$
and
$$
\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+5x^4+\dots
$$
Expansion:
$$
\begin{align}
\sum_{n=1}^\infty\frac{z^n}{(1-z^n)^2}
&=\sum_{n=1}^\infty\sum_{k=0}^\infty(k+1)z^{kn+n}\\
&=\sum_{n=1}^\infty\sum_{k=1}^\infty kz^{kn}\\
&=\sum_{k=1}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/83680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Convergence of $\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})},$ I'm in trouble with this limit. The numerator diverges positively, but I do not understand how to operate on the denominator.
$$\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})},$$
$$\lim_{n \to \infty} ... | Let's make a few comments.
*
*Note that the terms of the sequence are always defined: for $n\geq 0$, $3(n+2)^2$ is greater than $0$; and $\cos(n\pi/5)$ can never be equal to zero (you would need $n\pi/5$ to be an odd multiple of $\pi/2$, and this is impossible).
*If $a_n$ and $b_n$ both have limits as $n\to\infty$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$ Facing difficulty finding limit
$$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$
For starters I have trouble simplifying it
Which method would help in finding this limit?
| $$
\begin{eqnarray}
\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}=\lim \limits_{x\to \infty}\left(\frac{x-1+1}{x-1}\right)^{2x+1}
=\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{2x+1}\\= \lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}
=\lim \limits_{x\to \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/90324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maxima of bivariate function [1] Is there an easy way to formally prove that,
$$
2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y\ge -x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y$$
$${0<x,y<1}$$
without resorting to checking partial derivatives of the quotient formed by the two sides, and finding local maxima?
[2] Similar... | Your first question:
With a little manipulation you get that it is equivalent to
$$x^2((1-x)^2+1)+y^2((1-y)^2+1) \ge 2xy[(1-x)(1-y)+1].$$
This can be obtained from addition of two inequalities
$$x^2(1-x)^2+y^2(1-y)^2 \ge 2xy(1-x)(1-y)$$
$$x^2+y^2\ge 2xy.$$
Both of them are special cases of $a^2+b^2\ge 2ab$, which follo... | {
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"timestamp": "2023-03-29T00:00:00",
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Adding a different constant to numerator and denominator Suppose that $a$ is less than $b$ , $c$ is less than $d$.
What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?
| One nice thing to notice is that
$$
\frac{a}{b}=\frac{c}{d} \Leftrightarrow \frac{a}{b}=\frac{a+c}{b+d}
$$
no matter the values of $a$, $b$, $c$ and $d$. The $(\Rightarrow)$ is because $c=xa, d=xb$ for some $x$, so $\frac{a+c}{b+d}=\frac{a+xa}{b+xb}=\frac{a(1+x)}{b(1+x)}=\frac{a}{b}$. The other direction is similar.
Th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadratic forms and prime numbers in the sieve of Atkin I'm studying the theorems used in the paper which explains how the sieve of Atkin works, but I cannot understand a point.
For example, in the paper linked above, theorem 6.2 on page 1028 says that if $n$ is prime then the cardinality of the set which contains all ... | The main thing is that the norm of $s + t \omega$ is $s^2 + s t + t^2,$ which is a binary form that represents exactly the same numbers as $3x^2 + y^2.$
It is always true that, for an integer $k,$ the form $s^2 + s t + k t^2$ represents a superset of the numbers represented by $x^2 + (4k-1)y^2.$ For instance, with $k=... | {
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Finding the limit of a sequence $\lim _{n\to \infty} \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right)$ If there were a regular square root I would multiply the top by its adjacent and divide, but I've tried that with this problem and it doesn't work. Not sure what else to do have been stuck on it.
$$ \lim _{... | $$
\begin{align*}
\lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}-
\sqrt [3]{n} \right)
&= \lim _{n\to \infty } \sqrt [3]{n^2} \cdot \sqrt[3]{n} \left( \sqrt [3]{1+ \frac{1}{n}}-
1 \right)
\\ &= \lim _{n\to \infty } n \left( \sqrt [3]{1+ \frac{1}{n}}-
1 \right)
\\ &= \lim _{n\to \infty } \frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/92272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is every Mersenne prime of the form : $x^2+3 \cdot y^2$? How to prove or disprove following statement :
Conjecture :
Every Mersenne prime number can be uniquely written in the form : $x^2+3 \cdot y^2$ ,
where $\gcd(x,y)=1$ and $x,y \geq 0$
Since $M_p$ is an odd number it follows that : $M_p \equiv 1 \pmod 2$
Accordin... | (Outline of proof that, for prime $p\equiv 1\pmod 6$, there is one positive solution to $x^2+3y^2=p$.)
It helps to recall the Gaussian integer proof that, for a prime $p\equiv 1\pmod 4$, $x^2+y^2=p$ has an integer solution. It starts with the fact that there is an $a$ such that $a^2+1$ is divisible by $p$, then uses u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$
Is there any formula that tells this or why is it like that?
| \begin{equation}
\text{You can use De Moivre's identity:}
\end{equation}
\begin{equation}
\text{Let's Call:}\\\\
\end{equation}
\begin{equation}
\mathrm{z=\cos x+i \sin x}\\
\mathrm{\frac{1}{z}=\cos x-i \sin x}\\
\end{equation}
\begin{equation}
\text{Now subtracting both equations together, we get:}\\
\end{equation}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/97654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
The square of an integer is congruent to 0 or 1 mod 4 This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.
This problem is from assignment 6. The notes from this lecture can be found here.
a) P... | $$\begin{align}
x^2 \mod 4 &\equiv (x \mod 4)(x \mod 4) \pmod 4 \\
&\equiv \begin{cases}0^2 \mod 4 \\
1^2 \mod 4 \\
2^2 \mod 4 \\
3^2 \mod 4 \end{cases} \\
&\equiv \begin{cases}0 \mod 4 \\
1 \mod 4 \\
4 \mod 4 \\
9 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
If $n\ge 3$, $4^n \nmid 9^n-1$ Could anyone give me a hint to prove the following?
If $n\ge 3$, $4^n \nmid 9^n-1$
| Hint :
Try to prove using induction :
$1.$ $9^3 \not \equiv 1 \pmod {4^3}$
$2.$ suppose : $9^k \not \equiv 1 \pmod {4^k}$
$3.$ $9^k \not \equiv 1 \pmod {4^k} \Rightarrow 9^{k+1} \not \equiv 9 \pmod {4^k}$
So you have to prove :
$ 9^{k+1} \not \equiv 9 \pmod {4^k} \Rightarrow 9^{k+1} \not \equiv 1 \pmod {4^{k+1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/100393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Simple expressions for $\sum_{k=0}^n\cos(k\theta)$ and $\sum_{k=1}^n\sin(k\theta)$?
Possible Duplicate:
How can we sum up $\sin$ and $\cos$ series when the angles are in A.P?
I'm curious if there is a simple expression for
$$
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta
$$
and
$$
\sin\theta+\sin 2\theta+\cdots+\si... | Take the expression you have and multiply the numerator and denominator by $1-\bar{z}$, and using $z\bar z=1$:
$$\frac{1-z^{n+1}}{1-z} = \frac{1-z^{n+1}-\bar{z}+z^n}{2-(z+\bar z)}$$
But $z+\bar{z}=2\cos \theta$, so the real part of this expression is the real part of the numerator divided by $2-2\cos \theta$. But the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/102477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Conditional Probability Question Bowl A contains 6 red chips and 4 blue chips. Five chips are randomly chosen and transferred without replacement to Bowl B. One chip is drawn at random from Bowl B. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl... | There are $\frac{10!}{6!4!}$ (= 210) possible arrangements for the chips, and $\frac{5!}{2!3!}$ arrangements for the chips desired in bowl B. Any given arrangement of bowl B can occur for every corresponding arrangement in bowl A (also $\frac{5!}{2!3!}$ combinations)
The total number of possiblilities with the correct ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$.
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
| Here is a solution using induction:
Let $f(x)=x^3+6x^2+11x+6$
Since we want to see if it is divisible by 3 let us assume that $f(x)=3m$.
For the case where $x=0$, $f(0)=6$ which is divisible by 3.
Now that we have proved for one case let us prove for the case of $f(x+1)$
$$f(x+1)=(x+1)^3+6(x+1)^2+11(x+1)+6$$
$$= x^3+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 1
} |
roots of complex polynomial - tricks What tricks are there for calculating the roots of complex polynomials like
$$p(t) = (t+1)^6 - (t-1)^6$$
$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$$\left( \frac{t+1}{t-1} \right)^6 = 1$$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which b... | Notice that $t=1$ is not a root. Divide by $(t-1)^6$.
If $\omega$ is a root of $z^6 - 1$, then a root of the original equation is given by $\frac{t+1}{t-1} = \omega$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/105129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Generalization of Pythagorean triples Is it known whether for any natural number $n$, I can find (infinitely many?) nontrivial integer tuples $$(x_0,\ldots,x_n)$$ such that $$x_0^n + \cdots + x_{n-1}^n = x_n^n?$$
Obviously this is true for $n = 2$.
Thanks.
| These Pythagorean triples can appear in the most unexpected place.
If: $a^2+b^2=c^2$
Then alignment: $N_1^3+N_2^3+N_3^3+N_4^3+N_5^3=N_6^3$
$N_1=cp^2-3(a+b)ps+3cs^2$
$N_2=bp^2+3bps-3bs^2$
$N_3=ap^2+3aps-3as^2$
$N_4=-bp^2+3(2c-b)ps+3(3c-3a-2b)s^2$
$N_5=-ap^2+3(2c-a)ps+3(3c-2a-3b)s^2$
$N_6=cp^2+3(2c-a-b)ps+3(4c-3a-3b)s^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = ... | Let $\:\rm d = (n^3+1,\:n^2+2).\:$ Observe that $\rm \ d \in \{1,\:3,\:9\} \iff\ d\:|\:9\iff 9\equiv 0\pmod d\:.$
mod $\rm (n^3\!-a,n^2\!-b)\!:\ a^2 \equiv n^6 \equiv b^3\:$ so $\rm\:a=-1,\:b = -2\:\Rightarrow 1\equiv -8\:\Rightarrow\: 9\equiv 0\:. \ \ $ QED
Or, if you don't know congruence arithmetic, since $\rm\: x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
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How to get from $a\sqrt{1 + \frac{b^2}{a^2}}$ to $\sqrt{a^2 + b^2}$ I have the following expression: $a\sqrt{1 + \frac{b^2}{a^2}}$. If I plug this into Wolfram Alpha, it tells me that, if $a, b$ are positive, this equals $\sqrt{a^2 + b^2}$.
How do I get that result? I can't see how that could be done. Thanks
| $$a\sqrt{1 + \frac{b^2}{a^2}}$$
$$=a\sqrt{\frac{a^2 + b^2}{a^2}}$$
$$=a\frac{\sqrt{a^2 + b^2}}{|a|}$$
So when $a$ and $b$ are positive, $|a|=a$. Hence:
$$=\sqrt{a^2 + b^2}$$
Without the assumption:
$$\sqrt{a^2} =|a|=\begin{cases} a && a \geq 0\\ -a &&a < 0\\ \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right? Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right?
This is from my lecture notes which is used to solve:
But when $x = 0$, $(1 + 2^x + 3^x = 3) \gt (3^0 = 1)$? The thing is how do I choose which what expression should go on the left & right side?
| When $x=0$, the left side $3^0=1$, the center is $3$ as you say, and the right side is $3\cdot 3^0=3 \cdot 1=3$ so the center and right sides are equal. But you want this for large $x$, so could restrict the range to $x \gt 1$, say.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/111661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What are the possible values for $\gcd(a^2, b)$ if $\gcd(a, b) = 3$? I was looking back at my notes on number theory and I came across this question.
Let $a$, $b$ be positive integers such that $\gcd(a, b) = 3$. What are the possible values for $\gcd(a^2, b)$?
I know it has to do with their prime factorization decompos... | If $p$ is a prime, and $p|a^2$, then $p|a$; thus, if $p|a^2$ and $p|b$, then $p|a$ and $p|b$, hence $p|\gcd(a,b) = 3$. So $\gcd(a^2,b)$ must be a power of $3$.
Also, $3|a^2$ and $3|b$, so $3|\gcd(a^2,b)$; so $\gcd(a^2,b)$ is a multiple of $3$.
If $3^{2k}|a^2$, then $3^k|a$ (you can use prime factorization here); so if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Finding a simple expression for this series expansion without a piecewise definition I am doing some practice Calculus questions and I ran into the following problem which ended up having a reduction formula with a neat expansion that I was wondering how to express in terms of a series. Here it is: consider
$$
I_{n} =... | $$
\color{green}{I_n=\sum\limits_{i=2}^{n} (-1)^{n-i}\cdot i\cdot\left(\frac{\pi}{2} \right)^{i-1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving an algebraic identity using the axioms of field I am trying to prove (based on the axioms of field) that
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
So, my first thought was to use the distributive law to show that
$$(a-b)(a^2+ab+b^2)=(a-b)\cdot a^2+(a-b)\cdot ab+(a-b)\cdot b^2$$
And then continuing from this point.
My probl... | Indeed you need distributive, associative and commutative laws to prove your statement.
In fact:
$$\begin{split}
(a-b)(a^2+ab+b^2) &= (a+(-b))a^2 +(a+(-b))ab+(a+(-b))b^2\\
&= a^3 +(- b)a^2+a^2b+(-b)(ab)+ab^2+(-b)b^2\\
&= a^3 - ba^2+a^2b - b(ab) +ab^2-b^3\\
&= a^3 - a^2b+a^2b - (ba)b +ab^2-b^3\\
&= a^3 -(ab)b +ab^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Numbers are too large to show $65^{64}+64^{65}$ is not a prime I tried to find cycles of powers, but they are too big. Also $65^{n} \equiv 1(\text{mod}64)$, so I dont know how to use that.
| Hint $\rm\ \ x^4 +\: 64\: y^4\ =\ (x^2+ 8\:y^2)^2 - (4xy)^2\ =\ (x^2-4xy + 8y^2)\:(x^2+4xy+8y^2)$
Thus $\rm\ x^{64} + 64\: y^{64} =\ (x^{32} - 4 x^{16} y^{16} + 8 y^{32})\:(x^{32} - 4 x^{16} y^{16} + 8 y^{32})$
Below are some other factorizations which frequently prove useful for integer factorization. Aurifeuille, Le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
Is $ f(x) = \left\{ \begin{array}{lr} 0 & : x = 0 \\ e^{-1/x^{2}} & : x \neq 0 \end{array} \right. $ infinitely differentiable on all of $\mathbb{R}$? Can anyone explicitly verify that the function $
f(x) = \left\{
\begin{array}{lr}
0 & : x = 0 \\
e^{-1/x^{2}} & : x \neq 0
\end{array}
... | For $x\neq 0$ you get:
$$\begin{split}
f^\prime (x) &= \frac{2}{x^3}\ f(x)\\
f^{\prime \prime} (x) &= 2\left( \frac{2}{x^6} - \frac{3}{x^4}\right)\ f(x)\\
f^{\prime \prime \prime} (x) &= 4\left( \frac{2}{x^9} - \frac{9}{x^7} +\frac{6}{x^5} \right)\ f(x)
\end{split}$$
In the above equalities you can see a path, i.e.:
$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what... | ${\displaystyle{\frac{1}{9}}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Proof by induction:
For $n=1, {\displaystyle{\frac{1}{9}}(10^1+3 \cdot 4^1 + 5) = \frac{27}{9} = 3}$, so the result holds for $n=1$
Assume the result to be true for $n=m$, i.e. $\displaystyle{\frac{1}{9}(10^m+3 \cdot 4^m + 5)}$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 1
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How to prove a trigonometric identity $\tan(A)=\frac{\sin2A}{1+\cos 2A}$ Show that
$$
\tan(A)=\frac{\sin2A}{1+\cos 2A}
$$
I've tried a few methods, and it stumped my teacher.
| First, lets develop a couple of identities.
Given that $\sin 2A = 2\sin A\cos A$, and $\cos 2A = \cos^2A - \sin^2 A$ we have
$$\begin{array}{lll}
\tan 2A &=& \frac{\sin 2A}{\cos 2A}\\
&=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\\
&=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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Residue of a pole of order 6 I am in the process of computing an integral using the Cauchy residue theorem, and I am having a hard time computing the residue of a pole of high order.
Concretely, how would one compute the residue of the function $$f(z)=\frac{(z^6+1)^2}{az^6(z-a)(z-\frac{1}{a})}$$ at $z=0$?
Although it i... | $$g(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}}$$
we know:
$$(a+b)^n =a^n+\frac{n}{1!}a^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+...+b^n$$
$$ \text{As regards }: |a|<1 $$
Taylor series of f(z) is:
$$g(z)=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $\int\limits_0^{\frac{\pi}{2}} \frac{\sin(2nx)\sin(x)}{\cos(x)}\, dx$
How to evaluate
$$ \int\limits_0^{\frac{\pi}{2}} \frac{\sin(2nx)\sin(x)}{\cos(x)}\, dx $$
I don't know how to deal with it.
| Method 1. Let $I(n)$ denote the integral. Then by addition formula for sine and cosine,
$$\begin{align*}
I(n+1) + I(n)
&= \int_{0}^{\frac{\pi}{2}} \frac{[\sin((2n+2)x) + \sin(2nx)]\sin x}{\cos x} \; dx \\
&= \int_{0}^{\frac{\pi}{2}} 2\sin((2n+1)x) \sin x \; dx \\
&= \int_{0}^{\frac{\pi}{2}} [\cos(2nx) - \cos((2n+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Primes modulo which a given quadratic equation has roots Given a quadratic polynomial $ax^2 + bx + c$, with $a$, $b$ and $c$ being integers, is there a characterization of all primes $p$ for which the equation
$$ax^2 + bx + c \equiv 0 \pmod p$$
has solutions?
I have seen it mentioned that it follows from quadratic reci... | I never noticed this one before.
$$ x^3 - x - 1 \equiv 0 \pmod p $$
has one root for odd primes $p$ with $(-23|p) = -1.$
$$ x^3 - x - 1 \equiv 0 \pmod p $$
has three distinct roots for odd $p$ with $(-23|p) = 1$ and $p = u^2 + 23 v^2 $ in integers.
$$ x^3 - x - 1 \equiv 0 \pmod p $$
has no roots for odd $p$ with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx$? How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$
| To elaborate on Pantelis Damianou's answer
$$
\newcommand{\cis}{\operatorname{cis}}
\begin{align}
\int_0^\pi\frac{\cos(nx)}{5+4\cos(x)}\mathrm{d}x
&=\frac12\int_{-\pi}^\pi\frac{\cos(nx)}{5+4\cos(x)}\mathrm{d}x\\
&=\frac12\int_{-\pi}^\pi\frac{\cis(nx)}{5+2(\cis(x)+\cis(-x))}\mathrm{d}x\\
&=\frac12\int_{-\pi}^\pi\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to get rid of the integral in this equation $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$? How to get rid of the integral $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$ when $f(x)=x^2$?
| Summarising the comments, you'll get
$$
\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}f(t)\right)^2}dt}
=\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}t^2\right)^2}dt}
=\int\limits_{x_0}^{x}{\sqrt{1+4t^2}dt}
$$
To solve the last one substitute $t=\tan(u)/2$ and $dt=\sec^2(u)/2du$. Then $\sqrt{1+4t^2}= \sqrt{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Root Calculation by Hand Is it possible to calculate and find the solution of $ \; \large{105^{1/5}} \; $ without using a calculator? Could someone show me how to do that, please?
Well, when I use a Casio scientific calculator, I get this answer: $105^{1/5}\approx " 2.536517482 "$. With WolframAlpha, I can an even more... | Another way of doing this would be to use logarithm, just like Euler did:
$$
105^{1/5} = \mathrm{e}^{\tfrac{1}{5} \log (105)} = \mathrm{e}^{\tfrac{1}{5} \log (3)}
\cdot \mathrm{e}^{\tfrac{1}{5} \log (5)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (7)}
$$
Use $$\log(3) = \log\left(\frac{2+1}{2-1}\right) = \log\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/127310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "92",
"answer_count": 5,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.