Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}... | \begin{align}
\sin \alpha+\cos \alpha &= 0.2 \\
(\sin \alpha+\cos \alpha)^2 &= 0.04 \\
\sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos\alpha &= 0.04 \\
1+\sin 2\alpha &= 0.04 \\
\sin 2\alpha &= -0.96
\end{align}
and that would be straight forward to proceed
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Angles between two vertices on a dodecahedron Say $20$ points are placed across a spherical planet, and they are all spaced evenly, forming the vertices of a dodecahedron. I would like to calculate the distances between the points, but that requires me to find out the angles between the vertices.
From the origin of the... | As noted in Wikipedia's "Dodecahedron" entry, if $s$ is the length of an edge of a dodecahedron, and $r$ the radius of its circumsphere, then
$$r = s \frac{\sqrt{3}}{4}\left( 1 + \sqrt{5} \right)$$
So, if two points $A$ and $B$ are joined by and edge, and $O$ is the center of the dodecahedron, then $\triangle AOB$ is i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the number of real solutions of the following? $ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $ What is the number of real solutions of the following?
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
My solution:
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
... | To see where the infinite solutions come from, first note that:
$$
\sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1 \iff 1=|\sqrt{x-1}-2| + |\sqrt{x-1}-3|
$$
Now consider the case where $5\le x\le10$. This implies that:
$$
\sqrt{x-1}-2\ge\sqrt{5-1}-2=0 \qquad \text{and} \qquad\sqrt{x-1}-3\le\sqrt{10-1}-3=0
$$
Thus... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots. If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots.
MY attempt:
We can open and get a bi quadratic but that is two difficult to show that it has real roots.THere must be an easy way.!
| $$(2x^2-3x+1)(2x^2+5x+1)=(2x^2+x+1)^2-16x^2=9x^2\\\Rightarrow 2x^2+x+1=\pm5x\\ \Rightarrow 2x^2+ax+1=0 $$ where $a=-4\ \mbox{or}\ 6$. Hence the discriminant, $$\Delta=a^2-8>0$$ always. So the roots are real.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of cosines product Let $\displaystyle P_n:=\prod_{k=1}^n \cos\frac{\pi}{k+2}$. Evaluate $\displaystyle\lim_{n\to \infty} P_n$.
I've only shown that the limit is positive.
Let $\vartheta_k:=\pi/(k+2)$. We have $\log P_n=\sum_{k=1}^n \log\cos\vartheta_k$. Now,
$$\log\cos\vartheta_k=-\frac{1}{2}\log(1+\tan^2\varthe... | We may use the product for $\sin x$ and $\cos x$ to conclude that
\begin{align}
\prod_{n = 1}^\infty \cos\frac{x}{n} &= \prod_{n = 1}^\infty \prod_{k = 0}^\infty \left( 1 - \frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&= \prod_{k = 0}^\infty \prod_{n = 1}^\infty \left( 1-\frac{4x^2}{(2k + 1)^2 n^2 \pi^2} \right)\\
&= \p... | {
"language": "en",
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Find a vector $\mathbf x$ whose image under $T$ is $b$. I am having trouble with this question and how to get the answer.
With $T$ defined by $T(\mathbf x)=A\mathbf x$, find a vector $x$ whose image under $T$ is $b$.
$$
A = \begin{pmatrix}
1 & -3 & 2 \\
3 & -8 & 8 \\
0 & 1 & 2 \\
1 & 0 & 8
\end{pmatrix} \qquad,\qq... | Hint 1: If $A$ has $3$ columns, the dimension of $x$ must be $3$.
Hint 2: To check your result, compute $Ax$ and see if you got $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A question about determing a residue Let $f(z)$ be analytic at $z=w$ and have a pole at $z=a$.
How does one show that the residue of $\displaystyle\frac{f(z)}{w-z}$ at $z=a$ equals the singular/principal part of $f(z)$ evaluated at $z=w$?
For example, let $ \displaystyle f(z) = \frac{\cot z}{z^{2}} = \frac{1}{z^{3}} - ... | Good old-fashioned computation with Laurent series.
$$\begin{align}
\frac{f(z)}{w-z} &= \frac{f(z)}{(w-a) - (z-a)} = \frac{1}{w-a} f(z) \frac{1}{1 - \frac{z-a}{w-a}}\\
&= \frac{1}{w-a}f(z)\sum_{\nu = 0}^\infty \left(\frac{z-a}{w-a}\right)^\nu\\
&= \frac{1}{w-a}\left(\sum_{k = -m}^\infty a_k (z-a)^k\right)\sum_{\nu = 0}... | {
"language": "en",
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Sequence of continuous, integrable functions that are not dominated I am looking for a sequence of continuous functions ${f_n}: [0,1] \rightarrow [0,\infty]$ such that $\int f_n d\mu \rightarrow 0$, $f_n(x) \rightarrow 0$ for all $x$, but $f(x) = \sup f_n(x)$ is not in $L_1$.
The "rotating tower" functions with growing... | Let
$$f_n(x) = \begin{cases}\frac{1}{x} &, \frac{1}{n+1} \leqslant x < \frac{1}{n}\\
(n+1)2^{n+1}\left(x -\frac{1}{n+1} + \frac{1}{2^{n+1}}\right) &, \frac{1}{n+1} - \frac{1}{2^{n+1}} \leqslant x < \frac{1}{n+1}\\
n2^n\left(\frac1n + \frac{1}{2^n} - x\right) &, \frac{1}{n} < x < \frac1n + \frac{1}{2^n}\\
0 &, \text{ ot... | {
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Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be... | If you don't mind doing a lot of arithmetic, here's an "easy" way to solve things:
Note that $(49+20\sqrt[3]{6})^{3/2} = 788.401\ldots$. Since you're looking for positive integer solutions $a$ and $b$, one need at worst examine numbers from $1$ to $788$: That is, if $b\ge1$, then $\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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seeming ugly limit i want to compute the limit
$$\lim_{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}-\frac{x^5}{120}-\frac{x^6}{720}}{x^7}$$
Instead of doing some messy calculation, I think if there is some ingenious way to compute this limit, but i don't know how to do. thank you so much.
| Hints:
$$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=1+x+\frac{x^2}{2!}+\ldots+\frac{x^6}{6!}+\frac{x^7}{7!}+\ldots\implies$$
$$e^x-1-x-\ldots-\frac{x^6}{6!}=\frac{x^7}{7!}+\frac{x^8}{8!}+O(x^9)\implies$$
$$\frac{e^x-1-x-\ldots-\frac{x^6}{6!}}{x^7}=\frac1{7!}+\frac{x}{8!}+O(x^2)\xrightarrow[x\to 0]{}\;?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solutions in positive integers of $a+b\mid ab+1$ and $a-b\mid ab-1$ I am interested in a proof for the following claim. Suppose that for integers $a>b>1$ the following two conditions hold:
$$a+b\mid ab+1,$$ $$a-b\mid ab-1.$$
Then $\frac{a}{b}<\sqrt{3}$. Furthermore, is it possible to determine all positive integer solu... | To prove the bound, note first that necessarily $\gcd (a,b) = 1$, and then write
$$\begin{align}ab+1 &= (b-\gamma)(a+b)\\
\iff \gamma(a+b) &= b(a+b) - ab - 1 = b^2 -1\\
ab - 1 &= (b+\delta)(a-b)\\
\iff \delta(a-b) &= ab - 1 - b(a-b) = b^2-1.
\end{align}$$
So both, $a+b$ and $a-b$ divide $b^2-1$. Since $\gcd (a+b,a-b) =... | {
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Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\... | Another idea: Using the inequality $$\frac{1}{\sqrt{2n-1} +\sqrt{2n+1}}\gt\frac{1}{2\sqrt{2n}}, n\ge1$$ we get the folliwng chain of equities/inequalities:
$$\sum_{i = 1}^{4999}{\frac{1}{\sqrt{2i-1} +\sqrt{2i+1}}} \gt \sum_{i = 1}^{4999}{\frac{1}{2\sqrt{2i}}} = \frac{1}{2\sqrt{2}}\sum_{i = 1}^{4999}{\frac{1}{\sqrt{i}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
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Help solve this through complex numbers I have to do the following problem:
If $ax+cy+bz = X, cx+by+az=Y, bx+ay+cz=Z$, then show that
$(a^2+b^2+c^2-bc-ca-ab)(x^2+y^2+z^2-yz-zx-xy) = X^2+Y^2+Z^2-YZ-XZ-XY$.
It's easy enough through normal Algebra if I pair $X^2-XY = X(X-Y)$ etc., and collect the coefficients on both si... | HINT:
$$\sum_{\text{cyc}}(X^2-YZ)=X^2+Y^2+Z^2+(XY+YZ+ZX)(w^2+w)$$
$$=(X+Yw+Zw^2)(X+Yw^2+Zw)$$ where $w$ is a complex cube root of $1$
$$X+Yw+Zw^2=ax+cy+bz +w(cx+by+az)+w^2(bx+ay+cz)$$
$$=a(x+w^2y+wz)+bw^2(x+y^2w+wz)+cw(x+y^2w+wz)=(x+y^2w+wz)(a+bw^2+cw)$$
$$\text{Similarly, for } X+Yw^2+Zw$$
Alternatively, without usi... | {
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"timestamp": "2023-03-29T00:00:00",
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solve differential equation using given substitution Solve the following equation by making substitution
$$y=xz^n \text{ or } z=\frac{y}{x^n}$$
and choosing a convenient value of n.
$$\frac{dy}{dx}= \frac{2y}{x} +\frac{x^3}{y} +x\tan\frac{y}{x^2}$$
I thought it can be solved in 2 ways
1.making it exact differential equ... | Here is one approach (others may also be possible, but we will use your hint).
We are given:
$$\tag 1 \frac{dy}{dx}= \dfrac{2y}{x} +\dfrac{x^3}{y} +x \tan \frac{y}{x^2}$$
Lets choose the substitution:
$$\tag 2 z = \dfrac{y}{x^2} \rightarrow y = x^2 z$$
Differentiating $(2)$ yields:
$$\tag 3 \dfrac{dy}{dx} = 2x z + x^2 ... | {
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Solving $x \equiv 9 \pmod{11}, x \equiv 6 \pmod{13}, x \equiv 6 \pmod{12}, x \equiv 9 \pmod{15}$ $$x \equiv 9 \pmod{11}$$
$$x \equiv 6 \pmod{13}$$
$$x \equiv 6 \pmod{12}$$
$$x \equiv 9 \pmod{15}$$
Does this system have a solution? I want to solve this using the Chinese remainder theorem, but there's $\gcd (12,15)=3$. H... | A good method to deal with systems of congruence equations with non coprime modulos by hand is splitting them up according to their prime factorization.
A general approach:
*
*For each congruence, where the modulus is not a prime power, split the equation into several congruence equations with prime power modulus, e... | {
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"timestamp": "2023-03-29T00:00:00",
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Identify and put conic in standard form: $4x^2 - 16x + 3y^2 + 24y + 52 = 0$ How do I put this conic in standard form and identify it?
$$4x^2 - 16x + 3y^2 + 24y + 52 = 0$$
| Start by completing the square.
$$\begin{align} 4x^2 - 16 x + 3y^2 + 24 y + 52 & = (4x^2 - 16x \color{blue}{\bf + 16}) + (3y^2 + 24y \color{blue}{\bf + 48}) + 52 \color{blue}{\bf - 16 - 48} \\ \\ & = 4(x^2 -4x + 4) + 3(y^2 + 8 y + 16) - 12 = 0\\ \\ & = \cdots \end{align}$$
Added:
Factoring gives us: $$\begin{align} 4(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Comparing sums of surds without any aids Without using a calculator, how would you determine if terms of the form $\sum b_i\sqrt{a_i} $ are positive? (You may assume that $a_i, b_i$ are integers, though that need not be the case)
When there are 5 or fewer terms involved, we can try and split the terms and square both s... | This is answer is incomplete it only handles the case of six surds.
For six surds there is still way to do it in general using squaring. Suppose you want to check if the inequality :
$$ \sqrt{a} + \sqrt{b} + \sqrt{c} < \sqrt{a'} + \sqrt{b'} + \sqrt{c'} $$
is true. Squaring both sides and letting
$$ A = bc,\quad B=ac... | {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Identities Like $A \sin(x) + B \cos(y) = \cdots$ Are there any identities for trigonometric equations of the form:
$$A\sin(x) + B\sin(y) = \cdots$$
$$A\sin(x) + B\cos(y) = \cdots$$
$$A\cos(x) + B\cos(y) = \cdots$$
I can't find any mention of them anywhere, maybe there is a good reason why there aren't ide... | Since
$$
A \cos(a+b) = A \cos(a) \cos(b) - A \sin(a) \sin(b) \ \ \ \ \ \ (1) \\
B \cos(a-b) = B \cos(a) \cos(b) + B \sin(a) \sin(b) \ \ \ \ \ \ (2)
$$
(1) + (2) gives
$$
A \cos(x) + B \cos(y) = (A+B) \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + (B-A) \sin(\frac{x+y}{2}) \sin(\frac{x-y}{2})
$$
where
$$
x = a + b \\ y = a -... | {
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Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$ I am trying to solve the following integral
$$\int \frac{dx}{x\sqrt{x^2-1}}$$
I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then
\begin{align}
&\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\
&\int \frac{1}{x} \tex... | You had the "gist" of what you needed to do, but as others have noted, your substitution should yield the integrand $\dfrac{1}{u^2+1}$.
We have
$$\int \frac {dx}{x \sqrt{x^2 - 1}} = \int \frac{x\,dx}{x^2\sqrt{x^2-1}}$$
As you did, we let $\, u=\sqrt{x^2-1}$. Then $du=\frac{x}{\sqrt{x^2-1}}\,dx$, so $x\,dx=\sqrt{x^2 - ... | {
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What is the sum of the series? How to find the sum of the series $$\sum_{n=0}^\infty \frac {x^{3^n}+(x^{3^n})^2} {1-x^{3^{n+1}}}$$ under the assumptions $x >0,\,x\neq 1,$ in a closed form?
| Since $\displaystyle \frac{u+u^2}{1-u^3} = \frac{1}{1-u} - \frac{1}{1-u^3}$,
$$\sum_{n=0}^N \frac {x^{3^n}+(x^{3^n})^2} {1-x^{3^{n+1}}}
= \sum_{n=0}^N \left( \frac{1}{1-x^{3^{n}}} - \frac{1}{1-x^{3^{n+1}}}\right)
=\frac{1}{1-x^{3^0}} - \frac{1}{1-x^{3^{N+1}}}
$$
This implies
$$\sum_{n=0}^{\infty} \frac {x^{3^n}+(x^{3^n... | {
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Solving inequalities, simplifying radicals, and factoring. (Pre calculus) (Q.1) Solve for $x$ in $x^3 - 5x > 4x^2$
its a question in pre calculus for dummies workbook, chapter 2. The answer says:
then factor the quadratic: $x(x-5)(x+1)>0$. Set your factors equal to $0$ so you can find your key points.When you have them... | Just for the first Question, as you're given, set $p(x)=x^3-4x^2-5x$. We are asked to make $p(x)$ positive, so one can factor it to its basic parts as $$p(x)=x(x-5)(x+1)$$ and then follow the table below:
I think you can easily find out why that interval is taken. (-:
| {
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Prove that the sequence $(3n^2+4)/(2n^2+5) $ converges to $3/2$
Prove directly from the definition that the sequence $\left( \dfrac{3n^2+4}{2n^2+5} \right)$ converges to $\dfrac{3}{2}$.
I know that the definition of a limit of a sequence is $|a_n - L| < \varepsilon$ .
However I do not know how to prove this using thi... | Let $\epsilon>0$,
$$\left|u_n-\frac{3}{2}\right|=\left|\frac{3n^2+4}{2n^2+5}-\frac{3}{2}\right|=\frac{7}{4n^2+10}<\epsilon\iff 4n^2>\frac{7}{\epsilon}-10$$
so let $N=\frac{1}{2}\sqrt{\left|\frac{7}{\epsilon}-10\right|}$ then forall $n>N$ we have $\left|u_n-\frac{3}{2}\right|<\epsilon$ and we conclude.
| {
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Evaluating a double integral by using $\int\arctan (1+\sin 2t)\ dt/(1+\sin 2t)$ I am trying to evaluate the following integral:
$$
I:=\iint_D\frac{\mathrm dy\ \mathrm dx}{1+(x+y)^4}
$$
where $D := \{(x,y)\mid x^2+y^2\le1,x\ge0,y\ge0\}$.
I tried to evaluate it in the following way: first I rewrite it in terms of the pol... | Let $(\xi,\eta) = (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})$. In terms of $(\xi,\eta)$, the integral becomes:
$$\int_0^{\frac{1}{\sqrt{2}}} \frac{2\xi }{1 + 4\xi^4}d\xi + \int_{\frac{1}{\sqrt{2}}}^1 \frac{2\sqrt{1-\xi^2} }{1+4\xi^4}d\xi$$
For the $1^{st}$ integral, let $u = 2\xi^2$, we have:
$$\int_0^{\frac{1}{\sqrt{... | {
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Multiplying and simplifying expressions The expression is:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2}$$
What I did was subtract the exponents of the numerator to the exponents of the denominator. I did a cross multiplication too.
What confuses me is the whole numbers and how i should deal wi... | First, we'll express the product as a single fraction, and use the commutativity of multiplication to "rearrange" terms:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The intuition behind Trig substitutions in calculus I'm going through the MIT open calculus course, and in one of the lectures (19-28min marks) the professor uses the trig substitution $x = \tan \theta$ to find the integral of $\frac{dx}{x^2 \sqrt{1+x^2}}$.
His answer: $-\csc(\arctan x) + c$, which he shows is equival... | The basic "family" of trigonometric substitutions uses the definitions of trig functions to produce a ratio $ \ \frac{x}{a} = \ $ [trig function] $\theta \ \rightarrow \ x = a \cdot $ [trig function] $\theta \ $ , $ \ a \ $ being a constant. The legs and hypotenuse of a right triangle can produce three possibilities r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true? I found the following relational expression by using computer:
For any natural number $n$,
$$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor.$$
Note that $\lfloor... | Note that by the strict concavity of $\sqrt{x}$, Jensen's Inequality says
$$
\hspace{-1cm}\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}=\sqrt{25n+50}\tag{1}
$$
More precisely, Taylor's Formula with remainder says
$$
\begin{align}
\sqrt{n+2+x}
=\sqrt{n+2}+\frac{x}{2\sqrt{n+2}}-\int_0^x\frac{(x-t)\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 2
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$ How to prove the following equality?
$$\sum_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}.$$
| Another approach without using hidden beta and gamma functions where one gets to excercise manipulation of binomial coefficients and generating functions is to use
$$\prod_{k=1}^m \frac{1}{n+k} =
\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1}.$$
Now compute how often the fraction $1/q$ where $q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4... | Let $b_{n}:=\sum_{k=1}^{n-1}a_{k}$ for $n=2,3,\cdots$,
then $b_{n+1}-b_{n}=a_{n}$ for $n=2,3,\cdots$.
Then, the equation reads as $b_{n+1}-b_{n}=-\frac{1}{2}b_{n}+\frac{n}{4}$ for $n=2,3,\cdots$ with $b_{2}=a_{1}=\frac{1}{4}$. Rearraging the terms, we get $$\begin{cases}b_{n+1}-\frac{1}{2}b_{n}=\frac{n}{4},{\quad}n=2,3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of all 4-digit numbers formed by using digits 0, 2, 3, 5 and 8? Find the sum of all 4-digit numbers formed by using digits 0, 2, 3, 5 and 8.
Finding the total number of number is possible, but how can the sum be found?
| If you fix 8 as the last digit, you see that there are $4 \cdot 3 \cdot 2$ ways to complete the number. Thus, 8 appears 24 times as the last digit. By the same logic, if we enumerate all possible numbers using these 5 digits, each number appears 24 times in each of the 4 positions. That is, the digit 8 contributes $(24... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Integrate :$\int\frac{1}{\sqrt[3]{\tan(x)}}\, \mathrm dx$ How to integrate
$$\int\dfrac{1}{\sqrt[3]{\tan(x)}}\, \mathrm dx?$$
| Sub $u=\tan{x}$ and get
$$\int du \frac{u^{-1/3}}{1+u^2}$$
Then sub $u=y^3$ to get
$$3 \int dy \frac{y}{1+y^6} = \frac{3}{2} \int \frac{dv}{1+v^3} = \frac12 \int \frac{dv}{1+v} + \int \frac{dv}{1-v+v^2} - \frac12 \int dv \frac{v}{1-v+v^2} $$
Each of these integrals may be evaluated in turn.
$$\int \frac{dv}{1+v} = \log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Chinese remainder theorem :Algebraic solution I need help in a question:
It is required to find the smallest $4$-digit number that when divided by $12,15$, and $18$ leaves remainders $8,11$, and $14$ respectively. Here's how I've attempted:
Let the number be $a$, then $$a=12p+8 = 15q+11 = 18r+14$$
Hence, $p=(5q+1)/2$ a... | $$a = 12p+8 = 15q+11 = 18r+14$$
You should break these down into linear expressions with prime-power coefficients.
$$a = 12p+8 \implies a =
\left\{ \begin{array}{c}
4(3p+2) \\
3(4p+2)+2
\end{array}
\right \}$$
$$a = 15q+11 \implies a =
\left\{ \begin{array}{c}
3(5q+3)+2 \\
5(3q+2)+1
\end{array}
\right \}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Trigonometric Coincidence I Know that using Taylor Series, the formula of $\sin x$ is
$$x-x^3/3!+x^5/5!-x^7/7!\cdots,$$
and the unit of $x$ is radian (where $\pi/2$ is right angle).
However, the ratio of the circumference and the diameter of a circle is also $\pi$. Is it a coincidence? Or is there a proof?
| Let's start with your series and take a couple of derivatives:
$$
\begin{align}
v(x)&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\dots\\
u(x)=v'(x)&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\dots\\
u'(x)=v''(x)&=\hphantom{1}-\,\,x\,\,\,+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show that $\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$ Show that
$$\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$$
using 2 ways: the first using contour integration and the second using real analysis.
| Perform integration by parts,
$\begin{align} J&=\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx\\&=\left[\frac{1}{a}\ln\left(\text{arctanh}\left(\frac{ax}{2}\right)\right)x\right]_0^{\infty}-\int_0^{\infty}\frac{1}{a}\ln\left(\text{arctanh}\left(\frac{ax}{2}\right)\right)dx\\
&=-\frac{1}{a}\int_0^{\infty}\ln\left(\text{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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How to find an approximate values of rational function $f(x)$ for large $x$, neglecting $\frac{1}{x^4}$ and successive terms? This is the function that I want to find an approximate value for it neglecting $\displaystyle \frac{1}{x^4}$ and successive terms
$$
f(x)=\frac{25x}{(x-2)^2(x^2+1)}.
$$
| Since $\frac1{1+x}=1+O(x)$, we get
$$
\begin{align}
\frac{25x}{(x-2)^2(x^2+1)}
&=\frac{25}{x^3(1-\frac2x)^2(1+\frac1{x^2})}\\[9pt]
&=\frac{25}{x^3}\left(1+O\left(\frac1x\right)\right)^2\left(1+O\left(\frac1{x^2}\right)\right)\\[6pt]
&=\frac{25}{x^3}\left(1+O\left(\frac1x\right)\right)\\[6pt]
&=\frac{25}{x^3}+O\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric problem using basic trigonometry
If $x$ is a solution of the equation: $$\tan^3 x = \cos^2 x - \sin^2 x$$
Then what is the value of $\tan^2 x$?
This is the problem you are supposed to do it just with highschool trigonometry , but i can't manage to do it please help
Here are the possible answers:
$$a) ... | Hint: $$(1+\tan^2 x)^2 (\tan^2 x)^3 = \left((1+\tan^2 x) (\tan^3 x)\right)^2 = \left( \frac{1}{\cos^2 x} (\cos^2 x - \sin^2 x) \right)^2 = \left( 1 - \tan^2 x\right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Exponential algebra problem We need to solve for x:
$$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$
My proposed solution is below.
| Let's do this in another way:
$$3^3\cdot 2^{2x+1}=2^{3x-0.5}\cdot 3^{2x}$$
$$2^{-x+1.5}=3^{2x-3}$$
Let's take a log with base 2:
$$\log_{2}{2^{-x+1.5}}=\log_{2}{3^{2x-3}}$$
$${-x+1.5}=\log_{2}{3^{2x-3}}$$
Let's move to base 3:
$$-x+1.5=\frac{\log_3{3^{2x-3}}}{\log_3{2}}$$
$$-x+1.5=\frac{2x-3}{\log_3{2}}$$
$$-x(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find the square root of $(x^2 + 3x + 7)(x^2 + 5x + 3) + (x − 2)^2$ I want to find the square root of $$(x^2+3x + 7)(x^2+5x+3)+ (x −2)^2$$
First , I would like to know if it is really necessary to expand everything , because I think it is in the given form for a special reason.
Anyway I expanded and got $$x^4+8 x^3+... | As $x^2+3x + 7-(x^2+5x+3)=-2(x-2), x^2+5x+3=(x^2+3x +7)+2(x-2) $
$\implies (x^2+3x +7)(x^2+5x+3)+(x-2)^2$
$=(x^2+3x +7)\{x^2+3x +7+2(x-2)\}+(x-2)^2$
$=(x^2+3x +7)^2+2(x^2+3x +7)(x-2)+(x-2)^2$
$=\{(x^2+3x +7)+(x-2)\}^2$
Alternatively, if we $x^2+3x + 7=a, x^2+5x+3=b, x-2=c\implies b-a=2c$
$\implies (x^2+3x +7)(x^2+5x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$? I'm a ninth grader so please try to explain the answer in simple terms .
I cant fully understand the explanation in my book .
It just assumes that the expression that should be added has a degree of 1.
I apologize if t... | If you don't want to use polynomial long division, you can factor $x^2+2x-3$ by seeing it has roots at $1$ and $-3$, Then, you need to add a degree 1 polynomial to your polynomial such that the sum is divisible by $x-1$ and $x+3$. Those are relatively prime, so the sum will then be divisible by the product, which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Show that $ (n+1)(n+2)\cdots(2n)$ is divisible by $ 2^n$, but not by $ 2^{n+1}$ darij grinberg's note:
I. Niven, H. S. Zuckerman, H. L. Montgomery, An introduction to the theory of numbers, 5th edition 1991, §1.4, problem 4 (b)
Let $n$ be a nonnegative integer. Show that $ (n+1)(n+2)\cdots(2n)$ is divisible by $ 2^n... | Let's count directly:
The number of multiples $m_2$ of $2$ between $n+1$ and $2n$ is $\left\lfloor \frac {2n}2 \right\rfloor-\left\lfloor \frac {n}2 \right\rfloor$ (we take the multiples of $2$ up to $2n$ and subtract the number up to $n$).
The number of multiples $m_4$ of $4$ between $n+1$ and $2n$ is $\left\lfloor \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
How find this$\lim_{n\to\infty}n^2\left(n\sin{(2e\pi\cdot n!)}-2\pi\right)=\frac{2\pi(2\pi^2-3)}{3}$ show that
$$\lim_{n\to\infty}n^2\left(n\sin{(2e\pi\cdot n!)}-2\pi\right)=\dfrac{2\pi(2\pi^2-3)}{3}$$
we are kown that
$$\lim_{n\to\infty}n\sin{(2\pi e\cdot n!)}=2\pi$$
because we note
$$e=1+\dfrac{1}{1!}+\dfrac{1}{2... | Clearly the fractional part of $\mathrm{e} \cdot n!$ equals to
$$\begin{eqnarray}
\{\mathrm{e} \cdot n!\} &=& \left\{\sum_{k=0}^{n} \frac{n!}{k!} + \sum_{k=0}^\infty \frac{n!}{(n+1+k)!}\right\} =\left\{ \sum_{k=0}^\infty \frac{n!}{(n+1+k)!}\right\} = \left\{ \sum_{k=0}^\infty \frac{1}{(n+1)_{k+1}}\right\} \\
&=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$... | There are already a lot of good answers here, so I'm adding this one
primarily to dazzle people w/ my Mathematica diagram-creating skills.
As noted previously,
$x(t)=a \cos (t)$
$y(t)=b \sin (t)$
does parametrize an ellipse, but t is not the central angle. What
is the relation between t and the central angle?:
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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How to evaluate $\sqrt{5+2\sqrt{6}}$ + $\sqrt{8-2\sqrt{15}}$? My exams are approaching fast and I found this question in one of the unsolved sample papers.
I tried squaring the whole term but couldn't work out the answer.
I am a ninth grader so please try to explain in simple terms.
| It's worth noting that $(\sqrt a \pm \sqrt b)^2 = (a+b) \pm 2\sqrt{ab}$.
So, if the answer to $\sqrt{M \pm 2\sqrt N}$ is going to be of the form $
\sqrt a \pm \sqrt b$ then we need $N=ab$ and $M = a+b$.
$\sqrt{5+2\sqrt{6}}$
Note that $6 = 2 \times 3$ and $5 = 2 + 3$.
Hence $$(\sqrt 3 + \sqrt 2)^2 = 5+2 \sqrt 6$$
$\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the Determinant of a $3\times 3$ matrix. Show that:
$$
\begin{vmatrix}
x & a & b \\
x^2 & a^2 & b^2 \\
a+b & x+b & x+a \\
\end{vmatrix} = (b - a)(x-a)(x-b)(x+a+b)
$$
I tried expanding the whole matrix out, but it looks like a total mess. Does anyone have an idea how this could be simplified?
| Regard the matrix as a polynomial $f(x)$ in $x$.
Note that $a$ and $b$ are roots, because when you set $x = a$ you get that two of the columns are equal, and ditto for $x = b$.
Note that $- a - b$ is a root, because when you set $x = -a-b$ you get that two of the rows are equal.
So
$$
f(x) = c (x - a) (x - b) (x + a + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Bounds for $\binom{n}{cn}$ with $0 < c < 1$. Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$? I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn} \leq \left(\frac{e}{c}\right)^{cn}$.
| Stirling's Asymptotic Expansion, derived here, is
$$
n!=\sqrt{2\pi n}\,n^ne^{-n}\left(1+\frac1{12n}+\frac1{288n^2}-\frac{139}{51840n^3}-\frac{571}{2488320n^4}+O\left(\frac1{n^5}\right)\right)
$$
From which we get
$$
\begin{align}
&\frac{n!}{(cn)!((1-c)n)!}\\[6pt]
&=\frac{\left(c^c(1-c)^{1-c}\right)^{-n}}{\sqrt{2\pi c(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How find the $\angle B$
In $\Delta ABC$ such $I$ is incentre,and $$\angle A=80^{0},AI+IB=BC$$,
find the $\angle B$
my idea:let $AB=c,AC=b,BC=a$ then we have
$$\dfrac{AI}{ID}=\dfrac{AB}{AD}=\dfrac{AC}{DC}=\dfrac{AB+AC}{AD+DC}=\dfrac{b+c}{a}$$
and
$$AD^2=AB\cdot AC-BD\cdot DC=bc-BD\cdot DC$$
$$BD=\dfrac{ac}{b+c},DC=\dfr... | Hint: Apply sin rule to $BIC$, get $BC$ in terms of $BI$.
Hint: Apply sin rule to $BIA$, get $AI$ in terms of $BI$.
Now substitute these into $AI + BI = BC$, you will get a trigonometric equation in terms of $\angle IBC$. Solve it to determined that $\angle IBC = 20^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found t... | The statement is true for an arbitrary complex number $n$, with matrix logarithm. All the $\log$s below refer to the natural logarithm.
For convenience, let's set $$ A = \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}. $$
We know that if $k \in \mathbb N$ and $k \ge 2$, then $A^k = O$. Therefore,
$$ \exp A = \sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 4
} |
Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive ... | Let's substitute the variable: $x=y+\frac{(a+b+c)}{3}$
The equation will now look like
$$3y^3+(2a^2-2ab-2ac+2b^2-2bc+2c^2)y+\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}=0$$
Now we apply Cardano's method.
$$Q=\left(\frac{2a^2-2ab-2ac+2b^2-2bc+2c^2}{3}\right)^3+\left(\frac{\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}}{2}\right)^2=\frac{8}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Re-write $1 \cdot x$ to $x$. Given the following bi-directional re-write rules (where $1$ is a constant, $^{-1}$ is a unary operator, $\cdot$ is a binary operator, and $x,y,z$ are arbitrary terms):
$$\begin{align*}
x \cdot 1 &= x \\
x \cdot (y \cdot z) &= (x\cdot y) \cdot z \\
x \cdot x^{-1} &= 1
\end{align*}$$
we're a... | Yes, observe that:
\begin{align*}
1 \cdot x &= (x \cdot x^{-1}) \cdot x \\
&= x \cdot \left(x^{-1} \cdot x\right) \\
&= x \cdot \left((x^{-1} \cdot x) \cdot 1\right) \\
&= x \cdot \left((x^{-1} \cdot x) \cdot (x^{-1} \cdot (x^{-1})^{-1}) \right) \\
&= x \cdot \left(((x^{-1} \cdot x) \cdot x^{-1}) \cdot (x^{-1})^{-1} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? Let $a,b,c \in \mathbb{R}$, and $abc=1$. What is the simple(st) way to prove inequality
$$
a+b+c \le a^2+b^2+c^2.
$$
(Of course, it can be generalized to $n$ variables).
| We may assume that $a+b+c > 0$.
Then you can do this by using Cauchy-Schwarz and the power mean inequality as follows:
$$a+b+c \le \sqrt{a^2 + b^2 + c^2} \sqrt{3} \le \sqrt{a^2 + b^2 + c^2} \sqrt{a^2 + b^2 + c^2} = a^2 + b^2 + c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
How to test a matrix for inconsistency? I understand how to test two equations with two variables but what happens when I get up to 4 variables? I have no idea how to test that. For example with the matrix/system of equations
$x_1 +2 {x_2} + x_3 = 1$
$-2 x_1 - 4 {x_2} - x_3 = 0$
$5 {x_1} + 10 {x_2} + 3 {x_3} = 2$
$3 {x... | Try to solve the equation; and check whether the system has the solution or not.
$\begin{pmatrix}1 & 2 & 1 \\ -2 & -4 & -1 \\ 5 & 10 & 3 \\ 3 & 6 & 3 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 2 \\ 4\end{pmatrix} $, and multiply first row by 3 and substract to 4th row then
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why n! equals sum of some expression? Why n! equals sum of some expression?
Especially I need to know why this expression is true?
$$
n!= \left(\frac{n+1}{2}\right)^{p(n)} \; \prod_{j=0}^{q(n)}\sum_{i=0}^j(n-2i),
$$
Where
\begin{gather*}
p(n)=\frac{\cos(\pi n+\pi)+1}{2}\\\\
q(n)=\frac{2n+\cos(\pi n)-5}4
\end{gather*}
H... | Summary:
This formula is just a convoluted way of saying,
$$
n! = [n \cdot 1][(n-1) \cdot 2][(n-2) \cdot 3] \cdots
\left\{
\begin{array}{cc}
\left[ \frac{n+1}{2} \right] & n \text{ is odd} \\
\left[\frac{n}{2} \cdot \frac{n+2}{2} \right] & n \text{ is even}
\end{array}
\right.
$$
Explanation:
Note that by sum of an ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is acc... | One that's not been mentioned so far(?): knowing that
$$
0 < e^x =
1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots
$$
proves the inequality except for $-1 < x < 0$. But in that region
$$
e^x - (1+x) =
\frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots
$$
is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "113",
"answer_count": 27,
"answer_id": 7
} |
Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$ Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$.
I have absolutely no clue where to start and what to do, please provide complete proof and answer.
| Suppose that $\gcd(a+b,a^2+b^2)=d$
Then $d \mid a+b$ and $d \mid a^2+b^2$.
$d \mid (a+b)(a+b)-(a-b)(a+b)=2a^2$.
$d \mid (a+b)(a+b)+(a-b)(a+b)=2b^2$
$d \mid (2a^2,2b^2) = 2(a^2,b^2)=2(a,b)^2=2.1=2 \implies d=1$ or $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I integrate $\sec(x)$? My HW asks me to integrate $\sin(x)$, $\cos(x)$, $\tan(x)$, but when I get to $\sec(x)$, I'm stuck.
| $$\int \sec x dx$$
Using Weierstrass substitution,
$$\tan x=\tan(\frac{x}{2}+\frac{x}{2}) \iff \tan(\frac{x}{2}+\frac{x}{2})=\frac{2\tan\frac{x}{2}}{1-\tan^2(\frac{x}{2})}$$
$$ \tan(\frac{x}{2}+\frac{x}{2})=\frac{2\tan\frac{x}{2}}{1-\tan^2(\frac{x}{2})}\iff \tan x= \frac{2t}{1-t^2}$$
$$\tan(\frac{x}{2})=t \iff \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/506780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$ let $a,b,c>0$, show that
$$a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$
I know this
$$a+b+c\ge 3\sqrt[3]{abc}$$
so
$$\Longleftrightarrow 6\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$
But this maybe not true?
| Let $a=x^6$, $b=y^6$ and $c=z^6$, where $x$, $y$ and $z$ be positive numbers.
Hence, by Schur and AM-GM we obtain:
$$a+b+c+3\sqrt[3]{abc}=\sum_{cyc}(x^6+x^2y^2z^2)\geq\sum_{cyc}(x^4y^2+x^4z^2)=$$
$$=\sum_{cyc}(x^4y^2+y^4x^2)\geq2\sum_{cyc}x^3y^3=2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\tan x+\tan \left(x+\frac{\pi }{3}\right)+\tan \left(x+\frac{2\pi }{3}\right)=3\tan 3x$? The following is the equation.
\begin{eqnarray}
\tan x+\tan \left(x+\frac{\pi }{3}\right)+\tan \left(x+\frac{2\pi }{3}\right)&=&3\tan 3x\\
\tan x+\frac{\tan x+\tan \frac{\pi }{3}}{1-\tan x\space \tan \frac{\pi }{3}}+\... | Let $$\tan3x=\tan3A$$
$\implies 3x=n\pi+3A,x =\frac{n\pi}3+A$ where $n$ is any integer
We can set $x=A,\frac\pi3+A,\frac{2\pi}3+A$ (In fact, we can any three in-congruent values of $n\pmod 3$)
Now,
$$\tan3A=\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$$
On rearrangement, $$\tan^3x-3\tan3A\tan^2x-3\tan x+\tan3A=0$$ which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (... | Another way to see this is as follows:
Suppose $x^2+xy+y^2 <0$, then $xy<-x^2-y^2< 0$, hence $2xy<xy<-x^2-y^2$ and thus $x^2+2xy+y^2<0$, which is absurd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
trig equation $(\sqrt{\sqrt{2}+1})^{\sin(x)}+(\sqrt{\sqrt{2}-1})^{\sin(x)}=2$ Please help me to solve this trig equation.
$$(\sqrt{\sqrt{2}+1})^{\sin(x)}+(\sqrt{\sqrt{2}-1})^{\sin(x)}=2$$
| To solve
$$\left(\sqrt{\sqrt{2}+1}\right)^{\sin(x)}+\left(\sqrt{\sqrt{2}-1}\right)^{\sin(x)}=2,$$
let us rewrite as
$$\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)} + \left(\sqrt{2}-1\right)^{\frac{1}{2}\sin(x)}=2.$$
First consider the auxiliary equation
$$ y + \frac{1}{y} = 2.$$
You can verify this has only one solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sequence summation GRE question
This problem is giving me a lot of trouble. the only way i can think of doing it is to take the integeral a(k) from 1 to 100 but that is definitely not what i am supposed to do since its for the GRE and no knowledge of calculus is necessary. . . helpppp
| If you write out the first few terms of the series, you'll see that $$a_1+a_2+a_3+\cdots+a_{99}+a_{100}$$$$=\left(1-\frac{1}{2}\right)+ \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3} - \frac{1}{4} \right) \cdots+\left(\frac{1}{99} - \frac{1}{100}\right) + \left( \frac{1}{100} - \frac{1}{101} \right).$$
Rearrang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Find $\frac{a+b+c}{x+y+z}$ given $a^2+b^2+c^2$, $x^2+y^2+z^2$ and $ax+by+cz$. We are given $a^2+b^2+c^2=m$, $x^2+y^2+z^2=n$ and $ax+by+cz=p$ where $m,n$ and $p$ are known constants. Also, $a,b,c,x,y,z$ are non-negative numbers.
The question asks to find the value of $\dfrac{a+b+c}{x+y+z}$.
I have thought a lot about th... | Let $m=5/4,n=2,p=1$. We can produce a one parameter family of solutions to the three equations so that $(a+b+c)/(x+y+z)$ has infinitely many values for that family.
Let $a=\cos t,\ b=\sin t, c=1/2$ (so $a^2+b^2+c^2=5/4$ is satisfied.)
Let $x=\cos u,\ y=\sin u, z=1$ (so $x^2+y^2+z^2=2$ is satisfied.)
Using the differenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of $\sum_{n=1}^{\infty}\frac1{n^3}\frac{\sinh\pi n\sqrt2-\sin\pi n\sqrt2}{{\cosh\pi n\sqrt2}-\cos\pi n\sqrt2}=\frac{\pi^3}{18\sqrt2}$ Show that
$$\sum_{n=1}^{\infty}\frac{\sinh\big(\pi n\sqrt2\big)-\sin\big(\pi n\sqrt2\big)}{n^3\Big({\cosh\big(\pi n\sqrt2}\big)-\cos\big(\pi n\sqrt2\big)\Big)}=\frac{\pi^3}{18\sqr... | This simple and elegant solution relies on two results from Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. and/or from Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$.
both of which are derived from the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote $\mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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If $(x-8)\cdot (x-10) = 2^y$, where $x,y\in \mathbb{Z}$. then $(x,y)$ is (1) If $(x-8)\cdot (x-10) = 2^y$, where $x,y\in \mathbb{Z}$. Then the no. of ordered pairs of $(x,y)$
(2) If $x^4-6x^2+1 = 7\cdot 2^y$,where $x,y\in \mathbb{Z}$. Then the no. of ordered pairs of $(x,y)$
$\underline{\bf{My\; Try}}::$ for (1) one , ... | for $2$, you have $x^4-6x^2+(1-7.2^y)=0 \Rightarrow x^2= \frac{6\pm\sqrt{32-28.2^y}}{2}$
do you now see what should be $y$???
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding Jordan Basis of a matrix Having trouble finding the Jordan base (and hence $P$) for this matrix
$A = \begin{pmatrix}
15&-4\\ 49&-13
\end{pmatrix}$
I know that the eigenvalue is $1$, this gives an eigenvector $\begin{pmatrix}
2\\ 7
\end{pmatrix} $
Now to create the Jordan basis and find $P$ (of which its column... | Let
$$P=\left[ \begin{array}{cc}2 & a\\ 7 & b\end{array} \right]$$
You are meant to solve the equation $AP = PJ$, with
$$J=\left[ \begin{array}{cc}1 & 1\\ 0 & 1\end{array} \right].$$
$J$ has the previous form since its only eigenvalues $\lambda=1$ has an eigenspace of dimension $1$.
Expanding $AP = PJ$, you get the fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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Find the limit -> Infinity with radicals First guess to multiply by $x^{-1.4}$ so the radical in numerator with $x^7$ becomes 1 and other stuff becomes 0. But then denominator becomes $-\infty$. What is the right approach?
$$ \lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} $... | Putting $\frac1x=h$
$$\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}=\frac{(1+3h^7)^{\frac15}}{h^{\frac75}}-\frac{(2-h^3)^{\frac14}}{h^{\frac34}}=\frac{(1+3h^7)^{\frac15}-h^{\frac{13}{20}}(2-h^3)^{\frac14}}{h^{\frac75}}$$ as $\frac75-\frac34=\frac{7\cdot4-5\cdot3}{20}=\frac{13}{20}$
$$\sqrt[8]{x^7 + x^2 + 1} - x=\frac{(1+h^5+h^7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Standard inductive problem Question: Prove that $2^n \geq (n+1)^2$ for all $n \geq 6$.
I have tried to prove this below and I'm interested if my method was correct and if there is a simpler answer since my answer seems unnecessarily long for such a simple claim.
Inductive hypothesis $$2^n \geq (n+1)^2$$
We need to sho... | Except for showing a base case I think your method is good. Note that without a base case you do not have an induction, so make sure you pick e.g., $n=6$ and note that $2^6=64\ge (6+1)^2=49.$ You might consider trying to show the inductive hypothesis by showing that $n\ge 6\implies (n+1)^2\ge 2n+3$ and thus
$$2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/527305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem : If $\sin^2\theta = \frac{x^2+y^2+1}{2x}$ , $x$ must be .... Problem : If $\sin^2\theta = \frac{x^2+y^2+1}{2x}$ , $x$ must be
(a) $1$
(b) $-2$
(c) $-3$
(d) $ 2$
My approach :
Since $0 \leq \sin^2\theta \leq 1$
$\Rightarrow 0 \leq \frac{x^2+y^2+1}{2x} \leq 1 $
$\Rightarrow x^2 +y^2+1 -2x \leq 0$
$\Rightar... | Yes the approach is right, except for one small thing, you should not just assume x to be positive while transferring variables across inequality signs. However, in this case, in the first inequality condition, if x is negative, the condition becomes:
$x^2 + y^2 + 1 <= 0$
which is wrong
x has to be positive, so doesn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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probability regarding three people throwing a die There are 3 players, A, B, C, taking turns to roll a die in the order ABCABC....
What's the probability of A is the first to throw a 6, B is the second, and C is the third?
The answer said it's 216/1001, but I always got 125/1001
The way I did it was:
Let $X_1$= num... | You can avoid infinite sums altogether if you use a conditioning argument.
Let $X_1$, $X_2$, and $X_3$ be the times of the first six for each of the three players.
We look at the two man game first. Let $q=\mathbb{P}(X_2<X_3)$. By first step analysis, we have
$q={1\over 6}+\left({5\over 6}\right)^2q$, so $q={6\over 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/536420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Limit of $x_n/n$ for sequences of the form $x_{n+1}=x_n+1/x_n^p$
*
*Given $x_1 = 1, x_{n+1} = x_n + \frac{1}{x_n} (n\ge1)$, Prove whether the limit as follow exist or not. If so, find it
$$\lim_{n\to\infty}\frac{x_n}{n}$$
*Given $x_1 = 1, x_{n+1} = x_n + \frac{1}{\sqrt{x_n}} (n\ge1)$, Prove whether the limit as foll... | By the inspiration from @kedrigern, I solved the second one. Hope somebody can find a simple way to solve it:
Let $y_n = x_n^2$,then:
\begin{align*}
\ y_{n+1} &= x_{n+1}^2 = x_n + \frac{1}{\sqrt{x_n}}= x_n^2 + 2\sqrt{x_n} + \frac{1}{x_n}
\\&= y_n + 2\sqrt{x_n} + \frac{1}{x_n}
\\&=y_{n-1} + 2(\sqrt{x_n} + \sqrt{x_{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\... | Assume that $x,y,z\in\mathbb{R}^+$ and $x^2+y^2+z^2=3$. We want to prove:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq 3.$$
Since $f(w)=\frac{1}{w}$ is a convex function on $\mathbb{R}^+$, we have:
$$ \frac{x^2}{3}f(y)+\frac{y^2}{3}f(z)+\frac{z^2}{3}f(x)\geq f\left(\frac{x^2y+y^2 z+z^2 x}{3}\right), $$
so:
$$\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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time speed distance Two horses start simultaneously towards each other and meet after $3h 20 min$. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second $5 hours$ later than the second arrived at the departure of the first.
MY TRY::
Let spee... | First, let's identify what you actually want to solve for, which is $\frac{d}{b}$. Solve for $a$ in your first equation: $a = 3/10 d - b$ and substitute into the second equation
$$
\frac{d}{\frac{3}{10} d - b} - \frac{d}{b} = 5\\
db- d\left(\frac{3}{10} d - b\right) = 5b\left(\frac{3}{10} d - b\right)\\
d\left(2b - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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gaussian and mean curvatures I am trying to review, and learn about how to compute and gaussian and mean curvature. Given $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, how can I compute the gaussian and mean curvatures?
This is what I have so far,
$$K(u, v) = \frac{a^2 b^2 c^2}{[c^2 \sin^2(v) (a^2 \sin^2(u... | There is a much easier way to find the Gaussian curvature of ellipsoid without going through the tedious computation. I found this from Do Carmo's classic book of differential geometry (Ex.21, p.172-173).
First of all, we prove the following:
Let $M$ be a regular orientable surface with orientation $\mathbf{N}$ (i.e. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does $\sum_{n=1}^\infty \frac{1+5^n}{1+6^n}$ converge? Question
Use appropriate tests to decide whether the series converges or diverges
$$\sum_{n=1}^\infty \frac{1+5^n}{1+6^n}$$
I'm not sure how to complete this question. I have attempted to complete this with the ratio test, however this becomes very messy quickly.
... | It is not that messy...
$$\frac{a_{n+1}}{a_n} = \frac{1+5^{n+1}}{1+6^{n+1}}\frac{1+6^{n}}{1+5^{n}} = \frac{1+6^{n}}{1+6^{n+1}}\frac{1+5^{n+1}}{1+5^{n}} = \frac{6^n(6^{-n}+1)}{6^n(6^{-n}+6)}\frac{5^n(5^{-n}+5)}{5^n(5^{-n}+1)} = \frac{6^{-n}+1}{6^{-n}+6}\frac{5^{-n}+5}{5^{-n}+1}.$$
As $n\rightarrow\infty$, $6^{-n},5^{-n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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finding determinant as an function in given matrix Calculate the determinant of the following matrix as an explicit
function of $x$. (It is a polynomial in $x$. You are asked to find
all the coefficients.)
\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\
x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\
0 & 0 & 0 & x^{10} & x^{11}... | First, note that the 5th column is a multiple of the 4th column. That is,
\begin{bmatrix}
x^4\\
x^9\\
x^{11}\\
x^{13}\\
x^{15}\\
\end{bmatrix}
is $x$ times
\begin{bmatrix}
x^3\\
x^8\\
x^{10}\\
x^{12}\\
x^{14}\\
\end{bmatrix}.
Because the determinant of a matrix does not change when you subtract a multiple of one column... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to efficiently compute $17^{23} (\mod 31)$ by hand? I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 3... | Note that $17\equiv 3/2 \pmod {31}$. We can then calculate this.
$2^5 = 1$, so $2^{23}=8$.
$3^3 = -4$, so $3^6 = 16$. We then have $3^{24} = 16^4 = 2$, and dividing through by $8$, we get $8$. Divide this by $3$ to get $31+8 = 39$, divide by 3 to get $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How find this sum $\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}$ Find the sum
$$\sum_{n=1}^{\infty}n\sum_{k=2^{n - 1}}^{2^{n}\ -\ 1}\dfrac{1}{k(2k+1)(2k+2)}$$
My try:
note
$$\dfrac{1}{k(2k+1)(2k+2)}=\dfrac{2}{(2k)(2k+1)(2k+2)}=\left(\dfrac{1}{(2k)(2k+1)}-\dfrac{1}{(2k+1)(2k+2)}\right)$$
Then I ca... | Throughout, let $H_k\equiv \sum_{j=1}^k \frac{1}{j}$ and let $\gamma$ be the Euler-Mascheroni constant.
First, note
\begin{eqnarray}
\sum_{n=1}^{\infty} n \sum_{k=2^{n-1}}^{2^n-1} \frac{1}{k(2k+1)(2k+2)}
&=& \sum_{j=0}^{\infty} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)}.
\end{eqnarray}
Mathematica says
\begin{eqnarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/551111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Cauchy's Problem Question $x(x^2+1)dy-(3x^2+1)ydx= x(x^2+1)^2dx,\ \ y(1)=2$ I want to solve the following equation
$$x(x^2+1)dy-(3x^2+1)ydx= x(x^2+1)^2dx,\ \ y(1)=2$$
what I chose to do is to order the equation to $()dy+()dx=0$ then to find integrating factor. so what I did :
$$(x^2+x)dy=(x(x^2+1)^2+(3x^2+1)y)dx$$
$$(x... | Use the quotient rule; first, divide by $x^2 (x^2+1)^2 dx$:
$$\frac{x (x^2+1) y'-(3 x^2+1) y}{x^2 (x^2+1)^2} = \frac{1}{x}$$
The LHS may be rewritten:
$$\frac{d}{dx} \frac{y}{x (x^2+1)} = \frac{1}{x}$$
Integrate both sides with respect to $x$:
$$\frac{y}{x (x^2+1)} = \log{x} + C$$
where $C$ is an integration constant. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/551232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $7200a+720b+72c=1000x+340+y<10000$? What is the easiest way to solve $7200a+720b+72c=1000x+340+y<10000$ where all variables are one digit natural numbers?
Trial and error method seems to be tedious.
| It’s immediately clear that $a$ must be $1$, which implies that $x$ must be at least $7$.
$7200a+720b+72c$ is clearly divisible by $9$. The digits of $1000x+340+y$ are $x,3,4$, and $y$, so $x+3+4+y$ must be a multiple of $9$, so either $x=7$ and $y=4$, $x=8$ and $y=3$, or $x=9$ and $y=2$. Then $1000x+340+y$ is $7344$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integral $\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}\mathrm dx$ I need your assistance with evaluating the integral
$$\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}dx$$
I tried manual integration by parts, but it seemed to only complicate the i... | It is easy to see that the integral is equivalent to
$$
\begin{align*}
\int_0^\infty \frac{1}{x\sqrt{2}+\sqrt{2x^2+1}}\frac{\log x}{\sqrt{1+x^2}}dx &= \sqrt{2}\int_0^\infty \frac{\sqrt{x^2+\frac{1}{2}}-x}{\sqrt{1+x^2}}\log x\; dx\tag{1}
\end{align*}
$$
This integral is a special case of the following generalised equati... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 3,
"answer_id": 0
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The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$
| $$x-y = x^2 + y^2 - xy \Leftrightarrow \\
2x-2y = 2x^2 + 2y^2 - 2xy \Leftrightarrow \\
0= 2x^2 + 2y^2 - 2xy-2x+ 2y \Leftrightarrow \\
(x-y)^2+(x-1)^2+(y+1)^2=2$$
As $x,y$ are integers, there are only 2 possibilities for each bracket: $0$ or $1$. So two of the squares have to be $1$ and the third one must be $0$.
Second... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555235",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| The equation can be rewritten as:
$$x^2 \left(\frac y z - 1\right) + y^2 \left(\frac z x - 1\right) + z^2 \left(\frac x y - 1\right) > 0 $$
Then since I noticed that $x > 0, y > 0, z > 0$ is not sufficient to solve the problem, I did a change of variables to convert the problem into one of that form:
$$v = \frac x y - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Proving a 2nd order Mean-Value theorem
Let $f\in C^1([a,b])$ have 2nd-order derivative in $(a,b)$. Prove that there exists $c\in (a,b)$ such that
$$f(b)-2 f\left(\frac{a+b}{2}\right)+f(a)=\frac{1}{4} (b-a)^2 f''(c)$$
| Hint: As you say, generally $a+b\not\in[a,b]$, but $\frac{a+b}{2}$ divides $[a,b]$ to two. Consider, then, the function $g\in C^1([a,\frac{a+b}{2}])$ defined $g(x) = f(x+\frac{b-a}{2})-f(x)$.
Added (some more explicit remarks, please consider before reading everything)
*
*
One notes that $g$ indeed is (twice-)diffe... | {
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"url": "https://math.stackexchange.com/questions/555478",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expressing $1+a_1(b_1+a_2(b_2+a_3(b_3+a_4(b_4+a_5(\cdots)))))$ as an infinite continued fraction. Euler derived the following identity
$$
1+a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+\cdots= \cfrac{1}{
1- \cfrac{a_{1}}{
1+a_{1}- \cfrac{a_{2}}{
1+a_{2}- \cfrac{a_{3}}{
1+a_{3} - \ddots}}}}\;... | The answer is easier than I initially thought, $(3)$ can be expressed in the form of $(2)$ as follows:
$$
1+a_{1}b_{1}\left(1+\frac{a_{2}b_{2}}{b_{1}}\left(1+\frac{a_{3}b_{3}}{b_{2}}\left(1+\frac{a_{4}b_{4}}{b_{4}}(1+\cdots)\right)\right)\right)
$$
So the answer to the question is yes.
$$
1+a_{1}(b_{1}+a_{2}(b_{2}+a_3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Integration of $\int\frac{x^2-1}{\sqrt{x^4+1}} \, dx$ Integration of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+1}} \,dx$
$\bf{My\; Try}$:: Let $x^2=\tan \theta$ and $\displaystyle 2xdx = \sec^2 \theta \, d\theta\Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta$
$$
\begin{align}
& = \int\frac{\tan \t... |
Use binomial expansions of the expressions in respective integrals as given in the previous solution of the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to find $x^4+y^4+z^4$ from equation? Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
| solution 2: since
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2x^2y^2-2y^2z^2-2x^2z^2$$
and
$$(xy+yz+xz)^2=x^2y^2+x^2z^2+y^2z^2+2xyz(x+y+z)$$
since
$$xy+yz+xz=2,xyz=-\dfrac{2}{3}$$
so $$x^2y^2+y^2z^2+x^2z^2=8$$
so
$$x^4+y^4+z^4=5^2-2\cdot 8=9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/556726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if:
$$a^2 = b(b+c)$$
where $a, b, c$ are the sides opposite to $A, B, C$ respectively.
I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\... |
1) $\angle A=2\angle B$
angle bisector of $\angle A$ cut $BC$ at $D$ $\Longrightarrow$ $\dfrac{CD}{DB}=\dfrac{b}{c}$
thus, $CD=\dfrac{b}{b+c}a$
also, $\triangle CAD\sim\triangle CAB$ $\Longrightarrow$ $CA^{2}=CD\cdot CB$
hence, $b^{2}=\dfrac{b}{b+c}a\cdot a$ $\Longrightarrow$ $a^{2}=b(b+c)$
2) $a^{2}=b(b+c)$
angle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Prove that $\sqrt{x}$ is continuous on its domain $[0, \infty).$ Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.
Proof.
Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $s \neq 0.$ Let $\delta=2\sqrt{a}\epsilon.$ Then, $\forall x \in do... | We need to prove that for any point $a \in (0, \infty)$, for every $\varepsilon>0$ there exists a $\delta > 0$ such that $$|x-a|<\delta\implies|\sqrt{x}-\sqrt{a}|<\varepsilon.$$
So, to find a $\delta$, we turn to the inequality $|\sqrt{x}-\sqrt{a}|<\varepsilon.$ Since we want an expression involving $|x-a|$, multiply b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Find the maximum and minimum values of $A \cos t + B \sin t$ Let $A$ and $B$ be constants. Find the maximum and minimum values of $A \cos t + B \sin t$.
I differentiated the function and found the solution to it as follows:
$f'(x)= B \cos t - A \sin t$
$B \cos t - A \sin t = 0 $
$t = \cot^{-1}(\frac{A}{B})+\pi n$
Howe... | Let $\displaystyle C = \sqrt{A^2 + B^2}$.
Then $\displaystyle A \cos t + B \sin t = C \left(\frac{A}{\sqrt{A^2+B^2}} \cos t + \frac{B}{\sqrt{A^2+B^2}} \sin t\right)$, and we can visualize a right triangle with opposite $A$ and adjacent $B$ to notice that we can find an angle $\phi$ such that
$\displaystyle \frac{A}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What is the value of the integral $\int_{0}^{\infty} \frac{x \ln (1+x^{2})}{\sinh (\pi x)} \, dx $? Mathematica obtains
$$\int_0^{\infty}\frac{x}{\sinh(\pi x)}\ln(1+x^2) \ \mathrm{d}x=-\frac{3}{4}+\frac{\ln 2}{6}+9\ln(A)-\frac{\ln \pi}{2}$$ where $A$ is the Glaisher-Kinkelin constant.
A numerical approximation of the... | Cody's answer gave me the idea to look at $\displaystyle \int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt $.
First add the restriction $ \text{Re}(s) >1$.
Then
$$ \begin{align} &\int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\ &= \frac{1}{2}\int_{-\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/561736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
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Imaginary numbers and polynomials question I have a task which I do not understand:
Consider $w = \frac{z}{z^2+1}$ where $z = x + iy$, $y \not= 0$ and $z^2 + 1 \not= 0$.
Given that Im $w = 0$, show that $| z | = 1$.
Partial solution (thanks to @ABC and @aranya):
If I substitute $z$ with $x + iy$ then we have $w = \frac... | Saying that the imaginary part of $w$ is zero means $w=\bar{w}$, so
$$
\frac{z}{z^2+1}=\frac{\bar{z}}{\bar{z}^2+1}
$$
and therefore
$$
z\bar{z}^2+z=z^2\bar{z}+\bar{z}.
$$
This can be rewritten as
$$
z-\bar{z}=z^2\bar{z}-z\bar{z}^2
$$
and, collecting $z\bar{z}$ in the left-hand side,
$$
z-\bar{z}=z\bar{z}(z-\bar{z}).
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Determine (without using a calculator) which of the following is bigger: $1+\sqrt[3]{2}$ or $\sqrt[3]{12}$ I have encountered the following question in a highschool book in the subject of powers. and, it seems I can't solve it....
Determine (without using a calculator) which of the following is bigger: $1+\sqrt[3]{2}$... | After reading your helpfull answers, got something of my own:
$ (1+\sqrt[3]{2})<\sqrt[3]{12} \iff (1+\sqrt[3]{2})^3<12 \iff 1+3\sqrt[3]{2}+3\sqrt[3]{4} +2 < 12 \iff 3\sqrt[3]{2}+3\sqrt[3]{4} < 9 \iff \sqrt[3]{2}+\sqrt[3]{4} < 3 \iff 2+3\sqrt[3]{2}+3\sqrt[3]{4}+4 < 27 \iff 3\sqrt[3]{2}+3\sqrt[3]{4} < 21 \iff \sqrt[3]{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Recurrence sequence limit I would like to find the limit of
$$
\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1}, & n \ge 1
\end{cases}
$$
I tried to use this - $\lim \limits_{n\to\infty}a_n=\lim \limits_{n\to\infty}a_{n+1}=L$, so $L=\lim \limits_{n\to\infty}a_n\dfrac{n^2+2n}{n^2+2n+1}=L\cdot... | It's easy to see that $a_{n+1}=a_n\left[1-\frac{1}{(n+1)^2}\right]$ and observing that
$$
\begin{align}
a_2&=a_1\left(1-\tfrac{1}{2^2}\right)\\
a_3&=a_2\left(1-\tfrac{1}{3^2}\right)=a_1\left(1-\tfrac{1}{2^2}\right)\left(1-\tfrac{1}{3^2}\right)\\
&\vdots\\
a_n&=a_{n-1}\left(1-\tfrac{1}{n^2}\right)=a_1\left(1-\tfrac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Strange mistakes when calculate limits I have difficulties with calculating the following limits. W|A gives the correct answers for both of them:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \... | For the first problem, divide everything by x first. You have now terms which look like Sqrt[1+a] where "a" is small. Apply the Taylor series, that is to say that, close to a=0, Sqrt[1+a] can be approximated by (1 + a / 2 - a^2 / 8). Are you able to continue with this ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Using the ratio Test to see if a series converges or diverges I need to find if $\sum_{n=1}^\infty\dfrac{3^n}{2^n+1}$ converges or diverges. I am trying to use the ratio test, which gives $\dfrac{3(2^n)+1}{2(2^n)+1}$. I am struggling to rearrange this into a form where it is clear what it tends to as n tends to infinit... | $$\lim_{x \to \infty} |\frac{a_{n+1}}{a_n}|$$
So, $$|\frac{3^{n+1}}{2^{n+1} + 1} \cdot \frac{2^n + 1}{3^n}|$$
$$|\frac{3^n \cdot 3}{2^n \cdot 2 + 1} \cdot \frac{2^n + 1}{3^n}|$$
$$|\frac{3 \cdot 2^n + 3}{2 \cdot 2^n +1}|$$
$$\lim_{x\to \infty} \frac{3 \cdot 2^n + 3}{2 \cdot 2^n +1} = 3/2$$
Since the limit is greater th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the radius of convergence and the interval of convergence of the series $\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$ Series is:
$$\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$$
So, I understand that I use the ratio test to find r, but I can't simplify the equation to the point where I can do this. Here's where I am so f... | HINT: use the ratio test to find r.
We put
\begin{equation}
a_n=\frac{n}{n^3+1}
\end{equation}
and
\begin{equation}
\lim =\frac{a_{n+1}}{a_n}=1
\end{equation}
Then the series converges for $|x-4|<1$. If $|x-4|=1$, we have
\begin{equation}
\sum_{n=1}^\infty \frac{n}{n^3+1}
\end{equation}
and
\begin{equation}
\frac{n}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the four digit number? Find a four digit number which is an exact square such that the first two digits are the same and also its last two digits are also the same.
| Given $\overline{aabb}=1100a+11b=k^2$, consider mod $4$:
$$k\equiv 0,1,2,3 \pmod{4} \\
k^2\equiv 0,1 \pmod{4}\\
1100a+11b\equiv 3b\equiv 0,1 \pmod{4} \Rightarrow b=0,3,4,7,8 \ \ \ \ \ \ (1)$$
Also, the last digit of $k^2$ can be:
$$b=0,1,4,5,6,9 \ \ \ \ \ \ (2)$$
Hence, from $(1)$ and $(2)$:
$$b=0 \ \ \text{or} \ \ 4.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ I need help on a homework assignment. How to show that $\lim_{n\to\infty} \left(\dfrac{n^5}{3^n}\right) = 0$? We've been trying some things but we can't seem to find the answer.
| Let $a_n = \frac{n^5}{3^n}$. Then clearly $a_n >0$ for all $n \in \mathbb{N}$. The ratio of two consecutive terms of the sequence is:
$$
\frac{a_{n+1}}{a_n} = \frac{(n+1)^5}{3^{n+1}} \frac{3^n}{n^5} = \frac{1}{3} \left(1+\frac{1}{n}\right)^5
$$
For $n \geqslant 5$:
$$
\frac{a_{n+1}}{a_n} = \frac13\left(1+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Proof of Heron's Formula for the area of a triangle
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.
Could you please provide the proof of this formula?
Thank you in advance.
| A simple derivation exploits the cosine theorem. We have $\Delta=\frac{1}{2}ab\sin C$, hence
$$ 4\Delta^2 = a^2 b^2 \sin^2 C = a^2 b^2 (1-\cos C)(1+\cos C).\tag{1}$$
On the other hand, $ 2ab\cos C = a^2+b^2-c^2$, hence
$$ 2ab(1+\cos C) = (a+b)^2-c^2 = (a+b+c)(a+b-c), \tag{2}$$
$$ 2ab(1-\cos C) = c^2-(a-b)^2 = (a-b+c)(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$ Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$
How would I do this?
| $$
\begin{equation}
\begin{split}
\frac{z-1}{z+1} &= \frac{\cos\theta + i\sin\theta-1}{\cos\theta + i\sin\theta+1} \\
&= \frac{\cos\theta-1+i\sin\theta}{\cos\theta+1+i\sin\theta} \\
&=\frac{-2\sin^2\frac{\theta}{2} + i\sin\theta}{2\cos^2\frac{\theta}{2}+i\sin\theta} \\
&= \frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots$ Evaluate $$ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots$$
All i could do was to see that $$\frac{1}{3}=\frac{1}{2.1+1},\frac{1}{5}=\frac{1}{2.2+1},\frac{1}{7}=\frac{1}{2.3+1},\dots$$
SO,... | From this, for $|x|<1,$
$$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$
$$\text{and }\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4+\cdots$$
Now subtract
Observe that $$ 1 + \frac13\frac14+\frac15\frac1{4^2}+\frac17\frac1{4^3}+\cdots$$
$$=2\sum_{0\le r<\infty}\frac1{2r+1}\left(\frac12\right)^{2r+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/580143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
About the $\lim_{h\to0}\frac{\text{arcsec}\left(\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}\right)}{h}$ I want to prove that
$$\lim_{h\to0}\frac{\text{arcsec}(x+h)-\text{arcsec}(x)}{h}=\frac{1}{|x|\sqrt{x^2-1}}.$$
without using definition of derivative and by the following method:
Because
$$\text{arcsec}(p)-\text{arc... | A simple approach is to put $\text{arcsec}\, x = y$ so that $\sec y = x$ and $\text{arcsec}\, (x + h) = y + k$ so that $\sec(y + k) = x + h$. It should be clear that $k$ tends to zero with $h$. And then the desired limit is $\lim_{k \to 0}\dfrac{k}{\sec(y + k) - \sec y}$ which can be easily handled by changing $\sec$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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.