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How to evaluate $\int\sqrt{1-\tan x} \,dx$ without using up paper? $$\int\sqrt{1-\tan x} \, dx$$
is a very interesting integral. I attempted to evaluate it with the substitution $u^2=1-\tan x$ and then obtaining partial fractions. However, the coefficients are extremely complicated.
Is there an easier way to do it?
| Using your substitution $u^2=1-\tan x$:$$\int \sqrt{1-\tan x}dx=-\int\frac{2u^2}{(u^2-1)^2+1}du=-\int \frac{u^2+\sqrt 2}{u^4-2u^2+2}du-\int \frac{u^2-\sqrt 2}{u^4-2u^2+2}du$$
$$=-\int \frac{1+\frac{\sqrt 2}{u^2}}{\left(u-\frac{\sqrt{2}}{u}\right)^2-2+2\sqrt 2}du-\int \frac{1-\frac{\sqrt 2}{u^2}}{\left(u+\frac{\sqrt{2}}{u}\right)^2-2-2\sqrt 2}du$$
$$=-\frac{1}{\sqrt{2\sqrt 2-2}}\arctan\left(\frac{u-\frac{\sqrt 2}{u}}{\sqrt{2\sqrt 2-2}}\right)-\frac{1}{2\sqrt{2\sqrt 2+2}}\ln\left(\frac{u+\frac{\sqrt{2}}{u}-\sqrt{2+2\sqrt 2}}{{u+\frac{\sqrt{2}}{u}+\sqrt{2+2\sqrt 2}}}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why doesn't the substitution $x=it$ work in the integral $\int\frac{x^2}{{(x^2+1)}^2}dx$? For İntegral $$\int\frac{x^2}{{(x^2+1)}^2}dx$$ I used this substitution $x=it$
We have,
$$\begin{align}
\int\frac{x^2}{{(x^2+1)}^2}dx &=-i\int\frac{t^2}{{(t^2-1)}^2}dt \\
&=-\frac i4\left( \int\frac{1}{t-1}dt+\int\frac{1}{(t-1)^2}dt-\int\frac{1}{t+1}dt+\int\frac{1}{(t+1)^2}dt\right) \\
&=-\frac i4\left(\ln\left|\frac{t-1}{t+1}\right|-\frac{2t}{t^2-1}+C \right)
\end{align}$$
In the answer the function $\ln$ doesn't included.
My answer is wrong. Where is the mistake? I can not see.
| your result can be written in a simpler form as:
$$-\frac{i}{4}\left( \frac{2t}{{{t}^{2}}-1}+\ln \left( 1+t \right)-\ln \left( 1-t \right) \right)$$
now substitute $t=-ix$
$$\begin{align}
& =-\frac{i}{4}\left( \frac{2\left( -ix \right)}{{{\left( -ix \right)}^{2}}-1}+\ln \left( 1-ix \right)-\ln \left( 1+ix \right) \right) \\
& =-\frac{i}{4}\left( \frac{2ix}{{{x}^{2}}+1}+\ln \left( 1-ix \right)-\ln \left( 1+ix \right) \right) \\
& =-\frac{i}{4}\left( \frac{2ix}{{{x}^{2}}+1}+\left( \left( ix \right)-\frac{{{\left( ix \right)}^{2}}}{2}+\frac{{{\left( ix \right)}^{3}}}{3}-\cdots \right)+\left( \left( ix \right)+\frac{{{\left( ix \right)}^{2}}}{2}+\frac{{{\left( ix \right)}^{3}}}{3}-\cdots \right) \right) \\
& =-\frac{i}{4}\left( \frac{2ix}{{{x}^{2}}+1}+2\left( ix \right)+\frac{2{{\left( ix \right)}^{3}}}{3}+\frac{2{{\left( ix \right)}^{5}}}{5}+\cdots \right) \\
& =-\frac{i}{4}\left( \frac{2ix}{{{x}^{2}}+1}+2ix-\frac{2i{{x}^{3}}}{3}+\frac{2i{{x}^{5}}}{5}-\cdots \right) \\
& =\frac{1}{2}\left( \frac{x}{{{x}^{2}}+1}+x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\cdots \right) \\
& =\frac{1}{2}\left( \frac{x}{{{x}^{2}}+1}+{{\tan }^{-1}}\left( x \right) \right) \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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?
| $$\frac{1}{(x-x_1)(x-x_2)} = \frac{1}{(x - \frac{-3+\sqrt{13}}{2})(x-\frac{3+\sqrt{13}}{2})} = \frac{1}{(\frac{2x + 3-\sqrt{13}}{2})(\frac{2x -3-\sqrt{13}}{2})}=\frac{4}{(2x + 3-\sqrt{13})(2x - 3-\sqrt{13})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Jordan Decomposition of $\left(\begin{smallmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{smallmatrix}\right)$ over $\mathbb{F}_5$
Find the Jordan decomposition of
$$
A := \begin{pmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}
\in M_3(\mathbb{F}_5),
$$
where $\mathbb{F}_5$ is the field modulo 5.
What I've done so far
The characteristic polynomial is
\begin{equation}
P_A(t) = (4 - t)(1-t)(3-t) - (1-t)
= -t^3 + 8t^2-18t+1
\equiv 4t^3 + 3t^2 + 2t + 1 \mod5.
\end{equation}
Therefore, $\lambda = 1$ is a zero of $P_A$, since $4+3+2+1 = 10 \equiv 0 \mod 5$.
By polynomial division one obtains
$$
P_A(t)
= (t + 4)(4t^2 + 2t + 4)
= (t + 4)(t + 4) (4t + 1)
\equiv 4 (t + 4)^3
$$
Therefore $\lambda = 1$ is the only eigenvalue of $A$.
To find the eigenspace we calculate the kernel of $A + 4 E_3$ and obtain
$$
\text{span}\left( \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right)
$$
Since $(A + 4 E_3)^2 = 0$, the kernel of $(A + 4 E_3)^2$ is the whole space.
Now, I choose $v := (1, 0, 0) \in \text{ker}(A + 4 E_3)^2$ such that $v \not\in \text{ker}(A + 4 E_3)$.
We calculate $(A + 4E)v = (3, 0, 1)$ and then
$$
(A + 4E)
\begin{pmatrix} 3 \\ 0 \\ 1\end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},
$$
but the zero vector can't be a basis vector of our Jordan decomposition.
Have I made a mistake in my calculations?
| You want three independent vectors, $v_1,v_2,v_3$ that have the properties:
$$Av_1=v_1, Av_2=v_2+v_1, Av_3=v_3.\tag{1}$$
For $v_2=(1,0,0)^T$ not in the eigenspace, we get $v_1=Av_2-v_2=(3,0,1)^T$ [*] and then you need a $v_3$ which is in the eigenspace of $A$ but not a multiple of $v_1.$ We'll choose $v_3=(0,1,0)^T.$ Then if $$S=\begin{pmatrix}v_1&v_2&v_3\end{pmatrix}=\begin{pmatrix}3&1&0\\0&0&1\\1&0&0\end{pmatrix}$$ then we can show:
$$J=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}=S^{-1}AS$$
Since $Se_i=v_i$ and the equalities in (1) and and $S^{-1}v_i=e_i,$ so we get $$Je_1=e_1,Je_2=e_1+e_2,Je_3=e_3$$
Since $\det S = 1$ even in the integers, we can use Wolfram Alpha to invert $S$ over the integers and get $$S^{-1}=\begin{pmatrix}0&0&1\\1&0&-3\\0&1&0
\end{pmatrix}$$
[*] Note that since $(A-E_3)^2=0,$ you have $A^2-A=A-E_3$ and hence, when $v_1=Av_2-v_2,$ you have $Av_1=(A^2-A)v_2=(A-E_3)v_2=Av_2-v_2=v_1.$
We could have started with any $v_2$ not in the eigenspace. We'll always get some multiple of our original $v_1$ for $v_1.$ Then $v_3$ can be any of $20$ vectors in the eigenspace not a multiple of $v_1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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
| Obviously, $a,b,c\in \left[0,2\right]$ so $abc\leqslant 8$. Thus
$$abc = \sqrt[3]{abc}\cdot \sqrt[3]{(ab)(bc)(ca)} \leqslant 2\left(\frac{ab+bc+ca}{3}\right) \leqslant 2(ab+bc+ca).\tag{1}$$
Thus
$$ 4=a^2+b^2+c^2+abc \leqslant (a+b+c)^2.$$
Equality occurs when we have equality in (1), which happens only when both sides are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving $d = \frac{v^2}{2g} \left( 1+ \sqrt{ 1+ \frac{2g y_0}{v^2 \sin^2{\theta}} } \right)\sin{2\theta}$ for $\theta$
$$d = \frac{v^2}{2g} \left( 1+ \sqrt{ 1+ \frac{2g y_0}{v^2 \sin^2{\theta}} } \right)\sin{2\theta}$$
Need to find an equation for the angle.
This is the equation for the range of a projectile thrown from a height at an angle. I have tried to solve it myself and even used sites like Wolfram Alpha and Symbolab but to no avail. Could anybody please help?
The equation and derivation for the range of projectile launched at an angle from a height
| After a little messing around, I figured it out. Here it goes:
Given equation:
$$d = \frac{v^2}{2g} \left( 1+ \sqrt{ 1+ \frac{2g y_0}{v^2 \sin^2{\theta}} } \right)\sin{2\theta}$$
$$\implies \dfrac{2gd}{v^2}=(\sin 2\theta) +\left[\dfrac{(2\sin\theta\cos\theta)\sqrt{v^2\sin^2\theta+2gy_0}}{v\sin\theta}\right]$$
$$\implies \dfrac{2gd}{v^2}=2\cos\theta\left( \sin\theta+\sqrt{\sin^2\theta+\frac{2gy_0}{v^2}} \right)$$
$$\implies \dfrac{gd}{v^2}\sec\theta-\sin\theta=\sqrt{\sin^2\theta+\frac{2gy_0}{v^2}}$$
Let $\dfrac{gd}{v^2}=a$ and $\dfrac{2gy_0}{v^2}=b.$
$$\implies a\sec\theta-\sin\theta=\sqrt{\sin^2\theta+b}$$
Squaring on both sides,
$$\implies a^2\sec^2\theta+\sin^2\theta-2a\sec\theta\sin\theta=\sin^2
\theta+b$$
$$\implies a^2(1+\tan^2\theta)-2a\tan\theta-b=0$$
$$\implies a^2\tan^2\theta-2a\tan\theta+(a^2-b)=0$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve a system of non-linear equations How should I go around solving this system of non-linear equations?
$$x+\frac{1}{y} =2 \frac{1}{3}$$
$$y+\frac{1}{z}=2\frac{3}{4}$$
$$z+\frac{1}{x}=-3\frac{1}{2}$$
I managed to solve it using substitution (i.e. eliminating $x$ and $y$ in the first equation), but it seems like there might be an easier method, consider the equations appear symmetrical
| Hint: From the first equation we get
$$y=\frac{1}{\frac{7}{3}-x}$$ and from the last one $$z=-\frac{7}{2}-\frac{1}{x}$$ so we get
$$\frac{1}{\frac{7}{3}-x}+\frac{1}{-\frac{7}{2}-\frac{1}{x}}=\frac{11}{4}$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.
Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.
How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?
Any help or hint will be appreciated.
| $\Sigma \sin A\ge\Sigma\cos A\iff\Sigma\sin\left (A-\frac{\pi}{4}\right)\ge 0.$
If all are greater than $\frac{\pi}{4}$, it is trivially true.
let $A<\frac {\pi}{4}<B\le C$
Now using identity $\Sigma \sin x-\sin\Sigma x=4\prod\sin\left(\frac{x+y}{2}\right)$, we get:
$\Sigma\sin\left (A-\frac{\pi}{4}\right)=\frac{1}{\sqrt 2}+4\prod\cos \left (A+\frac{\pi}{4}\right)\ge \frac{1}{\sqrt 2}>0$
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve for x : $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$
Solve $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ for $x$
I started by multiplying both sides of the equation by $\frac{1}{2\sqrt{2}}$ to obtain
$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$
I am stuck here. Any hints on solving the R.H.S will be appreciated.
| square both sides, we get
$$2 \sin^2(x) + 6 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$
that is
$$2 + 4 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$
that is
$$2 + 4 \cos^2(x) + \sqrt{12}\sin(2x) = (1 + \sqrt{3})^2$$
that is
$$2 + 4 \frac{1}{1+\tan^2(x)} + \sqrt{12}\frac{2\tan x}{1+\tan^2(x)} = (1 + \sqrt{3})^2$$
Multiply everything by $1 + \tan^2(x)$
$$2(1+\tan^2(x)) + 4 +2 \sqrt{12}\tan x = (1 + \sqrt{3})^2(1+\tan^2(x)) $$
This is quadratic in $y = \tan(x)$
$$ay^2 + by + c = 0$$
where
\begin{align}
a &= 2- (1 + \sqrt{3})^2\\
b &=2\sqrt{12}\\
c &= 2 + 4 - (1 + \sqrt{3})^2
\end{align}
Solving you get something like
\begin{align}
y_1 &= 1.0000 \\
y_2 &= 0.2679
\end{align}
Now taking the inverse tangent of that you get
\begin{align}
x_1 &= \tan^{-1} y_1 =\tan^{-1} 1.0000 = 45^\circ\\
x_2 &= \tan^{-1} y_2 =\tan^{-1} 0.2679= 15^\circ
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
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.
| To evaluate $\int_0^\infty\frac{t^2+1}{t^4+\frac32 t^2+\frac32}$ we'll first write $t^4+\frac32 t^2+\frac32=(t^2+at+b)(t^2-at+b)$ with $a:=\sqrt{\frac32-\sqrt{6}},\,b:=\sqrt{\frac32}$. Hence $$\int_0^\infty\frac{t^2+1}{(t^2+at+b)(t^2-at+b)}=\frac{1}{2ab}\int_0^\infty\sum_\pm\frac{a\pm (1-b)t}{t^2\pm at+b}dt\\=\frac{1}{2ab}\left[\sum_\pm\left(\pm\frac{1-b}{2}\ln(t^2\pm at+b)+\frac{a(1+b)}{\sqrt{4b-a^2}}\arctan\frac{2t\pm a}{\sqrt{4b-a^2}}\right)\right]_0^\infty\\=\frac{1+b}{2b\sqrt{4b-a^2}}\left[\sum_\pm\left(\frac{\pi}{2}\mp\arctan\frac{a}{\sqrt{4b-a^2}}\right)\right]\\=\frac{\pi(1+b)}{2b\sqrt{4b-a^2}}=\frac{\pi(2+\sqrt{6})\sqrt{2\sqrt{6}+1}}{6\sqrt{23}}.$$(You'll want to double-check my arithmetic at the end.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $abc-a^2-b^2-c^2$ if $x,y,z$ are given in terms or $a,b,c$
if $$x=\frac{a}{b}+\frac{b}{a}$$
$$y=\frac{b}{c}+\frac{c}{b}$$
$$z=\frac{c}{a}+\frac{a}{c}$$
Then find
$$abc-a^2-b^2-c^2$$
As there are 6 variables in terms of 3 , we can fix any three without violating the conditions.
Say if $a=b=c=2$ then clearly the answer is$-4$ but how to find it in pure algebric way.
| It seems you want to find $xyz-x^2-y^2-z^2.$
If so:
$$xyz-\sum_{cyc}x^2=\prod_{cyc}\left(\frac{a}{b}+\frac{b}{a}\right)-\sum_{cyc}\left(\frac{a}{b}+\frac{b}{a}\right)^2=$$
$$=\frac{\prod\limits_{cyc}(a^2+b^2)}{a^2b^2c^2}-\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-6=$$
$$=\frac{\sum\limits_{cyc}\left(a^4b^2+b^4a^2+\frac{2}{3}a^2b^2c^2\right)}{a^2b^2c^2}-\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-6=2-6=-4.$$
In the full writing it's the following:
$$xyz-x^2-y^2-z^2=$$
$$=\frac{(a^2+b^2)(a^2+c^2)(b^2+c^2)}{a^2b^2c^2}-\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+6\right)=$$
$$=\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2+2a^2b^2c^2}{a^2b^2c^2}-$$
$$-\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+6\right)=$$
$$=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+2\right)-$$
$$-\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{b^2}+6\right)=2-6=-4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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CG Optimisation Function I need to prove the following statement:
Given the unique minimiser $\overline{x}$ of the function $f(x) := \frac{1}{2}<Ax,x> + <b,x> + c$ for some $c \in \mathbb{R}, b \in \mathbb{R}^n$ and a symmetric positive definite matrix $A$. Prove that $f(x) = f(\overline{x}) + \frac{1}{2}\vert\vert x-\overline{x} \vert\vert_A^2$ holds for all $x \in \mathbb{R}^n$
So far I got to this point:
$\begin{align*}
f(\overline{x}) + \frac{1}{2}\vert\vert{x - \overline{x}}\vert\vert_A^2 &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}((x-\overline{x})^TA(x-\overline{x})) \\
&= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}(x^TAx - x^TA\overline{x} - \overline{x}^TAx + \overline{x}^TA\overline{x}) \\
&= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^Tb + \frac{1}{2}\overline{x}^TA\overline{x} \\
&= \overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^Tb \\
&= \frac{1}{2}x^TAx - x^Tb + c + 2\overline{x}^Tb
\end{align*}$
How can I get $<b,x>$ from $2\overline{x}^Tb - x^Tb$?
| There is probably an error in your third equality, because you cannot get from $2\overline{x}^Tb - x^Tb$ to $<b,x>$ (since $\overline{x}$ depends on $A$). It goes well until the expression right before that:
$$\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^TA\overline{x},$$
but I do not see what the next step was that you took. If you plug in the expression for $\overline{x}$ (you can explicitly write the minimizer in terms of $A$ and $b$), you should arrive at the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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}
| We have
\begin{align} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+z)} &= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \sum_{k=0}^\infty \frac{1-(-1)^k}{(k+1)(k+z)} = \\
&= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \sum_{m=0}^\infty \frac{2}{(2m+2)(2m+1+z)} = \\
&= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \frac12 \sum_{m=0}^\infty \frac{1}{(m+1)(m+\frac{1+z}{2})} = \\
&= \frac{\psi(z)+\gamma}{z-1} - \frac12 \frac{\psi(\frac{1+z}{2})+\gamma}{\frac{1+z}{2}-1} = \\
&= \frac{\psi(z)-\psi(\frac{1+z}{2})}{z-1}\end{align}
where $\psi(z)$ is the digamma function and $\gamma$ is Euler-Mascheroni constant.
We have then
\begin{align} \int_0^1 \ln(1+x^n)dx &=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(n+nk+1)} = \\
&= \frac{1}{n} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+1+\frac{1}{n})} = \\&= \frac{1}{n} \frac{\psi(1+\frac{1}{n})-\psi(1+\frac{1}{2n})}{(1+\frac{1}{n})-1} = \\
&= \psi\big(1+\frac{1}{n}\big)-\psi\big(1+\frac{1}{2n}\big)\end{align}
It turns out that for any $n\in\mathbb N$ it can be expressed in terms of elementary functions. This is because
\begin{align} \psi\big(1+\frac{1}{n}\big) &= -\gamma + \sum_{k=0}^\infty\frac{\frac1n}{(k+1)(k+1+\frac1n)} = \\
&= -\gamma + \sum_{k=0}^\infty\Big(\frac{1}{k+1}-\frac{1}{k+1+\frac1n}\Big) = \\
&= -\gamma + \sum_{k=0}^\infty\int_0^1 (x^k-x^{k+\frac1n})dx = \\
&= -\gamma + \int_0^1 \frac{1-x^{\frac1n}}{1-x}dx = \\
&= -\gamma + \int_0^1 \frac{(1-y)ny^{n-1}}{1-y^n}dy = \\
&= -\gamma + \int_0^1 \frac{ny^{n-1}}{\prod_{k=1}^{n-1}(y-e^{i\frac{2\pi k }{n}})}dy \end{align}
and the last integral can be calculated using the partial fraction decomposition.
| {
"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": 0
} |
How to solve the recurrence relation $a_{1}=2, a_{n}=\frac{a_{n-1}+2}{2 a_{n-1}+1}(n \geq 2)$ with generating functions? There's already a way to solve it, called "fixed point method", that is, from the relation we define its characteristic equation as $x=\dfrac{x+2}{2x+1}$，then we have $x_1=1,x_2=-1$. So the following relation established:
$$
\frac{a_{n}-1}{a_{n}+1}=\frac{\frac{a_{n-1}+2}{2 a_{n-1}+1}-1}{\frac{a_{n-1}+2}{2 a_{n-1}+1}+1}=-\frac{1}{3} \cdot \frac{a_{n-1}-1}{a_{n-1}+1}
$$
It is obvious that $\displaystyle \frac{a_{n}-1}{a_{n}+1}=\frac{1}{3} \cdot\left(-\frac{1}{3}\right)^{n-1}$, and then we have $a_{n}=\dfrac{3^{n}-(-1)^{n}}{3^{n}+(-1)^{n}}$.
My question is, how to solve this kind of recurrence relations with generating functions? Also, "fixed points" can be applied to solving recurrences like $a_{n+1}=\dfrac{a_{n}^{2}+b}{2 a_{n}+d}$, which seems impossible to solve using generating functions.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{%
\left\{\begin{array}{rcl}
\ds{a_{1}} & \ds{=} & \ds{2}
\\
\ds{a_{n}} & \ds{=} &
\ds{{a_{n - 1} + 2 \over 2 a_{n - 1} + 1}\,,\quad n \geq 2}
\end{array}\right.}}$
$\ds{\large Another\ Method}$: Lets $\ds{a_{n} \equiv x_{n}/y_{n}}$ such that
\begin{align}
&\bbox[5px,#ffd]{a_{n} = {x_{n} \over y_{n}} =
{x_{n - 1}\,/y_{n - 1} + 2 \over 2x_{n - 1}\,/y_{n - 1} + 1} = {x_{n - 1} + 2y_{n - 1} \over 2x_{n - 1} + y_{n - 1}}}
\end{align}
and sets $\ds{\quad x_{n} = x_{n - 1} + 2y_{n - 1}\,,\quad
y_{n} = 2x_{n - 1} + y_{n - 1}}$ which is equivqlent to
\begin{align}
\pars{\begin{array}{c}\ds{x_{n}} \\ \ds{y_{n}}\end{array}}
& =
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}
\pars{\begin{array}{c}
\ds{x_{n - 1}} \\ \ds{y_{n - 1}}
\end{array}} =
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}^{2}
\pars{\begin{array}{c}
\ds{x_{n - 2}} \\ \ds{y_{n - 2}}
\end{array}}
\\[5mm] & = \cdots =
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}^{n - 1}
\pars{\begin{array}{c}
\ds{x_{1}} \\ \ds{y_{1}}
\end{array}}
\end{align}
The above matrix has eigenvalues $\ds{\lambda_{1} = 3}$ and $\ds{\lambda_{2} = -1}$ with,
respectively, orthonormal eigenvectors
$\ds{{\bf u}_{1} = {1 \over \root{2}}{1 \choose 1}}$ and
$\ds{{\bf u}_{2} = {1 \over \root{2}}
{-1 \choose \phantom{-}1}}$. Then,
\begin{align}
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}} & =
\sum_{j = 1}^{2}\lambda_{j}\,{\bf u}_{j}\,{\bf u}_{j}^{T}
\\[5mm] \mbox{and}\
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}^{n - 1} & =
\sum_{j = 1}^{2}\lambda_{j}^{n - 1}\,\,
{\bf u}_{j}\,{\bf u}_{j}^{T}
\\[2mm] & =
{ 1 \over 2}\pars{\begin{array}{cc}
\ds{3^{n -1} - \pars{-1}^{n}} & \ds{3^{n -1} + \pars{-1}^{n}}
\\
\ds{3^{n -1} + \pars{-1}^{n}} & \ds{3^{n -1} - \pars{-1}^{n}}
\end{array}}
\end{align}
Therefore,
\begin{align}
a_{n} & =
{\bracks{3^{n - 1} - \pars{-1}^{n}}\ \overbrace{x_{1}/y_{1}}^{\ds{= a_{1} = 2}}\ +\ 3^{n - 1} + \pars{-1}^{n} \over
\bracks{3^{n - 1} + \pars{-1}^{n}}x_{1}/y_{1} + 3^{n - 1} - \pars{-1}^{n}}
\\[5mm] & =
\bbx{3^{n} - \pars{-1}^{n} \over 3^{n} + \pars{-1}^{n}} \\ &
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Proof by induction for nth derivative Show the following hold by induction:
$$\frac {d^n}{dx^n}\frac {e^x - 1}{x} = (-1)^n \frac{n!}{x^{n+1}} \left( e^x \left(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}\right) - 1 \right)$$
Proof.
It's not hard to show the base case hold.
For inductive step, we can also write this as:
$$\frac {d^n}{dx^n}\frac {e^x - 1}{x} = (-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!} - (-1)^n \frac{n!}{x^{n+1}} $$
Take derivative on both side:
$$\frac {d^{n+1}}{dx^{n+1}}\frac {e^x - 1}{x} = \frac {d}{dx}(-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!} - \frac {d}{dx}(-1)^n \frac{n!}{x^{n+1}} $$
$$\frac {d^{n+1}}{dx^{n+1}}\frac {e^x - 1}{x} = \underbrace{\frac d{dx}(-1)^n\frac{n!}{x^{n+1}}e^x\sum_{k=0}^n(-1)^k\frac{x^k}{k!}}_? - \underbrace{(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}}_{hold} $$
Therefore, my question is for the first part, how do I show the following hold:
$$ \frac d{dx}(-1)^n\frac{n!}{x^{n+1}}e^x\sum_{k=0}^n(-1)^k\frac{x^k}{k!}
= (-1)^{n+1}\frac{(n+1)!}{x^{n+2}} e^x \sum_{k=0}^{n+1} (-1)^{k} \frac{x^k}{k!}$$
| It looks like you need to use the product rule here:
$$\frac {d}{dx}((-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!})$$
$$=(-1)^n \frac{d}{dx}(\frac{n!}{x^{n+1}}) e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} \frac{d}{dx}(e^x) \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} e^x \frac{d}{dx}(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!})$$
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}}e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=1}^{n} (-1)^{k} \frac{x^{k-1}}{(k-1)!}$$
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}}e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n-1} (-1)^{k+1} \frac{x^{k}}{k!}$$
Now notice that the last two terms cancel each other out except for the extra $n$th term in the middle sum. So this becomes
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}}e^x (-1)^{n} \frac{x^n}{n!}$$
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^{n+1} \frac{(n+1)!}{x^{n+2}}e^x (-1)^{n+1} \frac{x^{n+1}}{(n+1)!}$$
Here I just multiplied by $(-1)(-1)\frac{(n+1)}{(n+1)}\frac{x}{x}=1$ in the second term. Finally, we use the distributive property to finish it off.
$$(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\left(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}+(-1)^{n+1} \frac{x^{n+1}}{(n+1)!}\right)$$
$$(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n+1} (-1)^{k} \frac{x^k}{k!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral $\small \int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} \text{d}x$
Prove $$\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} \text{d}x = 8\pi$$
I tried using a modified version of the integrand and an origin-indented semicircular contour on the positive real half of the complex plane.
Despite my best efforts, I am unable to retrieve the desired integral, as the negative factor from the negative imaginary axis is preventing me from being able to simplify the integrals.
On top of that, the residue doesn't come anywhere close to the exact result.
Are there better methods that I am missing?
| Before we start I should mention that we will use some instances of Schroder's Integral which evaluates Gregory coefficients, namely:
$$\sf (-1)^{n-1}G_n=\int_0^\infty\frac{1}{(\pi^2+\ln^2 t)(1+t)^n}dt$$
But let's adjust things first.
$$\sf \color{orange}{I=\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} dx} \overset{x^2=t}=2\int_0^\infty \frac{(\pi\sqrt t-\ln t)^3}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$\sf =2\pi^3\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt-6\pi^2 \int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$\sf +6\pi \int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt-2\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$\sf =2\pi^3 I_1-6\pi^2I_2+6\pi I_3-2 I_4$$
Evaluation of $I_1$.
$$\tt \color{blue}{I_1=\int_0^\infty \frac{(1+t)-1}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)}dt-\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt}$$
$$\tt \color{blue}{=G_1+G_2=\frac12-\frac{1}{12}=\frac{5}{12}}$$
Evaluation of $I_2$.
I have solved this integral here.
$$\tt \color{red}{I_2=\int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt\overset{t=\frac{1}{x}}=-\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=\frac{\pi}{24}}$$
Evaluation of $I_3$.
$$\tt \color{purple}{I_3= \int_0^\infty \frac{(\pi^2 +\ln^2 t)-\pi^2}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{1}{(1+t)^2}dt-\pi^2\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt}$$
$$\tt \color{purple}{=1+\pi^2 G_2=1-\frac{\pi^2}{12}}$$
Evaluation of $I_4$.
Write $\ln^3 t= \ln t((\pi^2+\ln^2 t ) -\pi^2 )$ to get:
$$\tt \color{green}{I_4=\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}=\int_0^\infty \frac{\ln t}{(1+t)^2}\frac{dt}{\sqrt t}-\pi^2 \int_0^\infty \frac{\ln t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}}$$
$$\tt \color{green}{=-\pi+\pi^2 I_2=-\pi +\frac{\pi^3}{24}}$$
$$\require{cancel} \sf \Rightarrow \color{orange}I=\cancel{2\pi^3\color{blue}{\frac{5}{12}}} -\cancel{6\pi^2\color{red}{\frac{\pi}{24}}}+6\pi \left(\color{purple}{1-\cancel{\frac{\pi^2}{12}}}\right)-2\left(\color{green}{-\pi +\cancel{\frac{\pi^3}{24}}}\right)=\color{orange}{8\pi}$$
We actually only used $G_1$ and $G_2$ and I'm sure that we can evaluate them using elementary methods, atleast the first one is pretty easy, but for the second one we might have to strive a little.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Investigating this $\int_{0}^{a}\frac{(9-4\cos x)\cos x\cos(x/2)}{\sqrt{1+\sin(x/2)}}\mathrm dx$ Investigating this type of integral
$$\int_{0}^{a}\frac{(9-4\cos x)\cos x\cos(x/2)}{\sqrt{1+\sin(x/2)}}\mathrm dx=H(a)$$
I got the following results for:
$H(\pi)=2\sqrt{2}$,
$H(\pi/2)=2\sqrt{3}$,
$H(\pi/3)=2\sqrt{5}$ and
$H(2\pi/3)=\sqrt{21}$
expanded to... but I can't see it given me a better way of working it out.
$$\int_{0}^{a}\frac{9\cos x\cos^2(x/4)-9\cos x\sin^2(x/4)-4\cos^2 x\cos^2(x/4)+4\cos^2 x\sin^2(x/4)}{\sqrt{1+\sin(x/2)}}dx$$
My question is: What is the general formula for $H(a)?$
| Substitute $u=1+\sin(x/2)$ and then due to the fact that $\cos(x)=1-2(u-1)^2$ the integral breaks up into integrals of powers:
$$H(x)=2\int\frac{(9-4(1-2(u-1)^2))(1-2(u-1)^2)}{\sqrt{u}}du=-2\int({13\over\sqrt{u}} + 68 \sqrt{u} - 98 u^{3/2} + 64 u^{5/2} - 16 u^{7/2})du$$
which can easily be integrated for the result:
$$H(x)=-52 \sqrt{u} + 272 \frac{u^{3/2}}{3} - 392 \frac{u^{5/2}}{5} + 256 \frac{u^{7/2}}{7} -
64 \frac{u^{9/2}}{9}\Bigg|_{u=1+\sin(\frac{x}{2})}$$
and for the definite integral defined above
$$H(a)=-52 \sqrt{1+\sin(\frac{x}{2})} + 272 \frac{(1+\sin(\frac{x}{2}))^{3/2}}{3} - 392 \frac{(1+\sin(\frac{x}{2}))^{5/2}}{5} + 256 \frac{(1+\sin(\frac{x}{2}))^{7/2}}{7} -
64 \frac{(1+\sin(\frac{x}{2}))^{9/2}}{9}+\frac{3236}{315}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Given that $a$ and $b$ are integers satisfied $3 \mid ab(a + b) + 2$, prove that $9 \mid ab(a + b) + 2$.
Given that $a$ and $b$ are integers satisfy $3 \mid [ab(a + b) + 2]$, prove that $9 \mid [ab(a + b) + 2]$.
This problem is adapted from a recent competition. (The last few ones were and the next few ones will be.) I want to ask if the solution I have provided below is correct and if there are any other solutions which are more suitable more a test setting.
| We have that $$3\mid ab(a + b) + 2 \implies ab(a + b) \equiv 1\text{ (mod 3)}$$
$$\implies \left[ \begin{align} ab \equiv 1&\text{ (mod 3) and } a + b \equiv 1\text{ (mod 3)}\\ ab \equiv -1&\text{ (mod 3) and } a + b \equiv -1\text{ (mod 3)} \end{align} \right.$$
If $ab \equiv -1\text{ (mod 3)}$ then we have that $\left[ \begin{align} a \equiv 1&\text{ (mod 3) and } b \equiv -1\text{ (mod 3)}\\ a \equiv -1&\text{ (mod 3) and } b \equiv 1\text{ (mod 3)} \end{align} \right. \implies 3 | a + b$
(which is contradictory to $a + b \equiv 2\text{ (mod 3)}$).
$\implies ab \equiv 1\text{ (mod 3)}$ and $a + b \equiv 1\text{ (mod 3)} \implies a \equiv b \equiv 2\text{ (mod 3)}$.
Let $a = 3m - 1$ and $b = 3n - 1 \ (m, n \in \mathbb Z)$.
We have that $$ab(a + b) + 2 = (3m - 1)(3n - 1)[(3m - 1) + (3n - 1)] + 2$$
$$= [9mn - 3(m + n) + 1][3(m + n) - 2] + 2$$
$$= [3(m + n) - 1][9mn - 3(m + n) + 2] - 9mn + 2$$
Furthermore, $$[3(m + n) - 1][9mn - 3(m + n) + 2] + 2 = 9(3m^2n + 3mn^2 - m^2 - n^2 - 3mn + m + n)$$
which is divisible by $9$. That means $$[3(m + n) - 1][9mn - 3(m + n) + 2] - 9mn + 2$$
divisible by $9$ or $9 \mid [ab(a + b) + 2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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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.
| Suppose we are looking at the binomial coefficient $$\binom {2^n}m=\frac {(2^n)!}{m!(2^n-m)!}=\frac {2^n}m\cdot\frac {2^n-1}1\cdot\frac{2^n-2}2\dots \cdot \frac {2^n-(m-1)}{m-1}$$
(assuming $m\gt 1$: we have only the first term for $m=1$, and for $m=0$ the empty product or straightforwardly the value $1$)
Now suppose $2^n\gt r \gt 0$ and that $r=2^st$ where $t$ is odd, and we have $s\lt n$, then all the fractions in the product apart from the first are of the form $$\frac {2^n-r}r=\frac {2^n-2^st}{2^st}=\frac {2^{n-s}-t}{t}$$and this is a fraction with odd numerator and denominator. On the other hand with $\frac {2^n}m$ we have that $m$ is divisible by a lower power of $2$ than $2^n$ unless $m=2^n$. So we have a positive power of $2$ in the product.
This can be adapted for powers of any prime.
| {
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Nature of $\sum\left(\cos \frac{1}{n^\alpha}\right)^n$
How can I study the convergence of this series $$\sum\cos^n
\frac{1}{n^\alpha}$$ depending on $\alpha >0$
?
I found that $$\cos^n \frac{1}{n^\alpha} \sim \exp\left(\frac{-1}{2n^{2\alpha -1}}\right)$$
The case $2\alpha - 1 \geq 0$ can then be treated, as the series diverges.
But how can I exploit that same similar to treat the other case?
Thanks
| Using two facts, this one and this one, we have
$$1-\frac{x^2}{2}\leq \cos{x}\leq e^{-\frac{x^2}{2}}, x \in \left[0,\frac{\pi}{2}\right]$$
or, because $\frac{1}{n^{\alpha}}$ will be close to $0$ for suficiently large $n$
$$1-\frac{1}{2n^{2\alpha}}\leq
\cos{\frac{1}{n^{\alpha}}}\leq
\frac{1}{e^{\frac{1}{2n^{2\alpha}}}}$$
and, applying Bernoulli's inequality
$$1-\frac{n}{2n^{2\alpha}}\leq
\left(1-\frac{1}{2n^{2\alpha}}\right)^n\leq
\cos^n{\frac{1}{n^{\alpha}}}\leq
\frac{1}{e^{\frac{n}{2n^{2\alpha}}}}$$
Thus, for $2\alpha-1\geq0$
$$\lim\limits_{n\to\infty}\cos^n{\frac{1}{n^{\alpha}}} \ne 0$$
and the series doesn't converge.
For $2\alpha-1<0$ or $0<1-2\alpha$ and
$$0<\cos^n{\frac{1}{n^{\alpha}}}\leq
\frac{1}{e^{\frac{n}{2n^{2\alpha}}}}=
\left(\frac{1}{e}\right)^{\frac{n^{(1-2\alpha)}}{2}}$$
Using this limit (more details here)
$$\lim\limits_{n\to\infty}\frac{n^{1-2\alpha}}{\ln{n}}=\infty$$
for suficiently large $n$ we will have
$$n^{1-2\alpha}>4\ln{n}=\ln{n^4} \Rightarrow \\
e^{\frac{n^{1-2\alpha}}{2}}>e^{\frac{\ln{n^4}}{2}}=n^2 \Rightarrow \\
\left(\frac{1}{e}\right)^{\frac{n^{1-2\alpha}}{2}}<\frac{1}{n^2}$$
and by comparison test, the series converges.
| {
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How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.
| When partial fraction decomposition becomes a bit overwhelming, you can apply the Horowitz-Ostrogradsky algorithm ! [Manuel Bronstein - Symbolic Integration I]
It is very mechanical, only the calculation of $H$ is tedious, but the rest is quite easy.
So we start with $$\frac AD=\frac{1}{(x+1)(x+2)^2(x+3)^3}$$
$A=1$
$D=(x+1)(x+2)^2(x+3)^3$
$D\,'=(x+2)(x+3)^2(6x^2+22x+18)$
$D^-=\gcd(D,D\,')=(x+2)(x+3)^2$
$D^*=D/D^-=(x+1)(x+2)(x+3)$
$B=\sum_{i=0}^{\deg(D^-)-1}b_ix^i=b_0+b_1x+b_2x^2$
$C=\sum_{i=0}^{\deg(D^*)-1}c_ix^i=c_0+c_1x+c_2x^2$
And let's identify to the null polynomial $$\forall x:\quad H(x)=A-B\,'D^*+BD^*{D^{-}}'/D^--CD^-=0$$
$H(x)=-c_2x^5+(b_2-c_1-8c_2)x^4+(2b_1-2b_2-c_0-8c_1-21c_2)x^3+(3b_0+4b_1-15b_2-8c_0-21c_1-18c_2)x^2+(10b_0-4b_1-12b_2-21c_0-18c_1)x+(7b_0-6b_1-18c_0+1)$
This system solves to $\begin{cases}B=\frac 94x^2+\frac{25}{2}x+17\\C=\frac 94x+\frac 52\end{cases}$
And the formula says :
$$\int \frac AD\mathop{dx}=\frac{B}{D^-}+\int\frac{C}{D^*}\mathop{dx}=\frac{9x^2+50x+68}{4(x+2)(x+3)^2}+\int\frac{9x+10}{4(x+1)(x+2)(x+3)}\mathop{dx}$$
The last part is still solved by partial fraction decomposition, but is much simpler:
$\int=-\frac{17}8\ln(x+3)+2\ln(x+2)+\frac 18\ln(x+1)$
| {
"language": "en",
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Greatest common divisor proof attempt I am trying to prove the following assertion:
"If $a$ and $b$ are odd, then $(2a,2b)=(a+b, a-b)$".
$(x,y)$ denotes the greatest common divisor of $x$ and $y$. I am trying to prove it by showing that $(a,b) \vert (a+b, a-b)$ and $(a+b, a-b) \vert (a,b)$.
Denote $(a+b, a-b)$ by $d.$ Since $d \vert (a+b)$ and $d \vert (a-b)$ then $d \vert (a+b) + (a-b) = 2a$ and also $d \vert (a+b)-(a-b) = 2b$. This implies that $d \vert (2a,2b) = 2(a,b)$.
The other way of the proof is where I am struggling. Suppose $(a,b) = y$. Since $a$ and $b$ are odd we have that $a=2n+1$ and $b=2m+1$ for some integers $n,m.$ The sum $a+b$ and the difference $a-b$ is therefore even, and also we have that
$a=yr$ and $b=ys$ for some integers $r,s$ so $a+b = y(r+s)$ and $a-b = y(r-s).$ This implies that $y \vert (a+b)$ and $y \vert (a-b)$ which implies that $y \vert (a+b, a-b)$. However, I have not been able to show that $2y \vert (a+b, a-b)$. Any pointers in the right direction would be much appreciated!
| $ D=(a,b)$ so that $D|a,\ D|b,\ 2D=(2a,2b)$ and $D$ is odd
Hence $D|a+b,\ D|a-b$ so that $D|(a+b,a-b)$
Since $a+b,\ a-b$ are even, $2D|(a+b,a-b)$
When $D'=2kD|(a+b,a-b),\ k>1$, then
$$2kD=D'|(a+b)-(a-b)=2b
,\ kD|b$$
Similarly, $kD|a$ so that it is a contradiction to $D=(a,b)$
| {
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nth term of the sequence 1,2,3,5,7,9............ what will be the formula for finding the nth term of a series in which the difference between the terms increase by 1 after every k elements
For example (for k = 3) : 1,2,3,5,7,9,12,15,18........
(k=2) : 1,2,4,6,9,12,16,20,25......
i found some formula for smaller values of k (2,3,4,5)
for k=1 : floor(((2n+1)*(2n+1))/8)
k=2 :floor(((n+1)*(n+1))/4)
i tried to look for a pattern in these formula and i found for k>2:
if k is even : floor((n+k/2)^2)/(2*k))
if k is odd : floor(n+k/2)*(n+k/2+1)/(2*k))
but these do not hold true for k>=10
| For positive integers $n$ and $k$ we find: $$a_{nk}=\left(1+2+\cdots+n\right)k=\frac{1}{2}\left(n+1\right)nk$$
Then for $r\in\left\{ 1,2,\dots,k\right\} $ we find: $$a_{nk+r}=a_{nk}+\left(n+1\right)r=\frac{1}{2}\left(n+1\right)nk+\left(n+1\right)r=\frac{1}{2}\left(n+1\right)\left(nk+2r\right)$$
Setting $m=nk+r$ we find:
*
*$\lceil\frac{m}{k}\rceil=n+1$ and
*$m-k\lceil\frac{m}{k}\rceil+k=nk+r-nk-k+k=r$
so that: $$a_{m}=a_{nk+r}=\frac{1}{2}\lceil\frac{m}{k}\rceil\left(m+r\right)=\frac{1}{2}\lceil\frac{m}{k}\rceil\left(2m-k\lceil\frac{m}{k}\rceil+k\right)$$
| {
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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.
| Since $x^3\equiv_3 x$ for each integer $x$ we have $$3|a+b+c \iff a+b\equiv_3 -c$$ $$\iff (a+b)^3\equiv_3 (-c)^3$$
$$\iff a^3+3ab(a+b)+b^3\equiv_3 -c^3$$
$$\iff a^3+b^3+c^3\equiv_3 0 $$ $$\iff 3\mid a^3+b^3+c^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove limit doesn't exist $\lim\limits_{(x,y)\to(0,0)} \frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$ I need to show that limit doesn't exist:
$\lim\limits_{(x,y)\to(0,0)} \frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$
How can I show it?
| Polar coordinates:
$\dfrac{1-\cos (r^2)}{r^6 \cos^2 t \sin^2 t}= 4\dfrac{1- \cos (r^2)}{r^6 \sin^2(2t)}.$
$2t \not = kπ$, $k=0,1,2,3$.
Fix $t$, and consider
$A \dfrac{1-\cos (r^2)}{r^6}$, with $A=\dfrac{4}{\sin^2(2t)}>0$.
$A \lim_{r \rightarrow 0}\dfrac{1-\cos (r^2)}{r^6}=\infty.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\int_{a}^{b}\frac{\ln x ~ dx}{(x+a)(x+b)}=\frac{\ln ab}{2(b-a)} \ln \left(\frac{(a+b)^2}{4ab} \right), a, b>0.$ This integral is tabulated in Gradshetyn and Ryzhik.
The Mathematica gives nice results only for particular values of $a$ and $b$. Can one prove this result?
| Following MartinR's suggestion we show a slight generalization.
Let $r>0$ and $0<a<b$. By using the substitution $t=\frac{ab}{x}$, it follows that
$$\begin{align}
I_r&:=\int_{a/r}^{rb}\frac{\ln(x) }{(x+a)(x+b)}\,dx\\
&=\int_{rb}^{a/r}\frac{\ln(ab/t) }{(\frac{ab}{t}+a)(\frac{ab}{t}+b)}\cdot -\frac{ab}{t^2}dt\\
&=
\int_{a/r}^{rb}\frac{\ln (ab/t) }{(b+t)(a+t)}\,dt\\&=\ln (ab)\int_{a/r}^{rb}\frac{dt}{(b+t)(a+t)}\,dt-I.\end{align}$$
Hence
$$\begin{align}
I_r&=\frac{\ln (ab)}{2}\int_{a/r}^{rb}\frac{dt}{(b+t)(a+t)}\,dt\\
&=\frac{\ln (ab)}{2(b-a)}\int_{a/r}^{rb}\left(\frac{1}{a+t}-\frac{1}{b+t}\right)\,dt
\\&=\frac{\ln (ab)}{2(b-a)}\left[\ln\left(\frac{a+t}{b+t}\right)\right]_{a/r}^{rb}\\
&=
\frac{\ln (ab)}{2(b-a)}\ln\left(\frac{(a+rb)^2}{(1+r)^2ab}\right).
\end{align}$$
For $r=1$ we got the result. For $r\to +\infty$ we obtain
$$\int_0^{+\infty} \frac{ \ln(x)}{(x+a)(x+b)} dx=\frac{\ln (ab)\ln(b/a)}{2(b-a)}=\frac{\ln^2 (b)-\ln^2(a)}{2(b-a)}.$$
See Integral involving logarithm: $\int_0^\infty \frac{ \ln x}{(x+a)(x+b)} dx$.
P.S. Case $b=-1$ and $a>0$
$$\begin{align}
\int_0^1 \frac{\ln{x}}{\left(1-x\right)\left(x+a\right)}dx
&= \frac{1}{1+a}\int_0^1 \frac{\ln x}{1-x}dx - \frac{1}{1+a}\int_0^{1}\frac{\ln (x)}{x+a}dx \\
&= \frac{\pi^2}{6(1+a)}-\frac{\text{Li}_2(-1/a)}{1+a}
\end{align}$$
Moreover, by letting $t=1/x$
$$\begin{align}\int_1^{+\infty} \frac{\ln{x}}{\left(1-x\right)\left(x+a\right)}dx
&=\frac{1}{a}\int_0^{1} \frac{\ln{t}}{(1-t)(1/a+t)}dx\\
&=\frac{1}{a}\left(\frac{\pi^2}{6(1+1/a)}-\frac{\text{Li}_2(-a)}{1+1/a}\right)
\end{align}$$
Hence, after summing the two parts we get
$$\int_0^{+\infty} \frac{\ln{x}}{\left(1-x\right)\left(x+a\right)}dx=\frac{\pi^2}{3(1+a)}-\frac{\text{Li}_2(-1/a)+\text{Li}_2(-a)}{1+a}=\frac{\pi^2+\ln^2(a)}{2(a+1)}$$
where we used the identity $\text{Li}_2(-1/a)+\text{Li}_2(-a)=-\frac{\pi^2}{6}-\frac{\ln^2(a)}{2}$.
| {
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Prove $x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$ . Prove
$$x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$$
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
We have
$$(\,x+ y+ z\,)^{\,2}+ (\,-\,x+ y+ z\,)^{\,2}+ (\,x- y+ z\,)^{\,2}+ (\,x+ y- z\,)^{\,2}= 4(\,x^{\,2}+ y^{\,2}+ z^{\,2}\,)= 36$$
Or
$$\left ( z+ (\,y- x\,) \right )^{\,2}+ \left ( z- (\,y- x\,) \right )^{\,2}+ (\,3- 2\,z\,)^{\,2}= 27$$
Or
$$3\,z^{\,2}- 6\,z+ (\,y- x\,)^{\,2}= 9$$
Or
$$(\,y- x\,)^{\,2}= -\,3(\,z- 1\,)^{\,2}+ 12\leqq 12\,\therefore\,y- x\leqq |\,y- x\,|\leqq 2\sqrt{3}$$
Q.E.D. The equality condition occurs when $z= 1\,\therefore\,x+ y= 2\,\therefore\,x= 1- \sqrt{3},\,y= 1+ \sqrt{3}$.
Say it (Added)
The @user10354138's solution is so amazing, I try writing the inequality into the homonogeous form, then find $t\!=\!constant$ such that $3(\!y- z\!)^{\!2}\leqq 2t(\!x+ y+ z\!)^{\!2}+ 2(\!1- t\!)(\!x^{\!2}+ y^{\!2}+ z^{\!2}\!)$. That will lead to:
$${\rm discriminant}= 0\,\therefore\,t= -\,2,\,-\,\frac{1}{2},\,1$$
The coefficients of $y^{2}$ and $z^{2}$ both are negative there, I can't make the form like the solution as follow !
| Another using calculus:
$(x+y+z)^2 = x^2+y^2+z^2 +2(xy+yz+zx) \implies (xy+yz+zx)=0 \implies xy = -z(x+y)\implies xy=-z(3-z)$
$(y-x)^2 = (x+y)^2 -4xy$
Now
$(y-x)^2 = (3-z)^2 +4z(3-z) = 9+6z-3z^2$
Maximum of $(y-x)$ can be found by setting $\frac{d\sqrt{9+6z-3z^2}}{dz} = 0$
so $z = 1$
and Evaluating at $z=1$ $(y-x) \le2\sqrt{3}$
| {
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Proving $a = b = c$ under certain conditions For all real a, b, c, prove a = b = c if $$\frac{a^2+b^2+c^2}{3} = (\frac{a+b+c}{3})^2 $$
The first idea that came to mind would be to prove this inequality by contradiction. However, I am unsure about how to go about it. Would using the AM-GM inequality be useful? This is what it could possibly be simplified to make things easier:
$$ 2(a^2 + b^2 + c^2) = 2(ab + ac + bc) $$
Any help would be extremely appreciated.
| Write
$$ 2(a^2 + b^2 + c^2) = 2(ab + ac + bc) $$
As
$$ (a^2 + b^2 - 2ab) + (c^2 + b^2 - 2cb)+ (a^2 + c^2 - 2ac) = 0$$
Now since $ (a^2 + b^2 - 2ab) = (a-b)^2 $
The expression becomes
$$(a-b)^2 + (a-c)^2 + (c-b)^2 =0 $$
Which is only true if $$ a = b = c $$
| {
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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.
|
Hence, $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$
Don't forget $+2\pi n$.
Since the signs of $\cos\theta$ and $\sin\theta$ place $\theta$ in the fourth quadrant, only two candidates hold up. You can choose between them by testing whether $\cos\theta > \cos\frac{7\pi}{4}$
| {
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Summing cube roots in fractions I found this problem, and understand the solution, but do not understand why they made the first assumption.
The problem:
The first line of the solution says that:
The cube root of $1$ plus the cube root of $2$ plus the cube root of $4$ is a factor of $2-1$.
Why are you meant to assume this to solve the problem?
| They are pointing to the fact (it's not an assumption) that $$ (x^2+xy+y^2)(x-y) = x^3-y^3$$
that is
$$ \frac{1}{x^2+xy+y^2} = \frac{x-y}{x^3-y^3}$$
If you use that for $x=1$, $y=\sqrt[3]{2}$, you get
$$ \frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}} = \frac{\sqrt[3]{2}-1}{2-1} = \sqrt[3]{2}-1$$
To get the other two fractions use $x=\sqrt[3]{2}$, $y=\sqrt[3]{3}$ and $x=\sqrt[3]{3}$, $y=\sqrt[3]{4}$.
| {
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Functions satisfying: $f(f(x)^2+f(y))=xf(x)+y$ The problem is to find all the continuous functions $f:\mathbb{R}\to \mathbb{R}$ defined by :$f(f(x)^2+f(y))=xf(x)+y$
I'm trying my best to figure out a way to find the expression of this unknown function by plugging some numbers, but I could not. Thanks in advance for your help.
| The answers are $f(x) = \pm x$.
First, plug in $x = 0$ and $y = 0$, to get that $f(f(0)^2 + f(0)) = 0$. Therefore, let $x = f(0)^2 + f(0)$. Then, we get $f(f(y)) = y$. Then, plug in $y = f(0)^2 + f(0)$. We get that $f(0) = f(0)^2 + f(0)$, or that $f(0) = 0$.
Now, let $f(a) =1$. Clearly, $f(1) =a$ as well. Now, plugging $x = a, y = 1$ in to the original equation, we get that $f(a+1) = a+1$. However, plugging in $x = 1, y = 1$ gives us that $f(a^2 + a) = a + 1$, showing that $a^2 + a = a + 1$, which means that $a = \pm 1$.
Case 1: $f(1) = 1$. Let $x = 1$. Then, we get $f(1+f(y)) = 1+y$. Now, let $y = 1$. Then, we get that $f(f(x)^2 + 1) = xf(x)+1$. However, if, in the first equation, we let $y = xf(x)$, we get that $f(1+xf(x)) = 1 + xf(x)$, which means that $f(x)^2 = x(f(x))$ or that $f(x) = x$.
Proceed similarly in the second case.
| {
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Length of a perpendicular bisector to a line intersecting a curve The line $x+y-2=0$ intersects the curve $2x^2-y^2+2x+1=0$ at the points $A$ and $B$. The perpendicular bisector of the line $AB$ intersects the curve at the points $C$ and $D$. Find the length of the line $CD$ in the form $a\sqrt5$, where $a$ is an integer.
This was a question from a recent exam for IGCSE Additional Mathematics, and here are my steps:
$$y = -x+2$$
Substitute into the equation of the curve:
$$2x^2 - (-x+2)^2+2x+1=0$$
$$2x^2 - (x^2-4x+4) + 2x + 1 = 0$$
$$x^2+6x-3=0$$
$$x = \frac{-6 ± \sqrt{(6)^2-4(1)(-3)}}{2}$$
$$= 2\sqrt{3}-3, -2\sqrt{3}-3$$
$$y = -2 \sqrt{3} + 5, 2\sqrt{3}+5$$
The slope of $AB$ is $\frac{(-2\sqrt{3}+5) - (2 \sqrt{3} + 5)}{(2\sqrt{3}-3) - (-2\sqrt{3}-3)} = \frac{-4\sqrt3}{4\sqrt3} = -1$, so the slope of the perpendicular bisector is $1$.
The midpoint of $AB$ is $\left(\frac{(2\sqrt{3}-3) + (-2\sqrt{3}-3)}{2}, \frac{(-2\sqrt{3}+5) + (2 \sqrt{3} + 5)}{2} \right) = (-\frac{6}{2}, \frac{10}{2} = (-3, 5)$. The perpendicular bisector can be written in the form $y = x + c$, and with $x = -3, y=5$, we have $c = 8$. Therefore, $y = x+8$.
To find $C$ and $D$, we have:
$$2x^2-(x+8)^2 + 2x + 1 =0 $$
$$2x^2-(x^2+16x+64) + 2x+1=0$$
$$x^2-14x-63=0$$
$$x = \frac{-(-14) ± \sqrt{(-14)^2-4(1)(-63)}}{2}$$
$$= 7 + 4\sqrt7, 7 - 4\sqrt7$$
$$y = 15 + 4\sqrt7, 15 - 4\sqrt7$$
Therefore the distance between $C$ and $D$ is:
$$\sqrt{\left((7 + 4\sqrt7) - (7 - 4\sqrt7)\right)^2 + \left((15 + 4\sqrt7) - (15 - 4\sqrt7)\right)^2}$$
$$= \sqrt{(8\sqrt7)^2 + (8\sqrt7)^2}$$
$$= 8\sqrt{14}$$
but my answer is not in the form $a\sqrt5$.
Is my answer correct? If not, I would be happy to receive an alternate solution or a correction to my work.
| Sorry if this answer is too short.
I think it should have been $x+y+2=0$. That gives a distance that is a multiple of $\sqrt 5$.
I checked after your comment about it being $x-y+2=0$. That also gives a multiple of $\sqrt 5$ - the same, curiously, as for $x+y+2=0$.
On second thoughts, not that curious. The lines $x+y+2=0$ and $x-y+2=0$ are reflections of each other in the $x$-axis and the curve is symmetrical about the $x$-axis, too.
| {
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"source": "stackexchange",
"question_score": "1",
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First order differential equation with $y,y',$ and $\sqrt y$ I have been struggling with this equation:
$(x^2+1)y'-2xy=4\sqrt{(x^2+1)y}\arctan x$
I have tried with $y=z^m$ to make homogeneous equation, but I didn't get anything anything useful.
Left side also looks a lot like quotient rule, so I tried solving in that direction, but again it didn't work. Whatever I do I can't seem to get it to any standard form.
Thanks for help.
| Let $y = z^2$. We have:
$$ 2(x^2 + 1) z z' - 2x z^2 = 4 z \sqrt{x^2+1} \arctan(x)$$
$$ z' - \frac{xz}{x^2 + 1} = 2 \frac{1}{\sqrt{x^2+1}}\arctan(x)$$
Homogeneous equation:
$$z' = \frac{xz}{x^2+1} $$
Therefore, $z = c \sqrt{x^2+1}$
Variation of constant:
$$c' \sqrt{x^2+1} + c \frac{x}{\sqrt{x^2+1}} - c \frac{x}{\sqrt{x^2+1}} = 2 \frac{1}{\sqrt{x^2+1}}\arctan(x) $$
$$c' = \frac{2}{x^2+1}\arctan(x)$$
$$c = \arctan(x)^2 + a$$
$$z = \arctan(x)^2 \sqrt{x^2+1} + a\sqrt{x^2+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimize value of the function $a^2+b^2+c^2+2\sqrt{3abc}$ Let $a,b,c$ be the positive real numbers such that $a+b+c=1$. Find Minimize of $$P=a^2+b^2+c^2+2\sqrt{3abc}$$
WA says that $P$ gets only a local minimum. But i think it must be maximum value of $P$.
Then by AM-GM: $$\text{L.H.S}= a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}$$
$$\le a^2+b^2+c^2+2(ab+bc+ca)$$
$$=(a+b+c)^2=1$$
| Let $\psi(x)$ is symmetric function and have gradient on every point of its domain. If $x_i^* = x_j^*$ for any $i,j \in [1,n]$, then $\psi_{x_1}(x^*) = \psi_{x_2}(x^*) = \cdots = \psi_{x_n}(x^*)$.
Let $\xi_i = (x_1,\cdots,x_{i-1},x_{i+1},\cdots,x_n)$. if $x_i = \varphi(\xi_i)$ and $\nabla x_i = \pm \mathbf{1}$, then there is a Lagrange function $$L(\xi_i,\lambda) = \psi(\xi_i) + \lambda(\alpha - x_i)$$ and $$\nabla L = 0 \longrightarrow \nabla \psi(\xi_i) = \lambda \nabla x_i = \pm \lambda \mathbf{1} $$
where $\psi(\xi_i)$ is also symmetric function.
Therefore, if $\xi_i^* = c\mathbf{1}$ and $x_i = 0$ such that $\varphi(\xi_i^*) = 0$, then $\xi_i^*$ is critical point of $L(\xi_i),$ as well as $\psi(\xi_i)$.
Hence, $\vartheta_i = (x_1,\cdots,x_{i-1},0,x_{i+1},\cdots,x_n)$ is critical point of $\psi(x)$ for any $i \in [1,n]$.
The symmetric function $f(x) = x_1^2 + x_2^2 + x_3^2 + a\sqrt{x_1,x_2,x_3}$ where $x_1 + x_2 + x_3 = b$ thereby have critical point $$\vartheta_3 = \left(\frac{b}{2},\frac{b}{2},0\right)$$
and $$\min f(x) = f\left(\vartheta_3\right) = \frac{b^2}{2}.$$
For instance $b = 1$, it is $$\min f(x) = \frac{1}{2}.$$
| {
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Why are $\left(\begin{smallmatrix}0&1\\0&0\\\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0\\0&1\\\end{smallmatrix}\right)$ not similar?
Why are the matrices $\begin{pmatrix}0 & 1 \\0 & 0 \\\end{pmatrix}$ and $\begin{pmatrix}0 & 0 \\0 & 1 \\\end{pmatrix}$ not similar?
I can do a row operation. But on the other hand they don't have the same
characteristic polynomial. How does it relates to: $A$ is similar to $B$ if and only if $A$ and $B$ have the same canonical form ?
| Hint: Suppose they were equivalent then there had to exists a invertible matrix $S := \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ such that
\begin{equation} \tag{1}
S^{-1} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} S = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\end{equation}
holds.
We have
$
S^{-1}
= \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.
$
Therefore, $(1)$ becomes
\begin{align}
& \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\
\iff & \frac{1}{ad - bc} \begin{pmatrix} c d & d^2 \\ -c^2 & -c d \end{pmatrix}
= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\end{align}
This yields $cd = 0$ and $-cd = 1$, which clearly is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the remainder of the division of $a$ by $18$ knowing that $\gcd(a^{226} +4a +1, 54)=3$ Let's define $b:= a^{226} +4a +1$. We know that $b$ is odd because $54$ is even and the gcd is odd. But if $a$ were odd, $b$ would be even; so $a$ is also even.
We also know that $3\nmid a$ (since $3\nmid b$, if it did divide $a$, it would be a divisor of $1$, which is absurd.) Applying Fermat's theorem, we have
$$a^{226} + 4a + 1 = (a^2)^{113} + 4a + 1 \equiv 4a + 2 \equiv 0 \pmod 3$$
This means that $a \equiv 1 \pmod 3$. We infer the following congruences:
$$a \equiv 0 \pmod2 \\ a \equiv 1 \pmod 3$$
If I had a $\pmod 9$ instead of a $\pmod 3$ in the last congruence, I'd be able to aply the Chinese Remainder Theorem.
How can I bound the values of $r_{9}(a)$, given that $r_{3}(a)=1$?
| Let's try to find a solution where all steps can be performed without computer.
binomial expansion and reduction of the big power
You found $a=6c+4$. In the binomial expansion of $(6c+4)^{226}$ all powers of $6$ from the third on are divisible by $54$. $\binom{226}{2}=113\cdot 225$ is divisible by $9$, so that also the quadratic coefficient reduces to zero modulo $54$. Thus
$$
(6c+4)^{226}\equiv 226⋅(6c)⋅4^{225}+4^{226}=(113⋅3c+1)⋅4^{226}\pmod{54}
$$
reduction of the dyadic powers
Then with
$$
2^{18}=(64)^3\equiv 10^3\equiv -80\equiv 28\pmod{54}\implies 2^{18k+1}\equiv 2\pmod{54}
$$
we can reduce
$$
4^{226}=2^{25⋅18+2}\equiv 2^2=4\pmod{54}
$$
equivalent GCD conditions
The GCD identity then reduces to
$$
GCD\left((15c+1)⋅4+4⋅(6c+4)+1,54\right)=3\\
GCD(30c+21,54)=3\\ GCD(10c+7,18)=1
$$
which is satisfied for $c\not\equiv 2\pmod3$.
($10c+7$ is invertible $\bmod{18}$, the multiplicative group is $\{1,5,7,11,13,17\}+18\Bbb Z=\pm1+6\Bbb Z$, thus $10c\in \{0,4\}+6\Bbb Z$, $-2c\in \{0,-2\}+6\Bbb Z$ and consequently $c\in\{0,1\}+3\Bbb Z$.)
(new) conclusion $a\bmod{18}$
Combining the parametrizations together we get that either
*
*$a=6(3k)+4=18k+4$ giving $a\equiv 4\pmod{18}$ or
*$a=6(3k+1)+4=18k+10$ giving $a\equiv 10\pmod{18}$.
old conclusion $a\bmod8$
With $c=3k$ one gets $a=18k+4\equiv 2k+4\pmod 8$ and with $c=3k+1$ likewise $a=18k+10\equiv 2k+2\pmod 8$.
| {
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Differential equation $\frac{dy}{dx}+\frac{2y}{x}=\frac{4}{x^2}$ (need help)
Find the general solution of the differential equation:
$$\frac{dy}{dx}+\frac{2y}{x}=\frac{4}{x^2}$$
$$\frac{dy}{dx}x^2+2xy=4\tag1$$
$$x^2+\int 2xydx=\int 4dx\tag2$$
$$x^2+yx^2=4x+c\tag3$$
$$y=\frac{4x+c}{x^2+1}\tag4$$
The answer from the book is $$y=\frac{4}{x}+\frac{c}{x^2}$$
| You can solve firstly the homogeneous differential equation.
$\frac{dy}{dx}=-\frac{2y}{x}$
$\frac{dy}{y}=-\frac{2}{x}dx$
$\int \frac{dy}{y}=-\int\frac{2}{x}dx$
$\ln(y)=-2\cdot \ln(x)+c$
$\ln(y)=- \ln(x^2)+c$
$\ln(y)=\ln\left(\frac{1}{x^2}\right)+c$
$y_H=C\cdot \frac{1}{x^2}$
Variation of the constant.
$y_I=C(x)\cdot \frac{1}{x^2}$
Differentiating (product rule)
$y_I^{'}=C^{'}(x)\cdot \frac{1}{x^2}-C^{}(x)\cdot \frac{2}{x^3}$
Plugging into the differential equation
$C^{'}(x)\cdot \frac{1}{x^2}\underbrace{-C^{}(x)\cdot \frac{2}{x^3}+2\cdot C(x)\cdot \frac{1}{x^3}}_{=0}=\frac{4}{x^2}$
$C^{'}(x)\cdot \frac{1}{x^2}=\frac{4}{x^2}$
$C^{'}(x)=4\Rightarrow C(x)=4x$
Thus the solution is $y=y_H+y_I=C\cdot \frac{1}{x^2}+4x\cdot \frac{1}{x^2}$
$$y=\frac{C}{x^2}+ \frac{4}{x}$$
| {
"language": "en",
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Prove that $abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+ca(c+a-2)+a+b+c \le 2$ Prove that if the real numbers $a,b,c$ lie in the interval $[0,1]$,
then:
\begin{align*}
abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+ca(c+a-2)+a+b+c \le 2.
\end{align*}
I've tried this:
\begin{align*}
abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+
\end{align*}
\begin{align*}
+ ca(c+a-2)+(a+b+c-2) \le abc(a^2+b^2+c^2-2abc-2a-2b-2c)+
\end{align*}
\begin{align*}
+ab(2-2)+bc(2-2)+ca(2-2)+(3-2)=
\end{align*}
\begin{align*}
=abc(a^2+b^2+c^2-2abc-2a-2b-2c)+1
\end{align*}
but that $1$ in the last line ruins it even though that first term is clearly negative. Can anyone help me?
| $$\sum_{cyc}ab(a+b-2)\leq0$$ and
$$abc\sum_{cyc}(a^2-a)\leq0.$$
Thus, it's enough to prove that
$$a+b+c-abc(a+b+c)\leq2.$$
Now, let $a=\frac{1}{1+x},$ $b=\frac{1}{1+y}$ and $c=\frac{1}{1+z},$ where $x$, $y$ and $z$ are non-negatives.
Thus, we need to prove that
$$2(1+xyz)^2+\sum_{cyc}(3x^2y^2z+x^2y^2+4x^2yz+x^2y+x^2z+xy+x)\geq0,$$ which is obvious.
| {
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How to calculate the GCD? How to evaluate the following with the help of Mobius function ?
$$\displaystyle\sum_{i=1}^n \sum_{j=i+1}^n \sum_{k=j+1}^n \sum_{l=k+1}^n {gcd(i,j,k,l)^4} .$$
In other words, we have to select all possible quadruplets from (1 ton n), and then sum up their value with power 4.
Example:-
N=4 :
(1,2,3,4) : gcd(1,2,3,4)^4 = 1
Total sum = 1
For Second Case, let N=5 :
(1,2,3,4) : gcd(1,2,3,4)^4 = 1
(1,2,3,5) : gcd(1,2,3,5)^4 = 1
(1,2,4,5) : gcd(1,2,4,5)^4 = 1
(1,3,4,5) : gcd(1,3,4,5)^4 = 1
(2,3,4,5) : gcd(2,3,4,5)^4 = 1
Total sum = 1+1+1+1+1 = 5.
My approach is:
To calculate the number of quadruplets with gcd=2,3,4,...till n.
Say, the number of quadruplets with gcd=2 are x1, for gcd=3, its x2...and so on.
Now, l = $$\binom{n}{4}$$-(x1+x2+.......xn-1) = number of quadruplets with gcd=1.
Now, the final answer = l^4+x1^4+x2^4+.....
The only problem I have is how to calculate the number of quadruplets with gcd=x with the help of mobius function ?
| A quadruplet with gcd a multiple of $d$ occurs iff all $i,j,k,l$ are multiples of $d$, for which there are ${\lfloor n/d\rfloor\choose d}$ possibilities.
So you'll arrive at
$${n\choose 4}\cdot 1^4+{\lfloor n/2\rfloor\choose 4}\cdot (2^4-1^4) +{\lfloor n/3\rfloor\choose 4}\cdot (3^4-1^4)+{\lfloor n/4\rfloor\choose 4}\cdot (4^4-2^4)+\cdots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sqrt{5}BA \leq PA +PB+\sqrt{2}PC$ Let $ABC$ be an isosceles right triangle ($\angle B=90^o$) and a point $P$ in its plane. Prove the inequality $\sqrt{5}BA \leq PA +PB+\sqrt{2}PC$. Find all poins $X$ for which the equality holds.
It might be the most difficult problem I have ever met. First, I thought it has a relation to the lemma $\frac{PA}{a}+\frac{PB}{b}+\frac{PC}{c}\geq \sqrt 3$ . But no! The equality holds differently from the problem. I try to create a new isosceles right triangle with side AC but it does not work. Any idea for this, thank!
| Let $AB=1$, $A(1,0),$ $B(0,0)$, $C(0,1)$ and $P(x,y)$.
Thus, by Minkowski:
$$PA+PB+\sqrt2PC=\sqrt{(x-1)^2+y^2}+\sqrt{x^2+y^2}+\sqrt{2(x^2+(y-1)^2)}=$$
$$=\sqrt{(x-1)^2+(-y)^2}+\sqrt{(-y)^2+(-x)^2}+\sqrt{(-x+y-1)^2+(x+y-1)^2}\geq$$
$$\geq\sqrt{(x-1-y-x+y-1)^2+(-y-x+x+y-1)^2}=\sqrt5=\sqrt5BC.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274549",
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"source": "stackexchange",
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Under condition $2x^2 + y^2= 4$ for real numbers $x, y$, find the maximum and minimum value of $4x + y^2$. How can I solve this problem. I can find absolute maximum and minimum value of equation with given interval. But here, I don't know where should I start. Could you explain step by step?!
| $4x+y^2=4x +4-2x^2=$
$4+ 2x(2-x);$
Constraint:
$2x^2=4-y^2$;
$x_{1,2} =\pm \sqrt{2-y^2/2}$
$4-y^2 \ge 0$; or $|y| \le 2$; and
$-√2 \le x \le √2$.
Max & Min of
$Y:=4+4x-2x^2=$
$ -2(x^2-2x)+4= -2(x-1)^2+6$;
$\max (Y _x)= 6$; (cf. lab bhatthachargee).
$\min (Y_x)$ :
The parabola $Y=-2(x-1)^2 +6$ has vertex at $(1,6)$, and is opening downward.
Since $-√2 \le x \le √2$, we find
the minimum at the boundary point:
$x=-√2$:
$\min Y_x=-2(-√2-1)^2+6$.
| {
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Can we conclude $0^\circMy attempt
Based on the sine rule and the graph of $\sin A = k a$ (where $k$ is a constant) in interval $(0,\pi)$,
increasing $a$ up to $1/k$ will either
*
*increase $A$ up to $90^\circ$.
*decrease $A$ up to $90^\circ$.
So I cannot conclude that increasing $a$ will increase $A$.
Now I use the cosine rule (it is promising because the cosine is decreasing in the given interval).
\begin{align}
A &= \cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)\\
B &= \cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)\\
C &= \cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)\\
\end{align}
It is hard to show that $0^\circ<A\leq B\leq C<180^\circ$ for any $\triangle ABC$ with $0<a\leq b\leq c$. Could you show it?
It means that I need to show that
$$
-1<\frac{a^2+b^2-c^2}{2ab}\leq \frac{a^2+c^2-b^2}{2ac} \leq \frac{b^2+c^2-a^2}{2bc}<1
$$
for $0<a\leq b\leq c$.
| Let $a<b$ and $D\in AC$ such that $CD=BC.$
Thus, $$\measuredangle A<\measuredangle CDB=\measuredangle CBD<\measuredangle CBA.$$
| {
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Prove $\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}$
Prove
$$S=\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}.$$
I don't know how to evaluate this problem .
At first I used partial fraction but I got divergent series, so I used
$$\Gamma(n)=\int_0^{+\infty}t^{n-1}e^{-t}dt.$$
This yields
$$S = \sum_{n=1}^{\infty}\frac{\int_0^{+\infty}t^{n-1}e^{-t}dt}{(2n+1)(2n+2)(n-1)!}$$
Now i can exchange integral and sum.
But I don't know how to proceed.
| Since
$$\Gamma\left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{4^nn!}$$
We want to prove:
$$\sum_{n=1}^{\infty}\frac{\frac{(2n)!\sqrt{\pi}}{2^{2n}n!}}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}$$
Which is equivalent to:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n}n}{(2n+1)(2n+2)2^{2n}}=\frac{4-\pi}{4}$$
Consider the central binomial coefficient series:
$$\sum_{n=1}^{\infty} {2n \choose n} x^n=\frac{1}{\sqrt{1-4x}}$$
Substituting $x^2$ instead of $x$ we obtain:
$$\sum_{n=1}^{\infty} {2n \choose n} x^{2n}=\frac{1}{\sqrt{1-4x^2}}$$
Integrating twice we get:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n} x^{2n+2}}{(2n+1)(2n+2)}=\frac{\sqrt{1-4x^2}+2x\sin^{-1}(2x)-1}{4}$$
(We have to make sure that the value at zero of the left hand side and its derivative is zero, like on the right)
Now, dividing by $x^2$, taking derivative and multiplying by $\frac{1}{2}x$ we obtain:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n} nx^{2n}}{(2n+1)(2n+2)}=\frac{1-x\sin^{-1}(2x)-\sqrt{1-4x^2}}{4x^2}$$
Now we just plug in $x=\frac{1}{2}$ to obtain:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n} n}{(2n+1)(2n+2)2^{2n}}=\frac{1-\frac{1}{2}\sin^{-1}(2\cdot \frac{1}{2})-\sqrt{1-4(\frac{1}{2})^2}}{4(\frac{1}{2})^2}=\frac{4-\pi}{4}$$
As desired.
| {
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"url": "https://math.stackexchange.com/questions/3277383",
"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.
| Too long for comments.
Using another CAS, I have not been able to obtain the rhs (except for $m=2$) but numerically the results do agree with your conjecture (checked up to $m=20$).
Considering
$$f_m=\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n$$ running cases, what I obtained is
$$f_3=\frac{2 ^2}{3^2} \, _4F_3\left(1,1,\frac{4}{3},\frac{5}{3};\frac{3}{2},2,2;1\right)$$
$$f_4=\frac{3^3}{4^3} \,
_5F_4\left(1,1,\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{4}{3},\frac{5}{3},2,2;1
\right)$$
$$f_5=\frac{4^4}{5^4} \,
_6F_5\left(1,1,\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{5}{4},\frac
{6}{4},\frac{7}{4},2,2;1\right)$$
$$f_6=\frac{5^5}{6^5}\,
_7F_6\left(1,1,\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac
{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5},2,2;1\right)$$
$$f_7=\frac{6^6}{7^6}\,
_8F_7\left(1,1,\frac{8}{7},\frac{9}{7},\frac{10}{7},\frac{11}{7},\frac{12}{7},
\frac{13}{7};\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6},2,2;1
\right)$$ which, as written, reveal very clear patterns.
$$\color{blue}{f_m=\frac{(m-1)^{m-1}}{m^{m-1}}\,
_{m+1}F_m\left(1,1,\frac{m+1}m,\cdots,\frac{2m-1}m;\frac m{m-1},\cdots,\frac {2m-3}{m-1},2,2;1\right)}$$
Trying on Wolfram Cloud, I obtained the same results but no simplification at all. Surprising, isn't it ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| would solve it like this:
Hint: You want to split limits into two, in order for the main limit to be defined, two sub-limits need to for a start be undefined, and that will be achieved like this:
$$\lim_{x\to 2} \frac{\frac{\sqrt[3]x-\sqrt[3]2}{x-2}}{\frac{x^3-8}{x-2}}=\frac{l1}{l2}$$
then I will solve just $l1$ but you will get an idea:
$$l1=\lim_{x\to 2}\frac{\sqrt[3]x-\sqrt[3]2}{x-2} \cdot \frac{\sqrt[3]{x^2+\sqrt[3]{2x}+\sqrt[3]4}}{\sqrt[3]{x^2+\sqrt[3]{2x}+\sqrt[3]4}}= \frac{x-2}{(x-2)(\sqrt[3]{x^2+\sqrt[3]{2x}+\sqrt[3]4})}=\frac{1}{3 \cdot \sqrt[3]4}$$
Now you do the same for the $l2$, divide it, and it's done. THIS IS MY FIRST ANSWER!!! GIVING BACK TO THE COMMUNITY!!!
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$. Let $x$ be an odd natural number. Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$.
For the first part I would assume
$$x \equiv 1 \mod 2$$
$$x^2 \equiv 1^2 \mod 2^2$$
$$x^2 +2 \equiv 1^2+2 \mod 2^2$$
$$x^2 +2 \equiv 3 \mod 4$$
Can someone briefly tell me if this is correct and if it is a rule that squaring the number on the left means you square both numbers on the right?
Also I don't know how to start with the second part.
Note: This question is to do with rings in general.
| First, you can square both sides in the congruence relation but you must not square the number that you are "moding".
If $x$ is odd, then $x\equiv1$ or $3$ (mod $4$). So, by some calculations,
$$x^2\equiv1(\mod 4)$$
$$x^2+2=3(\mod 4)$$
Then, for the next part, we will use prove by contradiction.
First, we can see that all of the factors of $x^2+2$ is odd because $x^2+2$ is odd.
If $x^2+2$ has no prime factor which has a remainder of $3$ when divided by $4$, then $x^2+2$ should have only prime factors which has a remainder of $1$ when divided by $4$. As the product of any number of numbers that leaves a remainder of $1$ when divided by $4$ also has a remainder of $1$ when divided by $4$, so it contradicts that $x^2+2\equiv3 (\mod 4)$
$\therefore x^2+2 $ has at least one factor that leaves a remainder of $3$ when divided by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find coordinate points where the tangent line is horizontal for $f(x) = –\sin(8x) + 6\cos(4x) – 8x$ Problem:
Consider the function $f(x) = –\sin(8x) + 6\cos(4x) – 8x$ where $–\pi/4 < x < \pi/2$. Find the exact $x$-coordinates of the points on the graph of f at which there is a horizontal tangent line.
This is my attempt.
Find derivative
$$ f’(x) = -\cos(8x)\cdot8 – 6\sin(4x)\cdot4 – 8 $$
$$f’(x) = -8\cos(8x) – 24\sin(4x) – 8$$
$$ f’(x) = -8(\cos(8x) + 3\sin(4x))$$
$$f’(x) = 0 $$
$$0 = -8(\cos(8x) + 3\sin(4x) + 1)$$
Use double angle formula $1-\sin(4x)$ to replace $\cos(8x)$
$$0 = -8(1 – \sin^2(4x) + 3\sin(4x) + 1)$$
Let $u = \sin(4x)$
$$ 0 = 1 – u^2 + 3u + 1$$
$$ 0 = -u^2 + 3u + 2$$
Use quadratic formula to solve for $u$.
$$\sin(x) = (3\pm \sqrt{17}) / 2$$
Answer can only be negative due to out of bounds so $\sin(x) = (3-\sqrt17) / 2$
Multiply equation by $4$ because $\cos(4x) $
$$\sin(4x) = [4\cdot(3-\sqrt{17})] / 2$$
I tried multiplying by $4$ but $\sin$ becomes out of bounds
So I just tried to get the inverse of $\sin(x)$ to find the reference angle.
$$\sin^{-1}(\frac{3–\sqrt{17}}{2}) = -0.596\text{rad} = -34^\circ$$
$$x = (\pi + 0.596), (2\pi-0.596), (-0.596), (-\pi+0.596)$$
After graphing these tangent lines in a calculator, the points are close but not exact and thus are not horizontal. I've done other horizontal tangent line problems but this one, it seems I have to use the quadratic formula to find $\sin(x)$, rather than factoring.
Any advice on what I'm doing wrong would be greatly appreciated.
| Let $$g(x) = f(x/4) = -\sin 2x + 6 \cos x - 2x.$$ Then
$$\begin{align*}
g'(x) &= -2 \cos 2x - 6 \sin x - 2 \newline
&= -2 (1 - 2 \sin^2 x) - 6 \sin x - 2 \newline
&= 2 (2 \sin^2 x - 3 \sin x - 2) \newline
&= 2 (2 \sin x + 1)(\sin x - 2).
\end{align*}$$
Hence the critical points of $g$ occur when $\sin x = -1/2$ (it is not possible for $\sin x = 2$). There are four such $x \in (-\pi, 2\pi)$: $$x \in \left\{-\frac{5\pi}{6}, - \frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}.$$ Consequently, the critical points of $f$ occur when $\sin 4x = -1/2$, and for $-\pi/4 < x < \pi/2$, this occurs at $$x \in \left\{ -\frac{5\pi}{24}, - \frac{\pi}{24}, \frac{7\pi}{24}, \frac{11\pi}{24} \right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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compute the following integral in closed form : $\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2}(x)+8\cos^{2} x}dx$ Evaluate
$I=\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2} (x)+8\cos^{2} x}dx$
How can I starte in this hard integral , at first use $y=\frac{π}{2}$ but no result so I this use :
$y=\tan \frac{x}{2}$ then $dx=2\frac{1}{1+y^2}dy$
$x=2\arctan y$
$\cos x=\frac{1-y^2}{1+y^2}$ $&$ $\sin x=2\frac{y}{1+y^2}$
So :
$8\cos^{2} x+(1+\sqrt 2)sin^{2} x=\frac{8(1-y^2)^2+4(1+\sqrt 2)y^2}{(1+y^2)^2}$
Now I get $arctan$ integral
$I=2\int_0^{\infty}\frac{(1+y^2)\arctan y}{8(1-y^2)^2+4(1+\sqrt 2)y^2}dy$
But I don't know how to complete this work!
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int_{0}^{\pi/2}\frac{x}{\pars{1 + \root{2}}
\sin^{2}\pars{x} + 8\cos^{2}\pars{x}}\,\dd x:\ {\LARGE ?}}$.
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\pi/2}{x \over
\pars{1 + \root{2}}
\sin^{2}\pars{x} + 8\cos^{2}\pars{x}}\,\dd x}
\\[5mm] & =
\int_{0}^{\pi/2}{x \over
\pars{1 + \root{2}}
\bracks{1 - \cos\pars{2x}}/2 + 8\bracks{1 + \cos\pars{2x}}/2}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\pi/2}{2x \over
9 + \root{2}
+ \pars{7 - \root{2}} \cos\pars{2x}}\,2\,\dd x
\\[5mm] & =
{7 + \root{2} \over 94}
\int_{0}^{\pi}{x \over
a + \cos\pars{x}}\,\dd x
\end{align}
where $\ds{a \equiv {65 + 16\root{2} \over 47} > 1}$.
Then,
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\pi}{x \over
a + \cos\pars{x}}\,\dd x}
\\[5mm] = &\
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi}{-\ic\ln\pars{z} \over
a + \pars{z + 1/z}/2}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left. -2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi}{\ln\pars{z} \over
z^{2} + 2az + 1}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\tag{1}\label{1}
\end{align}
Roots of $\ds{z^{2} + 2az + 1 = 0}$ are given by
$\ds{r_{\pm} \equiv -a \pm \root{a^{2} - 1}}$ where $\ds{r_{-} < -1}$ and $\ds{-1 < r_{+} < 0}$.
\eqref{1} can be handle with the
"Polylogarithm Machinery".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integrating the trigonometric function So the problem goes as follows:-
$$\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx$$
My attempt is as follows:-
\begin{align*}\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx&=\int{{\cos^2x+2\sin x\cos x}\over{(2\cos x-\sin x)^2}}~dx\\
&=
\int{{(\cos x)(\cos x+2\sin x)}\over{(2\cos x-\sin x)^2}}~dx
\end{align*}
Now i could see that $\cos x+2\sin x $ is $-1$ times derivative of the denominator:-
$$\cos x+2\sin x=(-1){{d(2\cos x-\sin x)}\over{dx}}$$
And to handle the $\cos x$ term in the numerator i tried the following:-
$$2\cos x=t+\sin x \implies \cos x={{t+\sin x}\over{2}}$$
But the $\sin x$ term in the numerator is annoying.
| Recall that $\int -{f'\over f^2} dx= {1\over f}+C$, setting $f(x)=2 \cos x-\sin x $ your integral is ${1\over 2 \cos x-\sin x}$
| {
"language": "en",
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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?
| Hint: Multiply numerator and denominator by $$\sqrt{7\sqrt{3}+4\sqrt{5}}$$ and then by $$\sqrt{67}$$
For your work we get
$$\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\int_ {0}^ \infty (\frac{ \sin x}{x})^3 = A$, then $\int_ {0}^ \infty \frac{ x- \sin x}{x^3} = ??$
If $\int_ {0}^ \infty (\frac{ \sin x}{x})^3 = A$, then $\int_ {0}^ \infty \frac{ x- \sin x}{x^3} = k A $. Find the value of k
The main problem is how to get started as in the given integral , there is a term of $(\sin x)^3$, where as in the integral to be computed , there is no cube. Please help :)
Edit :I'm still in highschool so don't know how to compute this integral and that is why the first one was given
| As $\sin^3 x=(3\sin x-\sin 3x)/4$, we have
\begin{align}
A&=\int_{0}^{\infty}\frac{(3x-\sin 3x)-3(x-\sin x)}{4x^3}\,dx
\\ &=\frac{1}{4}\int_{0}^{\infty}\frac{3x-\sin 3x}{x^3}\,dx-\frac{3}{4}\int_{0}^{\infty}\frac{x-\sin x}{x^3}\,dx
\\ &=\frac{9}{4}\int_{0}^{\infty}\frac{y-\sin y}{y^3}\,dy-\frac{3}{4}\int_{0}^{\infty}\frac{y-\sin y}{y^3}\,dy
\\ &=\frac{3}{2}\int_{0}^{\infty}\frac{x-\sin x}{x^3}\,dx.
\end{align}
Thus $k=2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Derivation of general parametric equation of chord on ellipse How is the equation of a chord for the ellipse in parametric form given 2 points $P(a\cos\theta,b\sin\theta)$ and $Q(a\cos\varphi,b\sin\varphi)$ as $$bx\cos((\theta+\varphi)/2) + ay\sin((\theta+\varphi)/2) = ab\cos((\theta-\varphi)/2)$$ derived?
I am not sure how to proceed after using point-gradient form. Any help is greatly appreciated, thanks.
| Recall the two-point form of line
$$
\frac{y-y_P}{x-x_P}=\frac{y_Q-y_P}{x_Q-x_P}
$$
yields
$$
\frac{y-b\sin\theta}{x-a\cos\theta}=\frac{b\sin\varphi-b\sin\theta}{a\cos\varphi-a\cos\theta}.
$$
Clearing denominators,
$$
(y-b\sin\theta)(a\cos\varphi-a\cos\theta)=(b\sin\varphi-b\sin\theta)(x-a\cos\theta)
$$
and use sum-to-product formulae,
$$
-a\sin\frac{\varphi-\theta}2\sin\frac{\varphi+\theta}2(y-b\sin\theta)=b\cos\frac{\varphi+\theta}2\sin\frac{\varphi-\theta}2(x-a\cos\theta)
$$
Removing the common factor $\sin\frac{\varphi-\theta}2$ and rearranging,
\begin{align*}
b\cos\frac{\varphi+\theta}2x+a\sin\frac{\varphi+\theta}2y&=ab\left[\cos\theta\cos\frac{\varphi+\theta}2+\sin\frac{\varphi+\theta}2\sin\theta\right]\\
&=ab\cos\left(\theta-\frac{\varphi+\theta}2\right)\\
&=ab\cos\frac{\theta-\varphi}2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Writing $I + x(A+B)+ M_2x^2+M_3x^3+\cdots = e^{x(A+B)+x^2R_2+x^3R_3+\cdots}$ Suppose I have two complex matrices $A,B$, of the same dimension I am wondering if it possible to represent for $x \in \mathbb{R}$, $e^{x/2A}e^{xB}e^{x/2A} = e^{x(A+B) + x^2R2+x^3R_3+\cdots}$ for some matrices $R_2,R_3,\ldots$.
I know that $e^{x/2A}e^{xB}e^{x/2A} = (I+x/2A+x^2/8A +\cdots)(I+xB+x^2/2B + \cdots)(I+x/2A+x^2/8A + \cdots)= I + x(A+B)+ M_2x^2+M_3x^3+\cdots $ for some matrices $\{M_i\}_{i=2}^{\infty}$.
Can I then write
$I + x(A+B)+ M_2x^2+M_3x^3+\cdots = e^{x(A+B)+x^2R_2+x^3R_3+\cdots}$ for some matrices $\{R_i\}_{i=2}^{\infty}$ ?
| Just expand:
$$\log(I + A) = A - \frac{1}{2} A^2 + \frac{1}{3} A^3 + \ldots $$
(converging if $\|A\| < 1$)
so (say for $|z|<r$ where $\log(e^{Az/2} e^{Bz} e^{Az/2})$ is analytic)
$$ \eqalign{\log(I + M_1 z + M_2 z^2 + M_3 z^3 &+ \ldots) = (M_1 z + M_2 z^2 + M_3 z^3 + \ldots) \cr &- \frac{1}{2} (M_1 z + M_2 z^2 + \ldots)^2 + \frac{1}{3} (M_1 z + \ldots)^3 + \ldots\cr
&= M_1 z + \left(M_2 - \frac{M_1^2}{2}\right) z^2 + \left(M_3 + \frac{M_1M_2 + M_2 M_1 }{2} + \frac{M_1^3}{3}\right) z^3+\ldots}$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b,c>0$ and $2(a+b+c)=3$ prove $\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4$ $$\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4$$
With $a,b,c>0$ and $2(a+b+c)=3$.
First I use AM-GM inequality
$$\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4\left(\frac{abc}{\prod_{cyc}(b^2+2(b+1))^2)}\right)^{1/3}$$
I don't if my idea help me or no and can complete my problem ?
| $f(x)=\frac{1}{(x+1)^2+1}$ is convex for $x\geq 0$, so by Jensen:
$$\sum_{cyc}\frac{a}{b^2+2(b+1)}≥(a+b+c)\frac{1}{\left(\frac{ab+bc+ca}{a+b+c}+1\right)^2+1}\geq (a+b+c) \frac{1}{\left(\frac{a+b+c}{3}+1\right)^2+1}=\frac{6}{13}$$
Last inequality is true because of $ab+bc+ca\leq \frac{(a+b+c)^2}{3}$
| {
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"url": "https://math.stackexchange.com/questions/3290335",
"timestamp": "2023-03-29T00:00:00",
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Number of ways to reach ground floor in $n$ steps Given an integer $n$ which is the number of steps from first floor to ground floor in a building. We can either move $1$ step down, or $2$ step down, or $3$ step down. However, we may move $3$ steps down at most once. In other words, a $3$ step move can be done any time but only once. We have to find the number of ways to reach the ground floor.
I thought the solution is simply:
$f[n] = f[n-1]+f[n-2]+f[n-3]$
However, I am not getting the right answer. What could be possibly wrong?
| The problem can be attacked using generating functions (e.g. here).
We can take $3$ steps move after the $k$-th step. I.e. split the entire walk into first $k$ steps, one $3$ steps move and $n-k-3$ steps. Then, for the first part, the number of integer solutions of
$$x_1+2x_2=k, x_1\geq0,x_2\geq0 \tag{1}$$
is the number of ways to traverse those $k$ steps with $1$ or $2$ steps moves. For the last - the number of integer solutions of
$$x_3+2x_4=n-k-3, x_3\geq0,x_4\geq0 \tag{2}$$
is the number of ways to traverse the last $n-k-3$ steps with $1$ or $2$ steps moves.
All these for $k=0$ to $n-3$.
Generally, the number of integer solutions for $$x_1+2x_2=k, x_1\geq0,x_2\geq0 \tag{3}$$
is the coefficient of $x^k$ of the generating function
$$(1+x+x^2+...)(1+x^2+x^4+...+x^{2n}+...)=\frac{1}{1-x}\cdot \frac{1}{1-x^2}=\\
\frac{1}{2(1-x)^2} + \frac{1}{4(1-x)} + \frac{1}{4(1+x)}=...$$
which is
$$...=\frac{1}{2}\left(\sum\limits_{n=0}(n+1)x^n\right)+
\frac{1}{4}\left(\sum\limits_{n=0}x^n\right)+
\frac{1}{4}\left(\sum\limits_{n=0}(-1)^nx^n\right)=\\
\sum\limits_{n=0}\left(\frac{n+1}{2}+\frac{1+(-1)^n}{4}\right)x^n$$
and the coefficient is
$$\frac{k+1}{2}+\frac{1+(-1)^k}{4} \tag{4}$$
Back to $(1)$ and $(2)$ we have
$$\frac{k+1}{2}+\frac{1+(-1)^k}{4} \text{ and }
\frac{n-k-3+1}{2}+\frac{1+(-1)^{n-k-3}}{4}$$
or
$$\frac{k+1}{2}+\frac{1+(-1)^k}{4}+\frac{n-k-3+1}{2}+\frac{1+(-1)^{n-k-3}}{4}=\\
\frac{n-1}{2}+\frac{2+(-1)^k+(-1)^{n-k-3}}{4}=\\
\frac{n}{2}+\frac{(-1)^k+(-1)^{n-k-3}}{4}$$
and finally
$$\sum\limits_{k=0}^{n-3}\left(\frac{n}{2}+\frac{(-1)^k+(-1)^{n-k-3}}{4}\right)=\\
\frac{n(n-2)}{2}+\sum\limits_{k=0}^{n-3}\left(\frac{(-1)^k+(-1)^{n-k-3}}{4}\right)=\\
\frac{n(n-2)}{2}+\frac{1}{2}\left(\sum\limits_{k=0}^{n-3}(-1)^k\right) \tag{5}$$
See if you can simplify it any further.
| {
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"answer_count": 3,
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} |
Finding all elements in $x\in U_{143}$ such that $x^2=1 \pmod{143}$
Find all elements in $x\in U_{143}$ such that $x^2=1 \pmod{143}$
I am kind of stuck here. I know that $143=11*13$, and perhaps looking at $U_{11},U_{13}$ will help but I am unable to find a solution so far.
| $x^2 \equiv 1 \pmod{143}$ is equivalent to solving the following system:
\begin{align*}
x^2 & \equiv 1 \pmod{11}\\
x^2 & \equiv 1 \pmod{13}
\end{align*}
In general for a prime $p>2$, the congruence $x^2 \equiv 1 \pmod{p}$ has only two solutions, namely $x = 1,p-1$. The reason being if $p \mid (x-1)(x+1)$ then by the prime property either $p \mid x-1$ or $p \mid x+1$. However it cannot divide both (why?).
Thus the above system reduces to the following:
\begin{align*}
x & \equiv 1 \pmod{11} & x & \equiv 1 \pmod{11} & x & \equiv -1 \pmod{11} & x & \equiv -1 \pmod{11}\\
x & \equiv 1 \pmod{13} & x & \equiv -1 \pmod{13} & x & \equiv 1 \pmod{13} & x & \equiv -1 \pmod{13}
\end{align*}
There are quite a few ways to do this, one of them being Chinese remainder theorem.
We can solve it directly as well, for example for
\begin{align*}
x & \equiv 1 \pmod{11}\\
x & \equiv -1 \pmod{13}
\end{align*}
From the first equation we can say $x=11k+1$, then in the second equation we will have $11k+1 \equiv -1 \pmod{13}$, which is same as $11k \equiv 11 \pmod{13}$. Thus $k \equiv 1 \pmod{13}$ ($\because \, 11$ is invertible mod $13$). So we have $\color{red}{x=12}$ as a solution.
| {
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"answer_count": 4,
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} |
Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?
Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?
I first let $y= x^2 -20x +100$.
Then substitute it in the function -------> $f(x) = \sqrt{-y+500}+\sqrt{y-100}$.
Which means, $100\leq{y}\leq{500}$. I then tried to find values of $y$ that would make both radicals disappear. I found four values of $y$ that made $f(x)$ integers; $100, 500, 356,$ and $244$. I also checked their discriminants and all of them were greater than $0$, which means each value of $y$ means two solutions for $x$, meaning, there are $8$ elements in the range of $f$ are integers. But the correct answer is $9$ and I can't seem to find the last one.
| Let $z=y-300$. You want to solve $\sqrt{200-z}+\sqrt{200+z}=N,$ where $N$ is an integer.
Square both sides: $400+2\sqrt{200^2-z^2}=N^2$ or $\sqrt{200^2-z^2}=\dfrac{N^2-400}2,$
which means $0\le\dfrac{N^2-400}2\le200$ or $400\le N^2\le800.$
Note we must have $N\ge0$. Can you take it from here?
| {
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} |
Solving $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$ How do you solve $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$?
Working
in modulo $p$ we have $x^5 = 0 \pmod{p}$ and $x = 0$ which is the only solution in modulo $p$ since $a^{p} \equiv a \mod{p}$ for any prime $p$ and integer $a$. Taking this back we get $p^2y^5 = 0$ which implies $y=0$. Hence the only solution is $(x,y) = (0,0)$.
Edit/attempt 2:
Since $x = 0 \mod{p}$ we can let $x = pm$ where $m<x$ then we get $p^2 (m^5 p^3 + m^2 p y^3 + y^5) = 0 \implies m^5 p^3 + m^2 p y^3 + y^5$ since $p$ can't be zero. But now $y^5 = 0 \mod{p}$ ; suppose $y = pn$ where $n < y$. Then $p^3 (m^5 + m^2 n^3 p + n^5 p^2) =0$ and now $m^5 + m^2 n^3 p + n^5 p^2=0.$
| Above equation shown below:
$x^5+px^2y^3+p^2y^5 = 0$ -------$(1)$
Equation $(1)$ has numerical solution given below:
$(x, y, p)=(-6, 3, -8)$
| {
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} |
Prove $\int_0^1 \frac{\tanh^{-1} (\beta t) dt}{t\sqrt{(1-t)(1- \alpha t)}}=\log (a) \log (b)$ If we set:
$$\alpha= \frac{(ab-1)^2+(a-b)^2}{(ab+1)^2+(a+b)^2}$$
$$\beta= \frac{(ab+1)^2-(a+b)^2}{(ab+1)^2+(a+b)^2}$$
Then it follows that:
$$\int_0^1 \frac{\tanh^{-1} (\beta t) dt}{t\sqrt{(1-t)(1- \alpha t)}}=\log (a) \log (b)$$
I have derived this result in a very roundabout way, most of the details you can see in this post, however from the symmetry of it I suspect there may be better and more clear ways to prove it, which is why I'm asking a separate question.
Aside from the proof, I'm interested in deeper reasons or implications for this identity (if they exist) and some references to similar ones.
| We can solve this integral using only substitutions and integration by parts, as follows:
$$I:=\int_0^1 \frac{\operatorname{arctanh} (\beta t) }{t\sqrt{(1-t)(1- \alpha t)}}dt=\int_0^1 \frac{\operatorname{arctanh}(\beta t)}{t(1-t)}\sqrt{\frac{1-t}{1-\alpha t}}dt$$
$$\overset{\large \frac{1-t}{1-\alpha t}=x}=\int_0^1 \frac{\operatorname{arctanh}\left(\beta \frac{1-x}{1-\alpha x}\right)}{\sqrt x(1-x)}dx\overset{x=y^2}=2\int_0^1 \frac{\operatorname{arctanh}\left(\beta \frac{1-y^2}{1-\alpha y^2}\right)}{1-y^2}dy$$
$$\overset{\large y=\frac{1-x}{1+x}}=\int_0^1 \operatorname{arctanh}\left( \frac{4\beta x}{(1+x)^2-\alpha (1-x)^2}\right)\frac{dx}{x}=\frac12 \int_0^1 \ln\left(\frac{\left(ab+x\right)\left(\frac{1}{ab}+x\right)}{\left(\frac{a}{b}+x\right)\left(\frac{b}{a}+x\right)}\right)\frac{dx}{x}$$
$$\overset{IBP}=\frac12 \int_0^1 \ln x \left(\frac{1}{\frac{a}{b}+x}+\frac{1}{\frac{b}{a}+x}-\frac{1}{ab+x}-\frac{1}{\frac{1}{ab}+x}\right)dx$$
In each of the integral from above we will simplify the denominator using the substitution $x\to kx$, where $k$ is the constant found in each denominator.
$$\Rightarrow I=\frac12 \left(\int_0^\frac{b}{a}\frac{\ln\left(\frac{a}{b}x\right)}{1+x}dx+\int_0^\frac{a}{b}\frac{\ln\left(\frac{b}{a}x\right)}{1+x}dx-\int_0^\frac{1}{ab}\frac{\ln\left(ab x\right)}{1+x}dx-\int_0^{ab}\frac{\ln\left(\frac{x}{ab}\right)}{1+x}dx\right)$$
$$\small =\color{red}{\frac12} \left(\ln\left(\frac{a}{b}\right)\ln\left(1+\frac{b}{a}\right)+\ln\left(\frac{b}{a}\right)\ln\left(1+\frac{a}{b}\right)-\ln(ab)\ln\left(1+\frac{1}{ab}\right)-\ln\left(\frac{1}{ab}\right)\ln\left(1+ab\right)\right)$$
$$+\color{chocolate}{\frac12}\left(\int_0^\frac{b}{a}\frac{\ln x}{1+x}dx+\int_0^\frac{a}{b}\frac{\ln x}{1+x}dx-\int_0^\frac{1}{ab}\frac{\ln x}{1+x}dx-\int_0^{ab}\frac{\ln x}{1+x}dx\right)$$
We can also rewrite the four integrals from above as:
$$\color{blue}{\int_\frac{1}{ab}^\frac{b}{a}\frac{\ln x}{1+x}dx}+\int_{ab}^\frac{a}{b}\frac{\ln x}{1+x}dx\overset{\color{blue}{x\to \frac{1}{x}}}=\color{blue}{\int_{ab}^\frac{a}{b}\frac{\ln x}{x}dx-\int_{ab}^\frac{a}{b}\frac{\ln x}{1+x}dx}+\int_{ab}^\frac{a}{b}\frac{\ln x}{1+x}dx$$
$$=\int_{ab}^\frac{a}{b}\frac{\ln x}{x}dx=\frac{\ln^2 x}{2}\bigg|_{ab}^\frac{a}{b}=-2\ln a\ln b$$
So with some algebra for the first term we finally get:
$$I=\color{red}{\frac12}\left(4\ln a \ln b\right)+\color{chocolate}{\frac12}\left(-2\ln a \ln b\right)=\boxed{\ln a\ln b}$$
An alternative approach using Feynman's trick can be found here, which shows:
$$\int_0^1 \ln\left(\frac{\left(ab+x\right)\left(\frac{1}{ab}+x\right)}{\left(\frac{a}{b}+x\right)\left(\frac{b}{a}+x\right)}\right)\frac{dx}{x}=2\ln a\ln b$$
It might be useful in the future so I'll also mention that, since $\int_0^1 \frac{\ln x}{t+x}dx=\operatorname{Li}_2\left(-\frac{1}{t}\right)$ the following Dilogarithm identity arises from above:
$$\boxed{\operatorname{Li}_2\left(-\frac{a}{b}\right)+\operatorname{Li}_2\left(-\frac{b}{a}\right)-\operatorname{Li}_2\left(-ab\right)-\operatorname{Li}_2\left(-\frac{1}{ab}\right)=2\ln a\ln b;\ a,b>0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Number of nine digits numbers whose sum of the digits is even I am reading Mathematical Circle. Problem $48$ in chapter two says that
How many nine-digit numbers have an even sum of their digits?
I am trying in this way, that we can divide the problem in four cases.
*
*$1$ even digit and $8$ odd digits
*$3$ even digits and $6$ odd digits
*$5$ even digits and $4$ odd digits
*$7$ even digits and $2$ odd digits
For the first case we get $4 \cdot 5^8 +5\cdot 5^7 \cdot 5$ number of solution. Because if the even digit is placed in first place (left to right) then we get $4\cdot 5^8$ ways to write the number and if an odd digit is placed in first place then we get $5\cdot 5^7 \cdot 5$ ways to write the number. Similarly for the second case we get $4\cdot 5^8+5^9$ , for the third case we get $4\cdot 5^8+5^9$ and for the fourth case we get $4\cdot 5^8+5^9$ ways to write the number. So total number is $ 4 \cdot (4\cdot 5^8+5^9)$.
The answer is different. So Where I have made a mistake?
Thanks.
| Let $A$ be the set of nine digit numbers. Its size is $|A|=9\cdot10^8$: we're free to choose eight digits among all ten digits, but the most significant digit can't be zero.
Now consider the following map $f\colon A\to A$:
$$
f(x)=\begin{cases}
x-9 & x\equiv 9\pmod{10} \\[4px]
x+1 & x\not\equiv9\pmod{10}
\end{cases}
$$
The map is bijective. If $A_0$ and $A_1$ denote the subsets of $A$ consisting of numbers with even or, respectively, odd digit sum, we can see that $f$ induces bijective maps $f_{01}\colon A_0\to A_1$ and $f_{10}\colon A_1\to A_0$.
Therefore $|A_0|=|A|/2=45\cdot10^7$.
| {
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"answer_id": 3
} |
Distance From Center Of Circle To Intersection Of Diagonals Of A Cyclic Quadrilateral Given a cyclic quadrilateral $ABCD$ with length $AB=w$, $BC=x$, $CD=y$, and $DA=z$, compute the distance from the center of the circumscribing circle to the intersection point of the diagonals $AC$ and $BD$ in terms of $w,x,y,z$.
| I will use picture and the notations from
https://en.wikipedia.org/wiki/Cyclic_quadrilateral
so the cyclic quadrilateral under study is $ABCD$, it sides are $AB=a$, $BC=b$, $CD=c$, $DA=d$ (instead of $w,x,y,z$, which i cannot type in this context without error.)
The intersection of the diagonals is $P$, and the center of the circle $(ABCD)$ is $O$, we denote by $R$ the radius $R=OA=OB=OC=OD$.
When two letters are used in connection with metric relation, we mean always the corresponding lengths.
We will use the following facts and solve the issue:
*
*The power of the interior point $P$ is
$$PA\cdot PC=PB\cdot PD=R^2-OP^2\ .$$
*The relations of Brahmagupta and Parameshvara, and similar relations for the trigonometric functions of the angles in $A,B,C,D$, and of the angle $\theta$ between the diagonals.
*The sine theorem, for instance
$$
\begin{aligned}
\frac{PA}{DA}
&=
\frac
{\sin \widehat{PDA}}
{\sin \widehat{APD}}
=
\frac
{\sin \widehat{BDA}}
{\sin \theta}
=
\frac
{\sin \frac 12\widehat{BOA}}
{\sin \theta}
=
\frac {\frac{AB/2}R}
{\sin \theta}\ ,
\\[2mm]
&\qquad
\text{ which implies}
\\[2mm]
PA
&= \frac{AD\cdot AB}{2R\sin\theta}
= \frac{da}{2R\sin\theta}
\ ,\qquad\text{ and similarly}
\\
PC &= \frac{CD\cdot CB}{2R\sin\theta}
= \frac{cb}{2R\sin\theta}
\ .
\end{aligned}
$$
We are now in position to join the following relations:
$$
\begin{aligned}
R^2-OP^2
&=
PA\cdot PC
=\frac {abcd}{4R^2\sin^2\theta}\ ,
\\
4R^2
&=
\frac 14\cdot
\frac
{(ab+cd)(ac+bd)(ad+bc)}
{(s-a)(s-b)(s-c)(s-d)}\text{ (Parameshvara) with }
\\
s&=\frac 12(a+b+c+d)\ ,
\\[2mm]
&\qquad\text{ and we use now}
\\
\tan^2\frac\theta 2
&=
\frac{(s-b)(s-d)}{(s-a)(s-c)}\text{ to compute}
\\
\frac1{\cos^2\frac\theta 2}
&=
1+
\tan^2\frac\theta 2
=
\frac{(s-a)(s-c)+(s-b)(s-d)}{(s-a)(s-c)}
\sin^2\theta
\\
\cos^2\frac\theta 2
&=
\frac
{(s-a)(s-c)}
{(s-a)(s-c)+(s-b)(s-d)}
\\
\sin^2\frac\theta 2
&=
\frac
{(s-b)(s-d)}
{(s-a)(s-c)+(s-b)(s-d)}
\\
\sin^2\theta &=
4
\sin^2\frac\theta 2
\cos^2\frac\theta 2
\\
&=
4\cdot
\frac
{(s-a)(s-b)(s-c)(s-d)}
{(\; (s-a)(s-c)+(s-b)(s-d)\;)^2}\ ,
\\
4R^2\sin^2\theta
&=
\frac
{(ab+cd)(ac+bd)(ad+bc)}
{(\;(s-a)(s-c)+(s-b)(s-d)\;)^2}\ ,
\\
PA\cdot PC
&=\frac {abcd}{4R^2\sin^2\theta}
=
\frac
{abcd\;(\;(s-a)(s-c)+(s-b)(s-d)\;)^2}
{(ab+cd)(ac+bd)(ad+bc)}
\ ,\\[2mm]
&\qquad\text{ and finally}
\\
\color{blue}
{OP^2}
&=
R^2-PA\cdot PC
\\
&=
\color{blue}
{
\frac 1{16}
\cdot
\frac 1{(s-a)(s-b)(s-c)(s-d)}
\cdot
\frac
{(ac+bd)^2}
{(ab+cd)(ad+bc)}
\cdot
\Big(\
bd(a^2-c^2)^2+ac(b^2-d^2)^2
\ \Big)
}\ .
\end{aligned}
$$
Note:
At the last step we have used sage to factorize. Code and results:
sage: S.<a,b,c,d> = PolynomialRing(QQ)
sage: RR = 1/16 * (a*c+b*d)*(a*b+c*d)*(a*d+b*c) / (s-a) / (s-b) / (s-c) / (s-d)
sage: PAPC = a*b*c*d * ((s-a)*(s-c)+(s-b)*(s-d))^2 / ( (a*c+b*d)*(a*b+c*d)*(a*d+b*c) )
sage: factor( RR - PAPC )
-(a*b^4*c + a^4*b*d - 2*a^2*b*c^2*d + b*c^4*d
- 2*a*b^2*c*d^2 + a*c*d^4)
* (a*c + b*d)^2
/ ((a*b + c*d)*(b*c + a*d)
*(a + b + c - d)
*(a + b - c + d)
*(a - b + c + d)
*(a - b - c - d))
(Lines were manually broken.)
| {
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Evaluate $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$ The problem is as follows:
Find the value of $\textrm{H}$ which belongs to a certain vibration coming from a magnet.
$$H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$$
It was easy to spot that each term was related to multiples of two and three of the first angle. So I rewrote that equation like this:
$$H=\sec \frac{2\pi}{7}+\sec \frac{2\times 2\pi}{7}+\sec \frac{3\times 2\pi}{7}$$
One method which I tried was to transform the multiples of each angle into their equivalents as a single one as shown below:
$$\cos^{2}\omega=\frac{1+\cos 2\omega}{2}$$
$$\cos 2\omega= 2 \cos^{2}\omega - 1$$
$$\cos^{3}\omega=\frac{1}{4}\left(3cos\omega+\cos 3\omega \right)$$
$$\cos 3\omega = 4 \cos^{3}\omega - 3 cos\omega$$
Therefore by plugin these expressions into the above equation would become into (provided that secant function is expressed in terms of secant):
$$H=\frac{1}{\cos \frac{2\pi}{7}}+\frac{1}{2\cos^{2}\frac{2\pi}{7}-1}+\frac{1}{4\cos^{3}\frac{2\pi}{7}-3\cos\omega}$$
But from here on it looks convoluted or too algebraic to continue. My second guess was it could be related to sum to product identity but I couldn't find one for the secant.
Does it exist a shortcut or could it be that am I missing something? Can somebody help me to find the answer?
Can this problem be solved without requiring to use Euler's formulas?
| Using multiple angle formulas, we get
$$
\begin{align}
\cos(\theta)&=x\\
\cos(2\theta)&=2x^2-1\\
\cos(3\theta)&=4x^3-3x\\
\cos(4\theta)&=8x^4-8x^2+1
\end{align}\tag1
$$
Consider $\cos(3\theta)=\cos(4\theta)$, which happens when $3\theta+4\theta=2k\pi$ for some $k\in\mathbb{Z}$ (it also happens when $3\theta-4\theta=2k\pi$, but those cases are a subset). Thus,
$$
x=\cos\left(\frac{2k\pi}7\right)\implies8x^4-4x^3-8x^2+3x+1=0\tag2
$$
Since $k$ and $7-k$ give the same values for $\cos\left(\frac{2k\pi}7\right)$ and $k=0$ gives $\cos\left(\frac{2k\pi}7\right)=1$, if we divide $(2)$ by $x-1$, we get the polynomial satisfied by $x=\cos\left(\frac{2k\pi}7\right)$ for $k\in\{1,2,3\}$; that is,
$$
8x^3+4x^2-4x-1=0\tag3
$$
The polynomial satisfied by $x=\sec\left(\frac{2k\pi}7\right)$ for $k\in\{1,2,3\}$ is then
$$
x^3+4x^2-4x-8=0\tag4
$$
Vieta's formulas then give that
$$
\bbox[5px,border:2px solid #C0A000]{\sec\left(\frac{2\pi}7\right)+\sec\left(\frac{4\pi}7\right)+\sec\left(\frac{6\pi}7\right)=-4}\tag5
$$
Furthermore, they also give that
$$
\sec\left(\frac{2\pi}7\right)\sec\left(\frac{4\pi}7\right)\sec\left(\frac{6\pi}7\right)=8\tag6
$$
| {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Get ascending sum on 100% based on a number For each number I want to get the array of number ascending cumulative
For example :
$$100 \% = 45 \% + 55 \%$$
$$100 \% = 23.\overline{33} \% + 33.\overline{33} \% + 43.\overline{33} \%$$
$$100 \% = 10\% + 20\% + 30\% + 40\% $$
$$100 \% = a \%+b\%+c\%+d\%+e\%+f\%+g\%+h\% \ ??$$
What is the formula?
Thanks
| Yo seem to want to get $n$ numbers in arithmetic progression such that their sum is $100$, so mean $\frac{100}{n}$, and each of the $n-1$ steps is $10$
So the lowest value is $\frac{100}{n} - 10\frac{n-1}{2} = \frac{100}{n} -5n+5$
and the highest value is $\frac{100}{n} +10\frac{n-1}{2} = \frac{100}{n} +5n-5$
and with $n=8$ you get $$100 = -22.5 -12.5 -2.5 + 7.5+17.5+27.5+37.5+47.5$$
If you are not so fixed about the steps being $10$ and they are instead $d$ then the lowest value is $\frac{100}{n} - d\frac{n-1}{2}$ and the highest $\frac{100}{n} + d\frac{n-1}{2}$. For example with $n=8$ and $d=3$ you could have $100 = 2+5+8+11+14+17+20+23$.
| {
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"source": "stackexchange",
"question_score": "1",
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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 $x = 45^{\circ},225^{\circ}$
| From $$\sin(2x)=2\sin x\cos x$$
$$\sin x\cos x = \frac{1}{2} \Leftrightarrow \sin(2x)=1$$
So the answer is $x=\frac{\pi}{4}$ in $0<x<\pi.$
The general solution is $$x=\frac{(4n\pi+\pi)}{4}, n\in\mathbb{Z}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
Exercise II.2.2: Prove that $\lim \dfrac{n^3}{2^n} = 0$ I'm trying to solve Problem II.2.2 in textbook Analysis I by Amann/Escher.
Could you please verify whether my attempt on (b) and (c) is fine or it contains logical gaps/errors?
My attempt:
From binomial theorem, we have $$2^n = (1+1)^n = \sum_{k=0}^n {n \choose k} = {n \choose 4} + \sum_{k=0 \atop k\neq 4}^n {n \choose k} = n(n-1)(n-2)(n-3) + \sum_{k=0 \atop k\neq 4}^n {n \choose k}$$
As a result, $$\begin{aligned} \lim \dfrac{n^3}{2^n} &= \lim \dfrac{n^3}{n(n-1)(n-2)(n-3) + \sum_{k=0 \atop k\neq 4}^n {n \choose k}} \\ &= \lim \dfrac{1}{n \left( 1-\dfrac{1}{n} \right) \left(1-\dfrac{2}{n} \right) \left(1-\dfrac{3}{n} \right) + \dfrac{\sum_{k=0 \atop k\neq 4}^n {n \choose k}}{n^3}} \\ &= \lim \dfrac{1}{n + \dfrac{\sum_{k=0 \atop k\neq 4}^n {n \choose k}}{n^3}} = 0\end{aligned}$$
| Your idea is correct, but you can be more concrete and more simple:
It is $2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k}>\binom{n}{4}$ for $n$ big enough.
$\binom{n}{4}=\frac{n!}{4!(n-4)!}=\frac{n(n-1)(n-2)(n-3)}{24}<\frac{n^4}{24}$
We can stipulate like this:
$\lim_{n\to\infty} \frac{n^3}{2^n}\leq\lim_{n\to\infty} \frac{n^3}{n^4/24}=\lim_{n\to\infty} \frac{24}{n}=0$
Because you have certainly already proven, that $\lim_{n\to\infty} \frac1n=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.
| Hint.
A way to solve this DE is by making $z = y^2$ and then
$$
\frac 12 x z' - 3z = x^2
$$
which is a linear DE. Solving for $z$ we have
$$
z = y^2 = C_1 x^6-\frac 12x^2
$$
NOTE
The homogeneous part
$$
\frac 12 x z'_h -3z_h = 0
$$
is separable giving
$$
z_h = C_0 x^6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof: $\binom{x+y+n-1}{n} = \sum_{k=0}^{n} \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}$ I wanted to prove following equation
$\binom{x+y+n-1}{n} = \sum_{k=0}^{n} \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}$
Using Vandermonde's identity
$\binom{a+b}{t} = \sum_{k=0}^{t} \binom{a}{t-k} \binom{b}{k}$
whereas $a=x+n-k-1$ and $b=y+k-1$ but it doesn't add up.
Where is my mistake? How do i proof it?
| That's called "double convolution": we have
$$
\eqalign{
& \sum\limits_{0\, \le \,k\, \le \,n\,} {\left( \matrix{
a + k \cr
k \cr} \right)\left( \matrix{
b - k \cr
n - k \cr} \right)} \quad \left| \matrix{
\,a,b \in \mathbb C \hfill \cr
\;0 \le n \in \mathbb Z \hfill \cr} \right.\quad = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {\left( \matrix{
a + k \cr
k \cr} \right)\left( \matrix{
b - k \cr
n - k \cr} \right)} = \quad \quad (1) \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {\left( { - 1} \right)^{\,k} \left( \matrix{
- a - 1 \cr
k \cr} \right)\left( { - 1} \right)^{\,n - k} \left( \matrix{
n - b - 1 \cr
n - k \cr} \right)} = \quad \quad (2) \cr
& = \left( { - 1} \right)^{\,n} \left( \matrix{
n - a - b - 2 \cr
n \cr} \right) = \quad \quad (3) \cr
& = \left( \matrix{
a + b + 1 \cr
n \cr} \right) \quad \quad (4) \cr}
$$
where:
(1) we can omit the sum bounds because they are implicit in the binomials, this simplifies the algebraic manouvres;
(2) upper negation;
(3) Vandermonde convolution;
(4) upper negation again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$
| We have to prove that
$$\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(a)}+\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(b)}\geq 2$$ this is $$\ln\left(\frac{a+b}{2}\right)(\ln(a)+\ln(b))\geq 2\ln(a)\ln(b)$$ or
$$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq 2\ln(a)\ln(b)$$ Now we have by AM-GM:
$$\frac{a+b}{2}\geq \sqrt{ab}$$
taking the logarithm on both sides
$$\ln\left(\frac{a+b}{2}\right)\geq \frac{1}{2}\ln(ab)$$ now is
$$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq \frac{1}{2}(\ln(ab))^2$$
Now we have to show that $$(\ln(ab))^2\geq 4\ln(a)\ln(b)$$ this is true since
$$(\ln(a)+\ln(b))^2\geq 4\ln(a)\ln(b)$$
this is equivalent to
$$(\ln(a)-\ln(b))^2\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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$\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right) $, where $n \in \mathbb{N}$ $$
\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)
$$
where $n \in \mathbb{N}$.
In this question, what I thought was, since $n \to \infty$ and $\cos ^2x$ is periodic , all I need is actually the fractional part of this. And it easy to say that $n+1> \sqrt[3]{n^3+n^2+2n}> n$.
But evaluation of $\sqrt[3]{n^3+n^2+2n} - n$ , is getting tricky. I'm sure there must be a short way to solve it. Can someone help me with it?
| $\sqrt[3]{n^3 + n^2 + 2n} = n \sqrt[3]{1 + \frac{1}{n} + \frac{2}{n^2}}$
In the limit, this cube root $\rightarrow 1 + \frac{1}{3n}$, so this subexpression $\rightarrow n\pi + \pi/3$, so the argument to cosine goes to an integer multiple of $\pi$ plus $\pi/3$, and so the square of the cosine of such angles goes to $1/4$.
We use Newton's binomial formula.
\begin{align*}
&\lim_{n \rightarrow \infty} \cos^2\left(\pi \sqrt[3]{n^3 + n^2 + 2n}\right) \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left(n \pi \sqrt[3]{1 + \frac{1}{n} + \frac{2}{n^2}}\right) \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left( n \pi \sum_{k=0}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right) \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left( n \pi \left( \binom{1/3}{0} + \binom{1/3}{1} \left( \frac{1}{n} + \frac{2}{n^2} \right) + \sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right) \right) \\
&\qquad = \lim_{n \rightarrow \infty} \cos^2 \left( n \pi + \frac{\pi}{3} + \frac{2 \pi}{3n} + n \pi\sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right) \\
&\qquad = \lim_{n \rightarrow \infty} \left( \cos \left( n \pi + \frac{\pi}{3}\right) \cos \left( u \right) - \sin \left( n \pi + \frac{\pi}{3}\right) \sin \left( u \right)
\right)^2 \text{,}
\end{align*}
where $u = \frac{2 \pi}{3n} + n \pi\sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k$. By continuity, \begin{align*}
&\lim_{n \rightarrow \infty} \left( \cos \left( n \pi + \frac{\pi}{3}\right) \cos \left( u \right) - \sin \left( n \pi + \frac{\pi}{3}\right) \sin \left( u \right)
\right)^2 \\
&\qquad = \left( \left( \lim_{n \rightarrow \infty} \cos \left( n \pi + \frac{\pi}{3}\right) \right) \left( \lim_{n \rightarrow \infty}\cos \left( u \right) \right) - \left( \lim_{n \rightarrow \infty} \sin \left( n \pi + \frac{\pi}{3}\right) \right) \left( \lim_{n \rightarrow \infty} \sin \left( u \right) \right)
\right)^2 \\
&\qquad = \left( \left( \lim_{n \rightarrow \infty} \begin{cases} 1/2, & \text{$n$ even} \\ -1/2, & \text{$n$ odd} \end{cases} \right) \left( \lim_{n \rightarrow \infty}\cos \left( u \right) \right) - \left( \lim_{n \rightarrow \infty} \begin{cases} \sqrt{3}/2, & \text{$n$ even} \\ -\sqrt{3}/2, & \text{$n$ odd} \end{cases} \right) \left( \lim_{n \rightarrow \infty} \sin \left( u \right) \right) \right)^2 \text{.}
\end{align*}
So we examine
$$ \lim_{n \rightarrow \infty}\cos \left( u \right)
= \lim_{n \rightarrow \infty}\cos \left( \frac{2 \pi}{3n} + n \pi\sum_{k=2}^\infty \binom{1/3}{k} \left( \frac{1}{n} + \frac{2}{n^2} \right)^k \right)
\text{.} $$
First, $\left| \binom{1/3}{k} \right| \leq 1$ (in fact, we can tighten this to ${} \leq 1/3$). Second, for $n > 2$, $\frac{1}{n} + \frac{2}{n^2} < \frac{2}{n}$, so the sum is bounded using the geometric series $\frac{1}{3} \sum_{k=2}^\infty \left( \frac{2}{n} \right)^k = \frac{4}{3n(n-2)}$, which goes to $0$ as $n$ goes to $\infty$, as does $\frac{2 \pi}{3n}$. Consequently, again using continuity,
$$ \lim_{n \rightarrow \infty}\cos \left( u \right)
=\cos \left( \lim_{n \rightarrow \infty} u \right) = \cos 0 = 1 $$
and also
$$ \lim_{n \rightarrow \infty}\sin \left( u \right)
=\sin \left( \lim_{n \rightarrow \infty} u \right) = \sin 0 = 0 \text{.} $$
Therefore,
\begin{align*}
&\lim_{n \rightarrow \infty} \cos^2\left(\pi \sqrt[3]{n^3 + n^2 + 2n}\right) \\
&\qquad = \left( \left( \lim_{n \rightarrow \infty} \begin{cases} 1/2, & \text{$n$ even} \\ -1/2, & \text{$n$ odd} \end{cases} \right) \cdot 1 - \left( \lim_{n \rightarrow \infty} \begin{cases} \sqrt{3}/2, & \text{$n$ even} \\ -\sqrt{3}/2, & \text{$n$ odd} \end{cases} \right) \cdot 0 \right)^2 \\
&\qquad = 1/4 \text{.}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Substitution methods of first order linear eqautions [problem help] Okay so I am having some computational issues towards the end of the problems
Problem 1: $y' = (4x+y)^2$
$z = 4x + y$
$y = z - 4x$
$y' = \frac{dz}{dx} -4$
$z^2 = \frac{dz}{dx}-4$
$z^2 + 4 = \frac{dz}{dx}$
$\int \frac{dx}{x} = \int \frac{dz}{z^2+4}$
$ln \vert x \vert + c = \ln \vert z^2+4 \vert$
$ce^x = (4x+y)^2+4$
$y(x) = \sqrt{ce^x-4}-4x$
I don't know it seems like every single problem with this substitution is a nightmare computationally.
Second problem:
$x(x+y)y'+y(3x+y)=0$
$y' = \frac{-y(3x+y)}{x(x+y)}$
This is a homogeneous equation
$y' = -v \frac{(3x+xv)}{(x+xv)}$
$y' = -v \frac{(3+v)}{(1+v)}$
$v + \frac{dv}{dx} = \frac{-v^2-3v}{1+v}$
$x \frac{dv}{dx} = \frac{-v^2 -3v -v -v^2}{1+v}$
$x \frac{dv}{dx} = \frac{-2v^2-4v}{1+v}$
$\int \frac{1+v}{-2v(v-2)}dv = \int \frac{dx}{x}$
This a partial fractions, an ugly one at that:
$1+ v = \frac{A}{-2v}+\frac{B}{v-2}= A(v-2) -2Bv$
Now the only values of $v$ where I get any where is $v=2$ and $v = 0$ but I do not think these are viable because we get zeros in the fractions at this point.
| 1.)
Your integration is not correct.
$$
\int\frac{dz}{z^2+4}=\frac12\arctan(\frac z2)+C
$$
so that $z=2\tan(2(x-C))$, $y=2\tan(2(x-C))-4x$.
2.)
You want to compute the coefficients in the partial fraction decomposition
$$
-\frac{1+v}{2v(v-2)}=\frac{A}{v}+\frac{B}{v-2}
$$
To that end you multiply with the denominator to get
$$
1+v=-2A(v-2)-2Bv.
$$
You need to keep these two forms of the equation separate, do not mix them.
Indeed then inserting $v=0$ and $v=2$ gives $4A=1$ and $4B=-3$.
But, carefully check again the signs in your formulas
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Factoring $x^4 + 5x^3 + 4x^2 + 2x - 3$
So I have to factor this polynomial
$$x^4 + 5x^3 + 4x^2 + 2x - 3$$
I got $(x^2 + 2x -3)(x^2 + 3x + 1)$
but when I multiplied it, I got a different equation:
$$x^4 + 5x^3 + 4x^2 - 7x - 3$$
I don’t really understand how to find the factors. I only found numbers that could add up to 5 and multiply to -3 but they’re wrong. What did I do wrong and what can I do to fix this?
In the format of:
($x^2$+ __ x +__ ) ($x^2$ + __ x + __ )
or
$$(x^2 + ax + b) (x^2 + cx + d)$$
Also, how does comparing coefficients help factor this polynomial?
| $(x^2+ax+b)^2=x^4+2ax^3+a^2x^2+2bx^2+2abx+b^2$
we find:
$a=\frac{5}{2}$
$b=\frac{-9}{8}$
and we get following polynomial with these values:
$x^4+5x^3+4x^2-\frac{45}{8}x+\frac{81}{64}$
We compare this with given polynomial; if:
$x^4+5x^3+4x^2+2x-3=0$
then:
$(x^2+\frac{5}{4}x-\frac{9}{8})^2=\frac{273}{64}-\frac{61}{8}x$
This equation has two real roots ; $x_1≈-4.202$ and $x_2≈0.53$ and two complex roots ; $x_3≈ -0.66-0.95 i$ and $x_3≈ -0.66+0.95 i$, so the factorized form can be:
$(x-0.52)(x+4.2)(x+0.66+095 i)(x-0.66+095 i)$
You can get the same result from Wolfram.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 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 $\frac{-1}{13(13+h)}$?
| $$\frac{1}{a\pm b} \ne\frac{1}{a}\pm\frac{1}b$$
So, $$\frac{1}{13+h}-\frac{1}{13} = \frac{13-13-h}{(13+h)13}=-\frac{h}{13(13+h)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Integration of $ \int x^{2} \sqrt{2x-6} dx $
$$ \int x^{2} \sqrt{2x-6} dx = ?$$
My Attempt:
by partial integration
$$ \int x^{2} \sqrt{2x-6} dx = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \int x(2x-6)^{3/2}dx$$
continuing partial integration
$$ = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \left[ \frac{x(2x-6)^{5/2}}{5} - \int \frac{(2x-6)^{5/2}}{5} dx\right] $$
$$ = (x^{2}/3)(2x-6)^{3/2} - (2x/15)(2x-6)^{5/2} + (2/105)(2x-6)^{7/2} + C$$
Is this the correct and best/simplest answer?
Strangely, the multiple choices only include answers in the form:
$$ A(2x+6)^{7/2} + B(2x+6)^{5/2} + C(2x+6)^{3/2} + D$$
where $A,B,C,D$ are constants.
| Hint:Substituting $$t=\sqrt{2x-6}$$ then we get $$x=\frac{t^2+6}{2}$$ then we get
$$dx=tdt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find 3×3 orthogonal matrix satisfying the following conditions. Let $A$ be 3×3 orthogonal real matrix.
For $\mathbf{u}$ $= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \mathbf{v}$ $= \begin{pmatrix} 0 \\ 0 \\ \sqrt3 \end{pmatrix}, \mathbf{w}$ $= \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$,
$Au=v, Aw=-w$ and $detA=-1$
Find $A$.
I noted that $u, v, w$ are linearly independent. And I write
$A\begin{pmatrix}
1 & 0 & 1 \\
1 & 0 & -1 \\
1 & \sqrt3 & 0
\end{pmatrix}=\begin{pmatrix}
0 & a & -1 \\
0 & b & 1 \\
\sqrt3 & c & 0
\end{pmatrix}.$
Then $A=\begin{pmatrix}
0 & a & -1 \\
0 & b & 1 \\
\sqrt3 & c & 0
\end{pmatrix}\begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & 0 \\
-\frac{1}{2\sqrt3} & -\frac{1}{2\sqrt3} & \frac{1}{\sqrt3} \\
\frac{1}{2} & -\frac{1}{2} & 0
\end{pmatrix}$
Now, how can I find $a, b, c$??
Am I doing this right?
If there is a better solution, please help me.
| Your approach is fine; you can proceed by using the fact that $\det(A)=-1$ and $A$ is orthogonal. From
$$-1=\det(A)=\det\begin{pmatrix}
0 & a & -1 \\
0 & b & 1 \\
\sqrt3 & c & 0
\end{pmatrix}\det\begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & 0 \\
-\frac{1}{2\sqrt3} & -\frac{1}{2\sqrt3} & \frac{1}{\sqrt3} \\
\frac{1}{2} & -\frac{1}{2} & 0
\end{pmatrix},$$
you find that $a+b=-2$, so you can substitute $b=-(a+2)$. Now multiply the two matrices and choose $a$ and $c$ so that the matrix is orthogonal.
Alternatively, the fact that $Au=v$, i.e.
$$A\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}0\\0\\\sqrt{3}\end{pmatrix},$$
shows that the first two rows of $A$ are orthogonal to $u$. Because $A$ is an orthogonal matrix, it follows that the third row is orthogonal to the first two, so the third row is a scalar multiple of $u$. Its inner product with $u$ equals $\sqrt{3}$, so the third row must equal $\frac{u}{\sqrt{3}}$.
Can you proceed in this way with $Aw=-w$ to determine $A$?
| {
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"source": "stackexchange",
"question_score": "4",
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} |
Prove /disprove $M_2=\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}; a, b , c , d \in \mathbb{Q}\right\}$ is simple ring
Show that $$M_2=\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}; a, b , c , d \in \mathbb{Q}\right\}$$ is simple ring
My proof :
we know that $M_2$ is a ring under adition and matrix multiplication and has unity \begin{pmatrix} 1& 0 \\ 0 & 1 \end{pmatrix}. we can find two element $A $ and $B $ in $M_2$ such that $AB \neq \begin{pmatrix} 0& 0 \\ 0 & 0 \end{pmatrix}$. For example $ A=\begin{pmatrix} 1& 0 \\ 1 & 0 \end{pmatrix}$ and $ B= \begin{pmatrix} 1& 1 \\ 0 & 0 \end{pmatrix}$
If we show that $ M_2$ has no non trivial ideal , then $ M_2$ become a simple ring
Let $A$ be any ideal of $M_2$ . If $A= \{0\}$ , $0 $ being a $2 \times 2$ null matrix, then there is nothing to prove . let $A \neq\{0\}$. Then there exist a nonzero matrix $X \in A$ of the form $ X=\begin{pmatrix} a_{11}& a_{12} \\ a_{21} & a_{22} \end{pmatrix}$
Since $X$ is a nonzero matrix , atleast one the $4 $ entries in $X$ is nonzero . let $a_{12} \ne 0 \in \mathbb{Q}$
We choose four matrix in $M_2$ as follow
let $ P=\begin{pmatrix} 1& 0 \\ 0 & 0 \end{pmatrix}$,
$ Q=\begin{pmatrix} 0& 0 \\ 1 & 0 \end{pmatrix}$,
$ S=\begin{pmatrix} 0& 1 \\ 0 & 0 \end{pmatrix}$,$ T=\begin{pmatrix} 0& 0 \\ 0 & 1 \end{pmatrix}$
Now by doing multiplication $ PXQ=\begin{pmatrix} a_{12}& 0 \\ 0 & 0 \end{pmatrix}$ and $ SXT=\begin{pmatrix} 0& 0 \\ 0 & a_{12} \end{pmatrix}$
Since $X\in A$ and $A $ is an ideal of $M_2 $, therefore $PXQ + SXT \in A$
now $ \begin{pmatrix} a_{12}& 0 \\ 0 & 0 \end{pmatrix} +\begin{pmatrix} 0& 0 \\ 0 & a_{12} \end{pmatrix}=\begin{pmatrix} a_{12}& 0 \\ 0 & a_{12} \end{pmatrix} =K\in A$
since $a_{12} \neq 0 \in \mathbb{Q}$, $a_{12}^{-1} \in \mathbb{Q}$ that is $K^{-1} \in M_2$
Since $A$ is an ideal of $M_2$ that is $ KK^{-1} = I \in A$
Thus $A$ is an ideal of $M_2 $ containing the unity $ I $ it implies $A = M_2$
Hence prove that $M_2$ is simple ring
Is my proof is correct ?? yes/No
| $SXT=\begin{pmatrix} 0& 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a_{11}& a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} 0& 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} a_{21}& a_{22} \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0& 0 \\ 0 & 1\end{pmatrix}=\begin{pmatrix}0& a_{22} \\ 0 & 0 \end{pmatrix}$.
If you want the $a_{12}$ as the last entry of the matrix you can do $QXT$.
The rest is correct.
| {
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"source": "stackexchange",
"question_score": "3",
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A curiosity on a first three natural numbers Let's review a triple of numbers, $1, 2, 3$, it is a curiosity that
$$1+2+3 = 1\times2\times3 = 6$$
Are there another triples (or not necessary triples) such that their multiple equal to their sum?
And generalised pattern of such identities would be interesting and appreciated.
PS: Conjecture: Reviewing $t$ fold case of such numbers, they are seem to be the integer solutions of the equation
$$n(n+1)(n+2)\cdots(n+t) = \binom{t+1}{1}n + \binom{t+1}{2}$$
PSS: Integer solution (for consequent integers)
$$\prod_{k=0}^{2s} (n+k) = \sum_{k=0}^{2s} (n+k)$$
for $n=-s$. But these sums and products are 0.
PS3: Still we can easily find such combinations using the following pattern:
$$\prod_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1 \times \prod_{k=0}^t a_k = \left(\sum_{k=0}^t a_k\right)+\sum_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1$$
| If we know, that $A=1,B=2,C=3$ is a solution we can look for another solution with larger numbers by
$$(A+a)+(B+b)+(C+c) = (A+a)(B+b)(C+c) \\
-----------------------------\\
(1+a)+(2+b)+(3+c) = (1+a)(2+b)(3+c)\\
6+a+b+c = 6+ 2c+3b+6a+bc+3ab+2ac+abc\\
a+b+c = 2c+3b+6a+bc+3ab+2ac+abc\\
0 = c+2b+5a+bc+3ab+2ac+abc\\
$$
If no number $a,b,c$ is negative, all must be zero.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to flip all fractions in the power series for $\ln(1 + x)$? I am trying to evaluate this using power series:
$$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$
By using the power series for $\ln(1 + x)$, I have recognized that dividing through by $x$ and setting $x = -2$ will get you this:
$$1 + \frac{2}{2} + \frac{2^2}{3} + \frac{2^3}{4} + ..$$
This seems so close, but I can't seem to figure out how to flip each fraction so that it matches. How can I do this?
If I am on the completely wrong path and this is a coincidence, please point me in the right direction.
| You have $\sum_{n=1}^\infty \frac{n}{2^{n-1}}$ which is equal to $F'(\frac{1}{2})$, where $F(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$,
So we have $F'(x) = \frac{1}{(1-x)^2}$ and $F'(\frac{1}{2}) = 4$
Alternatively without derivatives (Axion004 idea) :
Let $S = \sum_{n=1}^\infty \frac{n}{2^{n-1}} = \sum_{n=0}^\infty \frac{n+1}{2^n} = \sum_{n=0}^\infty \frac{n}{2^n} + 2 = \frac{1}{2}\sum_{n=1}^\infty \frac{n} {2^{n-1}} + 2 = \frac{S}{2} + 2 $
So $ \frac{S}{2} = 2 $ and $S = 4$
| {
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Prove that $XX’,YY’$ and $ZZ’$ are concurrent at $T$
Let $\Delta ABC$ has median lines respectively be $AX,BY,CZ$. And $X',Y',Z'$ be respectively midpoint of bisectors $AA', BB', CC'$. Prove that $XX',YY'$ and $ZZ'$ are concurrent and assume that point is $T$ and prove that $T$ is in the straight line connecting the center of the inscribed circle and the Lemoine point of the triangle $\Delta ABC$.
My teacher said that we could use Vector to solve it but u habeco no idea about that approach. Help me.
| I will use barycentric coordinates of points, considered w.r.t. the given triangle $\Delta ABC$ with sides denoted by $a,b,c$. A point $P$ has then barycentric coordinates $(x,y,z)$, $x+y+z=1$, iff we have a relation of the form
$$
P = xA+yB +zC\ .
$$
(Using point affixes. This translates as the relation
$OP=xOA+yOB+zOC$ in terms of the vectors $OP;OA,OB,OC$ considered w.r.t. one (and any) fixed reference point $O$.)
Sometimes we use the notation $P(x:y:z)$ instead of $P(x,y,z)$, in this case the sum $x+y+z$ may or may not be normed to $1$, so this point is then explicitly $P(\ x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z)\ )$.
A good reference is
Max Schindler, Evan Chen, Barycentric Coordinates for the Impatient.
Note that the point $T$ is in the ETC
the point
X(37) = CROSSPOINT OF INCENTER AND CENTROID
and that some lines after this information it claims
X(37) lies on these lines:
1,6 2,75 3,975 7,241 ...
and so on. Being on the line through 1, 6 (i.e. through the incenter 1 = X(1), and the symmedian = Lemoine point 6 = X(6) ) is exactly our problem. On this line there are also some other remarcable triangle centers, for instance X(9)...
Let us compute now...
The mid points of the sides have homogeneous coordinates
$$
\begin{aligned}
X&(0:1:1)\ , \\
Y&(1:0:1)\ , \\
Z&(1:1:0)\ .
\end{aligned}
$$
For instance, $X$ is then $\left(0,\frac12,\frac12\right)$ after passing to normed coordinates, corresponding to
$X=0A+\frac 12B+\frac 12C=\frac 12(B+C)$.
The angle bisectors of the angles $A,B,C$ intersect the opposite sides in $A',B',C'$ with coordinates
$$
\begin{aligned}
A'&(0:b:c)\ , \\
B'&(a:0:c)\ , \\
C'&(a:b:0)\ .
\end{aligned}
$$
For instance, $A'=\frac b{b+c}B+\frac c{b+c}C$, thus a point on $BC$, and the ratio $BA':A'C=c:b$, as stipulated by the angle bisector theorem.
The intersection $I$ of the angle bisectors is $(a:b:c)$, because it is on each line $AA'$, $BB'$, $CC'$, because the determinant corresponding to the barycentric coordinates of the points $A$, $A'$, and $I$ (presumably $(a:b:c)$),
$$
\begin{vmatrix}
1 &0&0\\0&b&c\\a&b&c
\end{vmatrix}
=0
$$
vanishes, and the same happens by sinymmetry also for the triples $A,B',I$, and $C,C',I$.
The mid points of the angle bisectors are now
$$
\begin{aligned}
X' &=\frac 12(A+A')\\
&=\frac 12(1,0,0)+\frac 12\cdot\frac 1{b+c}(0,b,c)\\
&=\frac 1{2(b+c)}(b+c,b,c)\\
&= (b+c:b:c)\ , \\
Y'&=(a:a+c:c)\ ,\\
Z'&=(a:b:a+b)\ .
\end{aligned}
$$
The point of intersection of three lines $XX',YY',ZZ'$ is, after solving a system of linear equation (as in the check in the sequel):
$$
T = \Big(\ a(b+c),\ b(a+c),\ c(a+b)\ \Big)\ .
$$
To see that $T$ is on $XX'$ we compute the determinant with rows given by the barycentric coordinates of the three points,
$$
\begin{aligned}
\begin{vmatrix}
0 & 1 & 1\\
b+c & b & c\\
a(b+c) & b(a+c)& c(a+b)
\end{vmatrix}
&=
(b+c)
\begin{vmatrix}
0 & 1 & 1\\
1 & b & c\\
a & b(a+c)& c(a+b)
\end{vmatrix}
\\
&=
(b+c)
\begin{vmatrix}
0 & 1 & 1\\
1 & b & c\\
0 & b(a+c)-ab& c(a+b)-ac
\end{vmatrix}
\\
&=
bc(b+c)
\begin{vmatrix}
0 & 1 & 1\\
1 & b & c\\
0 & 1& 1
\end{vmatrix}
\\
&=0\ .
\end{aligned}
$$
By (cyclic) symmetry, the point $T$ also lies on $YY'$ and $ZZ'$.
We have to show finally that $T=X(37)$, as labeled in the Encyclopedia of Triangle Centers, ETC, is on the line through the incenter $X(1)=I(a:b:c)$, and the symmedian = Lemoine point $X(6)=L=(a^2:b^2:c^2)$, so we compute one more easy determinant, built using the barycentric coordinates of the points $I, L, T$:
$$
\begin{aligned}
\begin{vmatrix}
a & b & c\\
a^2 & b^2 & c^2 \\
a(b+c) & b(a+c)& c(a+b)
\end{vmatrix}
&=
abc
\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
b+c & a+c & a+b
\end{vmatrix}
\\
&=
abc
\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a+b+c & a+b+c & a+b+c
\end{vmatrix}
\\
&=
abc(a+b+c)
\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
1 & 1 & 1
\end{vmatrix}
\\
&=0\ .
\end{aligned}
$$
Note: This is an analytic solution. (A solution that is in the same time algebraic, barycentric, vectorial, analytic.. These attributes are in this case hard to distinguish, we are performing "blind geometry" without any (need of a) picture.)
If a synthetic solution is also wanted, i can try to type and insert it.
| {
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Solving $\frac53\left(3-\frac{x}{5}\right) = x$ Although I had already solved the following equation, I can't figure out why I fail when passing thru a $15$ denominator I get a bad result.
$$\frac53\cdot\left(3-\frac{x}{5}\right) = x$$
Transformation to get $15$ denominator:
$$\begin{align}
\frac{5\cdot 5(15\cdot 3-3x)}{15} &= \frac{15\cdot x}{15} \tag{1}\\[4pt]
25\cdot(45-3x) &= 15x \tag{2}\\
1125 - 75x &= 15x \tag{3}\\
1125 &= 90x \tag{4}\\[4pt]
x &= \frac{1125}{90} \tag{5}
\end{align}$$
Using alternative approaches, $x = 15/4 = 3.75$, which I believe is the right answer.
What am I doing wrong on the above try?
| Multiplying both by $$\frac{3}{5}$$ we get
$$3-\frac{x}{5}=\frac{3}{5}x$$ so $$3=\frac{4}{5}x$$ thus
$$x=\frac{15}{4}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many ordered triples $(x,y,z)$ of real numbers are there such that $x+y^2=z^3$, $x^2+y^3=z^4$, and $x^3+y^4=z^5$ How many ordered triples $(x,y,z)$ of real numbers are there such that
$x+y^2=z^3$
$x^2+y^3=z^4$
$x^3+y^4=z^5$?
Adding those equations together and factorizing gives you $x(1+x+x^2)+y^2(1+y+y^2)=z^3(1+z+z^2)$
Which is of the same form as the 1st equation, but I don’t know how to use this to my advantage. Hints, suggestions and solutions would be appreciated.
Taken from the 2018 BIMC
| Hint
From $z^3\cdot z^5=(z^4)^2$ we obtain $$(x+y^2)(x^3+y^4)=(x^2+y^3)^2$$which leads to $$xy^2(x-y)^2=0$$
| {
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Combinatorics generating functions of a series Hi I found that the generating function of a series $a_n$ is:
$$\frac{(1-x)(1+2x)}{(1+3x)(1-3x)}$$
I need to find a formula for $a_n$.
I tried some things and found that the generating function is equal to:
$$\frac{1}{3}\cdot (1+2x)\cdot( \frac{2}{1+3x} + \frac{1}{1-3x})$$
but I cant get any further than that.
| Your approach is fine. We see or obtain with polynomial division
\begin{align*}
1+2x=\frac{2}{3}(1+3x)+\frac{1}{3}\quad&\Longrightarrow\quad\frac{1+2x}{1+3x}=\frac{2}{3}+\frac{1}{3(1+3x)}\\
1+2x=-\frac{2}{3}(1-3x)+\frac{5}{3}\quad&\Longrightarrow\quad\frac{1+2x}{1-3x}=-\frac{2}{3}+\frac{5}{3(1-3x)}\tag{1}
\end{align*}
We start with OPs expression and obtain with (1) and the geometric series expansion
\begin{align*}
\color{blue}{\frac{1}{3}}&\color{blue}{(1+2x)\left( \frac{2}{1+3x} + \frac{1}{1-3x}\right)}\\
&=\frac{2}{3}\cdot\frac{1+2x}{1+3x}+\frac{1}{3}\cdot\frac{1+2x}{1-3x}\\
&=\frac{2}{3}\left(\frac{2}{3}+\frac{1}{3(1+3x)}\right)+\frac{1}{3}\left(-\frac{2}{3}+\frac{5}{3(1-3x)}\right)\\
&=\frac{2}{9}+\frac{2}{9}\sum_{n=0}^\infty(-3)^nx^n+\frac{5}{9}\sum_{n=0}^\infty 3^nx^n\\
&\,\color{blue}{=\frac{2}{9}+\sum_{n=0}^\infty\frac{1}{9}\left(2(-1)^n+5\right)3^nx^n}
\end{align*}
| {
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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?
| You can't multiply or divide and keep the inequalities the the same unless you know the term you are multiplying by is positive.
In 1) $\sqrt{x^2 + a^2} > b-(x+a)$ does not mean
$(\sqrt{x^2 + a^2})^2 > [b-(x+a)]^2$ unless $\sqrt{x^2 + a^2} > |b-(x+a)|$ and you don't know that.
And $x(2b-2a)>b^2-2ba$ does not mean
$x > \frac {b^2 - 2ba}{2b-2a}$ unless you know that $2b - 2a > 0$. And you do not know that.
And $(x+a)-b>-\sqrt{x^2+a^2}$ most certainly does NOT mean
$[(x+a)-b]^2 > (-\sqrt{x^2+a^2})^2$. That $-\sqrt{x^2 + a^2}$ is negative should have been a big tip off that if $(x+a) - b$ is positive we'd know absolutely nothing how the values compare and if $(x+a)-b < 0$ then this statement is completely false so $0 > (x+a)-b > -\sqrt{x^2 +a^2}\implies \sqrt{x^2 + a^2} > |(x+a)-b|\implies x^2 + a^2 > [(x+a)-b]^2$.
| {
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Limit development for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. I'm trying to solve this problem: Find a limit development of order
7 for the function $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. Where
we use the next definition:
Definition. Let $f\colon I\to\mathbb{R}$ be a function and $x_{0}\in I$. We
say that $f$ has a limit development of order $n$ in $x_{0}$ provided
that there exist $a_{0},a_{1},\ldots,a_{n}\in\mathbb{R}$ such that
for $x\in I$
$$
f(x)=a_{0}+a_{1}(x-x_{0})+\ldots+a_{n}(x-x_{0})^{n}+o((x-x_{0})^{n}) \text{ (small $o$)}.
$$
Using Taylor series for $\arctan x$ at 0, we have that $\arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots$,
and therefore
\begin{align*}
\arctan(x^{2}) & =x^{2}-\frac{(x^{2})^{3}}{3}+\frac{(x^{2})^{5}}{5}-\frac{(x^{2})^{7}}{7}+\ldots\\
& =x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots
\end{align*}
So, replacing for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$ I have
that:
\begin{align*}
f(x) & =\frac{x}{x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots}-\frac{1}{x}\\
& =\frac{x}{x\left(x-\frac{x^{5}}{3}+\frac{x^{9}}{5}-\frac{x^{13}}{7}+\ldots\right)}-\frac{1}{x}\\
& =\frac{1}{x}\cdot\left(1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots\right)^{-1}-\frac{1}{x}
\end{align*}
But I couldn't obtain the form that is necessary for the limit development
because I have that part with the inverse of $1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots$.
Could you help me or give me some suggestion?
Thanks.
| Just to check people's answers,
Wolfy says
$x^3/3 - (4 x^7)/45 + (44 x^{11})/945 + O(x^{13})
$.
| {
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Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $ Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $
I know I need to use induction for this proof, but it feels like a pretty complicated one.
Basis: For $n = 2$,
$$1^2 < \frac{8}3 < 1^2 + 2^2$$
Induction Hypothesis: Assume $P(n)$ holds for $n=k$, that is,
$$1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$$
We need to show that $P(n)$ also holds for $n=k+1$
Proof:
$$1^2+2^2+{...}+{(k)}^2=1^2+2^2+{...}+{(k-1)}^2+{k}^2$$
After this, I'm not sure how to use the assumed inequality to prove it because it's a less than inequality. If I could get a hint that'd be awesome.
| If you're allowed to use $\sum\limits_{k=1}^{n} k^2 =\frac{n(n+1)(2n+1)}{6} $,
$$1^2 + 2^2 + {…}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {…}+ {n}^2$$
because
$\dfrac{(n-1)n(n-\frac{1}{2})}{3} \lt \dfrac{n^3}{3}$
and
$\dfrac{n(n+1)(n+\frac{1}{2})}{3} \gt \dfrac{n^3}{3}$
| {
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"url": "https://math.stackexchange.com/questions/3342188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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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.
| Writing $A^n$ as $\begin{bmatrix}a^n & b_n\\ 0 & 1\end{bmatrix}$. Expanding $A^{n+1} = AA^n$ leads to a recurrence relation of the form:
$$b_{n+1} = a b_n + b$$
Since $b_1 = b$, solving the recurrence relation will lead to $$b_n = (a^{n-1}+ a^{n-2} + \cdots + 1)b = \begin{cases} \frac{a^n-1}{a-1} b, & a \ne 1\\ nb, & a = 1\end{cases}$$
| {
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"source": "stackexchange",
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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?
| Multiplying by $\sqrt{\cdot} + \sqrt{\cdot}$ at numerator and denomitor your get
$$\begin{aligned}f(x)&= \frac{4adx}{{\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}}}\\
\end{aligned}$$
And therefore $\lim\limits_{x \to \infty} f(x) = \frac{2ad}{\sqrt{a^2+b^2}}$ by pulling $x$ at the denominator as
$$\begin{aligned}\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}&=x\left(\sqrt{(a+ d/x)^2 + b^2} + \sqrt{(a - d/x)^2 + b^2}\right)\\
\end{aligned}$$ for $x>0$ and $\lim\limits_{x \to \infty} d/x =0$.
Easy then to get $\lim\limits_{x \to -\infty} f(x)$ as $f$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Comparing $a^b$ and $b^a$ when $b < e < a$
If $0 < b < e < a$, how can I determine whether $a^b$ or $b^a$ is greater?
I know this question has been asked before, but I want to solve this question by this method. It worked fine for first two cases, and I want to know whether I can apply it to the third case as well.
We can equivalently compare $(b^{1 \over b})^{ab}$ and $(a^{1 \over a})^{ab}$. So define $$f(x)=x^{ 1 \over x} .$$
Now, $$f'(x)={1 \over x} \cdot (x)^{{1 \over x}-1}+x^{1 \over x} \cdot \ln{x} \cdot{{(-1)} \over {x^2}} =\bigg(\frac{x^{ 1 \over x}}{x^2}\bigg)\cdot \bigg(1- \ln x\bigg)$$
So for $x < e$ , $f'(x)>0$ , and so $f(x)$ is increasing, and for $x > e$ , $f'(x)<0$ , and so f(x) is decreasing.
We also have $\lim_{x \to 0}x^{ 1 \over x}=0$ and $\lim_{x \to \infty}x^{ 1 \over x}=e^{\lim_{x \to \infty}\frac{\ln x }{x}}=1$, so our function will increase from $x=0$ to $x=e$ and then decrease, tending to $1$ as $x \to \infty$.
Case 1: $a>b>e$: $a^{1 \over a} < b^{1 \over b}$, so $a^{{ab} \over a} < b^{{ab} \over b}$ and $a^{b} < b^{a}$. For example, $4^5 > 5^4$.
Similarly:
Case 2: $a<b<e$: $a^{1 \over a} < b^{1 \over b}$, so $a^{{ab} \over a} < b^{{ab} \over b}$ or $a^{b} < b^{a}$ For example, $\sqrt2^{\sqrt3}$ < $\sqrt3^{\sqrt2}$.
Case 3: $b < e < a$.
Here's where I'm stuck. Can we apply same method to solve it?
| This is certainly the hardest of the three cases: In this case sometimes $a^b$ is larger, sometimes $b^a$ is.
Since $\log$ is increasing, it follows from your argument that we may as well work with the more tractable function $\log f(u) = \frac{\log u}{u}$, and so compare
$$\frac{\log a}{a} \qquad \textrm{and} \qquad \frac{\log b}{b} .$$
Again since $\log$ is increasing, $\log f$ is increasing on $(0, e)$ and decreasing on $(e, \infty)$.
As a toy example, let's consider the case $a = 3, b = 2$, so that we're comparing $2^3$ and $3^2$. Of course we can evaluate both expressions directly and compare, but we'd like to use a method that also applies when we can't compute by hand. If we notice that $$\frac{\log 2}{2} = \frac{\log 4}{4},$$ then since $3$ and $4$ are both greater than $e$, we've reduced to case (1):
$$\frac{\log 2}{2} = \frac{\log 4}{4} < \frac{\log 3}{3} .$$
We can generalize this method: For any $s > 0$ we have
$$\frac{\log \left[\left(1 + \frac{1}{s}\right)^s\right]}{\left(1 + \frac{1}{s}\right)^s} = \frac{\log \left[\left(1 + \frac{1}{s}\right)^{s + 1}\right]}{\left(1 + \frac{1}{s}\right)^{s + 1}}$$ (check this!). But $\left(1 + \frac{1}{s}\right)^s < e < \left(1 + \frac{1}{s}\right)^{s + 1}$, so choosing a bound for $b$ of the form $\left(1 + \frac{1}{s}\right)^s$, gives a bound $\frac{\log b}{b} > \frac{\log c}{c}$ or $\frac{\log b}{b} < \frac{\log c}{c}$ for some $c > e$. If we choose $s$ appropriately, then, we can compare $\frac{\log a}{a}, \frac{\log b}{b}$ as desired. (In our above example, we took $s = 1$.)
Example Consider $a = e + \frac{1}{2}, b = e - \frac{1}{2}$---it is not reasonable to try to compute $a^b$ and $b^a$ by hand! Using a fourth-order Taylor polynomial estimate for $\exp$ we find that $$e - \frac{1}{2} < \left(1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{24}\right) - \frac{1}{12} = \frac{9}{4},$$ and using a third-order Taylor polynomial estimate gives
$$e + \frac{1}{2} < \left(1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{6}\right) + \frac{1}{2} = \frac{10}{3} < \frac{27}{8}.$$
But taking $s = 2$ in the above identity gives $$\frac{\log \frac{9}{4}}{\frac{9}{4}} = \frac{\log \frac{27}{8}}{\frac{27}{8}},$$
and putting this all together gives
$$\frac{\log\left(e - \frac{1}{2}\right)}{\left(e - \frac{1}{2}\right)} < \frac{\log \frac{9}{4}}{\frac{9}{4}} = \frac{\log \frac{27}{8}}{\frac{27}{8}} < \frac{\log\left(e + \frac{1}{2}\right)}{\left(e + \frac{1}{2}\right)} ,$$
and we conclude
$$(e - 1)^{e + 1} < (e + 1)^{e - 1} .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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?
| Take your examples together, \begin{align*}
\int e^{2\theta}\sin(3\theta)d\theta
&=\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{2} \left(\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta\right)
\end{align*}
Substituting the integral for a variable, say $X$, gives you:
$$X=\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{2} \left(\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} X\right)$$
simplifying gives you:
$$X=\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{4}e^{2\theta}\cos(3\theta)- \frac{9}{4} X$$
so your answer is
$$X=\int e^{2\theta}\sin(3\theta)=\frac{4}{13}\left(\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{4}e^{2\theta}\cos(3\theta)\right)=\frac{e^{2\theta}\left(2\sin(3\theta)-3\cos(3\theta)\right)}{13}$$
and a simple derivative check shows this to be true. Note your first example last line, you are missing a $\frac{1}{2}$ on the right hand side.
| {
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"url": "https://math.stackexchange.com/questions/3351012",
"timestamp": "2023-03-29T00:00:00",
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Calculate the definite integral: The integral was: $$\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\frac{\pi + 4x^6}{1-\sin(|x|+\frac{\pi}{6})}$$
What I did was to identify that its an even function and write it as: $$2\int_{0}^{\frac{\pi}{6}}\frac{\pi + 4x^6}{1-\sin(x+\frac{\pi}{6})}$$
Then I wrote the denominator in terms of $\cos$ as $$\sin(x+\frac{\pi}{6})=\cos(\frac{\pi}{3}-x)$$ and then I applied the identity $1-\cos x=2\sin^2(\frac{x}{2})$ to finally get :
$$2\biggl[\biggl(\int_{0}^{\frac{\pi}{6}}\frac{\pi}{2\sin^2(\frac{\pi}{6}-\frac{x}{2})}dx\biggl)+\biggl(\int_{0}^{\frac{\pi}{6}}\frac{4x^6}{2\sin^2(\frac{\pi}{6}-\frac{x}{2})}dx\biggl)\biggl]$$
Now the first half is easy to get but how do I integrate the second one?
Can someone suggest any steps or perhaps an alternative way?
| Looking at the monster given by Mathematica in comments, consider
$$I=\int_{0}^{\frac{\pi}{6}}\frac{\pi + 4x^6}{1-\sin(x+\frac{\pi}{6})}\,dx=\pi\int_{0}^{\frac{\pi}{6}}\frac{dx}{1-\sin(x+\frac{\pi}{6})}+4\int_{0}^{\frac{\pi}{6}}\frac{x^6}{1-\sin(x+\frac{\pi}{6})}\,dx$$ The first one is simple.
For the second one, why not to try series expansion around $x=0$
$$\frac{1}{1-\sin(x+\frac{\pi}{6})}=2+2 \sqrt{3} x+5 x^2+\frac{11 x^3}{\sqrt{3}}+\frac{91 x^4}{12}+\frac{301 x^5}{20
\sqrt{3}}+O\left(x^6\right)$$ which, integrated termwise would give
$$\int_{0}^{\frac{\pi}{6}}\frac{x^6}{1-\sin(x+\frac{\pi}{6})}\,dx\approx 0.00902$$ while the numerical integration would give $\approx 0.00937$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $x_1, \cdots, x_n \in (0, 1)$ such that $\sum_{i \ne j}\frac{1}{x_i - x_j} = \frac{1}{1 - x_i} - \frac{1}{x_i}$, calculate $\sum_{i = 1}^nx_i$.
Given positive integer $n$ and $x_1, x_2, \cdots, x_{n - 1}, x_n \in (0, 1)$ such that for all $i = \overline{1, n}$, we always have that $$\large \sum_{i \ne j}\frac{1}{x_i - x_j} = \frac{1}{1 - x_i} - \frac{1}{x_i}$$. Calculate the value of $\displaystyle \sum_{i = 1}^nx_i$.
We have that
$$\sum_{j \ne 1}\frac{1}{x_1 - x_j} = \frac{1}{1 - x_1} - \frac{1}{x_1}$$
$$\sum_{j \ne 2}\frac{1}{x_2 - x_j} = \frac{1}{1 - x_2} - \frac{1}{x_2}$$
$$\vdots$$
$$\sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor - 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1} - x_j} = \frac{1}{1 - x_{\left\lfloor\frac{n}{2}\right\rfloor - 1}} - \frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1}}$$
$$\sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor - x_j} = \frac{1}{1 - x_\left\lfloor\frac{n}{2}\right\rfloor} - \frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor}$$
$$\implies \sum_{j \ne 1}\frac{1}{x_1 - x_j} + \sum_{j \ne 2}\frac{1}{x_2 - x_j} + \cdots + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor - 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1} - x_j} + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor - x_j}$$
$$ = \sum_{i = 1}^\left\lfloor\frac{n}{2}\right\rfloor\dfrac{1}{1 - x_i} - \sum_{i = 1}^\left\lfloor\frac{n}{2}\right\rfloor\dfrac{1}{x_i}$$
Similarly, $$\implies \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor + 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor + 1} - x_j} + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor + 2}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor + 2} - x_j} + \cdots + \sum_{j \ne n - 1}\frac{1}{x_{n - 1} - x_j} + \sum_{j \ne n}\frac{1}{x_n - x_j}$$
$$ = \sum_{i = \left\lfloor\frac{n}{2}\right\rfloor + 1}^n\dfrac{1}{1 - x_i} - \sum_{i = \left\lfloor\frac{n}{2}\right\rfloor + 1}^n\dfrac{1}{x_i}$$
But that's all I got.
| Let $P(x) = \prod_{k=1}^n (x-x_k) = x^n - Ax^{n-1} + \cdots$ and by
Vieta's formula, $\displaystyle\;A = \sum_{i=1}^n x_i$
In terms of $P(x)$, the horrible sum on LHS equals to
$$\sum_{j=1,\ne i}^n \frac{1}{x_i-x_j} = \frac12 \frac{P''(x_i)}{P'(x_i)}$$
This implies following rational function and hence its numerator vanishes at $n$ points $x_1,\ldots,x_n$.
$$\frac12\frac{P''(x)}{P'(x)} - \frac{1}{1-x} + \frac{1}{x} =
\frac{(4x-2)P'(x) + (x^2-x)P''(x)}{2x(x-1)P'(x)}
$$
The numerator is a polynomial of degree at most $n$. So it is a multiple of $P(x)$. Comparing coefficients of $x^n$, we obtain
$$(4x-2)P'(x) + (x^2-x)P''(x) - n(n+3)P(x) = 0$$
Throwing the expansion $P(x) = x^n - Ax^{n-1} + \cdots$ into above expression and extract the coefficient of $x^{n-1}$, we obtain
$$(2n+2)A - (n^2+n) = 0\quad\implies\quad \sum_{i=1}^n x_i = A = \frac{n}{2}$$
Update
For a solution without calculus, multiply both sides by $x_i(1-x_i)$ and sum over $i$. One obtain:
$$\sum_{i,j;j\ne i}\frac{x_i - x_i^2}{x_i - x_j} = \sum_{i=1}^n(2 x_i - 1) = 2A - n$$
Swapping the summation index $i,j$ in LHS and take averages, we obtain
$$\begin{align}
{\rm LHS} = \sum_{i,j;j\ne i}\frac{x_i - x_i^2}{x_i - x_j}
&= \frac12\sum_{i,j;j \ne i}\frac{(x_i - x_i^2) - (x_j - x_j^2)}{x_i - x_j}\\
&= \frac12\sum_{i,j;j \ne i}(1 - x_i - x_j)\\
&= \frac12\left[\sum_{1 \le i,j \le n}(1 - x_i - x_j) - \sum_{i=1}^n (1 - 2x_i)\right]\\
&= \frac12\left[(n^2 - 2nA) - (n - 2A)\right]
\end{align}
$$
Combine with RHS $= 2A - n$, we get
$$n^2 - 2nA - n + 2A = 4A - 2n
\iff n(n+1) = 2(n+1) A$$
This leads to $A = \frac{n}{2}$ again.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $A+B+C=180$, then $\frac{\sin2A+\sin2B+\sin2C}{\cos A+\cos B+\cos C-1}=8\cos\frac A2 \cos\frac B2 \cos\frac C2$ Then $2A+2B+2C =360$
So $$\sin 2C=-\sin(2A+2B)$$
Putting that in the equation
$$\frac{2\sin(A+B)\sin(A-B)-2\sin(A+B)\cos(A+B)}{\cos A+\cos B-\cos(A+B)+1}$$
$$\frac{2\sin(A+B)[\sin(A-B)-\cos(A+B)]}{\cos A+\cos B-\cos(A+B)+1}$$
I don’t know how to proceed. Please help me continue
| We will proceed by simplifying the numerator and denominator separately and repeatedly use some well-known formulae.
For simplifying the numerator: \begin{align} \sin 2A + \sin 2B + \sin 2C & = 2\sin(A+B)\cos(A-B) + 2\sin C \cos C \\ & = 2\sin(\pi - C) \cos(A - B) + 2\sin C \cos C \\ & = 2\sin C [\cos(A - B) + \cos(\pi - (A + B))] \\ & = 2\sin C[\cos(A - B) - \cos(A + B)] \\ & = 4\sin A \sin B \sin C\end{align}
For simplifying the denominator: \begin{align} \cos A +\cos B + \cos C - 1 & = 2\cos(\frac{A +B}{2}) \cos(\frac{A-B}{2}) - 2 \sin^2 \frac{C}{2} \\ & = 2\cos(\frac{\pi}{2} - \frac{C}{2})\cos(\frac{A-B}{2}) - 2\sin^2\frac{C}{2} \\ & = 2\sin \frac{C}{2}[\cos(\frac{A-B}{2}) - \sin \frac{C}{2}] \\ & =2\sin \frac{C}{2}[\cos(\frac{A-B}{2}) - \cos(\frac{A+B}{2})] \\ & =4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\end{align}
Can you take it from here?
| {
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Flaw in A.M.-G.M. inequality for finding the range of function. Let $f(x) =2^x+2^ {-x} +3^x+3^ {-x} +4$. The minimum value of this function is 8.
Which comes as
$\dfrac{2^x+\frac{1}{2^x}}{2}\ge \sqrt{2^x\cdot\frac{1}{2^x}}$
$2^x+2^{-x}\ge 2$
similarly,
$3^x+3^{-x}\ge 2$
Both these terms have minimum value at x=0
But what if we us the inequality as-
$\dfrac{2^x+2^{-x}+3^x+3^{-x}+4}{5}\ge\sqrt[5]4$
Hence the minimum value comes out to be $\sqrt[5]4$ which is less than 8.
I cannot see any flaw in both of the methods. What am I missing?
| There's no guarantee there ever has to be situation where the LHS value has to match the RHS value when using the AM-GM inequality. The inequality only states the $2$ may be equal, but the LHS is always at least the size of the RHS.
In your last equation, note that since $2^x + 2^{-x} \ge 2$ and $3^x + 3^{-x} \ge 2$, then $\frac{2^x + 2^{-x} + 3^x + 3^{-x} + 4}{5} \ge \frac{2 + 2 + 4}{5} = \frac{8}{5} \gt \sqrt[5]{4}$.
| {
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Find the point on the graph of $z= x^{2} +y^{2} +10$ nearest to the plane $x+2y-z=0$. Find the point on the graph of $z= x^{2} +y^{2} +10$ nearest to the plane $x+2y-z=0$.
So, any point on the given surface will be $(x,y,x^{2} +y^{2} +10)$ .
I need to minimize the function $(x +2y-x^{2}-y^{2}-10)/(\sqrt{6})$
The only critical point is $(1/2,1)$.
But this point gives maximum of the function.
How would I find the nearest point.
What I think is that, I should change the sign of the function, to keep the distance positive, then I'll get the same critical point, but the value will be minimum. So I'll get the nearest point.
In this case if I am asked to find the maximum distance, what it should be then$?$
| Note that you have to minimize the function $$ \frac{ |x +2y-x^{2}-y^{2}-10|}{\sqrt{6}} $$In your post, you does not take absolute value.
Observe that $$x^2 + y^2 -x - 2y + 10=(x-\frac{1}{2})^2 + (y-1)^2 + \frac{35}{4} $$ which is always positive. so $$ \frac{ |x +2y-x^{2}-y^{2}-10|}{\sqrt{6}} = \frac{ x^2+y^2-x-2y+10}{\sqrt{6}} \geq \frac{35}{4\sqrt6}
$$ It attains its minimum at $(x, y)=(\frac 12, 1)$.
| {
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How many $2020$-digit-numbers divisible by 3 satisfying the digits are $2, 4, 5, 7$ and digits $4, 5$ must not stand next to one another are there?
Determine how many $2020$-digit-numbers simultaneously satisfy these three constraints:
i/ every digit of the number is one of the following: $2, 4, 5, 7$;
ii/ digits $4$ and $5$ must not stand next to one another;
iii/ the number is divisible by $3$.
Let $x_n, y_n, z_n, t_n$ be the number of $n$-digit-numbers that end with $2, 4, 5, 7$ respectively $(n \in \mathbb Z^+)$.
We have that $x_n + y_n + z_n + t_n = S_n$.
(with $S_n$ being that number of $n$-digit-numbers that satisfy the first two conditions of the problem).
It can be seen that $t_n = x_n, y_n = z_n \implies \left\{ \begin{align} x_{n + 1} &= x_n + 2y_n\\ y_{n + 1} &= S_n\\ S_{n + 1} &= 2(x_n + y_n) \end{align} \right. \implies S_{n + 2} = 3S_{n + 1} + 2S_n$.
There exist constants $p$ and $q$ such that $S_n$ can always be expressed in the form of $p\alpha^n + q\beta^n$ for $\forall n \in \mathbb Z^+$ where $\alpha$ and $\beta$ are roots of the equation $x^2 - 3x - 2 = 0$.
Furthermore, it can be obtained that $S_1 = 4$ and $S_2 = 10$.
$\implies \left\{ \begin{align} \alpha + \beta &= 3\\ \alpha\beta &= - 2\\ p + q &= 4\\ p\alpha + q\beta &= 10 \end{align} \right.$ $\implies \left\{ \begin{align} (\alpha, \beta) = \left(\dfrac{3 + \sqrt{17}}{2}, \dfrac{3 - \sqrt{17}}{2}\right)\\ (p, q) = \left(2 + \dfrac{4}{\sqrt{17}}, 2 - \dfrac{4}{\sqrt{17}}\right)\end{align} \right.$
$$ \implies S_n = 2\left[\left(1 + \dfrac{2}{\sqrt{17}}\right) \cdot \left(\dfrac{3 + \sqrt{17}}{2}\right)^n + \left(1 + \dfrac{2}{\sqrt{17}}\right) \cdot \left(\dfrac{3 - \sqrt{17}}{2}\right)^n \right]$$
| This is an answer to your question about solving the recurrence relation.
The method of solving a linear recurrence relation such as yours is as follows.
Solve $x^2-3x-2=0$. This has roots $\frac{3+\sqrt {17}}{2}$ and $\frac{3-\sqrt {17}}{2}$.
Then $S_n=A\left(\frac{3+\sqrt {17}}{2}\right)^n+B\left(\frac{3-\sqrt {17}}{2}\right)^n$ for constants $A$ and $B$.
The constants can be found from the values of $S_1$ and $S_2$.
Obtaining your recurrence relation
There were some queries about your derivation of this. However, it can be obtained as follows.
$$x_{n+1}=t_{n+1}=S_n$$
$$y_{n+1}=S_n-z_n, z_{n+1}=S_n-y_n$$
Therefore
$$y_{n+1}+z_{n+1}=2S_n-y_n-z_n$$
$$S_{n+1}-2S_n=2S_n-(S_n-2S_{n-1})$$
$$S_{n+1}=3S_n+2S_{n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Write the expression so that the variable is only presented once and the exponents are positive. There is this problem
$\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}}$
and i know that the answer is
$\frac{2x^{\frac{5}{6}}}{3y}$
but i cant find the logic of how is solved, Where do the coefficients come from?
I understand that fractions are the result of multiplication of expressions, but I do not understand how all come together.
| You need to know some basic Exponential identities :
$$\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}}$$
$$=\frac{10}{15} \times \frac{x^{1/3}}{x^{-1/2}} \times \frac{y^{-1/4}}{y^{3/4}}$$
Using $\frac{x^a}{x^b} = x^{a-b}$
$$=\frac{5 \times 2}{5 \times 3} \times x^{1/3+1/2} \times y^{-1/4-3/4}$$
$$=\frac{2}{3} \times x^{5/6} \times y^{-1}$$
$$=\frac{5 \times 2}{5 \times 3} \times x^{5/6} \times y^{-1}$$
$$=\frac{2x^{\frac{5}{6}}}{3y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse function mod $$ y = 3x + 7 \pmod 4,\quad x,y \in \mathbb{Z}_4 $$
$x = 2, y =$ ??
Find the inverse function and verify the value of $y$.
I solved only the first question $y = 13\bmod 4 = 1 \bmod 4.
How to solve the second question for inverse?
| The only possibilities are:
$$(x,y) \in \{ (0,3),(1,2),(2,1),(3,0) \}$$
So the functions are as follows:
$$y=\left\{\begin{array}{c} 3 & x=0 \\ 2 & x=1 \\ 1 & x=2 \\ 0 & x=3 \\ \end{array}\right\}$$
$$x=\left\{\begin{array}{c} 3 & y=0 \\ 2 & y=1 \\ 1 & y=2 \\ 0 & y=3 \\ \end{array}\right\}$$
And if you need some formula, it is:
$$y \overset{4}{\equiv} 3x+7 \overset{4}{\equiv} -x+3 \Longrightarrow x+y \overset{4}{\equiv} 3 \Longrightarrow \left\{ \begin{array}{c} y \overset{4}{\equiv} 3-x \\ x \overset{4}{\equiv} 3-y \\ \end{array} \right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362684",
"timestamp": "2023-03-29T00:00:00",
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$\sum_{k=1} ^n (k^2 +1)k!=n(n+1)!$ I'm to prove this by mathematical induction:
Edited: I made a typo error.
$\sum_{k=1} ^n (k^2+1)k!=n(n+1)!$
I made the test and the rightside is true.
So I tested: $N+1$
$N(N+1)! + (N^2+1)N!$
$N(N+1)(N!)+(N^2+1)N!$
$N!\left[N(N+1)+N^2+1\right]$
$N! (N^2+N+N^2+1)$
$\sum_{k=1} ^ {N+1} (k^2+1)k!=$ $N!(2N^2+N+1)$
Is this a valid proof?
| $\sum _{k=1} ^{N+1} (k^2+1)k!$=$[(N+1)^2 +1]!+N(N+1)!\\
=(N+1)![(N^2+3N+2]\\
=(N+1)(N+1)(N+2)\\
=(N+1)(N+2)!$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the sequence $C_{n+1} = 1 + \frac{C_n}{C_n + 1}$ is monotonically increasing by induction I have a sequence
$$C_{n+1} = 1 + \frac{C_n}{C_n + 1}$$
With a base case $C_1 = 3/2$ and want to prove that it's monotonically increasing by induction.
Whenever I try to prove $C_n < C_{n+1} \implies C_{n+1} < C_{n+2}$
where $$C_{n+2} = 1 + \frac{2C_n + 1}{3C_n + 2} $$
I get to the inequality $$ \frac{C_n}{C_n + 1} < \frac{2C_n + 1}{(C_n + 1)^2} $$
but don't know how to get to the point where that implies that $C_{n+1} < C_{n+2}$
Thank
| rewrite as $$C_{n+1} = 1 + \frac{C_n}{C_n + 1}= \frac{2C_n+1}{C_n + 1}=2-\frac{1}{1+C_n}$$
now
$$C_{n+2} -C_{n+1}=2-\frac{1}{1+C_{n+1}}-(2-\frac{1}{1+C_{n}})=\\
\frac{1}{1+C_{n}}-\frac{1}{1+C_{n+1}}=
\\\frac{1+C_{n+1}-(1+C_{n})}{(1+C_{n+1})(1+C_{n})}=\\
\frac{C_{n+1}-C_{n}}{(1+C_{n+1})(1+C_{n})}=$$so now you can build your induction
$$ C_{n+1}>C_n \implies C_{n+2} > C_{n+1}$$because
$$C_{n+2} -C_{n+1}=\frac{C_{n+1}-C_{n}}{(1+C_{n+1})(1+C_{n})}=\frac{C_{n+1}-C_{n}}{(positive)(positive)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find coefficient of $x^2$ Let $C$ be the coefficient of $x^2$ in the expansion of the product
$$(1-x)(1+2x)(1-3x) \cdots (1+14x)(1-15x)$$
Find: $\lvert C \rvert $
| You may consider doing it manually.
A term of $x^2$ arises from the product of two factors with the $1$s from the rest, so by multiplying all the pairs of coefficients we get our answer.
Or, our answer is: $$|-1(2-3+4-5 \dots 14-15) + 2(-3+4-5-6\dots -13+14-15) \dots -13(14-15) +14(-15)|$$
$$=|-1(-7) + 2(6-15) - 3(-6) + 4(5-15) -5(-5) \dots -13(-1) +14(-15)|$$
$$=588$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find equations of all parabolas passing through points (0,3), (2,0) and tangential to $x+y=0$ I need to find equations of all parabolas, that pass through points $(0,3)$ and $(2,0)$ on affine plane $\mathbb{R}^{2}$ and tangent to line $x+y=0$ at origin.
I am trying to solve this problem using conformal geometry. Any hints?
| I don't know how to structure an argument using conformal geometry, so here's an alternative approach.
Using a similar strategy to this answer, we note that the equation for the conic through five points $P=(P_x, P_y)$, $Q=(Q_x,Q_y)$, $R=(R_x,R_y)$, $S=(S_x,S_y)$, $T=(T_x,T_y)$ is given by
$$\left|\begin{array}{c,c,c,c,c,c}
x^2 & y^2 & x y & x & y & 1 \\
P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\
Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\
R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\
S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\
T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\
\end{array}\right| = 0 \tag{$\star$}$$
Generalizing the problem at hand a bit, let's take $P:=(p^2,0)$ and $Q:=(0,q^2)$ (the squares are convenient later). Let $R:=(0,0)$. Point $S$ will be a vanishingly-small displacement of $R$ along the tangent line $x+y=0$; so $S := (s,-s)$. Finally, $T$ is some other point; say, $T = (t\cos\theta,t\sin\theta)$. Then $(\star)$ becomes, after dividing through by $s$ and then also taking $s\to 0$:
$$\begin{align}
0 &= x^2 q^2 t \sin\theta\cos\theta + y^2 p^2 t \sin\theta\cos\theta \\
&-x y\left( p^2 t \sin^2\theta + q^2 t \cos^2\theta - p^2 q^2 (\cos\theta+\sin\theta) \right) \\
&- x\,p^2 q^2 t \sin\theta\cos\theta \\
&- y\,p^2 q^2 t \sin\theta\cos\theta
\end{align} \tag{1}$$
Since our conic is specifically a parabola, we must have
$$(\text{coefficient of}\;xy)^2 = 4(\text{coefficient of}\;x^2)(\text{coefficient of}\;y^2) \tag{2}$$
This condition yields a quadratic in $t$, which we solve and simplify to give
$$t = \frac{p^2 q^2 (\cos\theta + \sin\theta)}{\left(p\sin\theta\pm q\cos\theta\,\right)^2}\tag{3}$$
Substituting back into $(1)$, we discard a common factor of $p^2q^2\sin\theta\cos\theta(\sin\theta+\cos\theta)/(p\sin\theta\pm q\cos\theta)$, leaving an equation that's independent of $T$ except for a sign ambiguity:
$$q^2 x^2 + p^2 y^2 \pm 2p q\,x y - p^2 q^2 x - p^2 q^2 y = 0 \tag{4}$$
which we can write as
$$\left(\frac{x}{p}\pm\frac{y}{q}\right)^2 = x+y
\tag{$\star\star$}$$
Substituting $p\to\sqrt{2}$ and $q\to\sqrt{3}$ is left as an exercise to the reader. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.