Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Is this simple proof about symmetric sums correct? I'm asked to prove
$$x \gt 0, y \gt 0, z \gt 0 \rightarrow$$
$$\left(\frac{x+y}{x+y+z}\right)^\frac{1}{2}+\left(\frac{x+z}{x+y+z}\right)^\frac{1}{2} + \left(\frac{y+z}{x+y+z}\right)^\frac{1}{2} \le 6^\frac{1}{2}$$
I rewrite the summands and say that it is sufficien... | You are almost there. Squaring both sides yields $$A+B+C+2(AB)^{1/2}+2(BC)^{1/2}+2(CA)^{1/2}\leq 6,$$ and so you only need to prove that $$(AB)^{1/2}+(BC)^{1/2}+(CA)^{1/2}\leq 2.$$ This inequality follows directly from the fact that $A+B+C=2$ along with Cauchy Schwarz.
Alternative Solution: Before squaring both sides... | {
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Calculate this sum $\sum \frac{1}{(4n+1)(4n+3)}$ I'm having troubles to calculate this sum: $\sum \frac{1}{(4n+1)(4n+3)}$. I'm trying to use telescopic series, without success:
$\sum \frac{1}{(4n+1)(4n+3)}=1/2\sum \frac{1}{(4n+1)}-\frac{1}{(4n+3)}$
I need help here
Thanks a lot
| $$\begin{align} \frac{1}{2} \sum_{k=0}^{\infty} \left ( \frac{1}{4 k+1} - \frac{1}{4 n+3} \right ) &= \frac{1}{2} \left ( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots \right )\\ &= \frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \\ &= \frac{\pi}{8} \end{align}$$
| {
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Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want ... | I assume right to left associativity, as is usually the case. Note that
$$(\alpha \beta)^{-1} = {\beta}^{-1} {\alpha}^{-1}, \text{ and}$$
$$ (a_1 a_2 \dots a_n)^{-1} = (a_1 a_n \dots a_2).$$
These are standard facts, and easily shown if you like to. Hence,
$$qpq^{-1} = (1 2 3)(4 5 6 8)(1 4 3 8 2)(1 2)(1 5)(4 8 6 5)(1 ... | {
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Least value of $a$ for which at least one solution exists? What is the least value of $a$ for which
$$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a$$
has atleast one solution in the interval $(0,\frac{\pi}{2})$?
I first calculate $f'(x)$ and put it equal to $0$ to find out the critical points.
This gives
$$\sin(x)=\frac{2}{3... | One possible approach: Find a common denominator, then :
$$\frac{4}{\sin x}+\frac{1}{1-\sin x}=a\iff \frac{4(1- \sin x) + \sin x}{\sin x - \sin^2x} = a$$ $$ \iff 4-3\sin x = a(\sin x - \sin^2 x)\tag{$\sin x \neq 0$}$$
Now write the equation as a quadratic equation in $\sin x$:
$$a\sin^2 x - (3 + a)\sin x + 4 = 0 $$
You... | {
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Verify that: $2\cot{x}/\tan{2}x = \csc^2x-2$ Verify the following:
$$\frac{2\cot{x}}{\tan{2}x} = \csc^2x-2\;.$$
| Hint:$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$
$$\frac{2\cot x}{\tan 2x}=\frac{1-\tan ^2x}{\tan ^2x}=\frac{1}{\tan ^2x}-1=\frac{cos^2x}{sin^2x}-1=$$$$=\frac{1-sin^2x}{sin^2x}-1= \csc^2x-2$$
| {
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Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that ... | Not sure if this is correct, but here goes anyway,
$$u_n=n\left(\int_0^1\frac{1}{(1+x)^2}dx-\frac 1n\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$$
Let $$\begin{align} A_k
&=\int_{\frac{k-1}{n}}^{\frac{k}{n}}\frac{1}{(1+x)^2}\mathrm{d}x -\frac{1}{n\left(1+\frac{k}{n}\right)^2} \\
& =\frac{1}{n\left(1+\frac{k}{n}\r... | {
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If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(... | The previous solution was wrong, because I used $\frac {da}{db} = -1$ instead. I've added an explanation of why we should have used $\frac {da}{db} = 1$.
The simplest approach I can think of, is to realize that you have a conic section, which is an ellipse.
Because you are interested in extreme values of $a-b = K$, th... | {
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A simultaneous system of equations Solve for $a,b,c$:
\begin{align}
2ab+a+2b=24\\
2bc+b+c=52\\
2ac+2c+a=74\\
\end{align}
Solving them simultaneously is leading to very difficult situation. Plz help.
| \begin{align}
(a+1)(2b+1)&=(2ab+a+2b)+1=25\\
(2b+1)(2c+1)&=2(bc+b+c)+1=105\\
(2c+1)(a+1)&=(2ca+2c+a)+1=75
\end{align}
So put $u=a+1,v=2b+1,w=2c+1$ and I think you'll get the answer. (Hint: consider $u^2=(uv)(uw)/(vw)$, et cetera)
| {
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a elementary number theory problem please help me to find $ a,b \in \Bbb Z $ such that $a|b$ and $ \forall z \in \Bbb Z $,we have $a+z|b+z$.
| If $a = b$ the result is trivial.
So I will assume that $a \ne b$.
If $a$ and $b$ are positive,
we must have $a < b$.
If we choose $z$ such that
$2(a+z) > (b+z)$
(i.e., $z > b-2a$),
then $1 < \frac{b+z}{a+z} < 2$
which contradicts $(a+z)|(b+z)$.
So $a$ and $b$ cannot both be positive.
If $a$ and $b$ are both negative,
... | {
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Introduction to Calculus equality I was reading through Apostol's Calculus where he has this equality, but I can't figure out how to prove this equality. I would appreciate a hint if you could help. Thank you.
$$
[(\sum_{k=1}^{n-1}k)+1]^3-(\sum_{k=1}^{n-1}k)^3=n^3-1^3
$$
Above shown to be incorrect should be:
$$
\sum_{... | I think there was a little problem with parentheses. You may intend
$$\sum_{k=1}^{n-1}(k+1)^3 -\sum_{k=1}^{n-1}k^3.$$
If so, it is just a question of expanding out the sums. The first is the sum of the cubes from $2^3$ to $n^3$, and the second is the sum of the cubes from $1^3$ to $(n-1)^3$. Equivalently, it is
$$\left... | {
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Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma... | Responding to the observation/comment by Random Variable from Thu Aug 8, what is missing above to turn this into a rigorous proof is an evaluation of the quantity
$$R(a,b) = \sum_{m=a}^b
\operatorname{Res}\left(\prod_{k=a}^b \frac{1}{k^2+x^2};x=im\right).$$
We put
$$g_1(x) = \prod_{k=a}^b \frac{1}{k+ix}
\quad \text{an... | {
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Help with following equality Could you please help me understand how this is?
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \,.$$
Thank you.
| $\text{L.H.S.}$
$=\displaystyle \frac 1 1-\frac 1 2+\frac 1 3 - \frac 1 4 +\cdots+\frac 1 {2n-1} - \frac 1 {2n}$
$\displaystyle=\frac 1 1+\frac 1 2+\frac 1 3 + \frac 1 4 +\cdots+\frac 1 {2n-1} + \frac 1 {2n}-2\Big(\frac 1 2+\frac 1 4+\cdots+\frac 1 {2n}\Big)$
$\displaystyle=\frac 1 1+\frac 1 2+\frac 1 3 + \frac 1 4 +\c... | {
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An integral related to the beta function: $\int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \, dx $ I came across an exercise in a textbook that says to show that $$ \int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \, dx = 2^{m+n-2} B(m,n), \ (m,n >0),$$ and then deduce that $$ \int_{-\frac{... | Let's prove the first part, you need change the variables.
*
*$y:=\frac{(1+x)^2}{2(1+x^2)}$, then $1-y=\frac{(1-x)^2}{2(1+x^2)}$, and $$\mathrm{d}y=\frac{1-x^2}{(1+x^2)^2}=2\cdot y^{\frac{1}{2}}\cdot (1-y)^{\frac{1}{2}}\cdot \frac{1}{1+x^2}\mathrm{d}x$$
Theorfore,
\begin{eqnarray}
&&\int_{-1}^{1}\frac{(1+x)^{2m-1}(1... | {
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Conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational Motivation
I am working on one of the questions from Hardy's Course of Pure Mathematics and was wondering if I could get some assistance on where to go next in my proof. I have attempted rearranging the expression in numerous ways from the step I am at... | Try to square $\sqrt{a+\sqrt b}+\sqrt{a-\sqrt b}$.
| {
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Number of non-negative integer solutions of an equation using generating functions Using convolution, find how many non-negative integer solutions of
$$\begin{aligned} x_1 + 5 x_2 &= n\\ x_1 + 5 x_2 &= 60\end{aligned}$$
Is anyone able to solve this problem or figure out the idea? Thank you so much.
| Here's a combinatorial interpretation: the number of solutions to the second equation is the number of ways to make 60 cents in change using 1 cent and 5 cent coins.
I assume that the second equation is a special case of the first, and that you want to know how to solve both. The method for $n=60$ illustrates the gene... | {
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How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$ My understanding would lead me to believe that:
$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$
But it actually equals: $\sqrt{1 + e^4}$
Can you explain why?
| Also you can use squaring both sides:
$$
\sqrt{4+4e^4}=2\sqrt{1+e^4}\\
4+4e^4=4(1+e^4)
$$
since $\sqrt{x^2}= \pm x$ and the expressions you start with is one of these two solutions (positive)
| {
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Column space of complex matrix Let $A=\begin{pmatrix}
1 & \alpha & \alpha^2\\
1 & \beta & \beta^2
\end{pmatrix}, \alpha,\beta\in\mathbb{C},\alpha\ne\beta$. I know that the column space of A is supposed to be $\mathbb{C}^2$, but I'm not sure how to get there.
My attempt:
The column space of $A$ is the set of ... | As $\alpha \neq \beta$, than the first two columns are linear independent, so the dimension of the image is at least $2$ that means your column space is $\mathbb{C}^2$.
If you like a constructive one more than make the following
$$ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} + (-\alpha) \cdot \begin{pmatrix} 1 \\ 1 ... | {
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$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ How to prove:$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ ?
| It is enough to check divisibility by $2$ and by $9$. Divisibility by $2$ is trivial, since $1^n+\cdots+9^n$ and $3(1^n+6^n+8^n)$ are both odd.
To show divisibility by $9$, we first check separately the case $n=1$. This is easy, the expression in that case is $0$. For $n\ge 2$, the terms $3^n$, $6^n$, and $9^n$ are di... | {
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How to calculate the asymptotic expansion of $\sum \sqrt{k}$? Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that
$$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$
hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$.
With some more calculations... | With Mathematica it very easy:
Series[Sum[Sqrt[k], {k, 1, n}], {n, Infinity, 2}]// TeXForm
$$\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta
\left(-\frac{1}{2}\right)+\frac{\sqrt{\frac{1}{n}}}{24}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$ Let $A,B,C\in \mathbb{R}$ with $\sin{A}+\sin{B}+\sin{C}=0$. Prove that
$$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$$
| The first idea that comes to mind it to express everything in terms of $\sin A,\ \sin B,\ \sin C$ (and the cosines). It turns out that it works:
Begin with $\cos (A - 2B ) = \cos A \cos 2B + \sin A \sin 2B $ (by the formula for $\cos (x+y)$). Now, get rid of the double angles: $\cos 2B = 1 - 2 \sin^2 B$ and $\sin 2B = ... | {
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Prove inequality: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$
I can do this by: $EAT^2$ (expand all of the thing)
*
*$(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2... | For non-negative variables also TL helps:
Let $a+b+c=3$ (we can assume it because our inequality is homogeneous).
Hence, we need to prove that
$$\sum_{cyc}(28a^4-(3-2a)^4-27)\geq0$$ or
$$\sum_{cyc}(a-1)(a^3+9a^2-9a+9)\geq0$$ or
$$\sum_{cyc}\left((a-1)(a^3+9a^2-9a+9)-10(a-1)\right)\geq0$$ or
$$\sum_{cyc}(a-1)^2(a^2+10a+... | {
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Triple Integral and symmetry The problem is as follows: Compute the intergal
$$I=\iiint_B \frac{x^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz,$$ where $B$ is the unit ball defined by $B=\{(x,y,z) \mid x^2+y^2+z^2 \leq 1\}$.
The official solution is tricky: The change of variable $(x,y,z) \mapsto (z,y,x)$ transforms the integral... | Forget about the Jacobian. From symmetry considerations it is obvious that
$$I:=\iiint_B \frac{x^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz=\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz\ .$$
Therefore $2 I=\int_B 1 \:dx\:dy\:dz= {\rm vol}(B)$ and $I={2\pi\over3}$.
| {
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Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am hav... | This here should work with $n \geq 1 $ :
$$s_n = \sum\limits_{n=1}^\infty \frac{1}{n^2}= \frac{1}{1} + \frac{1}{4} +\frac{1}{9} + \frac{1}{16}+ ...+ \frac{1}{n²} $$$$
b_n = \sum\limits_{n=1}^\infty \frac{1}{2^{n-1}} = \frac{1}{1} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8} + ...+ \frac{1}{2^{n-1}} $$
$b_n$ is directly com... | {
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Evaluate $\sum_{k=1}^{n}(k^2 \cdot (k+1)!)$ We have to evaluate the following:
$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$
Any hints ?
| Using $k^2 = (k+3)(k+2) - 5 (k+2) + 4$ we write $$f_k = k^2 (k+1)! = (k+3)! - 5 (k+2)! + 4 (k+1)! = G_{k+1} - G_k$$
where $G_k = (k+2)! - 4(k+1)!$, therefore
$$
\sum_{k=1}^n f_k = \sum_{k=1}^n \left(G_{k+1} - G_k\right) = G_{n+1} - G_1
$$
Since $G_1 = -2$, and $G_{n+1} = (n+3)! - 4(n+2)! = (n+2)! (n-1)$ we get
$$
... | {
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Series expansion $(1-\cos{x})^{-1}$ How do i get the series expansion $(1-\cos{x})^{-1} = \frac{2}{x^2}+\frac{1}{6}+\frac{x^2}{120}+o(x^4)$ ?
| $$\frac{1}{1 - x} = 1 + x + x^2 + O(x^3)$$
and so
$$\frac{1}{1 - \cos x}= \frac{1}{\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{6!}} = \frac{2}{x^2} \frac{1}{1 - (\frac{x^2}{12} - \frac{x^4}{360} + O(x^6))}$$
$$ = \frac{2}{x^2} \left( 1 + \left(\frac{x^2}{12} - \frac{x^4}{360}\right) + \frac{x^4}{144} + O(x^6) \right) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Euler's infinite product for the sine function and differential equation relation Euler's infinite product for the sine function
$$\displaystyle \sin( x) = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{\pi^2k^2} \right)$$
http://en.wikipedia.org/wiki/Basel_problem
We know that $\sin( x)$ satisfies $y''+y=0$ differentia... | The first sum is a known sum which I will not prove here:
$$\sum_{k=1}^{\infty} \frac{1}{\pi^2 k^2-x^2} = \frac{1}{2 x} \left ( \frac{1}{x}-\cot{x}\right)$$
The second sum, on the other hand, I could not find in a reference. You can, however, evaluate it using residues. That is,
$$\sum_{k=-\infty}^{\infty} \frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | maybe this following solution is by Vasc?
since use Cauchy-Schwarz inequality,we have
$$(a^3+3b)(a+3b)\ge (a^2+3b)^2$$
It suffices to show that
$$\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\ge 6$$
By Holder
$$\left(\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\right)^2[\sum_{cyc}(a^2+3b)(a+3b)]\ge[\sum_{cyc}(a^2+3b)]^3=[a^2+b^2+c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 0
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How can find this sequence $ a_{n+1}=a_{n}+na_{n-1},$ let $a_{n+1}=a_{n}+na_{n-1},a_{1}=1，a_{2}=2$.
find the $a_{n}=?$
my ideas: $\dfrac{a_{n+1}}{(n+1)!}=\dfrac{1}{n+1}\dfrac{a_{n}}{n!}+\dfrac{1}{n+1}\dfrac{a_{n-1}}{(n-1)!},$
and let $b_{n}=\dfrac{a_{n}}{n!}$,then we have
$(n+1)b_{n+1}=b_{n}+b_{n-1},b_{1}=1,b_{2}=1$,b... | If $f(x)=\sum_{n\ge 1}a_nx^{n+1}$ then
$$
x^3f'(x)+(x-1)f(x)+x^3+x^2=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Relation between area of a triangle on a sphere and plane We know area of a plane triangle $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$.
I was just thinking: let we have a triangle with arc length $a,b,c$ on a sphere of radius $r$, do we have any similar kind of formula for that spherical triangle? when ... |
Source: This Dr. Math Article
Novice here, so please excuse any mistakes. The ratio should be:
$$
\frac{ 180 \cdot \sqrt{ s(s-a)(s-b)(s-c) } }
{ 4 \cdot \pi \cdot R^2 \cdot \arctan \left(
\sqrt{ \tan \left( \frac{s}{2} \right) \cdot \tan \left( \frac{s-a}{2} \right) \cdot \tan \left( \frac{s-b}{2} \right) \cdot \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
What are some useful tricks/shortcuts for verifying trigonometric identities? What "tricks" are there that could help verify trigonometric identities?
For example one is:
$$a\cos\theta+b\sin\theta = \sqrt{a^2+b^2}\,\cos(\theta-\phi)$$
| Expand the cosine of the difference of angles on the right.
$$ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi$$
Now, collect terms so that the expression is a linear combination of $\sin \theta$ and $\cos \theta$, as in the expression on the left.
$$ \begin{align}
\sqrt{a^2 + b^2}\cos(\theta - \phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$ How to calculate following with out using L'Hospital rule
$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$
| Provided the following limits exist,
\begin{align*}
A &= \lim_{x \to -1^{+}}\left(\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{\sqrt{x+1}} \right)\\
&= \lim_{x\to -1^+}\left(\frac{\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\frac{\sqrt{x+1} - 0}{x + 1}} \right)\\
&= \frac{\lim_{x\to -1^+}\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Such a sequence is called Arithemtico-Geomteric Progression.
$$S_n=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } } $$
$$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i+1 } } }=\sum _{ i=2 }^{ n+1 }{ \frac { i-1 }{ { 2 }^{ i } } }$$
Subtracting
$$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { 1 }{ { 2 }^{ i }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\r... | To "discover" the proof, we typically work backwards. We "assume" ${|x+1|\over|2x+1|}<\epsilon$, and find what $\delta$ should be. Then we will have to rewrite the proof.
Let's think about the numerator and denominator separately. We want $|x+1|$ small, and intuitively, we want $|2x+1|$ large. So let's keep $2x+1$ awa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over the matrix. Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over $\mathbb{C}$ to \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}
| Elementarily, assume $$N = \begin{bmatrix}a & b\\c & d\end{bmatrix} $$
Then
$$0=N^2 = \begin{bmatrix}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix} $$
Now if $b=0$, you get
$$0=N^2 = \begin{bmatrix}a^2 & 0\\c(a+d) & d^2\end{bmatrix} $$
so that $a=d=0$, and the result follows.
Otherwise, you must conclude $a=-d$ so
$$0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Given $a+b+c=0$, simplify the following. I am here again to ask a question about an exercise I saw around but i'm having a lot of trouble with. I know the answer is 3abc, but as in many of my questions, I am interested in the why and how.
Given $a+b+c=0$, simplify:
$$\frac{(a^3-abc)^3+(c^3-abc)^3+(b^3-abc)^3}{(c^2-ab)(... | If you just sub in $c=-a-b$ then each factor of the denominator reduces to $(a^2+ab+b^2)$.
Each term of the numerator has a factor that is one of these denominator factors as well. For instance the first term from the numerator is $a^3(a^2-bc)^3$, which is $a^3(a^2+ab+b^2)^3$.
So we have
$$\begin{align}
\frac{a^3(a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/340169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Integral check $\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}$ $$\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}$$ Partial fraction decomposition of $\frac{6x+4}{x^4+3x^2+5}$ is of the following form: $$\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+1}$$ We need to find $A,B,C$ and $D$ \ $$\frac{6x+4}{x^4+3x^2+5}=\frac{Ax+B}{x^2+2}+\frac{Cx... | All the signs of all your letters are switched. This is because $A+A+C=A+0=A=6$, $B+B+D=B+0=B=4$, etc. Also, $$(x^2+1)(x^2+2)=x^4+3x^3+2\neq x^2+3x+5$$ Your polynomial has no real roots, so you'll have to do some manipulation in $\Bbb C$, as follows:
Let $x^2=u$. Then $$u^2+3u+5=0$$ has solutions $$u_1=\frac{-3+i\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/342198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
*
*Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
*If $0<\theta < \frac{\pi}{2} $ and $\sin 2\theta=\cos 3\theta~~$ then find the value of $\sin\theta$
| For the first one:
$$\lim_{x\to 0}\frac{5^x-4^x}{x} = \frac{\mathrm{d}(5^x-4^x)}{\mathrm{d}x}(0) = \frac{\mathrm{d}5^x}{\mathrm{d}x}(0)-\frac{\mathrm{d}4^x}{\mathrm{d}x}(0) = \log_e\frac{5}{4}$$
For the second:
$$\sin 2\frac{\pi}{10} + \cos 3\frac{\pi}{10} = 0.$$
The golden-ratio triangle has angles $\frac{\pi}{5}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$ I came across the following problem :
Let $a,b,c$ are non-zero real numbers .Then the minimum value of $a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}?$ This is a multiple choice question and the options are $0,6,3^2,6^2.$
I do not ... | $a^2+b^2+c^2+a^{-2}+b^{-2}+c^{-2}=(a-a^{-1})^2+(b-b^{-1})^2+(c-c^{-1})^2+6$, whence the minimum occurs when $a=a^{-1},b=b^{-1},c=c^{-1}$ and is $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Calculate volume in a 3D sort of space using cartesian coordinates Find the volume bounded by the cylinder $x^2 + y^2 = 1$, the planes $x=0, z=0, z=y$ and lies in the first octant. (where x, y, and z are all positive)
| Use a triple integral. We're simply trying to find the volume$-$there's no density involved or anything$-$so the important question is just to find the correct bounds.
Let's examine the $x$ variable first. $x$ can move from $0$ to $1$, since it's constrained below by $0$ and cannot be greater than $1$ (otherwise $x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the value of : $\lim_{x\to\infty}x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$ How can I show/explain the following limit?
$$\lim_{x\to\infty} \;x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$$
Some trivial transformation seems to be eluding me.
| A short calculation, using equivalents:
$$ x\Bigl(\sqrt{x^2-1}-\sqrt{x^2+1}\Bigr)=\frac{x\bigl((x^2-1)-(x^2+1)\bigr)}{\sqrt{x^2-12}+\sqrt{x^2+1}}= \frac{-2x}{\sqrt{x^2-1}+\sqrt{x^2+1}}.$$
Now, for $x>0$, we have
$$ \sqrt{x^2-1}+\sqrt{x^2+1}=x\biggl(\sqrt{1-\frac1{x^2}}+\sqrt{1+\frac 1{x^2}\biggr)}
\sim_{+\infty}2x, $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Struggling with an integral with trig substitution I've got another problem with my CalcII homework. The problem deals with trig substitution in the integral for integrals following this pattern: $\sqrt{a^2 + x^2}$. So, here's the problem:
$$\int_{-2}^2 \frac{\mathrm{d}x}{4 + x^2}$$
I graphed the function and because... | Let $x=2\tan\theta$
and $dx=2 \sec^2\theta$
$2\tan\theta = -2, 2$
$\theta = \frac{-\pi}{4}, \frac{\pi}{4}$
$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{2\sec^2\theta}{4+4\tan^2\theta}d\theta$
$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{2\sec^2\theta}{4\sec^2\theta}d\theta$
$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{1}{2} d\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Set of all triangles with two equal edges inscribed in a circle. Let $\Delta$ be the set of all triangles with two equal edges inscribed in a circle of radius $R$.
So, how do I show that:
1, The equilateral triangle in $\Delta$ is the one maximizing the area.
2, The equilateral triangle in $\Delta$ is the one maximizi... | We can prove that Isosceles Triangles have Two Equal Angles
Now using Law of Sines, $$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=2R$$
Let $A=B=x\implies C=\pi-2x,\sin C=\sin(\pi-2x)=\sin 2x$
So, $x=A>0$ and $\pi-2x=C>0\implies x<\frac\pi2\implies 0<x<\frac\pi2$
So, $a=b=2R\sin x,c=2R\sin2x$
So, the circumference ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
| Just double your expression and regroup using $\binom{i}{n} = \binom{n-i}{n}$ $$ 1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= \\ =\frac 1 2 \left(
(1 + (n+1))\cdot\binom{n}{0}+ (2 + n)\cdot\binom{n}{1}+(3 + (n-1))\cdot\binom{n}{2}+ \cdots+((n+1)+1)\cdot\binom{n}{n}\right)=
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Convergence of the infinite series $ \sum_{n = 1}^\infty \frac{1} {n^2 - x^2}$ How can I prove that for every $ x \notin \mathbb Z$ the series
$$ \sum_{n = 1}^\infty \frac{1} {n^2 - x^2}$$
converges uniformly in a neighborhood of $ x $?
| If $x \notin \mathbb{Z}$, then there exists some $\delta>0$ such that $(x-\delta,x+\delta) \cap \mathbb{Z} = \emptyset$. Then $s(y) = \sum_{n = 1}^\infty \frac{1} {n^2 - y^2}$ converges uniformly for $y \in B(x,\delta)$.
To see this, let $\epsilon>0$ and choose $N$ such that $N^2 > 2 (|x|+\delta)^2$. Then if $n \ge N$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Darts on a ruler probability If two points are selected at random on an interval from 0 to 1.5 inches, what is
the probability that the distance between them is less than or equal to 1/4"?
| The square $S=\{(x,y) | x,y \in [0, \frac{3}{2}] \}$ has area $(\frac{3}{2})^2$. The area $\Delta= \{(x,y) \in S \, |\, |x-y| > \frac{1}{4} \}$ can be easily rearranged to be a square with area $(\frac{3}{2}-\frac{1}{4})^2$. Hence the chance of landing in $S \setminus \Delta$ is $\frac{(\frac{3}{2})^2-(\frac{3}{2}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of the selected elements of matrix is $255$ A $5\times 10$ matrix is given:
$$\begin{pmatrix}
1 & 6 & 11 & 16 & 21 & 26 & 31 & 36 & 41 & 46\\
2 & 7 & 12 & 17 & 22 & 27 & 32 & 37 & 42 & 47\\
3 & 8 & 13 & 18 & 23 & 28 & 33 & 38 & 43 & 48\\
4 & 9 & 14 & 19 & 24 & 29 & 34 & 39 & 44 & 49\\
5 & 10& 15 & 20 & 25 & 30 & 35... | Your matrix is the sum of these two:
$$A = \begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2\\
3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3\\
4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4\\
5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5\\
\end{pmatrix}$$
and
$$B = \begin{pmatrix}
0 & 5 & 10 & 15 & 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352990",
"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots+ \frac{1}{\sqrt{24} + \sqrt{25}}$ Simplify$$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}.$$
I know you can solve this using gener... | Hint: Multiply top and bottom of $\dfrac{1}{\sqrt{k+1}+\sqrt{k}}$ by $\sqrt{k+1}-\sqrt{k}$, and watch the house of cards collapse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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question about inverse functions Functions $f$ and $g$ are defined by
$$ f: x\mapsto 2x+1$$
$$g: x \mapsto \dfrac{2x +1}{x+3}$$
(i) Solve the equation $gf(x) = x $
(iii) Show that the equation $g^{-1} (x) = x$ has no solutions
I need help with these.
| $$ f(x)=2x+1,g(x)= \frac{2x +1}{x+3}$$
$$g(f(x)) =\frac{2(2x+1) +1}{2x+1+3}= x$$
$$\frac{4x+3}{2x+4}= x,x\neq-2\Rightarrow2x^2+4x=4x+3,x^2=3/2,x=\pm\sqrt{3/2}$$
$$g(x)=y= \frac{2x +1}{x+3}$$
$$x= \frac{2y +1}{y+3},xy+3x=2y+1,y(x-2)=1-3x,y=\frac{1-3x}{x-2},x\neq2$$
$$g^{-1}(x)=\frac{1-3x}{x-2}=x\Rightarrow x^2+x-1=0,x=\... | {
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"url": "https://math.stackexchange.com/questions/353706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding all $\alpha$ such that a matrix is positive definite I have
$A = $
$
\left[\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right]
$
and I want to find all $\alpha$ such that $A$ is positive definite.
I tried
$ x^tAx = $
$
\left[\begin{array}{r}
x & y & z
\... | Maybe this helps:
A hermitian matrix is positive definite $\Leftrightarrow$ all leading principal minors are positive.
So $\left|\begin{array}{r}
2
\end{array}\right|$, $\left|\begin{array}{rr}
2 & \alpha \\
\alpha & 2
\end{array}\right|$ and $\left|\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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When is this rational fraction bounded? Let $\Omega=(0,\infty)^2$. For $\alpha \gt 0, \beta \gt 0$, define a function
$f_{\alpha,\beta}$ on $\Omega$ by putting
$$
f_{\alpha,\beta}(x,y)=\frac{(xy)^{\alpha}(1+x^2)(1+y^2)}{\big(xy(x+y)+1\big)^{\beta}}
$$
For which pairs $(\alpha,\beta)$ is $f_{\alpha,\beta}$ bounded ?
Pu... | The necessary and sufficient condition is $3\beta\ge 2\alpha+4$ and $\alpha\ge 2$.
When $x,y\ge 1$, you have seen $3\beta\ge 2\alpha+4$ is necessary. In this case, it is also sufficient. Note that $xy(x+y)+1\ge 2(xy)^{\frac{3}{2}}$, and when $x,y\ge 1$, $1+x^2\le 2x^2$, $1+y^2\le 2y^2$. It follows that
$$f_{\alpha,\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $z$ is a complex number of unit modulus and argument theta If $z$ is a complex number such that $|z|=1$ and $\text{arg} z=\theta$, then what is $$\text{arg}\frac{1 + z}{1+ \overline{z}}?$$
| $$ \frac{1+z}{1 + \bar z} = \frac{1+z}{1 + \bar z} \times \frac{1+z}{1 + z} = \frac{(1+z)^2}{|1+z|^2} $$
the argument should be $ 2 \arg (1 +z) = 2 \arctan \left( y \over x+1\right)$
Let $x = \cos \theta$ and $y = \sin \theta $, we have $\arctan \left( \frac{y}{x+1}\right) = \arctan \left( \frac{\sin \theta}{\cos \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
mixed permutations and combinations I have a problem that I am not too sure of. In a team of 16, there are 5 couples and 6 single people. In how many ways can at most 1 couple be chosen if 6 people are required to represent the team at a conference?
This is my solution: 6P6 (0 couples and 6 single people only) + 11C6 *... | The following approach is reasonably systematic. There will be lots of words, but at the end there will be a more or less compact formula.
We count first the teams that have no couple, then, in basically the same way, the teams that have $1$ couple. A couple, viewed as an entity, will be called a family. There are $5$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find minimum in a constrained two-variable inequation I would appreciate if somebody could help me with the following problem:
Q: find minimum
$$9a^2+9b^2+c^2$$
where $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$
| Another approach - the symmetry here suggests Purkiss Principle (conditions to be verified), so the extremum is attained when $a = b$.
So $c = (9-a^2) - 2a^2 = 9-3a^2$
and $9(a^2+b^2) + c^2 = 18a^2 + (9-3a^2)^2 = 9a^4 - 36a^2 + 81 = 9 (a^2 - 2)^2 + 45$
which is minimised when $a^2 = 2$ or $a = \pm \sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
| Technique 1:$$\sqrt{x^2y^2} = xy \le \dfrac{x^2 + y^2}{2}\le x^2 + y^2$$ Technique 2:$$(x - y)^2 \ge 0 \implies x^2 +y^2 - 2xy \ge 0 \implies x^2 + y^2 \ge 2xy\ge xy$$ Technique 3 (my favorite): There's a statement $\dfrac{y}{x} + \dfrac{x}{y} \ge 2$ with many classical proofs (which I'd not state here). We can write t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/357272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 25,
"answer_id": 2
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Diagonalizing symmetric 2x2 matrix If A is a $2\times2$ symmetric matrix ($A^T = A$) where $b$ does not equal zero ($a$'s are on the diagonal, $b$'s occupy the other $2$ spaces), find a matrix $X$ such that $X^T AX$ is diagonal.
What is the simple way to solve this problem (using orthogonal diagonalization intro linear... | You can also just go with the computations ;)
You easily get the two eigenvalues : $\lambda_1=a-b$ and $\lambda_2=a+b$ and the corresponding eigenvectors:
$V_1=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $V_2=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
you can also make them have norm 1. Then:
$V_1=\frac{1}{\sqrt{2}}\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this value:$\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$
Let $f(x)=\ln{x}-\dfrac{1}{x}+3$, and $a_{i}>0,i=1,2,3$, such that $a_{3}:a_{2}:a_{1}=e^2:e:1$.
Suppose $f(a_{1})+f(a_{2})+f(a_{3})=\dfrac{e^5-e^2}{1-e}$; what is the value of $\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$? (where $e=2.718\cdots, \ln{x}=\log_{e}{x}$)
| $\dfrac{e^2+1}{e}\quad $ B:$\dfrac{e^2+3}{e+1}\quad$ C :$\dfrac{e^2+5}{e+2}\quad$ D :$\dfrac{e^3+e+2}{e^2+1}$
this problem have nice solution? Thank you
my idea:$a_{3}=e^2*a_{1},a_{2}=e*a_{1}$
since $f(a_{1})+f(a_{2})+f(a_{3})=3\ln{a_{1}}+3-\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+9=\dfrac{e^5-e^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then w... | Squeeze principle can also help. Note that $$\lim_{n \to \infty}\frac{ 1\cdot 1 + 2\cdot 2 + 3 \cdot 3 + \cdots + n \cdot n}{n^{3}} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}}\leq \lim_{n\to\infty}\: \frac{2^{2}+3^{2}+4^{2}+\cdots +(n+1)^{2}}{n^{3}}$$
$$\Longrightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
| Have you tried dividing the original numerator and denominator each by $x^2 = \sqrt{x^4} = \sqrt[3]{x^6}\,$? This is a good example where algebraic manipulations are easier to use than is using L'Hopital.
$$
\lim_{x \to +\infty}\frac{\sqrt{x^4+1}}{\sqrt[\large 3]{x^6+1}}\cdot \frac {1/\sqrt{x^4}}{1/\sqrt[3]{x^6}} =\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Irrational equation, How to solve it? The equation
$$\sqrt[3]{x^2-1} + x = \sqrt{x^3-2}$$ has a solution $x = 3.$ How to solve this eqution?
| You have to rewrite this as
$$
\sqrt[3]{x^2-1}= \sqrt{x^3-2}-x
$$
and take the 3rd power of both members obtaining
$$
x^2-1=(x^3-2)\sqrt{x^3-2}-3x(x^3-2)+3x^2\sqrt{x^3-2}-x^3.
$$
Now, you can isolate the square root obtaining
$$
3x^4+x^3+x^2-6x-1=(x^3+3x^2-2)\sqrt{x^3-2}.
$$
I think from here you can go on by yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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I need to find the value of $a,b \in \mathbb R$ such that the given limit is true I am given that $\lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx=1$ need to find the value of $a,b \in \mathbb R$ such that the given limit is true. I was able to work the whole thing out, but I have a question about one step in my work. Ther... | Your assumption is correct; if $b\neq 2$, then $8-b^3\neq 0$, and hence the limit would either not exist or be infinite. But you know the limit is $1$.
A shorter way to do the first part is: $\sqrt[3]{8x^3+ax^2}-bx=x(\sqrt[3]{8+\frac{a}{x}}-b)$. The cube root approaches 2 as $x\rightarrow \infty$, so if $b\neq 2$, th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the limit of function - irrational function How can I find the following limit:
$$ \lim_{x \rightarrow -1 }\left(\frac{1+\sqrt[5]{x}}{1+\sqrt[7]{x}}\right)$$
| A useful principle when trying to find a limit is
"always expand around zero".
Since we want to see what happens
when $x \to -1$,
let $x = y-1$ so that
we can look at what happens when $y \to 0$.
The expression becomes
$\frac{1+\sqrt[5]{y-1}}{1+\sqrt[7]{y-1}}$.
For an odd value of $n$,
since $y-1 < 0$ (for small $y$),
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Finding coefficient of generating functions I have the equation
$$(1+x+x^2+\ldots+x^k+\ldots)(1+x^2+x^4+\ldots+x^{2k}+\ldots)(x^2+x^3)$$
how of I find the coefficent of $x^{24}$. I know to condense this down to
$$\frac1{1-x}\cdot\frac1{1-x^2}\cdot x(1+x)$$
but I don't know what to do after that
| Here's another way: the coefficient of $x^{24}$ is the number of integer solutions to $$a+b+c=24$$ where $a\ge0$; $b\ge0$ is even; and $c$ is $2$ or $3$.
If $c=2$, then we have $a+b=22$. There are $12$ values of $b$ (namely, $0,2,\dots,22$), and for each, a value of $a$. So, $12$ solutions so far.
If $c=3$, then $a+b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this equation by Fourier series? $$ y''+3y=\sin ^4 x ,\quad y=\frac{1}{8} +\frac{\cos2x}{2}-\frac{\cos4x}{104}.$$
Now the text book states the solution, but I don't know the process of solving this equation. I need your help!
| The first thing you should do is express $\sin^4{x}$ in terms of cosines:
$$\begin{align}\sin^4{x} &= \left ( \frac{1-\cos{2 x}}{2} \right )^2\\ &= \frac{1}{4} - \frac{1}{2} \cos{2 x} + \frac{1}{4} \cos^2{2 x}\\ &= \frac{1}{4} - \frac{1}{2} \cos{2 x} + \frac{1}{4} \left (\frac{1+\cos{4 x}}{2} \right )\\ &= \frac38 - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:
$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$
| Use the AM-GM inequality: For nonnegative $x,y$, we have $\displaystyle\sqrt{xy}\le\frac{x+y}2$. This follows immediately from expanding and rearranging the obvious inequality $(\sqrt x-\sqrt y)^2\ge0$.
Here, we have $$\begin{array}{rl} a\sqrt{1-b^2}+b\sqrt{1-a^2}&\le|a|\sqrt{1-b^2}+|b|\sqrt{1-a^2}\\&\le\frac{a^2+(1-b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Finding equations of lines I'm sitting an exam soon and have been revising for a while but I sometimes get stuck with questions that I know are simple but still have no idea about how to do them at all! Please explain to me how this is done. (Note: this is all the info I have about the question):
Let $L$ be the line wi... | a) The line $y_L=-x$ has a slope of $-1$, so a perpendicular line must have a slope of $-\frac{-1}{1}=1$; hence $y_P=x+C$, where $b=y_P(a)=a+C$ implies $C=b-a$. The lines intersect at $y_L(x_I)=-x_I=x_I+b-a=y_P(x_I)$; that is, at $x_I=\frac{a-b}{2}$. Thus the distance between the point $P=(a,b)$ and the line $L$ is giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fractional Trigonometric Integrands
*
*$$∫\frac{a\sin x+b\cos x+c}{d\sin x+e\cos x+f}dx$$
*$$∫\frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx$$
*$$∫\frac{dx}{a\sin x+\cos x}$$
What are the relations between the numerator in the denominator, and what is the general pattern to solve these type of questions?
| You can use the following way to calculate all of them once. Let
$$ A=\int \frac{\sin xdx}{d\sin x+e\cos x+f}, B=\int \frac{\cos xdx}{d\sin x+e\cos x+f}, C=\int \frac{dx}{d\sin x+e\cos x+f}. $$
It is easy to check
$$ dA+eB+fC=1, dB-eB=\ln|d\sin x+e\cos x+f|+Const. $$
So
$$ A=\frac{d(1-fC)}{d^2+e^2}-\frac{e\ln|d\sin x+e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$
My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This impl... | Two matrices $A$ and $B$ are similar if there exists a matrix $P$ such that $A=PBP^{-1}$.
The solutions to your equation are $x=2,3$. Thus, all matrices which satisfy your equation must be similar to $B=\begin{bmatrix}v_1&0\\0&v_2\end{bmatrix}$, where $v_1$ and $v_2$ are either $2$ or $3$.
Choosing $P=\begin{bmatrix}a&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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How to find the limit of these sequences? Let $\{a_n\}$ be a real-valued sequence such that $a_1 \geq 0$ and $$a_{n+1}=\ln(a_{n}+1)$$ for all $n\ge1$. How can we find the following limits?
$$\lim_{n\to \infty}na_n=?,$$ $$\lim_{n\to \infty}\frac{n(na_n-2)}{\ln n}=?$$
Thanks in advance.
| If you use Stolz–Cesàro theorem, it is much easier to get the limit. Note that $\lim_{n\to\infty}a_n=0$. By using Stolz–Cesàro theorem,
\begin{eqnarray*}
\lim_{n\to\infty}na_n&=&\lim_{n\to\infty}\frac{n}{\frac{1}{a_n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{a_{n+1}}-\frac{1}{a_n}}=\lim_{n\to\infty}\frac{a_na_{n+1}}{a_n-a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Inequality with mathematical expectations Let a random variable $X \ge 0.$ How to prove the inequality $EX^4EX^8 \le EX^3EX^9$?
| Let $p=\frac{6}{5}$ and $q=6$. Note that $\frac{1}{p}+\frac{1}{q}=1$, $X^4=X^{\frac{5}{2}}\cdot X^{\frac{3}{2}}$, $X^8=X^{\frac{1}{2}}\cdot X^{\frac{15}{2}}$, $\frac{5p}{2}=\frac{q}{2}=3$ and $\frac{3q}{2}=\frac{15p}{2}=9$. Therefore, by Hölder's inequality,
$$E X^4\le \left(E(X^{\frac{5}{2}})^p\right)^{\frac{1}{p}}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-... | There's a clever approach to this problem. In particular, $$(w-1)^2 \geq 0 \implies w^2 + 1 \geq 2w \implies w + \frac{1}{w} \geq 2$$
and this works for any $w$. Using this, you can show that one side is always at least $2$. Meanwhile, the other is at most $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How does one get the Bernoulli numbers via the generating function? Here is the definition:
Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$
I've tried to naively expand $\frac{x}{e^x-1}$ around $x_0=0$ and didn't quite understand what should be done w... | Expand $e^{x}$ around $x=0$ and then reexpand
\begin{align}\frac{1}{1+\frac{x}{2!}+\frac{x^2}{3!}+\ldots}=1-\left(\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\ldots\right)+\left(\frac{x}{2!}+\frac{x^2}{3!}+\ldots\right)^2-\left(\frac{x}{2!}+\ldots\right)^3+\ldots=\\
=1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\ldo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})... | Here because of a duplicate. Though it does seem like the original asker does pop in from time to time.
I'm not sure if the duplicate asker is aware of algebraic integers such as $$\theta = \frac{1}{2} + \frac{\sqrt{17}}{2},$$ which is a solution to $x^2 - x - 4$, though the duplicate asker is aware of the Minkowski bo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $ab=cd$ then $a+b+c+d $ is composite Let $a,b,c,d$ be natural numbers with $ab=cd$.
Prove that $a+b+c+d$ is composite.
I have my own solution for this (As posted) and i want to see if there is any other good proofs.
| From $ab=cd$, We may assume $a=\frac{cd}{b}$. So $M=a+b+c+d = \frac{cd}{b}+b+c+d = \frac{(b+c)(b+d)}{b}$ and so $bM=(b+c)(b+d)$ and $M|(b+c)(b+d)$. We assume that $M$ is not composite, so it is prime. Now we may know that either $b+c$ or $b+d$ is divisible by $M$. So $M\leq b+c$ or $M\leq b+d$ which both result in cont... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383394",
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"answer_count": 8,
"answer_id": 0
} |
Evaluating this integral : $ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $ The question :
$$ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $$
I tried dividing by $\cos^2 x$ and splitting the fraction.
That turned out to be complicated(Atleast for me!)
How do I proceed now?
| Let $\displaystyle I = \int\frac{1-7\cos^2 x}{\sin^7 x\cos^2 x} = \int\frac{\sin^2 x-6\cos^2 x}{\sin^7 x\cos^2 x}dx$
$\displaystyle I = \int\frac{\sin^7 x-6\sin^5 x\cos^2 x}{\sin^{12}x\cos^2 x}dx = -\int \bigg[\frac{1}{(\sin^ 6 x\cos x)^2}\bigg]'dx = -\frac{1}{\sin^6 x\cos x}+\mathcal{C} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Please correct my work, finding Eigenvector Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$.
$$A=
\begin{bmatrix}
+3 & +2\\
-2 & -3\\
\end{bmatrix}
$$
$$A-\ell I =
\begin{bmatrix}
+3-\ell & +2\\
... | The problem is that you rounded. The 0.009 in the second row is in fact a 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How find all this positive of $n$? Let $n$ be a number of the form $a^2+b^2$ with $a,b\in N^{+},(a,b)=1$, such that
every prime $P \lt \sqrt{n}$ satisfies $P|ab$. Find all such positive integers $n$?
I find $n=5$, and $n=13$,are there any others?
when $n=5$,then $a=2,b=1$, and $P=2<\sqrt{5}$
when $n=13$,then $a=3,b=2$... | First note that $\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \mid ab$. Thus $n=a^2+b^2 \geq 2ab \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p}$.
If $n \geq 26$, then $\sqrt{n}>5$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 5, p \text{prime}}{p}=60$. This implies that $\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (s... | The second integral in closed form (by computer algebra) equals
$$ \frac1{\sqrt{2(a_2-a_1)}} F\left(\text{asin}(\sqrt{1-a_2/a_1}), \sqrt{\frac{a_1-a_3}{a_1-a_2}}\right), $$
where $a_1$ is the real root and $a_{2,3}$ are the complex roots of $8x^3+x+7=0$, and $F(\phi,k)$ is the incomplete elliptic integral of the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
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$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$ I need help with calculating this integral:
$$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\rig... | This hypergeometric function is not an elementary function, but its inverse is - see Bring radical.
\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac12,\frac34,\frac54;\frac{x}{64}\right)\,dx \\
&=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 1,
"answer_id": 0
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Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $ Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$
Thanks!
| $$\text{Note that, we have }\dfrac{x^4}{x^2+1} = \dfrac{x^4-1}{x^2+1}+\dfrac1{x^2+1} = x^2-1 + \dfrac1{x^2+1}$$
$$\text{Hence, }\int \dfrac{x^4}{x^2+1} dx = \int (x^2-1) + \int \dfrac1{x^2+1} = \dfrac{x^3}3 - x + \arctan(x) + \text{ constant}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve the roots of a cubic polynomial? I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:
"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.
Any help would be appreciated.
| Let the roots be $a,a\cdot r,a\cdot r^2$
Using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$
and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$
On division, $ar=4$ as $a\cdot r\ne0$
$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.
Given the function
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
$5+x^4+y^4$.
First,... | Write $x^4=2x^4-x^4$ and similarly $y^4=2y^4-y^4$
$$(x^2-y^2)^2 \ge 0$$
$$x^4+y^4-2x^2y^2 \ge 0$$
$$2x^4-x^4+2y^4-y^4-2x^2y^2 \ge0$$
$$2x^4+2y^4 \ge x^4+y^4+2x^2y^2$$
$$x^4+y^4 \ge \dfrac{(x^2+y^2)^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$ When I'm trying to find the limit of $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :
$\dfrac{\sqrt{1-\cos x}}{\sin x} = \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)} = \dfrac{\sqr... | Now
$$\frac{\sqrt{1 - \cos(x)}}{\sin(x)} = \frac{\sqrt{2 \sin^2(x/2)}}{\sin(x)}
= \sqrt{2} \frac{ |\sin(x/2)|}{\sin(x)} = \sqrt{2} \frac{|x/2| + O(x^3)}{x + O(x^3)}\\
= \sqrt{2} \frac{\text{sgn}(x)/2 + O(x^2)}{1 + O(x^2)} = \text{sgn}(x)/\sqrt{2} + O(x^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I find all the solutions of $\sin^5x+\cos^3x=1$
Find all the solutions of $$\sin^5x+\cos^3x=1$$
Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.
| Use what you know about the magnitudes of $\sin x$ and $\cos x$.
$$
\begin{aligned}
\sin^5x+\cos^3x &\le |\sin^5x+\cos^3x| \\
&\le |\sin^5 x| + |\cos^3 x| \\
&\le |\sin^2 x| + |\cos^2 x| \\
&= \sin^2 x + \cos^2 x\\
&= 1
\end{aligned}
$$
The inequality where the exponents are changed is only satisfied if the individual ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Finding $y$ value of canonical ellipse. I have an ellipse:
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
This may be a simple question, but my mind plays tricks on me at the moment;
Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?
Hope someone can help me - thanks in advance :... | You can rearrange $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ to $$y^2 = b^2(\frac{x^2}{a^2} - 1)$$
With knowledge of $a,b$ and $x$ you can evaluate $y^2$ and $$y_{1,2} = \pm \sqrt{y^2} = \pm \sqrt{b^2(\frac{x^2}{a^2} - 1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Having trouble using eigenvectors to solve differential equations The question asked to solve $$\frac{dx}{dy} = \begin{pmatrix}
5 & 4 \\
-1 & 1\\
\end{pmatrix}x$$ ,where $$ x = \begin{pmatrix} x_1 \\
x_2 \\ \end{pmatrix}$$
I went ahead an found the determinant of matrix $$ |A - I\lambda| = \lam... | The characteristic polynomial is:
$$|A - \lambda I| = 0 \rightarrow \lambda^2-6 \lambda+9 = 0 \rightarrow \lambda_{1,2} = 3$$
Substituting in the first eigenvalue to find the first eigenvector:
$$[A - \lambda I]v_1 = 0 \rightarrow \begin{pmatrix}
2 & 4 \\-1 & -2\\\end{pmatrix}v_1 = 0$$
After RREF, for the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$ If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$
just give hint to start solution.
| Hint: $(x-3)^3\geq0,(y-5)^2\geq0,(z-4)^2\geq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ... | Here's one way for square roots:
$$(\sqrt{x+y}+\sqrt{x-y})^2 = 2x + 2\sqrt{x^2-y^2}.$$
Perhaps this is easier to evaluate. For cube roots:
$$
(\sqrt[3]{x+y}+\sqrt[3]{x-y})^3 = 2x + 3\sqrt[3]{(x+y)(x^2-y^2)}+3\sqrt[3]{(x-y)(x^2-y^2)}.
$$
Of course, to get the coefficients in all of these I am using the binomial theorem.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
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Find min of $IA + IB + IC +ID$ in tetrahedron $ABCD$ Let the point $I$ in tetrahedron $ABCD$. Find $\min\{IA + IB + IC + ID\}$.
I can't solve this problem, even in the case ABCD regular. Please help
|
I only can solve regular case, here is the solution:
$H$ and $F$ are one the plane $BCD$, $H,F$ is the foot of $A,I$,
$G$ is on $AH$ and $GH=IF=q,IG=FH=p,\angle FHB=\theta,BH=HD=HB=b,AH=h $
$AI=\sqrt{p^2+(h-q)^2},BF^2=p^2+b^2-2pbcos(\theta),FD^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta),FC^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd
If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.
Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Righta... | HINT:
$k^2-j^2+k-j=(k-j)(k+j+1)$
As $(k+j+1)-(k-j)=2j+1$ which is odd, they must be of opposite parity, exactly one of them must be divisible by $2$
Method 2:
If $a,b$ are odd, observe that one of $(a-b),(a+b)$ is divisible by $4,$ the other by $2$
Method 3:
$(2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1\equiv 1\pmod 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$ how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form
$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$
| We have for $x$ with $\cos x \ne 0$
\begin{align*}
\frac 1{\tan x + \sec x} &= \frac 1{\frac{\sin x}{\cos x} + \frac 1{\cos x}}\\
&= \frac{\cos x}{1 + \sin x}
\end{align*}
And hence
\[ \lim_{x \to \frac \pi 2} f(x) = \lim_{x\to\frac\pi 2} \frac{\cos x}{1 + \sin x} = \frac{\cos \frac \pi 2}{1 + \sin \frac \pi 2} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Generating functions combinatorical problem In how many ways can you choose $10$ balls, of a pile of balls containing $10$ identical blue balls, $5$ identical green balls and $5$ identical red balls?
My solution (not sure if correct, would like to have input):
Define generated function:
$$\begin{align}
A(x) & =(x^0+x^... | Here’s one elementary approach. As you see, it confirms your result.
Let $b,g$, and $r$ be the numbers of blue, green, and red balls chosen to make up a set of $10$ balls. You’re looking for the number of solutions in non-negative integers to the equation $$b+g+r=10\;,\tag{1}$$ subject to the condition that $g\le 5$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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A problem on matrices: Find the value of $k$
If $
\begin{bmatrix}
\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
\sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \\
\end{bmatrix}^k
=
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
$, then the least positiv... | I founded out that $$A^k=
\begin{bmatrix}
\cos \frac{k.2 \pi}{7} & -\sin \frac{k.2 \pi}{7} \\
\sin \frac{k.2 \pi}{7} & \cos \frac{k.2 \pi}{7} \\
\end{bmatrix}
$$
using appropriate trigonometric formulae. Now for $A=I$, $k$ should be equal to $7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Algebraic expression in its most simplified form I am trying to simplify the algebraic expression:
$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$
I am having trouble though. My current thoughts are:
$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$... | You did not distribute the term $(x - 3)$ in the denominator when you wrote:
$$\begin{align} & =\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x} \\ \\ & =\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} \end{align}$$
What would be correct is the following denominator:
$$\begin{align} & \quad\color{blue}{(x-3)}[x(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1... | You can replace the $\sqrt{n}$ on the RHS with another $\sqrt{n+1}$. Therefore you have $n\ge2\sqrt{n+1}$, or $n^2\ge4n+4$. $n^2-4n+4\ge8$, or $(n-2)^2\ge8$. The lowest integer solution to this is $5$, so $n\ge5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
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The Laplace operator. What will be the value of $\Delta\left(\frac 1{r^2}\right)$ if $r=|x|=\sqrt{x_1^2+x_2^2+x_3^2}$?
May be this can be determined using Green's formula.
| If we don't wanna use coordinate transform, it can be also done in a "hands-on" way:
$$\Delta u= \sum_{i}\frac{\partial^2 u}{\partial x_i^2},$$
where $$u = \frac{1}{r^2} = \frac{1}{\sum\limits_i x_i^2}= \frac{1}{x_1^2 + x_2^2 + x_3^2}.$$
We can compute it term by term:
$$
\frac{\partial u}{\partial x_1} = -\frac{1}{(\s... | {
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Evaluate : $\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$ Evaluate: $$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
First approach :
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4(1-\cos^2x)}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{\cos^2xdx}{4 - 3\cos^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \f... | You multiply and divide by $\sec^{4}(x)$ and see if it's working. The denominator becomes $$\sec^{2}(x) + 4 \tan^{2}(x) \sec^{2}(x)$$ Now use $1+\tan^{2}(x)=\sec^{2}(x)$.
So
*
*In numerator you have $\sec^{2}(x)$
*In denominator you have $\sec^{2}(x) \cdot \left(1+4\tan^{2}(x)\right) = (1+\tan^{2}(x))\cdot(1+4\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Algebra simplification in mathematical induction . I was proving some mathematical induction problems and came through an algebra expression that shows as follows:
$$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$
The final answer is supposed to be:
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I walked through every possible expansion; I com... | When we are given an expression of the form:
$$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$
We should recognize that this is a case of simply adding fractions which can also be referred to as rational expressions.
The first thing I would suggest is to rewrite this as a case of adding fractions:
$$\dfrac{k(k+1)(2k+1)}{6} + \d... | {
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac... | The generalized results for the even and odd cases of $$I_n=\int_0^\infty \frac{\log(1+x^n)}{(1+x^2)^2}dx
$$
are respectively as follows
\begin{align}
I_{2m} =& -\frac{m\pi}4+\frac{m\pi}2\ln2+\pi\sum_{k=1}^{[\frac m2]}\ln \cos\frac{(m-2k+1)\pi}{4m}\\
I_{2m+1} =& -\frac{(2m+1)\pi}8+\frac{(4m+1)\pi}8\ln2+\frac{(-1)^m G}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 2
} |
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't h... | Hint:$f^{-1}(f(x))=x$ where $f(x)=\frac{1-x}{1-2x}$,$(x\ne 1/2)$
Solution:Here $f^{-1}(x)=\frac{1-x}{1-2x}$.We know that $f^{-1}(f(x))=x$ and as $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}=f^{-1}(f(x))$ so it follows that $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}=f^{-1}(f(x))=x$ So we dont need to simplify.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
How to find out X in a trinomial How can I find out what X equals in this?
$$x^2 - 2x - 3 = 117$$
How would I get started? I'm truly stuck.
| A different way if you have not seen the quadratic formula yet.
Recall that $(x-1)^2 = x^2-2x+1$ and so
$$
x^2-2x-3 = \left(x^2-2x+1\right)-4=(x-1)^2-4
$$
and your equation $120=x^2-2x-3$ becomes equivalent to
$$
117=(x-1)^2-4
$$
so $(x-1)^2 = 121 = 11^2$. Therefore, $x-1 = 11$ or $x-1 = -11$.
In the first case, $x = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"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.