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Geometry Question Points on a Parabola We have the curve with equation $y^2=x+4$. Point $A$ is on the curve and has coordinates $(0,2)$, $B$ and $C$ are other points lying on the curve where the line segment $AB$ is perpendicular to $BC$. What values can the $y$-coordinate of the point $C$ be?
| Let $B(b^2-4,b)$ and $C(c^2-4,c)$, thus:
$$m_{AB}=\frac{2-b}{4-b^2}=\frac{1}{2-b}$$
$$m_{BC}=\frac{b-c}{b^2-c^2}=\frac{1}{b-c}$$
Perpendicularity condition:
$$m_{AB}=-\frac{1}{m_{BC}}\implies c-b=\frac{1}{2-b}\implies c=b+\frac{1}{2-b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5{x+3}+\frac 5{y+3}+\frac 5{z+3}-3\right]_{min}\Rightarrow \left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\left[ a+b+c\right]_{min}\Rightarrow a=b=c$
and from here, we get $$\frac 1{x+3}=\frac1{y+3}=\frac 1{z+3} \Rightarrow x=y=z$$
Finally, It must be $x=y=z=\frac 13$
and $$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}=\frac 32$$
Am I correct?
Note: I'm sorry for wrong mathematical symbols.
|
My solution:
$$[...]\quad \implies \quad\left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\color{red}{\left[ a+b+c\right]_{min}\Rightarrow a=b=c}$
$\color{red}{\style{font-family:inherit}{\text{That step}}}$ does not follow.
You presumably mean to use $a=\frac{1}{x+3}\,$, $b=\frac{1}{y+3}\,$, $c=\frac{1}{z+3}\,$. But "$\,\left[ a+b+c\right]_{min}\Rightarrow a=b=c\,$" only holds if the product $\,abc = \text{const}\,$ (and in that case it follows from AM-GM indeed). But here $\,abc=\frac{1}{(x+3)(y+3)(x+3)}\,$ which is obviously not constant, since all that's given is that $\,x+y+z=1\,$.
Another way, instead: by the harmonic mean inequality written for $(x+3), (y+3), (z+3)\,$:
$$
\begin{align}
\frac{3}{\cfrac 1{x+3}+\cfrac 1{y+3}+\cfrac 1{z+3}} &\le \frac{(x+3)+(y+3)+(z+3)}{3} = \frac{x+y+z+9}{3} = \frac{10}{3} \\[5px]
\implies \quad \cfrac 1{x+3}+\cfrac 1{y+3}+\cfrac 1{z+3} &\ge \frac{9}{10} \quad\quad \style{font-family:inherit}{\text{with equality iff}}\;\;x=y=z
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate: $-T_1^{2k}+T_2^{2k}-T_3^{2k}+T_4^{2k}-T_5^{2k}+T_6^{2k}-\cdots+T_{2n}^{2k}$ $T_n={n(n+1)\over 2}$, observing only the even terms of $T_n$ such as the series below
$$\color{red}{-1^2+3^2=8\times1^2}$$
$$\color{blue}{-1^2+3^2-6^2+10^2=8\times3^2}$$
$$\color{Red}{-1^2+3^2-6^2+10^2-15^2+21^2=8\times6^2}$$
$$\color{blue}{-1^2+3^2-6^2+10^2-15^2+21^2-28^2+36^2=8\times10^2}$$
$$-T_1^2+T_2^2-T_3^2+T_4^2-T_5^2+T_6^2-\cdots+T_{2n}^2$$
$$T_2^2-T_1^2+T_4^2-T_3^2+T_6^2-T_5^2-\cdots+T_{2n}^2-T_{2n-1}^2$$
$$8(1+2^3+3^3+\cdots n^3)$$
Well-known $1+2^3+3^3+\cdots +n^3=(1+2+3+\cdots n)^2=\left({n(n+1)\over 2}\right)^2=T_n^2$
so we have
$$8(1+2^3+3^3+\cdots n^3)=8T_n^2$$
My question is: Can we find a general formula for $$-T_1^{2k}+T_2^{2k}-T_3^{2k}+T_4^{2k}-T_5^{2k}+T_6^{2k}-\cdots+T_{2n}^{2k}?$$
$k:=1,2,3,...$
| The formula for $k=2$ is
$$
4 n^2 (n + 1)^2 (2 n^4 + 4 n^3 - 2 n + 1)
\\= 16 T_n^2 (8 T_n^2- 4 T_n + 1)
$$
The formula for $k=3$ is
$$
n^2 (n + 1)^2 (32 n^8 + 128 n^7 + 96 n^6 - 160 n^5 - 61 n^4 + 294 n^3 - 33 n^2 - 228 n + 114)
\\= 8 T_n^2 (256 T_n^4 - 384 T_n^3 + 390 T_n^2 - 228 T_n + 57)
$$
Thus, an interesting question is to express the sum as a polynomial in $T_n$, as in Faulhaber polynomials.
| {
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"url": "https://math.stackexchange.com/questions/2578915",
"timestamp": "2023-03-29T00:00:00",
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On $\sum_{n=0}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}+x$ and solvable Galois groups? In this comment, I hastily assumed that,
$$\sum_{\color{blue}{n=0}}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}$$
In fact, this is valid only for $b=1$ and, as Robert Israel pointed out, for general $b$ is missing a rational term. So for odd $b$ we have,
$$\begin{aligned}
\sum_{n=0}^\infty \frac{1}{n^2+n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{1^2-4c}\big)}{\sqrt{1^2-4c}}\\
\sum_{n=0}^\infty \frac{1}{n^2+3n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{3^2-4c}\big)}{\sqrt{3^2-4c}}-\frac1{c-2}\\
\sum_{n=0}^\infty \frac{1}{n^2+5n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{5^2-4c}\big)}{\sqrt{5^2-4c}}-\frac{2c-10}{(c-4)(c-6)}\\
\sum_{n=0}^\infty \frac{1}{n^2+7n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{7^2-4c}\big)}{\sqrt{7^2-4c}}-\frac{3c^2-56c+252}{(c-6)(c-10)(c-12)}\end{aligned}$$
But I'm having trouble finding $b=9$.
Questions:
*
*What is the rational term for $b=9$?
*For $b=2m+1$ and $m>0$, is it true that the rational term has form $\displaystyle \frac{P_1(c)}{P_2(c)}$ where $P_1(c)$ is a polynomial of degree $m-1$, while $P_2(c)$ has degree $m$?
*This may be like throwing a curveball, but do the equations $P_1(c)=0$ and $P_2(c)=0$ have solvable Galois groups?
| Well, if a quadratic polynomial $p(x)$ has a negative discriminant, $\sum_{n\in\mathbb{Z}}\frac{1}{p(n)}$ can be found through the Poisson summation formula or just by considering $\frac{d}{dx}\log(\cdot)$ applied to the Weierstrass product for the cosine function. For instance
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+9n+c}=4\sum_{n\in\mathbb{Z}}\frac{1}{(2n+9)^2+(4c-49)}=4\sum_{n\in\mathbb{Z}}\frac{1}{(2n+9)^2+\sqrt{4c-81}^2}$$
and assuming $d>0$
$$ \sum_{n\in\mathbb{Z}}\frac{1}{(2n+1)^2+d^2}=\frac{\pi}{2d}\,\tan\frac{\pi d}{2} $$
so by assuming $c>\frac{81}{4}$ we have
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+9n+c}=\frac{2\pi\tan\left(\pi\sqrt{c-\tfrac{81}{4}}\right)}{\sqrt{4c-81}} $$
and $p(n)=n(n+9)+c$ fulfills $p(n)=p(-n-9)$, so the LHS of the previous line can be written as
$$ 2\sum_{n\geq 0}\frac{1}{n^2+9n+c}+\sum_{k=1}^{8}\frac{1}{p(-k)}$$
such that
$$\sum_{n\geq 0}\frac{1}{n^2+9n+c}=\frac{\pi\tan\left(\pi\sqrt{c-\tfrac{81}{4}}\right)}{\sqrt{4c-81}}-\frac{4 \left(c^3-45c^2+654c-3044\right)}{(c-8)(c-14)(c-18)(c-20)}.$$
$P_2(c)$ has a pretty clear structure, I am not so confident about the Galois group of $P_1(c)$ over $\mathbb{Q}$, but the same manipulation applies in the general case.
| {
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"timestamp": "2023-03-29T00:00:00",
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check convergence $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $
check convergence of $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $
so for $x\le0$ : $e^{3x}\le1$ then we get $0\le\frac{e^{3x}}{1+x^2}\le\frac{1}{1+x^2}\le\frac{1}{x^2}$
and because $\int _{-\infty }^{-1}\:\frac{1}{x^2}$ is convergence we get $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $ is convergence
is that answer correct
thanks alot
| Well, you are correct, but your arguments could be a lot more precise.
Note that when: $x \in (-\infty,0), e^x < 1$. Hence, $$f(x) = \frac{e^x}{1+x^2} < \frac{1}{1+x^2} = g(x)$$
Hence, by the comparison test as: $$\int_{-\infty}^{0} \frac{1}{1+x^2}\, dx = \arctan x \bigg \lvert_{-\infty}^{0} = \frac{\pi}{2}$$ which converges. Hence, the original integral also converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581648",
"timestamp": "2023-03-29T00:00:00",
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Find the equation from the given equation in which leading term has c0-efficient $1$
Find the equation obtained from $5x^3-7x^2+8x-9=0$ into one in which the co-efficients of leading term is unity and the others are integers.
Below I am showing my attempt:
Applying the transformation $x\mapsto x+h$,then $f(x)=5x^3-7x^2+8x-9$ transforms into $f(x+h)=5(x+h)^3-7(x+h)^2+8(x+h)-9$
$ =5(x^3+h^3+3x^2h+3xh^2)-7(x^2+h^2+2xh)+8(x+h)-9$.
But this is not giving anything feasible as the leading term of $x$ must have co-efficient $1$.
How can I arrive at it?Should I use any alternative transformation.
| Then, divide it by $5$ to obtain: $$f’(x+h) = x^3+x^2(3h-\frac75) + x(3h^2-\frac{14h}{5}+\frac85) +(h^3-\frac{7h^2}{5}+\frac{8h}{5}-\frac95)$$
Now, note that $f’(x+h)$ is monic and the coefficients of every power of $x$ should $\in \mathbb{Z}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f_n(x) = \frac{x^2}{(1+x^2)^n}$ If $\sum_{n=0}^\infty$ converges pointwise, find it's limit.
Let $f_n(x) = \frac{x^2}{(1+x^2)^n}$ If $\sum_{n=0}^\infty f_n(x)$ converges pointwise, find it's limit. Determine if it converges uniformly.
It is clear that $0\le \frac{x^2}{(1+x^2)^n} < 1, \forall n \ge 1$.
With the assumption is does converge pointwise it appears that it is a geometric series so we have.
$\displaystyle \sum_{n = 0}^\infty f_n(x) = \sum_{n = 0}^\infty \frac{x^2}{(1+x^2)^n} = \frac{x^2}{1-\frac{1}{(1+x^2)}}$
Being geometric is seems that it would converge uniformly but I am having difficulty showing why that is the case.
| Note that we have
$$\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}=x^2\left(\frac{1-\frac{1}{(1+x^2)^{N+1}}}{1-\frac{1}{1+x^2}}\right)=1+x^2-\frac{1}{(1+x^2)^N}$$
For $x>0$, we have $\lim_{N\to \infty }\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}=1+x^2$.
Now, take $\epsilon=\frac1e$. Then,
$$\left|\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}-\left(1+x^2\right)\right|=\frac{1}{(1+x^2)^N}$$
Take $x=\frac{1}{\sqrt N}$. Then, note that
$$\lim_{N\to\infty}\frac{1}{(1+x^2)^N}=\lim_{N\to\infty}\frac{1}{\left(1+\frac1N\right)}=\frac1e$$
Therefore, the convergence is not uniform on $(0,\infty)$, but is uniform on $[\delta,\infty)$ for any $\delta>0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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limit of a trigonometric function $ \lim\limits_{x\to\pi/3} \frac{1 - 2\cos (x)}{\sin (3x)} $ compute the limit of
$$ \lim_{x\to \frac{\pi }{3} } \frac{1 - 2 \cos (x)}{\sin (3x)} $$
I would like to not do a translation with the change of variable $ t = x - \frac{\pi }{3} $
| Taking into account "I would like to not do a translation with the change of variable ...", let us use the Taylor series around $x=a$.
$$\cos(x)=\cos (a)-(x-a) \sin (a)-\frac{1}{2} (x-a)^2 \cos (a)+O\left((x-a)^3\right)$$
$$\sin(x)=sin (a)+(x-a) \cos (a)-\frac{1}{2} (x-a)^2 \sin (a)+O\left((x-a)^3\right)$$ So, using $a=\frac \pi 3$
$$\cos(x)=\frac{1}{2}-\frac{1}{2} \sqrt{3} \left(x-\frac{\pi }{3}\right)-\frac{1}{4}
\left(x-\frac{\pi }{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
$$1-2\cos(x)=\sqrt{3} \left(x-\frac{\pi }{3}\right)+\frac{1}{2} \left(x-\frac{\pi
}{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
$$\sin(3x)=-3 \left(x-\frac{\pi }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
$$\dfrac{1 - 2\cos (x)}{\sin (3x)}=\frac{\sqrt{3} \left(x-\frac{\pi }{3}\right)+\frac{1}{2} \left(x-\frac{\pi
}{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right) } {-3 \left(x-\frac{\pi }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^3\right) }$$ $$\dfrac{1 - 2\cos (x)}{\sin (3x)}=-\frac{1}{\sqrt{3}}-\frac{1}{6} \left(x-\frac{\pi
}{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^2\right)$$ which shows the limit and how it is approached.
| {
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"timestamp": "2023-03-29T00:00:00",
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To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\cos A } - 1$$
$$\Rightarrow \tan A + \cot A =\frac{ \sqrt 2 }{\cos A} -1 + \cot A $$
and the 2nd way as -
$$(\sin A + \cos A)^2 = 2$$
$$\sin ^2 A + \cos ^2 A + 2\sin A\cos A = 2 $$
$$\Rightarrow 2\sin A\cos A=1$$
$$\Rightarrow \sin A\cos A=\frac12$$
As we can see the first way is unable to give an answer in absolute Real Number, and the second way doesn't go even near to what is required to proof.
I know few trigonometry identities as per my textbook, those are
*
*$\sin^2 A + \cos^2 A = 1$
*$1 + \cot^2 A = \csc^2 A$
*$\tan^2A + 1 = \sec^2 A$
| Let's use $\sin^2A=\sqrt{1-\cos^2A}$
Equation transform into$$\sin A+\sqrt{1-\sin^2A}=\sqrt2$$
Which is quadratic equation and easy to solve
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $ \prod_{i=1}^n 3^{2^i} + \sum_{i=1}^n 3^{2^{i+1}-1} \prod_{j=i+1}^n 3^{2^j}$ I'm working on difference equations and got stuck on the algebra with this one:
$$ u_n = \prod_{i=1}^n 3^{2^i} + \sum_{i=1}^n 3^{2^{i+1}-1} \prod_{j=i+1}^n 3^{2^j}$$
I need to simplify this to just a function of n. Any help would be much appreciated.
|
We obtain
\begin{align*}
\color{blue}{\prod_{i=1}^n3^{2^i}}&\color{blue}{+\sum_{i=1}^n3^{2^{i+1}-1}\prod_{j=i+1}^n3^{2^j}}\\
&=3^{2^1+2^2+\cdots+2^n}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{i+1}+2^{i+2}+\cdots+2^{n}}\\
&=3^{2\left(2^n-1\right)}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{i+1}\left(2^{n-i}-1\right)}\\
&=3^{2\left(2^n-1\right)}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{n+1}-2^{i+1}}\\
&=3^{2\left(2^n-1\right)}+3^{2^{n+1}-1}\sum_{i=1}^n1\\
&=\frac{1}{9}3^{2^{n+1}}+\frac{1}{3}3^{2^{n+1}}n\\
&\color{blue}{=(3n+1)3^{2^{n+1}-2}}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given matrix times vector, find the inverse of the matrix times the same vector. Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.
My thought are the following:
Step 1:
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$
$V=8A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$
$\frac{1}{8} V=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$
Step 2:
Can I assume $AV=\lambda V=8*\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$ here? Hence, $V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$? Is this a correct step?
Step 3:
If step 2 is correct, then
$\frac{1}{8}V = \frac{1}{8} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} V=\begin{bmatrix} \frac{1}{8} \\ \frac{1}{8} \\ \frac{1}{8} \\\frac{1}{8}\end{bmatrix}$
Does the above inference sounds right?
| In general, no you can't assume $AV = \lambda V$. Not every vector is an eigenvector. Even if $V$ were an eigenvector, then this would still not give you a well-defined answer.
For one example, take $A = I$, the $4 \times 4$ identity matrix. Then $AV = V$, hence $V = \begin{bmatrix} 8 \\ 8 \\ 8 \\ 8 \end{bmatrix} \neq \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$.
As another example, take $A = 2I$. Then $AV = 2V$, hence $V = \begin{bmatrix} 4 \\ 4 \\ 4 \\ 4 \end{bmatrix} \neq \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$.
As you can see, there clearly needs to be more conditions placed on $A$ before this has a clear solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $(y+ x^3y + 2x^2)dx + (x + 4xy^4+ 8y^3)dy = 0$ Basically I tried to group the terms
ydx + xdy is a perfect differential but I couldn't think a method for ($x^3$y)dx + (4x$y^4$)dy . Also dividing by xy didn't help me.
| $$(y+ x^3y + 2x^2)\text{d}x + (x + 4xy^4+ 8y^3)\text{d}y = 0$$
$$\Rightarrow (y\text{d}x + x\text{d}y)+ x^3y\text{d}x + (2x^2\text{d}x) + 4xy^4\text{d}y+ (8y^3\text{d}y) = 0$$
$$\Rightarrow \text{d}(xy)+ (x^3y\text{d}x + 4xy^4\text{d}y) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$
$$\Rightarrow \text{d}(xy)+ xy(x^2\text{d}x + 4y^3\text{d}y) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$
$$\Rightarrow \text{d}(xy)+ xy(\frac{\text{d}(x^3)}3 + \text{d}(y^4)) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$
$$\Rightarrow \text{d}(xy)+ \frac{(xy+2)}3\text{d}(x^3) + (xy+2)\text{d}(y^4) = 0$$
$$\Rightarrow \text{d}(xy)+ (xy+2)(\frac{1}3\text{d}(x^3) + \text{d}(y^4)) = 0$$
$$\Rightarrow \frac{\text{d}(xy)}{(xy+2)}+ \frac{1}3\text{d}(x^3) + \text{d}(y^4) = 0$$
$$\Rightarrow \ln(xy+2)+ \frac{1}3x^3 + y^4 = C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| Let's try to spot a pattern.
$$f(x)=3+\frac x3=\frac13(3^2+x)$$ $$f(f(x))=f\left(\frac13(9+x)\right)=3+\frac{9+x}{9}=4+\frac x9=\frac1{3^2}(3^3+3^2+x)$$ $$f(f(f(x)))=f\left(\frac19(36+x)\right)=3+\frac{36+x}{27}=\frac{13}3+\frac x{27}=\frac1{3^3}(3^4+3^3+3^2+x)$$
So iterating $2018$ times, we have $$f(f(f(\cdots f(x))))=\frac1{3^{2018}}(3^{2019}+3^{2018}+...+3^2+x)$$ or $$3^1+3^0+3^{-1}+...+3^{-2016}+\frac x{3^{2018}}\approx 4.5+\frac{2018}{3^{2018}}$$ Now since $3^{2018}$ is much bigger than $2018$, that term could be negligible so we can conclude that $$\boxed{f(f(f(\cdots f(x))))=4.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 3
} |
Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 + 3 + 5 + \dots &= x\\
1 + 1 + 2 + 3 + \dots &= x\\
1 + 2 + 3 + 5 + 8 + \dots &= 2x \text{ (shifting and adding)}
\end{align}$$
We in fact get the same sequence. But the new sequence is one less than the original sequence. So the new sequence is $x-1$.
But $x-1=2x$ which implies that $x=-1$.
So $1+1+2+3+5+8+\dots=x$ which means...
$1+1+2+3+5+8+13+21+\dots=-1$
Is this right or wrong? Can someone please tell? Thanks...
| Since the terms of the series don't go to $0$, the series does not converge. Relatedly, $x=\infty$ also solves your equation $x-1=2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
simplification of trigonometric expression I search a simplification for
$$ \arcsin\biggl( \frac{1}{\sqrt 2} \frac{x}{\sqrt {(x-1)^2 +1}}\biggr)$$
any idea ? I tried composition, derivation and setting $x$ as a sine, cosine or a tangent in vain.
Thanks
| Let us substitute $x = 1 + \tan \theta $. We then get: $$E = \arcsin \left (\frac1 {\sqrt 2} \frac {1+\tan \theta}{\sqrt {\sec^2 \theta}} \right) $$ $$=\arcsin \left (\frac {1}{\sqrt 2} (\cos \theta + \sin \theta) \right) $$ $$=\arcsin \left (\sin \theta \cos \frac {\pi}{4} + \cos \theta \sin \frac {\pi}{4} \right) $$ $$=\arcsin \left (\sin (\theta + \frac {\pi}{4}) \right) $$ $$=\, ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Some infinite series (just for fun!) I have a few infinite series problems that I think MSE might enjoy, whose answers I already know:
$$\sum_{n=0}^\infty \frac{2^{n-2^n}}{1+2^{-2^n}}=\text{?}$$
$$\sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}=\text{?}$$
$$\sum_{n=0}^\infty \frac{3^n(1+2\cdot3^{-3^n})}{2\cosh(3^n\ln 3)+1}=\text{?}$$
After these all get answered, I will post explanations of how I came up with answers for them. Have fun!
| \begin{align}
\frac{1}{1-2}+S_1&=\lim_{N\to\infty}(\frac{1}{1-2}+\sum_{n=0}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}(\frac{1}{1-2}+\frac{1}{1+2}+\sum_{n=1}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}(\frac{2}{1-2^2}+\sum_{n=1}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}\frac{2^{N+1}}{1-2^{2^{N+1}}}\\
&=0
\end{align}
\begin{align}
\frac{1}{(4^{-2^{-1}}-4^{2^{-1}})^2}-S_2&=\frac{1}{(4^{-2^{-1}}-4^{2^{-1}})^2}-\sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}\\
&=\lim_{N\to\infty}(\frac{1}{(4^{-2^{-1}}-4^{2^{-1}})^2}-\frac{1}{(4^{-2^{-1}}+4^{2^{-1}})^2}-\sum_{n=1}^N \frac{4^{n}}{(4^{-2^{n-1}}+4^{2^{n-1}})^2})\\
&=\lim_{N\to\infty}(\frac{4}{(4^{-2^{0}}-4^{2^{0}})^2}-\sum_{n=1}^N \frac{4^{n}}{(4^{-2^{n-1}}+4^{2^{n-1}})^2})\\
&=\lim_{N\to\infty}\frac{4^{N+1}}{(4^{-2^{N}}-4^{2^{N}})^2}\\
&=0
\end{align}
Same thing for $S_3$. Note that $\cosh(x)=\frac{e^x+e^{-x}}{2}$. Just some calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find all ordered pairs such that $x + y^2 = 2$ and $y + x^2 = 2$ How can I find all ordered pairs $(x,y)$ such that $x + y^2 = 2$ and $y + x^2 = 2$?
A solution/explanation would be greatly appreciated.
If it helps, the solutions given in the textbook are
x=-2, y=-2; x=1, y=1; x= (1+root 5)/2, y=(1-root 5)/2; x= (1-root 5)/2, y=(1+root 5)/2
| Well $x+y^2=y+x^2$ so $x-y=x^2-y^2=(x-y)(x+y) $
If $x-y=0$ then $x=y$ and $x+x^2=2$. We can use the the quadratic equation to solve that.
If $x-y\ne 0$ then $1=(x+y) $ and $y=1-x $
So $x+y^2=x^2+y=x^2-x+1=2$ and we can use the quadratic equation to solve that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?
| We get $$\frac{a}{1+a}+\frac{b}{1+b}-\frac{a+b}{1+a+b}={\frac {ba \left( a+b+2 \right) }{ \left( 1+a \right) \left( 1+b
\right) \left( 1+a+b \right) }}
\geq 0$$
the numerator can be calculated as
$$a(1+b)(1+a+b)+b(1+a)(1+a+b)-(a+b)(1+a)(1+b)=...$$
we also have under the same conditions
$$\frac{a+b+c}{1+a+b+c}\le \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}$$ and so on ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Find the area above the $x-$ axis included between the parabola $y^2=ax$ and the circle $x^2+y^2=2ax$.
Find the area above the $x-$ axis included between the parabola $y^2=ax$ and the circle $x^2+y^2=2ax$.
The circle $x^2+y^2=2ax$ can be represented as $(x-a)^2+y^2=a^2$ which has centre at $(a,0)$ and radius $a$.
Since they both intersect at $(0,0)$ and $(a,a)$ so the area included will be
$\int _0^a \int _0^{\sqrt {ax}} dy dx-\int_0^a \int _0^{\sqrt {2ax-x^2}}dy dx$
But the problem is I am unable to compute the 2nd integral.
The answer to the first is $\frac{2}{3}a^{{2}}$.
Please help to solve it.
|
The graph is drawn taking $a=2$. What you need to notice is that the second integral is just $1/4$-th the area of the circle. So required area is $$\int_0^a\sqrt{ax}dx+\frac{\pi a^2}4=\frac{a^2}{3/2}+\frac{\pi a^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find values of x so that the matrix is not invertible Given
$$A=\begin{pmatrix}
5 & 6 & 6-x & 8 \\
2 & 2-x & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7
\end{pmatrix}\in\mathbb{R}^{4\times 4}$$
One must find all values of $x\in\mathbb{R}$ such that the matrix is not invertible.
I tried finding the determinant of $A$ and in the process got that $x = 4$ and $x = 6/192$. Therefore, $A1 = 1360; A2 = 816; A3 = 1146; A4 = 1088$. However when I calculate $\det(A) = 1360 - 816 + 1146 - 1088= 602$ and I do not get 0 so that I could prove it is not invertible.
| HINT
Let's calculate $\det A$ as a function of x and then set $\det A=0$.
Note that adding a scalar multiple of one column/row to another column/row does not change the value of the determinant. Thus you can simplify the matrix A before to perform the calculation.
You should obtain:
$$\det A=2(-13 x^2 + 19 x - 4)$$
and from here find the values for which $\det A=0$.
Notably
$$\det A=\begin{vmatrix}
5 & 6 & 6-x & 8 \\
2 & 2-x & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7
\end{vmatrix}=
\begin{vmatrix}
-1 & x & 6-x & 8 \\
x & -x & 2 & 8 \\
0 & 4 & 2 & 8 \\
-1 & -3 & 6 & 7
\end{vmatrix}=\\=-1\begin{vmatrix}
-x & 2 & 8 \\
4 & 2 & 8 \\
-3 & 6 & 7
\end{vmatrix}-x\begin{vmatrix}
x & 6-x & 8 \\
4 & 2 & 8 \\
-3 & 6 & 7
\end{vmatrix}+1\begin{vmatrix}
x & 6-x & 8 \\
-x & 2 & 8 \\
4 & 2 & 8 \\
\end{vmatrix}=\\=
-1\cdot(34x+136)-x\cdot(18x-72)+(128-8x^2)=-26x^2+38x-8 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $ab$ given the value of a limit If $$\lim_{x\to 0} \frac{ae^x -b}{x} = 2$$ then find $ab$.
I have tried everything that atleast I know about limits or algebra but I got not a single idea how to do this.
Please help
| We have
$$e^x = \sum_{k = 0}^\infty\frac{x^k}{k!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
by Taylor Series expansion of $e^x$, so
$$\lim_{x\to 0} \frac{ae^x -b}{x} = \lim_{x\to 0}\frac{a(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)-b}{x} = \lim_{x\to 0}\bigg(\frac{a}{x}+a+\frac{x}{2!}+...-\frac{b}{x}\bigg) = 2$$
Now, notice that as $x \to 0$, the terms $\frac{x}{2!}$, $\frac{x^2}{3!},...$ will all become $0$ and we are left with:
$$\lim_{x\to 0}\bigg(\frac{a-b}{x}+a\bigg) = 2$$
This is possible only when we have $a = b$ so that the limit $\lim_{x\to 0}\frac{a-b}{x}$ does not $\to \infty$ (notice that we have a real expression on RHS) and equal to $0$; and $a = 2$ so that the equation is satisfied. Therefore, $a = b = 2$ and $ab = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum\limits_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $
I've got in my assignment to show if the following series converges or diverges.
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$
Attempt:
\begin{align*}
\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) &=\sum_{n=1}^\infty \left( 1 + \frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...-1-\frac{1}{n}\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)
\end{align*}
At this point I'm lost. I tried using D'Alambert as follows:
$$\lim \frac{a_{n+1}}{a_n}=\lim\frac{\sqrt[n+1]{e}-1-\frac{1}{n+1}}{\sqrt[n]{e}-1-\frac{1}{n}}$$
which I tried to simplify with basic limit laws (hopefully correctly):
$$\lim\frac{\sqrt[n+1]{e}-1}{\sqrt[n]{e}-1} = 1$$
I don't know where to go from here. Thank you for all your help in advance.
| Just use comparison test. Using this Proving Schwarz derivative $\frac{f''(0)}{2} =\lim\limits_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule? you have $$\lim_{n\to\infty}{n^2}\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) =\lim_{x\to 0}\frac1{x^2}\left(e^x -1 -x\right) =\lim\limits_{x\to 0} \frac{\frac{e^x -1}{x}-1}{x}= \frac12$$
Hence, for n large enough we have $$ \left|n^2\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) -\frac1{2}\right|\le \frac14$$
That is $$ \frac1{4n^2}\le \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) \le \frac3{4n^2}$$
But $$\sum_{n=1}^\infty\frac1{n^2}<\infty$$
That is
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right)$$
converges too
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$ . Prove that $ABC$ is the equilateral triangle. For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$. Prove that $ABC$ is the equilateral triangle.
| With the standard notation, if $M_A$ is the midpoint of $BC$,
$$ AM_A^2 = \frac{2b^2+2c^2-a^2}{4},\quad AC^2 = b^2,\quad CM_A^2=\frac{a^2}{4} $$
hence if $\widehat{M_A A C}=30^\circ$ we have
$$ \cos 30^\circ = \frac{\sqrt{3}}{2} = \frac{AM_A^2+AC^2-CM_A^2}{2\cdot AM_A\cdot AC}=\frac{-a^2+3b^2+c^2}{4b\cdot AM_A} $$
and
$$ 2b\sqrt{3}\cdot AM_A = -a^2+3b^2+c^2 $$
$$ 2a\sqrt{3}\cdot BM_B = -b^2+3a^2+c^2 $$
lead to
$$ 3b^2(2b^2+2c^2-a^2)=(-a^2+3b^2+c^2)^2 $$
$$ 3a^2(2a^2+2c^2-b^2)=(-b^2+3a^2+c^2)^2 $$
or
$$ a^4+3b^4+c^4 = 3a^2 b^2+2a^2 c^2 $$
$$ b^4+3a^4+c^4 = 3a^2 b^2+2b^2 c^2 $$
so by difference $b^4-a^4 = c^2(a^2-b^2)$. If we assume $a\neq b$ we get $a^2+b^2+c^2=0$, which is absurd, hence $a=b$ and
$$ a^4+c^4 = 2a^2 c^2 $$
leads to $a=c$, too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator. Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $\sqrt{a} - \sqrt{b}$. I get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$. I am stuck here. How do I simplify this? Or is it that my method is wrong?
| $$\frac {1}{\sqrt n +\sqrt {n+1}} = \sqrt {n+1} - \sqrt {n}$$
Therefore the dsire sum is $ A-B$ where,
$$A= \sqrt 100+\sqrt 98 +....+\sqrt 2 = 338.047...$$ and
$$B=\sqrt 99+\sqrt 97 +....+\sqrt 1 = 333.415....$$
Thus [A-B]=4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Differentiating two trig functions. Check. I have two problems that I'd like a quick checkup on. I am trying to differentiate these two functions:
*
*$f(x) = x \cdot \cos x + 2\cdot \tan x$
and
*$f(x) = e^x \cdot \cos x$
Here are my attempts:
*
*Going to use a combo of sum and product rules.
$$f'(x) = \frac{d}{dx} (x\cdot cosx) + \frac{d}{dx} (2\cdot tanx)$$
$$= x\cdot(-sinx) + cosx + 2sec^2x$$
Is there anyway to simplify from here? It looks like I'm missing something...
*$$f'(x) = e^x\cdot sinx + cosx \cdot e^x$$
$$e^x(\sin x + \cos x)$$
Here's another check:
*Differentiate $y = sec\theta \cdot tan\theta$
$$= y' = sec\theta \cdot sec^2\theta + tan\theta \cdot sec\theta \cdot tan\theta$$
$$= sec^3\theta + tan^2\theta \cdot sec\theta$$
$$= sec\theta \cdot (sec^2\theta + tan^2\theta)$$
| Perhaps extending the steps will help. We have
$$f(x) = x\cos(x) + 2\tan(x)$$
Then
\begin{align*}
f'(x) &= \frac{d}{dx}\left(x\cos(x) + 2\tan(x)\right)\\
&= \frac{d}{dx}\bigg( x \bigg) \cos(x) + x\frac{d}{dx}\bigg( \cos(x) \bigg) + \frac{d}{dx}\bigg(2\tan(x) \bigg )\\
&= \cos(x) - x\sin(x) + 2\sec^2(x)
\end{align*}
For the second we have
$$f(x) = e^{x}\cos(x)$$
Then
\begin{align*}
f'(x) &= \frac{d}{dx}\bigg(e^{x}\cos(x)\bigg)\\
&= \frac{d}{dx}\bigg(e^{x}\bigg)\cos(x) + e^{x}\frac{d}{dx}\bigg(\cos(x)\bigg)\\
&= e^{x}\cos(x) - e^{x}\sin(x)\\
&= e^{x}\left(\cos(x) - \sin(x)\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $15\cdot x=27\pmod{18}$ EXERCISE:
Solve $15\cdot x\equiv 27\pmod{18}$
SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.
So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!
Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equiv 9\pmod{6}$
So, from now and then I can't understand how to proceed!
The book continue with this:
"We have a unique solution of $5\cdot x\equiv 9\pmod{6}$ which is":
$x=5\cdot 9 =45\equiv 3\pmod 6$ and $15\cdot 3\equiv 27\pmod{18}\equiv 9\pmod{18} $
The other solutions are:$(9,15)$. So all solutions are these: $(3,9,15)$
Can anyone explain me how the author of the book continue the exercise?I really can't understand how he ends up with these solutions!Is there another way to reach these solutions?
I would really appreciate a thorough explanation, since I've just started working on these type of exercises and I have to clear my mind on them.
Thanks for your time !
| It is possible to solve this problem using algebraic symbols or augmented matrices. For example the following uses augmented matrices to solve the problem:
$$\begin{array}{c|c|c|}
\hline
\text{x} & \text{y} & \text{c} & \text{s} & \text{t} \\
\hline
\text{15} & \text{-18} & \text{27} \\
& \text{+3} & \text{+3} & \text{-15} \\
& & & & \text{3} \\
\hline
\text{5} & \text{-6} & \text{9} \\
& \text{1} & \text{1} & \text{-5} \\
\hline
\text{1} & & \text{3} & \text{9} \\
& \text{1} & \text{1} & \text{-5}
\\
\hline
\end{array}$$
I have included animations for the construction of the augmented matrices in the PDF document "How I solved 15x≡27 (mod 18)" and a description of the "method" and its relation to the Euclidean Algorithm and the Extended Euclidean Algorithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to calculate this exponential integral? I am trying to calculate
$\int_0^{+\infty}\log_2(1+ax)\frac{1}{a}e^{-\frac{x}{a}}dx$
by using the integration by parts method like:
\begin{equation}
\begin{aligned}
\int_0^{+\infty}\log_2(1+ax)\frac{1}{a}e^{-\frac{x}{a}}dx &= -\int_0^{+\infty}\log_2(1+ax)(e^{-\frac{x}{a}})'dx\\
&=\log_2(1+ax)e^{-\frac{x}{a}}|_0^{+\infty} + \int_0^{+\infty}\frac{a}{\ln2(1+ax)}e^{-\frac{x}{a}}dx\\
&= \int_0^{+\infty}\frac{a}{\ln2(1+ax)}e^{-\frac{x}{a}}dx = \frac{a}{\ln2}\int_0^{+\infty}\frac{1}{(1+ax)}e^{-\frac{x}{a}}dx\\
&= \frac{a}{\ln2}\int_0^{+\infty} \frac{-\frac{1}{a^2}}{-\frac{1}{a^2}(1 + ax)}e^{-\frac{x}{a}}dx = \frac{-\frac{1}{a}}{\ln2}\int_0^{+\infty} \frac{1}{-\frac{1}{a^2} - \frac{x}{a}}e^{-\frac{x}{a}}dx\\
&= \frac{-\frac{1}{a}}{\ln2}\int_0^{+\infty}\frac{e^{-\frac{x}{a}}e^{-\frac{1}{a^2}}}{(-\frac{1}{a^2} - \frac{x}{a})e^{-\frac{1}{a^2}}}dx \\
&= \frac{-\frac{1}{a}e^{\frac{1}{a^2}}}{\ln2}\int_0^{+\infty}\frac{e^{-\frac{x}{a}-\frac{1}{a^2}}}{(-\frac{x}{a}-\frac{1}{a^2})}dx\\
\end{aligned}
\end{equation}
How can I further integrate the $\int_0^{+\infty}\frac{e^{-\frac{x}{a}-\frac{1}{a^2}}}{(-\frac{x}{a}-\frac{1}{a^2})}dx$ in order to get a closed form expression?
Thank you in advance!
| You are almost done. Continuing from where you left: $$I =\int_{0}^{\infty} \frac{e^{-x/a-1/a^2}}{-x/a-1/a^2}\, dx$$ Substituting $u = \frac{x} {a} + \frac1{a^2} \implies du = \frac1{a} \, dx$ gives us: $$I = a \int_{\frac1{a^2}}^{\infty} - \frac{e^{-u}}{u} \, du$$ which is the exponential integral $E_1(u)$.
Hope you can take it from here.
| {
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Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$ If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$
Try: $a+b+c=0,ab+bc+ca=1,abc=-1$
Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$
Could some help me to explain short way to calculate product,Thanks
| You already know the value of $(a-b)^2+(b-c)^2+(c-a)^2$. Use the fact that\begin{multline}(a-b)^2(b-c)^2(c-a)^2=-4 a b c (a+b+c)^3+(a b+c b+a c)^2 (a+b+c)^2+\\+18 a b c (a b+c b+a c) (a+b+c)-4 (a b+c b+a c)^3-27 a^2 b^2 c^2\end{multline}and that\begin{multline}(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=(a+b+c)^4+\\-6 (a b+c b+a c) (a+b+c)^2+9 (a b+c b+a c)^2\end{multline}to get the other coefficients. This is easy, since you know $a+b+c$, $ab+bc+ca$ and $abc$.
| {
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Proof verification: If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$ Would someone be willing to verify the following proof?
Theorem: Suppose $a, b \in \mathbb{Z}; n \in \mathbb{N}$. If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$.
Proof:
$a \equiv b (mod$ $n) \rightarrow xn = a - b; x \in \mathbb{Z}$
$\rightarrow a = xn + b$
Then, $a^3 = (xn)^3 + 3b(xn)^2 + 3b^2(xn) + b^3$.
This gives us $a^3 - b^3 = (xn)^3 + 3b(xn)^2 + 3b^2(xn) = n(x^3n^2 + 3bx^2n + 3b^2x)$
$n | n(x^3n^2 + 3bx^2n + 3b^2x) \rightarrow n | (a^3 - b^3)$
Therefore $a^3 \equiv b^3 (mod$ $n)$.
| Your proof is correct.
You could have stopped at $a^3=b^3+kn$, it doesn't matter whether $k$ has a complicated expression, if it's an integer, then $a^3\equiv b^3\pmod n$.
However, notice that you can factorize $xn$ actually, and that $xn=a-b$.
Thus $a^3-b^3\equiv(a-b)(a^2+ab+b^2)\equiv 0\pmod n$ since $(a-b)\equiv 0\pmod n$.
You are not forced to get back to definition everytime, try to use modular calculus directly as well, it is powerful.
| {
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Let $X$ and $Y$ be independent and identically distributed random variables, find $P(X \geq 2Y)$
Let $X$ and $Y$ be independent and identically distributed random variables with probability mass function $p(n)=\frac{1}{2^{n}}$. Find $P(X \geq 2Y)$.
My attempt: \begin{align} P(X \geq 2Y) & = \sum\limits_{y=1}^{\infty}\sum\limits_{x=2y}^{\infty}P(X) \\ & = \sum\limits_{y=1}^{\infty} \left( \frac{1}{2^{2y}}+\frac{1}{2^{2y+1}}+\frac{1}{2^{2y+2}} + \ldots \right) \\ & = \sum\limits_{y=1}^{\infty}\frac{1}{2^{2y}} \left( 2 \right) \\ & = \frac{2}{3}.\end{align}
$\frac{2}{3}$ however does not match the answer at the back. What am I doing wrong? Help!
(The answer is given to be in the interval $[0.27,0.3]$.)
| We have $$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \mathbb{P}(X \geq 2Y, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \mathbb{P}(X=x, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \frac{1}{2^x} \frac{1}{2^y}.$$
Thus,
$$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \frac{1}{2^y} \frac{2}{2^{2y}} = 2 \sum_{y \geq 1} \frac{1}{2^{3y}}= \frac{2}{7}.$$
| {
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Finding $\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$
find the :
$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$$
My Try :
$$\dfrac{1}{1+\dfrac{\sin 70^{\circ}}{\cos 70^{\circ}}}+\dfrac{1}{1+\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}}$$
$$\dfrac{\cos70^{\circ}}{\cos 70^{\circ}+\sin 70^{\circ}}+\dfrac{\cos20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}$$
now what do i do ?
|
See that $$\tan (x) = \frac{\sin x}{\cos x}=\frac{\cos (90^{\circ}-x)}{\sin (90^{\circ}-x)}=\frac{1}{\tan (90^{\circ}-x)}$$
what happen if you take $x=70$
$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}} = \dfrac{1}{1+\frac{1}{\tan 20^{\circ}}}+\dfrac{1}{1+\tan 20^{\circ}}= \dfrac{\tan 20^{\circ}}{1+\tan 20^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}} = 1$$
| {
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Find the centre of a circle given the equation of a tangent to the circle and the x coordinate of the centre The line with equation $2x+y-5=0$ is a tangent to the circle with equation $(x-3)^2 + (y-p)^2=5$
Find the two possible values of $p$.
| $$y=5-2x$$
Since this is a tangent, it touches the circle.
$$\begin{align}(x-3)^2+(y-p)^2&=5\\(x-3)^2+(5-2x-p)^2&=5\\x^2-6x+9+25+4x^2+p^2+2(-10x)+2(2px)+2(-5p)&=5\\5x^2+(4p-26)x+(p^2-10p+29)&=0\end{align}$$
We require that $$\boxed{b^2-4ac=0}$$
$$\begin{align}(4p-26)^2-20(p^2-10p+29)&=0\\4(2p-13)^2-20(p^2-10p+29)&=0\\4p^2-52p+169-5p^2+50p-145&=0\\p^2+2p-24&=0\\(p+6)(p-4)&=0\\p&=-6,4\end{align}$$
| {
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Limit: $\lim_{x\to2} \frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)$ Can someone help me understand how the solution for the following limit is $1/4$?
I've been trying to solve it but I always end up in a 'dead end' with an
indetermination. If someone could help me, that would be awesome.
$$\lim_{x\to2} \frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)$$
| Note that for $x\to 2$
$$\frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)=\frac{\sin\left(\frac{x-2}{x+2}\right)}{x-2}=\frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-2}{x+2}}\cdot\frac{1}{x+2}\to 1\cdot\frac14=\frac14$$
indeed
$$y=\frac{x-2}{x+2}\to 0 \quad \implies \frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-2}{x+2}}=\frac{\sin y}{y}\to1$$
| {
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Proving a limit through the delta-epsillon definition of a limit, discarding the delta upper bound Prove that the limit exists:$$\lim\limits_{x \to -3} \frac{x+5}{x+2} = -2$$
My answer is:
Let $\epsilon > 0$, there exists a $\delta=\boxed{min\{\frac{\epsilon}{6},\frac{1}{2}\}}$. Let $x$ such that $0<|x + 3| < \delta$,
$$|f(x) +2| = |\frac{x+5}{x+2} +2| = |\frac{3x+9}{x+2}| = 3 \cdot |\frac{x+3}{x+2}| < 3 \cdot \delta \cdot |\frac{1}{x+2}|$$
Now I need somehow to get rid of that $|\frac{1}{x+2}|$ so I could be able to compare the rightmost value to $\epsilon$ in order for me to show that the limit exists.
So I do the following: $$-\delta < x + 3 < \delta \implies -1-\delta<x+2<\delta-1 \implies \frac{1}{\delta-1} < \frac{1}{x+2} < \frac{1}{-1-\delta}$$
I choose to bound delta as the following $\delta \leq \frac{1}{2}$ and I get that $$-2 < \frac{1}{x+2} < -\frac{2}{3} \implies |\frac{1}{x+2}| < 2$$
So now we'll get $$ 3 \cdot \delta \cdot |\frac{1}{x+2}| < 6\cdot\delta = \epsilon \implies \delta=\frac{\epsilon}{6}$$
This concludes my proof. Now, it has been a few months since I last proved a basic limit like that, and I get every step of the way. However, something bugs me about the way I chose my $\delta$. What would happen if I didn't bound $\delta \leq \frac{1}{2}$? Can I just leave the delta as it is, write $|\frac{1}{x+2}|$ as a function of delta and from there compute epsilon? I can see why someone would bound delta, my guess is that it shortens the calculation of epsilon. Less $min$ and $max$, this way the proof stays cleaner. I never saw something like what I suggested and I was wondering if it is possible. Can't find an excuse why something like that wouldn't be applicable.
Suggestions are greatly appreciated.
| We can find the optimal value for $\delta$ in this way, let assume
$$ \left|\frac{x+5}{x+2} +2\right|<\epsilon \implies -2-\epsilon<\frac{x+5}{x+2}<-2+\epsilon \implies -2-\epsilon<1+\frac{3}{x+2}<-2+\epsilon \implies -3-\epsilon<\frac{3}{x+2}<-3+\epsilon \implies \frac{-3-\epsilon}{3}<\frac{1}{x+2}<\frac{-3+\epsilon}{3}\\\implies -\frac{3}{3-\epsilon}<x+2<-\frac{3}{3+\epsilon} \\\implies -\frac{3}{3-\epsilon}+1<x+3<-\frac{3}{3+\epsilon}+1 \\\implies \frac{-\epsilon}{3-\epsilon}<x+3<\frac{\epsilon}{3+\epsilon}\implies|x+3|<min\{\frac{\epsilon}{3+\epsilon},\frac{\epsilon}{3-\epsilon}\}=\frac{\epsilon}{3+\epsilon}=\delta$$
| {
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"Different form" of Ceva's theorem?
Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.
Now, the Art of Problem Solving site site reads one of the solutions as
Using a different form of Ceva's Theorem, we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$
I don't understand what this "different form" of Ceva's theorem is. How does one derive it?
| One way to prove it is to use Mass Points. We assign weights $a$ to vertex $A$, $b$ to $B$, $c$ to $C$. Then the weights of $F$ is $a+b$, of $D$ is $b+c$ of $E$ is $a+c$ and of $P$ is $a+b+c$.
Then we have that
$$\frac{AP}{PD} = \frac{b+c}{a} \implies \frac{AD}{PD} = \frac{a+b+c}{a} \implies \frac{PD}{AD} = \frac{a}{a+b+c}$$
Similarly:
$$\frac{PE}{BE} = \frac{b}{a+b+c} \quad \quad \frac{PF}{CF} = \frac{c}{a+b+c}$$
Adding this identities we have the wanted form.
Note that the method of mass points is another way of using Ceva's and Menelaus' Theorem. In fact all these identities can be proved using them, although it's a much more tedious process.
Here's a way to prove it purely by Ceva's and Menelaus Theorem.
Let $\frac{AF}{FB} = \frac ba$, $\frac{BD}{DC} = \frac cb$. Then by Ceva's Theorem we have that $\frac{CE}{EA} = \frac ac$. Now by Menelaus' Theorem on $B-P-E$ and $\triangle ADC$ we have that:
$$\frac{AP}{PD} = \frac{CB}{DB} \times \frac{EA}{CE} = \frac{b+c}{c} \times \frac ca = \frac{b+c}{a} \implies \frac{PD}{AD} = \frac{a}{a+b+c}$$
Similarly we get the other identities mentioned above. Hence the proof.
| {
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Sum of $\frac{C_r}{(r+1)(r+2)}$ from $r=0$ to $r=n$ I am trying to compute the following sum: $$\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}$$
Integrating $(1+x)^n = \sum_{r=0}^{n}C_rx^r$ with respect to $x$,
$$\int(1+x)^n dx=\int\sum_{r=0}^{n}C_rx^rdx$$
$$\implies \frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}$$
Again integrating with respect to $x$ from $0$ to $1$ yields,
$$\int_0^1\frac{(1+x)^{n+1}}{n+1}dx=\int_0^1\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}dx$$
$$\implies\bigg[\frac{(1+x)^{n+2}}{(n+1)(n+2)}\bigg]_0^1=\bigg[\sum_{r=0}^{n}C_r\frac{x^{r+2}}{(r+1)(r+2)}\bigg]_0^1$$
$$\implies\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}=\frac{2^{n+2}-1}{(n+1)(n+2)}$$
But it is written in my textbook that the answer is $$\frac{2^{n+2}-(n+3)}{(n+1)(n+2)}$$
So, where am I going wrong? Is it correct to use integral twice in such situations?
| We have to respect the integration constant when evaluating an indefinite integral.
When calculating $\int(1+x)^n\,dx = \int\sum_{r=0}^nC_rx^r\,dx$ we obtain
\begin{align*}
\int(1+x)^n\,dx&=\frac{(1+x)^{n+1}}{n+1}+K_1\\
\int\sum_{r=0}^nC_rx^r\,dx&=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}+K_2
\end{align*}
with $K_1,K_2$ constants.
It follows
\begin{align*}
\frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\color{blue}{+K}\tag{1}
\end{align*}
with $K=K_2-K_1$.
By setting $x=0$ in (1) we obtain
\begin{align*}
\color{blue}{\frac{1}{n+1}=K}
\end{align*}
We conclude
\begin{align*}
\frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\color{blue}{+\frac{1}{n+1}}\tag{2}
\end{align*}
Integrating (2) from $0$ to $1$ gives now the correct result
\begin{align*}
\color{blue}{\int_{0}^1\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}\,dx}&=\int_{0}^1\left(\frac{(1+x)^{n+1}}{n+1}-\frac{1}{n+1}\right)\,dx\\
&=\left[\frac{(1+x)^{n+2}}{(n+1)(n+2)}-\frac{x}{n+1}\right]_0^1\\
&=\left[\frac{2^{n+2}}{(n+1)(n+2)}-\frac{1}{n+1}\right]-\left[\frac{1}{(n+1)(n+2)}\right]\\
&\color{blue}{=\frac{2^{n+2}-(n+3)}{(n+1)(n+2)}}
\end{align*}
as expected.
| {
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Prove there exists no rational solutions $a,b,c$, and $d$ such that $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$ I want to prove there exists no rational solutions $a,b,c$, and $d$ such that $$a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$$I have proven there exists no rational solutions $a,b$, and $c$ such that $$a+b\sqrt2=\sqrt3$$ $$a + b\sqrt2 + c\sqrt3 = \sqrt6$$ $$a + b\sqrt2 + c\sqrt3 = \sqrt5$$ A hint would be much appreciated.
| $(b\sqrt{2}+c\sqrt{3})^2 = (\sqrt{5} - a - d\sqrt{6})^2\implies 2b^2+2bc\sqrt{6}+3c^2= 5+a^2+6d^2 - 2a\sqrt{5}-2d\sqrt{30}+2ad\sqrt{6}\implies m+r\sqrt{6}= q\sqrt{5}+t\sqrt{30}\implies (r\sqrt{6} - q\sqrt{5})^2 = (t\sqrt{30}-m)^2\implies 6r^2-2qr\sqrt{30}+5q^2= 30t^2-2tm\sqrt{30}+m^2\implies \sqrt{30} \in \mathbb{Q}$, a contradiction.
| {
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How to calculate $\int \frac{dx}{(a^2 + x^2)^2}$? I'm trying to use a trig substitution but I'm stuck. Here's what I did so far:
$$\int \frac{dx}{(a^2 + x^2)^2}$$
Let $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+sin^2\theta))^2} $$
$$\int \frac{a\cos \theta d\theta}{(a^2 \cos^2\theta)^2} =\int \frac{d\theta}{a^3cos^3\theta} $$
I don't know what to do anymore
| This could be done in a "general" way.
Denoting
$$
I_k=\int \frac{dx}{(x^2+a^2)^k},
$$
one has (for $k>1$)
$$
\begin{align}
I_k&=\frac{1}{a^2}\int\frac{(x^2+a^2)-x^2}{(x^2+a^2)^k}\ dx\\
&=\frac{1}{a^2} I_{k-1}-\frac{1}{a^2}\int\frac{x^2}{(x^2+a^2)^k}\ dx\\
&=\frac{1}{a^2} I_{k-1}+\frac{1}{2a^2(k-1)}\int x\ d\left(\frac{1}{(x^2+a^2)^{k-1}}\right)\\
&=\frac{1}{a^2} I_{k-1}+\frac{1}{2a^2(k-1)}\left[\frac{x}{(x^2+a^2)^{k-1}}-I_{k-1}\right],
\end{align}
$$
which gives
$$
I_k=\frac{x}{2a^2(k-1)(x^2+a^2)^{k-1}}+\frac{2k-3}{2a^2(k-1)}I_{k-1}.
$$
Now you can get $I_2$ by observing that
$$
I_1=\frac{1}{a}\arctan\frac{x}{a}+C.
$$
Notes: The recursive relation above allows one to calculate $I_k$ for any positive integer $k$ in principle, which can be used to calculate any indefinite integral of rational functions.
| {
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Question related to beta and gamma function I'm trying to derive the following integral.
$$\int_0^\infty \frac{x^8(1-x^6)}{(1+x)^{24}} \, dx.$$
What transformations can I use?
| In general
$$\int^{\infty}_{0}\frac{x^{u-1} }{(1+bx)^{m+1}}dx=\frac{1}{b^{u}}\beta (u,m+1-u).$$
Proof:
We make the transformation $y=\frac{bx}{1+bx}$ so
$$y-1 =\frac{bx}{1+bx} -1 = \frac{bx-1-bx}{1+bx} = \frac{-1}{1+bx}\Rightarrow $$ $$bx+1=\frac{1}{1-y} \Rightarrow (1-y)^{m+1}=\frac{1}{(1+bx)^{m+1}}.$$
We have,
$$bx+1=\frac{1}{1-y} \Rightarrow bx= \frac{1}{1-y} -1 = \frac{1-1+y}{1-y}= y (1-y)^{-1}$$
so $x=\frac{y(1-y)^{-1}}{b}$ then it follows $dy=b^{-1}(1-y)^{-2}dx$, and the desired integral is
$$\frac{1}{b^{u}}\int^{1}_{0}y^{u-1}(1-y)^{m-u}dy=\frac{1}{b^{u}}\beta (u,m+1-u). $$
Example:
In your case we take $b=1$, $m+1=24$, and take $x^8 (1-x^6)=x^8-x^{14}$, and the difference of the results
| {
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Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.
| Hint: $4 = 2^2$ and $8 = 2^3$.
Let $z = 2^x$. Then, using the hint, we can rewrite the original equation as \begin{align} 2^x + 4^x = 8^x &\implies 2^x + (2^2)^x = (2^3)^x \\ &\implies 2^x + (2^{x})^2 = (2^{x})^3 \\ &\implies z + z^2 = z^3. \end{align} Bringing all of the terms to one side of the equation, we obtain $$ 0 = z^3 - z^2 - z = z(z^2 - z -1). $$ One possible solution is $z = 0$, but this doesn't work, as $z = 2^x = 0$ has no real solutions. Dealing with the factor $z^2 - z - 1$, e.g. via the quadratic formula, we get $$ z = \frac{1 \pm \sqrt{1 - (4)(1)(-1)}}{2(1)} = \frac{1}{2} \pm \frac{\sqrt{5}}{2}. $$ Again, since $z = 2^x$, so $x = \log_2(z)$, and so $$ x = \log_2\left( \frac{1}{2} \pm \frac{\sqrt{5}}{2} \right). $$ However, $\log$ is undefined for negative numbers, so this reduces to only one real solution, namely $$ x = \log_2\left( \frac{1}{2} + \frac{\sqrt{5}}{2} \right). $$
Addendum: This is probably overkill for what is likely a precalculus problem, but I $\heartsuit$ the complex logarithm, so: just for funsies, note that we can deal with the logarithm of a negative number if we are willing to deal with complex numbers and some subtle issues in complex analysis. Let's suppose that $2^x = z$. Then, if we assume that $x = a+ib \in \mathbb{C}$ and that $z\in\mathbb{R}$, we have
$$2^x = \mathrm{e}^{\log(2) x} = \mathrm{e^{\log(2)a + i\log(2)b}} = \mathrm{e}^{\log|z| + i2k\pi}, $$
where $\log$ the real natural logarithm (or the principle branch of the complex logarithm, if you prefer), and $k\in\mathbb{Z}$. Equating real and imaginary parts, we get
$$ \begin{cases}
a = \frac{\log|z|}{\log(2)} \\
b = \frac{2\pi}{\log(2)}.
\end{cases}
$$
Therefore any number of the form
$$ \frac{\log|z|}{\log(2)} + i \frac{2k\pi}{\log(2)} $$
is a solution to the equation $2^x = z$, where $z$ is any nonzero real number (positive or negative).
In particular, we determined above that there are solutions corresponding to $$ z = \frac{1}{2} \pm \frac{\sqrt{5}}{2}. $$ Let's label these two solutions $$ \varphi = \frac{1}{2} + \frac{\sqrt{5}}{2} \qquad \text{and} \qquad \psi = \frac{1}{2} - \frac{\sqrt{5}}{2} = -\varphi^{-1}. $$ As $|\varphi| = \varphi$ and $|\psi| = \varphi^{-1}$, we may use the arguments above to conclude that the solutions to the original equation must then be elements of the set $$ \left\{ \frac{\log(\varphi)}{\log(2)} + i \frac{2k\pi}{\log(2)}, \frac{\log(\varphi^{-1})}{\log(2)} + i \frac{2k\pi}{\log(2)} \ \middle|\ k \in \mathbb{Z} \right\} = \left\{ \pm\frac{\log(\varphi)}{\log(2)} + i \frac{2k\pi}{\log(2)} \ \middle|\ k \in \mathbb{Z} \right\}. $$ There is a rather lovely kind of symmetry here, n'est-ce pas?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equiv b \text{ mod } 3 \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$
Besides, the prime factorization of $999=3^3\cdot37$, but I don't know how to go on.
I would appreciate any hints.
| $$999=a^3-b^3=(a-b)^3+3ab(a-b)\geq(a-b)^3,$$ which says $a-b$ is divided by $3$ and $a-b\leq9$ and we get not so many cases:
$a-b=3$ or $a-b=9$.
| {
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Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$ Find all solutions of the equation:
$z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$
I don't have any other idea than to guess them.
| HINT:
You can write the same equation as:
$$(z^2-2z+5)(z^2+2i\sqrt2 z-2)=0$$
Then you can find the roots of each brackets separately...
SOLUTION:
$$z=1-2i$$
$$z = 1+2i$$
$$z = -i\sqrt2$$
| {
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How to calculate $\int _\frac{3}{4}^{\frac{3}{2}} \sqrt{9 - 4x^2}$ $$\int _\frac{3}{4}^{\frac{3}{2}} \sqrt{9 - 4x^2}$$
First I set $x = \frac{3}{2}\sin\theta, dx = \frac{3}{2}\cos \theta d\theta$:
$$\int \sqrt{9 - 4x^2}dx = \int \sqrt{4\left(\frac{9}{4} - x^2\right)}dx = 2\int \sqrt{\left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\sin^2\theta}*\frac{3}{2}\cos \theta$$
$$3\int \frac{3}{2}\cos^2\theta d\theta = \frac{9}{2}\int \frac{\cos 2\theta + 1}{2}$$
$$\frac{9}{2}\left(\int \frac{1}{2} + \int \cos 2\theta \right) = \frac{9}{2}\left(\frac{1}{2}\theta + \frac{1}{4}\sin2\theta\right) = \frac{9}{2}\left(\frac{1}{2}\arcsin\left(\frac{2}{3}x\right) + \frac{1}{4}\sin\left(2\arcsin\left(\frac{2}{3}x\right)\right)\right)$$
The problem is evaluating the definite integral, it's getting way too messy. How do I go about this?
| When you substituted $x = \dfrac 32 \sin \theta$, you could have changed the limits of integral then, according to your new variable, i.e. $\theta$.
The new limits will be $$ \text{Upper limit :} \quad x = \frac{3}{4} =\frac 32 \sin \theta \implies \theta =\dfrac{\pi}{6}$$ and $$\text{Lower limit :} \quad x = \frac{3}{2} =\frac 32 \sin \theta \implies \theta =\dfrac{\pi}{2}$$
Now use these limits in the following expression you got, to evaluate the integral $$\rm I=\frac{9}{2}\left(\frac{1}{2}\theta + \frac{1}{4}\sin2\theta\right) \Bigg |_{\pi/6}^{\pi/2}$$
| {
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What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?
| For fun:
$\dfrac{\sqrt{x+2} -2}{(\sqrt{x+2})^2-4}=$
$\dfrac{\sqrt{x+2}-2}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}=$
$\dfrac{1}{\sqrt{x+2}+2}.$
The limit $x \rightarrow 2$ is?
| {
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Number of solution is twice $(x,y)$ Problem: Count the number of $2 \times 2$ matrices $A$ with $A^TA=-I$ in $Z_p$ for $p>2$.
Answer: if $p$ is an odd prime, the number of such matrices $A$ is twice the number of solutions $(x,y)$ to the congruence $x^2+y^2 \equiv -1 \pmod p$.
What's the reason behind "twice"?
| $$A^TA=-I$$
is equivalent to $A^{-1}=-A^T$ and hence implies $AA^T=-I$.
If $A=\begin{bmatrix}
a&b \\
c&d
\end{bmatrix}$ Then
$$A^TA=-I \Leftrightarrow AA^T=-I \mbox{ and }A^TA=-I \\
\begin{bmatrix}
a&b \\
c&d
\end{bmatrix}\begin{bmatrix}
a&c\\
b&d
\end{bmatrix}= \begin{bmatrix}
-1&0 \\
0&-1
\end{bmatrix} \mbox{ and } \begin{bmatrix}
a&c \\
b&d
\end{bmatrix}\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}= \begin{bmatrix}
-1&0 \\
0&-1
\end{bmatrix}\Leftrightarrow \\
a^2+b^2=-1 \\
c^2+d^2=-1 \\
ac+bd=0 \\
a^2+c^2=-1 \\
b^2+d^2=-1 \\
ab+cd=0
$$
Now
$$a^2+b^2=-1=a^2+c^2 \Rightarrow b^2=c^2 \Rightarrow b= \pm c\\
a^2+b^2=-1=b^2+d^2 \Rightarrow a^2=d^2 \Rightarrow a= \pm d \\
$$
Using $b =\pm c$ and $a=\pm d$ in
$$ac+bd=0 \\
ab+cd=0$$
you get that either $a=b=c=d=0$ or the signs in $b =\pm c$ and $a=\pm d$ are opposite.
Therefore, if you combine everything we got, the system above reduces to
$$a^2+b^2=-1$$
and, either $c=-b, d=a$, or $c=b, d=-a$.
| {
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Identity involving binomial coefficients I might have written this in a needlessly cumbersome way, but I want to prove that for odd positive integers $n$, $$\sum_{k\ odd}^{n}\binom{2n+1}{2k}=\begin{cases}
\binom{2^n+1}{2}, & \text{if}\ n\ \text{mod}\ 4 =1\\
\binom{2^n}{2}, & \text{if}\ n\ \text{mod}\ 4 =3
\end{cases}$$
I have tested these identities and they should hold in general. Thank you very much in advance!
Edit:
I found the following formula in the wikipedia article for binomial coefficients, under multisections of sums (https://en.wikipedia.org/wiki/Binomial_coefficient#Multisections_of_sums) which works nicely.$$\binom{n}{2}+\binom{n}{6}+\binom{n}{10}+\cdots=\frac{1}{2}(2^{n-1}-2^{\frac{n}{2}}\cos\frac{n\pi}{4})$$
It would be great if someone can provide a reference or proof for this. Or if there is a much faster way of getting at the same result then please ignore this entirely.
| Preliminaries
Note that if $n\mid m$, then
$$
\begin{align}
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}
&=\frac1n\sum_{k=0}^{n-1}1\\[6pt]
&=1
\end{align}
$$
and that if $n\nmid m$, then
$$
\begin{align}
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}
&=\frac1n\frac{e^{2\pi im}-1}{e^{2\pi im/n}-1}\\[6pt]
&=0
\end{align}
$$
Thus,
$$
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}=[\,n\mid m\,]
$$
where $[\dots]$ are Iverson Brackets.
Therefore, setting $n=4$, we get
$$
\begin{align}
[\,k\equiv0\pmod4\,]
&=1^k+i^k+(-1)^k+(-i)^k\\
[\,k\equiv2\pmod4\,]
&=1^{k-2}+i^{k-2}+(-1)^{k-2}+(-i)^{k-2}\\
&=1^k-i^k+(-1)^k-(-i)^k
\end{align}
$$
First Question
$$
\begin{align}
\sum_{k=0}^n\frac{\overbrace{1^k-i^k+(-1)^k-(-i)^k}^{[\,k\equiv2\pmod4\,]}}4\binom{2n+1}{k}
&=\frac{2^{2n+1}-(1+i)^{2n+1}+0^{2n+1}-(1-i)^{2n+1}}4\\
&=2^{2n-1}-2^{n-\frac12}\cos\left(\frac{(2n+1)\pi}4\right)\\[6pt]
&=2^{2n-1}-2^{n-1}\left(\cos\left(\frac{n\pi}2\right)-\sin\left(\frac{n\pi}2\right)\right)\\[6pt]
&=2^{2n-1}+2^{n-1}(-1)^{\left\lfloor\frac{n-1}2\right\rfloor}\\
&=\frac{2^n\left(2^n+(-1)^{\left\lfloor\frac{n-1}2\right\rfloor}\right)}2\\
&=\bbox[5px,border:2px solid #C0A000]{\left\{\begin{array}{}
\binom{2^n+1}{2}&\text{if }n\equiv1,2\pmod4\\
\binom{2^n}{2}&\text{if }n\equiv0,3\pmod4
\end{array}\right.}
\end{align}
$$
Second Question
$$
\begin{align}
\sum_{k=0}^n\frac{\overbrace{1^k-i^k+(-1)^k-(-i)^k}^{[\,k\equiv2\pmod4\,]}}4\binom{n}{k}
&=\frac{2^n-(1+i)^n+0^n-(1-i)^n}4\\
&=\bbox[5px,border:2px solid #C0A000]{2^{n-2}-2^{\frac n2-1}\cos\left(\frac{n\pi}4\right)+\frac{[n=0]}4}
\end{align}
$$
| {
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check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges How to check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges?.
$\begin{align}
\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k &=
\lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k \\
&= \lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7\frac{k}{k}-2\frac{1}{k}}{8\frac{k}{k}-3 \frac{\sqrt{k}}{k}}\right)}^k \\
&= \lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7-2\frac{1}{k}}{8-3 \frac{\sqrt{k}}{k}}\right)}^k \\
\end{align}$
Am I on the right track ? If so, how should I go on ? I need some help to do it without the root test.
| It should be
\begin{align*}
\lim_{k\rightarrow\infty}\left(\left(\dfrac{7k-2}{8k-3\sqrt{k}}\right)^{k}\right)^{1/k}=\dfrac{7}{8}<1,
\end{align*}
by root test, it is convergent.
| {
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Addition of $2$ Events
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(X + Y = k)$ for $2 \le k \le 2n$.
\begin{align}P(X + Y = k) &= \sum_{(x,y)\,:\,x+y=k} P(x, y) \\
&= \sum_{(x,y)\,:\,x+y=k} \frac{1}{n^2} \\
&= (k - 1)\frac{1}{n^2} \\
&= \frac{k-1}{n^2} \end{align}
When $k = 2: (1, 1)$
When $k = 3: (1, 2), (2, 1)$
When $k = 4: (1, 3), (3, 1), (2, 2)$
When $k = 5: (1, 4), (4, 1), (2, 3), (3, 2)$
$$\#(x, y) = k - 1$$
Textbook Answer:
$\frac{k-1}{n^2}\,\,\,$ for $\,\,\,2 \le k \le n+1$
$\frac{2n-k+1}{n^2}\,\,\,$ for $\,\,\,n+2 \le k \le 2n$
Why are there $2$ intervals being considered?
|
Why is there 2 intervals being considered?
Hint: Look what happens if you go "backwards" from $k=2n$ "down":
*
*$k=2n$: $(n,n)$
*$k=2n-1$: $(n,n-1)$, $(n-1, n)$
*$k=2n-2$: $(n,n-2)$, $(n-1, n-1)$, $(n-2, n)$
etc. What you've got will keep growing with growing $k$, but at some point the number of ways to get the sum $k$ starts shrinking. Thereby the second formula.
| {
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Nilpotent $4 × 4$ matrix
Find a nilpotent $4 × 4$ matrix with only $-1$ and $3$ as entries.
$\begin{bmatrix}
-1 & -1 & 3 & 3 \\
3 & -1 & 3 & -1 \\
3 & -1 & 3& -1 \\
-1 & -1 & 3 & 3
\end{bmatrix}$ is the matrix I came up with, but I don't think it works. Anyone have a better one?
| $$
\left(
\begin{array}{cccc}
3 & 3 & 3 & 3 \\
-1 & -1 & -1 & -1 \\
-1 & -1 & -1 & -1 \\
-1 & -1 & -1 & -1 \\
\end{array}
\right)
$$
| {
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How can we show that $y_n=z_n$?
Let $$y_{n+1}=2^n\sqrt{2(1-\sqrt{1-(\frac{y_n}{2^n})^2}})$$ and $$z_{n+1}=\frac{2z_n}{\sqrt{2(1+ \sqrt{1- (\frac{z_n}{2^n})^2}})}$$
$y_1=z_1=2$
How can one shows that $z_n=y_n$?
I have assumed that $y_n=z_n$ then if that implies $y_{n+1}=z_{n+1}$ then by induction they are equal. But I feel this is not a way to show that $y_n=z_n$
| see that, $$1-\sqrt{1-(\frac{y_n}{2^n})^2} = \frac{\left(1-\sqrt{1-(\frac{y_n}{2^n})^2}\right)\left(1+\sqrt{1-(\frac{y_n}{2^n})^2}\right)}{1+\sqrt{1-(\frac{y_n}{2^n})^2}} = \frac{(\frac{y_n}{2^n})^2}{1+\sqrt{1-(\frac{y_n}{2^n})^2}} $$
Hence $$y_{n+1}=2^n\sqrt{2\left(1-\sqrt{1-(\frac{y_n}{2^n})^2}\right)}=2^n\sqrt{\frac{\color{blue}4(\frac{y_n}{2^n})^2}{\color{blue}2(1+\sqrt{1-(\frac{y_n}{2^n})^2})}} =\frac{2y_n}{\sqrt{2(1+ \sqrt{1- (\frac{y_n}{2^n})^2}})}$$
The result easily follows since $y_1=z_1$
| {
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Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2}
+\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi}
.$$
which can be expressed as
$$D= \left[
\frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2}
\right]_0^{2\pi}
.$$
This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.
| It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have
$$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$
hence by differentiating both sides of $(1)$ with respect to $a$ we have
$$ \int_{0}^{2\pi}\frac{\cos(\theta)\,d\theta}{(1-a\cos\theta)^2}=\frac{2a\pi}{(1-a^2)^{3/2}},\qquad \int_{0}^{2\pi}\frac{d\theta}{(1-a\cos\theta)^2}= \frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$
and by differentiating again
$$ \int_{0}^{2\pi}\frac{\sin^2(\theta)\,d\theta}{(1-a\cos\theta)^3}=\frac{\pi}{(1-a^2)^{3/2}} \tag{3}$$
is simple to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
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A simple problem with a simple and a nonsimple solution I have this seemingly simple problem to solve. Its statement is straightforward:
"If $x+y+z=3,x^2+y^2+z^2=5, x^3+y^3+z^3=7$ show that $x^4+y^4+z^4=9,x^5+y^5+z^5\neq 11$".
There is a highschool approach, that is I expanded $(x+y+z)^2, (x+y+z)^3$ and managed to get $$xyz=-2/3, xy+yz+zx=2$$ etc, which I guess I could employ to expand $(x+y+z)^4, (x+y+z)^5$ and get some results (I guess).
The thing is I don't think this is the proper approach. Is there a better way than this tedious, long calculation (if it is a solution that is)
Say we could divide the polynomials
$$f=x^4+y^4+z^4,g=x^3+y^3+z^3, h=x^2+y^2+z^2,k=x+y+z$$
Wouldn't this give us what we want by using the remainder polynomial?
Thanks in advance for your time.
| If you are in the market for more sophisticated ways... you have found $x, y, z$ to be the roots of $P(t)=t^3-3t^2+2t+\frac23=0$. Now if $\alpha$ is a root of $P(t)$, we seek the minimal polynomial of $\alpha^m$ (for $m=4, 5$).
So if $P_4(r)$ is the polynomial with roots $x^4, y^4, z^4$, then it is given by the resultant (why?)
$$P_4(r) = Res(r-t^4, P(t))=r^3-9r^2+\tfrac{536}9r-\tfrac{16}{81}$$
so $x^4+y^4+z^4=9$.
$\parallel$ly
$$P_5(r) = Res(r-t^5, P(t))=-r^3+\tfrac{29}3r^2-\tfrac{496}3r-\tfrac{32}{243}$$
so $x^5+y^5+z^5=\frac{29}3 \neq 11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642114",
"timestamp": "2023-03-29T00:00:00",
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} |
Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.
| Suppose $x,y,z > 0$ satisfy the system
$$
\begin{cases}
x^2 + xy + y^2 = 3&\;\;\;(\text{eq}1)\\[4pt]
y^2 + yz + z^2 = 1&\;\;\;(\text{eq}2)\\[4pt]
z^2 + zx + x^2 = 4&\;\;\;(\text{eq}3)\\
\end{cases}
$$
From $(\text{eq}2)$, since $y,z > 0$, we get $y,z < 1$, hence from $(\text{eq}1)$, we must have $x > 1$.
Let $a = x + y + z$, and let $b = xy + yz + zx$.
The goal is to find the value of $b$.
Since $x,y,z > 0$, and $x > 1$, we get $a > 1$.
Identically, we have $x^2 + y^2 + z^2 = a^2 -2b$.
From the sum $(\text{eq}1)+ (\text{eq}2) +(\text{eq}3)$, we get
$$2(x^2 + y^2 + z^2) + (xy + yz + zx) = 8$$
hence
$$2a^2-3b=8\qquad(\text{eq}4)$$
Subtracting $(\text{eq}2)$ from $(\text{eq}1)$, we get
\begin{align*}
&(x^2 +xy + y^2) - (y^2 + yz + z^2) = 2\\[4pt]
\implies\;&(x^2 - z^2) + (xy - yz) = 2\\[4pt]
\implies\;&(x-z)(x+y+z) = 2\\[4pt]
\implies\;&x-z =\frac{2}{a}\\[4pt]
\implies\;&z^2-2zx+x^2 = \frac{4}{a^2}\qquad(\text{eq}5)\\[4pt]
\end{align*}
Subtracting $(\text{eq}2)$ from $(\text{eq}3)$, we get
\begin{align*}
&(z^2 +zx + x^2) - (y^2 + yz + z^2) = 3\\[4pt]
\implies\;&(x^2 - y^2) + (zx - yz) = 3\\[4pt]
\implies\;&(x-y)(x+y+z) = 3\\[4pt]
\implies\;&x-y =\frac{3}{a}\\[4pt]
\implies\;&x^2-2xy+y^2 = \frac{9}{a^2}\qquad(\text{eq}6)\\[4pt]
\end{align*}
Subtracting $(\text{eq}1)$ from $(\text{eq}3)$, we get
\begin{align*}
&(z^2 +zx + x^2) - (x^2 + xy + y^2) = 1\\[4pt]
\implies\;&(z^2 - y^2) + (zx - xy) = 1\\[4pt]
\implies\;&(z-y)(x+y+z) = 1\\[4pt]
\implies\;&z-y =\frac{1}{a}\\[4pt]
\implies\;&y^2-2yz+z^2 = \frac{1}{a^2}\qquad(\text{eq}7)\\[4pt]
\end{align*}
From the sum $(\text{eq}5)+ (\text{eq}6) +(\text{eq}7)$, we get
$$2(x^2 + y^2 + z^2) - 2(xy + yz + zx) = \frac{14}{a^2}$$
hence
$$a^2-3b=\frac{7}{a^2}\qquad(\text{eq}8)$$
Subtracting $(\text{eq}8)$ from $(\text{eq}4)$, we get
\begin{align*}
&(2a^2-3b) - (a^2-3b) = 8-\frac{7}{a^2}\\[4pt]
\implies\;&a^2 = 8-\frac{7}{a^2}\\[4pt]
\implies\;&a^4 - 8a^2 + 7=0\\[4pt]
\implies\;&(a^2-1)(a^2-7)=0\\[4pt]
\implies\;&a^2 = 7\qquad\text{[since $a > 1$]}\\[4pt]
\end{align*}
Then from $(\text{eq}4)$, we have
$$b = \frac{2a^2-8}{3}$$
hence
$$b = \frac{2(7)-8}{3} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 4
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
| By direct factorization:
$$
\begin{align}
P(x) &= 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} \\[5px]
&= 1 + x + x^2 + x^3 + x^4 + x^5 \\
&\quad\quad + x + x^2 + x^3 + x^4 + x^5 + x^6 \\
&\quad\quad\quad\quad + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\\
&\quad\quad\quad\quad\quad\quad \ldots \\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\[5px]
&= \color{blue}{1} .(1+ x + x^2 + x^3 + x^4 + x^5) \\
&\quad\quad + \color{blue}{x}\cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\
&\quad\quad\quad\quad + \color{blue}{x^2} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\
&\quad\quad\quad\quad\quad\quad \ldots \\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad + \color{blue}{x^5} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\[5px]
&= (\color{blue}{1 + x + x^2 + x^3 + x^4 + x^5}) \cdot (1 + x + x^2 + x^3 + x^4 + x^5)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$
I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$
Then I am getting stuck at ranges of $\theta$
| If you transform $(X,Y)\mapsto(R,\Theta)$ where $X=R\cos\Theta,Y=R\sin\Theta$,
you should end up with the joint density of $(R,\Theta)$ as $f_{R,\Theta}(r,\theta)=\dfrac{r}{2\pi}e^{-r^2/2}\mathbf1_{\{r>0,\,0<\theta<2\pi\}}$.
This implies $R$ and $\Theta$ are independent, where $R$ has the Rayleigh distribution and $\Theta\sim\mathcal{U}(0,2\pi)$.
Now changing variables $(R,\Theta)\mapsto(U,V)$ such that $U=R\sin(2\Theta),V=R\cos(2\Theta)$,
you should be able to show that $U$ and $V$ are independent $\mathcal{N}(0,1)$ variables.
Note that $U=\dfrac{2XY}{\sqrt{X^2+Y^2}}$ and $V=\dfrac{X^2-Y^2}{\sqrt{X^2+Y^2}}$ are independent, which in turn means that
$\dfrac{U}{2}=\dfrac{XY}{\sqrt{X^2+Y^2}}$ and $\dfrac{V}{2}=\dfrac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $\mathcal{N}(0,1/4)$ variables.
This is independent of the post above:
Joint density of $(X,Y)$ is $\displaystyle f_{X,Y}(x,y)=\frac{1}{2\pi}e^{-\frac{1}{2}(x^2+y^2)}\,\quad,(x,y)\in\mathbb{R^2}$
We transform $(X,Y)\mapsto(R,\Theta)\mapsto(U,V)$ where
$x=r\cos\theta\,,y=r\sin\theta$ and $u=\frac{r}{2}\sin(2\theta)\,,v=\frac{r}{2}\cos(2\theta)$
$(x,y)\in\mathbb{R^2}\implies r>0\,, 0<\theta<2\pi\implies (u,v)\in\mathbb{R^2}$.
Note that this transformation is not one to one.
Jacobian of the transformation is $J\left(\frac{x,y}{u,v}\right) = J\left(\frac{x,y}{r,\theta}\right)J\left(\frac{r,\theta}{u,v}\right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=r\times\frac{2}{r}=2$
Now $\left(U=\frac{XY}{\sqrt{X^2+Y^2}},V=\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}\right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Y\stackrel{\text{i.i.d.}}{\sim}\mathcal{N}(0,1)\iff -X,-Y\stackrel{\text{i.i.d.}}{\sim}\mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=\frac{1}{2\pi}e^{-\frac{1}{2} 4(u^2+v^2)}|J_1||J_2|\times 2$$
$$=\frac{1}{\sqrt{\frac{1}{4}}\sqrt{2\pi}}\exp\left(-\frac{u^2}{2\cdot\frac{1}{4}}\right)\cdot\frac{1}{\sqrt{\frac{1}{4}}\sqrt{2\pi}}\exp\left(-\frac{v^2}{2\cdot\frac{1}{4}}\right)\quad ,(u,v)\in\mathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density).
This implies $U$ and $V$ are independent $\mathcal{N}(0,1/4)$ variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin.
| $$\frac{\sin(a-b)}{\sin(a+b)}=\frac{\alpha/\beta-\beta/\alpha}
{\alpha\beta-1/(\alpha\beta)}=\frac{\alpha^2-\beta^2}{\alpha^2\beta^2-1}$$
where $\alpha=\exp(ia)$ and $\beta=\exp(ib)$. Set $\gamma=\exp(ic)$.
Then
$$(\alpha^2-\beta^2)(\alpha^2\gamma^2-1)(\beta^2\gamma^2-1)
+(\beta^2-\gamma^2)(\alpha^2\beta^2-1)(\alpha^2\gamma^2-1)
+(\gamma^2-\alpha^2)(\alpha^2\beta^2-1)(\beta^2\gamma^2-1)
=0.$$
This simplifies to
$$(\alpha^4-\beta^4)\gamma^2+(\beta^4-\gamma^4)\alpha^2
+(\gamma^4-\alpha^4)\beta^2=0.$$
But that factors as
$$(\alpha^2-\beta^2)(\beta^2-\gamma^2)(\gamma^2-\alpha^2)=0.$$
If say $\alpha^2=\beta^2$ then $a=b$, as $a$ and $b$ are acute angles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3)$ overlap, but $(2x+7)(x+3)$ does not!
| Compute $(2x+1)(x+3)$. We obtain $(2x^2+6x)+(x+3)=2x^2+7x+3$. So why do think it could be $(2x+7)(x+3)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.
| Let's play.
$\sin y
= 3\cos(x+y)\sin x
= 3(\cos x \cos y-\sin x \sin y)\sin x
$
Divide by
$\cos x \cos y$
$\dfrac{\tan y}{\cos x}
= 3(\cos x -\sin x \tan y)\tan x
= 3\cos x\tan x -3\sin x\tan x \tan y
$
$\tan y(\dfrac{1}{\cos x}+3\sin x\tan x)
= 3\cos x\tan x
=3\sin x
$
$\tan y(1+3\sin^2 x)
=3\sin x\cos x
$
$\tan y
=\dfrac{3\sin x\cos x}{1+3\sin^2 x}
=\dfrac{\frac32\sin(2 x)}{1+3(1-\cos(2x))/2}
=\dfrac{3\sin(2 x)}{2+3(1-\cos(2x))}
=\dfrac{3\sin(2 x)}{5-3\cos(2x)}
$
At this point,
it being about midnight here,
I threw this at Wolfy
which said that
its max value was
$\dfrac34$
at
$\pi n +\arctan(1/2)$.
So the max value is
$\dfrac34$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,120,240,300$ ? How do I find 240 and 300?
| General form of solution is :
$x=k\pi± \pi/3= k\times 180 ± 60$
$k=0 ⇒ x=0± 60 ⇒x=60,.. x=-60=300$
$k=1 ⇒ x = 180 ± 60 ⇒ x= 180+60=240, x=180-60=120$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $ Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $
Find the minimum value of $h(\theta)$
$h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} = 5 \sin (\theta + 53.13) + \sqrt{2} $
Minimum value -
$5\sin (\theta + 53.13) + \sqrt{2} = -5 $
Therefore min value is = $ -5/5 - \sqrt{2} $
Why am I wrong ? And how should I do this question..
| Minimum is attained when $\sin (\theta + 53.13) = -1$ that is
$$h_{min}=h(3\pi/2+k\pi)=5 \sin (3\pi/2+k\pi) + \sqrt{2}=-5+\sqrt{2}$$
for the same reason the maximum is attained when $\sin (\theta + 53.13) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\implies \frac{dy}{dx}={(\frac{-y}{x})}^{\frac{1}{3}}$.
Slope at $(p,q)$ will be ${(\frac{-q}{p})}^{\frac{1}{3}}$.
So equation of tangent at $(p,q);$ $y-q={(\frac{-q}{p})}^{\frac{1}{3}}(x-p)$.
How to find the length of tangent intercepted in between the axes from here?
| Now, if $x=0$ we obtain $y=q+\sqrt[3]{qp^2}$ and for $y=0$ we obtain $x=p+\sqrt[3]{pq^2}.$
Thus, $A\left(p+\sqrt[3]{pq^2},0\right)$ and $B\left(0,q+\sqrt[3]{qp^2}\right)$ they are intersection points of the tangent with $x$-axis and with $y$- axis respectively.
Id est, $$AB=\sqrt{\left(p+\sqrt[3]{pq^2}\right)^2+\left( q+\sqrt[3]{qp^2}\right)^2}=$$
$$=\sqrt{p^{\frac{2}{3}}\left(p^{\frac{2}{3}}+q^{\frac{2}{3}}\right)^2+q^{\frac{2}{3}}\left(q^{\frac{2}{3}}+p^{\frac{2}{3}}\right)^2}=
p^{\frac{2}{3}}+q^{\frac{2}{3}}=a^{\frac{2}{3}}.$$
| {
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"url": "https://math.stackexchange.com/questions/2658068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $3\mid(a^2+b^2)$,then $3\mid a$ and $ 3\mid b$ I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?
Any help is greatly appreciated. Thank you!
|
I am trying to prove this by contradiction.
Why?
A contradiction where the remainder is either 1 or 2 and $a^2$ and $b^2$ can be most combinations ... that's a lot to check.
But a direct proof requires the remainder to be exactly $0$ (one option) which requires $a^2$ and $b^2$ to have opposite (add to a multiple of three) which can only happen one way, is a lot less to check.
.....
Advice: Get use to using negative moduli. Checking remainders being $0$ or $\pm 1$ is a lot easier than checking remainders being $0,1$ or $2$.
Example: If the remainders of $a^2 + b^2=0$, then the remainder of $a^2 = -b^2$ is a lot easier to write then: the remainder of the remainder of $b^2$ is $2$ than the remainder of $a^2$ is $1$ and vice versa, but if one is $0$ they both are.
.......
Taking those in mind the proof practically writes itself!
If $a^2 + b^2 \equiv 0$ then $a^2 \equiv -b^2$. Which means either $a^2 = b^2 =0$ or $a^2 \equiv \pm 1$ which $b^2 \equiv \mp 1$. But $0^2 \equiv 0$ and $(\pm 1)^2 \equiv 1$ so $x^2 \equiv -1$ is impossible. So $a^2 \equiv b^2 \equiv 0$ and $a\equiv b \equiv 0$. i.e. $3$ divides both $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing minimal polynomial An example in chapter 14.2 in Dummit and Foote computes the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We consider the field $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, and note that this is the same field as $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Now, the other roots of the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ are the distinct conjugates of $\sqrt{2}+\sqrt{3}$ under the Galois group. The distincr conjugates are $\pm\sqrt{2}\pm\sqrt{3}$. The minimal polynomial is therefore $[x-(\sqrt{2}+\sqrt{3})][x-(\sqrt{2}-\sqrt{3})][x-(-\sqrt{2}+\sqrt{3})][x-(-\sqrt{2}-\sqrt{3})]$. The next sentence states that this is "quickly computed" to be the polynomial $x^4-10x^2+1$. My question is how is this quickly computed? I'm not sure of a way to do it other than tediously expanding it. Further, how do we know that $x^4-10x^2+1$ is, in fact, irreducible?
| if $x = \sqrt 2 + \sqrt 3,$ then $x^2 = 5 + 2 \sqrt 6.$ So $(x^2 - 5)^2 = 24,$ and $x^4 - 10 x^2 + 25 = 24,$ and
$$ x^4 - 10 x^2 + 1 = 0. $$
By the Gauss lemma, without any rational roots, the possible rational factorings are
$$ (x^2 + ax +1)(x^2 - ax + 1), $$
$$ (x^2 + ax -1)(x^2 - ax - 1). $$
Neither one gives integer $a.$ Either $2 - a^2 = -10$ or $-2 - a^2 = -10$
| {
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"url": "https://math.stackexchange.com/questions/2662642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - \sqrt{x-4} = \sqrt{x-1} $
Squaring both Sides
$\mapsto x - 2 \sqrt{x^2 - 16}= -1$
$\mapsto \frac{x+1}{2} = \sqrt{x^2 - 16}$
Squaring both sides
$ \mapsto \frac{x^2 + 2x + 1}{4} = x^2 - 16$
$\mapsto 3x^2 -2x - 65 = 0$
Solving the quadratic equation
$$x = 5 OR \frac{-13}{3}$$
Checking in the initial equation we can see that $5$ is a valid root. But the second value that is given in the key to the book is $4$. How do I obtain $4$?
| I think it should be like this when the equation gets divided by $\sqrt{x-4}$:
$(\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4}) - \sqrt{(x-4)(x-1)} = 0$
$\mapsto (\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$
$\therefore$ Either $\sqrt{x-4} = 0$ OR $(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Problem with generalized eigenvectors in a 3x3 matrix. I have this matrix:
$$
A= \begin{pmatrix}
0 & 1 & 1 \\
0 & 1 & 0 \\
-1 & 1 & 2 \\
\end{pmatrix}
$$
I have founded the eigenvalues:
$$\lambda_{1,2,3}=1$$
So
$$\lambda=1$$$$\mu=3$$
I'm expecting to have one eigenvector plus two generalized eigenvectors. But, proceding I have some troubles:
$$
A= \begin{pmatrix}
-1 & 1 & 1 \\
0 & 0 & 0 \\
-1 & 1 & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y\\
z\\
\end{pmatrix}
= \begin{pmatrix}
0 \\
0 \\
0 \\
\end{pmatrix}
$$
Which cleary brings to two equal equations:
$$-x+y+z=0$$
I don't know how I should proceed. I can find a solution by trying some values but I don't like this method.
Which is the best and secure method to solve this problem? Thank you very much.
| I like to keep things as matrices and column vectors, visual. Since $(A - I)^2 = 0$ but $A - I \neq 0,$ we can take the third column of $R$ (for "Right") as anything we like for which
$$ (A - I) w \neq 0. $$
I like
$$
w =
\left(
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right)
$$
The second column will be $v = (A-I)w,$ which is automatically an eigenvector (WHY??)
$$
v =
\left(
\begin{array}{c}
1 \\
0 \\
1
\end{array}
\right)
$$
For the first column we can take any eigenvector that is independent of $v,$ looking back at $A-I$ we can take
$$
u =
\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right)
$$
So, we have
$$
R =
\left(
\begin{array}{ccc}
1 & 1 & 0 \\
1 & 0 & 0 \\
0 & 1 & 1
\end{array}
\right)
$$
which has determinant $1,$ very helpful, and
$$ R^{-1} A R = J. $$
The direction that is actually useful is $R J R^{-1} = A.$ Useful for finding $e^A$ or $A^{100}$ or any $f(A)$ with $f$ analytic.
$$
R^{-1} =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & -1 & 0 \\
-1 & 1 & 1
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
1 & 1 & 0 \\
1 & 0 & 0 \\
0 & 1 & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & -1 & 0 \\
-1 & 1 & 1
\end{array}
\right) =
\left(
\begin{array}{rrr}
0 & 1 & 1 \\
0 & 1 & 0 \\
-1 & 1 & 2
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...
Proof
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| $$
By the triangle inequality
$$\frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| \leq \left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \tag{$\star$}$$
Note that for all $x\in \mathbb{R}$
$$\left|x\right| < 1 +x^2 \implies \left|x\right| < \left(1 +x^2\right)\left(1+y^2\right)$$
Therefore
$$\frac{\left|x\right|}{\left(1 +x^2\right)\left(1+y^2\right)} \leq 1$$
Applying this fact to $(\star)$ we see that
$$\left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \leq \left(1 + 1\right)\left|x-y\right| \leq 2\left|x - y \right|$$
Therefore $f$ is a Lipschitz function, which implies $f$ is uniformly continuous on $\mathbb{R}$.
Please comment on validity and/or readability, thank you.
| Lipschitz-continuity implies uniform continuity, we cannot disagree on that. On the other hand there is a slightly more efficient way for proving that $|f'|$ is bounded:
$$ f'(x) = f(x)\cdot \frac{d}{dx}\log f(x) = \frac{1}{1+x^2}\cdot \frac{-2x}{1+x^2}$$
where $\left|\frac{1}{1+x^2}\right|\leq 1$ is trivial and $\left|\frac{2x}{1+x^2}\right|\leq 1$ just a bit less (AM-GM), so $|f'|\leq 1$.
By using the weighted AM-GM inequality we have the optimal bound $|f'|\leq\frac{3}{8}\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to prove $e^x\ge \left(1+\frac xn\right)^n$ for any real numbers $x, n > 0$ Can someone provide a detailed proof? I saw a proof here
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\
&\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\
&=1
\end{align}$$
where in going from (1) to (2) we used Bernoulli's Inequality. Note
that (2) is valid whenever $n>−x$ or $x>−n$. Since $e_n(x)$
monotonically increases and is bounded above by $e^x$, then $$e^x\ge
\left(1+\frac xn\right)^n \tag 3$$ for all $n\ge 1$.
But I don't know how do we get $(1)$.
| \begin{align}
\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n} &= \frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^{n + 1}}\left(1+\frac xn\right) \\
&= \frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left( \frac{n + x}{n} \right)^{n + 1}}\left(1+\frac xn\right) \\
&= \left(1+\frac x{n+1}\right)^{n+1}\left( \frac{n}{n + x} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left(\left(1+\frac x{n+1}\right) \frac{n}{n + x} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n}{n + x} + \frac{nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n(n+1)}{(n+1)(n + x)} + \frac{nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n(n + 1) + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n(n + 1) + [x(n+1) - x(n+1)] + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{(n + x)(n + 1) - x(n+1) + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( 1 + \frac{- x(n+1) + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( 1 + \frac{- xn - x + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( 1 + \frac{-x}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
(Full) Solution to $\int \frac{1-x}{x^2+x+1}dx$ with steps.
*
*If we use substitution, we get:
$u=x^2+x+1$
$du=(2x+1)dx$
$(-1/2)du=(-x-1/2)dx=(-x+1-3/2)dx$
*Then the integral becomes:
$\int ((-du/2)/u) + 3/2\int dx/(x^2+x+1$)
But why isn't it just:
$\int ((-1/2du)-3/2)/u$ ?
And why do you use + 3/2 in the original answer and not -3/2, if -1/2du = (-x+1-3/2)dx ? I don't really understand step 2.
Edit: Please post a full solution to the question in the title: $\displaystyle \int \frac{1-x}{x^2+x+1}dx$ with all the steps.
| $\displaystyle \int\frac{1-x}{x^2+x+1}dx=\frac{3}{2}\int \frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{2}\int\frac{2x+1}{x^2+2x+1}dx$.
Let $\displaystyle x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$. Then $\displaystyle dx=\frac{\sqrt{3}}{2}\sec^2\theta d\theta$ and $\displaystyle \left(x+\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\sec^2\theta$. So,
$\displaystyle \int \frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}=\frac{2}{\sqrt{3}}\int d\theta=\theta+C=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$ Here's what I tried:
Let the roots be $a$, $b$, $c$ and $d$, $a+b=4$. Then,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrightarrow 4ab + 4cd = 20 \Longrightarrow ab + cd = 5$$
$$abcd = 5$$
I can't figure out how to proceed.
| Two of the roots sum to 4. All 4 roots sum to 8. This means that the other two roots must also sum to 4.
If this polynomial had rational roots, they would have to be in the set $\{\pm1,\pm5\}$
By what we have above we might try $x=-1, x = 5$ alas these do not work.
We could factor the polynomial like so:
$(x^2 - 4x + A)(x^2 -4x + B)\\
A+B = 5\\
AB = 5$
And solve for $A,B$
Or we might try
$(x -2 + a)(x-2-a)(x-2 + b)(x-2-b)$
Then substituting
$y = x-2\\x = y+2$
Into the original polynomial
$(y+2)^4 - 8(y+2)^3 + 21(y+2)^2 -20(y+2) + 5\\
y^4 - 3y^2+1\\
y^2 = \frac 32 \pm \sqrt 2\\
y = \pm \sqrt {\frac 32 \pm \sqrt 2}\\
x = 2\pm \sqrt {\frac 32 \pm \sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using summation by parts to evaluate an alternating sum I want to evaluate
$$
\sum_{k=0}^n (-1)^k \binom{n}{k} k.
$$
I tried summation by parts, i.e. the formula
$$
\sum_{k=0}^n (f(k+1) - f(k))g(k)
= f(n+1)g(n+1) - f(0)g(0) - \sum_{k=0}^n f(k+1) (g(k+1) - g(k))
$$
with $f(k+1) - f(k) = (-1)^k \binom{n}{k}$ and $g(k) = k$. As
$$
(-1)^{k+1}\binom{n-1}{k+1} - (-1)^k \binom{n-1}{k}
= (-1)^{k+1} \left( \binom{n-1}{k+1} + \binom{n-1}{k} \right)
= (-1)^{k+1} \binom{n}{k+1}
$$
we have $f(k) = (-1)^k \binom{n-1}{k}$. Plugging into the formula
\begin{align*}
\sum_{k=0}^n (-1)^k \binom{n}{k} k
& = (-1)^{n+1} \binom{n-1}{n+1} (n+1) - (-1)^0 \binom{n-1}{0}\cdot 0
- \sum_{k=0}^n (-1)^{k+1} \binom{n-1}{k+1} \\
& = \sum_{k=0}^n (-1)^{k} \binom{n-1}{k+1}.
\end{align*}
But for example if $n = 3$ then
$$
\sum_{k=0}^n (-1)^k \binom{n}{k} k = -3 + 6 -3 = 0
$$
but
$$
\sum_{k=0}^n (-1)^{k} \binom{n-1}{k+1}
= \binom{2}{1} - \binom{2}{2}
= 1
$$
which is not equal, but I cannot see whats wrong with the above derivation??
| Summation by parts is definitely an overkill, differentiation a lesser overkill. For any $n\geq 1$ we have
$$ k\binom{n}{k} = n\binom{n-1}{k-1} $$
and
$$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}k = \sum_{k=1}^{n}(-1)^k\binom{n}{k}k = -n\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}=-n\sum_{j=0}^{n-1}\binom{n-1}{j}(-1)^j $$
so your sum is constantly zero for any $n\geq 2$ and you just have to compute it by hand for $n=0$ and $n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the eigenvalues of block matrix $C$
Find the eigenvalues of $$C = \begin{bmatrix}\begin{array}{c|c} 0 & A\\ \hline A^T & 0\end{array}\end{bmatrix}$$ where $$A = \begin{bmatrix}
0&0&0&1&1&1&1\\0& 1& 1& 0& 1& 0& 1\\0 &1 &1 &1 &0 &1 &0\\1& 0& 1& 0& 0& 1& 1\\1 &0 &1& 1& 1& 0& 0\\1 &1 &0& 0& 1& 1& 0\\1 &1& 0& 1& 0& 0 &1
\end{bmatrix}$$
The characteristic polynomial of $C$ is $$\det(xI-C)=x^2-A^2$$
Will the eigenvalues of $A$ be related to those of $C$ in some way?
I am unable to proceed here.
| Since $\det(xI-C)=\det(x^2I-AA^T)$, the eigenvalues of $C$ are the singular values of $A$ and their negatives.
Now, note that $AA^T=2I+2ee^T$, where $e$ denotes the all-one vector. Hence the set of eigenvalues of $AA^T$ consists of a simple eigenvalue $2+2\times7=16$ and the eigenvalue $2$ of multiplicity $6$. It follows that the singular values of $A$ are $4$ and six copies of $\sqrt{2}$.
Hence the eigenvalues of $C$ are $+4,-4$, six copies of $\sqrt{2}$ and six copies of $-\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$ I've been dealing with the following series for a while now, without real progress.
$$\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$$
After using WolframAlpha, I know it converges to $30$, but I can't see how to calculate it by myself.
Any leads would be greatly appreciated!
| Substitute $2/3$ by a variable $x$. Then the series defined by
$$\sum_{k=1}^\infty k^2 x^k$$
converges locally uniformly for $|x|<1$, hence we can integrate and differentiate it termwise so
$$\sum_{k=1}^\infty k^2 x^k=x\sum_{k=1}^\infty k^2 x^{k-1}=x\sum_{k=1}^\infty k\frac{d}{dx}( x^k)=x\frac{d}{dx}\left(\sum_{k=1}^\infty kx^k\right)\\=x\frac{d}{dx}\left(x\sum_{k=1}^\infty k x^{k-1}\right)=x\frac{d}{dx}\left(x\sum_{k=1}^\infty \frac{d}{dx}(x^k)\right)=x\frac{d}{dx}\left(x\frac{d}{dx}\sum_{k=1}^\infty x^k\right)=\left[x\frac{d}{dx}\right]^2\frac{x}{1-x}\\=\left[x\frac{d}{dx}\right]\frac{x}{(1-x)^2}=x\left(\frac{1}{(1-x)^2}+\frac{2x}{(1-x)^3}\right)=\frac{x(1+x)}{(1-x)^3}$$
Hence
$$\sum_{k=1}^\infty k^2(2/3)^k=\left[\sum_{k=1}^\infty k^2 x^k\right]_{x=2/3}=\frac{x(1+x)}{(1-x)^3}\bigg|_{x=2/3}=\frac{10}9\cdot 3^3=3\cdot 10=30$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem Factorize and prove that
$$
\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2
$$
using factor theorem.
My Attempt:
$\Delta$ is a homogeneous symmetric polynomial of degree $6$.
When $(x-y)^2+(y-z)^2+(z-x)^2=0$, i.e. $x=y=z$
$$
\Delta=\begin{vmatrix}
0&0&0\\
0&0&0\\
0&0&0\\
\end{vmatrix}=0
$$
Thus, $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor.
How do I extract the other $(x-y)^2+(y-z)^2+(z-x)^2$ from $\Delta$ $\color{red}{?}$
Does this have anything to do with all rows (or columns) being zero when $(x-y)^2+(y-z)^2+(z-x)^2=0$ $\color{red}{?}$
If I can extract that then i think I know how to proceed. The remaining factor must be a homogeneous quadratic symmetric polynomial, i.e. $p(x,y,z)=a(x^2+y^2+z^2)+b(xy+yz+zx)$
$$
\Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.a(x^2+y^2+z^2)+b(xy+yz+zx)
$$
$$
\Delta(1,0,0)=\begin{vmatrix}
-1&0&0\\
0&0&-1\\
0&-1&0\\
\end{vmatrix}=1=4.a\implies a=\frac{1}{4}
$$
$$
\Delta(1,1,0)=\begin{vmatrix}
-1&-1&1\\
-1&1&-1\\
1&-1&-1\\
\end{vmatrix}=\begin{vmatrix}
0&0&1\\
-2&0&-1\\
0&-2&-1\\
\end{vmatrix}\\
=\begin{vmatrix}
-2&0\\
0&-2\\
\end{vmatrix}=4=4.(2a+b)=4(1/2+b)=2+4b\\
\implies b=\frac{1}{2}
$$
$$
\Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.\frac{1}{4}(x^2+y^2+z^2)+\frac{1}{2}(xy+yz+zx)\\
=\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.(x^2+y^2+z^2+2xy+2yz+2zx)\\
=\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2(x+y+z)^2
$$
Note:
I am trying to factorize the determinant using factor theorem given the fact that the determinant is a homogeneous symmetric polynomial of degree 6.
| Hint: Note that your matrix has the form
$$\pmatrix{a&b&c\cr b&c&a\cr c&a&b\cr }$$
Which has the determinant
$$3abc-c^3-b^3-a^3$$
which can again be factored into
$$-\left(a+b+c\right)\,\left(a^2+b^2+c^2-ab-bc-ac\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Surface integral of sphere within a paraboloid in spherical coordinates Find the surface area of the portion of the sphere $x^2 + y^2 + z^2 =3c^2$ within the paraboloid $2cz =x^2+y^2$ using spherical coordinates. ($c$ is a positive constant)
I've found it in cartesian coordinates and then polar coordinates by taking limits $r$ from $0$ to $\sqrt 2 c$ and $\theta$ from $0$ to $2\pi$ and got the answer $4\sqrt 3\pi c^3$.
But for the spherical coordinates I am getting the limits $\phi$ from $\arccos(\sqrt 3)$ to $\pi/2$ and $\theta$ from $2\pi$ to $0$ and the answer is $6\pi/\sqrt3 c^2$ which is $2c$ times less than in cartesian coordinates. What am I doing wrong?
| The limits for $\phi$ are wrong. Note that $\sqrt{3}>1$ and $\arccos(x)$ is defined in $[-1,1]$. Solving the equation $2cz+z^2=3c^2$ we get $z=c$ and $z=-3c$ (which is not acceptable because $x^2+y^2=2cz=-6c^2<0$). Hence $\phi$ goes from $0$ (north pole) to $\arccos(1/\sqrt{3})$. Therefore
$$S=3c^2\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\arccos(1/\sqrt{3})}\sin(\phi) d\phi d\theta=6\pi c^2[-\cos(\phi)]_0^{\arccos(1/\sqrt{3})}=2\pi c^2(3-\sqrt{3}).$$
P.S. Your result $4\sqrt 3\pi c^3$ seems to be a volume not a surface:
$$S=\iint_{x^2+y^2\leq 2c^2}\frac{\sqrt{3}c}{\sqrt{3c^2-x^2-y^2}}dx dy=
2\pi\sqrt{3}c\int_{\rho=0}^{\sqrt{2}c}\frac{\rho}{\sqrt{3c^2-\rho^2}}d\rho=2\pi c^2(3-\sqrt{3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ (Solve for $k$) Question:
$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$
Solve for $k$
My attempt:
$$1^2-2^2+3^2-4^2+…-2016^2+2017^2\\
\begin{align}= (1-2)(1+2)+(3-4)(3+4)+…+(2015-2016)(2015+2016)+2017^2 \end{align}
$$
What should I do next?
| A very simple way of seeing this is to write the following:
$$\begin{align} \\
1^2-2^2+\dots-2016^2+2017^2 &= 1+(3^2-2^2)+\dots+(2017^2-2016^2) \\
&=1+(3-2)(3+2)+(5-4)(5+4)+\dots+(2017-2016)(2016+2017) \\
&=1+2+3+...+2016+2017 \\
&=\frac{(2018)(2017)}{2}=2017k \\
\end{align}$$
Which means that $k=\frac{2018}{2}=1009$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When is $2^{2n+2}+2^{m+2}+1$ a perfect square? Let $m,n$ be natural numbers satisfying $m\leq 2n$. Is it true that $$2^{2n+2}+2^{m+2}+1$$ is a perfect square if and only if $m=n$?
What I have tried: Under the assumption $m<n$, I've tried to 'squeeze' the above number between two consecutive squares, implying it cannot be a perfect square. This works fine because (writing $P(m,n)=2^{2n+2}+2^{m+2}+1$), we have $(2\cdot2^n)^2<P(m,n)<(2\cdot2^n+1)^2$. But this method doesn't work when $\frac{m}{2}\leq n<m$, and I wonder if there is any pair $(m,n)$ with $m\ne n$ making $P(m,n)$ a perfect square.
Any advice is welcome.
| Suppose $m>n$ and $k$ is an integer such that $$(2^{n+1}+k)^2= 2^{2n+2}+2^{m+2}+1.$$ This means $$2^{n+2}k+k^2=2^{m+2}+1$$ so $$(k-1)(k+1)=k^2-1=2^{n+2}(2^{m-n}-k).$$ Note that clearly $k$ must be odd and $\gcd(k-1,k+1)=2$, so either $k-1$ or $k+1$ is divisible by $2^{n+1}$. Since $k>1$ (if $k=1$ we would have $m=n$), this means that $k\geq 2^{n+1}-1$. Plugging this into the second equation above gives $$2^{m+2}+1\geq 2^{n+2}(2^{n+1}-1)+(2^{n+1}-1)^2=2^{2n+3}+2^{2n+2}-2^{n+3}+1.$$ As long as $n\geq 1$ (the case $n=0$ is trivial) we have $2n+2\geq n+3$ so we can conclude $$2^{m+2}\geq 2^{2n+3}$$ and thus $m\geq 2n+1$.
Thus, if $n<m\leq 2n$, then no such integer $k$ can exist and $2^{2n+2}+2^{m+2}+1$ is not a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$
Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$
IMO 1962/4
My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.
$$
\begin{align}
\cos^22x&=(\cos^2x-\sin^2x)^2\\
&=\cos^4x+\sin^4x-2\sin^2\cos^2x\\
&=\cos^4+(1-\cos^2x)^2-2(1-\cos^2)\cos^2x\\
&=\cos^4+1-2\cos^2x+\cos^4x-2\cos^2x+2\cos^4x\\
&=4\cos^4x-4\cos^2x+1
\end{align}
$$
Without knowledge of other trigonometric identities, $\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\cos 3x=4\cos^3x-3\cos x$:
$$
\begin{align}
\cos^23x&=(4\cos^3x-3\cos x)^2\\
&=16\cos^6x+4\cos^2x-24\cos^4x
\end{align}
$$
Therefore, the original equation can be written as:
$$\cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x+4\cos^2x-24\cos^4x-1=0$$
$$-20\cos^4x+6\cos^2x+16\cos^6x=0$$
Letting $y=\cos x$, we now have a polynomial equation:
$$-20y^4+6y^2+16y^6=0$$
$$y^2(-20y^2+6y+16y^4)=0\Rightarrow y^2=0 \Rightarrow x=\cos^{-1}0=\bbox[yellow,10px]{90^\circ}$$
From one of the factors above, we let $z=y^2$, and we have the quadratic equation:
$$16z^2-20z+6=0\Rightarrow 8z^2-10z+3=0$$
$$(8z-6)(z-\frac12)=0\Rightarrow z=\frac34 \& \ z=\frac12$$
Since $z=y^2$ and $y=\cos x$ we have:
$$\biggl( y\rightarrow\pm\frac{\sqrt{3}}{2}, y\rightarrow\pm\frac{\sqrt{2}}2 \biggr)\Rightarrow \biggl(x\rightarrow\cos^{-1}\pm\frac{\sqrt{3}}{2},x\rightarrow\cos^{-1}\pm\frac{\sqrt{2}}2\biggr)$$
And thus the complete set of solution is:
$$\bbox[yellow, 5px]{90^\circ, 30^\circ, 150^\circ, 45^\circ, 135^\circ}$$
As I do not have the copy of the answers, I still hope you can verify the accuracy of my solution.
But more importantly...
Seeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation?
| You can shorten the argument by noting at the outset that
$$
\cos3x=4\cos^3x-3\cos x=(4\cos^2x-3)\cos x
$$
so if we set $y=\cos^2x$ we get the equation
$$
y+(2y-1)^2+y(4y-3)^2=1
$$
When we do the simplifications, we get
$$
2y(8y^2-10y+3)=0
$$
The roots of the quadratic factor are $3/4$ and $1/2$.
A different strategy is to note that $\cos x=(e^{ix}+e^{-ix})/2$, so the equation can be rewritten
$$
e^{2ix}+2+e^{-2ix}+e^{4ix}+2+e^{-4ix}+e^{6ix}+2+e^{-6ix}=4
$$
Setting $z=e^{2ix}$ we get
$$
2+z+z^2+z^3+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}=0
$$
or as well
$$
z^6+z^5+z^4+2z^3+z^2+z+1=0
$$
that can be rewritten (noting that $z\ne1$),
$$
\frac{z^7-1}{z-1}+z^3=0
$$
or $z^7+z^4-z^3-1=0$ that can be factored as
$$
(z^3+1)(z^4-1)=0
$$
Hence we get (discarding the spurious root $z=1$)
$$
2x=\begin{cases}
\dfrac{\pi}{3}+2k\pi \\[6px]
\pi+2k\pi \\[6px]
\dfrac{5\pi}{3}+2k\pi \\[12px]
\dfrac{\pi}{2}+2k\pi \\[6px]
\pi+2k\pi \\[6px]
\dfrac{3\pi}{2}+2k\pi
\end{cases}
\qquad\text{that is}\qquad
x=\begin{cases}
\dfrac{\pi}{6}+k\pi \\[6px]
\dfrac{\pi}{2}+k\pi \\[6px]
\dfrac{5\pi}{6}+k\pi \\[6px]
\dfrac{\pi}{4}+k\pi \\[6px]
\dfrac{3\pi}{4}+k\pi
\end{cases}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the numbers having a particular factor,an upper limit and a specific set of digits. Question :
How many numbers less than $50000$ can be formed which are multiples $6$ using the digits $0,1,2,3,4,5$?
My attempt :
First of all, for a number to be a multiple of $6$, it should be a multiple of $2$ as well as $3$. So, the number must be even as well as it's sum of digits should be a multiple of $3$. Next, the number should have only the digits $1,2,3,4$ in the first position as it has to be less that $50000$. But, I face a problem, If I put $2$ or $4$ in the first position, then I will have one less even digit for the last position and there would be cases where I only use the digits $0,1,2,4,5$ or $1,2,3,4,5$ so as to make the number divisible by $3$. So, is there an efficient way to cover all the cases and get the answer quickly?
All help will be appreciated.
| We put it in three cases as the last digit must be either $0;2$ or $4.$ (I assume that you can use each digit only once).
First case: The number is in the form $\overline{abcd0}$, which means the numbers $a,b,c,d$ are $4$ of the remaining numbers available $(1;2;3;4;5)$.
There are $5$ ways of picking four numbers from this set, we add $a,b,c,d$ along with $e$ for each pick:
$1+2+3+4+0=10$, not divisible by $3$.
$1+2+3+5+0=11$, not divisible by $3$.
$1+2+4+5+0=12$, divisible by $3$.
$1+3+4+5+0=13$, not divisible by $3$.
$2+3+4+5+0=14$, not divisible by $3$.
We conclude that $(a,b,c,d)$ must be a permutation of $(1;2;4;5)$ that satisfy $a<5$.
For each sub-case $a=1;a=2;a=4$, note that $(b,c,d)$ is a permutation of $3$ remaining numbers in the set, so each sub-case will have $3!=6$ possible outcomes for $b,c,d$.
In total, this first case (including $3$ sub-cases) should have $6 \times 3=18$ satisfied numbers.
Second case: The number is in the form $\overline{abcd2}$, which means the numbers $a,b,c,d$ are $4$ of the remaining numbers available $(0;1;3;4;5)$.
There are $5$ ways of picking four numbers from this set, we add $a,b,c,d$ along with $e$ for each pick:
$0+1+3+4+2=10$, not divisible by $3$.
$0+1+3+5+2=11$, not divisible by $3$.
$0+1+4+5+2=12$, divisible by $3$.
$0+3+4+5+2=14$, not divisible by $3$.
$1+3+4+5+2=15$, divisible by $3$.
We conclude that $(a,b,c,d)$ must be a permutation of $(0;1;4;5)$ or $(1;3;4;5)$ that satisfy $a<5$ and $a\ne 0$.
*
*Case 2.1: $(a,b,c,d)$ is a permutation of $(0;1;4;5)$ that satisfy $a<5$ and $a\ne 0$. There are two sub-cases for this one ($a=1;a=4$), plus each sub-case has $6$ possible outcomes for $b,c,d$, so there are $6 \times 2 =12$ satisfied numbers.
*Case 2.2: $(a,b,c,d)$ is a permutation of $(1;3;4;5)$ that satisfy $a<5$ and $a\ne 0$. There are three sub-cases for this one ($a=1;a=3;a=4$), plus each sub-case has $6$ possible outcomes for $b,c,d$, so there are $6 \times 3 =18$ satisfied numbers.
So the second case should have a total of $12+18=30$ satisfied numbers.
Third case: The number is in the form $\overline{abcd4}$, this is similar to the second case, you can do it by yourself, after finishing it you can check the answer below.
There are $78$ five-digit numbers less than $50000$ divisible by $6$ and can be formed by using the digits $(0;1;2;3;4;5)$
I think you should do similarly for the case one, two, three, four digit numbers, so here are two extra hints to do them:
The number of ways to choose $k$ objects from $n$ objects, which the order does not matter, is: $$\frac{n!}{k!(n-k)!}$$
For this problem the order of choosing numbers is important, so the number of permutations of $n$ distinct objects is $n!$.
I don't think this is a quick way, but it will surely help you to get the correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Induction $(n^3+5n)$ is divisible by $6$
Here is what I have done
Assume $n=k$ is true
$\sum_{1}^{k} k =\dfrac{(k^3+5k)}{6}$
Now assume $n=k+1$ is true
$\sum_{1}^{k+1} k+1 =\dfrac{(k+1)^3+5(k+1)}{6}$
Then now
$\dfrac{(k+1)^3+5(k+1)}{6}=\sum_{1}^{k} k + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + \dfrac{(6k+6)}{6}$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+11k+6)}{6}$
But the other side doesnt equate (LHS)
$\dfrac{(k^3+3k^2+3k+1)+5k+5}{6}$
$\dfrac{(k^3+3k^2+8k+6)}{6}\ne\dfrac{(k^3+11k+6)}{6}$
I hope my method was clear enough so you can see where I went wrong. It would be much more use to me if you solved it as I learn better from looking at solutions and then applying them to other questions.
| a one line proof $$n^3+6n-n=(n-1)n(n+1)+6n$$ ready!
| {
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"timestamp": "2023-03-29T00:00:00",
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Verfication of deduction made using the Cauchy-Schwarz inequality Is the following proof correct?
Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.
Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the euclidean product in $\textbf{6.4}$. Now let $a,b,c,d$ be arbitrary positive numbers and let $u = (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|)$ and $v = (\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})$ then by appealing to the Cauchy Schwartz inequality we have\begin{align*}
|\langle u,v\rangle| &= |4| = |\langle (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|),(\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})\rangle|\\
&\leq \sqrt{|\sqrt{a}|^2+|\sqrt{b}|^2+|\sqrt{c}|^2+|\sqrt{d}|^2}\cdot\sqrt{\frac{1}{|\sqrt{a}|^2}+\frac{1}{|\sqrt{b}|^2}+\frac{1}{|\sqrt{c}|^2}+\frac{1}{|\sqrt{d}|^2}}\\
&= \sqrt{(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})} = \|u\|\cdot\|v\|.
\end{align*}
Squaring both sides yields the required result.
NOTE The reference to $6.4$ above is simply to what is commonly understood to be the dot product.
$\blacksquare$
| Yes, your proof is fine. Why did you write $|\sqrt{a}|$ ? If $a$ is positive, we always have $\sqrt{a}>0$.
| {
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Find three rational numbers $a,b,c$ s.t. $b^2-a^2=c^2-b^2=5$. A rational numbers cannot have irrational as$\exists n \in \mathbb{Z}, \, \sqrt[n]{\frac xy}$, but the two equalities give: $b^2=\frac{(a^2+c^2)}{2} \implies b = \sqrt[2]{\frac {(a^2+c^2)}{2}}$.
To avoid this, need $4\mid a$, & $a=c$; so that if $\exists t=a/2, \, b = \sqrt[2]{\frac {(a^2+c^2)}{2}} => 2t$.
As an example, $a=c=8, t=4, b=8$.
I am giving an idea for a particular case only, & am unable to pursue.
| We can rewrite these numbers with common denominator $D$:
$a=\dfrac{A}{D}$, $\;b=\dfrac{B}{D}$, $\;c=\dfrac{C}{D}$. Then we'll get diophantine equation (system of diophantine equations):
$$
B^2-A^2=C^2-B^2=5D^2.\tag{1}
$$
The smallest solution of $(1)$ is $(A,B,C,D)=(31,41,49,12)$.
So, example of such three rational numbers $a,b,c$ is: $$a=\dfrac{31}{12},\quad b=\dfrac{41}{12}, \quad c=\dfrac{49}{12}.$$
Short algorithm (to find $A,B,C,D$) description:
*
*for each $C$ (from some range) consider $B = 1,2,\ldots,C-1$;
*
*if $2B^2-C^2$ is square number, then denote $A=\sqrt{2B^2-C^2}$;
*
*if $\dfrac{B^2-A^2}{5}$ is square number, then denote $D=\sqrt{\frac{B^2-A^2}{5}}$ and output $(A,B,C,D)$.
| {
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How to use integration by parts to solve an integral? I'm having trouble with solving this integral by parts. Here is what I have so far. Can anyone please help me out?
Solve the integral $\int x^3\sqrt{x^2-1}dx$ by parts, choosing u = $x^2$ and
$dv = x\sqrt{x^2-1} dx$
$du_1 = 2xdx$
$v_1$ = $\int x\sqrt{x^2-1}dx$ $ t = x^2-1$ $dt = 2x dx$
$v_1$ = $\frac{1}{2}\int \sqrt{t}dt$
$v_1$ = $\frac{1}{2}(\frac{2}{3}t^\frac{3}{2}) + c$
$v_1$ = $\frac{1}{3} (x^2-1)^\frac{3}{2} + c$
$uv - \int vdu$
= $\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \int 2x\frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$u_2 = 2x$ $dv_2 = \frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$du_2 = 2dx$
| You used integration by parts once. And you used substitution to find $v_1$. Use substitution again:
$$I=\int 2x\cdot \frac13(x^2-1)^{3/2}dx;$$
$$t=x^2-1; \ \ dt=2xdx;$$
$$I=\frac13 \int t^{3/2}dt=\frac13\cdot \frac25 \cdot t^{5/2}+C=\frac{2}{15}(x^2-1)^{5/2}+C.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \sin \frac{x}{2} - \cos \frac{x}{2}$
So, $\frac{dy}{dx} = \frac{1}{2} \cdot (\cos\frac{x}{2} + \sin\frac{x}{2})$.
But apparently this is wrong. The correct solution is:
$\frac{dy}{dx} = -\frac{1}{2}\cdot(\cos\frac{x}{2} + \sin\frac{x}{2})$.
So I want to know what I have done wrongly here. Why is my answer not right?
| Hint: $y = |\sin(x/2)-\cos(x/2)|$ and $|\sin(x/2)-\cos(x/2)|=\color{red}{-}(\sin(x/2)-\cos(x/2))$ for $0 \le x \le \pi/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{\pi/2}\frac{x}{(a^2\cos^2x+b^2\sin^2x)}dx$ The question is to evaluate this definite integral:
$$\int_{0}^{\pi/2}\frac{x}{(a^2\cos^2x+b^2\sin^2x)}dx$$
I know the definite integration with limits from $0$ to $\pi$, where we use the identity $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$, but it obviously cannot be used here. I also tried breaking the $0$ to $\pi$ one into two parts to no avail. Since the other one had such a simple solution, I can't wrap my head around this one, and would love some help!
| Write $\alpha = a/b$ and substitute $\tan x = \alpha t$ to obtain
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{a^2\cos^2 x + b^2\sin^2 x} \, dx
= \frac{1}{ab} \int_{0}^{\infty} \frac{\arctan(\alpha t)}{1 + t^2} \, dt
=: \frac{1}{ab}I(\alpha). $$
Then
\begin{align*}
I'(\alpha)
&= \int_{0}^{\infty} \frac{t}{(1+t^2)(1+\alpha^2 t^2)} \, dt \\
&\hspace{1.5em} = \left[ -\frac{1}{2(1-\alpha^2)} \log\left( \frac{1+\alpha^2 t^2}{1+t^2} \right) \right]_{0}^{\infty}
= -\frac{\log \alpha}{1 - \alpha^2}.
\end{align*}
Now it is easy to check that, if we write $\chi_2(z) = \int_{0}^{z} \frac{\operatorname{artanh} \xi}{\xi} \, d\xi$ for the Legendre chi function, then
$$ \frac{d}{d\alpha} \chi_2\left(\frac{1-\alpha}{1+\alpha}\right) = \frac{\log \alpha}{1-\alpha^2}. $$
So it follows that
$$ I(\alpha) = \chi_2(1) - \chi_2\left(\frac{1-\alpha}{1+\alpha}\right) = \frac{\pi^2}{8} + \chi_2\left(\frac{\alpha-1}{\alpha+1}\right). $$
Plugging this back and manipulating a bit, we obtain
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{a^2\cos^2 x + b^2\sin^2 x} \, dx
= \frac{1}{ab} \left( \frac{\pi^2}{8} + \chi_2\left(\frac{a-b}{a+b}\right) \right). $$
| {
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$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$ Related to solving $~~2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
I've tried some arithmetic to find something like
$a^x=b~\Longrightarrow~x=\log_{a}b$
But what I've found is that
$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
$2^8\cdot2^{x-5}+3^{x-5}=2^8\cdot2^{3(x-5)}+3^{2(x-5)}$
$256\cdot2^{x-5}+3^{x-5}=256\cdot(2^{x-5})^3+(3^{x-5})^2$
$256\cdot a +b=256\cdot a^3+b^2$
$256 \cdot a(a^2-1)+b(b-1)=0$
I don't know how to solve this equation above. Maybe it doesn't the better way to solve my question (tittle) or maybe it is wrong.
Can someone help me to solve that?
| $$0=2^{x+3}(2^{2x-10}-1)+3^{x-5}(3^{x-5}-1)=f(x)$$
As $2^{x+3},3^{x-5}>0$ for real $x,$
If $x-5>0$, then $f(x)>0$, and if $x-5<0$, then $f(x)<0$.
Since we wanted $f(x) = 0$, it must be true that $x - 5 \not< 0$ and $x - 5 \not> 0$. So $x - 5 = 0$, and thus $x = 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the following seminorm of $A$ Consider the following operators defined on $(\mathbb{C}^2,\|\cdot\|_2)$:
$$S = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$
I want to calculate $\|A\|_S$, with
$$\|A\|_S:=\inf\left\{c\geq 0; \quad\sqrt{\langle SAx,Ax\rangle} \leq c \sqrt{\langle Sx,x\rangle},\;\forall x \in \overline{\text{Im}(S)}\right\}.$$
My attempt: Note that $\overline{\text{Im}(S)}= \text{Im}(S)= \left\{ \begin{pmatrix} x \\ x \end{pmatrix} \ : \ x\in \mathbb{C} \right\}$. We compute
$$ \left\langle S A\begin{pmatrix} x \\ x \end{pmatrix}, A \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle
= \left\langle S \begin{pmatrix} x \\ 0 \end{pmatrix}, \begin{pmatrix} x \\ 0 \end{pmatrix} \right\rangle
= \left\langle \begin{pmatrix} x \\ x \end{pmatrix}, \begin{pmatrix} x \\ 0 \end{pmatrix} \right\rangle = \vert x \vert^2 \leq 4 \vert x \vert^2$$
$$= \left\langle \begin{pmatrix} 2x \\ 2x \end{pmatrix}, \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle
= \left\langle S \begin{pmatrix} x \\ x \end{pmatrix}, \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle.$$
Hence
$$\|A\|_S\leq 1.$$
| You basically did all the work. You just showed that $\forall x \in \mathbb{C}, \left < SA \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | A \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = |x|^2$ and $\left < S \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = 4|x|^2$. Now you have to find the smallest $c \in \mathbb{R_+^*}$ so that $|x| \leq 2c|x|$, which is just $\frac{1}{2}$.
| {
"language": "en",
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Solving an equation in $\mathbb N$ I am trying to solve the equation $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$
I have made the following progress:
1) $x, y z$ have to be larger than $1$
2) only one of x, y, z can be $2$; rest should be larger
3) WLOG I have assummed $2\leqslant x\leqslant y\leqslant z$. Knowing this, $x$ has to be smaller than $5$. How to find $y$ and $z$ for all cases of $x \in\{2, 3, 4, 5\}$ to be sure that I am finding all cases?
Thanks.
| This is not an exhaustive list. But it is at least a partial solution.
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$
We know that $\dfrac{1}{ab}=\dfrac{1}{a(a+b)} + \dfrac{1}{b(a+b)}$
Let $x=2$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{1}{10}.$
We can use $a=2$ and $b=5$ to get
$$\frac{3}{5} = \frac{1}{2}+\frac{1}{14}+\frac{1}{35}.$$
Let $x=3$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{4}{15}.$
We consider $\gcd(a,b)=1$. Then
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{4}{15}$.
It seems $a=1$, $b=3$, and $g=5$ will work. We get
$$\frac{3}{5} = \frac{1}{3}+\frac{1}{5}+\frac{1}{15}.$$
Let $x=4$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{7}{20}.$
We consider $\gcd(a,b)=1$. Then
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{7}{20}$.
No solution there. Next we try
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{14}{40}$.
Then $(a,b,g)=(4,10,1)$ gives us
$$\frac{3}{5} = \frac{1}{4}+\frac{1}{4}+\frac{1}{10}.$$
Let $x=5$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{2}{5}.$
We consider $\gcd(a,b)=1$. Then
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{2}{5}$.
Then $(a,b,g)=(1,1,5)$ gives us
$$\frac{3}{5} = \frac{1}{5}+\frac{1}{5}+\frac{1}{5}.$$
| {
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"answer_id": 0
} |
Partial Derivative of arctan Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$
Find $f_x(x,y)$
My attempt,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
But the given answer is $\frac{1}{x^2+1}$.
How?
| Alternatively, denote: $x=\tan a; y=\tan b$. Then:
$$z=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b}=a+b.$$
So:
$$z_x=a_x=(\arctan x)'=\frac{1}{1+x^2}.$$
| {
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"answer_count": 3,
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} |
Minimum value of $p^2+q^2+r^2+s^2$.
If $p,q,r,s>0$ and $(p+q)(q+r)(r+s)(s+p)=16$ and $pq+qr+rs+sp=2$.
Then minimum value of $(p^2+q^2+r^2+s^2)$
Try: using Cauchy Schwarz Inequality
$(p^2+q^2+r^2+s^2)(q^2+r^2+s^2+p^2)\geq (pq+qr+rs+sp)^2=4$
But answer given as $6$. Could some help me to solve it. Thanks
| Write $p^2 + q^2 + r^2 + s^2 = X$. Then, $$2X + 2(pq+qr+rs+sp) = (p+q)^2 + (q+r)^2 + (r+s)^2 + (s+p)^2 = 2X + 4$$.
Write $a = p+q, b = q+r, c= r+s, d = s+p$, then by AM-GM on $a^2,b^2,c^2,d^2$ we see that $a^2+b^2+c^2+d^2 \geq 4 \sqrt[4]{(abcd)^2} = 16$.
Therefore, $2X+4 \geq 16$, and $X \geq 6$.
Equality is attained, precisely when all terms in the AM-GM are equal i.e. $p=r,q=s$, in which case $p+q = 2$ and $pq = \frac 12$. Solve to get $p,q = 1 \pm \frac 1{\sqrt 2}$, at which indeed $X= 6$ is obtained. Hence, the minimum value is $X = 6$.
| {
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Sum of the first integer powers of $n$ up to k Pascal's triangle has a lot of interesting patterns in it; one of which is the triangular numbers and their extensions. Mathematically:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}$$
At first, we could guess that the next summation is:
$$\sum_{n=1}^kn^3 ?=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}$$
Yet this is off. However, it is off geometrically. Notice:
$$\left(\sum_{n=1}^kn^3\right)-\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}=error$$
$$k=1, r=0$$
$$k=2, r=0.25$$
$$k=3, r=1$$
$$k=4, r=2.5$$
$$k=5, r=5$$
$$k=6, r=8.75$$
...
Consider the ratios of the errors:
$$er(k)=\frac{r(k+1)}{r(k)}$$
$$k=1, r=udf$$
$$k=2, r=4$$
$$k=3, r=2.5$$
$$k=4, r=2$$
$$k=5, r=1.75$$
$$k=6, r=1.6$$
Then, rewriting the error as a function of n starting at k = 5:
$$1.75=2.5-\frac{1.5}{2}$$
$$1.6=2.5-\frac{1.5}{2}-\frac{1.5}{10}$$
$$1.5=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}$$
$$1.42857=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}-\frac{1.5}{21}$$
The denominators in the series are from pascals triangle: (3rd columns, or dependent again on the triangular numbers)
Then the total formula for equating the two is:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\right)-\left(\sum_{n=1}^kn^3\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)=0$$
Super interesting!
At least, I thought it was interesting how this the error is related back to the previous power's formula. Am I missing something obvious? Any input is greatly appreciated. (I'm not smart, so in the likely event I missed something obvious try not to be too harsh)
Update:
For the next power (4), I found the formula with trial and error:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\cdot\frac{\left(4k+1\right)}{5}\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)+\frac{1}{12}\left(k-1\right)k\left(k+1\right)k$$
Any ideas on power (5) and so on? I'll continue to try and generalize it.
| We are trying to find the correction polynomial of degree p $$a(k,p)$$ where
$$
a(k,p)=\sum_{n=1}^k(n^p)-\frac{k}{(p+1)!}\prod_{n=1}^{p}(nk+1)=\sum_{f=0}^p(a_fx^f)
$$
The correction polynomial was of degree three for the sum of cubes and degree four for the sum of tesseracts. Assuming that the polynomial degree is always p then, since we know what the polynomial's value should be at infinitely many points, we can turn the problem of finding the polynomial's coefficients into a p by p linear system. Maybe try to get some computer algebra system to solve the system and see if there is any obvious pattern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713657",
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"source": "stackexchange",
"question_score": "14",
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} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
| Observe we have
\begin{align}
I=\int^{\pi/2}_0 \frac{x^2}{\sin x}\ dx = \int^{\pi/2}_0 \frac{x^2}{\cos\left(\frac{\pi}{2}-x \right)}\ dx = \int^{\pi/2}_0 \frac{(\frac{\pi}{2}-u)^2}{\cos u}\ du.
\end{align}
Then using integration by parts, we see that
\begin{align}
I&=\left(\frac{\pi}{2}-u\right)^2\left\{\log\left|1 + \sin u\right|-\log|\cos u|\right\}\bigg|^{\pi/2}_0 + 2\int^{\pi/2}_0\left(\frac{\pi}{2}-u \right)\log|\sec u + \tan u|\ du\\
&= 2\pi \left(\frac{1}{2}\int_{0}^{\pi/2}\log|\sec u+\tan u|\ du \right)-\frac{7}{2}\left(\frac{4}{7}\int^{\pi/2}_0 u \log|\sec u+\tan u|\ du \right)\\
&= 2\pi G - \frac{7}{2}\zeta(3).
\end{align}
Here, I have used the facts that
\begin{align}
G= \frac{1}{2}\int_{0}^{\pi/2}\log|\sec u+\tan u|\ du
\end{align}
and
\begin{align}
\zeta(3) = \frac{4}{7}\int^{\pi/2}_0 u \log|\sec u+\tan u|\ du.
\end{align}
See here for reference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
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"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 5
} |
On the least prime in an arithmetic progression $a + nb$ where $a,b$ are distinct primes. Dirichlet's Theorem: there are infinitely many primes in every arithmetic progression $\{a + nd: n \geq 0\}$ for coprime $a, d$. Consider just the case where $a, d$ can take on prime values or $0$.
Then define $a \oplus a = 0$, $a \oplus 0 = 0 \oplus a = a$ and $a \oplus b$ for $a \neq b$ two prime numbers to be the smallest number in the above progression ($a + nb$) greater than $a$ that is a prime.
For instance $3 \oplus 5 = 3 + 2\cdot 5 = 13$
By Dirichlet's theorem there are infinitely many primes, so choose the smallest one greater than $a$!
Associativity:
$$
(3 \oplus 5) \oplus 7 = 13 \oplus 7 = 13 + 4 \cdot 7 = 41 \\
3 \oplus (5 \oplus 7) = 3 \oplus 19 = 41
$$
Does associativity hold at least when $a, b, c$ are distinct?
| Take $7\oplus 11=29$, so that $(7\oplus 11)\oplus 13=29\oplus 13=107$.
On the other hand we have $11\oplus 13=37$, and $7\oplus(11\oplus 13)=7\oplus 37=229$. So
$$
(7\oplus 11)\oplus 13=107\neq 229=7\oplus(11\oplus 13)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$ Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$
i am trying to solving integrating combination
since given that
$\cot (x^2+y^2)dy+xdx+ydy=0$
then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem
| Using polar coordinates, we see that
\begin{align}
dx =&\ \cos\theta dr -r\sin\theta d\theta,\\
dy =&\ \sin\theta dr + r\cos\theta d\theta
\end{align}
then we see that
\begin{align}
\cot(x^2+y^2)dy +xdx+ydy = (\cot(r^2)\sin\theta+r) dr+r\cot(r^2)\cos\theta d\theta =0.
\end{align}
Hence it follows
\begin{align}
\frac{d\theta}{dr} = - \frac{\cot(r^2)\sin\theta+ r}{r\cot(r^2)\cos\theta} = -\frac{\tan\theta}{r}-\tan(r^2)\sec\theta.
\end{align}
Lastly, using the fact that
\begin{align}
\frac{d (\sin\theta)}{dr} = \cos\theta\frac{d\theta}{dr} = -\frac{\sin\theta}{r}-\tan(r^2)
\end{align}
then we have obtain a linear equation
\begin{align}
u'+\frac{1}{r}u=-\tan(r^2) \ \ &\implies\ \ \frac{d}{dr}(ru) = -r\tan(r^2)\\
&\implies\ u= \frac{1}{2r}\log|\cos(r^2)|+\frac{c}{r}.
\end{align}
Finally, we have that
\begin{align}
\sin \theta =\frac{1}{2r}\log|\cos(r^2)|+\frac{c}{r} \ \implies \ y = \frac{1}{2}\log|\cos(x^2+y^2)|+C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How do I calculate the sum of $\sum_{n=0}^{\infty}{x^{n}\over n!}\cdot{{x^2\over n+2}}?$ How do I calculate the sum of $$\sum_{n=0}^{\infty}{x^{n+2}\over (n+2)n!}=\sum_{n=0}^{\infty}\color{red}{x^{n}\over n!}\cdot{{x^2\over n+2}}?\tag1$$
$$\sum_{n=0}^{\infty}{x^{n}\over n!}=e^x\tag2$$
| Hint:
$$\dfrac{x^{n+2}}{(n+2) n!}=\dfrac{(n+1)x^{n+2}}{(n+2)!}=\dfrac{(n+2-1)x^{n+2}}{(n+2)!}$$
$$=x\cdot\dfrac{x^{n+1}}{(n+1)!}-\dfrac{x^{n+2}}{(n+2)!}$$
$$\implies\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2) n!}=x\sum_{n=0}^\infty\dfrac{x^{n+1}}{(n+1)!}-\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)!}$$
Now use $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the minimum value of $\frac{a+b}{2} + \frac{2}{ab-b^{2}}$, with $a,b \in \mathbb{R}$, $a>b>0$ Find the minimum value of $$ \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
where $a,b \in \mathbb{R}$, $a>b>0$.
Attempt : The only method I knew was using partial derivatives.
Let
$$f(a,b) = \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
then the partial derivatives are
$$ f_{a} = \frac{1}{2} - \frac{2b}{(ab-b^{2})^{2}},\\
f_{b} = \frac{1}{2} - \frac{2(a-2b)}{(ab-b^{2})^{2}}.$$
Setting them to $0$ implies
$$ \frac{2b}{(ab-b^{2})^{2}} = \frac{2(a-2b)}{(ab-b^{2})^{2}},$$
so
$$ 3b = a $$
with $ ab - b^{2} \ne 0$.
Subtitute this to $f(a,b)$:
$$ f(b) = 2b + \frac{1}{b^{2}} \implies f'(b) = \frac{2b^{3} - 2}{b^{3}}.$$
so we must have $b^{3} = 1 \implies b = 1$, for $f'(b)=0$. So this means $a=3$. Then
$$f(a,b) = 3.$$
How to check whether this is valid minimum global of the expression, preferably without testing $D(a,b) = f_{aa} f_{bb} - (f_{ab})^{2}$?
How to solve another way without multivariable calculus? One way perhaps is by letting $a = kb, k > 1$.
| Denote $c = a - b > 0$, then$$
\frac{a + b}{2} + \frac{2}{ab - b^2} = b + \frac{c}{2} + \frac{2}{bc} \geqslant 3 \sqrt[3]{b · \frac{c}{2} · \frac{2}{bc}} = 3.
$$
The equality is achieved at $b = 1$, $c = 2$, i.e. $(a, b) = (3, 1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:
$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$
holds
I don't see room to use derivatives, since it seems a little messy to calculate the $\lim_{x\to 0}$ of $\frac{2\ln(\cos{x})}{x^2}$
(which, I think, is necessary in order to make usage of derivatives).
Any hints? I've already tried cross multiplying but it doesn't lead anywhere, unless I missed something.
| The Maclaurin series of the tangent function has all coefficients positive (see formula). Integrating we get that the Maclaurin series of the function $-\log(\cos x)$ has all coefficients positive. Up to order $12$ it is
$$-\log (\cos x) =\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31 x^{10}}{14175} + \frac{691 x^{12}}{935550} +
\mathcal{O}(x^{14})$$
So the inequality should probably read
$$-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) >0$$ or
$$ \frac{2 \log(\cos(x))}{x^2}<-1 - \frac{x^2}{6}$$
ADDED:
We can prove the inequality $-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) >0$ by taking the derivatives of the function $-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) $ up to order $5$. One can check easily that they all take value $0$ at $0$. Then one calculates:
$$\left(-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) \right)^{(5)}=4(5-\cos 2x)\sec^4 x \tan x>0$$ for $0<x<\pi/2$. From here one obtains the inequality . Note that we also get $\tan x>x + \frac{x^3}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
| HINT
We have
$$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x-\frac{9}{4}\right)^2-\frac{81}{8}+9=2\left(x-\frac{9}{4}\right)^2-\frac98$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Calculus max rectangle Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $x^2+y^2=a^2$ , and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $z=x^n+y^n$ . If $n=2$ , it is clear that $z$ has the constant value $a^2$ for all rectangles. If $n<2$, show that the square maximizes $z$, and if $n>2$, show that $z$ is maximized by a degenerate rectangle in which x or y is zero.
My approach was;
i took 1st and 2nd derivatives
1st: $nx^{n-1}+ny^{n-1}y'$
2nd: $[n(n-1)x^{n-2}+n(n-1)y^{n-2}y'+y''ny^{n-1}]$
And tried to put $\sqrt{a^2-x^2}$ to the $y$ then took the derivatives and that doesn't go anywhere too.
| By Lagrange multipliers: Let $g(x,y) = x^2+y^2$ and $f(x,y) = x^n+y^n$. First assume $x$ and $y$ are positive. The constraint is $g = a^2$. Since the case $n=2$ is done, assume $n \neq 2.$
We have $\nabla f = (nx^{n-1},ny^{n-1})$ and $\nabla g = (2x,2y).$ So the equation $\nabla f = \lambda \nabla g$ gives us
$$nx^{n-1} = \lambda 2x, \mbox{ and } ny^{n-1} = \lambda 2y.$$
Cancel an $x$ from the first equation and a $y$ from the second:
$$nx^{n-2} = \lambda 2, \mbox{ and } ny^{n-2} = \lambda 2.$$
Conclude that $x= y.$ Then $2x^2 = a^2$ and so $x= y = a/\sqrt{2}.$
So the critical points are $(a/\sqrt{2},a/\sqrt{2})$ and the points where either $x$ or $y$ equal zero. Then
$$f(a/\sqrt{2},a/\sqrt{2}) = 2\left(\frac{a}{\sqrt{2}}\right)^n$$
and
$f(0,y) = y^n$ which is maximized when $y=a$. Similarly for $f(x,0)$. The max will be which ever value is greater.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the last three digits of $19^{100}$ Find the last three digits of $19^{100}$
$19^{100}=361^{50}=(1+360)^{50}=\binom{50}{0}+\binom{50}{1}360+\binom{50}{2}360^2+\cdots$
When we divide it by $1000$, the remainder comes out to be $001$, so the last three digits must be $001$, but in my book, the answer is given as $801$. I dont know where I am wrong. Please help.
| Alternatively:
$$19^{100}=(20-1)^{100}=\underbrace{20^{100}-100\cdot 20^{99}+\cdots -\begin{pmatrix} 100\\ 3\end{pmatrix}\cdot 20^3}_{=1000A}+\begin{pmatrix} 100\\ 2\end{pmatrix}\cdot 20^2-\begin{pmatrix} 100 \\ 1\end{pmatrix}\cdot 20+\begin{pmatrix} 100 \\ 0\end{pmatrix}=$$
$$1000A+1978000+1.$$
Hence:
$$19^{100} \equiv 1 \pmod{1000}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Help with $\int \frac{x-\sqrt x}{x+\sqrt x} \, \mathrm d x$ I need help with this integral. I tried $$\int \frac{x-\sqrt x}{x+\sqrt x}\,\mathrm dx = \int \frac{(x-\sqrt x)^2}{x^2-x} \,\mathrm dx = \int \frac {x(x-2\sqrt x+1)}{x(x-1)}\,\mathrm dx = \int \frac {(x-2\sqrt x+1)}{(x-1)}\,\mathrm dx$$ but I don't know how to proceed? Thank you for your help.
| Before substitution:
$$\int \frac{x-\sqrt x}{x+\sqrt x} dx= \int \frac{\sqrt x -1}{\sqrt x+1} dx = \int \frac{\sqrt x +1-2}{\sqrt x+1} dx = \int \left(1- \frac{2}{\sqrt x+1} \right)dx$$ $$= x-2 \int \frac{1}{\sqrt x+1}dx$$
Then $x = (u-1)^2$:
$$\int \frac{1}{\sqrt x+1}dx = \int \frac{2(u-1)}{u} du = 2\int \left( 1- \frac{1}{u} \right) du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
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