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What is the equation of the circle having a radius of square root of 85, through (5,9) and (1,-7)? In solving this problem, I first solved the slope of the perpendicular bisector and I solved the midpoint of the two points. I don't know what to do next. I hope you can help me.
| Let center to be $(a,b)$. $$(a-5)^2+(b-9)^2=85, (a-1)^2+(b+7)^2=85$$
$$a^2-10a+b^2-18b=-21,a^2-2a+b^2+14b=35$$
Subtracting gives $$8a+32b=56\Rightarrow a+4b=7\Rightarrow a=7-4b$$
$$35=a^2-2a+b^2+14b=49-56b+16b^2-14+8b+b^2+14b=35+17b^2-34b\Rightarrow b^2-2b=0$$
$$b=0\Rightarrow a=7-4b=7; b=2 \Rightarrow a=7-4b=-1$$
$$(x-7)^2+y^2=85 \mathrm{\ or\ } (x+1)^2+(y-2)^2=85$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\forall a\ge 2, \exists n\ge 1$ such that $2n+1\mid a+n!$ Let $a\ge 2$ be an integer. Prove that $\exists n\ge 1$ such that $$2n+1\mid a+n! \text{ or }a-n!$$
I think we'll construct a solution for $n$ as a function of $a$. But I have no idea what that function looks like. I just took care of some trivial cases.
First if $a$ is a factorial number i.e. $a=k!$ then set $n=k$
If $a$ is of the form $k!+2k+1$ then set $n=k$ and observe that $$a-n!=2k+1$$
Note that if $a\equiv 1$ or $2$ $\pmod 3$ then set $n=1$,so we only have to deal with multiples of $3$ now .
| Note that if $2n+1$ is prime, then by Wilson's theorem:
$$-1 \equiv (2n)!\equiv (-1)^{n}(n!)^2 \mod (2n+1)$$
Thus $(n!)^2 \equiv (-1)^{n+1} \mod (2n+1)$
Thus if $2n+1$ is a prime of the form $1\text{mod}$ $ 4$ then $(n)!^2 \equiv -1 \mod(2n+1)$ (also some thanks to John Omielan).
Claim: For all $a \geq 2$, $a^2+1$ contains primes of the form $1 \mod 4$.
Proof:
By quadratic reciprocity, it contains no primes of the form $3\mod 4$. Hence, if it contains no primes of the form $1 \mod 4$ then it is a power of $2$. Thus we must have $a^2+1 = 2^{j}$ with $j \geq 2$ for some $a$. But $a^2+1 \equiv 1 \text{ or }2 \mod 4$, which is impossible.
Let $p_{a}$ be a prime $1 \mod 4$ that divides $a^2+1$. Note that $((\frac{p_{a}-1}{2})!)^2 \equiv -1 \mod p_{a}$ and, therefore;
$$p_{a}| a+\left(\frac{p_{a}-1}{2}\right)! \text{ or }a-\left(\frac{p_{a}-1}{2}\right)! $$
$$\Leftrightarrow \left(a+\left(\frac{p_{a}-1}{2}\right)!\right)\left(a-\left(\frac{p_{a}-1}{2}\right)!\right) \equiv 0 \mod p_{a}$$
$$\Leftrightarrow a^2+1 \equiv 0 \mod p_{a}.$$
The last statement is true. Thus $n = \frac{p_{a}-1}{2}$ does the trick.
| {
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A beautiful problem on the Pigeonhole Principle I recently solved this beautiful mathematical problem on Pigeonhole Principle from Romania TST 2000.
Let $S$ be the set of interior points of a sphere in three-dimensional space and let $D$ be the set of interior points of a disc on a plane. For any two points $X,$ and $Y$ in the space, let $d(X, Y)$ denote the distance between them. Determine if there is a function $f: S\rightarrow D$ such that
$d(A, B)\leq d(f(A), f(B))$ for any two points $A, B\in S.$
What are you initial thoughts on this problem? Do you think there is such a function or there isn't. I have given my solution to this problem below. You can also post a solution if you have another one :)
| We claim that such a function does not exist. To prove this claim, we will assume that there exists such a function $f$, and arrive at a contradiction.
Suppose that there exists such a function $f: S\rightarrow D,$ then we will consider a cube of side length $a$ that is completely in the sphere.
Fix a random positive integer $n$, we will partition the cube into exactly $n^3$ congruent cubes, such that these identical cubes determine $n^2$ identical squares on each face of the larger cube, and $n$ identical line segments on each edge of the larger cube.
Now, we consider the $(n+1)^3$ points that are the vertices of the $n^3$ identical cubes. Let these points be $S_1, S_2, \cdots, S_{n^3},$ we know that the distance between any two of these points is at least $\frac{a}{n}.$
Let $D_1, D_2, \cdots, D_{n^3}$ be the points on the disk $D,$ that the function $f$ maps $S_1, S_2, \cdots,$ and $S_{n^3}$ to, more specifically, let
$$D_i=f(S_i)$$
for all $i\in{1,2,\cdots,n^3}.$
Since our function $f$ has the property that, for any two points $A, B\in S,$ $d(A, B)\leq d(f(A), f(B))$ holds. Hence, the pairwise distances between the any of the $D_i$ are at least $\frac{a}{n}.$
This means that we can draw circles around each of those points of radii $\frac{a}{2n}$ such that none of the $(n+1)^3$ circles intersect. Let the radius of the disk $D$ be $r,$ we know that each of these circles is contained in the disk concentric with $D$ of a radius of $r+\frac{a}{2n}.$
All these $(n+1)^3$ circles are non-intersecting, and contained in the larger disk. the sum of their areas is lesser than the area of the larger disk. Hence, for all $n\in\mathbb{N},$ we have
\begin{align*}
& \pi\left(\frac{a}{n}\right)^2(n+1)^3\leq\pi\left(r+\frac{a}{n}\right)^2\\
\Rightarrow & \left(\frac{a}{n}\right)^2(n+1)^3\leq\left(r+\frac{a}{n}\right)^2\\
\Rightarrow & \left(\frac{a^2}{n^2}\right)(n+1)^2(n+1)\leq\left(r+\frac{a}{n}\right)^2\\
\Rightarrow & \left(\frac{(n+1)^2}{n^2}\right)(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2\\
\Rightarrow & \left(\frac{n+1}{n}\right)^2(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2\\
\Rightarrow & \left(1+\frac{1}{n}\right)^2(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2.
\end{align*}
Now we note that as $n$ tends to infinity, the term $\frac{1}{n}$ will tend to $0,$ and hence,
\begin{align*}
1+\frac{1}{n} &\rightarrow 1\\
\left(1+\frac{1}{n}\right)^2 &\rightarrow 1\\
\left(1+\frac{1}{n}\right)^2(n+1)a^2 &\rightarrow\infty.
\end{align*}
since $n+1\rightarrow\infty.$ Hence, the left side of the inequality tends to infinity, implying that since
$$\Rightarrow\left(1+\frac{1}{n}\right)^2(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2,$$
the right side will also tend to infinity. But note that as $n\rightarrow\infty,$
\begin{align*}
\frac{a}{n} &\rightarrow 0\\
\Rightarrow r+\frac{a}{n} &\rightarrow r\\
\Rightarrow \left(r+\frac{a}{n}\right)^2 &\rightarrow r^2.
\end{align*}
which is finite, hence, we have arrived a contradiction, and such a function $f$ does not exist.
| {
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Eigenvalue and spectral condition
Let $A=\begin{pmatrix} 1& 1 \\ a^2 &1 \end{pmatrix} \text{ with } a\in (0,\frac{1}{2}]$. Show $$cond_2(A)=||A||_2 \cdot ||A^{-1}||_2\leq 4(1-a^2)^{-1}$$ by first showing $||A||^2_2\leq||A||_1||A||_{\infty}$.
$||A||^2_2$ is the maximal eigenvalue of $A^TA$ and $||A||_1||A||_{\infty}=2\cdot2 = 4$. Prove $||A||^2_2\leq||A||_1||A||_{\infty}=4$ by contradiction: Suppose $\lambda_{max} >4$, then $$4<\lambda_{\ast}=\frac{\sqrt{a^8+2a^4+4a^2+1}+a^4+3}{2} \iff 5<\sqrt{a^8+\underbrace{2a^4}_{\leq \frac{1}{2}}+\underbrace{4a^2}_{\leq 1}+1}+\underbrace{a^4}_{\leq 1} \\ \leq \sqrt{4}+1 = 3$$ contradiction!
So I know $||A||^2_2\leq 4 \Rightarrow ||A||_2\leq 2$ but I need to show $cond_2(A) = \underbrace{||A||_2}_{\leq 2}||A^{-1}||_2\leq 4(1-a^2)^{-1}$. I don't know why I needed to show $||A||^2_2\leq||A||_1||A||_{\infty}$ first? How does it help? I still need the maximal eigenvalue of $(A^{-1})^TA^{-1}$. The eigenvalues are $$\lambda_1=\frac{\sqrt{a^4-2a^2+5}-a^2+1}{2\sqrt{a^4-2a^2+5}-4} \\ \lambda_2=\frac{\sqrt{a^4-2a^2+5}-a^2-1}{2\sqrt{a^4-2a^2+5}+4}$$ What am I supposed to do now? Thanks for any help!
| If $\lambda_1,..., \lambda_n$ are the eigenvalues of an invertible $n\times n$ matrix $M$, then $\frac{1}{\lambda_1},\dots,\frac{1}{\lambda_n}$ are the eigenvalues of $M^{-1}$. Moreover, the eigenvalues of $M^T$ are the same as the eigenvalues of $M$.
Putting this together, we have that the eigenvalues of $(A^{-1})^{T} A^{-1}$ are the reciprocals of the eigenvalues of $((A^{-1})^{T} A^{-1})^{-1} = AA^T$, which in turn are the same as the eigenvalues of $(AA^T)^T = A^T A$. Moreover, since the eigenvalues of $A^T A$ are necessarily non-negative, we get that $\|A^{-1}\|_2^2 = \frac1{\lambda_{\min}(A^T A)}$. Now, the determinant of a matrix is just the product of its eigenvalues, so from a direct computation:
$$(1 - a^2)^2 = \det(A^T A) = \lambda_{\max}(A^T A) \cdot\lambda_{\min}(A^T A) = \frac{\|A\|_2^2}{\|A^{-1}\|_2^2}.$$
Hence $\|A^{-1}\|_2 = \frac{\|A\|_2}{1 - a^2}$, and so $\text{cond}_2(A) = \frac{\|A\|_2^2}{1 - a^2} \le \frac{4}{1 - a^2}$.
| {
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How can $x=1$ be a solution to this equation? This is the equation I am supposed to solve:
$$\log_{3}(2x^{2} - x-1)\ -\ \log_{3}(\ x-1)\ = 2$$
The textbook gives the solutions, $x=1, x=4$, with the working out as shown below:
While I understand what they are doing, I don't completely understand why they work it out as they do. Specifically, why isn't the term on the left side in the second row simplified? In my working out, $$\log_{3}\left(\frac{2x^{2} - x-1}{x-1}\right) \implies \log_{3}\left(\frac{(2x+1)(x-1)}{x-1}\right) \implies \log_{3}(2x+1)$$
This equals $2$, allowing it to be simplified further,
\begin{align*}
\: & \log_{3}(2x+1) = 2 \\
\implies \: & 2x+1 = 3^2 = 9 \\
\implies \: & 2x = 8 \\
\implies \: & x = 4 \\
\end{align*}
What's more, $x$ can't even equal $1$, because if it does, both logs simplify to $log(0)$, which is undefined. I am wondering if there is some other concept I am not aware of, or is the book making a mistake?
| $x=1$ cannot be a solution because the equation is undefined for $x=1$. The equation in reals is only meaningful if $x-1 > 0$ and $2x^2-x-1>0$, which means $x>1$. In that case one may rewrite the equation as follows:
$$\begin{align}
\log_3(2x^2-x-1) - \log_3(x-1)
&= \log_3\big((2x+1)(x-1)\big) - \log_3(x-1) \\
&= \log_3(2x+1) + \log_3(x-1) - \log_3(x-1) \\
&= \log_3(2x+1) \\
&\stackrel!= 2
\end{align}$$
and thus $2x+1=9$ which has only one solution $x=4$.
As you see, the culprit is not a division by $x-1$ as clained in the comments, because it's still invalid if no division is present like above.
| {
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Is $\tan^{-1}(0)=\pi$ or is $\tan^{-1}(0)=0$? This question came in the Dhaka University admission exam 2006-7
Q) The value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$ is -
(a) $0$
(b) $\frac{\pi}{2}$
(c) $\pi$
(d) $2\pi$
My attempt:
$$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$$
$$=\tan^{-1}\frac{1+2+3-1\cdot2\cdot3}{1-1\cdot2-2\cdot3-1\cdot3}$$
$$=\tan^{-1}\frac{0}{-10}$$
$$=\tan^{-1}0$$
Now, as no range is mentioned, I should pick the angle that is within the principal range of $\tan^{-1}$: (a). (c) and (d) are also acceptable solutions, but they don't fall within the principal range. So, I'll go with (a).
However, my question bank says that $\tan^{-1}0=\pi$, so the answer is (b). I don't understand their reasoning.
Which is the correct option?
| Let's find the value of $\arctan 2 + \arctan 3$ first...
Using the formula $$\arctan x + \arctan y = \arctan \left(\dfrac {x+y}{1-xy}\right)$$ we have $$\arctan 2 + \arctan 3 = \arctan \left(\dfrac {2+3}{1-2 \cdot 3}\right) \\ = \arctan \dfrac {5}{-5} \\ = \arctan (-1) \\ = \dfrac {3 \pi}{4}$$ but since $\arctan 1 = \dfrac {\pi}{4}$, $$\arctan 1 + \arctan 2 + \arctan 3 = \dfrac {\pi}{4} + \dfrac {3 \pi}{4} = \pi$$ which is answer $(c)$.
| {
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How do you finish the equations for $S^2 + \frac{7}{4}S= \frac{1}{2}$? Complete the Square:
$$S^2 + \frac{7}{4}S = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S \_\_\_\_\_\_\_ = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S + \frac{49}{64} = \frac{81}{64}$$
THIS is where I get stuck. I know that the Square Root of $\frac{81}{64}$ is $\frac{9}{8}$, but how do I find the Square Root of $S^2 + \frac{7}{4}S + \frac{49}{64}$?
Quadratic Equation:
$$A=1,\ \ B=\frac{7}{4},\ \ C= \frac{-1}{2}$$
$$\frac{\frac{-7}{4} \pm
\sqrt{\frac{49}{16}} + 2}{2}$$
THIS is where I get stuck because I don't know what to do with the 2 in the discriminant.
| When you complete the square, assuming the coefficient on the quadratic term is $1$, you take the coefficient on the linear term and divide by two and then square it. That gives you two numbers,
$b/2$ and $b^2/4$. Don't throw away the $b/2$. See where it comes in:
$$x^2+bx = c$$
$$x^2+bx+\frac{b^2}{4} = c+\frac{b^2}{4}$$
$$\left(x+\frac{b}{2}\right)^2 = c+\frac{b^2}{4}.$$
So you took $7/4$, divided by two to get $7/8$. Keep track of the $7/8$. Then squared it to get $49/64$. So the left side factors as $(s+7/8)^2$. If you multiply it out, you should see why this is always true.
| {
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Range of $f(x) = x+ \frac {1}{x}$ for $x<0$ One method to do this is by re-writing the function:
$$ x + \frac{1}{x} = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 $$
Then adding and subtracting 2 from RHS:
$$ = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 - 2 + 2 $$
$$=(\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 - 2 \cdot (\sqrt{x}) \cdot \left(\frac{1}{\sqrt{x}}\right) + 2 $$
$$= \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 + 2 $$
Now, since $$\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \geq 0 $$
Then,
$$ \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 + 2 \geq 2$$
Now, $\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 + 2$ is equal to our original function, $f(x)$. So, we can say that $f(x) \geq 2 $, but only for $ x > 0 $, since for $x<0$, we will have square root of negative numbers.
My teacher said that using the fact that $$f(x) \geq 2; \space \forall \space x>0 $$ we can say that for any $x<0$:
$$ f(x) = -f(\vert{x}\vert) \qquad (1)$$
And then we can say that for negative values of $x$: $$f(x) \leq -2 \qquad (2)$$
This is what I don't understand. How can we go from (1) to (2)?
From my understanding, (1) simply means that $f(x)$ is an odd function, which can be easily shown. How we go from that to (2) is the part that is confusing me... Seems like I'm missing something basic, any help is appreciated!
| Since for if $x \neq 0, |x| > 0 \implies f(|x|) \ge 2 \implies f(x) = -f(|x|) \le -2$. This means $(1) \implies (2)$. Alternatively, set $x = -x' \implies x' > 0\implies f(x) = x+\dfrac{1}{x} = -\left(x'+\dfrac{1}{x'}\right)= -\left(\sqrt{x'}-\dfrac{1}{\sqrt{x'}}\right)^2-2\le -2$.
| {
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Solving the integral of $\cos^2x\sin^2 x$ Solving the integral of $\cos^2x\sin^2 x$:
My steps are: $(\cos x\sin x)^2=\left(\frac{\sin(2x)}{2}\right)^2$. Now we know that
$$\sin^2(\alpha)=\frac{1-\cos (2x)}{2}\iff\left(\frac{\sin(2x)}{2}\right)^2=\frac 14\sin^2(2x)=\frac 14\cdot \frac{1-\cos (4x)}{2}=\frac 18(1-\cos (4x))$$
Rewriting all the steps:
$$\int \sin ^2\left(x\right)\cos ^2\left(x\right)dx=\frac{1}{8}\left(x-\frac{1}{4}\sin \left(4x\right)\right)+k, \,\, k\in \Bbb R \tag 1$$
Is there another method to solve this integral (1)?
| You can excpress $\cos^2 x$ as $1 - \sin^2 x$ and hence your primitive becomes:
$$\int \sin^2 x d x - \int \sin^4 x dx$$
which comes to the same answer in the end.
Full working:
$$\int \sin^2 a x \cos^2 a x \, \mathrm d x$$
$$=\int \sin^2 a x (1 - \sin^2 a x) \, \mathrm d x$$
$$=\int \sin^2 a x \, \mathrm d x - \int \sin^4 a x \, \mathrm d x$$
$$=\frac x 2 - \frac {\sin 2 a x} {4 a} - \int \sin^4 a x \, \mathrm d x + C$$
$$= \frac x 2 - \frac {\sin 2 a x} {4 a} - \left ({\frac {3 x} 8 - \frac {\sin 2 a x} {4 a} + \frac {\sin 4 a x} {32 a} }\right) + C$$
$$= \frac x 8 - \frac {\sin 4 a x} {32 a} + C$$
In the above:
$$\int \sin^4 a x \, \mathrm d x = \int \left( {\frac {3 - 4 \cos 2 a x + \cos 4 a x} 8}\right) \, \mathrm d x$$
using the usual power reduction formula.
| {
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Evaluating $\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$ So, my friend has challenged me to solve the following integral
$\displaystyle \tag*{} I=\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$
I started by doing $x^2=t$, so
$\displaystyle \tag{1} I=\frac 12 \int_0^\infty \frac{\arctan (x) \ln(1+x^2)}{x(x^4+x^2+1)} \mathrm dx$
Now, I first tried to solve this generalized integral:
$\displaystyle \tag*{} \mathcal L(m,n,a) = \int_0^\infty \frac{\arctan(mx)\ln(1+n^2x^2)}{x(a^2+x^2)} \ \mathrm dx$
We have $\mathcal L(0,1,a)=0$ and $\mathcal L(1,0,a)=0$. We now differentiate $\mathcal L(m,n,a)$ w.r.t $m$ first and then w.r.t $n$. We get:
$\displaystyle \tag{2} \begin{align} \partial m \ \partial n \ \mathcal L(m,n,a) &= \int_0^\infty \frac{2nx^2}{(a^2+x^2)(1+m^2x^2)(1+n^2x^2)} \ \mathrm dx \\\\
&= \frac{\pi n}{(m+n)(am+1)(an+1)} \\\\
&= \frac{\pi}{2(am+1)(an+1)} , \ \ \text{via symmetry} \end{align}$
And now taking the double integeral gives
$\displaystyle \tag{3} \mathcal L(1,1,a) = \int_0^1\int_0^1 \frac{\pi}{2(am+1)(an+1)} \ \mathrm dm \ \mathrm dn$
Using the elementary result:
$\displaystyle \tag*{} \int_0^1 \frac{1}{ay+1} \ \mathrm dy = \frac{\log(a+1)}{a}$
We find the value of $(3)$ as
$\displaystyle \tag*{} \mathcal L(1,1,a) = \frac{\pi \log^2(a+1)}{2a^2}$
Finally, we conclude that
$\displaystyle \tag*{}\int_0^\infty \frac{\arctan(x)\ln(1+x^2)}{x(a^2+x^2)} \ \mathrm dx={\color{Red} { \frac{\pi \log^2(a+1)}{2a^2}}}:= {\color{Blue}{ f(a)}}$
Now by completing square and performing, we have:
$\displaystyle \tag*{} \begin{align}\frac{1}{x^4+x^2+1} &=\frac{1}{\left(x^2+\frac 12\right)^2+\frac 34 }\\ &= \frac{1}{\left(x^2+\frac 12 - \frac{i \sqrt 3}{2}\right)\left(x^2+\frac 12 + \frac{i \sqrt 3}{2}\right)} \\ &= \frac{2}{i \sqrt 3} \left(\frac{1}{\left(x^2+\frac 12 - \frac{i \sqrt 3}{2}\right)}-\frac{1}{\left(x^2+\frac 12 + \frac{i \sqrt 3}{2}\right)}\right)\end{align} $
Adding all the pieces together, we get
$\displaystyle \tag{4} I=\frac{1}{i \sqrt 3} \left(f\left(\sqrt{\frac{1-i\sqrt3}{2}}\right) - f\left(\sqrt{\frac{1+i\sqrt3}{2}}\right)\right)$
My question is to simplify $(4)$ (if all my steps are correct) and also is there any short ways (methods others than contour integration) to destroy this integral?
Thanks :).
| Note that
$$\Im\left(-\frac{2}{\sqrt{3}\left(x^2+\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)} \right) = \frac{1}{x^4+x^2+1}$$
Hence
$$ I=\frac 12 \int_0^\infty \frac{\arctan (x) \ln(1+x^2)}{x(x^4+x^2+1)} \mathrm dx = \Im \left( -\frac{1}{\sqrt{3}} \int_0^\infty \frac{\arctan (x) \ln(1+x^2)}{x\left(x^2+\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)} \mathrm dx \right) $$
which is just the imaginary part of
$$ -\frac{1}{\sqrt{3}}\int_0^\infty \frac{\arctan(x)\ln(1+x^2)}{x(a^2+x^2)} \ \mathrm dx={\color{Red} { -\frac{\pi \log^2(a+1)}{2\sqrt{3}a^2}}}$$
with $ \displaystyle a = \sqrt{\frac{1}{2}+\frac{i\sqrt{3}}{2}}$
Hence
\begin{align*} I =& \Im\left(-\frac{\pi \left(\frac{1}{2}-\frac{i\sqrt{3}}{2} \right)\log^2\left(1 + \sqrt{\frac{1}{2}+\frac{i\sqrt{3}}{2}}\right)}{2\sqrt{3}}\right) \\
=& \Im\left(-\frac{\pi \left(\frac{1}{2}-\frac{i\sqrt{3}}{2} \right)\left[\ln\left|1 + \sqrt{\frac{1}{2}+\frac{i\sqrt{3}}{2}}\right| + i\arg\left(1 + \sqrt{\frac{1}{2}+\frac{i\sqrt{3}}{2}}\right) \right]^2}{2\sqrt{3}}\right)\\
=& \Im\left(-\frac{\pi \left(\frac{1}{2}-\frac{i\sqrt{3}}{2} \right)\left[\frac{1}{2}\ln(2+\sqrt{3}) + i\frac{\pi}{12}\right]^2}{2\sqrt{3}}\right)\\
=& \Im\left(-\frac{\pi \left(\frac{1}{2}-\frac{i\sqrt{3}}{2} \right)\left[\frac{\log^2(2+\sqrt{3})}{4} + \frac{ \pi i\ln(2+\sqrt{3})}{12} - \frac{\pi^2}{144} \right]}{2\sqrt{3}}\right)\\
=& -\frac{\pi^2\ln(2+\sqrt{3})}{48\sqrt{3}}+ \frac{\pi\ln^2(2+\sqrt{3})}{16} -\frac{\pi^3}{576}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Determine area bounded by $(x+y)^4 = ax^2y$ Determine the area of region $U$ bounded by the graphic of the curve:
$$(x+y)^4 = ax^2y,\ a > 0 \quad \text{(loop in the first quadrant)}$$
I used polar coordinates and I arrived to:
$$\iint r\,dr\,d\theta = \frac{a^2}{2}\int_0^{\frac{\pi}{2}}
\frac{\cos^4 \theta \,\sin^2 \theta} {(\cos \theta + \sin \theta)^8}
\,d\theta$$
but I don't know how to compute this integral. Any hint for it?
| \begin{align}
&\int_0^{\pi/2}
\frac{\cos^4 \theta \,\sin^2 \theta} {(\cos \theta + \sin \theta)^8}
\,d\theta\\
=& \int_0^{\pi/2}
\frac{\tan^2 \theta \,\sec^2 \theta} {(1+ \tan \theta)^8}
\,d\theta
\overset{x=\tan\theta}=\int_0^{\infty}
\frac{x^2}{(1+x)^8}dx\\
=& \int_0^{\infty}
\left( \frac{1}{(1+x)^6}-\frac{2}{(1+x)^7}+ \frac{1}{(1+x)^8}\right) dx=\frac{1}{105}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove close form $\int_a^1 \frac{x}{\sqrt{x^2-a^2}}\ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}dx=\frac{\pi}{2}(1-a)$ I found this integral in a textbook (stated without proof),
$$J(a)=\int_a^1 \frac{x}{\sqrt{x^2-a^2}}\ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}dx=\frac{\pi}{2}(1-a)$$
with $0\le\,a\le\,1$.
I tried to confirm its validity.
For $a=1$,
$$J(1)=\int_1^1 ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}dx=0$$
For $a=-1$,
$$J(-1)=\int_{-1}^1 \frac{x}{\sqrt{x^2-1}}ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}dx\;=\pi
$$
Simplifying the expression in the logarithm,
$$J(-1)=\int_{-1}^1 \frac{x}{\sqrt{x^2-1}}ln\frac{1-\sqrt{1-x^2}}{x}dx$$
Applying integration by parts,
$$J(-1)=\int_{-1}^1 \frac{\sqrt{x^2\,-1}}{x\sqrt{1-x^2}}dx=i\int_{-1}^1\frac{1}{x}dx=iln(x)|_{-1}^1$$
$$J(-1)=iln(1)-iln(i)=i(-i\pi)=\pi$$
Could there be any way to prove the $J(a)$?
| Integrate by parts with $d\left( \sqrt{x^2-a^2}\right)= \frac{x}{\sqrt{x^2-a^2}}dx$
\begin{align}
&\int_a^1 \frac{x}{\sqrt{x^2-a^2}}\ln\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\ dx\\
=& \int_a^1\frac1x \sqrt{\frac{x^2-a^2}{1-x^2}}dx \overset{y^2=\frac{x^2-a^2}{1-x^2}}=\int_0^\infty
\frac{(1-a^2) y^2}{(y^2+a^2)(y^2+1)}dy=\frac\pi2(1-a)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What are the eigenvalues of $\begin{pmatrix} A & B \\ B &-A \end{pmatrix}$ in terms of $A$ and $B$? It is known that the set of eigenvalues of the following block matrix
$$ C = \begin{pmatrix} A & B \\ B & A \end{pmatrix} $$
is the union of the eigenvalues of the matrices $A + B$ and $A - B$. I am interested in the matrix of the following form
$$ C = \begin{pmatrix} A & B \\ B &-A \end{pmatrix} $$
Is there a description of the eigenvalues of $C$ in terms of $A$ and $B$?
Edit. If $AB=BA$, then we can do the following.
$$
C =
\begin{pmatrix}
A & B \\ B &-A \end{pmatrix}
\begin{pmatrix}
v \\ u \end{pmatrix} =\lambda \begin{pmatrix}
v \\ u \end{pmatrix}
$$
implies
$$
\left\{
\begin{array}{l}
Av+Bu=\lambda v \\
Bv-Au=\lambda u
\end{array}.
\right.
$$
By multiplying the first equation by $B$ and assuming $AB=BA$, we get
$$
(A^2+B^2)u=\lambda^2u.
$$
Therefore, $\lambda^2$ is an eigenvalue of $A^2+B^2$, what is discussed in the comments.
| Partial answer when $A$ has inverse $A^{-1}$. Use the formula for the determinant of block-partitioned matrices in terms of Schur's complement. If $(A-\lambda I)$ has an inverse then $\det(A-\lambda I)\neq 0$
\begin{align}
\det \left(\begin{bmatrix} A&B \\ B& -A \end{bmatrix}-\lambda\begin{bmatrix} I&0 \\ 0& I
\end{bmatrix} \right)
=&
\det \left(\begin{bmatrix} A-\lambda\cdot I&B \\ B& -A-\lambda\cdot I \end{bmatrix}\right)
\\
=&
\det(-A-\lambda I) \cdot \det\left(-(A-\lambda I) - B(A-\lambda I)^{-1} B\right)
\end{align}
Thus, the eigenvalues of $\begin{bmatrix} A&B\\ B&-A \end{bmatrix} $ would be between the eigenvalues of $-A$ and between the roots of the equation
$$
\det\left(-(A-\lambda I) - B(A-\lambda I)^{-1} B\right)=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
I need $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
$((2207^2 -2)^2 - 2)^2 -2)^2 ... - 2)$
quite big
But I realized I didn't have to simplify it, I just have to check if
$((((((2207^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2$ is divisible by 3
| Let $x$ be a root of $x^2 + 3x + 1$. Then $y = 1/x$ is also a root, since $$y^2 + 3y + 1 = (1/x)^2 + 3/x + 1 = \frac{1 + 3x + x^2}{x^2}$$ and $x \ne 0$. Moreover, for each positive integer $n$,
$$x^{n+1} + y^{n+1} = (x+y)(x^n + y^n) - xy(x^{n-1} + y^{n-1}).$$
Since $x + y = -3$ and $xy = 1$, it follows that
$$x^{n+1} + y^{n+1} = -3(x^n + y^n) - (x^{n-1} + y^{n-1}).$$ Taken modulo $3$, we see that if $f_n = x^n + y^n$,
$$f_{n+1} \equiv -f_{n-1} \pmod 3.$$
So for instance, with $f_0 = x^0 + y^0 = 2$, we have
$$f_2 \equiv -f_0 \pmod 3, \\ f_4 \equiv -f_2 \equiv f_0 \pmod 3, \\ f_6 \equiv -f_4 \equiv -f_0 \pmod 3, \\ \vdots \\f_{2048} \equiv (-1)^{2048/2} f_0 = 2 \pmod 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$ Find the integer solutions:
$$a+b+c=3d$$
$$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$
Attempt:
Notice that $a=b=c=d=1$ is a solution.
Other facts:
Notice that $a^{2} + b^{2} + c^{2} > 0$, so $4d^{2}-2d+1>0$. Notice that $4d^{2}-2d+1$ has negative discriminant: $D = -12$, and so it is always one sign, which is positive in this case. So we cannot narrow $d$-solution this way.
Next, since $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc)$, we get
$$ 9d^{2} - 2(ab+ac+bc) = 4d^{2}-2d+1 $$
$$ 5d^{2} + 2d -1 = 2(ab+ac+bc) $$
$5d^{2} + 2d - 1 = 0$ has solutions:
$$ d_{1,2} = \frac{-2 \pm \sqrt{24}}{10} = \frac{-1 \pm \sqrt{6}}{5}$$
and when $d > (-1 + \sqrt{6})/5$, OR, $d < (-1 - \sqrt{6})/5$ we have
$5d^{2}+2d-1 > 0$. And when $d$ is between the 2 roots we have $5d^{2}+2d-1<0$.
Then also nocitce that $(a+b+c)d= 3d^{2}$, and so
$$a^{2}+b^{2}+c^{2}-(a+b+c)d = (d-1)^{2}$$
Now if $d \ne 1$ we must have
$$ a^{2}+b^{2}+c^{2}-(a+b+c)d > 0$$
Another fact:
$$d = \frac{a+b+c}{3} \ge (abc)^{1/3} $$
by AM-GM.
Also notice $d < \max(a,b,c)$, because
$$a^{2}+b^{2}+c^{2} =d^{2} + d^{2} + d^{2} + (d-1)^{2}$$
| $a+b+c=3d, a^2+b^2+c^2=4d^2-2d+1$
Just set $(a, b, c)=(k_1d+l_1, k_2d+l_2, k_3d+l_3)$, since if we have quadratic polynomials for $a, b, c$, the 4th degree polynomials have to be in $a^2+b^2+c^2$.
$4d^2-2d+1=(k_1^2+k_2^2+k_3^2)d^2+2(k_1l_1+k_2l_2+k_3l_3)d+(l_1^2+l_2^2+l_3^2).$
$k_1^2+k_2^2+k_3^2=4, k_1l_1+k_2l_2+k_3l_3=-1, l_1^2+l_2^2+l_3^2=1.$
$k_1+k_2+k_3=3, l_1+l_2+l_3=0.$
By solving this, you can get:
$(k_1, k_2, k_3, l_1, l_2, l_3) \\=\Bigg(k_1, -\dfrac {\pm\sqrt{-3k_1^2+6k_1-1}+k_1-3} {2}, \dfrac {\pm\sqrt{-3k_1^2+6k_1-1}-k_1+3} {2}, 1-k_1, \dfrac {\pm\sqrt{-3k_1^2+6k_1-1}+k_1-1} {2}, -\dfrac {\pm\sqrt{-3k_1^2+6k_1-1}-k_1+1} {2}\Bigg)$.
So,$a=k_1d+(1-k_1) \\ b=-\dfrac {\pm\sqrt{-3k_1^2+6k_1-1}+k_1-3} {2}d+\dfrac {\pm \sqrt{-3k_1^2+6k_1-1}+k_1-1} {2} \\ c=\dfrac {\pm \sqrt{-3k_1^2+6k_1-1}-k_1+3} {2}d-\dfrac {\sqrt{-3k_1^2+6k_1-1}-k_1+1} {2}$.
For instance, $a=d, b=\dfrac {2-\sqrt{2}} {2}d+\dfrac {1} {\sqrt{2}}, c=\dfrac {2+\sqrt{2}}{2}d-\dfrac {1} {\sqrt{2}}.$
Since $a, b, c, d$ are all integers, we can't find the integers in this solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding $\lim _{n\to \infty}n\sin (2\pi \sqrt{1+n^2})$ Question:
Find $\lim \limits _{n\to \infty}n\sin \left (2\pi \sqrt{1+n^2}\right )$, where $n\in \mathbb{N}$.
Context:
This was a question given in my workbook, where it mentioned a hint that $n$ belonging to natural numbers is an important piece of information.
Approach:
I tried expanding the inner part using binomial expansion and then using Taylor series to expand the $\sin (\cdot )$ out, but I seem to keep hitting an indeterminate form. Help in understanding would be much appreciated!
| Note that $\sqrt{x^2+1}$ is very close to $x$. And $\sin$ has period $2\pi$.
Computations. As integer $x \to \infty$,
\begin{align}
\sqrt{x^2+1} &= x\sqrt{1+\frac{1}{x^2}}
= x\left(1+\frac{1}{2 x^2} + o(x^{-2})\right)
= x + \frac{1}{2 x} + o(x^{-1})
\\
\sin\left(2\pi\sqrt{x^2+1}\right) &=
\sin\left(2\pi x + \frac{\pi}{x} + o(x^{-1})\right) =
\sin\left(\frac{\pi}{x} + o(x^{-1})\right) = \frac{\pi}{x}+o(x^{-1})
\\
x\sin\left(2\pi\sqrt{x^2+1}\right) &= \pi + o(1)
\end{align}
So we conclude
$$
\lim_{x\to\infty\\x \in \mathbb N} x\sin\left(2\pi\sqrt{1+x^2}\right)
= \pi .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$, $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and I cant simplify the determinant in a good way with row operations.
I also tried solving $z$ from $g=1$. $z=1-x-y$.
$f(x,y,1-x-y)=xy-x^2y-xy^2$ restricted to $h(x,y)=2x^2+2y^2-2x-2y+2xy+1$ with Lagrange multiplier but I made no progress there either as the partial derivatives got too messy.
| $z = 1-x-y\implies f(x,y,z) = xyz = xy(1-x-y)=xy-xy(x+y)=g(x,y)$, subject to: $x^2+y^2+(1-x-y)^2 = 1$, or $2(x^2+y^2)-2x-2y+2xy=0$, or $x^2+y^2-x-y+xy=0$, or $(x+y)^2-xy-(x+y)=0\implies g(x,y) = xy(1-(x+y))=((x+y)^2-(x+y))(1-(x+y))=(x+y)^2-(x+y)^3-(x+y)+(x+y)^2 = -t^3+2t^2-t,t = x+y$. Observe that: $(x+y)^2 - (x+y) = xy \le \dfrac{(x+y)^2}{4}\implies t^2-t \le \dfrac{t^2}{4}\implies 3t^2-4t \le 0 \implies t(3t-4) \le 0\implies 0 \le t \le \dfrac{4}{3}$. Thus the problem boils down to finding the max and min of $h(t) = -t^3+2t^2-t, 0 \le t \le \dfrac{4}{3}$. You have: $h'(t) = -3t^2+4t-1 = 0\implies (-3t+1)(t-1) = 0\implies t = \dfrac{1}{3}, 1$. Evaluating $h$ at end points and at critical points: $h(0) = 0, h(\frac{4}{3})= -\dfrac{4}{27}, h(1) = 0, h(\frac{1}{3}) = -\dfrac{4}{27}$. This shows that the min value is $-\dfrac{4}{27}$, and the max value is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate the definite integral $I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}$ I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omissions), and (2) if there are any better/faster ways of evaluating this integral. Thanks!
Problem:
Evaluate the definite integral
$$I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}.$$
Attempted Solution:
We first note that the integrand is an even function, thus
$$I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)} = \frac{1}{2}\int_{-\pi}^{\pi} \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}. $$
Let $z=e^{i\theta} \implies d\theta = \frac{dz}{iz}$, then we have
$$\sin\theta =\frac{e^{i\theta} - e^{-i\theta}}{2i} =\frac{z-z^{-1}}{2i} \implies \sin^2\theta = \frac{z^2-2+z^{-2}}{-4} $$
and
$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z+z^{-1}}{2}. $$
We can now substitute such that for the unit circle $\gamma(\theta) = e^{i\theta}, \;\theta\in[-\pi,\pi]$, we have
\begin{align*}
\frac{1}{2}\int_{-\pi}^{\pi} \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)} &= \frac{1}{2}\oint_\gamma \frac{(-\frac{1}{4})(z^2-2+z^{-2})}{10-3(z+z^{-1})}\frac{dz}{iz} \\
&= \frac{i}{8} \oint_\gamma \frac{z^2-2+z^{-2}}{z(10-3z-3z^{-1})} dz \\
&= \frac{i}{8} \oint_\gamma \frac{z^4 - 2z^2 + 1}{z^2(10z-3z^2 -3)} dz \\
&= -\frac{i}{24} \oint_\gamma \frac{z^4-2z^2 + 1}{z^2(z^2-\frac{10}{3}z + 1)} dz \\
&= -\frac{i}{24} \oint_\gamma \frac{z^4-2z^2 + 1}{z^2(z-3)(z-\frac{1}{3})} dz
\end{align*}
It's clear that our integrand has a pole of order 2 at $z=0$, and two simple poles at $z=3$ and $z=1/3$. Using the Residue Theorem, we can evaluate this integral as $2\pi i$ times the sum of the residues at $z=0$ and $z=1/3$, disregarding $z=3$ since this pole is outside of the curve $\gamma$.
At the simple pole $z=1/3$, we calculate
\begin{align*}
\text{Res}(1/3) &= \lim_{z\to\frac{1}{3}} (z-\frac{1}{3})\left(-\frac{i}{24} \frac{z^4-2z^2 + 1}{z^2(z-3)(z-\frac{1}{3})} \right) \\
&= -\frac{i}{24}\lim_{z\to\frac{1}{3}} \left( \frac{z^4-2z^2 + 1}{z^2(z-3)} \right) \\
&= \frac{i}{9}
\end{align*}
The pole at $z=0$ is of order 2, therefore we calculate
\begin{align*}
\text{Res}(0) &= \lim_{z\to 0}\frac{d}{dz} z^2 \left(-\frac{i}{24} \frac{z^4-2z^2 + 1}{z^2(z-3)(z-\frac{1}{3})} \right) \\
&= -\frac{i}{24} \lim_{z\to 0}\frac{d}{dz} \left( \frac{z^4-2z^2 + 1}{(z-3)(z-\frac{1}{3})} \right) \\
&= -\frac{i}{24} \lim_{z\to 0} \frac{(4z^3-4z)(z-3)(z-\frac{1}{3})-(z^4-2z^2+1)(2z-\frac{10}{3})}{(z-3)^2(z-\frac{1}{3})^2}\\
&= \frac{-5i}{36}
\end{align*}
Finally, we calculate
$$
I=2\pi i\Big(\text{Res}(0) + \text{Res}(\frac{1}{3})\Big) = 2\pi i\Big(\frac{-5i}{36} + \frac{i}{9}\Big) = \frac{\pi}{18}
$$
| I think is perfect. Another way is to use the Weierstrass substitution $u=\tan(x/2)$. But is cumbersome and also your way is the apropiate way for a complex analysis course
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tried two ways but neither is very successful. Could someone please suggest a way through?
Attempt 1: I try to set up a generating function
$$G(x)=f(4)x^4+f(5)x^5+\dots=\sum_{k=4}^\infty f(k)x^k$$
Substituting in the recurrence (edit: noticed I forgot to add the +1)
$$\begin{align}G(x)&=x^4+\sum_{k=5}^\infty(1-\frac{4}{k})f(k-1)x^k \\
&=x^4+x\sum_{k=5}^\infty (1-\frac{4}{k}) f(k-1)x^{k-1} \\
&= x^4+x\sum_{k=4}^\infty (1-\frac{4}{k+1})f(k)x^k \\
&= x^4+xG(x)-x\sum_{k=4}^\infty \frac{4}{k+1}f(k)x^k
\end{align} $$
But I don't know how to express the last part in terms of G(x).
Attempt 2: Just brute force it:
With arbitrary $k$,
$$\begin{align}f(n) &= \frac{(n-4)(n-5) \dots (n-(k+3))}{n(n-1)\dots(n-(k-1))}f(n-k)\\
&\ \ + \left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots(n-(k+2))}{n\dots(n-(k-2))}\right)\end{align}$$
Setting $n-k=4 \implies k = n -4$ and using $f(4)=1$,
$$\begin{align} f(n) &= \frac{(n-4)!}{n!/4!}+\left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots2}{n \dots 6}\right) \\
&= {n \choose 4}^{-1}+\left(1+\frac{(n-4)!}{n!}\left(\frac{(n-1)!}{(n-5)!} + \frac{(n-2)!}{(n-6)!}+\dots +\frac{5!}{1!}\right)\right)
\\
&= {n \choose 4}^{-1}+\frac{(n-4)!}{n!}\sum_{k=1}^{n-5}\frac{(k+4)!}{k!} \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=1}^{n-5}(k+4)(k+3)(k+2)(k+1) \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=5}^{n-1}k(k-1)(k-2)(k-3)\end{align}$$
Empirically this seems to be correct, but even Mathematica refuses to simplify it all the way down. Surely there is a better way?
| This recurrence is linear so we can solve it as follows
$$
\cases{
f_n = f_n^h+f_n^p\\
f_n^h = \frac{n-4}{n}f_{n-1}^h\\
f_n^p = \frac{n-4}{n}f_{n-1}^p + 1\\
}
$$
The solution for the homogeneous part is easy to determine
$$
f_n^h = \frac{c_0}{n(n-1)(n-2)(n-3)}
$$
now assuming $f_n^p = \frac{c_0(n)}{n(n-1)(n-2)(n-3)}$ after substitution we have
$$
c_0(n)-c_0(n-1) = n(n-1)(n-2)(n-3)
$$
and solving
$$
c_0(n) = \frac 15(n+1)n(n-1)(n-2)(n-3)
$$
hence
$$
f_n^p = \frac 15(n+1)
$$
and
$$f_n = \frac{f_0}{n(n-1)(n-2)(n-3)} + \frac 15(n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\ln^3(\sin(x))$ using Fourier series
Show that $$\int_0^{\frac\pi2} \ln^3(\sin(x)) \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - \frac{3\pi}4 \zeta(3)$$
I have seen a method for this elsewhere, but I would specifically like to reproduce this result in the same way I've computed the similar integrals of $\ln(\cos(x))\ln(\sin(x))$ and $\ln^2(\sin(x))$, as shown here and here - which involves a "complicated series", as Jack puts it.
In particular, I want to use the Fourier series
$$f(x) = \ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$
Expanding the integrand yields
$$\begin{align*}
-f(x)^3 &= \ln^3(2) + 3 \ln^2(2) \sum_{a=1}^\infty \frac{\cos(2ax)}a \\ &\quad + 3\ln(2) \left(\sum_{a=1}^\infty \frac{\cos^2(2ax)}{a^2} + 2 \sum_{a\neq b} \frac{\cos(2ax) \cos(2bx)}{ab}\right) \\
& \quad + \left(\sum_{a=1}^\infty \frac{\cos^3(2ax)}{a^3} + 3 \sum_{a\neq b} \frac{\cos^2(2ax) \cos(2bx)}{a^2b} + 6 \sum_{a\neq b\neq c} \frac{\cos(2ax) \cos(2bx) \cos(2cx)}{abc}\right)
\end{align*}$$
In the integral, the series with $\cos(2ax)$ vanishes; by orthogonality, $\cos(2ax)\cos(2bx)$ vanishes; in combination of both of these facts, $\cos^2(2ax)\cos(2bx)$ and $\cos^3(2ax)$ also vanish. So the integral reduces to
$$\int_0^{\frac\pi2} f(x)^3 \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - 6 \sum_{a\neq b\neq c} \frac1{abc} \int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2cx) \, dx$$
In the remaining integral, I'm fairly sure that most of the terms integrate to $0$ using the orthogonality argument, except in the case of $a+b=c$,
$$\int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2(a+b)x) \, dx \\ = \int_0^{\frac\pi2} \frac{\cos(2(a-b)x)\cos(2(a+b)x) + \cos^2(2(a+b)x)}2 \, dx = \frac\pi8$$
ETA: If there are no other triples, then I should end up with
$$-6 \sum_{a\neq b\neq c} \frac1{abc} \int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2cx) \, dx = -\frac{3\pi}4 \sum_{a+b=c} \frac1{abc} = -\frac{3\pi}4 \zeta(3) \\ \implies \sum_{a\neq b} \frac1{ab(a+b)} = \zeta(3)$$
Now,
$$\begin{align*}
\sum_{a\neq b} \frac1{ab(a+b)} &= \sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)} - \frac12 \sum_{a=1}^\infty \frac1{a^3} \\[1ex]
&= 2 \sum_{a<b} \frac1{ab(a+b)} - \frac12 \zeta(3) \\[2ex]
\sum_{a<b} \frac1{ab(a+b)} &= \sum_{b=2}^\infty \frac1{b(b+1)} + \sum_{b=3}^\infty \frac1{2b(b+2)} + \sum_{b=4}^\infty \frac1{3b(b+3)} + \cdots \\[1ex]
&= \sum_{b=2}^\infty \left(\frac1b - \frac1{b+1}\right) + \frac14 \sum_{b=3}^\infty \left(\frac1b - \frac1{b+2}\right) + \frac19 \sum_{b=4}^\infty \left(\frac1b - \frac1{b+3}\right) + \cdots \\[1ex]
&= (H_2 - H_1) + \frac{H_4 - H_2}4 + \frac{H_6 - H_3}9 + \cdots \\[1ex]
&= \sum_{n=1}^\infty \frac{H_{2n} - H_n}{n^2} \\[1ex]
&= \sum_{n=1}^\infty \frac{H_{2n}}{n^2} - 2\zeta(3)
\end{align*}$$
where the last equality is due to (31), and it remains to show
$$\sum_{n=1}^\infty \frac{H_{2n}}{n^2} = \frac{11}4 \zeta(3)$$
Rewriting the sum as follows leads me to think there may be a hidden Cauchy product, but I have not been able to find a decomposition.
$$H_{2n}-H_n = \sum_{k=1}^{2n} \frac1k - \sum_{k=1}^n \frac1k = \sum_{k=n+1}^{2n} \frac1k = \sum_{k=1}^n \frac1{n+k} \\ \implies\sum_{n=1}^\infty \frac{H_{2n}-H_n}{n^2} = \sum_{n=1}^\infty \sum_{m=1}^n \frac1{n^2(n+m)}$$
| Some of the last several steps I took in the OP are unnecessary. The last obstacle is the sum
$$\sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)} = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac1{ab(a+b)} = \Gamma(3)\zeta(3) = 2\zeta(3)$$
Let
$$f(x) = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b}}{ab(a+b)}$$
Then for $|x|<1$,
$$\begin{align*}
f'(x) &= \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b-1}}{ab} \\[1ex]
x f'(x) &= \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b}}{ab} \\[1ex]
&= \sum_{a=1}^\infty \left(\frac{x^a}a \sum_{b=1}^\infty \frac{x^b}b\right) \\[1ex]
&= \ln^2(1-x) \\[2ex]
f'(x) &= \frac{\ln^2(1-x)}x \\[2ex]
f(x) &= \int_0^x \frac{\ln^2(1-y)}y \, dy
\end{align*}$$
and the result follows. (Incidentally, this is a proof I had come across on this site several months ago but am unable to track it down.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Finding roots of $\tan z=2i$ I'm trying to find the roots of $\tan(z) = 2i.$
At the moment I have
$$
\tan(z)=\frac{\sin 2x +i \sinh 2y }{\cos 2x+\cosh2y}=2i
$$
At a loss as to where I can go from here. Any advice would be appreciated.
| Recall that $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$, $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$, and $\tan(z) = \frac{\sin(z)}{\cos(z)} = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})}$. So your equation becomes:
$$\frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})} = 2i$$
$$\frac{e^{iz} - e^{-iz}}{e^{iz} + e^{-iz}} = -2$$
Make the substitution $w = e^{iz}$. Then:
$$\frac{w - \frac{1}{w}}{w + \frac{1}{w}} = -2$$
$$\frac{w^2 - 1}{w^2 + 1} = -2$$
$$w^2 - 1 = -2(w^2 + 1)$$
$$3w^2 + 1 = 0$$
This is a simple quadratic with the roots $w = \pm \frac{i\sqrt{3}}{3}$. Or in polar form, $w = \frac{\sqrt{3}}{3} \operatorname{cis}(\pm\frac{\pi}{2}) = \frac{\sqrt{3}}{3} e^{\pm i\pi/2}$. So,
$$e^{iz} = \frac{\sqrt{3}}{3} e^{\pm i\pi/2}$$
$$iz = \ln(\frac{\sqrt{3}}{3}) + i(2\pi k \pm \frac{\pi}{2}), k \in \mathbb{Z}$$
$$iz = \frac{-\ln 3}{2} + i(2\pi k \pm \frac{\pi}{2})$$
$$z = -i\frac{-\ln 3}{2} + 2\pi k \pm \frac{\pi}{2}$$
$$z = 2\pi k \pm \frac{\pi}{2} + i\frac{\ln 3}{2}$$
The real part can be simplified a little by noting that it's just the set of numbers $(n.5) \pi$ for any integer.
$$z = \pi(n + \frac{1}{2}) + i\frac{\ln 3}{2}, n \in \mathbb{Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Help learning and understanding polynomial factorizations A number theory book I'm reading used this factorization as a main step for a proof:
Given $m > n$, integers, $$\left( a^{2^{m}} - 1 \right) = \left( a^{2^{m-1}} + 1 \right)\left( a^{2^{m-2}} + 1 \right) \left( a^{2^{m-3}} + 1 \right) \cdots \left( a^{2^n} + 1 \right) \left( a^{2^n} - 1 \right)$$
Original from the book (in Portuguese):
I'm having a bit of trouble trying to understand how to derive it systematically (in a simple way, without lots of calculations).
If this a known common factorization? Im I missing something?
If so (or not) could one suggest source to learn such factorizations?
Sorry for the possibly relatively low effort question. I just tried to find material on such factorizations and could not find.
Thanks in advance.
| *
*
$(x^2 -1) = (x+1)(x-1)$
We know that through experience. We can explain it through "you see that $+$ in one term and the $-$ in the other... they cause all the middle terms in an expansion to cancel out". That is $(x+1)(x-1) = x(x-1) + 1(x-1) = x^2 \underbrace{\color{red}{- x + x}}_{\text{cancel out}} -1=x^2 -1$
*
*a) so we we replace $x$ we an power, say $x^k$ then it follows that $(x^{2k}-1)= (x^k +1)(x^k-1)$. (Because $x^{2k} =(x^k)^2$ and so $((\color{green}{x^k})^2 -1) = (\color{green}{x^k} +1)(\color{green}{x^k}-1)$
*$2^k$ is of course and even number (it is a power of two, after all). And if $2^k = 2n$ what is $n$? Well, obviously that means $n =\frac {2^k}2 = 2^{k-1}$. Increasing an exponent by $1$ will double the value and decreasing the exponent by one will have the number. This is no surprise. But what the heck does this have to do with 1. above?
*Since $2^k$ is always even. $x^{2^k}$ must be a perfect square. And what is $\sqrt{x^{2^k}}$? Well it must be $x^{\frac {2^k}2}$. And what is $\frac {2^k}2 =2^{k-1}$ so $\sqrt{x^{2^k}} = x^{2^{k-1}}$. This might intuitively not look right because the we are so removed in exponents of exponents but it makes sense. $\sqrt{x^{2^k}} = x^{\frac {2^k}2}=x^{2^{k-1}}$. Meanwhile $\sqrt{\sqrt{x^{2^k}}} = \sqrt{x^{2^{k-1}} }= x^{2^{k-2}}$ and so on.
*So $(x^{2^k} -1) = (x^{2^{k-1}} +1)(x^{2^{k-1}} -1)$ and as $x^{2^{k-1}}$ is also a perfect square we get:
$(x^{2^k} -1) = (x^{2^{k-1}} +1)(x^{2^{k-1}} -1)=$
$(x^{2^{k-1}} +1)(x^{2^{k-2} }+ 1)(x^{2^{k-2}}-1 =$
$(x^{2^{k-1}} +1)(x^{2^{k-2}} + 1)(x^{2^{k-3}} + 1)(x^{2^{k-3} } -1)=$
$...$
$(x^{2^{k-1}} +1)(x^{2^{k-2} }+ 1)(x^{2^{k-3}} + 1)(x^{2^{k-4}}+ 1)......(x^{2^{k-m}}+1)(x^{2^{k-m}} -1)=$
$...$
$(x^{2^{k-1}} +1)(x^{2^{k-2}} + 1)(x^{2^{k-3}} + 1)(x^{2^{k-4}}+ 1)......(x^{2^1}+1)(x^{2^{0}} +1)(x^{2^{0}}-1)=$
$(x^{2^{k-1}} +1)(x^{2^{k-2}} + 1)(x^{2^{k-3}} + 1)(x^{2^{k-4}}+ 1)......(x^{2}+1)(x +1)(x-1) $
Your result of $m>n$ that $a^{2^m} -1$ (you had a typo... you typed $a^{2^{m-1}} - 1$ instead) is just a step on the way:
$a^{2^m}-1 = (a^{2^{m-1}} +1)(a^{2^{m-1}} -1)=$
$a^{2^m}-1 = (a^{2^{m-1}} + 1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1) =$
$a^{2^m}-1 = (a^{2^{m-1}} + 1)(a^{2^{m-2}}+1)....(a^{2^{n+1}}+1)(a^{2^n} +1)(a^{2^n}-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$ To evaluate the integral
$$I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$$
I define $t=-1+\sqrt{\frac{4}{x}-3}$, then we have $x=\frac{4}{(t+1)^2+3}$ and
$$dx=\frac{-8(t+1)}{(t^2+2t+4)^2}dt$$
So we get
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\int_0^\infty \frac{\sqrt{t}(t+1)}{(t^2+2t+4)^2}\ dt$$
Again, let $s=\sqrt{t}$, then $dt=2sds$ and the integral becomes
$$8\int_0^\infty \frac{s(s^2+1) 2s}{(s^4+2s^2+4)^2}\ ds=16\int_0^\infty \frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}\ ds$$
Next, I write the integrand as
$$\frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}=\frac{1}{s^4+2s^2+4}-\frac{s^2+4}{(s^4+2s^2+4)^2}$$
so I have to evaluate
$$I_1=\int_0^\infty \frac{1}{s^4+2s^2+4}\ ds\ and\ I_2=\int_0^\infty \frac{s^2+4}{(s^4+2s^2+4)^2}\ ds$$
But when I evaluate the integral $I_1$, I get a weird result as follows:
$$I_1=\int_0^\infty \frac{1}{(s^2+2)^2-(\sqrt{2}s)^2}\ ds
=\int_0^\infty \frac{1}{(s^2+\sqrt{2}s+2)(s^2-\sqrt{2}s+2)}\ ds=\int_0^\infty \left(\frac{\frac{1}{4\sqrt{2}}s+\frac{1}{4}}{s^2+\sqrt{2}s+2}+\frac{\frac{-1}{4\sqrt{2}}s+\frac{1}{4}}{s^2-\sqrt{2}s+2}\right)\ ds$$
and if we separate two parts and evaluate the integrals, we would get two divergent improper integrals. So how can I find $I_1$?(Hope that the method is elementary and without complex analysis.)
Another question: I'm a beginner at learning complex analysis. I conjecture that we can evaluate the integral $I$ in complex analysis (or maybe not worked). Hope everybody can give me some hints or solutions with the method in complex analysis.
| To answer your question on evaluating $I_1$, partial fraction decomposition yields a rather tedious result to work with. Instead, there is a nice substitution one can make by taking advantage of the symmetry within the integral. Consider the general case
$$I\equiv\int\limits_0^{+\infty}\frac {\mathrm dt}{t^4+\alpha t^2+\beta^2}=\int\limits_0^{+\infty}\frac 1{t^2}\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}$$
Making the inverse substitution $t\mapsto\tfrac {\beta}t$ on the right-hand side, then we get that
$$\beta I=\int\limits_0^{+\infty}\frac {\beta}{t^2}\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}=\int\limits_0^{+\infty}\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}$$
Adding the two integrals together, then the left-hand side becomes $2\beta I$
\begin{align*}
I & =\frac 1{2\beta}\int\limits_0^{+\infty}\left(1+\frac {\beta}{t^2}\right)\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}\\ & =\frac 1{2\beta}\int\limits_{-\infty}^{+\infty}\frac {\mathrm dx}{x^2+\alpha+2\beta}\\ & =\frac {\pi}{2\beta\sqrt{\alpha+2\beta}}
\end{align*}
Where the substitution $x=t-\tfrac {\beta}t$ was made to get to the penultimate step. Your integral $I_1$ is the case when $\alpha=\beta=2$.
Another way of evaluating your original integral would be to start off directly with the expression
$$I\equiv\int\limits_0^{+\infty}\frac {t^2(t^2+1)}{(t^4+at^2+b^2)^2}\,\mathrm dt=\int\limits_0^{+\infty}\frac {t^2+1}{t^2}\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]^2}$$
In a similar fashion, enforce the substitution $t\mapsto\tfrac bt$ and clear the resulting nested fractions to obtain
$$bI=\int\limits_0^{+\infty}\frac {b(t^2+1)}{t^2}\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]^2}=\int\limits_0^{+\infty}\frac {t^2+b^2}{t^2}\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]}$$
Adding the two integrals together and observing that the numerator can be factored as $(b+1)(t^2+b)$ gives
\begin{align*}
I & =\frac {b+1}{2b}\int\limits_0^{+\infty}\left(1+\frac b{t^2}\right)\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]^2}\\ & =\frac {b+1}{2b}\int\limits_{-\infty}^{+\infty}\frac {\mathrm dx}{(x^2+a+2b)^2}\\ & =\frac {\pi(b+1)}{4b(a+2b)\sqrt{a+2b}}
\end{align*}
Where the result proved in the first-half of this answer was used. Substituting $a=b=2$ immediately gives the answer as
$$\int\limits_0^{+\infty}\frac {t^2(t^2+1)}{(t^4+2t^2+4)^2}\,\mathrm dt\color{blue}{=\frac {\pi}{16\sqrt 6}}$$
Your original integral is $16$ times the result in blue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Are there more examples of this method in calculus? One of my absolute favorite derivations in math is the calculation of the sum of the alternating harmonic series, $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$, which is done by considering the function $f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ and noticing that $f'(x) = 1 - x + x^2 - x^3 = \frac{1}{1+x}$ implying $f(x) = \ln(1+x)$ and thus the series converges to $\ln(1+1) = \ln(2)$.
The method here of deriving a formula for an unknown function by taking its derivative and gaining more information about the original function can also be used to derive $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. If $f(\theta) = \cos(\theta) + i\sin(\theta)$, then $f'(\theta) = if(\theta)$, which solves to $f(\theta) = e^{i\theta}$.
My question is, are there more examples of this technique, or is it a coincidence that it worked so well in these two cases?
| $(1+x^2)^{-1}=1-x^2+x^4-x^6+\dots$,
$\arctan x=\int(1+x^2)^{-1}\,dx=x-(x^3/3)+(x^5/5)-(x^7/7)+\cdots$
$\pi/4=\arctan1=1-(1/3)+(1/5)-(1/7)+\cdots$.
Another example (that has come up often on this site): show that
$$
\sum_1^{\infty}{n\over2^n}={1\over2}+{2\over4}+{3\over8}+{4\over16}+{5\over32}+\cdots=2.
$$
Solution: let $f(x)=\sum_1^{\infty}nx^n$, so we want to show $f(1/2)=2$. Then $x^{-1}f(x)=\sum_1^{\infty}nx^{n-1}$ is the derivative of $\sum_1^{\infty}x^n=x/(1-x)$. The derivative of $x/(1-x)$ is $(1-x)^{-2}$, so $f(x)=x(1-x)^{-2}$, and $f(1/2)=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the right algebra to show they are indeed equivalent.
Note I ran into this problem, trying to do: $2\cos(\pi/8)+2\sin(\pi/8)$, where
$$2\cos(\pi/8)=\sqrt{2+\sqrt{2}},$$
$$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}.$$
| Alternatively
We want
$\sqrt{2+\sqrt 2} + \sqrt {2-\sqrt 2}=\sqrt{something}$ so
$\sqrt{2+\sqrt 2} + \sqrt {2-\sqrt 2}=$
$\sqrt{(\sqrt{2+\sqrt 2} + \sqrt {2-\sqrt 2})^2}=$
$\sqrt{(2+\sqrt 2) + (2-\sqrt 2) + 2\sqrt{2+\sqrt 2}\sqrt {2-\sqrt 2}}=$
$\sqrt{4 + 2\sqrt{(2+\sqrt 2)(2-\sqrt 2)}}=$
$\sqrt{4 + 2\sqrt{4 - 2}}=$
$\sqrt{4 + 2\sqrt 2}=$
$\sqrt {2(2+\sqrt 2)}$
Or we could go the other way
We want some to get $\sqrt{2(2+\sqrt 2)} = \sqrt{\sqrt{2+\sqrt 2} + \sqrt{2-\sqrt 2}}$ so we try to get it into expresions that look closer and closer to it.
$\sqrt{2(2+\sqrt 2)} = \sqrt{(2+\sqrt 2) + (2 + \sqrt 2)}=$
$\sqrt{(2+\sqrt 2) +2\sqrt 2 +(2-\sqrt 2)}=$
$\sqrt{\sqrt{2+\sqrt 2}^2 + 2\sqrt 2+ {\sqrt{2-\sqrt 2}}^2}$
Now if we can prove that $\sqrt 2 = {\sqrt{2+\sqrt 2}}{\sqrt{2-\sqrt 2}}$ we'd be golden.
So side track: $ {\sqrt{2+\sqrt 2}}{\sqrt{2-\sqrt 2}}=$
${\sqrt{(2+\sqrt 2)(2-\sqrt 2}}=$
${\sqrt{2^2 - \sqrt 2^2}}=$
${\sqrt{4-2}} = \sqrt 2$.
so.... back on track:
$\sqrt{\sqrt{2+\sqrt 2}^2 + 2\sqrt 2+{\sqrt{2-\sqrt 2}}^2}=$
$\sqrt{\sqrt{2+\sqrt 2}^2 + 2\sqrt{2+\sqrt 2}{\sqrt{2-\sqrt 2}+ {\sqrt{2-\sqrt 2}}^2}}=$
$\sqrt{(\sqrt{2+\sqrt 2} + \sqrt{2-\sqrt 2})^2}=$
$\sqrt{2+\sqrt 2} + \sqrt{2-\sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$ without L'Hosptial rule. I have a belief that it is possible to evaluate any limit without using L'Hospital rule or series expansions. However the limit $$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$$ stumps me.
I thought at the beginning that is it nothing but dividing top and bottom by $x^2$ and then using the limits: $$\lim\limits_{x\to 0} \frac{\ln(1+x^2)}{x^2} = 1, \quad \quad \lim \limits_{x\to 0} \frac{\arcsin x}{x} = 1$$
And then I found $$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x} = 0$$
Which made the limit $\frac{0}{0}$. I have no clue how to split the limit into a product or a sum of existent limits. Is it possible to solve this without L'Hosptial? Is there is possible usage of the squeeze theorem or a geometric interpretation to compute this limit? The answer according to WolframAlpha is $\frac{5}{2e}$ and it is clear how to find the $\frac{1}{e}$ but not the $\frac{5}{2}$.
| The numerator is $\sim x^2(\cos(x)-1)+3x^4\sim \frac{5}{2}x^4$
The denominator is $\sim (e^{\sqrt[4]{1+4x^3+8x^4}}-e)x$
So you need to evaluate the limit for
$$\frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3}$$
You did $x^2$ instead of $x^3$, so you got $0$, and you can show this term
$$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e\cdot x^3$$
Series expansion $\sqrt[4]{1+y}=1+\frac{1}4y+O(y^2)$, substitute $y=4x^3+8x^4$
$$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e^{1+x^3+2x^4}-e=e\cdot(e^{x^3+2x^4}-1)\sim e\cdot x^3$$
Final answer:
$$L=\frac{5}{2e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
Furthermore, $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\
&=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\
&=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3)
\end{aligned}
$$
and $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$
| \begin{align}J&=\int_0^{\frac{\pi}{2}}x\ln(\sin x)dx\\
K&=\int_0^{\frac{\pi}{2}}x\ln(\cos x)dx\\
J+K&=\int_0^{\frac{\pi}{2}}x\ln(\sin x\cos x)dx\\
&\overset{y=\frac{\pi}{2}-x}=\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-y\right)\ln(\sin y\cos y)dy\\
&\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln(\sin y\cos y)dy\\
&=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin(2y)}{2}\right)dy\\
&\frac{\pi}{4}\underbrace{\int_0^{\frac{\pi}{2}}\ln\left(\sin(2y)\right)dy}_{z=2y}-\frac{\pi^2}{8}\ln 2\\
&=\frac{\pi}{8}\int_0^\pi \ln(\sin z)dz-\frac{\pi^2}{8}\ln 2\\
&=\frac{\pi}{8}\int_0^{\frac{\pi}{2}} \ln(\sin z)dz+\frac{\pi}{8}\underbrace{\int_{\frac{\pi}{2}}^\pi \ln(\sin z)dz}_{x=\pi-z}-\frac{\pi}{8}\ln 2\\
&=\frac{\pi}{4}\underbrace{\int_0^{\frac{\pi}{2}} \ln(\sin x)dx}_{=-\frac{\pi}{2}\ln 2}-\frac{\pi^2}{8}\ln 2=\boxed{-\frac{\pi^2}{4}\ln 2}\\
J-K&=\int_0^{\frac{\pi}{2}}x\ln(\tan x)dx\\
&\overset{t=\tan x}=\int_0^\infty \frac{\ln t\arctan t}{1+t^2}dt
\end{align}
And see:
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$
https://math.stackexchange.com/a/4500094/186817
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$. Why doesn't $x^2$ overpower $2x$ when $x\to\infty$, so that the answer is $0$ (instead of $1$)?
Evaluate $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$$
Hello!
I was solving this problem, and here is my approach:
Inside the square root $x^2$ overpowers $2x$ as $x \to \infty$, so $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x = \lim_{x \to \infty} \sqrt{x^2} - x = \lim_{x \to \infty} |x| - x = 0$$ which is wrong answer. Actual answer is $1$.
This is the way I have solved so many problems when $x \to \infty$, especially when the terms are in denominator.
For example, consider $$\lim_{x \to \infty} \frac{x^2 + x}{2x^2 + 4x + 10}$$
Now, we ignore the smaller power terms, and answer is $\frac{1}{2}$. Why can we ignore smaller powers here but not in the original question on top?
Sorry if this is a stupid question.
| Okay.... I'd say the issue of "overpowering" is that it doesn't work when there isn't something else acting an the thing doing the overpowering to something finite, so that the thing being overpowered is forced to be insignificant. If the thing doing the overpowering goes to infinity and is only countered by something else going to negative infinity so they cancel out, then the thing being overpowered isn't being forced to go to $0$-- it's merely going ... somewhere else. Just not as fast as the overpowering thing.
As $x\to \infty$ then the $\sqrt{x^2} \to \infty$ and a higher rate than $\sqrt{+2x} \to \infty$ but that doesn't mean it goes to $0$.
We have $\sqrt{x^2 + 2x}> \sqrt {x^2} = x$ so we must have $\sqrt{x^2+2x}-x = f(x) > 0$.
Now we do know that $f(x) << x$
and as $x\to SomethingREALLYhuge$
we will have $f(x)\to SomethingNOWHEREnearashuge$,
but there is nothing forcing it to zero. We have no idea at this point what $\lim_{x\to\infty}\sqrt{x^2 + 2x} -x = \lim_{x\to \infty} f(x)$ ought to be.
[Compare this to $\frac {x^5 + x^3}{2x^5 + 4x^2}$ where the $x^5$s "overpower" then $x^3$ and $4x^2$ and breezily say $\lim \frac {x^5 + x^3}{2x^5 + 4x^2}\approx \lim \frac {x^5}{2x^5} = \frac 12$. In this case we are dividing the $x^5$ by $x^5$ to "overpower to something finite". So as $x^5$ is being forced to something finite the $x^3$ and $x^2$ are forced to $\frac {finite}{x^2}$ and $\frac {finite}{x^3}$ and go to zero.]
====
Now to actually do this thing. $\sqrt{x^2+2x} = \sqrt{x^2 + 2x + 1 -1}= \sqrt{(x+1)^1 - 1}$.
Now we can say that as $x\to \infty$ that $(x+1)^2$ "overpowers" the $-1$. The difference between $x^2$ overpowering $-1$ to insignificance and $x^2$ overpowering $2x$ but not to insignificance, is that $-1$ is finite and gets insignificant when compared to very large $x$, whereas $2x$ is not finite.
$\lim \sqrt{x^2 + 2x} -x =\lim \sqrt{(x+1)^2-1} - x\approx \lim \sqrt{(x+1)^2} - x=\lim (x+1) -x = 1$.
.....
Still, informality is only for estimating and intuiting answers. To actually answer the question we must be more formal and proper.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculate the sum of $\left\lfloor \sqrt{k} \right\rfloor$ I'm trying to calculate for $n\in \mathbb{N}$ the following sum :
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor$.
I tried putting in the first terms, which gave me
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=(1+2+3+\cdots+n)+\left\lfloor \sqrt{2} \right\rfloor+\left\lfloor \sqrt{3} \right\rfloor+\left\lfloor \sqrt{5} \right\rfloor+\cdots+\left\lfloor \sqrt{n^2-1} \right\rfloor$
$\iff \sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=\frac{n(n+1)}{2}+\left\lfloor \sqrt{2} \right\rfloor+\left\lfloor \sqrt{3} \right\rfloor+\left\lfloor \sqrt{5} \right\rfloor+\cdots+\left\lfloor \sqrt{n^2-1} \right\rfloor$.
I've been trying to somehow find a pattern between the different integer parts of the irrational numbers just like I did with the integers but I fail to success.
Is there a trick to use here or is my take wrong ?
Thank you.
| Here's a proof using the benefits of the Iverson bracket:
\begin{align}
\sum_{k=1}^{n^2}\lfloor\sqrt{k}\rfloor &= \sum_{k=1}^{n^2}\sum_{j=1}^n[\;j \le \sqrt{k}\;]\\
&= \sum_{j=1}^n \sum_{k=1}^{n^2}[\;j \le \sqrt{k}\;]\\
&= \sum_{j=1}^n \sum_{k=1}^{n^2}[\;j^2 \le k\;]\\
&= \sum_{j=1}^n \sum_{k=j^2}^{n^2}1\\
&= \sum_{j=1}^n (n^2-j^2+1)\\
&= n^3 - \frac{1}{6}n(n+1)(2n+1)+n\\
&= \frac{1}{6}n(4n^2-3n+5)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that $\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}$ has a solution
Show that there exists some $\epsilon >0$ s.t. for all $\mathbf{A}\in \mathbb{R}^{2\times 2} $
with $|( \mathbf{A})_{i, j}| < \epsilon $ for all $i, j$ (let the space of all such matrices be $E$) the equation
$$
\begin{align*}
\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}
\end{align*}
$$
has some solution $\mathbf{X}\in \mathbb{R}^{2\times 2} $.
My approach would be to use the Banach fixed point theorem:
Consider the mapping
$$
\begin{align*}
\Phi \colon E \to \mathbb{R}^{2, 2}, \quad \Phi ( \mathbf{X})
= \mathbf{A} - \mathbf{X}^{2}
.\end{align*}
$$
So first we need to find an $\epsilon > 0$ s.t. $\Phi [ E]\subseteq E$. Since
$$
\begin{align*}
\begin{bmatrix}
a & b \\ c & d
\end{bmatrix} ^{2} = \begin{bmatrix}
a^{2} + cb & ab + db \\ ac + c d & b c + d ^{2}
\end{bmatrix}
\implies \forall i, j \in \{ 1, 2\}\colon ( \mathbf{X}^{2})_{i, j} < 2 \epsilon^{2}
\end{align*}
$$
choosing any $\epsilon < 1 / 2$ will assure that $\Phi ( \mathbf{X}^{2}) \in E$ for $\mathbf{X} \in E$.
Next we have to determine when $\Phi $ becomes a contraction. One has
$$
\begin{align*}
\left\| \Phi ( \mathbf{X}) - \Phi ( \mathbf{Y})\right\|_{\infty }
&=
\left\| \mathbf{X}^{2} - \mathbf{Y}^{2}\right\|_{\infty}
\\
&= \max_{i \in \{ 1, 2\}} \sum_{j = 1}^{2} \left| ( \mathbf{X}^{2})_{i, j}- ( \mathbf{Y}^{2})_{i, j} \right|
\\
&\leqslant \max_{i \in \{ 1, 2\}} \sum_{j = 1}^{2}\!\left( \left| ( \mathbf{X}^{2})_{i, j}\right|+ \left| ( \mathbf{Y}^{2})_{i, j} \right|
\right)
\\
& < 2 \!\left( 2 \epsilon ^{2} + 2 \epsilon ^{2}\right) = 8 \epsilon ^{2}
\end{align*}
$$
and
$\left\| \mathbf{X} - \mathbf{Y}\right\|_{\infty} < 4\epsilon $ meaning we simply have to choose some
$\epsilon $ satisfying
$$
\begin{align*}
2 \epsilon ^{2} < \epsilon \iff \epsilon < \frac{1}{2}
.\end{align*}
$$
First of all, is my approach correct? Secondly, are there more elegant solutions (this question was in the
context of an analysis course, meaning I didn't think about any elegant linear algebra solutions).
| Lemma: There exists $\epsilon>0$ such that if $B$ is a $2\times 2$ matrix with $\|B-\frac{1}{4}I \|_\infty<\epsilon$, then the equation $Y^2=B$ has a solution $Y$.
Proof: Since the eigenvalues of $S$ are given by
$$\lambda_{1,2}(S)=\frac{1}{2} \left(S_{11}+S_{22}\pm \sqrt{(S_{11}+S_{22})^2-4(S_{11}S_{22}-S_{12}S_{21})} \right ),$$ and $\frac{1}{4}I$ has positive eigenvalues, there exists a neighborhood of $\frac{1}{4}I$ such that every matrix in that neighborhood whose eigenvalues are real must have positive eigenvalues as well. In other words, there exists $\epsilon>0$ such that for every matrix $B$ with $\|B-\frac{1}{4}I\|_\infty<\epsilon$, either $B$ has non-real eigenvalues or $B$ has real positive eigenvalues.
Next, by Jordan decomposition, there exists a matrix $P$ such that $P^{-1}BP=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$ with $\lambda_1,\lambda_2>0$ (if $B$ has two positive eigenvalues) or $P^{-1}BP=t\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ for some $t>0$ and angle $\alpha$ (if $B$ has non-real eigenvalues). In either case, it is easy to see that there exists $Z$ with $Z^2=P^{-1}BP$. To be precise, in the former case, let $Z=\begin{bmatrix} \sqrt{\lambda_1} & 0 \\ 0 & \sqrt{\lambda_2}\end{bmatrix}$ and in the latter case, let $Z=\sqrt{t}\begin{bmatrix} \cos (\alpha/2) & -\sin (\alpha/2) \\ \sin (\alpha/2) & \cos (\alpha/2) \end{bmatrix}$. We let $Y=PZP^{-1}$ to get $Y^2=B$.
Using this lemma, now let $B=A+\frac{1}{4}I$. Hence there exists $Y$ with $Y^2=B$. Finally, let $X=Y-\frac{1}{2}I$ to get $X^2+X=A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How to find all the solutions to $a+2b+3c+\cdots+9i=45$ I want to find all the solutions to the equation $a+2b+3c+4d+5e+6f+7g+8h+9i=45$ where all of the variables can only be of $3$ values: $0,1,2$
I know there are a few solutions for example, set all of them to be $1$, or $a=1, b=2, c=0, d=1, e=0, f=2, g=1, h=1, i=1$.
I tried to turn them all into mod $3$ and I got:
$a+2b+d+2e+g+2h\equiv 0 \pmod 3$ but I don't know what to do after that.
I understand that there may be ~$7\times 10^{20}$ solutions
| As @robjohn mentioned , you can find the number of all solutions using generating functions. If the possible values for the variables are $0,1,2$ ,we can say that
*
*$a:0,1,2$ . Hence , its generating function is $$x^0+x^1+x^2 = \frac{1-x^3}{1-x}$$
*If "b" can take the values of $0,1,2$ , then $2b:0,1,4$ . Hence , its generating function is $$x^0+x^2+x^4 = \frac{1-x^6}{1-x^2}$$
*If "c" can take the values of $0,1,2$ , then $3c:0,3,6$ . Hence , its generating function is $$x^0+x^3+x^6 = \frac{1-x^9}{1-x^3}$$
*If "d" can take the values of $0,1,2$ , then $4d:0,4,8$ . Hence , its generating function is $$x^0+x^4+x^8 = \frac{1-x^{12}}{1-x^4}$$
*If "e" can take the values of $0,1,2$ , then $5e:0,5,10$ . Hence , its generating function is $$x^0+x^5+x^{10} = \frac{1-x^{15}}{1-x^5}$$
*If "f" can take the values of $0,1,2$ , then $6f:0,6,12$ . Hence , its generating function is $$x^0+x^6+x^{12} = \frac{1-x^{18}}{1-x^6}$$
*If "g" can take the values of $0,1,2$ , then $7g:0,7,14$ . Hence , its generating function is $$x^0+x^7+x^{14} = \frac{1-x^{21}}{1-x^7}$$
*If "h" can take the values of $0,1,2$ , then $8h:0,8,16$ . Hence , its generating function is $$x^0+x^8+x^{16} = \frac{1-x^{24}}{1-x^8}$$
*If "i" can take the values of $0,1,2$ , then $9i:0,9,18$ . Hence , its generating function is $$x^0+x^9+x^{18} = \frac{1-x^{27}}{1-x^9}$$
Now , find the coefficient of $x^{45}$ in the expansion of $$\bigg(\frac{1-x^{3}}{1-x}\bigg)\bigg(\frac{1-x^{6}}{1-x^2}\bigg)\bigg(\frac{1-x^{9}}{1-x^3}\bigg)\bigg(\frac{1-x^{12}}{1-x^4}\bigg)\bigg(\frac{1-x^{15}}{1-x^5}\bigg)\bigg(\frac{1-x^{18}}{1-x^6}\bigg)\bigg(\frac{1-x^{21}}{1-x^7}\bigg)\bigg(\frac{1-x^{24}}{1-x^8}\bigg)\bigg(\frac{1-x^{27}}{1-x^9}\bigg)$$
NOTE: You can use sofwares such as SageMath to find the coefficent of $x^{45}$ in the expansion , i generally use wolfram alpha but this expansion was long for wolfram
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$ What is the value of
$$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$$
I have already tried the two paths result and I think now that the limit does exist and it is equals to 0. But i really cannot prove this.
|
$$L=\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$$
$$L=\lim_{(x, y) \to (0, 0)} \frac{xy^2-x^3+x^3}{x^2-y^2}=\lim_{x \to0} (-x)+\lim_{(x, y) \to (0, 0)} \frac{x^3}{x^2-y^2}=\lim_{(x, y) \to (0, 0)} \frac{x^3}{x^2-y^2}$$
Let $y=x-ax^2$
$$L=\lim_{x \to 0} \frac{x^3}{2ax^3-a^2x^4}=\lim_{x \to 0} \frac{1}{2a-a^2x}=\frac{1}{2a}$$
So the limit depends on the value of $a$, hence the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$ From R. D. Sharma's Objective Mathematics,
Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$
My attempt:
By A.M.-G.M. inequality,
$$\begin{align}
(x+y)^2 &\geq 4(xy) \\
(y+z)^2 &\geq 4(yz) \\
(z+x)^2 &\geq 4(zx)
\end{align}$$
By multiplying by $xy$, $yz$, $zx$ in these equations respectively, gives
$$\begin{align}
xy(x+y)^2 &\geq 4(xy)^2 \\
yz(y+z)^2 &\geq 4(yz)^2 \\
zx(z+x)^2 &\geq 4(zx)^2 \\
\end{align}$$
Adding these equations, we get
$$ A \geq 4[(xy)^2 + (yz)^2 + (zx)^2] $$
Using AM GM inequality once again on RHS, we get
$$ A \geq 12(xyz)^{\tfrac43}$$
However, provided answer key says that minimum value is $4xyz$. I do not want a solution but only wants to know that is something wrong in my procedure? Why?
EDIT : This question has been identified as a possible duplicate of another question . However, readers will believe that my problem is different if they re-read the paragraph just above .
| An approach with symmetric functions:
$$A=xy(1-z)^2+xz(1-y)^2+yz(1-x)^2\\
\quad =xy+xz+yz-6xyz+xyz(x+y+z)\\
\quad =xy+xz+yz-5xyz$$
the symmetric $\quad xyz\quad$ in terms of $\quad x^3+y^3+z^3:$
$(x+y+z)^3=p_3+3(1-x)(1-y)(1-z)=\cdots \\
xyz=\frac 1 3(x^3+y^3+z^3)+(xy+xz+yz)-\frac 1 3$
note also $\quad xy+xz+yz=\frac 1 2\left[(x+y+z)^2-(x^2+y^2+z^2)\right]$
Then
$$A=xy+xz+yz-5xyz\\
=2(x^2+y^2+z^2)-\frac 5 3 (x^3+y^3+z^3)-\frac 1 3\\
\ge 2 \left(\frac 1 3\right)-\frac 5 3\left(\frac 1 9 \right)-\frac 1 3=\frac 4{27}$$
Given that $$\begin{aligned}\min (x^2+y^2+z^2)=\frac 1 3\\
\min(x^3+y^3+z^3)=\frac 1 9\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Calculating eigenvalues of a matrix I have this quadratic matrix:
\begin{align*}
A = \begin{bmatrix}
0 & -1 & -2\\
-1 & 0 & -2\\
-2 & -2 & -3
\end{bmatrix} \implies A - \lambda I = \begin{bmatrix}
0 - \lambda & -1 & -2\\
-1 & 0 - \lambda & -2\\
-2 & -2 & -3 - \lambda
\end{bmatrix}
\end{align*}
After forming the characteristic polynomial and using the Rule of Sarrus, I get this equation:
$$-\lambda^3 - 3\lambda^2 + 9\lambda - 5$$
However the correct form of this term would be:
$$-(5 + \lambda)(1 - \lambda)^2$$
So the eigenvalues are $1$ and $-5$.
The thing I do not understand, is how to form this equation to get the values. Which approach and rules can solve this?
Thank you very much.
| For matrices of $\mathbb{R}^{3 \times 3}$ and above, the method is to try to apply row operations and factorize. Avoid computing the determinant directly, or expanding the terms first.
*
*Apply row operations to $ |A - \lambda I|$.
$$
\begin{matrix}
\\
-R_1+R_2 \rightarrow\\
\\
\end{matrix}
\begin{vmatrix}
-\lambda & -1 & -2 \\
-1 & -\lambda & -2 \\
-2 & -2 & -3 -\lambda \\
\end{vmatrix}
\implies
\begin{vmatrix}
-\lambda & -1 & -2 \\
-(1-\lambda) & 1-\lambda & 0 \\
-2 & -2 & -3-\lambda \\
\end{vmatrix}
$$
*Find the characteristic polynomial from the reduced expression. We will exploit column $3$.
$$
\begin{align}
p(\lambda) = & -2 \bigl( 2(1-\lambda) + 2(1-\lambda) \bigl) + (-3-\lambda)\bigl( -\lambda(1-\lambda) -(1-\lambda) \bigl)\\
= & -8(1-\lambda) + (-3-\lambda)(1-\lambda)(-\lambda-1)\\
= & -8(1-\lambda) + (3+\lambda)(1-\lambda)(1+\lambda)\\
= & (1-\lambda)\bigl(-8 + (3+\lambda)(1+\lambda) \bigl) \\
= & (1-\lambda)(-8+\lambda^2+4\lambda+3)\\
= & (1-\lambda)(\lambda^2+4\lambda-5)\\
= & (1-\lambda)(\lambda+5)(\lambda-1)\\
= & -(1-\lambda)^2(\lambda+5)
\end{align}
$$
You may refer to a similar question I have answered.
Link: Finding Eigen Vectors and Respective Eigen Values for 3x3 Matrice
| {
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Apparent inconsistency in geometric product associativity With a little bit of work, I have proven to myself that the geometric product between three vectors is associative ($a$, $b$, and $c$ are 1-vectors):
$$\begin{aligned}(ab)c &= a(bc) \\ &= (b \cdot c) a - (a \cdot c) b + (a \cdot b) c + a \wedge b \wedge c.\end{aligned}$$
And, by extension, that it is consistent for four vectors ($a$, $b$, $c$, $d$) as:
$$\begin{aligned} (abc)d &= a(bcd) \\ &= (b \cdot c)(a \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot b)(c \cdot d) \\ &\quad+ (b \cdot c)a \wedge d - (a \cdot c)b \wedge d + (a \cdot b)c \wedge d \\ &\quad+ b \wedge c(a \cdot d) - a \wedge c(b \cdot d) + a \wedge b(c \cdot d) \\ &\quad+ a \wedge b \wedge c \wedge d\end{aligned}$$
However, the following, which I would expect to match the rules for associativity, is quite different:
$$\begin{aligned} (ab)(cd) &= (a \cdot b + a \wedge b)(c \cdot d + c \wedge d) \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + (a \wedge b)(c \wedge d) \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + (a \wedge b).(c \wedge d) + a \wedge b \wedge c \wedge d \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + a \cdot (b \cdot (c \wedge d)) + a \wedge b \wedge c \wedge d \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) \\ &\quad + (a \cdot d)(b \cdot c) - (a \cdot c)(b \cdot d) \\ &\quad + a \wedge b \wedge c \wedge d \\ &= (b \cdot c)(a \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot b)(c \cdot d) \\ &\quad + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) \\ &\quad + a \wedge b \wedge c \wedge d.\end{aligned}$$
Which, when compared with $a(bcd)$ and $(abc)d$ has all pure scalars and the quad-vector, but is missing four of the six scaled bi-vectors.
Is it that my understanding of associativity is incorrect in that $(ab)(cd)$ is not the same as $a(bcd)$ or $(abc)d$, did I make a mistake in my algebra, or is there something else going on?
My digging into this comes from the desire to understand $e_{12} e_{23}$ which is short-hand for $(e_1 \wedge e_2)(e_2 \wedge e_3)$ (note, geometric product for $()()$) where $e_1$, $e_2$, $e_3$ are orthonormal basis vectors and I have been unable to determine whether the result should be $0$ or $e_{13}$. Also, that $e_1 e_2 e_2 e_3$ comes up in my working out of geometric products when $e_1 e_2$ is used as shorthand for $e_1 \wedge e_2$.
Follow-up:
Many thanks to Somos for encouraging me to keep looking and to Peeter Joot for his hint about ${\langle ( \wedge )( \wedge ) \rangle}_2$ as the apparent circular definition made me dig into just what $(a \wedge b)(c \wedge d)$ looks like.
Extrapolating from appendix C.1 of Geometric Algebra for Computer Science (Dorst, Fontijne, Mann).
$$
(a \wedge b)(c \wedge d) = \frac{1}{4} (ab - ba)(cd - dc)
$$
This expands out into four permutations of $abcd$:
$$
abcd - abdc - bacd + badc
$$
I won't go into the full expansion, but it results in 12 grade-0 terms (4 of which cancel), 24 grade-2 terms (8 of which cancel), and 4 grade-4 terms. The terms that don't cancel add together, thus requiring the $\frac{1}{4}$. The terms that do cancel are exactly those terms in my $(ab)(cd)$ expansion remaining after removing $(a \wedge b)(c \wedge d)$.
Thus, while the geometric product is defined to be associative, this exercise shows that associativity in action, which is what I needed.
| It looks like you are missing a factor in your expansion of $ \left( {a \wedge b} \right) \left( {c \wedge d} \right) $. In general, a product of bivectors $ A B $, should have scalar, bivector, and quadvector components:
$$A B = A \cdot B + {\left\langle{{ A B }}\right\rangle}_{2} + A \wedge B,$$
so for $ A = a \wedge b, B = c \wedge d $, we have
$$ \left( { a \wedge b } \right) \left( { c \wedge d } \right) = \left( { a \wedge b } \right) \cdot \left( { c \wedge d } \right) + {\left\langle{{ \left( { a \wedge b } \right) \left( { c \wedge d } \right)}}\right\rangle}_{2} + \left( { a \wedge b } \right) \wedge \left( { c \wedge d } \right).$$
You can drop the braces in the final wedge, but that grade-two term should provide the missing bivector terms.
On your rationale.
If you only want to understand $(e_1 \wedge e_2)(e_2 \wedge e_3)$, it's helpful to revert to the GA axioms, two of which are:
*
*Associativity
*Contraction: $x^2 = x \cdot x$, for any vectors x.
With those, plus $x \wedge y = x y$, for vectors $x, y$ where $x \cdot y = 0$, you have
$$\begin{aligned}\left( {e_1 \wedge e_2} \right) \left( { e_2 \wedge e_3 } \right)&=\left( {e_1 e_2} \right) \left( { e_2 e_3 } \right) \\ &=e_1 e_2 e_2 e_3 \\ &=e_1 \left( { e_2 e_2 } \right) e_3 \\ &=e_1 e_3.\end{aligned}$$
As the dot and wedges of a pair of bivectors are the grade-0 and grade-4 terms respectively (by definition), you can immediately conclude that
$$\left( {e_1 \wedge e_2} \right) \cdot \left( { e_2 \wedge e_3 } \right) = 0,$$
$$\left( {e_1 \wedge e_2} \right) \wedge \left( { e_2 \wedge e_3 } \right) = 0,$$
and
$$ {\left\langle{{\left( {e_1 \wedge e_2} \right) \left( { e_2 \wedge e_3 } \right)}}\right\rangle}_{2} = e_1 e_3.$$
One expansion.
Here's one way to expand the product of four vectors, grouping the first and last pairs.
$$\begin{aligned}a b c d &=\left( { a \cdot b + a \wedge b } \right) \left( { c \cdot d + c \wedge d } \right) \\ &=\left( { a \cdot b } \right) \left( { c \cdot d } \right) \\ &+\quad\left( { a \cdot b } \right) \left( { c \wedge d } \right)+\left( { c \cdot d } \right) \left( { a \wedge b } \right) \\ &+\quad \left( { a \wedge b } \right)\left( { c \wedge d } \right).\end{aligned}$$
The first term of this bivector-bivector product can be expanded with application of $ a b = a \cdot b + a \wedge b $ in reverse
$$\begin{aligned}\left( { a \wedge b } \right)\left( { c \wedge d } \right)&=\left( { a b - a \cdot b } \right)\left( { c \wedge d } \right) \\ &=a b \left( { c \wedge d } \right) - \left( { a \cdot b } \right) \left( { c \wedge d } \right) \end{aligned}$$
The product of $ b $ with $ c \wedge d $ is
$$\begin{aligned}b \left( { c \wedge d } \right) &=b \cdot\left( { c \wedge d } \right) +b \wedge\left( { c \wedge d } \right) \\ &=\left( { b \cdot c } \right) d- \left( { b \cdot d } \right) c+ b \wedge c \wedge d.\end{aligned}$$
Multiplying this by $ a $ on the left, we have a scalar grade
$$ \left\langle{{ a b \left( { c \wedge d } \right) }}\right\rangle= \left( { b \cdot c } \right) \left( { a \cdot d } \right) - \left( { b \cdot d } \right) \left( { a \cdot c } \right),$$
a bivector grade
$$\begin{aligned}{\left\langle{{ a b \left( { c \wedge d } \right) }}\right\rangle}_{2}&=\left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right)+ a \cdot \left( { b \wedge c \wedge d } \right) \\ &=\left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right) \\ &\quad + \left( { a \cdot b } \right) \left( { c \wedge d } \right) - \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right),\end{aligned}$$
and a grade-four component
$${\left\langle{{ a b \left( { c \wedge d } \right) }}\right\rangle}_{4}= a \wedge b \wedge c \wedge d.$$
Putting all the pieces together, here are all the grades of the $ a b c d $ product
$$\begin{aligned} \left\langle{{( a b)( c d) }}\right\rangle &=\left( { a \cdot b } \right) \left( { c \cdot d } \right) +\left( { b \cdot c } \right) \left( { a \cdot d } \right) - \left( { b \cdot d } \right) \left( { a \cdot c } \right) \\ {\left\langle{{ (a b)( c d) }}\right\rangle}_{2} &= \left( { a \cdot b } \right) \left( { c \wedge d } \right)+ \left( { c \cdot d } \right) \left( { a \wedge b } \right) - \left( { a \cdot b } \right) \left( { c \wedge d } \right) \\ &\quad+ \left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right)+ \left( { a \cdot b } \right) \left( { c \wedge d } \right) \\ &\quad- \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right) \\ &= \left( { a \cdot b } \right) \left( { c \wedge d } \right)+ \left( { c \cdot d } \right) \left( { a \wedge b } \right) + \left( { b \cdot c } \right) \left( { a \wedge d } \right) \\ &\quad - \left( { b \cdot d } \right) \left( { a \wedge c } \right)- \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right) \\ {\left\langle{{ (a b)( c d) }}\right\rangle}_{4} &= a \wedge b \wedge c \wedge d.\end{aligned},$$
which matches your result, finishing your experimental proof of associative multiplication for a product of four vectors.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Gelfands Trigonometry $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ Trying Prove the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ using the figure provided in Gelfands trigonometry.
What I have so far
$\sin(\alpha - \beta) = \frac{CD}{AC} = \frac{PQ}{AC} = \frac{BQ}{AC} - \frac{BP}{AC}$
$\sin(\alpha) = \frac{BQ}{AB} \implies AB\sin(\alpha) = BQ$
$\sin(\alpha - \beta) = \frac{AB\sin(\alpha)}{AC} - \frac{BP}{AC}$
$\frac{AB}{AC} = \frac{1}{\cos(\beta)}$ #corrected
$\sin(\alpha - \beta) = \frac{\sin(\alpha)}{\cos(\beta)} - \frac{BP}{AC}$ # corrected
Im stuck on what to do with $\frac{BP}{AC}$. I've seen the posts here about the derivation of $\sin(\alpha + \beta)$ from the same diagram and I understand that proof perfectly well, but I am stuck on this one.
| Try using the whole angle $\alpha$ as in the proof for the sin and cos addition formulas:
$$\sin \alpha = \frac{BQ}{AB} = \frac{BP}{AB} + \frac{CD}{AB}$$
$$=\frac{BP}{BC} \frac{BC}{AB} + \frac{CD}{AC} \frac{AC}{AB}$$
$$=\cos(\alpha - \beta) \sin \beta + \sin(\alpha - \beta) \cos \beta$$
and also:
$$\cos \alpha = \frac{AQ}{AB} = \frac{AD}{AB} - \frac{PC}{AB}$$
$$= \frac{AD}{AC} \frac{AC}{AB} - \frac{PC}{BC} \frac{BC}{AB}$$
$$= \cos(\alpha - \beta) \cos \beta - \sin(\alpha - \beta) \sin \beta$$
Now let $\cos(\alpha - \beta) = x, \sin(\alpha - \beta) = y$, and we have some simultaneous equations:
$$x \cos \beta - y \sin \beta = \cos \alpha$$
$$x \sin \beta + y \cos \beta = \sin \alpha$$
This means that:
$\sin \beta(x \cos \beta - y \sin \beta) = \cos \alpha \sin \beta \tag{1}$
$\cos \beta(x \sin \beta + y \cos \beta) = \sin \alpha \cos \beta \tag{2}.$
Thus $(1) - (2)$ gives:
$$-y \sin^2 \beta - y \cos^2 \beta = \cos \alpha \sin \beta - \sin \alpha \cos \beta, -y = \cos \alpha \sin \beta - \sin \alpha \cos \beta.$$
Remembering what $y$ is, we have the identity for $\sin(\alpha - \beta)$.
Solving similarly for $\cos(\alpha - \beta)$ gives the desired identity.
| {
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Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$. Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$.
I had hoped that I could use a strategy similar to Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$, wherein the author found that $\lim_{n\to\infty}\sqrt{a_{n+1}}-\sqrt{a_n}=\frac{1}{2}$ and used the telescoping sum $\sum_{k=1}^{n-1}\sqrt{a_{k+1}}-\sqrt{a_k}\approx\frac{1}{2}n$ to approximate $\sqrt{a_n}$. However, $a_{n+1}^{\frac{1}{3}}-a_{n}^{\frac{1}{3}}=\frac{a_{n+1}-a_{n}}{a_{n+1}^{\frac{2}{3}}+a_{n+1}^{\frac{1}{3}}a_{n}^{\frac{1}{3}}+a_{n}^{\frac{2}{3}}}=
\frac{a_{n}^\frac{1}{3}}{a_{n+1}^{\frac{2}{3}}+a_{n+1}^{\frac{1}{3}}a_{n}^{\frac{1}{3}}+a_{n}^{\frac{2}{3}}}=
\frac{1}{\frac{a_{n+1}^{\frac{2}{3}}}{a_n^{\frac{1}{3}}}+a_{n+1}^{\frac{1}{3}}+a_{n}^{\frac{1}{3}}}
$ which converges to $0$. I also tried $a_{n+1}^q-a_n^q$ for different positive and negative values of $q$ to no avail.
If anyone has any hints, I'd appreciate it. Thanks.
| Hint: It suffices to show that there exists a real number $q$ such that $\lim_{n\rightarrow \infty}{{a_n}\over{n^q}}$ converges. Once this is done, call the value $p$, and then divide both sides by $p$.
| {
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Solving definite integral $\int_{0}^{1} \tan^{-1}(1-1/x)dx$ $$\int_{0}^{1} \tan^{-1}\left(1-\frac1x\right)dx$$
Here's what I have done so far. (the answer is given as $-\pi/4$)
Let
$$
I = \int_{0}^{1}\tan^{-1}\left(1-\frac1x\right)dx = \int_{0}^{1}\tan^{-1}\left(\frac{x-1}x\right)dx.
$$
Since, $\int_{0}^{1} f(x)dx = \int_{0}^{1} f(1-x)dx$ one has\begin{align}
I &= \int_{0}^{1}\tan^{-1}\left(1-\frac1{1-x}\right)dx\\
& = \int_{0}^{1}\tan^{-1}\left(\frac x{x-1}\right)dx\\
& = \int_{0}^{1}\frac\pi2-\cot^{-1}\left(\frac x{x-1}\right)dx\\
& = \int_{0}^{1}\frac\pi2-\tan^{-1}\left(\frac{x-1}x\right)dx\\
& = \frac\pi2 - I,
\end{align}
Hence, $I = \dfrac\pi4$. The given answer is $-\dfrac\pi4$. Where have I gone wrong?
| By integration by parts, one has
\begin{eqnarray}
\int_0^1 \arctan\left(1 - \frac1x\right) \, dx &=& -\int_0^1 \frac{x}{1+(1-\frac1x)^2}\frac{dx}{x^2}\\
&=& -\int_0^1 \frac{x}{2x^2-2x+1}dx\\
&=& \boxed{-\frac\pi4}
\end{eqnarray}
Update:
\begin{eqnarray}
\int_0^1 \frac{x}{2x^2-2x+1}dx&=&\int_0^1 \frac{2x}{(2x-1)^2+1}dx\\
&=&\int_0^1 \frac{2x-1}{(2x-1)^2+1}dx+\int_0^1 \frac{1}{(2x-1)^2+1}dx\\
&=&\frac{1}{4}\ln[(2x-1)^2+1]+\frac12\arctan(2x-1)\bigg|_0^1 \\
&=&\frac\pi4.
\end{eqnarray}
| {
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"question_score": "5",
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Two different absorbing matrices seemingly giving the same answer. Can someone please explain to me how these two absorbing matrices,
$
M_1 = \left[
\begin{array}{ccccc}
\frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & 0 \\
\frac{1}{3} & 0 & 0 & 0 & \frac{2}{3} \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right]^n
$
$
M_2 = \left[
\begin{array}{ccccc}
\frac{1}{3} & \frac{2}{3} & 0 & 0 & 0 \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\
\frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} & 0 \\
\frac{1}{3} & \frac{1}{3} & 0 & 0 & \frac{1}{3} \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right]^n
$
when enumerated, that the element in the 1st row and 5th column is seemingly always equal to one another?
As for some background, these matrices can be used to calculate the odds of getting 4 consecutive 1's or four 2's in a row when rolling a 3-sided die. So a die labelled $\{1, 2, 3\}$, you care if you get $\{..., 1, 1, 1, 1, ...\}$ or $\{..., 2, 2, 2, 2, ...\}$. $M_2$ is more intuitive in state-model design, whereas I accidentally stumbled upon $M_1$ as a valid solution.
Also, I could be wrong that the aforementioned two elements are equal $\forall \mathbb{Z^*}$, which I have tested exhaustively, but unable to generate a closed form for.
| Suppose the die faces are A, B, C.
(1,5) of the second matrix is as you said, the chance that within n steps you have seen four As in a row or four Bs in a row.
(1,5) of the first matrix is the chance that within n steps that you have seen three As in a row, then something not an A.
In the second matrix, you have a 2/3 chance of starting such a subsequence (from scratch), a 1/3 chance of continuing it (twice), and a 1/3 chance of finishing it.
In the first matrix, you have a 1/3 chance of starting such a subsequence (from scratch), a 1/3 chance of continuing it (twice), and a 2/3 chance of finishing it.
There some more complications because the chances of not being in a subsequence vs the chance of restarting a subsequence in the middle also has to balance, but that's the rough gist of it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to derive the polar equation of a circle.
I have learnt that the standard equation of a circle whose centre is at (a,b) is
$$(x-a)^2 + (y-b)^2 = c^2
$$
I am trying to derive the polar equation of this circle but I am unfortunately stuck.
$$\begin{align}(x-a)^2+(y-b)^2&=c^2\\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=c^2\\
r^2\cos^2\theta-2ra\cos\theta+a^2+r^2\sin^2\theta-2rb\sin\theta+b^2&=c^2\\
r(r-2a\cos\theta-2b\sin\theta)&=c^2-a^2-b^2\end{align}$$
The following is stated in James Stewart's book on Precalculus but there is no proof stated in this book. So I am trying to reach a similar result by beginning at the standard equation of a circle.
| By to reach a similar result I understood that you are interested in an unparametrized form.
We obtain polar equation by continuing the derivation to its logical conclusion.
$$ r(r-2a\cos\theta-2b\sin\theta)=c^2-a^2-b^2$$
Distance to center from origin $OC=\sqrt {a^2+b^2}$ and power of circle /squared tangent length
$$ a^2+b^2-c^2= OT^2=T^2 $$
$$ = r(r- 2a \cos\theta-2b \sin\theta)=OT^2$$
$$ = r(r- 2 \sqrt {a^2+b^2}\cdot ( \frac{a}{\sqrt {a^2+b^2}}\cos \theta + \frac{b}{\sqrt {a^2+b^2}}\sin \theta) )$$
$$ \alpha=\angle COx= \tan^{-1}\frac{b}{a}, \;\theta=\angle AOx\;$$
$$ =r(r- 2 \sqrt {a^2+b^2}( \cos\alpha \cos \theta + \sin \alpha\sin \theta) )$$
$$OT^2= r\left(r- 2 \sqrt {a^2+b^2} \cos(\alpha- \theta)\right) \tag 1 $$
This can be readily recognized as the property of circle with a constant product of segments made by intersecting chords. Drop perpendicular CM onto OA. M is mid-point of AB, OMC is a right triangle
$$OM=\frac{OA+OB}{2}= OC \cos (\alpha-\theta)$$
$$ OA=r, OB=r- 2\sqrt{a^2+b^2}\cos (\alpha-\theta)= r- 2 Q $$
$$Q=\sqrt{a^2+b^2}\cos (\alpha-\theta) $$
$$ OA\cdot OB=T^2 \;, r(r-2Q)=OT^2;\;\tag 2 $$
Solving the quadratic we obtain
$$ r= Q\pm\sqrt{Q^2-T^2} $$
The radius vector intersects a circle at two points $(A,B)$. Finally the required polar coordinate equations are
$$ r= \sqrt{a^2+b^2}\cos (\alpha-\theta) \pm \sqrt{c^2-(a^2+b^2)\sin^2 (\theta-\alpha)} \tag 3 $$
Positive sign for outer part, negative sign for inner part and radicals disappear at tangent point $r=T.$
It can be verified by trigonometry for any Circle radius $c$ centered at $(a,b)$ the radius must always be $r=T $ at point of tangency.I.e.,
$$ @\; \theta = \tan^{-1}\frac{b}{a} \pm \sin^{-1}\frac{c}{\sqrt{a^2+b^2}}, \; r=T\; \text{ always}$$
Extreme radii $ \; @ \theta= \alpha\; $ are
$$ \sqrt{a^2+b^2}\pm c $$
All the details are shown in the graph drawn for assumed constants
$$(a,b,c)=(4,3,2), r_{max}=7, r_{min}=3, T=\sqrt{21}\approx 4.5826 $$
A Cartesian $(r-\theta)$ plot is also given for a distorted circle in an alternate coordinate representation, as we are more familiar using Cartesian coordinates. It may be ignored at first reading, it is intended to identify four cardinal points.. ( two extreme radii, two points of tangency).
As an exercise you can make the necessary change for a negative power eccentric circle.
Actually I hope it is not the " general" polar equation given in the text-book. It pertains only to two particular cases when the circles pass through the origin, when centered on $(x,y)$ axes, $r_{min}=0 $ but is not about general eccentric circles placed in arbitrary quadrants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x^2\left(4+\frac{2}{x}+\frac{1}{x^2}-a^2-\frac{2ab}{x}-\frac{b^2}{x^2}\right)}{x^2\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
Applying limit on the numerator and denominator
$$\frac{\lim_{x \to \infty}\left(4-a^2+\frac{2-2ab}{x}+\frac{1-b^2}{x^2}\right)}{\lim_{x \to \infty}\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
It can be seen that the denominator tends to $0$
| A variation. We obtain
\begin{align*}
\color{blue}{-\frac{1}{2}}&=\lim_{x\to\infty}\left(\sqrt{4x^2+2x+1}-ax-b\right)\\
&=\lim_{x\to\infty}\left(\sqrt{\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}}-ax-b\right)\\
&=\lim_{y:=2x+\frac{1}{2}\to\infty}\left(\sqrt{y^2+\frac{3}{4}}-a\left(\frac{y}{2}-\frac{1}{4}\right)-b\right)\\
&\,\,\color{blue}{=\lim_{y\to\infty}\left(y\sqrt{1+\frac{3}{4y}}-\frac{a}{2}y+\frac{a}{4}-b\right)}\tag{1}\\
\end{align*}
*
*Since the terms with factor $y$ in (1) need to cancel we obtain $\color{blue}{a=2}$.
*Taking $a=2$ the constant terms in (1) give $\frac{2}{4}-b=-\frac{1}{2}$ and $\color{blue}{b=1}$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Geometric proof for derivative of $f(x)=\sqrt{x}$ Using a square of area $x$,
we obtain
$$dx = (2\sqrt{x})d\sqrt{x}+(d\sqrt{x})^2$$
$$\frac{d\sqrt{x}}{dx} = \frac{1}{2 \sqrt{x}} - \frac{(d\sqrt{x})^2}{2\sqrt{x}dx}$$
How do I show that $\frac{(d\sqrt{x})^2}{2\sqrt{x}dx} \rightarrow 0$ as $dx\rightarrow0$?
| By your geometric construction, and also by your first line, when $dx \rightarrow 0$, $d\sqrt x \rightarrow 0$.
$\frac {dx} {d\sqrt x} = 2\sqrt x + d\sqrt x$
So when $dx \rightarrow 0, \frac {dx} {d\sqrt x} \rightarrow 2\sqrt x$
So when $x\ne 0, \frac {(d\sqrt x)^2} {2\sqrt x dx}$ is equivalent when $dx \rightarrow 0$ to $\frac {d\sqrt x} {4x}$.
When $dx\rightarrow 0$, this $\rightarrow 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the range of a Quadratic divided by a Quadratic I need to find the range of a function:
$$f(x) = \frac{e^{2x} - e^x + 1}{e^{2x} + 3e^x - 7} $$
Substituting $e^x = t $ and equating $f(x)$ to $y$, I get
$$(y-1)t^2 + (3y+1)t - (7y+1) = 0$$
Since $t$ must be real and positive, I get the range of y to be $(-\infty, -\frac{1}{3})\bigcup(1, \infty) $.
Is this correct?
| Substitute $f(x)$ with $y$: $y=\frac{e^{2x}-e^x+1}{e^{2x}+3e^x-7}$.
Rewrite in terms of $e^x$:
$$e^{2x}y+3e^xy-7y=e^{2x}-e^x+1$$
$$(y-1)(e^x)^2+(3y+1)e^x-(7y+1)=0$$
This equation has roots (with respect to $x$), when there are roots with respect to $e^x$ (discriminant is non-negative) and at least one root is positive.
$D=(3y+1)^2-4(y-1)(-7y-1)=37y^2-18y-3\ge0$ or $y\in\left(-\infty, \frac{9-8\sqrt{3}}{37}\right]\cup\left[\frac{9+8\sqrt{3}}{37}, \infty\right)$
In this case $e^x=\frac{-(3y-1)\pm\sqrt{37y^2-18y-3}}{2(y-1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the biggest possible value of $\frac 1a+\frac 1b$ given that $2 <\frac ab < 7$ where $a$ and $b$ are distinct integers and $a + b = 10$. Let $a$ and $b$ be distinct integers that satisfy $2 < \frac ab < 7$. If $a + b = 10$, then what is the biggest possible value of $\frac 1a +\frac 1b$?
It's easy to show that $\frac 1a +\frac 1b=\frac {10}{ab}$ for some $a$ and $b$. Now we just need to find $ab$. However, without finding the interval solutions, how do we get $ab$ from $2 < \frac ab < 7$? Or maybe we don't need to find the value of $ab$? This is where I get stuck and I appreciate any help.
| If you assume that $a$ and $b$ are positive, then there are only nine possible combinations with $a + b = 10$, and of these, only two also meet $2 < \frac{a}{b} < 7$:
*
*$a = 7, b = 3 \implies \frac{1}{a} + \frac{1}{b} = \frac{10}{21} \approx 0.47619$
*$a = 8, b = 2 \implies \frac{1}{a} + \frac{1}{b} = \frac{5}{8} = 0.625$
Clearly, the latter is bigger.
But let's consider the possibility of negative numbers. We're given that $a + b = 10$, so let's just make the substitution $b = 10 - a$ everywhere so that there's only one variable to work with. Then your question becomes:
What is the biggest possible value of $\frac{1}{a} + \frac{1}{10-a}$ given that $2 < \frac{a}{10-a} < 7$?
Obviously $a \ne 10$.
If $a < 10$, then $10 - a$ is positive, so we can multiply the inequality by it without changing the order.
$$2(10 - a) < a < 7(10 - a)$$
$$20 - 2a < a \text{ and } a < 70 - 7a$$
$$20 < 3a \text{ and } 8a < 70$$
$$a > \frac{20}{3} = 6.666... \text{ and } a < \frac{70}{8} = 8.75$$
The only integer values of $a$ meeting this inequality are 7 and 8, which I've already worked out above.
Alternatively, if $a > 10$, then $10 - a$ is negative, so multiplying by it flips the order.
$$2(10 - a) > a > 7(10-a)$$
$$20 - 2a > a \text{ and } a > 70 - 7a$$
$$20 > 3a \text{ and } 8a > 70$$
$$a < \frac{20}{3} = 6.666... \text{ and } a > \frac{70}{8} = 8.75$$
But this is a contradiction, so this case adds no new possible values for $a$ (or $b$). So we still have $a \in \{ 7, 8 \}$. And as already shown, $a = 8$ maximizes the target expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed)
Solve the polynomial in closed form:
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$
WolframAlpha obviously failed.
I tried several ways:
*
*I tried the Rational Root Thereom, but there is no rational root.
*I tried possible factorisations
$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^2+a_1x+a_2)(x^4+a_3x^3+a_4x^2+a_5x+a_6)$$
and
$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^3+a_1x^2+a_2x+a_3)(x^3+a_4x+a_5)$$
But, the expansions are terrible!
| The problem can be rewritten as $$x((x-1)x+1)(x^3+3x-1)+1=0$$ Now solving this we get $6$ roots:
$1.$ $-0.1399-1.7659i$
$2.$ $-0.1399+1.7659i$
$3.$ $-0.04270-0.56305i$
$4.$ $-0.04270+0.56305i$
$5.$ $0.67660-0.73635i$
$6.$ $0.67660+0.73635i$
These are all approximations I got from a software. Strict factorisation $($writing expression only in terms of factors$)$ is not really possible in my opinion or you can use these approx roots for that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Connection between thinking of an ellipse as a squished circle and the formal definition If I think of an ellipse fundamentaly as a squished circle. For exemple, I have a initial circle $x^2+y^2=1$ and I morph into $x^2+ 2y^2=1$.
How can I see, intuitively, that for all points on the curve, the sum of the two distances to the focal points is a constant like the formal definition of an ellipse demands?
| What do we have?
$$x^2+2y^2=1$$
$$\frac{x^2}{1^2}+\frac{y^2}{(1/\sqrt{2})^2}=1$$
$$a=1, b=\frac{1}{\sqrt{2}}$$
$$c=\sqrt{a^2-b^2}=\frac{1}{\sqrt{2}}$$
We need to show
$$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}={\rm const}$$
$$x^2+2y^2=1\Rightarrow y^2=\frac{1-x^2}{2}$$
$$(x-c)^2+y^2=\frac{x^2-2x\sqrt{2}+2}{2}=\frac{(x-\sqrt{2})^2}{2}$$
$$(x+c)^2+y^2=\frac{x^2+2x\sqrt{2}+2}{2}=\frac{(x+\sqrt{2})^2}{2}$$
$$A=\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=\frac{|x-\sqrt{2}|+|x+\sqrt{2}|}{\sqrt{2}}$$
$$x^2=1-2y^2\leq 1\Rightarrow -1\leq x\leq 1\Rightarrow A=\frac{\sqrt{2}-x+x+\sqrt{2}}{\sqrt{2}}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Olympiad Math Problem About Simultaneous Equations Given the question:
*
*$x + y = 1$
*$x^2 + y^2 = 2$
*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substituting this into $2)$ we get a quadratic $2y^2 - 2y - 1 = 0$.
However, when we solve for $y$ we get two possible solutions. Hence how do we proceed from here? Why doesn't simultaneous equations work to solve the above problem?
| $$1=(x+y)^2= x^2+2xy+y^2 = 2+2xy$$$$ \implies xy = -1/2$$
$$4=(x^2+y^2)^2 =x^4+2x^2y^2 +y^4$$$$\implies x^4+y^4= 7/2$$
and $$2=(x+y)(x^2+y^2) = x^3+xy(x+y)+y^3$$ $$\implies x^3+y^3 = 5/2
$$
So
$$7/2=(x+y)(x^4+y^4) = x^5+y^5 + xy(x^3+y^3)$$ $$\implies x^5+y^5 =19/4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Computing the exact value of the integral $∫_0^∞ \tanh(2x)\ln(\tanh x)dx.$ LATEST EDIT
Thanks to @FDP’s alternative method and @Claude Leibovici’s generalisation on the integral. Meanwhile, I had found a formula for the integral with power n in general.
$$\boxed{I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^n}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1)} $$
and its proof is shown below.
Recently, I investigate the integral
$$∫_0^∞ \tanh(2x)\ln (\tanh x)dx,$$
using the substitution $y=\tanh x$.
$$
\begin{aligned}
I &=\int_0^{\infty} \frac{2 \tanh x}{1+\tanh ^2 x} \ln (\tanh x) d x \\
&=\int_0^{\infty} \frac{2 y \ln y}{1+y^2} \cdot \frac{d y}{2\left(1-y^2\right)} \\
&=\int_0^{\infty} \frac{y \ln y d y}{1-y^4} \\
&\stackrel{y^2\mapsto y}{=} \frac{1}{4} \int_0^{\infty} \frac{\ln y}{1-y^2} d y
\end{aligned}
$$
By my post,
$$
\begin{aligned}
&\int_0^{\infty} \frac{\ln y}{1-y^2} d y=-\frac{\pi^2}{4}
\end{aligned}
$$
We now conclude that
$$\boxed{∫_0^∞ \tanh(2x)\ln (\tanh x)dx= -\frac{\pi^2}{16}}$$
Is there any other method to evaluate the integral? Your comments and alternative methods are highly appreciated.
| If we consider
$$I_n=\int \tanh(nx)\,\log(\tanh(x))\,dx$$ there is a (nasty) antiderivative involving logarithms of complex arguments and polylogarithms.
Doing the same as you did and using multiple angles formulae
$$I_1=\int_0^1 \frac y{1-y^2}\log(y)\,dy=-\frac{\pi ^2}{24}$$
$$I_2=\int_0^1 \frac {2y}{1-y^4}\log(y)\,dy=-\frac{\pi ^2}{16}$$
$$I_3=\int_0^1 \frac{y \left(y^2+3\right) }{1-2y^2-3 y^4}\log (y)\,dy=-\frac{5 \pi ^2}{72}-\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{6}-\frac{\log ^2(3)}{12}$$
$$I_4=\int_0^1 \frac{4 y \left(y^2+1\right) }{\left(1-y^2\right) \left(1+6 y^2+y^4\right) }\log (y)\,dy=
-\frac{\pi ^2}{16}-\frac{1}{16} \log ^2\left(3-2 \sqrt{2}\right)$$
$$I_5=\int_0^1 \frac{y \left(y^4+10 y^2+5\right) }{1-9y^2+5y^4+5 y^6}\log (y)\,dy=$$
$$\frac{1}{120} \left(-6 \left(2 \text{Li}_2\left(-1+\frac{2}{\sqrt{5}}\right)-2 \text{Li}_2\left(-5+2 \sqrt{5}\right)+\log ^2\left(5+2 \sqrt{5}\right)\right)-7 \pi ^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the points of tangency given a 2 variable function and a normal vector? I'm given the two variable function $f(x,y) = 4x^2+7y^2+5xy+14$ and asked to find the (2) points on the surface where the vector $6\hat i + 69\hat j + 3\hat k$ is normal to the tangent plane. So far I have written a generalized linearization using the points $f(a,b)$: $$f(a,b) = 4a^2+7b^2+5ab+14$$ $$\Rightarrow z_0 = 4a^2+7b^2+5ab+14$$ $$ $$ $$\frac{\partial f}{\partial x} = 8x+5y$$ $$\Rightarrow x_0 = 8a+5b$$ $$ $$ $$\frac{\partial f}{\partial y} = 14y+5x$$ $$\Rightarrow y_0 = 14b+5a$$ For the following linearization: $$ $$ $$L(x,y) = 4a^2+7b^2+5ab+14+(8a+5b)(x-a)+(14b+5a)(y-b)$$ $$ $$ Rearranging this to be in the standard form of a plane: $$ $$ $$x(8a+5b)+y(14b+5a)-z = 4a^2+7b^2+5ab-14$$ $$ $$ If the goal is to find the two points where the given vector is normal to this plane I am at a loss. Any help from this point would be greatly appreciated. Edit: Fixed an error in the partial derivatives.
| $z = f(x,y) = 4x^2 + 7y^2+5xy + 14$
The normal to the tangent plane is along the gradient of $f$
$\nabla f = \begin{bmatrix} 8 x + 5 y \\ 14 y + 5 x \\-1 \end{bmatrix} $
If $\nabla f$ is parallel to $ (6, 69, 3) $, then their cross product is zero.
$ \bigg(\nabla f\bigg) \times (6, 69, 3) = \bigg( (14 y + 5x)(3) + 69 , 6 (-1) - 3 (8x + 5y) , (8x + 5y) (69) - 6(14 y + 5x) \bigg) $
The right hand side of the above simplifies to
$ (42 y + 15 x + 69, - 24 x - 15 y - 6, 522 x + 261 y \bigg) = 0$
Factoring $3$
$ (14 y + 5 x + 23, - 8 x - 5 y - 2, 174 x + 87 y) = 0 $
Solve the first two equations
$ 5 x + 14 y = -23 $
$ -8 x - 5 y = 2 $
You get
$ x = \dfrac{ (-23)(-5) - (2)(14) }{-25 + 112} = \dfrac{ 87}{87} = 1 $
$ y = \dfrac{-8 - 2}{5} = -2$
Plug that into the third equation
$ (1)(174) + (87)(-2) = 0 $
So the solution of the first two equation satisfies the third equation.
The solution point is $ x = 1, y = -2, z = 4(1)^2 + 7(-2)^2 +5(1)(-2) + 14 = 36 $
And therefore, the point of tangency is $(1, -2, 36) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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compute $\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\mod p$
Let $p$ be an odd prime. Compute $\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\mod p.$
For an odd prime p, let $S(p) = \sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}$. Note that we may assume $p\equiv 1, 5\mod 6,$ since for $p=3$, we have $S(p) = \sum_{k=0}^1 \dfrac{2}{(k!)^2(2-2k)!} = 1+2 = 3\equiv 0\mod p.$ Computing a few more values, we get $S(5) = 4!(1 + 1/(1!^2\cdot 2!) + 1/(2!^2 \cdot 0!)) \equiv 2\mod 5, S(7) = 6! (1 + 1/(1!^2 \cdot 4!) + 1/(2!^2 \cdot 2!) + 1/(3!^2 \cdot 0!)) \equiv 6\mod 7.$ All congruences in this problem will be modulo p for simplicity. First consider the case where $p\equiv 1\mod 6$ and write $p= 6r + 5$ for some $r\ge 0$. Then $(p-1)/2 = 3r+2.$
We have $S(p) = \sum_{k=0}^{(p-1)/2} \dfrac{(p-1)(p-2)\cdots (p-2k)}{(k!)^2} \equiv \sum_{k=0}^{(p-1)/2} \dfrac{(2k)!(-1)^{2k}}{(k!)^2} = \sum_{k=0}^{(p-1)/2} \dfrac{(2k)!}{(k!)^2}$
But I'm not sure how to simplify this result. Would the Binomial Theorem be useful?
| $$\begin{aligned}
(2k)!&=(2k)(2k-2)\cdots2\cdot(2k-1)(2k-3)\cdots1\\
&=k!2^k(2k-1)(2k-3)\cdots1\\
&=k!(4k-2)(4k-6)\cdots2\\
&\equiv_pk!(4k-2-2p)(4k-6-2p)\cdots(2-2p)\\
&=k!(-4)^k(\frac{p-1}2-k+1)(\frac{p-1}2-k+2)\cdots \frac{p-1}2
\end{aligned}$$
The idea of the transformation above comes from this comment.
$$\begin{aligned}
&\sum_{k=0}^{(p-1)/2} \dfrac{(p-1)!}{(k!)^2(p-1-2k)!}\\
&\equiv_p\sum_{k=0}^{\frac{p-1}2} \dfrac{(2k)!}{(k!)^2}\\
&=\sum_{k=0}^{\frac{p-1}2}\frac{k!(-4)^k({\frac{p-1}2}-k+1)({\frac{p-1}2}-k+2)\cdots {\frac{p-1}2}}{(k!)^2}\\
&=\sum_{k=0}^{\frac{p-1}2}(-4)^k\frac{({\frac{p-1}2}-k+1)({\frac{p-1}2}-k+2)\cdots {\frac{p-1}2}}{k!}\\
&=\sum_{k=0}^{\frac{p-1}2}(-4)^k{{\frac{p-1}2}\choose k}\\
&=(1+(-4))^{\frac{p-1}2}\\
&=(-3)^{\frac{p-1}2}\\
&=(-1)^{\frac{p-1}2}3^{\frac{p-1}2}\\
&\equiv_p(-1)^{\frac{p-1}2}\left({\frac {3}{p}}\right)\\
&=(-1)^{\frac{p-1}2}\left({\frac {p}{3}}\right)(-1)^{\frac{3-1}2\frac{p-1}2}\\
&=(-1)^{p-1}\left({\frac {p}{3}}\right)\\
&=\left({\frac {p}{3}}\right)\\
&=\begin{cases}
1 &\text{ if } p\equiv_31,\\
0 &\text{ if } p=3,\\
-1 &\text{ if }p\equiv_3-1\end{cases}
\end{aligned}$$
The place where $(1+(-4))^{\frac{p-1}2}$ appears uses the binomial theorem.
$(\frac{3}{p})$ and $(\frac{3}{p})$ are Legendre symbols.
The place where $(-1)^{\frac{3-1}2\frac{p-1}2}$ appears uses the law of quadratic reciprocity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$ I tried using integration by parts and could get to this point:
\begin{align}
\int \sqrt{x^2 + a^2}\mathrm{d}x &= x\sqrt{x^2+a^2} - \int x\frac{2x}{\sqrt{x^2 + a^2}}\mathrm{d}x
\\\\ &= x\sqrt{x^2+a^2} -2\int \sqrt{x^2 + a^2}\mathrm{d}x + \int \frac{2a^2}{\sqrt{x^2 + a^2}}\mathrm{d}x
\end{align}
then, I don't know how to get:
$$\int \frac{2a^2}{\sqrt{x^2 + a^2}}\mathrm{d}x$$
| Those types of integrals usually take advantage of some properties of the hyperbolic sine and cosine functions - $$\sinh t=\frac{e^t-e^{-t}}{2}\\\cosh t=\frac{e^t+e^{-t}}{2}\\(\sinh t)'=\cosh t\\(\cosh t)'=-\sinh t$$ We solve by plugging in $x=a\sinh t$: $$\int\sqrt{x^2+a^2}dx=\int a\sqrt{a^2\sinh^2 t+a^2}\cosh tdt=\int a^2\sqrt{\sinh^2 t+1}\cosh tdt=a^2\int\sqrt{\cosh^2 t}\cosh^2 tdt=a^2\int \cosh^2 tdt$$
Where the last equality is true since $\sinh^2 t+1=\cosh^2 t$. Another identity of the hyperbolic sine and cosine functions is: $$\cosh^2 t=\frac{\cosh 2t+1}{2}$$ So the integral becomes: $$a^2\int \cosh^2 tdt=a^2\int \frac{\cosh 2t+1}{2}dt=\frac{a^2}{2}\left(\frac{\sinh 2t}{2}+t\right)=\frac{a^2}{2}\left(\sinh t\cosh t+t\right)$$
Since $\sinh 2t=2\sinh t\cosh t$.
Finally, plugging $x$ back, we get:
$$x=a\sinh t\Rightarrow \cosh t=\frac{\sqrt{a^2\sinh^2 t+a^2}}{a^2}=\frac{\sqrt{x^2+a^2}}{a^2}$$
And also: $$x=a\sinh t\Rightarrow t=\text{arcsinh} \frac{x}{a}$$
Hence: $$\int\sqrt{x^2+a^2}dx=\frac{a^2}{2}\left(\frac{\sqrt{x^2+a^2}}{a^2}+\text{arcsinh} \frac{x}{a}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the sum of Arithmetico-geometric series It was asked to find the sum of the AGP series $\left\{\dfrac{e^{\frac{1}{k}}+2 e^{\frac{2}{k}}+3 e^{\frac{3}{k}}+\cdots+k e^{\frac{k}{k}}}{k^2}\right\}$
I tried solving it with the formula for sum of $n^{th}$ terms of AGP series $\frac{a-[a+(n-1) d] r^n}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1-r)^2}$ but my answer is not matching with the given solution.
The solution of this series was given as $-\dfrac{e^{\frac{1}{k}}(e-1)}{\left(e^{\frac{1}{k}}-1\right)^2}+\dfrac{k e^{1+ \frac{1}{k}}}{e^{\frac{1}{k}}-1}$
Thanks for any help.
| $S=e^{\frac{1}{k}}+2 e^{\frac{2}{k}}+3 e^{\frac{3}{k}}+\cdots+k e^{\frac{k}{k}}=e^{\frac{1}{k}}\left(1+2 e^{\frac{1}{k}}+3 e^{\frac{2}{k}}+\cdots+k e^{\frac{k-1}{k}}\right)$
Use your formula, set $a=1, d=1, r=e^{\frac{1}{k}}, n=k$
$S=e^{\frac{1}{k}}\left(\frac{a-[a+(n-1) d] r^n}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1-r)^2}\right)=e^{\frac{1}{k}}\left(\frac{1-ke}{1-e^{\frac{1}{k}}}+\frac{e^{\frac{1}{k}}-e}{(1-e^{\frac{1}{k}})^2}\right)$
$=e^{\frac{1}{k}}\left(\frac{ke}{e^{\frac{1}{k}}-1}+\frac{-1}{e^{\frac{1}{k}}-1}+\frac{e^{\frac{1}{k}}-e}{(e^{\frac{1}{k}}-1)^2}\right)=e^{\frac{1}{k}}\left(\frac{ke}{e^{\frac{1}{k}}-1}+\frac{1-e^\frac{1}{k}}{(e^{\frac{1}{k}}-1)^2}+\frac{e^{\frac{1}{k}}-e}{(e^{\frac{1}{k}}-1)^2}\right)$
$=e^{\frac{1}{k}}\left(\frac{ke}{e^{\frac{1}{k}}-1}+\frac{(1-e)}{(e^{\frac{1}{k}}-1)^2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the limit $\lim_{n\to \infty }\int_{1}^{n}(n^2+x^3)^{-1/2} dx$ \begin{align}
\lim_{n\to \infty }\int_{1}^{n}\frac{dx}{\sqrt{n^2+x^3}}& \overset{x=nt}{=}\int_{1/n}^{1}\frac{dt}{\sqrt{1+nt^3}} \\
& \overset{t^3=p}{=}\int_{1/n^3}^{1}\frac{1/3p^{-2/3}}{\sqrt{1+np}}dp\\
& \overset{pn=q}{=}\frac{n^{-1/3}}{3}\int_{1/n^2}^{n}\frac{t^{-2/3}}{\sqrt{t+1}}dt \\
& \overset{t=\mathrm{tg}^2\theta }{=} 2\int_{0}^{\pi/2}\sin^{-1/3}\theta\cos^{-2/3}\theta d\theta\\
& =\mathrm{B}\left ( \frac{1}{3},\frac{1}{6} \right ) \\
& =\frac{\Gamma \left ( \frac{1}{3} \right )\Gamma \left ( \frac{1}{6} \right )}{\sqrt{\pi}}
\end{align}
Question: Have I calculated the limit correctly?
Why am I asking? After replacing $x=nt$, the limit and the integral can be interchanged and the limit will tend towards zero and the answer will be $0$.
Am I solving it correctly?
| I did direct substitution $x=n^{\frac{2}{3}}\tan^{\frac{2}{3}}\theta$ and $dx=\frac{2}{3}n^{\frac{2}{3}}\tan^{-\frac{1}{3}}\theta\sec^2\theta d\theta$ and got
$$\lim_{n\rightarrow\infty}\int_{\arctan(n^{-2/3})}^{\arctan(n^{1/3})} \frac{\frac{2}{3}n^{\frac{2}{3}}\tan^{-\frac{1}{3}}\sec^2\theta}{n\sec\theta}d\theta=(\lim_{n\rightarrow\infty}n^{-\frac{2}{3}})(\int_{0}^{\pi/2}\sin^{-\frac{1}{3}}\theta\cos^{-\frac{2}{3}}\theta d\theta)=0\times B(\frac{1}{3},\frac{1}{6})=0.$$
As usual first I checked with WA: https://www.wolframalpha.com/input?i=int_1%5E100000000+1%2Fsqrt%28100000000%5E2%2Bx%5E3%29dx
It couldn't solve the indefinite integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex contour integral using Cauchy integral formula I am trying to get integral in contour $C=\left \{ 3\cos(t)+2i\sin(t) : 0\leq t\leq 2\pi \right \}$
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz$$
What I tried was using partial fractions and integrate them separately
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz =\oint_{C} \frac{1}{4(z-1)}dz +\oint_{C} \frac{1}{-4(z+1)}dz+\oint_{C} \frac{1}{2(z-1)^2}dz$$
For $\int_{C}^{} \frac{1}{4(z+1)}dz$ +$\int_{C}^{} \frac{1}{-4(z-1)}dz$, using Cauchy's integral fomula they are zero.
For $\int_{C}^{} \frac{1}{2(z-1)^2}dz$, I tried to use Cauchy's integral formula like:
$$\frac{1}{2}\int_{C}^{} \frac{\frac{1}{(z-1)}}{(z-1)}dz$$
but can't solve it... What am I missing??
| The contour is simply an ellipse $\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$ that encloses both the singularities viz. $z=-1$ and $z=1$ of $f(z)$.
Use deformation principle to see
$\oint_C f(z)dz=\oint_{C_1}\frac{z/(z+1)}{(z-1)^2}dz+\oint_{C_2}\frac{z/(z-1)^2}{(z+1)}dz$
where $C_1$ and $C_2$ are circles enclosing the singularities $z=1$ and $z=-1$ respectively.
$\oint_C f(z)dz=2\pi i\left [\frac{d}{dz}\left (\frac{z}{z+1} \right ) \right ]_{z=1}+2\pi i\left [ \frac{z}{(z-1)^2} \right ]_{z=-1}=2\pi i\times (-1/4)+2\pi i\times (1/4)=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove that the distance between two points is longer than one point in the normal and the other point in a plane? So, imagine $P$ is a plane with the normal $n$, and $A = (a1, a2, a3)$ is a point above the plane $P$ that does not lie on the plane. $B$ is a point collinear on the line $L$: $r = (a_1, a_2, a_3) + \lambda n$.
$C$ is a point on $P$ other than $B$.
How do I prove $\|A-C \| > \|A-B \|$ for all points $C \ne B$?
I tried drawing a diagram with points $A$ and $B$ in a line $L$ parallel to the normal, since the shortest distance from $A$ to $P$ is $L$ perpendicular. I drew a line from $B$ to $C$. Since the $L$ is parallel to the normal $n$, $L$ is perpendicular to the plane $P$.
Hence, this would be a right-angled triangle $ABC$.
Now, a hypothenuse is longer than the other two sides.
What's the best way to approach this question?
| You have $\mathbf{B} = \mathbf{A} + \lambda \mathbf{n} $ for some $\lambda$. Point $\mathbf{C}$ is in the plane and $\mathbf{C} \ne \mathbf{B}$, therefore,
$ \mathbf{n} \cdot (\mathbf{C} - \mathbf{B}) = 0 $
Now
$ \mathbf{A} - \mathbf{C} = (\mathbf{A} - \mathbf{B}) + (\mathbf{B} - \mathbf{C}) = -\lambda \mathbf{n} + \mathbf{CB} $
It follows from this that
$\begin{equation} \begin{split}
\| \mathbf{A - C} \|^2 &= (\mathbf{A - C}) \cdot (\mathbf{A - C} ) \\&= (-\lambda \mathbf{n} + \mathbf{CB} ) \cdot (-\lambda \mathbf{n} + \mathbf{CB}) \\
&= \lambda^2 \| \mathbf{n} \|^2 + \| \mathbf{CB} \|^2 \end{split} \end{equation} $
Because $\mathbf{n} \cdot \mathbf{CB} = 0 $
Therefore,
$\| \mathbf{A - C} \|^2 \gt \| \mathbf{B} - \mathbf{C} \|^2 $
because $\lambda^2 \| \mathbf{n} \|^2 \gt 0$ for $\lambda \ne 0 $
Taking the square root of both sides of the above inequality, we deduce that
$\| \mathbf{A - C} \| \gt \| \mathbf{B} - \mathbf{C} \| $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Value of a cyclic sum Let $a,b,c$ satisfy the equation $x^3+px^2+qx+r=0$. Is it possible to determine $\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}$ in terms of $p,q,r$? I stumbled upon this while thinking about an inequality problem. What i could do so far is this:
We have the relations $a+b+c=-p,ab+bc+ca=q,abc=-r$ from vietas relations. So by slightly rewriting the terms we can make them a little bit symmetric like $\frac{abc}{ac+bc}$. In this manner,the numerators become constant since $abc$ is constant. But what bothers me is we will still be left with the denominators which are not completely free of $a,b$ or $c$. And evalutating them by expansion seems to be a daunting task. Is there any clever way to approach this?
| Note that $\frac{ab}{a+b} = \frac{1}{\frac{1}{a}+\frac{1}{b}}$.
Suppose we have the monic cubic $f(x) = x^3+a_2x^2+a_1x+a_0$ with roots $\alpha, \beta, \gamma$. We can make use of the following two operations:
*
*$\frac{1}{a_0}x^3f(\frac{1}{x})$ is the unique monic cubic with roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$.
*$-f(\alpha+\beta+\gamma-x) = -f(-a_2-x)$ is the unique monic cubic with roots $\alpha+\beta,\,\alpha+\gamma,\,\beta+\gamma $.
Applying (1) to the original cubic:
$x^3+\frac{q}{r}x^2+\frac{p}{r}x+\frac{1}{r}$ has roots $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$.
Applying (2) to the result:
$$\left(x+\tfrac{q}{r}\right)^3 - \tfrac{q}{r}\left(x+\tfrac{q}{r}\right)^2 +\tfrac{p}{r}\left(x+\tfrac{q}{r}\right) - \tfrac{1}{r} = x^3+\tfrac{2q}{r}x^2+\tfrac{q^2+pr}{r^2} x + \tfrac{pq-r}{r^2}$$
has roots $\frac{1}{a}+\frac{1}{b},\, \frac{1}{a}+\frac{1}{c},\, \frac{1}{b}+\frac{1}{c}$.
Finally, applying (1) again gives
$$x^3+\tfrac{q^2+pr}{pq-r}x^2 + \tfrac{2qr}{pq-r}x+\tfrac{r^2}{pq-r}$$
which has roots $\frac{1}{\frac{1}{a}+\frac{1}{b}},\,\frac{1}{\frac{1}{a}+\frac{1}{c}},\,\frac{1}{\frac{1}{b}+\frac{1}{c}}$.
Your cyclic sum is the sum of the roots of this polynomial, which by Vieta is $$-\frac{q^2+pr}{pq-r}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2+B(x-3)(x+2) + C(x-3)}{(x-3)(x+2)} = \frac{Ax^2 + 4Ax + 4A + Bx^2 - Bx - 6B +Cx - 3C}{(x-3)(x+2)}$
Looking at the numerator: $x^2(A+B) + x(4A-B+C) + (4A-6B-3C)$
So, comparing coefficients:
$A+B=1$,
$4A-B+C=0$
$4A-6B-3C=0$
I am struggling to solve these 3 equations to find A,B,C
| $$B=1-A$$
Putting this in 2nd equation.
$$4A-1+A+C=0\\\implies C=1-5A$$
Now, we have both $B$ and $C$ in terms of $A$. Putting that in 3rd equation.
$$4A-6+6A-3+15A=0\\\implies 25A=9\\\implies A=\frac9{25}$$
| {
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"url": "https://math.stackexchange.com/questions/4555146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is ${n \choose n/2}$ approximated in this manner? In this question, the author asked how to get an asymptotic growth of ${n \choose n/2}$, to which the answer is as follows. First:
$$ {n \choose n/2} = \frac{n!}{(n/2)!(n- n/2)!} = \frac{n!}{(\tfrac{n}{2}!)^2} $$
Now, to find an upper bound on this, take the upper bound of Stirling's approximation for $n!$ and the lower bound for $\tfrac{n}{2}!$ which if I'm not mistaking gives the following:
$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} $$
However, the answer states this should give:
$$ {n \choose n/2} \leq \frac{e \cdot n^{n + \tfrac{1}{2}} e^{-n}}{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}}\Big )^2} $$
I'm confused as to how one would get from the former to the latter. Up to now I get stuck around here:
$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi n} \cdot n^n e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} \cdot \tfrac{n}{2}^\tfrac{n}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi} \cdot n^{n + \tfrac{1}{2}} e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2}
$$
Now, do the latter terms just disappear due to $\lim_{n \rightarrow \infty} \frac{e^{\tfrac{1}{12n}}}{\Big ( e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} = 0$? But in this case, why should there still be an $e$ term in the numerator? Also what happened to the $\sqrt{2 \pi}$ in the numerator? It cannot have canceled out with the one in the denominator since the latter is still there...
| Hint: It seems the answerer didn't use the inequality chain stated in the Wiki-page
as starting point, but started instead with the asymptotic formula
\begin{align*}
\color{blue}{n!\sim\left(\frac{n}{e}\right)^n\sqrt{2\pi n}}\tag{1}
\end{align*}
We obtain from (1) assuming even $n$:
\begin{align*}
\color{blue}{\left(\frac{n}{2}\right)!}&\sim\left(\frac{n/2}{e}\right)^{n/2}\sqrt{\pi n}=\left(\frac{n}{2e}\right)^{n/2}\sqrt{\pi n}\\
&=\left(\frac{n}{2}\right)^{n/2}e^{-n/2}\sqrt{\pi n}=\left(\frac{n}{2}\right)^{n/2}e^{-n/2}\sqrt{2\pi}\sqrt{\frac{n}{2}}\\
&\color{blue}{=\left(\frac{n}{2}\right)^{\frac{n+1}{2}}\sqrt{2\pi}e^{-n/2}}\tag{2}
\end{align*}
and squaring (2) gives the denominator of the expression under consideration
\begin{align*}
\binom{n}{n/2}\stackrel{?}{\leq}\frac{e\,n^{n+\frac{1}{2}}e^{-n}}{\left(\left(\frac{n}{2}\right)^{\frac{n+1}{2}}\sqrt{2\pi}e^{-n/2}\right)^2}
\end{align*}
From (1) and (2) we derive the asymptotic formula
\begin{align*}
\color{blue}{\binom{n}{n/2}}&\sim\frac{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}}{\left(\left(\frac{n}{2}\right)^{\frac{n+1}{2}}\sqrt{2\pi}\,e^{-n/2}\right)^2}\\
&=\frac{\sqrt{2\pi}\,n^{n+\frac{1}{2}}\,e^{-n}}{\left(\left(\frac{n}{2}\right)^{\frac{n+1}{2}}\sqrt{2\pi}\,e^{-n/2}\right)^2}\tag{$ \sqrt{2\pi}\leq e$}\\
&\,\,\color{blue}{\leq \frac{e\,n^{n+\frac{1}{2}}\,e^{-n}}{\left(\left(\frac{n}{2}\right)^{\frac{n+1}{2}}\sqrt{2\pi}\,e^{-n/2}\right)^2}}
\end{align*}
where the last inequality holds only if $n$ is sufficiently large.
| {
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"timestamp": "2023-03-29T00:00:00",
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finding a coefficient to obtain a geometric sequence I have the following:
$$u_0=2 \\
u_{n+1} = \frac{2}{3}(u_n)-n-\frac{8}{3}\\
v_n=u_n + x\cdot n -1$$
I need to find the value of $x$ in order for $v_n$ to be a geometric sequence, which mean that I need to compute the following :
$$\frac{v_{n+1}}{v_n}$$
edit:
thanks for the hints but when i solve the equation it give 2 roots
| The hard part is to show that $$ u_{n} = \left(\frac{2}{3}\right)^n - 3 \, n + 1.$$
The remaining part is
\begin{align}
v_{n} &= u_{n} + x \, n - 1 \\
&= \left(\frac{2}{3}\right)^n + (x - 3) \, n.
\end{align}
To show the form of $u_{n}$ compute the first few terms to develop a pattern and then check:
\begin{align}
f_{n} &= \frac{2}{3} \, u_{n} - n - \frac{8}{3} \\
&= \frac{2}{3} \, \left(\frac{2}{3}\right)^n - 2 \, n + \frac{2}{3} - n - \frac{8}{3} \\
&= \left(\frac{2}{3}\right)^{n+1} - 3 \, (n+1) + 1 \\
&= u_{n+1}.
\end{align}
With these components then the desired answer can quickly be obtained.
| {
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Approximations of $\sum_{i=0}^k\binom{2k+1}{i}p^i(1-p)^{k-i}$ Given a biased coin with a probability $p$ of coming up heads, what is the probability that strictly more than half of $2k+1$ flips turn up heads?
This probability can be modelled as
$$\sum_{i=k+1}^{2k+1}\binom{2k+1}{i}p^{i}(1-p)^{2k+1-i} = p^k\sum_{i=0}^{k}\binom{2k+1}{i}p^i(1-p)^{k-i}\qquad(1)$$
This last sum is quite close to the following identity
$$1 = (p + (1-p))^k = \sum_{i=0}^k\binom{k}{i}p^i(1-p)^{k-i} \qquad (2)$$
which leads me to think there is some way of approximating $(1)$ using $(2)$, but I am coming up short with good approaches. How might I approximate the aforementioned probability?
| It looks similar but it really isn't. The binomial identity sum is, of course, always $1$, whereas the probability you are interested in has behavior that differs dramatically depending on whether $p > \frac{1}{2}, p = \frac{1}{2}$, or $p < \frac{1}{2}$.
Instead we can use the Chernoff bound. Let $X \sim \text{Bin}(2k+1, p)$ be the number of heads that occur when we flip a coin $2k+1$ times whose bias is $p$. Then the Chernoff bound gives that if $p \le \frac{1}{2}$ then
$$\mathbb{P} \left( \frac{X}{2k+1} \ge \frac{1}{2} \right) \le \exp \left( -(2k+1) KL \left( \frac{1}{2}, p \right) \right)$$
where $KL \left( \frac{1}{2}, p \right) = - \frac{1}{2} \log 4p(1 - p)$ is a KL divergence. (Strictly speaking we want $\mathbb{P} \left( \frac{X}{2k+1} \ge \frac{k+1}{2k+1} \right)$ which will give a slightly better bound but it'll be messier to work with.) This gives
$$\boxed{ \mathbb{P} \left( \frac{X}{2k+1} \ge \frac{1}{2} \right) \le (4p(1-p))^{\frac{2k+1}{2} } }$$
which tells us that for $p < \frac{1}{2}$ the probability decays exponentially quickly in $n$, with the base of the exponential depending on $p$, such that the exponential decay is faster the smaller $p$ is, which should be fairly intuitive. For example if $p = \frac{1}{3}$ the base of the exponential is $\frac{4}{3} \cdot \frac{2}{3} = \frac{8}{9}$. Note that $4p(1 - p) \le 1$ and attains its maximum at $p = \frac{1}{2}$, which should also be fairly intuitive.
By substituting $1 - p$ for $p$, or by applying the Chernoff bound in the other direction, we similarly get that if $p \ge \frac{1}{2}$ then
$$\mathbb{P} \left( \frac{X}{2k+1} \le \frac{1}{2} \right) \le \exp \left( -(2k+1) KL \left( \frac{1}{2}, 1 - p \right) \right) = \exp \left( -(2k+1) KL \left( \frac{1}{2}, p \right) \right)$$
which now gives that if $p > \frac{1}{2}$ then
$$\boxed{ \mathbb{P} \left( \frac{X}{2k+1} > \frac{1}{2} \right) \ge 1 - (4p(1-p))^{\frac{2k+1}{2}} }.$$
So we get the opposite behavior: instead of a probability that decays exponentially we get a probability which is exponentially close to $1$, which should again be intuitive.
The case $p = \frac{1}{2}$ should be analyzed separately; both of the above bounds trivialize in this case but of course we can just exactly compute the answer, which is $\frac{1}{2}$ by symmetry.
| {
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} |
Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but since $(\sin x + \cos x) \in\left(0, \sqrt2\right]$ the maximum possible value of $(\sin x + \cos x)(6 - \sin x)$ can be $(6+ \dfrac{\sqrt2}{2}) * \sqrt2 = 1+6\sqrt2$ which is greater than $9$.
My second attempt was to find the derivative, but I couldn't find its roots.
| Proof 1: Note that $\sin x + \cos x = \sqrt 2 \, \sin (x + \pi/4) \le \sqrt 2$
and $6 - \sin x > 0$.
If $\sin x \ge 0$, we have
$(\sin x + \cos x)(6 - \sin x) \le \sqrt 2 \cdot (6 - \sin x)
\le \sqrt 2 \cdot 6 < 9$.
If $\sin x < 0$, we have
$\sin x + \cos x \le \cos x \le 1$ and thus $(\sin x + \cos x)(6 - \sin x)
\le 6 - \sin x \le 7 < 9$.
We are done.
$\phantom{2}$
Proof 2: $(a - b)^2 \ge 0$ yields $ab \le \frac{(a + b)^2}{4}$ for all real numbers $a, b$.
We have
\begin{align*}
(\sin x + \cos x)(6 - \sin x) &= \frac14 \cdot 4(\sin x + \cos x)\cdot (6 - \sin x)\\
&\le \frac14 \cdot \frac{(4\sin x + 4\cos x + 6 - \sin x)^2}{4}\\
&=\frac{(3\sin x + 4\cos x + 6)^2}{16} \\
&\le \frac{(5 + 6)^2}{16}\\
& = \frac{121}{16} = 7.5625
\end{align*}
where we have used $(3\sin x + 4\cos x)^2
\le (3^2 + 4^2)(\sin^2 x + \cos^2 x) = 25$ (C-S inequality)
to get $-5 \le 3 \sin x + 4\cos x \le 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Minimal polynomial of $\cos\left(\frac{\pi}{13}\right)$ Find the Minimal polynomial of $\cos\left(\frac{\pi}{13}\right)$.
My try:
Let $$13 \theta=\pi \implies 9\theta=\pi-4\theta$$
$$\implies \cos(9\theta)=-\cos(4\theta)$$
Now Let $x=\cos(\theta)$
$$\implies 4(4x^3-3x)^3-3(4x^3-3x)=-(2(2x^2-1)^2-1) $$
$$\implies 256 x^9-576 x^7+432 x^5+16x^4-120 x^3-16 x^2+9x+2=0$$
Now how to test that this is Minimal?
| Chebyshev polynomial of the first kind $T_p(x)$ satisfies $T_p(\frac{\zeta^{i}+\zeta^{-i}}{2})=1$ for all $0\leq i\leq p-1$ where $\zeta=e^{\frac{2\pi i}{p}}$.
So, the minimal polynomial of $\cos(\frac{2\pi i}{p})$ over integers is
$$m_p(x)=\sqrt{\frac{T_p(x)-1}{x-1}}.$$
But, I don't know its series expansion.
From WA: $m_{13}(x)=-1 + 6 x + 24 x^2 - 32 x^3 - 80 x^4 + 32 x^5 + 64 x^6$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $? Latest Edit
We are glad to see there are 4 alternative solutions which give the same closed form to the integral:
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$
where $\psi$ denotes the Digamma Function.
In the post, we found that
$$\int_0^{\infty} \frac{\ln x}{a^2+t^2} d x =\frac{\pi \ln a}{2 a }$$
Now I want to generalise the integral as
$$
I_n=\int_0^{+\infty} \frac{\ln t}{\left(a+t^2\right)^n} d t
$$
where $n\in N$.
Replacing a by $\sqrt{a}$ and differentiating both sides w.r.t. $a$ by $n$ times yields
$$
\begin{aligned}
&J(a)=\int_0^{\infty} \frac{\ln t}{a+t^2} d t=\frac{\pi}{4 \sqrt{a}} \ln a\\
&\frac{d^n}{d a^n}(J(a))=\frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) \\ & (-1)^n n ! \int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)} d t= \frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)\\& \boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{(-1)^n \pi}{4 n !} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)}
\end{aligned}
$$
By Wolfram-alpha, we have
$$
\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)=(-1)^n a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)\cdots (*)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
In particular, when $a=1$, we have
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} \left(\frac{1}{2}\right)_n\left(\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
For example,
$$$$
\begin{aligned}
\int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right)
=-\frac{23 \pi}{96}
\end{aligned}
$$
$$
My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?
| The generalized Leibniz rule gives
$$\newcommand{\d}{\mathrm{d}}
\frac{\d^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1}
= \sum_{k=0}^n \binom n k \left( \frac{\d^{n-k}}{\d x^{n-k}} \ln(x) \bigg|_{x=1} \right)\left( \frac{\d ^{k}}{\d x^{k}} x^{-1/2} \bigg|_{x=1} \right)$$
For the $(n-k)$th derivative of $\ln(x)$, it is easy to show by, e.g., induction that
$$\frac{\d ^{n-k}}{\d x^{n-k}} \ln(x)
= \frac{(n-k-1)! \cdot (-1)^{n-k-1}}{x^{n-k}}$$
Similarly,
$$\begin{align*}
\frac{\d ^k}{\d x^k} x^{-1/2}
&= \frac{-1}{2} \frac{-3}{2} \frac{-5}{2} \cdots \frac{ -(2k-1)}{2} x^{-k-1/2} \\
&= \frac{(-1)^k (2k-1)!!}{2^k} x^{-k-1/2}
\end{align*}$$
so
$$\begin{align*}
\frac{\d ^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1}
&= \sum_{k=0}^n \binom n k (n-k-1)! \cdot (-1)^{n-k-1} \cdot \frac{(-1)^k (2k-1)!!}{2^k} \\
&= (-1)^{n-1} \cdot \sum_{k=0}^n \binom n k (n-k-1)! \cdot \frac{(2k-1)!!}{2^k} \\
&= (-1)^{n-1} n! \cdot \sum_{k=0}^n \frac{1}{(n-k) \cdot k!} \cdot \frac{(2k-1)!!}{2^k} \\
\end{align*}$$
I don't know if there's a meaningful simplification beyond this.
| {
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"url": "https://math.stackexchange.com/questions/4562577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate the limit of a sequence by Riemann sums and mean value theorem Calculate:
$\displaystyle \lim_{n \rightarrow \infty} \left( \frac{n \pi}{4} - \left( \frac{n^2}{n^2+1^2} + \frac{n^2}{n^2+2^2} + \cdots \frac{n^2}{n^2+n^2} \right) \right)$.
I solved it by taking into account that $\displaystyle \int_0^{1} \frac{1}{1+x^2} \mathrm{d}x = \frac{\pi}{4}$ and let the given sequence be:
$a_n= \displaystyle \frac{n \pi}{4} - \left( \frac{n^2}{n^2+1^2} + \frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right)$
Let $f(x) = \frac{1}{1+x^2}$, then:
$a_n = \displaystyle \frac{n \pi}{4} - \sum_{i=1}^n \frac{1}{1+\left( \frac{i}{n} \right)^2} = n \int_0^{1} f(x) \mathrm{d}x - \sum_{i=1}^n f\left( \frac{i}{n} \right) = n \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} f(x) \mathrm{d}x - n \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} f\left( \frac{i}{n} \right) \mathrm{d}x = n \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} \left( f(x)- f\left( \frac{i}{n} \right) \right) \mathrm{d}x$
Using Mean Value Theorem and doing a lot of calculations, I finally get that the limit is $\displaystyle \frac{1}{4}$.
Is it correct? Is there an easier method to solve the problem?
| Let us use the following
Lemma: If $f\in C^1[0,1]$ then we have $$\lim_{n\to\infty} \sum_{k=1}^nf(k/n)-n\int_0^1 f(x) \, dx=\frac{f(1)-f(0)}{2}$$ Using $f(x) =1/(1+x^2)$ we get your desired limit as $1/4$. The lemma is proved using mean value theorem and matches your approach. An extension of the lemma is available here.
| {
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"url": "https://math.stackexchange.com/questions/4564940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $n$ be a fixed positive integer such that $\sin(\frac\pi{2n})+\cos(\frac\pi{2n})=\frac{\sqrt n}2$ then find $n$.
Let $n$ be a fixed positive integer such that $\sin(\frac\pi{2n})+\cos(\frac\pi{2n})=\frac{\sqrt n}2$ then find $n$.
This question already exists here, but the answer mentioned there is incorrect. That post says the answer is $4\lt n\lt 8$. But the answer is only $n=6$.
How to show this mathematically, without calculator?
My Attempt:
The given equation is $$\sin(\frac\pi4+\frac\pi{2n})=\frac{\sqrt n}{2\sqrt2}$$
Also, $$\frac\pi4\lt\frac\pi4+\frac\pi{2n}\le\frac\pi4+\frac\pi2\\\implies \frac1{\sqrt2}\le\sin(\frac\pi4+\frac\pi{2n})\le1\implies\frac1{\sqrt2}\le\frac{\sqrt n}{2\sqrt2}\le1\\\implies2\le\ n\le2\sqrt2\\\implies4\le n\le8$$
| I try to solve a more general equation $$
\sin \left(\frac{\pi}{2 x}\right)+\cos \left(\frac{\pi}{2 x}\right)=\frac{\sqrt{x}}{2} \textrm{ for }x\geq 1\tag{*} $$
Squaring both sides yields $$
\begin{gathered}
\sin ^2\left(\frac{\pi}{2 x}\right)+\cos^2\left(\frac{\pi}{2 x}\right)+2 \sin \left(\frac{\pi}{2 x}\right) \cos \left(\frac{\pi}{2 x}\right)=\frac{x}{4} \\
0<\sin \left(\frac{\pi}{x}\right)=\frac{x}{4}-1 <1\Rightarrow 4\leq x\le 8\cdots (**).
\end{gathered}
$$
Let’s consider the function $f(x)=\sin \left(\frac{\pi}{x}\right)-\frac{x}{4}+1, \textrm{ where }x\in[4,8].$
Since both $\sin \left(\frac{\pi}{x}\right)$ and $-\frac{x}{4}$ are strictly decreasing on $[4, 8]$ implies that $f(x)$ is strictly decreasing on $[4,8]$. Noticing that $f(\frac{\pi}{6})=\sin\left(\frac{\pi}{6}\right) -\frac{6}{4}+1=0$ from $(**)$$\Rightarrow f(x)\ne 0 $ for any $6\ne x\in [4,8],$ we can conclude that $x=6$ is the unique solution to the given equation $(*). $
$$
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all triples $(x,y,z)$ of positive integers such that $5(x²+2y²+z²)=2(5xy-yz+4xz)$... Find all triples of $(x,y,z)$ of positive integers such that $5(x^2+2y^2+z^2)=2(5xy-yz+4xz)$, and at least one of $x,y,z$ is prime
My approach: Put $x,y,z=2$ one by one then we get value as $(x,y,z)=(14/3,2,10/3)$ and $(14/5,6/5,2)$. Now since RHS is a multiple of 2 LHS should also be $0 \pmod 2$, so at least one of $x,y,z$ must be equal to $2.$
So there is no integral solution.
| Solving as a quadratic in $x$, we have
$$
x = \pm \frac{1}{5}\sqrt{-(5y-3z)^2} + y + \frac{4z}{5}
$$
Since $(5y-3z)^2$ is non-negative, the only possibility is $5y-3z = 0$ or
$$y = \frac{3z}{5}$$
This gives
$$
x = 0 + \frac{3z}{5} + \frac{4z}{5} = \frac{7z}{5}
$$
Now substitute $x$ and $y$ in terms of $z$ in the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C = \sec A + \sec A - \tan A$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{{{\sec }^2}A - {{\tan }^2}A}}{{\sec A + \tan A}}} \right) \Rightarrow \sqrt 3 \sin C = \sec A + \left( {\frac{1}{{\sec A + \tan A}}} \right)$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{1}{{\cos A}} + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{{1 + \sin A + 1 - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
$\sqrt 3 \sin C = \frac{{2 + \sin A - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
Not able to proceed further
| Use Weierstrass substitution
$$\sqrt3\sin C=\dfrac{2(1+t^2)-2t}{1-t^2}$$
$$\iff t^2(2+\sqrt3\sin C)-2t+2-\sqrt3\sin C=0$$
The discriminant $=2^2-4(2^2-3\sin^2C)=-12\cos^2C$ which is $\le0$
But we need $-12\cos^2C\ge0\implies12\cos^2C=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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On solutions to special eighth/fourth powers $a^8+b^4 = c^8+d^4$? By the Lander-Parkin conjecture, it is assumed there are no non-trivial integer solutions to either,
$$a^8+b^8+c^8 = d^8+e^8+f^8\tag1$$
$$a^8+b^8 = c^8+d^8\tag2$$
However, if we relaxed $(1)$ a bit, then,
$$1118^8 + 1937^8 + 2502045^4 = 455^8 + 1846^8 + 3200691^4\tag{3a}$$
and after removing the common factor $13^8$ (pointed out by Adam Bailey),
$$86^8 + 149^8 + 14805^4 = 35^8 + 142^8 + 18939^4\tag{3b}$$
and infinitely many other primitive solutions like it. Likewise, if we relaxed $(2)$ to either,
$$a^8+b^4 = c^8+d^4\tag4$$
$$a^8+b^4 = c^4+d^4\tag5$$
then are there non-trivial solutions to $(4)$ and $(5)$?
P.S. In response to Bailey's answer, I forgot to add that solutions must be primitive. Otherwise, $(5)$ is easily solved.
| Solutions of (5) can easily be obtained from solutions of the 4-2-2 equation. Given any integers $A,B,C,D$ such that:
$$A^4+B^4=C^4+D^4$$
we have:
$$A^8+(AB)^4=(AC)^4+(AD)^4$$
Taking for example the smallest 4-2-2 solution:
$$59^4+158^4=133^4+134^4$$
we have:
$$59^8+9322^4=7847^4+7906^4$$
Unfortunately there doesn't appear to be any similar way to infer solutions of (4).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{1}{(y+z) x^4} + \frac{1}{(x+z) y^4} + \frac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$, with $xyz=1$ How to prove that $\frac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$...
I've tried substituting $a=1/x$, $b=1/y,$ $c=1/z$, and also writing the 1 that is being divided into $(xyz)^4.$ In both of those cases I tried applying Titu's lemma, but didn't manage to do anything. In the first case, I've observed that it is similar to Nesbitt's inequality, just with an extra $a^2$,$b^2$ and $c^2$ multiplied
| By Holder and AM-GM we obtain:
$$\sum_{cyc}\frac{1}{(y+z)x^4}=\frac{\sum\limits_{cyc}\frac{1}{(y+z)x^4}\sum\limits_{cyc}(y+z)x\sum\limits_{cyc}1}{6(xy+xz+yz)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}\frac{1}{x}\right)^3}{6(xy+xz+yz)}=\frac{(xy+xz+yz)^2}{6}\geq\frac{9}{6}=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to verify that the roots are rational, we need to prove that the discriminant of this equation (which I'll denote with $D$) is greater than $0$. In other words:
$D=B^2-4AC>0$ where $B=c(3a^2+b^2)$, $A=abc^2$ and $C=3a^2-ab+b^2$ Therefore:
$D=[c(3a^2+b^2)]^2-4(abc^2)(3a^2-ab+b^2)$
$D=[c^2(9a^4+6a^2b^2+b^4)]+[c^2(-12a^3b+4a^2b^2-4ab^3)]$
$D=c^2(9a^4+10a^2b^2+b^4-12a^3b-4ab^3)$
I'm not quite sure where to proceed from here. And this certainly does not prove that the roots are real as this expression can be negative. Is there a way to proceed from here? Are there any better or alternative ways to answer this? Please share your approaches!
| Since this problem was written for high-school contestants, I also have my doubts as to how many of them would spot the factorability of the discriminant. I decided to take the approach of writing the quadratic polynomial in "vertex form", thus
$$ abc^2·x^2 \ + \ c·( \ 3a^2 + b^2 \ )·x \ + \ ( \ 3a^2 - ab + b^2 \ ) $$
$$ = \ \ abc^2 · \left[ \ x^2 \ + \ \frac{\overbrace{3a^2 \ + \ b^2}^{K}}{abc}·x \ \right] \ + \ ( \ K \ - \ ab \ ) $$
$$ = \ \ abc^2 · \left( \ x \ + \ \frac{K}{2abc} \ \right)^2 \ + \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4a^2b^2c^2}·abc^2 \ \right) $$
$$ = \ \ abc^2 · \left( \ x \ + \ \frac{K}{2abc} \ \right)^2 \ + \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \ , $$
with $ \ K \ > \ 0 \ $ and (necessarily) $ \ ab \ \neq \ 0 \ \ . \ $ (I am presuming, as it is not stated, that $ \ a \ , \ b \ , \ c \ $ are all real.)
The only sign dependence of significance in this expression is that of the recurrent factor $ \ ab \ \ : \ $ if $ \ ab \ > \ 0 \ \ $ (the parabola "opens upward"), then this quadratic polynomial has real zeroes if $ \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \le \ 0 \ \ ; \ $ if $ \ ab \ < \ 0 \ \ $ (the parabola "opens downward"), then the direction of this "coefficients" inequality is reversed. With $ \ ab \ > \ 0 \ , \ $ the term $ \ \frac{K^2}{4ab} \ $ is also positive, so it remains to determine whether $ \ K \ \le \ ab \ + \ \frac{K^2}{4ab} \ \ . \ $ Since both the numerator and denominator in these ratios are positive, we may apply the AM-GM inequality to show that indeed
$$ \frac{ ab \ + \ \frac{K^2}{4ab}}{2} \ \ \ge \ \ \sqrt{ab · \frac{K^2}{4ab} } \ \ \Rightarrow \ \ ab \ + \ \frac{K^2}{4ab} \ \ \ge \ \ 2·\sqrt{ \frac{K^2}{4} } \ = \ K \ \ . $$
Hence, $ \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \le \ 0 \ $ for this case. [This portion ends up being similar to Reza Rajaei's argument.] With $ \ ab \ < \ 0 \ \ , \ \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ $ is equivalent to a sum of three positive terms, hence $ \ \left( \ K \ - \ ab \ - \ \frac{K^2}{4ab} \ \right) \ \ge \ 0 \ \ . $
Therefore, the polynomial in question always has at least one real zero. (Other than that we must have $ \ c \ \neq \ 0 \ \ , \ $ the value of $ \ c \ $ is of no consequence for this conclusion.)
| {
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Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1.$
Let $n$ be a positive integer and let $z$ be a root of $x^n-x-1.$ Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1$.
Solution. Write $z = re^{i\theta}, r > 0$ as $0$ is obviously not a root. Then $$1+2r\cos \theta = |z+1|^2-r^2 = r^{2n} - r^2.$$ We need to show that $$2(r\cos\theta - 1/r \cos \theta) = 2 \operatorname{Re}(re^{i\theta} - 1/r e^{-i\theta}) > 1/r^2 - 1.$$ We know that $2r\cos \theta = r^{2n}-r^2 - 1\Rightarrow \cos \theta = \dfrac{r^{2n}-r^2-1}{2r}$ from above. So we need to show that $$r^{2n}-r^2 - 1 - \dfrac{r^{2n}-r^2-1}{r^2} > 1/r^2 - 1\Leftrightarrow r^{2n+2} - r^4 - r^2 - r^{2n} + r^2 + 1 - 1 + r^2>0 \\ \Longleftrightarrow r^{2n+2} - r^{2n} - r^4 + r^2 = r^2(r^{2n} - r^{2n-2} -r^2 + 1).$$
I don't think I've made any algebraic mistakes, so I was wondering how I can show the last quantity is positive?
| $n>1$ and your last parenthesis is $(r^2-1)(r^{2n-2}-1),$ so you just have to prove that $r\ne1.$
Let us prove it by contradiction. If $r=1$ then (by what you already proved) $\cos\theta=-\frac12$ hence $z=e^{\pm i2\pi/3},$ but such a $z$ satisfies $z+1=-z^2\ne z^n.$
| {
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"question_score": "2",
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Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
| $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$
$$x^2\left(1+2\sqrt {1-x^2}+1-x^2\right)=1-x^2$$
$$x^2\left(2+2\sqrt {1-x^2}-x^2\right)=1-x^2$$
Substitute $x^2=y$,
$$y\left(2+2\sqrt {1-y}-y\right)=1-y$$
$$2y+2y\sqrt {1-y}-y^2=1-y$$
$$2y\sqrt {1-y}=1-y+y^2-2y$$
$$4y^2(1-y)=(y^2-3y+1)^2$$
$$y^4-2y^3+7y^2-6y+1=0$$
$$(y^2-y+3)^2=8$$
$$y^2-y+3=\pm 2\sqrt 2$$
$$y = \frac {1}{2}\bigg(1 \pm \sqrt{8 \sqrt 2 - 11}\bigg)$$
$$x=\pm\sqrt {\frac {1}{2}\bigg(1 \pm \sqrt{8 \sqrt 2 - 11}\bigg)}$$
Hope this helps.
| {
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"source": "stackexchange",
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Encountered $\int_0^{\infty} \frac{x^3}{e^{bx}+1}dx $ in statistical mechanics, curious how to solve it According to Wolfram Alpha, this integral = $\frac{7\pi^4}{120 b^4}$. I tried solving it using the residue theorem, but there are infinite residues which I believe means you can't apply the theorem.
Edit: I forgot to mention that b>0.
| Using the power series
$$
\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n \text { for }| x |<1,
$$
we can transform the integrand into a power series too.
For $b>0$, we have
$$
\begin{aligned}
I & =\int_0^{\infty} \frac{x^3}{e^{b x}+1} d x \\
& =\int_0^{\infty} \frac{e^{-b x} x^3}{1+e^{-b x}} d x \\
& =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty} e^{-b x} \cdot x^3 e^{-b n x} d x \\
& =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty} x^3 e^{-b(n+1) x} d x
\end{aligned}
$$
Letting $y=b(n+1)x$ yields
$$
\begin{aligned}
\int_0^{\infty} x^3 e^{-b(n+1) x} d x
& =\frac{1}{b^4(n+1)^4} \int_0^{\infty} y^3 e^{-y} d y \\
& =\frac{1}{b^4(n+1)^4} \Gamma(4) \\
& =\frac{6}{b^4(n+1)^4}
\end{aligned}
$$
We can now conclude that
$$
\begin{aligned}
I & =\frac{6}{b^4} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^4} \\
& =\frac{6}{b^4}\left[\sum_{n=1}^{\infty} \frac{1}{n^4}-2 \sum_{n=1}^{\infty} \frac{1}{(2 n)^4}\right]\\& =\frac{21}{4b^4} \zeta(4)\\&=\frac{7\pi^4}{120b^4} \end{aligned}\\
$$
Furthermore, replacing the power $3$ of $x$ by $m\in R^+$ gives a more general integral for $b>0$,
$$\boxed{\int_0^{\infty} \frac{x^m}{e^{b x}+1} d x= \frac{\Gamma(m+1)}{b^{m+1} }\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^{m+1}}= \frac{\Gamma(m+1)}{b^{m+1}}\left(1-\frac{1}{2^{m}}\right)\zeta(m+1) }$$
| {
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Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$
Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$
This is my textbook contest exercise.
$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$
$$x=\big(\sqrt [4]{2-x}+1\big)^4$$
I know that
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
So, we will have some cube powers which introduce still $\sqrt [4]{\cdot}$ roots.
This makes more complicated the equation. How can I find the minimal polynomial using a more simple way?
| In this answer, I wanted to construct the trigonometric method that requires significantly less computation.
Since $0<x<2$, making the substitution $x=2\sin^2\theta$, where $0<\theta<\frac {\pi}{2}$, we have:
$$\begin{align}&\sqrt {\sin\theta}-\sqrt {\cos\theta}=\frac {1}{\sqrt [4]{2}}\\
\implies&\sin\theta+\cos\theta=\frac{1}{\sqrt 2}+\sqrt {2\sin (2\theta)}\\
\implies&2\sqrt {\sin (2\theta)}+\sin (2\theta)=\frac 12\end{align}$$
Then the substitution $\sqrt {\sin (2\theta)}=u$, leads to:
$$\begin{align}&2u^2+4u-1=0,~u>0\\
\implies &\sin2\theta=\left(\frac {\sqrt 6-2}{2}\right)^2\\
\implies &\theta =\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\end{align}$$
Thus, the final answer is:
$$\bbox[5px,border:2px solid #C0A000]{x=2\sin^2\left(\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\right)}$$
| {
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Integrate expression with 3 multiplications $\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}dx$ Specifically I want to integrate the following:
$$\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}$$
I do know that
$$\frac4{\pi}\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)} = \dfrac{-16\cos(n\pi)}{(4n^2-1)^2{\pi}}$$
I just have no idea how to integrate it. I tried using trig identities to simplify the expression something by parts could be used for, but I got a very wrong answer.
| Using $$\cos x \sin 2 n x=\frac{1}{2}[\sin (2 n+1) x+\sin (2 n-1) x],$$
we have
$$
\begin{aligned}
&\quad \int_0^{\frac{\pi}{2}} x \cos x \sin 2 n x d x\\& =\frac{1}{2}\left[\int_0^{\frac{\pi}{2}} x[\sin (2 n+1)+\sin (2 n-1) x] d x\right] \\
& =-\frac{1}{2} \int_0^{\frac{\pi}{2}} x d\left(\frac{\cos (2 n+1) x}{2 n+1}+\frac{\cos (2 n-1) x}{2 n-1}\right) \\
& =-\frac{1}{2}\left(x\left[\frac{\cos (2 n+1) x}{2 n+1}+\frac{\cos (2 n-1) x}{2 n-1}\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{\cos (2 n+1) x}{2 n+1} d x-\frac{\cos (2 n-1) x}{2 n-1} d x\right) \\
& =\frac{1}{2}\left[\frac{\sin (2 n+1) x}{(2 n+1)^2}+\frac{\sin (2 n-1) x}{(2 n-1)^2}\right]_0^{\frac{\pi}{2}}\\&= \frac{(-1)^n}{2\left(4n^2-1\right)^2}\left[(2 n-1)^2-(2 n+1)^2\right]\\&=-\frac{4 n \cos n \pi}{\left(4 n^2-1\right)^2}
\end{aligned}
$$
| {
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Finding remainder of dividing $3^{2^n}$ by $2^{n+3}$ Find the remainder of dividing $3^{2^n}$ by $2^{n+3}$
I was trying to use $ord_{2^{n+3}}(3)$, but i don't see a future in looking for a constant when I change the value of n. However, I was looking for a less advanced solution, is it possible?
| You could show by induction that the remainder is $2^{n+2}+1$.
Base case ($n=1$): $\;3^{2^n}=9=2^{n+2}+1\pmod {2^4=16}$.
Now assume $3^{2^n}\equiv 2^{n+2}+1\bmod 2^{n+3}$; i.e., $3^{2^n}=2^{n+2}+1+k\cdot2^{n+3}$ for some integer $k$.
Then $3^{2^{n+1}}=\left(3^{2^n}\right)^2=2^{2n+4}+1+k^2\cdot2^{2n+6}+2^{n+3}+k\cdot2^{2n+6}+k\cdot2^{n+4}$
$\equiv1+2^{(n+1)+2}\pmod{2^{(n+1)+3}}.$
| {
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Determinant of $A_{2n}$ Determine the determinant of $$A_{2n}:=\begin{pmatrix}
a &0 &\cdots & \cdots &0 &b \\
0& a & & & b&0 \\
\vdots &\vdots &\ddots &\ddots &\vdots &\vdots \\
\vdots&\vdots & \ddots & \ddots &\vdots &\vdots \\
0& b & & & a&0 \\
b&0 &\cdots &\cdots &0 &a
\end{pmatrix}\in \mathbb{R}^{2n \times 2n}$$
My initial thought is if we assume that it is $2 \times 2$, then we will get $a^2-b^2$. If it is a $4 \times 4$ one then we would get $(a^2-b^2)^2$. So would the determinant for a $2n \times 2n$ is $(a^2-b^2)^n$? But how to prove it mathematically?
| $\vert A_{2n}\vert=\begin{vmatrix} B_{n} & C_{n} \\ C_{n} & B_{n} \end{vmatrix}$, Where $B_{n}=aI_{n}$ and $C_{n}=bI_{n}$. Because $B_{n}$ and $C_{n}$ non singular, then
$\vert A_{2n}\vert=\vert B_{n}^2-C_{n}^2\vert=\vert a^2I_{n}-b^2I_{n}\vert=\vert (a^2-b^2)I_{n}\vert=(a^2-b^2)^n.1=(a^2-b^2)^n.$
$\blacksquare$
| {
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"source": "stackexchange",
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Evaluating the limit $\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$ How can I evaluate the following limit?
$$\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$$
To solve this limit, I divided the numerator and denominator by $x^5$ and I got
$$\lim_{x\to\infty}\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x$$
But I realized when $x\to\infty$ the expression is still indeterminate.
I tried changing the $x=\frac {1}{y}$ variable to be able to apply Lophital's rule, but I couldn't get a useful result.
| So this is the typical $1^\infty$ type-limit. To find the limit I always start by adding and subtracting $1$ inside the parenthesis (you are "apparentlly" doing nothing), like so $$\lim_{x\rightarrow \infty}\Bigg(1-1+\frac{x^5+\pi x^4+e}{x^5+e x^4+\pi}\Bigg)^x=\lim_{x\rightarrow \infty}\Bigg(1+\frac{(\pi-e)(x^4-1)}{x^5+e x^4+\pi}\Bigg)^x=(\star)$$
Since we know that $\lim_{x\rightarrow \infty}(1+\frac{1}{x})^x=e$, our priority is to set our limit in a similar way.
$$(\star)=\lim_{x\rightarrow \infty}\Bigg(1+\frac{1}{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg)^x=\lim_{x\rightarrow \infty}\Bigg(\Bigg[1+\frac{1}{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg]^{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg)^{x\cdot\frac{(\pi-e)(x^4-1)}{x^5+e x^4+\pi}}$$
Please go ahead and see by yourself that this last limit is equal to the first one you gave. If not, trust me. Now, we know that the part inside the parenthesis converges to $e$ $$\lim_{x\rightarrow \infty}\Bigg[1+\frac{1}{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg]^{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}=e$$ and our exponent converges to $\pi-e$ since $$\lim_{x\rightarrow \infty}x\cdot\frac{(e-\pi)(x^4-1)}{x^5+e x^4+\pi}=\pi-e$$
And so the limit is $e^{\pi-e}$. If you found this answer helpful please flag it as a correct answer. Thanks!!
| {
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Evaluating a simple looking integral While solving an problem of Reator design i have encountered a integral,
$$ I =\int \left(\frac{3x+1}{1-x}\right)^{1/2} dx$$
$0<x<1$
My Approach: Rationalization didn't helped much, so i have substituted $$\frac{3x+1}{1-x}=t $$
and $$dt =\frac{3(1-x)+(3x+1)}{(1-x)^2}dx= \frac{4}{(1-x)^2}dx $$
also $x$ would become $x = \frac{t-1}{t+3}$,
the integral would become
$$I = 2\int \sqrt{t}\left(\frac{t+1}{(t+3)^2}\right) dt $$
$$I =2\int \left[\frac{t^{3/2}}{(t+3)^2} + \frac{t^{1/2}}{(t+3)^2}\right]dt $$
From here i can't form the expression into some standard integrable form,how can i proceed further.
Hints are appreciated
| We may deal with the integral by a single substitution
$$
\sqrt{3} \tan \theta=\left(\frac{3 x+1}{1-x}\right)^{\frac{1}{2}},
$$
then
$$
\begin{aligned}
6 \tan \theta \sec ^2 \theta d \theta & =\frac{4}{(1-x)^2} d x \\
& =4\left(\frac{3+3 \tan ^2 \theta}{4}\right)^2 d x \\
& =\frac{9}{4} \sec ^4 \theta d x
\end{aligned}
$$
and the integral can be transformed into
$$
\begin{aligned}
I & =\frac{8}{\sqrt{3}} \int \frac{\tan ^2 \theta}{\sec ^2 \theta} d \theta \\
& =\frac{8}{\sqrt{3}}\int\frac{1-\cos 2 \theta}{2} d \theta \\
& =\frac{4}{\sqrt{3}}\left(\theta-\frac{\sin 2 \theta}{2}\right)+C \\
& =\frac{4}{\sqrt{3}}\arctan \sqrt{\frac{3 x+1}{3-3 x}}+\sqrt{\frac{3 x+1}{1-x}}(x-1)+ C
\end{aligned}
$$
| {
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Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to LHS but did not succeed in solving by this means.
Please provide a solution.
| Another way.
Since $(x+y+z)^2\geq3(xy+xz+yz),$ by C-S we obtain:
$$\sum_{cyc}\frac{a^2+bc}{b+c}=\sqrt{\left(\sum_{cyc}\frac{a^2+bc}{b+c}\right)^2}\geq\sqrt{3\sum_{cyc}\frac{(a^2+bc)(b^2+ac)}{(b+c)(a+c)}}=$$
$$=\sqrt{3(a^2+b^2+c^2)}\geq a+b+c.$$
The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq3\left(\frac{a^{\frac{20}{9}}+b^{\frac{20}{9}}+c^{\frac{20}{9}}}{3}\right)^{\frac{9}{20}}.$$
| {
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For $a+b+c+d=16$, are there always four integers $x, y, z, w$ that satisfy some conditions including $10\mid x+2y+3z+4w$?
For $a, b, c, d \in \Bbb{N}, a+b+c+d=16$, do four integers $x, y, z, w$ always exist that satisfy the following three conditions?
*
*$\ 0\leq x\leq a, \ 0\leq y\leq b, \ 0\leq z\leq c, \ 0\leq w\leq d.$
*$\ x+y+z+w=10.$
*$\ 10\mid x+2y+3z+4w.$
This was from the Math Competition in Korea, which ended today.
My attempt:
\begin{align}
\text{if} \; x\geq 10 \text{ or } &y\geq 10 \text{ or } z\geq 10 \text{ or } w\geq 10: \\
x\geq10; \; & (x, y, z, w) = (10, 0, 0, 0) \\
y\geq10; \; & (x, y, z, w) = (0, 10, 0, 0) \\
z\geq10; \; & (x, y, z, w) = (0, 0, 10, 0) \\
w\geq10; \; & (x, y, z, w) = (0, 0, 0, 10) \\
\ \\
\end{align}
Then I tried all kinds of $(x, y, z, w)$s which satisfy conditions 2 and 3, and the results were like this...
\begin{align}
& (10, 0, 0, 0) \\
& (1, 8, 1, 0), (2, 6, 2, 0), (2, 7, 0, 1), (3, 4, 3, 0), (3, 5, 1, 1), (4, 2, 4, 0), (4, 3, 2,1), (4, 4, 0, 2), (5, 0, 5, 0), (5, 1, 3, 1), (5, 2, 1, 2), (6, 0, 2, 2), (6, 1, 0, 3), (0, 10, 0, 0) \\
& (0, 1, 8, 1), (0, 2, 6, 2), (0, 3, 4, 3), (0, 4, 2, 4), (0, 5, 0, 5), (1, 0, 7, 2), (1, 1, 5, 3), (1, 2, 3, 4), (1, 3, 1, 5), (2, 0, 4, 4), (2, 1, 2, 5), (2, 2, 0, 6), (3, 0, 1, 6), (0, 0, 10, 0) \\
& (0, 0, 0, 10)
\end{align}
What else should I do here?
| zwim found by brute force that
no such $x,y,z,w$ exist for $(a,b,c,d)=(1,7,7,1)$.
Let me prove that fact by hand. Suppose integer $x,y,z,w$ satisfy condition 1 and 2 for $(a,b,c,d)=(1,7,7,1)$.
$x+2y+3z+4w=3(x+y+z+w)-2x-y+w\ge30-2\times1-7+0=21.$
$x+2y+3z+4w=2(x+y+z+w)-x+z+2w\le20-0+7+2\times1=29.$
The two inequalites above imply $x,y,z,w$ do not satisfy the condition 3.
| {
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Evaluate $\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$ using Gamma and Beta functions I need to evaluate the following integral, using Gamma and Beta functions:
$$\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$$
I tried to use partial fractions, but I am not sure if that is correct, since doing that I get that the integral is divergent.
Since
$\frac{1}{\left(x^2+1\right)\left(x+1\right)^2}=\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{\left(x+1\right)^2}\right)-\frac{1}{2}\left(\frac{x}{x^2+1}\right)$
Then,
$\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{\left(x^2+1\right)\left(x+1\right)^2}dx=\frac{1}{2}\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{x+1}dx}+\frac{1}{2}\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{\left(x+1\right)^2}dx}-\frac{1}{2}\int_{0}^{\infty}{\frac{x\sqrt[5]{x}}{x^2+1}dx}}$
But you can see that the first of this new integrals is divergent, and we only need one integral to be divergent to say that the whole integral diverges, but it is not true since the original integral is convergent, so there must be something wrong with this procedure.
| Substitute $x=\tan u$
$$\begin{align}
& =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\tan u}}{{{(\tan u+1)}^{2}}}du} \\
& =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\frac{\sin u}{\cos u}}}{{{(\frac{\sin u}{\cos u}+1)}^{2}}}du} \\
& =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\sin u}}{\sqrt[5]{\cos u}}\frac{{{\cos }^{2}}u}{{{(\sin u+\cos u)}^{2}}}du} \\
& =\int_{0}^{\pi /2}{{{\sin }^{1/5}}u{{\cos }^{9/5}}u\frac{1}{{{(1+2\sin u\cos u)}^{2}}}du} \\
\end{align}$$
Using $$\frac{1}{{{(1+y)}^{2}}}=\sum\nolimits_{n=0}^{\infty }{{{(-1)}^{n}}(1+n){{y}^{n}}}$$
$$\begin{align}
& =\int_{0}^{\pi /2}{\left( \sum\nolimits_{n=0}^{\infty }{{{(-2)}^{n}}(1+n){{\sin }^{1/5+n}}u{{\cos }^{9/5+n}}u} \right)du} \\
& =\sum\nolimits_{n=0}^{\infty }{{{(-2)}^{n}}(1+n)\left( \int_{0}^{\pi /2}{{{\sin }^{1/5+n}}u{{\cos }^{9/5+n}}udu} \right)} \\
& =\sum\nolimits_{n=0}^{\infty }{{{(-2)}^{n}}(1+n)\left( \frac{1}{2}Beta \left( n/2+3/5,n/2+7/5 \right) \right)} \\
\end{align}$$
Is it possible to continue ??????
| {
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Asymptotic expansion of the solution of $\tan x=x$. I am trying to find an asymptotic expansion of the soltion $x_n$ of the equation $\tan x=x$ in the intervall $I_n= \left]-\frac{\pi}{2}+n\pi , \frac{\pi}{2} + n\pi\right[$. I have showed that
$$x_n = n\pi + \frac{\pi}{2}-\frac{1}{n\pi}+o\left(\frac{1}{n}\right)$$ But I still need an other term, to find the result :
$$ x_n =n\pi + \frac{\pi}{2}-\frac{1}{n\pi}+\frac{1}{2\pi n^2}+o\left(\frac{1}{n^2}\right)$$ !
Any help is really appreciated !
NB: I have checked many similar responses in the website but I need to see the calculation and understand it ;) !
| Too long for a comment
Let's look for the solution in the form $x=\pi n+\frac{\pi}{2}-x_1+x_2+...$, where $x_2\ll x_1\ll1$.
We also suppose $\frac{x_2}{x_1}\sim x_1$
$$\tan x=x\,\Rightarrow\,\tan(\pi n+\frac{\pi}{2}-x_1+x_2+..)=\cot(x_1-x_2-..)=x=\pi n+\frac{\pi}{2}-x_1+x_2+..$$
$$\cot(x_1-x_2-..)=\frac{\cos()}{\sin()}=\frac{1-\frac{(x_1-x_2)^2}{2}-...}{(x_1-x_2-..)(1-\frac{(x_1-x_2)^2}{6}-..)}=\frac{1-\frac{(x_1-x_2)^2}{2}-...}{x_1(1-\frac{x_2-..}{x_1})(1-\frac{(x_1-x_2)^2}{6}-...)}$$
$$\frac{1}{x_1}\Big(1+\frac{x_2}{x_1}+...\Big)\Big(1-\frac{(x_1-x_2)^2}{3}-...\Big)=\pi n+\frac{\pi}{2}-x_1+x_2+...$$
$$\frac{1}{x_1}+\frac{x_2}{x_1^2}+...=\pi n+\frac{\pi}{2}-x_1+x_2+...$$
$$\frac{1}{x_1}=\pi n\,\Rightarrow\,x_1=\frac{1}{\pi n}$$
$$\frac{x_2}{x_1^2}=\frac{\pi}{2}\,\Rightarrow\, x_2=\frac{\pi}{2}x_1^2=\frac{1}{2\pi n^2}$$
$$x=\pi n+\frac{\pi}{2}-x_1+x_2+...=\pi n+\frac{\pi}{2}-\frac{1}{\pi n}+\frac{1}{2\pi n^2}+...$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x)=1+2\cos x +3 \sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Let $f(x)=1+2\cos x +3\sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Solution given as in book:
since $f(x)+f(x+\pi)=2 \implies \dfrac{1}{2}f(x)+\dfrac{1}{2}f(x+\pi)=1$
and after comparing with original equation we get $a=\dfrac{1}{2}, b=\dfrac{1}{2}, c=\pi$
My doubt: Isn't there other way to approach this problem as it is not obvious to do $f(x)+f(x+\pi)=2$ because this kind of step does not come to mind immediately.
| Using $\cos(\theta)=\sin\left(\theta+\frac{\pi}{2}\right)$, as well as several formulas for linear combinations of $\sin$ and $\cos$ (as suggested in Hypernova's comment), gives
$$\begin{equation}\begin{aligned}
1 &= af(x)+bf(x+c) \\
& = a(1+3\sin(x)+2\cos(x)) + b(1+3\sin(x+c)+2\cos(x+c)) \\
& = a + b + a\sqrt{13}\cos(x+\alpha) + b\sqrt{13}\cos(x+c+\alpha) \\
& = (a + b) + \sqrt{13}\left(a\sin\left(x+\alpha+\frac{\pi}{2}\right)+b\sin\left(x+\alpha+\frac{\pi}{2}+c\right)\right) \\
& = (a + b) + \sqrt{13}d\sin\left(x+\alpha+\frac{\pi}{2}+\beta\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
where $\alpha=\arctan\left(-\frac{3}{2}\right)$, $d^2=a^2+b^2+2ab\cos(c)$ and $\tan(\beta)=\frac{b\sin(c)}{a+b\cos(c)}$. Because \eqref{eq1A} must be true for all $x$, but with $\sin\left(x+\alpha+\frac{\pi}{2}+\beta\right)$ varying between $-1$ and $1$, we therefore require
$$a+b=1, \; d = 0 \tag{2}\label{eq2A}$$
Next, we have
$$0 \le (a-b)^2-d^2 = -2ab(1+\cos(c)) \tag{3}\label{eq3A}$$
$$1 = (a+b)^2-d^2 = 2ab(1-\cos(c)) \tag{4}\label{eq4A}$$
From \eqref{eq4A}, since $1 - \cos(c) \ge 0$, then $ab \gt 0$. Thus, using \eqref{eq3A}, $1 + \cos(c) \le 0$ so, with $1 + \cos(c) \ge 0$, we have
$$1 + \cos(c) = 0 \; \; \to \; \; \cos(c) = -1 \; \; \to \; \; c = (2k+1)\pi \tag{5}\label{eq5A}$$
for any integer $k$. This then means, from \eqref{eq3A}, that
$$0 = (a-b)^2 - d^2 \; \; \to \; \; a = b \tag{6}\label{eq6A}$$
Finally, we get
$$a + b = 1 \; \; \to \; \; a = b = \frac{1}{2} \tag{7}\label{eq7A}$$
| {
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Computing $\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$ So I was solving the Cengage Mathematics Calculus book for JEE Adv. and I came across this question in the examples $$\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$$
The solution given in the book uses the expansion of $\sin x$ in solving the question, whereas WolframAlpha used L'Hopital's Rule in solving it.
My first instinct to solving the question when I read it was like this-
$$ \lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}} $$
$$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \frac{1}{x^2} - \frac{x}{x^3} $$
$$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{x}{x^3} $$
$$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{1}{x^2} $$
$$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \frac{1}{x^2} $$
$$ \implies 1 \times 0 $$
$$ \implies 0 $$
I do not understand, where I am going wrong with this procedure. I would really appreciate it if someone can answer.
Thank you.
For reference here is the solution given in Cengage.
| The limit $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right)$$ is of the indeterminate form $\infty - \infty$ so
$$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right) \neq \lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2}\right) - \lim_{x\to 0}\left(\frac 1{x^2}\right)$$ because the RHS makes no sense.
| {
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Solving the functional equations $f(g(x)) = \sin x$ and $g(f(x)) = \cos x$
Solving the functional equations $f(g(x)) = \sin x$ and $g(f(x)) = \cos x$ .
I've managed to reduce an expression for $f(x)$ as:
$f(\cos x) = \sin(f(x))$
But I have no idea how to proceed. I would appreciate some help !
| (Too long for a comment.)
If we assume that $g(x)$ is nice enough to have a Taylor series, then there is no solution.
Going in the other direction (i.e. replacing $x$ with $g(x)$ in the cosine equation and reduce using the sine equation), we get
$$ g(\sin x) = \cos g(x). $$
Putting $g(x) = a_0 + a_1 x + a_2 x^2 + \text{etc.}$, we get
\begin{align}
a_0 &= \cos a_0 &&(1) \\
a_1 &= -a_1 S &&(x) \\
a_2 &= -\frac{{a_1}^2 C}{2} - a_2 S &&(x^2) \\
-\frac{a_1}{6} + a_3 &= \frac{{a_1}^3 S}{6} - a_1 a_2 C - a_3 S &&(x^3) \\
-\frac{a_2}{3} + a_4 &= \text{(terms with $a_1$, $a_2$, $a_3$)} - a_4 S &&(x^4) \\
\frac{a_1}{120} - \frac{a_3}{2} + a_5 &= \text{(terms with $a_1$, $a_2$, $a_3$, $a_4$)} - a_5 S &&(x^5) \\
\frac{2 a_2}{45} - \frac{2 a_4}{3} + a_6 &= \text{(terms with $a_1$, $a_2$, $a_3$, $a_4$, $a_5$)} - a_6 S &&(x^6) \\
&\text{etc.}
\end{align}
where $C = \cos a_0$ and $S = \sin a_0$.
We see that $a_0$ is the Dottie number, the unique real fixed point of the cosine function (so that $C = 0.739...$ and $S = 0.673...$). All subsequent coefficients vanish, as the equation for the coefficient of $x^n$ (for $n \ge 1$) reduces to
$$a_n = -a_n S.$$
Thus $g(x) = a_0$, and the sine equation becomes $f(a_0) = \sin x$, a contradiction.
This of course doesn't rule out the possibility of solutions where $g(x)$ doesn't have a Taylor series.
| {
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In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$, find $\frac{r}{R}$ in terms of $x, c, y$ This problem comes from a previous JEE Advanced examination, the problem is as follows:
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$ where $c$ is the third side, find the ratio of the inradius to the circumradius in terms of $x, c, y$
This is indeed a pretty challenging and tricky problem. Upon initial examination, spamming algebra seems to not lead anywhere. I found a solution using some angle chasing, which I'll share below, please share your own approaches as well!
Let $a+b=x$ and $ab=y$
We know that:
$$x^2-c^2=y$$
$$(a+b)^2-c^2=ab$$
$$a^2+b^2-c^2=-ab$$
Dividing by $2ab$ and exploiting the Law of Cosines, we can say that:
$$\frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}$$
Now, this implies that $\angle C=120^\circ$
Now, using the Law of Sines, we get:
$$\frac{1}{2R}=\frac{\sin{C}}{c}$$
$$\frac{c}{2R}=\frac{\sqrt3}{2}$$
$$R=\frac{c}{\sqrt3}$$
We also know that:
$$\Delta= \frac{1}{2}ab\sin{C}$$
$$\Delta=\frac{\sqrt{3}y}{4}$$
Further, we can note that:
$$s=\frac{a+b+c}{2}$$
$$s=\frac{x+c}{2}$$
Also, $r=\frac{\Delta}{s}$, therefore:
$$\frac{2\sqrt{3}y}{4(x+c)}=r$$
$$r=\frac{\sqrt{3}y}{2(x+c)}$$
Finally:
$$\frac{r}{R}=(\frac{\sqrt{3}y}{2(x+c)})(\frac{\sqrt{3}}{c})$$
$$\frac{r}{R}=\frac{3y}{2c(x+c)}$$
| Goku's solution with some geometric point of view:
*
*Dropping a perpendicular from $O$, the circumcenter, to side $c$, we have $\frac{c}{2}=R\sin C=R\sin120^{\circ}$ and hence $R=\frac{c}{\sqrt 3}.$
*Droping perpendiculars from the incenter $I$, to the sides $a,b,c$ we find:
$a=r\cot\frac B2+r\cot \frac C2=r\cot\frac B2+\frac{r}{\sqrt3}\tag1$
$b=r\cot\frac A2+r\cot \frac C2=r\cot\frac A2+\frac{r}{\sqrt3}\tag2$
$c=r\cot\frac A2+r\cot \frac B2.\tag3$
Now, $(1)+(2)-(3)$ gives $a+b=c+\frac{2r}{\sqrt 3}$ and hence $r=\frac{\sqrt 3}{2}(x-c)$.
From 1 and 2, $$\frac rR=\frac{\frac{\sqrt 3}{2}(x-c)}{\frac{c}{\sqrt3}}=\frac{3(x-c)}{2c}=\frac{3(x^2-c^2)}{2c(x+c)}=\frac{3y}{2c(x+c)}$$
which is, I mean the final expression I found as the answer is, not only consistent but looks exactly like Goku's final expression for the answer.
| {
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Remainder of a function in two variables Supposed a function in two variables is defined as $f(a,b) = \frac{a(a-1)b^{2}}{2}$, and $a,b \in N$ where a is fixed on a certain value while b runs from 1 to n.
For instance,
$$f(4,1) = \frac{4(4-1)1^{2}}{2} = 6$$
$$f(4,2) = \frac{4(4-1)2^{2}}{2} = 24$$
$$f(4,3) = \frac{4(4-1)3^{2}}{2} = 54$$
$$f(4,4) = \frac{4(4-1)4^{2}}{2} = 96$$
$$f(4,5) = \frac{4(4-1)5^{2}}{2} = 150$$
I noticed that when I take the modulo of this functions in terms of $a$, I get
$$f(4,1) (mod \;4) =2$$
$$f(4,2) (mod \;4) =0$$
$$f(4,3) (mod \;4) =2$$
$$f(4,4) (mod \;4) =0$$
$$f(4,5) (mod \;4) =2$$
But the case is different if $a$ is odd number
$$f(3,1) (mod\; 3) = 0$$
$$f(3,2) (mod\; 3) = 0$$
$$f(3,3) (mod\; 3) = 0$$
$$f(3,4) (mod\; 3) = 0$$
$$f(3,5) (mod\; 3) = 0$$
Observation:
*
*If $a$ is odd, the remainder is always 0
*If $a$ is even, the remainder is a series of $\frac {a}{2}, 0, \frac {a}{2}, 0, \frac {a}{2}, ...$
Can you give me any idea regarding this?
| When $a$ is odd, $\frac{a-1}2$ is an integer, so your fraction is $a\times\frac{a-1}2\times b^2$, which is an integer multiple of $a$, so it's equal to $0$ mod $a$.
When $a$ is even, $\frac a2$ is an integer, so your fraction is $\frac a2\times(a-1)b^2$. This is an integer multiple of $\frac a2$, so mod $a$ it can only be $0$ or $\frac a2$, depending on whether $(a-1)b^2$ is even or odd. Now $a-1$ is odd and $b$ alternates between odd and even as we increment it. As a result, $(a-1)b^2$ also alternates between odd and even, so your remainder alternates between $\frac a2$ and $0$.
| {
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What is the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true in $\mathbb{R}$? The question arose from the following proof:
$f(x)=x^3+x+1$ is injective:
Let $a,b\in \mathbb{R}$ and $f(a)=f(b)$:
$f(a)=f(b)\Leftrightarrow a^3+a+1=b^3+b+1\Leftrightarrow a^3+a=b^3+b \Rightarrow a=b$
Is $a^3+a=b^3+b\Rightarrow a=b$ mathematically rigorous enough? I don't know the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true?
| What do you mean by "rigorous enough"? No argument is in your step $a^3 + a = b^3 + b \Longrightarrow a = b$. Some reasoning should be included there.
Method 1. Use calculus to show $x^3 + x$ is an increasing function. That it should be an increasing function is obvious by staring at a graph of it, and calculus allows you to explain why the function is increasing.
Method 2. Use algebra to show $x^3 + x$ is an increasing function. If $x < y$, then we want to show $x^3 + x < y^3 + y$. Write $y = x + h$ for $h > 0$. Then
$$
y^3 + y = (x+h)^3 + (x+h) = x^3 + 3x^2h + 3xh^2 + h^3 + x+h,
$$
so
$$
(y^3 + y) - (x^3 + x) = 3x^2h + 3xh^2 + h^3 + h.
$$
We want to show the right side is positive for all $h > 0$, no matter what $x$ is. View the right side as a quadratic polynomial in $x$:
$$
(3h)x^2 + (3h^2)x + (h^3 + h).
$$
Its discriminant is $(3h^2)^2 - 4(3h)(h^3 + h) = -3h^4 - 12h^2 < 0$, so that quadratic polynomial in $x$ has no real roots: its values as a $x$ varies are either always positive or always negative. At $x = 0$ its value is $h^3 + h$, which is positive, so its values at all real $x$ are positive.
Method 3. Use algebra to show that if $a^3 + a = b^3 + b$ then $a = b$. Rewrite the equality as $a^3 - b^3 = b - a$. Factoring the left side,
$$
(a-b)(a^2 + ab + b^2) = b-a = -(a-b).
$$
If $a \not= b$ then the term $a-b$ is nonzero, so you can divide by it to get
$$
a^2 + ab + b^2 = -1.
$$
If $a$ or $b$ is $0$ then this equation becomes $b^2 = -1$ or $a^2 = -1$, which have no solution (in real numbers). If $a$ and $b$ are both nonzero and have the same sign, then all three terms on the left side are positive, so their sum is not $-1$. If $a$ and $b$ are both nonzero with opposite signs, then rewrite the left side:
$$
(a+b)^2 - ab = -1.
$$
Here $(a+b)^2 \geq 0$ and $-ab > 0$ (since $ab$ is negative due to the opposite signs), so again such an equation is impossible.
| {
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Proving that $\lim_\limits{x \to \infty} x\cdot \ln\left(1 + \frac{1}{x}\right) = 1$ using Taylor series We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.
To do this, when we expand $\ln(1 + \frac{1}{x})$, we should get a series like $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$, multiplying by $x$ would make it $1 + \frac{a_1}{x} + \frac{a_2}{x^2} + ...$ which is $1$ if $x \to \infty$.
The Taylor expansion of $f(x) = \ln(1 + \frac{1}{x})$ at some $a$ is $\sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6}(x-a)^3 + ...$.
$f'(a) = -\frac{1}{a^2 + a}$, $f'(a) \cdot (x-a) = - \frac{x-a}{a^2 + a}$
$f''(a) = \frac{2a + 1}{(a^2 + a)^2}$, $f''(a) \cdot (x-a) = \frac{(2a + 1)(x-a)}{(a^2 + a)^2}$
So the Taylor series begins like this:
$\ln(1 + \frac{1}{x}) = \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ...$
Since the first term doesn't involve $x$ and we want it to be $\frac{1}{x}$, we would like it to be $0$ and the next term to be $\frac{1}{x}$. To make the first term $0$, $a$ needs to be approaching infinity.
$\lim_\limits{a \to \infty} \left( \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right) = \lim_\limits{a \to \infty} \left(- \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right)$.
Now, this isn't in the form $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$ that we wanted.
Do we need to pick some other $a$?
I found https://math.stackexchange.com/a/1071689/1095885 which says
When x is very large, using Taylor
$\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$
The series $\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{x^3} + ...$, which would be perfect, seems to be a Laurent series though.
Edit: When substituting $u = \frac{1}{x}$ and taking the Taylor series of $\ln(1 + u)$, why don't we have to use the chain rule?
| It seems that you're making this a bit more complicated than is necessary.
For $\lvert u \rvert < 1$,
$$
\ln(1 + u) = u - \frac12 u^2 + \frac13 u^3 - \frac14 u^4
+ \cdots
$$
You can verify this Maclaurin series (Taylor series centered at $0$) by differentiating both sides to obtain the well-known geometric series
$$
\frac{1}{1+u} = 1 - u + u^2 - u^3 + \cdots
$$
and by verifying that the constant term vanishes by evaluating at $u=0$.
Now, by dividing through by $u$, we have the series
$$
\frac1u \ln(1 + u) = 1 - \frac12 u + \frac13 u^2 - \frac14 u^3
+ \cdots
$$
which has the same radius of convergence, using, say the Alternating Series Test.
As $x \to \infty$, we can certainly bound $\lvert x \rvert > 1$, so that
$\lvert u \rvert = \lvert \frac1x \rvert < 1$, hence
$$
x \cdot \ln\biggl(1 + \frac1x\biggr)
= 1 - \frac{1}{2x} + \frac{1}{3x^2} - \frac{1}{4x^3}
+ \cdots
$$
converges too for any such $x$.
It's easy to see that all but the constant term vanish as $x \to \infty$, revealing the limit.
| {
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Find all positive integers x, y Find all positive integers x, y such that $(x^2 + y)(x + y^2) = (xy)^3$
I get the problem from a practice problem set of a math camp which was organised for our national math olympiad.
I think that I can solve it if I multiply the left but after multipling I found no other way other than give up.
| If $x,y$ are both odd, then $(x^2+y)(x+y^2)$ is even and $(xy)^3$ is odd.
If one of $x,y$ is even and the other is odd, then $(x^2+y)(x+y^2)$ is odd and $(xy)^3$ is even.
So $x,y$ must both be even and therefore $\geq 2$. Hence $\dfrac{x^2}{y} \leq \dfrac{x^2}{2}$ and $1 < \dfrac{x^2}{2}$ so that:
$$\dfrac{x^2}{y} + 1 < \dfrac{x^2}{2}+\dfrac{x^2}{2} = x^2$$
$$x^2 + y < x^2y$$
Similar reasoning shows that $x+y^2 < xy^2$. Hence:
$$(x^2+y)(x+y^2) < (x^2y)(xy^2) = (xy)^3$$
So there are no solutions in positive integers.
| {
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Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
\sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1}, \\
\sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure these results are correct since WolframAlpha verifies them. I wonder if the first formula can be proved, and also the general version can be proved: Formula related to combinatorial number.
Below is proof for the second formula:
First, note that $r\binom{n}{r}=n\binom{n-1}{r-1}$, and $\sum_{i=0}^n\binom{n}{i}=2^n$. We thus have
\begin{aligned}
\sum_{i=0}^n i\binom{n}{i}&=2^{n-1}n, \\
\sum_{i=0}^n i^2\binom{n}{i}
&= \sum_{i=0}^n i(i-1)\binom{n}{i}+\sum_{i=0}^n i\binom{n}{i} \\
&= 2^{n-2}(n^2-n) + 2^{n-1} n \\
&= n(n+1)2^{n-2}.
\end{aligned}
Therefore,
\begin{aligned}
\sum_{i=0}^{2n}(n-i)^2\binom{2n}{i}
&= n^2\sum_{i=0}^{2n}\binom{2n}{i}-2n\sum_{i=0}^{2n}i\binom{2n}{i}+\sum_{i=0}^{2n}i^2 \binom{2n}{i} \\
&= 2^{2n}n^2-2^{2n+1}n^2+n(2n+1)2^{2n-1} \\
&= 2^{2n-1} n,
\end{aligned}
and thus
\begin{aligned}
\sum_{i=0}^n i^2 \binom{2n}{n-i}
= \sum_{i=0}^n (n-i)^2 \binom{2n}{i}
= \frac12 \sum_{i=0}^{2n} (n-i)^2 \binom{2n}{i}
= 2^{2n-2} n.
\end{aligned}
| I don't konw how to prove it, but for general case I found something:
Integral Representation and the Computation of Combinatorial Sums by G.P.EGORYCHEV Page77
says that:
verify the formula$\sum\limits_{i=1}^n i\left(\begin{array}{c}
2 n \\
n-i
\end{array}\right)=\frac{1}{2}(n+1)\left(\begin{array}{c}
2 n \\
n-1
\end{array}\right)$ with Mathematica
Sum[Binomial[i, 1]*Binomial[2*n, n - i], {i, 1, n}]
$$
\frac{1}{2} (n+1) \binom{2 n}{n-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Limits problem (unable to solve further) Struggling to solve this problem,
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$
My approach:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +...+ \frac{1}{6n}\right)$
= $\displaystyle\lim\limits_{n \to \infty} \int_{0}^{1}(x^{n-1} + x^{n} + x^{n-2}+...+ x^{6n-1}) dx$
= $\displaystyle\lim\limits_{n \to \infty}\int_{0}^{1}\left(x^{n-1} \cdot \frac{x^{5n+1} - 1}{x-1}\right)dx$
got stuck here and don't know how to solve it further (other approaches which are simpler would also help)
| Let define $\displaystyle H_n=1+\frac {1}{2}+ \frac {1}{3}...+\frac {1}{n}
$ more precisely is the sum up to n $H_n=\sum_{k=0}^n=\frac {1}{k}$.
Using Euler-Maclaurin formula $\displaystyle H_n=ln(n)+\gamma+\frac{1}{2n}-\epsilon_n$ where $\gamma\approx0.5772...$and $0\le\epsilon_n\le\frac{1}{8n^2}$
$\displaystyle H_{6n}=ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}$
Noting that your expression is equal to:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$=
$\displaystyle \lim\limits_{n \to \infty} H_{6n}-H_n$ =
$\lim\limits_{n \to \infty}$ ($ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-(ln(n)+\gamma+\frac{1}{2n}-\epsilon_n)$)=
$\lim\limits_{n \to \infty}$ ($ln(6)+ln(n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-ln(n)-\gamma-\frac{1}{2n}+\epsilon_n$)= $ln(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$. Given four positive real numbers $a,b,c,d$. It is given that $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$. Prove $abcd \geq 3$.
I've applied AM-HM inequality and got the result, $a^4+b^4+c^4+d^4 \geq 12$. Then by AM-GM, $a^4+b^4+c^4+d^4\geq 4abcd; 12\geq 4abcd \implies 3\geq abcd$.
But this is contradictory. Then I realized I was doing all wrong. Does anyone have method for this one?
| By AM-GM, $\frac{a^4}{1+a^4}=\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}\ge \frac{3}{\sqrt[3]{(1+b^4)(1+c^4)(1+d^4)}}$, etc. Multiplying all these together we have $\frac{(abcd)^4}{\prod (1+a^4)}\ge \frac{3^4}{\prod (1+a^4)}$, i.e. $abcd\ge 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2- \frac{2}{\sqrt{n+1}}$
Let $i\in\mathbb{N}$ $$\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{n+1}}$$
A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?
| I think it is a good idea to prove the given inequality with First Principle of Mathematical Induction.
1) Step-1:
To see if this inequality is true for $n=1$,
$$ \frac{1}{2^ \frac{3}{4}}>2- \sqrt{2}$$
Which is true (substitute $2^{ \frac{1}{4}}=k$ where and $k>1$ you will get result). Hence $P(1)$ is true.
2) Step-2
Assuming that $P(k-1)$ is true we'll prove that if $P(k)$ will be true.
$$\sum_{i=1}^{k-1} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{k}}$$
Adding $\frac{1}{({k²+k})^\frac{3}{4}}$ both sides,
$$\sum_{i=1}^{k} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{k}}+\frac{1}{{(k²+k})^\frac{3}{4}}$$
Now we have to prove
$$2-\frac{2}{\sqrt{k}}+\frac{1}{{(k²+k})^\frac{3}{4}}>2-\frac{2}{\sqrt{k+1}}$$
Or,
$$\frac{2(\sqrt{k+1}-\sqrt{k})}{\sqrt{k(k+1)}}<\frac{1}{{(k(k+1))^\frac{3}{4}}}$$
Note that $1=(\sqrt{k+1})^2-(\sqrt{k})^2$, After substitution at RHS numerator and cancelling we'll get,$$2<\frac{\sqrt{k+1}+\sqrt{k}}{(k(k+1))^{1/4}}$$
Or,
$$2< \frac{(k+1)^{\frac{1}{4}}}{k^{\frac{1}{4}}}+\frac{k^{\frac{1}{4}}}{(k+1)^{\frac{1}{4}}}$$
Which is a trivial result of A.M.$\geq$ G.M.
Hence we're done of second step, and also the problem.
Therefore our claim is true!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The real and imaginary parts of $\frac{z}{(1-z)^2}$ I have been trying to evaluate the real and imaginary parts of $\frac{z}{(1-z)^2}$ and I was wondering if my solution was correct, as it's quite not nice.
I first wrote out the z's as (x+iy)'s, turning the denominator into $1+x^2 -y^2 -2x +i(2y(1+x))$. Then I multiplied the expression by the complex conjugate of the denominator and did a lot of simplification to turn the expression into $\frac{x^3-2x^2+x+xy^2+2xy+i(-y^3+y-yx^2-2yx-2x^2)}{x^4+y^4-4x^3+2y^2+6x^2-4x+12y^2+2y^2x^2+1}$ . Obviously here you can just split the fraction to obtain the real and imaginary parts. It just seems quite heavy, and I was wondering if my process wasn't the best for this type of problem. Thank you!
| By $w=\frac{z}{(1-z)^2}$, for the real part we have
$$2\Re(w)=w+\bar w=\frac{z}{(1-z)^2}+\frac{\bar z}{(1-\bar z)^2}=\frac{z(1-\bar z)^2+\bar z(1-z)^2}{(1-2\Re(z)+|z|^2)^2}=$$
$$=\frac{2\Re (z)-4|z|^2+2\Re(z)|z|^2}{\left(1-2\Re(z)+|z|^2\right)^2} \implies \Re \left(\frac{z}{(1-z)^2}\right) =\frac{x-2(x^2+y^2)+x(x^2+y^2)}{\left(1-2x+x^2+y^2\right)^2}$$
and similarly for the imaginary part we obtain
$$2i\Im(w)=w-\bar w=\frac{z}{(1-z)^2}-\frac{\bar z}{(1-\bar z)^2}=\frac{z(1-\bar z)^2-\bar z(1-z)^2}{\left(1-2\Re(z)+|z|^2\right)^2}=$$
$$=\frac{2i\Im (z)-2i\Im(z)|z|^2}{\left(1-2\Re(z)+|z|^2\right)^2} \implies \Im \left(\frac{z}{(1-z)^2}\right) =\frac{y-y(x^2+y^2)}{\left(1-2x+x^2+y^2\right)^2}$$
A more effective way
$$\frac{z}{(1-z)^2}= \frac{z}{(1-z)^2}\frac{(1-\bar z)^2}{(1-\bar z)^2}=\frac{z-2z\bar z+z\bar z^2}{\left(1-z-\bar z+z\bar z\right)^2}=\frac{z-2|z|^2+\bar z|z|^2}{\left(1-2\Re (z)+|z|^2\right)^2}$$
from which the previous result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to solve $|z|^2 + z|z| + 2 = 2i Im(z)$? this is the process I follow:
*
*$z|z| = 2iImz -|z|^2- 2 $
*$z \sqrt{Re(z)^2 + Im(z)^2} = 2iy - (Re(z)^2 + Im(z)^2) - 2$
*$z\sqrt{x^2 + y^2} =2iy -(x^2+y^2)-2$
*$z=(2iy -(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
then i get: $Im(z)=2y/\sqrt{x^2 + y^2}$ and $Re(z)=(-(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
what should i do next?
| Write $z=re^{i\theta}=r(\cos\theta + i \sin\theta)$ with $r,\theta\in\Bbb{R}$; your equation becomes $r^2+r\times r(\cos\theta+i\sin\theta)+2=2ir\sin\theta$.
Identifying real and imaginary parts of the left & right members, you get:
$\begin{cases}
r^2(1+\cos\theta)+2=0&(1)\\[5pt]
r^2\sin\theta=2r\sin\theta&(2)
\end{cases}$
Now (1) implies that there is no solution to the problem… (Or did I make a mistake?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Splitting six friends into two pairs and two singles Six friends agree to meet at the hotel Acropolis in Athens. It happens that there are four hotels with the same name. Each of the six friends picks one hotel at random and goes there.
What is the probability that two friends end up alone and the rest four in pairs?
My solution:
Since we have 6 friends and 4 possible choices for each the sample space consists of $4^6$ possible events.
There are $\binom{6}{2}$ choices for the first pair and $4$ hotel choices. There are $\binom{4}{2}$ choices for the second pair and $3$ hotel choices. There $\binom{2}{1}$ choices for the first single person and $2$ hotel choices. The last person has only one hotel choice. Since we have four groups and we do not care about order, we have to divide our results by $4!$. Putting it all together:
$$
P(A) = \frac{\frac{4\binom{6}{2}3\binom{4}{2}2\binom{2}{1}1}{4!}}{4^6} \approx 0.0439
$$
The author finds
$$
P(A) = \frac{12\binom{6}{2}\binom{4}{2}}{4^6} \approx 0.2637
$$
Who is correct?
| For another approach :
We know that $2$ hotels will take pairs and the rest $2$ will take only an individual. So , select which hotels will take pairs by $C(4,2)$.The rest automatically take one person.
Now , determine who will be alone and who will be pairs by $$\binom{6}{1}\times\binom{5}{1}\times\binom{4}{2}\times\binom{2}{2}$$
Then , the answer is $$\frac{\binom{4}{2}\bigg(\binom{6}{1}\times\binom{5}{1}\times\binom{4}{2}\times\binom{2}{2}\bigg)}{4^6}=\frac{6 \times 30 \times 6}{4^6}=\frac{72 \times 15}{4^6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$ Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The given equation is equivalent to $$2^{2x}+\dfrac12\cdot2^{2x}=3^x\sqrt3+\dfrac{3^x}{\sqrt3}$$ which is $$\dfrac32\cdot2^{2x}=3^x\left(\sqrt3+\dfrac{1}{\sqrt3}\right)$$ The last equation can be written as $$\dfrac32\cdot2^{2x}=\dfrac{4}{\sqrt{3}}\cdot3^x,$$ or $$\dfrac{2^{2x}}{3^x}=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}\iff\left(\dfrac43\right)^x=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}$$ How do I find $x$ from here, as it is obviously a difficulty for me? What's the approach supposed to be?
| Since your last step then $(4/3)^{x}=(4/3)^{3/2}$ then $x=3/2$ as Sil noticed.
Alternatively, let $x$ be a real number, then
\begin{align*}
2^{2x}+2^{2x-1}&=3^{x+1/2}+3^{x-1/2}\\
\frac{1}{3^{x+1/2}}\left(2^{2x}+2^{2x-1} \right)&=\frac{1}{3^{x+1/2}}\left(3^{x+1/2}+3^{x-1/2} \right)\\
2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cdot 3^{-(x+1/2)}&=3^{x+1/2}\cdot 3^{-(x+1/2)}+3^{x-1/2}\cdot 3^{-(x+1/2)}\\
2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cdot 3^{-(x+1/2)}&=4/3\\
\underbrace{e^{\ln 2^{2x}}\cdot e^{\ln 3^{-(x+1/2)}}}_{(*)}+e^{\ln 2^{2x-1}}\cdot e^{\ln 3^{-(x+1/2)}}&=4/3
\end{align*}
For $(*)$ note that is just $e^{(2x)\ln 2-(x+1/2)\ln 3}:=t$, then the above equation came be written as $\frac{3t}{2}=\frac{4}{3}$, that is, $t=8/9$.
Now, we need to find $x$ into $8/9=e^{(2x)\ln 2-(x+1/2)\ln 3}$. Applying logarithm both sides we have $(2x)\ln 2-(x+1/2)\ln 3=\ln(8/9)$. Rewriting in order to find $x$ we have $(2\ln 2-\ln 3)x=1/2\ln 3+\ln(8/9)$. Thus ,
$$\boxed{x}=\frac{\frac{\ln 3}{2}+\ln\frac{8}{9}}{2\ln 2-\ln 3}=\boxed{\frac{\frac{1}{2}\ln 3+3\ln 2-2\ln 3 }{2\ln 2 -\ln 3}=\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Activation function defined as a convolution (proof verification) I would like to request a proof verification.
I am currently reading this paper, and I am trying to prove Eq. (3) from the paper.
The operator $\ast$ is the convolution operation and $\phi_n(x) = n\phi(nx)$ with $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$. (Gaussian function)
I have tried this tedious calculation 3 times, and could not get the same answer as the paper. Can anyone check my calculations? I have written the full calculation process below.
I want to calculate the following function defined by a convolution.
$$G(x, \alpha, n) = (\mathrm{LeakyReLU[\alpha]} \ast \phi_n)(x) = \int_{-\infty}^{\infty} \mathrm{LeakyReLU}[\alpha](x-y)\phi_n(y)\,d{y}.$$
Since
$$\mathrm{LeakyReLU}[\alpha](x-y) = \begin{cases} x - y & (y \leq x) \\ \alpha(x-y) & (y > x) \end{cases},$$
we split the integral into two parts.
$$
\begin{aligned}
G(x, \alpha, n) &= \int_{-\infty}^{x} (x-y)\phi_n(y)\,d{y} + \int_{x}^{\infty} \alpha(x-y)\phi_n(y)\,d{y} \\
&= \int_{-\infty}^{x} (x-y)n\phi(ny)\,d{y} + \int_{x}^{\infty} \alpha(x-y)n\phi(ny)\,d{y} \\
&= \frac{n}{\sqrt{2\pi}} \int_{-\infty}^{x} (x-y) e^{-\frac{n^2y^2}{2}}\,dy + \frac{\alpha n}{\sqrt{2\pi}} \int_{x}^{\infty} (x-y) e^{-\frac{n^2y^2}{2}}\,dy.
\end{aligned}
$$
Let each term be $(\ast)$ and $(\ast\ast)$ respectively. Then
$$
\begin{aligned}
(\ast) &= \frac{nx}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{n^2y^2}{2}} \,dy - \frac{n}{\sqrt{2\pi}} \int_{-\infty}^x ye^{-\frac{n^2y^2}{2}} \,dy\\
& = \frac{nx}{\sqrt{2\pi}} \int_{-\infty}^{\frac{nx}{\sqrt{2}}} e^{-t^2} \frac{\sqrt{2}}{n} \,dt - \frac{n}{\sqrt{2\pi}} \left[-\frac{1}{n^2} e^{-\frac{n^2y^2}{2}}\right]^x_{-\infty} \\
&=\frac{x}{2} \cdot \frac{2}{\sqrt{\pi}} \int_{-\infty}^{\frac{nx}{\sqrt{2}}} e^{-t^2} \,dt + \frac{n}{\sqrt{2\pi}}\cdot\frac{1}{n^2}e^{-\frac{n^2x^2}{2}} \\
&= \frac{x}{2} \left[\mathrm{erf}\left(\frac{nx}{\sqrt{2}} \right) + 1\right] + \frac{1}{n}\cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{n^2x^2}{2}}
\end{aligned}
$$
I substituted $\frac{ny}{\sqrt{2}} = t$ in the second line. The second part is pretty much the same calculation, with the additional $\alpha$ multiplied.
$$
\begin{aligned}
(\ast\ast) &= \alpha\frac{nx}{\sqrt{2\pi}} \int_x^{\infty} e^{-\frac{n^2y^2}{2}} \,dy - \alpha \frac{n}{\sqrt{2\pi}} \int_x^{\infty} ye^{-\frac{n^2y^2}{2}} \,dy \\
&= \alpha \frac{nx}{\sqrt{2\pi}} \int_{\frac{nx}{\sqrt{2}}}^{\infty} e^{-t^2} \frac{\sqrt{2}}{n} \,dt - \alpha \frac{n}{\sqrt{2\pi}} \left[-\frac{1}{n^2} e^{-\frac{n^2y^2}{2}}\right]_x^{\infty} \\
&=\frac{\alpha x}{2} \cdot \frac{2}{\sqrt{\pi}} \int^{\infty}_{\frac{nx}{\sqrt{2}}} e^{-t^2} \,dt - \frac{\alpha n}{\sqrt{2\pi}}\cdot\frac{1}{n^2}e^{-\frac{n^2x^2}{2}} \\
&=\frac{\alpha x}{2} \left[1 - \mathrm{erf}\left(\frac{nx}{\sqrt{2}}\right)\right] - \frac{\alpha}{n}\cdot \frac{1}{\sqrt{2\pi}}e^{-\frac{n^2x^2}{2}}.
\end{aligned}
$$
I also substituted $\frac{ny}{\sqrt{2}} = t$ in the second line. Adding the two results gives
$$
\begin{aligned}
G(x, \alpha, n) =&\, \frac{x}{2}\left[\mathrm{erf}\left(\frac{nx}{\sqrt{2}}\right) + 1\right] + \frac{1}{n}\cdot \frac{1}{\sqrt{2\pi}}e^{-\frac{n^2x^2}{2}} \\
&+ \frac{\alpha x}{2}\left[1 - \mathrm{erf}\left(\frac{nx}{\sqrt{2}}\right)\right] - \frac{\alpha}{n}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{n^2x^2}{2}} \\
=& \frac{1+\alpha}{2}x + \frac{1-\alpha}{2}x\,\mathrm{erf}\left(\frac{nx}{\sqrt{2}}\right) + \frac{1}{n}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{n^2x^2}{2}} - \frac{\alpha}{n}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{n^2x^2}{2}}
\end{aligned}
$$
My problem here is that the first 3 terms coincide with the result in the paper. But I have no idea where the last term came from.
The equations are tedious, so please understand my abuse of notation, and there may be some typos.
Thanks in advance for check this tedious calculation for me.
| You’re right; the paper is wrong. This is easiest to see if you choose $\alpha=1$. Then the convolution of $x$ with a Gaussian should indeed yield $x$, not an additional Gaussian term. You don’t need to evaluate any integrals for that; it follows by symmetry, since a convolution of a symmetric and an antisymmetric function must yield an antisymmetric function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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