Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Minimize Trace for non-symmetric matrix Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which
shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symme... | Yes, the program is convex for any $A$.
Let $X \in S^n$ (the space of $n\times n$ dimensional symmetric matrices). Then,
\begin{align*}
\textrm{trace}(AX) &= \textrm{trace}\left(A\left(\sum_{i=1}^n\nu_i x_i x_i^T\right)\right)\\
&= \textrm{trace}\left(\sum_{i=1}^n\nu_i A x_i x_i^T\right)\\
&= \sum_{i=1}^n\nu_i \textrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Inversions of a Permutation Show that the largest number of inversions of a permutation of $\{1,2,3,4,5,6,7,8\}$ equals $n(n-1)/2$. Determine the unique permutation with $n(n-1)/2$ inversions. Also determine all those permutations with one fewer inversion.
| $\{8,7,6,5,4,3,2,1\}$ has $8(8-1)/2$ inversions.
The number of inversions is one less in the following permutations: $\{7,8,6,5,4,3,2,1\}$,$\{8,6,7,5,4,3,2,1\}$,$\{8,7,5,6,4,3,2,1\}$,$\{8,7,6,4,5,3,2,1\}$,$\{8,7,6,5,3,4,2,1\}$,$\{8,7,6,5,4,2,3,1\}$,$\{8,7,6,5,4,3,1,2\}$. There are no others because we need only one pai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $(A+C)(AD+AD) + AC + C$ using Boolean algebra I have solved the equation like this:
(A + C)(AD + AD) + AC + C
=(A + C)(AD) + (A + C)(AD) + AC + C
=AAD + ACD + AAD + ACD + AC + C
=AAD + AAD + ACD + ACD + AC + C
=AAD + ACD + AC + C
=AAD + ACD + C(A+1)
=AAD + ACD + C
=AD + ACD + C
=AD(1+C) + C
=AD + C
Am I rig... | It looks just fine! Well done!
A bit more simply, $$\begin{align}(A+C)(AD+AD)+AC+C &= (A+C)AD+AC+C\\ &= AAD+CAD+AC+C\\ &= AD+(AD+A+1)C\\ &= AD+C\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How find this inequality$\sqrt{\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)}+1$ let $x,y,z>0$,show that
$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\dfrac{(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y... | As the inequality is homogeneous, we can normalise by $xyz=1$. Then we have:
$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\left(x^2+\frac{1}{x}\right)\left(y^2+\frac{1}{y}\right)\left(z^2+\frac{1}{z}\right)}$$
$$\sqrt{3+\sum_{cyc} \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is there an easy way to factor polynomials with two variables? On a recent precal test, I saw a question involving the following expression:
$$(x+1)^2-y^2$$
Which factored out into:
$$(x+y+1)(x-y+1)$$
This wasn't very hard, considering that it was already written as the difference of two squares. I then thought "this ... | I know this is nothing algorithmical but for simple polynomial $F(x,y)$ e.g. with integer coefficients it is an approach. It basically consists of plotting the zeros $Z(F) = \{(x,y) \in \mathbb R^2 | F(x,y)=0\}$.
If you consider two polynomials $F(x,y)$ and $G(x,y)$ then $Z(F\cdot G) = Z(F) \cup Z(G)$. That means if yo... | {
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"url": "https://math.stackexchange.com/questions/516618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Superposition of Sine and Cosine functions I was wondering this:
Let a and b be real numbers. Is it always possible to find real numbers c, d, and e such that $$a\sin(x)+b\cos(x)=c\cos\left(\frac{x+d}e\right)$$
Why is this the case?
Thanks.
| $$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}} \sin{x} + \dfrac{b}{\sqrt{a^2+b^2}} \cos{x} \right)\\
=\sqrt{a^2+b^2}\left(\sin{\varphi} \sin{x} + \cos{\varphi} \cos{x} \right)=\sqrt{a^2+b^2}\cos{(x-\varphi)},$$ where
$$\sin{\varphi} =\dfrac{a}{\sqrt{a^2+b^2}}, \\
\cos{\varphi} =\dfrac{b}{\sqrt{a^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/517822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many ways are there to distribute 8 teachers to 4 schools where each school must get at least 1 teacher? Additional details: the teachers are considered distinct from one another.
So here is what I thought:
1) Choose four teachers to go to each one of the schools: $\binom{8}{4}\cdot4!$
2) For each of those situatio... | Another way is using exponential generating functions.
The EGF of this problem is:
$\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+
\frac{x^7}{7!}+\frac{x^8}{8!}\right)^4$
to solve the problem we can use
$(e^x-1)^4$
for that expression. This expands out to
$ 1-4 e^x+6 e^{2 x}-4 e^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify $\frac{x^{2}-4x}{x^{3}+2x^{2}-5x-6}$ Simplify $$\frac{x^{2}-4x}{x^{3}+2x^{2}-5x-6}$$
Attempt.
I get $$\frac{-\frac{5}{6}}{x+1}+\frac{\frac{-4}{15}}{x+3}+\frac{\frac{21}{10}}{x-2}$$ but it's wrong.
| $\displaystyle \frac{x^3-4x}{x^3+2x^2-5x-6} = \frac{x\cdot(x-2)\cdot(x+2)}{(x+1)\cdot(x-2)\cdot(x+3)} = \frac{x.(x+2)}{(x+1)\cdot(x+3)}$
Now Using Partial fraction Method::
$\displaystyle \frac{x^2+2x}{(x+1)\cdot(x+3)} = \frac{A}{(x+1)}+\frac{B}{(x+3)}$
$\displaystyle x^2+2x = A(x+3)+B(x+1)$
Now Put $x+1 = 0\Leftrighta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
| Yes indeed, that's the way to go about it. Now, we have $$\begin{align} \lim_{x \to 2}\frac{x-2}{\sqrt{x^2+5}-3} & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{x^2 - 4} \\ \\ & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{(x - 2)(x+2)} \\ \\ & \overset{x\neq 2}{=} \lim_{x \to 2}\dfrac {\sqrt{x^2 + 5} + 3}{x+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove, that $\lim_{x \to \infty} 2^x \sin(\pi/(2^k))$ converges to $\pi$. Prove, that $$\lim_{x \to \infty} 2^x \sin\left(\frac \pi {2^x}\right)=\pi.$$
I've tried calculus to no avail.
I found out that every element of this sequence is the area of a regular rectangle with $2^{k+2}$ sides that's inside a circle whose r... | Using $\sin$'s MacLaurin series,
\begin{align}
\lim_{x\rightarrow\infty} \ 2^x\sin\frac{\pi}{2^x} &= \lim_{x\rightarrow\infty} \ 2^x\left(\frac{\pi}{2^x} - \frac{\pi^3}{3! \cdot 2^{3x}} + \frac{\pi^5}{5! \cdot 2^{5x} } + \ \ldots\right) \\
& = \lim_{x\rightarrow\infty} \ \pi - \frac{\pi^3}{3! \cdot 2^{2x}} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f(3x-1)=9x^2+6x-7$, determine $f(x)$ if $f(3x-1)=9x^2+6x-7$ determine all the $f(x)$ functions.
I tried in this way :
$t=3x-1 \Rightarrow x=(t+1)/3$
$f(t)=9(t+1)^2/9-6((t+1)/3)-7((t+1)/3)\ldots$
but unfortunately I get the original function.
Thanks in advance.
| Using your method, substituting $t = 3x - 1 \iff x = \dfrac {t+1}{3} $, we have $$\begin{align} f(t) & = \dfrac{9(t+1)^2}{9} + \dfrac{6(t+1)}{3} - 7 \\ \\ & = (t+1)^2 +2(t+1) - 7 \\ \\ &= t^2 + 4t - 4 \end{align}$$
Hence, $$f(x) = x^2 + 4x - 4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Is my answer correct? (And what's the name of the rule?)$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!}$ Want to know if I solved this problem correctly:
$$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!} =\lim_{n \to \infty} \frac{1 \cdot 2 \ldots(n-1)n(n+1)(n+2)(n+3) - 1 \cdot ... | You can more easily write the expression (cancelling as much as possible) as $$\frac {n+3}n-\frac 1{n(n+1)(n+2)}=1+\frac 3n-\frac 1{n(n+1)(n+2)}$$
Then the limit is very easy to see.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7 Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7.
I have that such an integer n has n=3 or 7(mod 10) but don't know where to go from there.
Then show that there are infinitely many prime numbers n ... | Every prime number is congruent to $1,3,7$, or $9$ modulo $10$. Here is a a multiplication table modulo $10$
\begin{array}{c|cccc}
\times & 1 & 3 & 7 & 9 \\
\hline
1 & 1 & 3 & 7 & 9 \\
3 & 3 & 9 & 1 & 7 \\
7 & 7 & 1 & 9 & 3 \\
9 & 9 & 7 & 3 & 1 \\
\end{array}
Note that every integer ending in $3$ or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Extended Euclidean Algorithm in $GF(2^8)$? I'm trying to understand how the S-boxes are produced in the AES algorithm. I know it starts by calculating the multiplicative inverse of each polynomial entry in $GF(2^8)$ using the extended euclidean algorithm. However I am having some trouble understanding how to perform th... | Using Rijndael's finite field, the reducing polynomial is $x^8+x^4+x^3+x+1$.
Suppose we want to compute the inverse of $x^5+1$ in this field. We want to solve the equation
$$
a(x^5+1)+b(x^8+x^4+x^3+x+1)=1
$$
I like to use the Euclid-Wallis Algorithm. Since we are dealing with polynomials, I will write things rotated by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529156",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Showing "$30$ divides $n^5-n$ for all $n\in\Bbb N$" using induction
Prove that $(n^5 - n)$ divides by $30$ for every $ n\in N$, using induction only.
How on earth do I do that? Thing is $(n^5 - n)$ can't be opened using any known formula...
| Let $f(m)=m^5-m$
$\displaystyle\implies f(1)=1^5-1=0$
Let $f(m)$ is divisible by $30$ for $m=n$
Now $\displaystyle f(n+1)-f(n)=(n+1)^5-(n+1)-(n^5-n)=5n^4+10n^3+10n^2+5n$
$\displaystyle=5n(n^3-n)+10(n^3-n)+15n^2+15n$
$\displaystyle=5(n+2)n(n+1)(n-1)+15n(n+1)$
Observe that $n(n+1)(n-1),$ being product of $3$ consecutive ... | {
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"url": "https://math.stackexchange.com/questions/529757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $A^2$ where A is matrix, how find A? Problem is simple.
Given $$A^2=\begin{bmatrix}13 & 9 &-9 \\ 0 & 4 & 0 \\ 12 & 12 & -8 \end{bmatrix}$$
How find $A$?
I think a method using eigenvalues and I find them.
But I can't find an actual $A$.
Is it right to use eigenvalues?
| Here is an answer obtained by simple inspection. It is not a generally applicable method, but every step below can be done very easily by hand. I have absolutely no intention to recommend this answer. I just want to see how far simple inspection can reach.
*
*Observe that the second row of $A$ has two off-diagonal z... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the orthogonal trajectories of the following family of curves: $x^3y - xy^3 = \alpha$ The answer is $-6x^2y^2 + x^4 + y^4 = \beta$, but I get $-3x^2y^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta$ and I can't seem to find my mistake. My work:
Differentiaing the given equation, we get
$$3x^2y + x^3\frac{dy}{dx} - y^3 ... | Since this is tagged complex analysis,
let $z = x+iy$ and let $f(x+iy) = x^3y - xy^3$.
Expressed in terms of $z$, this is $f(z) = \frac 1 {16i}\left((z+\bar{z})^3(z - \bar z) + (z + \bar z)(z - \bar z)^3\right) = \frac 1{8i}(z^4 - \bar z^4) = \frac 1 4 Im(z^4) = Im(z^4/4)$.
So the curves $x^3y-xy^3 = A$ are the pullba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the solution to the following non-homogenous recurrence relation: Find the solution to the following non-homogenous recurrence relation:
$a_{n+2} + a_{n+1} - 2a_n = n$ for $a_0 = 1$, $a_1 = -2$
I have found the homogenous part with the characteristic polynomial is
$a_n = C_1(-2)^n + C_2$ but don't know how to find... | One way is to find a difference equation that is solved by the function $n \mapsto n$. If $b_n = n$, we see that $b_{n+1}-b_n = 1$ and so $b_{n+2}-2 b_{n+1} + b_n = 0$, with $b_0 = 0, b_1 = 1$.
Now let $x_n= (a_{n+1}, a_n, b_{n+1}, b_n)^T$, then we are looking for a solution to $x_{n+1} = A x_n$ where $A = \begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Decomposition of an ideal as a product of two ideals How to show $$5\mathbb{Z}[\sqrt[3]{2}] = (5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1).$$
Firstly, I think that I can say that $$(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1)= (25,5(\sqrt[3]{2}+2),5((\sqrt[3]{2})^2+3\sqrt[3]{2}-1),5((\sqrt[3]{2})^2+\s... | anon gave you the elementary answer, but if we want to use more machinery, we can use the Dedekind-Kummer theorem.
Namely, to factor $5\mathcal{O}_K$ (where $K=\mathbb{Q}(\sqrt[3]{2})$) it suffices to factor $x^3-2$ in $\mathbb{F}_5[x]$. But, this factors as $(x+2)(x^2+3x+4)$. Thus, we know that
$$5\mathcal{O}_K=(5,\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice... | $$\sqrt[3]{x}=1+\frac{x-1}{3}-\frac{2(x-1)^2}{9(2!)}+\sum_{n=1}^\infty (-1)^{n-1} \frac{(2+3n)(2+3(n-1))(x-1) ^n)}{3n(n!)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 13,
"answer_id": 11
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Inversion applied on circles I'm studying for my exam and one of the questions I am stuck on is:
Show that under inversion in the unit circle a circle with centre C and radius $S$ inverts into a circle with centre $\frac{C}{C\overline{C} - s^2}$ and radius $\frac{s}{C\overline{C} - s^2}$.
And I know:
$$Z\overline{Z}-... | Inversion is
$$z\mapsto \frac1{\bar z}$$
Your equation for the circle is basically the squared form of
$$ \lvert z - C\rvert = s $$
Now combine them:
\begin{align*}
\left\lvert\frac1{\bar z}-C\right\rvert&=s \\
\left(\frac1{\bar z}-C\right)\left(\frac1{z}-\bar C\right)&=s^2 \\
\frac1{z\bar z}-\frac{C}{z}-\frac{\bar C}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541292",
"timestamp": "2023-03-29T00:00:00",
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Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\c... | A more generalised approach over my earlier post. This is not intended to be an exhaustive proof but an experimental one. Constructive comments are most welcome.
Let
$$\sqrt{y^2-x}-x=Ay^n\qquad \cdots (1)\\
\sqrt{x^2+y}-y=\frac {y^{1-n}}A \qquad \cdots (2)\\$$
such that the original equation $$\left(\sqrt{y^2-x}-x\ri... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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"answer_id": 5
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How prove this inequality $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{3^n}\right)\ge\frac{14}{25}$
show that
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$
My try:
I only prove following not strong inequality:
... | The product corresponds to QPochhammer[1/3,1/3,m] which is a decreasing function. For m going to infinity, its limit is QPochhammer[1/3,1/3] which is approximately 0.5601260779279490 as mentioned by MJD (14/25 = 0.56000).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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How prove this $ab|a^8+b^4+1$
show that: there exsit infinite $(a,b)$ such
$$ab|a^8+b^4+1$$
my try: let $a^8+b^4+1=kab,k\in N^{*}$
and I can't work,Thank you
| We show that for any solution $(a, b), a, b \in \mathbb{Z}^+$, we can get another solution $(a', b'), a', b' \in \mathbb{Z}^+$ such that $a'+b'>a+b$. This will imply that starting from the solution $(1, 1)$, we can get an infinite sequence of distinct solutions $(a_n, b_n)$, with $2=a_0+b_0<a_1+b_1<a_2+b_2< \ldots$.
Co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality in triangle Let $ABC$ be a triangle and $M$ a point on side $BC$. Denote $\alpha=\angle BAM$, $\beta=\angle CAM$. Is the following inequality true?
$$\sin \alpha \cdot (AM-AC)+\sin \beta \cdot (AM-AB) \leq 0.$$
| As I already stated in a comment, your question only makes sense if you assume $0\le\alpha,\beta\le\pi$. As a consequence you get $\sin\alpha\ge0,\sin\beta\ge0$ which will be very useful.
I use $a$ to denote the length $AB$ and $b$ to denote $AC$. Yes, the letters are a bit counter-intuitive with respect to the points ... | {
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Prime factorization, distinct primes Let $n=p^eq^f$ where $p$ and $q$ are distinct primes and $e$ and $f$ are positive integers.
Show that $n$ has $(e + 1)(f + 1)$ distinct factors in $N$, and that the sum of all these factors is
$(p^{e+1} −1)(q^{f+1} −1) / (p − 1)(q − 1)$
(You may assume the uniqueness of prime facto... | I'll do a warm up for you. Warm-ups help us understand the nature of a problem and how all of the relevant parts interact. Therefore when an exercise comes with warm-ups and you want to understand how to do the exercise, it's a good idea to try your hand at the warm-ups. Just for future reference. If you followed the d... | {
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A problem on finding dy/dx If $a+b+c=0$ and $$y=\frac{1}{x^b+x^{-c}+1}+\frac{1}{x^c+x^{-a}+1}+\frac{1}{x^a+x^{-b}+1}$$then $\frac{dy}{dx}$=?
The only way which I can think of solving this is by differentiating each term. However, is there a simpler way?
| $$y=\frac{1}{x^b+x^{a+b}+1}+\frac{1}{x^{-a-b}+x^{-a}+1}+\frac{1}{x^a+x^{-b}+1}$$
$$y=\frac{1}{x^b+x^{a+b}+1}+\frac{x^{a+b}}{1+x^{b}+x^{a+b}}+\frac{x^b}{x^{a+b}+1+x^b}$$
| {
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How to prove this inequality?$\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2$ $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2$$
$n$ is positive integer.
Thank you !
| Because $\frac 1t$ is concave up, its integral is overestimated by the trapezoidal approximation. In this light, we have
$$
\begin{align}
\ln 2 = \int_n^{2n}\frac 1t dt &<
\frac 12 \left( \frac 1{n} + \frac 1{n+1} \right) +
\frac 12 \left( \frac 1{n+1} + \frac 1{n+2} \right) + \cdots+
\frac 12 \left( \frac 1{2n-1} + ... | {
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Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ Is it true that for $x\in[0,2\pi]$ we have
$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$
How can I prove it?
For other intervals what is the value of above series if is convergent?
| Here there is a way using Fourier series. Consider the orthonormal basis of $L^{2}([0,2\pi])$, given by the functions
$$ e_{n}(x) = \frac{1}{\sqrt{2\pi}} e^{-i n x}, \quad n \in \mathbb{Z}. $$
Since for $x \in [0,2\pi]$ the indicator function $\mathbf{1}_{[0,x]}$ is in $L^{2}([0,2\pi])$, it can be expressed in its deve... | {
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"source": "stackexchange",
"question_score": "12",
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What is the maximum value of $4(\sin x)^2 + 3(\cos x)^2$ The question is: What is the maximum value of: $4\sin^2\theta + 3\cos^2\theta$
This is the way I did it:
$4\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3$
The max value of $\sin^2\theta$ is $1$, so the answer must b... | Here's a geometric "solution". With the curve $x(t)= \sqrt{3} \cos t$ and $y(t)=\sqrt{4} \sin t$, $0\leq t \leq 2 \pi$, we get an ellipse with major semiaxis of 2 along the y-axis and minor semiaxis of $\sqrt{3}$ along the x-axis. Then the distance squared to the origin $x(t)^2+y(t)^2=4\sin^2 t + 3 \cos^2 t$ must be ma... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Volume of a cuboid whose diagonal and surface area is known The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:
a) 125
b) 236
c) 361
d) 486
Solution:
$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$
How can i find volume from this information?
| Define
$$\begin{cases}
p_1 = & \ell + b + h\\
p_2 = & \ell^2 + b^2 + h^2\\
p_3 = & \ell^3 + b^3 + h^3
\end{cases}
\quad\text{ and }\quad
\begin{cases}
s_1 = & \ell + b + h\\
s_2 = & \ell b + \ell h + b h\\
s_3 = & \ell b h
\end{cases}
$$
We know $p_1 = s_1 = 19$ and $p_2 = 125$.
Apply $AM \ge GM$ to the three number... | {
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Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$I=\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$$
It also can be represented as
$$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\phi}}\mathrm d\phi$$
| Okay, finally I was able to prove it.
Step 0. Observations. In view of the following identity
$$ \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \, d\theta = 2 \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \right), $$
Vladimir's result suggests that there may exists a general formula connecting
$$ I(r, s) = \int_{0}^{\f... | {
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Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long wa... | Alternatively, take
$$ \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}, $$
and integrate three times with lower limit $0$, giving
$$ \begin{align*}
-\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} \\
x + (1-x)\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} \\
\frac{3}{4}x^2 - \frac{1}{2}x - \frac{1}{2} (1-x)^2 \log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560816",
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"source": "stackexchange",
"question_score": "18",
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Calculating squared reciprocals of arithmetic series Let $n>0$ be an integer. Is it possible to calculate the value of the sum
$$1+\frac1{(1+n)^2}+\frac1{(1+2n)^2}+\ldots$$?
| One can prove Marko Riedel's claim using a simple Taylor series argument. We have
$$
\sum_{k=1}^\infty \frac{1}{(kn+1)^2} = \sum_{k=1}^\infty \frac{1}{(kn)^2} \left(\frac{1}{(1+(kn)^{-1})^2}\right),
$$
as well as the Taylor series
$$
\frac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots,
$$
absolutely convergent for $|x| <... | {
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"source": "stackexchange",
"question_score": "8",
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Prove that if $ac + bc + c^2 < 0$ then equation (usual notation) has two roots We have $a, b, c$ real parameters, $a ≠ 0$.
Prove that $ax^2 + bx + c = 0$ has two different roots ($b^2-4ac > 0$), if $ac + bc + c^2 < 0$
| $ac+bc+c^2<0\Rightarrow -ac>bc+c^2$. Now $\Delta=b^2-4ac>b^2+4bc+4c^2=(b+2c)^2>0\Rightarrow$ there are $2$ distinct real roots.
| {
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Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2 + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS}... | If you want a short proof, there are few shorter than using difference calculus:
$$\sum_{k=1}^nk^2=\sum_1^{n+1}(k^{\underline{2}}+k^{\underline 1})\delta k=\frac{1}{3}k^{\underline{3}}+\frac{1}{2}k^{\underline{2}}\big|_1^{n+1}=\left(\frac{1}{3}(n+1)^{\underline{3}}+\frac{1}{2}(n+1)^{\underline{2}}\right)-(0+0)={n+1\cho... | {
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How do you prove the inequality $x^2+y^2+z^2-y(x+z)>0$? This actually comes out of a matrix multiplication of $x^TAx$, but when you multiply it out, you get the following.
$x^2+y^2+z^2-y(x+z)>0$
I just can't figure out how to actually prove that that inequality holds for all $x,y,z\neq0$. Also, note that this is for a... | Since $(x-y)^2 \geq 0$, it follows easily that $xy \leq \frac{1}{2}(x^2+y^2)$. Similarly, $yz \leq \frac{1}{2}(y^2+z^2)$. That gives you
$$x^2+y^2+z^2-y(x+z) \geq \frac{1}{2}(x^2+z^2). $$
The right-hand-side above is positive unless $x=z=0$. But if $x=z=0$, then
$$ x^2+y^2+z^2-y(x+z)=y^2,$$
and that will be posit... | {
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$a_1=1,a_{n+1}=\frac{n}{a_n}+\frac{a_n}{n}$. Prove that for $n\ge4$, $\lfloor{a_n^2}\rfloor=n$ Define a sequence $\left\lbrace a_{n}\right\rbrace$ by
$\displaystyle{a_{1} = 1\,,\ a_{n + 1} = {n \over a_n} + {a_n \over n}.\quad}$ Prove that for $n \geq 4,\,\,\left\lfloor a_{n}^{2}\right\rfloor=n$
The substitution $b_{n... | First,we use Mathematical induction have following inequality
$$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$
we easy prove this function $$f(x)=\dfrac{x}{n}+\dfrac{n}{x} $$ is decreasing on $(0,n)$
becasue $$f'(x)=-\dfrac{n}{x^2}+\dfrac{1}{n}\le 0$$
since $a_{1}=1,a_{2}=2,a_{3}=2$,so
$$\sqrt{3}\le a_{3}\le\dfr... | {
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Simplyfying the equation
It is given that, $a+b+c=0$ we will have to prove that,
$$\frac{1}{x^b + x^{-c}+1} + \frac{1}{x^a + x^{-b}+1} + \frac{1}{x^c +
x^{-a}+1} =1$$
What I have done is, $a=-(b+c)$ into the equation and also I have tried with multiplying $x^a, x^b x^c$ but it didn't look simple. What should be d... | Keeping the first term as is, I am trying to set the numerator of the rest two terms as parts of the denominator of the first term.
$$\frac1{x^a+x^{-c}+1}=\frac{x^b}{x^{b+a}+1+x^b}=\frac{x^b}{x^{-c}+1+x^b}\text{ as }b+a=-c$$
$$\frac1{x^c + x^{-a}+1}=\frac{x^{-c}}{1+x^{-(c+a)}+x^{-c}}=\frac{x^{-c}}{1+x^b+x^{-c}}\text{ a... | {
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"source": "stackexchange",
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,... | HINT:Use induction and
$$16^{n+1}-1=16(16^n-1)+15$$
| {
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$\lim_{x \to 0} \frac{x^n}{\cos\sin x -\cos x}=l$, find $n$ such that $l$ is non zero finite real Problem: Consider $\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$. The value of $n$ such that $l$ is non zero finite real is:
(a) $3$, (b) $4$, (c) $5$, (d) $6$
Solution: The given limit can be written as $$-\frac{1}{2... | Clearly we have
$\displaystyle \begin{aligned}l &= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{{\displaystyle \sin\left(\dfrac{\sin x + x}{2}\right)\sin\left(\dfrac{\sin x - x}{2}\right)}}\\
&= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{\left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{\sin x - x}{2}\right)}\cdot\dfrac{\dfrac{\s... | {
"language": "en",
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"source": "stackexchange",
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Integral $\int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}dx$ Could you suggest any ideas how to evaluate this integral? Is there a closed-form result?
$$\int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}dx$$
| This integral can be solved by brutal force. I will use Vladimir's suggestion as a starting point:
$$I = \int_{1}^{\infty} \frac{\log(x + \sqrt{x^{2} - 1}) \log(1 + \sqrt{x^{2} + 1})}{x^{2}} \, dx. $$
By the knowledge in trigonometry, we obtain
\begin{align*}
I
&= \int_{1}^{\infty} \frac{\operatorname{arcosh} x \cdot (... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim_{x\to 3}\frac{1}{x+1}=\frac{1}{4}$ using $\epsilon$-$\delta$ definition of a limit First of all, I'd just like to say I'm fairly new to proving limits using $\epsilon-\delta$ definition, so I apologize in advance if I ask a very obvious question, or make an elementary error.
Given $\displaystyle\lim_{x\to ... | To maximize a fraction, we want to minimize its denominator. With this in mind, given any $\epsilon>0$, let $\delta = \min\{1,12\epsilon\}$. Then if $0 < | x-3| <\delta$, notice that:
\begin{align*}
0<|x-3|<\delta \leq 1
&\implies -1 < x-3 < 1 \\
&\implies 3 < x+1 < 5 \\
&\implies 12 < 4(x+1) < 20 \\
&\implies \frac{1... | {
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"url": "https://math.stackexchange.com/questions/578646",
"timestamp": "2023-03-29T00:00:00",
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Solve a Linear Equation Please help to solve this problem.what is the answer of this equation. Is it any famous problem?
If $a,b,c\in\mathbb{C}$ such that
$$a+b\neq0,\quad b+c\neq0,\quad a+c\neq0\\a+b+c=-23,\quad ab+ac+bc=-324$$
Then what is the value of $x$ in the following equation?
$$\frac{x^2+2b^2+c^2}{a+b}+\frac{x... | HINT:
$$\frac{x+2b^2+c^2}{a+b}=\frac{x+a^2+b^2+c^2+b^2-a^2}{a+b}=\frac{x+a^2+b^2+c^2}{a+b}+b-a$$
and $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$$
| {
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What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? We were asked to determine the composition $f \circ f \circ f \circ \cdots \circ f $, $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Has anyone an idea?
| Well, try to find some pattern:
$$\begin{align*}f^2(x)=f(f(x))&=f\left(\frac x{\sqrt{1+x^2}}\right)=\frac{\frac x{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac x{\sqrt{1+2x^2}}\\
f^3(x)=f(f^2(x))&=f\left(\frac x{\sqrt{1+2x^2}}\right)=\frac{\frac x{\sqrt{1+2x^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}=\frac x{\sqrt{1+3x^2}}\l... | {
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"url": "https://math.stackexchange.com/questions/581037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using euclidean algo to find d (RSA encryption) The questions says "let p = 5, q =11, n = 55 tocient(n) = 40. e=7.
Use the Euclidean algo to find the value of d.
This is driving me crazy. Here's what I did:
40=5(7) +5
7= 1(5) + 2
5= 2(2) +1
2 = 2(1) + 0
So the GCD is 1.
Now to get 1 = de +f*tocient(n) to find d.
1 = 5... | Mathematically, it's fine for $d$ to be negative, but because positive powers don't require computing the inverse, you may wish to make it positive. To do so, add and subtract $7 \cdot 40$ from the rightmost side:
$$
\begin{align}
1 &= -17 \cdot 7 + 3 \cdot 40 \\
&= -17 \cdot 7 + \color{red}{40 \cdot 7} + 3 \cdot 40 - ... | {
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"url": "https://math.stackexchange.com/questions/581216",
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Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\in... | $$\text{From }\frac{n+1}{(n+2)!},$$
$$=\frac{n+2-1}{(n+2)!}=\frac1{(n+1)!}-\frac1{(n+2)!}$$
Can you identify the Telescoping series?
The survivor will be $$\frac1{2!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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$a^2+b^2+c^2 \geq 3 abc$ I have to solve this Prove that
$a^2+b^2+c^2 \geq 3 abc$
given that
$a+b+c \geq 3 abc$
I am stuck!
|
If $a,b,c\in \Bbb R$, and if $a+b+c\ge 3abc$, then: $$a^2+b^2+c^2\ge 3abc$$
Proof:
*
*Case 1:
If $abc\le 1$, then we use AM-GM inequality, and have:
$$a^2+b^2+c^2=|a|^2+|b|^2+|c|^2\ge 3\sqrt[3]{|abc|^2}\ge 3abc$$
this is true because
$$\Longleftrightarrow |abc|^2\ge (abc)^3$$
*Case 2:
If $abc\ge 1$, since: $$a+b+... | {
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How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$ I'm having a bit of trouble solving this integral: $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$
Here is my attempt at a solution:
I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}d... | Integrate $\int \frac{\sqrt{1-x}}{\sqrt x} dx$
let
$$
\begin{equation} \tag 1
x=t^2, \,\,\,\,\, dx=2t\, dt
\end{equation}
$$
now; $\int \frac{\sqrt{1-t^2}}{t} dt$
$$
\begin{equation} \tag{as $dt=\frac{dx}{2t}$}
\int 2 \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 2
2 \int \sqrt{1 - t^2} dt
\end{equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Cofactor expansion method for finding the determinant of a matrix Use the determinant properties to simplify the given matrix and show that $\det(A) = (x -
y)(x - z)(x - w)(y - z)(y - w)(z - w)$ for
$$A =
\begin{pmatrix}
1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3\\
1 & w & w^2 & w^3 \\
\end{pmatrix}
$$... | Here is a non-clever approach that is still better than brute force. Subtract the bottom row from each of the preceding rows. This does not change the determinant. The result is
$$\begin{pmatrix}
0 & x-w & x^2-w^2 & x^3-w^3 \\
0 & y-w & y^2-w^2 & y^3-w^3 \\
0 & z-w & z^2-w^2 & z^3-w^3 \\
1 & w & w^2 & w^3
\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for c... | You can solve this problem in a single swoop. Since $\alpha$ and $\beta$ are the roots of the polynomial $x^2+x-2$ therefore we must have $\alpha$=-2 and $\beta$=1(or vice versa). Hence
$$\left(\frac1\alpha-\frac1\beta\right)^2=(-\frac12-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving by induction that $\sum_{i=2}^n(i^2-i) = \frac{n(n^2-1)}{3}$ for all $n \ge 2$ Doing proof by induction exercises, everything was fine until I tried to do one with $\sum$
:
Prove that
$$\sum_{i=2}^n(i^2-i) = \frac{n(n^2-1)}{3}$$
holds for all $n \ge 2$.
Now, my real problem is converting an algebraic expressi... | MY approach:
first show that $\sum^n i = \frac{n(n+1)}{2} $ and $\sum^n i^2 = \frac{n(n+1)(2n +1 )}{6} $.
To show then, it is much easier the induction. So, therefore
$$ \sum^n (i^2 - i) = \sum^n i^2 - \sum^n i = \frac{n(n+1)(2n +1 )}{6} - \frac{n(n+1)}{2} = $$
$$ = \frac{n(n+1)(2n-1) - 3n(n+1)}{6} = \frac{(n+1)(2n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Writing $\frac{(\sqrt{2}+1)^{2n+1}+(\sqrt{2}-1)^{2n+1}}{2\sqrt{2}}, n\geq2$ as sum of two perfect squares I tried to show that
$$
{\left(\sqrt{2\,} + 1\right)^{2n+1} + \left(\sqrt{2\,} - 1\right)^{2n+1}
\over
2\,\sqrt{2\,}}\,,\qquad n\geq2
$$
is written as the sum of two perfect squares. We used Newton's binomial form... | Let us look at a quadratic whose roots are
$$(\sqrt2 + 1)^2 \text{ and }(\sqrt2 - 1)^2$$
i.e.,
$$3+2\sqrt2 \text{ and }3 - 2\sqrt2$$
The quadratic is $x^2-6x+1$. Now let us look at the recurrence
$$a_{n+1} = 6a_n - a_{n-1}$$
where $a_0 = 1$ and $a_1 = 5$. This gives the sequence you are after. Now note that
$$a_0 = 0^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the integral $\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$ As suggested in the title, I am given the following integral:
$$\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$$
I have tried to solve it several times, but I wind up with the wrong answer, although its close. I assume there is a blunder somewhere.
Applying lon... | You're off at the start, with your long division: in particular, the numerator of the remainder needs to be $5x - 1$.
$$\frac{2x^3+x^2+0\cdot x +1}{x^2+x-2} = 2x - 1 + \frac{5x - 1}{x^2+x-2} = 2x - 1 + \frac{5x-1}{(x+2)(x-1)}$$
Procedurally, your approach is just fine: once you correct for this mistake, you should obt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Generating function for division of $n$ into smaller subsets. I need to find the generating function for the number of ways of dividing $n$ into parts out of even numbers. The ways are only different if their parts are different, meaning that $2+1+2$ and $2+2+1$ are the same, because only their arrangmenent is differen... | If the parts can be repeated, this generating function can be written as
\begin{align}
F(x) & = (1 + x^{2} + x^{2+2} + \ldots)\cdot (1 + x^{4}+x^{4+4}+\ldots)\cdot (1 + x^{6}+x^{6+6}+\ldots)\cdot \ldots\\[0.7em]
& = \dfrac{1}{1-x^{2}}\cdot \dfrac{1}{1-x^{4}} \cdot \dfrac{1}{1-x^{6}} \cdot \ldots\\[0.9em]
& = \prod_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can Master Theorem be applied on any of these? 1) $T(n) = 6T(n/2) + 2^{3 \log(n)}$
2) $T(n) = 8T(n/2) + \frac{n^3}{(\log(n))^4}$
3) $T(n) = 9T(n/3) + n(\log(n))^3$
Can the complexity for these be calculated with the Master Theorem?
I am not sure how to decide upon which case they fit in.
| For problem three, lets solve the recurrence
$$T(n) = 9 T(\lfloor n/3\rfloor) + n \times (1+\lfloor\log_3 n\rfloor)^3$$
where we set $T(0)=0.$
Let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor\log_3 n\rfloor} d_k 3^k.$$
Then we get the exact formula
$$T(n) = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 9^j \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{... | How about: substitute $x=u^2.$ Then (dropping absolute values for now), you get
$$\int \frac{2 u}{1+u+ \sqrt{u^2+1}} du$$
Now, substitute $u=\tan \theta,$ to get
$$2 \int \frac{\tan \theta \sec^2 \theta}{1+\tan \theta + \sec \theta} d\theta =
\int \frac{\sin \theta}{\cos^2\theta (\sin \theta + \cos \theta + 1)} d\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Is $\lim_{n \to \infty}{\left(\frac{3}{1+\frac{1}{n}}\right)^n} = \lim_{n \to \infty}{3^n}$? I started out with the following series:
$\sum_{n=1}^{\infty}{\left(\frac{3}{n}\right)^nn!}$
The goal is to find out if the sequence converges or diverges. It went quite well, hopefully, until I got stuck here:
$\lim_{n \to \in... | Using the Ratio Test (expounding):
$$\frac{a_{n+1}}{a_n}= \frac{3^{n+1} \cdot (n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n \cdot n!} = \frac{3^n \cdot 3 \cdot(n+1)n! \cdot n^n}{(n+1)^n \cdot (n+1)^1 \cdot 3^n \cdot n!} = \frac{3 \cdot n^n}{(n+1)^n} = 3 \cdot \left(\frac{n}{n+1} \right)^n$$ $$= 3 \cdot \left(\frac{1}{1+1/n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How many integers between $100$ and $500$ are divisibly by both $6$ and $12$, but not by $9$? The numbers divisible by both $6$ and $12$ and just the numbers divisible by $12$, so:
$12 \times 9 = 108$
$12 \times 41 = 492$
$\frac {492-108}{12} = 32 + 1 = 33$ numbers divisible by $12$, and in turn div. by $6$
$\frac {495... | The number of integers divisible by $12$ between $[100,500]$ is $$N_1=\left\lfloor\frac{500}{12}\right\rfloor-\left\lfloor\frac{100-1}{12}\right\rfloor$$
As lcm$(12,9)=18,$ among these $N_1$ numbers, the number of integers that are also divisible by $9$ between $[100,500]$ will be $$N_2=\left\lfloor\frac{500}{18}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/603702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find value of integral: $I=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$ Find value of integral: $$I_1=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$$ and $$I_2=\int_0^{2\pi}\frac{dx}{(2+\sin x)^2}$$
I don't know how, i need a solution, please
| We can also use contour integration.
Let $z=e^{ix}$, then $\mathrm{d}x=-i\,\mathrm{d}z/z$ and $\cos(x)=\frac{z+1/z}{2}$
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2}
&=\oint\frac{-i\,\mathrm{d}{z}/z}{\left(2+\frac{z+1/z}{2}\right)^2}\\
&=\oint\frac{-4iz\,\mathrm{d}{z}}{(z^2+4z+1)^2}\tag{1}
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$2^5 \cdot a^b=2,5ab$ I came across this problem in an elementary number theory book, and I think I solved it. Well, the question is posed as
$2^5 \cdot 9^2 = 2,592$. Are there any other pairs $a,b \in \mathbb{Z}$ such that $2^5 \cdot a^b = 2,5ab$ is satisfied ($a,b$ are the tens and one representative, not a produc... | There is a 'simple' way of doing this. Notice that $2^5\cdot a^b=25ab$ are both integers. Therefore, since $a^b$ is an integer, so too must be
$$
a^b=\frac{25ab}{2^5}=\frac{25ab}{32}
$$
Now this can be done by hand very simply. We know that $2500<a^b<2600$. How many multiples of $32$ fit this inequality? This is simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/606584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $\int \frac{1}{(x+a)(x+b)}$ Evaluate:
$$\int \frac{1}{(x+a)(x+b)}$$
My attempt:
$$\int \frac{1}{(x+a)(x+b)} = \frac{A}{x+a} + \frac{B}{x+b}$$
$$1 = A(x+b) + B(x+a)$$
$$x = -b$$
$$1 = A(-b + b) + B(-b + a)$$
$$1 = B(-b + a)$$
$$x = -a$$
$$1 = A(-a + b) + B(-a + a)$$
$$1 = A(-a + b)$$
At this point I have no id... | $$\displaystyle {1\over (x+a)(x+b)}={1\over (b-a)}\left[{1\over x+a}-{1\over x+b}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Conic matrix and diagonalization If I have the conic $C$:
$$
5x^2 - 4xy + 8y^2 = 36
$$
How would I express it as a matrix of the form:
$$
\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}a&b/2\\b/2&c\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = k
$$
Also, how to find a unitary matrix $P$ such that
$$
P^* \begin{bmatri... | Let
$$
A\equiv\left[\begin{array}{cc}
a & \frac{1}{2}b\\
\frac{1}{2}b & c
\end{array}\right]
$$
so that
\begin{align*}
\left[\begin{array}{cc}
x & y\end{array}\right]A\left[\begin{array}{c}
x\\
y
\end{array}\right] & =\left[\begin{array}{cc}
x & y\end{array}\right]\left[\begin{array}{cc}
a & \frac{1}{2}b\\
\frac{1}{2}b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/608814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a weird recurrence relation using exponential generating functions and differential equations So the recurrence relation is:
$a(n) = (n-1)\cdot a(n-2) + (n-1)(n-2)\cdot a(n-3)$
I've tried several things, but the instructions are to use a differential equation and nothing I'm doing seems to work. Here's what I h... | Let $A(x)$ be the exponential generating function. For convenience I’ll write $a_n$ instead of $a(n)$.
$$\begin{align*}
A(x)&=\sum_{n\ge 0}a_n\frac{x^n}{n!}\\\\
&=\sum_{n\ge 0}(n-1)a_{n-2}\frac{x^n}{n!}+\sum_{n\ge 0}(n-1)(n-2)a_{n-3}\frac{x^n}{n!}\\\\
&=x^2\sum_{n\ge 0}\frac{(n-1)a_{n-2}}{n(n-1)}\cdot\frac{x^{n-2}}{(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}$ How can we prove that?
$$\sum_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}=\frac{1}{3}e^{\frac{-x}{2}}x\left(e^{\frac{3x}{2}}-2\sin\left(\frac{\pi+3\sqrt{3}x}{6}\right)\right).$$
I think if we write the taylor expansion of $\sin(u)$ and $e^u$, we can arrive from RH... | This is actually calling for Laplace transform.
Set $f(x)=\sum_{n=1}^\infty \frac{x^{3n}}{(3n-1)!}$ then
\begin{align}
F(s) &= \int_0^\infty \frac{f(x)}{x} \, e^{-sx} \, {\rm d}x \\
&= \sum_{n=1}^\infty \frac{1}{s^{3n}} \int_0^\infty \frac{u^{3n-1} \, e^{-u}}{(3n-1)!} \, {\rm d}u \\
&= \frac{1}{s^3-1}
\end{align}
Now
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Taylor series of $(1+x)\ln(1+x)$ in $x=0$ How to determine the Taylor series of $(1+x)\ln(1+x)$ in $x=0$?
My idea is finding the second derivative of the expression, which is $\frac{1}{1+x}$.
The Taylor series of this expression is $1-x+x^2-x^3$ and so on. If I integrate then this series I get $\frac{x^2}{2}-\frac{x^3... | You might want to use the fact that
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2} \left( (1+x) \ln(1+x) \right) =
\frac{\mathrm{d}}{\mathrm{d}x} \left( \ln(1+x) + 1 \right) = \frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n
$$
Integrating term-wise twice, and using initial conditions of 0 for the function, and 1 for the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/612166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n>0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-... | Writing ${\displaystyle {x_i \over 2 - x_i} = -1 + {2 \over 2 - x_i}}$, you have
$$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = -n + \frac{2}{2-x_1} + \frac{2}{2-x_2} +\ldots + \frac{2}{2-x_n}$$
So it suffices to show
$$ \frac{2}{2-x_1} + \frac{2}{2-x_2} +\ldots + \frac{2}{2-x_n} \geq n + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluating an Infinite Product Does anyone know how to evaluate the infinite product
$$
\left(1 - \frac{4}{1}\right) \prod_{k = 3}^{\infty} \left( 1 - \frac{4}{k^2} \right)
$$
| For $k \ge 3$:
$$\prod_{k=3}^{\infty} \left ( 1-\frac{4}{k^2}\right )$$
This is equal to
$$\frac{1\cdot 5}{3 \cdot 3} \frac{2\cdot 6}{4 \cdot 4} \frac{3\cdot 7}{5 \cdot 5} \cdots$$
With cancellations: note that only $1\cdot 2$ survives in the numerator, and a single $3 \cdot 4$ survives in the denominator. Thus, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Asymptotics of probability of Newton-Pepys problem In Newton-Pepys problem one is interested in probability $p_n$ of getting at least $n$ sixes in $6 n$ independent throws of regular 6-sided die.
The number of sixes $S_m$ obtained in $m$ throws follows a Binomial distribution $\operatorname{Bin}(m,p)$, where $p=\frac... | The probability $p_n$ has a representation in terms of regularized beta-function, which incidentally is the cumulative distribution function of beta distribution.
$$
p_n = \Pr(S_{6n} \geqslant n) = I_{p}\left(n, 5n+1\right) = \Pr(X < p ) = \Pr\left(X < \frac{1}{6} \right))
$$
where $X$ follows $\mathcal{Be}\left(n,5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry equation $\sin(x)+\cos(x)-\tan(x)=0.4$ There's some way to find $x$ here ?
$$\sin(x)+\cos(x)-\tan(x)=0.4$$
| This is a difficult problem to solve. If you write $\tan x$ as the quotient of $\sin x$ and $\cos x$ you get
$$\sin x + \cos x + \frac{\sin x}{\cos x} = \frac{2}{5}$$
It is possible to rearrange this formula, to get $\sin x$ as the subject:
$$\sin x = \frac{(2-5\cos x)\cos x}{5+5\cos x}$$
Squaring both sides, and using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
integer $a$ for which $x^3-3x+a = 0$ has $3$ distinct real roots in $x\in (0,1)$ If the equation $x^3-3x+a=0$ has $3$ distinct roots between $0$ and $1$ then the integer value of $a$ is
Can we solve it without Calculus.
$\bf{My\; Try}::$ Let $x=2\cos \phi$, where $-1\leq \cos \phi\leq 1\Rightarrow 0\leq \phi \leq \pi$(... | HINT:
If two roots are same we can write $$x^3-3x+a=(x-c)(x-b)^2$$
Compare the coefficients of the different powers of $x$ to determine $a,b,c,d$
If three roots are same we can write $$x^3-3x+a=(x-d)^3$$
Comparing the coefficients of $x^2, d=0\implies a=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Which is wrong in the proof? Consider the following steps:
\begin{align}
a &= x \\
a+a &= a+x && \text{[add }a\text{ to both sides]}\\
2a &= a+x && \text{[}a+a = 2a\text{]}\\
2a-2x &= a+x-2x && \text{[subtract }2x\text{ from both sides]}\\
2(a-x) &= a+x-2x && \text{[}2a-2x = 2(a-x)\text{]}\\
2(a-x) &... | $$\begin{align}
a & = x \\
a + a & = a + x \\
2a & = a+x \\
2a - 2x & = a + x - 2x\\
2 (a -x) &= a-x
\end{align}$$
All is OK so far but in the next step you are trying to divide $0$ by $0$ which is not allowed $\frac{0}{0}$ is undefined.
Note: $a = x$ so $a - x = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the minimum of a function of two variables
Find the smallest value of $\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}\quad \quad$ for $a,b$ real and positive.
What I've done so far:
Let $F(a,b)=\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab... | Hint: Consider vectors $X, A, B, Y$ where $|X| = 7$, $|A| = a$, $|B| = b$, $|Y| = \sqrt{50}$ and $\angle XOA = 45^\circ$, $\angle AOB = 45^\circ$, $\angle BOY = 45^\circ $.
How does your expression relate to these vectors? Think of distances. Apply Cosine rule.
Hence, the minimum is $ \sqrt{7^2 + (\sqrt{50})^2 - 2 \tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find five consecutive odd integers such that their sum is $55$. So my professor asked us to do an Olympiad exercise which says that the sum of five consecutive odd integers is $55$, find those integers. But I've never seen such an exercise so it is quite new and if I learn how then I will be able to solve similar exerc... | Hint: Choose one of those equations and solve it:
$$\begin{align}
& \text{The first integer:} & \qquad x+\color{blue}{x+2+x+4+x+6+x+8}=55 \\\,\\
& \text{The second integer:} & \qquad \color{red}{x-2}+x+\color{blue}{x+2+x+4+x+6}=55 \\\,\\
& \text{The third integer:} & \qquad \color{red}{x-4+x-2}+x+\color{blue}{x+2+x+4}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 10,
"answer_id": 8
} |
Finding integer solutions for $6x+15y+20z=1$ Problem: Find integers $x$, $y$, and $z$ that satisfy the equation $6x+15y+20z=1$.
I noticed that $\gcd(6,15)=3$, $\gcd(15,20)=5$, and that $\gcd(6,20)=2$. And of course $\gcd(6,15,2)=1$.
Of course I know to set one of the variables (x, y, or z) to be zero. But I want to kn... | Hint: Using Euclidean Algorithm you can find the gcd of the 3 numbers. Try using Euclidean Algorithm in the inverse direction some way.
gcd(a, b, c) = gcd(gcd(a, b), c)
$15=6 \times 2+3$ so $3= 1 \times 15-6 \times 2$ (+)
$20=6 \times 3+2$
$3=2 \times 1+1$
$1=3-2=3-20+6 \times 3=-20+7 \times 3=-20+7 \times 15-14 \times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Real integral by keyhole contour Evaluate
$$\int_0^\infty \frac{\log x \; dx}{x^{2} + 2x + 2}$$
by integrating a branch of $(\log z)^{2}/(z^{2} + 2z +2)$ along a keyhole contour.
The thing I have trouble with is why I should be examining the square of the log - I guess it has something to do with ln x in fact NOT being... | Here are some details of the calculation. Let the keyhole contour be oriented counterclockwise and have the slot on the positive real axis. Call the segment above the real axis $\Gamma_1$, the large circle of radius $R$ $\Gamma_2$, the segement below the real axis $\Gamma_3$ and the circle of radius $\epsilon$ around ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
$f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$, Then Max $\left\{f(x)-g(x)\right\}$ Let $f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$.
Find the Minimum of function $\left\{f(x)+g(x)\right\}$ and the maximum of function $\le... | If one observes carefully $f(x)$ is nothing but distance of the point $A(1,5)$ from an arbitrary point $P(x,x^2)$ lying on the curve $y=x^2$.
Similarly $g(x)$ is nothing but distance of the point $B(-2,-1)$ from any arbitrary point $P(x,x^2)$ on the curve $y=x^2$.
Now,$AB=3\sqrt{5}$
In other words we have been asked to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $m^{2} + n^{2} = \csc^{2}\left(\theta\right)$ $$
\mbox{If}\quad\left\lbrace%
\begin{array}{rcl}
m^{2} + m'^{2} + 2mm'\cos\left(\theta\right) & = & 1
\\
n^{2} + n'^{2} + 2nn'\cos\left(\theta\right) & = & 1
\\
mn + m'n' + (mn' + m'n)\cos\left(\theta\right) & = & 0
\end{array}\right\rbrace
\quad\mbox{then pr... | For convenience, let $a=m$, $b=m^{\prime}$, $x=n$, $y=n^{\prime}$; so
$a^2+b^2+2ab\cos\theta=1$, $\;\;x^2+y^2+2xy\cos\theta=1$, $\;\;ax+by+(ay+bx)\cos\theta=0$.
From the 3rd equation, $(a+b\cos\theta)x+(b+a\cos\theta)y=0$, so $
y=-\frac{a+b\cos\theta}{b+a\cos\theta}x$.
Substituting into the 2nd equation gives $x^2+\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
| Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$ for positive numbers $x,y,z$. We will show $M\le1$, which is equivalent to$$\sum_{cyc}\frac{x^3z^2}{x^2y^3+yz^4+x^3z^2}\le1$$and after full expansion, it is equivalent to$$\sum_{cyc}x^8 y^3 z^4 + x^7 y^2 z^6\le\sum_{cyc}x^9y^3z^3+x^7y^7z$$
By rearrangement in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How find this limit $\lim_{x\to 0^{+}}\int_{x}^{1}\frac{\ln{(1+t)}}{\sqrt{t}}dt$ Find this limit
$$\lim_{x\to 0^{+}}\left(\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt+\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt\right)$$
My try: since
$$\int_{0}^{x}(\sqrt{y+1}-1)dy=\dfrac{2}{3}(1+x)^{3/2}-x$$
... | Note that you can use integration by parts on $\displaystyle\int_x^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt$ with $u=\ln(1+t)$ and $\,dv=\dfrac{1}{\sqrt{t}}\,dt$ to get
$$\begin{aligned}\int_x^1\frac{\ln(1+t)}{\sqrt{t}}\,dt &= \left.\left[2\sqrt{t}\ln(1+t)\right]\right|_x^1-2\int_x^1 \frac{\sqrt{t}}{t+1}\,dt\\ &= 2\ln 2 - 2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Series of inverses of binomial coefficients Can you think of a simple way of proving that
$$
\sum_{n=k+1}^\infty \frac{1}{n \choose k}
$$
is rational for any $k \geq 2$?
Here's the background. Consider a series:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)}
$$
Elementary algebra gives us that:
$$
\sum_{n=1}^\infty \frac{1}{n(... |
We show the result stated by @IgorRivin is valid.
\begin{align*}
\sum_{n=k+1}^{\infty}\frac{1}{\binom{n}{k}}=\frac{1}{k-1}\qquad\qquad k\geq 2\tag{1}
\end{align*}
We start proving a related identity from which we can easily derive (1).
The following is valid for $N\geq 1,k\geq 1$
\begin{align*}
\sum_{n=1}^{N}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction: $10^n+3\times4^{n+2}+5$ is divisible by $9$ I dealt with this problem but I couldn't resolve.
Prove by induction that $10^n+3\times4^{n+2}+5$ is divisible by $9$ for all non-negative integers $n$.
| Let P(n) be the statement that $10^n+3\times 4^{n+2}+5$ is divisible by $9$. When $n=0$, $10^n+3\times 4^{n+2}+5=54$ is divisible by $9$, so we have $P(0)$.
Suppose we have $P(k)$ for some nonnegative integer $k$. Then $10^k+3\times 4^{k+2}+5$ is divisible by $9$. Now
$(10^{k+1}+3\times 4^{k+3}+5)-(10^k+3\times 4^{k+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding the limit of $\left(\dfrac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n$ I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\m... | By Taylor series we have:
$$\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}=\frac 1 3\left(3+\frac1 n(\log a +\log b+\log c)++o\left(\frac 1 n\right)\right)=1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)$$
so
$$\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\exp\left(n\log\left(1+\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
An inequality with $a,b,c>0$ Let $a,b,c>0$. Prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq\left(1+\sqrt{2}+\sqrt{3}\right)\cdot\left(\frac{1}{a+b\sqrt{2}+c\sqrt{3}}+\frac{1}{b+c\sqrt{2}+a\sqrt{3}}+\frac{1}{c+a\sqrt{2}+b\sqrt{3}}\right)$$
| It will suffice to show that the following (stronger) inequality holds
for positive $a,b,c,u,v,w$ :
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq
(u+v+w)\bigg(\frac{1}{au+bv+cw}+\frac{1}{av+bw+cu}+\frac{1}{aw+bu+cv}\bigg)
$$
The fonction $f : x \mapsto \frac{1}{x}$ is convex on $(0,\infty)$, so
$$
\begin{array}{lcl}
f(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/638833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $n,k\in\mathbb N$, solve $3^k-1=x^n$. If $n,k\in\mathbb N$, solve $$3^k-1=x^n$$
This seems like an interesting problem. I've currently tried a few things (one could try a lot of things in this case):
$1)$ $x^n\equiv -1\pmod 3\Rightarrow x\equiv -1\pmod 3\Rightarrow n=2r+1$, where $r\in\mathbb N\cup\{ 0\}$.
Which bri... | Since $n$ is odd, it has some odd prime divisor $p$, or is $1$, if it is $1$, we have an infinite family of trivial solutions. Since if we care of only prime $n$ we are done (why?), let's suppose $n=p$.
$$3^k=(x+1)(1-x+x^2-x^3\cdots)$$
And since $x\equiv 2 \pmod{3}$, the residues on the right parentheses $\pmod3$ look ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/642624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving convergence of a sequence Let the following recursively defined sequence:
$a_{n+1}=\frac{1}{2} a_n +2,$
$a_1=\dfrac{1}{2}$.
Prove that $a_n$ converges to 4 by subtracting 4 from both sides.
When I do that, I get:
$2(\frac{1}{2} a_{n+1} -2)=(\frac{1}{2} a_n -2)$, so $y=2y$,
which is true only for $0$. But I'm n... | Set $b_n=a_n-4$. Then $b_1=-7/2$ and
$$
b_{n+1}=a_{n+1}-4=\frac{1}{2}a_n+2-4=\frac{1}{2}(a_n-4)=\frac{1}{2}b_n.
$$
Thus
$$
b_n=\frac{b_{n-1}}{2}=\frac{b_{n-2}}{2^2}=\cdots=\frac{b_{1}}{2^{n-1}}=-\frac{7}{2^n},
$$
and finally
$$
a_n=4-\frac{7}{2^n}.
$$
Hence
$$
\lim_{n\to\infty} a_n=4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
find set of points for lots of triangulations I should find a set of $n$ points $Q$ in a plane, so that $t(Q)$ (the number of possible triangulations) the following equation holds:
$$t(Q) \ge 2^{n-2\sqrt{n}}$$
I solved the problem using the vertex points of a convex polygon. Wikipedia says, that $t(Q) = C_{n-2}$ and ... | How complex is too complex?
The central binomial coefficient $\displaystyle \binom{2n}{n}$ is bigger than $\displaystyle \frac{4^n}{2n+1}$, so
$\displaystyle t(Q) = C_{n-2} = \frac{1}{n-1} \binom{2(n-2)}{n-2} \geq \frac{1}{n-1} \frac{4^{n-2}}{2n - 3} = \frac{2^{n+2\sqrt{n} - 4}}{(n-1)(2n-3)} 2^{n-2\sqrt{n}}$
Since $2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot... | Hrm. How can we make $x^3 + y^3$ out of that?
What tools do we have available to us? We have $x+y$. We have $xy$. And we have arithmetic.
I don't know yet how to make $x^3 + y^3$ exactly. Let's focus on one part: let's make $x^3$, and then see what we can do with what's left over.
It's pretty easy to make $x^3$ by mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
Find the value of A, B and C in the identities. $6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Find the value of A, B and C.
I started it like this:
$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Solving the right hand side:
$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$
$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx ... | Hint $\ x=1\,\Rightarrow\,c = 6.\,$ Cancelling $\,x-1\,$ yields $6x^2-5x+1 = (Ax-1)(Bx-1)\ $ so $\,\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/654198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$
This is what I did:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^... | Let $\displaystyle-2=r\cos\phi,1=r\sin\phi$ where $r>0$ so $\displaystyle\frac\pi2<\phi<\pi$ and $\displaystyle\tan\phi=-\frac12$
So we have $$r\cos(2x-\phi)=r\sin\phi$$
$$\implies \cos(2x-\phi)=\cos\left(\frac\pi2-\phi\right)$$
$$\implies 2x-\phi=2n\pi\pm\left(\frac\pi2-\phi\right)$$ where $n$ is any integer
Consideri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Help with the algebra in for this number theory proof? For all $n\geq 1$, prove with mathematical induction
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$
So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for al... | This exercise shows that the sum of the reciprocals of the squares converges to something at most $2$; in fact, the series converges to $\frac{\pi^2}{6}$.
For $n\geq 1$, denote the statement in the exercise by
$$
S(n) : 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} \leq 2 - \frac{1}{n}.
$$
Base step ($n=1$): ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Simplify the surds below. Simplify this:
$10 + 2\sqrt5 - 4\sqrt2\sqrt5 - 4\sqrt2$
Help please! $:'($
It came out like this:
$10 + 2\sqrt5 -4\sqrt10 - 4\sqrt2$
How should I do the next step? Um stuck.
| Note that $4 \sqrt{10} = 4 \sqrt 2\cdot\sqrt 5$, and $10 = 2\sqrt 5\cdot \sqrt 5$.
Hint: FACTOR!
$$\begin{align} 10 + 2\sqrt5 - \color{red}{4\sqrt2}\sqrt5 - \color{red}{4\sqrt2} & = \color{blue}{2\sqrt 5}\cdot \sqrt 5 + \color{blue}{2\sqrt 5} - \color{red}{4\sqrt 2}(\sqrt 5 + 1)\\ \\ & = \color{blue}{2\sqrt 5}\color{pu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\int\limits_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum\limits_{n=1}^\infty \frac{1}{(n+a)^2}$ Prove that $$\int_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum_{n=1}^\infty \frac{1}{(n+a)^2}$$
I know that we have a product of $x^a$, $\displaystyle\sum_{n=0}^\infty x^n$, and $\displaystyle\sum_{n=0}^\infty \frac{(1-x)^n}{n}$, ... | $$
\begin{align}
&\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\left[\left(1-x^a\right)+\left(x-x^{a+1}\right)+\left(x^2-x^{a+2}\right)+\dots\right]\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\left[\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example.
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|... | First let us do it without algebra. Draw a number line. Then $|x+1|+|x-3|$ is the sum of the distances of $x$ from $-1$ and from $3$. It is clear that if $x$ is between $-1$ and $3$, the sum of the distances is $4$. The sum of the distances will be $6$ if we are $1$ unit to the right of $3$, or $1$ unit to the left of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Induction: $\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}$ $$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}, \quad \text{for } x \not \in \{0,-1,-2,\dots,-n\}$$
Can somebody... | Hint. Use that fact that the result is an identity for all $x$ (with the stated exceptions). So, assume the result is true for a specific $n$. Then we can say
$$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}\ ,$$
and also
$$\eqalign{&\frac{n!}{... | {
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How to prove this is a rational number I'm not sure how to prove this is a rational number $\frac{q}{m}$, can some one show me?
$$\frac{q}{m}=\frac{(\frac{1+\sqrt5}{2})^n - (\frac{1-\sqrt5}{2})^n}{\sqrt5}$$
| Lt me consider $$S(n) = 2\frac{ \left((a+b)^n-(a-b)^n\right)}{b}$$The binomial theorem will be applied to each term and the result will be expanded and simplified. So, the results are successively $$S(1) = 4$$ $$S(2) = 8 a$$ $$S(3)= 4\left(3 a^2+b^2\right)$$ $$S(4)= 16 a \left(a^2+b^2\right)$$ $$S(5)=4 \left(5 a^4+10 ... | {
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Closed form integral $\int_b^c \frac{x^2}{\sqrt{(x-a)(x-b)(c-x)(d-x)}} dx$ Is there a closed form expression for the definite integral $$I=\int_b^c \frac{x^2}{\sqrt{(x-a)(x-b)(c-x)(d-x)}} dx$$ for $a<b<c<d$?
Mathematica 9.0 can do it for special cases using elliptic integrals, so I suspect that there may be a more gen... | This integral is listed in Byrd and Friedman
\begin{equation}
I^{(2)}(a,b,c,d)=\frac{d^2}{\sqrt{(d-b)(c-a)}}\big\{2K(k)+4(b^2/a^2-1)\Pi(\alpha^2,k)+
\frac{\alpha^2E(k)+(k^2-\alpha^2)K(k) + (2\alpha^2k^2+2\alpha^2-\alpha^4-3k^2)\Pi(\alpha^2,k)}{(\alpha^2-1)(k^2-\alpha^2)}\bigg\}
\end{equation}
where $\alpha^2=(c-b)/(d-... | {
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If $a,b,c,d\in\mathbb N$ and $a^2+b^2\mid ac+bd$, can it be true that $\gcd(a^2+b^2,c^2+d^2)=1$? or $3$? or $74$?
If $a,b,c,d\in\mathbb N$ and $a^2+b^2\mid ac+bd$, can it be true that $\gcd(a^2+b^2,c^2+d^2)=1$? or $3$? or $74$?
That problem is complicated. I've tried some approaches, but they're useless. E.g. if $$\g... | We have the identity
$$
b^2\bigl((a^2+b^2)-(c^2+d^2)\bigr) = (a^2+b^2)(c^2-b^2) - (ac-bd)(ac+bd).
$$
Hence $(a^2+b^2) \mid (ac+bd)$ forces $(a^2+b^2) \mid b^2(c^2+d^2)$. Since $a^2+b^2 > b^2$ for all $a,b \in \mathbb{N}$, we must have $\gcd(a^2+b^2,c^2+d^2)>1$. Furthermore, $\gcd(a^2+b^2,b^2)$ must divide $a^2\!$, so ... | {
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Find the least nonnegative residue Find the least nonnegative residue of $5^{18} \mod 11$.
To do this I took $5^2 \equiv 3 \mod 11$. Then I did $(5^2)^5 \equiv 3^5 \mod 11$. And $3^5 \equiv 1 \mod 11$.
So now I have $5^{10} \equiv 1 \mod 11$. Then I multiplied both sides by $5^8$ to get $5^{18} \equiv 5^8 \mod 11$. ... | ${\rm mod}\ 11\!:\ 2^{18} 5^{18}\equiv 10^{18}\equiv (-1)^{18}\!\equiv 1,\ $ so $\ 5^{18}\!\equiv 2^{-18}\!\equiv 2^2 (2^5)^{-4}\equiv 2^2(-1)^{-4}\!\equiv 2^2$
| {
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"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give ... | I would start by factoring $x^3$ out from the first two terms and noticing the pattern in the result.
$$
\begin{split}
x^5+4x^3 + x^2 + 4 &= x^3 \left(x^2+4\right) + x^2+4 \\
&= \left(x^2+4\right)\left(x^3+1\right) \\
&= \left(x^2+4\right)(x+1)\left(x^2-x+1\right), \\
\end{split}
$... | {
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"answer_count": 4,
"answer_id": 2
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