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Solving differential equation using power series $y'' - xy' -y = 0$ given $y(0) = 1$ and $y'(0) = 0$ I think I made a mistake somewhere:
$$y'' -xy' - y = 0$$ given $y(0) = 1$ and $y'(0) = 0$
so we have:
$$y = \sum_{n=0}^\infty C_nx^n$$
$$y' = \sum_{n=1}^\infty nC_nx^{n-1}$$
$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
so subbing:
$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2} - \sum_{n=0}^\infty nC_nx^{n-1} - \sum_{n=0}^\infty C_nx^{n} = 0$$
$$y' = \sum_{n=0}^\infty (n+2)(n+1)C_{n+2}x^{n} - \sum_{n=0}^\infty (n+1)C_nx^n = 0$$
$$C_{n+2} = \frac{C_n}{n+2}$$
so a few terms:
$C_0 = C_0$ and $C_1 = C_1$ and $C_2 = \frac{C_0}{2}$ and $C_3 = \frac{C_1}{3}$ and $C_4 = \frac{C_2}{4} = \frac{C_0}{4 \cdot 2}$ and $c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$
so the even terms are: $\frac{C_0}{2^n \cdot n!}$
and the odd terms are: $\frac{C_1}{1 \cdot 3 \cdot 5 \cdot (2n-1)}$
and so $$y = \sum_{n=0}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n-1)} x^{2n-1}$$
but I'm stuck here. If $y(0) = 1$ ... doesn't the equation become 0? I feel like I've hit an impossible condition so I feel like I've made a mistake somewhere.
| You did well. Note that the initial conditions $y(0)=1$ and $y'(0)=0$ simply means $C_0=1$ and $C_1=0$.
Also, noting that $C_1=0$, your final solution is
$$y=\sum_{n=0}^\infty\frac{c_0}{2^n\cdot n!}x^{2n}=c_0+\sum_{n=1}^\infty\frac{c_0}{2^n\cdot n!}x^{2n}$$
which dosn't make the equation $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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What is $(N)+ (N-1) + (N -2) + \cdots + 1$ called? This is purely for figuring the name of a mathematical concept.
For example, $N \times (N-1) \times \cdots \times 1$ is called factorial.
Question: What is $N + (N-1) + (N -2) + \cdots + 1$ called?
| The sum of numbers from $1$ to $n$ is called a "Triangular number".
From Wikipedia:
The triangle numbers are given by the following explicit formulas:
$$T_n = \sum_{k=1}^n{k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}} = \binom{n + 1}{2},$$
So, the first triangular numbers are:
$
T_1 = 1\\
T_2 = 1 + 2 = 3\\
T_3 = 1 + 2 + 3 = 6\\
T_4 = 1 + 2 + 3 + 4 = 10\\
...etc
$
| {
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Show that $\sum_{j=0}^{n}{n+j\choose 2j}5^j...$ I am trying to show that, $$\sum_{j=0}^{n}{n+j\choose 2j}5^j=\frac{\phi^{-4n}}{1+\phi^4}+\frac{\phi^{4n}}{1+\phi^{-4}}$$
where $\phi=\frac{1+\sqrt{5}}{2}$ and $n\ge 0$
An atempt:
$$\sum_{j=0}^{n}{n+j\choose n-j}5^j$$
$$\sum_{j=0}^{n}\frac{(n+j)!}{(n-j)!(2j)!}\cdot5^j$$
$$\sum_{j=0}^{n}\frac{(n-1)(n-2)(n-3)\cdots(n-j+1)\times n(n+1)(n+2)(n+3)\cdots(n+j)}{(2j)!}\cdot5^j$$
$$\sum_{j=0}^{n}\frac{(n^2-1)(n^2-4)(n-9)\cdots(n^2-(j-1)^2)\times n(n+j)}{(2j)!}\cdot5^j$$
I can't simplify this sum, it seems like the method I am trying to do it is incorrect.
Help reqiured, thank you.
| We use the coefficient of operator $[z^j]$ to denote the coefficient of $z^j$ in a series. This way we can write for instance
\begin{align*}
[z^j](1+z)^n=\binom{n}{j}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^n}\color{blue}{\binom{n+j}{n-j}\,5^j}
&=\sum_{j=0}^n[z^{n-j}](1+z)^{n+j}5^{j}\\
&=[z^n](1+z)^n\sum_{j=0}^{\infty}\left(5z(1+z)\right)^j\tag{1.1}\\
&\,\,\color{blue}{=[z^n]\frac{(1+z)^n}{1-5z-5z^2}}\tag{1.2}
\end{align*}
We set in (1.1) the upper limit of the series to $\infty$ without changing anything, since powers of $z$ greater $n$ are ignored due to the coefficient of operator $[z^n]$. This way we can use the geometric series expansion in (1.2).
We now use a clever change of variables stated as rule 5 in section 1.2 of Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev. Rule 5 adapted for this special case is:
\begin{align*}
[z^n]A(z)f^{n}(z)
&=[y^n]\left.\frac{A(z)f(z)}{f(z)-zf^{\prime}(z)}\right|_{z=g(y)}\tag{2}
\end{align*}
Here we have $f(z)=(1+z)$ and $g=g(y)$ is the inverse function of
\begin{align*}
\frac{z}{f(z)}=\frac{z}{(1+z)}=y\tag{3}
\end{align*}
We obtain
\begin{align*}
&z=\frac{y}{1-y}\\
\end{align*}
We get from (1.2),(2) and (3)
\begin{align*}
\color{blue}{[z^n]\frac{(1+z)^n}{1-5z-5z^2}}&=[y^n]\left.\frac{1+z}{(1+z)-z}\,\frac{1}{1-5z-5z^2}\right|_{z=\frac{y}{1-y}}\\
&=[y^n]\left.\frac{1+z}{1-5z-5z^2}\right|_{z=\frac{y}{1-y}}\\
&=[y^n]\frac{1-y}{1-7y-y^2}\tag{4}\\
&=[y^n]\frac{1-y}{\left(y-\phi^4\right)\left(y-\phi^{-4}\right)}\tag{5}\\
&=[y^n]\left(\frac{1}{1+\phi^4}\,\frac{1}{1-\phi^{-4}y}+\frac{1}{1+\phi^{-4}}\,\frac{1}{1-\phi^{4}y}\right)\tag{6}\\
&\,\,\color{blue}{=\frac{\phi^{-4n}}{1+\phi^4}+\frac{\phi^{4n}}{1+\phi^{-4}}}
\end{align*}
and the claim follows.
Comment:
*
*In (4),(5) we find $y=\frac{1}{2}\left(7\pm3\sqrt{5}\right)$ are the roots of $1-7y-y^2=0$. We obtain
\begin{align*}
\phi^4&=\left(\frac{1}{2}\left(1+\sqrt{5}\right)\right)^4=\frac{1}{2}\left(7+3\sqrt{5}\right)\\
\phi^{-4}&=\frac{2}{7+3\sqrt{5}}=\frac{1}{2}\left(7-3\sqrt{5}\right)
\end{align*}
*In (6) we make a partial fraction decomposition.
*In (7) we select the coefficient of $y^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $ S_n$ is uniformly convergent ? Yes/NO Is $\displaystyle S_n = \sum_{n=1}^{\infty} 2^n \sin\frac{1}{3^nx}$ uniformly convergent on the interval $[1, \infty)$ ? True /false
My attempt : NO, the given series will not uniformly convergent on $[1, \infty)$ because the cauchy sequence criterion for uniform convergence fail
That is $$\vert S_{n +m}(x) - S_n(x) |= 2^{n+1} \sin \frac{1}{3^{n+1}x} + ....... + 2^{n+m} \sin \frac{1}{3^{n+m}x} \ge 2^{n+1} \frac{2}{ \pi} \frac{1}{3^{n+1} x} +....... + 2^{n+m}\frac{2}{\pi} \frac{1}{3^{n+m}x} \ge \frac{2^{n+1} {2}}{\pi3^{n+1} x}$$
Is this true ?
Any hints/solution will be appreciated
thanks u
| True. The main idea is that $\sin x \leq x$ for $x\geq 0.$ Hence, on the domain $[1,\infty),$
$$
\sum_{n=1}^\infty 2^n \sin\left(\frac{1}{3^n x}\right) \leq \sum_{n=1}^\infty 2^n \left(\frac{1}{3^n x}\right) \leq \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n,
$$
which is obviously convergent as a geometric series.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$
Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$
By factor theorem we know that $(x-a)$ is a factor of $f(x)$ if $f(a)=0$.
In this case, $f(x)=x^n -nx +n-1 \implies f(1)=0$
Hence we conclude that $(x-1)$ is a factor. From hereon, how can I say that $(x-1)^2$ is a factor?
Can we approach the problem without calculus approach? This problem was taken from a book of pre-calculus algebra.
| Here is another elementary way using the binomial theorem
*
*$(1+y)^n = \sum_{k=0}^n\binom{n}{k}y^k$
Set $\boxed{y=x-1}$ and note that
*
*$(x-1)^2$ is a factor of $p(x) = x^n -nx +n-1$ if and only if $y^2$ is a factor of $p(y+1)$
Hence,
\begin{eqnarray*} p(y+1)
& = & (1+y)^n - (1+y)n + n-1\\
& = & 1+ny +\sum_{k=2}^n\binom{n}{k}y^k -n-ny+n-1 \\
& = & y^2\sum_{k=2}^n\binom{n}{k}y^{k-2}
\end{eqnarray*}
Done.
| {
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Find the Minimum Value of $x^2+y^2$ Given:
$x+y=2\sin{a}-\cos{b}\\
xy=2\cos{a}+\sin{b}$
Find the minimum value of $x^2+y^2$.
Attempt:
$\begin{aligned}
x^2+y^2&=(x+y)^2-2xy\\
&=(2\sin{a}-\cos{b})^2-2(2\cos{a}+\sin{b})\\
&=4\sin^2{a}-4\sin{a}\cos{b}+\cos^2{b}-4\cos{a}-2\sin{b}
\end{aligned}$
$\begin{aligned}
f_{a}(a,b)&=0\\
8\sin{a}\cos{a}-4\cos{a}\cos{b}+4\sin{a}&=0\\
\cos{a}\cos{b}&=2\sin{a}\sin{b}+\sin{a}\\
\end{aligned}$
$\begin{aligned}
f_{b}(a,b)&=0\\
4\sin{a}\sin{b}-2\sin{b}\cos{b}-2\cos{b}&=0\\
2\sin{a}\sin{b}&=\sin{b}\cos{b}+\cos{b}\\
\end{aligned}$
Tried to plot it on a graph, the answer should be -6.
| It looks ugly, but fortunately, the $a$ terms can be isolated from both equations:
$$\tan a = \frac{\cos b}{2\sin b +1}$$
$$\sin a = \frac{\cos b(1+\sin b)}{2\sin b}$$
One of the options is to use $1+\tan^{-2} a = \sin^{-2} a$ to get an equation for $b$ where you can use $u=\sin b$ to make it a polynomial equation. However, you must be careful about cases where some of the terms are zero, for example if $\tan a=0$, you can't use $\tan^{-1}a $. It turns out that $a=0$ and $a=\pi$ allow you to find a few solutions directly. It these cases, both expressions above must be zero, so $\cos b=0$ is required (which gives you the solutions
$$a=0, b=\pm \tfrac{\pi}{2}$$
$$a=\pi, b=\pm \tfrac{\pi}{2}$$
Computing the values (and maybe checking second derivatives) shows you what these are. $a=0, b=+\tfrac{\pi}{2}$ turns out to be the global minimum, but to be sure, you should check the full polynomial equation that you get with full reconstruction for other, non-special values of $a$ and $b$.
Hint: it's a periodic function of $a$ and $b$, just plot it from $-\pi$ to $\pi$ in both directions.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sin(\theta) + \sin(5\theta) = \sin(3\theta)$ . Find number of solutions and the solutions for this equation in $[0,\pi]$ I tried to solve the equation by doing this-
$$2\sin(\frac{\theta+5\theta}{2})\cos(\frac{\theta-5\theta}{2})=\sin(3\theta)\\
2\sin(3\theta)\cos(2\theta) = \sin(3θ)\\
\cos(2θ) = \frac{1}{2}\\
1-2\sin^2(θ) = \frac{1}{2}\\
\sin^2(θ) = (\frac{1}{2})^2\\
∴ θ = nπ ± α\\
Answer = \frac{π}{6},\frac{5π}{6} $$
But in the solution there are 6 solutions and in step 3 instead of dividing $\sin(3θ)$ by $\sin(3θ)$ they have taken it as common and made "$\sin(3θ)(2\cos(2θ)-1)$"
| As we cannot divide by $0$, there are two possible cases:
(1) $\sin3\theta=0$ (i.e. $\theta=0$, $\frac \pi3$, $\frac {2\pi}3$ or $\pi$)
Then both $2\sin3\theta\cos2\theta$ and $\sin3\theta$ are $0$. The equation is satisfied. We have $4$ solutions in this case.
(2) $\sin3\theta\ne0$.
Then we can divide both sides of $2\sin3\theta\cos2\theta=\sin3\theta$ by $\sin3\theta$ as it is not zero. So we have $\cos2\theta=\frac12$ and hence $\theta=\frac \pi6$ or $\frac{5\pi}6$. There are $2$ solutions in this case.
| {
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Prove that $4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$ .
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$
I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution too, but not pretty . !
| Hint:
Rewrite this as
$\tan\frac{\pi}{7}(8\cos^2\frac\pi7 -1)=\sqrt7$
$\tan^2\frac{\pi}{7}(8\cos^2\frac\pi7 -1)^2=7$
$(1-\cos^2\frac{\pi}{7})(8\cos^2\frac\pi7 -1)^2=7\cos^2\frac\pi7$
$(1-\cos\frac{2\pi}{7})(4\cos\frac{2\pi}{7}+3)^2=7\cos\frac{2\pi}{7}+7$
$8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=0$
All we need is to prove that $\cos\frac{2\pi}7$ is a root of $8x^3+4x^2-4x-1=0$
Start from $\cos3(\frac{2\pi}{7})=\cos4(\frac{2\pi}{7})$ ...
| {
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Factorisation of $\frac{1}{u^2+3u-1}$ So, I need to factor the expression $\frac{1}{u^2+3u-1}$ First I find the roots $x_1=\frac{-3+\sqrt{13}}{2}$ and $x_2=\frac{-3-\sqrt{13}}{2}$ then I have $\frac{1}{(2x+3+\sqrt{13})(2x+3-\sqrt{13})}$ But on the factorisation calculator it states that there's 4 in numerator. So my question is why?
| $$\frac{1}{{{u}^{2}}+3u-1}=\frac{1}{{{u}^{2}}+3u+\frac{9}{4}-\frac{13}{4}}=\frac{1}{{{\left( u+\frac{3}{2} \right)}^{2}}-\frac{13}{4}}=\frac{4}{{{\left( 2u+3 \right)}^{2}}-13}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $5^n + 3^n - 2^{2n+1} > 0$ by induction I am not sure how to deal with the $-2^{2n+1}$ term.
I did the basis proof for n=1
I am stuck at this step:
$$
5^{k+1}+3^{k+1}-2^{2(k+1)+1} = 5\cdot 5^k + 3 \cdot 3^k -2^3 \cdot 2^{2k}
$$
Any advice guys?
| The base case is simple.
We now assume $5^k+3^k-2^{2k+1}>0$. Then multiplying by 4 yields $4\cdot 5^k + 4 \cdot 3^k - 4\cdot 2^{2k+1}>0$, which implies $5\cdot 5^k + 3\cdot 3^k -2^{2k+3} > 5^k - 3^k >0$ as required.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find minimum and maximum of $P=a+b+c$ Let $a,b,c\ge 0$ such that $a^2+b^2+c^2+abc=4$. Find minimum and maximum of $$P=a+b+c$$
+)Maximum: Let $x=\frac{2\sqrt{ab}}{\sqrt{\left(c+a\right)\left(c+b\right)}};y=\frac{2\sqrt{bc}}{\sqrt{\left(a+b\right)\left(a+c\right)}};z=\frac{2\sqrt{ca}}{\sqrt{\left(b+c\right)\left(b+a\right)}}$
And by AM-GM inequality i proved that $P\le 3$ when $a=b=c=1$
+)Minimum: I will prove $P\ge 2$ and the equality occurs when $a=b=0;c=2$ but failed. I tried substituting and uvw
| For the minimum, note $P^2 = 4+2(ab+bc+ca)-abc \geqslant 4+6(abc)^{2/3}-abc \geqslant 4$, as $abc \in [0, 1]$ from the obvious AM-GM $4 = a^2+b^2+c^2+abc \geqslant 4(abc)^{3/4}$ and $f(t)=6t^2-t^3$ is increasing in $t\in [0, 1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solve a trigonometric equation where one constant can be negative I previously posted a code-related questions on StackOverflow here but after doing some more investigation work into the code, I had an epiphany when looking at
the actual Newton-Raphson section because the formula used to represent $f(x)$ is a trigonometric equation of the following form:
$$f(x) = a\cdot\cos x + b\cdot\sin x - c$$
Now I've been trying to solve this for "$x$" using guidance from mathforum.org here and Quora here and I've got as far as the format at the end of Harshit Vyas' answer, and it looks like this:
$$C1\cdot\cos x + C2\cdot\sin x - C3 = 0$$
This equation/formula calculates the values of $C1$, $C2$ and $C3$ before it starts the iteration loop, so they become constants as far as the equation is concerned:
$$\sin x = \frac{C2 \cdot C3 \pm C1 \cdot \sqrt{C1^2 + C2^2 - C3^2}}{C1^2 + C2^2}$$
I can then get the value for $x$ with $\sin^{-1}$. My questions are as follows:
*
*Have I missed anything when I was solving for $x$?
*$C1$, $C2$ and $C3$ are calculated using sines and cosines of three
provided angles, and they can be negative due to the use of $-\cos$
and since a negative number has a complex square root, is there a
way to compensate for this?
04/06/2019 Edit to add:
I've gone back over the equation and rewritten according to the steps in the Quora article I linked to and this is what I get as my result (with C1=a, C2=b and C3=c, which are the actual coefficients was referring to earlier); apologies for the long list but I wanted to show my working step by step:
$$C1 \cdot cos(\beta) + C2 \cdot sin(\beta) - C3$$
$$C1 \cdot cos(\beta) + C2 \cdot sin(\beta) = C3$$
$$C1 \cdot cos(\beta) = C3 - C2 \cdot sin(\beta)$$
$$C1^2 \cdot cos^2(\beta) = C3^2 - 2 \cdot C2 \cdot C3 \cdot sin(\beta) + C2^2 \cdot sin^2(\beta)$$
$$C1^2 \cdot (1 - sin^2(\beta)) = 2 \cdot C2 \cdot C3 \cdot sin(\beta) + C2^2 + C2^2 \cdot sin^2(\beta)$$
$$(C1^2 + C2^2)sin^2(\beta) - 2 \cdot C2 \cdot C3 \cdot sin(\beta) + (C3^2 - C1^2) = 0$$
Let t = $sin(\beta)$ for simplicity:
$$(C1^2 + C2^2)t^2 + (-2 \cdot C2 \cdot C3)t + (C3^2 - C1^2) = 0$$
Let D = the discriminant of the formula:
$$D = (-2 \cdot C2 \cdot C3)^2 - 4(C1^2 + C2^2)(C3^2 - C1^2)$$
$$D = 4C2^2C3^2 - 4(C1^2C3^2 - C1^4 + C2^2C3^2 - C1^2C2^2)$$
$$D = 4C1^2 + 4C1^2C2^2 - 4C1^2C3^2$$
$$D = 4C1^2(C1^2 + C2^2 - C3^2)$$
$$t = \frac{-(-2 \cdot C2 \cdot C3 \pm \sqrt{D})}{2(C1^2 + C2^2)}$$
$$t = \frac{(2 \cdot C2 \cdot C3 \pm \ 2 \cdot C1 \sqrt{C1^2 + C2^2 - C3^2})}{C1^2 + C2^2}$$
$$t = sin(\beta) = \frac{(C2 \cdot C3 \pm C1 \sqrt{C1^2 + C2^2 - C3^2})}{C1^2 + C2^2}$$
$$\beta = sin^1 \left (\frac{C2 \cdot C3 \pm C1 \cdot ^{\sqrt{C1^2 + C2^2 - C3^2}}}{C1^2 + C2^2} \right )$$
Does this look like right or have I missed anything else out?
| Write your equation in the form $$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos(x)+\frac{b}{\sqrt{a^2+b^2}}\sin(x)-\frac{c}{\sqrt{a^2+b^2}}\right)$$
and define $$\sin(\phi)=\frac{a}{\sqrt{a^2+b^2}}$$
$$\cos(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$
| {
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Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can't really figure out how to solve the equation without expanding the brackets.
| Since the $x^2$ coefficient in your cubic is $a+b+c$, the quadratic factor is of the form $x^2+k$, with the roots being $\pm\sqrt{-k}$. The $x=0$ case gives $$k=\frac{abc+(b+c)(c+a)(a+b)}{a+b+c}=ab+bc+ca.$$In your case $k=-5$.
Simplifying $k$ as above looks like it requires tedious algebra, but things aren't as bad as they seem. It's the ratio of two fully symmetric polynomials in $a,\,b,\,c$, one of degree $3$, the other $1$. This doesn't prove on its own that $k$ is a polynomial; but if it is, it must be fully symmetric and of degree $2$, and hence proportional to $ab+bc+ca$. The case $a=b=c$ gives $k=\frac{9a^3}{3a}=3a^2$, so it'll have to be $ab+bc+ca$ itself. So it makes sense to double-check whether $(ab+bc+ca)(a+b+c)=abc+(b+c)(c+a)(a+b)$. But of course it does, because both sides are fully symmetric cubic functions, so they have a fixed ratio. Again, the case $a=b=c$ strengthens this to equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Rolling two dice: different probabilities Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$:
\begin{array}{c c c c c c}
\{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\
& \{2,2\}, & \{2,3\}, & \{2,4\}, & \{2,5\}, & \{2,6\},\\
& & \{3,3\}, & \{3,4\}, & \{3,5\}, & \{3,6\},\\
& & & \{4,4\}, & \{4,5\}, & \{4,6\},\\
& & & & \{5,5\}, & \{5,6\},\\
& & & & & \{6,6\}
\end{array}
but when I roll two different dice, then the probability of getting sum of 7 is $1/6$
\begin{array}{c c c c c c}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{array}
How this is possible?
| If you roll two identical dies, there are 21 outcomes, but they have different probabilities: probability to get $(1, 1)$ (or any outcome with the same numbers on both dies) is $\frac{1}{36}$, but probability to get $(1, 3)$ (or any outcome with different numbers) is $\frac{2}{36}$. So correct answer for two identical dies is $P((1, 6)) + P((2, 5)) + P((3, 4)) = \frac{6}{36} = \frac{1}{6}$ too.
Just having set of $n$ outcomes doesn't mean any outcome has probability of $\frac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine? In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:
$$a = ln(b + \sqrt{b^2 - 1})$$
$$e^a = b + \sqrt{b^2 - 1}$$
$${1 \over e^a} = {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = b + \sqrt{b^2 - 1} + {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b(b + \sqrt{b^2 - 1}) + \sqrt{b^2 - 1}(b + \sqrt{b^2 - 1}) + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b^2 + b\sqrt{b^2 - 1} + b\sqrt{b^2 - 1} + b^2 - 1 + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {2b^2 + 2b\sqrt{b^2 - 1} \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = 2b$$
This is all well and good, but for the life of me I can't see: how would someone ever know in advance that adding $b + \sqrt{b^2 - 1}$ to its own reciprocal would clean things up so nicely? Is this just a "special case" which needs to be memorized, or is there some general principle behind it which can be used when solving similar equations?
| Another "interesting" way to solve this equation, and which may be more easily explainable how to "see", might be to intuit that the right-hand side, with its sum of a number and a square root with a square of that number underneath it kind of resembles a quadratic formula:
$$b + \sqrt{b^2 - 1}$$
has kinda the form of
$$\frac{-B + \sqrt{B^2 - 4AC}}{2A}$$
Namely, note the
$$-B + \sqrt{B^2 - \cdots}$$
in particular, exactly rewritable with the right choices of $A$, $B$, and $C$ as
$$\frac{-(-b) + \sqrt{(-b)^2 - 4\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}}{2\left(\frac{1}{2}\right)}$$
(think: "How do I cancel the $4$ down to a $1$?" "Ooh, I can also end up removing the $2$ for 'free' by noting $4 = 2 \cdot 2$!" "Now what about the wee problem of $B$ being negative on the outside which doesn't quiiiiiite match?") which is the quadratic formula for the quadratic equation
$$\frac{1}{2} x^2 + (-b) x + \frac{1}{2} = 0$$
and then, since that quadratic formula was just the original right-hand side rewritten, and the quadratic formula gives "$x$", and the original equals $e^a$, we have $x = e^a$, which gives
$$\frac{1}{2} (e^a)^2 + (-b)(e^a) + \frac{1}{2} = 0$$
divide out $e^a$,
$$\frac{1}{2} (e^a) + (-b) + \frac{1}{2} (e^{-a}) = 0$$
and you can pull out $b$ yourself to get
$$b = \frac{e^a + e^{-a}}{2} = \cosh\ a$$.
Insofar as the original argument from the website is concerned, I suspect what the author did was that they already knew the form that $\cosh$ is supposed to take (who writing that wouldn't?) and just used that as a guide. Once you have $e^a$, you know you need to add $e^{-a}$ (i.e. $\frac{1}{e^a}$). The method above, though, is what you might do if you hypothetically never knew what $\cosh$ was supposed to look like and were deriving it afresh.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Definite integral of $1/(2\sin^4x + 3\cos^2x)$ I have $f = \frac 1 {(2\sin^4x + 3\cos^2x)}$ which area should be calculated from $0$ to $\frac{3\pi}2 $.
I noticed that $$\int_0^{\frac{3\pi}2} f \,dx= 3\int_0^{\frac{\pi}2} f \,dx$$
I tried to calculate this integral with the Weierstrass substitution:
$t = \tan{\frac x2}$
I got this integral:
$$6\int_0^1 \frac{(1+t^2)^3}{32t^4+3(1-t^4)^2} \,dt$$
My second try: I divided and multiplied on $\cos^4x$.
$$\int_0^{\frac{\pi}2} f \,dx = \int_0^{\frac{\pi}2} \frac1{(2\tan^4x + \frac3{\cos^2x})\cos^4x} \,dx$$
$t = \tan x, dt = \frac{dx}{\cos^2x}, \cos^2x = \frac1{\tan^2 x +1}$
I got:
$$ \int_0^\infty \frac{t^2+1}{2t^4 + 3t^2 + 3} \,dt$$
Nothing of this helps me to calculate the area by getting primitive function.
| The indefinite integral of your function is given by $$-{\frac {\ln \left( -\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}\tan
\left( x \right) +\sqrt {2}\sqrt {3}+2\, \left( \tan \left( x
\right) \right) ^{2} \right) \sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2
}}{120}}+{\frac {\ln \left( -\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}
\tan \left( x \right) +\sqrt {2}\sqrt {3}+2\, \left( \tan \left( x
\right) \right) ^{2} \right) \sqrt {3}\sqrt {2\,\sqrt {2}\sqrt {3}-3
}}{60}}-1/30\,{\frac {2\,\sqrt {2}\sqrt {3}-3}{\sqrt {4\,\sqrt {2}
\sqrt {3}+6}}\arctan \left( {\frac {-\sqrt {2\,\sqrt {2}\sqrt {3}-3}
\sqrt {2}+4\,\tan \left( x \right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}}
\right) }+1/30\,{\frac { \left( 2\,\sqrt {2}\sqrt {3}-3 \right)
\sqrt {2}\sqrt {3}}{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {
\frac {-\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x
\right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}} \right) }+1/3\,{\frac {
\sqrt {2}\sqrt {3}}{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {
\frac {-\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x
\right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}} \right) }+{\frac {\ln
\left( \sqrt {2}\sqrt {3}+\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}
\tan \left( x \right) +2\, \left( \tan \left( x \right) \right) ^{2}
\right) \sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}}{120}}-{\frac {\ln
\left( \sqrt {2}\sqrt {3}+\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}
\tan \left( x \right) +2\, \left( \tan \left( x \right) \right) ^{2}
\right) \sqrt {3}\sqrt {2\,\sqrt {2}\sqrt {3}-3}}{60}}-1/30\,{\frac {
2\,\sqrt {2}\sqrt {3}-3}{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan
\left( {\frac {\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan
\left( x \right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}} \right) }+1/30\,
{\frac { \left( 2\,\sqrt {2}\sqrt {3}-3 \right) \sqrt {2}\sqrt {3}}{
\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {\frac {\sqrt {2\,
\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x \right) }{\sqrt {4\,
\sqrt {2}\sqrt {3}+6}}} \right) }+1/3\,{\frac {\sqrt {2}\sqrt {3}}{
\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {\frac {\sqrt {2\,
\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x \right) }{\sqrt {4\,
\sqrt {2}\sqrt {3}+6}}} \right)}
$$
solvable by the tan-half angle substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Complicated rational number functional equation: $f \left( x + \frac{y}{x} \right) = f(x) + \frac{f(y)}{f(x)} + 2y$
Let $\mathbb{Q}^+$ denote the set of positive rational numbers. Let $f : \mathbb{Q}^+ \to \mathbb{Q}^+$ be a function such that
$$f \left( x + \frac{y}{x} \right) = f(x) + \frac{f(y)}{f(x)} + 2y$$
for all $x,$ $y \in \mathbb{Q}^+$. Find all possible values of $f \left( \frac{1}{3} \right)$.
If I substitute in $x=y$, then I get $f(x+1)-f(x)=2x+1$. This suggests that $f(x)=x^2$ works, and one possible value of $f \left( \frac{1}{3} \right)$ is $\frac{1}{9}$. Did I miss anything?
| From $f(x+1)-f(x)=2x+1$, we can deduce that $f(x)=x^2+f(1)-1$ for all $x\in\mathbb{Z}^+$.
Note that we also have $f(1+\frac x1)=f(1)+\frac{f(x)}{f(1)}+2x$.
Therefore, $f(x+1)=f(x)+2x+1=\dfrac{f(x)}{f(1)}+2x+f(1)$.
Hence, $f(x)\left(1-\dfrac{1}{f(1)}\right)=f(1)-1$ for $x\in\mathbb{Q}^+$.
As $f$ is not constantly zero, $f(1)=1$. So, $f(x)=x^2$ for $x\in\mathbb{Z}^+$.
$f(3+\frac{1}{3})=f(3)+\frac{f(1)}{f(3)}+2(1)=3^2+\frac{1}{3^2}+2=11+\frac19$
$f(2+\frac{1}{3})=f(3+\frac{1}{3})-2(2+\frac13)-1=5+\frac49$
$f(1+\frac{1}{3})=f(2+\frac{1}{3})-2(1+\frac13)-1=1+\frac79$
$f(\frac{1}{3})=f(1+\frac{1}{3})-2(\frac13)-1=\frac19$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Deceptive intersection of polar curves Consider the graphs of $r=\cos 2\theta$ and $r=\cos\theta -1$.
I would like to find all the intersections for the two graphs.
Equating both and solve I only manage to find $(-1,\frac{\pi}{2}),(-1,\frac{3\pi}{2}),(-\frac{1}{2},\frac{\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3})$.
However, there seem to be other intersections as well as the two graphs may not meet at the same value of $\theta$.
Is there a way to solve for the other points of intersection?
| $\cos 2\theta = \cos\theta - 1\\
2\cos^2 \theta - 1 = \cos\theta - 1\\
\cos\theta(2\cos\theta - 1) = 0\\
\theta = \frac {\pi}{2},\frac {3\pi}{2}, \frac {\pi}{3}, \frac {5\pi}{3}$
However, the curves will also overlap when $r(\theta)$ in one curve equals $-r(\theta + \pi)$ in the other.
$\cos 2\theta = \cos\theta + 1\\
2\cos^2 \theta - 1 = \cos\theta + 1\\
2\cos^2 \theta - \cos\theta - 2$
$\cos \theta = \frac {1 - \sqrt {17}}{4}$
Which will give two more points.
The curves $r = \cos \theta - 1$ and $r=\cos\theta + 1$ are identical curves.
And the curves intersect at $r = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Closed Form of $a_n = \int_0^1 \ln(1+x^n) dx$ I want to know the closed form of :
$$a_n = \int_0^1 \ln(1+x^n)dx, \quad \forall n \in \mathbb{N}$$
I found :
$$0<a_n<\frac{1}{n+1}, \quad \lim_{n\to\infty} a_n =0$$
I started from
\begin{align}
&\ln (1+x)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+1}}{k+1} \\
& \Rightarrow \ln(1+x^n)=\sum_{k=0}^\infty \frac{(-1)^k x^{nk+n}}{k+1} \\
& a_n = \int_0^1 \ln(1+x^n)dx = \sum_{k=0}^\infty \frac{(-1)^k}{k+1}\int_0^1 x^{n+nk}dx \\
&=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{n+nk+1} \\
&=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{(n+1)(k+1)-k}
\end{align}
But I stucked here. Is there any closed (or approximated) from exist?
These are some results for litte $n$ :
\begin{align}
& a_1 = 2\ln 2 - 1 \\
& a_2 = \ln2 - 2 + \frac{\pi}{2} \\
& a_3 = 2\ln 2 - 3 + \frac{\pi \sqrt{3}}{3} \\
\end{align}
| Integrate by parts
$$a_n=\int_0^1 \ln({x^n+1})dx= -n+\ln2+ \int_0^1 \frac n{x^n+1}dx$$
With the roots
$x_k= e^{i\frac{(2k-1)\pi}n} $ for $x^n+1=0$, integrate
\begin{align}
\int_0^1 \frac{n}{x^n+1}dx & = -\sum_{k=1}^n\int_0^1 \frac{x_k}{x-x_k} dx
= -\sum_{k=1}^nx_k\ln(1-x_k^{-1})\\
&= -\sum_{k=1}^n x_k \left[i \frac{n-2k+1}{2n}+ \ln \left(2 \sin\frac{(2k-1)\pi}{2n}\right) \right]
\end{align}
Per $\sum_{k=1}^nx^k=0$ and the symmetry of $x^k$ to arrive at the close-form
$$\int_0^1 \frac n{x^n+1}dx =\sum_{k=1}^{[\frac n2]} ( \pi-\theta_k )\sin\theta_k -\cos\theta_k \ln\sin^2\frac{\theta_k}2,\>\>\>
\theta_k= \frac{2k-1}{n} \pi$$
Listed below are the results for the first few $n’s$
\begin{align}
& a_2 = - 2 + \ln 2 + \frac\pi2\\
& a_3 = -3 +2\ln2 + \frac\pi{\sqrt3}\\
& a_4 = -4 +\ln 2+ \frac\pi{\sqrt2}+\sqrt2\ln(1+\sqrt2)\\
& a_5 = -5 +2\ln2 +\frac\pi{\sqrt{10}}\sqrt{5+\sqrt5}+\sqrt5\ln\frac{1+\sqrt5}2\\
& a_6 =-6+\ln2 + \pi +\sqrt3\ln(2+\sqrt3) \\
& a_7 =-7+\ln2 + \frac\pi2\csc\frac\pi7
+\frac{\ln\csc^2\frac\pi{14}}{\sec\frac\pi7}
+\frac{\ln\csc^2\frac{3\pi}{14}}{\sec\frac{3\pi}7}
+\frac{\ln\csc^2\frac{5\pi}{14}}{\sec\frac{5\pi}7}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test It's simple to evaluate the sum as follows
\begin{align*}
\sum_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)&=\lim_{n \to \infty}\sum_{k=1}^{n}\left[\left(\sqrt{k+2}-\sqrt{k+1}\right)-\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\sum_{k=1}^{n}\left(\sqrt{k+2}-\sqrt{k+1}\right)-\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\left(\sqrt{n+2}-\sqrt{2}\right)-\left(\sqrt{n+1}-1\right)\right]\\
&=\lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}+\sqrt{n+2}}+1-\sqrt{2}\right)\\
&=1-\sqrt{2}.
\end{align*}
But how to determine the convergence directly by comparison and without evaluating?
| \begin{align} \sqrt{n+2}-2\sqrt{n+1}+\sqrt{n} &= \frac{(\sqrt{n+2}+\sqrt{n})^2-(2\sqrt{n+1})^2}{\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n}} = \\
&= \frac{2(\sqrt{n+2}\sqrt{n}-n-1)}{\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n}} = \\
&= \frac{2\big((\sqrt{n+2}\sqrt{n})^2-(n+1)^2\big)}{(\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n})(\sqrt{n+2}\sqrt{n}+n+1)}= \\
&= \frac{-2}{(\sqrt{n+2}+2\sqrt{n+1}+\sqrt{n})(\sqrt{n+2}\sqrt{n}+n+1)}= \\
&= \frac{1}{n^{3/2}}\frac{-2}{(\sqrt{1+\frac2n}+2\sqrt{1+\frac1n}+1)(\sqrt{1+\frac2n}+1+\frac1n)}\end{align}
$$ \lim_{n\rightarrow\infty}\frac{-2}{(\sqrt{1+\frac2n}+2\sqrt{1+\frac1n}+1)(\sqrt{1+\frac2n}+1+\frac1n)} = -\frac14 $$
so, by comparison test
$$ \sum_{n=1}^\infty (\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}) \text{ is convergent} \Leftrightarrow \sum_{n=1}^\infty \frac{1}{n^{3/2}} \text{ is convergent} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Why is my solution to $ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 5y = 10\sin x $ wrong? This is what I did to find out the general solution :-
$$ (D^2 - 2D + 5)y = 10\sin x \\
\therefore \text{Auxiliary equation is } m^2 - 2m + 5 = 0 \\
\rightarrow m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4.1.5}}{2.1} \\
= \frac{2 \pm \sqrt{4 - 20}}{2} \\
= \frac{2 \pm 4i}{2} = 1 \pm 2i \\
\therefore \text{Complementary function = } e^x(C_1\cos 2x + C_2 \sin 2x) \\
\text{Now, particular integral } = \frac{10 \sin x}{D^2 - 2D + 5} \\
= \frac{10 \sin x}{-1^2 -2D + 5} = \frac{10 \sin x}{4 - 2D} = \frac{5 \sin x}{2 - D} \\
= \frac{5\sin x}{2(1-\frac{D}{2})} = \frac{5}{2}\left(1-\frac{D}{2}\right)^{-1}\sin x \\
= \frac{5}{2}\left(1 + \frac{D}{2} + ...\right)\sin x \\
= \frac{5}{2}\left(\sin x + \frac{D}{2}\sin x\right) = \frac{5}{2}\left(\sin x + \frac{\cos x}{2}\right) = \frac{5}{4}(2\sin x + \cos x) \\
\text{Now, } y = \text{CF + PI} \\
\therefore y = e^x(C_1\cos 2x + C_2 \sin 2x) + \frac{5}{4}(2\sin x + \cos x) $$
But the answer given in the book is :-
$$ y = e^x(C_1 \cos 2x + C_2 \sin 2x) + 2 \sin x + \cos x $$
| Here's how to get the right answer with your method.
After you render
$PI=10(1-\dfrac{D}{2})^{-1}(\sin x)$
you then apply a difference of squares factorization to get
$(1-\dfrac{D}{2})^{-1})=(1+\dfrac{D}{2})(1-\dfrac{D^2}{4})^{-1}$
and then observe that when $D$ operates on $\sin x$ then $D^2=-1$. So
$(1-\dfrac{D}{2})^{-1}=(1+\dfrac{D}{2})(1+\dfrac{1}{4})^{-1}=\color{blue}{\dfrac{4}{5}}(1+\dfrac{D}{2})$
thus getting the factor of $4/5$ you missed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $-\frac {x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$ where $f(x)= \frac{1-e^{-x}}{x}$ Let $$\begin{cases}
f(x)= \frac{1-e^{-x}}{x}, & \text{if $x > 0$} \\
f(0)=1
\end{cases}$$
I have to prove that $\forall x \in \Bbb{R_+}-\frac {x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$ so i can show that $f'_d(0)=-\frac{1}{2}$
I tried using the mean value inequality but it didn't get me anywhere.
| We know that Taylor series for $e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}...$ now $f(x)\approx \frac{1-(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24})}{x}=\frac{x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}}{x}=1-\frac{x}{2}+\frac{x^2}{6}-\frac{x^3}{24}$ thus $-\frac{x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\frac{3}{2}}}{8a^3}+\frac{3b^2\sqrt{b^2-4ac}}{8a^3}-\frac{3b\left(b^2-4ac\right)}{8a^3}$$ multiplying by $ 8a^3$ and simplifying we get $$4a^3=-b^3+3abc+\left(b^2-4ac\right)^{\frac{1}{2}}\left(b^2-ac\right)$$ Now $a$ , $b$ and $c$ are integers and the left hand side is an integer and so must the right hand side be and this happens only if the term $b^2-4ac$ is a perfect square. Let $b^2-4ac = m^2$. This implies that $$x=\frac{m-b}{2a}$$ Substituting $ x =2^{\frac{1}{3}}$ and multiplying by $2$ yields $$2^{\frac{4}{3}}=\frac{m-b}{a}$$ so $2=\left(\frac{m-b}{a}\right)^{\frac{3}{4}}$ which implies that all powers of two can be written in this form. choosing $ 2^4=16$ shows that $16^{\frac{1}{3}}=\frac{h}{a}\ $ where $h=m-b$. Now all I have to do is prove that $16^{\frac{1}{3}} = 2\ \left(2^{\frac{1}{3}}\right)$ is an irrational number. Now $h=2a\left(2^{\frac{1}{3}}\right)$. the number $a$ is the product of some primes and this product must contain a $\left(2^{\frac{r}{3}}\right)$ where $ r \equiv 2 mod(3)$ but that would mean that $r$ is not divisible by $3$ which means that the term $\left(2^{\frac{r}{3}}\right)$ can't exist in the prime factorization of any integer which shows that the initial assumption ( that $a$ , $b$ and $c$ are integers) is wrong.
The same argument works for $x=-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$ but the signs are different.
edit: Guys this is proof verification. please don't suggest alternative methods without mentioning why the proof is wrong.
| Hint:
$$-c^3=(ax^2+bx)^3=4a^3+2b^3+6ab(-c)$$
Observe that $c$ is even,
use https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Alternate way of computing the probability of being dealt a 13 card hand with 3 kings given that you have been dealt 2 kings We are dealt 13 cards from a standard 52 card deck. If $A$ is the event where we are dealt at least two kings and $B$ is the event where we are dealt at least 3 kings, we want to know $P(B|A)$. I believe the correct way to do this is to calculate the probability of being dealt a hand with each number of kings separately as follows:
$\displaystyle \frac{{4 \choose 3}{48 \choose 10} + {4 \choose 4}{48 \choose 9}}{{4 \choose 2}{48 \choose 11} + {4 \choose 3}{48 \choose 10} + {4 \choose 4}{48 \choose 9}} \approx .17$.
However, it also makes sense to me that if we know we have been dealt 2 kings, it doesn't matter where in our hand they are, the $P(B|A)$ should be the same as the probability of getting dealt either 1 or 2 kings in an 11 card hand from a 50 card deck with two kings in it as follows:
$\displaystyle \frac{{2 \choose 1}{48 \choose 10} + {2 \choose 2}{48 \choose 9}}{{50 \choose 11}} \approx .4$
(Or compute $1-p$, where $p$ is the probability of getting no kings in an 11 card hand from a deck with 50 cards and only 2 kings.)
What is the issue with my logic here?
| Your first answer is a correct application of Bayes Rule. If, as you say, $A$ is the event "at least 2 kings" and $B$ is the event "at least 3 kings", then
$$P(B|A)=\frac{P(A|B) \cdot P(B)}{P(A)}$$
$$P(A|B) = 1$$
$$P(A) = \frac{\begin{pmatrix} 4\\2 \end{pmatrix}\begin{pmatrix} 48\\11 \end{pmatrix}+\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48\\9
\end{pmatrix} }{\begin{pmatrix} 52\\13 \end{pmatrix}}$$
$$P(B) = \frac{\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48 \\9 \end{pmatrix} }{\begin{pmatrix} 52\\13 \end{pmatrix}}$$
$$P(B|A) = \frac{P(B)}{P(A)} = \frac{\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48\\9 \end{pmatrix} }{\begin{pmatrix} 4\\2 \end{pmatrix}\begin{pmatrix} 48\\11 \end{pmatrix}+\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48\\9 \end{pmatrix} } $$
Your second answer, if you were to try to "apply Bayes Rule" would show the falicy. What are your events "$A$" and "$B$" in this case?
You can also answer the question with the $1-p$ type argument, but again, you need to consider what the events are, and compute Bayes' rule accordingly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that ratio of two consecutive Fibonacci numbers to converges to golden ratio by induction $$\varphi = \frac{1 + \sqrt{5}}{2}$$
We want to prove that ratio of two consecutive Fibonacci numbers approaches $\varphi$ by induction and also utilizing Newton-Raphson method for approximating $\sqrt{5}$ as a rational number with relatively prime numerators and denominators.
Let us first define the Fibonacci Sequence and then write down what we want to prove using the symbolic notation.
$$\phi_1,\phi_2 = 1$$
$$\phi_{n+2} = \phi_{n+1} + \phi_{n}$$
$$1 \le n$$
$$n \to \infty \Rightarrow \frac{\phi_{n+1}}{\phi_{n}} \to \varphi$$
First we need rational approximations for the irrational number $\sqrt{5}$, so we can connect it up to integers:
$$x_{n} = \frac{a_{n}}{b_{n}}$$
$$x_{1} = \frac{2}{1} = \frac{a_1}{b_1}$$
Where the limit as $n$ goes to infinity is $\sqrt{5}$. So according to the Newton-Raphson method we can write that,
$$x_{n+1} = x_n - \frac{F(x_n)}{F'(x_n)} = \frac{x_n^2 + 5}{2x_n}$$
Substituting the $\frac{a_n}{b_n}$ and $\frac{a_{n+1}}{b_{n+1}}$ respectively to the places of $x_n$ and $x_{n+1}$ we will get the following rational number with integer numerator and denominator,
$$\frac{a_{n+1}}{b_{n+1}} = \frac{a_n^2 + 5b_n^2}{2a_nb_n}$$
So that
$$a_{n+1} = a_n^2 + 5b_n^2$$
$$b_{n+1} = 2a_nb_n$$
Let us define
$$\varphi_n = \frac{1 + \frac{a_n}{b_n}}{2} = \frac{a_n + b_n}{2b_n}$$
$$n \to \infty \Rightarrow \varphi_{n} \to \varphi$$
For $n = 1$ (the first case):
$$\varphi_1 = \frac{1 + \frac{2}{1}}{2} = \frac{3}{2}$$
Numerator and denominator are two consecutive Fibonacci numbers, respectively the $\phi_4$ and $\phi_3$. Now here goes our induction hypothesis for $1 \le n$:
if numerator and denominator of $\varphi_n$ are two consecutive Fibonacci numbers, respectively $\phi_{3\cdot2^{n - 1} + 1}$ and $\phi_{3\cdot2^{n - 1}}$ (as the $\phi_{4}$ and $\phi_{3}$ are), then the numerator and denominator of $\varphi_{n+1}$ will be again two consecutive Fibonacci numbers, respectively $\phi_{3\cdot2^{n} + 1}$ and $\phi_{3\cdot2^{n}}$.
$$\varphi_1 = \frac{3}{2} = \frac{\phi_4}{\phi_3} = \frac{\phi_{3\cdot2^{0} + 1}}{\phi_{3\cdot2^{0}}}$$
$$\varphi_2 = \frac{13}{8} = \frac{\phi_7}{\phi_6} = \frac{\phi_{3\cdot2^{1} + 1}}{\phi_{3\cdot2^{1}}}$$
$$\varphi_3 = \frac{233}{144} = \frac{\phi_{13}}{\phi_{12}} = \frac{\phi_{3\cdot2^{2} + 1}}{\phi_{3\cdot2^{2}}}$$
$$...$$
If we can prove that, we will be proved that ratio of two consecutive Fibonacci numbers approaches $\varphi$, of course if i have no mistake to here.
Here is our assumption:
$$\varphi_n = \frac{a_n + b_n}{2b_n} = \frac{\phi_{3\cdot2^{n - 1} + 1}}{\phi_{3\cdot2^{n - 1}}}$$
And here is what we want to prove (i rewritten and manipulated the statement using previous definitions of $a_{n+1}$ and $b_{n+1}$):
$$\varphi_{n+1} = \frac{(a_n + b_n)^2 + (2b_n)^2 }{(2b_n)^2} = \frac{\phi_{3\cdot2^{n} + 1}}{\phi_{3\cdot2^{n}}}$$
$$\varphi_{n+1} = \varphi_{n}^2 + 1 = \frac{\phi_{3\cdot2^{n} + 1}}{\phi_{3\cdot2^{n}}}$$
It gets really cumbersome for me after this point. Anyone could please give me hints or point out my mistakes if there is?
| With $\phi_n:=\dfrac{F_{n+1}}{F_n}$,
$$F_{n+2}=F_{n+1}+F_n\iff \phi_{n+1}=1+\frac1{\phi_{n}}.$$
Then if the sequence of $\phi_n$ converges, it converges to a root of
$$p=1+\frac1p.$$
As all terms are positive, convergence must be to the positive root.
As one can observe, the $\phi_n$ are alternating around $\phi$ and getting closer and closer. We can show that
$$|\phi_{n+1}-\phi|<|\phi_n-\phi|.$$
Indeed
$$|\phi_{n+1}-\phi|=\left|1+\frac1{\phi_{n}}-\phi\right|=\left|\frac1{\phi_{n}}-\frac1\phi\right|=\frac{|\phi_n-\phi|}{\phi_{n}\phi}<|\phi_n-\phi|.$$
As $\phi_n>1$, then the distance to $\phi$ is at least divided by $\phi$ (in fact by nearly $\phi^2$) on every iteration, and linear convergence is guaranteed.
| {
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"url": "https://math.stackexchange.com/questions/3252226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why is row $n = 2^x$ in Pascal's triangle have all even numbers except the $1$'s? In row $n = 2^x$, $x$ being a positive integer, in the Pascal's triangle, all entries except the two $1$'s in extreme left and right are even.
I tried to prove but I couldn't.
Here is my try:-
Every entry is of the form $\frac{(2^x)!}{(k!)([2^x]-k)!}$
I counted the no.of 2's in the prime factorisation of $(2^x)!$ in the following way:-
No.of multiples of $2 = \frac{2^x}2 = 2^{x-1}$
No.of multiples of $4 = \frac{2^x}{2^2}= 2^{x-2}$
Similarily upto no.of multiples of $2^x = 1$.
So total no.of 2's in prime factorisation is
$$
2^{x-1} + 2^{x-2} + \cdots + 2 + 1 = 2^x-1
$$
But I cannot prove that the no.of 2's in the denominator of each entry will be less than $2^{x-1}$.
Can I get some hints/help?
Thank you.
| First note that in row 2 of Pascal's triangle is $1,2,1$, so the statement is true for $n=2$. Now let $x$ be arbitrarily chosen and suppose the statement holds for $n=2^{x}$.
In particular this means that for
$$(1+x)^{2^{x}}=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n-1}x^{n-1}+a_{n}x^{n}$$
we have $a_{1},...,a_{n-1}$ are all even numbers and $a_{0}=a_{n}=1$. So
$$(1+x)^{2^{x+1}}=b_{0}+b_{1}x+b_{2}x^{2}+...+b_{2n-1}x^{2n-1}+b_{2n}x^{2n}$$
$$=(1+a_{1}x+a_{2}x^{2}+...+a_{n-1}x^{n-1}+x^{n})^{2}$$
$$=1+x^{2n}+\sum^{n-1}_{i=1}a_{i}^{2}x^{2i}+\sum_{0\leq i<j\leq n}2a_{i}a_{j}x^{i+j}$$
Since for $0\leq i<j\leq n$ we find that $0<i+j<2n$ we find that $b_{0}=b_{2n}=1$ and for $0<k<2n$ we find that $b_{k}$ is the sum of even numbers and therefore even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Matrix Equation $B^3+C^3=\begin{pmatrix}-1 & 1\\ 0 & -2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$B^3+C^3=\begin{pmatrix}-1 & 1\\ 0 & -2\end{pmatrix}?$$
I try to use $$X^2-Tr(X)X+det X\cdot I_2=0_2$$ but there are $B$ and $C$, I don't still obtain anything.
thanks.
| The easiest part. We assume that $B,C$ are upper-triangular.
Then $b_{1,1}^3+c_{1,1}^3=-1$; according to Fermat (up to order) $b_{1,1}=-1,c_{1,1}=0$.
And $b_{2,2}^3+c_{2,2}^3=-2$, that implies $b_{2,2}=c_{2,2}=-1$ (see $(a+1)^3-a^3$ when $a>0$).
By identification, we conclude that there is an integer $p$ s.t. (up to order)
$B=\begin{pmatrix}-1&p\\0&-1\end{pmatrix},C=\begin{pmatrix}0&1-3p\\0&-1\end{pmatrix}$.
EDIT. There are solutions $(B,C)$ in $M_2(\mathbb{R})$ s.t. $B^3,C^3$ are not upper-triangular; for example
$B\approx \begin{pmatrix}0&-2.0107\\1&-1.8503\end{pmatrix},C\approx\begin{pmatrix}1.1426&-2.7184\\1&0\end{pmatrix}$.
Thus the question is: can we find such matrices in $M_2(\mathbb{Z})$ ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$ over the integers.
Solve $$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$ over the integers.
You can probably guess by now... This problem is adapted from a recent competition.
If there are any other solutions, please post them below. I have provided one if you want to check out.
| Not suprisingly,
$$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$
$$\iff (x^2y^2 + y^3 + y^2) - (4x^2y + 4y^2 + 4y) + (4x^2 + 4y + 4) = 3$$
$$\iff y^2(x^2 + y + 1) - 4y(x^2 + y + 1) + 4(x^2 + y + 1) = 3$$
$$\iff (y^2 - 4y + 4)(x^2 + y + 1) = 3 \iff (y - 2)^2(x^2 + y + 1) = 3$$
$$\implies (y - 2)^2 \mid 3 \implies (y - 2)^2 \in \{\pm 1, \pm 3\}$$
Having said that, $(y - 2)^2$ is a perfect square. $\implies (y - 2)^2 = 1 \implies \left\{ \begin{align} x^2 + y + 1 = 3\\ y - 2 = \pm 1 \end{align} \right.$.
$$\implies x^2 = -(y - 2) = \mp 1 \implies x^2 = -(y - 2) = 1 \iff \left\{\begin{align} x = \pm 1\\ y = 1\end{align} \right.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
Question-Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
I was playing around with the formulae
$$(a+b+c)^3=a^3 +b^3+c^3+3(a+b)(b+c)(c+a)$$ and,
$$a^3 +b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
and noted that if $(a+b+c)\equiv0$(mod 6)$\implies a^3 +b^3+c^3\equiv3abc$(mod 6). Now I am not sure how to show $3abc\equiv 0$(mod6), and even doing that, we have only half of the proof because then we need to prove that the converse is also true.
| Now, use that $a+b+c$ is an even number.
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3257174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Least value of $ab-cd$ is
If $a,b,c,d\in\mathbb{R}$ and $a^2+b^2=c^2+d^2=4$
$ac+bd=0$. Then least value of $ab-cd$ is
Plan
$(ac+bd)^2+(ab-cd)^2=a^2c^2+b^2d^2+a^2b^2+c^2d^2\geq 0$
$a^2(b^2+c^2)+d^2(b^2+c^2)\geq 0$
How do i solve it Help me please
| Let $a=2\sin\alpha$, $b=2\cos\alpha$, $c=2\sin\beta$ and $d=2\cos\beta$.
Then $4\sin\alpha\sin\beta+4\cos\alpha\cos\beta=0$ and hence $\cos(\alpha-\beta)=0$. Therefore, $\alpha=(n+\frac12)\pi+\beta$.
\begin{align*}
ab-cd&=2\sin2\alpha-2\sin2\beta\\
&=4\cos(\alpha+\beta)\sin(\alpha-\beta)\\
&=4\cos\left(\left(n+\frac12\right)\pi+2\beta\right)\sin\left(\left(n+\frac12\right)\pi\right)
\end{align*}
The least possible value is $-4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is $ \dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}$?
How to evaluate the integral $$ \dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}?$$
Here $f(z)=\dfrac{1}{z \cos z}$ has poles at $0$ and $\frac{\pm π}{2}$ .
Residue at $z=0 $ is 1 and residues at remaining poles add up to give 0. So the integral using Cauchy integral formula is $2π(4-π)$.
I think I am wrong. How to get the integral?
| It's not true that residues add up to $0$. We have
$$ \cos z = \sin(\frac{\pi}{2}-z) = \sin(\frac{\pi}{2}+z)$$
so
$$ {\rm Res}_{z=0} \frac{1}{z\cos z} = \lim_{z\rightarrow 0}\frac{1}{\cos z} = 1$$
$$ {\rm Res}_{z=\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow \frac{\pi}{2}}\frac{(z-\frac{\pi}{2})}{z \sin(\frac{\pi}{2}-z)} = -\frac2\pi$$
$$ {\rm Res}_{z=-\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow -\frac{\pi}{2}}\frac{(z+\frac{\pi}{2})}{z \sin(\frac{\pi}{2}+z)} = -\frac2\pi$$
and $$ \sum {\rm Res} = 1 - \frac{4}{\pi} = \frac{\pi-4}{\pi}$$
so the final result is $$ \frac{i}{4-\pi}\cdot 2\pi i\frac{\pi-4}{\pi} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $? This comes from a bigger problem :-
$$ \text{Evaluate } \int\frac{dx}{1+x^4} $$
After making $ \int \frac {dx}{1+x^4} = \frac{dx}{(1+x^2)^2 - (\sqrt{2}x)^2} $ and then applying partial fraction method, I got :-
$$ \int \frac{dx}{1 + x^4} =\frac{1}{2 \sqrt{2}} \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} dx - \frac{1}{2 \sqrt{2}} \int \frac{x - \sqrt{2}}{x^2 - \sqrt{2}x + 1} dx $$
Now, to the first integral, I tried making a u-substitution:-
$$ \text{Let }x^2 + \sqrt{2}x + 1 = u \\
\frac{du}{dx} = 2x + \sqrt{2} \\
\implies du = (2x + \sqrt{2}) dx \\ $$
As you can see, it is not the same as the numerator, which is $$ (x + \sqrt{2}) dx $$
Any hints on how to proceed ?
| Hint: We have $$x^2+\sqrt{2}x+1=\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}$$
| {
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"question_score": "5",
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Given $\cos (\theta)$ and $\sin (\theta)$, find $2\theta$ I am working on my scholarship exam. I worked through almost final step but got my answer wrong. Could you please have a look?
If $\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ with
$0\leq\theta<2\pi$, it follows that $2\theta = ..... \pi$
What I have got is below:
$\sin(2\theta)=2\sin\theta\cos\theta$
Then, $\sin(2\theta)=-\frac{1}{\sqrt{2}}$
Hence, $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$ (quadrant 3 or 4)
$\theta=\frac{5\pi}{8}$ or $\frac{7\pi}{8}$
Since $\cos\theta$ is positive and $\sin\theta$ is negative, $\theta$ should be in quadrant 4 but my $\theta$'s are not. So I cannot use my $2\theta$ as a final answer.
However, the answer key provided is $\theta=\frac{15\pi}{4}$, Why do you think that is the case? How did they get to this answer? Please help.
| The given information has $\cos$ positive and $\sin$ negative so $\theta$ is in the 4th quadrant. Makes, $2\theta$ in the 8th quadrant that is $7\pi/4\leq 2\theta \leq 4\pi$; which angle in the 8th quadrant gives $\sin 2\theta = \frac{-1}{\sqrt 2}$? if you want to add more that $ \cos 2\theta = \frac{1}{\sqrt 2}$.
How to find angle the 8th quadrant, $\theta$ in 4th quadrant $+ 2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264029",
"timestamp": "2023-03-29T00:00:00",
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If $ax^2 + 2hxy + by^2 = 1$, prove that $d^2y/dx^2 = \frac{h^2 - ab}{ (hx + by)^3}$
If $ax^2 + 2hxy + by^2 = 1$, prove that $\frac{d^2y}{dx^2} = \frac{h^2-ab}{(hx + by)^3}$.
I did try to simplify it, by taking derivatives on both the sides, and taking the $\frac{dy}{dx}$ common, and applying the quotient rule.
The answer I get is $$\frac{d^2y}{dx^2} = 4ha + \frac{2ab}{(2h+b)^2}.$$
Edit 1: Sorry, it's now $4ha/(2h+b)^3$. I forgot to take the 2nd derivative the first time. But it's still not the answer.
| \begin{align*}
\frac{d}{dx}\left(ax^2+2hxy+by^2\right) & =0\\
2ax+ 2h\underbrace{\left(x \, \frac{dy}{dx}+y\right)}_{\text{product rule}} + 2by\frac{dy}{dx}&=0\\
\frac{dy}{dx}&=-\frac{ax+hy}{hx+by}\\
\frac{d^2y}{dx^2}&=-\left[\frac{(hx+by)\frac{d}{dx}(ax+hy)-(ax+hy)\frac{d}{dx}(hx+by)}{(hx+by)^2}\right]\\
\frac{d^2y}{dx^2}&=-\left[\frac{(hx+by)\left(a+h\frac{dy}{dx}\right)-(ax+hy)\left(h+b\frac{dy}{dx}\right)}{(hx+by)^2}\right]\\
\frac{d^2y}{dx^2}&=-\left[\frac{\left(h^2x-abx\right)\frac{dy}{dx}+aby-h^2y}{(hx+by)^2}\right]\\
\frac{d^2y}{dx^2}&=-(h^2-ab)\left[\frac{x\frac{dy}{dx}-y}{(hx+by)^2}\right]\\
\frac{d^2y}{dx^2}&=-(h^2-ab)\left[\frac{-x\left(\frac{ax+hy}{hx+by}\right)-y}{(hx+by)^2}\right]\\
\frac{d^2y}{dx^2}&=(h^2-ab)\left[\frac{\color{red}{ax^2+2hxy+by^2}}{(hx+by)^3}\right]\\
\frac{d^2y}{dx^2}&=(h^2-ab)\left[\frac{\color{red}{1}}{(hx+by)^3}\right]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality for optimization. I found this problem:
Find the minimum value of:
$f(a,b,c)=a+b+c+\frac{9a}{(a+2b+c)^2}+\frac{9b}{(b+2c+a)^2}+\frac{9c}{(c+2a+b)^2}$
Where $a,b,c >0$.
After looking for some minimum values, I think the lowest one is for the equality case $a=b=c=3/4$.Obviously,
$f(a,a,a)=3a+\frac{27}{16a} \geq 2\sqrt{\frac{81}{16}}=9/2$
I got inspired from the equality case to "substitute" $a=(a+b+c)/3$ and then having this inequality:
$f(a,b,c) \geq 3(\frac{a+b+c}{3})+\frac{27}{16((a+b+c)/3)}=3(\frac{a+b+c}{3})+\frac{81}{16(a+b+c)}$.
Subtracting on the both sides by $a+b+c$ and simplifying by $9$, we get
$\frac{a}{(a+2b+c)^2}+\frac{b}{(b+2c+a)^2}+\frac{c}{(c+2a+b)^2} \geq \frac{9}{16(a+b+c)}$
This last inequality is really the problem
I`m pretty sure it is true after giving some arbitrary values and also looking at cases when some variables tend to infinity.
Still, I cannot prove it. I tried Titu's lemma but end up with an invalid inequality.
I'd like to see a nice proof for this inequality.
| Let $a+2b+c = x $ etc.
$$
f(a,b,c)=\frac{x+y+z}{4}-\frac{9(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac{9z}{(x)^2}+\frac{9x}{(y)^2}+\frac{9y}{(z)^2}
$$
By the rearrangement inequality,
$\frac{9z}{(x)^2}+\frac{9x}{(y)^2}+\frac{9y}{(z)^2}\ge \frac{9x}{(x)^2}+\frac{9y}{(y)^2}+\frac{9z}{(z)^2} $ as the minimum occurs when numerators and denominators are sorted in the same order. So we have
$$
f(x,y,z)\ge\frac{x+y+z}{4}+ 9(\frac1{x}+\frac1{y}+\frac1{z})-\frac{9(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})
$$
Now this is a symmetric convex function in the variables, where the minimum occurs when all variables equal, i.e.
$$
f(x,y,z)\ge\frac{3x}{4}+ \frac{27}{x}-\frac{81}{4 x} = \frac{3x}{4}+ \frac{27}{ 4 x} = \frac{9}{4}(\frac{x}3 + \frac3{x})
$$
The minimum of this occurs at $x=3$ so
$$
f(x,y,z)\ge \frac{9}{2}
$$
Now indeed that happens, as you already computed, at $a=b=c=3/4$ (which corresponds to $x=y=z=3$), so the bound is tight and this is your minimum.
Considering your last inequality, you can proceed in the same fashion:
$$
\frac{a}{(a+2b+c)^2}+\frac{b}{(b+2c+a)^2}+\frac{c}{(c+2a+b)^2} \geq \frac{9}{16(a+b+c)}\\
\leftrightarrow
-\frac{(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac{z}{(x)^2}+\frac{x}{(y)^2}+\frac{y}{(z)^2}- \frac{9}{4(x+y+z)} \geq 0
$$
Now the LHS can be bound by rearrangement, and you have the stronger condition
$$
-\frac{(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac1{x}+\frac1{y}+\frac1{z}- \frac{9}{4(x+y+z)} \geq 0
$$
Again using symmetry, an even stronger condition is
$$
-\frac{9}{4 x}+\frac3{x}- \frac{3}{4x} \geq 0
$$
but the LHS is identically zero and so this holds. Again, verify that $a=b=c=3/4$ solves the starting inequality, which says that the starting inequality is tight.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding angle in the isosceles triangle There is the following triangle. I infer that the triangle is isosceles. But I cannot go further.
| $AN = 5$ (Pythagoras), so $AC = 10 = AB$, which is how you can conclude $\triangle ABC$ is isosceles.
$\angle BAC = \angle MAN = \arctan\frac 34$, so $\angle CBA = \frac 12(180^{\circ} - \arctan \frac 34)$
$\angle MBN = \arctan \frac 36 = \arctan \frac 12$
$\therefore x = \frac 12(180^{\circ} - \arctan \frac 34) - \arctan \frac 12$
and you get a very nice answer in degrees.
EDIT:
The answer is $45^{\circ} = \frac{\pi}{4}$, where the latter is expressed in radian measure.
This answer used trigonometry. It is being left intact because the OP found it useful. However, there was a request from the OP in comments to see if a "by-hand" calculator free solution can be obtained with my approach. Indeed it can, and here is how:
The major problems to resolve are to try to reduce the arctangent terms to single inverse trigonometric ratios.
Rearranging the above,
$\displaystyle x = 90^{\circ} - (\frac 12\arctan\frac 34 + \arctan\frac 12)$
$\displaystyle \tan x = \tan (90^{\circ} - (\frac 12\arctan\frac 34 + \arctan\frac 12))= \frac 1{\tan(\frac 12\arctan\frac 34 + \arctan\frac 12)}$
which uses the identity $\tan (90^{\circ} - \theta) = \cot \theta = \frac 1{\tan\theta}$
Let's now focus on this expression: $\frac 12\arctan\frac 34$. This can be expressed as the arctangent of simple fraction as follows:
First let's find the tangent of the above. Note that $\tan \frac 12\theta = \frac{\sin\theta}{1+\cos\theta}$, which can be looked up in a reference, or may be proved fairly easily by considering the individual half-angle identities for sine and cosine and dividing them.
Also note that if $\theta = \arctan \frac34$ then $\sin\theta = \frac 35$ and $\cos\theta = \frac 45$, which is quite trivial to show by considering the $3-4-5$ special right triangle.
Putting those together:
$\displaystyle \tan \frac 12\arctan\frac 34 = \frac{\sin (\arctan\frac 34)}{1+ \cos (\arctan\frac 34)} = \frac{\frac 35}{1+\frac 45} = \frac 13$
Therefore $\frac 12\arctan\frac 34 = \arctan \frac 13$
So $\displaystyle \tan x =\frac 1{\tan(\frac 12\arctan\frac 34 + \arctan\frac 12)} = \frac 1{\tan(\arctan \frac 13 + \arctan \frac 12)}$
Apply the angle addition formula for tangent to get:
$\displaystyle \tan x = \frac 1{\frac{\tan(\arctan \frac 13) + \tan(\arctan \frac 12)}{1- \tan(\arctan \frac 13)\tan(\arctan \frac 12)}} = \frac 1{\frac{\frac 13+\frac 12}{1- \frac 13\cdot \frac 12}} = \frac 11 = 1$
and the only acute angle solution for $\tan x = 1$ is $x = 45^{\circ} = \frac{\pi}{4}$.
I hope this is a more "satisfying" solution for the OP. As I said, it's possible to do it without a calculator, just harder.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the easiest and simplest way to bound $xy$ Let $-1\le x\le 2$ and $-2\le y\le 3$. I have to bound $xy$. I can do it case by case. E.g
Case1
\begin{cases}
0\le -x\le 1\\
0\le -y\le2
\end{cases} Taking the product term by term yields $\boxed{0\le xy\le 2}$
Case2
\begin{cases}
0\le -x\le 1\\
0\le y\le3
\end{cases} Again I take the product to have $\boxed{-3\le xy\le0}$
Case3
\begin{cases}
0\le x\le 2\\
0\le -y\le2
\end{cases} then $\boxed{-4\le xy\le 0}$
Case 4
\begin{cases}
0\le x\le 2\\
0\le y\le3
\end{cases} then $\boxed{0\le xy\le6}$
Combining all these stuffs gives finally $-4\le xy\le 6$.
Does someone have an alternative method, more simpler and not as long as this thing above.
| In this case, it is easy to see that the least value of the product $xy$ is either $-1×3$ or $-2×2.$ Similarly, the greatest value of this product in the rectangle is either $-1×-2$ or $2×3,$ so that we have $$-4\le xy\le 6.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum \frac {xy}{\sqrt{ (x^2+z^2)(y^2+z^2)}}\leq \frac {3}{2} $ Prove that $$\sum \frac {xy}{\sqrt{ (x^2+z^2)(y^2+z^2)}}\leq \frac {3}{2} $$
I tried to apply Cauchy Schwarz but then I obtain Nesbitt's inequality with the wrong sign.
| By AM-GM
$$\sum_{cyc}\frac{xy}{\sqrt{(x^2+z^2)(y^2+z^2)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{x^2}{x^2+z^2}+\frac{y^2}{y^2+z^2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{x^2}{x^2+z^2}+\frac{z^2}{z^2+x^2}\right)=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a finite sum Find the following sum :
$\frac{1}{1^2}+ \frac{1+2}{1^2 +2^2}+\frac{1+2+3}{1^2 +2^2 +3^2}+...+\frac{1+2+3+....n}{1^2 +2^2 +3^2 +....+n^2}$
I tried to unify the denominators but i got a complicated form so i could not go on ?
| We obtain using the finite summation formulas
\begin{align*}
\sum_{k=1}^nk&=\frac{1}{2}n(n+1)\qquad\text{and}\qquad\sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1)\\
\end{align*}
and the harmonic numbers $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$
\begin{align*}
\color{blue}{\sum_{j=1}^n\frac{1+2+\cdots+j}{1^2+2^2+\cdots+j^2}}
&=\sum_{j=1}^n\frac{\frac{1}{2}j(j+1)}{\frac{1}{6}j(j+1)(2j+1)}\\
&=3\sum_{j=1}^n\frac{1}{2j+1}\\
&=3\left(\sum_{j=1}^{2n+1}\frac{1}{j}-\sum_{j=1}^{n}\frac{1}{2j}\right)\\
&\,\,\color{blue}{=3\left(H_{2n+1}-\frac{1}{2}H_n\right)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
solve the equation in $\Bbb C $ \begin{array}{l}{\text {Solve in } \mathbb{C}}: \\ {x^{2}+\left(\frac{x}{x+1}\right)^{2}=3} \\ {\text { my try: }} \\ {x^{2}(x+1)^{2}+x^{2}=3(x+1)^{2}} \\ {x^{2}\left(x^{2}+2 x+1\right)-3\left(x^{2}+2 x+1\right)+x^{2}=0} \\ ({x^{2}-3 )\left(x^{2}+2 x+1\right)+x^{2}=0} \\ {x^{4}+2 x^{3}-x^{2}-6 x-3=0} \\ {\text { Now what should I do? }}\end{array}
| A strange solution that uses symmetry reasoning to transform the problem.
The argument $\frac{x}{1+x}$ stick out and can be set as a new variable. In fact, let's throw in a minus sign for good measure (especially if you recognize it as one of the functions that is inverse of itself).
$$u=-\frac{x}{1+x}\quad \Rightarrow \quad x=-\frac{u}{1+u}$$
If you put this in, you get the same equation for $u$, meaning that both $x$ and $u$ are solutions of this equation. Therefore, instead of one fourth order equation, we have two completely symmetric equations for two variables:
$$\begin{align}x^2+u^2&=3 \\
xu+u+x&=0\quad (\text{from }u=-\frac{x}{1+x})
\end{align}
$$
Adding twice the second equation to the first one gives us
$$(x+u)^2+2(x+u)-3=0$$
which is factorized by hand:
$$\boxed{x+u=\{-3,1\}}$$
Subtracting twice the second equation gives
$$(x-u)^2=3+2(u+x)$$
and using the just derived solutions,
$$\boxed{x-u=\pm\sqrt{\{-3,5\}}}$$
Averaging the two boxed results gives you all four solutions:
$$x=\frac{-3\pm i\sqrt{3}}{2}$$
$$x=\frac{1\pm \sqrt{5}}{2}$$
Note that computing $u$ must give you the same results (that was the premise of this method), and indeed, the $\pm$ in the second boxed equation ensures that swapping the sign just gives you the associated $u$ to each $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A conjecture formula: $\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)$ By the help of Mathematica numeral calculations, I find the following formula holds
$$\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)\quad ?$$
$m>1$ is a positive integer. But I can't prove it.
| This is hardly a starter, but the following representation showing some commonalities of LHS and RHS might be useful.
The RHS can be written as
\begin{align*}
m\log\left(\frac{m}{m-1}\right)&=m\log\left(\frac{1}{1-\frac{1}{m}}\right)\\
&=-m\log\left(1-\frac{1}{m}\right)\\
&\,\,\color{blue}{=m\sum_{n=1}^\infty \frac{1}{nm^n}}
\end{align*}
The LHS can be written as
\begin{align*}
\sum_{n=1}^\infty&\frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n\\
&=\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn}{n}\left(\frac{(m-1)^{m-1}}{m^{m-1}}\right)^n\\
&=\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn}{n}\left(1-\frac{1}{m}\right)^{n(m-1)}\\
&\,\,\color{blue}{=m\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn-1}{n-1}\left(1-\frac{1}{m}\right)^{n(m-1)}}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/3277540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$
Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$.
This is simple to check by listing and the only exceptions are for $N = 2,3,8$. I am looking for a simple proof if possible. This problem arises from solving polynomials with constraints on the range of coefficients.
| By definition of floor,
$$
\sqrt{N}-1\lt\left\lfloor\sqrt{N}\right\rfloor\le\sqrt{N}\tag1
$$
Therefore,
$$
\begin{align}
1\le\frac{N}{\left\lfloor\sqrt{N}\right\rfloor^2}&\lt\frac{N}{\left(\sqrt{N}-1\right)^2}\\
&=2-\frac{\left(\sqrt{N}-2\right)^2-2}{\left(\sqrt{N}-1\right)^2}\\[6pt]
&\lt2\tag2
\end{align}
$$
when $N\gt\left(2+\sqrt2\right)^2=6+4\sqrt2\doteq11.656854$, so we just need to verify $(2)$ for $N\in\{9,10,11\}$ by hand. Because $\frac99,\frac{10}9,\frac{11}9\lt2$, we have shown that for $N\gt8$,
$$
1\le\frac{N}{\left\lfloor\sqrt{N}\right\rfloor^2}\lt2\tag3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Develop into Laurent series around $-i$: $\frac{z +1}{(z-i)^2} .$ Develop into Laurent series around $-i$:
$$\frac{z +1}{(z-i)^2} .$$
I don't even know how to start. Any hint helps! Thanks.
| Since it's supposed to be around $z={-i}$, it means that it needs to have form
$$ \sum_{n=-\infty}^\infty c_n (z-(-i))^n = \sum_{n=-\infty}^\infty c_n (z+i)^n $$
Let's then start with writing the function in the form using $z+i$ expressions:
$$ \frac{z+1}{(z-i)^2}= \frac{(z+i)+(1-i)}{((z+i)-2i)^2}$$
Let $w=z+i$, $a=1-i$, $b=2i$. We need to find the expansion $$ \frac{w+a}{(w-b)^2} = \sum_{n=-\infty}^\infty c_n w^n$$
We'll do it by calculating the Laurent series of $\frac{1}{(w-b)^2}$ first and then multiplying it by $w+a$. There will be two expansions, one for $|w|<|b|$ and another for $|w|>|b|$.
For $|w|<|b|$ we have (from the formula on the sum of a geometric series)
$$\frac{1}{b-w} = \frac{1}{b} \frac{1}{1-\frac w b}= \frac1 b \sum_{n=0}^\infty\left( \frac{w}{b}\right)^n = \sum_{n=0}^\infty \frac{w^n}{b^{n+1}}$$
We then get $$ \frac{1}{(b-w)^2} = \frac{d}{dw} \frac{1}{b-w} = \frac{d}{dw} \sum_{n=0}^\infty \frac{w^n}{b^{n+1}} = \sum_{n=1}^\infty \frac{nw^{n-1}}{b^{n+1}} $$
(the term for $n=0$ can be skipped in the last sum, as it would be $0$ anyway). Finally we have
$$ \frac{w+a}{(b-w)^2} = \sum_{n=1}^\infty \frac{nw^{n-1}(w+a)}{b^{n+1}} = \sum_{n=1}^\infty \Big(\frac{nw^n}{b^{n+1}} + \frac{anw^{n-1}}{b^{n+1}}\Big) = \\
= \sum_{n=1}^\infty \frac{nw^n}{b^{n+1}} + \sum_{n=0}^\infty \frac{a(n+1)w^n}{b^{n+2}} = \frac{a}{b^2}+ \sum_{n=1}^\infty \Big(\frac{n}{b^{n+1}}+\frac{a(n+1)}{b^{n+2}}\Big)w^n$$
We have then (after substituting $w=z+i$, $a=1-i$, $b=2i$):
$$ \frac{z+1}{(z-i)^2}= \frac{1-i}{(2i)^2}+ \sum_{n=1}^\infty \Big(\frac{n}{(2i)^{n+1}}+\frac{(1-i)(n+1)}{(2i)^{n+2}}\Big)(z+i)^n = \\
= \frac{-1+i}{4}+ \sum_{n=1}^\infty \frac{(n+1)+i(n-1)}{(2i)^{n+2}}(z+i)^n$$
This expansion is valid for $|w|<|b|$, that is $|z+i|<2$.
The other expansion for $|w|>|b|$ we get by expanding first
$$\frac{1}{b-w} = -\frac{1}{w} \frac{1}{1-\frac b w}= -\frac1 w \sum_{n=0}^\infty\left( \frac{b}{w}\right)^n = \sum_{n=0}^\infty \frac{-b^n}{w^{n+1}} = -\sum_{n=-\infty}^{-1} b^{-n+1} w^n$$
Then we follow as before:
$$ \frac{1}{(b-w)^2} = \frac{d}{dw} \frac{1}{b-w} = -\sum_{n=-\infty}^{-1} nb^{-n+1} w^{n-1} $$
$$ \frac{w+a}{(b-w)^2} = -\sum_{n=-\infty}^{-1} nb^{-n+1} w^{n-1}(w+a) = -\sum_{n=-\infty}^{-1} nb^{-n+1} (w^n+aw^{n-1}) = \\ = -\sum_{n=-\infty}^{-1} nb^{-n+1} w^n-\sum_{n=-\infty}^{-2} (n+1)b^{-n}aw^n = \\
= w^{-1}-\sum_{n=-\infty}^{-2} \frac{nb+(n+1)a}{b^{n}}w^n$$
so
$$ \frac{z+1}{(z-i)^2}= (z+i)^{-1}-\sum_{n=-\infty}^{-2} \frac{(n+1)+i(n-1)}{(2i)^{n}}(z+i)^n$$
This expansion is valid for $|z+i|>2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find GCD of those two complex polynomials? I have a polynomial
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3$$
I want to check if this has repeated roots. To do so, I'll find greatest common divisor (euclidean algorithm) of $f(x)$ and its derivative $f'(x)$.
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3 = i(x^6-3x^4+3x^2-1)+(x^6+3x^4+3x^2+1)-8x^3$$
$$f'(x) = i(6x^5-12x^3+6x)+(6x^5+12x^3+6x)-24x^2$$
I know I should find the GCD as $(x-i)(x-1)$ but this is where I'm stuck. What would the next step be?
| It's best not to split the polynomials up using $i$, but collect terms in powers of $x$.
$$ \eqalign{f_0 = f(x) &= \left( 1+i \right) {x}^{6}+ \left( 3-3\,i \right) {x}^{4}-8\,{x}^{3}+
\left( 3+3\,i \right) {x}^{2}+1-i \cr
f_1 = f'(x) &= \left( 6+6\,i \right) {x}^{5}+ \left( 12-12\,i \right) {x}^{3}-24\,{x
}^{2}+ \left( 6+6\,i \right) x
}
$$
The remainder of $f_0$ on division by $f_1$ is
$$ f_2 = f_0 - (x/6) f_1 = \left( 1-i \right) {x}^{4}-4\,{x}^{3}+ \left( 2+2\,i \right) {x}^{2}+
1-i
$$
Then take remainder of $f_1$ on division by $f_2$, etc.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rationalize the denominator of $\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$ I have to rationalize the denominator of $A =
\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$. I multiplied the fraction by $\frac{\sqrt{7\sqrt{3}-4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$, and $A=\frac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$. Then I multiplied by $\frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}+4\sqrt{5}}$, and $A=\frac{7\sqrt{201}+4\sqrt{335}}{67}$. Can someone tell me if there is a better solution and whether I am right?
| For positive $a$ and $b$, we have $\sqrt{a}/\sqrt{b}=\sqrt{a/b}$. So you can concentrate on
$$
\frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}-4\sqrt{5}}=\frac{(7\sqrt{3}+4\sqrt{5})^2}{49\cdot3-15\cdot5}=\frac{(7\sqrt{3}+4\sqrt{5})^2}{67}
$$
Thus you're done with
$$
\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}=\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Exponential equations, solve for x I'm preparing for uni entrance exam. I've been struggling with this problem for about 90 minutes, tried everything I could think of. Can anybody explain how to solve for x step by step?
$$3^{\frac{x-1}{2}}-2^{\frac{x+1}{3}}=2^{\frac{x-2}{3}}+3^{\frac{x-3}{2}}$$
| Hint: Your equation can be written as: $$3^{\frac{x-1}{2}} - 3^{\frac{x-3}{2}} = 2^{\frac{x-2}{3}} + 2^{\frac{x+1}{3}}.$$ Because $$3^{\frac{x-1}{2}} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3+2}{2}} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2} +\frac{2}{2}} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2} +1} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2}}3^1 - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2}}(3-1),$$ and
$$2^{\frac{x-2}{3}} + 2^{\frac{x+1}{3}} = 2^{\frac{x-2}{3}} + 2^{\frac{x-2+3}{3}}= 2^{\frac{x-2}{3}} + 2^{\frac{x-2}{3}+\frac{3}{3}}=2^{\frac{x-2}{3}} + 2^{\frac{x-2}{3}+1}=2^{\frac{x-2}{3}} + 2^{\frac{x-2}{3}}2^1 = 2^{\frac{x-2}{3}}(1+2),$$
we can rewrite the first equation above as $$3^{\frac{x-3}{2}}(3-1) = 2^{\frac{x-2}{3}}(1+2).$$ Simplifying, we get $$3^{\frac{x-5}{2}} = 2^{\frac{x-5}{3}}.$$ Can you see why this implies $x=5$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial.....but it didnt help either.
| Write $$x^4 - 12 x^2 + 8 x + 3=q(x)(x^2 + kx + 1)$$
then $q$ must be quadratic, so $q(x)= ax^2+bx+c$. By expanding and comparing the coeficients you will get $k$ (clearly $a=1$ and $c=3$)...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Volume generated by rotating around y-axis, curve $y=x^3$ and the lines $y=0$ and $x=2$
Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$
I first found what $x=2$ would be in terms of y.
$$y= (2)^3
\\ = 8$$
And in terms of y, the original equation becomes:
$$y=x^3
\\ x= y^{1 \over 3}$$
So,
$$V= \int^8_0 (y^{1 \over 3})^2 \pi dy
\\ = \pi \int_0^8 y^{2\over 3}dy
\\ =\pi \bigg[ y^{5\over 3} ({3 \over 5}) \bigg]^8_0
\\ = (8^{5 \over 3})({3 \over 5})\pi
\\ = {96 \over 5} \pi$$
Therefore the answer is ${96 \over 5} \pi$ units $^3$
However, the answer is supposed to be ${64 \over 5} \pi$ units${^3}$
What went wrong?
| The shell method would be easiest to use here:
$$
V=2\pi\int_{0}^{2}x\cdot x^3\,dx=2\pi\int_{0}^{2}x^4\,dx=
2\pi\frac{x^5}{5}\bigg|_{0}^{2}=\\
\frac{2\pi}{5}\left(2^5-0^5\right)=
\frac{2\pi}{5}\cdot 32=\frac{64\pi}{5}\ \text{cubic units}.
$$
Or you can use the washer method which is going to be a bit messier. In that case, you're subtracting the volume of a cylinder of radius $2$ and height $8$ from the volume you get by revolving the curve $x=\sqrt[3]{y}$ around the $y$-axis:
$$
V=\int_{0}^{8}\left(\pi\cdot 2^2-\pi\left(\sqrt[3]{y}\right)^2\right)\,dy=
\pi\int_{0}^{8}\left(4-y^{\frac{2}{3}}\right)\,dy=\frac{64\pi}{5}\ \text{cubic units}.
$$
Wolfram Alpha check.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine, whether the matrix is diagonalizable or not Suppose, we have the following matrix $$A=\begin{pmatrix} 1 & 1 & 4\\ 2 & 2 & 1\\ 2 & 4 & 4\end{pmatrix}$$ over the field $\mathbb{Z}_5$.
To show, that $A$ is diagonalizable, we need to show, that the dimension of the sum of all eigenspaces equals the dimension of the matrix.
Therefore, we will calculate the eigenvalues, eigenvectors and get the eigenspaces. We need to calculate the characteristic polynomial with
$$\chi_A(\lambda)=\begin{vmatrix} 1-\lambda & 1 & 4\\ 2 & 2-\lambda & 1\\ 2 & 4 & 4-\lambda \end{vmatrix}=4\lambda^3+2\lambda^2+4,\quad \lambda\in\mathbb{Z}_5$$
In order to compute the eigenvalues, I will need to find the roots of $4\lambda^3+2\lambda^2+4$.
\begin{align}
4\lambda^3+2\lambda^2+4&=0 && \mid &+1\\
4\lambda^3+2\lambda^2&=1 && \mid &\\
\lambda^2(4\lambda+2)&=1
\end{align}
We have $\lambda_1=1,\lambda_2=4,\lambda_3=1$
Now, we insert them into $\chi_a(\lambda)$
$\lambda = 1$:
$$
\begin{pmatrix}
0 & 1 & 4\\
2 & 1 & 1\\
2 & 4 & 3
\end{pmatrix}
\iff\dots \iff
\begin{pmatrix}
2 & 1 & 1\\
0 & 1 & 4\\
0 & 0 & 0
\end{pmatrix}
$$
$L:=\{t\cdot(4,1,1)\mid t\in \mathbb{Z}_5\}$
One of the (infinity) eigenvectors of $\lambda = 1$ is $v=(4,1,1)$ and the eigenspace is $E_A(\lambda = 1):=\{t\cdot(4,1,1)\mid t\in \mathbb{Z}_5\}$
For $\lambda = 4$, I will get an eigenvector of $u=(0,0,0)$ which wouldn't work with $u\neq 0$ in the definition of eigenvalues/-vectors. Does that mean, the matrix is not diagonalizable?
| The characteristic polynomial of that matrix is
$$x^3-7x^2-14=x^3+3x^2+1=(x-1)(x+2)^2\pmod 5=(x-1)(x-3)^2\pmod5$$
so we only need the dimension of the eigenvalue $\;x=-2=3\pmod5\;$:
$$\begin{cases}3x+y+4z=0\\{}\\
2x+4y+z=0\\{}\\
2x+4y+z=0\end{cases}\implies x+2y+3z=0$$
and since this last equation is a plane (in $\;\left(\Bbb Z_5\right)^2)\;$ , the matrix is diagonalizable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The number of triples that sum to a constant Problem:
How many triples are there of the form $(x_0,x_1, x_2)$ where $x_0 \in I$, $x_1 \in I$, $x_2\in I$ $x_0 \geq 0$,
$x_1 >= 0$, $x_2 >= 0$ and $n = x_0 + x_1 + x_2$ where $n \in I$?
Answer:
Let $c(n)$ be the number of tuples we can have for a given $n$. For $n = 0$, the only valid triple is $(0,0,0)$, hence $c(0) = 1$.
For $c(1) = 3$, the set of valid triples is: $(0,0,1 ), (0,1,0), (0,0,1)$ Hence $c(1) = 3$.
For $c(2) = 6$, the set of valid triples is:
$$ (1,0,1 ), (0,1,1 ), (0,0,2 ), (1,1,0), (0,2,0), (0,0,2)$$
Hence $c(2) = 6$.
Using the information on this URL:
How many $k-$dimensional non-negative integer arrays $(x_1,\cdots,x_k)$ satisfies $x_1+x_2+\cdots+x_k\le n$
I find the answer to be:
$$ c(n) = {{n+3}\choose{3}} - {{n+2}\choose{3}} $$
$$ c(n) = \frac{(n+3)!}{3!n!} - \frac{(n+2)!}{3!(n-1)!} $$
$$ c(n) = \frac{(n+3)(n+2)(n+1) - (n+2)(n+1)(n)}{6} $$
$$ c(n) = \frac{(n+2)(n+1)(n+3 - n)}{6} $$
$$ c(n) = \frac{3(n+2)(n+1)}{6} $$
$$ c(n) = \frac{(n+2)(n+1)}{2} $$
Do I have that right?
Thanks,
Bob
| Yes - this is correct. This can also be found by defining a helper function $b(n)$, which counts the number of ways to write $n$ as a sum of two numbers. Clearly, $b(n) = n+1$, as the first value, $v$, can be anything from $0$ to $n$, while the second value is $n - v$.
The function $c(n)$ is then equal to $$\sum_{i = 0}^{n}b(n-i)$$ This is because the first value, $i$, can be anything from $0$ to $n$. The number of ways to write the remaining two numbers so that the total sum is $n$ is $b(n-i)$. Plugging in the formula finds $$c(n) = \sum_{i = 0}^{n} (n-i+1) = n(n+1) - \frac{n(n+1)}{2} + n+1 = \frac{(n+2)(n+1)}{2} $$
which is the same formula you arrived at.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a sum that seems telescopable Calculate the limit:
$$\lim\limits_{n\to\infty}\sum\limits_{k=6}^{n}\frac{k^3-12k^2+47k-60}{k^5-5k^3+4k}$$
All I managed to do was to write the inside term as $\frac{(k-3)(k-4)(k-5)}{k(k+1)(k+2)(2-k)(1-k)}$ which leads me to think of telescoping but i dont know how.
I think that comparing it to the analog product somehow makes sense or using the fact that $\frac{1}{(2-k)(1-k)} = \frac{1}{1-k} - \frac{1}{2-k}$ helps.
| Hint: Try to find the partial fraction decomposition of the rational function "inside". You will indeed find a sort of telescopic sum with a few first terms and then a few terms that tend to 0 when n tends to infinity
Hint 2: $\frac{k^3−12k^2+47k−60}{k(k-1)(k+1(k-2)(k+2)}=\frac{\left(-15\right)}{k}+\frac{\frac{13}{3}}{k+1}+\frac{\frac{59}{3}}{k-1}+\frac{\left(-\frac{97}{12}\right)}{k-2}+\frac{\left(-\frac{11}{12}\right)}{k+2}$
then: $\sum_{k=6}^n \frac{k^3−12k^2+47k−60}{k(k-1)(k+1(k-2)(k+2)} = \sum_{k=6}^n \frac{\left(-15\right)}{k}+\frac{\frac{13}{3}}{k+1}+\frac{\frac{59}{3}}{k-1}+\frac{\left(-\frac{97}{12}\right)}{k-2}+\frac{\left(-\frac{11}{12}\right)}{k+2}=
(-15)\sum_{k=6}^n \frac{1}{k} + (13/3) \sum_{k=6}^n \frac{1}{k+1} + (59/3)\sum_{k=6}^n \frac{1}{k-1}
+ (-97/12) \sum_{k=6}^n \frac{1}{k-2} + (-11/12) \sum_{k=6}^n \frac{1}{k+2}$
and I let you finish the job.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find triple summation rel in a closed form $S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$
Evaluate $\displaystyle S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$
My attempt :
Let $$A=\sum_{k=1}^{m}\frac{1}{k(k+1)}
=\sum_{k=1}^{m}\left( \frac1{k}-\frac1{k+1} \right) = \frac{m}{m+1}$$
and a second sum :
$$B=\sum_{m=1}^{n}\frac{1}{(m+1)^{2}}$$
from here how I can complete ??
| Just for your curiosity.
We could also have a cloased form for the partial sum
$$S_p=\sum_{n=1}^{p}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$$since
$$\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}=\frac{1}{(m+1)^2 n (n+1)}$$
$$\sum_{m=1}^{n}\frac{1}{(m+1)^2 n (n+1)}=\frac{\pi ^2-6-6 \psi ^{(1)}(n+2)}{6 n (n+1)}$$
$$S_p=\sum_{n=1}^{p}\frac{\pi ^2-6-6 \psi ^{(1)}(n+2)}{6 n (n+1)}=\frac{12 (p+1)-\pi ^2 (p+2)+6 (p+2) \psi ^{(1)}(p+2)}{6 (p+1)}$$
Now, using, for large values of $q$, the expansion
$$\psi ^{(1)}(q)=\frac{1}{q}+\frac{1}{2 q^2}+\frac{1}{6 q^3}+O\left(\frac{1}{q^5}\right)$$ and continuing with Taylor series for large values of $p$
$$S_p=\left(2-\frac{\pi ^2}{6}\right)+\frac{1-\frac{\pi ^2}{6}}{p}+\frac{\pi ^2-3}{6
p^2}-\frac{2+\pi ^2}{6 p^3}+\frac{10+\pi ^2}{6 p^4}+O\left(\frac{1}{p^5}\right)$$
For illustration purposes, using $p=10$, the exact value is
$$S_{10}=\frac{42308191}{140873040}\approx 0.300328$$ while the above truncated expansion gives
$$\frac{125690-10909 \pi ^2}{60000}\approx 0.300375$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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It is impossible for $(x-1)^2+x^2+(x+1)^2$ to be a perfect square Prove that it is impossible for three consecutive squares to sum to another perfect square.
I have tried for the three numbers $x-1$, $x$, and $x+1$.
| The smallest solution you may think of:
Note that $(x-1)^2+x^2+(x+1)^2=3x^2+2\equiv 2\pmod{3}$, but $$y^2\equiv 0,1\pmod{3}$$for all integers $y$. Therefore, $(x-1)^2+x^2+(x+1)^2$ can never be a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $(X,Y)$ be distributed over $[0,1]\times[0,1]$ according to $ f(x,y)=6xy^2$. Find $P(XY^3 \leq \frac{1}{2})$ . Let $(X,Y)$ be distributed over $[0,1]\times[0,1]$ according to $f(x,y)=6xy^2 $. Find $P(XY^3 \leq \frac{1}{2})$.
I think this the double integration of f (X,Y) but i'm confused in limits.
| Let $(X,Y)$ be a random vector with density function $f_{(X,Y)}:\mathbb R^2 \to \Bbb R,$ $f_{(X,Y)}(x,y) = 6xy^2 \chi_{[0,1]^2}(x,y) $.
Let $\mu $ be distribution of $(X,Y)$, that is for $A \in \mathcal B(\mathbb R^2) $ (borel set), $ \mu(A) = \int_A f_{(X,Y)}(x,y) d\lambda_2(x,y) $
By that we have $\mathbb P(XY^3 \leq \frac{1}{2}) = \mathbb P(\{ \omega \in \Omega : (X,Y)(\omega) \in B\}) = \mu(B) $, where
$B=\{(x,y) \in \mathbb R^2: x \leq \frac{1}{2y^3} \}$
So now we integrate:
$$\mu(B) = \int_B f_{(X,Y)}(x,y) d\lambda_2(x,y) = \int_0^\frac{1}{\sqrt[3]2}\int_0^1 6xy^2 dxdy + \int_\frac{1}{\sqrt[3]2}^1\int_0^\frac{1}{2y^3} 6xy^2 dxdy = $$
$$ = \frac{1}{2} + \frac{3}{4}\int_\frac{1}{\sqrt[3]2} ^1 \frac{dy}{y^4} = \frac{1}{2} - \frac{1}{4}(1 - 2) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$
So, $ \mathbb P(XY^3 \leq \frac{1}{2}) = \frac{3}{4} $
We had to find point $y= \frac{1}{\sqrt[3]{2}}$ (where $y^3 = \frac{1}{2}$), because $ x\leq \min\{\frac{1}{2y^3},1\} $ so for $y\in[0,\frac{1}{\sqrt[3]2}) $, $x \in [0,1)$, and for $y \in [\frac{1}{\sqrt[3]2},1] $, $x \in [0,\frac{1}{2y^3}]$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area between the following curves and their asymptotes. The curve is given by the polar equation $ r = a( \sec \theta + \cos \theta) $
At $\theta = \pi /2, r$ goes to infinity, so asymptotes should be at $\theta = \pi /2$
To find the area I integrated $2$ $ \frac{a^{2}(\sec \theta + \cos \theta)^{2}}{2} $ ($= r^{2}/2 \times 2$, since curve is symmetric about $x$ axis) with respect to $\theta$ from $0$ to infinity and got infinity.
But answer is given $5 \pi a^{2}/2$
Where did I go wrong$?$
| We are given the curve
$$ r = a( \sec \theta + \cos \theta) \tag{1}$$
but we are not given the asymptote.
It is clearly a vertical asymptote of the form $x=$constant, but what is the constant?
So what is happening to $x=r\cos\theta$ as $\theta\to\frac{\pi}{2}$?
We can find this out by multiplying equation (1) by $\cos\theta$ to obtain
$$ r\cos\theta = a( 1 + \cos^2 \theta) $$
Since $\cos(\pi/2)=0$ we see that as $\theta\to\frac{\pi}{2}$, $r\cos\theta\to a$.
So the vertical asymptote of the curve is $x=a$.
In polar coordinates this is $r\cos\theta=a$
So $r=a\sec\theta$ is the polar equation of the asymptote of the function.
So the area between the function and its asymptote can be found by the following:
$$ \int_0^{\pi/2} a^2(\sec\theta+\cos\theta)^2-a^2\sec^2\theta\,d\theta $$
We shall see that $\dfrac{5\pi a^2}{2}$ is not the correct answer.
\begin{eqnarray}
\int_0^{\pi/2} a^2(\sec\theta+\cos\theta)^2-a^2\sec^2\theta\,d\theta&=&
a^2\int_0^{\pi/2}2+\cos^2\theta\,d\theta\\
&=&a^2\int_0^{\pi/2}2+\frac{1+\cos2\theta}{2}\,d\theta\\
&=&a^2\left[\frac{5}{2}\theta+\frac{1}{4}\sin(2\theta)\right]_0^{\pi/2}\\
&=&\frac{5\pi a^2}{4}
\end{eqnarray}
| {
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The function $g: \Bbb{Z}_{10} → \Bbb{Z}_{10}$ is defined by $g(x) ≡ 7x + 2 \mod 10$. Find its inverse function. Solution: From $x = g(y) ≡ 7y + 2 \mod 10$, we obtain
$$y ≡ 7^{−1}(x−2)≡3(x−2) \mod 10$$.
Hence, the inverse function $g^{−1}: \Bbb{Z}_{10} → \Bbb{Z}_{10}$ is defined by $g^{−1}(x) ≡ 3(x − 2) \mod 10$.
I don't understand the steps on how $7^{−1}(x−2)$ is logically equivalent $3(x−2) \mod 10$. I'm missing something, but I can't figure it out. Thanks in advance.
| $7^{-1}\pmod {10}$ refers to the class $k$ so that $7*k\equiv 1 \pmod {10}$. If such a class exists.
An $a^{-1} \pmod n$ will exist if and only if $\gcd(a,n) = 1$.
So what is $7^{-1}\pmod {10}$. What is the solution to $7*k\equiv 1 \pmod {10}$.
We could use Bezout and Euclid's alogorithm to solve $7k + 10a = 1$ but these are such simple and small numbers it's easier to use brute force.
The multiples of $7$ are $7,14\equiv 4, 21\equiv 1, 28\equiv 8, 35\equiv 5, 42\equiv 2, 49\equiv 7, 56\equiv 6, 63\equiv 3, 70 \equiv 0$.
So $7*3 \equiv 21 \equiv 1 \pmod {10}$.
And $7^{-1} = 3\pmod{10}$.
.......
So if $y= g(x) = 7x+2\pmod {10}$ the $g^{-1}$ is the function so that $x =g^{-1}(g(x)) = g^{-1}(y)$ so to find $g^{-1}$ we solve $y = 7x+2\pmod {10}$ for $x$ in terms of $y$.
$y \equiv 7x + 2 \pmod{10}$
$7x \equiv y - 2 \pmod{10}$.
$7^{-1}*7x \equiv 7^{-1}(y-2)\pmod {10}$
$3*7x \equiv 3(y-2)\pmod{10}$
$21x\equiv 3(y-2)\pmod{10}$
$x\equiv 3(y-2)\pmod{10}$
In my a opinion we should distribute the $3(y-2)$
So $x \equiv 3y - 6 \pmod{10}$.
Furthermore I think we should, for aesthetic purposes mostly, consistently use the postive reduced residue system. $-6\equiv 4 \pmod {10}$
So
$x \equiv 3y + 4 \pmod {10}$
And so $g^{-1}(y) = 3y+4\pmod {10}$ or replacing one variable with another:
$g^{-1}(x) = 3x + 4\pmod {10}$
...
To verify this we wont to so if $g^{-1}(g(x)) = x$.
$3(7x+2) + 4 \equiv $
$21x + 6 + 4 \equiv $
$21x + 10 \equiv $
$x + 0 \equiv x\pmod{10}$.
And that $g(g^{-1}(x)) = x$.
$7(3x+4) + 2 \equiv $
$21x + 28 + 2 \equiv 21x + 30\equiv$
$x + 0 \equiv x \pmod {10}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Are there some known results for solving this kind of modular arithmetic problem? Preliminaries
Consider a number in the form:
$$x = a^n b - c,$$
where $a, b, c$ and $n$ are natural numbers. For which $n$ we can say
that $x$ is divisible by $d$ with $a$ and $d$ coprime?
WLOG, consider the following example with $a=2$, $c=1$ and $d = 3$.
\begin{array}{llll}
2^{n}b - 1 \equiv 0 &(\text{mod}~ 3) & (1)\\
2^{n}b - 1 + 3 \equiv 0 &(\text{mod}~ 3) & (\text{we add a multiple of $d=3$)}\\
2^{n}b + 2 \equiv 0 &(\text{mod}~ 3) & (\text{we divide by a number (2) non divisible by $d=3$)}\\
2^{n-1}b + 1 \equiv 0 &(\text{mod}~ 3) & (2) \\
2^{n-1}b + 1 - 3\equiv 0 &(\text{mod}~ 3) & (\text{we subtract a multiple of $d=3$)}\\
2^{n-1}b - 2\equiv 0 &(\text{mod}~ 3) & (\text{we divide by a number (2) non divisible by $d=3$)}\\
2^{n-2}b - 1\equiv 0 &(\text{mod}~ 3) & (3).
\end{array}
Reiterating this schema until I get $2^{n-n} = 2^0 = 1$, I conclude that
*
*If $n$ is even (see equation (1) and (3)), then I need to prove
$$b - 1\equiv 0 ~(\text{mod}~ 3). $$
*If $n$ is odd (see equation (2)), then I need to prove
$$b + 1\equiv 0 ~(\text{mod}~ 3). $$
Now, depending on $b$, I can find the answer to the initial question.
My question
I am wondering if the presented procedure has a name, or I can arrive to the conclusion by simply applying some known results that I'm missing.
| Since $2+1=3$, it follows that $2+1\equiv0\pmod3$; i.e., $2\equiv-1\pmod3$.
Therefore $2^n\equiv(-1)^n\pmod3$.
$(-1)^n$ is $1$ when $n$ is even and $-1$ when $n$ is odd.
Therefore $2^n\equiv1\pmod3$ when $n$ is even and $2^n\equiv-1\pmod3$ when $n$ is odd.
So if $2^nb-1\equiv0\pmod3$ then $b-1\equiv0$ when $n$ is even and $-b-1\equiv0$ when $n$ is odd;
i.e., $b\equiv1$ when $n$ is even and $b\equiv-1\equiv2\pmod3$ when $n$ is odd.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $x \in \mathbb R$ and $x>2$ then $y+ \frac{1}{y} = x$ will have real solution.
Prove that for real number $x$, if $x > 2$ then there is real number $y$ such that $y + \frac{1}{y} = x$
My attempt:
Rewriting equation, we have:
$$\tag1y + \frac{1}{y} = x$$
$$\tag2 \frac{y^2+ 1}{y} = x$$
$$\tag3 y^2+ 1 = yx$$
$$\tag4 y^2 - yx + 1 = 0$$
We know that for quadratic equations of the format $aq^2 + bq + c$ ($a,b,c$ are constants and q is variable), if discriminant is positive, then equation will have real solutions.
Let discriminant be denoted by $D$, then
$D = \sqrt{b^2- 4ac}$
Substituting our values into the equation, we have:
$D = \sqrt{x^2 - 4}$
It can be seen that if $x > 2$ then $D$ is positive, hence $y + \frac{1}{y} = x$ will have real solution.
Is it correct?
| Let $y >0$; $x>2$;
AM-GM:
$y+1/y \ge 2\sqrt{y\cdot 1/y}=2$;
Equality for $y=1/y$, i.e $y=1$;
Consider $f(y):=y+1/y$, $y \in [1,\infty)=:D$.
$f([1,\infty))= [2,\infty)$.
$f$ is continuos in its domain $D$,
MVT:
For given $x \in (2, \infty)$ there is a $y_0$ with
$f(y_0)=x$;
Hence a real solution.
Note: $f(1/y_0)=x$, is also a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why does $\int_{-\infty}^{+\infty }\arctan\left(\frac{1}{1+x^2}\right)dx$ have a real value when the indefinite integral uses $i$? WolframAlpha gives a real closed form for this definite integral:
$$\int_{-\infty}^{+\infty } \arctan\left(\dfrac{1}{1+x^2}\right)dx = \sqrt{2\left(\sqrt{2}-1\right)}\;\pi$$
Yet, the formula it gives for the indefinite integral uses $i$.
$$\int \tan^{-1}\left(\frac{1}{x^2 + 1}\right) dx = x \tan^{-1}\left(\frac{1}{x^2 + 1}\right)
+ 2 \left(
\frac{\tan^{-1}\left(\frac{x}{\sqrt{1 - i}}\right)}{(1 - i)^{3/2}}
+
\frac{\tan^{-1}\left(\frac{x}{\sqrt{1 + i}}\right)}{(1 + i)^{3/2}}
\right) + C$$
Why isn't the definite integral non-real?
| WolframAlpha, though handy, is not yet advanced to perform certain integrals satisfactorily. For less familiar integrands, it tends to fall back to complex variables, leading often to uninteresting, perhaps only symbolically useful, results.
The integral in question can be integrated in real variables explicitly
\begin{align}
\int \tan^{-1}\frac{1}{1+x^2}\ dx &
= x\tan^{-1}\frac{1}{1+x^2}\\
&\>\>\>+\sqrt{2(\sqrt{2}-1)}\tan^{-1}\frac{x^2-\sqrt{2}}{x\sqrt{2(\sqrt{2}+1)}} \\
& \>\>\>- \sqrt{2(\sqrt{2}+1)}\tanh^{-1}\frac{x^2+\sqrt{2}}{x\sqrt{2(\sqrt{2}-1)}}+C
\end{align}
which eludes WA due to limitations in its current algorithm.
| {
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integration of $\int_{-\pi}^{\pi}\cos(x)\frac{1}{\sqrt\pi}\sin(\frac{n(x+\pi)}{2})\frac{1}{\sqrt\pi}\sin(\frac{m(x+\pi)}{2})dx$ I'm working on solving the integration
$\int_{-\pi}^{\pi}\cos(x)\frac{1}{\sqrt\pi}\sin(\frac{n(x+\pi)}{2})\frac{1}{\sqrt\pi}\sin(\frac{m(x+\pi)}{2})dx$ and my results after using the integration by parts for different cases for the integers $n$ and $m$ as the following:
$$
\int_{-\pi}^{\pi}\cos(x)\frac{1}{\sqrt\pi}\sin(\frac{n(x+\pi)}{2})\frac{1}{\sqrt\pi}\sin(\frac{m(x+\pi)}{2})dx=
\begin{cases}
0,& \text{for } n\neq m \& (n-m)^2\neq4,\\
0,& \text{for } n=m\neq1,\\
\frac{1}{2},& \text{for }n=m=1,\\
-\frac{1}{2},& \text{for } (n-m)^2=4. \\
\end{cases}
$$
Is there any way to make sure I have the correct results?
appreciate any help:)
here is a case when $n=m\neq1:$
using the identity:
$sin(A)sin(B)=\frac{1}{2}[cos(A-B)-cos(A+B)]$
I get:$\int_{-\pi}^{\pi}\cos(x)\frac{1}{\sqrt\pi}\sin(\frac{n(x+\pi)}{2})\frac{1}{\sqrt\pi}\sin(\frac{m(x+\pi)}{2})dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}cos(x)cos(\frac{n-m}{2}(x+\pi))-cos(x)cos(\frac{n+m}{2}(x+\pi))dx$
using the integration by parts twice for the first term:
$\begin{align}\int_{-\pi}^{\pi}cos(x)cos(\frac{n-m}{2}(x+\pi))dx&=[\frac{2}{n-m}cos(x)sin(\frac{n-m}{2}(x+\pi))]_{-\pi}^{\pi}+\frac{2}{n-m}\int_{-\pi}^{\pi}sin(x)sin(\frac{n-m}{2}(x+\pi))dx\\&=(\frac{1}{(1-\frac{4}{(n-m)^2})})\bigg[[\frac{2}{n-m}cos(x)sin(\frac{n-m}{2}(x+\pi))]_{-\pi}^{\pi}-\frac{4}{(n-m)^2}[sin(x)cos(\frac{n-m}{2}(x+\pi))]_{-\pi}^{\pi}\bigg]\\&=0.\end{align}$
Similarly for the other term:
$\int_{-\pi}^{\pi}cos(x)cos(\frac{n+m}{2}(x+\pi))dx=0$
Hence for $n=m\neq1:$
$\int_{-\pi}^{\pi}\cos(x)\frac{1}{\sqrt\pi}\sin(\frac{n(x+\pi)}{2})\frac{1}{\sqrt\pi}\sin(\frac{m(x+\pi)}{2})dx=0$
| We have $$2\sin\frac{n(x+\pi)}2\sin\frac{m(x+\pi)}2=\cos\frac{(n-m)(x+\pi)}2-\cos\frac{(n+m)(x+\pi)}2$$ and \begin{align}2\cos\frac{(n\pm m)(x+\pi)}2\cos x&=\cos\frac{(n\pm m)(x+\pi)-x}2+\cos\frac{(n\pm m)(x+\pi)+x}2\\&=\cos\frac{(n\pm m-1)x+(n\pm m)\pi}2+\cos\frac{(n\pm m+1)x+(n\pm m)\pi}2\end{align} so $$\cos x\sin\frac{n(x+\pi)}2\sin\frac{m(x+\pi)}2\\=\\\frac14\left[\left(\cos\frac{(n-m-1)x+(n-m)\pi}2+\cos\frac{(n-m+1)x+(n-m)\pi}2\right)-\left(\cos\frac{(n+m-1)x+(n+m)\pi}2+\cos\frac{(n+m+1)x+(n+m)\pi}2\right)\right].$$ Thus $$\int_{-\pi}^\pi\cos x\sin\frac{n(x+\pi)}2\sin\frac{m(x+\pi)}2\,dx=\frac12\left[\frac{\sin\frac{(n-m-1)x+(n-m)\pi}2}{n-m-1}+\frac{\sin\frac{(n-m+1)x+(n-m)\pi}2}{n-m+1}-\frac{\sin\frac{(n+m-1)x+(n+m)\pi}2}{n+m-1}-\frac{\sin\frac{(n+m+1)x+(n+m)\pi}2}{n+m+1}\right]_{-\pi}^\pi=\frac12\left[\frac{\sin\frac{(2(n-m)-1)\pi}2}{n-m-1}+\frac{\sin\frac{(2(n-m)+1)\pi}2}{n-m+1}-\frac{\sin\frac{(2(n+m)-1)\pi}2}{n+m-1}-\frac{\sin\frac{(2(n+m)+1)\pi}2}{n+m+1}\right]-\frac12\left[\frac{1}{n-m-1}-\frac{1}{n-m+1}-\frac{1}{n+m-1}+\frac{1}{n+m+1}\right]$$ ...
| {
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The max Greatest Common Divisor of 2 digit decimal numbers Assume $a$ and $b$ are different one digit positive integer, both within range $[1,9]$. Then what is the possible maximum Greatest Common Divisor of two digits numbers $\overline{ab} ,\overline{ba}.$ ( $\overline{ab}$ means $a$ is the tens digit, $b$ is the ones digit. So $\overline{ab}=10a+b$).
A iterative way to iterate through $1-9$ of $a,b$ for 36 times can find the answer, I want to know any better mathematical way to solve the question. Thanks.
| Let $D=\gcd(\overline{ab},\overline{ba})$.
We have
$$\tag1 D\mid \overline{ab}-\overline{ba}=9(a-b),$$
$$\tag2 D\mid a\cdot \overline{ba}-b\cdot \overline{ab}=a^2-b^2=(a+b)(a-b),$$
$$\tag3 D\mid 10\cdot\overline{ab}-\overline{ba}=99a.$$
$$\tag4 D\mid 10\cdot\overline{ba}-\overline{ab}=99b,$$
$$\tag5 D\mid \overline{ab}+\overline{ba}=11(a+b).$$
From $(1)$, $D$ cannot be difvisible by any prime $>7$ (provided $a\ne b$), hence we can cast out $11$ in $(3)$, $(4)$, $(5)$, and combine these into
$$\gcd(a,b)\mid D\mid \gcd(9,a+b)\gcd(a,b). $$
Also from $(5)$, $D<18$ and in particular $27\nmid D$.
Then the power of $3$ occuring in $D$ is the same as that in $a+b$ (by the digit sum rules for divisibility by $3$ and $9$).
We conclude
$$ D=\frac{\gcd(a,b)\gcd(a+b,9)}{\gcd(a,b,9)}.$$
In order to maximize, we can make $a+b$ a multiple of $9$ only by making $a+b=9$. Then $\gcd(a,b)\mid 9$ and $D=\gcd(a+b,9)=9$.
Or we can make $a+b$ a multiple of $3$, but $3\nmid a,b$. Then $D=3\gcd(a,b)<9$.
So the maximal gcd is $9$ and happens with $18$ and $81$, $27$ and $72$, etc.
| {
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Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$ .
I am stuck here please help.
This image gives some hint, but I am unable to understand it.
| Hint: By the equation #(3) we get in #(1)
$$z-1/z-\frac{1}{z-\frac{1}{z}}=y$$
and with $z=y-\frac{1}{y}$ we get
$$3y^6-9y^4+6y^2-1=0$$
Multiplying all terms together gives
$$x^2(1-z^2)+y^2(1-z^2)+z^2(1-y^2)=0$$
substituting
$$xy=x^2-1$$
$$yz=y^2-1$$
$$xz=z^2-1$$
so we get
$$xyz(x+y+z)=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Constrained optimisation: Minimize $(x+\frac{1}{x})^2 +(y+\frac{1}{y})^2$ subject to the constrain $x+y=1$
Minimize $(x+\frac{1}{x})^2 +(y+\frac{1}{y})^2$ subject to the
constrain that $x+y=1$, where $x$ and $y$ are positive.
I used Lagrange-optimization and proved that
$x-\frac{1}{x^3} = y - \frac{1}{y^3}.$
But I don't seem to get the answer.
| Hint: Prove that $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq \frac{25}{2}$$
Your inequality is equivalent to
$$(x^2+y^2)(1+\frac{1}{(xy)^2})\geq \frac{17}{2}$$
Since we have $$x^2+y^2=1-2xy$$
we get
$$1-2xy+\frac{1}{(xy)^2}-\frac{2}{xy}\geq \frac{17}{2}$$
Substituting $$xy=t$$ the inequality above is equivalent
$$\frac{(1-4t)(2+4t+t^2)}{2t^2}\geq 0$$
This is true,since we get by AM-GM $\frac{1}{2}\geq \sqrt{xy}$
| {
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Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| Your solution is incorrect. You proved that $\sin x \cos x = \frac{1}{2}\implies\sin x =\pm\dfrac1{\sqrt2},$
but the reverse implication does not necessarily hold.
You need to look at which of your solutions satisfy the original equation. $135^o$ and $315^o$ do not.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Finding distribution of $\frac{1}{\sqrt X}$ when $X$ is uniform on $[1,9]$ I haven't been able to figure this question out and have been googling for about a good hour so thought I should post this here.
Let $X$ be a $\mathrm{uniform}(1,9)$ random variable, i.e. it has the probability density function
$$
f_X(x) =
\begin{cases}
1/8, & \text{if 1 $\le$ $x$ $\le$ 9,} \\
0, & \text{otherwise.}
\end{cases}
$$
Find the Cumulative Distribution function and Probability density function of
$$Y = \frac {1}{\sqrt{X}}.$$
My wrong attempt went as follows:
First I determined the range of $ Y $ to be $y \in [\frac13,1] $ by filling in the boundrairies 1 and 9 into $ Y $
Calculate the CDF by taking the close integral over the interval $ [1,x]$
$$ CDF: \int_1^x \frac18 du = \frac18(x-1)$$
Then I find CDF of $ Y $ $$ F_y = P(Y \le y ) = P( \frac{1}{\sqrt{X}} \le y ) = P(X \le \frac{1}{y^2} ) = \frac18(\frac{1}{y^2} -1) $$
But this solutions seems to be wrong since it doesn't add up to 1 when integrated over its range $ [\frac13,1]$
If someone could give me an explanaiting where I made an error that be great!
| Mistake is in the line $\mathbb P(\frac{1}{\sqrt{X}} \le y) = \mathbb P(X \le \frac{1}{y^2}),$ because due to $\sqrt{X},y > 0$, we have:
$ \frac{1}{\sqrt{X}} \le y \iff 1 \le \sqrt{X}y \iff \frac{1}{y} \le \sqrt{X} \iff \frac{1}{y^2} \le X,$ so it should be $P(\frac{1}{\sqrt{X}} \le y) = \mathbb P(X \ge \frac{1}{y^2})$
As you calculated $F_X(t) = \frac{x-1}{8}\chi_{[1,9]}(t) $ and correcting your error, we get $($for $y \in [\frac{1}{3},1])$:
$F_Y(y) = \mathbb P(\frac{1}{\sqrt{X}} \le y) = \mathbb P(X \ge \frac{1}{y^2}) = 1 - \mathbb P(X < \frac{1}{y^2}) = 1 - \mathbb P(X \le \frac{1}{y^2}) = 1 - F_X(\frac{1}{y^2}) = 1 - \frac{1}{8}(\frac{1}{y^2}-1) = \frac{1}{8}(9-\frac{1}{y^2}) $
So we get the CDF of $Y$: $F_Y(t) = \frac{1}{8}(9-\frac{1}{t^2})\chi_{[\frac{1}{3},1]}(t)$
Now PDF, as we know since CDF is continuosly differentiable (piecewise at least) it holds, that $f_Y(y) = F_Y'(y)$ for every point in which $F_Y'$ exists (we can put whatever we want in the rest, since $PDF$ is up to the set of measure zero).
Differentiating gives us PDF : $f_Y(y) = \frac{1}{4y^3} \chi_{(\frac{1}{3},1)}(y)$ (Note that you could have included points $\frac{1}{3}$ and/or $1$ in characteristic function.
| {
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Solving $(xy)y'= x^2+3y^2$ I am having a very frustrating time with the back book that says my answer is way off but to me everything looks fine:
\begin{align*}
(xy)y'&= x^2+3y^2\\
y' &= \frac{x^2}{xy} + \frac{3y^2}{xy}\\
y' &= \frac{x}{y} + \frac{3y}{x}\\
y' &= \frac{1}{v} + 3v\\
y' &= \frac{1 + 3v^2}{v}\\
v+\frac{dv}{dx}x &= \frac{1+3v^2}{v}\\
\frac{dv}{dx}x&= \frac{1+3v^2-v^2}{v}\\
\frac{dv}{dx}x &= \frac{1+2v^2}{v}\\
\int \frac{v}{2v^2+1}\,dv &= \int\frac{1}{x}\,dx\\
u &= 2v^2+1\\
du &= 4v\,dv\\
dv &= \frac{1}{4v}\,du\\
\int \frac{v}{u} \frac{1}{4v}\,du &= \int \frac{1}{x} \,dx\\
\int \frac{1}{4u}\,du &= \ln|x| + c\\
\frac{1}{4} \int \frac{1}{u}\,du &= \ln|x| +c\\
\frac{1}{4} \ln|2v^2 + 1| &= \ln |x| + c\\
\ln|2v^2 + 1|&= 4\ln|x|+c\\
2v^2 + 1 &= e^{4\ln|x|}e^c\\
2v^2 + 1 &= Cx^4\\
2v^2 &= Cx^4\\
v^2 &= Cx^4\\
\frac{y}{x} &= \sqrt{Cx^4}\\
y &= x\sqrt{Cx^4}
\end{align*}
However the book says the answer is $x^2 + 2y^2 = Cx^6.$ I am fairly sure there are no mistakes.
| Your solution simplifies to the cubical parabola $$y=kx^3.$$ Substituting this into the original equation with $k=1$ for simplicity shows something must be wrong despite your confidence to the contrary, for we have on the one hand $(xy)y'=x^4(3x^2)=3x^6,$ and on the other side $x^2+3y^2=x^2+3x^6,$ which are clearly not equal, off in fact by $x^2.$
The most suspicious operation is where you subsumed the $+1$ into the constant -- wrongly, it appears. There's no legitimate way you could have managed that!
| {
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Determining the range of $6^n+ 6^{-n} +3^n +3^{-n}+2$
I have to solve for range of the function $$6^n+ 6^{-n} +3^n +3^{-n}+2$$
The textbook solves it as
$$\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \tag{1}$$
i.e., $$(a-b)^2+(a-b)^2$$
which will always be greater than $6$, so the range is $(6,\infty)$ (since other two terms are squared). But, if we take
$$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2 \tag{2}$$
or $$\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 +2 \tag{3}$$
instead of $(1)$, we get that the range is $(-2,\infty)$ or $( 2,\infty)$, respectively.
So, how do we know what range is correct?
| If you'll take $$3^n+3^{-n}+6^n+6^{-n}+2=\left(\sqrt3^n+\sqrt3^{-n}\right)^2+\left(\sqrt6^n+\sqrt6^{-n}\right)^2-2\geq-2,$$ but the equality does not occur.
But in the following writing
$$3^n+3^{-n}+6^n+6^{-n}+2=\left(\sqrt3^n-\sqrt3^{-n}\right)^2+\left(\sqrt6^n-\sqrt6^{-n}\right)^2+6\geq6$$ the equality occurs for $n=0,$ which says that $6$ is a minimal value.
| {
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Prove inequality with logarithms:$\log_{a}{\frac{a+b}{2}} + \log_{b}{\frac{a+b}{2}} \ge 2$ Let $ a, b \in (1, \infty)$. Prove that:
$$\log_{a}{(\frac{a+b}{2})} + \log_{b}{(\frac{a+b}{2})} \ge 2$$
I tried switching the bases on each of the logaritm but I got stuck:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$
| By AM-GM we have $$\frac{a+b}2\geq \sqrt{ab} $$ and since $\log$ is an increasing function for bases $>1$ we have $$\log_a(\frac{a+b}2) +\log_b(\frac{a+b}2) \geq \log_a(\sqrt{ab}) +\log_b(\sqrt{ab}) =1/2(2+\log_a(b)+\log_b(a))$$
And we have that $\log_a(b) +\log_b(a) =\log_a(b) +\frac1{\log_a(b)} \geq 2$ because $x+\frac1 x \geq 2$
| {
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Series of $\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1)$ I am reading slides from my lecture and saw this equality:
$$ \frac{1}{2}(e^{3x} - e^{2x} - e^x + \color{red}{1}) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} $$
but in my opinion it should be
$$ \frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} + \color{red}{\frac{1}{2}} $$
where $ \color{red}{1}$ has been lost?
| You are correct, it should be
$$\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge 0}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!} + \frac{1}{2}.$$
Note that, since the left-hand side is zero for $x=0$, you may also write
$$\frac{1}{2}(e^{3x} - e^{2x} - e^x + 1) = \sum_{r \ge \color{red}{1}}\frac{1}{2}(3^r-2^r-1) \frac{x^r}{r!}$$
where starting index in the sum is now $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Simplify $\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $ Simplify
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $$
Attempt:
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} = \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} = \sqrt{10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})} $$
let $X =\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}}$, then
$$ X^{2} = 10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15}) $$
How to continue?
| Let $$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}$$
So $$\sqrt{x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}$$For $$\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{x}+\sqrt{y}+\sqrt{z}.$$We get $x+y+z=10, 2\sqrt{xy}=\sqrt{24}, 2\sqrt{yz}=\sqrt{40}, 2 \sqrt{xz}=\sqrt{60}.$ So we get $xy=6, yz=10, zx=15 \Rightarrow xyz=30, x=2,y=3, z=5.$
Hence the requiredsquare root is $$\sqrt{2}+\sqrt{3}+\sqrt{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive $a$, $b$, $c$ with $abc=1$, show that $\sum_{cyc}\sqrt{a^2-a+1}\geq a+b+c$
Let $a,b,c$ are positive number such that $abc=1$. Prove that:
$$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\;\geq\; a+b+c$$
This problem froms my Math teacher. I have attempted to let $$(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$$. The inequality is equivalent to: $\frac{\sqrt{x^2-xy+y^2}}{y}+\frac{\sqrt{y^2-yz+z^2}}{z}+\frac{\sqrt{x^2-xz+z^2}}{x}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$
Then, I tried to use AM-GM but I stucked on it.
| The hint:
Use the Mixing Variables method.
Indeed, we can use the beautiful Can's idea.
Since $$\prod\limits_{cyc}(a-1)^2=\prod_{cyc}((a-1)(b-1))\geq0,$$ we can assume that $$(a-1)(b-1)\geq0$$ or
$$a+b\leq1+ab=1+\frac{1}{c}.$$
Thus, by C-S:
\begin{align}
\sqrt{a^2-a+1}&+\sqrt{b^2-b+1}\\
&=\sqrt{a^2+b^2-a-b+2+2\sqrt{(a^2-a+1)(b^2-b+1)}}\\
&\geq\sqrt{a^2+b^2-a-b+2+2\sqrt{\left(\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\right)\left(\left(b-\frac{1}{2}\right)^2+\frac{3}{4}\right)}}\\
&\geq\sqrt{a^2+b^2-a-b+2+2\left(\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)+\frac{3}{4}\right)}\\
&=\sqrt{a^2+b^2-a-b+2+2ab-a-b+2}\\
&=\sqrt{(a+b)^2-2(a+b)+4}.
\end{align}
But $f(x)=\sqrt{x^2-2x+4}-x$ decreases, which says
$$\sum_{cyc}(\sqrt{a^2-a+1}-a)\geq f(a+b)+\sqrt{c^2-c+1}-c\geq$$
$$\geq f\left(1+\frac{1}{c}\right)+\sqrt{c^2-c+1}-c=\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c.$$
Id est, it's enough to prove that:
$$\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c\geq0$$ and the rest is smooth.
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Rate of change with $a(t)=\frac{1}{t+4}$ on $[9,9+h]$ I am working on an exercise to find the rate of change between points $[9, 9+h]$ with the function $a(t)=\frac{1}{t+4}$.
The solution provided is $\frac{-1}{13(13+h)}$ whereas I arrive at $\frac{\frac{1}{h}}{h}$.
My working:
$a(t_1)$ = $\frac{1}{9+4}$ = $\frac{1}{13}.$
$x(t_2)$ = $\frac{1}{9+h+4}$ = $\frac{1}{13+h}.$
The rate of change is: $\frac{a(t_2)-a(t_1)}{t_2-t_1}.$
So: $\dfrac{\frac{1}{13+h}-\frac{1}{13}}{9+h-9}$ = $\dfrac{\frac{1}{13}+\frac{1}{h}-\frac{1}{13}}{h}$ = $\dfrac{\frac{1}{h}}{h}.$
Where did I go wrong and how can I arrive at $\frac{-1}{13(13+h)}$?
| $$\frac{1}{13+h}\neq\frac{1}{13}+\frac{1}{h}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many mandrel wraps are required to get a desired strip length I have a machine that wraps a strip of aluminum with thickness $t$ on a mandrel. If I want a specified length $x$, how many mandrel (diameter $d$) revolutions $n$ are required?
That is:
$$x=\pi(d+(d+2t)+(d+4t)+\cdots+(d+2nt))$$
Solving for $n$. My brain went tilt.
| Notice you have $n+1$ total $d$'s. Now using the fact that $1+2+\cdots+n= \dfrac{n(n+1)}{2}$ (this is not obvious but is well known, for example here), we have
$$
\begin{aligned}
x&= \pi \,\big(d+(d+2t)+(d+4t)+\cdots+(d+2nt)\big)\\
x&= \pi \big( (d+d+\cdots+d) + (2t+4t+\cdots+2nt) \big) \\
x&= \pi \big( (n+1)d + 2t(1+2+\cdots+n) \big) \\
x&= \pi \left((n+1)d + 2t \cdot \dfrac{n(n+1)}{2} \right) \\
\frac{x}{\pi}&= (n+1)d + 2t \cdot \dfrac{n(n+1)}{2} \\
\frac{x}{\pi}- (n+1)d &= 2t \cdot \dfrac{n(n+1)}{2} \\
\dfrac{\frac{x}{\pi}- (n+1)d}{t}&= n(n+1)
\end{aligned}
$$
We have this odd (but straightforwardly calculated term) $\dfrac{\frac{x}{\pi}- (n+1)d}{t}$. Let $C=\dfrac{\frac{x}{\pi}- (n+1)d}{t}$. Then
$$
\begin{aligned}
C&= n(n+1) \\
C&= n^2+n \\
n^2+n-C&=0
\end{aligned}
$$
which is a quadratic eqation, so $n= \dfrac{-1 \pm \sqrt{1+4C}}{2}$. The problem now is that this term is probably not an exact integer (one probably could show that this is always the case) but $n$ is a count so it must have an integer value. But being a real world problem just round down or up, depending on the context called by this scenario - I do not know anything about mandrels.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be able to cancel anything out. I am also struggling with other methods like AM-GM since $a,b$ are not necessarily positive. Any help would be appreciated.
| Assuming that $a,b > 0$, you want to prove that:
$$a^6+3a^4b^2+3a^2b^4+b^6+32a^4+32a^3b+32ab^3+32b^4 \geq 32 a^4b+ 32ab^4.$$
Now, show that $$a^6+a^4b^2+a^4b^2+a^4b^2+16a^4+16a^4+16a^3b+16a^3b \geq 32a^4b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the range of $P= \frac{\sum_{1}^n x_i}{\prod _{1}^n (x_i^2+1)}$ When I solved the following problem.
Find the range of $B\equiv\dfrac{x+y}{(x^2+1)(y^2+1)}$ where $x,y \in\mathbb{R}$.
Solution.
Setting $x=\tan u$ and $y=\tan v$, expression is $\frac{\sin 2u+\sin 2v+\sin (2u+2v)}4$
And from there, it is not very complex to get upperbound when $u=v=\frac{\pi}6$ and opposite lowerbound.
Hence the result : $\boxed{\text{Range is }\left[-\frac{3\sqrt 3}8,\frac{3\sqrt 3}8\right]}$
But I don't how to solve the generalization problem:
Find the range of $P=\displaystyle \dfrac{\sum_{1}^n x_i}{\prod _{1}^n (x_i^2+1)}$ where $x_i \in\mathbb{R}$.
| Notice that:
*
*$P$ is continuous on $\mathbb R^n$.
*$P(x_1, \dotsc, x_n) \ge 0$ for any $x_1, \dotsc, x_n \ge 0$.
*$P(0, \dotsc, 0) = 0$.
*$P(x_1, \dotsc, x_n) \to 0$ as $(x_1, \dotsc, x_n) \to \infty$.
These imply that the image of $[0, \infty)^n$ under $P$ has the form $[0, M]$ for some $M \ge 0$.
Also,
*$P(-x_1, \dotsc, -x_n) = -P(x_1, \dotsc, x_n)$ for any $x_1, \dotsc, x_n \in \mathbb R$.
*$P(-\lvert x_1 \rvert, \dotsc, -\lvert x_n \rvert) \le P(x_1, \dotsc, x_n) \le P(\lvert x_1 \rvert, \dotsc, \lvert x_n \rvert)$ for any $x_1, \dotsc, x_n \in \mathbb R$.
These imply that the image of $(-\infty, 0]^n$ under $P$ is $[-M, 0]$ and that the image of the whole $\mathbb R^n$ under $P$ is thus $[-M, M]$.
Now, suppose $x_1, \dotsc, x_n \ge 0$ and $P(x_1, \dotsc, x_n) = M$. Since $M$ is a maximum for $P$, we have that
\begin{align*}
\frac \partial {\partial x_i} P & = \frac {1 \cdot \prod_j (x_j^2 + 1) - (\sum_j x_j) \cdot 2 x_i \prod_{j \neq i} (x_j^2 + 1)} {\prod_j (x_j^2 + 1)^2} \\
& = \frac {x_i^2 + 1 - 2(\sum_j x_j) x_i} {(x_i^2 + 1) \prod_j (x_j^2 + 1)} = 0
\end{align*}
Therefore, if we let $s = \sum_j x_j$, then all the $x_i$'s satisfy the same equation
$$x^2 - 2 s x+ 1 = 0$$
Since the equation has at least one solution, its discriminant must be nonnegative, i.e., $s^2 \ge 1$. Since we are assuming $x_1, \dotsc, x_n \ge 0$, we have $s \ge 0$, therefore $s \ge 1$.
Now, the solutions to the equation are $s + \sqrt{s^2 - 1}$ and $s - \sqrt{s^2 - 1}$. If $s = 1$, then they are both equal to $1$, and so all the $x_i$'s must be equal to $1$ (this can only happen if $n = 1$). If $s > 1$, then $\sqrt{s^2 - 1} > 0$ and $x_i$ can't be equal to the solution $s + \sqrt{s^2 - 1}$, because $x_i \le s < s + \sqrt{s^2 - 1}$. Therefore all the $x_i$'s must be equal to $s - \sqrt{s^2 - 1}$.
Since all the $x_i$'s are equal to the same $x$, we have that $s = n x$, and so $x$ satisfies the equation
$$x^2 - 2n x^2 + 1 = 0$$
that is,
$$x = \frac 1 {\sqrt {2n - 1}}$$
From this we can compute the maximum of $P$:
$$\sum_i x_i = \frac n {\sqrt {2n - 1}}$$
$$\prod_i (x_i^2 + 1) = \left (\frac 1 {2n - 1} + 1 \right )^n = \left ( \frac {2n} {2n - 1} \right )^n$$
and finally, the maximum is
$$M = \frac {\frac n {\sqrt {2n - 1}}} {\left ( \frac {2n} {2n - 1} \right )^n} = \frac { (2n - 1)^n } {2^n n^{n-1} \sqrt {2n - 1}}$$
The range of $P$ is then $[-M, M]$.
| {
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"url": "https://math.stackexchange.com/questions/3332716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find the formula for the convolution of the sequences (generating functions)
Find the formula for the convolution of the sequences:
$a_n=\begin {cases} 1 & 0\leq n \leq 4 \\ 0 & n \geq 5 \end{cases} $
$b_n = 1$ $\forall n \in \mathbb{N}$
What I've been doing:
In other words:
$a_n = 1,1,1,1,1,0,0,...$ and $b_n=1,1,1,1,...$
So that means that their generating functions are:
$a(x)=1+x+x^2+x^3+x^4=\sum _{i=0} ^{4} x^i$
$b(x)=1+x+x^2+x^3+...=\sum _{j=0} ^{\infty} x^i$
So using the convolution formula i.e: $b(x)a(x)=\sum _{j=0} ^{\infty}(\sum _{i=0} ^{4}a_ib_{j-i})x^4=\sum _{j=0} ^{\infty}(\sum _{i=0} ^{4}x^i(x^{i-j}))x^4=\sum _{j=0} ^{\infty}(\sum _{i=0} ^{4}x^{2i-j})x^4$
And I can't get past that nor do I know if what I did was correct... Can someone guide me?
| One approach is to simplify the generating functions $a(x)$ and $b(x)$, multiply them, and manipulate the resulting generating function for the convolution, as follows:
$$\begin{align}
a(x) &= \sum_{i=0}^4 x^i = 1+x+x^2+x^3+x^4 \\
b(x) &= \sum_{j=0}^\infty x^j = \frac{1}{1-x} \\
\sum_{k=0}^\infty \left(\sum_{i=0}^k a_i b_{k-i}\right) x^k &= a(x) b(x) \\
&= \frac{1+x+x^2+x^3+x^4}{1-x} \\
&= -4 - 3 x- 2 x^2 -x^3 + \frac{5}{1 - x} \\
&= -4 - 3 x- 2 x^2 -x^3 + 5\sum_{k=0}^\infty x^k \\
&= 1 +2 x+3 x^2 +4x^3 + 5\sum_{k=4}^\infty x^k \\
&= \sum_{k=0}^\infty \min(k+1,5) x^k.
\end{align}$$
So $$\sum_{i=0}^k a_i b_{k-i}= \min(k+1,5).$$
More generally, convolution of an arbitrary sequence $(a_i)_{i=0}^\infty$ with the constant 1 sequence yields the sequence of cumulative sums $(\sum_{i=0}^k a_i)_{k=0}^\infty$. You can use the same generating function approach as above or just compute directly:
$$
\sum_{i=0}^k a_i b_{k-i} = \sum_{i=0}^k a_i \cdot 1 = \sum_{i=0}^k a_i.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Can there be two inequalities? $$x+a+\sqrt{x^2+a^2}>b$$
I can consider this inequality for $a,b,x>0$.
1) $\sqrt{x^2+a^2}>b-(x+a)$
$$x^2+a^2>b^2-2b(x+a)+(x+a)^2$$
$$0>b^2-2b(x+a)+2ax$$
$$x(2b-2a)>b^2-2ba$$
$$x>\frac{b^2-2ba}{(2b-2a)}$$
2) $(x+a)-b>-\sqrt{x^2+a^2}$
$$x^2+a^2<b^2-2b(x+a)+(x+a)^2$$
$$0<b^2-2b(x+a)+2ax$$
$$x(2b-2a)<b^2-2ba$$
$$x<\frac{b^2-2ba}{(2b-2a)}$$
I get two inequalities. What is Im doing wrong?
| Supposing you want to solve for $x$ in $$\sqrt{x^2+a^2}>b-a-x,$$ note that since LHS is never negative, the inequality is automatically true whenever RHS is negative, that is, for all $b-a<x.$ Thus, suppose $b-a-x\ge 0,$ then you may square both sides to have $$x^2+a^2>(b-a-x)^2=b^2-2b(a+x)+(a+x)^2=b^2-2ab-2bx+a^2+2ax+x^2,$$ which simplifies to $$0>b^2-2ab-2bx+2ax=b^2-2ab-(2b-2a)x,$$ or $$(2b-2a)x>b^2-2ab=b(b-2a).$$ Now recall that we assumed that $b-a\ge x.$ Again we have three subsidiary cases, namely $a=b,a<b,a>b.$ You can check that in the first case we have no solutions. If we have $a<b,$ then we can divide both sides by $2b-2a$ without changing the order, to get $$x>\frac{b(b-2a)}{2(b-a)}.$$ In the final case this inequality is reversed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\frac{1}{\sin^2(x)\cos^2(x)}$ $$ I = \int\frac{1}{\sin^2(x)\cos^2(x)}$$
I have tried the following:
$$\sin^2x = \sin x \cdot \sin x = \frac{1}{2}(1 - \cos2x)$$
$$\cos^2x = \cos x \cdot \cos x = \frac{1}{2}(1 + \cos2x)$$
The integral becomes:
$$I = 4\int\frac{1}{1 - \cos^2(2x)} = 4\int\frac{1}{\sin^2(2x)}$$
Substitute $u = 2x \implies du = 2dx$
$$I = 2\int\frac{1}{\sin^2u} = -2\cot(u)$$
Plugging back x I get:
$$I = -2\cot(2x)$$
I tried plugging in the integral into an integral-calculator and the answer was: $\tan(x) - \cot(x)$. Can you help me identify what I did wrong?
| $\tan(x) - \cot(x) = \tan(x) -\frac{1}{\tan(x)} = \frac{\tan^2(x) - 1}{\tan(x)}$
As $\tan(2x) = \frac{2\tan(x)}{1- \tan^2(x)}$
This implies $\tan(x) - \cot(x) = -2\cot(2x)$
Ur correct, actually both r same
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 & a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{5}=\begin{pmatrix} a^5 & a^{4}b+a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
Continuing this way, we obtain
$A^{k}=\begin{pmatrix} a^k & (a^{k-2}+a^{k-3}+a^{k-4}+.....+1)b \\ 0 & 1 \end{pmatrix}$
I am stuck here! I was wondering if you could give me some hints to move further. I appreciate your time.
| Hint: Use Cayley–Hamilton: $A^2-(a+1)A+aI=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of function $f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$
I tried to calculate limit when $x$ goes to infinity for the following function
$$f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$$
where $a$, $b$, $d$ are some positive constants.
It's easy to see that terms before and after minus sign goes to infinity so that gives me indeterminate symbol. Is there some way to solve this problem?
| For $x\not=0$,
$$\begin{align}\sqrt{(xa \pm d)^2 + x^2 b^2} &=|x|\sqrt{\left(a \pm \frac{d}{x}\right)^2+ b^2}\\
&=|x|\sqrt{a^2+b^2}\left(1\pm\frac{2ad}{(a^2+b^2)x}+o(1/x)\right)^{1/2}\\
&=|x|\sqrt{a^2+b^2}\left(1\pm\frac{ad}{(a^2+b^2)x}+o(1/x)\right)\\
&=|x|\sqrt{a^2+b^2}\pm\frac{ad|x|/x}{\sqrt{a^2+b^2}}+o(1)
\end{align}$$
where we used the expansion at $t=0$, $(1+t)^{1/2}=1+t/2+o(t)$.
Now it should be easy to verify that
$$\lim\limits_{x \to +\infty} f(x) = \frac{2ad}{\sqrt{a^2+b^2}}\quad\text{and}\quad \lim\limits_{x \to -\infty} f(x) = -\frac{2ad}{\sqrt{a^2+b^2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility proof for numbers coprime to $6$ Can someone check on this? If it is wrong please refrain from telling me the answer.
Suppose $(6,a) = (6,b) = 1$. We wish to prove $24 \mid (a^2-b^2)$.
We can see that $(4,a) = (4,b) = 1$ since $2 \mid 4$ and $2 \mid 6$, so, $(24, a) = (24, b) = 1$. By Bezouts identity, any integer can be represented by a linear combination of two integers that are relatively prime, so let $a^2 - bx_2 = ax_1 + 24y_1$ and $b^2 - ax_1 = bx_2 + 24y_2$ for some $x_1,y_1,x_2,y_2 \in \mathbb{Z}$. Then $a^2 - b^2 = bx_2 - bx_2 + ax_1 - ax_2 + 24y_1 - 24y_2$. So $a^2 - b^2 = 24(y_1 - y_2)$. So $24 \mid (a^2 - b^2)$. $\square$
| Hints.
*
*The numbers are all of the form $6n \pm 1$
*If $n$ is odd, then $3n \pm 1$ is even.
*$(6n \pm 1)^2 - 1$ is a multiple of 24.
*$(6a \pm 1)^2 - (6b \pm 1)^2 = [(6a \pm 1)^2 - 1] - [(6b \pm 1)^2 - 1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Commutative matrices of a square matrix Question: Find the all matrice that commute with $B=\begin{pmatrix} b & 1 & 0 \\ 0 & b & 1 \\ 0 & 0 & b \end{pmatrix}$.
My work:
Let $A=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$.
Now from $AB=BA$ implies,
$\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix} . \begin{pmatrix} b & 1 & 0 \\ 0 & b & 1 \\ 0 & 0 & b \end{pmatrix}
=\begin{pmatrix} b & 1 & 0 \\ 0 & b & 1 \\ 0 & 0 & b \end{pmatrix}.\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$
Then,
$\begin{pmatrix} a_1b & a_1+bb_1 & b_1+c_1b \\ a_2b & a_2+bb_2 & b_2+c_2b \\ a_3b & a_3+bb_3 & b_3+c_3b\end{pmatrix} =\begin{pmatrix} ba_1+a_2 & bb_1+b_2 & c_1b+c_2 \\ ba_2+a_3 & bb_2+b_3 & c_2b+c_3 \\ ba_3 & bb_3 & bc_3\end{pmatrix}$
Then, solving this,
$ a_1b= ba_1+a_2 \implies a_2=0$
$a_1+bb_1=bb_1+b_2 \implies a_1=b_2$
$b_1+c_1b=c_1b+c_2 \implies b_1=c_2$
$a_2b=ba_2+a_3 \implies a_3=0$
$a_2+bb_2=bb_2+b_3 \implies a_2=b_3$
$b_2+c_2b=c_2b+c_3 \implies b_2=c_3$
$a_3b=ba_3$
$a_3+bb_3= bb_3 \implies a_3=0$
$b_3+c_3b= bc_3 \implies b_3=0$.
Then, I am not sure about following setting. Is it right or wrong?
$A=\begin{pmatrix} a_1 & b_1 & ? \\ 0 & a_1=b_2 & b_1 \\ 0 & 0 & c_3 \end{pmatrix}$.
I was wondering if you could help to resolve this issue. I appreciate your time.
| I believe there is a simpler way, a path which makes the calculations easier.
Write $B$ in the form
$B = \begin{bmatrix} b & 1 & 0 \\
0 & b & 1 \\ 0 & 0 & b \end {bmatrix} = bI + N, \tag 1$
where
$N = \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0 \end {bmatrix}; \tag 2$
then
$AB = BA \tag 3$
becomes
$A(bI + N) = (bI + N)A \Longrightarrow bA + AN = bA + NA \Longrightarrow AN = NA; \tag 4$
with
$A = \begin{bmatrix} a_1 & a_4 & a_7 \\
a_2 & a_5 & a_8 \\ a_3 & a_6 & a_9 \end {bmatrix}, \tag 5$
equation
$AN = NA \tag 6$
yields
$\begin{bmatrix} a_1 & a_4 & a_7 \\
a_2 & a_5 & a_8 \\ a_3 & a_6 & a_9 \end {bmatrix} \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0 \end {bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0 \end {bmatrix} \begin{bmatrix} a_1 & a_4 & a_7 \\
a_2 & a_5 & a_8 \\ a_3 & a_6 & a_9 \end {bmatrix}, \tag 7$
or
$\begin{bmatrix} 0 & a_1 & a_4 \\
0 & a_2 & a_5 \\ 0 & a_3 & a_6 \end {bmatrix} = \begin{bmatrix} a_2 & a_5 & a_8 \\ a_3 & a_6 & a_9 \\ 0 & 0 & 0 \end {bmatrix}; \tag 8$
comparing entries of these two matrices we find
$a_2 = a_3 = a_6 = 0, \tag 9$
$a_1 = a_5 = a_9, \; a_4 = a_8, \tag{10}$
and $a_7$ unconstrained (that is, arbitrary).
Thus we have $A$ taking the form
$A = \begin{bmatrix} a & c & d \\
0 & a & c \\ 0 & 0 & a \end {bmatrix} = aI + cN + dN^2, \tag{11}$
for
$N^2 = \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0 \end {bmatrix} \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\ 0 & 0 & 0 \end {bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\
0 & 0 & 0 \\ 0 & 0 & 0 \end {bmatrix}; \tag{12}$
again, $a$, $c$, and $d$ may be freely selected. It is evident that matrices of the form (12) commute with $B$, since each is a polynomial in $N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I solve $\int e^{2\theta} \sin(3\theta)\, d\theta$ with integration by parts? $\int e^{2\theta}\sin(3\theta)d\theta$ seems to be leading me in circles. The integral I get when I use integration by parts, $\int e^{2\theta}\cos(3\theta)d\theta$ just leads me back to $\int e^{2\theta}\sin(3\theta)d\theta$. I am not sure how to solve it.
My Steps:
$\int e^{2\theta}\sin(3\theta)d\theta$
Let $u = \sin(3\theta)$ and $dv=e^{2\theta}d\theta$
Then $du = 3\cos(3\theta)d\theta$ and $v = \frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta} \sin(3 \theta)d\theta &= \frac{1}{2} e^{2\theta}\sin(3\theta) - \int\frac{1}{2}e^{2\theta}3\cos(3\theta)d\theta\\
&=e^{2\theta}\sin(3\theta) - \frac{3}{2}\int e^{2\theta}\cos(3\theta)d\theta\\
\end{align*}
$\int e^{2\theta}\cos(3\theta)d\theta$
Let $u = \cos(3\theta)$ and $dv = e^{2\theta}d\theta$
Then $du = -3\sin(3\theta)d\theta$ and $v=\frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta}\cos(3\theta) &= \frac{1}{2}e^{2\theta}\cos(3\theta)-\int (\frac{1}{2}e^{2\theta}\cdot-3\sin(3\theta))d\theta\\
&=\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta
\end{align*}
So you can see I just keep going in circles. How can I break out of this loop?
| Another method is to predict the answer:
$$\int e^{2x}\sin(3x)dx=Ae^{2x}\sin (3x)+Be^{2x}\cos (3x)+C \Rightarrow \\
e^{2x}\sin (3x)=2Ae^{2x}\sin (3x)+3Ae^{2x}\cos (3x)+2Be^{2x}\cos (3x)-3Be^{2x}\sin (3x) \Rightarrow \\
\begin{cases} 2A-3B=1\\ 3A+2B=0\end{cases}\Rightarrow A=\frac2{13};B=-\frac3{13}$$
Hence, the final answer is:
$$\int e^{2x}\sin(3x)dx=\frac2{13}e^{2x}\sin (3x)-\frac3{13}e^{2x}\cos (3x)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3351012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve $\tan x =\sec 42^\circ +\sqrt{3}$ For the trigonometric equation,
$$\tan x =\sec 42^\circ+\sqrt{3}$$
Find the angle $x$, where $0<x<180^\circ$.
I tried to solve for an unknown angle $x$ in a geometry problem with a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached my calculator, entering the right-hand-side and arctan-ing it for $x$.
To my surprise, the angle $x$ comes out at exactly 72 degrees. I did not expect such a neat relationship. Then, I thought I should have solved the equation analytically for the whole-degree angle without the calculator. I spent a good amount of time already and was not able to derive it yet.
Either the equation is not as innocent as it looks, or a straightforward method just eludes me.
| Sorry, but I am unable to work with degrees.
If you look here
$$\sec \left(\frac{7 \pi }{30}\right)=\sqrt{8+2 \sqrt{5}-2 \sqrt{15+6 \sqrt{5}}}$$ and here
$$\tan \left(\frac{2 \pi }{5}\right)=\sqrt{5+2 \sqrt{5}}$$
Simplify
$$\left(\sqrt{8+2 \sqrt{5}-2 \sqrt{15+6 \sqrt{5}}}+\sqrt 3\right)^2=5+2 \sqrt{5}$$
I understand your surprise.
Edit
Thinking that this could not be the only one, I computed
$$R_k=\tan \left(\frac{(k+5) \pi}{30} \right)-\sec \left(\frac{k\pi }{30}\right)$$ for $k=1,\cdots,60$.
Here are the "funny" results (I hope I did not miss any)
$$\left(
\begin{array}{cc}
k & R_k \\
5 & \frac{1}{\sqrt{3}} \\
7 & \sqrt{3} \\
19 & \sqrt{3} \\
20 & 2-\frac{1}{\sqrt{3}} \\
25 & \frac{2}{\sqrt{3}} \\
30 & 1+\frac{1}{\sqrt{3}} \\
31 & \sqrt{3} \\
35 & \frac{5}{\sqrt{3}} \\
43 & \sqrt{3} \\
50 & -2-\frac{1}{\sqrt{3}} \\
55 & -\frac{2}{\sqrt{3}} \\
60 & -1+\frac{1}{\sqrt{3}}
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
What means $\mathbb{R}^{n}$ in the equation $\mathbb{R}^{n}$ = $\operatorname{Ran}(A)+ \operatorname{Ker}(A^T)$? What means $\mathbb{R}^{n}$ in the equation $\mathbb{R}^{n}=\operatorname{Ran}(A)+ \operatorname{Ker}(A^T)$? Is it the number of the columns of $A$ $\in \mathbb{R}^{m \times n}$? And what is the difference between row-rank and column-rank?
| A better question is: what does the right hand side mean?
$\operatorname{Ran}(A)$ is the column space—the span of the columns of $A$.
$\operatorname{Ker}(A^T)$ is the null space of $A^T$.
$+$ means to add together all pairs of vectors one from each space: $V + W = \{v + w : v \in V, w \in W\}$.
For example, let $A = \begin{pmatrix}1 & 1 & 2 \\ 0 & 1 & -1 \\ 1 & 2 & 1\end{pmatrix}$. Then
$$\operatorname{Ran}(A) = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\} = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right\}$$
(Use row reduction to find a basis.)
$$\operatorname{Ker}(A^T) = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\}$$
(Because I picked $A$ such that $(1,1,2) + (0,1,-1) = (1,2,1)$—you can also use row reduction to find this.)
And finally
\begin{align} \operatorname{Ran}(A) + \operatorname{Ker}(A^T) &= \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right\} + \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\} \\ &= \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\} = \mathbf{R}^3. \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Positive integer solutions to $ p+q^{n}=x^2$ Consider two prime numbers $p$ and $q$ such that
$$
p+q^2=r^2,
$$
and $r\in\mathbb N$. It is not difficult to figure out that for any $n\in\mathbb N$ and $x\in\mathbb N$ there are no solutions of
$$
p^{2}+q^{n}=x^2.
$$
What about
$$
p+q^{n}=x^2,
$$
with $n>2$? So far I was only able to prove that there are no positive integer solutions if $q=2$. The question is a variation of a problem in the Italian 2019 mathematical Olympiads.
| $p + q^2 = r^2$ means $p = r^2 - q^2 = (r+q)(r-q)$. Since $p$ is prime, this implies $r-q=1$ and $r+q=p$, i.e. $p=2q+1$.
If $n > 2$ is even,
$p + q^n = r^2$ means $p = (r+q^{n/2})(r-q^{n/2})$,
so $r=q^{n/2}+1$ and $p = r + q^{n/2} = 2 q^{n/2}+1$.
If $n$ is divisible by $4$, there can't be solutions with $q \ne 3$, as
$2 q^{n/2}+1$ would be divisible by $3$. For $n$ even but not divisible by $4$,
I would expect solutions to exist.
Some examples:
$$\eqalign{5 + 2^2 &= 3^2\cr
19 + 3^4 &= 10^2\cr
17 + 2^6 &= 9^2\cr
163 + 3^8 &= 82^2\cr
487 + 3^{10} &= 244^2\cr
1459 + 3^{12} &= 730^2\cr
257 + 2^{14} &= 129^2\cr
39367 + 3^{18} &= 19684^2\cr
5264950288664609755476607823 + 311^{22} &= 2632475144332304877738303912^2\cr
2441406251 + 5^{26} &= 1220703126^2\cr
65537 + 2^{30} &= 32769^2\cr
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $\sin\alpha+\sin\beta=a$ and $\cos\alpha-\cos\beta=b$, then what is $\tan(\frac{\alpha-\beta}{2})$? If I square both the equations
$$2+2\sin(\alpha-\beta)=a^2+b^2$$
$$\sin(\alpha-\beta)=\frac{a^2+b^2-2}{2}$$
Since $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$, then $$\sin(\alpha-\beta)=\frac{2\tan\frac{\alpha-\beta}{2}}{1+\left(\tan\frac{\alpha-\beta}{2}\right)^2}$$
It’s obviously the too long to solve, so is there a shorter way to do this?
| Since $\frac{a}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ while $-\frac{b}{2}=\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$, $\tan\frac{\alpha-\beta}{2}=-\frac{b}{a}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$ I recently saw on this site, the identity
$$\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$$
which I instantly wanted to prove.
I know that I can "reduce" the problem to the evaluation of $\cos\frac\pi{15},$ as the rest is easy with the use of the half-angle formula.
I know that $\cos$ obeys the 'nice' relation
$$\cos nx=T_n(\cos x)$$
where $$T_n(x)=\frac{n}2\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{n-k}{n-k\choose k}(2x)^{n-2k}.$$
Thus, setting $t=\cos\frac\pi{15}$,
$$T_{15}(t)=-1.$$
The only thing left to do is solve for $t$. We can narrow down our search to the values $0<t<1.$
I have never dealt with degree-$15$ polynomials before, so I was hoping one of you could help me out. Thanks!
| Staying with your approach (which could be made simpler as said in comments and answers), you have
$$T_{15}(x)=-15 x+560 x^3-6048 x^5+28800 x^7-70400 x^9+92160 x^{11}-61440 x^{13}+16384
x^{15}$$
but $T_{15}(x)+1$ can be factorized as
$$T_{15}(x)+1=(x+1) (2 x-1)^2 \left(4 x^2-2 x-1\right)^2 \left(16 x^4+8 x^3-16 x^2-8 x+1\right)^2$$ which means that you are left with the quartic
$$16 x^4+8 x^3-16 x^2-8 x+1=0$$ which can be solved with radicals. Its solutions are
$$\left\{x= \frac{1}{8} \left(-1-\sqrt{5}-\sqrt{6
\left(5-\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8}
\left(-1-\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}\right)\right\},\left\{x=
\frac{1}{8} \left(-1+\sqrt{5}-\sqrt{6
\left(5+\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8}
\left(-1+\sqrt{5}+\sqrt{6 \left(5+\sqrt{5}\right)}\right)\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integration using Trig Substitution I was given the following problem: $$\int\sqrt{1-7w^2}\ dw$$ I used the sin substitution - getting $w=\frac1{\sqrt{7}}\sin\theta$. I then needed to change the $dw$ to a $d\theta$, so I got this: $dw=\frac1{\sqrt{7}}\cos\theta\ d\theta$. My new problem looks like this: $$\int\sqrt{1-7(\frac1{\sqrt{7}}\sin\theta)^2}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$
Continuing, I get: $$\int\sqrt{1-\sin^2\theta}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$ $$=\int\sqrt{\cos^2\theta}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$$$=\int\frac1{\sqrt7}|\sin^2\theta|\cos\theta\ d\theta$$ If I now set $u=\sin\theta$ I get: $$\frac1{\sqrt7}\int u^2 \ du$$$$=\frac1{3\sqrt7}u^3$$$$=\frac{\sin\theta}{3\sqrt7}+c$$
This is not the correct answer. Why not? Where did I go wrong? What is the proper way to do a problem like this one?
| Let us consider the general case where $a > 0$, $b > 0$, $a^{2}-b^{2}x^{2} \geq 0$ and $a/b \leq 1$:
\begin{align*}
\int\sqrt{a^{2} - b^{2}x^{2}}\mathrm{d}x
\end{align*}
According to the substitution $\displaystyle x = \frac{a\sin(\theta)}{b}$, we get $\displaystyle\mathrm{d}x = \frac{a\cos(\theta)}{b}\mathrm{d}\theta$. Thus we have
\begin{align*}
\int\sqrt{a^{2}-b^{2}x^{2}}\mathrm{d}x & = a\int\sqrt{\displaystyle 1 - \left(\frac{bx}{a}\right)^{2}} = \frac{a^{2}}{b}\int\sqrt{1 - \sin^{2}(\theta)}\cos(\theta)\mathrm{d}\theta\\\\
& = \frac{a^{2}}{b}\int\cos^{2}(\theta)\mathrm{d}\theta = \frac{a^{2}}{b}\int\frac{\cos(2\theta) + 1}{2}\mathrm{d}\theta\\\\
& = \frac{a^{2}}{b}\left[\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right] = \left[\frac{x\sqrt{a^{2}-b^{2}x^{2}}}{2} + \frac{a^{2}\arcsin\left(\displaystyle\frac{bx}{a}\right)}{2b}\right]
\end{align*}
In your case, $a = 1$ and $b = \sqrt{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the Eigenvectors Consider the matrix given below:
$\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}.$
The eigenvalues for this matrix are
$\dfrac{1+\sqrt 5}{2},\dfrac{1-\sqrt 5}{2}.$
I am facing trouble finding the eigenvectors. Please help.
| Find eigenvalues from the characteristic polynomial
$\lambda^2-\lambda-1=(\lambda+(\sqrt(5)-1)/2)*(\lambda-(\sqrt(5)+1)/2)$
$\lambda_1=(-\sqrt(5)+1)/2$
$\lambda_2=(\sqrt(5)+1)/2$
For every λ we find its own vector(s):
$\lambda_1=(-\sqrt(5)+1)/2$
$A-\lambda_1I=\left(\begin{matrix}
\frac{\sqrt{5}+1}{2} & 1 \\
1 & \frac{\sqrt{5}-1}{2}
\end{matrix}\right)$
$Av=\lambda v$ 1
$ \Rightarrow (A-\lambda I)v=0$
So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: 2
$\left(\begin{matrix}
\frac{\sqrt{5}+1}{2} & 1 & 0 \\
1 & \frac{\sqrt{5}-1}{2} & 0
\end{matrix}\right)$
$\begin{matrix}
x_1 & +\frac{\sqrt{5}-1}{2}*x_2 & = & 0
\end{matrix}$
General Solution: $X=\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2}*x_2 \\
x_2
\end{matrix}\right)$
Let $x_2=1, v_1=\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2} \\
1
\end{matrix}\right)$
$\lambda_2=(\sqrt(5)+1)/2$
$A-\lambda_2I=\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2} & 1 \\
1 & \frac{-\sqrt{5}-1}{2}
\end{matrix}\right)$
$(A-\lambda I)v=0$ 1
So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination: 2
$\left(\begin{matrix}
\frac{-\sqrt{5}+1}{2} & 1 & 0 \\
1 & \frac{-\sqrt{5}-1}{2} & 0
\end{matrix}\right)$
$\begin{matrix}
x_1 & -\frac{\sqrt{5}+1}{2}*x_2 & = & 0
\end{matrix}$
General Solution: $X=\left(\begin{matrix}
\frac{\sqrt{5}+1}{2}*x_2 \\
x_2
\end{matrix}\right)$
Let $x_2=1, v_2=\left(\begin{matrix}
\frac{\sqrt{5}+1}{2} \\
1
\end{matrix}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solving $\sin 3\theta=1/2$ on the interval $[0,2\pi]$. I don't understand where some solutions came from. I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:
All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3\theta=1/2$ are
$$\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}$$
| Write out $3\theta$: $\frac{\pi}{6}$, $\frac{5\pi}{6}$, $\frac{13\pi}{6}$, $\frac{17\pi}{6}$, $\frac{25\pi}{6}$, $\frac{29\pi}{6}$. Now subtract $\frac{\pi}{6}$ from $\frac{13\pi}{6}$, and you get $\frac{12\pi}{6}=2\pi$. The sine function is periodic, with this period. You get all the other by adding $2\pi$ or $4\pi$ to the first solutions.
Maybe a more obvious way of thinking about the problem is to say $\alpha=3\theta$, and find all solutions of $\sin\alpha=\frac 12$ in the interval $[0,6\pi]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361862",
"timestamp": "2023-03-29T00:00:00",
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Trying to find $c_{1}$ , $c_{2}$ for Big theta notation. Need to find $c_{1},$ $c_{2}$ and $n_{0}$.
\begin{equation}
c_{1}n^3 \leq \frac{n^3}{100} - 100n^2 - 100n + 3\leq c_{2}n^3
\end{equation}
\begin{equation}
c_{1} \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
\begin{equation}
\frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
This is what I have done so far, but I am currently stuck do I need to chose c2 which is positive and greater than $\frac{100}{n} + \frac{3}{n^3}$? And for c1 , I need a c1, which is >0. How do I find these constants?
| As in polynomials when the variable goes to $\infty$ their behavior become like their greatest power term, so the inequality use this fact. But as $n$ becomes greater $c_1,c_2$ become more closer to $\frac{1}{100}$. Therefore if you know the bound of $n$, then you can far $c_1,c_2$ away from $\frac{1}{100}$ as it allows you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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first time a random walk hit the number $b$ In the book probability theory and random processes by Grimmet, I don't understand the last line of the proof of the probability that the first time that a random walk hits the number $b$.
I don't understand the last step where he shows that the line above is equals to $\frac{b}{n}P(S_n=b)$. the result is the same as the probability of getting to $b$ without revisiting the origin.
Here is the definition of $M_n$
| We have
$$
\mathbb{P} (S_k=x) = \binom{ k}{ \frac {k+x}{2} }p^\frac {k+x}{2}q^{k-\frac {k+x}{2}},
$$
so
\begin{alignat*}{2}
f_b(n) & = \binom{n-1}{\frac{n-1+b-1}{2}}p^{\frac{n-1+b-1}{2}+1}q^{\frac{n-1-b+1}{2}}-\binom{n-1}{\frac{n-1+b+1}{2}}p^{\frac{n-1+b+1}{2}}q^{\frac{n-1-b-1}{2}+1}\\
& = \left[\binom{n-1}{\frac{n+b}{2}-1}-\binom{n-1}{\frac{n+b}{2}}\right]p^{\frac{n+b}{2}}q^{\frac{n-b}{2}}\\
& =
\left[
\frac{(n-1)!}{\left(\frac{n+b}{2}-1\right)!\left(\frac{n-b}{2}\right)!}
-
\frac{(n-1)!}{\left(\frac{n+b}{2}\right)!\left(\frac{n-b}{2}-1\right)!}
\right]p^{\frac{n+b}{2}}q^{\frac{n-b}{2}}\\
& = \frac{(n-1)!}{\left(\frac{n-b}{2}\right)!\left(\frac{n+b}{2}\right)!}\left[\frac{n+b}{2}-\frac{n-b}{2}\right]p^{\frac{n+b}{2}}q^{\frac{n-b}{2}}\\
& = \frac{b}{n}\mathbb{P}(S_n = b).
\end{alignat*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why does $r = \frac 23 - \frac 12 T$ rather than $r = \frac 23$? I have the following question. Using Gaussian elimination, I got an answer different from the book's. Can anyone please tell me what I'm doing wrong?
At The Crispy Critter's Head Shop and Patchouli Emporium, along with their dried up weeds, sunflower seeds and astrological postcards, they sell an herbal tea blend. By weight,
*
*Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile.
*Type II herbal tea is 40% peppermint, 20% rose hips and 40% chamomile.
*Type III herbal tea is 35% peppermint, 30% rose hips and 35% chamomile.
How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile?
Here are my steps:
Matrix:
$(E1) \frac{3}{10}p +\frac{4}{10}r + \frac{3}{10}c = \frac{2}{3}$
$(E2) \frac{4}{10}p +\frac{2}{10}r + \frac{4}{10}c = \frac{2}{3}$
$(E3) \frac{35}{100}p +\frac{3}{10}r + \frac{35}{100}c = \frac{2}{3}$
I now make the leading coefficient in E1 a 1:
$(new E1) p +\frac{4}{3}r + c = \frac{20}{9}$
I eliminate p from E2:
$(E1) -\frac{4}{10}p -\frac{16}{30}r - \frac{4}{10}c = -\frac{8}{9}$
$(E2) \frac{4}{10}p +\frac{2}{10}r + \frac{4}{10}c = \frac{2}{3}$
and get new E2:
$(new E2) r = \frac{2}{3}$
I eliminate p from E3:
$(E3) -\frac{35}{100}p -\frac{140}{300}r - \frac{35}{100}c = -\frac{7}{9}$
$(E3) \frac{35}{100}p +\frac{90}{300}r + \frac{35}{100}c = \frac{6}{9}$
and get new E3:
$(new E3) r = \frac{2}{3}$
So I now have the following matrix:
$(new E1) p +\frac{4}{3}r + c = \frac{20}{9}$
$(new E2) r = \frac{2}{3}$
$(new E3) r = \frac{2}{3}$
Now this leaves c as a free variable so I set it equal to T and I get
c = t,
$r = \frac{2}{3}$, and $p = \frac{4}{3} - T $
But the book says the answer is c = t,
$r = \frac{2}{3} - \frac{1}{2}T$, and $p = \frac{4}{3} - \frac{1}{2}T $
Any ideas what I did wrong?
| My approach is similar to Eriins. First of all we have to define the variables:
a: amount of tea mixture type 1 in pounds
b: amount of tea mixture type 2 in pounds
c: amount of tea mixture type 3 in pounds
Next we have to notice that the equations are dependent. That means we can omit one equation (here: the third equation) and replace it by $a+b+c=2$
$$
\begin{eqnarray} \textrm{peppermint constraint} \\
\frac{3}{10}a + \frac{4}{10}b + \frac{35}{100}c = 2/3\\ \textrm{hips constraint} \\
\frac{4}{10}a + \frac{2}{10}b + \frac{3}{10}c = 2/3\\ \textrm{The sum of all three tea mixtures is 2 pounds} \\
a + b + c = 2\\
\end{eqnarray}
$$
The solution is $a=\frac43-\frac{c}2, b=\frac23-\frac{c}2, c=c$
Since we have the additional condition that $a,b,c\geq 0$, we can deduce
$c\leq \frac43$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$ Evaluate:
$$ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$$
My approach :
$$\lim_{x\to0}{\frac{1}{x^2}-\frac{\cos^2(x)}{\sin^2(x)}}$$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} $$
Using $$\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} $$
$$ \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
$$ \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
Applying L Hopital,
$$\lim_{x\to0}\frac{2\sin(x)\cos(x)}{2x}=1$$
But the actual answer is $\frac{2}{3}$. What am I doing wrong here?
| Although $\frac{\sin^2x}{x^2}\to1$, $\frac{\sin^2x}{x^2}\frac{1}{x^2}-\frac{\cos^2x}{x^2}$ isn't asymptotic to $\frac{1}{x^2}-\frac{\cos^2x}{x^2}$, because $$\lim_{x\to0}\left(\frac{\sin^2x}{x^2}\frac{1}{x^2}-\frac{1}{x^2}\right)=\lim_{x\to0}\frac{\frac{\sin^2x}{x^2}-1}{x^2}=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^2-1}{x^2}=-\frac13.$$The technique I'd advise is $$\cot x=\frac1x\frac{1-x^2/2+o(x^2)}{1-x^2/6+o(x^2)}=\frac1x\left(1-\frac13x^2+o(x^2)\right)\\\implies\frac{1}{x^2}-\cot^2x=\frac{1}{x^2}\left(\frac23x^2+o(x^2)\right)\to\frac23.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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what is $x$ for $\tan^23x = 2\sin^23x$ If $x$ = acute angle then find $x$ such that $\tan^23x = 2\sin^23x$.
So
$\tan3x = \sqrt2\sin3x$
$\frac{1}{\cos3x} = \sqrt2$
$3x = 45^{\circ}$
what are all the possibilies for $x$ ?
because the question asked for all possibilities of $x$
The options are: $180, 195, 120, 135, 360$
| \begin{align}
\tan^23x&=2\sin^23x\\
\tan^23x-2\sin^23x&=0\\
\sin^23x(\frac1{\cos^23x}-2)&=0\\
\sin^23x(\sec^23x-2)&=0
\end{align}
So we have either
\begin{align}
\sin^23x&=0\\
3x&=0^{\circ}, 180^{\circ}, 360^{\circ}\\
x&=0^{\circ}, 60^{\circ}, 120^{\circ}
\end{align}
or
\begin{align}
\sec^23x-2&=0\\
\sec^23x&=2\\
\sec3x&=\pm\sqrt2\\
3x&=45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}\\
x&=15^{\circ}, 45^{\circ}, 75^{\circ}, 105^{\circ}
\end{align}
From the above, the only admissible solutions are $\boxed{15^{\circ}, 45^{\circ}, 60^{\circ}, 75^{\circ}}$.
The sum of these angles is $\boxed{195^{\circ}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent? How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?
Taken from the 2008 IMC https://chiuchang.org/wp-content/uploads/sites/2/2018/02/2008-IWYMIC-Individual.x17381.pdf
my attempt
The number of ten digit numbers in which the digits are either 2 or 3 is $2^{10}$ and the numbers of ten digit numbers where there is no pair of adjacent numbers that are the same is $2$ E.g($2323232323$ & $3232323232$) and we know that the number of ten digit numbers consisting of pairs of adjacent $2$s and adjacent $3$s is the same due to symmetry therefore the answer would be $\frac{2^{10}-2}{2}=511$ however this doesn't taken into account the possibilities of having adjacent pairs of $3$s and $2$s in the same number.
| Suppose you have $k$ $3$s and it does not end with a $3$.
That the same thing as saying you have $k$ characters $32$ and $10-k-k=10-2k$ twos. So of the $10-2k + k=10-k$ characters you must choose $k$ spaces for the $k$ $32$ characters. There ${10-k \choose k}$ ways to do that.
Now suppose you have $k$ $3$s and it does end with a $3$.
If we just ignore the last place and put the $3$ in it, that is the same thing as saying you have $k-1$ characters $32$ to place and $10-k-(k-1)=11-2k$ $2$s to place. There are ${10-k\choose k-1}$ ways to do that.
So there are ${10-k \choose k}+ {10-k\choose k-1}$ ways to place $k$ threes.
Now we can have at most $5$ threes. (Any more and we won't have enough $2$s to go between all $3$s.)
So there are
$\sum_{k=0}^5 {10-k \choose k}+ {10-k\choose k-1}$ ways. (Assume ${10\choose -1} = 0$.... after all.... this would be the number of ways to choose $0$ threes and have a $3$ at the end which isn't possible.)
So $({10\choose 0}) + ({9\choose 1}+ {9\choose 0}) +({8\choose 2}+{8\choose 1}) + ({7\choose 3}+{7\choose 2}) + ({6\choose 4}+{6\choose3}) + ({5\choose 5} + {5\choose 4})=$
$1 + (9+1) + (\frac {8*7}2 + 8) + (\frac {7*6*5}6 -\frac {7*6}2) + (\frac {6*5*4*3}{24}+\frac{6*5*4}6) + (1 +5)=$
$1 + 10 +(28+8) + (35+21) + (15+20) + 6=$
$11+36+56+35+5=144$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Direct telescopic proof for sum of $1^2+2^2+...+n^2$ I was teaching $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$$ for the first I did direct telescopic proof like below
$$\sum_{k=1}^{n}k=\sum_{k=1}^{n}k(\frac{k+1}{2}-\frac{k-1}{2})=\sum_{k=1}^{n}(\frac{k(k+1)}{2}-\frac{k(k-1}{2}))=\\\sum_{k=1}^{n}(f(k)-f(k-1))=\frac{n(n+1)}{2}-0$$
for the second I did a classic proof $$\begin{align}
\sum_{k=1}^n k^2
& = \frac{1}{3}(n+1)^3 - \frac{1}{2}n(n+1) - \frac{1}{3}(n+1) \\
& = \frac{1}{6}(n+1) \left[ 2(n+1)^2 - 3n - 2\right] \\
& = \frac{1}{6}(n+1)(2n^2 +n) \\
& = \frac{1}{6}n(n+1)(2n+1)
\end{align}$$ some of the student asked for direct telescopic proof for the case...
I can't find this kind of proof. can anybody help me to find, or write this kind proving. \
I tried to rewrite $1=\frac{k+1}{2}-\frac{k-1}{2}$and I have
$$\sum_{k=1}^{n}k^2(\frac{k+1}{2}-\frac{k-1}{2})=$$ I can't go further more.
I promised to my students to try to find a direct proof Idea. Thanks for any help.
| Use binomial coefficients. Write $k^2=2\binom{k}{2}+\binom{k}{1}$ so $\sum_{k=1}^nk^2=\sum_{k=1}^n(a_k-a_{k-1})$ with $$a_n=2\binom{n+1}{3}+\binom{n+1}{2}=\frac13 n(n^2-1)+\frac12 n(n+1)=\frac16 n(n+1)(2n+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k} ≤ \frac{1}{\sqrt{2k+1}}$
Let's prove that the unequality is right for $n = k + 1$ as well, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$
I've spent a comparable amount of time trying to prove this point but looks like I can't grasp what to do.
| Note that if $$\frac{a}{b}<1 \Rightarrow \frac{a}{b} < \frac{a+x}{b+x}, x>0.~~~(1)$$
Let $$P_1=\frac{1}{2}~ \frac{3}{4}~ \frac{5}{6}....\frac{2k+1}{2k+2}.$$
Let us add 1, up and down in each fraction to get
$$P_2=\frac{2}{3} ~\frac{4}{5} ~\frac{6}{7}...\frac{2k+2}{2k+3}.$$
So By (1), $$P_1<P_2 \Rightarrow P_1^2 < P_1. P_2 \Rightarrow P_1^2 < \frac{1}{2k+3}$$$$\Rightarrow P_1 <\frac{1}{\sqrt{2k+3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why Does this Limit as V approaches Infinity Equal Zero
Given the following problem: $$\lim_{v\to\infty}\frac19(\ln|v-1|) - \frac19(\ln|v+8|)$$
Wouldn't this be undefined - as it would equal $\infty-\infty$?
However, Symbolab said it equaled zero.
I'm solving a larger problem and this is just one of the last segments. Please let me know why this would equal zero. Thank you.
| As an aside, there is some ambiguity in your notation. Some people would use $1/9(\ln|v-1|)$ to represent $\frac{1}{9 \ln |v-1|}$ and others to represent $\frac{1}{9} \ln |v-1|$. Given you comments about the form of the limit, you seem to mean the latter.
"$\infty - \infty$" is one of several indeterminate forms. We are responsible for recognizing indeterminate forms so that we can manipulate them to expose their limits. (Two other indeterminate forms are "$\frac{0}{0}$", which turns up in every derivative, and "$\infty \cdot 0$", which turns up in every integral.)
For "$\infty - \infty$", try exponentiation. They you are looking at $\frac{\mathrm{e}^\infty}{\mathrm{e}^\infty}$ for which there is some credible chance of cancellation. Here, \begin{align*}
\lim_{v \rightarrow \infty} &\frac{1}{9} \ln |v-1| - \frac{1}{9} \ln |v+8| \\
&= \lim_{v \rightarrow \infty} \ln \left( \exp \left( \frac{1}{9} \ln |v-1| - \frac{1}{9} \ln |v+8| \right) \right) \\
&= \lim_{v \rightarrow \infty} \ln \left(
\frac{ \exp \left( \frac{1}{9} \ln |v-1| \right) }
{ \exp \left( \frac{1}{9} \ln |v+8| \right) }
\right) \\
&= \lim_{v \rightarrow \infty} \ln \left(
\frac{ \exp \left( \ln |v-1| \right)^{1/9} }
{ \exp \left( \ln |v+8| \right)^{1/9} }
\right) \\
&= \lim_{v \rightarrow \infty} \ln \left(
\frac{ |v-1|^{1/9} }{ |v+8|^{1/9} }
\right) \\
&= \lim_{v \rightarrow \infty} \ln \left(
\left( \frac{|v-1|}{|v+8|} \right)^{1/9} \right) \\
&= \lim_{v \rightarrow \infty} \frac{1}{9} \ln \left( \frac{|v-1|}{|v+8|} \right) \\
&= \frac{1}{9} \lim_{v \rightarrow \infty} \ln \left( \frac{|v-1|}{|v+8|} \right) \text{.}
\end{align*}
We are taking a limit as $v \rightarrow \infty$, so we may assume that $v > 1$. Therefore, the arguments to both absolute values are positive and we can replace them with their arguments. \begin{align*}
\lim_{v \rightarrow \infty} &\frac{1}{9} \ln |v-1| - \frac{1}{9} \ln |v+8| \\
&= \frac{1}{9} \lim_{v \rightarrow \infty} \ln \left( \frac{v-1}{v+8} \cdot \frac{1/v}{1/v} \right) \\
&= \frac{1}{9} \lim_{v \rightarrow \infty} \ln \left( \frac{1-\frac{1}{v} }{1+\frac{8}{v}} \right) \\
&= \frac{1}{9} \ln \lim_{v \rightarrow \infty} \left( \frac{1-\frac{1}{v} }{1+\frac{8}{v}} \right) & [\text{$\ln$ continuous near $1$}] \\
&= \frac{1}{9} \ln 1 \\
&= \frac{1}{9} \cdot 0 \\
&= 0 \text{.}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
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