Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the n-th derivative of a rational function Given $$f : \left(\frac{-1}{2}, \frac{1}{2}\right) \to \mathbb{R} \ \ \ \ \ \
\ \ \ \ f(x) = \frac{1}{1-x-x^2}$$
I am trying to find an expression for $f^{n}(x)$ and prove it by induction afterwards.
I tried first splitting into partial fractions but it got messy, not sure if that's the right approach.
| $$
1-x-x^2 = -\left(x-\frac{-1+\sqrt 5} 2\right)\left(x-\frac{-1-\sqrt5} 2\right) = \left( \frac{-1+\sqrt 5} 2 - x \right) \left( \frac{1+\sqrt 5} 2 + x \right)
$$
$$
\frac A {\frac{-1+\sqrt 5} 2 - x} + \frac B {\frac{1+\sqrt5} 2 + x} = \frac 1{\left( \frac{-1+\sqrt 5} 2 - x \right) \left( \frac{1+\sqrt 5} 2 + x \right)}
$$
$$
A\left(\frac{1+\sqrt 5} 2 + x \right) + B\left( \frac{-1+\sqrt 5} 2 - x \right) = 0x + 1
$$
So $A-B=0$ and $(A+B) \left( \dfrac{\sqrt 5} 2 \right) = 1.$ Hence $A=B=\dfrac 1 {\sqrt 5}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Stuck calculating the derivative of $f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$. I have to calculate the derivative of this:
$$f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$$
But I'm stuck. This is the point where I have arrived:
$$f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$$
How can I simplify? I didn't include all the passages.
| You have $\frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$
splitting into 2 fractions gives ;
$\frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}+\frac{x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$
$=\frac1{x(\ln(10))}+\frac{x}{(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$
$ = \frac1{x(\ln(10))}+\frac{10^y}{\ln(10)(\sqrt{5-x^2})}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Rational Functional Equation: $3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$ Suppose $f(x)$ is a rational function such that
$3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$
for all $x \neq 0$. Find $f(-2)$.
I tried substituting different values of $x$ to get a system of equations to solve for $f(x)$, but this didn't work. How should I take this from here?
| Following lulu's comment, set $x=-2$ and $-\frac{1}{2}$ to find, respectively,
$$\begin{cases}3f(-\frac{1}{2})+\frac{2f(-2)}{-2}=4 \\ 3f(-2)+\frac{2f(-\frac{1}{2})}{-\frac{1}{2}}=\frac{1}{4}\end{cases}$$so that we have a system of simultaneous equations in $f(-2)$ and $f(-\frac{1}{2})$. Solving this system yields $f(2)=\frac{67}{20}. $
To motivation this substitution, notice that $f(x)=\frac{1}{x}$ is an involution. In fact, comparing the original functional equation with the equation formed by setting $x=\frac{1}{x}$ allows us to find the general solution for $f(x)$.
| {
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Simple algebra simplification I have
$ \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$
Which simplified to
$ -4rS^{2} \pm \sqrt{16r^{2}S^{4} +1}$
But I can’t seem to get this
| Let $$D = \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$$
Now
\begin{align}
D &= \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}\\
&=\frac{-8rS^{2}}{2} \pm \frac{\sqrt{64r^{2}S^{4} + 4}}{2} \\
&= -4rS^2 \pm \frac{\sqrt{64r^{2}S^{4} + 4}}{\sqrt{4}} \\
&= -4rS^2 \pm \sqrt{\frac{64r^{2}S^{4} + 4}{4}}\\
&=-4rS^2 \pm \sqrt{16r^{2}S^{4} + 1}
\end{align}
| {
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Proving Hermite's Identity in a different approach To prove
$$S=\left [x \right]+\left [x+\frac{1}{n} \right]+\left [x +\frac{2}{n}\right]+\cdots+\left [x +\frac{n-1}{n}\right]=\left [nx \right]$$
using and starting with $$x-1 \lt \left [x \right]\le x \tag{1}$$ we have
$$x+\frac{1}{n}-1 \lt \left [x +\frac{1}{n}\right] \le x+\frac{1}{n} \tag{2}$$
$$x+\frac{2}{n}-1 \lt \left [x +\frac{2}{n}\right] \le x+\frac{2}{n} \tag{3}$$
$$x+\frac{3}{n}-1 \lt \left [x +\frac{3}{n}\right] \le x+\frac{3}{n} \tag{4}$$
and so on till
$$x+\frac{n-1}{n}-1 \lt \left [x +\frac{n-1}{n}\right] \le x+\frac{n-1}{n} \tag{n}$$
Adding all we get
$$nx+\frac{n(n-3)}{2} \lt S \le nx+\frac{n(n-1)}{2}$$
Now how can we prove that between $nx+\frac{n(n-3)}{2}$ and $nx+\frac{n(n-1)}{2}$ there is only on integer which is $\left [nx \right]$?
| This method doesn't work.
Adding all of them we get $$nx+\frac{\frac{n(n-1)}2}{n}-n<S\le nx+\frac{\frac{n(n-1)}2}{n}$$ or $$nx-\frac{n+1}2<S\le nx+\frac{n-1}2\implies -\frac n2<S-nx+\frac12\le\frac n2$$ and this interval gets wider and wider as $n$ gets larger.
| {
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Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1+C_2$$
$$=
\left|
\begin{matrix}
(c+a)&(a-b)&a \\
(a+b)&(b-c)&b \\
(b+c)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1-C_3$$
$$=
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|$$
How do I complete the rest?
| You're close. Perform $C_2 \mapsto C_2 - C_3$. Then you have
$$- \begin{vmatrix} c & b & a \\ a & c & b \\ b & a & c\end{vmatrix}$$
Now apply the determinant formula,
$$-(c(c^2 - ba) - b(ac - b^2) + a(a^2 - cb)) = 3abc - a^3 - b^3 - c^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Calculate $ \lim_{n\to \infty} \frac{n}{\ln(n)} \int_{1}^{n^2} \frac{\ln(x)}{x^2+nx+n^2}\,dx $ For every $ n\in \mathbb{N} $ and positive $ x $ , $ x \neq 0 $ we consider the function $ f_{n}(x) = \frac{\ln(x)} {x^2+nx+n^2} $
Calculate
$ \lim_{n\to \infty} \frac{n}{\ln(n)} \int_{1}^{n^2} f_{n}(x) \,\mathrm dx$
The correct answer should be $ \frac{2\pi \sqrt{3}}{9} $
How to approach this using high-school techniques? The result suggests that we have to work with an arctangent function probably.
| Hint
With $x=ny$ you obtain:
$$n \int_1^{n^2} f_n(x) dx = \int_\frac{1}{n}^n \frac{\ln(ny)}{n^2(y^2+y+1)} n^2 dy=\int_\frac{1}{n}^n \frac{\ln(n)+\ln(y)}{(y^2+y+1)} dy$$
So:
$$\frac{n}{\ln(n)} \int_1^{n^2} f_n(x) dx= \int_\frac{1}{n}^n \frac{1}{(y^2+y+1)} dy+\frac{1}{\ln(n)} \int_\frac{1}{n}^n \frac{\ln(y)}{(y^2+y+1)} dy$$
As $$ \int_0^{+\infty} \frac{|\ln(y)|}{(y^2+y+1)} dy <+ \infty$$ you have:
$$\left|\frac{1}{\ln(n)} \int_\frac{1}{n}^n \frac{\ln(y)}{(y^2+y+1)} dy \right| \leq \frac{1}{\ln(n)} \int_0^{+\infty} \frac{\ln(y)}{(y^2+y+1)} dy \to 0$$
so:
$$\lim_{n \to +\infty} \frac{n}{\ln(n)} \int_1^{n^2} f_n(x) dx = \lim_{n \to + \infty} \int_\frac{1}{n}^n \frac{1}{1+y+y^2} dy =\int_0^{+\infty} \frac{1}{1+y+y^2} dy$$
and the last integral is exactly:
$$\frac{2 \pi \sqrt{3}}{9}$$
| {
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British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz? The question states:
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$
Prove that
$x^2yz + xy^2z + xyz^2 ≤ 1/3$
I have a proof of this relying on the fact that:
$x^2/3 +y^2/3 + z^2/3 \geq (x+y+z)^3/9 $ (A corollary of C-S I believe)
Is there an elementary proof without this fact (or C-S in general)?
| I found an alternative proof using the GM-AM inequality and a small observation.
Firstly, using the AM-GM inequality for x,y,z we get
$x^2yz + xy^2z + xyz^2 = (x+y+z)xyz = (x+y+z)\left(\sqrt[3]{xyz}\right)^3 \leq \frac{(x+y+z)^4}{27} = \frac{1}{3}\frac{(x+y+z)^4}{9}$
Now, notice that: $(x+y+z)^4 \leq 9(x^2 + y^2 + z^2)^2$. This is because we can square-root both sides to get:
$(x+y+z)^2 \leq 3(x^2 + y^2 + z^2) \Leftrightarrow 0 \leq (x-y)^2 + (y-z)^2 + (z-x)^2$
Hence, continuing the first line of the proof we get
$\frac{1}{3}\frac{(x+y+z)^4}{9} \leq \frac{1}{3}(x^2 + y^2 + z^2)^2 = \frac{1}{3}$
and we are done!
Edit: This line of reasoning nicely shows that the equality holds iff $x=y=z$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find two least composite numbers $n$ such that $n\mid2^n-2$ and $n\mid3^n-3$.
$n\mid2^n-2$
$n\mid3^n-3$.
It comes down to
$2^n \equiv 2 \mod n$
$3^n \equiv 3 \mod n$
But how can I solve this?
| As suggested in the comments, one clear path for the solution is to consider Carmichael numbers. The first two work: $561= 3 \cdot 11 \cdot 17$ and $1105= 5 \cdot 13 \cdot 17$.
The solutions $n \le 10000$ are
$$
\begin{align}
561&= 3 \cdot 11 \cdot 17 \\
1105&= 5 \cdot 13 \cdot 17 \\
1729&= 7 \cdot 13 \cdot 19 \\
2465&= 5 \cdot 17 \cdot 29 \\
2701&= 37 \cdot 73 \\
2821&= 7 \cdot 13 \cdot 31 \\
6601&= 7 \cdot 23 \cdot 41 \\
8911&= 7 \cdot 19 \cdot 67 \\
\end{align}
$$
Of these, $2701$ is not a Carmichael number.
| {
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About the relation $\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2$ It is well known that
$$\int\frac{1}{1+x^2}\,\mathrm{d}x=\tan^{-1}x+C \tag{1}$$
However, I integrated this differently and got an unusual result.
Suppose we make the substitution $x=\sinh\theta$ and $\mathrm{d}x=\cosh\theta\,\mathrm{d}\theta$ so the integral becomes $$\int\frac{\cosh\theta}{\cosh^2\theta}\,\mathrm{d}\theta=\int\frac{1}{\cosh\theta}\,\mathrm{d}\theta \tag{2}$$
By the definition of $\cosh\theta$, we can rewrite this as $$\int\frac{2e^\theta}{e^{2\theta}+1}\,\mathrm{d}\theta=2\tan^{-1}e^\theta+C \tag{3}$$
Using the fact that $e^\theta=\cosh\theta+\sinh\theta$, we get $e^\theta=x+\sqrt{1+x^2}$, so the answer is then $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C \tag{4}$$
Equating $(4)$ with $(1)$, we have $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C=\tan^{-1}x \tag{5}$$
Plugging in $x=0$, we find $C=-\frac\pi2$. We now have the following strange relationship
$$\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2 \tag{$\star$}$$
This leads me to wonder: Why is this true geometrically, and does this relationship extend into the complex plane?
| We need to be very careful with principal values ,
Let arccot$(x)=2y\implies x=\cot2y$ and $0<2y<\pi$
$x+\sqrt{1+x^2}=\dfrac{\cos2y+1}{\sin2y}=\cot y$ as $y\ne\dfrac\pi2$
$\implies2\tan^{-1}(x+\sqrt{1+x^2})=2\left(\dfrac\pi2-\cot^{-1}(x+\sqrt{1+x^2})\right)=\pi-2y$
and $\tan^{-1}x=\dfrac\pi2-\cot^{-1}x=\dfrac\pi2-2y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761206",
"timestamp": "2023-03-29T00:00:00",
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Express the given polar equation in simplest rectangular form: $r = 4 + 3cosθ$ Express the given polar equation in simplest rectangular form: $r = 4 + 3cosθ$
My attempt
Multiplying r on both sides, we get
$r^2=4r+3 rcos\theta$
since, $x=rcos\theta$ and $y=rsin\theta$
and $r^2=x^2+y^2$
we get $x^2+y^2=4r+3x$
can anyone please explain after this step..
| You have $r= 4+3\cos(\theta)$
$r^2 = 4r+3r\cos(\theta)$
we know that ;$x= r\cos(\theta)\qquad y = r\sin(\theta)$
$r^2 = x^2+y^2$
$\implies x^2+y^2 =\sqrt{x^2+y^2}+3x$
$\implies (x^2+y^2-3x)^2 = x^2+y^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring Quadratic Trinomials that don't simplify by common terms. $10x^2 + 13xy - 3y^2$
$(10x^2 + 2xy) + (-15xy - 3y^2)$
$2x(5x + y) + 3y(-3x - y)$
How would this factor out further?
| $10x^2+13xy-3y^2$
$=10(x^2+1.3xy-0.3y^2)$
$=10(x^2+2\times x \times 0.65y+0.4225y^2-0.7225y^2)$
$=10(x+0.65y)^2-(\sqrt{7.225}y)^2$
Then you can use the equality $a^2-b^2=(a-b)(a+b)$ to continue:
Next steps (spoilers):
$=\left(\sqrt{10}x+\sqrt{10}\times 0.65y+\dfrac{17\sqrt{10}}{20}y\right)\left(\sqrt{10}x+\sqrt{10}\times 0.65y-\dfrac{17\sqrt{10}}{20}y\right)$
Next step:
$=\sqrt{10}\left(x+ 0.65y+\dfrac{17}{20}y\right)\sqrt{10}\left(x+ 0.65y-\dfrac{17}{20}y\right)$
Finally:
$=10(x+1.5y)(x-0.2y)=(2x+3y)(5x-y)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Removing a discontinuity? How would you "remove the discontinuity" of $f$ ? In other words, how would you define $f(4)$ in order to make $f$ continuous at $x=4$?
$$f(x) = \dfrac{x^2-x-12}{x-4}$$
| You have $f(x)= \dfrac{x^2-x-12}{x-4}$. Notice that $x=4$ is not in the domain of the function since then you would be dividing by $0$. However, if $x \neq 4$, then we have
$$
\require{cancel}
f(x)= \dfrac{x^2-x-12}{x-4}= \dfrac{(x-4)(x+3)}{x-4}=\dfrac{\cancel{(x-4)}(x+3)}{\cancel{x-4}}= x+ 3
$$
Notice that $x+3$ gets 'close' to $4+3=7$ when $x$ is 'close' to $x=4$. Then if we want to define a function which is equal to $f(x)$ when $x \neq 4$, is defined at $x=4$, and is continuous, we have to choose
$$
g(x)=
\begin{cases}
\dfrac{x^2-x-12}{x-4}= x+3, & x\neq 4 \\
7, & x=4
\end{cases}
$$
| {
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How can one prove that $\sqrt{ 2} \cdot \sqrt{ 3} = \sqrt{ 6}$? I have already proved that $\sqrt{2}\cdot \sqrt{2}=2$ so I hope I can now use $\sqrt{3}\cdot \sqrt{3}=3$ and the same for 6.
The exercise comes from Stillwell: Mathematics and its History. The other exercises have not been complicated so probably the solution to proving $\sqrt{2}\cdot \sqrt{2}=2$ I borrowed from Martin K was way over the top. But it cannot be adapted, I think, to this one where the roots are not of the same number.
Here is my attempt:
$r^2=2, s^2=3$.
$r\cdot r=2$ implies $r=\sqrt{2}$ and $s\cdot s=3$
implies $s=\sqrt{3}$
I also know that $\sqrt{6}\cdot \sqrt{6}=6$.
$r^2\cdot s^2=6$ so $(rs)^2=6$
This implies that $(\sqrt{2}\sqrt{3})^2=\sqrt{6}\cdot \sqrt{6}$
Which I am not sure about, but hope, implies that $\sqrt{2}\sqrt{3}=\sqrt{6}$
| Note that
$$a=\sqrt{2}\implies a^2=2$$
$$b=\sqrt{3}\implies b^2=3$$
then
$$(ab)^2=6\implies ab=\sqrt{6}$$
| {
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replacing Inequalities I encountered a problem today:
Prove that:
$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2} \ge \frac{a+b+c}{3}$$
for all $a,b,c>0$
I used the RMS-AM inequality to replace the LHS with
$$\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{3}}$$
and replaced the RHS using AM-GM inequality
$$\frac{3abc}{a^2+b^2+c^2}$$
I can prove the new inequality, but does that mean I have proved the original inequality? I couldn't find another way to prove the original inequality except for expanding the terms and using scalar products. Thanks in advance!
| We can use the Chebyshov's inequality.
Since $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering, we obtain:
$$a^3+b^3+c^3=a^2\cdot a+b^2\cdot b+c^2\cdot c\geq\frac{1}{3}(a^2+b^2+c^2)(a+b+c).$$
Also, you can use PM and C-S.
Indeed, by PM $$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\geq\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which gives your
$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\geq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$
Thus, it's enough to prove that
$$\sqrt{3(a^2+b^2+c^2)}\geq a+b+c,$$ which is true by C-S:
$$\sqrt{3(a^2+b^2+c^2)}=\sqrt{(1+1+1)(a^2+b^2+c^2)}\geq\sqrt{(a+b+c)^2}=a+b+c.$$
| {
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Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!
BACKGROUND: The background of this question is that I tried to find this integral:
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
| Maybe instead of integrating you need just differentiating.
$$\frac{df}{dx}=\frac{(2x^2-x)(2x-1)-(x^2-x+1)(4x-1)}{denom^2}\\
=\frac{x^2-4x+1}{denom^2}$$
In the other hand:
$$\frac{dg}{dx}=\frac{(x-2)(2x-1)-(x^2-x+1)}{denom^2}\\
\ =\frac{x^2-4x+1}{denom^2}=\frac{df}{dx}$$
See that both functions have same derivates, which means they differ by the same value from $x$ to $x'$ , this only means they have a same growth rate and they have same difference all along the range of x axis.
| {
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"answer_count": 16,
"answer_id": 6
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What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $?
What is the value of
$\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?
The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (180-\theta)$, This reduces the problem as the products till $89^\circ$. But that doesn't help proceed.
Thanks in advanced.
| Let us use $\sin(1^\circ) = \sin(179^\circ) = \cos(89^\circ)$, etc., to rewrite the product as
$$\prod_{i=1}^{45} \cos^2 \left( \frac{\pi}{180} (2i - 1) \right).$$
Now, $\pm \cos \left( \frac{\pi}{180} (2i-1) \right)$ for $i = 1, \ldots, 45$ are roots of the polynomial $P_{180}(x) + 1$, where $P_n$ is the Chebyshev polynomial such that $P_n(\cos \theta) = \cos (n\theta)$. In fact, since $-1$ is the minimum possible value of $P_n(x)$ for $-1 \le x \le 1$, they are all double roots; and this accounts for all 180 roots of the polynomial. On the other hand, $P_{180}(x)$ has the form $2^{179} x^{180} + \cdots + 1$, so $P_{180}(x) + 1$ has the form $2^{179} x^{180} + \cdots + 2$. Therefore, the square of the product above is equal to the product of roots of this polynomial, which is $\frac{2}{2^{179}} = 2^{-178}$; and the original desired product is $2^{-89}$.
| {
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Find the parametrization of the curve resulting from intersection of a function and a curve I have the following function $f(x,y) = 2-x^2-4y^2$ and the surface $2x+4y+z-1 = 0.$
How do i go about finding the parametrization of the curve resulting from intersection of these surfaces? I see that $f(x,y)$ is the equation of an ellipsoid. I have tried to set $ f(x,y) = z$ and go from there but i cant seem to find any parameterization.
| We have
*
*$z=2-x^2-4y^2$
*$z=1-2x-4y$
then
$$2-x^2-4y^2=1-2x-4y\iff x^2-2x-1+4y^2-4y=0 \\\iff(x-1)^2+(2y-1)^2=3$$
thus we can take
*
*$x=\sqrt 3\cos \theta +1$
*$y=\frac{\sqrt 3}2\sin \theta +\frac12$
*$z=1-2x-4y=1-2\sqrt 3\cos \theta -2-2\sqrt 3\sin \theta -2=-3-2\sqrt 3(\cos\theta+\sin \theta)$
| {
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Failure to correctly apply the method of Variation of Parameters I use the Method of Variation of Parameters to find a particular solution to a given linear inhomogeneous ODE. Subsequently, I insert that particular solution and its derivatives into the inhomogeneous ODE. I find that, contradictorily, the particular solution is infact no solution. If there is somebody to tell me where there is an error in either my arithmetic or my logic, then I will thank you very much.
$$ y'' + y = \frac{1}{\cos x +1} $$
$$ y = y_p + y_h $$
$$ y_h = \, c_1 \cos x + c_2 \sin x $$
$$ y_p = \int_{x_0}^x \frac{\mathrm W_1 (t)}{\mathrm{W} (t)} \, \mathrm d t \, \cos x + \int_{x_0}^x \frac{\mathrm W_2 (t)}{\mathrm{W} (t)} \, \mathrm d t \, \sin x $$
$$ \mathrm W (t) = \left |
\begin{matrix}
\cos t & \sin t \\
- \sin t & \cos t
\end{matrix}
\right |
= 1 $$
$$
\mathrm W_1 (t) = \left |
\begin{matrix}
0 & \sin t \\
(1 + \cos t)^{-1} & \cos t
\end{matrix}
\right |
= \frac{-\sin t}{1+\cos t} \\
$$
$$
\mathrm W_2 (t) = \left |
\begin{matrix}
\cos t & 0 \\
- \sin t & (1 + \cos t)^{-1}
\end{matrix}
\right |
= \frac{\cos t}{1+\cos t}
$$
$$
\int_{x_0}^x \frac{-\sin t}{1+\cos t} \, \mathrm d t \, = \ln {\left | 1 + \cos x \right |}
$$
$$
\int_{x_0}^x \frac{\cos t}{1+\cos t} \, \mathrm d t \, = x - \tan \frac {x}{2}
$$
$$
\begin{align*}
y_p =& \cos x \ln |1 + \cos x| + \sin x \left( x - \tan \frac x 2 \right) \\
=& \cos x \ln |1 + \cos x| + x \sin x - (1 - \cos x)
\end{align*}
$$
$$
\begin{align*}
y'_p =& - \sin x \ln \left | 1 + \cos x \right | - \cos x (1 + \cos x)^{-1} \sin x + \sin x + x \cos x - \sin x \\
=& - \sin x \ln \left | 1 + \cos x \right | - \cos x (1 - \cos x) \sin^{-1} x + x \cos x \\
=& - \sin x \ln \left | 1 + \cos x \right | - \cot x + \cos^2 x \sin^{-1} x + x \cos x \\
\end{align*}
$$
$$
\begin{align*}
y''_p =& -\cos x \ln |1 + \cos x| + \sin^2 x (1+ \cos x)^{-1} + \sin^{-2} x - 2 \cot x \sin x \\
& \,\,\,\, -\cot^2 x \cos x + \cos x - x \sin x
\end{align*}
$$
Note that $ \sin^2 x (1+ \cos x)^{-1} = 1-\cos x $
$$ y''_p = -\cos x \ln |1 + \cos x | + 1 + \sin^{-2} x - 2 \cos x -\cot^2 x \cos x - x \sin x $$
$$
\begin{align*}
y''_p + y_p =& \sin^{-2} x -\cos x -\cot^2 x \cos x \\
=& - \cos x + \sin^{-2} x \, (1- \cos^3 x) \\
=& - \cos x + \sin^{-2} x \, (1- \frac{1}{4} \cos 3x + \frac{3}{4} \cos x) \\
=& - \cos x + \sin^{-2} x \, (1- \frac{1}{4} (\cos 2x \cos x - \sin 2x \sin x) + \frac{3}{4} \cos x) \\
=& - \cos x + \sin^{-2} x - \frac{1}{4} \cot x \sin^{-1} x \cos 2x + \frac{1}{4} \sin^{-1} x \sin 2x + \frac{3}{4} \cot x \sin^{-1} x \\
=& - \cos x + \sin^{-2} x + \frac{1}{4} \sin^{-1} x \sin 2x + \frac{1}{4} \sin^{-1} x \cot x (3 - \cos 2x) \\
\end{align*}
$$
Note that $ 3 - \cos 2x = 2+1 - \cos 2x = 2 + 2 \sin^2 x = 2 (1 + \sin^2 x) $
And that $ \sin 2x = 2 \sin x \cos x \Longrightarrow \frac{1}{4} \sin^{-1} x \sin 2x = \frac{1}{2} \cos x $
$$
\begin{align*}
y''_p + y_p =& - \cos x + \sin^{-2} x + \frac{1}{2} \cos x + \frac{1}{2} \cot x \sin^{-1} x (1 + \sin^2 x) \\
=& - \frac{1}{2} \cos x + \sin^{-2} x + \frac{1}{2} \cos x \sin^{-2} x + \frac{1}{2} \cos x \\
=& \sin^{-2} x (1+\frac{1}{2} \cos x) \\
\neq& \sin^{-2} x (1- \cos x) \\
=& \frac{1}{\cos x +1}
\end{align*}
$$
If the arithmetic that I have shown here is correct, then the function $y_p$ that I obtained by Variation of Parameters is not a solution to the given inhomogeneous ODE.
| Here is another method that help check your answer
Without Wronskian method
$$y'' + y = \frac{1}{\cos x +1}$$
$$\sin(x) y'' + y\sin(x) = \frac{\sin(x)}{\cos x +1}$$
$$\sin(x) y'' +y'\cos(x)-y'\cos(x)+ y\sin(x) = \frac{\sin(x)}{\cos x +1}$$
$$(\sin(x) y')'+(-y\cos(x))' = \frac{\sin(x)}{\cos x +1}$$
Integrate
$$\sin(x) y'+(-y\cos(x)) = \int \frac{\sin(x)}{\cos x +1}dx$$
$$\sin(x) y'-y\cos(x)= -\ln|\cos x +1|+K_1$$
Which is of first order
$$\frac y {\sin(x)}=-K_1 \cot(x) -\int \frac {\ln|\cos x +1|}{\sin^2(x)}dx$$
$$\frac y {\sin(x)}= -K_1 \cot(x) + \cot(x){\ln|\cos x +1|} +\int \frac {\cos(x)}{\cos x +1|}dx$$
$$ y = C_1 \cos(x) +\cos(x){\ln|\cos x +1|} +\sin(x)\int \frac {\cos(x)}{\cos x +1|}dx+K_2\sin(x)$$
$$ \boxed{y = C_1 \cos(x) + K_2\sin(x)+\cos(x){\ln|\cos x +1|} +\sin(x) ( x-\tan(x/2))}$$
Edit
You made a sign mistake when you linearised $\cos^3(x)$
$$cos^3(x)=\cos(3x)/4+3\cos(x)/4$$
You wrote this
$$\begin{align*}
y''_p + y_p =& \sin^{-2} x -\cos x -\cot^2 x \cos x \\
=& - \cos x + \sin^{-2} x \, (1- \cos^3 x) \\
=& - \cos x + \sin^{-2} x \, (1- \frac{1}{4} \cos 3x \color{red}{+ \frac{3}{4} \cos x)} \\
\end{align*}
$$
It should be
$$y''_p + y_p = - \cos x + \sin^{-2} x \, (1- \frac{1}{4} \cos 3x \color{blue}{- \frac{3}{4} \cos x)} $$
| {
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Show $\forall x \in \mathbb{R}^* \quad (1+x)^n \leq 2^{n-1}(1+x^n)$ Let $n\in N$ such that $n \geq 2$
$\forall x \in \mathbb{R}^* \quad (1+x)^n \leq 2^{n-1}(1+x^n)$
I tried induction but with no luck. On the other hand I said that $f(x)=x^n$ is convex then by Jenson inquality for $f$ with $ f((a+b)/2) \leq (f(a)+f(b))/2 $.
| By induction for $x \in [0,+\infty)$: start with $n=2$. You know that in general $(x-1)^2 \geq 0$. This means $x^2-2x+1 \geq 0$, so $2x \leq 1+x^2$, and adding $1+x^2$ to both sides you get
$$(1+x)^2 = 1+2x+x^2 \leq 2(1+x^2)$$
which is your base case. Now assume the formula works for $n\geq2$. Then
$$(1+x)^{n+1} = (1+x)(1+x)^n \leq 2^{n-1}(1+x)(1+x^n).$$
Now you want to prove that $2^{n-1}(1+x)(1+x^n) \leq 2^n(1+x^{n+1})$, which is equivalent to $1+x+x^n+x^{n+1} \leq 2+2x^{n+1}$, namely $x^{n+1} -x^n-x+1 \geq 0$. This is true since $x \geq 0$ and
$$x^{n+1} -x^n-x+1 = (1-x)-x^n(1-x) = (1-x^n)(1-x) = (1-x)^2(1+x+\dots+x^{n-1}) \geq 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorics- wheel of fortune In how may different ways can you color a wheel of fortune with five different colors, if it has 60 fields each containing a different value? Each color needs to be used at least once.
My answer is :
$$\frac{60 \cdot 59 \cdot 58 \cdot 57 \cdot 56}{5!} \cdot 5! \cdot 5^{55}$$ First I chose $5$ fields out of $60$ to color with $5$ different colors then I paired the colors and the fields by multiplying by $5!$, and then for the $55$ remaining fields I can choose any of the $5$ colors so that is $5^{55}$. Not sure if that's correct.
| If there were no restrictions, we could color the wheel in $5^{60}$ ways since there are five choices of color for each of the $60$ fields. From these, we must subtract those choices in which fewer than five colors are used.
There are $\binom{5}{k}$ ways to exclude exactly $k$ of the colors and $(5 - k)^{60}$ ways to color the wheel of fortune with the remaining $5 - k$ colors. By the Inclusion-Exclusion Principle, the number of ways to color the wheel with five colors so that each color is used at least once is
\begin{align*}
\sum_{k = 0}^{5} (-1)^k\binom{5}{k}(5 - k)^{60} & = \binom{5}{0}5^k - \binom{5}{1}4^{60} + \binom{5}{2}3^{60} - \binom{5}{3}2^{60} + \binom{5}{4}1^{60} - \binom{5}{5}0^{60}\\
& = 5^{60} - 5 \cdot 4^{60} + 10 \cdot 3^{60} - 10 \cdot 2^{60} + 5 \cdot 1^{60} - 0^{60}
\end{align*}
Your answer is larger than the $5^{60}$ possible ways to color the wheel. The problem is that you designated five particular fields to be of different colors. However, if red appears exactly $56$ times and each of the other colors appears once, there are $56$ ways to designate which of those $56$ fields is the designated red field, so you have counted that particular arrangement $56$ times. If each of the five colors appears $12$ times, there are $12$ ways to designate a particular field as the field of that color for each of the five colors, so you count such arrangements $12^5$ times.
| {
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Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use prime factorization.
The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}
$$
is a rational $\sqrt{2}$ with a smaller denominator.
The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.
I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the rational root theorem (which uses unique factorization
in its proof).
| Assume that $\sqrt[3]{2}=\dfrac{a}{b}$ with $a,b\in\mathbb{Z^+}$
We will have $\dfrac{a^3}{b^3}=2\Leftrightarrow a^3=2b^3$ $(*)$
The above implies that $a$ must be divisible by $2$ (if $a$ is odd then $a^3$ is odd), let $a=2a_1$ with $a_1\in\mathbb{N}$, then $a^3=2b^3\Leftrightarrow 8a_1^3=2b^3\Leftrightarrow4a_1^3=b^3$
The above implies that $b$ must be divisible by $2$ (if $b$ is odd then $b^3$ is odd), let $b=2b_1$ with $b_1\in\mathbb{N}$, then $4a_1^3=b^3\Leftrightarrow 4a_1^3=(2b_1)^3\Leftrightarrow 4a_1^3=8b_1^3\Leftrightarrow a_1^3=2b_1^3$.
This will continue for an infinite loop, let $n\in\mathbb{Z^+}$ and $a_n=2a_{n+1}$ then
$$a^3=2b^3\Leftrightarrow a_1^3=2b_1^3\Leftrightarrow a_2^3=2b_2^3\Leftrightarrow a_3^3=2b_3^3\Leftrightarrow\cdots\Leftrightarrow a_n^3=2b_n^3\Leftrightarrow a_{n+1}^3=2b_{n+1}^3$$
Because $a_{n+1}\in\mathbb{Z^+}$,
$$2|a_{n},b_{n}\Rightarrow2^2|a_{n-1},b_{n-1}\Rightarrow2^3|a_{n-2},b_{n-2}\Rightarrow\cdots\Rightarrow 2^{n+1}|a,b$$
$a,b$ must be divisible by $2^{n+1}$ for all $n\in\mathbb{N}$, this is only possible if $a=b=0$, which makes $\sqrt[3]{2}=\dfrac{a}{b}$ undefined.
By contradicition, $\sqrt[3]{2}$ is irrational.
Alternatively...
If $\sqrt[3]{2}$ is rational then it can be expressed in the form $\sqrt[3]{2}=\dfrac{a}{b}$ with $a,b\in\mathbb{Z^+}$ and $\dfrac{a}{b}$ is an irreducible fraction or $GCD(a,b)=1$.
Because we have proven that both $a$ and $b$ are even numbers, $\dfrac{a}{b}$ now becomes reducible, which also contradicts the condition above. This will be a much quicker conclusion compared to the "infinite loop" method.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why this value is not the second root of the equation? $\sqrt{x+3}=x-2$
Why $\frac{5}{2} - \frac{\sqrt{21}}{2}$ is not root?
There is only one restriction: $\sqrt{x+3}$, but $\frac{5}{2} - \frac{\sqrt{21}}{2} > 0$.
$x^2-4x+4=x+3$, $x^2-5x+1=0$, $D=25-4=21>0$
$D>0$, =>, $x = \frac{ 5 \pm \sqrt{21} }{ 2 }$
| That is because the irrational equation:
$$\sqrt A=B\iff A=B^2\;\textbf{ and }\, B\ge 0,$$
i.e., in the present case, $\;x\ge 2$, whereas $\;\dfrac{5-\sqrt{21}}2<\dfrac12$.
| {
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Find $7^{100} \pmod{167}$ using Fermat's Little Theorem I understand how to use Fermat's Little theorem but only on exponents that are larger than the mod, therefore, I do not know how to do this question please help, thank you!
| Since $\phi(167) = 166 = 2\cdot 83$, the possible orders of $7$ mod $167$ are $2$, $83$ and $166$. Clearly, the order is not $2$. If we compute the Legendre symbol
$$\left(\frac{7}{167}\right) = - \left(\frac{167}{7}\right) = -\left(\frac{-1}{7}\right) = 1,$$
we see that $7$ is a quadratic residue mod $167$, and so the order can't be $166.$ (A primitive root can't be a quadratic residue.)
Thus, the order of $7$ mod $167$ is $83.$ So we have
$$7^{100} \equiv 7^{17} \pmod{167}.$$
Next, notice that $7^3 = 343$ is pretty close to $2\cdot 167 = 334$, so we
have $7^3 \equiv 9 \pmod{167}.$ Also, $2\cdot 9^2 = 162 \equiv -5 \pmod{167}$. Then we calculate:
$$7^{17} \equiv (7^3)^5\cdot 49 \equiv 9^2\cdot 9^2 \cdot 9 \cdot 49 \equiv (80\cdot 9^2 +1\cdot 9^2)\cdot 9 \cdot 49 \equiv (40(-5)+81)\cdot 9 \cdot 49 $$
$$\equiv (-119)\cdot 9 \cdot 49 \equiv -119\cdot 441 \equiv 48 \cdot (-60) \equiv -2880 \equiv 126 \pmod{167}.$$
| {
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Calculate the determinant of this $5 \times 5$ matrix
Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha
& \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0
& 0 & 1 & a^2
& b^2 \\ 0 & 0 & 0 & a
& b \\ 0 & 0 & 0 & -b
& a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$
I have trouble solving the determinant.. But what is immediately visible are those zeroes in the matrix, just one more zero is needed such that this matrix is a triangular matrix (the element $a_{54}$ must be zero for this but it is $-b$ instead). If it was zero we could just multiply the diagonal and the product would be our determinant.
I have tried various ways to form this matrix such that $a_{54}$ is zero but the way I formed harmed the matrix and I got a wrong determinant as solution : /
As example, I have multiplied row $4$ with $b$, multiply row $5$ with $a$ and then add row $4$ to row $5$. Because I multiplied row $5$ with $a$, I need to divide the determinant by $a$ at the end.
So then I have the matrix
$$\begin{pmatrix}
\sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab\\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2\\
0 & 0 & 1 & a^2 & b^2\\
0 & 0 & 0 & a & b\\
0 & 0 & 0 & 0 & a^2+b^2
\end{pmatrix}$$
$$\text{Thus }\det = \frac{\sin\alpha \cdot \sin\alpha \cdot 1 \cdot a \cdot (a^2+b^2)}{a}=\sin^2\alpha \cdot (a^2+b^2)$$
But this is wrong and I don't see how to get the correct determinant...?
| $$\det \left[\begin{array}{cc|c|cc}
\sin (\alpha) & \cos (\alpha) & a \,\sin (\alpha) & b\,\cos (\alpha) & a b \\
-\cos (\alpha) & \sin (\alpha) & - a^2 \sin (\alpha) & b^2 \cos (\alpha) & a^2 b^2\\ \hline
0 & 0 & 1 & a^2 & b^2\\ \hline
0 & 0 & 0 & a & b \\
0 & 0 & 0 & -b & a\end{array}\right] = \\ = \underbrace{\det \begin{bmatrix}
\sin (\alpha) & \cos (\alpha)\\ -\cos (\alpha) & \sin (\alpha)\end{bmatrix}}_{= 1} \cdot \det (1) \cdot \det \begin{bmatrix}
a & b\\ -b & a\end{bmatrix} = a^2 + b^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Reason for substitution : $a=z + \frac{1}{z}$. Have read regarding the substitution $a = z + \frac{1}{z}$ to get the factorization of $z^6+z^5+z^4+z^3 +z^2+z+1$ to get the form $a^3+a^2-2a -1=0$ in the book by Erickson martin, titled: Beautiful mathematics, on page #$58$ as shown below.
I am unable to get the process for the division, i.e. how to divide by $a$ the given polynomial. I mean that $a = \frac{z^2+1}{z}$ cannot divide $z^6+z^5+z^4+z^3 +z^2+z+1$.
For $z^5-1=0\implies (z-1)(z^4+z^3+z^2+z+1)=0$, want to use the same logic of symmetry for $z+z^4= z+\frac1z=a'$, but am hindered by the inability to divide $(z^4+z^3+z^2+z+1)$ by $a'$.
| I don't understand why you are referring to division. All you have to do is substitute
$$a=z+\frac1z$$
into
$$a^3+a^2-2a-1$$
and simplify. You should get
$$\frac{z^6+z^5+z^4+z^3+z^2+z+1}{z^3}\ .$$
(I believe the $z^6+z^5+z^4+z^3+z^2+1$ in your question is wrong.) Then you have
$$z^6+z^5+z^4+z^3+z^2+z+1=0\quad\hbox{if and only if}\quad a^3+a^2-2a-1=0\ .$$
| {
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Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| $$
\begin{aligned}
\int \frac{1}{\left(x^{2}+1\right)^{2}} d x &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{x^{2}+1}\right) \\
& \stackrel{IBP}{=} -\frac{1}{2 x\left(x^{2}+1\right)}-\frac{1}{2} \int \frac{1}{x^{2}\left(x^{2}+1\right)} d x \\
&=-\frac{1}{2 x\left(x^{2}+1\right)}-\frac{1}{2} \int\left(\frac{1}{x^{2}}-\frac{1}{x^{2}+1}\right) d x \\
&=-\frac{1}{2 x\left(x^{2}+1\right)}+\frac{1}{2 x}+\frac{1}{2} \arctan x+C \\
&=\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2} \arctan x+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just tried to work out the bottom of the first fraction which I found to be $\frac{(3-x)(4-x)-x}{(3-x)(4-x)}$ now taking the reciprocal gives me $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$.
I did try factorising the bottom to get to $\frac{(3-x)(4-x)}{(x-6)(x-2)}$ and then using partial fractions gets me to $\frac{-15}{2(x-6)}+\frac{1}{2(x-2)}$ which is definitely not where I want to be.
I feel like I need to work from $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$ and somehow pull out a $1$ here but not entirely sure how.
Would really appreciate if anyone could help me.
Thank you.
| The first expression can be written as
$$\frac{(3-x)(4-x)}{(3-x)(4-x)-x(4-x)+x(3-x)} = \frac{12-7x+x^2}{12-8x+x^2}.$$
The second can be written as
\begin{align}
1+\frac{1}{2}\cdot\frac{6-x-3(2-x)}{(2-x)(6-x)} & = \frac{24-16x+2x^2+6-x-6+3x}{2(12-8x+x^2)} = \frac{24-14x+2x^2}{2(12-8x+x^2)} \\
& = \frac{12-7x+x^2}{12-8x+x^2}.\end{align}
Since these expressions make sense for $\lvert x \rvert < 1$, then the left hand sides are equal, and your claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality with a condition If $a,b,c$ positives and $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} =3,$ I have to prove $\frac {1}{2 a^2+b^2} +\frac {1}{2 b^2+c^2} +\frac {1}{2 c^2+a^2} \le 1.$
Since we have $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} =3$, then implying Titu's lemma, we have:
$ \frac{2}{a^2} +\frac{1}{b^2} =\frac{1}{a^2} +\frac{1}{a^2} +\frac{1}{b^2} \ge \frac {9}{2 a^2+b^2},$
so:
$ \frac {1}{2 a^2+b^2} +\frac {1}{2 b^2+c^2} +\frac {1}{2 c^2+a^2} \le \frac{1}{3} ( \frac {1}{a^2} +\frac {1}{b^2} +\frac {1}{c^2}).$
It is enough to prove:
$\frac {1}{a^2} +\frac {1}{b^2} +\frac {1}{c^2} \le 3= \frac{1}{3} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
and I stuck. Thank you
| By AM-GM $$1-\sum_{cyc}\frac{1}{2a^2+b^2}=\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\sum_{cyc}\frac{1}{2a^2+b^2}=$$
$$=\frac{1}{9}\sum_{cyc}\left(\frac{1}{a^2}+\frac{2}{ab}-\frac{9}{2a^2+b^2}\right)=\sum_{cyc}\frac{4a^3+2ab^2+b^3-7a^2b}{9a^2b(2a^2+b^2)}\geq$$
$$\geq\sum_{cyc}\frac{7\sqrt[7]{(a^3)^4(ab^2)^2b^3}-7a^2b}{9a^2b(2a^2+b^2)}=0.$$
Also, we can use C-S and AM-GM:
$$\frac{1}{a^2}+\frac{2}{ab}=\frac{1}{a^2}+\frac{4}{2ab}\geq\frac{(1+2)^2}{a^2+2ab}\geq\frac{9}{a^2+2\cdot\frac{a^2+b^2}{2}}=\frac{9}{2a^2+b^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^3+b^3=10^5$ has no positive integer solution.
Prove that there do not exist two positive integers, $a$ and $b$, such that $$a^3+b^3=100\,000$$
I tried to use congruence modulo $7$ and some other modulo, but it does not seem to work. If possible, please prove it with modular congruences.
| $$a^3+b^3=(a+b)(a^2-ab+b^2).$$
So if $a^3+b^3=10^5$, the only possible prime factors of $a^2-ab+b^2$
are $p=2$ and $p=5$.
The quadratic form $x^2-xy+y^2$ is irreducible modulo $2$ and modulo $5$
so if $2\mid(a^2-ab+b^2)$ then $2\mid a$ and $2\mid b$. Likewise if
$5\mid(a^2-ab+b^2)$ then $5\mid a$ and $5\mid b$. If
$$a^3+b^3=(a+b)(a^2-ab+b^2)=10^5$$
then $(a,b)=(ra',rb')$ where $r^2\mid 10^5$ and $a'^2-a'b'+b'^2=1$.
As $a$ and $b$ are positive, this means $a'=b'=1$. Therefore $a=b=r$,
and $a^3+b^3=2r^3\ne10^5$.
This argument rules out $a^3+b^3$ equalling any power of $10$
for $a$, $b$ integers $>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Integrate $\int \frac{dx}{\sqrt{\sin^3x\sin(x+\alpha)}}$
Solve the indefinite Integration $\int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx$
$$
\int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx=\int \frac{1}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx\\=\frac{1}{\sin\alpha}\int \frac{\sin(x+\alpha-x)}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx=\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos x-\cos(x+\alpha)\sin{x}}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx\\
=\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos x}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx-\frac{1}{\sin\alpha}\int\frac{\cos(x+\alpha)\sin{x}}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx
$$
Is it possible to proceed further and complete the integration or what is the right substitution to find the solution ?
Note: I'm looking for a simple way to solve this, unlike here Finding indefinite integral $\int{dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$.
| Divide by $\sin^2 x$ on numerator and denominator and use the identity $\sin(A+B) = \sin A \cos B+\cos A \sin B $ to get to following:
$$\int\frac{\csc^2 x \, dx}{\sqrt{\cos a + \sin a\cot x}}$$
Noting that $\cot'(x) = -\csc^2 x$, it is straightforward now. The final antiderivative is:
$$-\frac{2 \sqrt{\cos a + \sin a \cot x}}{\sin a}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $f(x) = \frac{x^3+\cos x}{6}$ I am trying to calculate the derivative of $f$ using the product rule and quotient rule respectively.
However, I am getting different results for the product rule and quotient rule.
Did I made any mistake along the way?
$$f(x) = \frac{(x^3 + \cos x)}{6}$$
using the product rule (multiple by 1/6 instead of divide by 6)
$$f'(x) = \frac {1}{6} \frac{d}{dx}[x^3 + \cos x]$$
$$ = \frac {1}{6}(3x^2-\sin x)$$
using the quotient rule
$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2} $$
$$ = \frac {6(3x^2-\sin x)-(x^3+\cos x)}{36}$$
$$ = \frac{(3x^2-\sin x-x^3-\cos x)}{6}$$
| Note that by quotient rule we have
$$f'(x) = \frac {6 \frac {d}{dx}[x^3+\cos x]-(x^3+\cos x)\frac {d}{dx}[6]}{6^2}=\frac {6 (3x^2-\sin x)-0}{6^2}=\frac {1}{6}(3x^2-\sin x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the Laurent series in $1<|z|<2$ ring I just started Laurant series, I have this function
$$f(z) = {z^4 + 1 \over {(z-1)(z+2)}}$$ in $1<|z|<2$ ring, I know the theorems,but still cant figure it out, Any help with idea or transformation , welcome!
| Since $z^4+1=(z^2-z+3)(z-1)(z+2)-5z+7$, you have\begin{align}\frac{z^4+1}{(z-1)(z+2)}&=z^2-z+3+\frac{-5z+7}{(z-1)(z+2)}\\&=z^2-z+3+\frac2{3(z-1)}-\frac{17}{3(z+2)}.\end{align}On the other hand, since $1<|z|<2$, you have$$\frac1{z-1}=-\frac1{1-z}=z^{-1}+z^{-2}+z^{-3}+\cdots$$and$$\frac1{z+2}=\frac12\cdot\frac1{1+\frac z2}=\frac12-\frac z{2^2}+\frac{z^2}{2^3}-\frac{z^3}{2^4}+\cdots$$Now, put it all together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bounding a somewhat complicated integral (exponential of a polynomial) I am interested in bounding the following integral, where $a>0$ is a constant:
$$\int\limits_0^a \exp\left(\left(x^2 - \frac{2a^4+3}{4a^2}\right)^2\right) dx$$
I first conjectured that
$$\int\limits_0^a \exp\left(\left(x^2 - \frac{ba^4+c}{a^2}\right)^2 \right) dx \le a^{-1}\exp \left( \left( \frac{ba^4+c}{a^2}\right)^2 \right)$$
when $c \ge 0$ and $b > 0$. However, this is false (for example, take $c = 0$, $b = 9/20$, and plug in $a = 2$). However, this inequality does appear to hold when $b \ge \frac{1}{2}$, from the numerical tests I have done. I am not sure how to prove it, however. The expressions that I have seen for exponentials of quartics are quite complicated and do not easily admit an upper bound (see for instance Exponential of a Quartic). Does anyone have any ideas?
[Note: In fact, what I'm really interested in is the regime $a \rightarrow \infty$, but so far the upper bound I have guessed appears to hold up for all $a > 0$.]
Updated wording: I am not just looking for any bound on this integral, but I indeed would like to prove the bound that I have proposed above. This was not made clear in my original post; I apologize for that.
| Another approach I can think of that could provide more insight would be computing a Taylor Remainder for the expansion around $a=0$ of:
$$
F(a) = \int_0^a f(a,x)dx = \int_0^a \exp(g(a,x))dx
$$
$$
F(a) = F(0) + F'(a) + F''(\xi)/2
$$
Naturally $F(0) = 0$. So:
For that, we can use Leibniz Integral Rule:
$$
\frac{dF}{da} |_{x=a} = f(a,a) \times 1 + \int_0^a \frac{\partial f(a,x)}{\partial a}dx = W(a)
$$
Computing $f(a,a)$:
$$
f(a,a) = \exp \left(\frac{(1-b)a^4 -c}{a^2} \right)^2
$$
Now the partial derivative:
$$
\frac{\partial f(a,x)}{\partial a} = f(a,x) \times \frac{\partial g(a,x)}{\partial a} = f(a,x) \times 2 \left( \left( x^2 - \frac{ba^4+c}{a^2}\right) \left( -2ab +2\frac{c}{a^3}\right) \right)
$$
Well... I'll just use Mathematica:
$$
W(a) = \int_0^a \frac{\left(a^2-3\right) e^{\left(\frac{3}{4 a^4}-\frac{1}{2 a^2}+x^2\right)^2}
\left(4 a^4 x^2-2 a^2+3\right)}{2 a^9} \, dx+e^{\frac{\left(4 a^6-2 a^2+3\right)^2}{16
a^8}}
$$
The Remainder term will look like this:
$$
\frac{e^{\left(\frac{3}{4 \xi ^4}+\xi ^2-\frac{1}{2 \xi ^2}\right)^2} \left(\xi ^6+\xi
^2-3\right) \left(4 \xi ^6-2 \xi ^2+3\right)}{\xi ^9}+\int_0^{\xi }
\frac{e^{\left(\frac{3-2 \xi ^2}{4 \xi ^4}+x^2\right)^2} \left(-162 \xi ^2-24 \xi ^{14}
x^2-36 \xi ^6 \left(8 x^2+1\right)+9 \xi ^4 \left(24 x^2+13\right)+4 \xi ^{12} \left(4
x^4+30 x^2+5\right)-2 \xi ^{10} \left(48 x^4+8 x^2+63\right)+2 \xi ^8 \left(72 x^4+60
x^2+83\right)+81\right)}{4 \xi ^{18}} \, dx
$$
So, I call it quits on this attempt, but I do admit it is starting to look plausible, but for a very limited range on $a$, such that $a^{-1}$ is actually a big number and thus $a^{-1} K $ for big enough $K$ can be an actual bound... But thinking like this...
Yeah... Nope, I don't think this would work...
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the coefficient of $x^n$ in the expansion of $(1+x+x^2+x^3+\dots)^4$? I have just come across a question (in a past exam paper for a module that I will be taking soon) which asks what the coefficient of $x^n$ is in the expansion of
$$
(1+x+x^2+x^3+x^4+\dots)^4
$$
Can anyone give me an idea as to how this might be done? Using Wolfram Alpha to expand the brackets and then by inspecting the coefficient of each power manually, I was able to deduce that the coefficient of $n^\text{th}$ term is given by $C(n+2, 3)$. How might I have been able to work this out for myself?
| For $-1 < x < 1$, let
$$f(x) = (1+x+x^2+x^3+x^4+\cdots)^4$$
and for each nonnegative integer $n$, let $f^{(n)}(x)$ denote the $n$-th derivative of $f(x)$.
Summing the inner geometric series, we get $f(x) = (1-x)^{-4}$, and by a routine induction, we get
$$f^{(n)}(x)=n!{\small{\binom{n+3}{3}}}(1-x)^{-n-4}$$
Hence, the coefficient of $x^n$ in the power series expansion of $f(x)$ is
$$\frac{f^{(n)}(0)}{n!}=\frac{n!{\large{\binom{n+3}{3}}}(1-0)^{-n-4}}{n!}={\small{\binom{n+3}{3}}}$$
To show the induction . . .
For $n=0$,
\begin{align*}
f^{(0)}(x)
&=f(x)\\[4pt]
&=(1-x)^{-4}\\[4pt]
&=0!{\small{\binom{0+3}{3}}}(1-x)^{-0-4}\\[4pt]
\end{align*}
so the base case $n=0$ is verified.
Next, assume
$$f^{(n)}(x)=n!{\small{\binom{n+3}{3}}}(1-x)^{-n-4}$$
holds for some nonnegative integer $n$.
\begin{align*}
\text{Then}\;\;f^{(n+1)}(x)&=\frac{d}{dx}\left(n!{\small{\binom{n+3}{3}}}(1-x)^{-n-4}\right)\\[4pt]
&=n!{\small{\binom{n+3}{3}}}(n+4)(1-x)^{-n-5}\\[4pt]
&=n!\left(\frac{(n+3)(n+2)(n+1)}{6}\right)(n+4)(1-x)^{-n-5}\\[4pt]
&=(n+1)!\left(\frac{(n+4)(n+3)(n+2)}{6}\right)(1-x)^{-n-5}\\[4pt]
&=(n+1)!{\small{\binom{n+4}{3}}}(1-x)^{-n-5}\\[4pt]
\end{align*}
which completes the induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a trigonometric identity by integration. By considering, $$\int \sin^n (x)\cos^3(x)dx$$
Prove that:
$$ \frac{ \sin^8(x)}{8} - \frac{ \sin^6(x)}{6} + \frac{1}{24} =
\frac{\cos^8(x)}{8} - \frac{ \cos^6(x)}{3} + \frac{ \cos^4(x)}{4}$$
I have integrated the suggested integral successfully, that being:
$$ \frac{\sin^{n+1}(x)}{n+1} - \frac { \sin^{n+3}(x)}{n+3} +c $$
I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand $\sin^2(x) = (1-\cos^2(x))$ but this seems far too tedious.
| Hint:
$$I=\int\sin^{2m+1}x\cos^{2n+1}x\ dx=\int(1-\cos^2x)^m\cos^{2n+1}x\sin x dx$$
Set $\cos x=u$
Again,
$$I=\int(1-\sin^2x)^n\sin^{2m+1}x\cos x\ dx$$
Set $\sin x=v$
Here $2n+1=3$
Set $2m+2n+2=8,6,4$ etc.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$ To find the general solution $(x,y)$ satisfying the system of equations
\begin{align} 5(\sin x + \sin y) &= 1 \\ 5(\sin 2x + \sin 2y) &= 1
\end{align}
I applied $\sin C + \sin D$ rule and then divided these two equations, then I am stuck at
$$\cos\frac{x-y}{2} = 2\cos \frac{x+y}{2}\cos(x-y).$$
I do not know what to do further.
| Suppose you have instead
$$
e^{ix}+e^{iy} = a\\
e^{2ix}+e^{2iy} = b
$$
then
$$
e^{2ix}+e^{2iy}+2e^{ix}e^{iy} = a^2\Rightarrow 2e^{ix}e^{iy} =a^2-b
$$
and now
$$
e^{ix}+e^{iy} = a\\
e^{ix}e^{iy} =\frac{a^2-b}{2}
$$
etc.
NOTE
Another way is to follow the trig. identities
$$
\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) = \frac a2\\
\sin(x+y)\cos(x-y) = \frac a2
$$
now calling
$$
\frac{x+y}{2} = u\\
\frac{x-y}{2} = v
$$
$$
\sin^2u\cos^2v = (1+\cos(2u))(1-\cos(2v)) = a^2
$$
or
$$
1+\cos(2v)-\cos(2u)-\cos(2u)\cos(2v) = a^2\\
\sin(2u)\cos(2v) = \frac a2
$$
and
$$
\sin(2u) = \frac a2\left(\frac{\cos(2u)-1}{1-a^2+\cos(2u)}\right)
$$
and now calling $\cos(2u) = z$ we have
$$
\left(\frac 2a\right)^2\left(\frac{z-1}{1-a^2+z}\right)^2=1-z^2
$$
This quartic should be solved to obtain the solutions. Attached a plot showing the intersections between the curves
$$
\sin x + \sin y = \frac 15 \;\;\mbox{in red}\\
\sin(2x)+\sin(2y) = \frac 15\;\;\mbox{in blue}
$$
| {
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"url": "https://math.stackexchange.com/questions/2797777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove the theorem of Nicomachus by induction Prove the theorem of Nicomachus(AD.100) by induction:
$$
1^3 = 1,\ 2^3 = 3+ 5,\ 3^3 = 7 + 9 + 11,\ 4^3 = 13 + 15 + 17 + 19,\ ...
$$
My approach: from looking at the above pattern you can tell there is something of the following sort:
$$ 2^{n-1} + q = n ^3,$$
where $q$ is odd s.t. $q = 2k + 1.$
| Notice that by adding your equations you get
\begin{align*}
1^3&=1\\
1^3+2^3&=1+3+5\\
1^3+2^3+3^3&=1+3+5+7+9+11\\
1^3+2^3+3^3+4^3&=1+3+5+7+9+11+13+15+17+19
\end{align*}
And notice also that these equations imply the equations you want. (You get them simply by subtracting the two consecutive equations.)
So you actually want to prove that sum of the first $n$ cubes is the same as the sume of the first $1+2+\dots+n=\frac{n(n+1)}2$ odd numbers.
$$1^3+2^3+\dots+n^3 = \sum_{k=1}^{n(n+1)/2} (2k-1)$$
If you also know that the sum of the first $n$ odd numbers gives a square, this is the same as
$$1^3+2^3+\dots+n^3 = \left(\frac{n(n+1)}2\right)^2$$
See also:
*
*Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
*Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| Just to give a slightly different approach, note that
$$a^2=\left(x^2+{1\over x^2}\right)^2=x^4+{1\over x^4}+2\implies x^4+{1\over x^4}=a^2-2$$
and
$$a^3=\left(x^2+{1\over x^2}\right)^3=x^6+{1\over x^6}+3\left(x^2+{1\over x^2}\right)\implies x^6+{1\over x^6}=a^3-3a$$
From these we find that
$$\left(x^5+{1\over x^5}\right)\left(x+{1\over x}\right)=x^6+{1\over x^6}+x^4+{1\over x^4}=a^3-3a+a^2-2=(a^2-a-1)(a+2)$$
Finally,
$$\left(x+{1\over x}\right)^2=x^2+{1\over x^2}+2=a+2\implies x+{1\over x}=\sqrt{a+2}$$
(using the assumption $x,a\gt0$), so we have
$$x^5+{1\over x^5}=(a^2-a-1)\sqrt{a+2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Angle between planes with x, y and z variables Problem: Find the angle between the planes $3x-y+z-5=0$ and $x+2y+2z+2=0$.
We have been thought 1 formula for solving angles which is : Angle $= Arctan(\frac{m1-m2}{1+m1m2})$ but that is for parallel lines only and for equation with x and y only.
How to answer this kind of problem? Thank you.
| Finding the angle between two planes is similar to finding the angle between two vector (normal of the planes). The first plane is $3x-y+z-5=0$ has normal $\overrightarrow n_0 =\langle 3, -1, 1 \rangle$. The second plane is $x+2y+2z+2=0$ has normal $\overrightarrow n_1 =\langle 1, 2, 2 \rangle$.
To find the angle between two vectors we use the formula
$$cos(\theta)= \frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}$$
Insert A as $\overrightarrow n_0$ and B as $\overrightarrow n_1$.
$$\overrightarrow n_0 \bullet \overrightarrow n_1=3*1+(-1)*(2)+1(2) = 3$$
$$|\overrightarrow n_0| = \sqrt{3^2+(-1)^2+1^2}=\sqrt{11}$$
$$|\overrightarrow n_1| = \sqrt{1^2+2^2+2^2}=3$$
So,
$$cos(\theta)= \frac{3}{\sqrt{11}*3}=\frac{1}{\sqrt{11}}$$
Therefore
$$\theta= \arccos(\frac{1}{\sqrt{11}}) \approx 1.26451896 rad$$
Relevant Find the angle between two planes using their normal vectors
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
Decompose initial value problem into semi-simple and nilpotent matrices Consider the initial value problem:
$\begin{pmatrix}
\dot x(t) \\
\dot y(t) \\
\dot z(t)
\end{pmatrix}= A \vec x(t) = \begin{pmatrix}
-1 & 1 & 1 \\
0 & -1 & 4 \\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
x(t) \\
y(t) \\
z(t)
\end{pmatrix}, \: \: \: \begin{pmatrix}
x(0) \\
y(0) \\
z(0)
\end{pmatrix}=\begin{pmatrix}
x_0 \\
y_0 \\
z_0
\end{pmatrix}$
(i) Determine the eigenvalues and generalised eigenvectors of $A$.
(ii) Decompose $A$ into a semisimple matrix $S$ and a nipotent matrix $N$ such that $A=S+N$.
(a) first determine the semisimple part $S$ of $A$.
(b) then determine the nilpotent part $N$ of $A$ and show that $N^2=0$.
My attempt:
Since $A$ is an upper triangular matrix we have eigenvalues $\lambda=\{-1,-1,1\}$
Taking $\lambda_3=1, (A-\lambda I)v=\begin{pmatrix}
-2 & 1 & 1 \\
0 & -2 & 4 \\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=\begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix}$
So $-2y+4z=0$ and $-2x+y+z=0$
and I've ended up with eigenvector $v_3=(3,4,2)^T$
I get the feeling this is wrong because I tried the rest of the question and don't get $N^2=0$
Any help?
| Consider the initial value problem:
$$\vec x'(t) = A \vec x= \begin{pmatrix}
-1 & 1 & 1 \\
0 & -1 & 4 \\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}, ~~~~\vec x(0) =\begin{pmatrix}
x_0 \\
y_0 \\
z_0
\end{pmatrix}$$
Since $A$ is upper triangular, the two eigenvalues can be read off the main diagonal as $\lambda_{1,2} = -1, 1$.
$\lambda_1 = -1$ has multiplicity $n_1 = 2$ and $\lambda_2 = 1$ has multiplicity $n_2 = 1$.
The generalized eigenspace associated with $\lambda_1$ is $E_1 = \ker(A - \lambda_1I)^2 = \ker(A + I)^2$. Find
$$(A + I)^2 = \begin{pmatrix}
0 & 0 & 6 \\
0 & 0 & 8 \\
0 & 0 & 4 \\
\end{pmatrix}$$
A choice for generalized eigenvectors spanning $E_1$ is $v_1 = (1,0,0)^T$ and $v_2 = (0,1,0)^T$.
The generalized eigenspace associated with $\lambda_2$ is $E_2 = \ker(A-I)$. Find
$$(A - I) = \begin{pmatrix}
-2 & 1 & 1 \\
0 & -2 & 4 \\
0 & 0 & 0 \\
\end{pmatrix}$$
Let $v_3 = (3, 4, 2)^T$. The transformation matrix is
$$P = (~v_1~|~v_2~|~v_3~) = \begin{pmatrix}
1 & 0 & 3 \\
0 & 1 & 4 \\
0 & 0 & 2 \\
\end{pmatrix} ~\mbox{and}~ P^{-1} = \begin{pmatrix}
1 & 0 & -\dfrac{3}{2} \\
0 & 1 & -2 \\
0 & 0 & \dfrac{1}{2} \\
\end{pmatrix}$$
We are now ready to find $S$ and $N$ using $A = S + N$.
$$S = P\Lambda P^{-1}, ~\mbox{where}~\Lambda = \mbox{diag}(-1,-1,1)$$
Obtain $S = \begin{pmatrix}
-1 & 0 & 3 \\
0 & -1 & 4 \\
0 & 0 & 1 \\
\end{pmatrix}$ and $N = A - S = \begin{pmatrix}
0 & 1 & -2 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$. It's easy to verify that $N^2 = 0$.
The matrix exponential is given by $$e^{tA} = e^{t(S+N)} = e^{tS}e^{tN} = Pe^{t\Lambda}P^{-1} \sum_{k=0}^{m-1}\dfrac{{tN}^k}{k!}$$
Since $N^2$ yields a zero vector, all the higher terms in the matrix exponential will be zero. We have $e^{t\Lambda} = \mbox{diag}(e^{-t},e^{-t},e^t)$ and
$$e^{t A} = e^{t(S+N)} = e^{tS}e^{tN} = P e^{t \Lambda}P^{-1}(I + t N) = \begin{pmatrix}
e^{-t} & t e^{-t} & -2 t e^{-t}+\dfrac{3 e^t}{2}-\dfrac{3 e^{-t}}{2} ~~\\
0 & e^{-t} & -2 e^{-t}+2 e^t ~~\\
0 & 0 & e^t \\
\end{pmatrix}$$
The solution of the initial value problem is
$$x(t) = e^{t A} x_0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of $3 \times 3$ symmetric matrices with entries five ones and four zeros which are Non Singular Find Number of $3 \times 3$ symmetric matrices with entries five ones and four zeros which are Non Singular
My try:
The only possibilities of symmetric matrices with given entries are as follows:
$1.$ With diagonal entries $1,1,1$ one of which is
$$\begin{bmatrix}
1 & 0 &0 \\
0&1 &1 \\
0&1 & 1
\end{bmatrix}$$ and Non Diagonal entries can be Permuted in $3$ ways
$2.$ With Diagonal entries $0,0,1$ one of which is
\begin{bmatrix}
0 & 1 &1 \\
1&0 &0 \\
1&0 & 1
\end{bmatrix}
Here Number of Matrices possible are $3 \times 3=9$
Any clue of how to check Non singular matrices among these $12$ matrices?
| Let us remember that $3\times 3$ matrix $A=(a_{ij})$ is no-singular if, only if,
\begin{align}
\det (A)=
&+a_{11}\cdot ( a_{22} \cdot a_{33} - a_{23}\cdot a_{32}) \\
&-a_{12}\cdot ( a_{21} \cdot a_{33} - a_{31}\cdot a_{23})\\
&+a_{13}\cdot ( a_{21} \cdot a_{32} - a_{31}\cdot a_{22})\neq 0
\end{align}
Affirmation 1: If $a_{11}=1$ then $a_{22}\cdot a_{33} - a_{23}\cdot a_{32}\neq 0$ is non-zero if, and only if,
$$
-a_{12}\cdot ( a_{21} \cdot a_{33} - a_{31}\cdot a_{23}) +a_{13}\cdot ( a_{21} \cdot a_{32} - a_{31}\cdot a_{22})=0.
$$
Affirmation 2: If $a_{11}=1$ then $a_{22}\cdot a_{33} - a_{23}\cdot a_{32}\neq 0$ if, and only if,
$3$ of the $4$ entries are equal to $1$ and exactly $1$ entry equals $0$. And it is easy to see that there are $4$ ways for this to happen under these conditions.
Doing this process for the remaining $2$ products
$$
-a_{12}\cdot ( a_{21} \cdot a_{33} - a_{31}\cdot a_{23})
\qquad
+a_{13}\cdot ( a_{21} \cdot a_{32} - a_{31}\cdot a_{22})
$$
gives a total of $4+4+4=12$ non-sigular matrices of the required type.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$ I am trying to find the partial fraction of:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$
I started off with:
$$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$
After that I get the following equations:
$A+B = 0$; $-2A+2B =1$
Giving: $A=-B$, and secondly $4B=1$, Hence:
$A=-\frac{1}{4}$ and $B=\frac{1}{4}$
But is this correct?
| You want numerators of degrees lower than your denominators. Let's note that $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{1}{4}(\frac{1}{s^2-2s+2}-\frac{1}{s^2+2s+2}).$$ Complex numbers are required to obtain partial fractions with linear denominators, viz. $$\frac{1}{s^2\pm 2s+2}=\frac{1}{(s\pm 1)^2+1}=\frac{i}{2}(\frac{1}{s\pm 1 +i}-\frac{1}{s\pm 1 -i}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$f(x-y)=f(x)f(y)$
If
$f(x-y) = f(x)f(y) $ (Call this equations (1) )
for all $x$ and $y$, and $f(x)\not = 0$ for any $x$, find all possible values for $f(1997)$.
My attempt:
Substitute $x=y=0$ in $(1)$ to obtain $f(0) = 1$. If you now substitute $x=y=1$ in (1), you get $f(1)^2 = f(0) \implies f(1) = \pm 1$.
Now if $f(1)=1$, then you can see that $f(2)=1$ as well (by substituting $x=2,y=1$ in $(1)$ ). We can extend this to all natural numbers
(assume that $f(k)=1 \forall k\in \mathbb{N} \leq n$:
\begin{align*}
f((k+1)-1) &= f(k+1)f(1) \\
f(k) &= f(k+1)f(1) \\
f(k+1) &= 1
\end{align*}
On the other hand, if $f(1) = -1$, then (as before) $f(2) = 1$. This implies
\begin{align*}
f(3-1) &= f(3)f(1) \\
f(2) &= f(3)f(1) \\
f(3) &= -1
\end{align*}
This makes us think that $f(n) = (-1)^n \forall n \in \mathbb{N}$. Indeed, this is the case (assume that it holds for all $k\leq n$):
\begin{align*}
f((k+1)-1) &= f(k+1) f(1) \\
f(k) &= f(k+1) f(1) \\
(-1)^k &= f(k+1) f(1) \\
f(k+1) &= (-1)^{k-1} = (-1)^{k+1}
\end{align*}
Therefore, all the possible values for $f(1977)$ are $\{1,-1\}$.
But apparently, $f(1977) = -1$ is not given in the solution to this problem (this problem is from Richard Rusczyk's and Mathew Crawford's Intermediate Algebra):
Setting $y=x$, we get $f(0) = f(x)^2)$. Setting $x=0$, we get $f(0) = f(0)^2$, so $f(0)[f(0) - 1] = 0$. Hence, $f(0) = 0$ or $f(0) = 1$. But $f(x)$ is never equal to $0$, so $f(0) = 1$. Then $f(x)^2 = 1$ for all x, so $f(x) = -1$ or $f(x) = -1$ for all $x$.
Setting $x=2y$ in the given equation, we get $f(y) = f(2y)f(y)$. Since $f(y)$ is nonzero, we may divide both sides by $f(y)$ to get $f(2y)=1$. Since this holds for all $y$, in particular, we have $f(1977) = \boxed{1}$.
What did I miss in my solution?
Also, the fact that they did not mention the domain and co-domain of $f$ annoys me.
| You start with $f(1) = -1$, then you get $f(n) = (-1)^n$.
But, the assumption is for all $x$ and $y$. Then you have $f(\frac{1}{2}) = f(1)f(\frac{1}{2}) = -f(\frac{1}{2})$ or $f(\frac{1}{2}) = 0$. Note that $f(x) \neq 0$ for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n→∞}\left(\frac{\sqrt{n^2+n}-1}n\right)^{2\sqrt{n^2+n}-1}$ I have the limit
$$\lim_{n\to\infty}\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}.$$
I simplified the limit by using $(f(x)-1)\cdot g(x)$, where $f(x)= \sqrt{n^2+n}-1$ and $g(x) = 2\sqrt{n^2+n}- 1$ since it is in the form of $1^\infty$. I ended up getting $$\frac{(3+2n)(n-(\sqrt{n^2-n})+1}{n},$$ which tends to $\infty$.
Please help how to proceed further or a different way to solve it.
| $$
\left(\sqrt{1+\frac{1}{n}}-\frac{1}{n}\right)^{2n\left(\sqrt{1+\frac{1}{n}}-\frac{1}{2n}\right)} = \left(\sqrt{1+\frac{1}{n}}-\frac{1}{n}\right)^{2n+O(\frac{1}{n})}
$$
hence
$$
\lim_{n\to\infty}\left(\sqrt{1+\frac{1}{n}}-\frac{1}{n}\right)^{2n+O(\frac{1}{n})} = e^{-1}
$$
because
$$
\lim_{y\to 0}\left(\sqrt{1-y}-y\right)^{\frac{2}{y}} = \lim_{y\to 0}\left(1-\frac{y}{2}+O(y^2)\right)^{\frac{2}{y}} = e^{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way.
I eventually get $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$
And I am stuck...
| $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} =\frac {\frac{2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\cos^2\frac{\alpha}2}}{\frac{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}}{\cos^2\frac{\alpha}2}}=\frac{2\tan\frac{\alpha}2}{\tan^2\frac{\alpha}2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Trigonometric equations solutions within a range I wanted to ask a question about solving the following trigonometric equation:
$$ \tan\left(\theta + \frac{\pi}{6}\right) = \tan (\pi - \theta)$$
in the interval $0 ≤ \theta ≤ \pi$
This is the following route I took.
$$ \theta + \frac{\pi}{6} = \pi - \theta$$
$$ 2 \theta = \frac{5 \pi}{6}$$
$$ \theta = \frac{5 \pi}{12}$$
Now, using the interval above, I then clarified that there are no further solutions as the next solution would be $+ \pi$ so would be $\frac{17 \pi}{12}$ and hence out of this range.
However, I apparently made a slight "error". According to a friend in the lecture hall today, there is another solution, $\frac{11 \pi}{12}$ as follows:
$$ 2 \theta = \frac{5 \pi}{6}, \frac{11 \pi}{6}$$
$$ \theta = \frac{5 \pi}{12}, \frac{11 \pi}{12}$$
because in the range $0 ≤ 2 \theta ≤ \pi$, the result $\frac{11 \pi}{6}$ is possible.
I'm confused. How can it be that when I solved for $\theta = \frac{5 \pi}{12}$ there were no more solutions available in the interval but using $2 \theta = \frac{5 \pi}{6}$ there were solutions possible?
| The point is that you are adding the period $\pi$ to $\theta$ but it is wrong indeed the following holds
$$\tan x=\tan y \iff x=y+k\pi$$
and thus we have that
$$\tan\left(\theta + \frac{\pi}{6}\right) = \tan \left(\pi - \theta\right)\iff \theta + \frac{\pi}{6} = \pi - \theta+k\pi$$
$$\iff 2\theta=\frac56\pi+k\pi\iff\theta=\frac5{12}\pi+k\frac \pi 2$$
therefore
*
*$\theta_1=\frac5{12}\pi$
*$\theta_2=\frac5{12}\pi+\frac \pi 2=\frac {11} {12}\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $32\mid(a^2+3)(a^2+7)$ If $a$ be an odd integer prove that $32\mid(a^2+3)(a^2+7)$
These are not three or two consecutive integers so they are not divisible by $3$ or $2$ also mathematical induction can not be applied. Any idea how to solve it
| Use mod $4$, meaning if $a = 1 \pmod 4\implies a^2+3= 0\pmod 4, a^2+7 = 0\pmod 8\implies (a^2+3)(a^2+7)= 0 \pmod {32}$ . If $a = 3\pmod 4\implies a^2+3 = 0\pmod 4, a^2+7 = 0\pmod 8\implies (a^2+3)(a^2+7) = 0\pmod {32}$ . Either case gives the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Are the expressions involving the nth root and the natural logarithm equivalent? Are the following expressions equivalent?
Expression 1:
$1 - \frac{1}{\sqrt[n]{1+n}}, n\in Z^+ $
Expression 2:
$ \frac{ln(1+n)} {n}$
Under what set of conditions is it possible, if at all?
| As Richard Ambler suspected by his comment, may be you are considering the case of very large values of $n$ and the limit when $n\to \infty$.
If we use expansions
$$A=(n+1)^{\frac{1}{n}}\implies \log(A)={\frac{1}{n}}\log(n+1)={\frac{1}{n}}\left(\log(n)+\log\left(1+\frac 1n\right)\right)$$
$$\log(A)=\frac{\log \left({n}\right)}{n}+\frac{1}{n^2}-\frac{1}{2
n^3}+O\left(\frac{1}{n^4}\right)$$
$$A=e^{\log(A)}=1+\frac{\log \left({n}\right)}{n}+\frac{\log
^2\left({n}\right)+2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$1-\frac 1A=1 - \frac{1}{\sqrt[n]{1+n}}=\frac{\log \left({n}\right)}{n}+\frac{1-\frac{1}{2} \log
^2\left({n}\right)}{n^2}+O\left(\frac{1}{n^3}\right)$$ On the other side, doing the same,
$$B=\frac{\log (n+1)}{n}=\frac{\log \left({n}\right)}{n}+\frac{1}{n^2}-\frac{1}{2
n^3}+O\left(\frac{1}{n^4}\right)$$
$$1 - \frac{1}{\sqrt[n]{1+n}}-\frac{\log (n+1)}{n}=-\frac{\log ^2\left({n}\right)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
So, if you want to know for which $n$, you have
$$\left|-(n+1)^{-1/n}-\frac{\log (n+1)}{n}+1\right| \leq \epsilon$$, you need to solve for $n$
$$\frac{\log ^2\left({n}\right)}{2 n^2}=\epsilon\implies n=-\frac{W_{-1}\left(- \sqrt{2\epsilon }\right)}{ \sqrt{2\epsilon }}$$ where appears Lambert function.
For more conveniency, let $\epsilon =10^{-k}$ and compute the corresponding rounded value of $n$
$$\left(
\begin{array}{cc}
k & n \\
2 & 22 \\
3 & 104 \\
4 & 429 \\
5 & 1658 \\
6 & 6171 \\
7 & 22398 \\
8 & 79814 \\
9 & 280500 \\
10 & 975123 \\
11 & 3360257 \\
12 & 11495782
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expanding with fractional powers I've been given the following integral to evaluate
$$\int_0^1(3x^3 - x^2 + 2x - 4)\frac{1}{\sqrt{x^2 - 3x + 2}} dx.$$
Can I expand fractional powers?
| You should complete the square to obtain
$$ x^2-3x+2=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} $$
then let $x-\frac{3}{2}=\frac{1}{2}\sec(t)$, that is, $x=\frac{3}{2}+\frac{1}{2}\sec(t)$. Then $dx=\frac{1}{2}\sec(t)\tan(t)\,dt$ and
$$ \sqrt{x^2-3x+2}=\pm\frac{1}{2}\tan(t) $$
For $0\le x\le1$, $\sec(t)<0$ and $\sec^{-1}(-3)\le t\le\pi$, so you will need
$$ \sqrt{x^2-3x+2}=-\frac{1}{2}\tan(t) $$
Upon substitution into the original, you obtain an integral whose terms are powers of the secant function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determining coefficients of a trigonometric function with given tangent lines I am given the following problem:
Find a function of the form $f(x)=a+b\cos{(cx)}$ that is tangent to the line $y=1$ at the point $(0,1)$ and tangent to the line $y=x+\dfrac{3}{2}-\dfrac{\pi}{4}$ at the point $\left(\dfrac{\pi}{4},\dfrac{3}{2}\right)$.
I do not know what I am doing wrong here. We have that $f(0)=1$ so that $$a+b\cos{(0)}=a+b=1.$$ Also, we have that$$f\left(\dfrac{\pi}{4}\right)=\dfrac{3}{2}=a+b\cos{\left(\dfrac{c\pi}{4}\right)}.$$ Taking derivatives, we know that $f'(0)=0=-bc\sin{0},$ and so this is unhelpful in solving the system. Finaly, we have that$$f'\left(\dfrac{\pi}{4}\right)=1=-bc\sin{\left(\dfrac{c\pi}{4}\right)}.$$ Therefore we have a system of three equations with three unknowns and it should be solvable:\begin{align*}
&a+b=1\\
&a+b\cos{\left(\frac{c\pi}{4}\right)}=\frac{3}{2}\\
&bc\sin{\left(\frac{c\pi}{4}\right)}=-1
\end{align*}
So my initial thought here was to look at$$f\left(\frac{\pi}{2}\right)=a+b\cos\left(\frac{2c\pi}{4}\right)=a+b\cos\left(2\cdot\frac{c\pi}{34}\right)$$ and use double angle formulas. So
\begin{align*}f\left(\frac{\pi}{2}\right)&=1-b+b\left[2\cos^2{\left(\frac{c\pi}{4}\right)-1}\right]\\&=1-2b+2b\cos{\left(\frac{c\pi}{4}\right)}\cos{\left(\frac{c\pi}{4}\right)}\\&=1-2b+\frac{2}{b}\left(\frac{3}{2}-a\right)^2\\&=1-2b+\frac{2}{b}\left(\frac{1}{2}+b\right)^2\\&=1-2b+\frac{2}{b}\left(\frac{1}{4}+b+b^2\right)\\&=1-2b+\frac{1}{2b}+2+2b\\&=3+\frac{1}{2b}\end{align*}
And so this does not seem very helpful since I cannot solve for $b$ without knowing $f\left(\dfrac{\pi}{2}\right)$. I could use an alternate double angle formula, such as $\cos(2x)=1-2\sin^2{(2x)}$, and get that
$$f\left(\frac{\pi}{2}\right)=1-b+b\left[1-2\sin^2{\frac{c\pi}{4}}\right]=1-\frac{2}{bc^2}.$$
Setting them equal gives a solution for one variable, but again, this does not seem to help. I know I am missing something very basic here, but not sure what.
| Since$$
\begin{cases}
a + b = 1\\
a + b \cos \dfrac{πc}{4} = \dfrac{3}{2}\\
-bc \sin \dfrac{πc}{4} = 1
\end{cases} \Longleftrightarrow
\begin{cases}
b \cos \dfrac{πc}{4} = b + \dfrac{1}{2}\\
b \sin \dfrac{πc}{4} = -\dfrac{1}{c}\\
a = 1 - b
\end{cases}, \tag{1}
$$
and$$
\left( b + \frac{1}{2} \right)^2 + \left( -\frac{1}{c} \right)^2 = b^2 \Longrightarrow b = -\frac{1}{c^2} - \frac{1}{4},
$$
then$$
(1) \Longleftrightarrow
\begin{cases}
\cos \dfrac{πc}{4} = -\dfrac{c^2 - 4}{c^2 + 4},\ \sin \dfrac{πc}{4} = \dfrac{4c}{c^2 + 4} & \qquad (2)\\
a = \dfrac{1}{c^2} + \dfrac{5}{4},\ b = -\dfrac{1}{c^2} - \dfrac{1}{4}
\end{cases}.
$$
WA shows that (2) may have infinitely many solutions, but solutions can only be numerically obtained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimizer of $A=12a+13b+11c$
Let $a \ge 2$, $b \ge 5$ and $c \ge 5$ such that $2a^2+b^2+c^2=69$. Find the minimizer of $$A=12a+13b+11c$$
My try: Use Lagrange multipliers
We have: $L\left(a;b;c;\lambda \right)=12a+13b+11c+\lambda \left(2a^2+b^2+c^2-69\right)$
$$\begin{cases}
\frac{\partial L}{\partial a}=4a\lambda +12=0\Leftrightarrow \lambda =-\frac{3}{a} & \\
\frac{\partial L}{\partial b}=2b\lambda +13 =0\Leftrightarrow \lambda =-\frac{13}{2b} \\
\frac{\partial L}{\partial c}=2c\lambda +11=0\Leftrightarrow \lambda =-\frac{11}{2c} &
\end{cases} $$
Or $$\begin{cases}
\frac{3}{a}=\frac{13}{2b}=\frac{11}{2c} & \\ 2a^2+b^2+c^2=69 &
\end{cases} $$
But $\text{Min}A=155$ occurs when $a=2;b=5;c=6$. Help me!
| Let $a=2$, $b=5$ and $c=6$.
Hence, $A=155.$
We'll prove that it's a minimal value.
Indeed, let $a=u+2$, $b=v+5$ and $c=w+5$.
Hence, the condition gives
$$2u^2+v^2+w^2+8u+10v+10w=11$$ and we need to prove that
$$12u+13v+11w\geq11.$$
For the proof we can use the Contradiction method.
Indeed, let $12u+13v+11w<11$, $u=kx$, $v=ky$ and $w=kz$,
where $k>0$ and $12x+13y+11z=11.$
Thus, $k<1$ and
$$11=2u^2+v^2+w^2+8u+10v+10w=k^2(2x^2+y^2+z^2)+k(8x+10y+10z)<$$
$$<2x^2+y^2+z^2+8x+10y+10z,$$
which is a contradiction because we'll prove now that
$$2x^2+y^2+z^2+8x+10y+10z\leq11.$$
Indeed, we need to prove that
$$2x^2+y^2+z^2+\frac{(8x+10y+10z)(12x+13y+11z)}{11}\leq\frac{(12x+13y+11z)^2}{11}$$ or
$$13x^2+14y^2+44xy+28xz+23yz\geq0,$$ which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\sum_n(2n - \sqrt{n^2+1}-\sqrt{n^2-1})$? How to compute $\sum_n (2n - \sqrt{n^2+1}-\sqrt{n^2-1})$?
I tried two ways:
1. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= n - \sqrt{n^2+1} + n -\sqrt{n^2-1} \\
&= \frac{1}{n+\sqrt{n^2-1}}-\frac{1}{n-\sqrt{n^2+1}},
\end{align*}
but I don't know how to do later.
2. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= 2n - \frac{(\sqrt{n^2+1} + \sqrt{n^2-1})}{1} \\
&= 2n - \frac{2}{\sqrt{n^2+1} - \sqrt{n^2-1}},
\end{align*}
but I don't know how to do later too.
| Starting with Sangchul Lee's integral representation (which is a consequence of the Laplace transform)
$$ S = \int_{0}^{+\infty}\frac{I_1(x)-J_1(x)}{x(e^x-1)}\,dx = \frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi}\frac{e^{-x\cos\theta}-e^{ix\cos\theta}}{e^x-1}\sin^2\theta\,d\theta\,dx \tag{1}$$
and applying Fubini's theorem we get
$$ S = \frac{1}{\pi}\int_{0}^{\pi}\left[\psi(1-i\cos\theta)-\psi(1+\cos\theta)\right]\sin^2\theta\,d\theta \tag{2}$$
where
$$\begin{eqnarray*} S &=& \frac{1}{2\pi}\int_{0}^{\pi}\left[\psi(1-i\cos\theta)+\psi(1+i\cos\theta)-2\psi(1-\cos\theta)\right]\sin^2\theta\,d\theta\\&=&\frac{1}{\pi}\int_{0}^{\pi}\left[\log\Gamma(1-\cos\theta)-\text{Im}\,\log\Gamma(1-i\cos\theta)\right]\cos\theta\,d\theta \tag{3}\end{eqnarray*}$$
allows an efficient numerical evaluation of $S$ through standard integration techniques (composite Simpson's rule or Gaussian quadrature):
$$ S \approx 0.6369740582412\tag{4} $$
but I do not believe that $S$ has a simple closed form in terms of standard mathematical constants.
We have a similar situation here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the following limit I have been working hard on this limit:
$$(n^6 + n^5 + n^4)^{1/6} - n$$
as n goes to infinity. I have no clue of what i need to do. I always get indeterminante forms.
| $$ \left( n + \frac{1}{6} \right)^6 < n^6 + n^5 + n^4 < \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 $$
$$ $$
$$ $$
$$ \left( n + \frac{1}{6} \right)^6 = n^6 + n^5 + \frac{5 n^4}{12} + \frac{5n^3}{54} + \frac{5n^2}{432} + \frac{n}{1296} + \frac{1}{46656} $$
Note that, for $n \geq 1,$ this is smaller than $n^6 + n^5 + n^4$
$$ $$
$$ \left( n + \frac{1}{6} \right)^6 = n^6 + n^5 + \frac{5 n^4}{12} \;+ \;\mbox{lower degree} $$
$$ \left( n + \frac{1}{6} \right)^5 = n^5 + \frac{5 n^4}{6} \;+ \;\mbox{lower degree} $$
$$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = \left( n + \frac{1}{6} \right)^6 + 6 \left( n + \frac{1}{6} \right)^5 \; \frac{7}{72n} + \mbox{lower degree} $$
$$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = \left( n + \frac{1}{6} \right)^6 + \frac{7}{12n} \; \left( n + \frac{1}{6} \right)^5 \;+ \mbox{lower degree} $$
$$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = n^6 + n^5 + \frac{5 n^4}{12}+ \mbox{lower degree} + \frac{7 n^4}{12}+ \mbox{lower degree} $$
$$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = n^6 + n^5 + n^4 + \mbox{lower degree} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Are there uncountably many $A\in M_3 (\mathbb {R})$ such that $A^8=I $? I'm working on the following problem:
Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then
*
*the minimal polynomial of $A$ can only be of degree $2$.
*the minimal polynomial of $A$ can only be of degree $3$.
*either $A = I$ or $ A = -I$.
*there are uncountably many such $A$.
By taking $A=I $ we can eliminate options (1) & (2). For the minimal polynomial of $A$ in that case is of degree $1$. Now take
$$A =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}$$
Then $A^8=I $ but $A$ is neither $I$ nor $-I$. So option (3) is eliminated. Now I don't know how to proceed about with option (4). Can someone help me please? Thanks in advance.
| Let $O(\theta)$ be the $2 \times 2$ orthogonal matrix
$O(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}; \tag 1$
set
$P = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}; \tag 2$
then
$O^{-1}(\theta) = O^T(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}; \tag 3$
we have
$O^T(\theta) P O(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$
$= \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2 \cos \theta \sin \theta \\ 2 \cos \theta \sin \theta & \sin^2 \theta - \cos^2 \theta \end{bmatrix}$
$= \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ \sin 2\theta & -\cos 2 \theta \end{bmatrix}, \tag 4$
which is clearly an uncountable family of $2 \times 2$ matrices; furthermore,
$(O^T(\theta) P O(\theta))^2 = O^T(\theta) P O(\theta) O^T(\theta) P O(\theta) = O^T(\theta) P I P O(\theta)$
$= O^T(\theta)P^2 O(\theta) = O^T(\theta)I O(\theta) = I, \tag 5$
whence
$(O^T(\theta) P O(\theta))^8 = I \tag 6$
as well; now set
$A(\theta) = \begin{bmatrix} O^T(\theta) P O(\theta) & 0 \\ 0 & -1 \end{bmatrix}; \tag 7$
then
$A^8(\theta) = \begin{bmatrix} (O^T(\theta) P O(\theta))^8 & 0 \\ 0 & (-1)^8 \end{bmatrix} = \begin{bmatrix} I & 0 \\ 0 & 1 \end{bmatrix} = I, \tag 8$
and the family $\{ A(\theta)\mid \theta \in \Bbb R \}$ is also uncountable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
A geometric inequality for a triangle ABC I have to prove that:
$ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} \ge 4,$
for the sides $a,b,c$ of the triangle $ABC,$ $ w_a, w_b, w_c $ the angle bisectors and $s$ its semiperemeter. We have for the angle bisectors of the triangle $ABC$:
$ w_a = \frac {2}{b+c} \sqrt{b c s (s-a)},$ $ w_b = \frac {2}{c+a} \sqrt{c a s (s-b)},$ $ w_c = \frac {2}{a+b} \sqrt{a b s (s-c)}.$
Then, we have:
$ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} = \frac {1}{4 s} [ \frac { a^2 (b+c)^2}{b c (s-a)} + \frac { b^2 (a+c)^2}{a c (s-b)} + \frac { c^2 (a+b)^2}{a b (s-c)}].$
So, we have to prove
$ \frac{ a^2 (b+c)^2}{b c (s-a)} + \frac { b^2 (a+c)^2}{a c (s-b)} + \frac { c^2 (a+b)^2}{a b (s-c)} \ge 16 s $
Then I stuck, thank you.
| Taking from where you left off...
$LHS \ge \sum 4\cdot\dfrac{a^2}{s-a}\ge 4\cdot\dfrac{(a+b+c)^2}{(s-a)+(s-b)+(s-c)}= 4\cdot\dfrac{4s^2}{s} = 16s = RHS$ , by employing AM-GM and CS inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
| Though there are already 6 wonderful solutions, I want to share mine with you now. Wish that you can enjoy it.
Letting $x=\cos \theta$ yields
$$
I=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\cos \theta+\sin \theta)}{\cos \theta}\sin \theta d \theta
$$
and letting $x=\sin \theta$ yields $$
I=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\sin \theta+\cos \theta)}{\sin \theta} \cos \theta d \theta
$$
Combining them gives $$
\begin{aligned}
2 I &=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\sin \theta+\cos \theta)}{\sin \theta \cos \theta} d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\sin 2 \theta)}{\sin 2 \theta} d \theta \\
&\stackrel{2\theta\mapsto\theta}{=} \frac{1}{2} \int_{0}^{\pi} \frac{\ln (1+\sin \theta)}{\sin \theta} d \alpha \\
&=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\sin \theta)}{\sin \theta} d \theta \quad \textrm{( By symmetry)}\\
&\stackrel{\theta\mapsto \frac{\pi}{2} -\theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta \\
&=\frac{\pi^{2}}{8}
\end{aligned}
$$
Putting $a=0$ in my answer, we get $2 I=\dfrac{\pi^{2}}{8}$ and hence $\boxed{I=\frac{\pi^{2}}{16}}$.
:|D Wish you enjoy my solution!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 6
} |
If $a+b+c$ divides the product $abc$, then is $(a,b,c)$ a Pythagorean Triple? Firstly, I will define what Pythagorean Triples are for those who do not know.
Definition:
A Pythagorean Triple is a group of three integers $a$, $b$ and $c$ such that $a^2+b^2=c^2$, since the Pythagorean Theorem asserts that for any $90^\circ$ (right-angle) triangle $ABC$ with sides $a$, $b$ and $c$, one will always have the equation, $a^2+b^2=c^2$.
I was looking at Pythagorean Triples and noticed another property apart from how $a^2+b^2=c^2$. Here are the first $30$ Pythagorean Triples $(a,b,c)$ ordered from smallest to greatest value, i.e. $$(a,b,c)\qquad\text{ s.t. }\qquad a<b<c.\tag*{$\big(\text{s.t. = such that}\big)$}$$
I noticed that $a^2=(c+b)(c-b)$, but that is trivial since $$\begin{align}a^2&=(c+b)(c-b)\tag{given} \\ &=c^2-b^2 \\ \Leftrightarrow\,\,\,\, a^2+b^2&=c^2.\end{align}$$
However, I also noticed that by having "$u\mid v$" be read as "$u$ divides $v$", it appears that $$a+b+c\mid abc.$$ For example, $(a,b,c)=(3,4,5)$ is a classic Pythagorean Triple; $3^2+4^2=5^2$.
Also, $$\begin{align}3+4+5&=12 \\ \& \quad3\times 4\times 5 &= 60. \\ \\ 12 &\,\mid 60 \\ \Leftrightarrow \,\,\,\,3+4+5&\,\mid 3\times 4\times 5.\end{align}$$ This, I cannot prove to be true $-$ but I tested with all the $30$ Pythagorean Triples above, and I have come across no counter-example. Is there a proof? I do not know where to begin myself.
Conjecture:
Given three positive integers $a$, $b$ and $c$, if $a < b<c$ and $a^2+b^2=c^2$, then $$a+b+c\mid abc.$$
Thank you in advance.
Edit:
My conjecture was originally the other way round; i.e. if $a+b+c\mid abc$ then $a^2+b^2=c^2$. But $6$ is a counter-example, namely because it is a Perfect Number.
| Alternatively:
$$\frac{abc}{a+b+c}=\frac{abc(a+b-c)}{(a+b+c)(a+b-c)}=\frac{abc(a+b-c)}{2ab}=\frac{c(a+b-c)}{2},$$
which is a positive integer for both cases: $a,b,c$ are all even; $a,c$ are odd and $b$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$ Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$
Angle between curves at a point is given by the angle between their tangent planes at the point.
$$f(x,y,z):= 3x +z^2 -xy^2z$$
$\nabla f(1,-2,1) = \langle 3-y^2z,-2xyz,2z-xy^2\rangle_{(1,-2,1)} = \langle -1,4,-2 \rangle$
Equation of tangent plane to $xy^2z = 3x + z^2 $ will be
$-x + 4y -2z + d=0$ Putting $(1,-2,1)$ and solving for $d$ we have
$$x-4y+2z=11 \; \; \; (1)$$
Also, a nice way to write equation of tangent plane to curve $ax^2 + by^2 + cz^2 + 2ux + 2vy + 2wz + d =0$ at $P(x_0,y_0,z_0)$ would be:
$$ax\cdot x_0 + by\cdot y_0 + cz\cdot z_0 + u(x+x_0) + v(y+y_0) + w(z+z_0) +d=0$$
Hence tangent plane to $3x^2-y^2+2z=1$ will be $3x(1) -y(-2)+(z+1)=1$
$$ \Rightarrow 3x +2y+z=0 \; \; \; (2)$$
One of the reason for posting this is, how'd you follow this method to write equation of tangent plane to the first curve $xy^2z = 3x + z^2 $? Basically, is it possible to extend this method to equations where degree is greater than $2$ and contains terms such as $xyz$?
Now, angle between tangent planes is angle between their normals,
Direction ratios of normal to $(1)$ and $(2)$ respectively are
$a=\langle 1,-4,2\rangle$ and $b=\langle 3,2,1\rangle $
$\Rightarrow \theta= \arccos(\frac{a\cdot b}{|a||b|}) = \arccos(\frac{-3}{7\sqrt{6}})$
Problem here is, how'd I know if this is acute or not, i.e, if this is the answer that I was looking for ?
| We have
$$
\vec n_1 = \frac{\nabla (x y^2 z - 3 x + z^2)}{||\nabla(x y^2 z - 3 x + z^2)||} \\
\vec n_2 = \frac{\nabla (3 x^2 - y^2 + 2 z - 1)}{||\nabla(3 x^2 - y^2 + 2 z - 1)||}
$$
and the sought angle is
$$
\varphi = \min(|\arccos(\pm<\vec n_1,\vec n_2>)|) = \arccos(\frac{1}{\sqrt{742}})\approx 88^{\circ}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Using Lagrange Multiplier to prove identity Show that the maximum and minimum values of the function $u=x^2+y^2+xy$, where $ax^2+by^2=ab\ (a>b>0)$ are given by $$4(u-a)(u-b)=ab$$
My attempt- using Lagrange Multiplier method, $$F(x,y)= (x^2+y^2+xy)-\lambda(ax^2+by^2-ab)$$
$$dF= (2x+y-2a\lambda x)dx+(2y+x-2b\lambda y)dy$$
Equating to zero to get,
$$y/x= 2(a\lambda -1); y/x= 1/(2(b\lambda-1))$$
$$\implies 4(a\lambda-1)(b\lambda-1)=1$$
$$\implies 4(a^2\lambda-a)(b^2\lambda-b)=ab$$
I am stuck here. Finding $(x_0,y_0)$ which finds extrema for $u$ and then substituting to prove $$4(u-a)(u-b)=ab$$ is very tedious. Can someone help on how I should go about it?
|
Because the function $ \ u(x,y) \ = \ x^2 + xy + y^2 \ $ and the constraint ellipse $ \ ax^2 + by^2 = ab \ \ , $ with $ \ a > b > 0 \ $ have symmetry about the origin, we should expect that the points at which the level curves of the function (which are themselves rotated ellipses [marked in red in the graph above]) are tangent to the constraint curve [in blue] will occur in pairs $ \ (\pm x \ , \ \pm y) \ $ and $ \ (\pm x \ , \ \mp y) \ \ . $ The former pair lie on the line through the origin $ \ y = mx \ \ , \ m > 0 \ \ $ and the latter on the line $ \ y = nx \ \ , \ n < 0 \ \ \ , $ which permits us to describe these tangent points as $ \ (\pm x_{+} \ , \ \pm mx_{+}) \ $ and $ \ (\pm x_{-} \ , \ \mp nx_{-}) \ \ . $ We will identify the two level curves which represent the maximum and minimum values of our function on the constraint ellipse as
$$ u \ = \ c_{+} \ = \ x_{+}^2 \ + \ x_{+}y_{+} \ + \ y_{+}^2 \ = \ (1 + m + m^2)·x_{+}^2 \ \ \ \text{and} \ \ \ u \ = \ c_{-} \ = \ (1 + n + n^2)·x_{-}^2 \ \ . $$
We wish to show that these extremal values $ \ c_{+} \ $ and $ \ c_{-} \ $ satisfy the equation $ \ 4·(u-a)·(u - b) \ = \ ab \ \ , $ which implies the quadratic equation $ \ u^2 \ - \ (a + b) u \ + \ \frac34 ab \ = \ 0 \ \ . $ Applying Viete's relations, we will endeavor to demonstrate here that
$$ c_{+} \ + \ c_{-} \ = \ a + b \ \ \ \text{and} \ \ \ c_{+} \ · \ c_{-} \ = \ \frac34 ab \ \ . $$
The Lagrange equations for this extremization are
$$ 2x \ + \ y \ \ = \ \ \lambda \ · \ 2ax \ \ , \ \ x \ + \ 2y \ \ = \ \ \lambda \ · \ 2by \ \ , $$
as you have determined. We may write this as a system of two equations
$$ 2x · (1 - \lambda·a) \ + \ y \ \ = \ \ 0 \ \ , \ \ x \ + \ 2y · (1 - \lambda·b) \ \ = \ \ 0 \ \ . $$
The tangent level curves do not contact the constrain ellipse on either coordinate axis, so we "reject" the "trivial" solutions $ \ x = 0 \ , \ y = 0 \ $ and instead seek those solutions for which the "coefficient determinant" is
$$ \ 2 · (1 - \lambda·a) · 2 · (1 - \lambda·b) \ - \ 1 \ \ = \ \ 4ab · \lambda^2 \ - \ 4·(a+b) · \lambda \ + \ 3 \ \ = \ \ 0 \ \ . $$
We obtain
$$ \lambda \ \ = \ \ \frac{4·(a+b) \ \pm \ \sqrt{16·(a+b)^2 \ - \ 4·4ab·3}}{2 \ · \ 4ab} \ \ = \ \ \frac{ (a+b) \ \pm \ \sqrt{ a^2 \ - \ ab \ + \ b^2}}{2 ab} \ \ . $$
[We will label this discriminant $ \ \mathbf{D} \ = \ a^2 - ab + b^2 \ $ in what follows.]
Inserting these into the two linear equations above leads us to
$$ y \ \ = \ \ 2 · \left(a \ · \ \frac{ (a+b) \ \pm \ \sqrt{\mathbf{D}}}{2 ab} \ - \ 1 \right) x \ \ = \ \ \left( \frac{ (a-b) \ \pm \ \sqrt{\mathbf{D}}}{ b} \right) x \ \ , $$
$$ y \ \ = \ \ \left( \frac{1}{2 · ( b · \frac{ (a+b) \ \pm \ \sqrt{\mathbf{D}}}{2 ab} \ - \ 1 )} \right) \ x \ \ . $$
(We will spare the reader the algebraic manipulation in a few places along the way; here, it will show that the second equation is redundant.) From this, we find the equations of the two lines through the origin passing through the tangent points have the slopes $ \ m \ , \ n \ \ = \ \ \frac{ (a-b) \ \pm \ \sqrt{\mathbf{D}}}{ b} \ \ ; $ we observe that we indeed have $ \ m > 0 \ $ and $ \ n < 0 \ \ . $
Using this result in the constraint ellipse equation, we obtain
$$ ax_{+}^2 \ + \ b(mx_{+})^2 \ \ = \ \ ab \ \ \Rightarrow \ \ x_{+}^2 \ \ = \ \ \frac{ab}{a \ + \ m^2·b} \ \ = \ \ \frac{b}{2} \ · \ \frac{\mathbf{D} \ - \ (a-b) · \sqrt{\mathbf{D}} }{\mathbf{D}} \ \ , $$
and, similarly, $ \ x_{-}^2 \ \ = \ \ \frac{ab}{a \ + \ n^2·b} \ \ = \ \ \frac{b}{2} \ · \ \frac{\mathbf{D} \ + \ (a-b) · \sqrt{\mathbf{D}} }{\mathbf{D}} \ \ . $
With further algebra, we find
$$ m^2 \ , \ n^2 \ \ = \ \ \frac{(2a^2 - 3ab + 2b^2) \ \pm \ 2·(a-b)·\sqrt{\mathbf{D}}}{b^2} $$
$$ \Rightarrow \ \ 1 + m + m^2 \ , \ 1 + n + n^2 \ \ = \ \ \frac{2\mathbf{D} \ \pm \ (2a-b)·\sqrt{\mathbf{D}}}{b^2} \ \ , $$
and, finally, that
$$ c_{+} \ , \ c_{-} \ \ = \ \ (1 + m + m^2)·x_{+}^2 \ \ , \ \ (1 + n + n^2)·x_{-}^2 $$
$$ = \ \ \frac{[2\mathbf{D} \ \pm \ (2a-b)·\sqrt{\mathbf{D}}] \ · \ [\mathbf{D} \ \mp \ (a-b) · \sqrt{\mathbf{D}}] }{2·b·\mathbf{D}} $$
$$ = \ \ \frac{ 2·(a^2-ab+b^2)^2 \ - \ (a-b)·(2a-b)·(a^2-ab+b^2) \ \pm \ (2a-b-2a+2b)·\mathbf{D}\sqrt{\mathbf{D}} }{2·b·\mathbf{D}} $$
$$ = \ \ \frac{ (2a^2-2ab+2b^2-2a^2+3ab-b^2)·(a^2-ab+b^2) \ \pm \ b ·\mathbf{D}\sqrt{\mathbf{D}} }{2·b·\mathbf{D}} $$
$$ = \ \ \frac{ (b^2+ab )·\mathbf{D} \ \pm \ b ·\mathbf{D}\sqrt{\mathbf{D}} }{2·b·\mathbf{D}} \ \ = \ \ \frac{ (a+b ) \ \pm \ \sqrt{\mathbf{D}} }{2 } \ \ , $$
agreeing with the result of Oldboy's approach.
We now see directly that $ \ c_{+} \ + \ c_{-} \ = \ a + b \ $ and that
$$ c_{+} \ · \ c_{-} \ \ = \ \ \frac{(a + b)^2 \ - \ \mathbf{D}}{2 · 2} \ \ = \ \ \frac{(a^2 + 2ab + b^2) \ - \ (a^2 - ab + b^2)}{4} \ \ = \ \ \frac{3ab}{4} \ \ . $$
Hence, $ \ u = c_{+} \ $ and $ \ u = c_{-} \ $ satisfy the relation $ \ 4·(u-a)·(u - b) \ = \ ab \ \ . $
So the Lagrange method does provide a means of establishing the given equation, but the interpretation of the solution to the Lagrange equations is a bit different here from that for typical extremization problems.
For the graph above, the values $ \ a = 7 \ , \ b = 4 \ $ are used. Applying the formulas we produced in our discussion, we have $ \ \mathbf{D} \ = \ 7^2 - 7·4 + 4^2 \ = \ 37 \ \ , $ the slopes of the lines through the origin are $ \ m \ , \ n \ \ = \ \ \frac{ 3 \ \pm \ \sqrt{37}}{4} \ \approx \ 2.271 \ , \ -0.7707 \ \ , $ and the extremal values of our function are
$$ c_{+} \ , \ c_{-} \ \ = \ \ \frac{ 11 \ \pm \ \sqrt{ 37 } }{2 } \ \ \approx \ \ 8.5414 \ \ , \ \ 2.4587 \ \ . $$
(We can also find the coordinates of the tangent points of these level curves, but they are not called for in this problem.)
| {
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Problem related to quadratic equation If $x$ is a real number then what is the greatest value of
$$ f(x)=2(a-x)\left(x+\sqrt{x^2+b^2}\right)\;\;? $$
I cannot figure out how to convert this into a quadratic equation to find out the greatest value of $f(x).$
From what I've read, the answer is given by $a^2+b^2$.
| $$f(x)=2(a-x)\left(x+\sqrt{x^2+b^2}\right)$$
$$f'(x)=2 (a-x) \left(1+\frac{x}{\sqrt{x^2+b^2}}\right)-2 \left(x+\sqrt{x^2+b^2}\right)$$
$$f'(x)=2(a-x)\frac{x+\sqrt{x^2+b^2}}{\sqrt{x^2+b^2}}-2 \left(x+\sqrt{x^2+b^2}\right)$$
$$f'(x)=2\left(x+\sqrt{x^2+b^2}\right) \left(\frac{a-x}{\sqrt{x^2+b^2}}-1 \right)$$ So, you need to solve
$$\frac{a-x}{\sqrt{x^2+b^2}}=1\implies \frac{(a-x)^2}{x^2+b^2}=1$$ which makes your quadratic equation which, in fact, is linear
$$\frac{(a-x)^2}{x^2+b^2}=1 \implies (a^2-b^2)-2 a x=0$$ I am sure that you can take it from here.
| {
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Calculating the length of a curve I must calculate the length of the curve $$\frac{x^2}{72} - 9 \ln(x)$$ for $$9 \leq x \leq 9 e.$$
We know that formula för length is to integrate $$\sqrt{1+f'^2(x)}dx$$ between the two points of interest.
I calculated the derivata to be:
$$\frac{x}{36} - \frac{9}{x}$$
And squared:
$$\frac{x^2}{36^2} - 2 \cdot \frac{x}{36}\cdot\frac{9}{x} + \frac{9^2}{x^2} = $$
$$\frac{x^2}{36^2} - \frac{1}2 + \frac{9^2}{x^2}$$
The integral is then:
$$\int_9^{9e} \sqrt{1+\frac{x^2}{36^2} - \frac{1}2+ \frac{9^2}{x^2}} dx$$
$$\int_9^{9e} \sqrt{\dfrac{36^2x^2+x^4- 748x^2+ 9^2 \cdot 36^2}{36^2 \cdot x^2}} dx$$
$$\int_9^{9e} \sqrt{\dfrac{x^4 + 748x^2+ 9^2 \cdot 36^2}{36^2 \cdot x^2}} dx.$$
And I have no idea how to integrate that. The examples we have in the course is with trigonometric functions with a replacement with $$\tan^2(x)$$ but it is not applicable here...
| There is a nice substitution.
Let $$\frac{x}{36} - \frac{9}{x}=t \implies x=18 \left(\sqrt{t^2+1}+t\right)\implies dx=18 \left(\frac{t}{\sqrt{t^2+1}}+1\right)$$ which make
$$\int \sqrt{1+\left(\frac{x}{36}-\frac{9}{x}\right)^2}\,dx=18\int \left(t+\sqrt{t^2+1}\right)\,dt$$ which seems to be easy to compute.
| {
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How to prove $2e^x>x^3+3x$ holds for all $x \in \mathbb{R}$? Problem
Prove that $$2e^x>x^3+3x, \forall x \in \mathbb{R}.$$
Notes
It's clear that the inequality holds for all $x \leq 0$. Hence, we only need to reserch the situation when $x>0.$
Let $f(x)=2e^x-x^3-3x,(x>0).$ Here, if you use Tayor's formula of $e^x$, namely,$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots,$$ and plug it into $f(x)$,you'll have $$f(x)=2-x+x^2-\frac{2}{3}x^3+\cdots.$$ This will perhaps give nothing helpful. If you consider applying the derivative, $$f'(x)=2e^x-3x^2-3.$$ You can hardly find its zero-point, though there exists one root for $f'(x)=0$,say,$x_0=2.0715\cdots$ such that $f(x)$ reaches its minimum value,say, $$f(x)\geq f(x_0)=0.76990\cdots>0.$$
It's not easy to obtain these results without computer. Is there another tricky solution for this? Thanks in advance.
| For $x\leq0$ it's obvious.
But for $x>0$,
$$2e^x-x^3-3x
>2\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)-x^3-3x=$$
$$=\frac{1}{60}(x^5+5x^4-40x^3+60x^2-60x+120)=$$
$$=\frac{1}{60}\left(\left(x-\frac{5}{2}\right)^2\left(x^3+10x^2+\frac{15}{4}x\right)+\frac{5}{16}(52x^2-267x+384)\right)>0.$$
These coefficients we can get by the following way.
By calculator we can get that the polynomial $x^5+5x^4-40x^3+60x^2-60x+120$ for $x\geq0$ gets a minimal value around $x=\frac{5}{2}.$
Now, we'll choose $a$, $b$ and $c$ such that
$$x^5+5x^4-40x^3+60x^2-60x+120=\left(x-\frac{5}{2}\right)^2(x^3+ax^2+bx+c)+Ax^2+Bx+C,$$ $$x^3+ax^2+bx+c>0$$ and $$Ax^2+Bx+C\geq0$$ for $x\geq0.$
Indeed, we need $$x^5+5x^4-40x^3=\left(x^2-5x+\frac{25}{4}\right)(x^3+ax^2+bx+c)+Dx^2+Ex+F,$$
which gives the following system:
$$5=a-5,$$
$$-40=b-5a+\frac{25}{4},$$ which gives $a=10$ and $b=\frac{15}{4}.$
Now, we know that
$$x^5+5x^4-40x^3+60x^2-60x+120-\left(x-\tfrac{5}{2}\right)^2\left(x^3+10x^2+\tfrac{15}{4}x+c\right)=Ax^2+Bx+C$$
and it remains to choose a value of $c$, for which
$$Ax^2+Bx+C>0$$ and we see that $c=0$ is valid.
| {
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Proof that squares are divisible by 3 when their sum is In this proof, they write $3|a^2+b^2 \implies 3|a$, $3|b$. I tried using the same proof used to prove $3|a^2 \implies 3|a$, where $3$ being prime and writing $a^2 = a\cdot a$ suggests that $a$ is divisible by $3$. I'm not sure how to prove the $3|a^2+b^2$ case, though.
E9. There is no quadruple of positive integers $(x, y, z, u)$ satisfying $$x^2 + y^2 = 3(z^2 + u^2).$$
Solution. Suppose there is such a quadruple. We choose the solution with the smallest $x^2 + y^2$. Let $(a, b, c, d)$ be the chosen solution. Then $$a^2 + b^2 = 3(c^2 + d^2) \implies 3|a^2 + b^2 \implies 3|a, 3|b \implies a = 3a_1, b = 3b_1,\\a^2 + b^2 = 9(a^2_1 + b^2_1) = 3(c^2 + d^2) \implies c^2 + d^2 = 3(a^2_1 + b^2_1).$$
We have found a new solution $(c, d, a_1, b_1)$ with $c^2 + d^2 \lt a^2 + b^2$. Contradiction.
We have used the fact that $3|a^2 + b^2 \implies 3|a, 3|b$. Show this yourself. We will return to similar examples when treating infinite descent.
| It probably isn't the best solution, but you could try using congruence.
Since 3 is a pretty small number, you can test each case for
$$a,b\equiv 0,1,2\pmod 3$$
And for each one check if
$$a² + b² \equiv 0 \pmod 3 $$
It gives you (all results given modulo 3):
$$
\begin{matrix}
a & b & a^2+b² \\
0 & 0 & 0 \\
0 & 1 & 1 \\
0 & 2 & 1 \\
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 2 \\
2 & 0 & 1 \\
2 & 1 & 2 \\
2 & 2 & 2 \\
\end{matrix}
$$
As you can see:
$$a² + b² \equiv 0 \pmod 3 $$ iff $$a \equiv 0\pmod 3 \land b\equiv 0\pmod 3 $$
| {
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Show that $2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}$ is an increasing function I would like to show that the following function
\begin{align}
f_a(x)=2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}
\end{align}
is an increasing function in $x$ for $x \ge 0$ for any fixed $a>0$.
I did some simulations but not sure how to show a proof for this. I wanted to point out that, from simulation, it seems that $\frac{\Gamma((a+1)x)}{\Gamma(x)}$ can be decreasing for values $x=0$.
We can attempt this by showing that the derivative of a logarithm of $f_a(x)$ is positive. Let
\begin{align}
g_a(x)= \log (f_a(x))
\end{align}
(here log is base e) and the derivative of $g_a(x)$ is given by
\begin{align}
\frac{d}{dx} g_a(x)&= \frac{d}{dx} \left( ax \log(2)+ \log (\Gamma( (a+1)x))- \log (\Gamma( x)) \right)\\
&=a \log(2) + (a+1)\psi((a+1)x)-\psi(x)
\end{align}
where $\psi(x)$ is a digamma function.
Now it remains to show the following inequality for the difference of digamma functions
\begin{align}
(a+1)\psi((a+1)x)-\psi(x) \ge - a \log(2) .
\end{align}
| We have that
$$
r_a(x) = {{\Gamma \left( {\left( {a + 1} \right)x} \right)} \over {\Gamma \left( x \right)}}
= {{\Gamma \left( {x + a\,x} \right)} \over {\Gamma \left( x \right)}} = x^{\,\overline {\,a\,x\,} }
$$
where $x^{\,\overline {\,y\,} } $ denotes the Rising Factorial.
Now, the Rising Factorial is defined (for $x$ and $y$ real and also complex) as
$$
h(x,y) = x^{\,\overline {\,y\,\,} } = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}} = \prod\nolimits_{\;k\, = \,\,0}^{\,y} {\left( {x + k} \right)}
$$
where the last term denotes the Indefinite Product, computed for $k$ ranging between the indicated bounds.
So we can write $f_a(x)$ as
$$ \bbox[lightyellow] {
f_{\,a} (x) = 2^{\,a\,x} {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}
= \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {2\left( {x + k} \right)} = 2^{\,a\,x} x^{\,a\,x} \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {\left( {1 + k/x} \right)}
} \tag{1}$$
Concerning the derivative of $r_a(x)$, since
$$
\left\{ \matrix{
{\partial \over {\partial x}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\left( {\psi \left( {x + y} \right)
- \psi \left( x \right)} \right) \hfill \cr
{\partial \over {\partial y}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\psi \left( {x + y} \right) \hfill \cr} \right.
$$
then, as you already found
$$
\eqalign{
& {d \over {dx}}r_a(x) = {\partial \over {\partial x}}h(x,y) + {\partial \over {\partial y}}h(x,y){d \over {dx}}y = \cr
& = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {\left( {a + 1} \right)\psi \left( {x + ax} \right)
- \psi \left( x \right)} \right) = \cr
& = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {a\psi \left( {x + ax} \right) + \left( {\psi \left( {x + ax} \right)
- \psi \left( x \right)} \right)} \right) \cr}
$$
where, for $0<x$ (and $0<a$) $\Gamma(x+ax)/\Gamma(x)$ is clearly positive.
However, while $\psi(x+ax)-\psi(x)$ is also positive since $\psi(x)$ is increasing in that range, $a\psi(a+ax)$ introduces a negative term for lower $x$.
To determine the limit of $r_a'(x)$ as $x \to 0^+$, let's consider the series development of
$$ \bbox[lightyellow] {
\left\{ \matrix{
\ln \Gamma (cx) = \ln \left( {{1 \over {cx}}} \right) - \gamma cx + O\left( {x^{\,2} } \right) \hfill \cr
\psi (cx) = - {1 \over {cx}} - \gamma + {{\pi ^{\,2} } \over 6}cx + O\left( {x^{\,2} } \right) = \hfill \cr
= - {1 \over {cx}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + cx}}} \right)} \hfill \cr} \right.
} \tag{2}$$
Therefore
$$
\eqalign{
& \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {d \over {dx}}r_a(x) = \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\mathop {\lim }\limits_{x\; \to \;0^{\, + } } \left( {\left( {a + 1} \right)\psi \left( {x + ax} \right) - \psi \left( x \right)} \right) = \cr
& = {1 \over {a + 1}}\left( { - \gamma a} \right) \cr}
$$
which is negative for $0<a$.
Concerning $f_a(x)$ instead
$$
\eqalign{
& {d \over {dx}}f_{\,a} (x) = \;2^{\,a\,x} a\ln 2r_{\,a} (x) + 2^{\,a\,x} {d \over {dx}}r_{\,a} (x) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \left( {r_{\,a} (x)} \right)} \right) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \Gamma (x + ax) - {d \over {dx}}\ln \Gamma (x)} \right) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x)} \right) \cr}
$$
(which is the equation you already found)
and
$$ \bbox[lightyellow] {
\eqalign{
& \mathop {\lim }\limits_{x\; \to \;0^{\, + } } f_{\,a} '(x) = 1\left( {\left( {a\ln 2} \right){1 \over {a + 1}} - {{\gamma a} \over {a + 1}}} \right) = \cr
& = {a \over {a + 1}}\left( {\ln 2 - \gamma } \right) = {a \over {a + 1}}0.1159 \cdots \cr}
} \tag{3}$$
Proceeding with the development of the derivative above
$$ \bbox[lightyellow] {
\eqalign{
& {d \over {dx}}f_{\,a} (x)\;\mathop /\limits_{} \;\left( {2^{\,a\,x} r_{\,a} (x)} \right) = \cr
& = a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x) = \cr
& = a\ln 2 + \left( {a + 1} \right)\left( { - {1 \over {\left( {a + 1} \right)x}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)} } \right) - \left( { - {1 \over x} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {1 \over {k + 1 + x}}} \right)} } \right) = \cr
& = \left( {a\ln 2 - {1 \over x} - \left( {a + 1} \right)\gamma + {1 \over x} + \gamma } \right)
+ \left( {a + 1} \right)\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)}
- \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + x}}} \right)} = \cr
& = a\left( {\ln 2 - \gamma } \right) + \sum\limits_{0\, \le \,k} {\left( {{a \over {k + 1}} + {1 \over {k + 1 + x}}
- {{\left( {a + 1} \right)} \over {k + 1 + \left( {a + 1} \right)x}}} \right)} = \cr
& = a\left( {\ln 2 - \gamma } \right) + a\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}} \right)} \cr}
} \tag{4}$$
and
$$ \bbox[lightyellow] {
\eqalign{
& 0 \le {1 \over {k + 1}} - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}} = \cr
& = {1 \over {k + 1}} - {1 \over {\left( {1 + x/\left( {k + 1} \right)} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}\quad \left| {\;0 \le x,k} \right. \cr}
} \tag{5}$$
therefore your thesis is demonstrated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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} |
Prove that the equation $ xyz=xy+xz+yz+x+y+z$ has finite numbers of natural solutions. Prove that the equation
$$\begin{equation}
xyz=xy+xz+yz+x+y+z\tag{*}
\end{equation}$$
has finitely many natural solutions.
My attempt. Suppose that $x \leq y \leq z$ then
$$
xyz\leq xz+xz+yz+3z \implies xy \leq 2x+y+3 \implies x \leq \frac{y+3}{y-2}
$$
Since $$x \leq \frac{y+3}{y-2}=\frac{y-2+5}{y-2}=1+\frac{5}{y-2} \implies (x-1)(y-2) \leq 5,$$
we can conclude that the last inequality has finite number of natural solutions, so there are only finite numbers for $x,y.$
For any fixed $x,y$ the equation (*) reduces to linear equation for $z$ with finite numbers of solutions.
Is it ok?
Is there any other solutions?
| Suppose
$$xyz = xy + xz + yz + x + y + z $$
is a natural solution with $$x \leq y \leq z$$ then
$z = \frac {xy + x + y}{xy - x - y - 1}= 1 + \frac {2x + 2y + 1}{xy - x-y -1}$
So it must be $$xy - x - y -1\le 2x + 2y + 1$$ or
$$xy \le 3x + 3y + 2$$ which actually means at least $x$ is smaller than 8.
Likewise we can conclude that $y$ is smaller than 8. Now the $$z = \frac {xy + x + y}{xy - x - y - 1}= 1 + \frac {2x + 2y + 1}{xy - x-y -1}$$ gives us finitely many possibilities for $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $ 2\arcsin\left(\sinh(x)\right) + \arccos\left( 2 - \cosh(2x) \right) = 0 $
$$ 2\arcsin\big( \sinh(x) \big) + \arccos\big( 2 - \cosh(2x) \big) = 0 $$
If this is true, then for what values of $x$? (Clearly, the $\arcsin(\cdot)$ and $\arccos(\cdot)$ are not valid for every real argument.)
| I suppose a typo; changing the $+$ to $-$ makes
$$2\arcsin\big( \sinh(x) \big) \color{red}{-} \arccos\big( 2 - \cosh(2x) \big) = 0$$ true for all $x \geq 0$.
Composing Taylor series built at $x=0$ and assuming $x >0$, we have
$$2\arcsin\big( \sinh(x) \big)=2 x+\frac{2 x^3}{3}+\frac{x^5}{3}+\frac{79 x^7}{315}+\frac{493
x^9}{2268}+O\left(x^{11}\right)$$
$$\arccos\big( 2 - \cosh(2x) \big)=2 x+\frac{2 x^3}{3}+\frac{x^5}{3}+\frac{79 x^7}{315}+\frac{493
x^9}{2268}+O\left(x^{11}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Check if $f(x,y)=\frac{1}{\cos(x)\cos(y)}$ is integrable on $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)^2$ I started like this:
$$f(x,y)=\frac{1}{\cos(x)\cos(y)}$$
Notice $f(-x,-y)=f(x,y)$ so and $f(x,y)$ is continuous on the square $A=\left(\frac{-\pi}{2},\frac{\pi}{2}\right)^2$ so if $f$ is integrable then $\int_A df =4\int_0^{\pi/2}\int_0^{\pi/2} f(x,y) \,dx\,dy$.
I wanted to bound $f$ from below because I have a feeling it is divergent so $\cos(x)\leq \frac{4x}{\pi}+2$ so $$f(x,y)\geq \frac{1}{\left(-\frac{4x}{\pi}+2\right)\left(-\frac{4y}{\pi}+2\right)}=\frac{1}{\frac{16}{\pi^2}xy-\frac{8}{\pi}(x+y)+4}\geq\frac{1}{xy+4}$$ and here I am stuck because this bound it didnt give me too much. So I need different one that goes to zero as $x$ approaches $\pi/2$ and it cannot be linear in $x$ because cos is convex.
| We have
\begin{align}
\int_0^{\pi/2} \frac{dx}{\cos x} &= \int_0^{\pi/2} \frac{\cos x}{\cos^2 x}\,dx \\
&= \int_0^{\pi/2} \frac{\cos x}{1-\sin^2 x}\,dx \\
&= \begin{bmatrix} t = \sin x \\ dt = \cos x\,dx\end{bmatrix}\\
&= \int_0^{1} \frac{dt}{1-t^2}\\
&= \frac12\int_0^1 \frac{dt}{1-t} + \frac12\int_0^1 \frac{dt}{1+t}\\
&= \left[-\frac12 \ln(1-t) + \frac12 \ln(1+t)\right]_0^1\\
&= \frac12 \ln \left(\frac{1+t}{1-t}\right)\Bigg|_0^1\\
&= +\infty
\end{align}
so $$\int_{\left(\frac{-\pi}{2},\frac{\pi}{2}\right)^2} \frac{dx}{\cos x\cos y} = 4\int_0^{\pi/2} \left(\int_0^{\pi/2} \frac{dx}{\cos x}\right)\frac{dy}{\cos y} = +\infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\cos(\frac{2\pi}{5})$ in terms of $\sin(\frac{\pi}{5})$ I intend to calculate $\cos\left(\frac{2\pi}{5}\right)$ via the formula
$$\sin\left(\frac{\pi}{5}\right)=\sqrt{\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}}$$
This could be expressed as:
$\sqrt{\frac{5-\sqrt{5}}{8}}=\sqrt{\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}}$. Squaring both sides and manipulate a bit, this will give $\cos\left(\frac{2\pi}{5}\right)=-\frac{1-\sqrt{5}}{4}$. However, the true result is $\frac{\sqrt{5}-1}{4}$. What is wrong with my derivation?
| You shouldn't have to square or manipulate whatever, if you used the relevant one of the duplication formulæ:
$$\cos 2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ I am just wondering how to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ (This one is taken from Wolfram BTW)
Clearly, $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ can be rewritten as $\frac{1}{2(\sqrt[4]{\frac{3}{4}}-\sqrt[4]{\frac{1}{4}})}$, but what should I do next?
I am a little inexperienced in manipulating surds, so some helps are very much appreciated
Thank you!
| Start without the irrelevant $1/2$ and note that $\sqrt[4]{4} = \sqrt{2}$, so: $$\frac{1}{\sqrt[4]{\frac{3}{4}}-\sqrt[4]{\frac{1}{4}}} = \frac{\sqrt{2}}{\sqrt[4]{3} -1}.$$ Now try to rationalize the denominator in the standard fashion (using $(x-y)\cdot(x+y) = x^2 - y^2$) by multiplying with $\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1}$ to get
$$ \frac{\sqrt{2}}{\sqrt[4]{3} -1}\cdot\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1} = \frac{\sqrt{2}(\sqrt[4]{3}+1)}{\sqrt{3} -1}.$$
That gets us half-way to a rationalized denominator, so finish the job by multiplying with $\frac{\sqrt{3}+1}{\sqrt{3}+1}$:
$$\frac{\sqrt{2}(\sqrt[4]{3}+1)}{\sqrt{3} -1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}= \frac{\sqrt{2}(\sqrt{3}+1)(\sqrt[4]{3}+1)}{2}.$$
Now that we have a rational denominator, the basic idea is to rewrite the numerator as the square root of the square of its previous value, expand the squares, combine terms and see where we are:
$$\frac{\sqrt{2}(\sqrt{3}+1)(\sqrt[4]{3}+1)}{2} = \sqrt{\frac{2}{4}(\sqrt{3}+1)^2 (\sqrt[4]{3}+1)^2} = \sqrt{\frac{2}{4}(3+2\sqrt{3}+1)(\sqrt{3} +2\sqrt[4]{3}+1)} = \sqrt{\frac{2}{4}(10+6\sqrt{3} +4(2+\sqrt{3})\sqrt[4]{3})} = \sqrt{5 + 3\sqrt{3} + 2(2+\sqrt{3})\sqrt[4]{3}}.$$
You can see that already we have part of the required answer. We use the same trick of taking the square root of the square (thereby leaving everything the same) one more time on the last term to get rid of the pesky fourth root. We then expand the square and collect terms as usual:
$$\sqrt{5 + 3\sqrt{3} + 2(2+\sqrt{3})\sqrt[4]{3}} = \sqrt{5 + 3\sqrt{3} + 2\sqrt{(2+\sqrt{3})^2\sqrt{3}}}= \sqrt{5 + 3\sqrt{3} + 2\sqrt{(4+4\sqrt{3} +3)\sqrt{3}}}= \sqrt{5 + 3\sqrt{3} + 2\sqrt{4\sqrt{3} +12 +3\sqrt{3}}} = \sqrt{5 + 3\sqrt{3} + 2\sqrt{12 +7\sqrt{3}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of the inequality of sums N.B: before downvoting this question without having finished reading it please leave a comment explaining why, so I can improve and ask better question in future, thanks.
In a problem I've to proof that
$$\sum_{x_i\in A}{x_i}=1$$
After some effort
I found the follow identity
$$\sum_{x_i\in A}{x_i}=\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}$$
Whit
$a\ge1, b\ge1, 1\le n\le a+b$
I'd like to bound this sum between itself calculated for $n=1$ and $n=a+b$ in this way:
$$\sum_{k=max(0,1-b)}^{min(a,1)}{\frac{\binom{a}{k}\binom{b}{1-k}}{\binom{b+a}{1}}}\le\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}\le\sum_{k=max(0,a+b-b)}^{min(a,a+b)}{\frac{\binom{a}{k}\binom{b}{a+b-k}}{\binom{b+a}{a+b}}}$$
If this inequality is true then
$$1\le\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}\le1$$
So
$$\sum_{k=max(0,n-b)}^{min(a,n)}{\frac{\binom{a}{k}\binom{b}{n-k}}{\binom{b+a}{n}}}=1$$
Because
$$\sum_{k=max(0,1-b)}^{min(a,1)}{\frac{\binom{a}{k}\binom{b}{1-k}}{\binom{b+a}{1}}}=\sum_{k=0}^{1}{\frac{\binom{a}{k}\binom{b}{1-k}}{\binom{b+a}{1}}}={\frac{\binom{a}{0}\binom{b}{1}}{\binom{b+a}{1}}}+{\frac{\binom{a}{1}\binom{b}{0}}{\binom{b+a}{1}}}=\frac{a+b}{b+a}=1$$
and
$$\sum_{k=max(0,a+b-b)}^{min(a,a+b)}{\frac{\binom{a}{k}\binom{b}{a+b-k}}{\binom{b+a}{a+b}}}=\sum_{k=a}^{a}{\frac{\binom{a}{k}\binom{b}{a+b-k}}{\binom{b+a}{a+b}}}=\binom{a}{a}\binom{b}{a+b-a}=1$$
The problem is that i can't check this inequality my myself, can anyone explain me how to do it?
| All you need is a double counting argument. If we select $n$ objects from a set of $a+b$ objects there are $\binom{a+b}{n}$ ways to do this. On the other hand, if we pick $n$ objects from $a+b$ objects by selecting some $k$ objects from $a$ and the remainder from $b$, then if we sum over all possible values of $k$, we count the number of ways of selecting $n$ objects from $a+b$ objects. Thus
$$
\binom{a+b}{n}=\sum_{k=\max(0,n-b)}^{\min(a,n)} \binom{a}{k}\binom{b}{n-k}.
$$
Now divide both sides by $\binom{a+b}{n}$ and use the identity you have already proved to get your result.
| {
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"url": "https://math.stackexchange.com/questions/2838621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Shortest distance from ellipse to line What is the shortest distance between the ellipse
$$\frac{x^2}{4}+y^2=1$$
and the line $y=\frac{-\sqrt{3}}{2}x+8?$
I tried solving it by using a line that is tangent to the ellipse and parallel to the line given: $y=\frac{-\sqrt{3}}{2}x+b$
Then plugging the y back into the ellipse equation:
$x^2-b\sqrt{3}x+(b^2-1)=0$
Which means the discriminant would have to equal 0 for the line to be tangent, right?
$3b^2-4b^2+4=0$
$b=2$
But when I plug the y-intercept, (0,2), into a point to line distance formula I get
$$\frac{|0*\sqrt{3}+2*2-16|}{\sqrt{3+4}}$$
Which is $\frac{12 \sqrt{7}}{7}$, but the answer key says it's $\frac{3\sqrt{13}}{2}$
What am I doing wrong? I don't mind learning any other methods, but I haven't learned any calculus yet, so I would like to do the problem without it. Thank you.
| I think there is a typo in the answer.
Note that the shortest distance is always perpendicular distance.
Let $P$ be a point parallel to the given line $y=\frac{-\sqrt{3}}{2}x+8$.
Then the slope at $P$ is
$$\frac{dy}{dx}=-\frac{x}{4y}\mbox{------(1)}$$
Since the given line and tangent line are parallel, the slopes are equal.
$$\mbox{Slope of the given line is } \sqrt{3}x+2y=16 \mbox{ is }\frac{-\sqrt{3}}{2}$$
$$-\frac{x}{4y}=\frac{-\sqrt{3}}{2}$$
$$x=\sqrt{3}y$$
Now substitute the $x=\sqrt{3}y$ in the ellipse $x^2+4y^2=4$ and we get
$$12y^2+4y^2=4$$
$$y=\pm\frac12$$
Clearly $\left(\sqrt{3},-\dfrac12\right)$ is closer to the line.
Distance$=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=\dfrac{12\sqrt{7}}{7}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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How to find solution to this ODE with powerseries I am working on solving this ODE
$$ (1+x^2)y'' + xy'-y =0 $$
with $1. y(0)=0,y'(0)=1 \\
2. y(0)=1, y'(0)=1$
with $$ y= \sum_{n=0}^{ \infty} a_n x^n , y'= \sum_{n=1}^{ \infty}na_n x^{n-1}, y''=\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}$$
plugged into the equation follows:
$(1+x^2) \sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}+x \sum_{n=1}^{ \infty}na_n x^{n-1}-\sum_{n=0}^{ \infty} a_n x^n =0 $
$ \leftrightarrow \sum_{n=2}^{ \infty} n(n-1)a_n x^{n}+\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}+ \sum_{n=1}^{ \infty} na_n x^{n}-\sum_{n=0}^{ \infty} a_n x^n =0$
$ \leftrightarrow 2a_2-a_0 + \sum_{n=1}^{ \infty} (n(n-1)a_n + (n+2)(n+1)a_{n+2}+ na_n - a_n) x^n = 0 $
you get that
$ a_{n+2}= \frac{a_n(1-n^2)}{(n+2)(n+1)} $
let now be $ a_0=1 $, $a_1=0 $ so that $a_2= \frac12, $( $2a_2 - a_0 = 0 \leftrightarrow a_2= \frac12 )$
for uneven $n$ such as $a_3, a_5,..$ it equals zero.
but for even $n$ i can not find any regularity:
$ n=2 , a_4 = \frac{a_2 (-3)}{12}= -\frac{1}{8}$
$ n=4 , a_6= \frac{a_4 (-15)}{30}= \frac{1}{16} $
$ n=6 , a_8= \frac{a_6 (-35)}{56}= - \frac{5}{128} $
i am totally stuck and do not know how to proceed :(
I appreciate any help of you guys !
| Note that by binomial theorem $n^2-1=(n-1)(n+1)$ so that the recursion cancels to
$$
a_{n+2}=-\frac{n-1}{n+2}a_n=\frac{(n-1)(n-3)}{(n+2)n}a_{n-2}=...
$$
For the even subsequence one can write
$$
a_{2(k+1)}=-\frac{2k(2k-1)}{4(k+1)k}a_{2k}=(-1)^{k}\frac{(2k)(2k-1)\cdots2\cdot1}{4^{k}(k+1)k^2(k-1)^2\cdots2^2\cdot1}a_2
\\=(-1)^k\frac{(2k)!}{2^{2k+1}(k+1)!k!}a_0
=\frac{(-1)^k}{2^{2k+1}(k+1)}\binom{2k}{k}a_0
$$
etc.
| {
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"timestamp": "2023-03-29T00:00:00",
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integral $\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$ Evaluate $$\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$$
My Attempt
$$I_{1}=\int_{0}^{1}\left\lfloor{\frac{2}{x}} \right\rfloor dx$$
Put $x=2t$
$$I_{1}=\int_{0}^{\frac{1}{2}}2\left\lfloor{\frac{1}{t}} \right\rfloor dt=\int_{0}^{\frac{1}{2}}2\left\lfloor{\frac{1}{x}}\right\rfloor dx$$
let $$I_{2}=\int_{0}^{1} \left\lfloor{\frac{1}{x}} \right\rfloor dx=\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx+\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx$$
Given integral
$$
\begin{align}
\int_{0}^{1}\left(\left\lfloor{\frac{2}{x}}\right\rfloor-2\left\lfloor{\frac{1}{x}}\right\rfloor \right)dx&=I_{1}-2I_{2}\\
&=2\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx-2\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx-2\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx\\
&=-2\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx\\
&=(-2)(1)=-2.
\end{align}
$$
But answer given is $\ln(\frac{4}{e})$
What mistake am I making?
| The function $\lfloor \frac{2}{x} \rfloor$ is unbounded near $x = 0$ so if you try to split your integral, at best you can hope to interpret $I_1 = \int_0^1 \lfloor \frac{2}{x} \rfloor \, dx$ as an improper integral. Unfortunately, this integral actually diverges (it behaves like $\int_0^1 \frac{1}{x} \, dx$) and so is $I_2$ (this also follows from the calculations below) so you can't do all the manipulations as you are subtracting $+\infty$ from $+\infty$...
Let's see how the integrand (which I will denote by $f(x)$) behaves. Divide the interval $\left( 0, 1 \right]$ into intervals
$$\cdots, \left( \frac{2}{k+1}, \frac{2}{k} \right],\cdots,\left( \frac{2}{4}, \frac{2}{3} \right], \left( \frac{2}{3}, 1 \right]. $$
*
*If $x \in \left( \frac{2}{k+1}, \frac{2}{k} \right]$ then $\frac{2}{x} \in [k,k+1)$ so $\lfloor \frac{2}{x} \rfloor = k$.
*If $x \in \left( \frac{2}{k+1}, \frac{2}{k} \right]$ then $\frac{1}{x} \in [\frac{k}{2},\frac{k+1}{2})$. Now, if $k$ is even then $2\lfloor \frac{1}{x} \rfloor = 2 \frac{k}{2} = k$ while if $k$ is odd then $2 \lfloor \frac{1}{x} \rfloor = 2\frac{k-1}{2} = k - 1$.
*Hence, we have for all $x \in (0,1]$
$$ f(x) = \lfloor \frac{2}{x} \rfloor - 2 \lfloor \frac{1}{x} \rfloor = \begin{cases} 1 & x \in \left( \frac{1}{l+1}, \frac{2}{2l + 1} \right], l \geq 1, \\
0 & \textrm{otherwise}. \end{cases} $$
Hence,
$$ \int_0^1 f(x) \, dx = \lim_{n \to \infty} \int_{\frac{2}{2n+1}}^1 f(x) \, dx = \lim_{n \to \infty} \left( \sum_{l=1}^n \left( \frac{2}{2l+1} - \frac{1}{l+1} \right) \right) \\
= 2 \cdot \sum_{l=1}^{\infty} \left( \frac{1}{2l+1} - \frac{1}{2l+2} \right) = 2 \sum_{n=3}^{\infty} \frac{(-1)^{n+1}}{n} = 2 \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} - \left( 1 - \frac{1}{2} \right) \right) = 2 \left( \ln(2) - \frac{1}{2} \right) = 2 \ln(2) - 1 = \ln(4) - \ln(e) = \ln \left( \frac{4}{e} \right) $$
where I used the Taylor series of $\ln(1 + x)$ which converges at $x = 2$ to $\ln(2)$ to evaluate the infinite sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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The set of integers $n$ expressible as $n=x^2+xy+y^2$
Let $S$ be the set of integers $n$, such there exist integers $x,y$ with $$n=x^2+xy+y^2$$
Is the implication $$a,b\in S\implies ab\in S$$ true? If yes, how can I prove it?
I worked out $$n\in S\iff 4n\in S$$ and $$n\in S\iff 3n\in S$$
I tried two approaches. The first is to express $$(a^2+ab+b^2)(c^2+cd+d^2)$$ in the form $$f^2+fg+g^2$$ with polynomials $f,g$ with integer coefficients. I however could not find suitable $f$ and $g$.
The second approach is based on $$x^2+xy+y^2=\frac{(2x+y)^2+3y^2}{4}$$ If we have $n=x^2+xy+y^2$ , we have $u^2+3v^2=4n$ for some integers $u,v$ with equal parity. The main problem of this approach is to consider the equal parity.
Any ideas ?
| Note that
\begin{align*}
&\qquad16(a^2+ab+b^2)(c^2+cd+d^2)\\&=\left((2a+b)^2+3b^2\right)\left((2c+d)^2+3d^2\right)\\&=\left(\frac{(2a+b)(2c+d)+3bd}{2}\right)^2+3\left(\frac{d(2a+b)-b(2c+d)}{2}\right)^2
\end{align*}
| {
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Find a close formula using generating functions I want to find the sum:
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}$$
I did it by using the binomial theorem and I got
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
= 2^{-n} \sum \limits_{k=0}^n {n \choose k}2^k
= \left(\frac{1}{2}\right)^n3^n$$
I would like to know how to do it using generating functions. I have tried some calculations but I am stuck. This is what I have so far.
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
=\sum \limits_{n\ge0} \sum \limits_{k\ge0} {n \choose k} 2^{k-n} x^n
=\sum \limits_{k\ge0} 2^k \sum \limits_{n\ge0} {n \choose k}\left(\frac{x}{2}\right)^n$$
| The binomial coefficient in $\sum \limits_{k=0}^n \binom{n}{k}2^{k-n}$ indicates that we could use the Cauchy-product of exponential generating functions. We have
\begin{align*}
A(x)=\sum_{k=0}^\infty a_k\frac{x^k}{k!},\ \ B(x)=\sum_{l=0}^\infty b_l\frac{x^l}{l!}\quad\longleftrightarrow\quad
A(x)B(x)=\sum_{n=0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_kb_{n-k}\right)\frac{x^n}{n!}
\end{align*}
We obtain
\begin{align*}
\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}2^{k-n}}\right)\frac{x^n}{n!}
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}1^k\left(\frac{1}{2}\right)^{n-k}\right)\frac{x^n}{n!}\\
&=\left(\sum_{k=0}^\infty\frac{x^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{\left(\frac{x}{2}\right)^l}{l!}\right)\\
&=e^xe^{\frac{x}{2}}\\
&=e^{\frac{3}{2}x}\\
&=\sum_{n=0}^\infty\color{blue}{\left(\frac{3}{2}\right)^n}\frac{x^n}{n!}
\end{align*}
and the claim follows.
Besides the more common method above we also have the possibility to use ordinary generating functions. We can apply a transformation known as Euler transform of a series
\begin{align*}
A(x)=\sum_{n= 0}^\infty a_nx^n\quad\longrightarrow\quad
\frac{1}{1-x}A\left(\frac{x}{1-x}\right)=\sum_{n= 0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_k\right)x^n
\end{align*}
This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev (see Lemma 1).
We obtain
\begin{align*}
\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}2^{k-n}}\right)x^n
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}2^k\right)\left(\frac{x}{2}\right)^n\\
&=\frac{1}{1-\frac{x}{2}}\sum_{n=0}^\infty2^n\left(\frac{\frac{x}{2}}{1-\frac{x}{2}}\right)^n\\
&=\frac{1}{1-\frac{x}{2}}\cdot\frac{1}{1-2\left(\frac{\frac{x}{2}}{1-\frac{x}{2}}\right)}\\
&=\frac{1}{1-\frac{3}{2}x}\\
&=\sum_{n=0}^\infty\color{blue}{\left(\frac{3}{2}\right)^n}x^n
\end{align*}
and the claim follows.
| {
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Are $1+p^3+p^6$ and $1+p^4+p^8$ coprime? What are the primes $p$ for which $1+p^3+p^6$ and $1+p^4+p^8$ are coprime?
I know it is true for $p=2$ and $p=3$ and not true for any $p \equiv 1 \mod 6$. I conjecture that it true for all primes $p \equiv 5 \mod 6$.
Any counterexample $> 10^8$.
This is relevant to OEIS sequence A046685.
| This answer is no more than an expansion of the very first comment by Daniel Schepler (since hed did not write it as an answer himself).
Since we are interested in the "pointwise" GCD (greatest common divisor) of $f=x^6+x^3+1$ and $g=x^8+x^4+1$ for particular whole numbers $x$, it is a good idea to start with the GCD within the graded ring $\mathbb{Z}[x]$ of these two polynomials. If we even use the extended Euclidean algorithm, we get a Bézout identity:
$$(x^6 - x^5 - 2x^3 - x + 1)(x^6+x^3+1) + (-x^4 + x^3 + x + 2)(x^8+x^4+1)=3$$
See machine-generated answer by Will Jaggy/Community wiki for some details. Can also be found by PARI/GP with gcdext(x^6+x^3+1,x^8+x^4+1) (and multiplying the output by 3 to get rid of fractions).
This Bézout-type identity shows that for any $x\in\mathbb{N}$, the GCD of $x^6+x^3+1$ and $x^8+x^4+1$ is also a (positive) divisor of $3$. Therefore the GCD will be one or three.
Let $x\in\mathbb{N}$ be given. Let us consider all cases for $x$ modulo three. If $x\equiv +1 \pmod 3$, both $x^6+x^3+1$ and $x^8+x^4+1$ are clearly zero modulo three, which means that $\gcd(x^6+x^3+1, x^8+x^4+1)$ is really $3$ in that case. If $x\equiv 0 \pmod 3$ or $x\equiv -1 \pmod 3$, then $x^6+x^3+1 \equiv 0+1 \not\equiv 0 \pmod 3$, so three does not divide $x^6+x^3+1$, so the GCD has to be one in those cases, which was what you wanted to prove.
Nowhere did we use the property that $x$ is a prime number.
| {
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How would you prove that $\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$? How would you prove that $\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$?
Regards!
| After a little simplification,
$$t+2\frac{2t}{1-t^2}+4\frac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}+8\dfrac{1-\left(\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}\right)^2}{2\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}}
=\frac1t.$$
| {
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Computing a residue To find the value of $\int_0^\pi \frac{\sin^2 \theta}{a + \cos \theta} \, d\theta$ with $a > 1$ using residue theory, I am trying to compute the residue at the simple pole $z=-a+\sqrt{a^2-1}$ inside the unit disc. I tried both ways:
\begin{align}
\operatorname{Res}\left(\frac{z^4-2z^2+1}{z^4+2az^3+z^2},-a+\sqrt{a^2-1}\right) &= \frac{z^4-2z^2+1}{\frac d{dz}(z^4+2az^3+z^2)}\Bigg\vert_{z=-a+\sqrt{a^2-1}} \\
&= \frac{z^4-2z^2+1}{4z^3+6az+2z}\Bigg\vert_{z=-a+\sqrt{a^2-1}}
\end{align}
and
\begin{align}
\operatorname{Res}\left(\frac{z^4-2z^2+1}{z^4+2az^3+z^2},-a+\sqrt{a^2-1}\right) &= \lim_{z \to -a+\sqrt{a^2-1}} (z-(-a+\sqrt{a^2-1}))\frac{z^4-2z^2+1}{z^4+2az^3+z^2} \\
&= \lim_{z \to -a+\sqrt{a^2-1}} \frac{z^4-2z^2+1}{z^2(z-(-a-\sqrt{a^2-1}))}.
\end{align}
Is there a clean way to finding this particular residue? Because the methods I tried above require chugging through long arithmetic calculations.
| Note that we can simplify your analysis by writing
$$\begin{align}
\frac{\sin^2(\theta)}{a+\cos(\theta)}=\frac{1-a^2}{a+\cos(\theta)}+a-\cos(\theta)
\end{align}$$
The integral of the term $a-\cos(\theta)$ is trivial. If we wish to use contour integration for the term $\displaystyle \frac{1-a^2}{a+\cos(\theta)}$, we can proceed to write for $a>1$
$$\begin{align}
\int_0^\pi \frac{1-a^2}{a+\cos(\theta)}\,d\theta&=i(a^2-1) \oint_{|z|=1} \frac{1}{z^2+2az+1}\,dz\\\\
&=2\pi (1-a^2)\text{Res}\left(\frac{1}{z^2+2az+1}, z=-a+\sqrt{a^2-1}\right)\\\\
&=-\pi\sqrt{a^2-1}
\end{align}$$
Hence, we find that
$$\int_0^\pi \frac{\sin^2(\theta)}{a+\cos(\theta)}\,d\theta=\pi(a-\sqrt{a^2-1})$$
| {
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Range of $a$ in Trigonometric equation
If the inequality $\sin^2 x+a\cos x+a^2>1+\cos x$ hold for any $x\in \mathbb{R}.$ Then range of $a$ is
Try: $1-\cos^2 x+a\cos x+a^2-1-\cos x>0$
$$\cos^2 x+(1-a)\cos x-a^2<0$$
$$4\cos^2 x+4(1-a)\cos x-4a^2<0$$
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
Could some help me how to solve further, Thanks
| Your approach is fine.
You already got
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
which is equivalent to
$$\bigg(2\cos x+(1-a)\bigg)^2\lt 5a^2-2a+1\tag1$$
Since $5a^2-2a+1$ is always positive, $(1)$ is equivalent to
$$-\sqrt{5a^2-2a+1}\lt 2\cos x+1-a\lt \sqrt{5a^2-2a+1}$$
which is equivalent to
$$\frac 12\left(a-1-\sqrt{5a^2-2a+1}\right)\lt \cos x\lt\frac 12\left(a-1+\sqrt{5a^2-2a+1}\right)\tag2$$
The necessary and sufficient condition that $(2)$ holds for any $x\in\mathbb R$ is
$$\frac 12\left(a-1-\sqrt{5a^2-2a+1}\right)\lt -1\quad\text{and}\quad 1\lt \frac 12\left(a-1+\sqrt{5a^2-2a+1}\right),$$
i.e.
$$\sqrt{5a^2-2a+1}\gt a+1\quad \text{and}\quad \sqrt{5a^2-2a+1}\gt -a+3\tag3$$
If $a\lt -1$, then
$$(3)\iff 5a^2-2a+1\gt (-a+3)^2\iff a\in (-\infty, -2)\cup (1,\infty)$$
So, in this case, we have $a\in (-\infty,-2)$.
If $-1\le a\lt 3$, then
$$\begin{align}(3)&\iff 5a^2-2a+1\gt (a+1)^2\quad\text{and}\quad 5a^2-2a+1\gt (-a+3)^2\\\\&
\iff a\in (-\infty, -2)\cup (1,\infty)\end{align}$$
So, in this case, we have $a\in (1,3)$.
If $a\ge 3$, then
$$(3)\iff 5a^2-2a+1\gt (a+1)^2\iff a\in (-\infty,0)\cup (1,\infty)$$
So, in this case, we have $a\in [3,\infty)$.
Therefore, the range of $a$ is
$$a\in (-\infty,-2)\cup (1,3)\cup [3,\infty),$$
i.e.
$$a\in (-\infty, -2)\cup (1,\infty)$$
Another approach
First, let us prove the following lemma :
Lemma 1 :
If the inequality holds for any $x\in\mathbb R$, then $a$ has to, at least, satisfy $a\lt -2$ or $a\gt 1$.
Proof :
The inequality
$$\sin^2 x+a\cos x+a^2>1+\cos x$$
holds for $x=0$, so we have to have
$$a+a^2\gt 2,$$
i.e.
$$a\lt -2\quad\text{or}\quad a\gt 1\qquad\square$$
Next, let us prove the following lemma :
Lemma 2 :
For any $a$ satisfying $a\lt -2$ or $a\gt 1$, the inequality holds for any $x\in\mathbb R$.
Proof :
The inequality can be written as
$$t^2+(1-a)t-a^2\lt 0\tag4$$
where $t=\cos x$.
So, we want to prove that
$$a\in (-\infty, -2)\cup (1,\infty)\implies \text{$(4)$ holds for any $t$ satisfying $|t|\le 1$}\tag5$$
Let $f(t)$ be the LHS of $(4)$, and let us consider the graph of $y=f(t)$.
If $a\in (-\infty, -2)\cup (1,\infty)$, then we have
$$f(-1)=-a^2+a\lt 0\qquad \text{and}\qquad f(1)=-a^2-a+2\lt 0$$
Since the graph of $y=f(t)$ is an upward parabola, it follows that if $a\in (-\infty, -2)\cup (1,\infty)$, then $f(t)\lt 0$ for any $t$ satisfying $|t|\le 1.\quad\square$
If follows from Lemma 1 and Lemma 2 that the range of $a$ is
$$a\in (-\infty, -2)\cup (1,\infty)$$
| {
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What is the probability that exactly one box remains empty Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $\frac37$ (mcq)
Can you tell me how should I proceed then?
| Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$\begin{align}
S_1 &= \binom{4}{1} \left( \frac{3}{4} \right)^5 \\
S_2 &= \binom{4}{2} \left( \frac{2}{4} \right)^5 \\
S_3 &= \binom{4}{3} \left( \frac{1}{4} \right)^5 \\
S_4 &= 0
\end{align}$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_{[m]} = S_m - \binom{m+1}{m} S_{m+1} + \binom{m+2}{m} S_{m+2} - \dots \pm \binom{N}{m} S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_{[1]} = S_1 - \binom{2}{1} S_2 + \binom{3}{1} S_3 - \binom{4}{1} S_4= \boxed{0.585938}$$
| {
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Number of elements in a multiplicative group
Given $n$ the square of a prime odd number. How many elements $\bar x
∈$ $Z^*_n$ with $\bar x ^2$ = $\bar 1$?
Given $n$ the product of two distinct prime numbers. How many elements
$\bar x ∈$ $Z^*_n$ with $\bar x ^2$ = $\bar 1$?
Not knowing the value of $n$, how can I find out that cardinality? Is there any theorem that I should follow to induce $n$?
| For the first question: Let $n = p^2$. We're interested in $x$ such that $x^1 \equiv 1 \pmod {p^2}$. That is $p^2 | x^2 - 1$. We see that this is then equivalent to $p^2 | (x - 1)(x + 1)$. We have three cases:
$p|(x - 1)$ and $p|(x + 1)$:
If $p|(x - 1)$ and $p|(x + 1)$ then $p|(x + 1 - (x -1 )) = 2$. Since $p$ is an odd prime, this is not possible, so this case cannot happen.
$p^2|(x - 1)$:
In this case $x \equiv 1 \pmod p^2$.
$p^2|(x + 1)$:
In this case $x \equiv p^2-1 \pmod p^2$.
Thus if $n = p^2$ for an odd prime there are exactly two solutions to $x^2 \equiv 1 \pmod n$, namely $1$ and $p^2 - 1$.
Next suppose $n = pq$ for distinct primes $p$ and $q$. Again we can arrive at the equation $pq|(x + 1)(x - 1)$. We then have four cases:
$p|(x + 1)$, $q|(x - 1)$:
In this case we have $x + 1 = kp$ and $x - 1 = mq$. That is $x \equiv 1 \pmod q$ and $x \equiv -1 \pmod p$. Note that by the Chinese Remainder Theorem, this has a unique solution modulo $pq$.
$p|(x - 1)$, $q|(x + 1)$:
As above this congruence has a unique solution modulo $pq$ by the Chinese Remainder Theorem.
$pq|(x + 1)$:
In this case $x \equiv pq - 1 \pmod{pq}$
$pq|(x - 1)$:
In this case $x \equiv 1 \pmod{pq}$
Next note that one $x$ cannot satisfy any two of the cases above. This is because if such an $x$ did exist then $p|(x + 1)$, $p|(x - 1)$ and $q|(x + 1)$ and $q|(x - 1)$. As in part one, this would for $p = q = 2$, which is a contradiction. Thus there are exactly 4 solutions to $x^2 \equiv 1 \pmod n$ if $n$ is the product of two distinct primes.
| {
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Evaluating $\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx=\frac{4}{3}\pi\;$ with residue? Wolfam Alpha says that $$\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx=\frac{4}{3}\pi$$
I want to calculate this on my own, but have problems.Would be great, if you could tell me where and what it is.
$$\begin{align}\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx & =\operatorname{Re} \Biggr(\int_0^{2\pi}\frac{4e^{ix}}{5-4e^{ix}}dx \Biggl) \\
& = \operatorname{Re} \Biggr(\int_0^{2\pi}\frac{4z}{(5-4z)iz}dz \Biggl)\\
& = \operatorname{Re} \Biggr(-i\int_0^{2\pi}\frac{4}{(5-4z)}dz \Biggl) \\
& = \operatorname{Re} \Biggr(-i\Bigg[\ln(5-4z)\Bigg]_0^{2\pi}\Bigg) \\
& = \operatorname{Re} \Biggr(-i\ln(\frac{5-8\pi}{5})\Bigg)\\
\end{align}$$
I guess that I can't calculate the integral in this way, right ?
I guess that I can't calculate the integral in this way (4), right ?
| This is not the right way of calculating.$$\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx=\int_{|z|=1}\frac{2z+2z^{-1}}{5-2z-2z^{-1}}\dfrac{dz}{iz}=\int_{|z|=1}\frac{2z^2+2}{5z-2z^2-2}\dfrac{dz}{iz}$$The singularities are in $2,\dfrac{1}{2},0$ where $$\operatorname{Re}z_0=-1\\\operatorname{Re}z_{\dfrac{1}{2}}=\dfrac{5}{3}$$therefore $$I=\dfrac{1}{i}2\pi i\left(-1+\dfrac{5}{3}\right)=\pi\dfrac{4}{3}$$
| {
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What is $\int\frac{x^4}{1+ e^x} dx$? Here's the integral I have,
$$ \displaystyle\int\dfrac{x^4}{1+ e^x} dx $$
I tried the usual methods I know, but I failed miserably.
How would you all approach this problem?
| $\int\frac {x^4}{1+e^x} \ dx\\
\int\frac {x^4e^{-x}}{1+e^{-x}} \ dx$
Converting to a geometric series: $\frac {y}{1+y} = \sum_\limits{n=1}^\infty (-y)^n$
$\int x^4\sum_\limits{n=1}^\infty (-1)^ne^{-nx} \ dx$
considering just one term $\int x^4e^{-nx} = (\frac {x^4}{n} + \frac {4x^3}{n^2} + \frac {12x^2}{n^2} + \frac {24x}{n^3} + \frac {24}{n^4}) e^{-nx}$
$\sum_\limits{n=1}^{\infty}(-1)^n(\frac {x^4}{n} + \frac {4x^3}{n^2} + \frac {12x^2}{n^2} + \frac {24x}{n^3} + \frac {24}{n^4}) e^{-nx}$
It is worth noting
$\frac{1}{\Gamma(s)} \int_0^{\infty} \frac {x^{s-1}}{e^x -1} \ dx = \zeta(s)$
| {
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Sums of consecutive powers of 3 being perfect squares I was recently considering the puzzle of finding consecutive (integer) powers of 3 that sum to a square. It's not hard to show that this can be reduced to finding values of $n$ such that
$S_3(n) = \displaystyle\sum_{i=0}^n 3^i\tag*{}$
is either a square or one more than a square. It's not hard to find three such values: $S_3(0) = 1$, $S_3(2) = 4$ and $S_3(5) = 121$. However, I've verified numerically that for no other values of $n$ up to $10,000$ is $S_3(n)$ either a square or one more than a square.
How would one go about either proving that these are the only such values for $n$ or finding other values (other than brute force search, assuming there are more values to find)?
The question could obviously be generalized to bases other than 3. When is
$S_b(n) = \displaystyle\sum_{i=0}^nb^i\tag*{}$
either a square or one more than a square. Here are the results of my experimenting for low values of $b$ and $0<n\le10,000$.
$\begin{array}{c|l}b&n\\\hline2\\3&3^0+3^1 = 2^2\\&3^0+3^1+3^2+3^3+3^4 = 11^2\\4&4^0+4^1 = 2^2+1\\5\\6\\7&7^0+7^1+7^2+7^3 = 20^2\\8&8^0+8^1 = 3^2\\9&9^0+9^1 = 3^2+1\end{array}\tag*{}$
| We have
$$
\sum_{i=0}^n 3^i = \frac{3^{n+1}-1}{2}
$$
Therefore, if $\sum_{i=0}^n 3^i=a^2$, then we have
$$
3^{n+1} = 1 + 2a^2
$$
It follows from a result of Nagell that this equation has no solutions for $n\ge 2$ except $a=11$ and $n=4$. ($242+1 = 243$). The remaining cases - $n=0$ and $n=1$ - both give rise to easy solutions.
Note: Nagell's result is actually stronger, and replaces the '$3$' in the above expression with an arbitrary positive integer $y$: i.e., it states that there is no solution to $y^{n+1}=1+2a^2$ for $n\ge 2$ except $y=3,n=4,a=11$. I don't know if there is an easier proof of this special case.
Update: You also asked about the 'one more than a square' case, which we can prove using more elementary means. In this case, we have $\sum_{i=0}^n 3^i = a^2 + 1$ and so
$$
3^{n+1} = 3 + 2a^2
$$
We know that the number $9$ does not divide the right hand side of this equation: indeed, if $a$ is a multiple of $3$, then $2a^2$ is a multiple of $9$ and therefore $3 + 2a^2$ is not a multiple of nine. Conversely, if $a$ is not a multiple of $3$, then $3+2a^2$ cannot even be a multiple of $3$.
The only possible value of $3^{n+1}$ that is not a multiple of $9$ is $3$ (when $n=0$). From this it follows that $a=0$ too.
Nagell, Trygve, Contributions to the theory of a category of Diophantine equations of the second degree with two unknowns, Nova Acta Soc. Sci. Upsal., Ser. IV 16, No. 2, 38 p. (1954). ZBL0057.28304.
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} |
Work out $\int_{0}^{\pi/2}\cos^{2n}(x)\ln\sin(x) dx$ $$F(n)=\large \int_{0}^{\pi/2}\cos^{2n}(x)\ln\sin(x)\mathrm dx$$
$$\cos^{2n}(x)=\frac{1}{2^{2n}}{2n \choose n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}{2n \choose k}\cos\left[2\left(n-k\right)x\right]$$
$$I=\frac{1}{2^{2n}}{2n \choose n}\int_{0}^{\pi/2}\ln\sin(x)\mathrm dx+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}{2n \choose k}\int_{0}^{\pi/2}\cos\left[2\left(n-k\right)x\right]\ln\sin(x)\mathrm dx$$
$$I=-\frac{\pi \ln(2)}{2^{2n+1}}{2n \choose n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}{2n \choose k}\int_{0}^{\pi/2}\cos\left[2\left(n-k\right)x\right]\ln\sin(x)\mathrm dx$$
I can't continue.
I would like to evaluate $F(n)$
| You can let $\cos^2 (x) = t$ to obtain
\begin{align}
F(n) &= \frac{1}{4} \int \limits_0^1 t^{n-\frac{1}{2}} (1-t)^{-\frac{1}{2}} \ln(1-t) \, \mathrm{d} t \\
&= \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d}s} \int \limits_0^1 t^{n-\frac{1}{2}} (1-t)^{s - 1} \, \mathrm{d} t ~\Bigg \lvert_{s=\frac{1}{2}} \\
&= \frac{1}{4} \frac{\mathrm{d}}{\mathrm{d}s} \operatorname{B} \left(n+\frac{1}{2},s\right)\\
&= -\frac{\Gamma\left(n+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{4\Gamma(n+1)} \left[\psi (n+1) - \psi \left(\frac{1}{2}\right)\right] \, .
\end{align}
For $n \in \mathbb{N}_0$ the special values $\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)! \sqrt{\pi}}{4^n n!}$ , $\psi \left(\frac{1}{2}\right) = - \gamma - 2 \ln(2)$ and $\psi(n+1) = H_n - \gamma$ yield
$$F(n) = - \frac{\pi}{4} \frac{(2n-1)!!}{(2n)!!}[H_n + 2 \ln(2)] \, . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Pattern in factorising monic Quadratics I have noticed a pattern while factoring quadratic expressions in the form: $x^2+(a+1)x+a$ ($a$ is a constant). For example,
*
*$x^2+3x+2=(x+2)(x+1)$
*$x^2+92x+91=(x+91)(x+1)$
*$x^2+5x+4=(x+4)(x+1)$
The pattern which can be assumed here is:
$$x^2+(a+1)x+a=(x+a)(x+1)$$
My question is: How do you prove this identity?
| A couple of ways other than direct factorization:
*
*by the quadratic formula the roots of $\,x^2+(a+1)x+a\,$ are $$\,x_{1,2}=\dfrac{-(a+1)\pm\sqrt{(a+1)^2-4a}}{2}=\dfrac{-(a+1)\pm\sqrt{(a-1)^2}}{2}=\begin{cases}-1 \\-a\end{cases}\,$$and therefore the quadratic factors as $\,(x-x_1)(x-x_2)=(x+1)(x+a)\,$;
*two quadratics that have equal values at three different points are identical (since their difference would be a polynomial of degree $\le 2$ with $3$ roots), so it is enough to verify that $\,x^2+(a+1)x+a=(x+1)(x+a)\,$ for three values of $x$, for example $x=0,-1,-a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Reduction of Order Leads to Non-Elementary Integral
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
\begin{align}
\frac{dw}{dx}(x^2+x)-w(x^2+1)&=0 \\
\frac{dw}{dx}&=\frac{(x^2+x)^{-1}(x^2+1)}{w^{-1}} \\
\text{ln}(w)&=\int \frac{x^2+1}{x^2+x} \ dx \\
\text{ln}(w)&=\int 1-\frac{1}{x}+\frac{2}{x+1} \ dx \\
\text{ln}(w)&=x+\text{ln}\left(\frac{(x+1)^2}{x}\right)+C \\
w&=C_1\frac{e^x(x+1)^2}{x} \\
v&=C_1\int \frac{e^x(x+1)^2}{x} \ dx
\end{align}
Where $$C_1\int \frac{e^x}{x} \ dx$$ cannot be solved. How do I find $v$?
| $$xu''-(x+1)u'+u=0$$
$$x(u''-u')-(u'-u)=0$$
Let $v=u'-u$,
$$xv'-v=0$$
\begin{align}
xv' &= v \\
\frac{dv}{dx} &= \frac{v}{x} \\
\ln v &= \ln x + C \\
v &= Ax \\
u'-u &= Ax \\
u &= \dfrac{1}{\exp\left(-\int dx\right)}\int Ax\exp\left(-\int dx\right) dx \\
&= Ae^x\int xe^{-x} dx \\
&= Ae^x \left[-xe^{-x}-\int-e^{-x}dx\right] \\
&= Ae^x \left[-(x+1)e^{-x}+B\right] \\
&= k_1(x+1)+k_2e^x \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find no. of non negative integer solutions of $a+2b+3c = 200$ I want to find number of the non negative integer solutions of $a+2b+3c = 200$
(answer can contain $a,b,c = 0$)
Example one of the solutions is $a = 100, b = 50, c = 0$
I used stars and bars trick and here i get is
$\dfrac{\dbinom{200+2}{2}}{2 \times 3}$ but it is incorrect
any type of math is welcomed
what I have learnt from stars and bars technique is to find the solutions of $x+y = n , x,y,n > 0$ are positive integers but I am confused with equation of type $ax+by = n$ where $a , b> 1$
| Using generating functions, this would be the coefficient of $x^{200}$ in
$$f(x) = \frac{1}{1-x} \cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3} = \frac{1}{(1-x)^3 (1+x) (1+x+x^2)}.$$
Now, according to Maxima, if you calculate a partial fractions expansion of $f(x)$ you should get:
$$f(x) = \frac{2+x}{9(1+x+x^2)} + \frac{1}{8(1+x)} + \frac{17}{72(1-x)} + \frac{1}{4(1-x)^2} + \frac{1}{6(1-x)^3} = \\
\frac{2-x-x^2}{9(1-x^3)} + \frac{1}{8(1+x)} + \frac{17}{72(1-x)} + \frac{1}{4(1-x)^2} + \frac{1}{6(1-x)^3}.$$
Therefore, the coefficient of $x^{200}$ would be
$$-\frac{1}{9} + \frac{1}{8} + \frac{17}{72} + \frac{1}{4} \binom{201}{1} + \frac{1}{6} \binom{202}{2} = 3434.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Shorter Way to Solve $\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$ The limit to find is
$$\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$\lim_{x\to 0}\frac{(x-\ln(1+x))(x^2+x\ln(x+1)+\ln^2(x+1))}{(\sin x-x)(\sin x+x)}$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?
| Taylor Series around $0$ of $\ln(1+x)$:
$$\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$$
$$\ln^3(1+x) = x^3 -\frac{3}{2}x^4 + O(x^5)$$
Taylor Series around $0$ of $\sin(x)$:
$$\sin(x) = x - \frac{x^3}{6} + O(x^5)$$
$$\sin^2(x) = x^2 - \frac{x^4}{3} + O(x^6)$$
So
$$\frac{x^3 - \ln^3(1+x) }{\sin^2(x) - x^2} \simeq \frac{x^3 - x^3 +\frac{3}{2}x^4}{x^2 - \frac{x^4}{3} - x^2} = \frac{\frac{3}{2}x^4}{- \frac{x^4}{3}} = - \frac{9}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Showing $\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$
I wish to show that if $z$ is real, then
$$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$
I have shown this result, although my inequality is the wrong way around.
I considered
\begin{align}
|z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(triangle inequality)} \\
&=|z|^2+1 \\ \\
\Rightarrow |z^2+1|&\leq |z|^2+1 \\
\frac{1}{|z^2+1|}&\geq\frac{1}{|z|^2+1} \\
\frac{|e^{iz}|}{|z^2+1|}&\geq\frac{|e^{iz}|}{|z|^2+1} \\
\left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{e^{-\text{Im}(z)}}{|z|^2+1} \\
\left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{1}{|z|^2+1} \ \ \ \ \ \ \text{(if $z$ is real $\Rightarrow$ Im$(z)=0$)} \\
\end{align}
Where did I go wrong?
Also, I wonder, would this inequality still hold if $z$ was not real?
| If $z$ is real: $|e^{iz}|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{1}{|z|^2+1}.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^{iz}|
=e^{-y}$ is huge, and so
$$\left|\frac{e^{iz}}{z^2+1}\right|\gg\frac{1}{|z|^2+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the probability of the following event? Let $X_1, X_2, X_3, X_4$ be independent Bernoulli random variables. Then
\begin{align}
Pr[X_i=1]=Pr[X_i=0]=1/2.
\end{align}
I want to compute the following probability
\begin{align}
Pr( X_1+X_2+X_3=2, X_2+X_4=1 ).
\end{align}
My solution: Suppose that $X_1+X_2+X_3=2$ and $X_2+X_4=1$. Then $(X_2, X_4)=(0,1)$ or $(1,0)$. The probabilities of $(X_2, X_4)=(0,1)$ and $(1,0)$ are both $1/2 \times 1/2 = 1/4$. If $(X_2, X_4)=(0,1)$, then $X_1+X_2+X_3=X_1+X_3=2$. Therefore $X_1 = X_3 =1$. This happens with probability $1/4$. If $(X_2, X_4)=(1,0)$, then $X_1+X_2+X_3=X_1+1+X_3=2$. Therefore $(X_1, X_3) \in \{(1,0), (0,1)\}$. This happens with probability $1/4$. Therefore
\begin{align}
Pr( X_1+X_2+X_3=2, X_2+X_4=1 ) = 1/16+2/16=3/16.
\end{align}
Is this correct? Thank you very much.
| Just add as a supplementary technique:
You may also try to use the law of total probability with conditioning on $X_2$, since only $X_2$ are in common of the two events, then make use of the independence:
$$ \begin{align} &\Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1\} \\
=& \Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1|X_2 = 0\}\Pr\{X_2 = 0\} \\
&+ \Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1|X_2 = 1\}\Pr\{X_2 = 1\} \\
=& \Pr\{X_1 + X_3 = 2, X_4 = 1|X_2 = 0\}\Pr\{X_2 = 0\} \\
&+ \Pr\{X_1 + X_3 = 1, X_4 = 0|X_2 = 1\}\Pr\{X_2 = 1\} \\
=& \Pr\{X_1 + X_3 = 2\}\Pr\{X_4 = 1\}\Pr\{X_2 = 0\} + \\
&+ \Pr\{X_1 + X_3 = 1\}\Pr\{X_4 = 0\}\Pr\{X_2 = 1\} \\
=& \frac {1} {4}\times \frac {1} {2} \times \frac {1} {2} + \frac {1} {2} \times \frac {1} {2} \times \frac {1} {2} \\
=& \frac {3} {16}
\end{align}$$
So essentially you have counted the cases, just like what you have did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$
Or $\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$
Then i can't prove $\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$
I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QM\ge AM+GM$ with $n=2$ so is that true for $n=i$?
| An alternative approach, let
$$
u = \frac{a + b}{2} \quad \text{and} \quad v = ab
$$
so that the inequality becomes
$$
\frac{v}{u} + \sqrt{2u^2 - v} \ge \sqrt{v} + u.
$$
Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to
$$
x \ge \sqrt{1 + x} - \sqrt{1 - x} \quad (\text{for}\,\, x \le 0)
$$
which is shown fairly easily. The condition comes from
$$
u = 1 \implies b = 2 - a \implies x = a(2 - a) - 1 = -(a-1)^2 \le 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
| Yet an other solution. It is based on the fact that
$$ f = x^4+x^3+x^2+x+1\in\Bbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d)\ ,\qquad a,b,c,d\in\Bbb F_7\ .$$
Let $u\ne 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal,
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $a\leftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1)\ .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $\Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)
Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $b\ne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True
One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $\Bbb F_7$. Then there is a root in $\Bbb F_{7^2}=\Bbb F_{49}$, so the polynomials $f$ and $x^{49}-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f )
1
sage: gcd( x^(7^4) - x, f )
x^4 + x^3 + x^2 + x + 1
For this, one can also compute easily $(x^{49}-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^{49}-x,f)=(x^{49}-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Integral Resulting in Hypergeometric Function Any tips on how to prove this result?
$$
\int \frac{x}{x^K + c} dx = \frac{x^2 {}_2F_1 \left(1,\frac{2}{K};\frac{K+2}{K};-\frac{x^K}{c} \right)}{2c},
$$
where $${}_2F_1 (a, b;c;z)$$ is the hypergeometric function.
| Assuming $K,c>0$ , we can use the geometric series to find for $|x|^K < c$
$$ \int \limits_0^x \frac{t}{t^K + c} \, \mathrm{d} t = \frac{1}{c} \sum \limits_{n=0}^\infty \left(-\frac{1}{c}\right)^n \int \limits_0^x t^{K n +1} \, \mathrm{d}t = \frac{x^2}{c} \sum \limits_{n=0}^\infty \frac{1}{Kn+2} \left(-\frac{x^K}{c}\right)^n \, .$$
Since
$$ \frac{1}{K n +2} = \frac{1}{2} \frac{\frac{2}{K}}{\frac{2}{K} +n} = \frac{1}{2} \frac{\Gamma\left(\frac{2}{K}+1\right)}{\Gamma\left(\frac{2}{K}\right)} \frac{\Gamma\left(\frac{2}{K}+n\right)}{\Gamma\left(\frac{2}{K}+1+n\right)} \, ,$$
we get
$$ \int \limits_0^x \frac{t}{t^K + c} \, \mathrm{d} t = \frac{x^2}{2 c} \sum \limits_{n=0}^\infty \frac{\Gamma(1+n)}{\Gamma(1)} \frac{\Gamma\left(\frac{2}{K}+n\right)}{\Gamma\left(\frac{2}{K}\right)} \frac{\Gamma\left(\frac{2}{K}+1\right)}{\Gamma\left(\frac{2}{K}+1+n\right)} \frac{\left(-x^K/c\right)^n}{n!} \, .$$
Now note that the sum on the right-hand side is exactly the definition of the hypergeometric function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all matrices that commute with $A$
Given $$A = \begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.
Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to find other $B's$ using above Nilpotency of $A-3I$?
| By just writing out the matrix multiplication and simplifying you get:
\begin{align*}
AB &= BA\\
\begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}
\begin{bmatrix}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33} \\
\end{bmatrix}
&=
\begin{bmatrix}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33} \\
\end{bmatrix}
\begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}\\
\begin{bmatrix}
3b_{11} + b_{21} & 3b_{12} + b_{22} & 3b_{13}+b_{23} \\
3b_{21} + b_{31} & 3b_{22} + b_{32} & 3b_{23}+b_{33} \\
3b_{31} & 3b_{32} & 3b_{33}
\end{bmatrix}
&=
\begin{bmatrix}
3b_{11} & b_{11}+3b_{12} & b_{12} + 3b_{13} \\
3b_{21} & b_{21}+3b_{22} & b_{22} + 3b_{23} \\
3b_{31} & b_{31}+3b_{32} & b_{32} + 3b_{33}
\end{bmatrix} \\
\begin{bmatrix}
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33} \\
0 & 0 & 0
\end{bmatrix}
&=
\begin{bmatrix}
0 & b_{11} & b_{12} \\
0 & b_{21} & b_{22} \\
0 & b_{31} & b_{32}
\end{bmatrix}
\end{align*}
Hence, $b_{21},b_{31},b_{32}=0$, $b_{11}=b_{22}=b_{33}$ and $b_{12}=b_{23}$, confirming the solutions are exactly those given by Robert.
| {
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"url": "https://math.stackexchange.com/questions/2884923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve the equation $\frac{\sqrt {3}}{2}\sin(x) -\cos x=\cos^2x$ Solve the equation $\frac{\sqrt {3}}{2}\sin x-\cos x=\cos^2x$
My approach $\cos^2x=1-\sin^2x $
$\frac{\sqrt {3}}{2}\sin x-\cos^2x =\cos x$
$\frac{\sqrt {3}}{2}\sin x+\sin^2x -1=\cos x$
$\sqrt {3}\sin x+2\sin^2x-2=2\cos x$
Though the equation comes in form of $\sin x$ from here onward after squaring still not getting the answer.
| Just to flesh out A. Pongrácz's answer, let $c=\cos x,\,s=\sin x$ so $\frac{s\sqrt{3}}{2}=c(1+c)$ and $3(1-c^2)=4c^2(1+c)^2$. After some rearrangment, $(c+1)(c-\frac{1}{2})(2c^2+3c+3)=0$, with the quadratic factor lacking real roots. We must be careful with the signs of $c,\,s$. One solution is $c=-1,\,s=0$; the other is $c=\frac{1}{2},\,s=\frac{\sqrt{3}}{2}$. In other words, the real $x$ allowed are $x=\pi(2k+1),\,x=\frac{\pi (1+6k)}{3}$ for $k\in\mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving system of linear equations involving 3x3 matrix using adjoint matrix
Solve the following system using the adjoint matrix.
$$2x+4y-10z=-2$$
$$3x+9y-21z=0$$
$$x+5y-12z=1$$
Now, I have tried to solve it, and I got the determinant of matrix $A$ is equal to zero. What is the solution of the above system? Please help me. Thank you.
| The system of equation can be expressed in the matrix form as:
$$\begin{pmatrix}2&4&-10\\ 3&9&-21\\1&5&-12\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-2\\0\\1\end{pmatrix} \iff AX=B.$$
The formula to find $X$:
$$AX=B \iff A^{-1}AX=A^{-1}B \iff X=A^{-1}B=\frac{\text{adj}A}{|A|}B.$$
The determinant of $A$ (performing row operations):
$$|A|=\begin{vmatrix}2&4&-10\\ 3&9&-21\\1&5&-12\end{vmatrix}=\begin{vmatrix}0&-6&14\\ 0&-6&15\\1&5&-12\end{vmatrix}=\begin{vmatrix}0&0&-1\\ 0&-6&15\\1&5&-12\end{vmatrix}=-6.$$
The adjoint of $A$:
$$\text{adj}A=(C_{ij})^{\text{T}}=\begin{pmatrix}9(-12)-(-21)5&-(4(-12)-(-10)5)&4(-21)-(-10)9\\ -(3(-12)-(-21)1)&2(-12)-(-10)1&-(2(-21)-(-10)3)\\3\cdot 5-9\cdot 1&-(2\cdot 5-4\cdot 1)&2\cdot 9-4\cdot 3\end{pmatrix}=\\
=\begin{pmatrix}-3&-2&6\\ 15&-14&12\\6&-6&6\end{pmatrix}$$
Can you calculate $X$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Counting Conjugacy Classes: $A^6 = I$ This question is inspired by this earlier post. The question that I'm interested in is as follows:
Let $S= \{A \in \mathbb{Q}^{k \times k} : A^6 = I \text{ and }A^n \ne I \text{ for any }0 < n < 6\}$. How many orbits under conjugation by $GL_k(\mathbb{Q})$ does $S$ contain?
Per the argument that I've given in the linked post, it suffices to list representatives in rational canonical form. Thus, it is equivalent to enumerate the multisets of irreducible polynomials $\{p_1,\dots,p_n\}$ (which are the polynomials corresponding to companion matrix blocks) each of which divides $x^6-1$, whose degrees sum to $k$, and whose lcm divides $x^6 - 1$ but no polynomial of the form $x^n - 1$ for $n < 6$.
Of course, there is only one such multiset in the case of $k=2$, hence the answer to the linked question. As $k$ grows things get trickier. For $k=5$, we have
$$
\{x^2 - x + 1, x \pm 1, x \pm 1, x\pm 1\}\\
\{x^2 + x + 1, x+1, x\pm 1, x\pm 1\}\\
\{x^2 - x + 1, x^2 \pm x + 1 ,x \pm 1\}\\
\{x^2 + x + 1, x^2 + x + 1, x+1\}\\
$$
For a total of $14$ conjugacy classes.
How might one approach this problem in general?
| Please read my comments under the OP's question. In this answer, I would like to only address the case $m=6$, that is, the situation the OP asks for. I do not know how to deal with the general exponent $m$.
As we have established, the characteristic polynomial involved in each indecomposable block of $A$ is one of the following four cyclotomic polynomials: $$\Phi_1(x)=x-1\,,\,\,\Phi_2(x)=x+1\,,\,\,\Phi_3(x)=x^2+x+1\,,\text{ and }\Phi_6(x)=x^2-x+1\,.$$ Now, $A$ must have at least one indecomposable block of size $2\times 2$. Let $a$ and $b$ denote the number of $2$-by-$2$ indecomposable blocks of $A$ and the number of $1$-by-$1$ blocks of $A$. Then, $a\geq 1$ and $2a+b=k$. Denote by $J_1,J_2,\ldots,J_a$ the $2$-by-$2$ blocks of $A$, whereas $K_1,K_2,\ldots,K_b$ the $1$-by-$1$ blocks. We may assume that $J_1,J_2,\ldots,J_p$ have $\Phi_6$ as the characteristic polynomial, whereas $J_{p+1},J_{p+2},\ldots,J_a$ have $\Phi_3$ as the characteristic polynomial. Similarly, suppose that $K_1,K_2,\ldots,K_q$ have $\Phi_2$ as the characteristic polynomial, whilst $K_{q+1},K_{q+2},\ldots,K_b$ have $\Phi_1$ as the characteristic polynomial.
If $p\geq 1$, then there is no restriction on $q$. Thus, for each fixed $a\in\Biggl\{1,2,\ldots,\left\lfloor\dfrac{k}{2}\right\rfloor\Biggr\}$, there are $a$ ways to choose $p\geq 1$ and $b+1=k-2a+1$ ways to choose $q$. Thus, for $p\geq 1$, there are
$$\begin{align}\sum_{a=1}^{\left\lfloor\frac{k}2\right\rfloor}\,a(k-2a+1)&=(k-1)\,\sum_{a=1}^{\left\lfloor\frac{k}2\right\rfloor}\,a-4\,\sum_{a=1}^{\left\lfloor\frac{k}2\right\rfloor}\,\frac{a(a-1)}{2}
\\
&=\frac{k-1}{2}\,\left\lfloor\frac{k}{2}\right\rfloor\,\left(\left\lfloor\frac{k}{2}\right\rfloor+1\right)-\frac{2}{3}\,\left\lfloor\frac{k}{2}\right\rfloor\,\left(\left\lfloor\frac{k}{2}\right\rfloor+1\right)\,\left(\left\lfloor\frac{k}{2}\right\rfloor-1\right)
\end{align}$$
corresponding conjugacy classes.
If $b=0$, then $q\geq1$ is required. Thus, for each fixed $a\in\Biggl\{1,2,\ldots,\left\lfloor\dfrac{k}{2}\right\rfloor\Biggr\}$, there are $b=k-2a$ ways to choose $q\geq 1$. Hence, for $p=0$, there are
$$\sum_{a=1}^{\left\lfloor\frac{k}2\right\rfloor}\,(k-2a)=k\,\left\lfloor\frac{k}{2}\right\rfloor-\left\lfloor\frac{k}{2}\right\rfloor\,\left(\left\lfloor\frac{k}{2}\right\rfloor+1\right)$$
corresponding conjugacy classes.
This shows that
$$N(k,6)=\frac{1}{6}\,\left\lfloor\frac{k}{2}\right\rfloor\,\left(3\,(k-3)\,\left\lfloor\frac{k}{2}\right\rfloor-4\,\left\lfloor\frac{k}{2}\right\rfloor^2+9\,k-5\right)$$
is the total number of conjugacy classes. We have
$$N(1,6)=0\,,\,\,N(2,6)=1\,,\,\,N(3,6)=3\,,\,\,N(4,6)=7\,,\text{ and }N(5,6)=12\,.$$
(The OP miscounted something. From his list, there should be $12$ distinct conjugacy classes for $k=5$.)
Note that
$$\frac{(k-1)(k^2+10k-3)}{24}\leq N(k,6)\leq \frac{k(k-1)(k+10)}{24}\,.$$
The left-hand side is an equality iff $k$ is an odd positive integer. The right-hand side is an equality iff $k=1$ or $k$ is an even positive integer.
| {
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"url": "https://math.stackexchange.com/questions/2893357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluate $\int (3+4\sin x)^{-2} dx$
Evaluate: $\int\frac{dx}{(3+4\sin x)^2}$
My attempt: I have tried to express the integrand in terms of $\tan x$ and $\sec x$ but there was no use since the substitution $\tan x=z$ is of no use after that. I also tried to use Weierstrass substitution but i got a very complicated algebraic expression. Please help.
| By integration by parts, we have
$$
\begin{aligned}
\int \frac{d x}{(3+4 \sin x)^{2}} =&-\frac{1}{4} \int \frac{1}{\cos x} d\left(\frac{1}{3+4 \sin x}\right) \\
=&-\frac{1}{4 \cos x(3+4 \sin x)}+\frac{1}{4} \underbrace{\int \frac{\sec x \tan x}{3+4 \sin x}}_{J} d x
\end{aligned}
$$
$$
J=\int \frac{\sec x \tan x}{3+4 \sin x} d x= \int \frac{\sin x}{(1+\sin x)(1-\sin x)(3+4 \sin x)} d x
$$
Resolving the integrand of the last integral yields $$
J=\frac{1}{2} \underbrace{ \int \frac{d x}{1+\sin x} }_{K} +\frac{1}{14}\underbrace{\int \frac{d x}{1-\sin x}}_{L}-\frac{12}{7} \underbrace{\int \frac{d x}{4 \sin x+3}}_{M}
$$
$$
\begin{aligned}
K&=\int \frac{1-\sin x}{\cos ^{2} x} d x=\int\left(\sec ^{2} x-\tan x \sec x\right) d x=\tan x-\sec x+C_{1}
\end{aligned}
$$
Similarly,
$$
L=\int \frac{1+\sin x}{\cos ^{2} x} d x=\tan x+\sec x+C_{2}
$$
Letting $t=\tan \frac{x}{2}$, $$
\begin{aligned}
M &=\int \frac{1}{4\left(\frac{2 t}{1+t^{2}}\right)+3} \cdot \frac{2 d t}{1+t^{2}}=2 \int \frac{d t}{3 t^{2}+8 t+3} =\frac{2 \sqrt{7}}{7} \ln \left|\frac{3 t-\sqrt{7}+4}{3 t+\sqrt{7}+4}\right|+C_{3}
\end{aligned}
$$
Now we can conclude that $$
\begin{aligned}
I=&-\frac{1}{4 \cos x(3+4 \sin x)}+\frac{\tan x}{7}-\frac{3 \sec x}{28} +\frac{6 \sqrt{7}}{49} \ln \left|\frac{3 \tan \frac{x}{2}+\sqrt{7}+4}{3 \tan \frac{x}{2}-\sqrt{7}+4}\right|+C\\=& \frac{6 \sqrt{7}}{49} \ln \left|\frac{3 \tan \frac{x}{2}+\sqrt{7}+4}{3 \tan \frac{x}{2}-\sqrt{7}+4}\right|-\frac{4 \cos x}{7(3+4 \sin x)}+C
\end{aligned}
$$
| {
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"url": "https://math.stackexchange.com/questions/2894007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.