Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Implicit differentiations of same equation with two different answers:Why? I've an equation that I like to use for implicit differentiation. The equation is:$x^2 = \frac{(x+2y)}{(x-2y)}$
I used two different methods but got two different answers for same equation.
Can anyone kindly tell me where I am wrong? Why am I getting two different answers?
The first method I used is:
$$
\begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ 2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\ 2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\ 2x(x - 2y)^2 =& x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\ 2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\ \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\ \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{ [after simplification] } \end{align*}
$$
Then I used the same equation but multiplied both sides by $(x-2y)$. So here's the second procedure:
$$
\begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ x^2 (x-2y) =& x + 2y \\ x^3 - 2x^2y - x - 2y =& 0 \\ 3x^2 - 2x^2 \frac{dy}{dx} - 4xy - 1 - 2 \frac{dy}{dx} =& 0 \\ -2x^2 \frac{dy}{dx} - 2 \frac{dy}{dx} =& -3x^2 + 4xy + 1 \\ \frac{dy}{dx}(2x^2 + 2) =& 3x^2 - 4xy - 1 \\ \frac{dy}{dx} =& \frac{3x^2 - 4xy - 1}{2x^2 + 2}\end{align*}
$$
| They may not be different. Remember, there's an equation relating $x$ and $y$. If you use that equation, you might be able to get one of the answers to look like the other one.
| {
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"url": "https://math.stackexchange.com/questions/191226",
"timestamp": "2023-03-29T00:00:00",
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Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| If $x,y,z$ are nonnegative reals, then
$x^2+y^2+z^2\ge xy+yz+zx$ (with equality iff $x=y=z$), hence
$3(x^2+y^2+z^2)\ge x(x+y+z)+y(y+z+x)+z(z+x+y) = (x+y+z)^2$
(with equality iff $x=y=z$).
Letting $x=a+b, y=b+c, z=a+c$, we find $x+y+z=6$ and
$3(a+b)^2+3(b+c)^2+3(c+a)^2 \ge 36$.
Note that $(a+b)^2+(b+c)^2+(c+a)^2= 2(a^2+b^2+c^2+ab+bc+ca)$ so that we actually showed
$$ a^2+b^2+c^2+ab+bc+ca\ge 6$$
with equality iff $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
System of 3 linear congruences
Find all solutions: $$\begin{cases} x\equiv 39 \mod(189) \\ x\equiv 25 \mod(539) \\ x\equiv 399 \mod(1089) \end{cases}$$
But $189=3^3\cdot7$, $539=11\cdot7^2$ and $1089=3^2\cdot 11^2$, so I can't use here Chinese remainder theorem.
| You can find a system of congruences equivalent to the original system, but with pairwise relatively prime moduli.
For example, the first congruence is equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 39\pmod{7}$.
The second original congruence is equivalent to the system $x\equiv 25\pmod{11}$ $x\equiv 25\pmod{7^2}$.
Note that from the first congruence we had $x\equiv 4\pmod{7}$. If $ \equiv 25\pmod{7^2}$ then $x\equiv 4\pmod{7}$, our first two original congruences are equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 25\pmod{7^2}$, $x\equiv 25\pmod{11}$.
The third original congruence can be similarly dealt with. When we are through we end up with a system with moduli $3^3$, $7^2$, and $11$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$ I need to find an expression for $n$th derivative of $f(x) = e^{x^2}$. Really need help.
| Let $y=e^{x^2}$. We want to find $\frac{d^n y}{dx^n}$.
Note that $\ln y = x^2$ and so $\frac{1}{y} \frac{dy}{dx} = 2x\iff \frac{dy}{dx}=2xy$.
Differentiating again yields:
$\frac{d^2y}{dx^2}=2x\frac{dy}{dx}+2y$,
$\frac{d^3 y}{dx^3} = 2x \frac{d^2 y}{dx^2} + 4\frac{dy}{dx}$,
$\frac{d^4 y}{dx^4} = 2x \frac{d^3 y}{dx^3} + 6\frac{d^2 y}{dx^2}$.
We might guess that $\frac{d^n y}{dx^n} = 2x \frac{d^{n-1} y}{dx^{n-1}}+2(n-1) \frac{d^{n-2} y}{dx^{n-2}}$, which is easily proved by straightforward induction (we leave this as an exercise to the reader. ;) )
Let $u_k=\frac{d^k y}{dx^k}$ for easier notation. We thus have established
$u_n = 2x u_{n-1}+2(n-1)u_{n-2}$.
It is easy to find that the coefficient of $x^n y$ in $u_n$ is $2^n$, but I'm not seeing any nice way of solving the recursion at the moment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Positive Definite Matrix Problem Suppose that the Symmetric Matrix
$$B=\left(
\begin{array}{cc}
\alpha & a^T \\
a & A
\end{array}
\right)$$
of order $n+1$ is positive definitive.
(a) Show that the scalar $\alpha$ must be positive and the $n*n$ matrix $A$ must be positive definitive.
(b) What is the Cholesky factorization of $B$ in terms of $\alpha$, $a$, and the Cholesky factorization of $A$?
| I also find a solution for my problem:
$$B=\left(
\begin{array}{cc}
\alpha & a^T \\
a & A
\end{array}
\right)$$
is a symmetric matrix of order $n+1$ positive definite.
We will show that $\alpha$ is positive and $A$ is a $n\times n$ positive definite matrix.
$B$ is positive definite, so
$$\exists\;v=\left(
\begin{array}{c}
v_1 \\
\tilde{0}
\end{array}
\right)$$
a non-zero $(n+1)\times1$ vector such that $v^TBv$ is positive ($\tilde{0}$ denote $n$ zero elements of $v$).
Consequently, we have
$$\left(
\begin{array}{cc}
v_1 & \tilde{0}
\end{array}
\right)\left(
\begin{array}{cc}
\alpha & a^T \\
a & A
\end{array}
\right)\left(
\begin{array}{c}
v_1 \\
\tilde{0}
\end{array}
\right)=\left(
\begin{array}{cc}
v_1 \alpha & v_1 a^T
\end{array}
\right)\left(
\begin{array}{c}
v_1 \\
\tilde{0}
\end{array}
\right)=v_1 \alpha v_1=v_1^2\alpha >0$$
Because $B$ is positive definite the $v^T B v$ should be positive which in this case forces $\alpha$ to be positive.
Again, considering that
$$\exists\;v=\left(
\begin{array}{c}
0 \\
\tilde{v}
\end{array}
\right)$$
a non-zero $(n+1)\times1$ vector such that $v^T B v$ is positive ($\tilde{v}$ denote a non-zero $n\times1$ vector).
We can write our formula again:
$\left(
\begin{array}{cc}
0 & \tilde{v}^T
\end{array}
\right)\left(
\begin{array}{cc}
\alpha & a^T \\
a & A
\end{array}
\right)\left(
\begin{array}{c}
0 \\
\tilde{v}
\end{array}
\right)=\left(
\begin{array}{cc}
\tilde{v}^Ta & \tilde{v}^TA
\end{array}
\right)\left(
\begin{array}{c}
0 \\
\tilde{v}
\end{array}
\right)=\tilde{v}^TA \tilde{v}>0$
again, because $B$ is positive definite, $v^TB v$ should be positive and it will force $A$ to be positive definite.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the limit of $\sum\limits_{k=1}^{n} \left(\frac{k}{n^2}\right)^{1+k/n^2}$ when $n\to\infty$
Compute the limit
$$\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\frac{k}{n^2} + 1}$$
At a first look, I only thought of Riemann sums, but I don't see how I may apply it. What else could I do? I need some hints, suggestions.
| Let $x: = \frac {k} {n^2}$. Then, we have $$\lim_{n\to\infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)} = \lim_{n\to\infty} n^2 \int_{1/n^2}^{1/n} x^{x + 1}\, dx.$$ Trapezoidal rule (in quadrature) gives $$\int_{1/n^2}^{1/n} x^{x + 1}\, dx = \frac {1} {2} \left(\frac {1} {n} - \frac {1} {n^2} \right) \left(\left(\frac {1} {n}\right)^{1 + \frac {1} {n}} + \left(\frac {1} {n^2}\right)^{1 + \frac {1} {n^2}}\right) + R_n,$$ where $R_n \to 0$.
This is $$= \frac {1} {2 n^2} (n - 1) \left (\left(\frac {1} {n}\right)^{1 + \frac {1} {n}} + \left(\frac {1} {n}\right)^{2 + \frac {2} {n^2}}\right) + R_n = \frac {n - 1} {2 n^3} \left (\left(\frac {1} {n}\right)^{\frac {1} {n}} + \left(\frac {1} {n}\right)^{1 + \frac {2} {n^2}}\right) + R_n.$$ Since $$\left (\left(\frac {1} {n}\right)^{\frac {1} {n}} + \left(\frac {1} {n}\right)^{1 + \frac {2} {n^2}}\right) \to 1 \qquad \text{and} \qquad R_n \to 0,$$ we have $$\int_{1/n^2}^{1/n} x^{x + 1}\, dx \to \frac {n - 1} {2 n^3}.$$ Hence, $$\sum_{k=1}^{n} \left(\frac{k}{n^2}\right)^{\left(\frac{k}{n^2} + 1\right)} \to \frac {n - 1} {2 n} \to \frac {1} {2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$? Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$
| I think that if this exercise can be solved in a simple fashion (without heavy
computation), then this should be the way to approach it (otherwise, it is
just a mindless computation which just requires to apply some algorithm like
the, so-called, Weierstrass substitution and teaches you nothing).
So, since
$$
\sin^{3}x+\cos^{3}x=\left( \sin x+\cos x\right) \left( \sin^{2}x-\sin x\cos
x+\cos^{2}x\right) =\left( \sin x+\cos x\right) \left( 1-\sin x\cos
x\right) ,
$$
then we try to express $\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}$ as follows (if
possible), in order to be able to (easily) compute ${\displaystyle\int}\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}\;\mathrm{d}x$:
\begin{align*}
\frac{\sin^{3}x}{\sin^{3}x+\cos^{3}x} & =A+B\cdot\frac{\left( \sin x+\cos
x\right) ^{\prime}}{\sin x+\cos x}+C\cdot\frac{\left( 1-\sin x\cos x\right)
^{\prime}}{1-\sin x\cos x}=\\
& =A+B\cdot\frac{\cos x-\sin x}{\sin x+\cos x}+C\cdot\frac{\sin^{2}x-\cos
^{2}x}{1-\sin x\cos x}
\end{align*}
From here we obtain that
$$
\sin^{3}x=A\left( \sin^{3}x+\cos^{3}x\right) +B\left( \cos x-\sin x\right)
\left( 1-\sin x\cos x\right) +C\left( \sin x+\cos x\right) \left(
\sin^{2}x-\cos^{2}x\right)
$$
hence
\begin{align*}
0 & =(A+C-1)\sin^{3}x+(A-C)\cos^{3}x+B\left( \cos x-\sin x\right)
-(B+C)\sin x\cos^{2}x+(B+C)\sin^{2}x\cos x\\
& =(A+C-1)\sin^{3}x+(A-C)\cos^{3}x+B\left( \cos x-\sin x\right) -(B+C)\sin
x(1-\sin^{2}x)+(B+C)(1-\cos^{2}x)\cos x\\
& =(A+B+2C-1)\sin^{3}x+(A-B-2C)\cos^{3}x+(2B+C)\left( \cos x-\sin x\right)
\end{align*}
so
$$
\left\{
\begin{array}
[c]{r}
A+B+2C=1\\
A-B-2C=0\\
2B+C=0
\end{array}
\right.
$$
which has the (unique) solution
$$
\left\{
\begin{array}
[c]{l}
A=\frac{1}{2}\\
B=-\frac{1}{6}\\
C=\frac{1}{3}
\end{array}
\right.
$$
hence
\begin{align*}
{\displaystyle\int}\dfrac{\sin^{3}x}{\sin^{3}x+\cos^{3}x}\;\mathrm{d}x &=Ax+B\log\left\vert
\sin x+\cos x\right\vert +C\log\left\vert 1-\sin x\cos x\right\vert
+\text{some constant}\\
&=\frac{x}{2}-\frac{\log\left\vert \sin x+\cos x\right\vert }{6}+\frac
{\log\left\vert 1-\sin x\cos x\right\vert }{3}+\text{some constant}
\end{align*}
Let's hope I didn't make any mistake in my calculations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof the logaritmic identity Please help me proving the basic logarithmic identity $\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$
| Let $\gcd(a,b) = 1$ and $x = a^nb$ where $a,b,n \in \mathbb{Z}$, then
$$\begin{aligned}
\log_a x &= n + \frac{\ln b}{\ln a} = n + \frac{\ln b}{\ln(b+1)}\frac{\ln(b+1)}{\ln(b+2)}\cdots \frac{\ln(b+(m-1))}{\ln(b+m)}\frac{\ln (b+m)}{\ln a} \newline
&= n + \log_{b+1} b\log_{b+2}{b+1}\cdots \log_{b+m} (b + (m-1)) \ln_{a} (b+m).
\end{aligned}$$
Take case that $a = 3$ and $b = 4$, $n = 1$ and $m = 2$, then $\log_3 12 = 1 + \log_5 4 \log_6 5 \log_3 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/199805",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int \cos(x) \sqrt{\sin(2 x)} dx$
Evaluate the following indefinite integral:
$$\int \cos(x) \sqrt{\sin(2 x)} dx$$
Only hint I have is from W|A that expresses the integral in terms of a hypergeometric function and it looks rather ugly.
Can we solve it in a simpler way and get a nicer form? Thanks.
| Alternatively, rewrite
$$I=\sqrt{2}\int(1-(\sin x)^2)^\frac{1}{4}(\sin{x})^\frac{1}{2}d(\sin{x})\\=\sqrt{2}\int(1-t^2)^{\frac{1}{4}}t^\frac{1}{2}dt\\=\sqrt{2}\int\left(\frac{1}{t^2}-1\right)^\frac{1}{4}tdt \\=\frac{1}{\sqrt{2}}\int\left(\frac{1}{z}-1\right)^\frac{1}{4}dz$$
Now let
$$\frac{1}{z}-1=u^2$$
$$-\frac{dz}{z^2}=2udu$$
$$dz=-\frac{2udu}{(1+u^2)^2}=d\left(\frac{1}{1+u^2}\right)$$
$$\sqrt{2}I=\int u^\frac{1}{2}d\left(\frac{1}{1+u^2}\right)=\frac{u^\frac{1}{2}}{1+u^2}-\int\frac{d\left(u^\frac{1}{2}\right)}{1+u^2}$$
Where the last integral is equivalently $\int\frac{dv}{1+v^4}$ to which there exist various approaches.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
| You know that $$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$ and $$(a^2-b^2)=(a-b)(a+b)$$ so if $a=\sqrt{3+x}$ and $b=2$, we have $(a^2-b^2)=(\sqrt{3+x}-2)(\sqrt{3+x}+2)=3+x-4$. The same calculation can be done for another identity by taking $x=\sqrt[3]{7+x}$ and $y=2$. In fact you have $$(\sqrt[3]{7+x}-2)\big((\sqrt[3]{7+x})^2+2\sqrt[3]{7+x}+2^2\big)=(\sqrt[3]{7+x})^3-2^3=7+x-8$$ Now multiply your fraction to $\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}=1$ and $\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}=1$ simultaneously, so:
$$\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}\times\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}$$ $$=\frac{3+x-4}{7+x-8}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ $$=\frac{x-1}{x-1}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2} $$ which is equal to $$\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ near $x=1$. Now, I think taking the limit when $x$ tends to $1$ is so easy. It is $3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a spanning set for a null space I have a matrix $A$ that is like this:
\begin{equation}
A = \pmatrix{
1 & 2 & -3 & 1 & 5 \\
1 & 3 & -1 & 4 & -2 \\
1 & 1 & -5 & -2 & 12 \\
1 & 4 & 1 & 7 & -7
}
\end{equation}
The question is: Find a set of $5\times 1$ matrices whose linear span is the null space of $A$.
I did Gauss-Jordan and I got the matrix down to:
\begin{equation}
\pmatrix{
1 & 2 & -3 & 1 & 5 \\
0 & -1 & -2 & -3 & -7 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -2
}
\end{equation}
But doesn't that mean the matrix is inconsistent therefore there is no linear span? Any ideas? Thanks
| Your reduced matrix is correct. First you need to characterize the set of vectors $x$ that satisfy $A x = 0$. This set is called the null space or kernel, and I use the standard notation $\ker A$.
The reduction process above corresponds to pre-multiplying $A$ by an invertible matrix $G$ such that $G A = \tilde{A}$, where $\tilde{A}$ is the reduced matrix above. Since $G$ is invertible, you have $Ax = 0 $ iff $\tilde{A}x = 0$, or in other words, $\ker A = \ker \tilde{A}$. So we can focus on finding $\ker \tilde{A}$, since the matrix has a nicer form.
Suppose $ \tilde{A}\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5} = 0$. Then you can immediately see that we must have $x_5 = 0$. The third row tells us nothing. And the first two rows can be re-written as
$\pmatrix{ 1 && 2 \\ 0 && 1} \pmatrix{x_1 \\ x_2}+\pmatrix{ -3 && 1 \\ 2 && 3} \pmatrix{x_3 \\ x_4} = 0$. Since $\pmatrix{ 1 && 2 \\ 0 && 1}^{-1} = \pmatrix{ 1 && -2 \\ 0 && 1}$, we get
$\pmatrix{x_1 \\ x_2} = \pmatrix{ 7 && 5 \\ -2 && -3} \pmatrix{x_3 \\ x_4}$.
This tells us that if we select $x_3, x_4$, then $x_1, x_2$ are completely determined. Thus $ \tilde{A}x = 0$ iff $x_5 = 0$, $x_3, x_4$ are arbitrary, and $x_1,x_2$ given by the above formula, or in other words $\ker A = \{\pmatrix{7 x_3+5 x_4 \\ -2x_3-3 x_4 \\ x_3 \\ x_4 \\ 0} | x_3, x_4 \text{arbitrary} \}$.
Now you want to find a more convenient way of expressing this. Note that we can write $\pmatrix{7 x_3+5 x_4 \\ -2x_3-3 x_4 \\ x_3 \\ x_4 \\ 0} = x_3 \pmatrix{7 \\ -2 \\ 1 \\ 0 \\ 0} + x_4 \pmatrix{5 \\ -3 \\ 0 \\ 1 \\ 0}$. So we can write $\ker A = \text{sp} \{\pmatrix{7 \\ -2 \\ 1 \\ 0 \\ 0}, \pmatrix{5 \\ -3 \\ 0 \\ 1 \\ 0} \}$. That is, the null space of $A$ is the span of these two vectors.
| {
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Arithmetic Mean & Geometric Mean Question 1: if the arithmetic mean of two numbers is twice of their geometric mean, their ratio of sum of numbers to the difference of numbers equals?
Question 2: if the quadratic equation:
$(b^2+c^2)x2-2(a+b)cx+(c^2+a^2)=0$
has equal roots then?what is its AP & GP?
Question 3: If the expansion of:
$(1+x)^{50}$
let S be the sum of the coefficient of the odd power of x, then S will be?
Please help with this problems in brief.
-Thanks.
| $1.$ We can even do it without the quadratic formula. We have $a+b=4\sqrt{ab}$ and therefore
$$(a+b)^2=16ab.$$
Also,
$$(a-b)^2=(a+b)^2-4ab=(a+b)^2-\frac{1}{4}(a+b)^2=\frac{3}{4}(a+b)^2.$$
Thus
$$\frac{(a+b)^2}{(a-b)^2}=\frac{4}{3},$$
and therefore
$$\frac{a+b}{a-b}=\pm\frac{2}{\sqrt{3}}.$$
$2.$ The roots are equal precisely if the discriminant is $0$, that is, if
$$4(a+b)^2c^2-4(b^2+c^2)(c^2+a^2)=0.$$
Divide by $4$, expand everything, do the obvious cancellations.
We get
$$2abc^2=c^4+a^2b^2,$$
which can be rewritten as
$$(c^2-ab)^2=0.$$
We conclude that $ab=c^2$. We cannot have $c=0$ and $b=0$, else we would not have a quadratic equation. We conclude that the sequence $b,c,a$ is a three-term geometric sequence. If $a\ne 0$, then $a,c,b$ is also a three-term geometric sequence.
$3.$ I recommend that you look at the solution by robjohn.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
| Let $s_n=\sum_{k=0}^nk^3$; your generating function for these numbers will be $$f(x)=\sum_{n\ge 0}s_nx^n\;.$$
You know that the sequence satisfies the recurrence $s_n=s_{n-1}+n^3$. Multiply this recurrence by $x^n$ and sum over $n\ge 0$:
$$\sum_{n\ge 0}s_nx^n=\sum_{n\ge 0}s_{n-1}x^n+\sum_{n\ge 0}n^3x^n\tag{1}\;.$$
The lefthand side of $(1)$ is $f(x)$. We assume that $s_n=0$ for all $n<0$, so we can rewrite $(1)$ as $$f(x)=x\sum_{n\ge 0}s_{n-1}x^{n-1}+\sum_{n\ge 0}n^3x^n=x\sum_{n\ge 0}s_nx^n+\sum_{n\ge 0}n^3x^n=xf(x)+\sum_{n\ge 0}n^3x^n$$ and see that $$f(x)=\frac1{1-x}\sum_{n\ge 0}n^3x^n\;.\tag{2}$$
To deal with the summation in $(2)$, start with $$\frac1{1-x}=\sum_{n\ge 0}x^n\;.$$ Differentiate and multiply by $x$ to get $$\frac{x}{(1-x)^2}=\sum_{n\ge0}nx^n\;.$$ Repeat: $$\frac{x(1+x)}{(1-x)^3}=\sum_{n\ge0}n^2x^n\;.$$ And one more time: $$\frac{x(1+4x+x^2)}{(1-x)^4}=\sum_{n\ge 0}n^3x^n\;.$$ Thus,
$$f(x)=\frac{x+4x^2+x^3}{(1-x)^5}\;.$$ Now decompose $f$ into partial fractions:
$$f(x)=-\frac1{(1-x)^2}+\frac7{(1-x)^3}-\frac{12}{(1-x)^4}+\frac6{(1-x)^5}\;.$$
Finally, you need to know some standard generating functions. In particular, you need to know that $$\frac1{(1-x)^k}=\sum_{n\ge 0}\binom{n+k-1}{k-1}x^n\;.$$ With that you get finally that
$$\begin{align*}
f(x)&=\sum_{n\ge 0}\left(-\binom{n+1}1+7\binom{n+2}2-12\binom{n+3}3+6\binom{n+4}4\right)x^n\\
&=\sum_{n\ge 0}\frac14\left(n^4+2n^3+n^2\right)x^n
\end{align*}$$
and therefore that $$s_n=\frac14\left(n^4+2n^3+n^2\right)=\left(\frac{n(n+1)}2\right)^2\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/209268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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A Differential Equation with Trigonometric Coefficients Suppose we have the following second-order differential equation:
$\cos^2(x)y'' -\sin(x)y' + y = 0$
How do we determine its general solution? I couldn't even guess a particular solution; all my efforts led nowhere. I started off with something like $y = Ae^{B\cos(x)}$ but needless to say, that also looked like a dead-end.
| This is a linear ODE of trigonometric function coefficients. The current approach of solving it is to transform it to a linear ODE of polynomial function coefficients first.
Let $u=\cos x$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-(\sin x)\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-(\sin x)\dfrac{dy}{du}\right)=-(\sin x)\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)-(\cos x)\dfrac{dy}{du}=-(\sin x)\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}-(\cos x)\dfrac{dy}{du}=-(\sin x)\dfrac{d^2y}{du^2}(-\sin x)-(\cos x)\dfrac{dy}{du}=(\sin^2x)\dfrac{d^2y}{du^2}-(\cos x)\dfrac{dy}{du}$
$\therefore(\cos^2x)\left((\sin^2x)\dfrac{d^2y}{du^2}-(\cos x)\dfrac{dy}{du}\right)+(\sin^2 x)\dfrac{dy}{du}+y=0$
$(\cos^2x)(1-\cos^2x)\dfrac{d^2y}{du^2}+(1-\cos^2 x-\cos^3x)\dfrac{dy}{du}+y=0$
$u^2(1-u^2)\dfrac{d^2y}{du^2}+(1-u^2-u^3)\dfrac{dy}{du}+y=0$
$u^2(u^2-1)\dfrac{d^2y}{du^2}+(u^3+u^2-1)\dfrac{dy}{du}-y=0$
$\dfrac{d^2y}{du^2}+\dfrac{u^3+u^2-1}{u^2(u^2-1)}\dfrac{dy}{du}-\dfrac{y}{u^2(u^2-1)}=0$
$\dfrac{d^2y}{du^2}+\left(\dfrac{1}{u^2}+\dfrac{u}{u^2-1}\right)\dfrac{dy}{du}+\left(\dfrac{1}{u^2}-\dfrac{1}{u^2-1}\right)y=0$
| {
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"source": "stackexchange",
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Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$$
Finding $\cos(2x)^2$ frequency:
$$\cos(4x)=2\cos^2(2x)-1$$
$$\cos^2(2x)=\frac{\cos(4x)+1}{2}$$
$$f_1=\frac{2\pi}{4}=\frac{\pi}{2}$$
Finding $\sin(x)^2$ frequency:
$$\cos(2x)=1-2\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$f_2=\frac{2\pi}{2}=\pi$$
$$F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_1=\frac{\pi}{2}\times2=\pi$$
Finding $cos(x)^4$ frequency (I'll call it F2):
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
$$\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$$
Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1):
$$f_3=\frac{2\pi}{2}=\pi$$
$$F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_2=\frac{\pi}{2}\times2=\pi$$
Finding $y$'s frequency:
$$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$
$$F_y=\pi\times1=\pi$$
| I won't pretend that I would have thought of this without seeing the answer but we can look at the problem geometrically. This is more of a nice illustration than a genuine 'proof by pictures'.
We know firstly that the curve $x^4+y^4=a$ is symmetric about the $x$- and $y$-axes:
Suppose that $f(\theta)=\sin^4\theta+\cos^4\theta=a$ for some $a$ in the range of $f$. Now we know that $\sin^2\theta+\cos^2\theta$ is equal to one. Hence we can place the point $(\cos\theta,\sin\theta)$ on the intersection of the unit circle and the curve $x^4+y^4=a$ as shown:
Now as the curve has $\pi/2$-rotation symmetry, the point $(\cos(\theta+\pi/2),\sin(\theta+\pi/2))$ is also on the curve $x^4+y^4=a$ and we are done.
| {
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When is $1^5 + 2^5 + \ldots + n^5$ a square? When is $1^5 + 2^5 + \ldots + n^5$ a square? I found that this happens sometimes: $n=13$ gives $1001^2$, $n=133$ gives $9712992^2$ and $n=1321$ gives $942162299^2$.
I feel that the identity$$\displaystyle\sum_{i=1}^n i^5 = \tfrac{1}{12}[2n^6+6n^5+5n^4-n^2]$$ will be useful, since it's all square powers except one.. but I see no way to connect that.
| If $$a=1+2+3+\cdots+n$$ then
$$
1^5+2^5+3^5+\cdots+n^5 = \frac{4a^3-a^2}{3},
$$
so now the question is: when is that a square?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "30",
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Probability of rolling 5 dice? Say you picked $\{4 5 2 6 5\}$ to show up in the dice rolls. You win if you have 3, 4, or 5 numbers in the correct sequence.
*
*3 numbers in correct sequence would be; $\{4 5 2\}$ OR $\{5 2 6\}$ OR $\{2 6 5\}$
*4 numbers in the correct sequence would be; $\{4 5 2 6\}$ OR $\{5 2 6 5\}$
*5 numbers in the correct sequence would be; $\{4 5 2 6 5\}$
What would be the probability of winning? How would I calculate it?
| 3 numbers correct: $(\frac{5}{6})^2\cdot\frac{1}{6^3}\cdot\binom{5}{2}$
4 numbers correct: $\frac{5}{6}\cdot\frac{1}{6^4}\cdot\binom{5}{1}$
5 numbers correct: $\frac{1}{6^5}$
total: $\frac{1}{6^5}\bigg(1+5\cdot\binom{5}{1}+5^2\cdot\binom{5}{2}\bigg)=\frac{276}{7776}=\frac{23}{648}$
| {
"language": "en",
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"source": "stackexchange",
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Expected Number of Successes in a Sample
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $10$ calculators being broken, multiply the probabilities with the respective $0$ - $10$ and add those together?
For example:
\begin{align}
{10 \choose 0} \cdot \left(\frac{1}{10}\right)^0 \cdot \left(\frac{9}{10}\right)^{10} &= 0.3487 \tag{0}, \\
{10 \choose 1} \cdot \left(\frac{1}{10}\right)^1 \cdot \left(\frac{9}{10}\right)^9 &= 0.3874 \tag{1}, \\
\vdots & \\
{10 \choose 10} \cdot \left(\frac{1}{10}\right)^{10} \cdot \left(\frac{9}{10}\right)^0 &= 1 \cdot 10^{-10}. \tag{10}
\end{align}
Adding them together:
\begin{align}
E[X] = (0) \cdot 0 + (1) \cdot 1 + \ldots + (10) \cdot 10.
\end{align}
Or is it enough to calculate
\begin{align}
E[X] &= (1+2+3+4+5+6+7+8+9+10) \cdot \left(\frac{1}{10}\right).
\end{align}
Thanks.
| The answer should be
$\left({20 \choose 1}{180 \choose 9} + 2{20 \choose 2}{180 \choose 8} + \cdots
+ 10{20 \choose 10}{180 \choose 0} \right) / {200 \choose 10}$, where ${n \choose k}$ denotes the binomial coefficient ${n \choose k} = \frac{n!}{k!(n-k)!}$.
This follows because the probability $p_k$ that you your sample has $k$ broken calculators is ${20 \choose k}{180 \choose 10-k} / {200 \choose 10}$. Why? Because the denominator is counting the total number of ways you could sample $10$ calculators from a set of $200$, and the numerator is counting the total number of such subsets which have $k$ broken calculators, because it's equivalent to sampling $k$ calculators from the set of broken ones and $10-k$ calculators from the set of non-broken ones.
The expected value is therefor $p_1 + 2p_2 + 3p_3 + \cdots + 10p_{10}$ which is equal to the number I have on top.
| {
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"source": "stackexchange",
"question_score": "2",
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What is $\int\frac{dx}{x+\sqrt{1-x^2}}$?
Possible Duplicate:
Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$
$$\int\frac{dx}{x+\sqrt{1-x^2}}$$
I used $x = \sin(a),\ dx = \cos (a)\,da$.
However, I am getting suck after that.
I can't seem to find what to do after $$ \int \frac{\cos(a)\,da}{\sin(a)+\cos(a)}$$
Help please.
| Hint or starting point:
Nothing immediate jumps out to me when I see the integral
$$\int \frac{\cos \theta}{\sin \theta + \cos\theta} d\theta$$
However, when integrating rational function of trig functions, we can always use the Weierstrass substitution.
So, let $t = \tan \left(\frac{\theta}{2}\right), dt = \frac{1}{2}\sec^2\left(\frac{\theta}{2}\right) \ dx.$
With the Weierstrass sub, we have that:
$$\sin \theta = \frac{2t}{1+t^2}, \cos\theta = \frac{1-t^2}{1+t^2}, d\theta = \frac{2 \ dt}{1+t^2}$$
This allows us to rewrite our integral as:
$$\int\frac{2(1-t^2) \ dt}{(t^2+1)^2 \left(\frac{2t}{t^2+1} + \frac{1-t^2}{t^2+1}\right) }$$
Clean up the denominator to get:
$$2\int\frac{(t^2-1) \ dt}{t^4 - 2t^3 - 2t - 1}$$
From here, looks like partial fractions will do it. Note the denominator factors as
$$(t^2+1)(t^2-2t-1)$$
Second thought
Write the original integral as
$$\int \frac{1}{1+\cot \theta} d\theta$$
Let $u = \cot \theta, du = -(1+u^2) d\theta$. Then we have:
$$-\int\frac{1}{(1+u^2)(u^2+1)} \ du $$
This is a nicer partial fraction decomposition to work with and is less work in getting there compared to my original thinking.
| {
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"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Process for $(k+1)^3$? I've mentioned in previous questions that I have a hard time with simplifying algebraic equations.
For this equation, I assumed we would first put the equation into easier to read form:
$(k+1)(k+1)(k+1)$
Then apply FOIL to the first two:
EDIT: $(k^2+2k+1)(k+1)$
Although this is where I'm stuck. Any help is appreciated.
With all your help my answer is:
$k^3+3k^2+3k+1$
Thanks.
| Use the distributive law (after correcting your first multiplication):
$$(k^2+2k+1)(\color{red}{k}+\color{blue}{1})=(k^2+2k+1)\cdot\color{red}{k}+(k^2+2k+1)\cdot\color{blue}{1}\;.$$
Now $(k^2+2k+1)\cdot k=k^3+2k^2+k$, and $(k^2+2k+1)\cdot 1=k^2+2k+1$, so all that’s left is to add the two polynomials:
$$\begin{align*}
(k^3+2k^2+k)+(k^2+2k+1)&=k^3+(2k^2+k^2)+(k+2k)+1\\
&=k^3+3k^2+3k+1\;.
\end{align*}$$
You can also organize this calculation like a pencil-and-paper multiplication:
$$\begin{array}{r}
&&k^2&+&2k&+&1\\
&&&&k&+&1\\ \hline
&&k^2&+&2k&+&1\\
k^3&+&2k^2&+&k\\ \hline
k^2&+&3k^2&+&3k&+&1
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Concecutive last zeroes in expansion of $100!$
Possible Duplicate:
Highest power of a prime $p$ dividing $N!$
In decimal form, the number $100!$ ends in how many consecutive zeroes. I am thinking of the factorization of $100!$ but I am stuck. I try to count them and since there are 10, 20, 30,..., 100, there are at least 11 zeros. How should I proceed.
| \begin{align}
2\text{ goes into }100 & & 50\text{ times} \\
2^2\text{ goes into }100 & & 25\text{ times} \\
2^3\text{ goes into }100 & & 12\text{ times} \\
2^4\text{ goes into }100 & & 6\text{ times} \\
2^5\text{ goes into }100 & & 2\text{ times} \\
2^6\text{ goes into }100 & & 1\text{ time} \\
\end{align}
$$
50+25+12+6+2+1 = 96.
$$
Thus $2^{96}$ divides $100!$ and $2^{97}$ does not.
\begin{align}
5\text{ goes into }100 & & 20\text{ times} \\
5^2\text{ goes into }100 & & 4\text{ times}
\end{align}
Thus $5^{24}$ divides $100!$ and $5^{25}$ does not.
$$\min\{96,24\}=24.$$
So $(2\cdot5)^{24}$ divides $100!$ and $(2\cdot5)^{25}$ does not.
| {
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"question_score": "3",
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How to find solutions of $x^2-3y^2=-2$? According to MathWorld,
Pentagonal Triangular Number: A number which is simultaneously a pentagonal number $P_n$ and triangular number $T_m$. Such numbers exist when
$$\frac{1}{2}n(3n-1)=\frac{1}{2}m(m+1).$$
Completing the square gives
$$(6n-1)^2-3(2m+1)^2=-2.$$
Substituting $x=6n-1$ and $y=2m+1$ gives the Pell-like quadratic Diophantine equation
$$x^2-3y^2=-2,$$
which has solutions $(x,y)=(5,3),(19,11),(71,41),(265,153), \ldots$.
However, it does not state how these solutions for $(x,y)$ were obtained.
I know that the solution $(5,3)$ can be obtained by observing that $1$ is both a pentagonal and a triangular number.
Does obtaining the other solutions simply involve trial-and-error? Or is there a way to obtain these solutions?
| $$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
-1 \\
1
\end{array}
\right) \; = \;
\left( \begin{array}{c}
1 \\
1
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
1 \\
1
\end{array}
\right) \; = \;
\left( \begin{array}{c}
5 \\
3
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
5 \\
3
\end{array}
\right) \; = \;
\left( \begin{array}{c}
19 \\
11
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
19 \\
11
\end{array}
\right) \; = \;
\left( \begin{array}{c}
71 \\
41
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
71 \\
41
\end{array}
\right) \; = \;
\left( \begin{array}{c}
265 \\
153
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
265 \\
153
\end{array}
\right) \; = \;
\left( \begin{array}{c}
989 \\
571
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
\left( \begin{array}{c}
989 \\
571
\end{array}
\right) \; = \;
\left( \begin{array}{c}
3691 \\
2131
\end{array}
\right),
$$
EDIT, March 2016: From the stuff with the matrix above, we can use the Cayley-Hamilton theorem to give separate linear recurrences for $x$ and for $y.$ Just these:
$$ x_{k+2} = 4 x_{k+1} - x_k, $$
$$ y_{k+2} = 4 y_{k+1} - y_k. $$ The $x$ sequence is
$$ 1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, \ldots $$ while the $y$ sequence is
$$ 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, \ldots $$
Well. The theorem of Lagrange is that all values of the quadratic form (that are primitively represented) occur as output of the neighboring forms method, the same as doing continued fractions, if they are below $\frac{1}{2} \; \sqrt \Delta$ in absolute value, where in this case $\Delta = 12.$ So half the square root of that is $\sqrt 3,$ and $2$ is larger than this. This means that, while $-2$ is permitted to show up by the continued fraction method, it is possible that unexpected representations may occur. However, one may check with Conway's topograph method from The Sensual Quadratic Form and confirm that all appearances of $-2$ are along the "river" itself, meaning the simplest possible collection, as I illustrate with the matrix multiplications above. For your viewing pleasure, the topograph for $x^2 - 3 y^2,$ with a fair amount of detail:
=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=
Oh, well. The $-2$ at coordinates $(5,3)$ goes in the lower right open space, while the $-2$ at coordinates $(-5,3)$ goes in the lower left open space. If you think about it long enough, each edge in the infinite tree, including the little blue numbered arrow and the value on either side, is an indefinite quadratic form equivalent to $\langle 1,0,-3 \rangle,$ but is also an element in $PSL_2 \mathbb Z$ given by a little 2 by 2 matrix using the two column vectors in green.
Note that the automorph $$
\left( \begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}
\right)
$$ is visible as a pair of column vectors corresponding once again to $\langle 1,0,-3 \rangle,$ as, indeed, it must.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Find monic gcd($x^4+x^3+x+1$, $x^6+x^5+x^4+x+1$) in $\mathbb Z_{2}$ My working so far using the Euclidean algorithm and polynomial long division (which I won't fully show here)
$x^6+x^5+x^4x+1$ = $(x^2+1) \times (x^4+x^3+x+1) + (-2x^3-x^2)$
and $(-2x^3-x^2) \equiv (x^2)$ in $\mathbb Z_{2}[x]$
So a non-monic gcd would be $x^2$? Should I keep going?
$x^4+x^3+x+1 = (x^2+x) \times (x^2) + (x+1)$
I'm a little confused now - when do I stop?
The answer is supposed to be 1 but I'm not seeing where 1 comes from
| As wj32 stated, all I needed to do was to keep going one more step:
$x^2=(x+1)(x+1)+1$
$(x+1)^2$ is $(x^2+2x+1)$ but of course $2x \equiv 0$ in $\mathbb Z_{2}$
| {
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Integration of a trigonometric function I am having some difficulties with the calculation of the following integral. Can somebody help me please?
$$\int \frac{dx}{1+a\cos x},\text{ for }0<a<1$$
Thank you in advance
| Substitute, $t = \tan \left(\dfrac x2\right)$. So $x = 2\tan^{-1}t$ and $dx = \dfrac{2dt}{1+t^2}$. And the integral becomes,
$$\begin{align*}\int\dfrac{\dfrac{2}{1+t^2}}{1+\dfrac{a(1-t^2)}{1+t^2}}dt
&=\dfrac2{(1+a)}\int\dfrac1{1+\left[t\dfrac{\sqrt{1-a}}{\sqrt{1+a}}\right]^2}dt
&\color{blue}{u =\left[t\frac{\sqrt{1-a}}{\sqrt{1+a}}\right]\Rightarrow dt = \frac{\sqrt{1+a}}{\sqrt{1-a}}du }\\
&=\dfrac2{(1+a)}\frac{\sqrt{1+a}}{\sqrt{1-a}}\int\dfrac1{1+u^2}du
&\color{blue}{\int\dfrac1{1+u^2}du= \tan^{-1}u}\\
\\&=\dfrac2{(1+a)}\frac{\sqrt{1+a}}{\sqrt{1-a}}\tan^{-1}u\\\\
&=\dfrac2{\sqrt{1-a^2}}\tan^{-1}u&\color{blue}{u =\left[t\frac{\sqrt{1-a}}{\sqrt{1+a}}\right]}\\ \\
&=\dfrac{2\tan^{-1}\left[\frac{\sqrt{1-a}}{\sqrt{1+a}}\cdot\tan \left(\dfrac x2\right)\right]}{\sqrt{1-a^2}}\\\\
\end{align*}$$
$$\displaystyle\Large\therefore \boxed{\int \dfrac1{1+acosx}dx =\dfrac{2\tan^{-1}\left[\frac{\sqrt{1-a}}{\sqrt{1+a}}\cdot\tan \left(\dfrac x2\right)\right]}{\sqrt{1-a^2}} }$$
| {
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"source": "stackexchange",
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some problems related to primes I would like to learn the following:
a) Prove that the equation $1 + x + x^2 = py$ has integer solutions for infinitely many primes $p$.
b) Twin primes are those difference by 2. Show that 5 is the only prime belonging to two such pairs. Show also that there is a one-to-one correspondence between twin primes and positive integers n such that $d(n^2-1) = 4$, where as above $d(k)$ stands for the number of divisors of $k$.
| a) The proof is a modification of the "Euclid" proof that there are infinitely many primes. Let $p_1=3$. Putting $x=1$ and $y=1$, we see that $1+x+x^2=py$ has a solution.
Now suppose that we have found $n$ primes $p_1,p_2,\dots,p_n$ such that $1+x+x^2=p_iy$ has a solution for any $i$ from $1$ to $n$. We exhibit a new prime $p_{n+1}$ such that $1+x+x^2=p_{n+1}y$ has a solution.
Consider the number $1+(p_1p_2\cdots p_n)+(p_1p_2\cdots p_n)^2$. This is an integer $\gt 1$, so has a prime divisor $p_{n+1}$. Any such $p_{n+1}$ must be distinct from $p_1,p_2,\dots,p_n$. This is because if $p_{n+1}=p_i$, where $1\le i\le n$, then $p_i$ divides $(p_1p_2\cdots p_n)+(p_1p_2\cdots p_n)^2$, so cannot divide $1+ (p_1p_2\cdots p_n)+(p_1p_2\cdots p_n)^2$.
Thus we have found a new prime $p_{n+1}$ such that $1+x+x^2=p_{n+1}y$ has a solution.
b) The number $5$ belongs to the pairs $(3,5)$ and $(5,7)$. It is clear that $2$ does not belong to any pair, and $3$ belongs to only $1$ pair. We show that any prime $p\ge 7$ cannot belong to more than one pair. Suppose to the contrary that $p$ does. Then $p-2$, $p$, and $p+2$ are prime. Note that if $x$ is any integer, then one of $x-2$, $x$, or $x+2$ is divisible by $3$. A number divisible by $3$ and greater than $3$ cannot be prime.
For the second part of the question, we ask when $d(n^2-1)=4$. If $n$ is even, then $n-1$ and $n+1$ are relatively prime, so $d((n-1)(n+1))=d(n-1)d(n+1)$. But $1$ is the only $k$ such that $d(k)=1$. So we must have $d(n-1)=d(n+1)=2$. That forces the pair $(n-1,n+1)$ to be a pair of twin primes.
Now examine the case $n$ odd. Then $(n-1)(n+1)$ is divisible by $8$, so has the divisors $1$, $2$, $4$ and $8$. Since $d(n^2-1)=4$, it can have no others. Thus $n^2-1=8$, giving $n=3$.
So the set $A$ of numbers $n-1$ such that $d(n^2-1)=4$ consist of $2$ plus the smaller primes in a twin prime pair. This set can be easily put in one to one correspondence with the set $B$ of smaller primes in twin prime pairs. Just list the two sets as $a_1,a_2,\dots$ and $b_1,b_2,\dots$ and map $a_i$ to $b_i$.
But the one-to-one correspondence bit has absolutely nothing to do with twin primes. For any two infinite sets of natural numbers can be put in one to one correspondence. So to prove there is a one to one correspondence, we need not have done any analysis of the $n$ such that $d(n^2-1)=4$.
| {
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How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives:
\begin{align*}
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
3^7 \mod 5 = 2 \\\\
3^8 \mod 5 = 1 \\
3^9 \mod 5 = 3 \\
3^{10} \mod 5 = 4 \\
3^{11} \mod 5 = 2 \\\\
3^{12} \mod 5 = 1 \\
3^{13} \mod 5 = 3 \\
...
\end{align*}
As I see, there's a period $1, 3, 4, 2$. And intuitively, it's easy to predict that $3^{45357} \mod 5 = 3$. Unfortunately, I'm almost unfamilar with modular math. Could you give me strict analytic explaination of this thing?
| The pattern you've seen here is that $3^{4k+1}\;\text{mod}\:5=3$ for integers $k\ge 0$. You can prove this inductively from the base case, and the fact that $3^4=81$ is equal to $1$ modulo $5$. Hence, since $4\cdot 11339+1=45357$, then your prediction holds.
| {
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Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
| $\displaystyle \lim_{n \rightarrow \infty} \left( \sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 + 1} \right) = \lim_{n \rightarrow \infty} \left\{
n \left[
\left(
1 + \frac 1n
\right)^{\frac 13} - \left(
1 + \frac 1{n^3}
\right)^{\frac 13}
\right]
\right\} = \\
\displaystyle \lim_{n \rightarrow \infty} \left[
n \left(
1 + \frac 1{3n} - 1 - \frac 1{3n^3}
\right)
\right] = \lim_{n \rightarrow \infty} \left(
\frac 13 - \frac 1{3n^2}
\right) = \frac 13$
| {
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Geometry Prove - two perpendicular lines in a circle In a circle of radius r, two lines (AB and CD) are perpendicular to each other and meet at X.
Show that:
| Let $O$ be the center of the circle, and let $M$ and $N$ be the midpoints of segments $AB$ and $CD$, respectively. Write $a:=|XA|$, $b:=|XB|$, $c:=|XC|$, $d:=|XD|$, $m := |OM|$, $n:=|ON|$.
Then (noting that $OMXN$ is a rectangle),
$$\begin{align}
|AM| = \frac{1}{2}(a+b) &\qquad |ON|=|MX| = \frac{1}{2}|a-b| \\[5pt]
|CN| = \frac{1}{2}(c+d) &\qquad |OM|=|NX| = \frac{1}{2}|c-d|
\end{align}$$
Now, $\triangle AMO$ and $\triangle CNO$ are right triangles with hypotenuses $|AO| = |CO| = r$, so that
$$\begin{align}
r^2 &= |AM|^2 + |OM|^2 = \frac{1}{4}(a+b)^2 + \frac{1}{4}(c-d)^2 \\[5pt]
r^2 &= |CN|^2 + |ON|^2 = \frac{1}{4}(c+d)^2 + \frac{1}{4}(a-b)^2
\end{align}$$
Adding these equations causes the "$ab$" and "$cd$" terms to cancel; multiplying through by $2$ gives
$$4r^2 = a^2 + b^2 + c^2 + d^2$$
| {
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Find $x$ such that $\arctan(3/2)+\dots=\arctan x$ Find $x$ such that
$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$
| You may use the fact if $x > 0$, the argument of the complex number $x+iy$ is $\tan^{-1} \left(\frac{x}{y}\right)$. So if you compute the product
$$(2+3i)(4+5i)(2-5i)(3-8i) = 920 - 531i$$
you get that
$$\tan^{-1}\left(\frac{3}{2}\right) + \tan^{-1}\left(\frac{5}{4}\right) + \tan^{-1}\left(-\frac{5}{2}\right) + \tan^{-1}\left(-\frac{8}{3}\right) \equiv \tan^{-1}\left(-\frac{531}{920}\right) \mod 2\pi$$
By computing both side on a calculator, you can check this actually an equality.
| {
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Find the maxium $\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$ 1) $a, b, c$ are triangle edges's length such that $abc = 1$. Find max:
$$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$$
My idea: $$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}=\frac{abc}{a^3(b+c)}+\frac{abc}{b^{3}(c+a)}+\frac{abc}{c^3(a+b)}$$
Then use AM-GM ? I just can find min: $$\frac{1}{a^{3}\left ( b+c \right )}+\frac{1}{b^{3}\left ( c+a \right )}+\frac{1}{c^{3}\left ( a+b \right )}\geq \frac{3}{2}$$
2) Find for $x$, $y$, $z$ such that $\left\{\begin{matrix}
xy + 2(x+y)=0\\
\ yz + 2(y+z)=-3\\
zx + 2(z+x)=5
\end{matrix}\right.$
(Some one should edit my post: correct grammar...)
| For $0 < \epsilon < 1$ and $a = \epsilon^2$, $b = c = \frac 1 \epsilon$ we have
$$
abc = 1\\
a = \epsilon^2 \leq \frac 2 \epsilon = b + c \\
b = \frac 1 \epsilon \leq \frac 1 \epsilon + \epsilon^2 = a + c \\
c = \frac 1 \epsilon \leq \frac 1 \epsilon + \epsilon^2 = a + b
$$
All question conditions are satisfied.
Now
$$
I := \frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b} \geq
\frac{bc}{a^{2}b+a^{2}c} = \frac 1 {a^3(b +c)} = \frac 1 {2\epsilon^5}
$$
so for arbitrarily small $\epsilon$ we get arbitrarily great $I$.
| {
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Is this series convergent or divergent? Kindly asking, what can I do about series
$$
\left(\frac{1}{3}\right)^2+\left(\frac{1\times 4}{3\times 6}\right)^2+\left(\frac{1\times 4\times 7}{3\times 6\times 9}\right)^2+...+\left(\frac{1\times 4\times 7\times...\times (3n-2)}{3\times 6\times 9\times...\times3n}\right)^2+...$$
Indeed, the ratio test fails. Thank you.
| Let $a_n$ be the $n$th term.
The ratio of successive terms is
$$\frac{a_{n+1}}{a_n} = \frac{(3n+1)^2}{9(n+1)^2} = 1 - \frac{4}{3n} + O\left(\frac{1}{n^2}\right).$$
Thus, the series converges by Raabe's test, which tells us that if
$$\left|\frac{a_{n+1}}{a_{n}}\right| \sim 1 - \frac{s}{n} \hspace{5ex}(n\to\infty),$$
then the series converges absolutely if $s>1$, diverges if $s<1$, and may converge or diverge if $s=1$.
| {
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Factor $1+x+x^2+x^3+...+x^{14}$ In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving
$1+x+x^2+x^3 = 0 \to $
$1+x(1+x+x^2) = 0 \to $
$x(1+x+x^2) = -1$
which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$.
Now I'm asked to factor $1+x+x^2+x^3+...+x^{14}$ for $x \in \mathbb{R}$ and I'm a bit stuck. Again, we have the implication $x(1+x+x^2+...+x^{13}) = -1$, for which $x=-1$ is a solution, so again we have a factor $x+1$. But now I cannot apply guesswork to determining the rest of the factors, so I feel there is some kind of conclusion I can draw about the powers (perhaps their parity) to solve this problem?
| Using the fact
$$ x^n-1 = (x-1)(1+x+x^2+\dots+x^{n-1}),$$
our polynomial can be written in the form
$$ 1+x+x^2+x^3+…+x^{14} = \frac{x^{15}-1}{x-1}. $$
Now, we can find the roots of $ x^{15} - 1 $ using the complex variable tecniques
$$ x^{15}=1=e^{i2k\pi} \implies x = e^{\frac{i2k\pi}{15}},\quad k=0,1,2,\dots,14. $$
So, our polynomial can be written as
$$ 1+x+x^2+\dots+x^{14} =(x-e^{\frac{i2\pi}{15}})(x-e^{\frac{i4\pi}{15}})\dots (x-e^{\frac{i28\pi}{15}})$$
$$ = \Pi_{m=1}^{14}(x-e^{\frac{i2m\pi}{15}}). $$
Note that,
$$ e^{i\theta}= cos(\theta)+i\sin(\theta) $$
$$ e^{2k\pi i} = 1,\quad k\in \mathbb{Z}. $$
| {
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Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$
Possible Duplicate:
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?
How can I calculate precise value of that series: $\sum\limits_{i=0}^{n-1} i2^i$ ?
So far, I tried to differentiate the $ \sum\limits_{i=0}^{n} 2^i = 2^{i-1} - 1 $ series, but by result $2^n(n+2)$ isn't correct according to Wolfram ($2^n (n-2)+2$).
| Here is another way to do this
Consider the polynomial $$\begin{align}&P(x)=\sum^{n-1}_{i=0} \ i\ \cdot \ x^i= 0x^0 +1x^1+2x^2+3x^3+\cdots +(n-1)\ x^{n-1}\\&Q(x)=\cfrac{P(x)}{x}=1x^0+2x^1+3x^2+\cdots+(n-1)\ x^{n-2}, \quad \quad x \ne 0\\ &\int Q(x)\ \text d x = cx^0+x^1+x^2+x^3+\cdots+ x^{n-1}=c+\sum^{n-1}_{i=1}x^i \\ &\text{by geometric series, we have} \int Q(x)\ \text d x =c+\cfrac{x(1-x^{n-1})}{1-x}\\ &\text{we then differentiate back to have } Q(x)= \cfrac{(n-1) x^{n+1}-n x^n+x}{x(x-1)^2 }\\ &\text{and at last } P(x)=x\cdot Q(x) =\cfrac{(n-1) x^{n+1}-n x^n+x}{(x-1)^2 }\\ &\text{we compute $P(2)$ to get the desired result }\\ &P(2)=\cfrac{(n-1) 2^{n+1}-n\cdot 2^n+2}{(2-1)^2 }=(n-1) 2^{n+1}-n\cdot 2^n+2\ \ =\ \ 2^n(2(n-1)-n)+2\\ &\sum\limits_{i=0}^{n-1} i\cdot 2^i=P(2)=2^n(n-2) +2 \end{align}$$
| {
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Combinations of 5-digit numbers I'm just studying for finals here and wanted some confirmation that I'm doing things right.
So if we have 5-digit decimal numbers, there are:
*
*$10^5 - 9*8*7*6*5$ numbers with no $0$,
*$10^5 - 9*8*7*6*5$ numbers with no $1$, and
*$10^5-2(9*8*7*6*5)+8*7*6*5*4$ numbers with no $0$ or $1$.
Am I correct? Your help is so greatly appreciated.
| *
*A $5$-digit number with no $0$ can have any of the other $9$ digits in each of its $5$ places, so there are $9^5$ such numbers. Your calculation would be right if the digits of the number were required to be distinct, but they’re not. The number of $5$-digit numbers with at least one $0$ is therefore $9\cdot10^5-9^5=9\left(10^4-9^4\right)$.
*A $5$-digit with no $1$ can have any of the $8$ digits $2,3,4,5,6,7,8,9$ in the first position, and any of the $9$ digits other than $1$ in each of the remaining $4$ positions, so there are $8\cdot9^4$ such numbers. There are then $9\cdot10^4-8\cdot9^4$ $5$-digit numbers with at least one $1$.
*A $5$-digit with no $0$ and no $1$ can have any of the other $8$ digits in each of its $5$ places, so there are $8^5$ such numbers. The number of $5$-digit numbers with at least one $0$ and at least one $1$ is then $9\cdot10^4-9^5-8\cdot9^4+8^5=9\cdot10^4-17\cdot9^4+8^5$.
Note: I’m assuming throughout that a number is not permitted to begin with $0$.
| {
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Find the direct expression of $f(t)$ I am trying to find the direct expression of the function $f(t)$ given by $$f(t)= \int_{1}^\infty \frac{\arctan (tx)}{x^2\sqrt{x^2-1}}dx$$
It's hard for me to calculate the integration directly.Should I try the method of interchanging $\frac{d}{dx}$ with$\int$ or$\int$ with $\int$ ? Thanks for help.
| Presumably this was intended to be done with pen and paper, so here is what they might have had in mind. First note that convergence is excellent, so we may differentiate with respect to $t$ to obtain
$$ f'(t) = \int_1^\infty \frac{x}{1+t^2x^2} \frac{1}{x^2\sqrt{x^2-1}} dx.$$
Now let $x^2-1 = u^2$ so that $x\, dx = u\, du$ and
$$ f'(t) = \int_0^\infty \frac{u}{1+t^2(u^2+1)} \frac{1}{(u^2+1)\sqrt{u^2}} du =
\int_0^\infty \frac{1}{1+t^2(u^2+1)} \frac{1}{u^2+1} du.$$
Putting $$ g(u) = \frac{1}{2} \frac{1}{1+t^2(u^2+1)} \frac{1}{u^2+1}
\quad \text{we seek to evaluate} \quad
\int_{-\infty}^\infty g(u) du.$$
This may be done with the Cauchy Residue theorem applied to a half-circle contour in the upper half plane of radius $R$ (the contribution of the circle vanishes in the limit), giving
$$ 2 \pi i
\left(\operatorname{Res}_{u=i} g(u) +
\operatorname{Res}_{u=\sqrt{-\frac{1}{t^2}-1}} g(u) \right).$$
These poles are both simple and we obtain
$$ 2\pi i \left(- \frac{1}{4} i - \frac{1}{4} \frac{1}{\sqrt{-\frac{1}{t^2}-1}}\right) =
\frac{\pi}{2} - i \frac{\pi}{2}\frac{1}{\sqrt{-\frac{1}{t^2}-1}} =
\frac{\pi}{2} - \frac{\pi}{2}\frac{t}{\sqrt{t^2+1}} \quad( t\geqslant 0)$$
Integrating we have
$$ f(t) = C + \frac{\pi}{2} t - \frac{\pi}{2} \sqrt{t^2+1}.$$
Finally, to determine $C$, note that $f(0) = 0$ or
$$ 0 = C - \frac{\pi}{2} $$ and hence $C = \frac{\pi}{2}$ giving the result
$$ f(t) = \frac{\pi}{2} ( 1 + t - \sqrt{t^2+1} ).$$
When $t<0,$same calculation shows
$$ f(t)= \frac{\pi}{2}(-1 + t +\sqrt{t^2+1})$$
| {
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea.
Thanks:)
| Here $\prod_{cyc}$ refers to the cyclic product of $x, y, z$.
Let $a=x^2, b=y^2, c=z^2$ for positives $x, y, z$. Then by the Cauchy-Schwarz inequality, we have:
$$LHS^2=\prod_{cyc}[(x^4+y^2z^2)(x^2z^2+y^4)]\ge \prod_{cyc}(x^3z+y^3z)^2=x^2y^2z^2\prod_{cyc}(x^3+y^3)^2$$
Then, by Power-Mean and AM-GM:
$$x^2y^2z^2\prod_{cyc}(x^3+y^3)^2\ge x^2y^2z^2\prod_{cyc}[\frac{(x^2+y^2)^3}{2}]\ge x^2y^2z^2\prod_{cyc}[(x^2+y^2)^2xy]=RHS^2$$
Thus $LHS^2\ge RHS^2$, and both sides are positive so $LHS\ge RHS$ as desired.
Sidenote: There is also a proof by direct expansion in $a, b, c$: upon expanding and rearranging the inequality becomes $\sum_{cyc}(a-b)^2(\frac{a+b}{2})(c^3+abc)\ge 0$, which is clear.
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"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Algebraic manipulation of binomial theorem Prove, by algebraic manipulation, that:
\[ {{2n} \choose {n}} + {{2n} \choose {n+1}}={1\over2} {{2n+2} \choose {n+1}} \]
| $$\binom m r+\binom m{r+1}=\frac{m!}{(m-r)! r!}+\frac{m!}{(m-r-1)! (r+1)!}=\frac{r+1}{m+1}\frac{(m+1)!}{\{(m+1)-(r+1)\}! (r+1)!}+\frac{m-r}{m+1}\frac{(m+1)!}{\{m-(r+1)\}! (r+1)!}$$
$$=\binom{m+1}{r+1}\cdot\left(\frac{r+1}{m+1}+\frac{m-r}{m+1}\right)=\binom{m+1}{r+1}$$
Putting $m=2n,r=n,$
$$\binom {2n} n+\binom {2n}{n+1}=\binom{2n+1}{n+1}=\frac{(2n+1)!}{n!(n+1)!}=\frac1 2 \frac{(2n+2)\cdot (2n+1)!}{(n+1)\cdot n!(n+1)!}=\frac1 2\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac1 2\binom{2n+2}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded.
Thank you!
| Let $\arctan x=y\Leftrightarrow x=\tan y$. Then,
$$\sin^2 y+\cos^2 y=1\Leftrightarrow \tan^2 y+1=\frac{1}{\cos^2 y}\Leftrightarrow \frac{1}{x^2+1}=1-\sin^2 y\Leftrightarrow \sin^2 y=\frac{x^2}{x^2+1}$$
and so $$\sin y= \frac{x}{\sqrt{1+x^2}}\Rightarrow \arctan x=y=\arcsin \frac{x}{\sqrt{1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Prove that the equation $y^2=x^3-73$ has no integer solutions Prove that there are no integers $x,y$ such that $y^2=x^3-73$.
Thank you.
| You should send the equation into the congruence modulo $4$, for that according to Division Algorithm, (or Complete Residue System modulo 4) we have 4 possibilities for arbitrary number $x$.
$$ \text{If} \ x \equiv 0 \ \text{or} \ 2 \pmod{4} \ \ \text{then} \ \ y^2\equiv 0 - 73 \equiv 3 \equiv -1 \pmod{4} $$
So as we know in the congruence modulo $4$, it's not possible for a squared number (a number raised to the power of $2$) to be $-1$, all squared numbers are congruent to either $1$ or $0$ modulo $4$.
$$\text{If} \ \ x\equiv 3 \pmod{4}\ \ \text{then} \ \ y^2\equiv -1 -73\equiv2 \pmod{4} \ \ $$
And again it's not possible. The last part is $ x\equiv1\pmod{4} $ which is a little tricky! Look upon our equation and increase both side of it by $100$, the new equation will be: $$y^2+100=x^3+27 $$ By some algebraic factorising reformulate it like this:
$$y^2+100=(x+3)(x^2-3x+9)\qquad (1)$$
If we take the single factor $(x^2-3x+9)$ and notice that $$x^2-3x+9 \equiv 1-3+9\equiv-1 \pmod{4}$$
Our game is started from now on, there must be an odd prime like $q$ in which, $q\equiv -1 \pmod{4}$ and $q \mid x^2-3x+9$.
Actually that's very convenient and there is an odd prime like $q$ because if we show $x^2-3x+9$ as the product of $p_1,p_2,...,p_k $ (note that it's possible for a pair or more than two of them like $p_i$ and $p_j$ to have $p_i=p_j$ and this will cover $p_1^\alpha p_2^\beta ...$ the true Unique-Prime-Factorization representation of numbers according to Fundamental Theorem of Arithmetic) and neither every $p_i$ can be congruent to $1$ modulo $4$ nor the number of primes which are congruent to $-1$ is even.
We show that the second part must be true, first we know every odd prime should be congruent to either $1$ or $-1$ modulo $4$ and also:
$$ x^2-3x+9 = p_1p_2...p_k \equiv 1\times 1\times ... (-1)\times (-1)\times ...\equiv (-1)^n \pmod{4}$$ So it's obvious for $n$ to be odd. Let's get back to our found $q$ and our customized equation (1).
$$ y^2+100\equiv (x+3)(x^2-3x+9) \equiv 0 \pmod{q}$$
Therefore:
$$y^2 \equiv -100 \pmod{q}$$
$100=2^2\times 5^2$ and both $2$ and $5$ are primes and their remainders when they're divided by 4 are 2 and 1. So $2 \nmid q, q\nmid 5$ or better say $\text{gcd}(5,q)=\text{gcd}(2,q)=\text{gcd}(2\times 5,q)=1$, therefore there is an inverse of $10$ module $q$ and let denote it with $10\ '$, and multiply both sides of the congruence by $(10\ ')^2$:
$$ (10\ ')^2y^2\equiv -1 \pmod{q} $$
And we turn $10\ 'y$ into a new variable like $z$, therefore:
$$z^2\equiv -1 \pmod{q}$$ This congruence has solution if and only if $q\equiv 1 \pmod{4}$ (according to a theorem which is resulted in by famous Wilson's Theorem), but our poor $q$ is congruent to $-1$ modulo $4$, and eventually we proved that $y^2=x^3-73$ has no solution among integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| We have $\gcd(a+b,a^2-ab+b^2)=\gcd(a+b,3ab)$.
If $p$ is a prime with $p|\gcd(a+b,3ab)$, then $p=3$ or $p|a$ or $p|b$.
If $p|a$, then $p\nmid b$, hence $p\nmid a+b$. Similarly, if $p|b$, then $b\nmid a+b$.
Thus we conclude $p=3$.
But then if $9|\gcd(a+b,3ab)$, clearly $3|a$ or $3|b$ and again $3\nmid a+b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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simplify summation of factorial (random walk) I suspect that the expression
$$\sum_{n=0}^N \frac{(N-2n)^2}{n!(N-n)!}$$
simplifies to
$$\frac{2^N}{(N-1)!}$$
But I cannot find the intermediate steps. Can someone give me a hint how I can deduce this result?
(The expression comes up when calculating the average final position after a random walk of $N$ steps.)
| In the course of this post, you can see how to prove that $$\sum_{n=0}^Nn\binom{N}{n}=N2^{N-1}.\tag{1}$$ Two other nice facts we'll use are $$\sum_{n=0}^N\binom{N}{n}=2^N\tag{2}$$ and $$\binom{N+1}{n}=\binom{N}{n}+\binom{N}{n-1}.\tag{3}$$ From these facts, along with $\binom{N}{N+1}=0$, we see that $$\begin{align}\sum_{n=0}^{N+1}n^2\binom{N+1}{n} &= \sum_{n=0}^{N+1}n^2\binom{N}{n} + \sum_{n=0}^{N+1}n^2\binom{N}{n-1}\\ &= \sum_{n=0}^Nn^2\binom{N}{n} + \sum_{n=1}^{N+1}n^2\binom{N}{n-1}\\ &= \sum_{n=0}^Nn^2\binom{N}{n} + \sum_{n=0}^N(n+1)^2\binom{N}{n}\\ &= 2\sum_{n=0}^Nn^2\binom{N}{n} + 2\sum_{n=0}^Nn\binom{N}{n}+\sum_{n=0}^N\binom{N}{n}\\ &= 2\sum_{n=0}^Nn^2\binom{N}{n} + N2^N+2^N\\ &= (N+1)2^N+2\sum_{n=0}^Nn^2\binom{N}{n}.\end{align}$$ Note that $\sum_{n=0}^0n^2\binom{0}{n}=0=0(0+1)2^{0-2}$, and if $\sum_{n=0}^Nn^2\binom{N}{n}=N(N+1)2^{N-2}$, then by the above work, $$\begin{align}\sum_{n=0}^{N+1}n^2\binom{N+1}{n} &= (N+1)2^N+2\sum_{n=0}^Nn^2\binom{N}{n}\\ &= (N+1)2^N+N(N+1)2^{N-1}\\ &= (N+1)(N+2)2^{N-1},\end{align}$$ so inductively, we have the identity $$\sum_{n=0}^Nn^2\binom{N}{n}=N(N+1)2^{N-2}\tag{4}$$ for all $N\geq 0$.
Now, using $(1)$ $(2)$ and $(4)$, we can see that $$\begin{align}N!\sum_{n=0}^N\frac{(N-2n)^2}{n!(N-n)!} &= \sum_{n=0}^N(N-2n)^2\binom{N}{n}\\ &= N^2\sum_{n=0}^N\binom{N}{n}-4N\sum_{n=0}^Nn\binom{N}{n}+4\sum_{n=0}^Nn^2\binom{N}{n}\\ &= N^22^N-N^22^{N+1}+N(N+1)2^N\\ &= N2^N,\end{align}$$ whence dividing by $N!$ yields $$\sum_{n=0}^N\frac{(N-2n)^2}{n!(N-n)!} = \frac{2^N}{(N-1)!},$$ as desired.
| {
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"url": "https://math.stackexchange.com/questions/258882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
| One reason the result looks difficult is because it seems to involve two variables.
But if you divide both sides by $y^n$, it can be rewritten as
$$ \eqalign{\frac{x^n}{y^n} - 1 &= \left(\frac{x - y}{y}\right) \left( \frac{x^{n-1} + x^{n-2} y + \ldots + x y^{n-2} + y^{n-1}}{y^{n-1}}\right)\cr
&= \left(\frac{x}{y} - 1 \right) \left(\frac{x^{n-1}}{y^{n-1}} + \frac{x^{n-2}}{y^{n-2}} + \ldots + \frac{x}{y} + 1 \right)} $$
With $x/y = r$, this says
$$ r^n - 1 = (r - 1)(r^{n-1} + r^{n-2} + \ldots + r + 1)$$
Now $$ r (r^{n-1} + r^{n-2} + \ldots + r + 1) = r^{n} + r^{n-1} + \ldots + r^2 + r$$
Subtract $1(r^{n-1} + \ldots + r + 1) = r^{n-1} + \ldots + r + 1$ and you see that the terms in $r, r^2, \ldots, r^{n-1}$ all cancel, leaving
$$ (r - 1)(r^{n-1} + r^{n-2} + \ldots + r + 1) = r^n - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/260362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with Inequality Given that $x, y, z$ are nonnegative real numbers such that :
$$x^2 + y^2 + z^2 + xyz = 4$$
Prove that
$0 ≤ xy + yz + zx − xyz ≤ 2$
| use this $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
and $$x^2+y^2+z^2+xyz=4$$
then we set
$$x=2\cos{A},y=2\cos{B},z=2\cos{C}$$
$$\Longleftrightarrow 0\le 4(\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{A}\cos{C})-8\cos{A}\cos{B}\cos{C}\le 2$$
and since
$$\cos{A}\cos{B}\cos{C}=\dfrac{s^2-(2R+r)^2}{4R^2},\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{C}\cos{A}=\dfrac{s^2+r^2-4R^2}{4R^2}$$
then
$$\Longleftrightarrow 0\le\dfrac{s^2+r^2-4R^2}{R^2}-\dfrac{2}{R^2}\left(s^2-(2R+r)^2\right)\le 2$$
and left hand it suffices that
$$s^2+r^2-4R^2-2s^2+2(2R+r)^2\ge 0$$
$$\Longleftrightarrow 3r^2+8Rr+4R^2\ge s^2$$
use the Gerrseten inequality:
$$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$
then
$$\Longleftrightarrow 4R^2+4Rr+3r^2\le 3r^2+4R^2+8Rr$$
$$\Longleftrightarrow 4Rr\ge 0$$is obvious.
and the Right Hand,
$$\Longleftrightarrow s^2+r^2-4R^2-2(s^2-(2R+r)^2)\le 2R^2 $$
$$\Longleftrightarrow 2R^2+8Rr+3r^2\le s^2$$
since
$$s^2\ge r(16R-5r)$$,
it follow that
$$r(16R-5r)\ge 2R^2+8Rr+3r^2$$
$$\Longleftrightarrow (R-2r)^2\ge 0$$
is obivous.
| {
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"url": "https://math.stackexchange.com/questions/262608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A property of finite field of order $2^n$ Suppose $a$ and $b$ are elements of a finite field of order $2^n$ with $n$ odd and $a^2+ab+b^2=0$. Is it necessary that both $a$ and $b$ must be zero ?
I understand that the field has characteristic $2$ but don't know how to use the fact that $n$ is odd, please help.
| If $a=b$,
then $a^2+ab+b^2=3a^2=a^2=0$ and $a=0=b$,
we are done.
Suppose that $a\neq b$.
Observe that $0=(a-b)(a^2+ab+b^2)=a^3-b^3$.
Thus, $a^3=b^3$.
We claim that $a=0$ and $b=0$.
If $a\neq 0$,
then $(a^{-1}b)^3=1$ and the multiplicative order of $a^{-1}b$
in the multiplicative group $F-\{0\}$ is $1$
because $3\not\mid |F-\{0\}|=2^n-1$.
Hence, $a^{-1}b=1$ and $a=b$,
a contradiction.
Therefore, $a=0$.
By a similar argument, we have $b=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Eccentricity from the rotating ellipse formula I see that from a normal ellipse formula, we can acquire the eccentricity via this formula here.
However, for this formula (1):
$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$
When parameter $B = 0$, we would have normal ellipse, and the formula from the link above can be used.
But when $B ≠ 0$, we will have a tilting ellipse, and its eccentricity will change as well. In fact, if we are to find the current eccentricity of the given formula (1), what would be the formula for the eccentricity in this case?
| If $S(p,q)$ be a focus, $L: ax+by+c=0$ be the directrix and $e$ be the eccentricity of a conic with $P(x,y)$ is any point on the conic,
$$e=\frac{\sqrt{(x-p)^2+(y-q)^2}}{\frac{|ax+by+c|}{\sqrt{a^2+b^2}}}$$
On simplification, $$\{a^2(1-e^2)+b^2\}x^2-2abe^2xy+\{a^2+(1-e^2)b^2\}y^2+()x+()y+()=0$$
Comparing with the given equation, $a^2(1-e^2)+b^2=A, a^2+(1-e^2)b^2=C$ and $-2abe^2=B$
So, we have three unknowns $a,b,e$ with three equations which can be solved easily.
On elimination of $a,b,$ we get $$(B^2-4AC)e^4-4e^2\{B^2+(A-C)^2\}+4\{B^2+(A-C)^2\}=0$$
The value of discriminant is $4^2\{B^2+(A-C)^2\}^2-4(B^2-4AC)4\{B^2+(A-C)^2\}$
$=16\{B^2+(A-C)^2\}\{B^2+(A-C)^2-(B^2-4AC)\}=16(A+C)^2\{B^2+(A-C)^2\}$
So, $e^2=\frac{4\{B^2+(A-C)^2\}\pm4(A+C)\sqrt{B^2+(A-C)^2}}{2(B^2-4AC)}=\frac{2\sqrt{B^2+(A-C)^2}\{\sqrt{B^2+(A-C)^2}\pm(A+C)\}}{B^2-4AC}$
For the '-' sign, this is the same value we reached at in the 1st answer as $B^2-4AC=B^2+(A-C)^2-(A+C)^2=\{\sqrt{B^2+(A-C)^2}+A+C\}\{\sqrt{B^2+(A-C)^2}-A-C\}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate, $\int_{0}^{\pi/2}\sin (\tan\theta) \mathrm{d\theta}$? Integrate, $$\int_{0}^{\frac{\pi}{2}}\sin (\tan\theta) \mathrm{d\theta}$$
| Making a change of variables $u=\tan(\theta)$:
$$
\int_0^{\pi/2} \sin\left(\tan \theta\right) \mathrm{d} \theta = \int_0^\infty \frac{\sin(u)}{1+u^2} \mathrm{d} u \tag{1}
$$
In order to evaluate this we use the technique of Mellin transform.
*
*Evaluate the Mellin transforms of $\sin(u)$ and $\left(1+u^2\right)^{-1}$:
$$
\mathcal{M}_s(\sin(u))= \int_0^\infty u^{s-1} \sin(u) \mathrm{d} u = \frac{\sqrt{\pi}}{2} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)}
$$
defined in the strip $-1<\Re(s)<1$. Applying the inverse transform:
$$
\sin(u) = \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} \frac{\sqrt{\pi}}{2} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} u^{-s} \mathrm{d} s \tag{2}
$$
where $\gamma$ is an arbitrary real constant such that $-1<\gamma<1$.
The Mellin transform of the rational function $(1+u^2)^{-1}$ reads
$$
\mathcal{M}_s\left(\frac{1}{1+u^2}\right)= \int_0^\infty \frac{u^{s-1}}{1+u^2}\mathrm{d} u = \frac{1}{2} \Gamma\left(\frac{s}{2}\right) \Gamma\left(1-\frac{s}{2}\right) \tag{3}
$$
and is defined in the strip $0<\Re(s)<2$.
*Now substituting $(2)$ into $(1)$, choosing $\gamma$ such that $-1<\gamma<0$, and interchanging the order of integration:
$$\begin{eqnarray}
\int_0^\infty \frac{\sin(u)}{1+u^2} \mathrm{d} u &=& \frac{1}{2\pi i} \frac{\sqrt{\pi}}{2} \int_{\gamma-i \infty}^{\gamma+i \infty} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} \left( \int_0^\infty \frac{u^{-s}}{1+u^2} \mathrm{d}u \right) \mathrm{d} s \\ &\stackrel{(2)}{=}& \frac{1}{2\pi i} \frac{\sqrt{\pi}}{2} \int_{\gamma-i \infty}^{\gamma+i \infty} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} \left(\frac{1}{2} \Gamma\left(\frac{1-s}{2}\right) \Gamma\left(1-\frac{1-s}{2}\right) \right)\mathrm{d} s \\
&=& \frac{1}{2\pi i} \frac{\sqrt{\pi}}{4} \int_{\gamma-i \infty}^{\gamma+i \infty} \Gamma\left(\frac{1}{2}-\frac{s}{2}\right) \frac{\Gamma\left(\frac{1}{2}+\frac{s}{2}\right)^2}{\Gamma\left(1-\frac{s}{2}\right)} \left(\frac{1}{2}\right)^{-s} \mathrm{d}s \\
&\stackrel{\text{reflection}}{=}& \frac{1}{2\pi i} \frac{\pi}{2} \int_{\gamma - i \infty}^{\gamma+i \infty} \Gamma(s) \tan\left(\frac{\pi}{2} s \right) \mathrm{d}s
\end{eqnarray}
$$
The latter integral can be evaluated as a sum over residues at poles to the left of the integration contour, situated at odd negative integers. The poles of $\Gamma(s)$ are even non-positive integers are canceled by zeros of the tangent function. The poles at the odd negative integers are double poles:
$$
\operatorname{Res}_{s=-2k-1} \left(\frac{\pi}{2} \Gamma\left(s\right) \tan\left(\frac{\pi}{2} s\right) \right) = \frac{1}{\Gamma\left(2k+2\right)} \psi\left(2k+2\right)
$$
where $\psi(x)$ is the digamma function.
Thus we have the result:
$$\begin{eqnarray}
\int_0^\infty \frac{\sin u}{1+u^2} \mathrm{d} u &=& \sum_{k=0}^\infty \frac{\psi(2k+2)}{\Gamma(2k+2)} = \frac{1}{2} \left( \sum_{k=0}^\infty \frac{\psi(k+1)}{\Gamma(k+1)} - \sum_{k=0}^\infty (-1)^k \frac{\psi(k+1)}{\Gamma(k+1)}\right) \\ &=&
\frac{1}{2} \left(f(1) - f(-1)\right) = \frac{\exp(-1)}{2} \operatorname{Ei}(1) - \frac{\exp(1)}{2} \operatorname{Ei}(-1)
\end{eqnarray}
$$
where $\operatorname{Ei}(x)$ denotes the exponential integral special function.
We now proceed to prove that, for real non-zero $x$:
$$
f(x) = \sum_{k=0}^\infty x^k \frac{\gamma(k+1)}{\Gamma(k+1)} = \mathrm{e}^{x} \left( \frac{1}{2} \log(x^2) - \operatorname{Ei}(-x) \right)$$
Indeed, denoting the Euler-Mascheroni constant as $C$, and using $H_n = -\sum_{k=1}^n \frac{(-1)^k}{k} \binom{n}{k}$:
$$\begin{eqnarray}
\sum_{k=0}^\infty x^k \frac{\gamma(k+1)}{\Gamma(k+1)} &=& \sum_{k=0}^\infty x^k \frac{-C +H_{k}}{k!} = - C \mathrm{e}^{x} - \sum_{k=1}^\infty \frac{x^k}{k!} \sum_{m=1}^k \frac{(-1)^m}{m} \binom{k}{m} \\ &=&
-C \mathrm{e}^{x} - \sum_{m=0}^\infty \frac{x^m}{m!} \sum_{k=1}^\infty \frac{(-x)^k}{k \cdot k!} = -C \mathrm{e}^{x} - \mathrm{e}^{x} \left( \operatorname{Ei}(-x) - \log |x| -C \right) \\ &=& \mathrm{e}^{x} \left( \log |x| - \operatorname{Ei}(-x) \right)
\end{eqnarray}$$
| {
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Solving an Inequality Involving a Fraction I have the following inequality:
$$ \frac{x-1}{x+2} \geq 0.$$
I solved it pretty fast:
$$\begin{align}
\frac{x-1}{x+2} +1 & \geq 1\\\\
\left(\frac{x-1}{x+2} + 1\right)\cdot(x+2) & \geq 1 \cdot (x+2)\\\\
x-1 + 1\cdot(x+2) & \geq 1\cdot (x+2)\\\\
2x + 1 & \geq x+2\\\\
x + 1 & \geq 2\\\\
x & \geq 1
\end{align}$$
But that is not the only solution, the other solution is $x < -2$. How do I get to this solution?
| First note that when you multiply by a negative number the inequality changes in sign. In general, if $\dfrac{a}{b} > 0$, we have $a>0, b>0$ or $a<0, b<0$. Hence, we get that
$$(x-1) > 0, \,\,\,\,\, (x+2) > 0 \text{ i.e. } x>1$$
or
$$(x-1) < 0, \,\,\,\,\, (x+2) < 0 \text{ i.e. }x < -2$$
If $\dfrac{x-1}{x+2} = 0$, then $x=1$. Hence, the solution is $$x \in (-\infty,-2) \cup [1,\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How is $\sqrt{n^2+2n}-\lfloor\sqrt{n^2+2n}\rfloor=\frac{2}{\sqrt{1+\frac{2}{n}}+1}$. I have trouble seeing how
$$\sqrt{n^2+2n}-\lfloor\sqrt{n^2+2n}\rfloor=\frac{2}{\sqrt{1+\frac{2}{n}}+1}.$$
I can't see where to start even.
| Assuming that $n$ is a non-negative integer, we know that
$$ n^2 \leq n^2 + 2n < n^2 + 2n + 1 = (n+1)^2$$
Hence, this show that $\lfloor \sqrt{n^2 + 2n} \rfloor = n$, which simplifies the LHS to
$$ \sqrt{n^2+2n} - n.$$
Now let's work on simplifying the RHS, multiplying the numerator and denominator by $n$, we get
$$ \frac {2}{ \sqrt{1+ \frac {2}{n}} + 1} \times \frac {n}{n} = \frac {2n} {\sqrt{n^2 +2n} +n}$$
Let's also rationalize the denominator, by multiplying throughout by $\frac {\sqrt{n^2+2n} -n}{\sqrt{n^2+2n}-n}$ to obtain
$$\frac {2n} {\sqrt{n^2 +2n} +n} \times \frac {\sqrt{n^2+2n} -n}{\sqrt{n^2+2n}-n} = \frac {(2n) ( \sqrt{n^2+2n} - n)}{(n^2+2n) - (n^2)} = \sqrt{n^2+2n} - n$$
Hence, LHS = RHS.
Note: The statement is not true for negative integers, since the first inequality doesn't hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/272854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of a trigonometric inequality Does anyone know the proof of the following inequality
$$\sin(A)\sin(B)\sin(C)\le\left(\frac{3\sqrt{3}}{2\pi}\right)^3ABC$$
where $A,B,C$ are the vertex angles of a triangle.
| First note that $\log(\sin(x)/x)$ is concave for $0 \leq x,y,z < \pi$. Hence, we have that
$$\dfrac{\log(\sin(x)/x) + \log(\sin(y)/y) + \log(\sin(z)/z)}3 \leq \log \left( \dfrac{\sin((x+y+z)/3)}{(x+y+z)/3} \right)$$
Now you should be able to finish it off. Essentially same as Jensen's inequality if we consider the function $\log \left(\dfrac{x}{\sin(x)}\right)$.
Move your cursor over the gray area for the complete answer.
Taking $x+y+z = \pi$, we get that $$\dfrac{\log(\sin(x)/x) + \log(\sin(y)/y) + \log(\sin(z)/z)}3 \leq \log \left( \dfrac{3\sqrt{3}}{2 \pi} \right)$$Hence,$$\log \left(\dfrac{\sin(x) \sin(y) \sin(z)}{xyz}\right) \leq \log \left( \dfrac{3\sqrt{3}}{2 \pi} \right)^3$$Hence,$$\sin(x) \sin(y) \sin(z) \leq \left( \dfrac{3\sqrt{3}}{2 \pi} \right)^3 xyz$$
| {
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If $r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\}$ , find $ r_{\max}$ Studying differential equations I came cross through this:
Let $ \displaystyle{ r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\} } $, where $ a,b >0$. Find $ r_{ \max} $.
Here is what I did:
Fix $ a> 0$ and define $ \displaystyle{ g(b) = \frac{b}{ a+ a^2 b^3} ,\quad b>0 }$. Then we have that $ \displaystyle{ g'(b) = \frac{ a(1-2ab^3)}{ (a+ a^2 b^3)^2} }$.
Thus, $g$ has maximum at $ \displaystyle{b_0= \frac{1}{ \sqrt[3]{2a} }}$ which gives $ \displaystyle{ g(b) \leq g\left( \frac{1}{ \sqrt[3]{2a}} \right) = \frac{2}{ 3a \sqrt[3]{2a} }}$
So, $ \displaystyle{ r = \min \{ a , \frac{2}{ 3a \sqrt[3]{2a} } \}}$. I think that to find $a$ which gives $ r_{\max}$ I have to solve the equation $ \displaystyle{ a=\frac{2}{ 3a \sqrt[3]{2a}} }$.
Is this correct and why....?
Any ideas?
Thank's in advance!
edit: I edit the title.
| The maximum is when $$a=(4/27)^{1/7}\;\; \text{or}\;\; a=.76125127398793390138765377686933423...$$
$a$ is increasing as a increases. The other formula decreases as $a$ increases
$$a = 2^{\frac{2}7}/3^{\frac{3}7}$$ is when both sides are equal. Any change in a in either direction will cause one of them (and therefore the minimum) to decrease.
Your work up to that point was correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\sum_{m>n=1}^{\infty} \frac{1}{m!n!}$ Compute the series
$$\sum_{m>n=1}^{\infty} \frac{1}{m!n!}$$
| We have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{ m!\,n!} = \left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)\left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)=(e-1)^2$$
We must substract from that the "diagonal" terms: $$\sum_{n=1}^{\infty} \left(\frac{1}{n!}\right)^2 = I_0(2) - 1 \approx 2.2795853-1$$ (ref ; $I_0(\cdot)$ is the modified Bessel function of the first kind) and divide by two to get the desired region. Hence the result is
$$\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty} \frac{1}{ m!\,n!}=\frac{(e-1)^2 - (I_0(2)-1)}{2}= \frac{e^2-I_0}{2}+1-e \approx 0.83645357$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the only root of the equation $z-\sin(z)$ in the unit disk is zero. Prove that the only root of the equation $z-\sin(z)$ in the unit disk is $z=0$.
My first thought is Rouche's Theorem, but I don't know any bounds on $|\sin(z)|$. Suggestions?
| Not the most elegant solution, but here's something.
First of all, $f(z) = z - \sin z$ has a triple zero at $z=0$. Define
$$g(z) = \frac{z-\sin z}{z^3} = \frac{1}{3!} + \underbrace{\left(- \frac{z^2}{5!} + \frac{z^4}{7!} - \cdots \right)}_{h(z)}. $$
If $|z| < 1$, then by the triangle inequality,
\begin{align}
|h(z)| &\le \frac{1}{5!} + \frac{1}{7!} + \frac{1}{9!} + \cdots \\
&= \frac{1}{5!} \left( 1 + \frac{1}{6\cdot 7} + \frac{1}{6\cdot 7 \cdot 8 \cdot 9} + \cdots \right) \\
&\le \frac{1}{5!}\left(1 + \frac{1}{42} + \frac{1}{42^2} + \cdots + \right) \\
&= \frac{1}{5!}\cdot \frac{42}{41} < \frac{1}{3!}
\end{align}
Hence $g$ has no zero on the unit disc, which shows that the only zero for $f$ is the triple zero at the origin. (You can get by with a cruder estimate of $h$, of course.)
Added In fact, the same method shows that $z=0$ is the only zero on a much larger disc than the unit disc. If $|z| < 6$, for example, the final estimate for $h$ will be $|h(z)| < 7/120$ which is still less than $1/3!$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Differential equation involving matrices I want to show that
$ Y(t)=\cos(At)$ and $Y(t)=\sin(At)$ satisfy the equation
$$Y''+A^2Y=0 $$
subjected to the initial conditions $Y(0)=I, Y'(0)=0$ and $Y(0)=0, Y'(0)=I$ respictively
where $A$ is an constant $n \times n$ matrix..
| I found the shifting of indices to be highly confusing so I propose a method that does not rely on that.
We will first show that $Y(t)=\cos (At)$ solves the initial value problem $Y''+A^2 Y=0$ where $Y(0)=I$ and $Y'(0)=0$. Because $\cos At$ is defined as the infinite sum $\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k}t^{2k}}{(2k)!}$ and differentiation is linear we can differentiate term by term to get:
\begin{eqnarray*}
Y''(t)&=&\frac{d^2}{dt^2}\left[ I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots \right]\\
&=&\left[ -2\frac{A^2}{2!} +4\cdot 3 \frac{A^4 t^2}{4!} -6\cdot 5 \frac{A^6 t^4}{6!} +\ldots \right]\\
&=&-A^2\left[ I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots \right]\\
&=& -A^2 \sum\limits_{k=0}^{\infty} (-1)^k \frac{A^{2k}t^{2k} }{(2k)!}=-A^2Y(t)
\end{eqnarray*}
So $Y(t)$ satisfies $ Y''+A^2Y=0$. Also $Y(0)=\cos(0)=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{0^{2k}}{(2k)!}=I$ (The $0^{2k}$ in the summation is the zero matrix $[A\cdot t]_{t=0}=0$ and $0^0=I$, all higher powers of $0$ are of course $0$). And for $Y'(0)$ we find
\begin{eqnarray*}
Y'(0)&=&\frac{d}{dt}\left[ I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots \right] _{t=0}\\
&=& \left[ -2\frac{A^20}{2!}+4\frac{A^40^3}{4!}-6\frac{A^60^5}{6!}+\ldots \right] =0
\end{eqnarray*}
(here the $0^k$ is just the scalar $t=0$ )
For $Y(t)=\sin(At)$ with $\sin(At):=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!}$ we have:
\begin{eqnarray*}
Y''(t)&=&\frac{d^2}{dt^2}\left[ At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots \right]\\
&=&\left[ -3\cdot 2\frac{A^3t}{3!} +5\cdot 4 \frac{A^5 t^3}{5!} -5\cdot 6 \frac{A^7 t^5}{7!} +\ldots \right]\\
&=&-A^2\left[ At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots \right]\\
&=& -A^2 \sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!}=-A^2Y(t)
\end{eqnarray*}
So $Y(t)=\sin(At)$ also satisfies $Y''(t)+A^2Y(t)=0$. Here however $Y(0)=\sin( 0)=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{0^{2k+1}}{(2k+1)!}=0$ (again using $0$ as the zero matrix in the summation, and now the lowest power of $0$ is $0^1$ so here we have an infinite sum of zeros). And
\begin{eqnarray}
Y'(0)&=&\frac{d}{dt}\left[ At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots \right] _{t=0}\\
&=& \left[ A -3\frac{A^30^2}{3!}+5\frac{A^50^4}{5!}-7\frac{A^70^6}{7!}+\ldots \right] =A
\end{eqnarray}
I am guessing this is what you meant because even if we consider a scalar $A$ then $\left[\frac{d}{dt}\sin(At)\right]_{t=0}=A\cos(0)=A$.
| {
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"url": "https://math.stackexchange.com/questions/278938",
"timestamp": "2023-03-29T00:00:00",
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gambling probability problem We are given a fair coin. We start out with 5 dollars. We keep tossing the coin. If the outcome is different than the previous one, we are awarded another 5 dollars. However, we do not get anything if the outcome is the same as the previous one. Let's say we toss the coin X times in the long run. How much do we expect to have in the end?
| This problem may be solved using generating functions which is admittedly more powerful than strictly necessary in this case, but it does illustrate the method, which is worth knowing. Let $p_A(n)$ be a polynomial in $u$ such that $[u^k] p_A(n)$ is the probability of having a capital of $k$ dollars and the last coin came up heads and similarly $p_B(n)$ for tails. Then we have the following system of equations
$$ p_A(n) = \frac{1}{2} p_A(n-1) + \frac{1}{2} u^5 p_B(n-1) \\
p_B(n) = \frac{1}{2} u^5 p_A(n-1) + \frac{1}{2} p_B(n-1)$$
with initial conditions
$$ p_A(0) = \frac{1}{2} u^5 \quad \text{and} \quad p_B(0) = \frac{1}{2} u^5.$$
Now introduce the generating functions
$$ P_A(z) = \sum_{n\ge 0} p_A(n) z^n \quad \text{and} \quad
P_B(z) = \sum_{n\ge 0} p_B(n) z^n.$$
Multiply the recurrences by $z^n:$
$$ p_A(n) z^n = z \frac{1}{2} z^{n-1} p_A(n-1) + z \frac{1}{2} u^5 z^{n-1} p_B(n-1) \\
p_B(n) z^n = z \frac{1}{2} u^5 z^{n-1} p_A(n-1) + z \frac{1}{2} z^{n-1} p_B(n-1)$$
Sum over $n$ to obtain the following system of equations
$$ P_A(z) - \frac{1}{2} u^5 = \frac{1}{2}z P_A(z) + \frac{1}{2}u^5 z P_B(z) \\
P_B(z) - \frac{1}{2} u^5 = \frac{1}{2}u^5 z P_A(z) + \frac{1}{2}z P_B(z).$$
The solution to this system is
$$ P_A(z)=-{\frac {{u}^{5}}{{u}^{5}z-2+z}}\\
P_B(z) =-{\frac {{u}^{5}}{{u}^{5}z-2+z}}.$$
The generating function for the expected capital after $n$ steps is then given by
$$ f(z) = \left( \frac{d}{du} (P_A(z) + P_B(z)) \right)_{u=1} =
-10\, \left( 2\,z-2 \right) ^{-1}+10\,{\frac {z}{ \left( 2\,z-2 \right) ^{2}}}.$$
Now we have $$[z^n] f(z) = 5+5/2\,n,$$
which is precisely what was obtained above.
We can also compute the variance, which is where the generating functions start to shine. First compute the expected factorial moment $Q(Q-1)$ where $Q$ is the capital. This has generating function
$$ g(z) = \left( \frac{d}{du} \frac{d}{du} (P_A(z) + P_B(z)) \right)_{u=1} =
-40\, \left( 2\,z-2 \right) ^{-1}+140\,{\frac {z}{ \left( 2\,z-2 \right) ^{2}}}-100\,{\frac {{z}^{2}}{\left( 2\,z-2 \right) ^{3}}}
.$$ This gives $$[z^n] g(z) = 20+{\frac {115}{4}}\,n+{\frac {25}{4}}\,{n}^{2}.$$
The variance can now be obtained from
$$Var[Q] = E[Q^2] - E[Q]^2 = E[Q(Q-1)] + E[Q] - E[Q]^2,$$
which gives
$$Var[Q] = {\frac {25}{4}}\,n.$$
We could in fact compute all factorial moments if desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of strings of length n formed by $\{0,1,2\}$ such that $1$ and $2$ do not occur successively. What is the number of ways of forming a string of length $n$ from the set $\{0,1,2\}$ such that $1$ and $2$ do not occur successively.
| Let's use generating functions (Wilf's "generationfunctinology explains the technique)to solve this one. Call $a_n$, $b_n$, $c_n$ the number of strings of length $n$ ending in 0, 1, 2 respectively. By considering the ways of getting strings of length $n + 1$ by appending a digit to sequences of length $n$, this leads to the recurrences (remember to take care not to get adyacent 1, 2):
$$
a_{n + 1} = a_n + b_n + c_n \\
b_{n + 1} = a_n + b_n \\
c_{n + 1} = a_n + c_n
$$
Also, $a_0 = b_0 = c_0 = 1$. We are interested in $f(n) = a_n + b_n + c_n = a_{n + 1}$. Define ordinary generating functions:
$$
A(z) = \sum_{n \ge 0} a_n z^n \\
B(z) = \sum_{n \ge 0} b_n z^n \\
C(z) = \sum_{n \ge 0} c_n z^n
$$
By the properties of ordinary generating functions:
$$
\frac{A(z) - a_0}{z} = A(z) + B(z) + C(z) \\
\frac{B(z) - b_0}{z} = A(z) + B(z) \\
\frac{C(z) - c_0}{z} = A(z) + C(z)
$$
Your tame computer algebra package solves this system of equations:
$$
A(z) = \frac{1 + z}{1 - 2 z - z^2} \\
B(z) = C(z) = \frac{1}{1 - 2 z - z^2}
$$
We are interested in $a_{n + 1}$:
$$
A(z) = \frac{1}{2 (1 + \sqrt{2})} \cdot \frac{1}{1 - (1 + \sqrt{2})^{-1} z} + \frac{1}{2 (1 - \sqrt{2})} \cdot \frac{1}{1 - (1 - \sqrt{2})^{-1} z}
$$
This gives:
$$
a_n = \frac{(1 + \sqrt{2})^{n - 1}}{2} + \frac{(1 - \sqrt{2})^{n - 1}}{2}
$$
The final answer is that there are:
$$
f(n) = a_{n + 1} = \frac{(1 + \sqrt{2})^n}{2} + \frac{(1 - \sqrt{2})^n}{2}
$$ strings in total of length $n$.
| {
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Which of the following statements are true for the sides of a triangle? Which of the following statements are true?
(a) if $a,b$ and $c$ are the sides of a triangle , then $(ab+bc+ca)/(a^2+b^2+c^2)≥(1/2)$
(b) if $a,b$ and $c$ are the sides of a triangle , then $(ab+bc+ca)/(a^2+b^2+c^2)≤ 1$
(c) both statements above are true for all triples $(a,b,c)$ of strictly positive real numbers.
How can I do this?
My thoughts:
(b) true , (a) true, (c) false
| $$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
$$\implies a^2+b^2+c^2\ge ab+bc+ca\iff \frac{ab+bc+ca}{a^2+b^2+c^2}\le1$$
If $c\ge a-b, c^2\ge (a-b)^2=a^2+b^2-2ab$ which is true for any triangle
Similarly, $a^2\ge b^2+c^2-2bc$
and $b^2\ge c^2+a^2-2ca$
Adding we get, $$a^2+b^2+c^2\ge 2(a^2+b^2+c^2)-2(ab+bc+ca)$$
$$\iff 2(ab+bc+ca)\ge a^2+b^2+c^2\iff \frac{ab+bc+ca}{a^2+b^2+c^2}\ge \frac12$$
So, $(a),(b)$ are true, not $(c)$ as we need the difference of any two of $a,b,c$ to be $\ge$ the other.
| {
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| It is enough to show that for $t\geqslant0$,
$$f(t) = \log(3)-\log\left(t^3+t+1\right)+\tfrac23\left(t^2-1\right)
\geqslant 0$$
as the inequality in question is merely $f(a)+f(b)+f(c)\geqslant0$.
But $f’(t)=0$ only when $t=1$ where it has a minimum, so $f(t)\geqslant f(1)=0$.
| {
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What is wrong with this problem We know that:
$(a^n)^m=a^{nm}$
From this we have:
$-3^3=[(-3)^2]^\frac{3}{2}=(3^2)^\frac{3}{2}=27$
Find what's wrong
| The "law of exponents" that you cite:
$$\large (a^n)^m=a^{nm}$$
applies PROVIDED $\bf{a \gt 0}$.
Here, though, we have $\,\bf{a = -3 \lt 0}$, and hence:
$$-3^3=[(-3)^2]^{\large\frac{3}{2}}=(3^2)^{\large\frac{3}{2}}=27\quad \Longleftarrow \;\text{ False}.$$
$$-3^3 = -(3^3) = -[(3^2)^{\large \frac{3}{2}}] = -(9^{\large \frac{3}{2}}) = -27\quad \Longleftarrow \; \text{ True}$$
But why the trouble? It is a very straightforward computation:
$$-3^3 = -(3^3) = -(3\cdot 3\cdot 3) = -27$$
| {
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Improper integral with log and absolute value: $\int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx$ How do you show that $$ \int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx = \frac{\pi}{2} \log (\tanh 1)\, ?$$
I know that the integral converges since $\log |\tan x| = \frac{1}{2} \log(\tan^{2}x)$, and $\log (\tan^{2} x)$ behaves like $2 (-1)^{n} \log \left(x-\frac{n \pi}{2} \right)$ near $x= \frac{n \pi}{2}$. Thus the singularities at $x= \frac{n \pi}{2}$ are integrable.
| Denote the evaluated integral as $I$, then $I$ may be rewritten as
$$I=\frac{1}{2}\int_0^\infty \frac{\ln \sin^2 x}{1+x^2}\,dx-\frac{1}{2}\int_0^\infty \frac{\ln \cos^2 x}{1+x^2}\,dx$$
Using Fourier series representations of $\ln \sin^2 \theta$ and $\ln \cos^2 \theta$,
$$\ln \sin^2 \theta=-2\ln2-2\sum_{k=1}^\infty \frac{\cos2k\theta}{k}$$
and
$$\ln \cos^2 \theta=-2\ln2+2\sum_{k=1}^\infty (-1)^{k+1}\frac{\cos2k\theta}{k}$$
also note that
$$\int_0^\infty\frac{\cos ax}{1+x^2}\,dx=\frac{\pi e^{-a}}{2}$$
then
$$\begin{align}I&=-\sum_{k=1}^\infty \frac{1}{k}\int_0^\infty\frac{\cos2kx}{1+x^2}\,dx-\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\int_0^\infty\frac{\cos2kx}{1+x^2}\,dx\\&=-\pi\sum_{k=1}^\infty \frac{e^{-2k}}{2k}-\pi\sum_{k=1}^\infty (-1)^{k+1}\frac{e^{-2k}}{2k}\\&=\frac{\pi}{2}\ln\left(1-e^{-2}\right)-\frac{\pi}{2}\ln\left(1+e^{-2}\right)\\&=\frac{\pi}{2}\ln\left(\frac{1-e^{-2}}{1+e^{-2}}\right)\\&=\frac{\pi}{2}\ln\left(\tanh 1\right)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Real part of complex z must be zero
Show that the real part of any solution of $(z+1)^{100}=(z-1)^{100}$
must be zero.
I know that I must use the fact $e^{2\pi ki}$ somewhere but not sure exactly how or where to use it.
| If $z=x+iy$ and if we put the value of $z$ in $|z+1|=|z-1|$ which has been derived in other answers,
$$|x+iy+1|=|x+iy-1|\implies (x+1)^2+y^2=(x-1)^2+y^2$$
$$\implies (x+1)^2-(x-1)^2=0\implies4\cdot x\cdot1=0\implies x=0$$
A little generalization:
Let $|z-w|=|z+w|$
If $z=x+iy,w=a+ib$ where $x,y,a,b$ are real and $a\cdot b\ne0$
$$(x-a)^2+(y-b)^2=(x+a)^2+(y+b)^2\implies x\cdot a+y\cdot b=0$$
If $b=0\implies w=a,x\cdot a=0\implies x=0$ as $a\ne0\implies z=iy$ (In our case, $a=1$)
If $a=0\implies w=ib,y\cdot b=0\implies y=0$ as $b\ne0\implies z=x$
Geometrically, $$\arg(z)-\arg(w)=\arctan\frac yx-\arctan\frac ba$$
$$=\arctan \left(\frac{\frac yx-\frac ba}{1+\frac yx\cdot\frac ba}\right)=\arctan\left(\frac{ay-bx}{x\cdot a+y\cdot b}\right)=\frac\pi2 $$
$$\implies z\perp w $$
If $b=0,\arg(w)=\arctan 0=0$ or $\pi,$ so $w$ is parallel to $X$ axis, $z$ being perpendicular to $w,$ will be parallel to $Y$ axis.
Observe that, $\arg(z)=\frac\pi2+0=\frac\pi2$ or $\frac\pi2+\pi\equiv-\frac\pi2$
Similarly for $a=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| As $-1\leq\sin x\leq1\implies |\sin x-1|=1-\sin x$, $|\sin x-2|=2-\sin x$, $|\sin x-3|=3-\sin x$, $|\sin x+1|=1+\sin x$
Thus the given expression becomes, $1-\sin x+2-\sin x+3-\sin x+1+\sin x=7-2\sin x$ which is minimum when $\sin x $ is maximum and equal to $1$ which gives minimum value equal to $5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Solve without using L'Hôpital's rule? Is it possible to solve this without using L'Hôpital's rule?
$$
\lim_{x\to 0} \Big(\frac{3x+1}{x}-\frac{1}{\sin x}\Big)
$$
I tried to solve it but I got stuck at the $\frac{1}{\sin x}$ part.
$$
=\lim_{x\to 0}\frac{3+x^{-1}}{1}-\lim_{x\to 0}\frac{1}{\sin x}
$$
since $\lim_{x\to 0}\frac{1}{\sin x}$ does not exist.
| Let $\displaystyle L=\lim_{x \to 0} \frac{\sin x - x}{x^2}$. Replacing $x$ by $2y$, we obtain:
$
\begin{align*}
L & = \lim_{y \to 0} \frac{\sin 2y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2 \sin y + 2 \sin y - 2y}{4y^2} \\
& = \lim_{y \to 0} \left( \frac{2 \sin y(\cos y-1)}{4y^2}+\frac{1}{2} \lim_{y \to 0} \frac{\sin y - y}{y^2}\right) \\
& = \frac{1}{2} \lim_{y \to 0} \sin y \cfrac{-2\sin^2\cfrac{y}{2}}{y^2}+\frac{1}{2}L \\
& = -\frac{1}{4} \lim_{y \to 0} \sin y \cfrac{\sin^2\cfrac{y}{2}}{\cfrac{y^2}{4}}+\frac{1}{2}L,
\end{align*}
$
which implies that $L=0$.
Hence,
$
\begin{align*}
\lim_{x\to 0} \left(\frac{3x+1}{x}-\frac{1}{\sin x}\right) & = \lim_{x\to 0} \left(3+\frac{1}{x}-\frac{1}{\sin x}\right) = \lim_{x\to 0} \left(3+\frac{\sin x - x}{x \sin x}\right) \\
& = \lim_{x\to 0} \left(3+\frac{\sin x - x}{x^2} \cdot \frac{x}{\sin x} \right) = 3+0 \cdot 1 \\
& = \boxed{3},
\end{align*}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Definite integral $\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{\sin 2x}}\,\mathrm{d} x$ $$\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{\sin 2x}}\,\mathrm{d} x$$
I'm pretty sure I can finish it after finding the anti derivative. I tried changing the denominator to $2\sin x \cos x$ and subbing $u$ as $\sin x$.
$$\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{2\sin x\cos x}}\,\mathrm{d} x$$
| After a little algebra, we get
$$ \int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} dx \: \frac{\sin{x}}{\sqrt{\cos{x} \sqrt{1-\cos^2{x}}}}$$
Substitute $u=\cos{x}$:
$$\int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} = \frac{1}{\sqrt{2}} \int_0^1 du \: u^{-1/2} (1-u^2)^{-1/4} $$
Now substitute $v = u^2$:
$$\begin{align} \int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} &= \frac{1}{2 \sqrt{2}} \int_0^1 dv \: v^{-3/4} (1-v)^{-1/4} \\ &= \frac{1}{2 \sqrt{2}} \Gamma \left( \frac{1}{4} \right ) \Gamma \left( \frac{3}{4} \right ) \\ &= \frac{\pi}{2} \end{align} $$
That last step comes from the relation
$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
First week in Linear Algebra, need some help on this simple problem If you have
$5x + 2y + z = 0$
$2x + y = 0$ and you're asked to solve using back-substitution how would you go about doing it?
Initially I thought just simply the following:
$x + \frac{2}{5} y + \frac{1}{5} z = 0$ (divide by $5$)
$2x + y = 0$
$\displaystyle x + \frac{2}{5} y + \frac{1}{5} z = 0$
$\displaystyle 2(x + \frac{2}{5} y + \frac{1}{5} z) + \frac{1}{5} y - \frac{2}{5} z = 0$ (sub method I was taught in class)
But after simplifying I realized I basically just made the problem worse because I ended with:
$\displaystyle x + \frac{2}{5} y + \frac{1}{5} z = 0$
$\displaystyle \frac{1}{5} y + \frac{2}{5} z = 0$
So I pretty much still have 3 unknowns. Any suggestions or hints?
| I will write up what I wrote above into a solution:
Note that we begin with
$$5x + 2y + z = 0 $$
$$2x + y = 0$$
Subtracting twice the second from the first we get $x+z=0$, therefore, $z=-x$. Therefore, we get solutions of the form $y=-2x$ and $z=-x$ where $x$ can be anything. This means that all solutions are on the line $$(x,-2x,-x)$$ and anything on this line is a solution.
Note since that there are less equations than variables there cannot be a unique solution, which is why we get a line of solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is my Fourier Series computation done correctly? See my fourier series calculation of this function if you please!
$
f(t)=\left\{\begin{array}{ll} 0, & \text{for } \ -\pi<t<0 \\
1, & \text{for } \ 0 < t < \pi\end{array}\right. .
$
I start by calculation of the coefficient $a_0,a_k,b_k$: $$a_0=\frac{1}{\pi} \int\limits_{0}^{\pi}1\mathrm dt=1 ;$$ $$ a_k=\frac{1}{\pi}\int\limits_{0}^{\pi}\sin(kt)\mathrm dt=\frac{1-\cos(\pi k)}{\pi k} = \frac{1-(-1)^{k}}{\pi k};$$ $$ b_k= \frac{1}{\pi}\int\limits_{0}^{\pi}\cos(kt)\mathrm dt = \frac{\sin(\pi t)}{k} = 0$$
We will get: $f(t) \sim \frac{1}{2} + \sum\limits_{k=0}^{\infty}\frac{1-(-1)^{k}}{\pi k}\sin(kt) = \frac{1}{2} + \sum\limits_{n=0}^{\infty}\frac{2}{\pi (2n+1)}\sin((2n+1)t) $
How we can find relationship : $\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$ from this fourier series ?
| $$\begin{equation*} % a_0
f(t)=\begin{cases}0,&-\pi<t<0\\1,&0<t<\pi \end{cases}
\end{equation*}$$
\begin{equation*}a=-\pi\qquad l=\pi\qquad a+2l=\pi\qquad\qquad\qquad\end{equation*}
$$\begin{equation*}S(t)=\frac{a_0}2+\sum\limits_{k=1}^\infty a_k\cos\left(\frac{k\pi x}l\right)+b_k\sin\left(\frac{k\pi t}l\right)\\
\\
a_0=\frac1l\int\limits_{a}^{a+2l}f(t)\mathrm dt\qquad a_k=\frac1l\int\limits_{a}^{a+2l}f(t)\cos(t)\mathrm dt\qquad b_k=\frac1l\int\limits_{a}^{a+2l}f(t)\sin(t)\mathrm dt\\
\\\end{equation*}$$
\begin{align*} % a_0
a_0&=\frac1\pi\int\limits_{-\pi}^\pi f(t)\,\text dt\\
&=\frac1\pi\int\limits_{0}^\pi\,\text dt\\
&=\frac t\pi \Big|_0^\pi\\
&=\frac{\pi-0}\pi\\
&=1\\
\end{align*}
\begin{align*} %a_k
a_k&=\frac1\pi\int\limits_{-\pi}^\pi f(t)\cos(kt)\,\text dt\\
&=\frac1\pi\int\limits_{0}^\pi \cos(kt)\,\text dt\\
&=\frac1\pi\frac{\sin(kt)}k\Big|_0^\pi\\
&=\frac{\sin(k\pi)}{\pi k}\\
&=0
\end{align*}
\begin{align*} % b_k
b_k&=\frac1\pi\int\limits_{-\pi}^\pi f(t)\sin(kt)\,\text dt\\
&=\frac1\pi\int\limits_{0}^\pi\sin(kt)\,\text dt\\
&=\frac{-\cos(kt)}{\pi k}\Big|_0^\pi\\
&=\frac{\cos(0)-\cos(k\pi)}{k\pi }\\
&=\frac{1-(-1)^k}{k\pi }\\
&=\frac{1+(-1)^{k+1}}{k\pi }\\
\end{align*}
\begin{align*} % S(t)
S(t)&=\frac{1}2+\sum\limits_{k=1}^\infty \frac{1+(-1)^{k+1}}{k\pi }\sin\left({k t}\right)\\
&=\frac{1}2+\frac2\pi\sum\limits_{k=1,3,5,\dots}^\infty\frac{\sin({k t})}k\\
&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)t\big)}{2r-1}\\
\\
&=\frac12+\frac2\pi\left(\sin(t)+\frac{\sin(3t)}3+\frac{\sin(5t)}5+\dots\right)
\end{align*}
\begin{align*} % S(t)
S(t)
&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)t\big)}{2r-1}\\
\\
\\% S(π/2)
S\left(\tfrac\pi2\right)
&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\
1&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\
\frac\pi4&=\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\
\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find real roots? $4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$ How to find real roots?
$$4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$$
I try LMVT but vary difficult
| Trying to piece together the above comments. First, the equation can be rewritten:
$$
\frac{5^x - 4^x}{5 - 4} = \frac{6^{x^2} - 5^{x^2}}{6 - 5}
$$
By LMVT, there exist constants $4 \le \alpha \le 5$ and $5 \le \beta \le 6$
such that:
$$
\begin{align*}
\frac{5^x - 4^x}{5 - 4} &= x \alpha^{x - 1} \\
\frac{6^{x^2} - 5^{x^2}}{6 - 5} &= x^2 \beta^{x^2 - 1}
\end{align*}
$$
Equating them results in:
$$
\begin{align*}
\alpha^{x - 1} &= x \beta^{x^2 - 1} \\
\frac{1}{x} &= \frac{\beta^{x^2 - 1}}{\alpha^{x - 1}} \\
- \ln x &= x^2 \ln \beta - x \ln \alpha - \ln \frac{\beta}{\alpha}
\end{align*}
$$
All constants are positive.
But here I'm stuck.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding all matrices $B$ such that $AB=BA$ for a fixed matrix $A$ Let
$$ A=\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 2
\end{pmatrix}
$$
Find all matrices $B$ such that $AB=BA$.
Attempt at solution: I can show that $A$ is invertible so its inverse must be one of the elements. But how do I go about showing there are more of them? or not?. I can set set up the unknown matrix to be a matrix with 9 unknowns and then (at least in principle) try to solve or this system. But I think this is not a very productive way to do this. If these were $2 \times 2$ matrces that would be ok. How should I proceed? Any hints?
Thanks for your time and help.
| $$A=\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 2
\end{pmatrix}$$
$$A=\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
-3 & -1 & 1
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
0 & 0 & 2
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 1
\end{pmatrix}$$
This is the diagonalization of $A$, $A=PDP^{-1}$. Any diagonal matrix in place of $D$ gives a matrix that commutes with $A$. Using $B =PD_0P^{-1}$ :
\begin{align}
AB &= PDP^{-1}PD_0P^{-1} \\
&= PDD_0P^{-1} & \text{($P^{-1}P = I$)}\\
& = PD_0DP^{-1} &\text{(since diagonal matrices commute)}\\
& = PD_0P^{-1}PDP^{-1} \\
& = BA
\end{align}
Lets now look at the full variable form:
$$\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
-3 & -1 & 1
\end{pmatrix}\begin{pmatrix}
\lambda_1 & 0 & 0 \\
0& \lambda_2 & 0 \\
0 & 0 & \lambda_3
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 1
\end{pmatrix}
= \begin{pmatrix}
\lambda_1 & 0 & 0 \\
0& \lambda_2 & 0 \\
-3\lambda_1 & -\lambda_2 & \lambda_3
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 1
\end{pmatrix}$$
$$\begin{pmatrix}
\lambda_1 & 0 & 0 \\
0& \lambda_2 & 0 \\
-3\lambda_1 & -\lambda_2 & \lambda_3
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0& 1 & 0 \\
3 & 1 & 1
\end{pmatrix}= \begin{pmatrix}
\lambda_1 & 0 & 0 \\
0& \lambda_2 & 0 \\
-3\lambda_1+3\lambda_3 & -\lambda_2 + \lambda_3& \lambda_3
\end{pmatrix}$$
This gives $B$ as a function of the three variables $\lambda_1, \lambda_2$, and $\lambda_3$:
$$B=\begin{pmatrix}
\lambda_1 & 0 & 0 \\
0& \lambda_2 & 0 \\
-3\lambda_1+3\lambda_3 & -\lambda_2 + \lambda_3& \lambda_3
\end{pmatrix}$$
And this general $B$ commutes with $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/296857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Determinant of block matrix Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.
\begin{align}
D=\begin{vmatrix}
a_{11} &a_{12} & \cdots & a_{1k} &0 &\cdots &0 \\
a_{21}& a_{22}& \cdots & a_{2k} & 0 &\cdots &0 \\
\vdots& \vdots & & \vdots & \vdots & &\vdots\\
a_{k1} & a_{k2} & \cdots & a_{kk} & 0 &\cdots & 0\\
c_{11}& c_{12} & \cdots& c_{1k} & b_{11} & \cdots & b_{1m}\\
\vdots& \vdots & & \vdots & b_{21}&\cdots & b_{2m}\\
c_{m1}& c_{m2} & \cdots & c_{mk} & b_{m1}& \cdots & b_{mm}
\end{vmatrix}
\end{align}
If I have got a determinant $$D_1=
\begin{vmatrix}
0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\
0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\
b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
$$
Then $D_1$ is equal to $(-1)^{k \times m}$$D$.
I know that the existence of the factor -1 is due to the interchange of 2 row, but i have a question on that $k \times m$.In my book,it said that i have to do $k\times m$ times row operations to transform $D_1$ into $D$.However,i thought only k times is needed for $D_1$ transform into $D$.
If i have done 1 times row operation for $D_1$
$$D_1=
\begin{vmatrix}
0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\
0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\
b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
=\begin{vmatrix}
a_{11} &\cdots &0&0 &a_{12} & \cdots & a_{1k} \\
a_{21} &\cdots &0 &0& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
a_{k1} &\cdots & 0& 0& a_{k2} & \cdots & a_{kk} \\
c_{11} & \cdots & b_{1m}&b_{11} & c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
c_{m1}& \cdots & b_{mm}&b_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
$$
Correct me if i have made any mistakes
| Apply Laplace expansion http://en.wikipedia.org/wiki/Laplace_expansion.
you can directly calculate $D_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve for $ x$, $-\frac{1}{2}x^2 + 2x + 5 = 0$ I'm having trouble solving this equation for $x$:
$$-\frac{1}{2}x^2 + 2x + 5 = 0$$
What's the steps to take to solve it?
Thanks.
| If you multiply
$$-\frac{1}{2}x^2 + 2x + 5 = 0\tag{1}$$
by $-2$ you get $$x^2-4x-10=0$$ Using the quadratic formula: $$ax^2 + bx + c = 0 \iff x = \frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right),$$ where in this case, $a = 1,\;b= -4,\; c = -10$ we have $$x=\frac 12 (4 \pm \sqrt {16+40})=2 \pm \sqrt {14}\tag{2}$$
Walking through the simplification of $(2)$
$$x = \frac12 (4 \pm \sqrt{16 + 40}) \;=\; \frac12(4 \pm \sqrt{56})\; = \;\frac12 (4 \pm \sqrt{4\cdot 14}) \;=\; \frac12 (4 \pm \sqrt{4}\sqrt{14})\;$$ $$ = \;\frac12 (4 \pm 2\sqrt{14})\;=\; \frac 12 \cdot 4 \pm \frac12 \cdot 2\sqrt{14}\;\; = \;\;2 \pm \sqrt{14}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the exact value of $ \cos(36^\circ) $? The problem reads as follows:
Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of
$$\cos(36^\circ)$$
it says that you may find useful the following identities:
$$\cos^2 t+\sin^2 t = 1,\\
\sin 2t = 2\sin t\cos t,\\
\sin 3t = 3\sin t - 4\sin^3 t.
$$
Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?
Thanks in advance :)
| Let $t=\frac{\pi}{5}$ (so $t$ is $36^\circ$). Since $108=180-72$, we have $3t=\pi-2t$ and therefore
$$\sin(3t)=\sin(\pi-2t).$$
But $\sin(\pi-2t)=\sin(2t)=2\sin t\cos t$.
Also, by the identity you were given, $\sin(3t)=3\sin t-4\sin^3 t$. Thus
$$3\sin t-4\sin^3 t=2\sin t\cos t.$$
But $\sin t\ne 0$, so we can cancel a $\sin t$ and obtain
$$3-4\sin^2 t=2\cos t.$$
Substitute $1-\cos^2 t$ for $\sin^2 t$ and simplify a bit. We get
$$4\cos^2 t-2\cos t-1=0.$$
Use the Quadratic Formula to solve this quadratic equation for $\cos t$, rejecting the negative root. We get
$$\cos t=\frac{2+\sqrt{20}}{8}.$$
We can simplify this to $\dfrac{1+\sqrt{5}}{4}.$
| {
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"url": "https://math.stackexchange.com/questions/300081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there a closed form or a better solution? This is another integral in the book "irresistible integral"
I can find that:
$$\int_0^\infty \frac{1}{\left( x^4 +2ax^2+1 \right)^{m+1}} \, \text{d}x =\frac{{{\left( -1 \right)}^m}\sqrt{2}\pi }{4\left( m! \right)}\left. \left( \frac{\text{d}^m}{\text{d}x^m} \sqrt{x}\sqrt{a+\sqrt{x}} \right) \right|_{x=1}$$
Do we have a better closed form? or better solution?
My working:
Let $$I(b)=\int_0^\infty \frac{1}{x^4+2ax^2+b}\,\text{d}x$$
then $$I^{(m)}(b)=m!(-1)^m\int_0^\infty \frac{1}{(x^4+2ax^2+b)^{m+1}} \, \text{d}x$$
Also $$I(b)=\int_0^\infty \frac{1}{(x^2+\alpha)(x^2+\beta)}\,\text{d}x$$
where $$\alpha=a+\sqrt{a^2-b}>0,\ \ \beta=a-\sqrt{a^2-b}>0$$
$$I(b)=\frac{1}{\beta-\alpha}\int_0^\infty \left(\frac{1}{x^2+\alpha}-\frac{1}{x^2+\beta}\right)$$
$$=\frac{\pi}{2(\beta-\alpha)}\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)$$
$$=\frac{\pi}{4\sqrt{a^2-b}}\frac{\sqrt{2}\sqrt{a-\sqrt{b}}}{\sqrt{b}} = \frac{\sqrt{2}\pi}{4\sqrt{b}\sqrt{a+\sqrt{b}}}$$
Which follows that $$\int_0^\infty \frac{1}{( x^4+2ax^2+1 )^{m+1}}\text{d}x =\frac{(-1)^m\sqrt{2}\pi }{4m!} \left. \left( \frac{\text{d}^m}{\text{d}x^m}\sqrt{x}\sqrt{a+\sqrt{x}} \right) \right|_{x=1}$$
| For every $a\gt-1$, consider
$$
I_m(a)=\int_0^\infty \frac{\mathrm dx}{\left(x^4+2ax^2+1\right)^{m+1}},$$
then, for every $|t|$ small enough,
$$
\sum_{m=0}^{+\infty}I_m(a)t^m=\frac{\pi}{2\sqrt2\sqrt{1-t}\sqrt{a+\sqrt{1-t}}}.
$$
To deduce a closed form formula for each $I_m(a)$ from this expression does not seem obvious. However, one sees readily that the radius of convergence of the series in the LHS is $1-a^2$ if $-1\lt a\leqslant0$ and $1$ if $a\geqslant0$. Thus, when $m\to\infty$, $m^{-1}\log I_m(a)\to-\log(1-a^2)$ if $-1\lt a\leqslant0$ and $m^{-1}\log I_m(a)\to0$ if $a\geqslant0$.
| {
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"url": "https://math.stackexchange.com/questions/300217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| Hint: calculate the remainder each of the first few terms contributes. You should be able to make a reasonable hypothesis, then confirm it. What is the remainder when $(2k+1)^2$ is divided by $8?$ How many terms are in the sum?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Theory of Partial Fraction Decomposition The function in question that I want to decompose is $$\dfrac{8x^3 + 7}{(x+1)(2x+1)^3}$$
I had the idea to to break this down into:
$$\dfrac{A}{x+1} + \dfrac{Bx^2 +Cx + D}{(2x+1)^3} + \dfrac{Ex + F}{(2x+1)^2} + \dfrac{G}{2x+1}$$
Well this turns into a really messy system of 4 equations and 7 variables.
Is there an easier way to decompose the function? Can I possibly eliminate any of these variables at the outset?
| You only need to break it down into $\displaystyle\frac{A}{x+1}$ and $\displaystyle\frac{Bx^2 + Cx + D}{(2x+1)^3}.$ What you end up with is
$\begin{align*}
8A + B &= 8\\
12A + B +C &= 0\\
6A + C + D &= 0\\
A + D &= 7
\end{align*}$
Solving gives $A = 1, B = 0, C = -12, D =6.$
Note that this is equivalent to lab bhattacharjee's decomposition.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculating simple limit: $\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}$ I am trying to calculate this limit:
$$
\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}
$$
What I get is $\frac{4}{3}$, however, according to Wolfram Alpha, it should be 1. What am I missing here?
| $$ \frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\frac{(x+2)(3x-1)}{(x+2)(4x+1)}=\frac{(3x-1)}{(4x+1)}\text{ if }x+2\ne0$$
As $x\to-2,x\ne-2\implies x+2\ne0$
So, $$\lim_{x\to-2}\frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\lim_{x\to-2}\frac{(3x-1)}{(4x+1)}=\frac{3(-2)-1}{4(-2)+1}=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{a}{b^2+5}+ \frac{b}{c^2+5} + \frac{c}{a^2+5} \le \frac 12$ Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$
I have an ugly solution for this solution.
| First ,we can get rid of denominators, by noticing that
$\frac{1}{x^2+5} \leq \frac{4-x}{18}$ for any $x\in [0,2]$
(indeed, we have
$$
(4-x)(x^2+5)-18=((x-1)^2)(2-x) \geq 0 \tag{1}
$$
So it will suffice to show that the number
$$
f(a,b,c)=a(4-b)+b(4-c)+c(4-a) \tag{2}
$$
is $\leq 9$. Note that $f(a,b,c)=4(a+b+c)-(ab+ac+bc)$ is fully symmetric
in $a,b,c$.
We need a “lower-dimensional” result :
Lemma. Let $\alpha,\beta$ be two positive numbers such that $\alpha^3+\beta^3=2$.
Then $\alpha(3-\beta)+\beta(3-\alpha) \leq 4$, with equality iff $\alpha=\beta=1$.
Proof of lemma The proposed inequality is equivalent to
$\beta(3-2\alpha) \leq 4-3\alpha$, or $(2-\alpha^3)(3-2\alpha)^3 \leq (4-3\alpha)^3$. Now
$$
(4-3\alpha)^3-(2-\alpha^3)(3-2\alpha)^3=(\alpha-1)^4 \bigg(\frac{42}{25}+
\big(\frac{13}{10}-\alpha\big)\big(8\alpha+\frac{32}{5}\big)\bigg),
$$
which concludes the proof of the lemma.
Now let $a,b,c$ be positive numbers with $a^3+b^3+c^3=1$. Set $t=\big(\frac{b^3+c^3}{2}\big)^{\frac{1}{3}}$, so that $b^3+c^3=2t^3$. Then the numbers
$\alpha=\frac{b}{t}$ and $\beta=\frac{c}{t}$ satisfy the hypotheses of the lemma ; we deduce
$$
b(3t-c)+c(3t-b) \leq 4t^2, \ \text{with equality iff } \ b=c=t \tag{3}
$$
This means that
$$
f(a,b,c) \leq f(a,t,t), \ \text{with equality iff } \ b=c=t \tag{4}
$$
The set $K=\lbrace (a,b,c) \in [0,+\infty[^3 | a^3+b^3+c^3=1\rbrace$ is compact, so
the continuous map $f$ attains its maximum on $K$ at some point $(a_0,b_0,c_0)$. Then (4)
shows that we must have $b_0=c_0$, and by symmetry $a_0=b_0=c_0=1$, qed.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to evaluate this rational limit Evaluate:
$$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6}$$
| Exploiting the Taylor series of a function at the point $x=0$, we have
$$ \frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln(1+x)+x^4-x^6}$$
$$=\frac{(-3\sqrt{x}+x^3+(x^6-\frac{x^{12}}{3!}-\dots )))( (1+4\sqrt{x}+\dots)-1)}{4 (x+\frac{x^2}{2}\dots)+x^4-x^6}$$
$$ \sim \frac{(-3\sqrt{x})(4\sqrt{x})}{4x} = -3. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
| $$\begin{align}
\gcd(a+b, a^2 + b^2)
&= \gcd(a+b, a^2 + b^2 - a(a+b))
\\&= \gcd(a+b, b^2 - ab)
\\&= \gcd(a+b, b^2 - ab + b(a+b))
\\&= \gcd(a+b, 2b^2)
\end{align}
$$
Now, $\gcd(a+b,b) = \gcd(a,b) = 1$, so we can get rid of the factors of $b$ and have
$$ \gcd(a+b, a^2 + b^2) = \gcd(a+b, 2) $$
The strategy I used was still the basic idea of the Euclidean algorithm; since I couldn't compare numeric values, I instead simplified by working to eliminate the variable $a$, starting with the largest power of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
} |
Complex Integration : $\int_1^{1+i}\frac{1}{1+z^2}dz$
Integrate alonf the line segment from $z=1$ to $z=1+i$ :
$$\int_1^{1+i}\frac{1}{1+z^2}dz$$
If I integrate, it is just the identity $tan^{-1}z$, but the answer to this question is $$\frac{\pi}{4}-\frac{1}{2}\arctan2+\frac{i}{4}\log5$$ which I don't understand how they got?
| The integral can be parameterized by writing $z=1+it$:
$$\int_0^1 \frac{i}{(1+ti)^2+1}dt = \int_0^1 \frac{2 i + 2 t - i t^2}{4+t^4}dt$$
$$ = \int_0^1 \frac{2 t}{t^4+4} + i\int_0^1 \frac{2-t^2}{t^4+4}dt$$
$$ = \frac{i}{2} \arctan(1/2) - \frac{\log 5}{4} =
\frac{\pi}{4}-\frac{1}{2}\arctan2+\frac{i}{4}\log5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$ How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$?
For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is $(-\infty, -6]$ which is not my answer.
| Consider the families of quadratic functions $f_a(x)=(a+4)x^2-2ax+2a-6 $ we needs to find functions with their graph below $x$ axis these functions has maximum if $a+4<0$ and don't intersect the $x$ axis if discriminant is negative $D<0$ so we nedd to find solution of system of innequalities
$$\begin{matrix}
a+4<0\\
D=4a^2-4(a+4)(2a-6)<0
\end{matrix}$$
$$\begin{matrix}a<-4 \\ 4a^2-8(a+4)(a-3)<0
\end{matrix}$$
$$\begin{matrix}a<-4 \\a^2-2a+24<0\end{matrix}$$
$$\begin{matrix}a<-4 \\a^2+2a-24>0\end{matrix}$$
$$\begin{matrix}a<-4\\a^2+6a-4a-24>0\end{matrix}$$
$$\begin{matrix}a<-4,\\(a-4)(a+6)>0\end{matrix}$$
$a+6<0\iff a\in(-\infty,-6)$ is solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
help for proving an equation by induction For this equation:
$$-1^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)$$
how can I prove this by induction?
When I set $n = 1$ for the base case I got:
$$-1^3 + (-3)^3 + (-5)^3 + \ldots + (-3)^3 = -28$$
but am having trouble with the following inductive steps
| First, as the power of each summand is odd, I'd write the equation as
$$-1^3-3^3-5^3-\ldots-(2n+1)^3=-(n+1)^2(2n^2+4n+1)\Longleftrightarrow$$
$$1^3+3^3+5^3+\ldots+(2n+1)^3=(n+1)^2(2n^2+4n+1)$$
The base case is
$$n=1:\;\;\;\;\;1^3+3^3=28\stackrel ?=(1+1)^2(2+4+1)=28\ldots\ldots good$$
Assume for $\,n\,$ and show for $\,n+1\,$:
$$1^3+3^3+\ldots+(2n+1)^3+(2n+3)^3\stackrel{\text{Ind. hypothesis}}=(n+1)^2(2n^2+4n+1)+(2n+3)^3=\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/312405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
| Hint: use Parseval/Plancherel theorem on $(\sin{x}/x)^2$.
That is, the FT of $(\sin{x}/x)^2$ is
$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$
Plancherel/Parseval says that
$$\int_{-\infty}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{1}{2 \pi} \int_{-2 }^{2 } dk \: \pi^2 \left ( 1 - \frac{|k|}{2} \right )^2 = \frac{\pi}{2} \frac{4}{3} = \frac{2 \pi}{3}$$
$$\therefore \: \int_{0}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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Evaluating $\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$ $$I(a)=\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$$ I'd like to evaluate the integral by differentiating with respect to parameter $a$ but no success yet. Seems impossible. What would the other options?
Edit: A hypothetical closed form solution:
$$I(a)=\frac{\pi}{4}(1-a)e^{-a}$$
| A special case is easy, using $4 \sin^3(x) = 3 \sin(x) - \sin(3x)$:
$$
I(0) = \int_0^\infty \frac{\sin^3(x)}{x} \mathrm{d} x = \frac{3}{4} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x - \frac{1}{4} \int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x = \frac{\pi}{4}
$$
It is easy to see that $I^\prime(a)$ is singular at $a=0$.
In order to evaluate $I(a)$ numerically some massaging is needed:
$$ \begin{eqnarray}
I(a) &=& \int_0^\infty \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \int_{-\infty}^\infty \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \sum_{n=-\infty}^\infty \int_{-\frac{\pi}{2} + \pi n }^{\frac{\pi}{2} + \pi n} \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x}
\\ &=& \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sum_{n=-\infty}^\infty (-1)^n \sin^3(y) \cos(a \tan y) \frac{\mathrm{d}y}{y+\pi n} \\
&\stackrel{t =\tan(u)}{=} & \frac{1}{2} \int_{-\infty}^\infty \frac{u^3}{(1+u^2)^{5/2}} \cos(a u) f(u) \mathrm{d} u = \int_{0}^\infty \frac{u^3}{(1+u^2)^{5/2}} \cos(a u) f(u) \mathrm{d} u
\end{eqnarray}
$$
where $f(u)$ is defined by the sum:
$$
\begin{eqnarray}
f(u) &=& \sum_{n=-\infty}^\infty \frac{(-1)^n}{\pi n + \arctan(u)} = \frac{1}{\arctan(u)} - 2 \arctan(u) \sum_{n=1}^\infty \frac{(-1)^n}{\pi^2 n^2 - \arctan(u)^2} \\
&=& \frac{1}{\pi} \left( \Phi \left(-1,1,\frac{\arctan(u)}{\pi }\right)+\Phi
\left(-1,1,1-\frac{\arctan(u)}{\pi }\right) \right)
\end{eqnarray}
$$
where $\Phi(z,s,a)$ denotes the Lerch transcendent. With this we can plot $I(a)$ using numerical methods:
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "18",
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$F:= \{ a+b\sqrt{7} \mid a,b \in \mathbb{Q} \}$ closed under addition, subtraction, multiplication, and division I am in my math class and I came across this problem on my past midterm. How can we prove that $F:=\{ a+b\sqrt{7} \mid a,b \in \mathbb{Q} \} $ is closed under addition, subtraction, multiplication, and division by a nonzero number in the set?
| The only somewhat tricky one is division. Let $x=a+b\sqrt{7}$, where $a+b\sqrt{7}\ne 0$. Note that $a-b\sqrt{7}\ne 0$, since $\sqrt{7}$ is irrational. (This needs proof.)
Then
$$\frac{1}{x}=\frac{a-b\sqrt{7}}{(a-b\sqrt{7})(a+b\sqrt{7})}=\frac{a}{a^2-7b^2}+\frac{-b}{a^2-7b^2}\sqrt{7}.$$
Remark: To show that if $a$ and $b$ are not both $0$, then $a-b\sqrt{7}\ne 0$, maybe proceed as follows. Suppose to the contrary that $a-b\sqrt{7}=0$. By multiplying through by a suitable non-zero integer if necessary, we can assume that $a$ and $b$ are integers. We can also assume without loss of generality that the integers $a$ and $b$ have no common factor $\gt 1$. We obtain $a=b\sqrt{7}$, and therefore $a^2=7b^2$, Then $7$ divides $a^2$, but since $7$ is prime, $7$ divides $a$. Thus $a=7c$ for some $c$, and therefore $7c^2=b^2$. It follows that $7$ divides $b$, contradicting the fact that $a$ and $b$ have no common factor greater than $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Infinite Series :$ \sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2}$ Prove that:
$$\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2} = \frac{-\pi^{\frac{3}{2}}}{12}\left( \pi^2+6\gamma(1-2\log 2)-12\log 2\right)$$
where $\gamma$ is Euler-Mascheroni Constant and $\psi(z)$ is the Digamma Function.
| Consider the series
\begin{align}
S(x,y) = \sum_{n=0}^{\infty} \frac{\Gamma(n+x)}{n! \ (n+y+1)^{2}}.
\end{align}
This series can be evaluated as follows
\begin{align}
S(x,y) &= \sum_{n=0}^{\infty} \frac{\Gamma(x) \ (x)_{n}}{\Gamma(2) \ n!} \ \int_{0}^{\infty} e^{-(n+y+1)t} \ t \ dt \\
&= \Gamma(x) \ \int_{0}^{\infty} e^{-(y+1)t} \left( \sum_{n=0}^{\infty} \frac{\Gamma(x) \ (x)_{n} \ e^{-nt}}{n!} \right) \ t \ dt \\
&= \Gamma(x) \ \int_{0}^{\infty} e^{-(y+1)t} \ t \ (1-e^{-t})^{-x} \ dt \\
&= - \Gamma(x) \ \int_{0}^{1} u^{y} (1-u)^{-x} \ \ln(u) \ du \\
&= - \Gamma(x) \partial_{y} B(y+1, 1-x) \\
S(x,y) &= - \Gamma(x) \ B(y+1, 1-x) \ \left[ \psi(y+1) - \psi(2+y-x) \right].
\end{align}
which yields
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma(n+x)}{n! \ (n+y+1)^{2}} = - \Gamma(x) \ B(y+1, 1-x) \ \left[ \psi(y+1) - \psi(2+y-x) \right].
\end{align}
Taking the derivative with respect to $x$ leads to
\begin{align}
\partial_{x} S(x,y) &= - \Gamma(x) \ B(y+1,1-x) \ \left[ \psi_{1}(2+y-x) + (\psi(y+1) - \psi(2+y-x)) \right. \\
& \hspace{15mm} \left. \cdot(\psi(2+y-x) - \psi(1-x) + \psi(x)) \right]
\end{align}
which yields
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma(n+x) \ \psi(n+x)}{n! \ (n+y+1)^{2}} &=
- \Gamma(x) \ B(y+1,1-x) \ \left[ \psi_{1}(2+y-x) \right. \\
& \hspace{15mm} \left. + (\psi(y+1) - \psi(2+y-x)) \cdot(\psi(2+y-x) - \psi(1-x) + \psi(x)) \right].
\end{align}
Taking $x=y=1/2$ leads to the two series
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2} \right)}{n! \ \left(n+\frac{3}{2}\right)^{2}} = \frac{\pi^{3/2}}{2} ( 2 \ln 2 -1).
\end{align}
and
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2}\right) \ \psi\left(n+\frac{1}{2}\right)}{n! \ \left(n+ \frac{3}{2} \right)^{2}} &=
- \frac{\pi^{3/2}}{12} \ \left[ \pi^{2} + 6 \gamma (1 - 2 \ln 2 ) - 12 \ln 2 \right].
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
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Exponential equation, $(3+2\sqrt2)^x+1=6(\sqrt2+1)^x$ http://i.stack.imgur.com/YP2Ha.png
$$(3+2\sqrt2)^x+1=6(\sqrt2+1)^x \qquad\qquad x\in\mathbb{R}$$
I managed to find one of the solutions (x=2), but I got stuck. I would really appreciate a step by step solution. Thanks in advance :)
| $$(3+2^{1/2})^x+1=6((2^{1/2}+1)^x)$$
$$(2^{1/2}+1)^{2x}+1=6((2^{1/2}+1)^x)$$ taking$(2^{1/2}+1)^x=y$ we get
$$y^2-6y+1=0$$ from there
$$y_{1,2}=\frac{6\pm\sqrt{32}}{2}=\frac{6\pm4\sqrt{2}}{2}=3\pm 2\sqrt2$$ or
$$(2^{1/2}+1)^x=3+2\sqrt2=(\sqrt2+1)^2\Rightarrow x_1=2$$
$$(2^{1/2}+1)^x=3-2\sqrt2=(\sqrt2+1)^{-2}\Rightarrow x_2=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof: Two non identical circles have at most 2 same points I'm struggeling with an analytic proof for the fact, that two different circles have at most 2 same points. (I try to solve it analytical, because geometrical I already prooved it).
I tried to start with the equations
$r_1^2=(x-a_1)^2+(y-b_1)^2$ and $r_2^2=(x-a_2)^2+(y-b_2)^2$, further
$r_1^2=x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2$ and the same for the 2nd equation.
Then I get
\begin{eqnarray*}
r_1^2&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2\\
r_2^2&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2\\
0&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2\\
0&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Rightarrow& x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2=x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Leftrightarrow& 0=-2xa_1+a_1^2-2xb_1+b_1^2-r_1^2+2xa_2-a_2^2+2yb_2-b_2^2+r_2^2
\end{eqnarray*}
But now I don't know how to move on. Can someone give a hint?
| Write the circle equation as:
$$
x^2 + y^2 + \alpha x + \beta y + \gamma = 0
$$
This equation has three unknowns, so three points are sufficient to uniquely identify a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/326654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate the double integral $$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x^2+y^2+\sin(\pi(x^2+y^2)))\,dy\,dx$$
*sorry if the mathjax is off, I'm new at it.
Anyways, I can use the properties of double integrals to make it
$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x^2+y^2 \,dy\,dx + \int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sin(\pi(x^2+y^2))\,dy\,dx$$
From there I can solve the first double integral expression but I'm not sure on the second double integral expression:
$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sin(\pi(x^2+y^2))\,dy\,dx$$
A nudge in the right direction would be appreciated on how to solve this second expression.
| Convert the integral into polar coordinates. Note that the region you are integrating is the semi-circle of radius lying on the right half plane. Hence, the integral becomes
$$\int_{-\pi/2}^{\pi/2} \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr \,d \theta = \pi \cdot \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$$
All we need to evaluate now is $\displaystyle \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$.
\begin{align}
\int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr & = \int_0^3 r^3 dr + \int_0^3 r \sin( \pi r^2) \,dr = \left. \dfrac{r^4}4 \right \vert_{r=0}^{r=3} + \int_0^3 \dfrac{d( \cos(\pi r^2))}{-2 \pi}\\
& = \dfrac{81}4 - \dfrac{\cos(9 \pi) - \cos(0)}{2 \pi} = \dfrac{81}4 + \dfrac1{\pi}
\end{align}
Hence, the integral is $$\pi \left(\dfrac{81}4 + \dfrac1{\pi}\right) = 1 + \dfrac{{81}\pi}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
Consider an equilateral triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Rotate $P$ clockwise along $A$ by $60^o$ to $B'$, along $B$ by $60^o$ to $C'$, along $C$ by $60^o$ to $A'$
We note that,
$${\triangle APB} \cong ${\triangle ACB'}$$
$${\triangle APC} \cong ${\triangle A'CB}$$
$${\triangle BPC} \cong ${\triangle AC'B}$$
Thus we have, the area of the hexagon $AB'CA'BC'$
$$=2.{\triangle ABC}$$
We also observe that
${\triangle APB'}$, ${\triangle BPC'}$ and ${\triangle A'PC}$ are equilateral triangles
${\triangle PCB'} \cong {\triangle AC'P} \cong {\triangle A'BP} = RightAngle {\triangle}$
Thus we can say
$$2.{\triangle ABC} = [AB'CA'BC']$$
$$= {\triangle APB'} + {\triangle BPC'} + {\triangle A'PC} + 3.{\triangle PCB'}$$
$$= 12^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4} + 3.\frac{1}{2}(5)(12)$$
$$=169\frac{\sqrt3}{2} + 90$$
$${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then
$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
$$\frac{\sqrt3}{4}a^2 = 169\frac{\sqrt3}{4} + 45$$
$$a^2 = 169 + 45\frac{4}{\sqrt3}$$
$$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$
Simplifying
$$a = \sqrt{169 + 60\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How to simplify square root $$
\sqrt[3]{a}(\sqrt[3]{a^2}-\sqrt[3]{a^5})
$$
How can this be simplified? I can't find anything for doing the subtraction.
| *
*Distribute $\;\sqrt[\large 3]{a}\;$ over the difference,
*use the property that $\;\sqrt[\large a]{b}\cdot \sqrt[\large a]{c} =
\sqrt[\large a]{b\cdot c},\;$ and
*remember that $\sqrt[\Large a]{b^a} = b$
$$
\begin{align} \sqrt[\Large 3]{a}\left(\sqrt[\Large 3]{a^2}-\sqrt[\Large 3]{a^5}\right) & = \sqrt[\Large 3]{a}\cdot \sqrt[\Large 3]{a^2} - \sqrt[\Large 3]{a}\cdot \sqrt[\Large3]{a^5} \\ \\
& = \sqrt[\Large 3]{a\cdot a^2} - \sqrt[\Large a]{a\cdot a^5} \\ \\
& = \sqrt[\Large 3]{a^3} - \sqrt[\Large 3]{a^6} \\ \\
& = a - \sqrt[\Large 3]{(a^2)^3} \\ \\
& = a \;\;- \;\;a^2 \\ \\
&
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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For $x > -1$ proof that $ \arctan x + \arctan\frac{1-x}{1+x} = \frac{\pi}{4} $ For $x > -1$ proof that $\arctan x + \arctan\dfrac{1-x}{1+x} = \dfrac{\pi}{4} $
I have no idea how to approach this, some kind of help would be greatly appreciated!
edit: Thank you all!
| Let $\arctan(x) = a$. We then have
\begin{align}
\arctan(x) + \arctan \left(\dfrac{1-x}{1+x}\right) & = a + \arctan \left(\dfrac{1-\tan(a)}{1+\tan(a)}\right)\\
& = a + \arctan\left(\dfrac{\tan(\pi/4) - \tan(a)}{1+\tan(\pi/4) \cdot \tan(a)}\right)\\
& = a + \arctan \left(\tan(\pi/4-a)\right)
\end{align}
Note that since $x>-1$, we have $a = \arctan(x) \in \left(-\dfrac{\pi}4, \dfrac{\pi}2 \right)$. Hence, $$\dfrac{\pi}4 - a \in \left(-\dfrac{\pi}4, \dfrac{\pi}2 \right) \implies \arctan \left(\tan(\pi/4-a)\right) = \dfrac{\pi}4-a$$
This gives us
\begin{align}
\arctan(x) + \arctan \left(\dfrac{1-x}{1+x}\right)& = a + \arctan \left(\tan(\pi/4-a)\right)\\
& = a + \dfrac{\pi}4 - a = \dfrac{\pi}4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that :
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ .
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $
What to do next?
| As $A+B=\pi-C\implies \frac{A+B}2=\frac\pi2-\frac C2$
So, $$\tan\left(\frac{A+B}2\right)=\tan\left(\frac\pi2-\frac C2\right)$$
$$\frac{\tan \frac A2+\tan \frac B2}{1-\tan \frac A2\tan \frac B2}=\cot\frac C2=\frac1{\tan\frac C2} $$
$$\sum \tan \frac A2\tan \frac B2=1$$
Now, $$\sum (\tan \frac A2-\tan \frac B2)^2\ge0$$
$$2\left(\tan ^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)\ge2\sum \tan \frac A2\tan \frac B2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Solve in the positive real number the system of equation. Solve in the positive real numbers, the system of equations :
$$(2x)^{2013} + (2y)^{2013} + (2z)^{2013} =3 $$
and $$xy + yz + zx + 2xyz = 1.$$
| Let $x=\frac{a}{2}, y=\frac{b}{2}, z=\frac{c}{2}$, so
$$ab+bc+ca+abc=4, a^{2013}+b^{2013}+c^{2013}=3$$
Now by A.M.$\geq$G.M. $4=ab+bc+ca+abc \geq 3\sqrt[3]{a^2b^2c^2}+abc$. Let $x=\sqrt[3]{abc}$, so that $4 \geq 3x^2+x^3$.
Factorising, we get $(x-1)(x+2)^2 \leq 0$, so $x \leq 1$. Thus $abc \leq 1$, so $ab+ac+bc=4-abc \geq 3$.
Now by power mean inequality and the inequality $a^2+b^2+c^2 \geq ab+ac+bc$ (which is equivalent to $\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2) \geq 0$), we get
$$1=\sqrt[2013]{\frac{a^{2013}+b^{2013}+c^{2013}}{3}} \geq \sqrt{\frac{a^2+b^2+c^2}{3}} \geq \sqrt{\frac{ab+bc+ca}{3}} \geq 1$$
Since equality holds, we must have $a=b=c$, so $3=3a^{2013}$ and we get $a=b=c=1$.
Therefore the only solution in positive reals is $(x, y, z)=(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4.\binom{n}{r-2}+........+(-1)^r.(r+1).(r+2).\binom{n}{0}$
Now after that how can i calculate it
Thanks
| We shall use the combinatorial identity
$$\sum_{j=0}^{k}{(-1)^j\binom{n}{j}}=(-1)^k\binom{n-1}{k}$$
This can be proven easily by induction, and there is also probably some combinatorial argument why it holds. We shall use the equivalent form
$$\sum_{j=0}^{k}{(-1)^{k-j}\binom{n}{j}}=\binom{n-1}{k}$$
Now $(r-k+1)(r-k+2)=k(k-1)-(2r+2)k+(r^2+3r+2)$, so
\begin{align}
& \sum_{k=0}^{r}{(-1)^k(k+1)(k+2)\binom{n}{r-k}} \\
&=\sum_{k=0}^{r}{(-1)^{r-k}(r-k+1)(r-k+2)\binom{n}{k}} \\
& =\sum_{k=2}^{r}{(-1)^{r-k}k(k-1)\binom{n}{k}}-(2r+2)\sum_{k=1}^{r}{(-1)^{r-k}k\binom{n}{k}}+(r^2+3r+2)\sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}}
\end{align}
We have
\begin{align}
\sum_{k=2}^{r}{(-1)^{r-k}k(k-1)\binom{n}{k}} & =\sum_{k=2}^{r}{(-1)^{(r-2)-(k-2)}n(n-1)\binom{n-2}{k-2}} \\
& =n(n-1)\sum_{k=0}^{r-2}{(-1)^{(r-2)-k}\binom{n-2}{k}} \\
& =n(n-1)\binom{n-3}{r-2} \\
& =\frac{r(r-1)(n-r)}{n-2}\binom{n}{r}
\end{align}
\begin{align}
\sum_{k=1}^{r}{(-1)^{r-k}k\binom{n}{k}} & =\sum_{k=1}^{r}{(-1)^{(r-1)-(k-1)}n\binom{n-1}{k-1}} \\
& =n\sum_{k=0}^{r-1}{(-1)^{(r-1)-k}\binom{n-1}{k}} \\
& =n\binom{n-2}{r-1} \\
& =\frac{r(n-r)}{n-1}\binom{n}{r}
\end{align}
\begin{align}
\sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}} & =\binom{n-1}{r} \\
& =\frac{n-r}{n}\binom{n}{r}
\end{align}
Thus
\begin{align}
& \sum_{k=0}^{r}{(-1)^k(k+1)(k+2)\binom{n}{r-k}} \\
& =\frac{r(r-1)(n-r)}{n-2}\binom{n}{r}-(2r+2)\frac{r(n-r)}{n-1}\binom{n}{r}+(r^2+3r+2)\frac{n-r}{n}\binom{n}{r} \\
& =\binom{n}{r}\frac{(n-r)(r(r-1)n(n-1)-(2r+2)rn(n-2)+(r^2+3r+2)(n-1)(n-2))}{n(n-1)(n-2)} \\
& =\binom{n}{r}\frac{(n-r)(2r^2+(6-4n)r+(2n^2-6n+4))}{n(n-1)(n-2)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of pairwise coprime triples of positive integers (a,b,c) with aFind the number of pairwise coprime triples of positive integers $a,b,c$ with $a\lt b\lt c$ such that
a|bc−31, b|ca−31, c|ab−31
Details and assumptions:
The notation n∣m means that $n$ is a divisor of $m$.
Clarification: $(ab−31)$,$(ac−31)$, and $(bc−31)$ may be zero or negative.
Any help will be appreciated.
Thanks
| Hint: $a, b, c$ all divide $ab+ac+bc-31$, so since they are pairwise coprime, $$abc \mid (ab+ac+bc-31)$$
Continuation: Note that the above is equivalent to the given conditions.
Consider 3 cases, where $ab+ac+bc-31$ is $=0, <0, >0$.
Case 1: $ab+ac+bc-31=0$, then $31=ab+ac+bc \geq a(a+1)+a(a+2)+(a+1)(a+2)=3a^2+6a+2$, so $a \leq 2$.
If $a=1$, then $c>b>1$, and $(b+1)(c+1)=32$. We have $3 \leq b+1<\sqrt{32}$ so $b+1=4, c+1=8$ and $(a, b, c)=(1, 3, 7)$.
If $a=2$, then $c>b>2$ and $(b+2)(c+2)=35$. We have $5 \leq b+2<\sqrt{35}$ so $b+2=5, c+2=7$ and $(a, b, c)=(2, 3, 5)$.
Case 2: $ab+ac+bc-31<0$, then $abc \leq |ab+ac+bc-31|=31-ab-ac-bc$.
If $a \geq 2$, then $b \geq 3, c \geq 4$, so $31 \geq abc+ab+ac+bc \geq 2(3)(4)+2(3)+2(4)+3(4)>31$, a contradiction.
Thus $a=1$, so $c>b>1$ and $2bc+b+c \leq 31$.
If $b \geq 3$, then $c \geq 4$, so $31 \geq 2bc+b+c \geq 2(3)(4)+(3)+(4)=31$, so equality holds and we have $(a, b, c)=(1, 3, 4)$. Checking, this works.
Otherwise $b=2$, so $5c \leq 29$, giving $c=3, 4, 5$. However we also have $c \mid ab-31=-29$, so this gives no solution.
Case 3: $ab+ac+bc-31>0$, then $abc \leq ab+ac+bc-31$.
If $a \geq 3$, then $abc \geq 3bc \geq ab+ac+bc>ab+ac+bc-31 \geq abc$, a contradiction.
Thus $a=1, 2$. If $a=2$, then $c>b>2$ and $2bc \leq 2b+2c+bc-31$, so $bc+31 \leq 2b+2c$, so $(b-2)(c-2)+27 \leq 0$, a contradiction.
Thus $a=1$, so $bc \leq b+c+bc-31$, so $b+c \geq 31$. Also $bc \mid (b+c+bc-31)$, so $bc \mid (b+c-31)$. If $b+c=31$, we have $(a, b, c)=(1, 2, 29), (1, 3, 28), \ldots, (1, 15, 16)$. Otherwise $bc \leq b+c-31$, so $(b-1)(c-1)+30 \leq 0$, a contradiction.
To conclude, the solutions are $(1, 3, 7), (2, 3, 5), (1, 3, 4), (1, 2, 29), (1, 3, 28), \ldots, (1, 15, 16)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorize $3m^4-6m^3+14m^2-6m+11$ I have this expression:
$3m^4-6m^3+14m^2-6m+11=0$ and I want to factorize it in $(m^2+1)(3m^2-6m+11)$.
How can I do it? Thanks for any help!
| $3m^4-6m^3+14m^2-6m+11$
$=3m^4-6m^3+11m^2+3m^2-6m+11$
$=m^2(3m^2-6m+11)+(3m^2-6m+11)$
$=(m^2+1)(3m^2-6m+11)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof: How to prove $n$ is odd if $n^2 + 3$ is even New to the whole proof thing.
Trying to figure out that, for all integers $n$, if $n^2 + 3$ is even, then $n$ is odd.
Thank you for the help.
| If $n^2 + 3$ is even then $n^2 + 3 = 2k$ , where $k$ is an integer.
Thus, $n^2 = 2k - 3 = 2k - 4 + 1 = 2(k-2) + 1$, where $k$ is an integer, which means $k-2$ is an integer.
Now, this implies that $n^2$ is odd.
Assume $n$ is even.
Then $n$ can be written in the form $2m$, where $m$ is an integer.
Thus, $n^2 = (2m)^2 = 4m^2 = 2(2m^2)$.
We know $2m^2$ is an integer.
Thus $n^2$ is even.
But this gives us a contradiction because we already have established that $n^2$ is odd.
Therefore $n$ is odd.
| {
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How do I determine the intersection of span A and span B? Consider the following 2 sets of vectors in $\mathbb R^4$: $A = \{v_1, v_2, v_3\}, B = \{w_1, w_2, w_3\}$. You are given that $A$ is a set of linearly independent vectors and that $B$ is a set of linearly independent vectors.
Let
$v_1 = (3,1,4,1), v_2 = (5,9,2,6), v_3 = (5,3,5,8), w_1 = (9,7,9,3), w_2 = (2,3,6,4), w_3 = (6,2,8,4)$.
(Wrote them sideways when it should be top to bottom in brackets but left to right shown)
Determine the intersection of span$\,A$ and span$\,B$ and write your answer as the span of a set of linearly independent vectors.
I'm really lost in class. Please show steps and answers that I can learn. Please help...
Thank you
| The intersection will have dimension 2 or 3 so we can control our results. As your vectors doesn't look very nice we will use the gauss algorithm to make them sweeter. Sweet vectors are vectors with a lot of zeroes. When we calculate $$a_1=v_2-v_3=\begin{pmatrix} 5-5 \\ 9-3 \\ 2-5 \\ 6-8\\ \end{pmatrix} = \begin{pmatrix} 0 \\ 6 \\ -3 \\ -2 \\ \end{pmatrix}$$
Calculating
$$a_2=3 v_2 -5 v_1= \begin{pmatrix} 15-15 \\ 27-5 \\ 6-20 \\ 18-5\\ \end{pmatrix}=
\begin{pmatrix} 0 \\ 22 \\-14\\ 13 \end{pmatrix}$$
When we calculate now
$$a_3=3a_2-11 a_1= \begin{pmatrix} 0 \\ 66-66\\ -42+33\\ 39+22\\ \end{pmatrix}= \begin{pmatrix}
0 \\ 0 \\-9 \\ 61
\end{pmatrix}$$
Now we calculate
$$a_4=3a_1 -a_3=\begin{pmatrix} 0 \\ 18\\ -9+9\\ -2-61 \\ \end{pmatrix}=\begin{pmatrix} 0 \\18 \\ 0 \\ 63\\ \end{pmatrix}=9\cdot \begin{pmatrix} 0\\ 2 \\ 0 \\ 7 \\\end{pmatrix}$$
Simplify $v_1$ like this too and the other set, and it will give you an easy (easier) system of equations.
Mathematica gives me those reduced echolons forms
$$A=\left(
\begin{array}{cccc}
1 & 0 & 0 & \frac{191}{18} \\
0 & 1 & 0 & -\frac{67}{18} \\
0 & 0 & 1 & -\frac{61}{9} \\
\end{array}
\right)\qquad B=
\left(
\begin{array}{cccc}
1 & 0 & 0 & -\frac{28}{61} \\
0 & 1 & 0 & -\frac{6}{61} \\
0 & 0 & 1 & \frac{53}{61} \\
\end{array}
\right)
$$
where every row is a vector.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/341462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Complete induction:figure out the last step I tried to solve this task on complete induction. I am sorry for the form, if you see a $6$
it means: a divisible through six. Could anyone please check if I have done right? Because at the end, I don´t know how to show that 4n³+2n²+3n is divisible through six.
Please show by complete induction, that for every $n\in\mathbb N$: $(n²)² -n³ -n²+n$ is divisible through $6$
1) $\frac{(1²)²-1³ -1²+1 }{6}=\frac06= 0$ true statement
2) inductive hypothesis:
$\frac{(n²)²-n³-n²+n }{6}$
3) inductive step:
$n\gt n+1$
$$\frac{((n+1)²)²-(n+1)³-(n+1)²+(n+1)}{6}=$$
$$\frac{(n+1)(n+1)(n+1)(n+1)-(n+1)(n+1)(n+1)-(n+1)(n+1)+(n+1)}{6}=$$
$$\frac{(n²+2n+1)(n²+2n+1)–(n²+2n+1)(n+1)-(n²+2n+1)+(n+1)}{6}=$$
$$\frac{(n²)²+ 2n³+n² +2n³+4n²+2n+1–(n²+2n+1)(n+1)-(n²+2n+1)+(n+1)}{6}=$$
$$\frac{(n²)²+4n³+5n²+2n+1-(n³+n²+2n²+2n+n+1)-(n²+2n+1)+(n+1)}{ 6}=$$
$$4(n²)²-n³-n²-2n$$
=inductive
hypothesis
$+ 4n³+2n²+3n $ ??? (divisible through six?)
| If you want to use induction, you are looking to assume $n^4-n^3-n^2+n$ is a multiple of $6$ and prove $(n+1)^4-(n+1)^3-(n+1)^2+n$ is. So $$((n+1)^4-(n+1)^3-(n+1)^2+n+1)-(n^4-n^3-n^2+n)=4(n+1)^3+6(n+1)^2+4(n+1)+1-3(n+1)^2-3(n+1)-1-2(n+1)-1+1=4n^3+3n^2-n+=n(n+1)(4n-1)$$
One of the first two terms is even and one of the three is a multiple of $3$, so the difference $f(n+1)-f(n)$ is divisible by $6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/342046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$ Show that
$$\begin{aligned} \int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2}
\right)dx &= \frac{7\pi^3}{108} \\ \int_0^{\pi/3}x\log^2
\left(2\sin\frac{x}{2} \right)dx &= \frac{17\pi^4}{6480}\end{aligned}$$
*
*I can solve $\displaystyle \int_0^\pi \log^2 \left(2\sin \frac{x}{2}
\right)dx $ but I don't know what to do if the limits are from $0$ to $\pi/3$.
*I have no idea what to do if the integrand contains an $x$.
*I feel that the Polylogarithm function will be involved however I don't know how it can be implemented here.
It would be really great if someone could take the initiative to prove these.
| Here's another approach for evaluating the one without the $x$ in front.
First notice that it's equivalent to showing that $$\int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx = \frac{7 \pi^{3}}{216}. $$
Using the principal branch of the logarithm and assuming that $0 < x < \pi$, we have $$ \begin{align} \log(1-e^{2ix}) &= \log (e^{-ix}-e^{ix}) + \log(e^{ix}) \\ &= \log(-2i \sin x) + ix \\ &= \log(2 \sin x) - \frac{i \pi}{2} + ix. \end{align}$$
Squaring both sides and integrating, $$\int_{0}^{\pi /6} \left(\log(2 \sin x) - \frac{i \pi}{2} + ix \right)^{2} \ dx = \int_{0}^{\pi /6} \log^{2} (1-e^{2ix}) \ dx . $$
Then equating the real parts on both sides of the equation, we get
$$\begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \int_{0}^{\pi/6} \left(x- \frac{\pi}{2} \right)^{2} \ dx + \text{Re} \int_{0}^{\pi /6} \log^{2}(1-e^{2ix}) \ dx \\ &= \frac{19 \pi^{3}}{648} +\text{Re} \int_{C} \log^{2}(1-z) \frac{dz}{2iz} \\ &=\frac{19 \pi^{3}}{648} + \frac{1}{2} \ \text{Im} \int_{C} \frac{\log^{2}(1-z)}{z} \ dz \end{align}$$
where $C$ is the portion of the unit circle from $z=1$ to $z=e^{ \pi i /3}$.
But since $\frac{\log^{2}(1-z)}{z}$ is analytic for $\text{Re}(z) <1$,
$$ \begin{align} \int_{C} \frac{\log^{2}(1-z)}{z} \ dz &= \int_{1}^{e^{\pi i /3}} \frac{\log^{2}(1-z)}{z} \ dz . \end{align} $$
Then integrating by parts twice, we get
$$ \begin{align} \text{Im} \int_{1}^{e^{\pi i /3}} \frac{\log^{2}(1-z)}{z} \ dz &= \text{Im} \ \log^{2}(1-z) \log(z) \Bigg|^{e^{\pi i /3}}_{1} + 2 \ \text{Im} \int_{1}^{e^{\pi i /3}} \frac{\log(1-z) \log (z)}{1-z} \ dz \\ &= \text{Im} \ \log^{2}(e^{-\pi i /3}) \log(e^{\pi i /3}) + 2 \ \text{Im} \ \log(1-z) \text{Li}_{2}(1-z) \Bigg|^{e^{\pi i / 3}}_{1} \\ &+ 2 \ \text{Im} \int_{1}^{e^{\pi i / 3}} \frac{\text{Li}_{2}(1-z)}{1-z} \ dz \\ &=- \frac{ \pi^3}{27} - \frac{2 \pi }{3}\text{Im} \ i \ \text{Li}_{2} (e^{- \pi i /3}) - 2 \ \text{Im} \ \text{Li}_{3}(1-z) \Bigg|^{e^{\pi i/3}}_{1} \\ &= - \frac{ \pi^3}{27} - \frac{2 \pi }{3}\text{Im} \ i \ \text{Li}_{2} (e^{- \pi i /3}) - 2 \ \text{Im} \ \text{Li}_{3}(e^{ -\pi i /3}) \\ &= - \frac{\pi^3}{27} - \frac{2 \pi }{3} \sum_{n=1}^{\infty} \frac{\cos (n \pi /3)}{n^{2}} +2 \sum_{n=1}^{\infty} \frac{\sin (n \pi /3)}{n^3}. \end{align}$$
Integrating both sides of the Fourier series $$\sum_{n=1}^{\infty} \frac{\sin (k \theta)}{k} = \frac{\pi - \theta}{2} \ , \ 0 < \theta < 2 \pi$$
we get
$$\sum_{n=1}^{\infty} \frac{\cos (k \theta)}{k^{2}} = \frac{\theta^{2}}{4} - \frac{\pi \theta}{2} + \frac{\pi^{2}}{6} .$$
And integrating a second time, $$ \sum_{n=1}^{\infty} \frac{\sin (k \theta)}{k^{3}} = \frac{\theta^{3}}{12} - \frac{\pi \theta^{2}}{4} + \frac{\pi^{2} \theta}{6}.$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{\cos (n \pi /3)}{n^{2}} = \frac{\pi^{2}}{36} $$
and $$ \sum_{n=1}^{\infty} \frac{\sin (n \pi /3)}{n^{3}} = \frac{5 \pi^{3}}{162}. $$
So finally we have
$$ \begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \frac{19 \pi^{3}}{648} + \frac{1}{2} \left[ - \frac{ \pi^{3}}{27} - \frac{2 \pi }{3} \left(\frac{\pi^{2}}{36} \right) + 2 \left( \frac{5 \pi^{3}}{162} \right) \right] \\ &= \frac{7 \pi^{3}}{216} . \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/342777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 3,
"answer_id": 0
} |
logarithm problem - four tuple How many distinct four tuple (a,b,c,d) of rational numbers are there with $a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7=2005$
Can we proceed like this :
Using $\log a +\log b = \log(ab)$ and $m\log a = \log a^m$
$\Rightarrow \log_{10}2^a \cdot 3^b \cdot 5^c \cdot 7^d = 2005$
Please guide how to proceed further..
| Hint: $10^{2005}=2^a \cdot 3^b \cdot 5^c \cdot 7^d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/343811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.