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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 $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
| \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
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451199",
"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 dodecahedron, how would I find the angle between two adjacent vertices on the same face, and the angle between two vertices on the same face but not connected by an edge?
| 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 isosceles with legs $r$ and base $s$; applying the Law of Cosines to its vertex angle, we have ...
$$s^2 = r^2 + r^2 - 2 r\cdot r\cos\angle AOB = 2 r^2 \left( 1 - \cos\angle AOB \right) = 4 r^2 \sin^2\frac{1}{2}\angle AOB$$
so that
$$\sin\frac{1}{2}\angle AOB = \frac{s}{2r} = \frac{2s}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6}$$
whence
$$\angle AOB = 2 \arcsin \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6} = 41.8103\dots^\circ$$
If $A$ and $C$ are non-adjacent vertices on a face, then $d := |AC|$ is a diagonal of a regular pentagon with side length $s$. Thus,
$$d = \frac{s}{2}\left( 1 + \sqrt{5} \right)$$
Just as above, we can compute
$$\sin\frac{1}{2}\angle AOC = \frac{d}{2r} = \frac{s\left(1+\sqrt{5}\right)}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}}{3}$$
whence
$$\angle AOC = 2 \arcsin \frac{\sqrt{3}}{3} = 70.5288\dots^\circ$$
(You may recognize this as the central angle between adjacent vertices of a cube. It's often helpful to realize that a dodecahedron's face diagonals form the edges of a family of cubes, as shown in the Wikipedia entry. Moreover, one can think of constructing a dodecahedron by taking a cube and pitching a pup-tent on each face, where a triangular tent face and a quadrilateral tent face form a regular pentagon.)
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
$$ \implies \sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1 $$
$$ \implies (\sqrt{x-1}-2) + (\sqrt{x-1}-3) = 1 $$
$$ \implies \sqrt{x-1} = 3$$
So, $ x = 10$ is the only solution.
But the answer key (and Wolfram alpha too) says there are infinite number of solutions to this equation. Where I am going wrong?
| 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 we obtain:
$$
1=|\sqrt{x-1}-2| + |\sqrt{x-1}-3| = (\sqrt{x-1}-2) - (\sqrt{x-1}-3) = 1
$$
Since we obtained a tautology ($1=1$), it follows that any $x\in[5,10]$ satisfies the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452927",
"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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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,
$$\log\cos\vartheta_k=-\frac{1}{2}\log(1+\tan^2\vartheta_k)>-\frac{1}{2}\tan^2\vartheta_k=-\frac{1}{2}\frac{\sin^2\vartheta_k}{\cos^2\vartheta_k}>-2\vartheta_k^2.$$
Therefore we can write that
$$\log P_n>-2\pi^2\sum_{k=1}^n \frac{1}{(k+2)^2}>-2\pi^2\int_2^{\infty}\frac{dx}{x^2}=-\pi^2$$
i.e. $P_n>e^{-\pi^2}$.
| 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)\\
&= \prod_{k = 0}^\infty \frac{\sin\frac{2x}{2k + 1}}{\frac{2x}{2k + 1}}
\end{align}
which could be used to approximate your product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 $b$.
$$
A = \begin{pmatrix}
1 & -3 & 2 \\
3 & -8 & 8 \\
0 & 1 & 2 \\
1 & 0 & 8
\end{pmatrix} \qquad,\qquad
b = \begin{pmatrix}
1 \\
6 \\
3 \\
10
\end{pmatrix}
$$
What I have done so far is that I've combined the two matrices into a augmented matrix. And row reduced it to get:
$$\begin{pmatrix}
1 & -3 & 2 & 1 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
So does this just mean that the answer to the question is $\mathbf x = \begin{pmatrix}
1 \\ 3 \\ 0 \\ 0 \end{pmatrix} $??
| 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 $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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}} - \frac{1}{3z} + O(z)$.
Then $ \displaystyle\text{Res} \Big[ \frac{\cot z}{z^{2}(2-z)},0 \Big] = \frac{1}{2^{3}} - \frac{1}{3(2)} = - \frac{1}{24}$.
It came up in a proof of the Mittag-Leffler partial fractions expansion theorem.
| Good old-fashioned computation with Laurent series.
$$\begin{align}
\frac{f(z)}{w-z} &= \frac{f(z)}{(w-a) - (z-a)} = \frac{1}{w-a} f(z) \frac{1}{1 - \frac{z-a}{w-a}}\\
&= \frac{1}{w-a}f(z)\sum_{\nu = 0}^\infty \left(\frac{z-a}{w-a}\right)^\nu\\
&= \frac{1}{w-a}\left(\sum_{k = -m}^\infty a_k (z-a)^k\right)\sum_{\nu = 0}^\infty \left(\frac{z-a}{w-a}\right)^\nu\\
&= \frac{1}{w-a} \sum_{s = -m}^\infty \left(\sum_{k = -m}^s\frac{a_k}{(w-a)^{s-k}}\right)(z-a)^s
\end{align}$$
The residue is the coefficient of $(z-a)^{-1}$, that is
$$\frac{1}{w-a}\sum_{k = -m}^{-1} \frac{a_k}{(w-a)^{-1-k}} = \sum_{k=-m}^{-1} \frac{a_k}{(w-a)^{-k}} = \sum_{k=-m}^{-1} a_k(w-a)^k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/455162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 heights seem to not converge to zero. A function that rises over reducing intervals, such as $f_n(x) = \sqrt{n} {\bf 1}_{[0,\frac{1}{n})}$ wouldn't satisfy the last condition…
Would you have any suggestions on how I could find such functions?
| Let
$$f_n(x) = \begin{cases}\frac{1}{x} &, \frac{1}{n+1} \leqslant x < \frac{1}{n}\\
(n+1)2^{n+1}\left(x -\frac{1}{n+1} + \frac{1}{2^{n+1}}\right) &, \frac{1}{n+1} - \frac{1}{2^{n+1}} \leqslant x < \frac{1}{n+1}\\
n2^n\left(\frac1n + \frac{1}{2^n} - x\right) &, \frac{1}{n} < x < \frac1n + \frac{1}{2^n}\\
0 &, \text{ otherwise}. \end{cases}$$
Then $f_n \to 0$ pointwise, $\int_0^1 f_n(x)\,dx = \log (1 + 1/n) + \frac{n}{2^{n+1}} + \frac{n+1}{2^{n+2}} \to 0$, but $\sup_n f_n(x) = \frac1x \notin L^1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456599",
"timestamp": "2023-03-29T00:00:00",
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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 written as $(p+q)^2$
| 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 $\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b}-1 =(49+20\sqrt[3]{6})^{1/2}$, hence $1\le a\le788$. In doing so, you will presumably stumble across the solution $(48,288)$ and its reverse, and no others.
You can streamline this a bit by looking for solutions with $a\le b$ so that
$$2\sqrt[3]{a}\le\sqrt[3]{a}+\sqrt[3]{b} = 1+(49+20\sqrt[3]{6})^{1/2}=10.238\ldots$$
leads to $a\le134$. Whether you can easily streamline it further is less clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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",
"url": "https://math.stackexchange.com/questions/457146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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 solutions in this case ?
| 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 $\gcd (a+b,a-b) = \gcd(a+b,2) \in \{1,\,2\}$, we have
$$\frac{(a+b)(a-b)}{\gcd(a+b,a-b)} \mid b^2-1 \Rightarrow \frac{a^2-b^2}{2} \leqslant b^2-1 \Rightarrow a^2 \leqslant 3b^2 - 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457289",
"timestamp": "2023-03-29T00:00:00",
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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
$$\frac{\sqrt{3}-1}{2} +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$
Now by taking two as common and separating the positive and negative terms gives
$$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$
Can we do like this please suggest. Thanks.
| 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:
$$\sum_{i = 1}^{4999}{\frac{1}{\sqrt{2i-1} +\sqrt{2i+1}}} \gt \sum_{i = 1}^{4999}{\frac{1}{2\sqrt{2i}}} = \frac{1}{2\sqrt{2}}\sum_{i = 1}^{4999}{\frac{1}{\sqrt{i}}} \ge \frac{1}{2\sqrt{2}}\frac{4999}{\sqrt{\frac{\sum_{i = 1}^{4999}{i}}{4999}}} = \frac{1}{2\sqrt{2}} \frac{4999}{\sqrt{2500}} \approx 35.35$$
The last inequality is obtained by using the Root mean square-Harmonic mean inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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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 sides, but the chapter is of complex numbers and more specifically, cube roots of unity. How do I solve this using that idea?
The best I've come up with is to replace $-1$ with $\omega+\omega^2$, but I later seem to apply the reverse and thus, render the use of $\omega$ pointless. Some guidance, please!
| 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 using Complex Numbers,
$$X^2-YZ=(ax+cy+bz)^2-(cx+by+az)(bx+ay+cz)$$
$$=(a^2-bc)x^2+(c^2-ab)y^2+(b^2-ca)z^2+(2ca-b^2-ca)xy+(2ab-c^2-bc)zx+(2bc-ab-a^2)yz$$
$$\implies\sum_{\text{cyc}}(X^2-YZ)=(x^2+y^2+z^2-xy-yz-zx)(a^2+b^2+c^2-ab-bc-ca)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464333",
"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 equation
2.making it linear differential equation
But I do not understand when i will do substitution.
Please help me solve this question.
| 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:
$$\tag 3 \dfrac{dy}{dx} = 2x z + x^2 \dfrac{dz}{dx}$$
Substituting $(2)$ into $(1)$ yields:
$$\tag 4 \frac{dy}{dx}= \dfrac{2 x^2 z}{x} +\dfrac{x^3}{x^2 z} +x \tan \frac{x^2 z}{x^2} = 2 x z + \dfrac{x}{z} +x \tan z$$
Now, since we have two expressions for $\dfrac{dy}{dx}$, we can equate $(3)$ and $(4)$, yielding:
$$2x z + x^2 \dfrac{dz}{dx} = 2 x z + \dfrac{x}{z} +x \tan z$$
Simplifying, yields:
$$x^2 \dfrac{dz}{dx} = \dfrac{x}{z} +x \tan z$$
Simplifying yields:
$$\dfrac{dz}{dx} = \dfrac{z \tan z + 1}{x z}$$
This can be written as:
$$\tag 5 \displaystyle \int\dfrac{z ~ dz}{z \tan z + 1} = \int \dfrac{dx}{x}$$
Now, you can integrate each side of $(5)$ and then substitute in for $z$ and you are done.
Upon integrating, you get:
$$ \ln(z \sin z + \cos z) = \ln x + c$$
You have an expression for $z$, substitute and you can simplify a bit.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\gcd (12,15)=3$. How can I deal with this?
| 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.g.: $$x \equiv 6 \pmod{12} \Leftrightarrow \begin{cases}x \equiv 0 \pmod{3} \\
x \equiv 2 \pmod{2^2}\end{cases} $$
*Group the equations according to the base prime of the modulus.
*Whenever you have more than one equation for a prime, choose only the/an equation with the highest prime power.
*Then, compare the other equations to the choosen one. They either contradict the first one or are redundant (they are implied by the first one). Examine them one by one and either cut them off or conclude that there are no solutions.
Example:
$$x\equiv 1\pmod{2} \tag{1}$$
$$x\equiv 3\pmod{8} \tag{2}$$
$$x\equiv 3\pmod{8} \tag{3}$$
$$x\equiv 1\pmod{4} \tag{4}$$
Choose equation (2). (2) obviously implies (1). So, we can ignore (1). Same for (3). (4) however contradicts (1), so there are no solutions to this system. If we instead had $x\equiv 3\pmod{4}$ in (4), then this would be a redundancy, too, and the whole system would simplify to $x\equiv 3\pmod{8}$.
In your case:
$$\begin{align*}x \equiv 9 \pmod{11} &\Leftrightarrow x \equiv 9 \pmod{11}\\
x \equiv 6 \pmod{13} &\Leftrightarrow x \equiv 6 \pmod{13} \\
x \equiv 6 \pmod{12} &\Leftrightarrow \begin{cases}x \equiv 0 \pmod{3} \\
x \equiv 2 \pmod{4}\end{cases} \\
x \equiv 9 \pmod{15} &\Leftrightarrow \begin{cases} x \equiv 0 \pmod{3} \\
x \equiv 4 \pmod{5}\end{cases} \end{align*}$$
Only $3$ occurs twice, and the equations are obviously identical. So you are left with five equations with pairwise coprime modulos and you can apply CRT.
| {
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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: $$\begin{align} 4(x^2 -4x + 4) + 3(y^2 + 8 y + 16) - 12 & = 0\\ \\ 4(x - 2)^2 + 3(y + 4)^2 &= 12\end{align}$$
Now divide through by $12$.
$$\dfrac{(x-2)^2}{3} + \frac{(y+4)^2}{4} = 1 \iff \left(\dfrac{x-2}{\sqrt 3}\right)^2 + \left(\frac{y+4}{2}\right)^2 = 1$$
This equation describes an ellipse with center $(2, -4)$.
| {
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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 sides, to reduce the number of surds that are involved. For example, to determine if $$\sqrt{2} - \sqrt{3} - \sqrt{5} + \sqrt{7} > 0, $$ we can square both sides of $\sqrt{2} + \sqrt{7} > \sqrt{3}+\sqrt{5} $ to obtain $$9 + 2 \sqrt{14} > 8 + 2 \sqrt{15}.$$
Repeated squaring eventually resolves this question, as the number of surds are reduced.
However, when there are more than 6 terms involved, then repeated squaring need not necessarily reduce the terms that are involved.
E.g. How would you determine if
$$\sqrt{2} - \sqrt{3} + \sqrt{5} - \sqrt{7} - \sqrt{11} + \sqrt{13} < 0 $$
I can think of several approaches
*
*There are special cases, which allow us to apply Jensen's inequality. However, this gives a somewhat restrictive condition on the set of values.
*Show that $$ \sqrt{2} + \sqrt{5} + \sqrt{13} < 7.26 < \sqrt{3} + \sqrt{7} + \sqrt{11} $$
However, it might not be feasible to guess what the middle number is, unless you already had a calculator.
*Calculate the surds to the appropriate level of precision (e.g. use Taylor expansion). This could be a little annoying.
Do you have any other suggestions?
| 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
$$ A = bc,\quad B=ac,\quad C=ab,\quad K = a^2+b^2+c^2,\quad \dots $$
we get an expression of the form:
$$ K + 2\sqrt{A} + 2\sqrt{B} + 2\sqrt{C} < K' + 2\sqrt{A'} + 2\sqrt{B'} + 2\sqrt{C'}\quad\quad (*)$$
now square again, you get another expression of the same shape
$$ K_1 + (4K+2a) \sqrt{A} + (4K+2b)\sqrt{B} + (4K+2c)\sqrt{C} <\\ K'_1 + (4K'+2a') \sqrt{A'} + (4K'+2b')\sqrt{B'} + (4K'+2c')\sqrt{C'} \quad\quad(**)$$
Now if $X,Y$ and $\theta$ are positive numbers and $X \ge \theta /2$, $Y \ge \theta /2$ then as it is easy to check
$$ X^2-\theta X < Y^2 - \theta X \quad\quad \text{if and only if}\quad\quad X < Y $$
letting $X$ equal to the left hand side of (*), $Y$ equal to the right hand side and $\theta = 2K+a$, (assuming $a$ is the smallest of $a,b,c,a',b',c'$) as we have trivially
$$ X = K + 2 \sqrt{A} + \dots > \frac{2K+a}{2} = \frac{\theta}2 $$
and the same for $Y$, we can subtract $2K+a$ times (*) from (**) and obtain an inequality equivalent to the original but with one surd less.
With some care I think you can manage to extend this to positive and negative coefficients. But I can't see how to extend it in general for larger number of surds. For example if you have four surds in one side then squaring twice you get at most seven surds
$$ \sqrt{ab}, \sqrt{ac}, \sqrt{ad}, \sqrt{bc}, \sqrt{bd}, \sqrt{cd},\sqrt{abcd} $$
if you manage somehow to reduce the number of surds but leave more than three surds, then squaring again a couple of times recovers all the surds, so you need to reduce four surds in a single blow.
By the way in your example squaring the inequality we see that it is equivalent to
$$ 20 + 2 \sqrt{10} + 2\sqrt{26}+2\sqrt{65} < 21 + 2 \sqrt{21}+2\sqrt{33}+2\sqrt{77} $$
in this case you have finished as every term in the left is smaller than the corresponding term in the right.
| {
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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 identities for these? Thanks!
| 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
$$
x = a + b \\ y = a - b
$$
substitute
$$
Q = (A+B) \cos(\frac{x-y}{2}) \\
R = (B-A) \sin(\frac{x-y}{2}) \\
P = \frac{x+y}{2}
$$
then
$$
A \cos(x) + B \cos(y) = Q \cos P + R \sin P = \sqrt{Q^2+R^2} \cos(P-\phi)
$$
where
$$
\sin \phi = \frac{R}{\sqrt{Q^2+R^2}} \\
\cos \phi = \frac{Q}{\sqrt{Q^2+R^2}}
$$
| {
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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} \text{d}u \\
&\int \frac{1}{\sqrt{u^2+1}} \text{d}u\\
\end{align}
Now, this is where I am having trouble. How can I evaluate that? Please provide only hints
Thanks!
EDIT:
The problem specifically states that one must use substitution with $u = \sqrt{x^2-1}$. This problem is from the coursera course for Single Variable Calculus.
| 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 $x\,dx=\sqrt{x^2 - 1}\,du = u \,du$.
Note that $$u = \sqrt{x^2 - 1} \implies u^2 = x^2 - 1 \iff x^2 = u^2 + 1 $$
So substituting gives us $$\int \frac{x\,dx}{x^2\sqrt{x^2-1}} = \int \dfrac{u \,du}{(u^2 + 1)u} = \int \frac {du}{u^2 + 1}$$
Now, we can use trigonometric substitution, and given a denominator of the form $u^2 + 1$, put $u = \tan \theta$. This gives us: $$\int \frac {du}{u^2 + 1} = \arctan(u) + C = \arctan(\sqrt{x^2 - 1}) + C$$
| {
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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
$$\sum_{n=0}^{\infty} \frac {x^{3^n}+(x^{3^n})^2} {1-x^{3^{n+1}}} = \frac{1}{1-x} - \lim_{N\to\infty}\frac{1}{1-x^{3^{N+1}}}
= \frac{1}{1-x} -
\begin{cases}
1, & |x| < 1\\
\frac12 & x = -1\\
\text{undefined}, & x = 1\\
0, & |x| > 1\\
\end{cases}$$
| {
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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, put these points on a number line and plug in test numbers form each possible section to determine whether the factor would be positive or negative. Then, given that you're looking for positive solution, think about the possibilities: (+)(+)(+) = +,
++- = - , -+- = +, --- = -. Therefore, your solution is $-1 < x < 0$ or $x > 5$.
so i know how he got the $x>5$ but i don't get the $-1 < x < 0$ cuz it suppose to be $x > -1$ and what these possibilities have to do with the solution ? please explain to me in details.
(Q.2) solve for x in $x^\frac{5}{3} - 6x = x^\frac{4}{3}$
same book. The answer says: Next, factor out an $x$ from each term: $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$. The resulting expression is similar to $y^3(y^2 - y - 6)$, which factors into $y^3(y+2)(y-3)$.Similarly, you can factor $x(x^\frac{2}{3} - x^\frac{1}{3} - 6)$ into $x(x^\frac{1}{3} + 2)(x^\frac{1}{3} - 3) = 0$
I don't get how did he factor the main equation and $x^\frac{5}{3}$ became $x^\frac{2}{3}$ and $x^\frac{4}{3}$ became $x^\frac{1}{3}$. I know how to factor like this but with numbers not fractions. And also how this expression $x(x^\frac{2}{3} - x^\frac{1}{3} - 6) = 0$ is similar to that $y^3(y^2 - y - 6)$. I'm solving for x so why he got the y into the answer now ?! (weird)
(Q.3) simplify $\frac{8}{4^\frac{2}{3}}$
same book. The answers says: change it to $\frac{8}{\sqrt[3]{4^2}}$ ( i get this one ) but then he said multiply the numerator and denominator by one more cube root of 4 . so how i can multiply $\frac{8}{\sqrt[3]{4^2}}$ to $\frac{8}{ \sqrt[3]{4}}$ and get $4$ ? isn't supposed to be $\frac{8}{\sqrt[3]{4^2}}*\frac{8}{\sqrt[3]{4^2}}$equals $4$ ?
| 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 this definition.
Any help is kindly appreciated.
| 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 polar coordinate:
$$
I = \iint \frac{r\ \mathrm dr\ \mathrm d\theta}{1+r^4(\cos\theta+\sin\theta)^4} = \int_0^{\pi/2}\mathrm d\theta\int_0^1\frac{\mathrm d(r^2)}{1+(\cos\theta+\sin\theta)^4(r^2)^2}.
$$
Then, I used the well-known technique to get
$$
I = \int_0^{\pi/2}\frac{\arctan (1+\sin 2\theta)}{1+\sin 2\theta}\mathrm d\theta.
$$
The problem is how to evaluate this. I could get a simpler formula by substituting $1+\sin 2\theta$, but I would not proceed further.
I would be grateful if you give a clue (not necessarily a complete solution).
| 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:
$$\int_0^{\frac{1}{\sqrt{2}}} \frac{2\xi}{1 + 4\xi^4} d\xi = \frac{1}{2}\int_0^1 \frac{du}{1+u^2} = \frac{\pi}{8}$$
For the $2^{nd}$ integral, let $\cos\theta = \xi$ and then let $t = \tan\theta$, we have:
$$\int_{\frac{1}{\sqrt{2}}}^1 \frac{2\sqrt{1-\xi^2} d\xi}{1+4\xi^4}
= \int_0^{\frac{\pi}{4}} \frac{2\sin(\theta)^2 d\theta}{1 + 4\cos(\theta)^4}
= \int_0^{1} \frac{2 \left(\frac{t^2}{1+t^2}\right)\left(\frac{dt}{1+t^2}\right) }{1 + 4\left(\frac{1}{1+t^2}\right)^2}
= \int_0^1 \frac{2t^2 dt}{t^4+ 2t^2 + 5}
$$
Since
$$\frac{2t^2}{t^4 + 2t^2 + 5} = \frac{2t^2}{(t^2+1)^2+4}
=\frac{t^2}{2i}\left(\frac{1}{t^2+1-2i} - \frac{1}{t^2+1+2i}\right)
=\frac{i}{2}\left(\frac{1-2i}{t^2+1-2i} - \frac{1+2i}{t^2+1+2i}\right)
$$
We can evaluate the $2^{nd}$ integral and get following ugly expression:
$$\frac{i}{2} \left(
\sqrt{1-2i} \tan^{-1}\left(\frac{1}{\sqrt{1-2i}}\right)
- \sqrt{1+2i} \tan^{-1}\left(\frac{1}{\sqrt{1+2i}}\right)
\right)$$
Since $\sqrt{1-2i} = \frac{\varphi-i}{\sqrt{\varphi}}$ where $\varphi$ is the golden ratio,
the original integral can be expressed as:
$$\begin{align}&\frac{\pi}{8} + \frac{i}{2\sqrt{\varphi}}\left(
(\varphi - i)\tan^{-1}\left(\frac{\varphi + i}{\sqrt{5\varphi}}\right)
- (\varphi + i)\tan^{-1}\left(\frac{\varphi - i}{\sqrt{5\varphi}}\right)
\right)\\
= & \frac{\pi}{8} + \frac{1}{2\sqrt{\varphi}}\left(
\tan^{-1}(\sqrt{\varphi}^3)
-\varphi \tanh^{-1}(\frac{1}{\sqrt{\varphi}^3})
\right)
\end{align}$$
Numerically, the integral $\sim 0.3926990817+0.1021715030 = 0.4948705847$.
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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 with the negatives.
Please clear it for me.
| First, we'll express the product as a single fraction, and use the commutativity of multiplication to "rearrange" terms:
$$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(2\cdot \color{red}{\bf 8})\;a^2b^2c^2\;xy^2z^5}$$
Now, we can cancel common integer factors that appear in both numerator and denominator, and multiply the remaining integer factors:
$$\frac{(3\cdot \color{red}{\bf 8})\cdot( 3\cdot \color{blue}{\bf 5})\;a^4b^2c^3\; x^3y^3 z^3}{(5\cdot \color{blue}{\bf 5})\cdot(2\cdot \color{red}{\bf 8})\;a^2b^2c^2\;xy^2z^5}= \frac{9a^4b^2c^3x^3y^3z^3}{10\;a^2 b^2 c^2 x^2z^5}$$
Now, using the fact that $\dfrac {k^n}{k^m} = k^{n-m}$ for $n \geq m$, and that $\dfrac {k^n}{k^m} = \dfrac 1{k^{m-n}}$ for $n \lt m$, then simplifying, we have:
$$\begin{align} \frac{9a^4b^2c^3x^3y^3z^3}{10\;a^2 b^2 c^2 x^2z^5} & = \frac{9 a^{4-2}b^{2-2}c^{3-2}x^{3-1}y^{3-2}}{10z^{5-3}}\\ \\ & = \frac{9a^2b^0cx^2y}{10z^2} \\ \\ & = \frac{9a^2cx^2y}{10z^2}\end{align}$$
Putting all steps together, we have $$\frac{24a^4b^2c^3}{25xy^2z^5} \cdot \frac{15x^3y^3z^3}{16a^2b^2c^2} = \frac{9a^2cx^2y}{10z^2}$$
| {
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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 equivalent to $-\frac{1+x^2}{x} + c$ by drawing a right triangle on the blackboard.
I get the math behind each step of it, but I can't wrap my head around why that equivalence works. We just used an arbitrary $x = \tan \theta$ substitution, where $\theta$ moves differently than x does, and the expression $-\frac{1+x^2}{x} + c$ by itself doesn't know anything about trigonometry. But I type both into Excel for a bunch of different x values, and obviously they are equivalent.
I guess I'm not really sure what my question is here, but I could just use some perspective. It just seems like substituting ANY function in for x then integrating it shouldn't work, especially when crossing into polar coordinates.
| 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 relating square roots of the sum or difference of two squared terms:
$ \sqrt{x^2 + a^2} \ $ can only be the hypotenuse of a right triangle with legs of lengths $ \ x \ $ and $ \ a \ $ . It is the usual practice to make $ \ x \ $ the leg opposite the angle $ \ \theta \ $ and $ \ a \ $ the "adjacent" leg, which produces the ratio $ \ \tan \theta = \frac{x}{a} \ $ , as described above. The associated differential is then $ \ dx = a \sec^2 \theta \ d\theta \ $ ; the radical is then given by $ \frac{\sqrt{x^2 + a^2}}{a} = \sec \theta \ \rightarrow \ \sqrt{x^2 + a^2} = a \sec \theta \ $ .
The other arrangements involve differences of squared terms, which requires the radical to be the length of one of the triangle's legs. We have either
$ \sqrt{x^2 - a^2} \ $ , which makes $ \ x \ $ the hypotenuse; generally, the side adjacent to $ \theta \ $ is chosen as the leg $ \ a \ $ , leading us to $ \ \sec \theta = \frac{x}{a} \ \rightarrow \ dx = a \sec \theta \tan \theta \ d\theta \ $ and $ \frac{\sqrt{x^2 + a^2}}{a} = \tan \theta \ \rightarrow \ \sqrt{x^2 + a^2} = a \tan \theta \ $
or
$ \sqrt{a^2 - x^2} \ $ , making $ \ a \ $ the hypotenuse; the side opposite to $ \theta \ $ is usually chosen to be the leg $ \ x \ $ , giving $ \ \sin \theta = \frac{x}{a} \ \rightarrow \ dx = a \cos \theta \ d\theta \ $ and $ \frac{\sqrt{x^2 + a^2}}{a} = \cos \theta \ \rightarrow \ \sqrt{x^2 + a^2} = a \cos \theta \ $ .
$$\\$$
Returning to your integral, the "tangent substitution" (with $ \ a \ = 1 \ $ ) produces
$$\int \ \frac{1 \cdot \sec^2 \theta \ d\theta}{1^2 \cdot \tan^2 \theta \ \cdot \ 1 \cdot \sec \theta} \ = \ \int \ \frac{\cos \theta \ d\theta}{ \sin^2 \theta } \ , $$
which can now be completed through a "$v-$subsitution" , $ \ v = \sin \theta \ $ , yielding the result
$$\int \ v^{-2} \ dv \ = \ -v^{-1} \ + \ C \ \rightarrow \ -\frac{1}{\sin \theta} \ + \ C \ \ \text{or} \ \ -\csc \theta \ + \ C \ . $$
[Maybe this was already clear to you, but you also seemed to be asking about the rationale for the choice of particular trig functions...]
Referring back to the associated right triangle, the side opposite $ \ \theta \ $ is $ \ x \ $ and the hypotenuse is $ \sqrt{x^2 + 1^2} \ $ , so the "back substitution" to return to a function of $ \ x \ $ gives us $ \ -\frac{\sqrt{x^2 + 1}}{x} \ + \ C \ $ . (I believe you have omitted a radical in your expression; a check against WolframAlpha confirms this...)
Were we to simply construct a right triangle with the legs and hypotenuse described (and thus $ \ \theta = \arctan \frac{x}{1} \ $ ) and ask for the expression representing $ \ -\csc \theta \ $ , we would obtain this same result.
| {
"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 $\lfloor x\rfloor$ is the largest integer not greater than $x$.
I can neither prove this nor find any counterexample even by using computer.
Could you show me how to prove this? Or could you get an counterexample?
Update: I've just asked a related question.
Generalization of $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$
| 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
$$
\begin{align}
\sqrt{n+2+x}
=\sqrt{n+2}+\frac{x}{2\sqrt{n+2}}-\int_0^x\frac{(x-t)\,\mathrm{d}t}{4\sqrt{n+2+t}^3}\tag{2}
\end{align}
$$
Summing $(2)$ for $x\in\{-2,-1,0,1,2\}$ yields
$$
\hspace{-1cm}5\sqrt{n+2}-\frac5{4n^{3/2}}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}\tag{3}
$$
Since
$$
\begin{align}
\sqrt{25n+50}-\sqrt{25n+49}
&=\frac1{\sqrt{25n+50}+\sqrt{25n+49}}\\
&\gt\frac1{2\sqrt{25n+50}}\\
&\gt\frac5{4n^{3/2}}\quad\text{for }n\ge14\tag{4}
\end{align}
$$
we get that for $n\ge14$,
$$
\hspace{-5mm}\sqrt{25n+49}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt\sqrt{25n+50}\tag{5}
$$
$(5)$ says that for $n\ge14$,
$$
\left\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right\rfloor
=\left\lfloor\sqrt{25n+49}\right\rfloor\tag{6}
$$
It is simple to verify $(6)$ for $1\le n\lt14$ (it is false for $n=0$).
| {
"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 $q\ge m+1$ occurs on the right side and with what coefficients. It occurrs for the first time in the term for $n=q-m$ and for the last time in $n=q-1,$ so its coefficient is (keeping in mind that $k = q-n-1$)
$$\sum_{n=q-m}^{q-1} (-1)^{q-n-1} \binom{m-1}{q-n-1}.$$
Put $p=q-n-1$ in the above sum to get
$$\sum_{p=q-(q-m)-1}^{q-(q-1)-1} (-1)^p \binom{m-1}{p} =
\sum_{p=m-1}^0 (-1)^p \binom{m-1}{p} = \sum_{p=0}^{m-1} (-1)^p \binom{m-1}{p} = 0,$$
so for $q\ge m+1$ the sum telescopes and the contribution of the value $1/q$ is zero.
The leftover contribution is
$$\sum_{q=2}^m \frac{1}{q}\sum_{n=1}^{q-1} (-1)^{q-n-1} \binom{m-1}{q-n-1}
= \sum_{q=2}^m \frac{1}{q}\sum_{n=0}^{q-2} (-1)^n \binom{m-1}{n}$$
To conclude we ask where $\binom{m-1}{n}$ appears and for what values of $q$, getting
$$ \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} \sum_{q=2+n}^m \frac{1}{q}
= \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} \left(H_m - H_{n+1}\right).$$
Split the sum in two to obtain first,
$$H_m \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} =
H_m (0 - (-1)^{m-1} ) = (-1)^m H_m$$
and second,
$$ \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} H_{n+1} =
\sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1} - (-1)^{m-1} H_m =
\sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1} + (-1)^m H_m$$
so the difference between the two is
$$- \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1}.$$
This is the binomial transform of $H_{n+1}$ and since
$$\sum_{n\ge 1} H_{n+1} z^n = f(z) = \frac{1}{z} \frac{1}{1-z} \log \frac{1}{1-z}$$
the sum has generating function
$$\frac{1}{1-z} f\left(\frac{z}{z-1}\right) =
\frac{1}{z} (1-z) \log \frac{1}{1-z} = \left(\frac{1}{z}-1\right) \log \frac{1}{1-z}
\\= \sum_{m\ge 1} \left(\frac{1}{m+1}-\frac{1}{m}\right) z^m
= -\sum_{m\ge 1} \frac{1}{m(m+1)} z^m.$$
Observe that we need the coefficient of $z^{m-1}.$ It follows that the original sum has value
$$ \frac{1}{(m-1)!} \frac{1}{(m-1)m} = \frac{1}{(m-1)m!}$$
as was to be shown.
I will prove the partial fraction identity in an additional answer.
| {
"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 $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4=\frac{15}{32}...$ So it seems the pattern is $a_n = \frac{2^{n}-1}{2^{n+1}}$, but I have been unable to show this algebraically. Here is what I tried:
$$\begin{align*}
2a_n &= \frac{n}{2} - \sum_{k=1}^{n-1}a_k\\
a_n + \sum_{k=1}^n a_k &= \frac{n}{2}\\
a_{n-1} + \sum_{k=1}^{n-1} a_k &= \frac{n-1}{2} \\
2a_n - a_{n-1} & = \frac{1}{2} \\
a_n = \dfrac{2a_{n-1} + 1}{4}
\end{align*}$$
I am so close, I can taste the closed form. Can someone nudge me in the right direction without giving too much away?
| 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 $$\begin{cases}b_{n+1}-\frac{1}{2}b_{n}=\frac{n}{4},{\quad}n=2,3,\cdots,\\ b_{2}=\frac{1}{4}.\end{cases}$$
Let $\mu_{n}:=2^{n}$ for $n=2,3,\cdots$.
Multiplying both sides of the equation by $\mu_{n+1}$, we get
$$\mu_{n+1}b_{n+1}-\mu_{n}b_{n}=\mu_{n}\frac{n}{2}.$$
Summing this from $2$ to $(n-1)$ for $n=2,3,\cdots$, we get
$$\begin{aligned}&\underbrace{\sum_{k=2}^{n-1}[\mu_{k+1}b_{k+1}-\mu_{k}b_{k}]}_{\text{telescoping sum}}=\frac{1}{2}\sum_{k=2}^{n-1}k2^{k}\\ &{\implies}\mu_{n}b_{n}-\mu_{2}b_{2}=\frac{1}{2}2^{n}(n-2)\\ &{\implies}b_{n}=\frac{1}{2^{n}}+\frac{1}{2}(n-2),\end{aligned}$$
where we have used the fact that $\mu_{2}b_{2}=1$.
Then, the desired solution is $$a_{n}=b_{n+1}-b_{n}=\frac{1}{2}\bigg(1-\frac{1}{2^{n}}\bigg),{\quad}n=2,3,\cdots.\tag*{$\blacksquare$}$$
| {
"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 $(24 \cdot 8 + 240 \cdot 8 + 2400 \cdot 8 + 24000 \cdot 8)$. In total, we have $$(0 + 2 + 3 + 5 + 8)(24 + 240 + 2400 + 24000) = 479952$$ as our total sum.
Update: In case 4-digit numbers cannot start with 0, then we have overcounted. Now we have to subtract the amount by which we overcounted, which is found by answering: "What is the sum of all 3-digit numbers formed by using digits 2,3,5, and 8?" Now if 8 appears as the last digit, then there are 6 ways to complete the number, so 8 contributes $(6\cdot 8 + 60 \cdot 8 + 600 \cdot 8)$. In total, we have
$$(2 + 3 + 5 + 8)(6 + 60 + 600) = 11988.$$
Subtracting this from the above gives us 467964.
| {
"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.
$$\int \frac{dv}{1+v} = \log{(1+v)}$$
$$\int \frac{dv}{1-v+v^2} = \int \frac{dv}{(v-1/2)^2+3/4} = \frac{2}{\sqrt{3}} \arctan{\frac{2 v-1}{\sqrt{3}}}$$
$$\int dv \frac{v}{1-v+v^2} = \int dv \frac{v-1/2}{(v-1/2)^2+3/4} +\frac12 \int \frac{dv}{(v-1/2)^2+3/4} = \\\frac12 \log{[(v-1/2)^2+3/4]} + \frac{1}{\sqrt{3}} \arctan{\frac{2 v-1}{\sqrt{3}}}$$
I get, putting this all together,
$$\int dx \, (\tan{x} )^{-1/3} = \frac12 \log{\frac{1+v}{1-v+v^2}} + \frac{\sqrt{3}}{2} \arctan{\frac{2 v-1}{\sqrt{3}}} +C$$
where $v=(\tan{x})^{-2/3}$.
| {
"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, $p=(5q+1)/2$ and $r=(5q-1)/6$
So, $a=15q+11$
Now if I put $q=67$, $a=1016$ (wrong answer because $r$ is not an integer) .
So where did I go wrong in the algebraic method?
| $$a = 12p+8 = 15q+11 = 18r+14$$
You should break these down into linear expressions with prime-power coefficients.
$$a = 12p+8 \implies a =
\left\{ \begin{array}{c}
4(3p+2) \\
3(4p+2)+2
\end{array}
\right \}$$
$$a = 15q+11 \implies a =
\left\{ \begin{array}{c}
3(5q+3)+2 \\
5(3q+2)+1
\end{array}
\right \}$$
$$a = 18r+14 \implies a =
\left\{ \begin{array}{c}
2(9r+7) \\
9(2r+1)+5
\end{array}
\right \}$$
Some of these can be combined.
$$a =
\left\{ \begin{array}{c}
4(3p+2) \\
2(9r+7)
\end{array}
\right \}
\implies a = 4u$$
$$a =
\left\{ \begin{array}{c}
3(4p+2)+2 \\
3(5q+3)+2 \\
9(2r+1)+5
\end{array}
\right \}
\implies a = 9v+5$$
$$a=5(3q+2)+1 \implies a = 5w+1$
$$\left\{ \begin{array}{l}
a=4u \\
a=9v+5 \\
a=5w+1
\end{array}
\right \}
\implies
\left\{ \begin{array}{l}
45a=180u \\
20a=180v+100 \\
36a=180w+36
\end{array}
\right \}
\implies
\left\{ \begin{array}{rcl}
45a &=& 180u \\
-80a &=& 180(-4v)-400 \\
36a &=& 180w+36
\end{array}
\right \}
$$
Adding the three equations, we get
$$ n= 180x + 176$$
Since the smallest $3$-digit number is $100$, the smallest $3$-digit number of the form $ n= 180x + 176$ must be between $100$ and $100+180-1=279$
\begin{array}{c}
100 &\le &180x+176 &\le &279 \\
-76 &\le &180x &\le &103 \\
-\dfrac{76}{180} &\le &180x &\le &\dfrac{103}{180} \\
0 &\le &180x &\lt &1 \\
&&x &= &0
\end{array}
So $n=176$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Trigonometric Coincidence I Know that using Taylor Series, the formula of $\sin x$ is
$$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!}-\dots\\[6pt]
&=-v(x)
\end{align}
$$
That is,
$$
\begin{align}
v'&=u\\
u'&=-v
\end{align}\tag{1}
$$
Furthermore,
$$
\begin{align}
u(0)&=1\\
v(0)&=0
\end{align}\tag{2}
$$
Consider the derivative of $u^2+v^2$ :
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(u^2+v^2\right)
&=2uu'+2vv'\\
&=-2uv+2uv\\[6pt]
&=0
\end{align}
$$
Therefore, $u^2+v^2$ is constant. Since $u(0)^2+v(0)^2=1$, we must have
$$
u^2+v^2=1\tag{3}
$$
which, by $(1)$, also says that
$$
u^{\prime\hspace{1pt}2}+v^{\prime\hspace{1pt}2}=1\tag{4}
$$
$(3)$ says that $(u,v)$ lies on the unit circle and $(4)$ says that $(u,v)$ moves at unit speed (where $x$ is time); that is, the length of the arc swept out below is $x$:
$\hspace{35mm}$
Thus, if we measure the angle at the center of a unit circle subtended by an arc as the length of that arc, we have the position of the point on the circle at angle $x$ is $(u(x),v(x))$. No coincidence.
These functions are more commonly called the sine and cosine of the angle $x$:
$$
\begin{align}
\cos(x)&=u(x)\\
\sin(x)&=v(x)
\end{align}\tag{5}
$$
The circumference of a unit circle is $2\pi$, so a quarter circle (subtending a right angle) is $\pi/2$. Thus,
$$
(\cos(\pi/2),\sin(\pi/2))=(0,1)\tag{6}
$$
A half circle (subtending a straight angle) is $\pi$; therefore,
$$
(\cos(\pi),\sin(\pi))=(-1,0)\tag{7}
$$
$\hspace{8mm}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show that $\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$ Show that
$$\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$$
using 2 ways: the first using contour integration and the second using real analysis.
| Perform integration by parts,
$\begin{align} J&=\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx\\&=\left[\frac{1}{a}\ln\left(\text{arctanh}\left(\frac{ax}{2}\right)\right)x\right]_0^{\infty}-\int_0^{\infty}\frac{1}{a}\ln\left(\text{arctanh}\left(\frac{ax}{2}\right)\right)dx\\
&=-\frac{1}{a}\int_0^{\infty}\ln\left(\text{arctanh}\left(\frac{ax}{2}\right)\right)dx\\
\end{align}$
Perform the change of variable $y=\text{arctanh}\left(\frac{ax}{2}\right)$,
$\begin{align}J&=-\frac{2}{a^2}\int_0^1\frac{\ln x}{1-x^2}dx\\
&=\frac{2}{a^2}\int_0^1\frac{x\ln x}{1-x^2}dx-\frac{2}{a^2}\int_0^1\frac{\ln x}{1-x}dx
\end{align}$
In the first intégral perform the change of variable $y=2x$,
$\begin{align}J&=\frac{1}{2a^2}\int_0^1\frac{\ln x}{1-x}dx-\frac{2}{a^2}\int_0^1\frac{\ln x}{1-x}dx\\
&=-\frac{3}{2a^2}\int_0^1\frac{\ln x}{1-x}dx\\
&=\frac{3}{2a^2}\zeta(2)\\
&=\frac{1}{4a^2}\pi^2
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
How to find an approximate values of rational function $f(x)$ for large $x$, neglecting $\frac{1}{x^4}$ and successive terms? This is the function that I want to find an approximate value for it neglecting $\displaystyle \frac{1}{x^4}$ and successive terms
$$
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(\frac1{x^4}\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric problem using basic trigonometry
If $x$ is a solution of the equation: $$\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) \sqrt{2}-1, b) \sqrt{2}+1, c) \sqrt{3}-1, d) \sqrt{3}+1, e)\sqrt{2}+3$$
| Hint: $$(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{2}{\log_32})=-1.5-\frac{3}{\log_32}$$
$$x=1.5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the square root of $(x^2 + 3x + 7)(x^2 + 5x + 3) + (x − 2)^2$ I want to find the square root of $$(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+26 x^2+40 x+25$$ and after that also rational roots test is of no use because there are no real factors.
Please help.
| As $x^2+3x + 7-(x^2+5x+3)=-2(x-2), x^2+5x+3=(x^2+3x +7)+2(x-2) $
$\implies (x^2+3x +7)(x^2+5x+3)+(x-2)^2$
$=(x^2+3x +7)\{x^2+3x +7+2(x-2)\}+(x-2)^2$
$=(x^2+3x +7)^2+2(x^2+3x +7)(x-2)+(x-2)^2$
$=\{(x^2+3x +7)+(x-2)\}^2$
Alternatively, if we $x^2+3x + 7=a, x^2+5x+3=b, x-2=c\implies b-a=2c$
$\implies (x^2+3x +7)(x^2+5x+3)+(x-2)^2=a\cdot b +c^2$
$=a(a+2c)+c^2=a^2+c^2+2ca=(a+c)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$? I'm a ninth grader so please try to explain the answer in simple terms .
I cant fully understand the explanation in my book .
It just assumes that the expression that should be added has a degree of 1.
I apologize if this question is too simple or just stupid but this is a genuine doubt.
| If you don't want to use polynomial long division, you can factor $x^2+2x-3$ by seeing it has roots at $1$ and $-3$, Then, you need to add a degree 1 polynomial to your polynomial such that the sum is divisible by $x-1$ and $x+3$. Those are relatively prime, so the sum will then be divisible by the product, which is $x^2+2x-3$. So,
$$
\begin{aligned}
x^2 + 2x -3 &\mid x^4+2x^3−2x^2+x−1+ax+b\ \mbox{if}\\
(x-1)(x+3) &\mid x^4+2x^3−2x^2+x−1+ax+b\\
P(1) &= P(-3) = 0\ \mbox{where}\\
P(x) &= x^4+2x^3−2x^2+x−1+ax+b\\
P(1) &= a + b + 1 = 0,\\
P(-3) &= -3a + b + 5 = 0\\
a + b &= -1\\
-3a + b &= -5\\
a&=1\\
b&=-2
\end{aligned}
$$
Added
The poster asked why I needed a degree $1$ polynomial. Remember that integers have this property: when I divide integer $a$ by integer $b$, I get an equation:
$$
a = bq + r,\quad 0\le r < |b|.
$$
The absolute value is what we call a Euclidean norm, and division has the remainder have its norm smaller than the dividend. The integers $\mathbb{Z}$ are called a Euclidean ring.
Polynomials are also a Euclidean ring, where the norm is just the degree:
$$
a(x) = b(x)q(x) + r(x),\quad 0\le \deg r < \deg b.
$$
So, if you had found an arbitrary solution $p(x)$ to your problem, then you could divide $p$ by $x^2+2x-3$ and get a remainder of degree less than $2$, and that would be unique. So, use $ax+b$ so you have only two integers to find.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Show that $ (n+1)(n+2)\cdots(2n)$ is divisible by $ 2^n$, but not by $ 2^{n+1}$ darij grinberg's note:
I. Niven, H. S. Zuckerman, H. L. Montgomery, An introduction to the theory of numbers, 5th edition 1991, §1.4, problem 4 (b)
Let $n$ be a nonnegative integer. Show that $ (n+1)(n+2)\cdots(2n)$ is divisible by $ 2^n$, but not by $ 2^{n+1}$.
I have no idea how to prove this. Can anyone help me through the proof. Thanks.
| Let's count directly:
The number of multiples $m_2$ of $2$ between $n+1$ and $2n$ is $\left\lfloor \frac {2n}2 \right\rfloor-\left\lfloor \frac {n}2 \right\rfloor$ (we take the multiples of $2$ up to $2n$ and subtract the number up to $n$).
The number of multiples $m_4$ of $4$ between $n+1$ and $2n$ is $\left\lfloor \frac {2n}4 \right\rfloor-\left\lfloor \frac {n}4 \right\rfloor$
The number of multiples $m_8$ of $8$ between $n+1$ and $2n$ is $\left\lfloor \frac {2n}8 \right\rfloor-\left\lfloor \frac {n}8 \right\rfloor$
Now consider $$\sum_{r=1}^\infty m_{2^r}$$
This counts each multiple of $2$. Every multiple of $4$ is counted twice - as a multiple of $2$ and a multiple of $4$, and every multiple of $2^r$ is counted $r$ times. Also, when $2^r\gt 2n$ we have $m_{2^r}=0$ so the sum is finite (we could calculate a finite upper limit).
Adding the $m_{2^r}$ we see that every term apart from the first cancels, because $\left\lfloor \frac {2n}{2^{r+1}} \right\rfloor=\left\lfloor \frac {n}{2^{r}} \right\rfloor$ leaving the exact power of $2$ which divides the product as $\left\lfloor \frac {2n}{2} \right\rfloor=n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
How find this$\lim_{n\to\infty}n^2\left(n\sin{(2e\pi\cdot n!)}-2\pi\right)=\frac{2\pi(2\pi^2-3)}{3}$ show that
$$\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!}+\cdots+\dfrac{1}{n!}+\dfrac{1}{(n+1)!}+O(\dfrac{1}{(n+1)!})$$
then
$$2e\pi\cdot n! =2k\pi+\dfrac{2\pi}{n+1}+o(\dfrac{1}{n+1}))$$
for this problem we
$$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}+\dfrac{1}{(n+1)!}+\dfrac{1}{(n+2)!}+o(\dfrac{1}{(n+2)!})$$
so
$$2\pi en!=2k\pi+\dfrac{2\pi}{n+1}+\dfrac{2\pi}{(n+1)(n+2)}+o(\dfrac{1}{(n+1)(n+2)})$$
and use
$$\sin{x}=x-\dfrac{x^3}{6}+o(x^3)$$
But I can't work
| Clearly the fractional part of $\mathrm{e} \cdot n!$ equals to
$$\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\} \\
&=& \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \mathcal{o}\left(n^{-3}\right)
\end{eqnarray}
$$
Hence
$$
n \cdot \sin\left(2 \pi \mathcal{e} n!\right) = 2 \pi - \frac{2 \pi}{n^2} \left(1 + \frac{2 \pi^2}{3}\right) + \mathcal{o}\left(n^{-3}\right)
$$
Hence
$$
\lim_{n \to \infty} n^2 \left( \sin\left(2 \pi \cdot \mathcal{e}\cdot n!\right) - 2 \pi \right) = - 2 \pi \left(1 + \frac{2 \pi^2}{3}\right) \approx -47.6248875475793{\color
\gray{467109}}$$
Here is a numerical confirmation:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating
$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$
So we should be able to find the area of the ellipse by
$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$
$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$
$$=\pi\frac{a^2+b^2}{2}.$$
First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.
What am I doing wrong?
| 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?:
Since y is bSin[t] and x is aCos[t], we have:
$\tan (\theta )=\frac{b \sin (t)}{a \cos (t)}$
or
$\tan (\theta )=\frac{b \tan (t)}{a}$
Solving for t, we have:
$t(\theta )=\tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$
We now reparametrize using theta:
$x(\theta )=a \cos (t(\theta ))$
$y(\theta )=b \sin (t(\theta ))$
which ultimately simplifies to:
$x(\theta)=\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$
$y(\theta)=\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$
Note that, under the new parametrization, $y(\theta)/x(\theta) =
tan(\theta)$ as desired.
To compute area, we need $r^2$ which is $x^2+y^2$, or:
$r(\theta )^2 = (\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2+
(\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2$
(note that we could take the square root to get r, but we don't really need it)
The above ultimately simplifies to:
$r(\theta)^2 =
\frac{1}{\frac{\cos ^2(\theta )}{a^2}+\frac{\sin ^2(\theta )}{b^2}}$
Now, we can integrate $r^2/2$ to find the area:
$A(\theta) = (\int_0^\theta
\frac{1}{\frac{\cos ^2(x )}{a^2}+\frac{\sin ^2(x )}{b^2}} \, dx)/2$
which yields:
$A(\theta) = \frac{1}{2} a b \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$
good for $0\leq \theta <\frac{\pi }{2}$
Interestingly, it doesn't work for $\theta =\frac{\pi }{2}$ so we
can't test the obvious case without using a limit:
$\lim_{\theta \to \frac{\pi }{2}} \, \frac{1}{2} a b \tan
^{-1}\left(\frac{a \tan (\theta )}{b}\right)$
which gives us $a*b*Pi/4$ as expected.
| {
"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
} |
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$$
$\sqrt{8-2\sqrt{15}}$
Note that $15 = 5 \times 3$ and $8 = 5 + 3$.
Hence $$(\sqrt 5 - \sqrt 3)^2 = 8 - 2 \sqrt{15}$$
\begin{align}
\sqrt{5+2\sqrt{6}} + \sqrt{8-2\sqrt{15}}
&= (\sqrt 3 + \sqrt 2) + (\sqrt 5 - \sqrt 3) \\
&= \sqrt 5 + \sqrt 2
\end{align}
Added 1/18/2021
Another way of looking at $\sqrt{M + 2\sqrt N} = \sqrt a + \sqrt b$.
\begin{align}
\sqrt{M + 2\sqrt N} &= \sqrt a + \sqrt b \\
M + 2 \sqrt N &= (a+b) + 2\sqrt{ab} \\
\hline
M &= a + b \\
N &= ab \\
\hline
x^2 - Mx + N &= (x-a)(x-b) \\
\{a,b\} &= \dfrac{M \pm \sqrt{M^2-4N}}{2}
\end{align}
So if $\sqrt{5 + 2\sqrt 6} = \sqrt a + \sqrt b$
$$\{a,b\} = \left\{\dfrac{5 \pm \sqrt{5^2 - 4(6)}}{2}\right\} = \{3,2\}$$
$$x^2-5x+6 = (x-2)(x-3)$$
So
$$\sqrt{5 + 2\sqrt 6} = \sqrt 2 + \sqrt 3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the Determinant of a $3\times 3$ matrix. Show that:
$$
\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 + b)
$$
for some constant $c$.
To compute $c$, note that you obtain $x^{3}$ only from the following term of the expansion with respect to the first column
$$
x^{2} \cdot \begin{vmatrix}a&b\\x+b&x+a\end{vmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Bounds for $\binom{n}{cn}$ with $0 < c < 1$. Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$? I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn} \leq \left(\frac{e}{c}\right)^{cn}$.
| Stirling's Asymptotic Expansion, derived here, is
$$
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-c)n}}\small\left(1-\frac{1-c+c^2}{12c(1-c)}\frac1n+\frac{1-2c+3c^2-2c^3+c^4}{288c^2(1-c)^2}\frac1{n^2}+O\left(\frac1{n^3}\right)\right)
\end{align}
$$
Absolute Bounds
For $n\ge1$,
$$
\sqrt{2\pi n}\,n^ne^{-n}\left(1+\frac1{12n}\right)\le n!\le\sqrt{2\pi n}\,n^ne^{-n}\left(1+\frac1{12n}+\frac1{288n^2}\right)
$$
which gives
$$
\sqrt{2\pi n}\,n^ne^{-n}\le n!\le\frac{313}{288}\sqrt{2\pi n}\,n^ne^{-n}
$$
From which we get, for $1\le cn\le n-1$,
$$
\left(\frac{288}{313}\right)^2\frac{\left(c^c(1-c)^{1-c}\right)^{-n}}{\sqrt{2\pi c(1-c)n}}
\le\binom{n}{cn}
\le\frac{313}{288}\frac{\left(c^c(1-c)^{1-c}\right)^{-n}}{\sqrt{2\pi c(1-c)n}}
$$
Note that
$$
\frac{e}{\sqrt{2\pi}}\doteq1.0844\lt1.0868\doteq\frac{313}{288}
$$
so this is close to Hagen von Eitzen's answer.
The bound mentioned in Hagen's answer is based on the fact that the ratio
$$
\frac{\Gamma(n)}{\sqrt{2\pi n}\,n^ne^{-n}}
$$
is decreasing in $n$. Thus, the greatest ratio is for $n=1$ and therefore,
$$
\sqrt{2\pi n}\,n^ne^{-n}\le n!\le\frac{e}{\sqrt{2\pi}}\sqrt{2\pi n}\,n^ne^{-n}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How find the $\angle B$
In $\Delta ABC$ such $I$ is incentre,and $$\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=\dfrac{ab}{b+c}$$
$$\Longrightarrow AD=\sqrt{bc-\dfrac{a^2bc}{(b+c)^2}}$$
so
$$AI=\dfrac{b+c}{a+b+c}AD$$
and
$$BI=\dfrac{a+c}{a+b+c}BE$$
so
$$AI+BI=BC\Longleftrightarrow \dfrac{b+c}{a+b+c}AD+\dfrac{a+c}{a+b+c}BE=a$$
then I have ugly,and can't work,Thank you someone can have other methods.
| Hint: Apply sin rule to $BIC$, get $BC$ in terms of $BI$.
Hint: Apply sin rule to $BIA$, get $AI$ in terms of $BI$.
Now substitute these into $AI + BI = BC$, you will get a trigonometric equation in terms of $\angle IBC$. Solve it to determined that $\angle IBC = 20^\circ$.
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found the pattern:
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$
But I have yet to figure out how to prove it algebraically.
Suggestions?
| The statement is true for an arbitrary complex number $n$, with matrix logarithm. All the $\log$s below refer to the natural logarithm.
For convenience, let's set $$ 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}^\infty \frac{1}{k!}A^k = I + A. $$
All eigenvalues of $A$ are zero, so $A$ is the principal logarithm.
$$ \log \left( I + A \right) = A. $$
This leads to
$$ \log \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$
And
$$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \exp \begin{pmatrix} 0 & n \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}. $$
| {
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"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
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Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive real roots $=3$
But are they distinct $?$
Answer :- number of distinct real roots $ =1$
| Let's substitute the variable: $x=y+\frac{(a+b+c)}{3}$
The equation will now look like
$$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}{27}(a^2-ab-ac+b^2-bc+c^2)^3+\frac{1}{324}(a+b-2c)^2(a-2b+c)^2(b-2a+c)^2$$
Then the answer is given depending on the sign of Q.
If $Q>0$ there is only one real root
If $Q<0$ there are three distinct real roots
If $Q=0$ there are two distinct real roots.
| {
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Re-write $1 \cdot x$ to $x$. Given the following bi-directional re-write rules (where $1$ is a constant, $^{-1}$ is a unary operator, $\cdot$ is a binary operator, and $x,y,z$ are arbitrary terms):
$$\begin{align*}
x \cdot 1 &= x \\
x \cdot (y \cdot z) &= (x\cdot y) \cdot z \\
x \cdot x^{-1} &= 1
\end{align*}$$
we're asked to prove that $1 \cdot x = x$ (ie, there is a chain $t_0 \to t_1 \to t_3 \to \cdots \to t_n$ with $t_0 = 1\cdot x$, $t_n = x$, and $t_i \to t_{i+1}$ meaning one of the 3 equations above re-writes $t_i$ to $t_{i+1}$ (in either direction)).
After staring at this for a while I'm beginning to doubt whether or not this is possible... can anyone a) confirm this is indeed possible and b) potentially nudge me in the right direction?
| Yes, observe that:
\begin{align*}
1 \cdot x &= (x \cdot x^{-1}) \cdot x \\
&= x \cdot \left(x^{-1} \cdot x\right) \\
&= x \cdot \left((x^{-1} \cdot x) \cdot 1\right) \\
&= x \cdot \left((x^{-1} \cdot x) \cdot (x^{-1} \cdot (x^{-1})^{-1}) \right) \\
&= x \cdot \left(((x^{-1} \cdot x) \cdot x^{-1}) \cdot (x^{-1})^{-1} \right) \\
&= x \cdot \left((x^{-1} \cdot (x \cdot x^{-1})) \cdot (x^{-1})^{-1} \right) \\
&= x \cdot \left((x^{-1} \cdot 1) \cdot (x^{-1})^{-1} \right) \\
&= x \cdot \left(x^{-1}\cdot (x^{-1})^{-1} \right) \\
&= x \cdot 1 \\
&= x \\
\end{align*}
| {
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How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? Let $a,b,c \in \mathbb{R}$, and $abc=1$. What is the simple(st) way to prove inequality
$$
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$$
| {
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} |
How to test a matrix for inconsistency? I understand how to test two equations with two variables but what happens when I get up to 4 variables? I have no idea how to test that. For example with the matrix/system of equations
$x_1 +2 {x_2} + x_3 = 1$
$-2 x_1 - 4 {x_2} - x_3 = 0$
$5 {x_1} + 10 {x_2} + 3 {x_3} = 2$
$3 {x_1} + 6 {x_2} + 3 {x_3} = 4$
How do I find out if it is inconsistent?
| Try to solve the equation; and check whether the system has the solution or not.
$\begin{pmatrix}1 & 2 & 1 \\ -2 & -4 & -1 \\ 5 & 10 & 3 \\ 3 & 6 & 3 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 2 \\ 4\end{pmatrix} $, and multiply first row by 3 and substract to 4th row then
$\begin{pmatrix}1 & 2 & 1 \\ -2 & -4 & -1 \\ 5 & 10 & 3 \\ 0&0&0 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 2 \\ 1\end{pmatrix} $
which says that $0 = 1$ in the 4th row, which is inconsistent.
| {
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Why n! equals sum of some expression? Why n! equals sum of some expression?
Especially I need to know why this expression is true?
$$
n!= \left(\frac{n+1}{2}\right)^{p(n)} \; \prod_{j=0}^{q(n)}\sum_{i=0}^j(n-2i),
$$
Where
\begin{gather*}
p(n)=\frac{\cos(\pi n+\pi)+1}{2}\\\\
q(n)=\frac{2n+\cos(\pi n)-5}4
\end{gather*}
How to prove this equality?
Is it true?
| Summary:
This formula is just a convoluted way of saying,
$$
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.
$$
Explanation:
Note that by sum of an arithmetic sequence,
$$
\sum_{i = 0}^j (n - 2i) = (n - j)(j+1)
$$
Also,
$$
\frac{2n + \cos(\pi n) - 5}{4}
=
\frac{2n + (-1)^n - 5}{4}
=
\left\{
\begin{array}{cc}
\frac{n - 3}{2} & n \text{ is odd}\\
\frac{n - 2}{2} & n \text{ is even}
\end{array}
\right.
$$
Finally,
$$
{\left(\frac{n+1}{2}\right)}^{\left(\frac{\cos(\pi n + \pi)+1}{2}\right)}
=
{\left(\frac{n+1}{2}\right)}^{\left(\frac{(-1)^{n+1}+1}{2}\right)}
=
\left\{
\begin{array}{cc}
\frac{n + 1}{2} & n \text{ is odd}\\
1 & n \text{ is even}
\end{array}
\right.
$$
Thus, if $n$ is odd,
\begin{align*}
{\left(\frac{n+1}{2}\right)}^{\left(\frac{\cos(\pi n + \pi)+1}{2}\right)} \prod_{j=0}^\frac{2 n +\cos(\pi n) - 5}{4}\sum_{i=0}^j(n-2i)
&=
\frac{n+1}{2} \prod_{j=0}^{\frac{n - 3}{2}} (n - j)(j+1) \\
&= \frac{n+1}{2} [n \cdot 1][(n-1) \cdot 2] \cdots \left[ \left( \frac{n + 3}{2} \right) \left( \frac{n-1}{2} \right) \right] \\
&= n!
\end{align*}
And if $n$ is even,
\begin{align*}
{\left(\frac{n+1}{2}\right)}^{\left(\frac{\cos(\pi n + \pi)+1}{2}\right)} \prod_{j=0}^\frac{2 n +\cos(\pi n) - 5}{4}\sum_{i=0}^j(n-2i)
&=
1 \cdot \prod_{j=0}^{\frac{n - 2}{2}} (n - j)(j+1) \\
&= [n \cdot 1][(n-1) \cdot 2] \cdots \left[ \left( \frac{n + 2}{2} \right) \left( \frac{n}{2} \right) \right] \\
&= n!
\end{align*}
So indeed the formula holds for any $n$.
| {
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Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.
| One that's not been mentioned so far(?): knowing that
$$
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
$$
is an alternating series whose terms decrease in absolute value
and start out positive. Therefore it is positive by the usual argument:
group the terms as
$$
e^x - (1+x) =
\left( \frac{x^2}{2} + \frac{x^3}{3!} \right) +
\left( \frac{x^4}{4!} + \frac{x^5}{5!} \right) + \cdots
$$
and observe that each combined term is positive, QED.
(This actually works for $-3 < x < 0$, but you still want to use
$e^x > 0$ to prove the inequality for very negative $x$.)
| {
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Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$ Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$.
I have absolutely no clue where to start and what to do, please provide complete proof and answer.
| Suppose that $\gcd(a+b,a^2+b^2)=d$
Then $d \mid a+b$ and $d \mid a^2+b^2$.
$d \mid (a+b)(a+b)-(a-b)(a+b)=2a^2$.
$d \mid (a+b)(a+b)+(a-b)(a+b)=2b^2$
$d \mid (2a^2,2b^2) = 2(a^2,b^2)=2(a,b)^2=2.1=2 \implies d=1$ or $2$.
| {
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How do I integrate $\sec(x)$? My HW asks me to integrate $\sin(x)$, $\cos(x)$, $\tan(x)$, but when I get to $\sec(x)$, I'm stuck.
| $$\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}{2}\sec^2(\frac{x}{2})dx=dt$$
$$\frac{1}{2}\sec^2(\frac{x}{2})dx=dt \iff dx= 2\cos^2\frac{x}{2}dt$$
$$dx= 2\cos^2\frac{x}{2}dt \iff dx=\frac{2}{(1+t^2)}dt$$
$$\cos(x)=\cos(\frac{x}{2}+\frac{x}{2}) \iff \cos( \frac{x}{2}+ \frac{x}{2})= \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})$$
$$\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{1}{t^2+1}-\frac{t^2}{t^2+1} \iff \frac{1-t^2}{t^2+1}=\cos x$$
$$\frac{1-t^2}{t^2+1}=\cos x \iff \sec x=\frac{1+t^2}{1-t^2} $$
$$\int \sec x dx=\int \frac{1+t^2}{1-t^2}. \frac{2}{(1+t^2)}dt \iff \int \frac{1+t^2}{1-t^2}. \frac{2}{(1+t^2)}dt=\int \frac{2}{1-t^2}dt$$
Partial fractions!
$$\frac{2}{1-t^2}\equiv \frac{A}{1-t}+\frac{B}{1+t}\iff 2\equiv A(1+t)+B(1-t)$$
Solution is $t=1, A=1,t=-1,B=1$
$$\int \frac{1}{1+t}+\frac{1}{1-t}dt=\ln|1+t|-\ln|1-t|+c \iff \ln|\frac{1+t}{1-t}|.\ln|\frac{1+t}{1+t}|+c=\ln|\frac{(1+t)^2}{1-t^2}|+c$$
$$\ln|\frac{(1+t)^2}{1-t^2}|+c=\ln|\frac{t^2+2t+1}{1-t^2}|+c \iff \ln|\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}|$$
So,
$$\ln|\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}| \iff \ln|\sec x + \tan x|$$
This is a standard integral for $\sec x$
Conclusion!
$$\int \sec xdx=\ln|\sec x+\tan x|+c$$
| {
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How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$ let $a,b,c>0$, show that
$$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}).$$
Done!
| {
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How to prove $\tan x+\tan \left(x+\frac{\pi }{3}\right)+\tan \left(x+\frac{2\pi }{3}\right)=3\tan 3x$? The following is the equation.
\begin{eqnarray}
\tan x+\tan \left(x+\frac{\pi }{3}\right)+\tan \left(x+\frac{2\pi }{3}\right)&=&3\tan 3x\\
\tan x+\frac{\tan x+\tan \frac{\pi }{3}}{1-\tan x\space \tan \frac{\pi }{3}}+\frac{\tan x+\tan \frac{2\pi }{3}}{1-\tan x\space \tan \frac{2\pi }{3}} &=& 3\tan 3x\\
\end{eqnarray}
I think that :
\begin{eqnarray}
\tan \frac{2\pi }{3}=-\tan \frac{\pi }{3}\\
\end{eqnarray}
Therefore,
\begin{eqnarray} \tan x+\frac{\tan x+\tan \frac{\pi }{3}}{1-\tan x\space \tan \frac{\pi }{3}}+\frac{\tan x-\tan \frac{\pi }{3}}{1+\tan x\space \tan \frac{\pi }{3}}&=&3\tan 3x\\ \end{eqnarray}
For me, it's too complex to continue...
Maybe there are something wrong.
Are there any other methods to prove the answer easily?
Thank you for your attention.
| Let $$\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$$ which is a cubic equation in $\tan x$
Now, apply Vieta's formula to find
$\sum \tan x=3\tan3A,\prod\tan x=-\tan3A $
| {
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It becomes $2xy - xy \ge 0$, then $xy \ge 0$.
How is this a contradiction?
I think I'm missing some key point.
| Another way to see this is as follows:
Suppose $x^2+xy+y^2 <0$, then $xy<-x^2-y^2< 0$, hence $2xy<xy<-x^2-y^2$ and thus $x^2+2xy+y^2<0$, which is absurd.
| {
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trig equation $(\sqrt{\sqrt{2}+1})^{\sin(x)}+(\sqrt{\sqrt{2}-1})^{\sin(x)}=2$ Please help me to solve this trig equation.
$$(\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.$$
You can verify this has only one solution, $y=1$.
Next, notice
$$ \frac{1}{\sqrt{2}+1} = \sqrt{2}-1.
$$
By laws of exponents, our original equation is then
$$\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)} +
\frac{1}{\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)}}=2,
$$
which we recognize as the auxiliary equation with
$y=\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)}$.
Therefore the solution to our original equation has
$$
\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)}=1.
$$
Since $\sqrt{2}+1>1$, we must hve
$$ \frac{1}{2} \sin(x) =0.$$
The solutions are the integer multiples of $\pi$.
| {
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Sequence summation GRE question
This problem is giving me a lot of trouble. the only way i can think of doing it is to take the integeral a(k) from 1 to 100 but that is definitely not what i am supposed to do since its for the GRE and no knowledge of calculus is necessary. . . helpppp
| If you write out the first few terms of the series, you'll see that $$a_1+a_2+a_3+\cdots+a_{99}+a_{100}$$$$=\left(1-\frac{1}{2}\right)+ \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3} - \frac{1}{4} \right) \cdots+\left(\frac{1}{99} - \frac{1}{100}\right) + \left( \frac{1}{100} - \frac{1}{101} \right).$$
Rearranging the parentheses yields $$1 + \left(-\frac{1}{2} + \frac{1}{2}\right) + \left(-\frac{1}{3} + \frac{1}{3}\right) + \cdots +\left(-\frac{1}{100} + \frac{1}{100} \right) - \frac{1}{101}$$ $$=1-\frac{1}{101} = \frac{100}{101}.$$
You can confirm that the general formula for the $n$th sum is $$\sum_{k=1}^{n}\frac{1}{k} - \frac{1}{k+1}=1-\frac{1}{n+1}$$ by induction, since it is clearly true for $n=1$, and then assuming that it holds for $n$, we can see that $$\sum_{k=1}^{n+1}\frac{1}{k} - \frac{1}{k+1}=\sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{n+1} - \frac{1}{n+2} \right)$$
so from our inductive hypothesis we get
$$\left(1-\frac{1}{n+1}\right)+\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$
$$=1-\frac{1}{n+2}.$$
| {
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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 this problem, but I can't solve it.
I can write $(a+b+c)^2$ as $a^2+b^2+c^2+2(ab+bc+ca)=m+2(ab+bc+ca)$, but then I don't know how to determine $ab+bc+ca$. I'm also not certain how to use the equation $ax+by+cz=p$ to help me in finding the given expression's value.
Any help would be much appreciated.
| 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 difference formula for cosine, the third equation becomes
$$\cos(t-u)+1/2=1,$$
which holds as long as say $t-u=\pi/3$. So put $t=u+\pi/3$ [with also $0<u<\pi/6$ to keep $t,u$ in the first quadrant so that $a,b,x,y$ are positive] and consider the quantity $(a+b+c)/(x+y+z)$, which becomes
$$\frac{\cos(u+\pi/3)+\sin(u+\pi/3)+1/2}{\cos u + \sin u + 1},$$
which is a smooth nonconstant function of $u$ and so takes on infinitely many values.
| {
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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
$$\sum_{n=1}^{\infty}\frac{\sinh\big(\pi n\sqrt2\big)-\sin\big(\pi n\sqrt2\big)}{n^3\Big({\cosh\big(\pi n\sqrt2}\big)-\cos\big(\pi n\sqrt2\big)\Big)}=\frac{\pi^3}{18\sqrt2}$$
I have no hint as to how to even start.
| 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 $\mathbb{W}=\mathbb{Z}/\{0\}$) :
$$\sum_{n\in\mathbb{W}}\frac{1}{n^2+z^2}=\frac{\pi\coth\pi z}{z}-\frac{1}{z^2}$$
valid for every complex $z$ exept the poles. The two later results from the links are in $\mathbb{W}$ formalism :
$$\begin{align}&\sum_{m\in\mathbb{W}}\frac{1}{(m-a)^2+b^2}=\frac{\pi\mathbin{\color{blue}{\sinh2\pi b}}}{b\left(\cosh2\pi b-\cos2\pi a\right)}-\frac{1}{a^2+b^2}\tag{$\varphi$}\\ \\
&\sum_{m\in\mathbb{W}} \frac{1}{m^4+c^4} = \frac{\pi}{\sqrt{2} \, c^3} \frac{\mathbin{\color{green}{\sinh\left(\sqrt{2} \pi c\right)+\sin\left(\sqrt{2} \pi c\right)}}}{\cosh\left(\sqrt{2} \pi c\right) -\cos\left(\sqrt{2} \pi c\right)}-\frac{1}{c^4}\tag{$\nu$}\end{align}$$
from these two formulae we are able by some simple manipulations obtain the sum $S$, for which we have:
$$S=\frac{1}{2}\sum_{n\in\mathbb{W}}\frac{1}{n^3}\frac{\mathbin{\color{red}{\sinh\left(\sqrt{2} \pi n\right)-\sin\left(\sqrt{2} \pi n\right)}}}{\cosh\left(\sqrt{2} \pi n\right) -\cos\left(\sqrt{2} \pi n\right)}$$
Note that
$$\mathbin{\color{red}{\sinh\left(\sqrt{2} \pi n\right)-\sin\left(\sqrt{2} \pi n\right)}}=2\mathbin{\color{blue}{\sinh\left(\sqrt{2} \pi n\right)}}-\left[\mathbin{\color{green}{\sinh\left(\sqrt{2} \pi n\right)+\sin\left(\sqrt{2} \pi n\right)}}\right]$$
i.e., this suggest using $2\varphi-\nu$. Choosing $a=b=n/\sqrt2$ in $\varphi$ and $c=n$ in $\nu$ we have for our sum:
$$\begin{align}
S &=\frac{1}{2\pi\sqrt{2}}\sum_{n\in\mathbb{W}}\frac{2}{n^3}\left(\frac{1}{n}+\sum_{m\in\mathbb{W}}\frac{1}{(m-n/\sqrt2)^2+n^2/2}\right)-\frac{1}{n^3}\left(\frac{2}{n}+\sum_{m\in\mathbb{W}}\frac{2n^3}{m^4+n^4}\right)\\ \\
&=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\frac{1}{m^2-mn\sqrt{2}+n^2}-\frac{1}{m^4+n^4}\tag{1} \\ \\
&=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\frac{m^2+n^2}{m^4+n^4}-\frac{1}{m^4+n^4}\tag{2} \\ \\
&=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{m^2}{n^2}\frac{m^2}{m^4+n^4}\tag{3}\\ \\
&=\frac{1}{\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{2}\left(\frac{m^2}{n^2}+\frac{n^2}{m^2}\right)\frac{1}{m^4+n^4}\tag{4}\\
&=\frac{1}{2\pi\sqrt{2}}\sum_{n,m\in\mathbb{W}^2}\frac{1}{m^2n^2}=\frac{1}{2\pi\sqrt{2}}\left(\sum_{n\in\mathbb{W}}\frac{1}{n^2}\right)^2=\frac{2}{\pi\sqrt{2}}\zeta^2(2)=\frac{\pi^2}{18\sqrt2}\tag{5}
\end{align}$$
Explanations:
$(1)$ terms $2/n^4$ cancel each other; $(m-n/\sqrt2)^2+n^2/2=m^2-mn\sqrt{2}+n^2$
$(2)$ on the summation domain, lattice $(n,m)\in \mathbb{W}\times\mathbb{W}$ there is a symmetry $\sum_{n,m}=\sum_{-n,m}$, ergo by taking "average" $\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\frac{1}{m^2-mn\sqrt{2}+n^2}\!=\!\frac{1}{2}\!\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2}\left(\frac{1}{m^2-mn\sqrt{2}+n^2}+\frac{1}{m^2+mn\sqrt{2}+n^2}\right)$
$(3)$ uncomfortable terms $1/(n^4+m^4)$ cancels too
$(4)$ another symmetry = changing order of summation $n\longleftrightarrow m$
$(5)$ also $\zeta(2)=\sum_{n=1}^{\infty}1/n^2=\pi^2/6$
| {
"language": "en",
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"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 , Given $(x-8)\cdot (x-10) = 2^y$. and $x,y\in\mathbb{Z}$
So $(x-8)=2^m\Rightarrow x= 8+2^m$ and $(x-10) = 2^n\Rightarrow x=10+2^n$and $m+n=y$
Now $10+2^n-8-2^m = x-x = 0\Rightarrow 2^m-2^n = 2$
Now I did not understand How can i solve after that
Help me
Thanks
| 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",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 columns will consist of the two basis vectors) I'm aware that I need to find $v_1$ s.t $(A-I)v_1 = 0$, so $v_1 = \begin{pmatrix}
2\\ 7
\end{pmatrix}$. Now to find $v_2$ I need to do $ker(A-I)^2$ but $(A-I)^2 = 0$ so any non zero vector is in the kernel right$?$ So why doesn't it work if I choose $v_2 = \begin{pmatrix}
1\\ 0
\end{pmatrix}$ $?$
| 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 following non-trivial equations:
$$\left\{ \begin{array}{rcl}15a - 4b & = & a+2\\49a-13b & = & b+7\end{array} \right.$$
$$\left\{ \begin{array}{rcl}14a - 4b & = & 2\\49a-14b & = & 7\end{array} \right.$$
These equations are linearly dependent, so you can choose to solve the first one:
$$\left\{ \begin{array}{rcl}a & = & \frac{1+2k}{7}\\b & = & k\end{array} \right.$$
If you fix $k$, say $k=0$, then
$$P=\left[ \begin{array}{cc}2 & \frac{1}{7}\\ 7 & 0\end{array} \right]$$
Note that $$det(P(k))=\left[ \begin{array}{cc}2 & \frac{1+2k}{7}\\ 7 & k\end{array} \right] = -1 \neq 0 ~\forall k,$$ so the choice of $k$ is arbitrary.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $\frac75-\frac34=\frac{7\cdot4-5\cdot3}{20}=\frac{13}{20}$
$$\sqrt[8]{x^7 + x^2 + 1} - x=\frac{(1+h^5+h^7)^{\frac18}}{h^{\frac78}}-\frac1h=\frac{h^{\frac18}(1+h^5+h^7)^{\frac18}-1}h$$
as $1-\frac78=\frac18$
$$\implies\lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x}=\lim_{h\to0} \frac{(1+3h^7)^{\frac15}-h^{\frac{13}{20}}(2-h^3)^{\frac14}}{h^{\frac75}}\cdot \frac h{h^{\frac18}(1+h^5+h^7)^{\frac18}-1}$$
$$=\lim_{h\to0} \frac{(1+3h^7)^{\frac15}-h^{\frac{13}{20}}(2-h^3)^{\frac14}}{h^{\frac25}}\cdot \frac 1{h^{\frac18}(1+h^5+h^7)^{\frac18}-1}\left(\text{ as }\frac75-1=\frac25\right)$$
$$=\frac10\cdot\frac1{-1}=-\infty$$
| {
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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 show that $2^{n+1} \geq (n+2)^2$ or alternatively
$$2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$$
claim $$\frac{(n+2)^2}{(n+1)^2} <2, \forall n \geq 6$$
notice that it is true for $n=6$ and $$\frac{\frac{(n+2)^2}{(n+1)^2}}{\frac{((n+1)+2)^2}{((n+1)+1)^2}}<1$$ so $\frac{(n+2)^2}{(n+1)^2}$ is decreasing as $n$ gets larger so we have proven the above claim.
So we have that for all $n \geq 6$ we have $2^n \geq (n+1)^2$ and by our induction hypothesis and $\frac{(n+2)^2}{(n+1)^2} <2$ so we can conclude that $2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$.
| 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
$$2\cdot 2^n\ge 2(n+1)^2 \ge (n+1)^2+2n+3=(n+2)^2$$
where $2n+3$ is the difference between $(n+1)^2$ and $(n+2)^2$.
| {
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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 $
$\Rightarrow x^2 +y^2+1 -2x \leq 0$
$\Rightarrow (x-1)^2 +y^2 =0$
$\Rightarrow x =1 $
Therefore option is (a)
Is it correct? please suggest thanks.
| 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't really matter.
| {
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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 $X_1$= number of rolls till A throw a 6
$X_2$= number of rolls till B throw a 6
$X_3$= number of rolls till C throw a 6
$P(X_1=i)=\frac{1}{6} (\frac{5}{6})^{i-1}$
$P(X_2=j)=\frac{1}{6} (\frac{5}{6})^{j-1}$
$P(X_3=k)=\frac{1}{6} (\frac{5}{6})^{k-1}$
therefore, $P$(A is the first to throw a 6, B is the second, and C is the third)
$=P(X_1 < X_2 < X_3) $
= $\sum_{i<j<k}^{ } (\frac{1}{6})^3 (\frac{5}{6} )^{(i-1)+(j-1)+(k-1)} $
$= (\frac{1}{6})^3 \sum_{i<j}^{ } (\frac{5}{6} )^{(i-1)+(j-1)} \sum_{k=j+1}^{\infty } (\frac{5}{6} )^{k-1} $
$ = (\frac{1}{6})^3 \sum_{i<j}^{ } (\frac{5}{6} )^{(i-1)+(j-1)} \frac{(\frac{5}{6})^{j}} {1-\frac{5}{6}} $
$ =(\frac{1}{6})^2 \sum_{i<j}^{ } (\frac{5}{6} )^{(i-1)+(j-1)+j}$
$ =(\frac{1}{6})^2 \sum_{i}^{ }(\frac{5}{6} )^{(i-1)}\sum_{j=i+1}^{\infty} (\frac{5}{6} )^{2j-1}$
$ =(\frac{1}{6})^2 \sum_{i}^{ }(\frac{5}{6} )^{(i-1)} \frac{(\frac{5}{6})^{2i+1}} {1-\frac{5}{6}} $
$ =(\frac{1}{6}) \sum_{i}^{ }(\frac{5}{6} )^{(i-1)+2i+1}$
$ =(\frac{1}{6}) \sum_{i=1}^{\infty }(\frac{5}{6} )^{3i}$
$=\frac{125}{546}$
Could anyone help me check where went wrong? Thanks so much!
| 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 $q={6\over 11}$.
For the full three man game, let $p=\mathbb{P}(X_1<X_2<X_3)$,
First step analysis gives $p={1\over 6}q+\left({5\over 6}\right)^3p$,
or ${91\over 216} p={1\over 11}$. Finally, this means $p={216\over 1001}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 follow exist or not. If so, find it
$$\lim_{n\to\infty}\frac{x_n}{n}$$
In both cases, $x_n$ is increasing, so I tried to get an upper bound on $x_n$ (possibly depending on $n$) to apply a squeeze theorem; but failed.
| 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-1}}) +(\frac{1}{x_n}+\frac{1}{x_{n-1}})
\\&\vdots
\\&= y_1 + 2\sum_{i=1}^n\sqrt{x_i} + \sum_{i=1}^n\frac{1}{x_i}
\end{align*}
(1)Since $1\le x_n \le n$. then:
\begin{align*}
\ y_{n+1} &= 1 + 2\sum_{i=1}^n\sqrt{x_i} + \sum_{i=1}^n\frac{1}{x_i}
\\&\ge 1+2\sum_{i=1}^n1+\sum_{i=1}^n\frac{1}{x_i}
\\&=1+2n+\sum_{i=1}^n\frac{1}{x_i}
\\&\ge n
\end{align*}
Hence：
$$ \frac{1}{x_{n+1}} \le \frac{1}{\sqrt{n}}~~That~~is: \frac{1}{x_{n}} \le \frac{1}{\sqrt{n-1}}(n\ge 2)$$
(2)Since $1\le x_n \le n$,then：
$$\sqrt{x_n} \le \sqrt{n}$$
(3)The upper bound of $y_{n+1}$：
\begin{align*}
\ 0\le y_{n+1}&= 1+2\sum_{i=1}^n\sqrt{x_i}+\sum_{i=1}^n\frac{1}{x_i}
\\&\le 1+2\sum_{i=1}^n\sqrt{i}+\sum_{i=1}^n\frac{1}{x_i}~~~~~~~~~~~~~~~~~~~~~\ldots\ldots from~(2)
\\&= 1+2\sum_{i=1}^n\sqrt{i}+(1+\sum_{i=2}^n\frac{1}{x_i})
\\&\le 1+2\sum_{i=1}^n\sqrt{i}+(1+\sum_{i=2}^n\frac{1}{\sqrt{i-1}})~~~\ldots\ldots from~(1)
\\&= 1+2(\sqrt{n}+\sum_{i=1}^{n-1}\sqrt{i})+(2+\sum_{i=3}^n\frac{1}{\sqrt{i-1}})
\\&\le 1+2(\sqrt{n} + \int_1^n\sqrt{x}dx)+(2+\int_2^n\frac{1}{\sqrt{x-1}}dx)
\\&=2\sqrt{n}+2\sqrt{n-1}+\frac{4n^\frac{3}{2}-1}{3}
\end{align*}
Simplify,we have：
$$0\le y_{n+1} \le 2\sqrt{n}+2\sqrt{n-1}+\frac{4n^\frac{3}{2}-1}{3} $$
That is：$$0 \le \frac{x_{n+1}}{n} \le \frac{\sqrt{2\sqrt{n}+2\sqrt{n-1}+\frac{4n^\frac{3}{2}-1}{3}}}{n}$$
By the squeeze theorem：
$$ \lim_{n \rightarrow \infty}\frac{x_n}{n}=0 $$
| {
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"url": "https://math.stackexchange.com/questions/537240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$ but I don't obtain anything.
| 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:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{9}{x^2y+y^2 z+z^2 x},$$
and it is sufficient to prove $x^2y+y^2z+z^2x\leq 3$. In virtue of the Cauchy-Schwarz inequality,
$$x^2 y + y^2 z + z^2 y \leq \sqrt{(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2y^2)},$$
so it is sufficient to prove $x^2y^2+y^2z^2+z^2y^2\leq 3$, that is equivalent to $x^4+y^4+z^4\geq 3$, that is trivial in virtue of the Cauchy-Schwarz inequality, again.
| {
"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 speed of 1st be a kmph and 2nd be b kmph
Let the distance between A and B be d km
d = 10a/3 + 10b/3
and
d/a - d/b = 5
now i cant solve it. :(
Spoiler: The answer is $10$ hours.
| 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
$$
\frac{d}{\frac{3}{10} d - b} - \frac{d}{b} = 5\\
db- d\left(\frac{3}{10} d - b\right) = 5b\left(\frac{3}{10} d - b\right)\\
d\left(2b - \frac{3}{10}d \right) = \frac{3}{2}bd - 5b^2 \\
\frac{3}{10}d^2- \frac{1}{2}d b -5b^2 = 0
$$
then, dividing by $b^2$
$$
\frac{3}{10}\left(\frac{d}{b}\right)^2 - \frac{1}{2}\frac{d}{b} - 5 =0 \\
3\left(\frac{d}{b}\right)^2 - 5\frac{d}{b} - 50 =0
$$
which is a quadratic in the variable you want to solve for.
| {
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"timestamp": "2023-03-29T00:00:00",
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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,
$$K(u, v) = \frac{a^2 b^2 c^2}{[c^2 \sin^2(v) (a^2 \sin^2(u)+b^2 \cos^2(u))+a^2 b^2 \cos^2(v)]^2}$$
Please help me out.
| 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. a unit normal vector field). Let $f$ be a nowhere-zero smooth function on $M$. Let $p\in M$ and $\mathbf{v_1}$,$\mathbf{v_2}\in T_p(M)$ such that $\mathbf{v_1}$ and $\mathbf{v_2}$ are orthonormal, and that $\mathbf{v_1} \times\mathbf{v_2}=\mathbf{N}$. Then the Gaussian curvature of $M$ at $p$ is
\begin{align}
K=\frac{<d(f\mathbf{N})(\mathbf{v_1})\times d(f\mathbf{N})(\mathbf{v_2}),f\mathbf{N}>}{f^3}
\end{align}
where $d(f\mathbf{N})$ is the differential of $f\mathbf{N}$ at $p$.
To prove the above proposition, one has to start with the right-hand side of the equality and resort to the definition of differential, followed by the use of product rule of usual differentiation. After that, plug it into the expression and perform algebraic manipulation to finally cancel the function $f$. At last, consider $K$ as the determinant of the shape operator, and express every remaining vector in terms of the orthonormal basis $\mathbf{v_1}$,$\mathbf{v_2}$ in order to finish the proof.
Now, consider $M$ as the ellipsoid $\displaystyle g(x,y,z):=\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$. Notice that the gradient of the function $h$ is normal to the tangent plane of $M$ at each point. So we have
\begin{align}
&\nabla{h}=2(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}) &|\nabla{h}|=2\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}=2f
\end{align}
where we let $f$ to be the restrction of the function
\begin{align}
\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}
\end{align}
to $M$. Then we obtain
\begin{align}
N=\frac{\nabla{h}}{|\nabla{h}|}=\frac{1}{f} (\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2})
\end{align}
It is easy to check that
\begin{align}
d(f\mathbf{N})=
\begin{pmatrix}
\frac{1}{a^2} & 0 & 0 \\
0 & \frac{1}{b^2} & 0 \\
0 & 0 & \frac{1}{c^2} \\
\end{pmatrix}
\end{align}
Let $\mathbf{u}$ and $\mathbf{v}$ be an orthonormal basis for $T_pM$ such that $\mathbf{u}\times\mathbf{v}=\mathbf{N}$. Then we obtain
\begin{align}
&d(f\mathbf{N})(\mathbf{u})=(\frac{u_1}{a^2},\frac{u_2}{b^2},\frac{u_3}{c^2}) &d(f\mathbf{N})(\mathbf{v})=(\frac{v_1}{a^2},\frac{v_2}{b^2},\frac{v_3}{c^2})
\end{align}
Using the formula,
\begin{align}
K&=\frac{1}{f^3a^2b^2c^2}\begin{vmatrix}
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3 \\
x & y & z \\
\end{vmatrix}\\
&=\frac{1}{f^3a^2b^2c^2}<\mathbf{N},\mathbf{p}> \\
&=\frac{1}{f^4a^2b^2c^2}(\frac{x^2}{a^2},\frac{y^2}{b^2},\frac{z^2}{c^2}) \\
&=\frac{1}{f^4a^2b^2c^2}
\end{align}
at point $\mathbf{p}=(x,y,z)\in{M}$
| {
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"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.
If it is the right method, can someone show me how to do it, otherwise how would you figure this out?
$$\lim_{n\to\infty} \left(\frac{1 +5^{n+1}}{1+6^{n+1}} \frac{1+6^n}{1+5^n}\right)$$
| 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 $n\rightarrow\infty$, $6^{-n},5^{-n}\rightarrow 0$, and thus the limit of the ratio test is $\frac{5}{6}$. Conclude that ...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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}\\
0 & 0 & 0 & x^{12} & x^{13}\\
0 & 0 & 0 & x^{14} & x^{15}
\end{bmatrix}
Can someone help me with this question?
| 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 from another column, we get that that matrix has the same determinant as that of
\begin{bmatrix}
1 & x & x^2 & x^3 & 0\\
x^5 & x^6 & x^7 & x^8 & 0\\
0 & 0 & 0 & x^{10} & 0\\
0 & 0 & 0 & x^{12} & 0\\
0 & 0 & 0 & x^{14} & 0\\
\end{bmatrix}
and you can easily tell that the determinant of that matrix is $0$.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
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 ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 31) \equiv 7 (\mod 31)$, etc.). While this makes the computation easier, I'm thinking there might be a better way to do this.
| 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 $13$.
| {
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"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 can't
| 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{eqnarray}
\sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)} = -\frac{1}{2}\psi(2^j) + \psi(2^j+\frac{1}{2}) - \frac{1}{2} \psi(2^j+1)
\end{eqnarray}
where $\psi$ is the digamma function. To compute this without as much help from softare, use the original poster's expression for the summand and the fact that
$$
\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}
$$
for any $z\in\mathbb{C}$ as long as $z\ne -1,-2,-3,\ldots$.
Using explicit formulas (see Wikipedia) for $\psi(m)$ and $\psi(m+\frac{1}{2})$ where $m$ is an integer,
\begin{eqnarray}
-\frac{1}{2}\psi(2^j) + \psi(2^j+\frac{1}{2}) - \frac{1}{2} \psi(2^j+1) &=&
-\frac{1}{2}H_{2^j-1} - \frac{1}{2} H_{2^j} - 2 \log 2 + \sum_{k=1}^{2^j} \frac{2}{2k-1} \\
&=& -\frac{1}{2}H_{2^j-1} - \frac{1}{2} H_{2^j} - 2 \log 2 + 2(H_{2^{j+1}} - \frac{1}{2} H_{2^j}) \\
&=& -\frac{1}{2}H_{2^j-1} - \frac{3}{2} H_{2^j} - 2 \log 2 + 2 H_{2^{j+1}} \\
&=& -\frac{1}{2}(H_{2^j} -\frac{1}{2^j}) - \frac{3}{2} H_{2^j} - 2 \log 2 + 2 H_{2^{j+1}} \\
&=& \frac{1}{2^{j+1}} - 2 \log 2 + 2 (H_{2^{j+1}} - H_{2^j}).
\end{eqnarray}
Observe
\begin{eqnarray}
\sum_{j=0}^m \frac{1}{2^{j+1}} - 2 \log 2 + 2 (H_{2^{j+1}} - H_{2^j})
&=& 1 - 2^{-(m+1)} + 2\left(H_{2^{m+1}}- \log 2^{m+1}\right) - 2H_1.
\end{eqnarray}
It follows
\begin{eqnarray}
\sum_{j=0}^{\infty} \sum_{k=2^j}^{\infty} \frac{1}{k(2k+1)(2k+2)} &=& \lim_{m\to\infty} 1 - 2^{-(m+1)} + 2\left(H_{2^{m+1}}- \log 2^{m+1}\right) - 2H_1 \\
&=& 2\gamma - 1.
\end{eqnarray}
| {
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"answer_count": 3,
"answer_id": 0
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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 :
$$(x^2+x)dy=(x(x^2+1)^2+(3x^2+1)y)dx$$
$$(x^2+x)dy=(x(x^4+2x^2+1)+(3x^2y+y))dx$$
$$(x^2+x)dy=(x^5+2x^3+x+3x^2y+y)dx$$
$$(x^2+x)dy+(-x^5-2x^3-x-3x^2y-y)dx$$
set $Q(x,y)=x^2+x,\ \ P(x,y)=-x^5-2x^3-x-3x^2y-y$
$\frac{dQ}{dx}\neq \frac{dP}{dy}$ so I need to find integration factor.
$$\frac{\frac{dP}{dy}-\frac{dQ}{dx}}{Q}=\frac{-3x^2-2x-2}{x^2+x}$$
this is the right way? I need some advice how to continue, thanks.
| 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. Using $y(1)=2$ implies that $C=1$. Therefore
$$y(x) = x (x^2+1) (1+\log{x})$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $7344$, $8343$, or $9342$. Only the first of these is divisible by $72$: it’s $72\cdot102$.
It follows that $100a+10b+c=102$; if $a,b$, and $c$ must be single digits, this implies that $a=1$, $b=0$, and $c=2$. Thus, it’s impossible to meet the conditions: if $a,b,c,x$, and $y$ are all single digits, one of them must be $0$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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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 integrand more. I also tried to evaluate it with a CAS, but it was not able to handle it.
| 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 equation:
$$\begin{align*}\mathcal{I}(k) :&= \int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx \\ &= E'(k)-\left(\frac{1+k^2}{2} \right)K'(k)+\left(k^2 K'(k)-E'(k) \right)\frac{\log k}{2}+\log 2-1 \tag{2}\end{align*}$$
where $K'(k)$ and $E'(k)$ are complementary elliptic integrals of the first and second kind respectively.
Putting $k=\frac{1}{\sqrt{2}}$ in equation $(2)$,
$$
\begin{align*}
\mathcal{I}\left(\frac{1}{\sqrt{2}}\right)&=E'\left(\frac{1}{\sqrt{2} }\right)-\frac{3}{4}K'\left(\frac{1}{\sqrt{2}} \right)-\left\{\frac{1}{2} K'\left(\frac{1}{\sqrt{2}} \right)-E'\left(\frac{1}{\sqrt{2}} \right)\right\}\frac{\log 2}{4}+\log 2-1
\end{align*}
$$
Using the special values,
$$
\begin{align*}
E'\left(\frac{1}{\sqrt2} \right) &= \frac{\Gamma\left(\frac{3}{4} \right)^2}{2\sqrt\pi}+\frac{\sqrt{\pi^3}}{4\Gamma\left(\frac{3}{4} \right)^2}\\
K'\left(\frac{1}{\sqrt2} \right) &= \frac{\sqrt{\pi^3}}{2\Gamma\left(\frac{3}{4} \right)^2}
\end{align*}
$$
we get
$$
\mathcal{I}\left(\frac{1}{\sqrt{2}}\right)=\frac{1+\log\sqrt[4]2}{2\sqrt{\,\pi}}\Gamma\left(\frac34\right)^2-\frac{\sqrt{\,\pi^3}}8\Gamma\left(\frac34\right)^{-2}+(\log 2-1)\, \tag{3}
$$
Putting this in equation $(1)$, we get the answer that Cleo posted.
How to prove Equation $(2)$?
We begin with Proposition 7.1 of "The integrals in Gradshteyn and Ryzhik:
Part 16" by Boettner and Moll.
$$\int_0^\infty \frac{\log x}{\sqrt{(1+x^2)(m^2+x^2)}}dx = \frac{1}{2}K'(m)\log m$$
Multiplying both sides by $m$ and integrating from $0$ to $k$:
$$
\begin{align*}
\int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx &= \frac{1}{2}\int_0^k m K'(m)\log(m)\; dm
\end{align*}
$$
The result follows since
$$\begin{align*} \int m K'(m)\log(m)\; dm &= 2E'(m)-\left(1+m^2 \right)K'(m)+\left(m^2 K'(m)-E'(m) \right)\log m\\ &\quad +\text{constant} \tag{4}
\end{align*}$$
One can verify equation $(4)$ easily by differentiating both sides with respect to $m$ and using the identities
$$
\begin{align*}
\frac{dE'(k)}{dk}&= \frac{k}{k^{'2}}(K'(k)-E'(k))\\
\frac{dK'(k)}{dk}&= \frac{k^2 K'(k)-E^{'}(k)}{kk^{'2}}
\end{align*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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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 one leads to
$$(x-y)^2+(x-1)^2+(y-1)^2=2$$
| {
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"source": "stackexchange",
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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:
$$v = \frac x y - 1, v > 0$$
$$u = \frac y z - 1, u > 0$$
$$\frac {x^2} {y^2} \left(\frac y z - 1\right) + \left(\frac z x - 1\right) + \frac {z^2} {y^2} \left(\frac x y - 1\right) > 0 $$
$$(v+1)^2u + (v+1)^{-1}(u+1)^{-1} - 1 + (u + 1)^{-2}\left((v+1)(u+1)-1\right) > 0$$
Then I put it into a computer to simplify the fraction because no way I want to do that by hand...
$$\frac{\left( {u}^{3}+2\,{u}^{2}+u\right) \,{v}^{3}+\left( 3\,{u}^{3}+6\,{u}^{2}+4\,u+1\right) \,{v}^{2}+\left( 3\,{u}^{3}+5\,{u}^{2}+3\,u\right) \,v+{u}^{3}+{u}^{2}+u}{\left( {u}^{2}+2\,u+1\right) \,v+{u}^{2}+2\,u+1} > 0$$
That's my first time solving this kind of problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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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-)differentiable on $(a,\frac{a+b}{2})$ and $$g^\prime(x) = f^\prime(x+\frac{b-a}{2})-f^\prime(x).$$
*
One also notes $g(\frac{a+b}{2}) - g(a) = f(b)-2f(\frac{a+b}{2})+f(a)$.
*
The mean value theorem implies that there exists some $c\in(a,\frac{a+b}{2})$ such that $$g(\frac{a+b}{2}) - g(a) = \frac{b-a}{2}g^\prime(c) = \frac{b-a}{2}\left(f^\prime(c+\frac{b-a}{2})-f^\prime(c)\right).$$
*This last part completes the problem:
Another application of the mean value theorem for $f^\prime$ over the interval $[c,c+\frac{b-a}{2}]\subset[a,b]$, yields $\tilde{c}\in(c,c+\frac{b-a}{2})$ such that $$f^\prime(c+\frac{b-a}{2})-f^\prime(c) = \frac{b-a}{2}f^{\prime\prime}(\tilde{c}).$$
| {
"language": "en",
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"source": "stackexchange",
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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
$$
1+a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+\cdots= \cfrac{1}{
1- \cfrac{a_{1}}{
1+a_{1}- \cfrac{a_{2}}{
1+a_{2}- \cfrac{a_{3}}{
1+a_{3} - \ddots}}}}\;\;\;\;\;\;\;(1)
$$
where the LHS can be expressed as
$$
1+a_1(1+a_2(1+a_3(1+a_4(1+a_5(\cdots)))))\;\;\;\;\;\;\;\;\;\;\;\;(2)
$$
Now suppose that we replace those $1$'s with $b_{i}$'s we get
$$
1+a_{1}(b_{1}+a_{2}(b_{2}+a_3(b_{3}+a_4(b_{4}+a_5(\cdots)))))\;\;\;\;\;\;\;\;\;\;\;\;(3)
$$
My question is: Is it possible to express $(3)$ as an infinite continued fraction?
Thanks.
| 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.
$$
1+a_{1}(b_{1}+a_{2}(b_{2}+a_3(b_{3}+a_4(\cdots)))= \cfrac{1}{
1- \cfrac{a_{1}b_{1}}{
1+a_{1}b_{1}- \cfrac{\frac{a_{2}b_{2}}{b_{1}}}{
1+\frac{a_{2}b_{2}}{b_{1}}- \cfrac{\frac{a_{3}b_{3}}{b_{2}}}{
1+\frac{a_{3}b_{3}}{b_{2}} - \ddots}}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 $\displaystyle 2xdx = \sec^2 \theta \, d\theta\Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta$
$$
\begin{align}
& = \int\frac{\tan \theta - 1}{\sec \theta}\cdot \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta = \frac{1}{2}\int \frac{\left(\tan \theta - 1\right)\cdot \sec \theta}{\sqrt{\tan \theta}} \, d\theta \\
& = \frac{1}{2}\int \left(\sqrt{\tan \theta}-\sqrt{\cot \theta}\right)\cdot \sec \theta \, d\theta
\end{align}
$$
Now i did not understand how can i solve it
Help me
Thanks
|
Use binomial expansions of the expressions in respective integrals as given in the previous solution of the problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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",
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} |
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 $2\angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof.
From the Law of Sines,
$$a = 2R\sin A = 2R\sin (2B) = 4R\sin B\cos B$$
$$b = 2R\sin B$$
$$c = 2R\sin C = 2R\sin(180 - 3B) = 2R\sin(3B) = 2R(\sin B\cos(2B) + \sin(2B)\cos B)$$
$$=2R(\sin B(1 - 2\sin^2 B) +2\sin B\cos^2 B) = 2R(\sin B -2\sin^3 B + 2\sin B(1 - \sin^2B))$$
$$=\boxed{2R(3\sin B - 4\sin^3 B)}$$
Now,
$$\implies b(b+c) = 2R\sin B[2R\sin B + 2R(3\sin B - 4\sin^3 B)]$$
$$=4R^2\sin^2 B(1 + 3 - 4\sin^2 B)$$
$$=16R^2\sin^2 B\cos^2 B = a^2$$
Now, to prove the converse:
$$c = 2R\sin C = 2R\sin (180 - (A + B)) = 2R\sin(A+B) = 2R\sin A\cos B + 2R\sin B\cos A$$
$$a^2 = b(b+c)$$
$$\implies 4R^2\sin^2 A = 2R\sin B(2R\sin B + 2R\sin A\cos B + 2R\sin B\cos) $$
$$ = 4R^2\sin B(\sin B + \sin A\cos B + \sin B\cos A)$$
I have no idea how to proceed from here. I tried replacing $\sin A$ with $\sqrt{1 - \cos^2 B}$, but that doesn't yield any useful results.
|
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 bisector of $\angle A$ cut $BC$ at $D$ $\Longrightarrow$ $\dfrac{CD}{DB}=\dfrac{b}{c}$ $\Longrightarrow$ $CD=\dfrac{b}{b+c}a$
$a^{2}=b(b+c)$ $\Longrightarrow$ $\dfrac{a}{b}=\dfrac{b+c}{a}=\dfrac{b}{\dfrac{b}{b+c}a}$ $\Longrightarrow$ $\dfrac{BC}{CA}=\dfrac{CA}{CD}$ and $\angle DCA=\angle ACB$
thus, $\triangle ACD\sim\triangle ACB$ $\Longrightarrow$ $\angle CAD=\angle CBA$ $\Longrightarrow$ $\angle A=2\angle B$
| {
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"url": "https://math.stackexchange.com/questions/557704",
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"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
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, $\forall x \in dom,$ and $\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\epsilon.$
Can I do this?
| 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 by the conjugate to remove the square roots.
$$|\sqrt{x}-\sqrt{a}|<\varepsilon\implies|\sqrt{x}-\sqrt{a}|\cdot|\sqrt{x}+\sqrt{a}|<\varepsilon\cdot|\sqrt{x}+\sqrt{a}|$$
$$|x-a|<\varepsilon\cdot |\sqrt{x}+\sqrt{a}|. \tag{1}$$
Now, if you require that $|x-a|<1$, then it follows that $x-a<1$, so $a - 1<x<a+1$, and therefore that $\sqrt{x}<\sqrt{a+1}.$ Therefore, $\sqrt{x}+\sqrt{a}<\sqrt{a+1}+\sqrt{a},$ which combined with $(1)$ tells us that $$|x-a|<\varepsilon(\sqrt{a+1}+\sqrt{a}).$$
So, let $\delta = \mathrm{min}(1,\ \varepsilon(\sqrt{a+1}+\sqrt{a}))$. This proves that $f(x) = \sqrt{x}$ is continuous on $(0, \infty)$. $\square$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
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} |
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 $
$t = \cot^{-1}(\frac{A}{B})+\pi n$
However, I got stuck here on how to formulate the minimum and maximum points. Any explanation would be appreciated.
| 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
$\displaystyle \frac{A}{\sqrt{A^2+B^2}} = \sin \phi$
and
$\displaystyle \frac{B}{\sqrt{A^2+B^2}} = \cos \phi$.
This means that $\displaystyle A \cos t + B \sin t = C (\sin \phi \cos t + \cos \phi \sin t) = C \sin (\phi + t)$. Now we can easily see that the maximum is $C = \sqrt{A^2+B^2}$ and the minimum is $-C = -\sqrt{A^2+B^2}$.
And of course, the maximum is reached when $\displaystyle \phi + t = \frac{\pi}{2}$, and the minimum when $\displaystyle \phi + t = \frac{3\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560711",
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"source": "stackexchange",
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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 integral strongly suggests that this is incorrect. What is the correct value?
| 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
$$ \begin{align} &\int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\ &= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\&= \frac{1}{2} \int_{-\infty}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s} \sinh \pi t} \, dt \\ &= \frac{1}{2} \, \text{Im} \, \text{PV} \int_{-\infty}^{\infty} \frac{1}{(1-it)^{s} \sinh \pi t} \, dt \\ &= \frac{1}{2} \, \text{Im} \left(\pi i \ \text{Res}\left[\frac{1}{(1-iz)^{s} \sinh \pi z},0\right] + 2 \pi i \sum_{n=1}^{\infty} \text{Res} \left[\frac{1}{(1-iz)^{s} \sinh \pi z},in \right] \right) \\ &= \frac{1}{2} \text{Im} \left(\pi i \left(\frac{1}{\pi} \right) + 2 \pi i \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\pi (1+n)^{s}} \right) \\ &= \frac{1}{2} \, \text{Im} \Big(i + 2i \big( (1-2^{1-s}) \zeta(s)-1\big) \Big) \\& = (1-2^{1-s})\zeta(s) - \frac{1}{2}. \end{align}$$
By analytic continuation, the result is valid for all complex values of $s$.
Differentiating under the integral sign and letting $s=-1$ we get
$$ \begin{align} \frac{1}{2} \int_{0}^{\infty} \frac{t \log(1+t^{2})}{\sinh \pi t} \, dt + \int_{0}^{\infty} \frac{\arctan t}{\sinh \pi t} \, dt &= 2^{1-s} \log 2 \ \zeta(s) + (1-2^{1-s}) \zeta'(s) \Bigg|_{s=-1} \\ &= 4 \log (2) \left(-\frac{1}{12} \right) - 3 \zeta'(-1) \\ &= - \frac{\log 2}{3} - 3 \zeta'(-1) . \end{align}$$
So we need to evaluate $ \displaystyle \int_{0}^{\infty} \frac{\arctan t}{\sinh \pi t} \, dt $.
Let $ \displaystyle I(z) = \int_{0}^{\infty} \frac{\arctan \frac{x}{z}}{\sinh \pi x} \, dx $.
Then
$$ \begin{align} I(z) &= \int_{0}^{\infty} \frac{1}{\sinh \pi x} \int_{0}^{\infty} \frac{\sin tx}{t} e^{-zt} \, dt \, dx \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin t x}{\sinh \pi x} \, dx \, dt \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{e^{-zt}}{t} \tanh \left( \frac{t}{2} \right) \, dt . \end{align} $$
Now differentiate under the integral sign.
$$ \begin{align} I'(z) &= - \frac{1}{2} \int_{0}^{\infty} \tanh \left(\frac{t}{2} \right) e^{-zt} \, dt \\ &= - \frac{1}{2} \int_{0}^{\infty} \frac{1-e^{-t}}{1+e^{-t}} e^{-zt} \, dt \\ &= - \frac{1}{2} \int_{0}^{\infty} \left( -1 + 2 \sum_{n=0}^{\infty} (-1)^{n} e^{-tn}\right) e^{-zt} \, dt \\ &= \frac{1}{2z} - \sum_{n=0}^{\infty} (-1)^{n} \frac{1}{z+n} \\ &= \frac{1}{2z} - \frac{1}{2} \psi \left(\frac{z+1}{2} \right) + \frac{1}{2} \psi \left( \frac{z}{2} \right) \tag{1} \end{align}$$
Integrating back and using Stirling's formula to determine the value of the constant of integration, we find
$$ I(z) = \frac{\ln z}{2} - \log \Gamma \left(\frac{z+1}{2} \right) + \log \Gamma \left(\frac{z}{2} \right) - \frac{\log 2}{2} .$$
So
$$ \int_{0}^{\infty} \frac{\arctan x}{\sinh \pi x} \, dx = I(1) = \frac{\log \pi}{2} - \frac{\log 2}{2} .$$
Combining this result with the first result we have
$$ \begin{align} \int_{0}^{\infty} \frac{x \log(1+x^{2})}{\sinh \pi x} \, dx &= 2 \left(-\frac{\log 2}{3} - 3 \zeta'(-1) - \frac{\log \pi}{2} + \frac{\log 2}{2} \right) \\&=\frac{\log 2}{3} - \log \pi - 6 \zeta'(-1) \\ &= \frac{\log 2}{3} - \log \pi - \frac{1}{2} + 6 \log A \\ &\approx 0.0788460364 . \end{align}$$
$ $
$(1)$ http://mathworld.wolfram.com/DigammaFunction.html (5)
| {
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"url": "https://math.stackexchange.com/questions/561736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
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 $w = \frac{x + iy}{x^2+2xyi-y^2+1}$ or written slightly different $w= \frac{x + iy}{x^2-y^2+1+2xyi}$ or $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}$.
In this format we can multiply with complex conjugate $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}\cdot\frac{(x^2-y^2+1)-(2xyi)}{(x^2-y^2+1)-(2xyi)}$.
Then we get $w=\frac{x^3+xy^2-x^2yi-y^3i+x+iy}{(x^2+^2+1)^2-(2xyi)^2} = \frac{(x^3+xy^2+x)+(-x^2yi-y^3i+iy)}{(x^2+^2+1)^2+(2xy)^2} = \frac{(x^3+xy^2+x)+i(-x^2y-y^3+y)}{(x^2+^2+1)^2+(2xy)^2}$.
And as stated above, imaginary part of $w = 0$ meaning $\frac{-x^2y-y^3+y}{(x^2+^2+1)^2+(2xy)^2} = 0$.
As denominator can't be zero, means nominator is zero $-x^2y-y^3+y=0$. Or $y\cdot(-x^2-y^2+1)=0$. As $y\not=0$ implies $-x^2-y^2+1=0$ or $x^2+y^2=1$.
What now?
| 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,
$$
z-\bar{z}=z\bar{z}(z-\bar{z}).
$$
The hypothesis $y\ne0$ means $z\ne\bar{z}$ and so we get
$$
1=z\bar{z},
$$
that is, $|z|^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562873",
"timestamp": "2023-03-29T00:00:00",
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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: $1+\sqrt[3]{2}$ or $\sqrt[3]{12}$
Any ideas?
Thank you!
Shir
| 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} + \sqrt[3]{4} < 7 $ and the last inequality is true since $\sqrt[3]{2}<2$ and $\sqrt[3]{4}<4$
Thank you all for your answers!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/566605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $L=\lim \limits_{n\to\infty}a_n\dfrac{n^2+2n}{n^2+2n+1}=L\cdot1$ What does this result mean? Where did I make a mistake?
| It's easy to see that $a_{n+1}=a_n\left[1-\frac{1}{(n+1)^2}\right]$ and observing that
$$
\begin{align}
a_2&=a_1\left(1-\tfrac{1}{2^2}\right)\\
a_3&=a_2\left(1-\tfrac{1}{3^2}\right)=a_1\left(1-\tfrac{1}{2^2}\right)\left(1-\tfrac{1}{3^2}\right)\\
&\vdots\\
a_n&=a_{n-1}\left(1-\tfrac{1}{n^2}\right)=a_1\left(1-\tfrac{1}{2^2}\right)\left(1-\tfrac{1}{3^2}\right)\cdots\left(1-\tfrac{1}{n^2}\right)=a_1\prod_{k=2}^n\left(1-\tfrac{1}{k^2}\right)
\end{align}
$$
that is
$$
a_n=\tfrac{3}{4}\prod_{k=2}^n\left(1-\tfrac{1}{k^2}\right).
$$
It's easy to see that
$$
\prod_{k=2}^n\left(1-\tfrac{1}{k^2}\right)=\tfrac{(n+1)}{2n}\to\frac{1}{2}\qquad\text{for }n\to\infty
$$
and finally
$$
\lim_{n\to\infty}a_n=\frac{3}{8}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \\
= \lim_{x \to +\infty} x\left(\sqrt{1+ \frac {\sqrt x}{x}} + \sqrt{1 - \frac{\sqrt x}{x}} - 2\right) = \left[t = \frac 1x\right] = \lim_{t \to 0+} \frac{\sqrt{1+\sqrt t} - 1 +\sqrt{1-\sqrt t} - 1}{t} =
\left[c = \sqrt t\right] = \lim_{c \to 0+} \frac{\sqrt{1+c} - 1 +\sqrt{1-c} - 1}{c^2} = \frac 12c + \left(-\frac 12 c\right) = 0
$$
It's because:
$$
\lim_{c \to 0+} \frac{\sqrt{1+c}-1}{c} = \left|k = \sqrt{1+c}\right| = \lim_{k \to 1+} \frac{k-1}{k^2-1} = \lim_{k \to 1+} \frac{1}{k+1} = \frac 12
$$
The second part is similar to it. But the answer is -$\frac 14$.
And the second limit is:
$$
\lim_{x \to 0} \left(\frac{e^x - \sin x}{\sqrt{1 - 2x} + \log(x+1)}\right)^\cfrac{1}{x^2} = \exp\left(\lim_{x \to 0} \frac{1}{x^2}\left(\frac{e^x-\sin x}{\sqrt{1-2x} + \log(x+1)}-1\right)\right) =
\exp\left(\lim_{x \to 0}\frac{e^x - \sin x - \sqrt{1-2x} - \log(x+1)}{x^2} \cdot \frac{1}{\sqrt{1-2x}+\log(x+1)}\right) = \exp\left(\lim_{x \to 0} \left(\frac{e^x-1}{x \cdot x} - \frac{\sin x}{x \cdot x} - \frac{\sqrt{1-2x}-1}{x \cdot x} - \frac{\log(x+1)}{x \cdot x}\right)\right) = \exp\left(\lim_{x \to 0} \left(\frac 1x - \frac 1x + \frac 1x - \frac 1x\right)\right) = e^0 = 1
$$
The answer is $e^\frac 32$. Where am I wrong?
| 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 ?
| {
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"timestamp": "2023-03-29T00:00:00",
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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 infinity, or is it just an inconclusive test?
| $$\lim_{x \to \infty} |\frac{a_{n+1}}{a_n}|$$
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 than $1$ the series diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 far:
$$
\frac{(n+1)(x-4)^{n+1}}{(n+1)^3+1}\bullet\frac{n^3+1}{n(x-4)^n}
$$
which gets me to
$$
\frac{(n+1)(x-4)(n^3+1)}{n[(n+1)^3+1]}
$$
Which is where I'm stuck. How can I simplify this further so I can find R? Do I just need to multiply this out and look at the coefficients of the highest powers of n?
| 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^3+1}\sim \frac{1}{n^2}
\end{equation}
Then
\begin{equation}
\sum_{n=1}^\infty \frac{n}{n^3+1}
\end{equation}
converges, and the original function series converges for $|x-4|\leq 1$.
| {
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"url": "https://math.stackexchange.com/questions/571073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
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 $(2)$:
$$b=0 \ \ \text{or} \ \ 4.$$
And:
$$k^2=1100a+11b=11(100a+b)=11^2\cdot 9a+11(a+b) \Rightarrow a+b\equiv 0 \pmod{11}.$$
So, $\overline{aabb}=7744$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"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 $n \geqslant 5$:
$$
\frac{a_{n+1}}{a_n} = \frac13\left(1+\frac{1}{n}\right)^5 \leqslant \frac{1}{3} \left(1+\frac{1}{5}\right)^5 < 1
$$
Hence for $n\geqslant 5$ the sequence is decreasing, and since sequence elements are positive the sequence converges to this lower bound, i.e. zero.
| {
"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
$$ 2ab(1+\cos C) = (a+b)^2-c^2 = (a+b+c)(a+b-c), \tag{2}$$
$$ 2ab(1-\cos C) = c^2-(a-b)^2 = (a-b+c)(-a+b+c),\tag{3}$$
and by multiplying $(2)$ and $(3)$ and exploiting $(1)$
$$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)\tag{4} $$
which is equivalent to
$$ \Delta = \sqrt{s(s-a)(s-b)(s-c)}\tag{5} $$
as wanted.
Alternative derivation: by considering the circumcenter $O$ and its distances from the sides we have
$$2\Delta = R\sum_{cyc}a\cos A,\qquad 16\Delta^2 = \sum_{cyc}a^2\cdot 2bc(\cos A)=\sum_{cyc}a^2(b^2+c^2-a^2)\tag{6}$$
through $4R\Delta=abc$ and the cosine theorem. By rearranging the RHS of $(6)$
$$ 16\Delta^2 = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)\tag{7} $$
immediately follows.
| {
"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?
| $$
\begin{equation}
\begin{split}
\frac{z-1}{z+1} &= \frac{\cos\theta + i\sin\theta-1}{\cos\theta + i\sin\theta+1} \\
&= \frac{\cos\theta-1+i\sin\theta}{\cos\theta+1+i\sin\theta} \\
&=\frac{-2\sin^2\frac{\theta}{2} + i\sin\theta}{2\cos^2\frac{\theta}{2}+i\sin\theta} \\
&= \frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}\cdot\frac{-\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}} \\
&=\tan\frac{\theta}{2}\cdot i\cdot \frac{i\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}\\
&= i\tan\frac{\theta}{2}
\end{split}
\end{equation} \\
$$
Clearly the above expression is purely imaginary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"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, we should be able to write
$$ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots=1+\sum_{n=1}^{\infty}\frac{1}{2n+1}.\frac{1}{4^n}$$
This is what i could simplify the given question to...
But, this is doing me no good.. I could not go any further..
I would be thankful if some one can help me to clear this.
I would request users who are trying to help me to just give necessary hints but not post it as answer.
Thank You.
| 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
$$\text{arcsec}(p)-\text{arcsec}(q)=\text{arcsec}\left(\frac{pq}{1+\sqrt{p^2-1}\sqrt{q^2-1}}\right),p,q>0\,\text{or}\,p,q<0$$
I arrived at the following limit
$$\lim_{h\to0}\frac{\text{arcsec}\left(\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}\right)}{h}\tag{*}$$
For evaluating recent limit I wrote it's as follows
$$\lim_{h\to0}\frac{\text{arcsec}\left(\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}\right)}{\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}}\times\lim_{h\to0}\frac{\frac{x^2+xh}{1+\sqrt{(x+h)^2-1}\sqrt{x^2-1}}}{h}$$
but it converts to $0\times\infty$.
Please, help me for evaluating $(*)$.
Also for $\arcsin(x)$ by using this method see here.
| 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 $\sec$ into $1/\cos$. Some people may find it as essentially equivalent to the rule for differentiation of inverse function.
In that case another approach is to use the following relation $$\text{arcsec}\, x = \arccos \left(\frac{1}{x}\right) = \frac{\pi}{2} - \arcsin\left(\frac{1}{x}\right)$$ Using this we get
$\displaystyle \begin{aligned}L &= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{1}{x}\right) - \arcsin\left(\dfrac{1}{x + h}\right)}{h}\\
&= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{1}{x}\sqrt{1 - \dfrac{1}{(x + h)^{2}}} - \dfrac{1}{x + h}\sqrt{1 - \dfrac{1}{x^{2}}}\right)}{h}\\
&= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}\right)}{h}\\
&= \lim_{h \to 0}\dfrac{\arcsin\left(\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}\right)}{\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}}\cdot\dfrac{\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{x(x + h)}}{h}\\
&= \lim_{h \to 0}1\cdot\dfrac{\sqrt{(x + h)^{2} - 1} - \sqrt{x^{2} - 1}}{xh(x + h)}\text{ (because }\lim_{y \to 0}\dfrac{\arcsin y}{y} = 1\text{)}\\
&= \lim_{h \to 0}\dfrac{(x + h)^{2} - x^{2}}{xh(x + h)\left\{\sqrt{(x + h)^{2} - 1} + \sqrt{x^{2} - 1}\right\}}\\
&= \lim_{h \to 0}\dfrac{h(2x + h)}{xh(x + h)\left\{\sqrt{(x + h)^{2} - 1} + \sqrt{x^{2} - 1}\right\}}\\
&= \frac{2x}{x\cdot x\cdot 2\sqrt{x^{2} - 1}} = \frac{1}{x\sqrt{x^{2} - 1}}\end{aligned}$
| {
"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
} |
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