Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proof for Inequality Can somebody tell me what is the name of the inequality:
\begin{equation} \sum_{t=1}^T \frac{1}{\sqrt{t}} \leq 2\sqrt{T} \end{equation}
or any hint/link how to prove above?
Thanks.
| We will prove that for every $n \in \mathbb Z_{\gt 0}$ we have
\begin{equation*} \tag{1}
\sum_{j=1}^n \frac{1}{\sqrt{j}} \lt 2 \sqrt{n}.
\end{equation*}
We proceed by Mathematical Induction.
Base case. The inequality is true for $n=1$ since $1 < 2$.
Inductive step. Let $k \in \mathbb Z_{\gt 0}$ be arbitrary. Assume that
\begin{equation*} \tag{2}
\sum_{j=1}^k \frac{1}{\sqrt{j}} \lt 2 \sqrt{k}.
\end{equation*}
This is the inductive hypothesis. We need to prove that
\begin{equation*} \tag{3}
\sum_{j=1}^{k+1} \frac{1}{\sqrt{j}} \lt 2 \sqrt{k+1}.
\end{equation*}
Before proceeding with the proof of (3), we notice that
$4k(k+1) < (2 k + 1)^2.$ Consequently, $2 \sqrt{k}\sqrt{k+1} < 2 k + 1$, and therefore,
\begin{equation*} \tag{4}
2 \sqrt{k}\sqrt{k+1} + 1 < 2(k + 1).
\end{equation*}
Now we proceed with the proof of (3).
\begin{alignat*}{2}
\sum_{j=1}^{k+1} \frac{1}{\sqrt{j}} & \lt 2 \sqrt{k} + \frac{1}{\sqrt{k+1}} & \qquad & \text{by (2)} \\
& = \frac{2 \sqrt{k} \sqrt{k+1} + 1}{\sqrt{k+1}} & & \\
& \lt \frac{2 (k+1)}{\sqrt{k+1}} & & \text{by (4)} \\
& = 2 \sqrt{k+1}. & &
\end{alignat*}
This proves that (2) implies (3) and completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solving $a = \sqrt{b + x} + \sqrt{c + x}$ for $x$ I'm trying to solve a very simple looking square root equation but nothing seems to work. The equation has this form (solve for $x$):
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
Squaring both sides obviously doesn't help since it will still give me a square root. Rearranging and then squaring doesn't help either.
The problem looks very simple to me but I have no idea on how to approach this.
| $$\begin{align}a=\sqrt{b+x}+\sqrt{c+x}&\Rightarrow \sqrt{b+x}=a-\sqrt{c+x}\\&\Rightarrow b+x=a^2-2a\sqrt{c+x}+c+x\\&\Rightarrow 2a\sqrt{c+x}=a^2+c-b\\&\Rightarrow 4a^2(c+x)=(a^2+c-b)^2\\&\Rightarrow x=\frac{(a^2+c-b)^2-4a^2c}{4a^2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Basic Logarithm question - I can't get both answers from quadratic Here's the Question :
If $xy$ = $64$ and $\log_x y + \log_y x = \frac{5}{2}$, find $x$ and $y$
I can get this to
$$log_x y + \frac{1}{\log_x y} \frac{5}{2}$$
let $\log_x y = N$
$$N + \frac{1}{N} = \frac{5}{2}$$
Multiply by 2
$$2N + \frac{2}{N} = 5$$
Multiply by N
$$2N^2 + 2 = 5N$$
$$2N^2 - 5N + 2 = 0$$
$$(2N - 1)(N - 2)$$
Giving :
$$N = \frac{1}{2}$$
$$N = 2$$
Therefore :
$$\log_x y = \frac{1}{2} $$
$$\log_x y = 2$$
Giving
$$x^2 = y$$
$$x^{\frac{1}{2}} = y$$
Part of the original question :
$$xy = 64$$
As $x^2 = y$
$$x * x * x = 64$$
$$x^3 = 64$$
Therefore:
$$x = 4$$
$$y = 16$$
I can't seem to solve for $y = x^{\frac{1}{2}}$ though
Solving for $x^{\frac{1}{2}} = y$
$$x^{\frac{1}{2}} * x^{\frac{1}{2}} = 64$$
$$x^{\frac{1}{2} + \frac{1}{2}} = 64$$
$$x= 64$$
$$xy= 64$$
$$64y= 64$$
Therefore
$$x = 64$$
$$y = 1$$
This is wrong though.
Answer :
$$(4,16) or (16,4)$$
I don't see how they got the second part. The first part makes sense but I'm not able to solve for $y = x^{\frac{1}{2}}$
| $$\log_x(y)+\log_y(x)=\frac{\ln(y)}{\ln(x)}+\frac{\ln(x)}{\ln(y)}=\frac{\ln^2(x)+\ln^2(y)}{\ln(x)\ln(y)}=\frac{(\ln(x)+\ln(y))^2}{\ln(x)\ln(y)}-2=\frac{\ln^2(xy)}{\ln(x)\ln(y)}-2$$
Then if $xy=64$,
$$\log_x(y)+\log_y(x)=\frac{5}{2}\implies \frac{\ln^2(64)}{\ln(x)\ln(y)}=\frac{9}{2}\implies \frac{2\ln^2(64)}{9}=\ln(x)\ln(y)\underset{xy=64}{=}\ln(x)\ln(\frac{64}{x})=-\ln^2(x)+\ln(x)\ln(64)$$
and thus, if you set $X=\ln(x)$ you get
$$X^2-\ln(64)X+\frac{2\ln^2(64)}{9}=0.$$
$$\Delta =\ln^2(64)-\frac{8}{9}\ln^2(64)=\frac{1}{9}\ln^2(64)$$
Therefore $$X=\frac{\ln(64)\pm\frac{1}{3}\ln(64)}{2}$$
and thus $X=\frac{1}{3}\ln(64)=\ln(\sqrt[3]{64})=\ln(4)$ ou $X=\frac{2}{3}\ln(64)=\ln(\sqrt[3]{64^2})=\ln(16).$
Therefore $x=4$ and $y=16$ or $x=16$ and $y=4$. Finally, we conclude that $(x,y)=(4,16)$ or $(x,y)=(16,4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Solve for real value of $x$: $|x^2 -2x -3| > |x^2 +7x -13|$ Here I have a question:
Solve for real value of $x$:
$$|x^2 -2x -3| > |x^2 +7x -13|$$
I got the answer as $x = (-\infty, \frac{1}{4}(-5-3\sqrt{17}))$ and $x=(\frac{10}{9},\frac{1}{4}(3\sqrt{17}-5)$
Please verify it if it is correct or not. Thanks
| Brute force (as I am wont to do):
Look at where $(x^2-2x-3)^2 - (x^2+7x-13)^2 = -(9x-10)(2x^2+5x-16)$ is strictly positive.
The zeroes are ${10 \over 9}$ and ${1 \over 4} (-5 \pm 3 \sqrt{17})$.
Since the leading coefficient is $-1$, we see that the answer is
$(-\infty,{1 \over 4} (-5 - 3 \sqrt{17})) \cup ({10 \over 9}, {1 \over 4} (-5 + 3 \sqrt{17}))$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$ After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this?
| the easier way to do this is to derive the chebyshev recurrence relation for $$a_n = \cos nt, x = \cos t$$ by the sum to product formula. that is, $$a_{n+2} +a_n = \cos (nt+2t) + \cos(nt) = 2\cos t \cos(nt) = 2xa_n, a_0 = 1, a_1 = x. $$
now we can compute $$a_2 = 2x^2 - 1, a_3 = 2x(2x^2-1) - x=4x^3 - 3x,\\a_4=8x^4-6x^2-(2x^2-1) = 8x^4-8x^2 + 1\\a_5 = 16x^5-16x^3+2x-(4x^3-3x) = 16x^5-20x^3+5x\\a_6 = 32x^6-40x^4+10x^2-(8x^4-8x^2+1)=32x^6-48x^4+18x^2-1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Finding a Basis for this subspace Set $V=\mathbb{R}^{2x3}$ and let $U$ be a subspace of $V$ defined by:
\begin{equation*}
U=\{B=(b_{ij})\in V\mid b_{11} + b_{12} + b_{13} = -4(b_{21} + b_{22} + b_{23})\}.
\end{equation*}
I would greatly appreciate it if someone could help me understand how to find a basis for $U$ and the dimension of $U$.
| Note that $B\in U$ if and only if
\begin{align*}
B
&=
\begin{bmatrix}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23}
\end{bmatrix} \\
&=
\begin{bmatrix}
-b_{12}-b_{13}-4\,b_{21} -4\,b_{22}-4\,b_{23} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23}
\end{bmatrix} \\
&= b_{12}\underbrace{\begin{bmatrix}
-1 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}}_{=u_1}
+
b_{13}
\underbrace{
\begin{bmatrix}
-1 & 0 & 1 \\
0&0&0
\end{bmatrix}}_{=u_2}
+b_{21}
\underbrace{
\begin{bmatrix}
-4 & 0 & 0 \\
1&0&0
\end{bmatrix}}_{=u_3} \\
&\qquad\qquad\qquad\qquad\qquad\qquad+
b_{22}
\underbrace{
\begin{bmatrix}
-4 & 0 & 0 \\
0&1&0
\end{bmatrix}}_{=u_4}
+
b_{23}
\underbrace{
\begin{bmatrix}
-4 & 0 & 0 \\
0&0&1
\end{bmatrix}}_{=u_5}
\end{align*}
This proves that $\{u_1,u_2,u_3,u_4,u_5\}$ is a basis for $U$. In particular, $\dim U=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Continued fraction for $\int_{0}^{\infty}(e^{-xt}/\cosh t)\,dt$ In one of the comments to a question I posted on MSE, I got this wonderful continued fraction $$\int_{0}^{\infty}\frac{e^{-xt}}{\cosh t}\,dt = \frac{1}{x}\genfrac{}{}{0pt}{}{}{+}\frac{1^{2}}{x}\genfrac{}{}{0pt}{}{}{+}\frac{2^{2}}{x}\genfrac{}{}{0pt}{}{}{+}\frac{3^{2}}{x}\genfrac{}{}{0pt}{}{}{+\dots}\tag{1}$$ Another article here (search for "Conjecture 1") mentions that this is a conjecture for $x \geq 1$. I believe this formula $(1)$ should be true whenever both sides of the equation are defined.
Let me know if there is an elementary proof available for $(1)$.
Update: As mentioned in this comment the integral in $(1)$ can be easily shown to be equal to an infinite series as follows
\begin{align}
\int_{0}^{\infty}\frac{e^{-xt}}{\cosh t}\,dt &= 2\int_{0}^{\infty}\frac{e^{-xt}}{e^{t} + e^{-t}}\,dt\notag\\
&= 2\int_{0}^{1}\frac{v^{x}}{1 + v^{2}}\,dv \text{ (putting }v = e^{-t})\notag\\
&= 2\int_{0}^{1}\sum_{k = 0}^{\infty}(-1)^{k}v^{x + 2k}\,dv\notag\\
&= 2\sum_{k = 0}^{\infty}(-1)^{k}\int_{0}^{1}v^{x + 2k}\,dv\notag\\
&= 2\sum_{k = 0}^{\infty}\frac{(-1)^{k}}{x + 2k + 1}\notag
\end{align}
Also it is mentioned there that there is a general result by Rogers which converts the above sum into the continued fraction on the right of $(1)$. If there are any references to this result of Rogers, it would be of great help here.
Further Update: It turns out that the continued fraction expansion mentioned above is quite helpful in approximating $\pi$.
| In addition to the nice answer by @metamorphy, one can also transform the series into the continued fraction by expressing it as an hypergeometric function
\begin{align}
S(x)&=2\sum_{k = 0}^{\infty}\frac{(-1)^{k}}{x + 2k + 1}\\
&=\frac{1}{x+1}\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right)
\end{align}
(see below).
Then the ratio
\begin{align}
\frac{1}{S(x)}&=(x+1)\frac{1}{\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right)}\\
&=(x+1)\frac{\,_2F_1\left( 0,0;\frac{x+1}{2};\frac12 \right)}{\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right)}\\
&=2\frac{\mathbf F\left( 0,0;\frac{x+1}{2};\frac12 \right)}{\mathbf F\left( 1,1;\frac{x+3}{2};\frac12 \right)}
\end{align}
where the regularized HG functions are introduced.
This ratio can be expressed as a continued fraction:
\begin{equation}
\frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a+1,b+1;c+1;z\right)}={x
_{0}+\cfrac{y_{1}}{x_{1}+\cfrac{y_{2}}{x_{2}+\cfrac{y_{3}}{x_{3}+\cdots}}}}
\end{equation}
where
\begin{align}
x_{n}&=c+n-(a+b+2n+1)z\\
y_{n}&=(a+n)(b+n)z(1-z)
\end{align}
Here, $a=b=0,c=\frac{x+1}{2},z=1/2$. Then
\begin{align}
x_n&=\frac x2\\
y_n&=\frac{n^2}{4}
\end{align}
We obtain
\begin{align}
\frac{1}{2S(x)}&={\frac x2+\cfrac{1^2/4}{x/2+\cfrac{2^2/4}{x/2+\cfrac{3^2/4}{x/2+\cdots}}}}\\
&={\frac x2+\cfrac{1^2/2}{x+\cfrac{2^2}{x+\cfrac{3^2}{x+\cdots}}}}
\end{align}
and thus
\begin{equation}
S(x)=\cfrac1{ x+\cfrac{1^2}{x+\cfrac{2^2}{x+\cfrac{3^2}{x+\cdots}}}}
\end{equation}
which is the desired result.
The hypergeometric form of the series can be obtained by denoting the general term of the series
\begin{equation}
c_k=\frac{2(-1)^{k}}{x + 2k + 1}
\end{equation}
the successive ratio of the terms is
\begin{align}
\frac{c_{k+1}}{c_k}&=-\frac{x+2k+1}{x+2k+3}\\
&=\frac{(k+1)(k+\frac{x+1}{2})}{k+\frac{x+3}{2}}\frac{-1}{k+1}
\end{align}
with $c_0=\frac{2}{x+1}$, the series writes
\begin{equation}
S(x)=\frac{2}{x+1}\,_2F_1\left( 1,\frac{x+1}{2};\frac{x+3}{2};-1 \right)
\end{equation}
Now, the transformation
\begin{equation}
\,_2F_1(a,b;c;z)=(1-z)^{-a}\,_2F_1\left(a,c-b;c;\frac z{z-1}\right)
\end{equation}
gives the expression
\begin{equation}
S(x)=\frac{1}{x+1}\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right)
\end{equation}
which is also tabulated here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Are the following expression equivalent? I need to check wether this expressions are equal but I haven't yet learned enough about the floor function to tell (I also
$$ \frac 1 4 \left (5-\cos \left(n \frac \pi 2 \right)-2 (-1)^n \left(1+ \left \lfloor\frac 1 4 (-2+n) \right \rfloor - \left\lfloor \frac n 4 \right \rfloor \right)-\sin \left(n \frac \pi 2 \right)\right )$$
And
$$ \frac 5 4 - \frac 1 4 (-1)^n- \frac 1 2 \sin \left(n \frac \pi2 \right)$$
For all $n\in \Bbb{N}$.
| You need to think about what happens if $n$ is even, and then if $n$ is odd. For example, if $n$ is even, the second expression is clearly $1$, since $\sin\left(n\frac{\pi}{2}\right) = 0$. For the first expression, we get
$$\frac 1 4 \left (5-(\pm 1)-2\left(1+ \left \lfloor\frac 1 4 (-2+n) \right \rfloor - \left\lfloor \frac n 4 \right \rfloor \right)\right ),$$
where we have $+1$ if $n\equiv 0\mod{4}$ and $-1$ if $n\equiv 2\mod{4}$. But if $n\equiv 0\mod{4}$, then $\left \lfloor\frac 1 4 (-2+n) \right \rfloor - \left\lfloor \frac n 4 \right \rfloor = -1$, so we get $\frac{1}{4}(5-1) = 1$, while if $n\equiv 2\mod{4}$, then $\left \lfloor\frac 1 4 (-2+n) \right \rfloor - \left\lfloor \frac n 4 \right \rfloor = 0$ and we get $\frac{1}{4}(5+1-2) = 1$.
The analysis when $n$ is odd is presumably similar.
In general, when you are trying to analyze an expression like this involving the floor function, you are going to have to consider cases. The fact that you have trig functions whose arguments are multiples of $\frac{\pi}{2}$ should lead you even further in that direction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An identity involving the Bessel function of the first kind $J_0$ Today, computing $\int_{0}^{\pi/2}\sin^2(\sin^2 x)\,dx$, I found an interesting identity:
$$\sum_{k\geq 0}\frac{(-1)^k(4k)!}{(2k)!^3 4^k}=\cos(1)\cdot\sum_{k\geq 0}\frac{(-1)^k}{k!^2 4^k}.$$
How would you prove it?
| This answer is due to Giulio Francot, a friend of mine. His solution is extremely close to my original manipulations.
We have:
\begin{eqnarray*}
\int\limits_{0}^{\frac{\pi }{2}}\sin ^{2}(\sin ^{2}(x))dx
&=&\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (2\sin ^{2}(x))}{2}dx \\
&=&\frac{
\pi }{4}-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}\sum\limits_{k\geq 0}
\frac{(-1)^{k}4^{k}}{(2k)!}\sin ^{4k}(x)dx \\
&=&\frac{\pi }{4}-\frac{1}{2}\sum\limits_{k\geq 0}\frac{(-1)^{k}4^{k}}{(2k)!
}\int\limits_{0}^{\frac{\pi }{2}}\sin ^{4k}(x)dx \\
&=&\frac{\pi }{4}-\frac{1}{2}
\sum\limits_{k\geq 0}\frac{(-1)^{k}4^{k}}{(2k)!}\frac{(4k)!}{
(4^{k}(2k)!)^{2}}\frac{\pi }{2} \\
&=&\frac{\pi }{4}-\frac{\pi }{4}\sum\limits_{k\geq 0}\frac{(-1)^{k}}{
((2k)!)^{3}}\frac{(4k)!}{4^{k}}.\tag{1}
\end{eqnarray*}
On the other hand:
\begin{eqnarray*}
\int\limits_{0}^{\frac{\pi }{2}}\sin ^{2}(\sin ^{2}(x))dx
&=&\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (2\sin ^{2}(x))}{2}
dx=\int\limits_{0}^{\frac{\pi }{2}}\frac{1-\cos (1-\cos (2x))}{2}dx \\
&=&\frac{\pi }{4}-\int\limits_{0}^{\frac{\pi }{2}}\frac{\cos (1)\cos (\cos
(2x))}{2}dx-\int\limits_{0}^{\frac{\pi }{2}}\frac{\sin (1)\sin (\cos (2x))}{
2}dx \\
&=&\frac{\pi }{4}-\frac{\cos (1)}{2}\int\limits_{0}^{\frac{\pi }{2}}\cos
(\cos (2x))dx=\frac{\pi }{4}-\frac{\cos (1)}{4}\int\limits_{0}^{\pi }\cos
(\cos (x))dx \\
&=&\frac{\pi }{4}-\frac{\cos (1)\pi }{4}J_{0}(1).\tag{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving that $\cos(2\pi/n)$ is algebraic I want to prove this without using any of the properties about the field of algebraic numbers (specifically that it is one). Essentially I just want to find a polynomial for which $\cos\frac{2\pi}{n}$ is a root.
I know roots of unity and De Moivre's theorem is clearly going to be important here but I just can't see how to actually construct the polynomial from these facts.
| Euler's Formula implies
$$
\left[\cos\left(\frac{2\pi}n\right)+i\sin\left(\frac{2\pi}n\right)\right]^n=1\tag{1}
$$
The Binomial Theorem says
$$
\begin{align}
1
&=\sum_{k=0}^ni^k\binom{n}{k}\cos^{n-k}\!\left(\frac{2\pi}n\right)\sin^k\!\left(\frac{2\pi}n\right)\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\sin^{2k}\!\left(\frac{2\pi}n\right)\\
&+i\sum_{k=0}^{\lfloor(n-1)/2\rfloor}(-1)^k\binom{n}{2k+1}\cos^{n-2k-1}\!\left(\frac{2\pi}n\right)\sin^{2k+1}\!\left(\frac{2\pi}n\right)\tag{2}
\end{align}
$$
Looking at the real part of $(2)$ yields
$$
\begin{align}
1
&=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\sin^{2k}\!\left(\frac{2\pi}n\right)\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\left[1-\cos^2\!\left(\frac{2\pi}n\right)\right]^k\\
&=\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m\cos^{n-2m}\!\left(\frac{2\pi}n\right)\sum_{k=m}^{\lfloor n/2\rfloor}\binom{n}{2k}\binom{k}{m}\tag{3}
\end{align}
$$
Equation $(3)$ gives a polynomial equation whose root is $\cos\!\left(\frac{2\pi}n\right)$. Furthermore, the coefficient of $\cos^n\!\left(\frac{2\pi}n\right)$ is $2^{n-1}$, which implies that $2\cos\!\left(\frac{2\pi}n\right)$ is an algebraic integer. This polynomial also has a root of $1$, so it is reducible.
Example: Using $n=7$ in $(3)$, we have that $\cos\!\left(\frac{2\pi}7\right)$ satisfies
$$
64x^7-112x^5+56x^3-7x-1=0\tag{4}
$$
Note that $1$ is also a root of $(4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial derivatives and finding the maximum, is there a simple argument that imply this inequality (perhaps using symmetry somehow?).
Thanks !
| We have $$\frac{(x+y)^2}{x^2+xy+y^2} = \frac{(x+y)^2}{\frac{3}{4}(x+y)^2 + \frac{1}{4}(x-y)^2}\leq \frac{(x+y)^2}{\frac{3}{4}(x+y)^2}=\frac{4}{3}$$
The inequality holds since squares are non-negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to distribute $x \times \sqrt{x^2 + 1}$ How do I distribute the x in this problem?
How do I "gain access" so to say. Does it become $x^\frac{1}{2}(x^2 + 1)^\frac{1}{2}$ or $(x^3+x)^\frac{1}{2}$ ?
Or do I need to even do that in order to integrate $\int x(x^2+1)^\frac{1}{2}$ ?
| Note that
$$\sqrt{x^2}=|x|.$$
If $x\ge 0$, then $x=|x|=\sqrt{x^2}$ leads $$x\sqrt{x^2+1}=\sqrt{x^2}\cdot\sqrt{x^2+1}=\sqrt{x^2(x^2+1)}.$$
If $x\lt 0$, then $x=-|x|=-\sqrt{x^2}$ leads$$x\sqrt{x^2+1}=-\sqrt{x^2}\cdot\sqrt{x^2+1}=-\sqrt{x^2(x^2+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| Tennenbaum gave a geometric proof of the irrationality of $\sqrt{2}$:
The large square has side length $a$, while the light purple/blue square has side length $b$, with $a, b$ positive integers such that $({a\over b})^2=2$. But then it's easy to see that the blue square has twice the area of the pink square - that is, $({2b-a\over a-b})^2=2$. Since the numerator and denominator are each positive integers integers, and are less than $a$ and $b$ respectively, we have an infinite descent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 1
} |
sum of series $\sum_{k = 0}^n (-1/4)^k \cos((\pi/4) k)$ What is the sum of series $$\sum_{k=0}^n \left(-\frac14\right)^k \cdot \cos \left(\frac{k\pi}4\right) \text{ ?}$$
I was trying to see over some inputs its behavior: (to see if cos has some interesting information) for $n=1: (1/2)^{0.5}$, $n=2:0$, $n=3$: $-(1/2)^{0.5}$ $n=4: -1$, $n=5:-(1/2)^{0.5}$, $n=6$: $(3/4)^{0.5},\ldots$ but with no conclusions. It reminds a bit of fourier series, but it isn't a function of $x$.
| Consider the series
\begin{align}
S_{n}(x) = \sum_{k=0}^{n} e^{\frac{i k \pi}{4}} x^{k}
\end{align}
It can quickly be determined that
\begin{align}
S_{n}(x) = \frac{1 - e^{- \pi i/4} x - e^{(n+1) \pi i/4} x^{n+1} + e^{n \pi i/4} x^{n+2} }{ 1 - \sqrt{2} x + x^{2}}.
\end{align}
Taking the real part leads to, where $Re(S_{n}(x))$ is the desired series,
\begin{align}
\sum_{k=0}^{n} \cos\left(\frac{k \pi}{4} \right) \, x^{k} = \frac{1 - \cos(\pi/4) x - \cos((n+1) \pi/4) x^{n+1} + \cos(n \pi/4) x^{n+2} }{ 1 - \sqrt{2} x + x^{2}}.
\end{align}
Taking the imaginary part leads to
\begin{align}
\sum_{k=0}^{n} \sin\left(\frac{k \pi}{4} \right) \, x^{k} = - \frac{x}{\sqrt{2}} \, \frac{1 + \sqrt{2} \left(\sin\left(\frac{(n+1)\pi}{4} \right)
- \sin\left(\frac{n\pi}{4} \right) x \, \right) x^{n+1} }{ 1 - \sqrt{2} x + x^{2}}.
\end{align}
If $x \to \frac{1}{x}$ then the series becomes
\begin{align}
\sum_{k=0}^{n} \cos\left(\frac{k \pi}{4} \right) \, \left(\frac{1}{x}\right)^{k} = \frac{x^{n+1} \left(x - \frac{1}{\sqrt{2}} \right) - x \cos\left(\frac{(n+1) \pi}{4}\right) + \cos\left(\frac{n \pi}{4}\right)}{
(1 - \sqrt{2} x + x^{2}) \, x^{n} }.
\end{align}
When $x = -1/4$ the result becomes
\begin{align}
\sum_{k=0}^{n} \cos\left( \frac{k \pi}{4} \right) \, \left( \frac{-1}{4} \right)^{k} = \frac{1}{17 + 2 \sqrt{2}} \, \left[ 16 + 2 \sqrt{2} + \frac{(-1)^{n}}{4^{n}} \left( 4 \cos\left( \frac{(n+1) \pi}{4} \right) + \cos\left( \frac{n \pi}{4} \right) \right) \right]
\end{align}
and
\begin{align}
\sum_{k=0}^{n} \sin\left( \frac{k \pi}{4} \right) \, \left( \frac{-1}{4} \right)^{k} = \frac{1}{8 + 17 \sqrt{2}} \, \left[ 2 \sqrt{2} + \frac{(-1)^{n}}{4^{n}} \left( 4 \sin\left( \frac{(n+1) \pi}{4} \right) + \sin\left( \frac{n \pi}{4} \right) \right) \right].
\end{align}
For the case $x = -4$ the series become
\begin{align}
\sum_{k=0}^{n} \cos\left( \frac{k \pi}{4} \right) \, (-4)^{k} = \frac{1}{17 + 4 \sqrt{2}} \, \left[ 1 + 2 \sqrt{2} + (-4)^{n} \left( \cos\left( \frac{(n+1) \pi}{4} \right) + 4 \cos\left( \frac{n \pi}{4} \right) \right) \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Need a closed formula for the generating function $x/(1+x+x^2)$. I used partial fractions but the obtained formula is only correct for the first two elements.
$\dfrac{x}{(1+x+x^2)}=\dfrac{x}{(1+a_1x)(1+a_2x)}=\dfrac{A_1}{(1+a_1x)+A_2(1+a_2x)}$
$x=\dfrac{-1 \pm\sqrt3i}{2}$
Then let $a_1=\dfrac{-1+\sqrt3i}{2}$ and $a_2=\dfrac{-1-\sqrt3i}{2}$
and then calculated $A_1$ and $A_2$ to obtain the formula
$a_n=\dfrac{(-1)^n}{\sqrt3i} \cdot \left(\left(\dfrac{-1-\sqrt3i}{2}\right)^n-\left(\dfrac{-1+\sqrt3i}{2}\right)^n\right)$
The first three elements become $0$, $1$, $1$, but the answer from the Taylor series is $0$, $1$, $-1$.
| Partial fractions are not the most efficient way here.
Our expression is $\frac{x(1-x)}{1-x^3}$. Expand $\frac{1}{1-x^3}$ using $\frac{1}{1-t}=1+t+t^2+t^3+t^4+\cdots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the derivative to nth order How to find $$\frac{d^ny}{dx^n}$$
of $$y=\frac{x}{lnx-1}$$
Appreciated advance
| By applying the quotient rule we have $$\frac{dy}{dx} = \frac{1 \times (\ln x -1)- \frac{1}{x}x}{(\ln x - 1)^2} = \color{blue}{\frac{\ln x -2}{(\ln x - 1)^2}}$$
Now you can apply the quotient rule again to get $$\frac{d^2y}{dx^2} = \frac{\frac{1}{x} \times (\ln x -1)^2 - 2(\ln x -1) \frac{1}{x} \times (\ln x - 2)}{(\ln x - 1)^4} =\frac{\frac{1}{x}(\ln x -1)((\ln x-1) -2(\ln x -2))}{(\ln x - 1)^4} = \frac{\frac{1}{x}((\ln x-1) -2(\ln x -2))}{(\ln x - 1)^3}=\frac{\frac{1}{x}(\ln x -2\ln x+3)}{(\ln x - 1)^3}=\frac{(-\ln x+3)}{x(\ln x - 1)^3} = \color{blue}{\frac{-(\ln x-3)}{x(\ln x - 1)^3}}$$
Now do you see a pattern developing it is $$\color{blue}{\frac{d^ny}{dx^n}=\frac{(-1)^{n+1}(\ln x -(n+1))}{x^{n-1}(\ln x - 1)^{n+1}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry to the 24th power How can I find the value of
$$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$
Specifically, is there some easy method that I am overlooking?
| A note: A general formula that is applicable is
\begin{align}
\sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right].
\end{align}
When $n=6$ this reduces to
\begin{align}
\sin^{24}\left(\frac{\pi}{24}\right) + \cos^{24}\left( \frac{\pi}{24}\right) = \frac{1}{4^{11}} \left[ \sum_{r=0}^{5} \binom{24}{2r} \cos\left(1 - \frac{r}{6} \right) \pi \, + \frac{1}{2} \binom{24}{12} \right].
\end{align}
The remaining details have been given by Micah's solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 0
} |
Find the area between $y=x; y=2x; y=1/x; y=3/x$ Find the area between $y=x; y=2x; y=1/x; y=3/x$ using the substitution $(x,y)=(\frac{u}{v},uv)$
I made a sketch but don't see how I can make use of the substitution given, to help.
| $x=\frac{u}{v}, y=uv\implies u^2=xy\;$ and $\;v^2=\frac{y}{x}$, so
$1\le u^2\le3\;, \;1\le v^2\le2 \implies 1\le u\le\sqrt{3},\;1\le v\le\sqrt{2}$.
Since $\displaystyle\begin{vmatrix}x_u &x_v\\y_u & y_v\end{vmatrix}=\begin{vmatrix}\frac{1}{v} &-\frac{u}{v^2}\\v & u\end{vmatrix}=\frac{2u}{v}$,
$\displaystyle A=\int_1^{\sqrt{2}}\int_1^{\sqrt{3}}\frac{2u}{v}dudv=\int_1^{\sqrt{2}}\frac{2}{v}dv=\big[2\ln v\big]_1^{\sqrt{2}}=2\ln\sqrt{2}=\ln 2$.
Alternatively, $xy=k\implies r^2\sin\theta\cos\theta=k\implies r^2=\frac{k}{\sin\theta\cos\theta}=2k\csc 2\theta$,
so using polar coordinates gives $\displaystyle A=\int_{\pi/4}^{\tan^{-1}2}\frac{1}{2}\left(6\csc 2\theta-2\csc2\theta\right)d\theta=\int_{\pi/4}^{\tan^{-1}2}2\csc2\theta d\theta$
$\displaystyle=\int_{\pi/4}^{\tan^{-1}2} \frac{2}{2\sin\theta\cos\theta}d\theta=\int_{\pi/4}^{\tan^{-1}2}\frac{\sec^2\theta}{\tan\theta}d\theta=\big[\ln(\tan\theta)\big]_{\pi/4}^{\tan^{-1}2}=\ln 2-\ln1=\ln 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find Maxima and Minima Sir, help me to find the maxima, minima & saddle point.
$$f(x,y)=x^2-x^2y+2xy^2+4y$$
here $$\begin{align}f_x(x,y) & =2x-2xy+2y^2
\\ f_{xx}(x,y) & =2-2y
\\ f_y(x,y) & =-x^2+4xy+4
\\ f_{yy}(x,y) & =4x
\\ f_{xy}(x,y) & =-2x+4y
\end{align}$$
For critical point I put $f_x=0$ & $f_y=0$, but fail to subsitute $y=(x^2-4)/4x$ to find the value of $x$ in the equation $-x^2+4xy+4=0$
| setting $$f_x = 0 \to x = \frac{y^2}{y-1} \, or\, y = 1 $$ in the case $y = 1, x = \pm 2\sqrt 2$ and in the other case we have $$0 = f_y \to 0 = -\frac{y^4}{(y-1)^2} + \frac{4y^3}{y-1} + 4 \to -y^4+4y^3(y-1)+4(y-1)^2=3y^4-4y^3+4(y-1)^2 $$ the last equation has two roots $$y = 0.634, 1.2828 , x = \frac{y^2}{y-1} \text{ and }x = \pm 2\sqrt 2, y = \pm 1.$$
you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expanding an Integrand I am trying to solve this integral but it looks like I do not get the right result. Can you please help me?
$$\int{(t-4)(t+2)^{\frac 4 5}}dt$$
I set $u=t+2$, so I get $\int{(u-6)u^{\frac 4 5}} \Bbb d t$ and then the solution I get is $$\frac 5 {14} (t+2)^{\frac {14} 5}-\frac{10} 3 (t+2)^{\frac 9 5} .$$
Can you please tell me where I am wrong?
| $$\begin{align}
\int{(t-4)(t+2)^{\frac {4}{5}}}dt
&=\int({t(t+2)^{\frac {4}{5}}-4(t+2)^{4/5}})dt\\
&=\int{t(t+2)^{\frac {4}{5}}}dt-4\int{(t+2)^{4/5}}dt\\
\end{align}$$
Subtitute $u=t+2$ and $du=dt$
$$\begin{align}
&=\int{(u-2)u^{4/5}}du-4\int{(t+2)^{4/5}}dt\\
&=\int{(u^{9/5}-2u^{4/5}})du-4\int{(t+2)^{4/5}}dt\\
&=\int{u^{9/5}}du-2\int{u^{4/5}}du-4\int{(t+2)^{4/5}}dt\\
&=\frac{5}{42}(t+2)^{9/5}(3t-22)+c
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For what value of $\alpha$ is the limit $\lim{x\to 0} \frac{\arctan(x) \cdot \log(\sin(x)) - x \cdot \log(x) }{x^\alpha}$ finite and non-zero? My brother asked me this question and I found it very difficult.
For which value of $\alpha$ is the limit
$$
\lim_{x\to0} \frac{\arctan(x) \cdot \log(\sin(x)) - x \cdot \log(x) }{x^\alpha}
$$ finite and different from zero? The thing that confuses me is that I can't use McLaurin for $\log x$, because it is not defined in 0.
Is the question really hard or am I missing something?
| It is possible to arrive to the same conclusion as that of Mark Fischler,
without using Taylor series. First note that
\begin{equation*}
\log (\sin x)=\log \left( x\frac{\sin x}{x}\right) =\log x+\log \left( \frac{%
\sin x}{x}\right) =\log x+\log \left( 1+\frac{\sin x-x}{x}\right) .
\end{equation*}
Then,
\begin{eqnarray*}
\frac{\arctan x\log (\sin x)-x\log x}{x^{\alpha }} &=&\frac{\arctan x\left(
\log x+\log \left( 1+\frac{\sin x-x}{x}\right) \right) -x\log x}{x^{\alpha }}
\\
&=&\frac{\left( \arctan x-x\right) \log x+\arctan x\log \left( 1+\frac{\sin
x-x}{x}\right) }{x^{\alpha }} \\
&=&\left( \frac{\arctan x-x}{x^{3}}\right) \left( \frac{\log x}{x^{\alpha -3}%
}\right) \\
&&+\left( x^{3-\alpha }\right) \left( \frac{\arctan x}{x}\right) \left(
\frac{\sin x-x}{x^{3}}\right) \left( \frac{\log \left( 1+\left( \frac{\sin
x-x}{x^{3}}\right) x^{2}\right) }{\left( \frac{\sin x-x}{x^{3}}\right) x^{2}}%
\right)
\end{eqnarray*}
Using standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \
\lim_{x\rightarrow 0}\frac{\arctan x}{x}=1,\ \ \ \ \ \ \ \
\lim_{x\rightarrow 0}\frac{\arctan x-x}{x^{3}}=-\frac{1}{3},\ \ \ \
\lim_{u\rightarrow 0}\frac{\log (1+u)}{u}=1 \\
\lim_{x\rightarrow 0^{+}}\frac{\log x}{x^{\alpha -3}} &=&\left\{
\begin{array}{ccc}
0 & if & \alpha <3 \\
-\infty & & \alpha =3 \\
+\infty & & \alpha >3%
\end{array}%
\right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\lim_{x\rightarrow 0^{+}}x^{3-\alpha }=\left\{
\begin{array}{ccc}
0 & if & \alpha <3 \\
1 & & \alpha =3 \\
+\infty & & \alpha >3%
\end{array}%
\right.
\end{eqnarray*}
it follows that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\arctan x\log (\sin x)-x\log x}{x^{\alpha }}%
=\left\{
\begin{array}{ccc}
\left( -\frac{1}{3}\right) \left( 0\right) +\left( 0\right) \left( 1\right)
\left( -\frac{1}{6}\right) \left( 1\right) =0 & if & \alpha >3 \\
\left( -\frac{1}{3}\right) \left( -\infty \right) +\left( 1\right) \left(
1\right) \left( -\frac{1}{6}\right) \left( 1\right) =+\infty & if & \alpha
=3 \\
\left( -\frac{1}{3}\right) \left( +\infty \right) +\left( +\infty \right)
\left( 1\right) \left( -\frac{1}{6}\right) \left( 1\right) =-\infty & if &
\alpha <3.%
\end{array}%
\right.
\end{equation*}
For any real value of $\alpha ,$ the limit is either infinite or
zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\large{\frac{7x}{2x-1} +\frac{5x-2}{2x+2}}$ Simplify the rational expression.
$$\frac{7x}{2x-1} + \frac{5x-2}{2x+2}$$
My work:
\begin{eqnarray*} \frac{7x}{2x-1} + \frac{5x-2}{2x+2}
&=& \frac{7x(2x+2)+(5x-2)(2x-1)}{(2x-1)(2x+2)} \\ &=& \frac{7x(2x+2)+(5x-2)(2x-1)}{(2x-1)(2x+2)}.\end{eqnarray*}
Is this correct?
| \begin{eqnarray*} \frac{7x}{2x-1} + \frac{5x-2}{2x+2}
&=& \frac{7x(2x+2)+(5x-2)(2x-1)}{(2x-1)(2x+2)} \\ &=& \frac{7x(2x+2)+(5x-2)(2x-1)}{(2x-1)(2x+2)}\\
&=&\frac{14x^2+14x+10x^2-5x-4x+2}{(2x-1)(2x+2)} \\
&=& \frac{24x^2+5x+2}{(2x-1)(2x+2)}\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solving inequalities with fractions on both sides Solve this inequality: $\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5}$.
My solution:
Multiply $(x+5)^2$ on both sides:
$\frac{(x^2 -2)(x+5)^2}{2} = \frac{x^4 + 10x^3 + 23x^2 -20x -50}{2}$
$(6x^2 -8x - 1)(x+5) = 6x^3 + 22x^2 -41x - 5 $
$\frac{x^4 + 10x^3 + 23x^2 -20x -50}{2} < 6x^3 + 22x^2 -41x - 5$
$0.5x^4 - x^3 - 10.5x^2 + 31x - 20 < 0 $
I don't think doing this way will lead to solving the question....
| This is very similar to Meshal's approach, but you can avoid having to factor a quartic. It gives the general idea for a slightly different approach as well.
Move all terms to the same side:
$$\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5} \iff \frac{x^2 -2}{2} - \frac{6x^2 -8x - 1}{x+5} < 0$$
Now put everything on a common denominator and factor the cubic in the numerator:
$$\frac{x^3-7 x^2+14 x-8}{2(x+5)} < 0 \overset{\;\color{green}{(*)}}{\iff} \frac{(x-1) (x-2) (x-4)}{2(x+5)} < 0$$
You can make a sign table of this fraction which leads immediately to the solution set and/or follow a scheme like Meshal's.
You can also notice that all roots (three in the numerator and one in the denominator) are simple roots (i.e. their multiplicity is 1) and thus sign changes occur in each root.
These roots are in increasing order $-5$, $1$, $2$ and $4$ and clearly the inequality is not satisfied for $x<-5$ (check with a value or reason for $x \to -\infty$); so it is satisfied for $x \in (-5,1) \cup (2,4)$.
To factor in step $\color{green}{(*)}$, you can follow general algorithms or group as follows:
$$\begin{align}
\color{blue}{x^3}\color{red}{-7 x^2+14 x}\color{blue}{-8}
& = \color{blue}{(x-2)(x^2+2x+4)}\color{red}{-7x(x-2)} \\
& = (x-2)(x^2-5x+4) \\
& = (x-2)(x-1)(x-4)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\lim\limits_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}$
How to calculate
$$\lim_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}?$$
I know the result is $1/(6e)$.
| In the same spirit as other answers but trying to get some more information $$A=\left(\frac{\sin(x)}x\right)^{\frac 1{x^2}}$$ $$\log(A)=\frac 1{x^2}\,\log\left(\frac{\sin(x)}x\right)$$ Now, using Taylor $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\frac{\sin(x)}x=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^5\right)$$ Now, using $$\log(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O\left(y^4\right)$$ replace $y$ by $(\frac{x^2}{6}-\frac{x^4}{120}+\cdots)$, expand just keeping the low powers of $x$; this will give $$\log\left(\frac{\sin(x)}x\right)=-\frac{x^2}{6}-\frac{x^4}{180}+\cdots$$ $$\log(A)=-\frac{1}{6}-\frac{x^2}{180}+\cdots$$ So, $$A=e^{-\frac{1}{6}-\frac{x^2}{180}+\cdots}=e^{-\frac{1}{6}}e^{-\frac{x^2}{180}}=e^{-\frac{1}{6}}(1-\frac{x^2}{180})$$ which shos the limit and how it is reached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation of product of $m+1$ alternate numbers We know that
$$\begin{align}
\sum_{r=1}^n \prod_{j=0}^m (r+j)&=\sum_{r=1}^n r(r+1)(r+2)\cdots(r+m)\\
&=(m+1)!\sum_{r=1}^n\binom {r+m}{m+1}\\
&=(m+1)!\binom {n+m+1}{m+2}\\
&=\frac {(n+m+1)^\underline{m+2}}{m+2}\end{align}$$
but is there a similar closed form for
$$\sum_{r=1}^n \prod_{j=0}^m (r+2j)=\sum_{r=1}^n r(r+2)(r+4)\cdots (r+2m)$$
?
Edit 1
Here's a summary of results from wolframlpha for the first few values of $m$ (please excuse the messy alignment).
From the summary we observe some interesting patterns as follows:
*
*all results have factors $n,(n+1)$
*all results have a factor of $\frac 1{2(m+2)}$ or $\frac 1{m+2}$ where the product of the other factors results in a $2n^{m+2}$ term or $n^{m+2}$ term respectively
*for $m=2$: $(n+4),(n+5)$ are factors
*for $m=4$: $(n+8),(n+9)$ are factors
*for $m=6$: $(n+12),(n+13)$ are factors
*for $m=8$: $(n+16),(n+17)$ are factors
*this implies that for even $m$: $(n+2m), (n+2m+1)$ are factors
Perhaps the observations above might be helpful in working out the solution.
The solution could then possibly be of the form:
$$\begin{cases}
\dfrac {n(n+1)(n+2m)(n+2m+1)\cdot f(n,m)}{2(m+2)}\qquad \text{for even $m$}\\
\dfrac {n(n+1)\cdot g(n,m)}{2(m+2)}\qquad \qquad\qquad\qquad\qquad\text {for odd $m$}
\end{cases}$$
| \begin{align*}\sum_{r=1}^n r(r+2)(r+4)...(r+2m)&=\sum_{1\le r=2k\le n}r(r+2)(r+4)...(r+2m)\\&+\sum_{1\le r=2k+1\le n}r(r+2)(r+4)...(r+2m) \\ &=\underbrace{\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{(2k+2m)!!}{(2k-2)!!}}_{S_1}+\underbrace{\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\frac{(2k+2m+1)!!}{(2k-1)!!}}_{S_2}\end{align*}
where: $\quad (2x)!!=2^x .x!\quad $ and $\quad (2x-1)!!=\dfrac{(2x)!}{(2x)!!}$
Note that: $\quad 0!!=2^0.0!=\frac{0!}{0!!}=(-1)!!=1$
\begin{align*} S_1&=(2m+2)!!+\frac{1}{2m+4}\sum_{k=2}^{\left\lfloor\frac{n}{2}\right\rfloor}\Delta\left(\frac{(2m+2k)!!}{(2k-4)!!}\right)\\ &=(2m+2)!!+\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n}{2}\right\rfloor+2\right)!!}{\left(2\left\lfloor\frac{n}{2}\right\rfloor-2\right)!!}-\frac{1}{2m+4}\cdot\frac{(2m+4)!!}{0!!}\\&=\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n}{2}\right\rfloor+2\right)!!}{\left(2\left\lfloor\frac{n}{2}\right\rfloor-2\right)!!}\end{align*}
\begin{align*} S_2&=(2m+1)!!+\frac{1}{2m+4}\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\Delta\left(\frac{(2m+2k+1)!!}{(2k-3)!!}\right)\\ &=(2m+1)!!+\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n+1}{2}\right\rfloor+1\right)!!}{\left(2\left\lfloor\frac{n+1}{2}\right\rfloor-3\right)!!}-\frac{1}{2m+4}\cdot\frac{(2m+3)!!}{(-1)!!}\\&=\frac{(2m+1)!!}{2m+4}+\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n+1}{2}\right\rfloor+1\right)!!}{\left(2\left\lfloor\frac{n+1}{2}\right\rfloor-3\right)!!}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Integration consquence How does it follow from $$\int_0^\infty \left(\frac{1}{a^2+x^2} \right) dx= \frac{\pi}{2a}$$ that $$\int_0^\infty \left(\frac{1}{\left({a^2+x^2}\right) ^2 } \right) dx= \frac{\pi}{4a^3}\text{ ?}$$
| An alternate method would be to use integration by parts:
$\displaystyle\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx=\frac{1}{a^2}\left[\int_0^{\infty}\frac{1}{x^2+a^2}dx-\int_0^{\infty}\frac{x^2}{(x^2+a^2)^2}dx\right]$,
so letting $\displaystyle u=x, \;dv=\frac{x}{(x^2+a^2)^2}dx, \;du=dx, \;v=-\frac{1}{2(x^2+a^2)}dx$ yields
$\displaystyle\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx=\frac{1}{a^2}\left[\int_0^{\infty}\frac{1}{x^2+a^2}dx+\left[\frac{x}{2(x^2+a^2)}\right]_0^{\infty}-\frac{1}{2}\int_0^{\infty}\frac{1}{x^2+a^2}dx\right]$
$\displaystyle\hspace{1.3 in}=\frac{1}{a^2}\left[\frac{\pi}{2a}+0-\frac{1}{2}\cdot\frac{\pi}{2a}\right]=\frac{1}{a^2}\cdot\frac{\pi}{4a}=\frac{\pi}{4a^3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $\cos2a$ and $\cos b$, find $a+b$. High School Level Mathematics:
If $\cos(2\alpha) = -\frac{63}{65}$ for $\alpha \in (0,\frac{\pi}{2})$ and $\cos(\beta) = \frac{7}{\sqrt{130}}$ for $\beta \in (0,\frac{\pi}{2})$ then, without using a calculator, what is $\alpha + \beta$?
The answer is ${3\pi\over 4}$ but I need steps.
| Notice $$\cos \alpha=\sqrt{\frac{1+\cos 2\alpha}{2}} \quad (\forall \space 0<\alpha<\frac{\pi}{2})$$ $$=\sqrt{\frac{1-\frac{63}{65}}{2}}=\sqrt{\frac{2}{2\times65}}=\color{blue}{\frac{1}{\sqrt{65}}}$$ $$\implies \sin\alpha=\sqrt{1-\cos^2\alpha} \quad (\forall \space 0<\alpha<\frac{\pi}{2})$$ $$=\sqrt{1-\left(\frac{1}{\sqrt{65}}\right)^2}=\color{blue}{\frac{8}{\sqrt{65}}}$$
$$\implies \sin\beta=\sqrt{1-\cos^2\beta} \quad (\forall \space 0<\beta<\frac{\pi}{2})$$ $$=\sqrt{1-\left(\frac{7}{\sqrt{130}}\right)^2}=\color{blue}{\frac{9}{\sqrt{130}}}$$ Now, we have $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ Substitute all the values in the above expression, we get $$\sin(\alpha+\beta)=\left(\frac{8}{\sqrt{65}}\right)\left(\frac{7}{\sqrt{130}}\right)+\left(\frac{1}{\sqrt{65}}\right)\left(\frac{9}{\sqrt{130}}\right)$$$$=\frac{65}{\sqrt{65\times130}}=\sqrt{\frac{65}{130}}=\frac{1}{\sqrt{2}}$$ $$\implies \alpha+\beta=\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \quad \text{or} \quad \pi-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) $$
By the given values of $\alpha$ & $\beta$ we have $\color{blue}{0<(\alpha+\beta)<\pi}$
$$\color{blue}{\alpha+\beta}=\pi-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) =\pi-\frac{\pi}{4}$$$$\color{blue}{=\frac{3\pi}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality $4+xy+yz+zx \ge 7xyz$ Let $x,y,z$ be non negative real numbers such that $x+y+z=3$, prove the following inequality:
$$4+xy+yz+zx \ge 7xyz$$
I tried MV and taking out one variable but I got nothing
| Alternative approach:
We have that $1=\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$, therefore $xyz\leq 1$.
Since $xyz\leq 1$, we have $4 \geq 4xyz$.
Also, by AM-HM, we have:
$$1=\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Therefore
$$\frac{1}{3} \geq \frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
$$3 \leq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
$$3xyz \leq yz+xz+xy$$
Adding the latter inequality and $4 \geq 4xyz$ gives the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding extrema. Find the minimum distance between point $M(0,-2)$ and points $(x,y)$ such that: $y=\frac{16}{\sqrt{3}\,x^{3}}-2$ for $x>0$ .
I used the formula for distance between two points in a plane to get: $$d=\sqrt{x^{2}+\frac{256}{3x^{6}}}$$ And this is where I cannot come up with how to proceed. I tried calculus but the first derivative of $d(x)$ is fairly ugly expression...
A few techniques on how to handle problems on maxima and minima with(out) using calculus would be really useful.
| We give a non-calculus approach. It is in my opinion a fair bit harder than the calculus way.
We want to minimize $x^2+\frac{256}{3x^6}$, or equivalently
$$\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{256}{3x^6}.$$
By the arithmetic mean geometric mean inequality (AM/GM) we have
$$\frac{1}{4}\left( \frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{256}{3x^6} \right)\ge \sqrt[4]{\frac{x^2}{3}\cdot \frac{x^2}{3}\cdot\frac{x^2}{3}\cdot \frac{256}{3x^6}}\tag{1}$$
with equality when all the terms on the left are equal. The right-hand side of (1) is $\frac{4}{3}$. Equality is achieved when $\frac{x^2}{3}=\frac{256}{3x^6}$, that is, when $x=\pm 2$.
We conclude that the square of the distance has minimum value $\frac{16}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve in positive integers: $5^x 7^y +4=3^z$ Solve in positive integers: $5^x 7^y +4=3^z$.
I tried to solve it with log but I couldn't complete.
| Rewrite this as
$$3^z-4=5^x7^y$$
Then we see, because $x>0$, that
$$3^z-4\equiv 0\mod 5 \iff 3^z\equiv -1\mod 5$$
so $z=2j$ is even.
Then we have $(3^j-2)(3^j+2)=5^x7^y$. Since the gcd of these two divides $4$--because $(3^j-2)+(3^j+2)=4$--we have that they are coprime (any larger factor would then be a factor of $5^x7^y$ which is odd), i.e. one of the following two cases hold
$$\begin{cases} 3^j+2=5^x\quad ,\quad 3^j-2=7^y \\ 3^j+2=7^y\quad ,\quad3^j-2=5^x\end{cases}.$$
In the second case we get $7^y\equiv 2\mod 3$ which is impossible, so we are in case $1$. Then $3^j=7^y+2$ so $j\equiv 2\mod 6$ which recovers our original conclusion, that $z=12\ell+4\equiv 0\mod 4$, which is a contradiction as this implies
$$-3\equiv 3^z-4\equiv 5^x7^y\equiv 0\mod 5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Find the maximum possible area of a certain right triangle I want to find the maximum possible area of a right triangle with hypotenuse $=10$.
My approach so far: let $x,y$ be the lengths of the two sides adjacent to the right angle; then $$100=x^2+y^2$$
Area $=\frac{xy}{2}$, so by the substitution method I got equality. But my result is wrong (probably I made a mistake in the equality), could someone show what to do?
|
The area of triangle $ABC$ is just $\Delta ABO + \Delta BOC$. If $\angle BOC = \theta$, the area is:
$$\frac{1}{2} r^2 \sin \theta + \frac{1}{2}r^2 \sin(180 - \theta) = \frac{1}{2}r^2 \big( \sin \theta + \sin(180 - \theta) \big) = \frac{1}{2}r^2 (2 \sin \theta) = r^2 \sin \theta$$
where we have used the identity $\sin \theta = \sin(180 - \theta)$.
The maximum value of $\sin \theta$ is $1$, so the maximum area is $r^2$. In this case, $r = 5$ so the maximum area is $25$.
This occurs when $\theta = \frac{\pi}{2}$ or $90º$, so $\Delta ABC$ must in fact be isoceles.
Alternatively, if $A = (-5,0)$, $B = (5 \cos t, 5 \sin t)$, $C = (5,0)$, then the area of $\Delta ABC$ using the shoelace formula is:
$$
\begin{vmatrix}
5 & 0 \\
5 \cos t & 5 \sin t \\
-5 & 0 \\
5 & 0 \\
\end{vmatrix}
$$
so the area is: $$\frac{1}{2} \big((25 \sin t + 0 + 0) - (0 + -25 \sin t + 0) \big) = 25 \sin t$$
which takes the maximum value $25(1)$ when $t = \frac{\pi}{2}$, again.
The area of triangle $ABC$ is can also be computed using determinants:
$$
\begin{vmatrix}
5 & 0 & 1 \\
5 \cos t & 5 \sin t & 1 \\
-5 & 0 & 1 \\
\end{vmatrix}
$$
which when computed, gives:
$$\frac{1}{2} \left((25 \sin t + 0 + 0) - (-25 \sin t + 0 + 0) \right) = 25 \sin t.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of $\frac{1}{\sqrt{x^8+1}} $ How can we integrate $\frac{1}{\sqrt{x^8+1}}$ thanks for help.
I couldn't find any way. I tried to factor, substituted with $\tan(a)$ but it's not working :'(
| Given $\displaystyle \int \frac{1}{\sqrt{x^8+1}}dx\;,$ Now Let $x^4 = t\;,$ Then $\displaystyle 4x^3dx = dt\Rightarrow dx = \frac{1}{4x^3}dt = \frac{1}{4}t^{-\frac{3}{4}}dt$
So Integral Convert into $$\displaystyle \frac{1}{4}\int t^{-\frac{3}{4}}\cdot \left(1+t^2\right)^{-\frac{1}{2}}dt$$
Now Using Integrals of differential binomials and Chebyshev’s criterion
Integration of $$\displaystyle \int x^{m}\cdot (a+bx^n)^{p}dx$$ in terms of elementry function is Possible only, When
$$\displaystyle p\;,\frac{m+1}{n}\;,\frac{m+1}{n}+p$$ is an Integer, Where $m,n,p\in \mathbb{Q}$.
So above we have $\displaystyle m=-\frac{3}{4}\notin \mathbb{Z}$ and $\displaystyle \frac{m+1}{n} = \frac{\left(-\frac{3}{4}+1\right)}{2} = \frac{1}{8}\notin \mathbb{Z}$ and
$\displaystyle \frac{m+1}{n}+p = \frac{1}{8}-\frac{1}{2} = -\frac{3}{8}\notin \mathbb{Z}$
So We can Not express the Integral $$\displaystyle \int\frac{1}{\sqrt{x^8+1}}dx$$ in terms of Elementry function..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating 2 integrals in polylogarithmic functions Are we aware of any nice way of calculating these $2$ integrals?
$$i) \space \int_0^1 \frac{\text{Li}_2\left(x-x^2\right)}{x^2-x+1} \, dx$$
$$ii)\space \int_0^1 \frac{\text{Li}_2\left(x-x^2\right)}{\left(x^2-x+1\right)^2} \, dx$$
If considering in the first integral that
$$\frac{1}{{x^2-x+1}}=\frac{1}{i \sqrt{3}}\left(\frac{1}{ \left(x-\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\right)}-\frac{1}{ \left(x-\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)\right)}\right)$$ then it seems Mathematica can handle the resulting integrals.
It's also easy to note that $$\frac{1}{i \sqrt{3}}\int_0^1 \frac{(1-2x)\log \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) \log \left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{ x(1-x)} \, dx=0$$
and then for the remaining part Mathematica yields some shorter result in polylogarithms.
So, we have that
$$ \int_0^1 \frac{\text{Li}_2\left(x-x^2\right)}{x^2-x+1} \, dx$$
$$=\frac{i}{\sqrt{3}}\int_0^1 \frac{ (1-2 x) \log ^2\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{ x-x^2} \, dx-\frac{i}{\sqrt{3}}\int_0^1 \frac{ (1-2 x) \log ^2\left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}{ x-x^2} \, dx$$
$$=\frac{2i}{\sqrt{3}}\int_0^1 \frac{ (1-2 x) \log ^2\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{ x-x^2} \, dx$$
where the last equality is simply got by using the variable change $x\mapsto 1-x$.
EDIT: Using the idea of dilogarithm reflection formula, we get that
$$\int_0^1 \frac{\text{Li}_2\left(x^2-x+1\right)}{x^2-x+1} \, dx=\frac{2}{\sqrt{3}} \int_{-\pi/6}^{\pi/6} \text{Li}_2\left(\frac{3 \sec ^2(x)}{4}\right) \, dx$$
$$=\frac{4}{\sqrt{3}} \int_{-\pi/6}^{\pi/6} \text{Li}_2\left(-\frac{\sqrt{3}}{2} \sec (x)\right) \, dx+\frac{4}{\sqrt{3}} \int_{-\pi/6}^{\pi/6} \text{Li}_2\left(\frac{\sqrt{3}}{2} \sec (x)\right) \, dx$$
but not sure yet if it might lead to an elegant solution.
| A first step. Let:
$$ I_k = \int_{0}^{1}\frac{(x-x^2)^k}{1-x+x^2}\,dx.$$
We have $I_0=\frac{2\pi}{3\sqrt{3}}, I_1=-1+\frac{2\pi}{3\sqrt{3}}$ and:
$$ I_{n+1} = \int_{0}^{1}\frac{(x-x^2)(x-x^2)^n}{1-x+x^2}\,dx = I_n-\int_{0}^{1}x^n(1-x)^n\,dx = I_n - \frac{\Gamma(n+1)^2}{\Gamma(2n+2)}$$
hence:
$$ I_{n}=\frac{2\pi}{3\sqrt{3}}-\sum_{k=0}^{n-1}\frac{1}{(2k+1)\binom{2k}{k}}\tag{1}$$
and:
$$\begin{eqnarray*} \mathcal{J}_1=\int_{0}^{1}\frac{\text{Li}_2(x-x^2)}{1-x+x^2}\,dx &=& \frac{\pi^3}{9\sqrt{3}}-\sum_{n\geq 1}\frac{1}{n^2}\sum_{k=1}^{n}\frac{\Gamma(k)^2}{\Gamma(2k)}\\&=&\frac{\pi^3}{9\sqrt{3}}-\sum_{k\geq 1}\frac{\Gamma(k)^2}{\Gamma(2k)}\left(\zeta(2)-H^{(2)}_{k-1}\right)\\&=&\sum_{k\geq 1}\frac{\Gamma(k)^2}{\Gamma(2k)}H_{k-1}^{(2)}.\tag{2}\end{eqnarray*}$$
Maybe it is useful to recall that
$$\frac{\Gamma(k)^2}{\Gamma(2k)} = [z^{2k-1}]\frac{2\arcsin(z/2)}{\sqrt{1-z^2/4}},\qquad H_{k-1}^{(2)}=[z^k]\frac{z\cdot\text{Li}_2(z)}{1-z}.$$
According to (a carefully driven) Mathematica, we simply have:
$$ \mathcal{J}_1 = \frac{\pi^3}{81\sqrt{3}},\tag{3}$$
quite astonishing. I bet that the reflection formula
$$ \text{Li}_2(x)+\text{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x)$$
and the other dilogarithm functional identities are behind this conspiracy, just cannot grasp how, at the moment.
As a matter of fact, that is not too difficult. If we complete the OP's manipulations with integration by parts, we are left to computing the two "conjugated" integrals:
$$ \int_{0}^{1}\frac{\log(x)\log(x-\omega)}{x-\omega}\,dx,\qquad \int_{0}^{1}\frac{\log(1-x)\log(x-\omega)}{x-\omega}\,dx$$
where the first one, for instance, is just:
$$ \frac{i}{81}\left(2\pi^3-27\pi\,\text{Li}_2(\overline{\omega})+54i\,\zeta(3) \right)$$
by the residue theorem. As an alternative, we can deal with $(2)$ through the Flajolet-Salvy technique. In a similar fashion,
$$\begin{eqnarray*} \mathcal{J}_2 &=& \int_{0}^{1}\frac{\text{Li}_2(x-x^2)}{(1-x+x^2)^2}\,dx\\&=&3+\frac{\pi^2}{9}-\frac{16}{3}\log 2-\frac{32}{3\sqrt{3}}\int_{0}^{1}\frac{x}{1-x^2}\arctan\left(\frac{x}{\sqrt{3}}\right)\log\left(\frac{1+x}{2}\right)\,dx.\end{eqnarray*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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A polynomial that satisfies $x^pf(1-x) + (1-x)^pf(x) = 1$ The context of this question is the construction of the Daubechies wavelet.
$f$ is a polynomial of degree $p-1$ which satisfies the equation:
$$
x^pf(1-x) + (1-x)^pf(x) = 1 \tag{1}
$$
Since
$$
f(x) = \frac{1}{(1-x)^p} - \frac{x^p}{(1-x)^p} f(1-x)
$$
and
$$
\frac{1}{(1-x)^p} = \sum_{k=0}^{\infty} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k
$$
it is argued that
$$
f(x) = \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k + O(x^p) = \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k
$$
where higher order terms are zero because $f$ is degree $p-1$.
My problem is that I can't seem to verify that $(1)$ holds with this $f$. I've tried expanding the $(1-x)^p$ and $(1-x)^k$ terms. This turns into a mess. I've tried doing induction on $p$. Again, I get nowhere. I've tried other approaches with no success. I feel like I'm missing something simple!
Here is an example of one of my attempts:
\begin{align*}
x^pf(1-x) + (1-x)^pf(x) &= x^p \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} (1-x)^k + (1-x)^p \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k \\
&= \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix}\left( x^p(1-x)^k + x^k(1-x)^p \right) \\
&= \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix}\left( x^{p-k}[x(1-x)]^k + [x(1-x)]^k(1-x)^{p-k} \right)
\end{align*}
I get to this point, and I can't help but thinking of using the Binomial Theorem, but I cannot seem to manipulate the factorial terms in nice way.
| I've found a method of validation which uses a system-of-equations type approach to find the coefficients of the polynomial which satisfies:
$$
x^pf(1-x) + (1-x)^pf(x) = 1 \tag{1}
$$
I then use induction to show that these coefficients are equal to the ones given above. It's somewhat tedious. I typed it up in markdown some time ago, and I thought I'd post it here for historical reasons, given the popularity of the Daubechies wavelet.
Let $g(x) = (1-x)^p f(x)$. This implies $f(x) = \frac{g(x)}{(1-x)^p}$. Since $f$ is a polynomial, this implies $g$ has a zero of order $p$ at $x=1$. So, $g^{(n)}(1) = 0, \;n=0,1,\ldots,p-1$. We can make the factorization:
$$
g(x) = (1-x)^p \sum_{j=0}^{p-1} a_j x^j \tag{2}
$$
It's clear that the coefficients, $a_j$, are the coefficients of $f$. How do we find them? Notice that differentiating condition $(1)$ $n$ times yields:
$$
g^{(n)}(x) + (-1)^n g^{(n)}(1-x) = \delta[n]
$$
where $\delta[n]$ is one only if $n=0$ and $1$ otherwise. Plugging in $x=1$, we find:
$$
g^{(n)}(0) = \delta[n], \; n=0,1,\ldots,p-1
$$
This yields $p$ linear equations. Thus, to have a unique solution, there must be exactly $p$ unknowns, which justifies why $f$ should be degree $p-1$.
Using $(2)$ and $n=0$, it's immediate that $a_0=1$. On the other hand, viewing $g(x) = r(x)s(x)$ with $r(x) = (1-x)^p$ and $s(x) = \sum_j= a_j x^j$, one can apply the general Leibniz rule to find:
$$
g^{(n)}(x) = \sum_{l=0}^n \sum_{j=n-l}^{p-1} \begin{pmatrix} n \\ l \end{pmatrix} \frac{p!}{(p-l)!} \frac{j!}{(j-(n-l))!} (-1)^l a_j x^{j-(n-l)} (1-x)^{p-1}
$$
Thus,
$$
g^{(n)}(0) = \sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} \frac{p!(n-1)!}{(p-l)!} (-1)^l a_{n-l}
$$
This yields explicitly the set of $p$ linear equations:
$$
\sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} \frac{p!(n-1)!}{(p-l)!} (-1)^l a_{n-l} = \delta[n], \; n=0,1,\ldots,p-1 \tag{3}
$$
This forms a triangular linear system is where the diagonal is $1$, so the polynomial, $f$, satisfying $(1)$ exists and is unique. It is not difficult manipulate $(3)$ into a recurrence beginning with $a_0$. However, it turns out that there is nicer closed form expression for the coefficients.
Proposition
$$
a_n = \begin{pmatrix} p-1+n \\ n \end{pmatrix} \tag{4}
$$
Proof
Clearly, $a_0=1$ using the above. We use induction to show that $(3)$ holds for $n>0$ precisely when we use $(4)$. That is, we show:
$$
\sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} \frac{p!(n-l)!}{(p-l)!} (-1)^l \begin{pmatrix} p-1+n-l \\ n-l \end{pmatrix} = \sum_{l=0}^n \begin{pmatrix} p \\ l \end{pmatrix} (-1)^l \frac{(p-1+n-1)!}{(p-1)!} = 0
$$
holds for $n>0$. The base case holds:
$$
\sum_{l=0}^1 \begin{pmatrix} p \\ l \end{pmatrix} (-1)^l \frac{(p-1)!}{(p-1)!} = 0
$$
Next, we show that
$$
\sum_{l=0}^{n+1} \begin{pmatrix} n+1 \\ l \end{pmatrix} (-1)^l \frac{(p-l+n)!}{(p-l)!} = 0 \tag{5}
$$
under the induction hypothesis. First, split the sum:
$$
(5) = \sum_{l=0}^n \begin{pmatrix} n+1 \\ l \end{pmatrix} (-1)^l \frac{(p-l+n)!}{(p-l)!} + (-1)^{n+1} \frac{(p-1)!}{(p-n-1)!} =: (a) + (b)
$$
Next, we use an identity sometimes called Pascal's rule to split the binomial term in $(a)$:
$$
(a) = \sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l+n)!}{(p-l)!} + \sum_{l=1}^n \begin{pmatrix} n \\ l-1 \end{pmatrix} (-1)^l \frac{(p-l+n)!}{(p-l)!} =: (c) + (d)
$$
First, we focus on $(c)$:
\begin{align*}
(c) &= \sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l+n-1)!}{(p-l)!} ((p+n) - l) \\
&= (p+n) \sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l+n-1)!}{(p-l)!} + \sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l+n-1)!}{(p-l)!} (-l) \\
&= 0 + \sum_{l=0}^n \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l+n-1)!}{(p-l)!} (-l)
\end{align*}
The first term is $0$ due to the induction hypothesis. Next, let's look at $(d)$:
\begin{align*}
(d) &= \sum_{l=1}^n \begin{pmatrix} n \\ l-1 \end{pmatrix} (-1)^l \frac{(p-l+n)!}{(p-l)!} \\
&= -\sum_{l=0}^{n-1} \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l)(p-l+n-1)!}{(p-l)!} \\
&= -\sum_{l=0}^{n} \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l)(p-l+n-1)!}{(p-l)!} + (-1)^n \frac{(p-n)(p-1)!}{(p-n)!} \\
&= 0 + \sum_{l=0}^{n} \begin{pmatrix} n \\ l \end{pmatrix} (-1)^l \frac{(p-l+n-1)!}{(p-l)!}l + (-1)^n \frac{(p-n)(p-1)!}{(p-n)!}
\end{align*}
where the last step is again due the induction hypothesis. Wrapping up, we have that:
$$
(c) + (d) = (-1)^n \frac{(p-1)!}{(p-n-1)!}
$$
So that:
$$
(5) = (a) + (b) = 0
$$
which completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$
If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$
I don't know how to find all possible values.
I tried making the lower of the two roots as negative. That does not seem to work out. As $4>0$ the curve will be concave up. There can be numerous cases possible.
| Examining the extrema of $f(x)=4x^2+ax+a-3$ which is a family of upwards open parabolas:
$$
f'(x) = 8 x + a = 0 \Rightarrow x = -\frac{1}{8}a \\
f''(x) = 8 > 0
$$
So we have a local minimum at $x = -(1/8)a$. There the function has the value:
$$
f(-(1/8)a)
= \frac{1}{16}a^2 - \frac{1}{8}a^2 + a - 3
= -\frac{1}{16}a^2 + a - 3
$$
The minima regarding different $a$ values lie on the curve
$$
m(x) = -4x^2-8x -3
$$
(Large image version)
From
$$
0 = -\frac{1}{16} a^2 + a - 3 \Rightarrow \\
a^2 = 16 a - 48 \Rightarrow \\
(a - 8)^2 = 64 - 48 = 16 \Rightarrow \\
a = 8 \pm 4
$$
we get that the minimum is non-negative for $a \in [4, 12]$ and there we will find no negative value for $f$ at all.
So we have a chance to find negative $f$ values for negative argument only for $a \in I = \mathbb{R} \setminus [4,12]$, where the minimum occurs for a negative function value.
For positive $a\in I$ the minimum is at $x_m = -(1/8) a < 0$, so those $a$ are valid.
For negative $a \in I$ for $x = -1/2 < 0$ we have
$f(-1/2) = 1 - a/2 + a - 3 = a/2 - 2 < 0$, so those $a$ are valid as well.
This gives $I = \mathbb{R} \setminus [4,12]$ as set of all feasible $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
How do i find all integers $y$ such that $y^3 = 3x^2+3x+7$, where $x$ is also an integer? I have tried to find all integers $y$ such that $$y^3 = 3x^2+3x+7$$, where
$x$ is also an integer but i didn't succed only i guess that no integer $y$
$x$satisfied that equation so i would like to solve to find all integers
satisfiing it ?
Thank you for any help .
| Try to take mod $3 \rightarrow y^3 \equiv 1 \pmod {3} \rightarrow y \equiv 1 \pmod {3} \implies y^3 \equiv 1 \pmod {9}$
Therefore $3x^2 + 3x + 7 \equiv 1 \pmod {9} \rightarrow 3x^2 + 3x \equiv 3 \pmod {9} \rightarrow x^2 + x \equiv 1 \pmod {3}$
This is impossible. If $x \equiv 1 \pmod {3}, x^2 + x \equiv 2 \pmod {3}$ If $x \equiv 2 \pmod{3}, x^2 + x \equiv 0 \pmod{3}$ and obviously if $x \equiv 0 \pmod {3}, x^2 + x \equiv 0 \pmod{3}$
Therefore no such x and y exist where $x$ and y are both integers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I evaluate $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$? How do i evaluate this limit : $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+............+\frac{1}{\sqrt{n}}\right)$$ ?
Thank you for any help .
| \begin{align}
&\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}}\right) \\
= \ &\frac{1}{n}\left(\sqrt{\frac{n}{1}} + \sqrt{\frac{n}{2}} + \ldots + \sqrt{\frac{n}{n}}\right) \\
= \ &\frac{1}{n}\left(\frac{1}{\sqrt{\frac{1}{n}}} + \frac{1}{\sqrt{\frac{2}{n}}} + \ldots + \frac{1}{\sqrt{\frac{n}{n}}}\right) \\
\rightarrow \ &\int_0^1 \frac{1}{\sqrt{x}} \mathrm{d}x = 2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Inverse laplace transform with square completion I need to find the inverse laplace of this : $$\frac{s+2}{s^2+2s+5}$$
I know that completing the square should help me to solve this so I get
$$\frac{s+2}{(s+1)^2+4}$$
Then separating this equation i get
$$\frac{s}{(s+1)^2+2^2} + \frac{2}{(s+1)^2+2^2}$$
I can find the inverse Laplace for the second part easily but I can't find it for
$$\frac{s}{(s+1)^2+2^2}$$
What am I not understanding here?
| Just another approach. The zeroes of $s^2+2s+5$ occur at $s=-1\pm 2i$, and:
$$\text{Res}\left(\frac{s+2}{s^2+2s+5},s=-1\pm 2i\right)=\frac{1}{2}\pm \frac{i}{4},\tag{1}$$
so:
$$ f(s) = \frac{s+2}{s^2+2s+5} = \frac{\frac{1}{2}+\frac{i}{4}}{s-(-1+2i)}+\frac{\frac{1}{2}-\frac{i}{4}}{s-(-1-2i)}\tag{2} $$
and since $\mathcal{L}^{-1}\left(\frac{1}{s-\alpha}\right)=e^{\alpha x}$ we have:
$$\begin{eqnarray*}\mathcal{L}^{-1}(f(s)) &=& \left(\frac{1}{2}+\frac{i}{4}\right) e^{(-1+2i)x}+\left(\frac{1}{2}-\frac{i}{4}\right) e^{(-1-2i)x}\\&=&\color{red}{e^{-x}\cos(2x)+\frac{1}{2}e^{-x}\sin(2x)}.\tag{3}\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is :$\sqrt{i\pi+\sqrt{i\pi+\sqrt{i\pi+\sqrt\cdots}}}$ irrational or transcendental or real number? Is there someone who can show me if :$$\sqrt{i\pi+\sqrt{i\pi+\sqrt{i\pi+\sqrt\cdots}}}$$ is irrational or real or transcendental number ?
Thank you for any help
| Let's assume this is a complex number $z=x+iy$ in the first quadrant. Then $z+i\pi=z^2$ implies $$x+i(y+\pi)=(x+iy)^2 = x^2-y^2 +2ixy$$ so
$$ x = x^2 - y^2,\ \ \ y + \pi = 2xy.$$
From the second equation we get $x = \frac{y+\pi}{2y}$, which plugging into the first gives
$$ \frac{y+\pi}{2y} = \frac{y^2+\pi^2+2\pi y}{4y^2} - 2y^2,$$
from which
$$2y(y+\pi) = y^2+\pi^2 + 2\pi y - 4y^4 \Longrightarrow 4y^4 + y^2 -\pi^2=0.$$
Then using the assumption that $y>0$ we obtain
$$ y^2 = \frac{-1 + \sqrt{1+16 \pi^2}}{8} \Longrightarrow y = \frac{\sqrt{-1+\sqrt{1+16\pi^2}}}{2\sqrt{2}}.$$
Computing $x$ using $x= (y+\pi)/2y$ and the above
$$ x = \frac{1}{2}+\pi\sqrt{\frac{2}{-1+\sqrt{1+16\pi^2}}},$$
so
$$ z = \frac{1}{2}+\pi\sqrt{\frac{2}{-1+\sqrt{1+16\pi^2}}} + \frac{i}{4}\sqrt{-2+2\sqrt{1+16\pi^2}},$$
which is transcendental. The case of the other quadrants should be similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
} |
A limit problem: $\lim\limits_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$ I need help in solving the limit below:
$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$$
What I've done is to simplify the upper part to: $$\frac{2^{n+1}-1}{2^n}$$
Any hints or solutions will be greatly appreciated.
| Here are the steps
$$ \lim\limits_{n\to\infty} \frac{\sum\limits_{k=0}^n \frac{1}{2^k}}{\sum\limits_{k=0}^n \frac{1}{3^k}} = \lim\limits_{n\to\infty} \frac{2-\frac1{2^n}}{\frac12\left(3-\frac1{3^n}\right)}$$
$$= \frac{2-0}{\frac12\left(3-0\right)} = \frac2{\frac32} = \frac43$$
Study geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What happens to the following limit when $b\in [0, 1)$ and $b > 1$? Problem:
Find all $a, b$ which make the following statement true: $\lim_{x\to 0}\frac{\exp{\left(\sin ax\right)} - \cos x}{x^b} = \frac{1}{2}$
Attempted solution:
Firstly, let's notice that if $b < 0$ then the limit is $0$ no matter what the value of $a$ is.
Secondly, let's rewrite the fraction using the Maclaurin Series for $\exp$ and $\cos$:
$$
\frac{1 + \sin(ax) + \frac{1}{2!}(\sin^2 ax) + \frac{1}{3!}\sin^3(ax) + \dots+ (-1)(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots)}{x^b} = \\
\frac{\sin(ax) + \frac{1}{2!}(\sin^2(ax) + x^2) + \frac{1}{3!}\sin^3(ax) + \frac{1}{4!}(\sin^4(ax) - x^4) + \dots}{x^b}
$$
From this representation it is clear that if $b = 1$ and $a = 0.5$:
$$
\begin{split}
&\lim_{x\to 0}\frac{\sin(0.5 x) + \frac{1}{2!}(\sin^2(0.5 x) + x^2) + \frac{1}{3!}\sin^3(0.5x) + \frac{1}{4!}(\sin^4(0.5x) - x^4) + \dots}{x} &= \\
&\lim_{x\to 0}\frac{\sin(0.5 x)}{x} + 0 &= \frac{1}{2}
\end{split}
$$
At this point let's remember that $\lim_{x\to 0}\sin x = \lim_{x\to 0} x$ so the fraction may be rewritten like so:
$$
\lim_{x\to 0}\frac{ax + \frac{1}{2!}((ax)^2 + x^2) + \frac{1}{3!}(ax)^3 + \frac{1}{4!}((ax)^4 - x^4) + \dots}{x^b}
$$
which makes this limit $0$ if $b\in[0, 1)$ and $\infty$ if $b > 1$.
| If $b=0$ then the limit is $0$ independently of $a$.
Note that no matter how $a$ and $b>0$ are, this limit is of the type $\frac 0 0$, therefore let us use l'Hospital's theorem on it, replacing the numerator and the denominator by their derivatives, respectively: $\lim \limits _{x \to 0} \frac {a \Bbb e ^{\sin ax} \cos ax + \sin x} {b x^{b-1}} = \lim \limits _{x \to 0} \frac {\cos ax} b \frac {a \Bbb e ^{\sin ax} + \tan x} {x^{b-1}} = \frac 1 b \lim \limits _{x \to 0} \frac {a \Bbb e ^{\sin ax} + \tan x} {x^{b-1}}$. Note that the numerator tends to $a$.
If $b<1$ then $x$ must be positive (what would $(-1) ^{0.5}$ mean?), so the denominator tends to $\infty$, so the whole limit is $0$. If $b=1$ then the denominator is constant $1$ and the limit is $\frac a b$. Finally, if $b>1$ you must have again $x>0$ and the limit is $\frac 1 b \frac 1 {0^+} = \infty$.
Therefore, the only possibility is $a=\frac 1 2, b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the cubic equation of roots $α, β, γ$. Taken from Fitzpatrick $4$ unit course textbook. The question says:
If the cubic equation $\ ax^3+bx^2+cx+d$ has roots $α, β, γ$. Find the cubic equation who's roots are $α^2, β^2, γ^2$
I keep getting a $±$ sign that I can't get rid of. The answer in the back is $x(ax+c)^2=(bx+d)^2$
Thanks for any help.
| Notice that if $$x^3 + \frac{b}{a} x^2 + \frac{c}{a}x + \frac{d}{a} = (x-\alpha)(x-\beta)(x-\gamma),$$ then
$$
x^3 + \frac{b}{a} x^2 + \frac{c}{a}x + \frac{d}{a} = x^3 - (\alpha + \beta + \gamma) x^2 + (\alpha \beta + \alpha \gamma + \beta \gamma) x - (\alpha \beta \gamma).
$$
Therefore
\begin{align*}
\alpha \beta \gamma &= -\frac{d}{a}\\
\alpha \beta + \alpha \gamma + \beta \gamma &= \;\;\frac{c}{a}\\
\alpha + \beta + \gamma &= - \frac{b}{a}.
\end{align*}
Now the polynomial we would like to find is $(x - \alpha^2)(x - \beta^2)(x - \gamma^2).$ Expanding this out we see this is equal to
$$
x^3 - (\alpha^2 + \beta^2 + \gamma^2)x^2 + (\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \beta^2 \gamma^2) x - (\alpha^2 \beta^2 \gamma^2).
$$
To find these expressions in terms of $a,b,c,$ and $d$, we just fiddle with the above three equations.
The easiest value to find is $\alpha^2 \beta^2 \gamma^2$. Clearly this is just $d^2/a^2$. To find $\alpha^2 + \beta^2 + \gamma^2$, observe that
$$
(\alpha + \beta + \gamma)^2 = (\alpha^2 + \beta^2 + \gamma^2) + 2(\alpha \beta + \alpha \gamma + \beta \gamma).
$$
Therefore
$$
\alpha^2 + \beta^2 + \gamma^2 = \frac{b^2}{a^2} - 2 \frac{c}{a}.
$$
Finally, to calculate $\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \beta^2 \gamma^2$, we observe that
$$
(\alpha \beta + \alpha \gamma + \alpha \beta)^2 = (\alpha^2 \beta^2+\alpha^2 \gamma^2 + \beta^2 \gamma^2) + 2(\alpha \beta \gamma)(\alpha + \beta + \gamma)
$$
so the value in question is equal to
$$
\frac{c^2-2bd}{a^2}.
$$
Putting this altogether, the polynomial we are looking for must be
$$
x^3 - \left(\frac{b^2 - 2ac}{a^2}\right)x^2 + \left(\frac{c^2-2bd}{a^2}\right) x - \frac{d^2}{a^2}.
$$
This polynomial will have the same zeros if we multiply by a constant, so we multiply by $a^2$ to get the polynomial
\begin{align*}
x^3 - (b^2 - 2ac) x^2 + (c^2 - 2bd)x - d^2 &= (x^3 + 2ac x^2 + c^2x) - (b^2x^2 + 2bd x + d^2) \\
&= x(ax+c)^2 - (bx+d)^2,
\end{align*}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$ If $$x+y+z=7$$ and $$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}$$ Find the value of $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$$
I tried but I got nothing
| Multiply both sides of $\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = \frac{7}{10}$ by $x+y+z$. Then it becomes $\frac{x+y+z}{x+y} + \frac{x+y+z}{y+z} + \frac{x+y+z}{z+x} = \frac{49}{10}$. Then you can simplify $\frac{x+y+z}{x+y}$ into $1+\frac{z}{x+y}$, and so on, getting
$3+\frac{z}{x+y} + \frac{x}{y+z} + \frac{y}{z+x} = \frac{49}{10}$. Now subtract 3 from both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to integrate $\int \frac{1} {{\sin x+\sec^2x}}{d}x$ I am a newbie in learning topic of integration. My friend asked me to find indefinite integral shown below
$$y=\int \frac{1} {{\sin(x)+\sec^2(x)}} \, \mathrm{d}x \tag 1$$
What I tried until now is the substitution
$m=\sin(x)$ and
$$\frac{\textrm{d}m}{\textrm{d}x}=\cos(x)$$
Now, converting equation $(1)$ in terms of $m$ to get
$$y=\int \frac{(1-m^2)^{1/2}} {{1+m(1-m^2)}} \, \mathrm{d}m$$
But, as you can see, it became more complicated than the original equation $(1)$. So, can anybody help me to integrate this integral?
| Rewrite the integral as
\begin{align}
I=&\int \frac{1} {{\sin x+\sec^2x}}dx\\
=&\int \frac{\sin^2x-1} {{\sin^3x-\sin x-1}}dx\\
=&\ \frac1{2a+3}\int \frac1{\sin x -a}+ \frac{2(a+1)\sin x+a(2a+1)} {{\sin^2x+a\sin x+a^{-1}}}\ dx\\
\end{align}
where $a$ satisfies the cubic equation $a^3-a-1=0$, given by
$$a =\sqrt[3]{\frac12+\frac12\sqrt{\frac{23}{27}}}+ \sqrt[3]{\frac12-\frac12\sqrt{\frac{23}{27}}}\approx 1.3247
$$
Then, integrate with the substitution $t=\tan(\frac\pi4-\frac x2)$ to obtain
\begin{align}
I =&\ \frac2{2a+3}\int \frac1{a^3t^2+a^{-4}}-\frac{a^8+a^{-6}\ t^2} {a^{-3}\ t^4-2a^{-5}\ t^2 +a^4}\ dt\\
=& \ \frac2{2a+3}\bigg(
\sqrt{a}\tan^{-1}(a^{7/2}t)
- \frac{1+a^{21/2}}{2a^2\sqrt{2(a^{11/2}-1)}}\tan^{-1}\frac{at-{a^{9/2}}t^{-1}}{\sqrt{2(a^{11/2}-1)}}\\
&\hspace{3cm}+ \frac{1-a^{21/2}}{2a^2\sqrt{2(a^{11/2}+1)}}\coth^{-1}\frac{at+{a^{9/2}}t^{-1}}{\sqrt{2(a^{11/2}+1)}}\ \bigg)
\end{align}
As an example, the definition integral below evaluates to
$$\int_{-\frac\pi2}^{\frac\pi2} \frac{1} {{\sin x+\sec^2x}}dx
=\frac\pi{2a+3}\bigg(\ \frac{a^{21/2}+1}{a^2\sqrt{2(a^{11/2}-1)}}-\sqrt a \bigg)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
} |
A problem in Algebra If $a+b+c+d=0=a^7+b^7+c^7+d^7$,
Then prove $a(a+b)(a+c)(a+d)=0$ and all are real numbers.
I am confused and don't even know which to tag for help.
| Yeah, I saw the duplicate question. But here is another useful application of the AM-GM inequality(didn't think of it, did you)
Tricky factorization:
$$0=-(a^7+b^7+c^7+d^7)$$$$=7(b+c)(c+d)(d+a)E$$ where $E=(b^2+c^2+d^2+bc+cd+da)^2+bcd(b+c+d)$. However, $$4E=((b+c)^2+(c+d)^2+(b+d)^2)^2)-4abcd$$$$=((b+a)^2+(c+a)^2+(a+d)^2)^2)-4abcd$$$$=(3a^2+2a(b+c+d)+b^2+c^2+d^2)^2-4abcd$$ The last expression reduces to $$(a^2+b^2+c^2+d^2)^2-4abcd$$ However we see that $$(a^2+b^2+c^2+d^2)^2-4abcd \geq 12|abcd| \geq 0$$ from the AM-GM inequality. We conclude that $4E \geq 0$ with equality if and only if $|a|=|b|=|c|=|d|=0$. Otherwise one of $b+c,b+d,c+d$ must be $0$. Hence, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve: $-\frac{1}{\sqrt{2}} < \sin \theta + \cos \theta < \frac{1}{\sqrt{2}}$ The question is:
Solve $$-\frac{1}{\sqrt{2}} \lt \sin\theta + \cos\theta < \frac{1}{\sqrt{2}}$$ for values of $\theta$ between $0^\circ$ and $180^\circ$.
I realized that:
$$\begin{align}
-\frac{1}{\sqrt{2}} < \sin\theta + \cosθ &< \frac{1}{\sqrt{2}} \\[4pt]
\left|\sin\theta + \cosθ\right| &< \frac{1}{\sqrt{2}} \\[4pt]
\left(\sin\theta + \cos\theta\right)^2 &< \frac12 \\[4pt]
\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta &< \frac12 \\[4pt]
1 + 2\sin\theta\cos\theta &< \frac12 \\[4pt]
1 + \sin 2\theta &< \frac12 \\[4pt]
\sin 2\theta &< -\frac12
\end{align}$$
But I don't know where to go from here. Can someone help me figure out how to get to the answer in the book: $105^\circ < θ < 165^\circ$. Thanks!
| From that point you can do
$$\sin2\theta<-\frac12$$
Since $\theta$ is between $0$ and $180^o$, $2\theta$ is between $0$ and $360^o$. Then,
$$210^o<2\theta<330^o$$
Now, dividing by $2$, you obtain your book's result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to solve this 2-D integration? How to solve
$$I=\int_{[0,1]^2}\frac{dxdy}{(1+xy)(1+x^2)}$$
I've tried using the diffeomorphism $(x,y)=(u,v/u)$ from $\text{int}\{(u,v)\mid 0\le u\le 1,0\le v\le u\}$ to $\text{int}[0,1]^2$, but since the Jaccobian determinant becomes unbounded at the boundary, the transformation leads to an improper integral, which I think makes it invalid.
If I integrate directly over the original domain, the result will be too complicated to calculate.
I can't come up with another diffeomorphism. Is there another method to go on?
| I don't know if this is the slickest way, but at least it works. First decompose into partial fractions with respect to $x$:
$$
\frac{1}{(1+xy)(1+x^2)} = \frac{1}{1+y^2} \left( \frac{y^2}{1+xy} + \frac{1-xy}{1+x^2} \right)
.
$$
Rewrite this (using $y^2=(1+y^2)-1$ as)
$$
\frac{1}{1+xy}
- \frac{1}{(1+xy)(1+y^2)}
+ \frac{1}{1+x^2} \cdot \frac{1}{1+y^2}
- \frac{x}{1+x^2} \cdot \frac{y}{1+y^2}
.
$$
The last two terms are easy to integrate over the square, since $x$ and $y$ separate.
The integral of the second term equals the integral that we're trying to compute (call it $I$), by symmetry.
The integral of the first term is trickier, but we can expand $(1+xy)^{-1}=1 - x y + x^2 y^2 - x^3 y^3 + \dotsb$, which upon termwise integration gives the known series $1-1/2^2+1/3^2-1/4^2+\dotsb = \pi^2/12$. One could probably also get this in the same way as the integral of $(1-xy)^{-1}$ over the unit square is computed in Chapter 8 of Proofs from THE BOOK by Aigner & Ziegler, namely by setting $x=u-v$ and $y=u+v$.
All combined, we get
$$
I = \frac{\pi^2}{12} - I + \left( \frac{\pi}{4} \right)^2 - \left( \frac{\ln 2}{2} \right)^2
,
$$
so
$$
I=\frac{7 \pi^2}{96} - \frac{\ln^2 2}{8}
.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$ confusion $$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$$
=$$x^2+3\ln \left(x-3\right)+2\ln \left(x+1\right)+C$$
Where did the $x^2$ come from?
What I did:
Partial fraction then integrate, got the answer without $x^2$ term.
|
$$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$$
$$=\int\bigg(2x+\frac{3}{x-3}+\frac{2}{x+1}\bigg)dx$$
$$=2\int\frac{1}{x+1}dx+3\int\frac{1}{x-3}dx+2\int xdx$$
$$\boxed{\color{red}{=\color{blue}{x^2}+2 \ln|x+1|-3 \ln|x-3|+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
$
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
| Answer to question:
If we take $x=3$, we see that the equality doesn't hold, thus answers a, b and c are wrong. The only option left is d.
In case 1 you made an error, this equality doesn't hold:
$$\frac{x^2−5x+12}{x−3}≤0$$
if you take $x=5$ we see that
$$\frac{5^2−5\cdot 5+12}{5−3} = \frac{25−25+12}{2} = \frac{12}{2}=6\geq 0$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's what I did: \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2} \\ = \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2 (4-6/x+7/x^2)} \\ \\ = \lim_{x \to \infty} \frac{\sqrt{x^2(1/x+1/x^2)}(1-\sqrt{x^2(2/x+3/x^2)}}{x(4-6/x+7/x^2)} \\ \\ \lim_{x \to \infty} \frac{\sqrt{1/x+1/x^2}(1-x \sqrt{2/x+3/x^2})}{4-6/x+7/x^2}
\end{align*} If I now evaluate this limit, everything in the numerator goes to zero except $1$. And the denominator leaves me with $4$. So I thought the answer should be $1/4$?
| Your first step of dividing both numerator and denominator by $x^2$ was a good idea. The denominator then converges to 4.
Now, you can write for the numerator:
$$ \frac{x\sqrt{x+1}(1-\sqrt{2x+3})}{x^2}= \frac{x\sqrt{x+1}}{x\sqrt{x}}\cdot\frac{1-\sqrt{2x+3}}{\sqrt{x}} $$
by splitting $x^2$ into $x\sqrt{x}$ and $\sqrt{x}$.
The first term apparently goes to 1, while the second terms converges to $-\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$a^4+b^4+c^4+d^4 \neq 2^{2011}$ Prove (elementary, meaning no high level theorems used) that there can not exist 4 prime numbers a,b,c,d $\geq$ 7 such that
\begin{equation}a^4+b^4+c^4+d^4=2^{2011}\end{equation}
I tried the following:
The last digit l(d) of the fourth power of a prime number is 1 or 5.
The last digit of $2^{2011}$ is 8.
But there exists a combination that holds true:
\begin{equation}l(a)=1,l(b)=1,l(c)=1,l(d)=5\end{equation}
\begin{equation}1+1+1+5=8\end{equation}
| If $n$ is odd, $n^4\equiv 1\pmod{16}$, hence if $a,b,c,d$ are prime numbers and at least one of them is odd,
$$ a^4+b^4+c^4+d^4 \in \{1,2,3,4\}\pmod{16}.$$
However, $2^{2011}\equiv 0\pmod{16}$ and $2^4+2^4+2^4+2^4$ is way smaller than $2^{2011}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Probability question about coin flipping If I flip an unbiased coin an infinite number of times, what is the probability that, at some point, the number of heads will be twice the number of tails?
I have already tried making a branching probability tree, but I just get overwhelmed and it seems to go on forever... I don't know how to reconcile it.
| Empirically it seems to be $$\dfrac{3}{8^1} + \dfrac{6}{8^2} + \dfrac{21}{8^3} + \dfrac{90}{8^4} + \dfrac{429}{8^5} + \dfrac{2184}{8^6} + \dfrac{11628}{8^7} + \dfrac{63954}{8^8} + \dfrac{360525}{8^9} + \dfrac{2072070}{8^{10}} + \dfrac{12096045}{8^{11}} + \dfrac{71524440}{8^{12}} + \dfrac{427496076}{8^{13}} + \dfrac{2578547760}{8^{14}} + \dfrac{15675792072}{8^{15}} + \dfrac{95951017602}{8^{16}} + \cdots $$
which seems to be about $0.573$.
Added: As a sum it seems to be $\displaystyle \sum_{n=1}^{\infty} \dfrac{2}{8^n(3n-1)} {3n \choose n}$ which is $\frac{3}{4}(3-\sqrt{5})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Solving following quartic equation Solve in $\mathbb{R}$ :
$$(x^2+2)^2+8x^2=6x (x^2+2) $$
My try: I tried to make the graph by calculating values for $x=1, 2, 3, 4$ and I found that the function is positive at $x=0$ but negative at $x=1$, so the graph must have crossed the $x$ axis, thus there will be a root between $0$ and $1$ and similarly this was the case between $3$ and $4$, but I was unable to solve it precisely. I was also unable to find about the other two roots.
What is the way to solve it just by algebra or rough plotting with the help of pen and paper and not using any computational tools?
| Given that
$$(x^2+2)^2+8x^2=6x(x^2+2)$$
Put $x^2+2=y$. Then
$$y^2+8x^2=6xy$$
$$8x^2-6xy+y^2=0$$
$$8x^2-4xy-2xy+y^2=0$$
$$4x(2x-y)-y(2x-y)=0$$
$$(4x-y)(2x-y)=0$$
$$(4x-x^2-2)(2x-x^2-2)=0$$
$$(x^2-4x+2)(x^2-2x+2)=0$$
Now we have
$$x^2-4x+2=0$$
This is quadratic eq
$$x=\frac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}$$
$$x=\frac{4 \pm 2\sqrt{2}}{2}$$
$$ x=2 \pm \sqrt{2}$$
Similarly we have second eq
$$x^2-2x+2=0$$
Bu quadratic formula
$$x=\frac{-(-2) \pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}$$
$$x=\frac{2 \pm \sqrt{4(-1)}}{2}$$
$$x=\frac{2 \pm 2\sqrt{-1}}{2}$$
$$x=1 \pm i$$
As $\sqrt{-1}=i$.
So all roots of "$x$" we have
$$x=2 \pm \sqrt{2} , 1 \pm i $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Epsilon-delta proof of $\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2}$ I have been doing $\varepsilon$-$\delta$ proofs for fun and I challenged myself to prove $$\displaystyle\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2},\quad a,b\in\mathbb{R}$$
The definition says: We say that $\displaystyle\lim_{x\to\infty}f(x)=l$ if for any positive number $\varepsilon$ we can find a positive number $N$ (depending on $\varepsilon$ in general) such that $|f(x)-l|<\varepsilon$ whenever $x>N$.
So I started with: $\left|\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}\right|<\varepsilon$ whenever $x>N$.
Manipulating the first inequatlity
\begin{gather*}
-\varepsilon<\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}<\varepsilon\\
-\varepsilon+\dfrac{a+b}{2}<\sqrt{(x+a)(x+b)}-x<\varepsilon+\frac{a+b}{2}
\end{gather*}
At this point I thought about adding $x$, squaring the expressions and then expanding them. I did it and I got: $$\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2-a\varepsilon-b\varepsilon-2x\varepsilon+\varepsilon^2<x^2+ax+bx+ab<\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2+a\varepsilon+b\varepsilon+2x\varepsilon+\varepsilon^2$$
And here I'm not sure how to proceed. Am I on the right track?
Thanks for any help / hints.
Note: There might be other ways to prove this, but I'd like to do it using just algebra if possible, even if it's not the best method.
| The problem is that the $\epsilon$-method requires the result of the limit...
A "look-alike" method without knowing the result of the limit...
Given
$$
\lim_{x \rightarrow \infty} \left( \sqrt{ ( x + a ) ( x + b ) } - x \right).
$$
Let
$$
\sqrt{ ( x + a ) ( x + b ) } - x = \ell.
$$
Rewrite this as
$$
( x + a ) ( x + b ) = ( x + \ell)^2.
$$
So
$$
x^2 + ( a + b ) x + a b = x^2 + 2 \ell x + \ell^2,
$$
or
$$
( a + b - 2 \ell ) x = \ell^2 - a b.
$$
So we can find $x$ for a given $\ell$, using
$$
x = \frac{ \ell^2 - a b }{ a + b - 2 \ell },
$$
and for the limit, we find $x \rightarrow \infty$.
We clearly see that
$$
\lim_{\ell \rightarrow \frac{a+b}{2}} x
= \lim_{\ell \rightarrow \frac{a+b}{2}} \frac{ \ell^2 - a b }{ a + b - 2 \ell } = \infty,
$$
consequently, we obtain
$$
\lim_{x \rightarrow \infty} \left( \sqrt{ ( x + a ) ( x + b ) } - x \right) = \frac{a+b}{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Tangent line on open interval using the mean value theorem I am having problems understanding how to solve the following question:
Find all points in the open interval $(a, b)$ where the tangent line to $y = f(x)$ is parallel to the secant line joining $(a, f(a))$ and $(b, f(b))$ when $f(x) = x^5 − 5x + 1$ with domain $[−1, 1]$.
I have no problems understanding how to solve this equation at a given point using the mean value theorem, but am having problems understanding how to solve this equation on an open interval.
Link to Desmos Graph of my problem
My current solution is
$$
y= (f'(m_2))(x-m_2)+f(m_2)
$$
where $m_2 = \left(\frac{m+5}{5}\right)^{1/4}$ and $m = \frac{f(a)-f(b)}{a-b}$ which is the slope of the secant line.
Please any help would be much appreciated as I do not have the solution and am unsure of the correct solution.
| The tangent line to $y=f(x)$ at a point $x\in [a,b]$ is given by the derivative $f'(x)$. We will assume that $a<b$. For
$$f=x^5-5x+1 \tag 1$$
the derivative is given by
$$f'(x)=5x^4-5\tag 2$$
We wish to find all points for which
$$f'(x)=\frac{f(b)-f(a)}{b-a}\tag 3$$
The Mean Value Theorem assures that there exists such a point. Thus, we substitute $(1)$ and $(2)$ into $(3)$ to obtain
$$\begin{align}
5x^4-5&=\frac{(b^5-a^5)-5(b-a)}{b-a}\\\\
&=b^4+ab^3+a^2b^2+a^3b+a^4-5\\\\
x^4&=\frac{b^4+ab^3+a^2b^2+a^3b+a^4}{5}\\\\
x^4&=\frac{b^4}{5}\left(\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1\right)\tag 4
\end{align}$$
It is easy to show that polynomial $x^4+x^3+x^2+x+1>0$ for all $x$. Thus, the real solutions are given by
$$x=\pm|b|\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4} \tag 5$$
We now analyze carefully the solutions as given by $(5)$.
CASE 1: $0<a<b$ or $a<b<0$.
For $0<a<x<b$ or $a<x<b<0$, then
$$x=b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}$$
CASE 2: $a<0<x<b$.
$$x=b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}$$
occurs when $a/b$ satisfies
$$\frac ab <\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}<1$$
which occurs for $r<\frac ab <0$, where
$$r=\frac{1}{12}\left(-3-\frac{(15)^{2/3}}{(4\sqrt{6}-9)^{1/3}}+(15(4\sqrt{6}-9)^{1/3}\right)\approx. -0.605829586188268$$
is the negative, real-valued root of the equation $4y^4-y^3-y^2-y-1=0$.
CASE 3: $a<x<0<b$.
$$x=-b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}$$
occurs when $a/b$ satisfies
$$\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}<-\frac ab$$
which occurs for $\frac ab <r$, where
SUMMARY:
The point $x$, where the slope of the tangent line to the curve $y=x^5-5x+1$ is equal to the slope of the secant line on the interval $[a,b]$, is given by
$$x=
\begin{cases}
b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}, & ab>0\,\,\text{or}\,\,a<0<b,\,\,\text{and}\,\,\frac{a}{b}>r\\\\
-b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}, & a<0<b,\,\,\text{and}\,\,\frac{a}{b}<r
\end{cases}
$$
where
$$r=\frac{1}{12}\left(-3-\frac{(15)^{2/3}}{(4\sqrt{6}-9)^{1/3}}+(15(4\sqrt{6}-9)^{1/3}\right)\approx. -0.605829586188268$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$ If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$.
I tried to substitute the equation 1 into equation 2 so that I can find the value of $k$ or $h$, but it did not work as the base is not the same (I cannot compare to find the answer)
How can I find the answer?
| We can see that $2^h4^{h-k}=\frac{20}{9}$. On the other hand $2^{2h-k}=\frac{4}{3}$. So $\frac{4^{h-k}}{2^{k-h}}=\frac{5}{3}$. This implies $2^{3h-3k}=\frac{5}{3}$. Since $2^k=\frac{3}{4}2^{2h}$ we have $2^{3k}=\frac{27}{64}2^{6h}$. Therefore, $\frac{2^{3h}}{2^{3k}}=\frac{64}{27}\frac{2^{3h}}{2^{6h}}=\frac{64}{27}2^{-3h}=\frac{5}{3}$. This implies $(2^h)^3=64/45$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The exact value of $\cot\frac{7\pi}{6}$? I am working on a trigonometry question at the moment and am very stuck. I have looked through all the tips to solving it and I cant seem to come up with the right answer. The problem is
What is exact value of
$$\cot \left(\frac{7\pi}{6}\right)? $$
| We have $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{\cos \left(\frac{7\pi}{6}\right)}{\sin \left(\frac{7\pi}{6}\right)} \equiv \frac{-\cos \left(\frac{\pi}{6}\right)}{-\sin \left(\frac{\pi}{6}\right)} = \sqrt{3}$$
You can easily see this using a "CAST" diagram to reduce $\cos \left(\frac{7\pi}{6}\right)$and $\sin \left(\frac{7\pi}{6}\right)$ to standard results.
As per @Scientifica's comment, an easier method would be to simply note that $\tan$ is $\pi$-periodic so that shifting its argument by $\pi$ will yield no change to it's value, or: $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{1}{\tan \left(\frac{7\pi}{6} - \pi\right)} = \frac{1}{\tan \left(\frac{\pi}{6}\right)} = \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| For the special case $x=\dfrac12$:
If you accept that $1+x+x^2+\dotsb=\dfrac1{1-x}$, the same picture works — just move the horizontal and vertical lines. Instead of them being at $1,1\frac12,1\frac34,\dotsb,2$, you should put them at $1,1+x,1+x+x^2,\dotsb,\dfrac1{1-x}$. The area of the square is then $\left(\dfrac1{1-x}\right){}^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 11
} |
Solve $\cos3x - 18\cos x +10 =0$ I want to solve
$$\cos 3x - 18\cos x +10 =0 $$
I tried:
1) Replacing $\cos 3x$ to $\cos^3x - 3\cos x$
2) Replacing $\cos x$ to $t$ we get:
$$t^3 - 21t +10 = 0$$
So we get cubic equation. But I can't solve it.
| Given $$cos3x - 18cosx +10 =0 $$ $$\implies 4\cos^3x-3\cos x-12\cos x+10 = 0$$ Factorizing the cubic equation as follows $$\left(\cos x-\frac{1}{2}\right) \left(\cos x-2\right)\left(\cos x+\frac{5}{2}\right)= 0$$ $$\text{if}\ \cos x-\frac{1}{2}=0 \implies \cos x=\frac{1}{2}\iff x=2n\pi\pm\frac{\pi}{3} $$ $$\text{if}\ \cos x-2=0 \implies \cos x\neq 2$$$$\text{if}\ \cos x+\frac{5}{2}=0 \implies \cos x\neq \frac{-5}{2}$$ Hence, the general solution for $x$ is $$x=2n\pi\pm \frac{\pi}{3}$$ Where, $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Nonnegative Integer solutions of $x+y-xy=0$ I would like to see other methods, besides the one I use here to find all the nonnegative integer solutions of an equation like $$x+y-xy=0$$.
This is the one I used:
First we note that for $x=1$ we get $1+y-y=0$, but $1+y-y=1$ so $x\neq 1$.
Then as $x\neq 1$ we have $y=\frac{x}{x-1}=1+\frac{1}{x-1}$. Furthermore $y(0)=0$ and $y(2)=2$ so $x=y=0$ and $x=y=2$ are solutions. Then if $x>2$, $x-1>1>0$ so as $x-1>1$ we have $\frac{1}{x-1}<1$ and as $x-1>0$ we have $\frac{1}{x-1}>0$ so $0<\frac{1}{x-1}<1$ for $x>2$, and so $1<1+\frac{1}{x-1}<2$ for $x>2$. As $y=1+\frac{1}{x-1}$, we get that for $x>2$, $1<y<2$. As there is no integer between $1$ and $2$, for $x>2$, $y$ can't be a integer. So the only nonnegative solutions of the equations are $x=y=0$ and $x=y=2$.
How else can I prove this, using less theory (For example, I think it can be proven using only Peano and a little more)?
| We have $$x+y-xy=0$$ $$\implies x+y=xy$$ $$\implies \text{sum of two non-negative integers}=\text{product of two non-negative integers}$$ For non-negative integer solution of the above equation, we have $x=0$, & $y=0$
Notice, $$\frac{x}{xy}+\frac{y}{xy}=\frac{xy}{xy}$$ $$\implies \frac{1}{x}+\frac{1}{y}=1$$ $$\implies \text{sum of inverse of two non-negative integers}=1$$ For non-negative integer solution of the above equation, we have $x=2$, & $y=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
find $\left( \frac{x}{x+y} \right)^{2007} + \left( \frac{y}{x+y} \right)^{2007}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
if $x, y$ are non-zero numbers satisfying $x^2 + xy + y^2 = 0$, find the value of $$\left(\frac{x}{x+y} \right)^{2007} + \left(\frac{y}{x+y} \right)^{2007}$$
(A). $2$ (B). $1$ (C). $0$ (D). $-1$ (E). $-2$
expanding it would give us $$ \frac { x^{2007} + y^{2007}} {(x+y)^{2007}}$$
how do I calculate this? Very appreciate for all of those who had helped me
| Hint:
We have:
$$A = \left(\frac{x}{x+y} \right)^{2007} + \left(\frac{y}{x+y} \right)^{2007}$$
$$ = \left( \frac{x}{y} \right)^{1003} \frac{x}{x+y} + \left( \frac{y}{x} \right)^{1003}\frac{y}{x+y} $$
$$ = - \left[ \left( \frac{x}{y} \right)^{1002}+ \left( \frac{y}{x} \right)^{1002} \right]$$
From the condition $x^2 + xy + y^2 =0$, we have
$$\left( \frac{x}{y} \right)^2 + \left( \frac{x}{y} \right) + 1 =0 $$
or,
$$\left( \frac{x}{y} \right)^3 = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to rewrite $\frac{d}{d(x+c)}$? I would like to know how to rewrite the following equations:
$$
\frac{d (f(x))}{d(x+c)} =0\\
\frac{d^2 (f(x))}{d(x+c)^2} =0\\
$$
Here $x$ is a variable, $c$ is a constant and $f(x)$ is a function of x.
I would also like to know the reasoning behind the answer.
| Use the chain rule. Define $u = x + c$ then use the fact that $$\frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du}$$ where the $\cdot$ represents any function, so
$$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$
It also follows that
$$
\begin{array}{rcll}
\frac{d^2f}{dx^2} &=& \frac{d}{dx} (\frac{df}{dx}) &\quad\mbox{definition of 2nd derivative} \\
&=& \frac{d}{dx} \big(\frac{du}{dx} \frac{df}{du}\big) &\quad\mbox{using the result above} \\
&=& \frac{d}{dx} \big(\frac{du}{dx} \big)\frac{df}{du} + \frac{du}{dx} \frac{d}{dx} \big(\frac{df}{du}\big) &\quad\mbox{using the rule for the derivative of a product} \\
&=& \frac{d^2u}{dx^2} \frac{df}{du} + \frac{du}{dx} \frac{du}{dx} \frac{d}{du} \big(\frac{df}{du}\big) &\quad\mbox{using various results from above} \\
&=& \frac{d^2u}{dx^2} \frac{df}{du} + (\frac{du}{dx})^2 \frac{d^2f}{du^2} &\quad\mbox{simplifying}
\end{array}
$$
So, in summary,
$$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$
and
$$\frac{d^2f}{dx^2} = \frac{d^2u}{dx^2} \frac{df}{du} + (\frac{du}{dx})^2 \frac{d^2f}{du^2}$$
The above is correct and valid for any $u(x)$ but it's written in a somewhat backwards way. You already have $\frac{df}{dx}$ and $\frac{d^2f}{dx^2}$ and want to find $\frac{df}{du}$ and $\frac{d^2f}{du^2}$. Well, that's simple algebra now to get those from the above. It's even simpler with the specific example of $u = x + c$ because, then, $\frac{du}{dx} = 1$ and $\frac{d^2u}{dx^2} = 0$, so
$$\frac{df}{du} = \frac{df}{dx}$$
and
$$\frac{d^2f}{du^2} = \frac{d^2f}{dx^2}$$
for that particular example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$? How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$ ?
Please let me know the fastest method you know of for such type of problems. Is there any way other than finding the left hand and right hand derivative using the concept of limits (That makes it huge)? Can intuition be used in any way?
| $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\\$ then
$$f(x)=\sqrt{(\sqrt{x-2}+\sqrt{2})^2}+\sqrt{(\sqrt{x-2}-\sqrt{2})^2}=\\|\sqrt{x-2}+\sqrt{2}|+|\sqrt{x-2}-\sqrt{2}|$$ now note that $ x\geq 2$ so ,when $$x \rightarrow 4^+$$
$$f(x)=\sqrt{x-2}+\sqrt{2} +\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\\f'_{4^+}=2\frac{1}{2\sqrt{x-2}}=\frac{1}{\sqrt{2}}$$
when $$x \rightarrow 4^-$$
$$f(x)=\sqrt{x-2}+\sqrt{2} -(\sqrt{x-2}-\sqrt{2})=2\sqrt{2}\\f'_{4^-}=0$$
so it is not differentiable at $x=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
new plane equation after transformation of coordinates I have a plane equation $ax + by + cz + d = 0$ w.r.t to a particular coordinate frame.
this coordinate frame w.r.t to the world coordinate frame is
$$\begin{vmatrix}
r_1 & r_2 & r_3 & x_1 \\
r_4 & r_5 & r_6 & y_1 \\
r_7 & r_8 & r_9 & z_1
\end{vmatrix}$$
now, what is the equation of plane in the world coordinate frame?
| The equation of the plane can be created from the dot product of two homogeneous vectors
$$ ax+by+cz+d = 0$$
$$ \begin{pmatrix} x\\y\\z\\1 \end{pmatrix}^\intercal \begin{pmatrix} a\\b\\c\\d \end{pmatrix} =0 $$
The world coordinates of the local $(x,y,z,1)$ are
$$ \begin{pmatrix} u \\ v \\ w \\ 1 \end{pmatrix} =
\begin{vmatrix}
r_1 & r_2 & r_3 & x_1 \\
r_4 & r_5 & r_6 & y_1 \\
r_7 & r_8 & r_9 & z_1 \\
0 & 0 & 0 & 1
\end{vmatrix} \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} $$
Taking the inverse you find that the coefficients $(A,B,C,D)$ of the plane $Au+Bv+Cw+D=0$ are
$$ \begin{pmatrix} A \\ B \\ C\\ D \end{pmatrix} = \begin{vmatrix}
r_1 & r_2 & r_3 & x_1 \\
r_4 & r_5 & r_6 & y_1 \\
r_7 & r_8 & r_9 & z_1 \\
0 & 0 & 0 & 1
\end{vmatrix}^{-\intercal} \begin{pmatrix} a\\b\\c\\d \end{pmatrix} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Question based of orthocenter distance from angular points In an acute angled triangle ABC,$\angle A=20^\circ $,let D,E,F be the feet of altitudes through A,B,C respectively and H is the orthocenter of $\bigtriangleup ABC $.Find $\frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}$
Since $AH=2R cosA,AD=2R cos A+2R cos B cos C$
$\frac{AH}{AD}=\frac{2R cos A}{2R cos A+2R cosB cos C}=\frac{cos A}{cos A+cosB cos C}$
$BH=2R cosB,BE=2R cos B+2R cos A cos C$
$\frac{BH}{BE}=\frac{2R cos B}{2R cos B+2R cosA cos C}=\frac{cos B}{cos B+cosA cos C}$
$CH=2R cosC,CF=2R cos C+2R cos A cos B$
$\frac{CH}{CF}=\frac{2R cos C}{2R cos C+2R cosA cos B}=\frac{cos C}{cos C+cosA cos B}$
but since we have only A given,not B and C.How will we find these ratios?
| Here's one solution using mass point geometry.
Assign mass points as following: $aA, bB, cC$ and let $(a+b+c)H$ be the center of the mass. Then from this it follows that points D, E, F have the following masses: $b+c, c+a, a+b$, respectively. Then from the mass points definition we have:
$$\frac{AH}{HD} = \frac{b+c}{a} \implies \frac{HD}{AH} + 1 = \frac{a}{b+c} + 1 \implies \frac{AD}{AH} = \frac{b+c+a}{b+c} \implies \frac{AH}{AD} = \frac{b+c}{a+b+c}$$
Simularly:
$$\frac{CH}{CF} = \frac{a+b}{a+b+c} \quad \text{and} \quad \frac{BH}{BE} = \frac{a+c}{a+b+c}$$
Adding them we get:
$$\frac{AH}{AD} + \frac{CH}{CF} + \frac{BH}{BE} = \frac{2(a+b+c)}{a+b+c} = 2$$
The fact that one of the angle is $20$ degrees is redundant.
If you're new to mass point geometry you can easily prove this by using Menelaus and Ceva Theorem. Actually the mass point geometry is an implicit and quicker way of using the Menalaus and Ceva Theorem, because if you want to use them explicitly, you need to find the right combination of lines and triangles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$ Please help me to evaluate this integral in a closed form:
$$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$
Using integration by parts I found that it could be expressed through integrals of elementary functions:
$$I_1=\int_0^\infty\log\left(1-e^{-\pi x}\right)\log\left(1+x^2\right)dx$$
$$I_2=\int_0^\infty x\log\left(1-e^{-\pi x}\right)\arctan x\,dx$$
| 1. Solution
Here is another solution: Notice that
\begin{align*}
&\int_{0}^{\infty} \operatorname{Li}_2(e^{-\pi x}) \arctan x \, dx \\
&\qquad = \overbrace{\left[ -\tfrac{1}{\pi} \operatorname{Li}_3(e^{-\pi x}) \arctan x \right]_{0}^{\infty}}^{=0} + \int_{0}^{\infty} \frac{\operatorname{Li}_3(e^{-\pi x})}{\pi(1 + x^2)} \, dx \\
&\qquad = \sum_{n=1}^{\infty} \frac{1}{n^3} \int_{0}^{\infty} \frac{e^{-n\pi x}}{\pi(1+x^2)} \, dx \\
&\qquad = \sum_{n=1}^{\infty} \frac{1}{n^2} \int_{0}^{\infty} \frac{e^{-x}}{(\pi n)^2 + x^2} \, dx \qquad (n\pi x \mapsto x) \\
&\qquad = \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{3x^2} - \frac{\coth x}{x^3} \right) e^{-x} \, dx
\end{align*}
In order to evaluate the last integral, we introduce the function $I(s)$ defined by
$$ I(s) = \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{3x^2} - \frac{\coth x}{x^3} \right) x^{s} e^{-x} \, dx. $$
Then $I(s)$ is analytic for $\Re(s) > -1$ and our integral can be written as $I(0)$. Now assume for a moment that $\Re(s) > 3$. Then
\begin{align*}
I(s)
&= \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{x^3} + \frac{1}{3x^2} - \frac{2}{x^3(1 - e^{-2x})} \right) x^{s} e^{-x} \, dx \\
&= \frac{\pi^2}{2} \left( \Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) - 2 \sum_{n=0}^{\infty} \frac{\Gamma(s-2)}{(2n+1)^{s-2}} \right) \\
&= \frac{\pi^2}{2} \left( \Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) - 2 \Gamma(s-2)(1 - 2^{2-s})\zeta(s-2) \right).
\end{align*}
By the principle of analytic continuation, this relation continues to hold on $\Re(s) > -1$. So we only need to take limit as $s \to 0$. To this end, we treat two parts separately:
*
*It is easy to check that $\Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) \to \frac{1}{9}$ as $s \to 0$.
*Using the functional equation of $\zeta(s)$ and the reflection formula for $\Gamma(s)$, we have
$$ 2 \Gamma(s-2)(1 - 2^{2-s})\zeta(s-2) = \frac{(1 - 2^{s-2})\pi^{s-2}}{\cos(\pi s/2)} \zeta(3-s). $$
Taking $s \to 0$, this converges to $\frac{3}{4\pi^2}\zeta(3)$.
Combining these two observations, we have
$$ I(0) = \frac{\pi^2}{18} - \frac{3}{8}\zeta(3). $$
2. Generalization
We can generalize the problem by considering the family of integrals
$$ I_n = \int_{0}^{\infty} \operatorname{Li}_{n}(e^{-\pi x}) \arctan x \, dx. $$
When $n$ is even, we can use the same technique as in our solution to obtain
Claim. For $m = 0, 1, 2, \cdots$ we have
$$ I_{2m} = \frac{(-1)^m \pi^{2m}}{2} \sum_{k=0}^{2m+1} \frac{2^k B_k H_{2m+1-k}}{k!(2m+1-k)!} - \frac{1}{2}\eta(2m+1), \tag{1} $$
where $(B_k)$ are the Bernoulli numbers, $(H_k)$ are the harmonic numbers, and
$$\eta(s) = (1-2^{1-s})\zeta(s)$$
is the Dirichlet eta function.
Remark 1. Following @Marco Cantarini's calculation, we have
$$ I_n = \sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!(2k+1)} \eta(n-2k) - \frac{1}{2}\eta(n+1). \tag{2} $$
This reduces to a finite summation when $n = 2m$ is even:
$$ I_{2m} = (-1)^m \pi^{2m} \sum_{k=0}^{2m} \frac{(1-2^{k-1})B_k}{(2m+1-k)!(2m+1-k)(k!)} - \frac{1}{2}\eta(2m+1). \tag{3} $$
So we have two different representations for $I_{2m}$.
Remark 2. Using (1), we find that for $|z| < 1$,
$$ \sum_{m=0}^{\infty} I_{2m}z^{2m} = \frac{1}{2}\left( \frac{\operatorname{Si}(\pi z)}{\sin (\pi z)} - \int_{0}^{\infty} \frac{\cosh(z t)}{e^t + 1} \, dt \right). \tag{4} $$
Thus in principle, we can find the values of $I_{2m}$ by differentiating (4) $2m$ times.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Elegant solution for $\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy$ I have the following integral: $\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy$
I already know the solution, but it needs three substitutions. Is there a simpler, more elegant way to go about this?
| $$\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy=$$
(Substitute $u=\sin(y)$ and $du=\cos(y)dy$):
$$\int\frac{1}{u^2+u-6}du=$$
$$\int\frac{1}{\left(u+\frac{1}{2}\right)^2-\frac{25}{4}}du=$$
(Substitute $s=u+\frac{1}{2}$ and $ds=du$):
$$\int\frac{1}{s^2-\frac{25}{4}}ds=$$
$$\int-\frac{4}{25(1-\frac{4s^2}{25})}ds=$$
$$-\frac{4}{25}\int\frac{1}{1-\frac{4s^2}{25}}ds=$$
(Substitute $p=\frac{2s}{5}$ and $dp=\frac{2}{5}ds$):
$$-\frac{2}{5}\int\frac{1}{1-p^2}dp=$$
$$-\frac{2}{5}\tanh^{-1}\left(p\right)+C=$$
$$-\frac{2}{5}\tanh^{-1}\left(\frac{2s}{5}\right)+C=$$
$$-\frac{2}{5}\tanh^{-1}\left(\frac{2\left(u+\frac{1}{2}\right)}{5}\right)+C=$$
$$-\frac{2}{5}\tanh^{-1}\left(\frac{2\left(\sin(y)+\frac{1}{2}\right)}{5}\right)+C=$$
$$-\frac{2}{5}\tanh^{-1}\left(\frac{2\sin(y)+1}{5}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of Squares in terms of Sum of Integers We know that the sum of squares can be expressed as a multiple of the sum of integers as follows: $$\begin{align}
\sum_{r=1}^n r^2
&=\frac 16 n(n+1)(2n+1)\\
&=\frac {2n+1}3\cdot \frac {n(n+1)}2\\
&=\frac {2n+1}3\sum_{r=1}^nr\end{align}$$
Is there a simple direct proof to express the sum of squares as $\dfrac {2n+1}3$ multiplied by the sum of integers, without first deriving the formula for the sum of squares and then breaking it down as shown above?
| Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$\begin{align}
\sum_{r=1}^n r^2&=(n+1)^3-1-\sum_{r=1}^n (r+1)\left((r+1)^2-r^2\right)\\\\
&=(n+1)^3-1-\sum_{r=1}^n (r+1)\left(2r+1\right)\\\\
&=(n+1)^3-1-2\sum_{r=1}^n r^2-3\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\\\
3\sum_{r=1}^n r^2&=(n+1)^3-1-3\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\\\
\sum_{r=1}^n r^2&=\frac{n(n+1)}{2}\frac{2(n+2)}{3}-\sum_{r=1}^{n}r\\\\
&=\frac{n(n+1)}{2}\left(\frac{2(n+2)}{3}-1\right)\\\\
&=\frac{n(n+1)}{2}\frac{2n+1}{3}
\end{align}$$
ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts. Here, we let $f_k=g_k=k$ and write
$$\begin{align}
\sum_{k=1}^{n}k^2&=n\left(\frac{n(n+1)}{2}\right)-\sum_{k=0}^{n-1}\sum_{\ell=0}^{k} k\\\\
&=n\left(\frac{n(n+1)}{2}\right)-\sum_{k =0}^{n-1}\frac{k(k+1)}{2}\\\\
&=(n+1)\left(\frac{n(n+1)}{2}\right)-\frac12\sum_{k =1}^n k(k+1)\\\\
\frac32\sum_{k=1}^{n}k^2&=(n+1)\left(\frac{n(n+1)}{2}\right)-\frac12\sum_{k =1}^n k\\\\
\sum_{k=1}^{n}k^2&=\frac23 (n+1)\left(\frac{n(n+1)}{2}\right)-\frac13\sum_{k =1}^n\,k\\\\
&=\frac13(2n+1)\left(\frac{n(n+1)}{2}\right)
\end{align}$$
again as expected!
| {
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Absolute Min and Max of $f(x, y)=x^2+4y^2-2x^2y+4$ Using Partial Derivatives Consider this problem:
Find the absolute minimum and absolute maximum of $f(x, y)=x^2+4y^2-2x^2y+4$ on the rectangle given by
$-1\leq x\leq1$ and $-1\leq y\leq1$
I solved this problem using partial derivatives:
I got $f_x=2x-4xy=0$
And $f_y=8y-2x^2=0$
Solving simultaneously gives the following critical points: $(0,0), (0,\frac{1}{2}), (\sqrt2,\frac{1}{2}) $
And with the restriction: $-1\leq x\leq1$ and $-1\leq y\leq1$
The critical points becomes: $(0,0), (0,\frac{1}{2}) $
$f(0,0)=4 $ and $f(0,\frac{1}{2}) =5$
Now the quagmire is how to use partial derivatives (probably of second order) to determine which is the absolute minimum and maximum.
Plotting a graph of $z=x^2+4y^2-2x^2y+4$ suggests that:
$(0,0,4) $ is the absolute minimum and
$(0,\frac{1}{2},5) $ the maximum
But I need to be sure I'm on the right path. I also need to know how to use partial derivatives to differentiate a minimum critical point from a maximum critical point.
| Hint:
Since $2x-4xy=0\implies 2x(1-2y)=0\implies x=0$ or $y=\frac{1}{2}$ and
$8y-2x^2=0\implies x^2=4y$, either $x=0, y=0$ or $y=\frac{1}{2}, x=\pm\sqrt{2}$.
Therefore $(0,0)$ is the only critical point in the interior of the region.
On the left and right edges of the rectangle, $g(y)=4y^2-2y+5$ and $g^{\prime}(y)=8y-2=0$ if $y=\frac{1}{4}$.
On the top edge, $k(x)=-x^2+8$ so $k^{\prime}(x)=-2x=0$ if $x=0$.
On the bottom edge, $l(x)=3x^2+8$ so $l^{\prime}(x)=6x=0$ if $x=0$.
Now you just need to compare the values of the function at the points
$(0,0)$ and $(1, \frac{1}{4}),\; (-1, \frac{1}{4}),\; (0,1), \;(0,-1), \;(-1, 1), \;(-1, -1),\; (1, -1),\; (1,1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed-forms of the integrals $\int_0^1 K(\sqrt{k})^2 \, dk$, $\int_0^1 E(\sqrt{k})^2 \, dk$ and $\int_0^1 K(\sqrt{k}) E(\sqrt{k}) \, dk$ Let denote $K$ and $E$ the complete elliptic integral of the first and second kind.
The integrand $K(\sqrt{k})$ and $E(\sqrt{k})$ has a closed-form antiderivative in term of $K(\sqrt{k})$ and $E(\sqrt{k})$, so we know that
$$
\int_0^1 K\left(\sqrt{k}\right) \, dk = 2,
$$
and
$$
\int_0^1 E\left(\sqrt{k}\right) \, dk = \frac{4}{3}.
$$
I couldn't find closed-form antiderivatives to the integrals $\int K(\sqrt{k})^2 \, dk$, $\int E(\sqrt{k})^2 \, dk$, $\int E(\sqrt{k})K(\sqrt{k}) \, dk$, but I've conjectured, that
$$\begin{align}
\int_0^1 K\left(\sqrt{k}\right)^2 \, dk &\stackrel{?}{=} \frac{7}{2}\zeta(3),\\
\int_0^1 E\left(\sqrt{k}\right)^2 \, dk &\stackrel{?}{=} \frac{7}{8}\zeta(3)+\frac{3}{4},\\
\int_0^1 K\left(\sqrt{k}\right)E\left(\sqrt{k}\right) \, dk &\stackrel{?}{=} \frac{7}{4}\zeta(3)+\frac{1}{2}.
\end{align}$$
How could we prove this closed-forms? It would be nice to see some references to these integrals.
| We have
$$ K(\sqrt{k}) = \int_{0}^{1}\frac{dt}{\sqrt{1-t^2}\sqrt{1-k t^2}}\tag{1} $$
hence:
$$ K(\sqrt{k})^2 = \iint_{(0,1)^2}\frac{dt\,ds}{\sqrt{1-t^2}\sqrt{1-s^2}\sqrt{(1-kt^2)(1-ks^2)}}\tag{2}$$
and since:
$$\begin{eqnarray*} \int_{0}^{1}\frac{dk}{\sqrt{(1-ks^2)(1-kt^2)}}&=&\frac{1}{st}\int_{0}^{st}\frac{dk}{\sqrt{1-\left(\frac{s}{t}+\frac{t}{s}\right)k+k^2}}\\&=&\frac{1}{st}\,\left.\log\left(2k-\left(\frac{s}{t}+\frac{t}{s}+2\sqrt{k^2-\left(\frac{s}{t}+\frac{t}{s}\right)k+1}\right)\right)\right|_{0}^{st}\\&=&\frac{1}{st}\,\log\left(\frac{t^2+s^2-2t^2 s^2-2st\sqrt{(1-s^2)(1-t^2)}}{(s-t)^2}\right)\tag{3}\end{eqnarray*}$$
it follows that:
$$ \int_{0}^{1}K(\sqrt{k})^2\,dk = \iint_{\left(0,\frac{\pi}{2}\right)^2}\log\left[\frac{\sin^2(\phi-\theta)}{(\sin\phi-\sin\theta)^2}\right]\cot(\phi)\cot(\theta)\,d\phi\,d\theta$$
and now we may use a change of coordinates and the Fourier series of $\log\sin$.
An interesting chance is also given by exploiting the expansion of $K(k)$ with respect to the base of $L^2(0,1)$ given by the shifted Legendre polynomials. We have:
$$ K(k) = 2\sum_{n\geq 0}\frac{P_n(2k-1)}{2n+1}\tag{4} $$
and since:
$$ \int_{0}^{1}P_n(2\sqrt{k}-1)^2\,dk = \frac{1}{2n+1},$$
$$ \int_{0}^{1}P_n(2\sqrt{k}-1)P_{n+1}(2\sqrt{k}-1)\,dk=\frac{n+1}{(2n+1)(2n+3)}\tag{5}$$
we have:
$$\begin{eqnarray*} \int_{0}^{1}K(\sqrt{k})^2\,dk&=&4\sum_{n\geq 0}\frac{1}{(2n+1)^3}+8\sum_{n\geq 0}\frac{n+1}{(2n+1)^2(2n+3)^2}\\&=&\color{red}{\frac{7}{2}\,\zeta(3)+1}.\tag{6}\end{eqnarray*}$$
This is the same approach used by Zhou to prove similar identities.
| {
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"url": "https://math.stackexchange.com/questions/1384080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Find the approximate square root of $(a^4-a^{-4})/(a^2-a^{-2})$ Find the approximate square root of
$$\dfrac{\left( \dfrac{12}{5}\right)^4 - \left( \dfrac{5}{12}\right)^4 }{\left( \dfrac{12}{5}\right)^2 - \left( \dfrac{5}{12}\right)^2}$$
I tried using the formula for $(a^4-b^4)$ and $(a^2-b^2)$. Then cancelled the common $(a-b)$, substituted the values and simplified. Didn't work.
Answer is 13
Can someone tell me how to solve it with steps of possible?
It will be a great help
| Let $x= \frac{12}{5}$ to obtain
\begin{align}
f(x) &= \frac{x^{4} + \frac{1}{x^{4}}}{x^{2} + \frac{1}{x^{2}}} = \frac{x^{8}-1}{x^{2} \, (x^{4}-1)} = \frac{x^{4} + 1}{x^{2}} = x^{2} + \frac{1}{x^{2}} = \left(x + \frac{1}{x}\right)^{2} - 2
\end{align}
Taking the square root of both sides leads to
\begin{align}
\sqrt{f(x)} = \sqrt{x^{2}+ \frac{1}{x^{2}}} = x \, \sqrt{1 + \frac{1}{x^{4}}} = x \, \left( 1 + \frac{1}{2 \, x^{4}} - \frac{1}{8 \, x^{8}} + \mathcal{O}\left(\frac{1}{x^{10}}\right) \right) \approx x
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex number z satisfies both the inequality $|z-ai|=a+4$ and the inequality $|z-2|<1$ The number of integral values of $a$ for which at least one complex number z satisfies both the inequality $|z-ai|=a+4$ and the inequality $|z-2|<1$.
I supposed $z=x+iy$ and put in both equations,
$|x+iy-ai|=a+4\Rightarrow \sqrt{x^2+(y-a)^2}=a+4$$\Rightarrow x^2+y^2-2ya+a^2=a^2+8a+16...(1)$
$|x+iy-2|<1\Rightarrow \sqrt{(x-2)^2+y^2}<1.....(2) $
$\Rightarrow (x-2)^2+y^2<1$
$\Rightarrow x^2+y^2-4x+4<1......(2)$
How will i solve further to get integer values of $a$?
| HINT....Assuming $a$ is real, for a geometrical interpretation: the first equation represent the point on a circle centre $(0,a)$ and radius $a+4$, and the second one represents the points inside the circle centre$(2,0)$, radius $1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability of rolling a dice 8 times before all numbers are shown. What is the probability of having to roll a (six sided) dice at least 8 times before you get to see all of the numbers at least once?
I don't really have a clue how to work this out.
Edit: If we are trying to find the number of situations where all of the numbers are shown, for seven rolls, a favorable outcome would be one in which there are two numbers the same, for example: 1123456. These numbers can be arranged in 7!/2!5! ways, and there are 6 different numbers that could be repeated.
So the probability of getting all 6 numbers with 7 throws is (6*7!/2!5!)/6^7 = 126/279936.
Is that right?
Then 1 minus this is the probability?
Edit: prob. of not getting all six numbers with seven rolls = 1-prob of getting all six numbers with seven rolls
Six numbers same with seven rolls means one number repeated twice 7!/2! ways of doing this for each repeated number
6*(7!/2!) is number of ways of getting all six numbers with seven rolls. Total number of outcomes 6^7.
Prob of getting all six numbers with seven rolls = (6*(7!/2!))/6^7 = 0.054
Prob of not getting all six numbers with seven rolls (=prob of needing at least 8 rolls to get all numbers) = 1-0.054 = 0.946
| (Edit: Answer is for the probability of seeing all the numbers for $n$ rolls)
For exactly $n$ rolls, the problem can be solved using a markov chain
\begin{align*}
A&=
\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0\cr 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0\cr 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0\cr 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0\cr 0 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}\cr 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix}
\end{align*}
where the rows and columns are the number of faces of the die seen.
The probability of seeing all the faces in exactly 8 rolls is $(A^8)[0,6]$, which is $\dfrac{665}{5832}\approx 0.11402606310014$
For any $n$, it can be found by finding the generating function $G(z)$ and in turn finding the coefficient of $z^n$
\begin{align*}
\mathbb{P}(n) = 1-\frac{20}{2^n}+15\left(\frac{2}{3}\right)^n+\frac{15}{3^n}-6\left(\frac{5}{6}\right)^n-\frac{6}{6^n}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $\cot\angle MAB + \cot\angle MBC + \cot\angle MCA \geq 3\sqrt{3}$ Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $$\cot\measuredangle MAB + \cot\measuredangle MBC + \cot\measuredangle MCA \geq 3\sqrt{3}$$
| $\cot A+\cot B+\cot C\ge \sqrt3 \ $ and $\ \sin A\cdot \sin B\cdot \sin C\le \dfrac {3\sqrt3} 8$
are well known inequalities
$(1+\dfrac {\alpha} {\beta} +\dfrac {\beta} {\gamma})\cot A+(1+\dfrac {\beta} {\gamma} +\dfrac {\gamma} {\alpha})\cot B+(1+\dfrac {\gamma} {\alpha} +\dfrac {\alpha} {\beta})\cot C= \\ $
$\displaystyle\sum_{cyc}\cot A+\sum_{cyc}\dfrac {\alpha} {\beta} \cdot \dfrac {\sin B} {\sin A\cdot \sin C} \ge \sqrt3 +3\dfrac 1 {\left(\sin A\cdot \sin B\cdot \sin C\right)^{\frac 1 3}}\ge \sqrt3 +2\sqrt3=3\sqrt3$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$
I showed that in
$$ \forall x \in [0,\frac{\pi}{4}] \quad \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{\pi}{4}} \,dx = 1 -\frac{\sqrt2}{2} $$
but it's not good enough
| Using the fact that $\sin(x)\le x$ for $x\ge0$,
$$
\begin{align}
\int_0^{\pi/4}\frac{1-\cos(x)}x\,\mathrm{d}x
&=\int_0^{\pi/4}\frac{2\sin^2(x/2)}x\,\mathrm{d}x\\
&=\int_0^{\pi/4}\frac{\sin(x/2)}{x/2}\cdot\sin(x/2)\,\mathrm{d}x\\
&\le\int_0^{\pi/4}1\cdot\frac x2\,\mathrm{d}x\\[4pt]
&=\frac{\pi^2}{64}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$, prove that $ x^2+y^2=1$. If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$,
prove that $$ x^2+y^2=1$$
I tried taking sine of both the sides, I only come to this result: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$
| As $\arccos u+\arcsin u=\dfrac\pi2 $
from the given relation, $\arccos x+\arccos y=\dfrac\pi2$
Taking cosine in both sides, $$xy-\sqrt{(1-x^2)(1-y^2)}=0\iff xy=\sqrt{(1-x^2)(1-y^2)}$$
Now square both sides.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{n\to \infty}(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n})^{1/n}$ Find:
$$\lim_{n\to \infty}\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n}$$
I was wondering,is there a formula for multiplication of sines,when angles are in arithmetic progression(likewise there is formula for summation of sines).I also tried this question by taking log of both sides and then applying L Hospital rule.But it is becoming messy.Can someone please assist me in solving this question?
| $\bf{My\; Solution::}$ USing $\bf{n^{th}}$ root of Unity, $\displaystyle$ So Let $x= (1)^{\frac{1}{n}}\Rightarrow x^n - 1=0$
So $$x^n-1 = (x-1)\cdot (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})\;,$$
Where $$\displaystyle \alpha^r = \cos\left(\frac{2r\pi}{n}\right)+i\sin \left(\frac{2r\pi}{n}\right)\;,$$ and $r=0,1,2,3,.....(n-1)$
So $$\displaystyle \frac{x^n-1}{x-1} = (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})$$
Now Using The formula
$$\bullet\; x^n-1 = (x-1)\cdot \left(x^{n-1}+x^{n-2}+x^{n-3}+........+x^2+x+1\right)$$
So $$\displaystyle \left(x^{n-1}+x^{n-2}+x^{n-3}+........+x^2+x+1\right) = (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})$$
Now Put $x=1$ in above equation and taking Modulus on both side, We get
$$\displaystyle 1 = \left|(1-\alpha)\cdot (1-\alpha^2)\cdot.........(1-\alpha^{n-1})\right| = \left|(1-\alpha)\right|\cdot \left|(1-\alpha^2)\right|\cdot .........\left|(1-\alpha^{n-1})\right|$$
So we get
$$\displaystyle n= \left|1-\cos\left(\frac{2\pi}{n}\right)-i\sin \left(\frac{2\pi}{n}\right)\right|\cdot \left|1-\cos\left(\frac{4\pi}{n}\right)-i\sin \left(\frac{4\pi}{n}\right)\right|.........\left|1-\cos\left(\frac{2(n-1)\pi}{n}\right)-i\sin \left(\frac{2(n-1)\pi}{n}\right)\right|$$
So $$\displaystyle n = 2\sin\left(\frac{\pi}{n}\right)\cdot 2\sin\left(\frac{2\pi}{n} \right)....2\sin\left(\frac{(n-1)\pi}{n}\right)$$
So $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Now Replace $n\rightarrow 2n\;,$ We get
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
So $$\displaystyle \lim_{n\rightarrow \infty}\left[\sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)\right]^{\frac{1}{n}} = \lim_{n\rightarrow \infty}\left(\frac{2n}{2^{2n-1}}\right)^{\frac{1}{n}}$$
$$\displaystyle = \frac{\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}}{\lim_{n\rightarrow \infty}(2)^{2-\frac{1}{n}}} = \frac{1}{2^2}$$
For Calculation of $$\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}$$
Let $$\displaystyle L=\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}\Rightarrow \ln(L)=\lim_{n\rightarrow \infty}\frac{\ln(2n)}{n}$$
Now Using $$\bf{L,Hopital\; Rule}\;,$$ we get $$\ln(L)=0\Rightarrow L=e^{0}=1$$
I did not Understand Where I have Done Wrong, Plz anyone explain me, Thanks
| {
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Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots.
I can see two ways of proving it.
The first one is to notice, that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. It has the only root 1. And it is not the root of $x^4 + x^3 + x^2 + x + 1$. So, $x^4 + x^3 + x^2 + x + 1$ does not have roots.
Another way is to solve it as a palindromic polynomial. It does not have roots.
But is there any way to directly manipulate the expression to show that it is always greater than zero?
| Assuming that exists $a
$ such that $$a^{4}+a^{3}+a^{2}+a+1=0.
$$ Now it's obvious that $a\neq0
$. So we have $$0=\frac{\left(a^{4}+a^{3}+a^{2}+a+1\right)}{a}=a^{3}+a^{2}+a+1+\frac{1}{a}
$$ so $$a^{3}+a^{2}+a+1+\frac{1}{a}=a^{4}+a^{3}+a^{2}+a+1\Longrightarrow\frac{1}{a}=a^{4}\Rightarrow1=a^{5}
$$ so we have $$a=1
$$ and this is absurd. Now since $f\left(x\right)=x^{4}+x^{3}+x^{2}+x+1
$ is a continuous function, to see if it is positive it is sufficient see what happens at one point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad \quad
\boldsymbol{2.)}\ a^{2}+b^{2}-c^{2}\\
\boldsymbol{3.)}\ a^{2}+b^{2} \quad \quad \quad \quad
\boldsymbol{4.)}\ a^{2}+b^{2}+c^{2}$
By observation if all the three value
$\dfrac{m-a^{2}}{b^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=1$
Then $m=a^{2}+b^{2}+c^{2}$
I want to know if their is any other simple and short method other than the stated observation .
I have studied maths up to $12$th grade.
| the given equation is equivalent to $$-\left(a^4+3 a^2 b^2+3 a^2 c^2+b^4+3 b^2 c^2+c^4\right)
\left(a^2+b^2+c^2-m\right)=0$$ thus we obtain $$m=a^2+b^2+c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to solve the trigonometric equation $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$ the question is:
Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.
Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarly solving the above equation it comes out to be:
$$\sqrt2 u^2-u-\sqrt2=0=(u-\sqrt2)(\sqrt2u+1)=0 \implies \sin x+\cos x=\sqrt2 \quad(1)$$
and
$$\sin x+\cos x= \frac{-1} {\sqrt2}\quad (2)$$
so solving the results differently i got the answers:
$$x=2n\pi + \frac{\pi}4, 2n\pi +\frac{7\pi}{12}, 2n\pi-\frac{\pi}{12}$$
but the answers are:
$$x=2n\pi + \frac{\pi}4, 2n\pi -\frac{5\pi}{12}, 2n\pi+\frac{11\pi}{12}$$
I divided the eq(2) by $\sqrt2$
| Notice, we have $$\sin x-\cos x-2\sqrt2\sin x\cos x=0$$
$$\sin x-\cos x=2\sqrt2\sin x\cos x$$
$$\sin x-\cos x=\sqrt2\sin 2x$$
$$\frac{1}{\sqrt2}\sin x-\frac{1}{\sqrt2}\cos x=\sin 2x$$
$$\sin 2x=\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}$$
$$\sin 2x=\sin \left(x-\frac{\pi}{4}\right)$$
Writing the general solution , we get
$$2x=2n\pi+\left(x-\frac{\pi}{4}\right)\iff x=\frac{(8n-1)\pi}{4}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\frac{(8n-1)\pi}{4}}}$$
or $$2x=2n\pi+\pi-\left(x-\frac{\pi}{4}\right)\iff x=\frac{(8n+5)\pi}{12}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\frac{(8n+5)\pi}{12}}}$$
Where, $n$ is any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Exponential Simultaneous Equations Solve the following simultaneous equations:
$$2^x + 2^y = 10$$
$$x + y = 4$$
Looking at it, it is obvious that the answers are $(3,1)$ and $(1,3)$, however, I was wondering if they could be solved algebraically. Here's my approach:
$$2^x + 2^{4-x} = 10$$
$$2^x + \frac{(2^4)}{(2^x)} = 10$$
$$2^x + \frac {16}{2^x} = 10$$
And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.
| Notice that:
\begin{align*}
2^x+\frac{16}{2^x}={}&10 \\
(2^x)^2+16={}&10\cdot 2^x \\
(2^x)^2-10\cdot 2^x+16={}&0 \\
(2^x)^2-8\cdot 2^x-2\cdot 2^x+16={}&0
\end{align*}
By factorizing we get
\begin{align*}
&\phantom{2^x-8}(2^x-8)(2^x-2)=0 \iff{} \\
&{}\iff 2^x-8=0\vee2^x-2=0 \iff{} \\
&{}\iff 2^x=8\vee2^x=2\iff {} \\
&{}\iff2^x=2^3\vee2^x=2^1\iff{} \\
&{}\iff x=3\vee x=1.
\end{align*}
Hence, we get that the solutions are:
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=1,\ 3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
How do people perform mental arithmetic for complicated expressions?
This is the famous picture "Mental Arithmetic. In the Public School of S. Rachinsky." by the Russian artist Nikolay Bogdanov-Belsky.
The problem on the blackboard is:
$$
\dfrac{10^{2} + 11^{2} + 12^{2} + 13^{2} + 14^{2}}{365}
$$
The answer is easy using paper and pencil: $2$.
However, as the name of the picture implies, the expression ought be simplified only mentally.
My questions:
*
*Are there general mental calculation techniques useful for performing basic arithmetic and exponents?
*Or is there some trick which works in this case?
*If so, what is the class of problems this trick can be applied to?
| $$\color{Green}{n^2=1+3+5+\cdots+(2n-1)}$$
Therefore
$$\begin{align}
10^2+11^2+12^2+13^2+14^2
&=(1\cdot+19)+(1\cdot+21)+(1\cdot+23)+(1\cdot+25)+(1\cdot+27)\\
&=5\times(1\cdot+19)+(4\times21)+(3\times23)+(2\times25)+27
\end{align}$$
Also, $1+3+\cdot+19=10^2=100.$
Hence
$$10^2+11^2+12^2+13^2+14^2=500+84+69+50+27=730$$
$$\color{Green}{10^2+11^2+12^2+13^2+14^2=2\times365.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "183",
"answer_count": 20,
"answer_id": 4
} |
strange fibonacci recurrence As it is well known fibonacci numbers satisfy the recurrence relation $$F_{n}=F_{n-1}+F_{n-2}$$ with initial conditions $F_{0}=0$ and $F_{1}=1$.
While playing around with numbers,I noticed the following recurrence relation for the fibonacci numbers
$$F_{n}=4F_{n+1}+F_{n-4}-(2F_{n+2}+F_{n-2})$$
Valid for $n\geq 4$
Can anyone enjoy the challenge of proving this simple relation?
| It can be seen that:
$$F_{n-4} = F_{n-2} - F_{n-3} = - F_{n-1} + 2 F_{n-2} = 2 F_{n} - 3 F_{n-1} = - 3 F_{n+1} + 5 F_{n},$$
$$ F_{n-2} = F_{n} - F_{n-1} = - F_{n+1} + 2 F_{n}$$
and leads to:
\begin{align}
4F_{n+1}+F_{n-4}-(a F_{n+2}+F_{n-2}) &= 4 F_{n+1} - 3 F_{n+1} + 5 F_{n} - a F_{n+2} + F_{n+1} - 2 F_{n} \\
&= 2 F_{n+1} - a F_{n+2} + 3 F_{n}\\
&= (2 - a) F_{n+2} + F_{n}.
\end{align}
When $a =2$ this reduces to
$$ 4 F_{n+1} + F_{n-4} - 2 F_{n+2} - F_{n-2} = F_{n}.$$
When $a = 1$ this reduces to
$$ 4 F_{n+1} + F_{n-4} - F_{n+2} - F_{n-2} = L_{n+1}$$
where $L_{n}$ are the Lucas numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\frac{\sqrt{x^2+1}}{x^2}$ How to integrate: $$\int \frac{\sqrt{x^2+1}}{x^2}\text dx$$
I've typed this into wolframalpha as a result I received something with inverse hyperbolic sine. Is it possible to have it without this hyperbolic sine? If so, how to do that?
| Let $$ I= \displaystyle \int \frac{\sqrt{x^2+1}}{x^2}dx = \int \frac{\sqrt{x^2+1}}{x^2} \times \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}dx $$
$$\displaystyle I = \int\frac{x^2+1}{x^2\sqrt{x^2+1}}dx = \int\frac{1}{\sqrt{x^2+1}}dx+\int \frac{1}{x^2\sqrt{x^2+1}}dx$$
Now in second Integral , Put $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get $$\displaystyle I = \ln\left|x+\sqrt{x^2+1}\right|+J$$
So Integral $J$ Convert into $$\displaystyle -\int \frac{t}{\sqrt{t^2+1}}dt\;,$$ Now Substitute $t^2+1=u^2\;,$ Then $tdt = udu$
So Second Integral $J$ into $$\displaystyle -\int \frac{udu}{u} = -u=-\sqrt{t^2+1}=-\frac{\sqrt{x^2+1}}{x}$$
So $$\displaystyle I = \frac{\sqrt{x^2+1}}{x^2}dx= \ln\left|x+\sqrt{x^2+1}\right|-\frac{\sqrt{x^2+1}}{x}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to compute $\int_0^1 \frac{1}{1+x^4}\;dx$? I am having problems trying to compute $$\int_0^1 \frac{1}{1+x^4}\;dx$$
Wolfram alpha gives an answer $$\frac{\pi + 2 \coth^{-1}(\sqrt{2})}{4 \sqrt{2}}$$
| Notice, $$\int_{0}^{1}\frac{dx}{1+x^4}$$
$$=\int_{0}^{1}\frac{\frac{1}{x^2}dx}{x^2+\frac{1}{x^2}}$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\frac{2}{x^2}dx}{x^2+\frac{1}{x^2}}$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)dx}{x^2+\frac{1}{x^2}}$$
$$=\frac{1}{2}\left[\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)dx}{x^2+\frac{1}{x^2}}-\int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)dx}{x^2+\frac{1}{x^2}}\right]$$
$$=\frac{1}{2}\left[\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)dx}{\left(x-\frac{1}{x}\right)^2+2}-\int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)^2-2}\right]$$
$$=\frac{1}{2}\left[\int_{0}^{1}\frac{\left(1+\frac{1}{x^2}\right)dx}{\left(x-\frac{1}{x}\right)^2+(\sqrt{2})^2}-\int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)^2-(\sqrt{2})^2}\right]$$
$$=\frac{1}{2}\left[\frac{1}{\sqrt 2}\left\{\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt 2}\right)\right\}_{0}^{1}-\left\{\frac{1}{2\sqrt 2}\ln\left|\frac{x+\frac{1}{x}-\sqrt 2}{x+\frac{1}{x}+\sqrt 2}\right|\right\}_{0}^{1}\right]$$
$$=\frac{1}{2}\left[\frac{1}{\sqrt 2}\left\{\tan^{-1}\left(0\right)-\tan^{-1}\left(\infty\right)\right\}-\frac{1}{2\sqrt 2}\left\{\ln\left|\frac{2-\sqrt 2}{2+\sqrt 2}\right|-\ln|1|\right\}\right]$$
$$=\frac{1}{2}\left[-\frac{\pi}{2\sqrt 2}-\frac{1}{\sqrt 2}\ln\left|\frac{2-\sqrt 2}{2}\right|\right]$$
$$=-\frac{\pi}{4\sqrt 2}-\frac{1}{2\sqrt 2}\ln\left|\frac{2-\sqrt 2}{2}\right|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum_{k=1}^{n} \frac1{\sin^2 \frac{\left( 2k-1\right)\pi}{4n+2}}=2n\left( n+1\right)$ Prove that $$\frac{1}{\sin^{2}\frac{\pi }{4k+2}}+\frac{1}{\sin^{2}\frac{3\pi }{4k+2}}+\frac{1}{\sin^{2}\frac{5\pi }{4k+2}}+\cdots+\frac{1}{\sin^{2}\frac{(2k-1)\pi }{4k+2}}=2k(k+1)$$
| Using $(7)$ from this answer, we get
$$
\begin{align}
\sum_{k=1}^n\frac1{\sin^2\left(\frac{2k-1}{4n+2}\pi\right)}
&=\sum_{k=1}^n\csc^2\left(\frac{2k-1}{2n+1}\frac\pi2\right)\tag{1}\\
&=\sum_{k=1}^n\sec^2\left(\frac{2n-2k+2}{2n+1}\frac\pi2\right)\tag{2}\\
&=\sum_{k=1}^n\sec^2\left(\frac{k}{2n+1}\pi\right)\tag{3}\\
&=n+\sum_{k=1}^n\tan^2\left(\frac{k}{2n+1}\pi\right)\tag{4}\\[4pt]
&=n+n(2n+1)\tag{5}\\[12pt]
&=2n(n+1)\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\frac1{\sin(x)}=\csc(x)$
$(2)$: $\csc(x)=\sec\left(\frac\pi2-x\right)$
$(3)$: substitute $k\mapsto n+1-k$
$(4)$: $\sec^2(x)=1+\tan^2(x)$
$(5)$: use $(7)$ from this answer
$(6)$: simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the family of functions $f(x),g(x)$ such that $f/g=x$ Suppose we have two functions $f(x)$ and $g(x)$, defined on $x\geq1$, we know
$$f(1)=g(1)=1$$
$$g,f>0$$
$$\frac{\text{d} f}{\text{d} x}\geq0,\quad\frac{\text{d} g}{\text{d} x}\leq0$$
and
$$\frac{f}{g}=x$$
for $x\geq1$.
My approach was this:
Clearly
$$f=xg$$
So we only need to define a general solution for $g$. Let us look for what exponent values, $n$, hold if we let
$$g=x^{-n}$$
giving
$$f=x^{1-n}$$
thefore we find that
$$0\geq g'=-nx^{-n-1}\Rightarrow n\geq 0$$
$$0\leq f'=(1-n)x^{-n}\Rightarrow 1\geq n$$
Therefore
$n\in[0,1]$. Are there any other possible solutions not of this form? I have tried a power series expansion, however this isn't greatly helpful as I have shown these functions behave nicely with non-integer exponents.
| Since $f(x)=xg(x)$, then $f'(x)=xg'(x)+g(x)$ and since $f'(x)\ge0$ then $ xg'(x)+g(x) \ge0$ which implies that $xg'(x)\ge-g(x)$......(*)
Now since $x\ge 1$ and $g(x)\ge 0$ so that $xg(x)\ge0$, therefore from (*) we have, $\frac{g'(x)}{g(x)}\ge-\frac{1}{x} $$\Rightarrow$ $\ln g(x)\ge -\ln x $ $\Rightarrow$ $g(x) \ge\frac{1}{x}$.
Define a function $g(x)=\frac{a}{bx+c}$, where, $a=b+c$; $a,b,c>0$; $b+c\ge1$.
Then, clearly $g(x)= \frac{a}{bx+c}\ge\frac{1}{x}$, since $bx+c\ge x$ for all $x\ge 1$, we have $\frac{a}{bx+c}=\frac{b+c}{bx+c}\ge\frac{1}{x}$
$\Rightarrow $ $bx+cx \ge bx+c \Rightarrow x \ge 1$.
Now, at $x=1$, $g(1)=\frac{a}{b+c}=1$ iff $a=b+c$. Also, $g'(x)=-\frac{ab}{(bx+c)^2}\le 0$ iff $ab>0$
Note:
$ab<0$ is impossible case since when $ab<0$ $\Rightarrow$ $(b+c)b<0$ but $b+c\ge 1$ so that $b<0$ in this case $bx+c\ge bx+(1-b) = b(x-1)+1$ would be greater that $x$ iff $b(x-1)>x-1$ $\Rightarrow$ $b\ge 1$ which is a contradicts the assumption that ($b<0$).
Now, since $f(x)=xg(x)$, then we define $f$ to be $f(x)=\frac{ax}{bx+c}$, where $a,b,c$ are defined above.
$$f(x)=\frac{(b+c)x}{bx+c}\Rightarrow f(1)=1.$$ Also, $f'(x) = \frac{(bx+c)a-abx}{(bx+c)^2}$ would be $\ge$ than $0$ for all $x\ge 1$ iff $(bx+c)a-abx\ge 0$ $\Rightarrow$ $abx+ac-abx\ge 0$ $\Rightarrow$$ac=(b+c)c\ge c \ge 0$, (since $b+c\ge 1$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I complete this proof the cantor set does not meet $\left(\frac{3s+1}{3^k}, \frac{3s+2}{3^k}\right)$? We define the cantor set $K$ as the set of all ternary numbers $0.{a_1a_2a_3\cdots}$ such that $a_i = {0, 2}$ for all $i$, i.e. no digit is allowed to be 1, and the first one is zero.
Here is my proof in progress.
Theorem For all $x \in K$, either $x \leq \frac{3s + 1}{3^k}$ or $x \geq \frac{3s + 2}{3^k}$
Proof We proceed by induction on $k$.
(Base Case) For $k = 1$ we show that $x \in K$ implies $x \leq s + \frac{1}{3}$ or $x \geq s + \frac{2}{3}$ for all natural $s$. By case analysis on the first digit, either $x \leq \frac{1}{3}$ or $x \geq \frac{2}{3}$. Notice $x \leq \frac{1}{3}$ implies $x \leq \frac{1}{3} + s$ for all $s$. For $s \geq 1$, $s + \frac{2}{3} \geq \frac{5}{3} > 1$ and so there is no $x \in K$ such that $x \geq s + \frac{2}{3}$.
(Inductive case)
Suppose that $x = 0.a_1a_2\cdots$. If $a_1 = 0$, then $3x \in K$ so by the inductive hypothesis we have two cases. The first case is $3x \leq \frac{3s + 1} {3^k}$ for all $s$, so $x \leq \frac{3s+1}{3^{k + 1}}$ for all $s$. The second case is that $3x \geq \frac{3s + 2}{3^k}$ for all $s$, so $x \geq \frac{3s + 2}{3^{k + 1}}$ for all $s$.
Now here is where I get stuck. If $a_1 = 2$, then $3x - 2 \in K$, so we can apply the inductive hypothesis and similar reasoning to obtain that, for all $s$ either $x \leq \frac{3s + 1}{3^{k + 1}} + \frac{2}{3}$ or $x \geq \frac{3s+2}{3^{k + 1}} + \frac{2}{3}$. The problem is of course that rescaling the interval $[\frac{2}{3}, 1]$ to $[0, 1]$ "shifts" the $s$ value when compared to the intervals excluded in the previous step of the induction. I have an intuition that if $a_1 = 1$ it suffices to consider the case where $\frac{3s + 1}{3^{k + 1}} \geq \frac{2}{3}$, as the other cases are covered by the case where $a_1 = 0$, but I have no idea how to formalise this intuition. So could someone give me hints as to how to complete this proof, or at least how to formalise the aforementioned intuition?
| $\textbf{Hint:}$
If $a_1=2$, then $x\ge\frac{2}{3}$, so $x\ge\frac{3s+2}{3^{k+1}}$ whenever
$\frac{3s+2}{3^{k+1}}\le\frac{2}{3}\iff 3s+2\le2\big(3^k\big)\iff 3s+3\le2\big(3^k\big)\iff s\le2\big(3^{k-1}\big)-1$.
Therefore we only need to consider values of $s$ satisfying $2\big(3^{k-1}\big)\le s<3^k$, and
$\displaystyle \frac{3s+1}{3^{k+1}}<x<\frac{3s+2}{3^{k+1}}\iff \frac{3s-2\big(3^k\big)+1}{3^k}<3x-2<\frac{3s-2\big(3^k\big)+2}{3^k}$
| {
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"url": "https://math.stackexchange.com/questions/1397259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given $P(x) = x^4+ax^3+bx^2+cx+d$ such $P'(0)=0.$ If $P(-1)
Given $P(x) = x^4+ax^3+bx^2+cx+d$ such that $x=0$ is the only root of $P'(x) = 0.$
If $P(-1)<P(1)\;,$ Then in interval $\left[-1,1\right],$
$\bf{Options::}$
$(a)\;\; P(-1)$ is the minimum and $P(1)$ is the maximum of $P(x).$
$(b)\;\; P(-1)$ is not minimum and $P(1)$ is the maximum of $P(x).$
$(c)\;\; P(-1)$ is the minimum and $P(1)$ is not the maximum of $P(x).$
$(d)\;\;$ Neither $P(-1)$ is the minimum nor$P(1)$ is the maximum of $P(x).$
$\bf{My\; Try::}$ Given $P(x) = x^4+ax^3+bx^2+cx+d\;,$ Then $P'(x) = 4x^3+3ax^2+2bx+c$
Now Given $P'(0) = 0\Rightarrow c=0$
Now Given $P(-1)<P(1)\Rightarrow 1-a+b+d<1+a+b+d\Rightarrow a>0$
Now I did not Understand How can I solve it, Help me
thanks
| Let for example $a=1$, $b=10$ $d=1$:
$a=1$, $b=-10$ $d=1$:
$a=1$, $b=1$ $d=1$:
$a=-1$, $b=1$ $d=1$:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b,c>0$ so that $a+b+c=1$...
Let $a,b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6\geq 2\sqrt{2}\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )$$
What I think is we should replacing $1-a ,1-b ,1-c$ with $b+c,c+a, a+b$ respectively on the right hand side, I have no idea if this is true, any help will be appraciated.
| $$\begin{align}\\&\frac ab+\frac ba+\frac bc+\frac cb+\frac ca+\frac ac+6-2\sqrt 2\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\\&=\frac{b+c}{a}+\frac{a+c}{b}+\frac{b+a}{c}+6-2\sqrt 2\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\\&=\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6-2\sqrt 2\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\\&=\frac{1-a}{a}-2\sqrt{2}\sqrt{\frac{1-a}{a}}+2+\frac{1-b}{b}-2\sqrt{2}\sqrt{\frac{1-b}{b}}+2+\frac{1-c}{c}-2\sqrt 2\sqrt{\frac{1-c}{c}}+2\\&=\left(\sqrt{\frac{1-a}{a}}-\sqrt 2\right)^2+\left(\sqrt{\frac{1-b}{b}}-\sqrt 2\right)^2+\left(\sqrt{\frac{1-c}{c}}-\sqrt 2\right)^2\ge 0\end{align}$$
| {
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$\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$ $\int_{0}^{\frac{\sqrt{2}-1}{2}}\frac{dx}{(2x+1)\sqrt{x^2+x}}$
This is in the form of $\frac{1}{linear\sqrt{quadratic}}$.I put $x=\frac{1}{t}$
$\int_{\frac{2}{\sqrt2-1}}^{\infty}\frac{dt}{(2+t)\sqrt{t+1}}$Then put $t+1=p^2$
From now,it got complicated.Its answer is $\frac{\pi}{4}$.Answer is elusive.
| Let $\sqrt{x^2+x}=u$
$\implies\dfrac{2x+1}{2\sqrt{x^2+x}}dx=du$ and $(2x+1)^2=4u^2+1$
$$\implies\int\dfrac{dx}{(2x+1)\sqrt{x^2+x}}=\int\dfrac{(2x+1)dx}{(2x+1)^2\sqrt{x^2+x}}=\int\dfrac{2du}{4u^2+1}=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The behavior of quadratic formula in the limit $a\to 0$ $$\lim\limits_{a\to0}\frac{-b+\sqrt{b^2-4ac}}{2a}$$
$b$ and $c$ are constants. As $a$ approaches $0$, what does the formula approach?
Example:
$$\lim\limits_{a\to0}\frac{-5+\sqrt{5^2-4·0.001·3}}{2·0.001}\approx-0.6$$
| $$\begin{align*}
\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}
&= \lim_{a\to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{b+\sqrt{b^2-4ac}}{b+\sqrt{b^2-4ac}}\right)\\
&= \lim_{a\to 0}\frac{-b^2+(b^2-4ac)}{2a(b+\sqrt{b^2-4ac})}\\
&= \lim_{a\to 0}\frac{-2c}{b+\sqrt{b^2-4ac}}\\
&= \frac{-2c}{b+|b|}
\end{align*}$$
If $b>0$, then you get $-\dfrac cb$, the root of the linear equation hinted in another answer; but
if $b<0$, then the limit does not exist, because
*
*the larger one of the two quadratic roots becomes $+\infty$ when $a\to 0^+$;
*the smaller one of the two quadratic roots becomes $-\infty$ when $a\to 0^-$;
| {
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If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand it now.
| Starting from
$$3x^2-2x+7=0\\$$
$$x^2-\frac{2}{3}x+\frac{7}{3}=0\\$$
$$x^2-2\cdot\frac{1}{3}\cdot x+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{7}{3}=0\\$$
$$\left(x-\frac{1}{3}\right)^2+\frac{21-1}{9}=0\\$$
$$ \left(x-\frac{1}{3}\right)^2=\frac{-20}{9}$$
| {
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inequality involving $x$, $x^3$,$\sin(x),\cos(x)$ Let $x \in \left[0,\dfrac {\pi} 2 \right]$. Prove the inequality
$$6x \ge 6\sin x +x^3 \cdot \cos x$$
there is nice solution using Taylor expansion. Is there other one?
| (EDIT) Let $f(x) = 6 x - 6 \sin(x) - x^3 \cos(x)$.
$f''(x) = 6 (\sin(x) - x \cos(x)) + 6 x^2 \sin(x) + x^3 \cos(x)$.
Now $\sin(x) - x \cos(x) = \cos(x) (\tan(x) - x) \ge 0$ for $x \in (0,\pi/2)$ while $x^2 \sin(x) \ge 0$ and $x^3 \cos(x) \ge 0$ in this interval as well, so $f''(x) \ge 0$ for $x \in [0,\pi/2]$.
But $f(0)=0$ and $f'(0) = 0$, so $$f(x) = f(0) + f'(0) x + \int_0^x (x-t) f''(t)\; dt \ge 0$$
for $0 \le x \le \pi/2$.
| {
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orthogonal polynomials in (-1,1) with a modified weight function I would like to find the orthogonal polynomial system $ \{ P_n(x), n \in \textbf{Z} \} $ corresponding to the weight function $ w(x) = \frac { 1} {\sqrt{1-x^2} (1+\sqrt{1-x^2} ) } $ defined on the interval $(-1,1)$. Noticing that this $w(x)$ is a modified Chebyshev weight function, I am just wondering whether the solution to this problem studied previously. I am not sure about it. Your responses are very much appreciated.
| The integral $ \int_{x=-1}^{+1} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2}) } = \int_{x=-1}^{0} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2} )} + \int_{x=0}^{1} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2} )} $.
Using substitutions $ s \rightarrow -\sqrt{1-x^2}$ and $t \rightarrow\sqrt{1-x^2} $ respectively in to the first and second integrals on the right-hand side gives
$ \int_{x=-1}^{+1} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2}) } =
- \int_{s=-1}^{0} \frac{ds}{(1+s)^{ \frac{1}{2}} (1-s)^{\frac{3}{2}}} +
\int_{t=0}^{1} \frac{dt}{(1+t)^{ \frac{3}{2}} (1-t)^{\frac{1}{2}}} $ .
Thus, the resulting polynomial space appears to be
$L_2[(-1,1),w(x)] =L_2[(-1,0),w_1] \cup L_2[(0,1),w_2] $.
Where
$ w(x) = \frac{1} { \sqrt{1-x^2} (1+\sqrt{1-x^2} ) }$ ,
$w_1(x) = \frac{-1} { (1+x)^{1/2} (1-x)^{3/2 } } $ and
$w_2(x) = \frac{1} { (1+x)^{3/2} (1-x)^{1/2 } } $.
So two different beta distributions for the half-intervals of $(-1,1)$, leading to two different orthogonal polynomial sequences there.
| {
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Correctness of proof for the convergence of a series Does the following series converge?
$\sum_{n=1}^\infty \frac{5^n + (-1)^n}{2^n3^n}$
What I've done so far.
$0\leq \sum_{n=1}^\infty \frac{5^n + (-1)^n}{2^n3^n} \leq \sum_{n=1}^\infty \frac{5^n + 1^n}{6^n} = \sum_{n=1}^\infty \left(\frac{5}{6}\right)^n + \sum_{n=1}^\infty \left(\frac{1}{6}\right)^n$
We know that last two series both converge, since they are an instance of a geometric series with both $|q|<1$.
Now from this, am I allowed to conclude that the original series converges? What stops it from alternating between 0 and the sum of the values of the geometric series?
| Notice, we have $$\sum_{n=1}^{\infty}\frac{5^n+(-1)^n}{2^n\cdot 3^n}$$ $$=\sum_{n=1}^{\infty}\frac{5^n+(-1)^n}{6^n}$$
$$=\sum_{n=1}^{\infty} \frac{5^n}{6^n}+\sum_{n=1}^{\infty}\frac{(-1)^n}{6^n}$$
$$=\sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n+\sum_{n=1}^{\infty}\left(\frac{-1}{6}\right)^n$$
$$=\left(\frac{5}{6}+\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^3+\ldots\right)+\left(\frac{-1}{6}+\left(\frac{-1}{6}\right)^2+\left(\frac{-1}{6}\right)^3+\ldots\right)$$ $$=\left(\frac{\frac{5}{6}}{1-\frac{5}{6}}\right)+\left(\frac{-\frac{1}{6}}{1-\left(-\frac{1}{6}\right)}\right)$$
$$=\left(\frac{\frac{5}{6}}{\frac{1}{6}}\right)+\left(\frac{-\frac{1}{6}}{\frac{7}{6}}\right)$$
$$=5-\frac{1}{7}=\frac{35-1}{7}$$ $$=\frac{34}{7}$$ Above is a finite value. Hence the series converges.
| {
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Number of reals such that $\frac{1}{x-f(1)}+\frac{2}{x-f(2)}+\frac{3}{x-f(3)}=0$
If $f(x)= x^3+3x^2+6x+2009$ and $g(x)=\dfrac{1}{x-f(1)}+\dfrac{2}{x-f(2)}+\dfrac{3}{x-f(3)}.$
Then number of real solutions of $g(x)=0$ is
$\bf{My\; Try::}$ Given $f(x) = x^3+3x^2+6x+2009\;,$ Now
$$f'(x) = 3x^2+6x+6 = 3[(x+1)^2+1]>0\forall x\in \mathbb{R}$$
So $f(x)$ is an Strictly Increasing function.
So $f(1)<f(2)<f(3)\;,$ Now Let $f(1) = a$ and $f(2) = b$ and $f(3) = c\;,$
Then we get $a<b<c$
Now $$\displaystyle g(x) = \frac{1}{x-a}+\frac{2}{x-b}+\frac{3}{x-c}.$$
$$\displaystyle = \frac{(x-b)\cdot (x-c)+2(x-c)\cdot (x-a)+3\cdot (x-a)\cdot (x-b)}{(x-a)\cdot (x-b)\cdot (x-c)}$$
Now How can i solve after that, help me
Thanks
| Two solutions. It is enough to see that
$$
g^\prime(x)=-\sum_{i=1}^3 \frac{i}{(x-f(i))^2}<0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed-form of $\int_0^1 \operatorname{Li}_3\left(1-x^2\right) dx$ By using dilogarithm functional equations we can show that
$$
\int_0^1 \operatorname{Li}_2\left(1-x^2\right)\,dx = \frac{\pi^2}{2}-4,
$$
where $\operatorname{Li}_2$ is the dilogarithm function.
Could we evaluate in closed-form the following integral?
$$
I = \int_0^1 \operatorname{Li}_3\left(1-x^2\right)\,dx,
$$
where $\operatorname{Li}_3$ is the trilogarithm function.
A related integral with known closed-form is
$$\int_0^1 \operatorname{Li}_3\left(\frac{1}{x^2}\right)\,dx = \zeta(3)+\frac{\pi^2}{3}-8\ln2 - 4\pi\,i,$$
where $\zeta$ is the Riemann zeta function.
| Expanding my comment: the substitution $1-x^2\mapsto x$, followed by expanding the trilogarithm and keeping in mind Legrende's duplication formula $B(n+1,\frac12)=2^{2n+1}B(n+1,n+1)$, we arrive at $$I=\sum_{n=1}^{\infty} \frac{2^{2n}}{n^3(2n+1)\binom{2n}{n}}=\sum_{n=1}^{\infty}\frac{2^{2n}}{n^3\binom{2n}{n}}-\sum_{n=1}^{\infty}\frac{2^{2n+1}}{n^2(2n+1)\binom{2n}{n}}\\=4\int_0^1\frac{\arcsin^2x}{x}\,dx-4\int_0^1\arcsin^2x \,dx$$
where I used the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}=2\arcsin^2x$.
The second integral is easily evaluated by IBP twice: $$\int_0^1\arcsin^2x \,dx=\int_0^{\frac{\pi}{2}}x^2\cos x \,dx=\frac{\pi^2}{4}-2$$
The first integral may be evaluated by IBP and using that $\displaystyle -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}$:
$$\begin{align} \int_0^1\frac{\arcsin^2x}{x}\,dx\\&=\int_0^{\frac{\pi}{2}}x^2\cot x \,dx\\&=-2\int_0^{\frac{\pi}{2}}x\ln\sin x \,dx\\
&=2\int_0^{\frac{\pi}{2}}x\left(\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}\right)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{2}}x\cos(2nx)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\frac1{4n^2}((-1)^n-1)\\&=\frac{\pi^2}{4}\ln2-\frac{7}{8}\zeta(3).
\end{align}$$
| {
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Proving: $\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.$ $$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0\end{equation}$$
I found this inequality in this paper: http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf (Equation (3)).
How exactly can I prove it? I tried induction but to no avail...
| Let $f(x) = \frac{2x}{x+2}$, $g(x) = \ln(1+x)$, $h(x) = \frac{x^2 + 2x}{2x+2}$.
$f(0) = 0$, $g(0) = 0$, $h(0) = 0$.
$f' = \frac{2(x+2)-2x}{(x+2)^2} = \frac{4}{(x+2)^2}$, $g' = \frac{1}{1+x}$, $h'= \frac{(2x+2)(2x+2)-2(x^2+2x)}{(2x+2)^2}=\frac{2x^2+4x+4}{(2x+2)^2}$
Now it is easy to check that
$$f'=\frac{4}{(x+2)^2} \leq g' = \frac{1}{1+x} \leq h'=\frac{2x^2+4x+4}{(2x+2)^2}$$
| {
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If $f’(x) = \sin x + (\sin4x)(\cos x)$, then $f’(2x^2 + \pi/2) $is? If $$f'(x) = \sin x + \sin4x \cdot \cos x,$$ then $$f'(2x^2 + \pi/2)$$ is?
Given answer: $$4x\cos(2x^2) – 4x\sin(8x^2) \sin(2x^2)$$
I tried and I'm getting the answer as $\cos(2x^2) - \sin(8x^2)\sin(2x^2)$
| If $$f(x) = \sin x+\sin 4x\cdot \cos x\;,$$ Then put $$\displaystyle x = 2x^2+\frac{\pi}{2}$$
So $$\displaystyle f\left(2x^2+\frac{\pi}{2}\right) = \sin \left(2x^2+\frac{\pi}{2}\right)+\sin \left(8x^2+4\pi\right)\cdot \cos \left(2x^2+\frac{\pi}{2}\right)$$
So $$\displaystyle f\left(2x^2+\frac{\pi}{2}\right) = \cos 2x^2-\sin 4x^2\cdot \sin 2x^2$$
Above we have used $$\displaystyle \bullet\; \sin \left(\frac{\pi}{2}+\phi\right) = \cos \phi$$ and $$\displaystyle \bullet\; \cos \left(\frac{\pi}{2}+\phi\right) = -\sin \phi$$
and $$\displaystyle \bullet\; \sin (2\pi+\phi) = \sin \phi$$
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