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I somehow deduced that $\tan x=\iota$ for any real value of $x$ by equating the value of $\tan(\frac{\pi}{2}+x)$ obtained using two identities. Let's assume that we're familiar with the identity : $\tan \Bigg (\dfrac{\pi}{2} + x \Bigg ) = -\cot x$ which we have derived using the unit circle.
I was trying to equate the values of $\tan \Bigg ( \dfrac{\pi}{2} + x \Bigg )$ obtained using the above mentioned identity and the compound angle identity and I got a weird result. Have a look :
$$\tan \Bigg ( \dfrac{\pi}{2} + x \Bigg ) = \dfrac{\tan\dfrac{\pi}{2} + \tan x}{1 - \tan \dfrac{\pi}{2} \tan x}$$
For the sake of simplicity, let us assume that $\tan \dfrac{\pi}{2} = a$ and $\tan x = b$.
$$ \therefore \tan \Bigg ( \dfrac{\pi}{2} + x \Bigg ) = \dfrac{a + b}{1 - ab} \implies -\cot x = \dfrac{a + b}{1 - ab}$$
Also,
$$-\cot x = \dfrac{-1}{\tan x} = \dfrac{-1}{b}$$
$$ {\color{red} {\therefore \dfrac{-1}{b} = \dfrac{a + b}{1 - ab} \implies -1 + ab = ab + b^2 \implies -1 = b^2}}$$
This leads us to :
$$\tan x = b = \sqrt{-1} = \iota$$
which is not true.
So, what went wrong here?
I think that the ${\color{red}{\text{highlighted part}}}$ was wrong because while cross-multiplying, I automatically made the assumption that $1 - ab$ has a real value which won't be the case if $\tan \Bigg ( \dfrac{\pi}{2} \Bigg )$ doesn't have a real value (which is actually the case as $\tan \Bigg ( \dfrac{\pi}{2} \Bigg ) = \dfrac{\sin \Bigg ( \dfrac{\pi}{2} \Bigg )}{\cos \Bigg ( \dfrac{\pi}{2} \Bigg )} = \dfrac{1}{0}$ which does not have a real value and approaches $\infty$)
Was this the mistake I made?
Thanks!
| This is a cool "paradox," hadn't seen it before!
Even before the red line, the identity
$$\tan \Bigg ( \dfrac{\pi}{2} + x \Bigg ) = \dfrac{\tan\dfrac{\pi}{2} + \tan x}{1 - \tan \dfrac{\pi}{2} \tan x}$$
is objectionable.
This isn't true-- or, more accurately, it isn't even grammatically correct, since $\frac{\pi}{2}$ isn't in the domain of the tangent function.
Similarly, when you let $a = \tan\big(\frac{\pi}{2}\big)$ you are saying something grammatically incorrect, and so you can't expect to do formal algebraic manipulations with $a$ and receive something meaningful. It might be instructive to replace your identity with the identity
$$\tan \Bigg ( \dfrac{\pi}{2} + x \Bigg ) = \lim_{y \to \frac{\pi}{2}} \dfrac{\tan y + \tan x}{1 - \tan y \tan x}$$
(which is valid for any $x$ not an integer multiple of $\pi$), and see what happens.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's going on with this Wolfram Alpha matrix multiplication? I was computing the matrix exponential of
$$X = \begin{pmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$
by diagonalization. Wolfram Alpha and Symbolab both solve this problem correctly. After the diagonalization and taking the exponential of the diagonal matrix, the only thing left to do is to compute the product
$$e^X = Pe^DP^{-1} = \begin{pmatrix}1&-1&1\\ 1&0&-2\\ 1&1&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&e&0\\ 0&0&e^3\end{pmatrix}\begin{pmatrix}\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\ -\frac{1}{2}&0&\frac{1}{2}\\ \frac{1}{6}&-\frac{1}{3}&\frac{1}{6}\end{pmatrix},$$
and this is where I ran into a problem. Symbolab computes this multiplication and obtains the value of $e^X$ that I was expecting (and that Wolfram Alpha and Symbolab both reported earlier), that is,
$$e^x = \begin{pmatrix}\frac{2+3e+e^3}{6}&\frac{-e^3+1}{3}&\frac{2-3e+e^3}{6}\\ \frac{-e^3+1}{3}&\frac{2e^3+1}{3}&\frac{-e^3+1}{3}\\ \frac{2-3e+e^3}{6}&\frac{-e^3+1}{3}&\frac{2+3e+e^3}{6}\end{pmatrix}.$$
But when I tried the same multiplication on Wolfram Alpha, all of the entries were off by $1/3$rd. I checked to make sure that I had typed everything in right, checked what Wolfram Alpha gave for the inverse to see if it matched, etc., and everything else seemed right.
Am I doing something wrong? Does Wolfram Alpha not know how to multiply matrices or am I just making a simple mistake somewhere?
| As user1551 pointed out in a comment, the answer was as simple as a missing $1$: If
$$D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}, \quad \textrm{then} \quad e^D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^3 \end{pmatrix}, \quad \textrm{not} \quad e^D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & e^3 \end{pmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find points that are on two elliptic curves in $F_p$ I have some basic cryptography question but I don't know if my calculations are not overcomplicated and if there's some simpler solution. Following algorithm is explained here.
Problem
There are two elliptic curves in $F_5$ with equations:
a) $y^2 = x^3 + 2x + 1 \pmod{5}$
b) $y^2 = x^3 + x + 1 \pmod{5}$
Find points that are in these two eliptic curves.
My approach is to determine points on these eliptic curves separately and then check for duplicates.
Determining quadratic residue in $\mod{5}$
For every value in $\mod{5}$ calculate square for it, so
$
(\pm 1)^2 \pmod{5} = 1 \\
(\pm 2)^2 \pmod{5} = 4 \\
(\pm 3)^2 \pmod{5} = 9 \pmod{5} = 4 \\
(\pm 4)^2 \pmod{5} = 16 \pmod{5} = 1 \\
$
So our quadratic residue set is $QR = \{1, 4 \}$
Determining points in a)
Equation is:
$
y^2 = x^3 + 2x + 1 \pmod{5}
$
Let's build table:
*
*In column $x$ we have all possible $x$.
*In second column we calculate curve equation.
*In third column we check if value is quadratic residue. So we check if its in $QR$. There is also option to check this with Euler theorem. For $y=3$ we check if $y^{(p-1)/2} \equiv 1 \pmod{p}$ so $3^2 \equiv 1 \pmod{5}$ is false because $3^2 \pmod{5} = 4$.
*For all quadratic residues we calculate square roots. We can get them from step where we determined quadratic residues. For example we have $1$ for $1^2$ and $4^2$. So $1$ and $4$ are square roots of $1$.
So points on this elliptic curve are:
$(0, 1)$, $(0, 4)$, $(1, 2)$, $(1, 3)$, $(3, 2)$, $(3, 3)$
Determining points in b)
Equation is:
$
y^2 = x^3 + x + 1 \pmod{5}
$
Let's build table:
So points on this eliptic curve are:
$(0, 1)$, $(0, 4)$, $(2, 1)$, $(2, 4)$, $(3, 1)$, $(3, 4)$, $(4, 2)$, $(4, 3)$.
Result
Points $(0, 1)$ and $(0, 4)$ are on these two functions.
Checking if correct
I've found website that generate points for given EC. Solution that I've found is correct. Is there faster way to found these points?
EC a)
EC b)
| Since you are working on the same field for both the curves. To get the common points $(x_0,y_0)$ that lie on both the curves, you just need to to equate the following (think in terms of intersection of two curves)
\begin{align*}
y_0^2 =x_0^3+2x_0+1 &\equiv x_0^3+x_0+1 \pmod{5}\\
x_0 & \equiv 0 \pmod{5}.
\end{align*}
Thus $x_0=0$. Now this gives us that $y_0^2 \equiv 1 \pmod{5}$, which has only two solutions, namely $y_0=1,4$. Thus the common points are $(0,1)$ and $(0,4)$ and $\color{blue}{\text{the point at infinity } \mathcal{O}}$.
Note: If the task was to find the points on each curve separately, then what you did would be a good approach.
| {
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How to find area of rectangle inscribed in ellipse. In an ellipse $4x^2+9y^2=144$ inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis.
Longer side which is parallel to major axis, relates to the shorter sides as $3:2$. Find area of rectangle.
I can find the values of $a$ and $b$ as
$$\frac{4x^2}{144}+\frac{9y^2}{144}=1$$
$$\frac{x^2}{6^2}+\frac{y^2}{4^2}=1$$
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, gives
$a=6$ & $b=4$. From here I have no idea how to solve further?
| We're lucky that the ellipse is centered at the origin. :)
In this case, the inscribed rectangle is also centered at the origin. If $P = (x, y)$ is the vertex of the inscribed rectangle at the first quadrant, then the smaller rectangle spanned by the origin and point $(x, y)$ is similar to the inscribed rectangle.
Each side of the inscribed rectangle is $2$ times that of the smaller rectangle. Hence, the area of the inscribed rectangle is $2^2 = 4$ times that of the smaller rectangle. The sides of the smaller rectangle are $x$ and $y$ respectively so we have $x \colon y = 3 \colon 2$. This means we have $2x = 3y$.
Plugging it into the equation of the ellipse, we have
\begin{align}
4x^2 + 9y^2 &= 144 \\
(2x)^2 + (3y)^2 &= 144 \\
(2x)^2 + (2x)^2 &= 144 \\
8x^2 &= 144 \\
x^2 &= 18 \Rightarrow x = \pm3\sqrt{2}
\end{align}
Since $(x, y)$ is in the first quadrant, we have $x = 3\sqrt{2}$ and hence $y = 2\sqrt{2}$. The area of the smaller rectangle is $xy = 12$, so the area of the inscribed rectangle is $4xy = 4 \cdot 12 = 48$.
Update
I have incorporated @zwim's comment into the answer. Yeah, it's more natural this way, thanks @zwim! :) Also I have corrected an error in the computation of $x$ and $y$ in my previous answer.
| {
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Question about finding the value of an infinite sum What is the value of:
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}?$$
I realised that
$$\sum_{n=2,4,6,8,...} \frac{1}{n^2} + \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 1 } \frac{1}{4n^2}+\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \frac{3}{4}\frac{\pi^2}{6} \qquad \Rightarrow \qquad \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \frac{1}{8} \cdot \pi^2$$
So
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2} + \sum_{k=0}^\infty \frac{1}{(4k+3)^2} = \frac{\pi^2}{8}$$
But I cannot find the value of the second summation.
Any suggestions?
| Recall the definition of Catalan's Constant:
$$K=\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}- \sum_{k=0}^{\infty} \frac 1 {(4k+3)^2}$$
Combining that with the equation you have already derived we immediately get $$\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}=\frac K2+\frac {\pi^2}{16}$$
| {
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Continuity of this Piecewise function $f:\mathbb{R}^2\to \mathbb{R}$ I have shown that $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(0,0) = 0$ and $\displaystyle f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}$ if $(x,y)\ne (0,0)$ isn't differentiable at $(0,0)$, now I'm trying to show whether it is continuous or not.
My attempt: I must show that $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x|y|}{\sqrt{x^2+y^2}}=f(0,0) = 0$. But $x^2+y^2-2|xy| = (|x|-|y|)^{2}\ge 0$ so $\displaystyle|xy|\le \frac{x^2+y^2}{2}$. So $\displaystyle\left|\frac{x|y|}{\sqrt{x^2+y^2}}\right| =\frac{|xy|}{\sqrt{x^2+y^2}}\le \frac{x^2+y^2}{\sqrt{x^2+y^2}} = (x^2+y^2)^{1/2}$ and once $(x^2+y^2)^{1/2}\to0$ as $(x,y)\to(0,0),$ we would have it.
Is this correct?
| Let us consider the directional derivative of $f$ in the direction $v = (a,b)$ at the point $(0,0)$. We have
\begin{align*}
D_{v}f(0,0) & = \lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t}\\\\
& = \lim_{t\to 0}\frac{ta|tb|}{t\sqrt{t^{2}a^{2}+t^{2}b^{2}}} = \frac{a|b|}{\sqrt{a^{2}+b^{2}}}
\end{align*}
On the other hand, its partial derivatives are given by
\begin{align*}
\begin{cases}
\displaystyle\frac{\partial f}{\partial x}(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(1,0)) - f(0,0)}{t} = 0\\\\
\displaystyle\frac{\partial f}{\partial y}(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(0,1)) - f(0,0)}{t} = 0
\end{cases}
\end{align*}
Finally, if $f$ was differentiable, we should have that
\begin{align*}
D_{v}f(0,0) = a\frac{\partial f}{\partial x}(0,0) + b\frac{\partial f}{\partial y}(0,0) = a\times 0 + b\times 0 = 0
\end{align*}
which is different from the previous result. Thus $f$ is not differentiable at $(0,0)$.
EDIT
Your approach is fine. Here I propose another way to solve it.
To begin with, notice that $|y| = \sqrt{y^{2}} \leq \sqrt{x^{2} + y^{2}}$. Consequently, one has
\begin{align*}
\frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1 \Rightarrow \frac{|xy|}{\sqrt{x^{2}+y^{2}}} \leq |x| \xrightarrow{(x,y)\rightarrow(0,0)} 0
\end{align*}
Hence $f$ is continuous at $(0,0)$.
| {
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Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplified I got
$$\int\dfrac{2-2\tan^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\tan^2\dfrac{x}{2}+6\tan\dfrac{x}{2}-5}dx$$
substituted $\tan^2\dfrac x2=\sec^2\dfrac x2-1$
$$\int\dfrac{4-2\sec^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\left(\tan\dfrac{x}{2}+\dfrac{3}{5}\right)^2-\dfrac{34}{5}}dx$$
I can't eliminate $\tan\frac x2$ term in numerator. I think I am not in right direction. your help to solve this integral is appreciated. thank in advance
| hint
Dividing by $\cos(x)$ and putting $\tan(x)=t$, it becomes
$$\int \frac{(2-t)dt}{(5+3t)(1+t^2)}$$
$$=\frac{1}{34}\int \Bigl(\frac{33}{5+3t}+\frac{-11t+7}{t^2+1}\Bigr)dt$$
$$=\frac{11}{34}\int \frac{3dt}{5+3t}-\frac{11}{68}\int \frac{2tdt}{t^2+1}+$$
$$\frac{7}{34}\int \frac{dt}{t^2+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Parametrization the curve of intersection of sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$. I've seen similar posts here but none of the answers helped me. I am trying to parametrize a curve of intersection of a (top half $z>0$) sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.
I tried $$x=\frac{1}{2}\cos t $$ $$y=\frac{1}{2}+\frac{1}{2}\sin t$$ $$z=\sqrt{5-\left(\frac{1}{2}\cos t\right)^2-\left(\frac{1}{2}+\frac{1}{2}\sin t\right)^2}=\sqrt{\frac{9}{2}-\frac{1}{2}\sin t}$$ for $t \in (0,2\pi)$, but I don't think it is correct. Even if it was, is there a better (more simple) approach?
Note: I need to find the circulation of a field $F=(y+z,x-z,0)$ over this curve so I need a good parametrizatian so that I was able to integrate it.
| Sphere
$$ x^2 +y^2 + z^2= 5 \tag1 $$
simplify given second Cylinder equation
$$ x^2+y^2 -y=0 $$
Subtract
$$z^2+y = 5 \; \tag2 $$
To satisfy this relation now introduce parameter $t$
$$z^2 = 5 \cos^2 t,\;\tag3 $$
so that
$$ y = 5 \sin ^2 t \tag4 $$
From (1) and (4)
$$ z^2= 5 -x^2 -25 \sin ^4 t \tag 5 $$
Equate RHS of $z^2$ from (3) and (5) simplify to find parametrization for $x$
$$ x = \sqrt{5 \sin^2 t (1-5 \sin^2 t )}\tag6$$
From(3)
$$ z= \sqrt 5 \cos t \tag 7$$
So parametrization for three coordinates
$$ (x,y,z)=\big( \sqrt{5 \sin^2 t (1-5 \sin^2 t )}, 5 \sin ^2 t ,\sqrt 5 \cos t \big) \tag8$$
where positive sign is taken before radical signs for top hemisphere.
| {
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Geometry question: Find the area of blue-shared area inside this isosceles See below, looks a bit interesting, but I cannot find a solution.
I think a starting point might be the similarity of the lower white triangle and the larger triangle composed of lower blue, pink, and lower right triangle though..
| Expanding-upon a comment, as requested.
Let the triangle have vertex angle $2\theta$, let $a\leq b$ be the sides of the squares, and let $t$ denote the target area. We immediately find
$$\begin{align}
a+a\cot\theta = |PQ| = b+b\tan2\theta \tag{1} \\
t = \frac12\cdot a\cdot a\cot\theta + \frac12\cdot b \cdot b\tan2\theta \tag{2}
\end{align}$$
From here, "all we have to do" is eliminate $\theta$ from $(1)$ and $(2)$. One way to do that here is first to treat the equations as a linear system in $\cot\theta$ and $\tan2\theta$; solving gives
$$\cot\theta = \frac{2t-b(a-b)}{a (a + b)} \qquad
\tan2\theta = \frac{2t+a(a-b)}{b (a + b)} \tag{3}$$
Reciprocating $\cot\theta$ into $\tan\theta$, and substituting into the double-angle identity $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$, and "simplifying" gives this relation among $t$, $a$, and $b$:
$$\begin{align}
0
&= 8 t^3 \\
&+ 4 t^2 (a - b)(a - 2 b) \\
&- 2 t\phantom{^2}(a^2 + b^2)(a^2 + 6 a b - b^2) \\
&- \phantom{2}a (a - b) (a^4 + 2 a^3 b - 2 a^2 b^2 - 2 a b^3 - 3 b^4)
\end{align} \tag{$\star$}$$
In the particular case of $a=7$ and $b=16$, $(\star)$ reduces into a factorable form and we have
$$(2t-339) (4t^2+1128t+49371) = 0 \quad\to\quad t = \frac{339}{2} \tag{$\star$'}$$
which gives our answer. $\square$
As an alternative to the above approach with $(3)$ we could re-write $(1)$ in terms of $\tan\theta$ alone:
$$(a-b)\tan^3\theta + (a+2b)\tan^2\theta-(a-b)\tan\theta-a=0 \tag{4}$$
For $a=7$ and $b=16$, this reduces to
$$(3\tan\theta-1) (3\tan^2\theta-12\tan\theta-7) =0 \tag{5}$$
Therefore (casually ignoring a positive root of the second factor),
$$\tan\theta = \frac13\qquad\to\qquad\tan2\theta=\frac{3}{4}\qquad\stackrel{\text{by (1)}}{\to}\qquad t = \frac{339}{2}\tag{6}$$
(See the edit history that also rewrites $(2)$ in terms of $\tan\theta$ alone before eliminating $\tan\theta$ from the system to get $(\star)$.)
An advantage of this approach is that $(4)$ is "more-immediate" to factor. Also, we see explicitly that $\tan\theta$ and $\tan2\theta$ are nice values. Knowing $\tan\theta=1/3$ facilitates drawing an accurate figure.
More importantly, $\tan2\theta=3/4$ reveals a non-obvious connection to the $3$-$4$-$5$ triangle. I can't help thinking that there's a clever way to detect and exploit it.
Yet-another way to proceed is to seek the distance $p:=|PQ|$. (@Aretino's answer uses another approach to this.) Clearly,
$$\tan\theta = \frac{a}{p-a} \quad \tan2\theta=\frac{p-b}{b} \qquad\stackrel{\text{dbl ang}}{\to}\qquad p^3 - p^2(2 a + b) + 2 a^2 b =0 \tag{7}$$
(The nicest cubic yet!) For $a=7$ and $b=16$, this reduces to
$$(p-28) (p^2-2p-56)=0 \quad\to\quad p = 28 \tag{8}$$
from which the result follows thusly:
$$t = \frac12a(p-a)+\frac12b(p-b) = \cdots = \frac{339}{2} \tag{9}$$
This solution seems like it might be the most accessible to middle-schoolers, although it, too, ignores the underlying $3$-$4$-$5$ triangle. (Perhaps I'm over-estimating the significance of that triangle!)
| {
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Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my lights go off.
Any leads? Thanks in advance.
| By conjugates $\sqrt{a+b\pm\sqrt{2ab+b^2}}=\sqrt{x}\pm\sqrt{y}$ i.e. $a+b=x+y,\,2ab+b^2=4xy$, so $x,\,y$ are roots of $t^2-(a+b)t+ab/2+b^2/4=(t-a-b/2)(t-b/2)$.
| {
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Can this inductive proof that $\sum_{i=0}^n2^{2i+1}=\frac23(4^n-1)$ be simplified? The general structure of equations I've used for the inductive step for proofs with a summation is something like:
We'll prove that $\sum_{i = 0}^{n + 1} (\text{something}) = (\text{closed form expression})$
\begin{align}
\sum_{i = 0}^{n + 1} (\text{something}) &= \sum_{i = 0}^n (\text{something}) + \text{last term} &\\
&= [\text{expression via I.H.}] + \text{last term} &\\
&= \text{do some work...} &\\
&= \text{some more work...} &\\
&= (\text{finally reach the closed form expression we want})
\end{align}
This structure is very nice, since the equation is one-sided, and very easy to follow. However I solved a problem that I couldn't solve with this one-sided structure, and I had to substitute the LHS with the closed form expression I'm trying to prove, so I could use some of its terms to simplify the RHS. This is fine and valid, but I'd like to know if there's a simpler way to perform this proof that doesn't employ the substitution you see below:
In other words, I couldn't figure out how to simplify $\frac{2}{3}(4^n - 1) + 2^{2n + 1}$ to get $\frac{2}{3}(4^{n + 1} - 1)$. The farthest I got was:
\begin{align}
&= \frac{2}{3}(4^n - 1) + 2^{2n + 1} &\\
&= \frac{2}{3}(4^n - 1) + 2\cdot 2^{2n} &\\
&= \frac{2}{3}(4^n - 1) + 2\cdot 4^n &\\
&= \frac{2}{3}(4^n - 1) + \frac{3 \cdot 2\cdot 4^n}{3} &\\
&= \frac{2}{3}(4^n - 1 + 3 \cdot 4^n) &\\
\end{align}
| Note that $2^{2n+1}=2\cdot 2^{2n}=2\cdot 4^n$. Now,
$$\frac{2}{3}(4^n-1)+2^{2n+1}=\frac{2}{3}(4^n-1)+2\cdot 4^n =\frac{8}{3}\cdot 4^n -\frac{2}{3} = \frac{2}{3}(4^{n+1}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multiplicate a polynomial by square and third roots?
| Considering $$y= \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$ compose Taylor series and use binomial expansions
$$\sqrt{n^2+1}=n+\frac{1}{2 n}+\cdots$$
$$\sqrt{n^2+n}=n+\frac{1}{2}-\frac{1}{8 n}+\cdots$$
$$\sqrt[3]{n^3+1}=n+\frac{1}{3 n^2}+\cdots$$
$$\sqrt[3]{n^3+n^2+1}=n+\frac{1}{3}-\frac{1}{9 n}+\cdots$$
$$y=\frac {-\frac{1}{2}+\frac{5}{8 n}+\cdots } {-\frac{1}{3}+\frac{1}{9 n}+\cdots }\to \frac 32$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a=7!$, $b=_{13}P_k$, $\frac{ab}{\operatorname{lcm}(a,b)}=120$, find the $k$
Question:
$a=7!$, $b=_{13}P_k$, $\dfrac{ab}{\operatorname{lcm}(a,b)}=120$, then find the $k$.
My attempts:
$$\frac{ab}{\operatorname{lcm}(a,b)}=\gcd{(a,b)}=120$$
$\gcd{(7!, _{13}P_k)}=120 \Longrightarrow \begin{cases} \dfrac{13!}{(13-k)!×5!}\in \mathbb {Z^+} \\ \dfrac{13!}{(13-k)!×5!} \mod 2≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 3≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 7≠0 \end{cases} \Longrightarrow \begin{cases}
\dfrac{13×12×\cdots (13-k+1)!}{120} \in \mathbb {Z^+} \\
\dfrac{13!}{(13-k)!×5!} \mod 2≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 3≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 7≠0 \end{cases} \Longrightarrow 13-k+1=10 \Longrightarrow k=4$
Is my solution correct? Do I have any missing or unnecessary/unneeded steps?
| I think it'd be simpler to do this:
$\frac {ab}{\operatorname{lcm}(a,b)}=\gcd(a,b)$ so we have
$\gcd(7!, b) = 120$
So $\gcd(2^4\cdot 3^2\cdot 5\cdot 7, b) = 2^3\cdot 3\cdot 5$.
So $3|b$ but $9\not \mid b$ and $8|b$ but $16\not\mid b$ and $5|b$ and $7\not \mid b$.
$b =13*(13-1)*....*(13-(k-1))= \prod_{m=13-k+1}^{13} m$. Let $L = 13-k+1$ If $L \le 9$ then $9|b$. So $L>9$. If $10 < L < 15$ then $5\not \mid b$ so $L\le 10$. So $L = 10$.
So $13- k+1=10$ and $k=4$.
But we must verify that $L=10$ is acceptable namely that $3|b$ (which is does as $3|12$) that $9\not \mid b$ (which it doesn't as $12$ is the only multiple of $3$ between $10$ and $13$ inclusively) that $8|b$ but $16\not \mid b$ (as $2|10$ and $4|12$ and $10, 12$ are the only two even numbers between $10$ and $13$ that is fine) and that $5|b$ (which as $5|10$ it does) and the $7\not \mid b$ (which as there are no multiples of $7$ between $10$ and $13$ it does not.
| {
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"source": "stackexchange",
"question_score": "1",
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Show that if $|z| < 1$, then $\displaystyle\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots$ Show that if $|z| < 1$, then
\begin{align*}
\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots
\end{align*}
converges
MY ATTEMPT
Let us start by noticing that
\begin{align*}
\frac{z}{1-z} - \frac{z}{1-z} + \frac{z}{1+z} = \frac{z}{1-z} + \frac{z - z^{2} -z - z^{2}}{1-z^{2}} = \frac{z}{1-z} - \frac{2z^{2}}{1-z^{2}}
\end{align*}
Similarly, we have that
\begin{align*}
\frac{z}{1-z} - \frac{2z^{2}}{1-z^{2}} + \frac{2z^{2}}{1+z^{2}} = \frac{z}{1-z} + \frac{2z^{2} - 2z^{4} - 2z^{2} - 2z^{4}}{1-z^{4}} = \frac{z}{1-z} - \frac{4z^{4}}{1-z^{4}}
\end{align*}
Based on this pattern, it can be proven by induction the given series is the same as
\begin{align*}
\frac{z}{1-z} = z + z^{2} + z^{3} + \ldots = \sum_{n=1}^{\infty}z^{n}
\end{align*}
which converges, since it is the geometric series inside the circle $|z| < 1$
(we may apply the ratio test, for instance).
Could someone tell me if my approach is correct? Any contribution is appreciated.
| So far, you've found
$$
\sum_{k=0}^{n-1} \frac{2^{k}z^{2^k}}{1+z^{2^k}} = \frac{z}{1-z} - \frac{2^{n}z^{2^{n}}}{1-z^{2^{n}}}.
$$
As you said, you'll need to prove this by induction. To finish the proof, you need to show
$$
\lim_{n\rightarrow\infty}\frac{2^n z^{2^n}}{1-z^{2^n}} =0.
$$
You can then combine these to get
$$
\lim_{n\rightarrow\infty}\sum_{k=0}^{n} \frac{2^{k}z^{2^k}}{1+z^{2^k}} = \frac{z}{1-z},
$$
which proves convergence.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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distributing different balls into different boxes I have this question:
*
*Find in how many ways we can distribute $10$ similar balls to $4$ different boxes such that in no box there are exactly $3$ balls.
*Find in how many ways we can distribute $10$ different balls to $4$ different boxes such that no box has exactly $3$ balls
For the first part, I did the inclusion-exclusion principle. I used
$|U− (A_1 \cup A2 \cup A_3)| = S_0-S_1+S_2-S_3$.
Firstly, $U = D(4,10) = {13 \choose 10}$, $S_1 = 4 \cdot {9 \choose7}$ because we put $3$ balls in one box, $S2= {4 \choose 2} \cdot {5 \choose 4}$ because now we have $6$ balls in $2$ boxes, $S_3 = {4 \choose 3} \cdot {1 \choose 1}$. We get that $S_0- S_1+S_2-S_3=166$.
For the second part of the question, which is the part I couldn't get through, I thought about the inclusion-exclusion principle same as before, but this time $U = 4^{10}$, $S_1 = 4 \cdot {10 \choose 3} \cdot 3^7$ because we need to put $3$ balls in the first box so we have ${10 \choose 3}$ options and then there are $7$ different balls left for $3$ different boxes so we go $3^7$, and there are $4$ more options as well.
I wanted to know if my way of thinking is right? and if I should keep the second question as I started it. Just felt wrong and lost too much guessing in it ..
Thanks to any helpers!
|
In how many ways can we distribute $10$ indistinguishable balls to $4$ different boxes such that no box receives exactly three balls?
Your strategy is correct. However,
$$\binom{13}{10} - \binom{4}{1}\binom{9}{7} + \binom{4}{2}\binom{5}{4} - \binom{4}{3}\binom{1}{1} = 286 - 144 + 30 - 4 = 168$$
In how many ways can we distribute $10$ different balls to $4$ different boxes such that no box has exactly three balls?
What you have done so far is correct.
There are four ways to distribute each of the $10$ balls, so there are $4^{10}$ ways to distribute the balls without restriction.
From these, we must subtract those cases in which one or more of the boxes receives exactly three balls.
A box receives exactly three balls: There are four ways to select the box which receives exactly three balls, $\binom{10}{3}$ ways to select which three balls that box receives, and $3^7$ ways to distribute the remaining seven balls to the remaining three boxes. Thus, there are
$$\binom{4}{1}\binom{10}{3}3^7$$
such distributions.
Two boxes each receive exactly two balls: There are $\binom{4}{2}$ ways to select which two boxes receive exactly three balls, $\binom{10}{3}$ ways to select which three balls are placed in the leftmost of those boxes, $\binom{7}{3}$ ways to select which three of the remaining seven balls are placed in the other box which is selected to receive exactly three balls, and $2^4$ ways to distribute the remaining four balls to the remaining two boxes. Hence, there are
$$\binom{4}{2}\binom{10}{3}\binom{7}{3}2^4$$
such distributions.
Three boxes each receive exactly three balls: There are $\binom{4}{3}$ ways to select which three boxes receive exactly three balls, $\binom{10}{3}$ ways to select which three balls are placed in the leftmost of those boxes, $\binom{7}{3}$ ways to select three of the remaining seven balls is placed in the middle of those boxes, $\binom{4}{3}$ ways to select which three of the remaining four balls is placed in the rightmost of those boxes, and one way to place the remaining ball in the remaining box. Hence, there are
$$\binom{4}{3}\binom{10}{3}\binom{7}{3}\binom{4}{3}1^1$$
such distributions.
By the Inclusion-Exclusion Principle, the number of ways ten distinct balls can be distributed to four distinct boxes so that no box receives exactly three balls is
$$4^{10} - \binom{4}{1}\binom{10}{3}3^7 + \binom{4}{2}\binom{10}{3}\binom{7}{3}2^4 - \binom{4}{3}\binom{10}{3}\binom{7}{3}\binom{4}{3}1^1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all $a$ such that $y=\log_\frac{1}{\sqrt3} (x-2a) = \log_3(x-2a^3-3a^2) $
Find all values of parameter $a\in \mathbb{Z}$ such that $$y= \log_\frac{1}{\sqrt3} (x-2a)$$ $$and$$ $$y = \log_3(x-2a^3-3a^2)$$ intersect at points with whole coordinates.
This is what I did:
$$\log_\frac{1}{\sqrt3}(x-2a) = \frac{log_3(x-2a)}{\log_33^{-1/2}}$$
$$\frac{\log_3(x-2a)}{\log_33^{-1/2}} = \log_3(x-2a^3-3a^2)$$
$$-2\log_3(x-2a) = \log_3(x-2a^3-3a^2)$$
$$\log_3\frac{1}{(x-2a)^2} = \log_3(x-2a^3-3a^2)$$
$$\frac{1}{(x-2a)^2} = (x-2a^3-3a^2)$$
I got up to this point, not sure how to proceed.
| After getting a common base, $3$, and exponentiating, we have
$$
\frac{1}{(x-2 a)^2} = -3 a^2 - 2 a^3 + x
$$Since we are requiring $a,x$ to be whole numbers, the RHS is an integer which means the LHS must be as well: then $x-2a=\pm 1$. However, since these came from logarithms, only the case $x-2a=1$ is allowed. Then the equation becomes $1=-3a^2-2a^3+1+2a$, whose solutions are $a=\{-2,0,1/2\}$.
So in summary, the point $(a,x)=(0,1)$ is the only whole number solution , and there is also the solution $(a,x)=(-2,-3)$ if one admits negative integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Evaluating $\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx$ How can i evaluate this integral, maybe differentiation under the integral sign? i started expressing the integral as the following,
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$
But i dont know how to keep going, ill appreciate any solutions or hints.
| I dont think Feynman's trick would work best here, following your path:
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\underbrace{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx}_{x=\frac{1-t}{1+t}}\:$$
$$=\int _0^1\frac{\ln \left(x^3-1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x-1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+3\right)}{x+1}\:dx+2\int _0^1\frac{\ln \left(x+1\right)}{x+1}\:dx$$
$$-\sum _{k=1}^{\infty }\frac{1}{k}\int _0^1x^{3k-1}\:dx\:+\sum _{k=1}^{\infty }\frac{1}{k}\:\int _0^1x^{k-1}\:dx-\int _0^1\frac{\ln \left(x^2+3\right)}{x+1}\:dx+\ln ^2\left(2\right)$$
To solve that remaining integral you can use the identity i derived here
So,
$$=\frac{2\zeta \left(2\right)}{3}-(-\frac{\ln ^2\left(3\right)}{4}-\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}-\frac{\ln ^2\left(4\right)}{4}+\frac{\ln \left(3\right)\ln \left(4\right)}{2}-\arctan ^2\left(\sqrt{\frac{1}{3}}\right)+\ln \left(2\right)\ln \left(4\right))+\ln ^2\left(2\right)$$
$$\frac{\pi ^2}{9}+\frac{\ln ^2\left(3\right)}{4}+\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}+\ln ^2\left(2\right)-\ln \left(3\right)\ln \left(2\right)+\frac{\pi ^2}{36}-2\ln ^2\left(2\right)+\ln ^2\left(2\right)$$
So your integral's solution is,
$$\boxed{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\frac{5\pi ^2}{36}+\frac{\ln ^2\left(3\right)}{4}+\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}-\ln \left(3\right)\ln \left(2\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the relation of the Infinite power series with these fraction series? Take this infinite series:
$$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{2i+1}{4i} = 1 + \frac{3}{4} + \frac{3\times5}{4\times8} + \frac{3\times5\times7}{4\times8\times12} + ....$$
We want to find the sum of this series. I didn't know how to solve this. But when I went to look at the solution, they compared this series with the infinite power series
$$P = (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ......$$
for some real $n$ and $x$. Equating the corresponding terms ($nx = \frac{3}{4}$ and $\frac{n(n-1)}{2}x^2 = \frac{3\times5}{4\times8}$) they found $n=-\frac{3}{2}$ and $x=-\frac{1}{2}$. So that the sum is simply $2^\frac{3}{2}$. And when I checked the infinite power series $P$ plugging these values of $x$ and $n$, it really turn out to be the series $S$. Now, I don't understand why this comparison works.
Let's generalise this thing. Say, $S$ is given by
$$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{ai+b}{di}$$
for some positive integers $a, b$ and $d$ with $b < a$, and it is guaranteed (given) that the series converges. Let $P$ be the same as above. Now, can anyone say if $S$, as defined just now, can always be compared with the series $P$, that is, are there always some real $n$ and $x$ such that $S = P?$
| Note that setting $nx=\frac{3}{4}$ and $\frac{n(n-1)}{2!}=\frac{3\cdot 5}{4\cdot 8}$ in order to get values for $n$ and $x$ is just a clever approach, but not a proof that $P=S$.
To make it a proof we have to additionally verify that the general terms evaluated at $n=\frac{3}{2}$ and $x=-\frac{1}{2}$ are equal.
Let's write $S$ and $P$ with general term. We have
\begin{align*}
S&=1+\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\cdots+
\color{blue}{\frac{3\cdot 5\cdots (2k+1)}{4\cdot 8\cdots (4k)}}+\cdots\tag{1}\\
P&=(1+x)^n\\
&=1+nx+\frac{n(n-1)}{2!}x^2+\cdots+\color{blue}{\frac{n(n-1)\cdots(n-k+1)}{k!}x^k}+\cdots\tag{2}
\end{align*}
We simplify the general term of $S$ somewhat
\begin{align*}
\frac{3\cdot 5\cdots (2k+1)}{4\cdot 8\cdots (4k)}&=\frac{1}{4^k}\cdot\frac{3\cdot 5\cdots (2k+1)}{1\cdot 2\cdots k}\\
&\,\,\color{blue}{=\frac{1}{4^kk!}\prod_{j=1}^k(2j+1)}
\end{align*}
The general term of $P$ evaluated $n=\frac{3}{2}$ and $x=-\frac{1}{2}$:
\begin{align*}
&\left.\frac{n(n-1)\cdots(n-k+1)}{k!}x^k\right|_{n=-\frac{3}{2},x=-\frac{1}{2}}\\
&\qquad\qquad=\frac{1}{k!}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{3}{2}-k+1\right)\left(-\frac{1}{2}\right)^k\\
&\qquad\qquad=\frac{1}{k!}\frac{(-1)^k}{2^k}\left(3\right)\left(5\right)\cdots\left(3+2k-2\right)\left(-\frac{1}{2}\right)^k\\
&\qquad\qquad\,\,\color{blue}{=\frac{1}{4^kk!}\prod_{j=1}^k(2j+1)}
\end{align*}
Since both terms are equal we can now conclude that $S=P$.
Note: In order to generalise this approach, we could start to study and compare general terms accordingly.
[Add-on (2020-07-08)]: This add-on is based on OP who has successfully calculated the generalisation in comments to this answer.
The general ($k$-th) term of $S = 1 + \sum_{n=1}^\infty\prod_{j=1}^n\frac{aj+b}{dj}$ is (provided $a\ne 0$):
\begin{align*}
\prod_{j=1}^k\frac{aj+b}{dj}&=\frac{1}{d^kk!}\prod_{j=1}^k(aj+b)\\
&=\left(\frac{a}{d}\right)^k\frac{1}{k!}\prod_{j=1}^k\left(j+\frac{b}{a}\right)\tag{3}
\end{align*}
Since the general term of $P=(1+x)^n$ is
\begin{align*}
\frac{x^k}{k!}\prod_{j=1}^k(n-j+1)&=\frac{(-x)^k}{k!}\prod_{j=1}^{k}\left(j-\left(n+1\right)\right)\tag{4}
\end{align*}
we obtain by comparison of (3) with (4):
\begin{align*}
x=-\frac{a}{d}\qquad\qquad n=-\left(1+\frac{b}{a}\right)
\end{align*}
We conclude, providing $|x|=\left|\frac{a}{d}\right|<1$ to assure convergence of the binomial series:
\begin{align*}
\color{blue}{S }= 1 + \sum_{n=1}^\infty\prod_{j=1}^n\frac{aj+b}{dj}\color{blue}{=\left(1-\frac{a}{d}\right)^{-\left(1+\frac{b}{a}\right)}}
\end{align*}
Hints:
*
*Alternatively to OPs generalisation we can also recall that $P=(1+x)^n$ is some kind of reference series used to derive $S$. We can turn the tables and play with different settings of $x$ and $n$ (respecting $|x|<1$) and check which different general terms and series $S$ we obtain this way.
*A technical aspect. We have to be aware of arithmetic precedence rules and write $\prod_{j=1}^k\color{blue}{(}aj+b\color{blue}{)}$ using brackets, since we have
\begin{align*}
\prod_{j=1}^kaj+b&=\left(\prod_{j=1}^kaj\right)+b=a^kk!+b\\
\prod_{j=1}^kb+aj&=\left(\prod_{j=1}^kb\right)+aj=b^k+aj\\
\prod_{j=1}^k\left(aj+b\right)&=(a+b)(2a+b)\cdots(ka+b)
\end{align*}
You might want to see this answer for more information regarding precedence rules.
| {
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Curve tangent line A curve with the equation $x^2-x+1$ has two tangent lines $a$ and $b$ that intersects at $x=1$, what is $y$? can I determine $y$ when $a$ and $b$ are perpendicular, or the gradient of $a$ and $b$ are $-1$
I tried using the second point, but knowing that they have the same gradient means they won't intersect, so it cannot be determined from that
from the first point, I know that the gradient of the tangent lines must be $2x-1$, and the equation for line is $y=mx+b$
if both are perpendicular, then $m_a × m_b = -1$. so does that mean I could make it $(2x-1)(2x-1)=-1$? because both lines have the same gradient?
I tried by making the gradient for line $a = (2x-1)$ and gradient for line $b -1/(2x-1)$. but they intersect at $x=1$
I'm stuck right there, is there a way of using either one to determine the y which both lines intersect?
| Let $A$ be the $x$-coordinate of the point of tangency of the line $a$, then its equation is
$$y=(2A-1)(x-A)+A^2-A+1.$$
Similarly for $b$,
$$y=(2B-1)(x-B)+B^2-B+1.$$
Therefore for $x=1$ we have
$$(2A-1)(1-A)+A^2-A+1=y=(2B-1)(1-B)+B^2-B+1$$
and we finally find
$$(A-B)(A+B-2)=0.$$
It follows that if $a$ and $b$ are distinct lines then $A+B=2$.
Therefore we have infinite pairs of lines $a$ and $b$ that satisfy the given conditions. The $y$-coordinate of their intersection point is $A(2-A)=AB$ (it depends on the pair of lines).
| {
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} |
Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easily solved using the substitution $y=\cos(x)$.
What if the power of sine is even?
| $$\sin^4x=(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)$$
$$1+\cos^2x=(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)$$
and so your integral becomes:
$$\int\frac{\sin^4x}{1+\cos^2x}dx=\int\frac{(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)}{(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)}dx=-\int1+\sin^2xdx+\int\frac{2+\sin^2x}{2-\sin^2x}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
How to find $f\circ g$ when $f(y)$ and $g(x)$ are given? $f(y)=\frac4{y-2},\;g(x)=\frac5{3x-1}$
Find the graph of $f\circ g$?
I am stuck at this step. Since $f(x)$ is not given.
$f\circ g\;=\;f(g(x))\\f(\frac5{3x-1})=\frac4{y-2}$
I am thinking whether $f^{-1}(y)=f(x)$?
| Variables are just place holders.
If $f(y)=\frac4{y-2}$ then $f(x) = \frac 4{x-2}$ and $f(c) = \frac 4{c-2}$ and $f(\text{fred the tap-dancing goat}) = \frac 4{\text{fred the tap-dancing goat}-2}$
And $f(g(x)) = \frac 4{g(x)-2}$.
So $f\circ g$ is the function so that $f\circ g(x) = f(g(x)) = \frac 4{g(x)-2} = \frac 4{\frac 5{3x-1}- 2}$
So just graph the graph $y = \frac 4{\frac 5{3x-1}- 2}$
Which can be rewritten as
$y=\frac {4(3x-1)}{5 - 2(3x-1)};3x -1 \ne 0$ or
$y = \frac {12x-4}{-6x+7}; x\ne \frac 13$ or as
$y = \frac {12x-14 + 10}{-6x+7}= -2- \frac {10}{6x-7}; x\ne \frac 13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3752936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
| Substitute $v=u^2+1$ so that $du=\frac{1}{2u}dv$ to turn the integral into:
$$\int{\frac{u(v-1)}{2uv^3}} dv$$
$$\frac{1}{2}\int{\frac{v-1}{v^3}} dv$$
$$=\frac{1}{2}(\int{\frac{1}{v^2} dv}-\int{\frac{1}{v^3}} dv)$$
$$=\frac{1}{2}(\frac{1}{2v^2}-\frac{1}{v})+C$$
$$=\frac{1}{4v^2}-\frac{1}{2v}+C$$
Substitute $v=u^2+1$ and simplify to get your answer:
$$\dfrac{1}{4(u^2+1)^2}-\dfrac{1}{2(u^2+1)}+c$$
$$=-\dfrac{2u^2+1}{4\left(u^2+1\right)^2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 1
} |
Unclear problem with $n$-th power matrix and limit
Find $$\lim\limits_{n \to \infty} \frac{A_n}{D_n}$$ where
$$\begin{pmatrix}
19 & -48 \\
8 & -21 \\
\end{pmatrix} ^{\! n} =
\begin{pmatrix}
A_n & B_n \\
C_n & D_n \\
\end{pmatrix}$$
$n$ - is the power of a matrix, but what is $A_n, B_n, C_n, D_n$ then? Is it a corresponding element of a matrix in the $n$-th power?
How is this type of problem called? And what is the way to solve that problem?
| $$A:=\begin{pmatrix}19&-48\\8&-21\end{pmatrix}=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}-5&0\\0&3\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=PDP^{-1}$$
So $A^n=PDP^{-1}PDP^{-1}\cdots PDP^{-1}=PD^nP^{-1}$
That is, $$A^n=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}(-5)^n&0\\0&3^n\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=\begin{pmatrix}-2 (-5)^n + 3^{1 + n}& 6 (-5)^n - 2\times 3^{1 + n}\\-(-5)^n + 3^n&
3 (-5)^n - 2\times 3^n\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct. Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct.
I'm puzzled on how to solve this problem, any help is appreciated. There was hints about turning the exponents into fractions and picking fractions between : $3^x < 3^\sqrt3 <3^y$
Then I simplified: $x< \sqrt3<y$
$x^2< 3<y^2$
$\sqrt2^2<3<\sqrt4^2$
So $x=\sqrt2$ and $y=\sqrt4=2$
$3^\sqrt2 < 3^\sqrt3 <3^2$
| Use the (easily verified) inequalities $\sqrt3\lt1.75=7/4$ and $\sqrt{7/4}\lt4/3$ to show that
$$3^\sqrt3\lt3^{1.75}=3\cdot3^{1/2}\cdot3^{1/4}\lt3\cdot{7\over4}\cdot{4\over3}=7$$
and $\sqrt3\gt1.7$ to show
$$3^\sqrt3\gt3^{1.7}=3\cdot3^{7/10}$$
Now show $3^7=3^4\cdot3^3=81\cdot27\gt1600\gt1024=2^{10}$ to conclude $3^{7/10}\gt2$ and thus
$$6=3\cdot2\lt3\cdot3^{7/10}=3^{1.7}\lt3^\sqrt3$$
Putting these together, we have $\lfloor3^\sqrt3\rfloor=6$.
Just to be thorough, let's verify the asserted inequalities:
$$\sqrt3\lt7/4\iff3\lt49/16\iff3\cdot16\lt49\\
\sqrt{7/4}\lt4/3\iff7/4\lt16/9\iff7\cdot9\lt4\cdot16\\
\sqrt3\gt1.7\iff3\gt1.7^2=2.89$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Let $ABC$ be a triangle and $M$ be the midpoint of $BC$. Squares $ABQP$ and $ACYX$ are erected. Show that $PX = 2AM$.
$\textbf{Question:}$ Let $ABC$ be a triangle and $M$ be the midpoint of $BC$. Squares $ABQP$ and $ACYX$ are erected. Show that $PX = 2AM$.
I could solve this problem using computational techniques but I am looking for purely synthetic solution. I tried drawing some extra midpoints, connected them. But still couldn't find the solution.
Any kind of hint or full solution both are appreciated.
| Let's first give names to the sides of $\overset{\Delta}{PAX}$ triangle: $|PA| = k$, $|AX| = m$ and $|PX| = b$. Also for $\overset{\Delta}{ABC}$ triangle: $|AB| = k$, $|AC| = m$, $|BC| = 2l$ and the length of the median $|AM| = a$. Writing law of cosines for $\overset{\Delta}{PAX}$ triangle for angle $\widehat{PAX}$, which is the opposite of angle $\widehat{A}$ in $\overset{\Delta}{ABC}$ triangle, along with the fact that $\widehat{PAB} = 90^{\circ}$ and $\widehat{XAC} = 90^{\circ}$ ($\widehat{PAX} + \widehat{A} + \widehat{PAB} + \widehat{XAC} = 360^{\circ}$), we have:
\begin{align*}
b^{2} &= k^{2} + m^{2} - 2\ k\ m\ cos (180^{\circ} - \widehat{A}) \\
b^{2} &= k^{2} + m^{2} + 2\ k\ m\ cos (\widehat{A}) \tag{1}
\end{align*}
Then write the same law for $ABC$ triangle along with Apollonius's theorem and we have:
\begin{align*}
(2l)^{2} &= k^{2} + m^{2} - 2\ k\ m\ cos(\widehat{A}) \\
4 l^{2} &= k^{2} + m^{2} - 2\ k\ m\ cos(\widehat{A}) \tag{2}
\end{align*}
\begin{align*}
k^{2} + m^{2} = 2(l^{2} + a^{2}) \tag{3}
\end{align*}
Summing $(1)$ and $(2)$ and using the right-hand side of $(3)$ in place of $k^{2} + m^{2}$ we get:
\begin{align*}
b^{2} + 4 l^{2} &= 2 (k^{2} + m^{2}) \\
b^{2} + 4 l^{2} &= 4 (l^{2} + a^{2}) \\
b^{2} &= 4 a^{2} \\
b &= 2a \implies |PX| = 2 |AM|
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760238",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
How to solve a non-linear system of 3 equations involving $x^2$, $y^2$ and $xy$ only? Let $b_1, b_2,b_3$ be real constants, how to solve the following non-linear system?
\begin{align}
(-3x^2 + y^2 -6xy) + b_1 &= 0 \qquad\text{(L1)}\\
(\phantom{-} x^2 -3 y^2 -6xy) + b_2 &= 0 \qquad\text{(L2)}\\
(\phantom{.}3x^2 + 3 y^2 -2xy) + b_3 &= 0 \qquad\text{(L3)}
\end{align}
I try (L1)-(L2) which yields: $y^2 = x^2 + \frac{b2-b1}{4}$.
Then, (L2)-3*(L3) leads to this intermediate results:
\begin{align}
x^2 &= \frac{3b_1-2b_2-12b_3}{20}\\
y^2 &= \frac{-2b_1+3b_2-12b_3}{20}
\end{align}
At this point, can I conclude the following statement?
\begin{align}
|x| &= \sqrt{\frac{3b_1-2b_2-12b_3}{20}}\\
|y| &= \sqrt{\frac{-2b_1+3b_2-12b_3}{20}}
\end{align}
What if $(3b_1-2b_2-12b_3) <0$ or $(-2b_1+3b_2-12b_3)<0$?
Does it mean that the system has no solution?
If I plug the previous results of $x^2$ and $y^2$ in (L1), (L2) and (L3), then I obtain:
\begin{align}
xy &= -\frac{1}{40}(3b_1 + 3b_2 + 8b_3)\\
xy &= -\frac{1}{40}(3b_1 + 3b_2 - 52b_3)\\
\end{align}
which is valid only if $b_3 = 0$.
In this case $xy = -\frac{3}{40}(b_1 + b_2)$.
Can I conclude something about the sign of $x$ and $y$?
| Since you have three equations and two unknowns, I should define $z=xy$ to make
$$\begin{align}
(-3x^2 + y^2 -6z) + a &= 0 \tag 1\\
(\phantom{-} x^2 -3 y^2 -6z) + b &= 0 \tag 2\\
(\phantom{.}3x^2 + 3 y^2 -2z) + c &= 0 \tag 3
\end{align}$$
So, four solutions
$$x=\pm\frac{\sqrt{3 a-2 b-3 c}}{2 \sqrt{5}}\qquad y=\pm \frac{\sqrt{3b-2 a-3 c}}{2 \sqrt{5}}\qquad z=\frac{3 a+3 b+2 c}{40}$$
So, as you noticed, if we are in the real domain, this gives already two conditions $3a-2b-3c \geq 0$ and $3b-2a-3c>0$. If these ared fulfilled, then $z=xy$ gives a relation between $a$, $b$ and $c$, but no more.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Motivation confusion in floor and ceiling function algebra From HMMT https://hmmt-archive.s3.amazonaws.com/tournaments/2019/nov/team/solutions.pdf:
Compute the sum of all positive real numbers $x \le 5$ satisfying $$x=\frac{\left\lceil{x^2}\right\rceil+\left\lceil x\right\rceil\cdot\left\lfloor x\right\rfloor}{\left\lceil x\right\rceil+\left\lfloor x\right\rfloor}.$$
The given solution says the following.
Note that all integer $x$ work. If $x$ is not an integer then suppose $n < x < n+ 1$. Then $x = n+\frac{k}{2n+1}$, where $n$ is an integer and $1 \le k \le 2n$ is also an integer, since the denominator of the fraction on the right hand side is $2n + 1$. We now show that all x of this form work.
Note that $$x^2=n^2+\frac{2nk}{2n+1}+\left (\frac{k}{2n+1}\right)^2=n^2+k-\frac{k}{2n+1}+\left (\frac{k}{2n+1}\right )^2.$$ For $\frac{k}{2n+1}$ between $0$ and $1$, $-\frac{k}{2n+1}+(\frac{k}{2n+1})^2$ is between $-\frac{1}{4}$ and $0$, so we have $n^2+k-1<x^2\le n^2+k$, and $\lceil x^2\rceil=n^2+k$.
Then, $$\frac{\left\lceil{x^2}\right\rceil+\left\lceil x\right\rceil\cdot\left\lfloor x\right\rfloor}{\left\lceil x\right\rceil+\left\lfloor x\right\rfloor}=\frac{n^2+k+n\cdot (n+1)}{2n+1}=n+\frac{k}{2n+1}=x,$$ so all $x$ of this form work...
To complete the solution there is more, but those parts I believe would be irrelevant to my question, although please comment to correct me if there is more context required, or click on the link, as there's only a paragraph more.
I don't understand, in the first paragraph, why $x=n+\frac{k}{2n+1}$ follows from $n < x < n+1$. More specifically, what is the motivation behind writing this? It seems that the paragraph after this claim is a proof that all $x$ of this form work, but in that proof, this fact is used. So then why do we have to "show that all $x$ of this form work", if it is used in the proof (and thus claimed without proof)?
| Since all floors and ceilings are integral, it is clear that the right hand side is a rational number. Because $n < x < n+1$, we have $\left\lfloor{x}\right\rfloor = n$ and $\left\lceil{x}\right\rceil = n+1$. Therefore the denominator of the right hand side is $(n+1) + n = 2n+1$. By hypothesis, this equals the left hand side which is $x$. Therefore, $x$ is a rational number whose denominator is $2n+1$ (which is not necessarily in reduced form). Since $n < x < n+1$, it follows that $x = n + \frac{k}{2n+1}$ where $1 \le k \le 2n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the parabola of the form $y=ax^2+b$ which best fits the points $(1,0), (4,4), (5,8)$ by minimizing the sum of squares, $S$, given by
$$S=(a+b)^2 + (16a+b−4)^2 + (25a+b−8)^2$$
My work so far
The variables are a and b, so we set
$$\frac{∂S}{∂a}=2(a+b)+32(16a+b−4)+50(25a+b−8)=0$$
and
$$\frac{∂S}{∂b}=2(a+b)+2(16a+b−4)+2(25a+b−8)=0.$$
Collecting terms, we get
$$1764a+84b−528=0$$
and
$$84a+6b−24 = 0,$$
At this point, I'm stuck. I checked the solution in the textbook, but was unsure how they were calculated. What would be the first step? Would it be using the least squares approximation?
Solution
and solving for a and b gives
$$a=\frac{16}{49}$$
and
$$b=\frac{-4}{7}$$
Since there is only one critical point and S is unbounded as $a,b → ∞$, this critical point is the global minimum. Therefore, the best fitting parabola is
$$y=\frac{16}{49}$$
and
$$x=\frac{-4}{7}$$
| Given set of $n$ points $(x_i,y_y)$, $i=1,\dots,n$,
the parameters of the best fit parabola $y(x)=a_2x^2+a_1 x+a0$
can be found as:
\begin{align}
\begin{bmatrix}
a_0 \\ a_1 \\ a_2
\end{bmatrix}
&=
\begin{bmatrix}
n & s_x & s_{x^2}\\
s_x & s_{x^2} & s_{x^3}\\
s_{x^2} & s_{x^3} & s_{x^4}
\end{bmatrix}
^{-1}
\cdot
\begin{bmatrix}
s_y \\ s_{xy} \\ s_{x^2y}
\end{bmatrix}
.
\end{align}
where
\begin{align}
s_{x}&=\sum_{i=1}^n x_i
,\quad
s_{x^2}=\sum_{i=1}^n x_i^2
,\quad
s_{x^3}=\sum_{i=1}^n x_i^3
,\quad
s_{x^4}=\sum_{i=1}^n x_i^4
,\\
s_{y}&=\sum_{i=1}^n y_i
,\quad
s_{xy}=\sum_{i=1}^n x_i y_i
,\quad
s_{x^2y}=\sum_{i=1}^n x_i^2y_i
.
\end{align}
For the set of three points $(1,0),(4,4),(5,8)$,
$a_0=\tfrac43$,
$a_1=-2$,
$a_2=\tfrac23$.
In a special case where the parameter $a_1$ is forsd to be zero,
the solution simplifies
by eliminating all corresponding items
to
\begin{align}
\begin{bmatrix}
a_0 \\ a_2
\end{bmatrix}
&=
\begin{bmatrix}
n & s_{x^2}\\
s_{x^2} & s_{x^4}
\end{bmatrix}
^{-1}
\cdot
\begin{bmatrix}
s_y \\ s_{x^2y}
\end{bmatrix}
.
\end{align}
For the set of three points $(1,0),(4,4),(5,8)$,
we have $n=3$,
$x=[1,4,5]^{\mathsf{T}}$,
$y=[0,4,8]^{\mathsf{T}}$,
$s_{x^2}=42$,
$s_{x^4}=882$,
$s_{y}=12$,
$s_{x^2y}=264$,
and
\begin{align}
\begin{bmatrix}
a_0 \\ a_2
\end{bmatrix}
&=
\begin{bmatrix}
3 & 42 \\
42 & 882
\end{bmatrix}
^{-1}
\cdot
\begin{bmatrix}
12 \\ 264
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 & -1/21 \\
-1/21 & 1/294
\end{bmatrix}
\cdot
\begin{bmatrix}
12 \\ 264
\end{bmatrix}
=
\begin{bmatrix}
-4/7 \\ 16/49
\end{bmatrix}
,
\end{align}
so
$a_0=-\tfrac47$,
$a_2=\tfrac{16}{49}$
and
\begin{align}
y(x)&=\tfrac{16}{49}\,x^2-\tfrac47
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
| You can solve it by using induction formula: $\color{blue}{\int \frac{dt}{(t^2+a^2)^n}=\frac{t}{2(n-1)a^2(t^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)a^2}\int \frac{dt}{(t^2+a^2)^{n-1}}}$ as follows
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$$$=\int \dfrac{d(x-2)}{((x-2)^2+3^2)^2}$$
$$=\frac{x-2}{2\cdot 3^2((x-2)^2+3^2)}+\frac{1}{2\cdot 3^2}\int \frac{d(x-2)}{(x-2)^2+3^2}$$
$$=\frac{x-2}{18(x^2-4x+13)}+\frac{1}{18}\left(\frac13\tan^{-1}\left(\frac{x-2}{3}\right) \right)+C$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{x-2}{18(x^2-4x+13)}+\frac{1}{54}\tan^{-1}\left(\frac{x-2}{3}\right)+C}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ I want to prove
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1
$$
First question:
Isn't this wrong? Shouldn't it be
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2
$$
Or
$$
a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3
$$
And the proof:
I start backwards.
\begin{align}
a+b-2&\sqrt{ab}\geq 0 \tag 4 \\
(\sqrt a)^2+(\sqrt b)^2-2&\sqrt{ab}\geq 0 \tag 5 \\
(\sqrt a)^2+(\sqrt b)^2-2&\sqrt{a}\cdot \sqrt{b}\geq 0 \tag 6
\end{align}
Let $x= \sqrt a$ and $y=\sqrt b$, so we see that
\begin{align}
x^2+y^2-2xy&\geq 0 \tag 7\\
(x-y)^2\geq 0 \tag 8
\end{align}
So finally we have
$$
(\sqrt a- \sqrt b)^2 \geq 0 \tag 9
$$
If $a=b=0$ we have $0\geq 0$ which is true for $0=0$ ($0>0$ is always false).
For $a>0$ and $b>0$ the square $(\sqrt a- \sqrt b)^2$ is always strictly positive.
And also, $a<0$ and $b<0$ is not valid because $\sqrt a$ and $\sqrt b$ are only defined for $a\geq 0$ and $b\geq 0$.
| You want to prove $a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0$.
You can use $a^2 > 0$ for all $a \in \mathbb{R}$ as an assumption and segue into (9), with the restriction that $a,b > 0$. Then, instead of working backwards, you're working forwards and the proof is nice.
Rewrite your proof so that it begins with the assumptions, then (9) becomes (7) with $x = \sqrt{a}$ and $y= \sqrt{b}$ which then becomes (4), and finally conclude your proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.
But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems impossible to bring it in $y = f(x)$ form.
*
*$ \frac{\sqrt{4-3y-y^2}}{(y+4)}= x $
*$ 4 - 3y - y^2 = x^2(y^2 + 8y + 16) \iff (x^2-1)y^2+y(-8x-3)+4(1-4x) = 0$
Is there a trivial methodology that I am missing or is it indeed impossible to inverse some relations?
| \begin{align*}
\sqrt{4-3y-y^2} &= x(y+4)\\
\sqrt{(1-y)(y+4)} &= x(y+4)\Rightarrow y\in[-4,1],x\in[0,+\infty)\\
\sqrt{y+4}\left(\sqrt{1-y}-x\sqrt{y+4}\right) &=0\Rightarrow \boxed{y=-4}\\
\sqrt{1-y}&=x\sqrt{y+4}\\
\sqrt{\frac{5}{y+4}-1}&=x\\
\boxed{y=\frac5{x^2+1}-4}\in(-4,1]\,\forall\,x\in[0,+\infty)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving for positive reals: $abcd=1$, $a+b+c+d=28$, $ac+bc+cd+da+ac+bd=82/3$ $$a,b,c,d \in \mathbb{R}^{+}$$
$$ a+b+c+d=28$$
$$ ab+bc+cd+da+ac+bd=\frac{82}{3} $$
$$ abcd = 1 $$
One can also look for the roots of polynomial
$$\begin{align}
f(x) &= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+acd+bcd)x + 1
\end{align}$$
and $f(x)$ has no negative roots... but how else do I proceed?
There is a trivial solution $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 27$. We just need to prove it's unique.
| [Here's an unsatisfactory answer as I don't expect the discriminant to be calculated in a contest-math setting.]
Let $ C = abc + bcd + cda + dab > 0$.
The equation $ x^4 - 28x^3 + \frac{82}{3} x^2 - Cx + 1 =0 $ has 4 positive real roots.
The discriminant (Thanks Wolfram) is
$$\Delta = -5565901568/81 + (82093312 C)/9 + (13588640 C^2)/27 - 74032 C^3 - 27 C^4 \\
= -9 ( C - \frac{244}{27} ) ^2(3C^2 + 8290C + 93488). $$
Since the equation has 4 real roots, the discriminant is non-negative.
The only positive value of $c$ which makes $\Delta$ non-negative is $ c = \frac{244}{27}$.
Hence, the solution is uniquely determined (up to permutation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\frac{a-ax^3/2-3bx^2}{(1+x^3)^{3/2}}$$
Finally by setting $a=2,b=0$, we get $$I(x)=\frac{2x}{(1+x^3)^{1/2}}+C$$
The question is: How to do it otherswise?
| Split the integral and integrate by parts:
$$\begin{align}I = \int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx &\equiv\int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \int\dfrac{x^2}{\left(1 + x^3\right)^{3/2}}x\,\mathrm dx \\ &= \int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \left(-\dfrac{2x}{3\left(1 + x^3\right)^{1/2}} + \dfrac23\int\dfrac{1 + x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx\right) \\ &= \int\dfrac{6 - 2 - 2x^3}{3\left(1 + x^3\right)^{3/2}}\,\mathrm dx + \dfrac{2x}{3\sqrt{1 + x^3}} \\ &= \dfrac23\int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx + \dfrac{2x}{3\sqrt{1 + x^3}} \\ &= \dfrac23I + \dfrac{2x}{3\sqrt{1 + x^3}} \\ \implies I &= \dfrac{2x}{\sqrt{1 + x^3}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding $\lim_{n\to\infty}\frac1{n^3}\sum_{k=1}^{n-1}\frac{\sin\frac{(2k-1)\pi}{2n}}{\cos^2\frac{(k-1)\pi}{2n}\cos^2\frac{k\pi}{2n}}$
For all $n\ge 1$, let
$$
a_{n}=\sum_{k=1}^{n-1} \frac{\sin \left(\frac{(2 k-1) \pi}{2 n}\right)}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right) \cos ^{2}\left(\frac{k \pi}{2 n}\right)}
$$
Find $\displaystyle\lim_{n\to\infty}\frac{a_n}{n^3}$.
This is a problem from the 2019 Putnam competition.
The official solutions use two different strategies; one reduces the expression as a telescoping sum and the other uses the asymptotic property of sine near zero. The two community-wiki answers below essentially elaborate on the official solutions.
Remark. I am curious if one can write $\displaystyle\frac{a_n}{n^3}$ as a Riemann sum so that one can write the limit as an integral. Naively, the fraction looks very much like $\frac{1}{n}\sum_{k=1}^{n-1}\cdots$, which is in the setting of Riemann sums. (This only serves as a comment, not a requirement for solving the problem.)
| This solution is based on one of the official solutions for the problem.
The dummy variable $k$ is a dummy variable in $a_n$. Since
\begin{align}
\sum_{k=1}^{n-1}f(k;n) &= f(1;n)+f(2;n)+\cdots+f(n-1;n)\\
&= f(n-1;n) + f(n-2;n) +\cdots f(n-(n-1);n)\\
&= \sum_{k=1}^{n-1}f(n-k;n)
\end{align}
we can substitute $n-k$ for $k$ to obtain
$$
a_{n}=\sum_{k=1}^{n-1} \frac{\sin \left(\frac{(2 k+1) \pi}{2 n}\right)}{\sin ^{2}\left(\frac{(k+1) \pi}{2 n}\right) \sin ^{2}\left(\frac{k \pi}{2 n}\right)}
$$
We then use the estimate
$$
\frac{\sin x}{x} = 1+O(x^2)\quad \text{as } x\to 0
$$
to rewrite the summand as
$$
\frac{\left(\frac{(2 k-1) \pi}{2 n}\right)}{\left(\frac{(k+1) \pi}{2 n}\right)^{2}\left(\frac{k \pi}{2 n}\right)^{2}}\left(1+O\left(\frac{k^{2}}{n^{2}}\right)\right)
$$
which simplifies to
$$
\frac{8(2 k-1) n^{3}}{k^{2}(k+1)^{2} \pi^{3}}+O\left(\frac{n}{k}\right)
$$
Consequently,
$$
\begin{aligned} \frac{a_{n}}{n^{3}} &=\sum_{k=1}^{n-1}\left(\frac{8(2 k-1)}{k^{2}(k+1)^{2} \pi^{3}}+O\left(\frac{1}{k n^{2}}\right)\right) \\
&=\left[\frac{8}{\pi^{3}} \sum_{k=1}^{n-1} \frac{(2 k-1)}{k^{2}(k+1)^{2}}\right]
+O\left(\frac{\log n}{n^{2}}\right) \end{aligned}
$$
Finally, note that
$$
\sum_{k=1}^{n-1} \frac{(2 k-1)}{k^{2}(k+1)^{2}}=\sum_{k=1}^{n-1}\left(\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}\right)=1-\frac{1}{n^{2}}\;.
$$
It is now easy to finish the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2}$? Note: Similar questions have been asked here and here, but this is quite different.
I am trying to evaluate
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} \quad (1)$$
I re-wrote the fraction as $$ \frac{2n+1}{2n(n+1)^2} = \frac1{2(n+1)} \cdot \frac{2n+1}{n(n+1)}= \frac1{2(n+1)} \left( \frac1n + \frac1{n+1} \right) = \frac1{2} \left( \frac1{n(n+1)} + \frac1{(n+1)^2} \right) = \frac12 \left( \left( \frac1n -\frac1{(n+1)} \right) + \frac1{(n+1)^2} \right)$$
Hence
$$(1) = \frac12 \sum_{n=1}^{\infty}\: \left( \frac1n -\frac1{(n+1)} \right) + \frac12 \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac12\lim_{n \to \infty}1-\frac1{n+1}+\frac{\pi^2}{12} = \frac{\pi^2}{12} + \frac12$$
Therefore
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} = \frac{\pi^2}{12} + \frac12$$
I am unsure about $$ \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac{\pi^2}{12} $$
We know the basic p-series $$ \sum_{n=1}^{\infty}\:\frac1{n^2} = \frac{\pi^2}{6} $$
Is this solution correct?
| 1). $\frac{2n+1}{2n(n+1)^2}=\frac{n+n+1}{2n(n+1)^2}=\frac{1}{2(n+1)^2}+\frac{1}{2n(n+1)}$
So
\begin{align*}
\sum_{n\geq1}{}\frac{2n+1}{2n(n+1)^2}=\frac{1}{2}\sum_{n\geq1}\frac{1}{(n+1)^2}+\frac{1}{2}\sum_{n\geq1}{}\frac{1}{n(n+1)}\\
=\frac{1}{2}(\zeta(2)-1)+\frac{1}{2}\sum_{n\geq1}{}\frac{1}{n}-\frac{1}{n+1}\\
=\frac{1}{2}(\zeta(2)-1)+\frac{1}{2}\\
=\frac{(\pi)^2}{12}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$, find $y^2+2y$
If $$y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$$ what is $y^2+2y$?
Attempt:
We know that for negative and fractional indices,
$$(1+x)^n = 1 + nx + n(n-1)/2!\cdot x^2 + n(n-1)(n-2)/3!\cdot x^3 + \cdots$$
Rewriting the series in question, we get:
$$\frac{2}{5} \left(1 + \frac{1\cdot3}{2!}\cdot \frac{2}{5}+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^2+\cdots\right)$$
I know this looks like the binomial expansion above, but I have no idea how to proceed further.
| You can calculate $y$ using a "well known" binomial series expansion for
$$(1-4x)^{-\frac 12} = \sum_{n=0}^{\infty}\binom{2n}nx^n$$
To see this rewrite the coefficients
$$\prod_{k=1}^n\frac{2k-1}{k} = \prod_{k=1}^n\frac{(2k-1)\cdot 2k}{k\cdot 2k}= \frac 1{2^n}\binom{2n}n$$
So in your case the series becomes
$$y = \sum_{\color{blue}{n=1}}^{\infty}\frac 1{2^n}\binom{2n}n \frac{2^n}{5^n} = \sum_{\color{blue}{n=1}}^{\infty} \binom{2n}n\frac 1{5^n}$$ $$ = \left.\frac 1{\sqrt{1-4x}}-1\right|_{x = \frac 15} = \sqrt 5 - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to prove the solution.
I start by pointing out that it has the form
\begin{equation}
\dfrac{dy}{dx} + P(x)y = Q(x)y^3
\end{equation}
where
\begin{equation}
P(x) = \dfrac{1}{2x}, \qquad Q(x) = -\frac{1}{3x}
\end{equation}
Rewriting y(x) as
\begin{equation}
y(x) = (3x)^{\frac{1}{2}} (2x + 3c)^{-\frac{1}{2}}
\end{equation}
Getting rid of that square root, I'll need it later on to simplify things
\begin{equation}
[y(x)]^2 = 3x(2x + 3c)^{-1}
\end{equation}
Calculating dy/dx
\begin{align}
\dfrac{dy}{dx} &= \frac{1}{2}(3x)^{-\frac{1}{2}}(3)(2x + 3c)^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)(2x + 3c)^{-\frac{3}{2}}(2)(3x)^{\frac{1}{2}} \\
&= \frac{3}{2}(3x)^{-\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} - (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{3}{2}} \\
&= (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= y \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= \dfrac{y}{2x} - y(2x + 3c)^{-1} \\
&= \dfrac{y}{2x} - y\left(\dfrac{y^2}{3x}\right) \\
&= \dfrac{y}{2x} - \dfrac{y^3}{3x}
\end{align}
Finally
\begin{align}
\dfrac{dy}{dx} + P(x)y &= \dfrac{y}{2x} - \dfrac{y^3}{3x} + \dfrac{y}{2x} \\
&= \dfrac{y}{x} - \dfrac{y^3}{3x}
\end{align}
which obviously isn't the same as equation 2. I don't know where I screwed up.
| @kira1985 you should not add the p(x).y on both side, here you have to find the answer .
But you are using the answer it self to solve the problem .As the answer is unknown to have to just find the value of the P(x) and Q(x).
which is
/ = /2 − (^3)/3
P(x) = - 1/(2x)
Q(x) = - 1/(3x)
Is the write answer for the given question, please check your question once more.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this statement. "There are not odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2=(x+y)^2$." I did try to prove this by contradiction(I guess it works). So I suppose this statement is false but I have confused about the meaning of it.
If this statement is false it means "There are odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2=(x+y)^2$." or "There are odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2\neq (x+y)^2$." or.....
Is proof by contradiction a good method? and could you help me how to do next from the statement that is supposed to be false?
Thank you very much.
| Write $x, y,$ and $z$ as $2a+1, 2b+1,$ and $2c+1$ respectively.
Then, the equation becomes $(2(a-c))^2+(2(b-c))^2=(2(a+b)+2)^2$, or $4(a-c)^2+4(b-c)^2=(2a+2b+2)^2$.
Expanding both sides, one gets $4(a^2-2ac+c^2)+4(b^2-2bc+c^2)=4a^2+4b^2+4+8ab+8a+8b$.
Dividing both sides by $4$, one gets $a^2-2ac+c^2+b^2-2bc+c^2=a^2+b^2+1+2ab+2a+2b$.
The $a^2$ and $b^2$ terms can then be cancelled, giving $-2ac+c^2-2bc+c^2=1+2ab+2a+2b$.
This gives a contradiction, because the LHS is even, while the RHS is odd.
| {
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"url": "https://math.stackexchange.com/questions/3776798",
"timestamp": "2023-03-29T00:00:00",
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Show that the inequality $\left|\int_{0}^{1} f(x)\,dx\right| \leq \frac{1}{12}$ holds for certain initial conditions
Given that a function $f$ has a continuous second derivative on the interval $[0,1]$, $f(0)=f(1)=0$, and $|f''(x)|\leq 1$, show that $$\left|\int_{0}^{1}f(x)\,dx\right|\leq \frac{1}{12}\,.$$
My attempt: This looks to be a maximization/minimization problem. Since the largest value $f''(x)$ can take on is $1$, then the first case will be to assume $f''(x)=1$. This is because it is the maximum concavity and covers the most amount of area from $[0,1]$ while still maintaining the given conditions.
Edit: Because of the MVT and Rolle's Theorem, there exists extrema on the interval $[0,1]$ satisfying $f'(c)=0$ for some $c\in[0,1]$. These extrema could occur at endpoints.
Then $f'(x)=x+b$ and $f(x)=\frac{x^2}{2}+bx+c$. Since $f(0)=0$, then $c=0$ and $f(1)=0$, then $b=-\frac{1}{2}$. Remark: Any function with a continuous, constant second derivative will be of the form $ax^2+bx+c$ and in this case, $a=-b$ and $c=0$. Now, $$\begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(x^2-x)\,dx\\&=\frac{1}{2}\bigg[\frac{x^3}{3}-\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=-\frac{1}{12}\end{align*}$$
Next, we assume that $f''(x)=-1$ and repeating the process yields $$ \begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(-x^2+x)\,dx\\&=\frac{1}{2}\bigg[\frac{-x^3}{3}+\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=\frac{1}{12}\end{align*}$$ Thus we have shown that at the upper and lower bounds for $f''(x)$ that $\frac{-1}{12}\leq\int_{0}^{1}f(x)\,dx\leq \frac{1}{12}
\Longleftrightarrow \left|\int_{0}^{1}f(x)\,dx\right|\leq\frac{1}{12}$ because $f''(x)$ is continuous on $[0,1]$.
I was wondering if this was 'rigorous' enough to be considered a full proof and solution to the problem.
| Use $\text{Taylor}$ we can get:
$$
f\left( 0 \right) =f\left( x \right) -xf'\left( x \right) +\frac{f''\left( \xi _1 \right)}{2}x^2,\xi _1\in \left( 0,x \right)
$$
$$
f\left( 1 \right) =f\left( x \right) +\left( 1-x \right) f'\left( x \right) +\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2,\xi _2\in \left( x,1 \right)
$$
so we can get:
$$0=2f\left( x \right) +\left( 1-2x \right) f'\left( x \right) +\frac{f''\left( \xi _1 \right)}{2}x^2+\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2$$
i.e.$$2f\left( x \right) =\left( 2x-1 \right) f'\left( x \right) -\frac{f''\left( \xi _1 \right)}{2}x^2-\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2$$
then we can get:
$$2\int_0^1{f\left( x \right) \text{d}x}=\int_0^1{\left[ \left( 2x-1 \right) f'\left( x \right) -\frac{f''\left( \xi _1 \right)}{2}x^2-\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2 \right] \text{d}x}$$
after simplification,we can get:$$4\int_0^1{f\left( x \right) \text{d}x}=-\int_0^1{\left[ \frac{f''\left( \xi _1 \right)}{2}x^2+\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2 \right] \text{d}x}$$
so$$4\left| \int_0^1{f\left( x \right) \text{d}x} \right|\leqslant \frac{1}{2}\int_0^1{\left( 2x^2-2x+1 \right) \text{d}x}=\frac{1}{3}$$
i.e.$$\left| \int_0^1{f\left( x \right) \text{d}x} \right|\leqslant \frac{1}{12}$$
| {
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"url": "https://math.stackexchange.com/questions/3777387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ for a natural number $ n \geq 2$? Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ holds for a natural number $ n \geq 2$?
My attempt at solving:
Using $(a+b)^2=a^2+2ab+b^2$; if $(a+b)^2=a^2+b^2$, $2ab=0$ therefore $a$ and/or $b$ must be $0$.
If $a$ and/or $b$ is $0$ then $a^2$ and/or $b^2$ will be $0$.
Therefore $(a+b)^2$ can never equal $a^2+b^2$.
| For even $n$ you are right. For odd $n$ we have also $a+b=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral of $\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$ $$\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$$
My approach:
Since it is easy to evaluate $\int{\sec^2x}$ , integration by parts seems like a viable option.
Let $$I_n=\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}}$$
$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\int{\frac{\sec x \tan x}{(\sec x+\tan x)^\frac{9}{2}}dx}$$
Evaluating the new integral again using by parts yields
$$\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}+\frac{9}{2}\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}\,dx}$$
$$=\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2} I_n$$
Plugging it back, we obtain
$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{81}{4}I_n $$
$$\frac{-77}{4}I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$$
This obviously doesn't match with bprp's answer. Help!
Edit:
How do I convert my answer to the answer obtained by him, if mine is correct
| $-77I_n=4 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 18 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$
$-77I_n= 11 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} - 7 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 11 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$
$-77I_n= 11 \times \frac{\sec x+\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x - \tan x}{(\sec x+\tan x)^\frac{9}{2}}$
First part is straightforward. The second part can be simplified multiplying both numerator and denominator by $(\sec x+\tan x)$. We know, $(\sec x+\tan x) (\sec x-\tan x) = 1$.
You can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The number $85^9 - 21^9 + 6^9$ is divisible by an integer between 2000 and 3000. Compute that integer.
The number $85^9 - 21^9 + 6^9$ is divisible by an integer between 2000 and 3000. Compute that integer.
I tried taking the expression mod 5, 7, etc. getting a few congruences, but nothing that led to an answer between 2000 and 3000. Note that I am not allowed to use a calculator or a computer.
Euler's Theorem might also lead to an answer.
| To elaborate on the trial division approach, we can observe that $$85^9 - 21^9 = (85-21)(85^8 + 85^7 21^1 + \cdots + 21^8), \tag{1}$$ thus this part is divisible by $85 - 21 = 64 = 2^6$. Then $6^9 = 2^9 \cdot 3^9$, hence we know $2^6$ divides the entire number $N = 85^9 - 21^9 + 6^9$. We know $2$ does not divide the second factor in the difference of ninth powers in equation $(1)$, because all of the terms in that factor are odd, and there are an odd number of terms. Therefore $2^6$ is the largest power of $2$ that divides $N$.
As for modulo $3$, the first term $85^9$ is not divisible by $3$ since $85$ is not divisible by $3$, but the other two terms are, so we are done there.
For modulo $5$, the first term is clearly divisible. The second term is $-1$ mod $5$ since $21 \equiv 1$, hence $-21^9 \equiv -1$. The third term has $6 \equiv 1$. So $N$ is divisible by $5$. If we check mod $7$, then we won't need to consider mod $5^2$.
For modulo $7$, $85 \equiv 1$ and $6 \equiv -1$, and $21 \equiv 0$; thus by similar reasoning as above, $N$ is divisible by $7$, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question:
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.
My try:
$$\begin{align}\\
&x^2+y^2-4x=0\\
&\implies (x-2)^2+y^2=4\\
&\implies |z-2|=2\\
\end{align}\\
$$
Now, $w=\frac{2z+3}{z-4}$
$$\begin{align}\\
&\frac w2=\frac{2z+3}{2z-8}\\
&\implies\require{cancel}\frac{w}{w-2}=\frac{2z+3}{\cancel{2z}+3-\cancel{2z}+8}\\
&\implies\frac{w}{w-2}=\frac{2z+3}{11}\\
&\implies\frac{2z}{11}=\frac{w}{w-2}-\frac{3}{11}\\
&\implies\frac{2z}{\cancel{11}}=\frac{8w+6}{\cancel{11}(w-2)}\\
&\implies z=\frac{4w+3}{w-2}\\
&\implies z-2=\frac{2w+7}{w-2}\\
\end{align}\\
$$
$$\therefore\left|\frac{2w+7}{w-2}\right|=2\\
\implies 2w+7=2w-4
$$
Now, what to do? Where is my fault? Or how to do it? Is there any other possible ways?
| You can't replace modulus like that in the last line ( because complex numbers with equal magnitude may not be actually equal. For e.g., $2i$ and $-2i$ have magnitude $2$ but aren't equal.)
Use $w = u+iv$
So, $|2(u+iv)+7| = 2|(u+iv)-2|$
Evaluating and squaring both sides,
$\Rightarrow (2u+7)^2 + 4v^2 = 4(u-2)^2 + 4v^2 $
$\Rightarrow 4u^2 + 28u +49 = 4u^2 - 16u + 16 $
$\Rightarrow 44u + 33 = 0 $
$\Rightarrow 4u +3 =0 $
| {
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"url": "https://math.stackexchange.com/questions/3781558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Inverse Laplace Transform via Circuit Analysis [HELP] Inverse Laplace Transform $\frac{1}{s^2 + \sqrt{2}s + 1}$
so what I did it changed the denominator to complete the square format which is $\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$, then I can solve for $s$, it will make it as
$$
\left(\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i\right)
\left(\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i\right)
$$
So now, to the sheet of paper is to do Partial Fraction Decomposition of this which is absurd to me because of complex roots it has:
$$
\frac{1}{s^2 + s\sqrt{2} + 1}
= \frac{1}{\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}}
$$
Partial Fraction of Complex root will be
$$
\frac{K}{\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i}
+ \frac{K^*}{\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i}
$$
to follow the formula sheet.
which I got my K = -$i\frac{\sqrt{2}}{2}$ and $K^*$ = $i\frac{\sqrt{2}}{2}$
the problem I get is magnitude and $\theta$ is undefined it makes no sense at all.
| Let us denote by $a$ the value $1/\sqrt 2$ to have an easy typing. The the partial fraction decomposition over complex numbers of the given fraction is
indeed
$$
\begin{aligned}
\frac{1}{s^2 + \sqrt{2}s + 1}
&=
\frac{1}{(s^2 + 2as + a^2)+a^2}
=
\frac1{2ia}\left(\frac 1{s+a-ia}-\frac 1{s+a+ia}\right)
\\
&=\frac K{s+a-ia}-\frac {K^*}{s+a+ia}
\ .
\\[3mm]
\text{Here:} &
\\
K
&=\frac 1{2ia}
=-\frac 1{2a}i
=\frac 1{2a}\left(\cos\left(-\frac\pi2\right) +i\sin\left(-\frac\pi2\right)\right)
\\
&=a\left(\cos\left(-\frac\pi2\right) +i\sin\left(-\frac\pi2\right)\right)\ ,
\\
|K| &= a\ ,\\
\theta &=-\frac \pi 2\ .
\end{aligned}
$$
The inverse Laplace transform is thus using the table:
$$
2ae^{-at}\cos(at+\theta)
=
2ae^{-at}\sin(at)
=
\sqrt 2\cdot e^{-t/\sqrt 2}\sin(t/\sqrt 2)
\ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $|\frac{1}{x-5}| \leq 1$ then condition $(1)$ and $(2)$ will not hold. So wouldn't the range be $x < 4$ and $5 \leq x$ ? I know this is not right since the answer should be $ 5\frac{1}{3} \leq x$ and $x < 4 \frac{2}{3}$ ... could someone provide a solution?
| Hints : by Cauchy Hadamard test , range will be $ |x-5|\gt\frac{1}{3} \implies x\gt \frac{16}{3},x\lt \frac{14}{3} $.
Only equality $x=\frac{16}{3} $ is possible because series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} $ is converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \times 14$
*$4 \times 14$
I think we have
$$\det \left[\begin{array}{ll}6 & 2 \\ 2 & 6\end{array}\right] = 4 \times 8$$
$$\det \left[\begin{array}{lll}6 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 6\end{array}\right] = 4^{2} \times10$$
and for any $n$ we have $\det(A_n)= 4^{n-1} \times (6+2×(n-1))$ so "1" is true.
| Your generalization, with an obvious but unstated definition of $A_n$, can be proved as follows. Define $E_n$ as the $n\times n$ matrix whose entry are all $1$s. You can show its only nonzero eigenvalue is $2n$, and the associated eigenspace is $1$-dimensional. By the rank-nullity theorem, the eigenvalue $0$ has multiplicity $n-1$. So $A_n=4I_n+2E_n$ has eigenvalues $4,\,2n+4$ with respective multiplicities $n-1,\,1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{x\to 0} \frac 1x -\frac{2}{e^{2x}-1}$ Factorizing the expression
$$\lim_{x\to 0} \frac{e^{2x}-1-2x}{x(e^{2x}-1)}$$
$$=\lim_{x\to 0} \frac{e^{2x}-(1+2x)}{(x)(e^x-1)(e^x+1)}$$
How do I proceed?
| Another approach using the Hopital's rule:
$$\lim_{x \to 0}\left(- \frac{2}{e^{2 x} - 1} + \frac{1}{x}\right)= \lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)}$$
$$\lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)} \stackrel{\,\,(0/0) \\\text{Hopital}}{=} \lim_{x \to 0} \frac{2 e^{2 x} - 2}{2 x e^{2 x} + e^{2 x} - 1}$$
$$\left(2 \lim_{x \to 0} \frac{e^{2 x} - 1}{2 x e^{2 x} + e^{2 x} - 1}\right)\stackrel{\,\,(0/0) \\\text{Hopital}}{=} 2 \lim_{x \to 0} \frac{2 e^{2 x}}{4 x e^{2 x} + 4 e^{2 x}}=\lim_{x \to 0} \frac{4e^{2x}}{4e^{2x}\left(x + 1\right)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
$p\equiv 1\pmod 4\Rightarrow p=a^2+b^2$ and $p\equiv 1\pmod 8\Rightarrow p=a^2+2b^2$, what about for $p\equiv 1\pmod {2^n}$ in general Primes $p$ with $p\equiv 1\pmod 4$ can be written as $p=a^2+b^2$ for some integers $a,b$. For $p\equiv 1\pmod 8$ we have $p=a^2+2b^2$. Can primes that satisfy $p\equiv 1\pmod{2^n}$ for $n>3$ be written in a similar form -- for example $p=a^2+4b^2$ for $n=4$?
| Let prime $p=a^2+2^{n-2}\cdot b^2\equiv1\pmod{2^n}$, where $n>4$ and $a,b$ is integers.
Then sequence $(n, min(p))$=(5,97), (6,193), (7,257), (8,257), (9,18433), (10,18433), (11,18433), (12,65537), (13,1097729), (14,65537), (15,1179649), (16,65537), (17,1179649), (18,26214401), (19,117964801), (20,26214401), (21,169869313), (22,104857601), ....
gp-code:
nminp()=
{
for(n=5,100,
forprime(p=3,10^9,
m= 2^n;
if(Mod(p,m)==1,
if(#thue('x^2+m/4, p),
print1("("n","p"), ");
break()
)
)
)
)
};
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
$$ \Rightarrow 2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq (b+c-2a)^3 $$
But couldn't take it any further, but it looks as if Hölder's Inequality may help. And also I don't think that it is a "warm-up" problem.
Any help will be appreciated.
| Using algebra.
Consider that we look for the minimum value of
$$F=a^3+b^3+c^3-3abc -2\left({b+c\over 2}-a\right)^3$$
$$\frac{\partial F}{\partial a}=3 a^2+6 \left(\frac{b+c}{2}-a\right)^2-3 b c \tag 1$$
$$\frac{\partial F}{\partial b}=3b^2-3 \left(\frac{b+c}{2}-a\right)^2-3 a c\tag 2$$
$$\frac{\partial F}{\partial c}=3c^2-3 \left(\frac{b+c}{2}-a\right)^2-3 a b\tag 3$$
Using $(3)$ the only possible value of $c$ is
$$c=\frac{1}{3} \left(2 \sqrt{4 a^2-a b+b^2}-2 a+b\right)\tag 4$$ Plug in $(2)$ and solve for $b$; the only possible solutions are
$$b_1=a \qquad \text{and} \qquad b_2=\frac{\sqrt{6}-1}{2} a\tag 5$$
$$b=b_1\implies\frac{\partial F}{\partial a}=0$$
$$b=b_2\implies\frac{\partial F}{\partial a}=\frac{1}{12} \left(359-140 \sqrt{6}\right) a^2$$ which must be rejected.
So $b=a$ and $c=a$ then $a=b=c$ and $F=0\,\, \forall a >0$
| {
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"url": "https://math.stackexchange.com/questions/3785807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$ How to evaluate:
$$\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$$
Can somebody help me handle it?
I have no idea what to do.
This is my attempt:
$$\cos24^\circ-\cos(60^\circ+24^\circ)-\cos12^\circ+\sin (12^\circ+30^\circ)$$
| $$-\cos84^\circ-\cos96^\circ$$
$$-\cos12^\circ=\cos192^\circ=\cos168^\circ$$
$$\sin42^\circ=\cos(?)$$
As $\cos5x=-\dfrac12$ for $x=120^\circ,240^\circ\equiv-120^\circ,480^\circ\equiv120^\circ,960^\circ\equiv-120^\circ\pmod{360^\circ}$
Now $\cos5x+\cos x=2\cos3x\cos2x$
$\cos5x=2(4\cos^3x-3\cos x)(2\cos^2x-1)=?$
So, the roots of $$16c^5-20c^3+16c+\dfrac12=0$$ are $5x=360^\circ n+120^\circ\iff x=72^\circ n+24^\circ; -2\le n\le2$
$$\sum_{n=-2}^2\cos(72^\circ n+24^\circ)=\dfrac0{16}$$
$$\implies\cos(-120)^\circ+\cos(-48^\circ)+\cos24^\circ+\cos96^\circ+\cos168^\circ=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
| By Chebyshev's sum inequality (https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality),
we have
$$a^n\cdot a + b^n\cdot b + c^n \cdot c
\ge \frac{1}{3}(a^n + b^n + c^n)(a+b+c).$$
Thus,
$$\frac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n + b^n + c^n} \ge \frac{a+b+c}{3} \ge \sqrt[3]{abc}.$$
We are done.
| {
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"url": "https://math.stackexchange.com/questions/3787573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that $8^{\frac{1}{3}}$ is $(8+i0)^{\frac{1}{3}}$
And, $r = 8$
And, $8\cos \theta = 8$ and $\theta = 0$.
So, $8^{\frac{1}{3}}\operatorname{cis} 0^\circ = 2\times (1+0)=2$
I just got only $2$. Where and how others $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ come from?
| Here,
$$\begin{align*}
8^{1/3} &= (|8|e^{2\pi kj})^\frac{1}{3}, k = 0,1,2\\
&= |8|^\frac{1}{3} e^{\frac{2}{3}\pi kj}, k = 0,1,2\\
&= 2 e^{\frac{2}{3}\pi kj}, k = 0,1,2\\
\end{align*}$$
so,for $k=1$,$k=2$ we get $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$
Or take:
$$8^{1/3}=x$$
Then we get,
$$(x-2)(x^2+2x+4)=0$$
Then we get our desired roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Using termwise differentiation and integration find the sum $\sum\limits_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + ... +(n+1)x^n$ Using termwise differentiation and integration find the sum $$\sum\limits_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + \cdots +(n+1)x^n$$
The answer can be achevied simply by setting $S = \sum\limits_{n=0}^\infty (n+1)x^n$ and evaluating $S(1-x) = \sum\limits_{n=1}^\infty x^n = \frac{1}{1-x}$. Hence $S = \frac{1}{(1-x)^2}$. However, this approach does not use termwise approach:
$$f(x) = 1 + 2x + 3x^2 + \cdots + (n+1)x^n$$
$$f'(x) = 2 + 6x + \cdots + n(n+1)x^{n-1}$$
Not sure how to proceed after this point using that approach. Could someone provide a solution? Thanks.
| We have
$$\sum_{n\geq 0}(n+1)x^n=\left(\sum_{n\geq 0}x^{n+1}\right)'=\left(\frac{x}{1-x}\right)'=\frac{1}{(1-x)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791891",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
How were amplitudes of the $\cos$ and $\sin$ chosen? I don't understand why we use $\displaystyle\sqrt{1^2+\left(\frac{1}{2}\right)^2}$ in the below transformation. Can someone help to explain?
from
$$f(x)=\frac{3}{5}-\frac{3}{5}e^t\left(\cos(2t)+\frac{1}{2}\sin(2t)\right)$$
transform to
$$f(x)=\frac{3}{5}-\frac{3}{5}\sqrt{1^2+\left(\frac{1}{2}\right)^2}e^t\left(\frac{1}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}\cos(2t)+\frac{\frac{1}{2}}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}\sin(2t)\right)$$
let $\displaystyle\frac{1}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}=\cos\phi$ and $\displaystyle\frac{\frac{1}{2}}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}=\sin\phi$,
$$f(x)=\frac{3}{5}-\frac{3}{5}\sqrt{1^2+\left(\frac{1}{2}\right)^2}e^t(\cos\phi\cos(2t)+\sin\phi\sin(2t))$$
| This is a way to normalize the vector $v=(a,b)=\left(1,\frac12\right)$ that is
$$|v|=\sqrt{a^2+b^2} \implies \hat v=\frac{v}{|v|}$$
has length equal to $1$ and this allows to perform the subsequent transformation for $\cos \phi$ and $\sin \phi$.
| {
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"url": "https://math.stackexchange.com/questions/3793748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
| $$A=x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=x + \frac{2x^{3}}{3} - \frac{2x^3}{3}\left(1+\frac{1}{x^2}\right)^{3/2}$$ For the very last term, let $\frac{1}{x^2}=\epsilon$ and use the binomial expansion
$$(1+\epsilon)^{3/2}=1+\frac{3 \epsilon }{2}+\frac{3 \epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ Replace $\epsilon$ by $\frac{1}{x^2}$ to make
$$\left(1+\frac{1}{x^2}\right)^{3/2}=1+\frac{3}{2 x^2}+\frac{3}{8 x^4}+O\left(\frac{1}{x^6}\right)$$
$$A=x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=x + \frac{2x^{3}}{3} - \frac{2x^3}{3}\left(1+\frac{3}{2 x^2}+\frac{3}{8 x^4}+O\left(\frac{1}{x^6}\right)\right)$$
$$A=-\frac{1}{4 x}+O\left(\frac{1}{x^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way? The following problem
$$\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)\tag1$$
was already solved in this solution.
The question here is how to prove $(1)$ by utilizing the Fourier series of
$$\tan x\ln(\sin x)=-\sum_{n=1}^\infty\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac1n\right)\sin(2nx)$$
$$=-\sum_{n=1}^\infty\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\sin(2nx),\quad 0<x<\frac{\pi}{2}$$
I wonder what kind of clever manipulation we need to do to create the integrand in $(1)$. I am sure it would be an amazing solution.
Thank you in advance.
This Fourier series can be found in the book, Almost Impossible Integrals, Sums and series, page $243$, Eq$(3.281)$.
| From here we have
$$\frac23\arcsin^4x=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}=\sum_{n=1}^\infty\frac{H_{n}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}-\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^4{2n\choose n}}$$
Set $x=1$ we get
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=\sum_{n=1}^\infty\frac{4^nH_{n}^{(2)}}{n^2{2n\choose n}}-\frac{15}{4}\zeta(4)\tag1$$
In this question we showed
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}+12\ln^2(2)\zeta(2)\tag2$$
Adding $(1)$ and $(2)$ yields
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{4}\zeta(4)-\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}$$
By using the Fourier series of $\tan x\ln(\sin x)$, we showed in this solution:
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$
substitute this result we get
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\frac{19}{4}\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\tag3$$
Now we use the well-known series expansion of $\arcsin^2 x$:
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{4^n x^{2n}}{n^2{2n\choose n}}$$
Multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{2n-1}\ln xdx=\frac{1}{4n^2}$ we get
$$\frac18\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=-\int_0^1\frac{\ln x\arcsin^2(x)}{x}dx$$
$$\overset{IBP}{=}\int_0^1\frac{\ln^2x\arcsin(x)}{\sqrt{1-x^2}}dx\overset{x=\sin\theta}{=}\int_0^{\pi/2}x\ln^2(\sin x)dx\tag4$$
From $(3)$ and $(4)$ we obtain
$$\int_0^{\pi/2} x\ln^2(\sin x)dx=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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} |
Why is $|-a+\sqrt{a^2-1}|<1<|-a-\sqrt{a^2-1}|$ (where $a>1$) true? Why does $|-a+\sqrt{a^2-1}|<1<|-a-\sqrt{a^2-1}|$ (where $a>1$) hold? I understand that $a>1 \implies 1<|-a-\sqrt{a^2-1}|$ and that $|-a+\sqrt{a^2-1}|<|-a-\sqrt{a^2-1}|$
But I can't see why $a>1 \implies |-a+\sqrt{a^2-1}|<1$.
Does anyone see why? Thank you.
| $$\begin{align}\left|-a-\sqrt{a^2-1}\right|=\left|-\left(a+\sqrt{a^2-1}\right)\right|=\left|a+\sqrt{a^2-1}\right|>a>1\end{align}$$
$$\begin{align}\left|-a+\sqrt{a^2-1}\right|=\left|-\left(a-\sqrt{a^2-1}\right)\right|=\left|a-\sqrt{a^2-1}\right|=\dfrac{1}{a+\sqrt{a^2-1}}<1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving limit using the definition of a limit Heres the question (provided for context) about proving a limit by using the definition of a limit:
$\mathop {\lim }\limits_{x \to 4} \left({x^2} + x - 11\right) = 9$
So, let’s get started. Let $\varepsilon > 0$ be any number then we need to find a number $\delta > 0$ so that the following will be true.
$\left| {\left( {{x^2} + x - 11} \right) - 9} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 4} \right| < \delta$
simplifying a bit
$\left| {\left( {{x^2} + x - 11} \right) - 9} \right| = \left| {{x^2}
+ x - 20} \right| = \left| {\left( {x + 5} \right)\left( {x - 4} \right)} \right| = \left| {x + 5} \right|\left| {x - 4} \right| < \varepsilon$
if, by some chance, we can show that $\left| {x + 5} \right| < K$ for some number $K$ then, we’ll have the following
$\left| {x + 5} \right|\left| {x - 4} \right| < K\left| {x - 4} \right|$
If we now assume that what we really want to show is $K\left| {x - 4} \right| < \varepsilon$ instead of $\left| {x + 5} \right|\left| {x - 4} \right| < \varepsilon$ we get the following,
$\left| {x - 4} \right| < \frac{\varepsilon }{K}$
All this is based on the assumption that we can show $\left| {x + 5} \right| < K$ for some $K$. To do this we assume that whatever $x$ is it must be close to $x=4$ since we are working with a limit. So lets assume that $x$ is within a distance of one of $x=4$. In terms of inequality we can assume
$\left| {x - 4} \right| < 1$
Starting with removing the absolute value bars we have
$- 1 < x - 4 < 1\hspace{0.5in} \Rightarrow \hspace{0.5in}3 < x < 5$
If we now add 5 to all parts of this inequality we get,
$8 < x + 5 < 10$
Now, since $x + 5 > 8 > 0$ (the positive part is important here) we can say that, provided $\left| {x - 4} \right| < 1$ we know that $x + 5 = \left| {x + 5} \right|$. Or, if take the double inequality above we have,
$8 < x + 5 = \left| {x + 5} \right| < 10$ $\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\left| {x + 5} \right| < 10\hspace{0.5in} \Rightarrow \hspace{0.5in}K = 10$
So, provided $\left| {x - 4} \right| < 1$ we can see that $\left| {x + 5} \right| < 10$ which in turn gives us,
$\left| {x - 4} \right| < \frac{\varepsilon }{K} = \frac{\varepsilon }{{10}}$
source
How did we go from this double inequality $8 < x + 5 < 10$ to this $8 < x + 5 = \left| {x + 5} \right| < 10$. From what I understand $|{x + 5}| < 10 $ can also be written as $-10<x + 5<10$, then the authors statement:
$8 < x + 5 = \left| {x + 5} \right| < 10$
should not be true since it is excluding a part of the interval that $|{x + 5}| < 10 $ includes (the interval which can be seen when the absolute value inequality is expanded into a double inequality i.e. $-10<x + 5<10$)
| You are working with the assumption $|x-4|<1$, that means $3<x<5$.
So $x+5=|x+5|$ because $x+5$ is always positive when $x \in (3,5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find continuous $f$ such that $f(x+1)=f(x)+f(\frac{1}{x})$
Find all real continuous functions that verifies :
$$f(x+1)=f(x)+f\left(\frac{1}{x}\right) \ \ \ \ \ \ (x\neq 0) $$
I found this result $\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$ and I tried to study the behaviour of the function $g$ defined as $g(x)=\frac{x}{x-1}$ and compare it with $x$ in order to use fixed point theorem but it won't work.
I need a hint and thanks.
| Some results
Assume that $f$ is defined and continuous on all of $\mathbb{R}$ even if the equation $f(x+1) = f(x) + f(\frac{1}{x})$ is not defined for $x=0.$
From the given identity
$
f(x+1) = f(x) + f(\frac1x)
$
we get
$
f(\frac1x) = f(x+1) - f(x) \to f(1) - f(0)
$
as $x\to0$. Thus $\lim_{R\to\infty}f(R) = \lim_{R\to\infty}f(-R) = f(1) - f(0).$
Furthermore, we have
$$\begin{align}
f(2) &= f(1+1) = f(1) + f(\frac{1}{1}) = f(1) + f(1) \\
f(3) &= f(2+1) = f(2) + f(\frac{1}{2}) = f(1) + f(1) + f(\frac{1}{2}) \\
f(4) &= f(3+1) = f(3) + f(\frac{1}{3}) = f(1) + f(1) + f(\frac{1}{2}) + f(\frac{1}{3}) \\
\vdots \\
f(n) &= f(1) + \sum_{k=1}^{n-1} f(\frac{1}{k}) \\
\end{align}$$
Taking limits as $n\to\infty$ gives
$
\lim_{n\to\infty} f(n) = f(1) + \sum_{k=1}^{\infty} f(\frac{1}{k})
,
$
but since $\lim_{n\to\infty} f(n) = f(1) - f(0)$ this implies
$$
f(0) = -\sum_{k=1}^{\infty} f(\frac{1}{k})
.
$$
However, the convergence of the series implies that $|f(\frac1k)|\to 0$ as $k\to\infty.$ By continuity of $f$ this means that $f(0)=\lim_{k\to\infty}f(\frac1k)=0.$ And from that follows that
$
\sum_{k=1}^{\infty} f(\frac{1}{k}) = 0
$
and that
$
\lim_{R\to\infty}f(\pm R) = f(1)
.
$
We also get that
$$
f(0) = f((-1)+1) = f(-1) + f(\frac1{-1}) = 2 f(-1)
,
$$
i.e. $f(-1) = 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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} |
If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| We can't bluntly take $\sin^{-1}(\sin 2y) = 2y$ and so with $\tan^{-1}(\tan 2y)$, because we don't know the value of $2y$ and the range in which it lies.
So, substitute directly.
$f(-10) = \sin^{-1}\left(\dfrac{-20}{101}\right)+\tan^{-1}\left(\dfrac{20}{99}\right) = -\sin^{-1}\left(\dfrac{20}{101}\right)+\tan^{-1}\left(\dfrac{20}{99}\right)$
Now let, $\tan z = \dfrac{20}{99} = \dfrac{20/101}{99/101}\Rightarrow \sin z =\dfrac{20}{101} \Rightarrow z = \sin^{-1}\left(\dfrac{20}{101}\right)$
(Here $0<\tan^{-1}\left(\dfrac{20}{99}\right)<\dfrac{\pi}2$)
So, we have $f(-10) = -\sin^{-1}\left(\dfrac{20}{101}\right) + \sin^{-1}\left(\dfrac{20}{101}\right) = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics? How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics?
My attempt.
These are equations with which I begin:
$\dfrac{dx}{ds} = y+u $;
$\dfrac{dy}{ds} = x+u $;
$\dfrac{du}{ds} = x+y $.
However, I am stuck, because I can not solve the equations for $\dfrac{dx}{ds}$ and $\dfrac{dy}{ds}$ because we have a dependence on $u$.
Thanks for any help.
| Your differential equations are linear, with constant coefficients, for which there are well-established methods of solution. First write them in matrix form as:
$$
\pmatrix{\frac{dx}{ds}\\
\frac{dy}{ds}\\
\frac{du}{ds}}=\pmatrix{0&1&1\\
1&0&1\\
1&1&0}\pmatrix{x\\y\\u}\ .
$$
Next, find the eigenvalues of the matrix $\ \pmatrix{0&1&1\\1&0&1\\1&1&0}\ $, which are $\ -1,-1,2\ $, and a complete set of eigenvectors, which can be taken to be
$$
\pmatrix{1\\-1\\0}, \pmatrix{0\\-1\\1}\ \text{ and }\ \pmatrix{1\\1\\1}\ .
$$
These allow you to diagonalise the matrix thus
$$
\pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{0&1&1\\1&0&1\\1&1&0} \pmatrix{1&0&1\\-1&-1&1\\0&1&1}=\pmatrix{-1&0&0\\0&-1&0\\0&0&2}\ .
$$
This, if you put
$$
z= \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{x\\y\\u}\ ,
$$
you can write your differential equations in the form
\begin{align}
\frac{dz}{ds}=& \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{0&1&1\\1&0&1\\1&1&0} \pmatrix{1&0&1\\-1&-1&1\\0&1&1}z\\
=& \pmatrix{-1&0&0\\0&-1&0\\0&0&2}z\ ,
\end{align}
which have solutions
\begin{align}
z_1&=Ae^{-s}\\
z_2&=Be^{-s}\\
z_3&=Ce^{2s}\ ,
\end{align}
and so we get
\begin{align}
\pmatrix{x\\y\\u}&= \pmatrix{1&0&1\\-1&-1&1\\0&1&1}z\ \text{, or}\\
\\
x&=Ae^{-s}+Ce^{2s}\\
y&=Ce^{2s}-(A+B) e^{-s}\ ,\text{ and}\\
u&=B e^{-s}+Ce^{2s}\ .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}$ given $n$ is a positive integer and $0 < x \lt 1$. Problem Statement: If $n$ is a positive integer and $0 < x \lt 1$, show that
$$ \frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}.$$
My Solution:
$$
\frac{ 1- x^{n+1} }{n+1} \lt \frac{1-x^n}{n} ~~~~\text{is true} \\
\text{if}~~~~ \frac{n}{1-x^n} \lt \frac{n+1}{1- x^{n+1} }$$
If we see the LHS we find that it is of the form of sum of a geometric series with the first term $n$ and the common ration $x^n$ (which is less than 1) similarly the RHS represents the sum of a geometric series with the first term as $n+1$ and common ratio $x^{n+1}$ (which is less than $x^n$, that is less than 1)
Now, my point is that the series represented by the LHS have the first term lesser than the first term of the series represented by the RHS, and the series represented by LHS decreases fastly in comparison to the series represented by RHS (because the common ratio $x^{n+1} \lt x^n$), hence the sum of series represented by the RHS is greater than the sum of the series represented by the LHS.
Is my solution and reasoning correct?
| Another option:
$$
\frac{1-x^n}{n} = \int_x^1 t^{n-1} \, dt
$$
and the integrand $t^{n-1}$ decreases strictly on $(x, 1)$ if $n$ increases.
Yet another option: The inequality
$$
\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}
$$
is equivalent to
$$
x^n < \frac{1}{n+1}x^0 + \frac{n}{n+1}x^{n+1}
$$
and that is true because the function $f(t) = x^t$ is strictly convex for fixed $x \in (0, 1)$. Graphically: The slope of the secant
$$
\frac{x^n-1}{n} = \frac{f(n)-f(0)}{n-0}
$$
increases with increasing $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Compute the area of a surface, encountering a strange integral compute the area of this yellow surface, which is actually a paraboloid:
$x^2+y^2=2az$(yellow one), cutted by $(x^2+y^2)^2=2a^2xy$(blue one)
To compute the part be surrounded in the blue surface, use polor coordinates:
\begin{cases}
x=r\cos\theta\\
y=r\sin\theta\\
z=\frac{r^2}{2a}\\
\end{cases}
use Gauss efficient to compute the area:
\begin{cases}
E=(\frac{\partial x}{\partial r})^2+(\frac{\partial y}{\partial r})^2+(\frac{\partial z}{\partial r})^2&=1+\frac{r}{a}\\
F=\frac{\partial x}{\partial r}\frac{\partial x}{\partial\theta}+\frac{\partial y}{\partial r}\frac{\partial y}{\partial\theta}+\frac{\partial z}{\partial r}\frac{\partial z}{\partial\theta}&=0\\
G=(\frac{\partial x}{\partial\theta})^2+(\frac{\partial y}{\partial\theta})^2+(\frac{\partial z}{\partial\theta})^2&=r^2\\
\end{cases}
$$S=\iint\limits_D\sqrt{EG-F^2}drd\theta$$
where D becomes $\left\{ (r,\theta )|\theta \in \left[ 0,\frac{\pi}{2} \right] \cup \left[ \pi ,\frac{3}{2}\pi \right] ,\mathrm{r}\in \left[ 0,\mathrm{a}\sqrt{\sin 2\theta} \right] \right\}$ since the blue surface can be written as $r^2=a^2\sin2\theta$.
I can't find anything wrong until now, but this intergal is extreme complex, and Mathematica give me a non elementary solution. However, the standard solution given by the text book to this question is$\frac{20-3 \pi}{9} a^{2}$
How can i get that answer? Or what's wrong with my method?
| I am coming from assumption, that to obtain desired answer you should have cylinder $(x^2+y^2)^2=2a^2xy$ in place of brought in title $x^2+y^2=2a^2xy$ (blue one).
Then we will have $\sqrt{1+z'_x{^2}+z'_y{^2}} = \frac{1}{a}\sqrt{a^2+x^2+y^2}$. After writing double integral it is enough to use usual polar coordinates, so we will have
$$S=\frac{4}{a}\int\limits_{0}^{\frac{\pi}{4}}\,d\phi \int\limits_{0}^{a\sqrt{\sin 2\phi}}r\sqrt{a^2+r^2}\,dr = \frac{4}{3} a^2 \left( \int\limits_{0}^{\frac{\pi}{4}}(\sin \phi +\cos \phi)^3\,d \phi - \frac{\pi}{4} \right ) =\\
=\frac{4}{3} a^2 \left( 2\sqrt{2} \int\limits_{0}^{\frac{\pi}{4}} \sin^3 \left( \phi+\frac{\pi}{4} \right )\,d\phi - \frac{\pi}{4}\right ) =\\
=\frac{4}{3} a^2 \left(2\sqrt{2} \int\limits_{0}^{\frac{\pi}{4}} \left( \cos^2 \left( \phi+\frac{\pi}{4} \right )-1 \right )\,d \left( \cos \left( \phi+\frac{\pi}{4} \right )-1 \right ) - \frac{\pi}{4}\right ) = \frac{a^2}{9}(20-3\pi)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$
| You used the inverse of $x_i$ instead of the inverse of $\frac{M}{m_i}$. So for example,
$$
9 \cdot \frac{88}{11} \cdot 5 + 1 \cdot \frac{88}{8} \cdot 1 \equiv 0 + 1 \cdot 11 \cdot 1 \equiv 3 \not \equiv 1\pmod{8}.
$$
If you use the inverse of $\frac{M}{m_i}$ instead, you would have
$$
x_1 \cdot \frac{88}{11} \cdot 8^{-1} + x_2 \cdot \frac{88}{8} \cdot 11^{-1} \equiv 0 + 1 \cdot 11 \cdot 11^{-1} \equiv 1\pmod{8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
recurrence relation where $c_n = c_{n-1} + 2c_{n-2}$
A sequence $(c_n)$ is defined recursively as follows: $c_0 = 1, c_1 = 1, $ and $c_n = c_{n-1} + 2c_{n-2}$ for $n\geq 2$. We use $[x^n]g(x)$ to denote the coefficient of $x^n$ of the polynomial $g(x).$ Show that $c_{2n} = [x^{2n}] \dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+\dfrac{1}{1+x-2x^2}\right)$ and that $\sum_{n\geq 0} c_{2n}x^n = \dfrac{1-2x}{1-5x+4x^2}.$ From this, one can deduce that $c_{2n} = 5c_{2n-2} - 4c_{2n-4}.$ Obtain a similar equation for $c_{2n}, c_{2n-4}$ and $c_{2n-8}.$
I know that $\dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+\dfrac{1}{1+x-2x^2}\right) = \dfrac{1-2x^2}{1-5x^2+4x^4}$ and so if I show that $c_{2n} = [x^{2n}] \dfrac{1-2x^2}{1-5x^2+4x^4},$ I can replace $x^2$ with $x$ and get that $\sum_{n\geq 0} c_{2n}x^n = \dfrac{1-2x}{1-5x+4x^2}$. But I'm not sure how to show that $c_{2n} = [x^{2n}] \dfrac{1-2x^2}{1-5x^2+4x^4}.$ I don't think I'll need to compute the exact coefficient and it does not seem useful to manipulate the recurrence equation by substituting $n$ with $2n$. I tried showing that $(1-5x^2+4x^4) \sum_{n\geq 0} c_{2n} x^{2n} = 1-2x^2.$ Matching coefficients results in $c_0+(c_2-5c_0)x^2 + \sum_{n\geq 0} (c_{2n}-5c_{2n-2}-2c_{2n-4})x^{2n} = 1-2x^2,$ but it seems I'd need to prove something like $c_{2n}-5c_{2n-2}-2c_{2n-4}.$ I think figuring out how to come up with $\dfrac{1-2x}{1-5x+4x^2}$ should help me obtain a similar equation relating $c_{2n}, c_{2n-4}, c_{2n-8}$
| Your attempt to equate coefficients in
$$(1-5x^2+4x^4)\sum_{n\ge 0}c_{2n}x^{2n}=1-2x^2$$
will work if you do it correctly. When you multiply out the lefthand side you should get
$$c_0+(c_2-5c_0)x^2+\sum_{n\ge 2}(c_{2n}-5c_{2n-2}+4c_{2n-4})x^{2n}=1-2x^2\,,$$
and you know that $c_0=1$ and $c_2=3$, so this reduces to
$$\sum_{n\ge 2}(c_{2n}-5c_{2n-2}+4c_{2n-4})x^{2n}=0\,.$$
This means that $c_{2n}-5c_{2n-2}+4c_{2n-4}\equiv 0$ for $n\ge 2$ and gives you the recurrence $c_{2n}=5c_{2n-2}-4c_{2n-4}$. And from that you can work backwards to get
$$\sum_{n\ge 0}c_{2n}x^{2n}=\frac{1-2x^2}{1-5x^2+4x^4}$$
and the rest. There is probably some perfectly reasonable way to get their first expression for $c_{2n}$ and work forwards, but I’m not seeing it at the moment. Unfortunately, that means that what I did above may not adapt easily to let you solve the last part of the problem. I can, however, offer an alternative approach.
It’s easy enough simply to solve the recurrence and get an explicit formula for $c_n$. We have the recurrence $c_n=c_{n-1}+2c_{n-2}+[n=0]$, where the last term is an Iverson bracket added to make the recurrence valid for all $n$ if we assume that $c_n=0$ for $n<0$. Multiplying through by $x$ and summing over $n$ we have
$$g(x)=xg(x)+2x^2g(x)+1$$
and hence
$$g(x)=\frac1{1-x-2x^2}=\frac1{(1+x)(1-2x)}\,.$$
Decomposing this into partial fractions, we get
$$\begin{align*}
g(x)&=\frac13\left(\frac2{1-2x}+\frac1{1+x}\right)\\
&=\frac13\left(2\sum_{n\ge 0}2^nx^n+\sum_{n\ge 0}(-1)^nx^n\right)\\
&=\frac13\sum_{n\ge 0}\left(2^{n+1}+(-1)^n\right)x^n\,,
\end{align*}$$
so that $c_{2n}=\frac{2^{2n+1}+1}3\,.$ It is now easy to verify that
$$\begin{align*}
5c_{2n-2}-4c_{2n-4}&=\frac{5\cdot 2^{2n-1}+5-4\cdot 2^{2n-3}-4}3\\
&=\frac{20\cdot2^{2n-3}-4\cdot 2^{2n-3}+1}3\\
&=\frac{2^{2n+1}+1}3\\
&=c_{2n}\,.
\end{align*}$$
If you use the closed form to write out $c_{2n},c_{2n-4}$, and $c_{2n-8}$ and tinker a bit along the lines of that last calculation of mine, you should be able to come up with the coefficients $\alpha$ and $\beta$ for the recurrence $c_{2n}=\alpha c_{2n-4}+\beta c_{2n-8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding an expression and proof for $\sqrt{1-4x}$ This is similar to $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{2n\choose n}x^n$$
However I want to find an expression the same way for $$\sqrt{1-4x}$$ rather than $$\frac{1}{\sqrt{1-4x}}$$
Here's my thoughts so far:
$$(1-4x)^\frac{1}{2}=\sum_{n\geq0}{\frac{1}{2}\choose n}(-4)^nx^n$$
but this is where I get stuck.
| If the representation
\begin{align*}
\frac{1}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n
\end{align*}
for $|x|<\frac{1}{4}$ is supposed to be known, we can also use
\begin{align*}
\sqrt{1-4x}&=\frac{1-4x}{\sqrt{1-4x}}\\
&=\sum_{n \geq 0}\binom{2n}{n}x^n-4\sum_{n\geq 0}\binom{2n}{n}x^{n+1}\\
&=1+\sum_{n\geq 1}\left(\binom{2n}{n}-4\binom{2n-2}{n-1}\right)x^n
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the inverse of $A$ Let
$$
A=
\begin{bmatrix}
n & n_1 & n_2 & \cdots & n_s & 0 \\
n_1 & n_1 & 0 & \cdots & 0 & 1 \\
n_2 & 0 & n_2 & \cdots & 0 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n_s & 0 & 0 & \cdots & n_s & 1 \\
0 & 1 & 1 & \cdots & 1 & 0
\end{bmatrix},
$$
where $n=\sum_{i=1}^sn_i$.
My questions: What is $A^{-1}$? Does it have an elegant expression?
My attempts: I tried some small matrices and found that $A^{-1}$ has the following form:
$$
A^{-1}=\begin{bmatrix}
\cdots & \cdots & \cdots & \cdots & \cdots & -\frac{1}{s} \\
\cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\
\cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\
-\frac{1}{s} & \frac{1}{s} & \frac{1}{s} & \cdots & \frac{1}{s} & 0
\end{bmatrix}
$$
While I faild to find patterns in those "dots" parts.
Thanks a lot.
| Partial Answer: Here's a strategy you might find helpful.
Let $M$ denote the submatrix of $A$ given by
$$
M = \pmatrix{n & n_1 & n_2 & \cdots & n_s \\
n_1 & n_1 & 0 & \cdots & 0 \\
n_2 & 0 & n_2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n_s & 0 & 0 & \cdots & n_s}.
$$
we see that $M$ can be expressed in the form
$$
M = \sum_{i=1}^s (e_1 + e_{1+i})(e_1 + e_{1+i})^T,
$$
where $e_1,e_2,\dots,e_{s+1}$ denotes the canonical basis of $\Bbb R^n$. This gives us the decomposition $M = BDB^T$, where
$$
B = \pmatrix{e_1 + e_2 & e_1 + e_3 & \cdots & e_1 + e_{s+1}} =
\pmatrix{1&1&\cdots&1\\1\\&1\\ && \ddots \\&&&1},
\quad D = \pmatrix{n_1 \\ & \ddots \\ && n_s}.
$$
So, the matrix $A$ can be written in the form
$$
A = \pmatrix{BDB^T & x\\ x^T & 0},
$$
where $x = (0,1,1,\dots,1)^T$.
Let $C$ denote the matrix
$$
C = \pmatrix{1&-1&\cdots & -1\\
&1\\
&&\ddots\\
&&&1}.
$$
We note that
$$
CB = \pmatrix{0\\&I_{s}}.
$$
With that in mind, we find that
$$
\tilde C A \tilde C ^T = \overbrace{\pmatrix{C&0\\0 & 1}}^{\tilde C} \pmatrix{BDB^T & x\\ x^T & 0} \pmatrix{C&0\\0 & 1}^T =
\pmatrix{D & Cx\\ (Cx)^T & 0}.
$$
It would suffice to find the inverse of this "nicer" matrix $\tilde C A \tilde C ^T$, then compute
$$
A^{-1} = \tilde C^T[\tilde C A \tilde C ^T]^{-1}\tilde C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expected number of days before all magical seeds become apple trees This is a question that I came across recently.
At the end of day $0$, $6$ magical seeds are planted. On each day following it, each seed has a chance to magically transform into an apple tree with a probability of $\frac{1}{2}$. The outcomes of seeds are independent of each another.
What is the expected number of days for all six seed to have become apple trees?
My solution:
$E(n)$ - number of expected days from the point that there is only n seed(s) left.
So, $E(1)$ - number of expected days for the last seed to grow.
$E(1) = 1 + \frac{1}{2} E(1) \,or \,E(1) = 2$. This we anyway know from a coin flip analogy.
$E(2) = 1 + \frac{2}{4} E(1) + \frac{1}{4} E(2) \,or\, E(2) = \frac{8}{3}$.
This comes from the fact that if at the end of a day, two seeds are left, I have 3 possible events - i) both seeds become trees the next day ($+1$ day). ii) one seed becomes tree and one seed is left (probability $\frac{2}{4}$). So we further add expected number of days for $E(1)$. iii) None of the seeds become a tree (probability $\frac{1}{4}$). So we add further expected number of days for $E(2)$.
Similarly, $E(3) = 1 + \frac{3}{8} E(1) + \frac{3}{8} E(2) + \frac{1}{8} E(3)$
$E(4) = 1 + \frac{4}{16} E(1) + \frac{6}{16} E(2) + \frac{4}{16} E(3) + \frac{1}{16} E(4)$
$E(4) = 1 + \frac{4}{16} E(1) + \frac{6}{16} E(2) + \frac{4}{16} E(3) + \frac{1}{16} E(4)$
$E(5) = 1 + \frac{5}{32} E(1) + \frac{10}{32} E(2) + \frac{10}{32} E(3) + \frac{5}{32} E(4) + \frac{1}{32} E(5)$
$E(6) = 1 + \frac{6}{64} E(1) + \frac{15}{64} E(2) + \frac{20}{64} E(3) + \frac{15}{64} E(4) + \frac{6}{64} E(5) + \frac{1}{64} E(6)$
This gives me an answer of $E(6) = \frac{55160}{13671}$. However the answer given is $(\log_2 6)$. I do not understand how the answer got into $\log$. When I calculate both, they are not same values.
Also, are there more generic and faster methods that I could use to get to the answer?
| Let $X_i$ the time of growth of the $i$th seed, $X$ the time until all grow.
$X_i\sim Geom (0.5)$ so $\Pr(X_i\leq t)=1-0.5^t$ for $t=1,2,\ldots$.
$X$ is the maximum of $X_1,\ldots,X_6$ so $\Pr(X\leq t)=\prod\limits_{i=1}^6\Pr(X_i\leq t)=(1-0.5^t)^6$ and $\Pr(X=t)=\Pr(X\leq t)-\Pr(X\leq t-1)=(1-0.5^t)^6-(1-0.5^{t-1})^6$
Finally, $E(X)=\sum\limits_{t=1}^\infty t\Pr(X=t)=\sum\limits_{t=1}^\infty t\left((1-0.5^t)^6-(1-0.5^{t-1})^6\right)=4.034$ (used Wolfram; I guess there are general methods to compute it) so I guess you are correct and I don't understand the $\log_2(6)$ either.
It could come from an argument like "each day half of the seeds become trees, so we need $\log_2(6)$ days for all to turn trees" but I don't see why this is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of multivariable function $f(x,y) = {(x^2+y^2)}^{x^2y^2}$ $$f(x,y) = {(x^2+y^2)}^{x^2y^2}$$
I need to find the limit at (0,0) point
I applied the exponent rule and got $$e^{x^2y^2ln(x^2+y^2)}$$
and now with chain rule, I need to find the limit of $${x^2y^2ln(x^2+y^2)}$$
and how? :D
There isn't L'Hôpital's rule for multivariable function, right?
| For $0 < x^2+y^2<1$ we have
$$1 \geqslant (x^2+y^2)^{x^2y^2} \geqslant (x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}$$
and
$$\lim\limits_{x \to 0 \\y \to 0}(x^2+y^2)^{\frac{1}{4}(x^2+y^2)^2}=\lim\limits_{t \to 0+}t^{\frac{1}{4}t^2} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Diophantine equation: $x^4+4=py^4$
find all primes p for which $x^4+4=py^4$ is solvable for integers.
My try: I started out with $p=2$ . LHS must be even, thus $(x^4+4)\mod(16)=4$ similarly
$(2y^4)\mod (16)=${$0,2$}. Thus for $p=2$ there are no solutions.
Now I tried factoring i.e $x^4+4=(x^2+2x+2)(x^2-2x+2)=py^4.$ I dont know what to do next
Any ideas??
| @WhatsUp has the answer, but I wanted to show just for fun how to systematically derive the other case $(X,Y,P)=(0,\pm 1,4)$. It doesn't answer the question because $P=4$ is not prime, but to those who ask "How can we derive the other solution without some kind of trial and error?" well here it is.
Consider the equation $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$
Let $pq=4$ so that $$(p^2-q^2)^2+64=(p^2+q^2)^2$$
Let $p=x^2+y^2$ and $q=x^2-y^2$ so that $p^2-q^2=(2xy)^2$ and $p^2+q^2=2(x^2+y^2)^2$ then $$(2xy)^4+64=4(x^2+y^2)^4$$ or $$(xy)^4+4=\frac 14(x^2+y^2)^4$$
The coefficient is rational, but we want it to be an integer. The requirement is that $x^2+y^2$ is even. So let $x=2s$ and $y=2t$ so it follows $$(4st)^4+4=4(s^2+t^2)^4.$$
Remember the constraint $pq=4$. Since $p=x^2+y^2$ and $q=x^2-y^2$ then it follows $(s^2+t^2)(s^2-t^2)=1$. Since $x$ and $y$ are integers then $s$ and $t$ are integers and so $s^2\pm t^2$ are also integers. Therefore $s^2+t^2=s^2-t^2=1$ or $2s^2=2$ or $s=\pm 1$ and thus $t=0$.
Therefore we derive $(X,Y,P)=(0,\pm 1, 4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of non negative integer solutions of $x+y+2z=20$
The number of non negative integer solutions of $x+y+2z=20$ is
Finding coefficient of $x^{20}$ in
$$\begin{align}
&\left(x^0+x^1+\dots+x^{20}\right)^2\left(x^0+x^1+\dots+x^{10}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{11}}{1-x}\right)\\
=&\left(1-x^{21}\right)^2(1-x)^{-3}\left(1-x^{11}\right)
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-3}-x^{11}(1-x)^{-3}$$
Or, coefficient of $x^{20}$ in $(1-x)^{-3}-$ coefficient of $x^9$ in $(1-x)^{-3}=\binom{22}{20}-\binom{11}{9}=176$
The asnwer is given as $121$. What's my mistake?
EDIT (after seeing @lulu's comment):
Finding coefficient of $x^{20}$ in $$\begin{align}
&\left(x^0+x^1+...+x^{20}\right)^2\left(x^0+x^2+...+x^{20}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{22}}{1-x^2}\right)\\
=&\left(1-x^{21})^2(1-x)^{-2}(1-x^{22})(1-x^2\right)^{-1}
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-2}(1-x^2)^{-1}$$
Not able to proceed next.
| If $z=0$, there are 21 ways $x+y=20$.
If $z=1$, there are 19 ways $x+y=18$.
$$\vdots\hspace{4in}$$
If $z=9$ there are 3 ways $x+y=2$.
If $z=10$ ther is 1 way $x+y =0.$
$$\begin{aligned}\text{number of ways } &= 1+3+\cdots+ 19+21 \\ &= \sum_{k=1}^{11} (2k-1)\\ &=2\sum_{k=1}^{11}k - 11 \\ &= 2\cdot 66 -11 \\ &= 121.\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Infinitely many solutions of the equation $\frac{x+1}{y}+\frac{y+1}{x} = 4$ Prove that there exists infinitely many positive integer solutions in $(x,y)$ to the equation :
$$\frac{x+1}{y} + \frac{y+1}{x} = 4$$
| $$\frac{x+1}{y} + \frac{y+1}{x} = 4$$
$$x^2+x+y^2+y=4xy$$
$$(x+y)^2+(x+y)=6xy$$
Let $X=x+y$ and $Y=x-y$. Then $X^2+X=\dfrac32\left(X^2-Y^2\right)$ or $(X-1)^2-3Y^2=1$.
That's a Pell equation, which has infinitely many solutions:
$X-1=$$1,2,7,26,97,...$ and $Y=$$0,1,4,15,56,...$.
Click on the blue numbers to see more. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$ For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$
$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$
Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$
$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$
I wish to find a proof with $a:\neq {\rm mid}\left \{ a, b, c \right \},$ or another proof$?$
Actually$,$ I also found a proof true for all $a,b,c \in \Big[\dfrac{1}{3},3\Big],$ but very ugly.
After clearing the denominators$,$ need to prove$:$
$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$
but we have$:$
$$f=\dfrac{1}{32} \left( 3-a \right) \left( 3-b \right) \Big( c-\dfrac{1}{3} \Big) +
\left( 3-a \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \left(
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right) \left( 3-c
\right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right) \left( 3-c
\right) \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right) \left( c-\dfrac{1}{3}
\right) \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) + \left( 3-b \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
\left( 3-b \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
\left( 3-a \right) \left( c-\dfrac{1}{3} \right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
\right) \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
\left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right) \left(
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right) \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{2}$$
So we are done.
If you want to check my decomposition$,$ please see the text here.
| I found a better estimation
$$ (a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant \frac{209}{9}.$$
Equality occur when $a=b=3,\,c=\frac 13$ or $a=b=\frac 13,\,c=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.
In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.
Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.
$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$
is what I have got so far, and from here I can’t continue.
Am I on the right direction? Any suggestions or hints will be much appreciated.
| Thanks for all the answers you guys provided, and after spending a large amount of time wandering in this question and understanding these answers, I came up with my solution which I think is valid.
By AM-GM:
$x^2+xy+y^2\\=\frac{x(x+y)}{2}+\frac{y(x+y)}{2}+\frac{x^2+y^2}{2}\\ \geq3\cdot \sqrt[3]{\frac{xy(x+y)^2}{4}\cdot\frac{x^2+y^2}{2}}\\\geq 3\cdot \sqrt[3]{\frac{x^2y^2z^2}{4}}$
Therefore $(x^2+y^2+z^2)^3\geq8\cdot(3\cdot \sqrt[3]{\frac{x^2y^2z^2}{4}} )^3=54x^2y^2z^2$
Now, since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ and $x+y+z=0$,
we have $6(x^3+y^3+z^3)^2=6(3xyz)^2=54x^2y^2z^2$
Hence the inequality is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| Form condition we get $c = \sqrt{a^2-ab+b^2},$ therefore
$$(a-c)(b-c)=a^2+b^2-c(a+b)=a^2+b^2-(a+b)\sqrt{a^2-ab+b^2}$$
$$=-\frac{ab(a-b)^2}{a^2+b^2+(a+b)\sqrt{a^2-ab+b^2}} \leqslant 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Solve differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$ I need to solve this differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$.
My attempt
I found that $y = x^2$ is a solution. Then I tried to put $y = x^2f(x)$, and solved this way:
$$2xf(x) + x^2f'(x) = xf^2(x) + 2xf(x) - x \implies x^2f'(x) = xf^2(x) - x \implies$$
$$xf'(x) = f^2(x) - 1 \implies \frac{df(x)}{f^2(x) - 1} = \frac{dx}{x} \implies$$
$$\frac{1}{2}\ln\left|\frac{f(x) - 1}{f(x) + 1}\right| = \ln|x| + C_* \implies \frac{f(x) - 1}{f(x) + 1} = Cx^2 \implies$$
$$f(x) = \frac{1 + Cx^2}{1 - Cx^2}$$
And we lost a solution $f(x) = -1$. So finally we have
$$y = x^2\frac{1 + Cx^2}{1 - Cx^2}, y = -x^2$$
Now I have 3 questions:
$\quad 1)$ Is my solution correct? I'm not sure that all solutions were found.
$\quad 2)$ When can we use particular solution to find all other solutions? I mean doing something like $y = g(x)h(x)$, where $g(x)$ is a particular solution.
$\quad 3)$ Is there an easier method to solve this equation?
| With a slightly different approach,
$$\frac{y'}x=\left(\frac y{x^2}+1\right)^2-2$$
indeed hints the change of variable
$$y=x^2z,$$ which makes the equation separable,
$$\frac{2xz+x^2z'}x=(z+1)^2-2,$$
$$\frac{z'}{z^2-1}=\frac1x$$
and
$$\text{artanh } z=\log Cx,$$
$$z=\tanh \log Cx.$$
$$y=x^2\sqrt{\frac{1+Cx}{1-Cx}}.$$
There remains to discuss the special case $z=\pm1$, such that $z'=0$, giving the two extra solutions
$$y=\pm x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding all real $(a,b,c)$ satisfying $a+b+c=\frac1{a}+\frac1{b}+\frac1{c}$ and $a^2+b^2+c^2=\frac1{a^2}+\frac1{b^2}+\frac1{c^2}$ I have been trying to find my error in the following question for a while, but am yet to succeed:
Find all triples $(a,b,c)$ of real numbers that satisfy the system of equations:
$$\begin{align}
a+b+c&=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\[6pt]
a^2+b^2+c^2&=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
\end{align}$$
I attempted to do it in the following way:
$(a+b+c)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}$
$a^2+b^2+c^2+2ab+2bc+2ac=\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
$ab+bc+ac=\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}$
From which we have a solution $(|a|, |b|, |c|)=(1,1,1)$
$(a-\frac{1}{a})^2=(\frac{1}{b}-b+\frac{1}{c}-c)^2$
$a^2-\frac{1}{a}^2=\frac{1}{b^2}+b^2+\frac{1}{c^2}+\frac{2}{bc}-\frac{2c}{b}-\frac{2b}{c}+2bc-2$
So we have: $\frac{1}{b^2}+b^2+\frac{1}{c^2}+c^2+\frac{2}{bc}-\frac{2c}{b}-\frac{2b}{c}+2bc-2=\frac{1}{b^2}-b^2+\frac{1}{c^2}-c^2$(combining the second equation in the statement with the one above)
$b^2+c^2+\frac{1}{bc}+bc-\frac{c}{b}-\frac{b}{c}=2$
How can I use the greatest portion possible of what I have done thus far to reach solve the question. My intuition tells me that we will have to use inequalities, however I can't fathom to see how.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Thus, our conditions give: $$uw^3=v^2$$ and $$(9u^2-6v^2)w^6=9v^4-6uw^3.$$
Now, if $v^2=0$ so $u=0,$ which gives $$ab-(a+b)^2=0$$ or $$a^2-ab+b^2=0,$$ which gives $a=b=0,$ which is impossible.
Thus, $u\neq0$, $v^2\neq0$ and $w^3\neq0$, which gives
$$(3u^2-2v^2)\frac{v^4}{u^2}=3v^4-2uw^3$$ or
$$v^6=u^3w^3$$ or
$$v^2=uw,$$ which with $v^2=uw^3$ gives $w^2=1$ and from here $w^3=1$ or $w^3=-1.$
For $w^3=1$ we obtain that $a$, $b$ and $c$ are roots of the equation:
$$x^3-3ux^2+3ux-1=0$$ or
$$(x-1)(x^2+x+1-3ux)=0,$$ which has three real roots for any real $u$ such that $$(1-3u)^2-4\geq0$$ or $$u\in\left(-\infty,-\frac{1}{3}\right]\cup[1,+\infty).$$
Id est, in this case we obtain the following solution: $\left(1,t,\frac{1}{t}\right)$ and any symmetric permutations of this.
The case $w^3=-1$ is a similar.
Also, by your work: $$(ab+ac+bc)abc=a+b+c$$ and by the given $$(a+b+c)abc=ab+ac+bc$$ and since easy to show that $a+b+c=ab+ac+bc=0$ is impossible (see my post),
we obtain $a^2b^2c^2=1.$
After this see my solution again.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$. This is Exercise 14 on page 110 of Analysis I by Amann and Escher.
The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.
If I attempt this, I get the following:
\begin{align*}
X^4 - 2X^3 - X^2 + 2X + 1 &= 0\\
\Rightarrow X^2 - 2X - 1 + \frac{2}{X} + \frac{1}{X^2} &= 0.
\end{align*}
My problem is that I don't understand how to make the suggested substitution. I'm wondering if there is something obvious I'm missing.
I appreciate any help.
| $y^2=x^2-2+1/x^2$, so the expression after multiplying by $1/x^2$ can be rewritten
$$y^2-2y+1=(y-1)^2=0$$
So $y=1$, or $x-1/x=1$ or $x^2-x-1=0$. Solving this quadratic gives $x=\frac{1\pm\sqrt5}2$, and it is easy to check that $x=0$ doesn't satisfy the original equation, so we have found all solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?
| $$\frac{(x-3)(x+5)}{ |x+5|} = (x-3)\frac{(x+5)}{ |x+5|}=(x-3)\text{sign}(x+5)$$
As $x \to -\infty$, then $\text{sign}(x+5)=-1$, So $(x-3)\text{sign}(x+5)=-x+3 \to \infty$.
Addition.
If we take limit point $x=-5$, then of course we come from infinite limit to non existence of limit. First multiplier $x-3 \to -8, x \to -5$, but second as $\text{sign}(x+5)$ have different one-sided limits $\pm 1.$ So $\mp 8$ for product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$ I have to evaluate the following limit
$$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$$
I tried to divide tan in $\frac{sin}{cos}$ or with Hopital but I can't understand where I'm making mistakes.
The final result is:
$\frac{b}{1-ab}$ if $ab \ne 1$
$- \infty$ if $ab=1$ and $a>0$
$+ \infty$ if $ab=1$ and $a<0$
| We have that by $\arctan x+\arctan \frac 1x= \pm \frac \pi 2$
$$x \tan \left(xa+ \arctan \frac{b}{x}\right)=x \tan \left(\pm\frac\pi 2+xa-\arctan \frac{x}{b}\right)=$$
$$=-\frac{x}{ \tan \left(xa-\arctan \frac{x}{b}\right)}=-\frac{x}{ xa-\arctan \frac{x}{b}} \frac{xa-\arctan \frac{x}{b}}{ \tan \left(xa-\arctan \frac{x}{b}\right)}$$
and since by standard limits
$$\frac{xa-\arctan \frac{x}{b}}{ \tan \left(xa-\arctan \frac{x}{b}\right)} \to 1$$
we only need to consider the first term that is for $ab\ne -1$
$$-\frac{x}{ xa-\arctan \frac{x}{b}}=-\frac{1}{ a-\frac1b\frac{\arctan \frac{x}{b}}{\frac x b}} \to\frac{b}{1-ab}$$
by standard limits and for $ab= 1$ by Taylor's expansion $\arctan (x)=x-\frac13x^3+o(x^3)$
$$-\frac{x}{ xa-\arctan \frac{x}{b}}=-\frac{x}{ xa-\arctan (xa)}=-\frac{x}{ xa-xa+\frac13 (xa)^3+o(x^3)}=$$
$$=-\frac{1}{ \frac13 x^2a^3+o(x^2)}\to -\operatorname{sign}(a) \cdot \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0$ How to show that
$$\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2}=0,$$ where $x,y,z>0$.
My attempt:
$$||(x,y,z)|| < \delta \implies |x|, |y|, |z| < \delta$$
$$\left | \frac{xyz}{x^2+y^2+z^2} \right | < \left | \frac{xyz}{x^2}\right | < \frac{\delta^3}{x^2}.$$
Now, I do not know how to proceed, and I think my attempt might be wrong.
| Note that $|x|\le \sqrt{x^2+y^2+z^2}$, $|y|\le \sqrt{x^2+y^2+z^2}$, and $|z|\le \sqrt{x^2+y^2+z^2}$.
Hence, given $\varepsilon>0$,
$$ \left| \frac{xyz}{x^2+y^2+z^2}\right|\le \sqrt{x^2+y^2+z^2}<\varepsilon$$
whenever $\sqrt{x^2+y^2+x^2}<\delta=\varepsilon$.
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Can I solve $\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2}$ by converting to polar coordinates? Is it correct to solve this problem like this?
$$\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2} $$
$$\lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}$$
$$\frac{x^2+y^2}{x^2y^2}=\lim_{r\to\ 0} \frac{1}{r^2\cos^2\theta\sin^2\theta}=\infty\implies \lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}=0$$
| You don't have to make it more complicated. Simply observe that $x^2 \le x^2+x^2y^2+y^2\implies \dfrac{x^2y^2}{x^2+x^2y^2+y^2} \le y^2\implies \text{limit} = 0$ .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z+x) \geqslant 8xyz $$
Also, I got
$$ \frac{x+y+z+(x+y+z)}{4} \geqslant \bigg[ xyz(x+y+z) \bigg] ^{1/4} $$
$$ \therefore x+y+z \geqslant 2 $$
But, I am stuck here. Any hints ?
| Setting $t=xyz(x+y+z),$ then $xy+yz+zx \geqslant \sqrt{3t}=\sqrt 3.$ Using known inequality
$$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx).$$
We have
$$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx)$$
$$\geqslant \frac{8}{9} \cdot \sqrt{3(xy+yz+zx)} \cdot (xy+yz+zx) = \frac{8}{9}\sqrt{3(xy+yz+zx)^3}$$
$$\geqslant \frac{8}{9}\sqrt{3\left(\sqrt{3}\right)^3} = \frac{8\sqrt[4]{3}}{3}.$$
Equality occur when $x=y=z=\frac{1}{\sqrt[4]{3}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\iint_{D} y^{3}\,dx\,dy$ $$\iint_{D} y^{3}\,dx\,dy$$ where $D$ is the domain between $x^2 + y^2 = 6$ circle and the parabola $y=x^2$
Edit: Also, I got the intersection of the curves $(\sqrt2,2)$ and $(\sqrt-3, -3)$
I draw the curves but I got confused at the bounds of the integrals.
| Double-check your work finding the points of intersection, as they should be $\left(-\sqrt{2},2\right)$ and $\left(\sqrt{2},2\right)$. These give $x=-\sqrt{2}$ and $x=\sqrt{2}$ as your lower and upper bounds for $x$, respectively.
Looking at the plot provided, it is clear that the lower and upper bounds for $y$ will be the parabola $y=x^2$ and the upper branch of the circle $x^2+y^2=6$ ($y=\sqrt{6-x^2}$), respectively. Therefore,
\begin{align*}
\iint_{D}y^3 dA &= \int_{-\sqrt{2}}^{\sqrt{2}}\left(\int_{x^2}^{\sqrt{6-x^2}}y^3 dy\right)dx\\
&= \int_{-\sqrt{2}}^{\sqrt{2}}\left(\frac{\left(\sqrt{6-x^2}\right)^4}{4}-\frac{\left(x^2\right)^4}{4}\right)dx\\
&= \frac{1}{4}\int_{-\sqrt{2}}^{\sqrt{2}}\left((6-x^2)^2-x^8\right)dx\\
&= \frac{1}{2}\int_{0}^{\sqrt{2}}(6-12x^2+x^4-x^8)dx\\
&= \frac{1}{2}\left(6\sqrt{2}-4\left(\sqrt{2}\right)^3+\frac{\left(\sqrt{2}\right)^5}{5}-\frac{\left(\sqrt{2}\right)^9}{9}\right)\\
&= \frac{608\sqrt{2}}{45}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Choose four integers from six integers such that $1\cdot \alpha_1 + 2\cdot \alpha_2 + 3\cdot \alpha_3 + 4\cdot \alpha_4 = 5\cdot \beta$. Could you help me with the following task?
Prove that for four numbers form a set $S = \{ \alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6 \}$ of distinct integers it is possible to choose four numbers of $S$ such that $1\cdot \alpha_1 + 2\cdot \alpha_2 + 3\cdot \alpha_3 + 4\cdot \alpha_4 = 5\cdot \beta$, where $\beta \in \mathbb{Z}$
At first, I have tried to edit the expression:
$$\begin{matrix}
1\cdot \alpha_1 + 2\cdot \alpha_2 + 3\cdot \alpha_3 + 4\cdot \alpha_4 &=& 5\cdot \beta \\
2\cdot \alpha_2 + 3\cdot \alpha_3 + 1\cdot \alpha_1 &=& 5\cdot \beta - 5\cdot \alpha_4 +1\cdot \alpha_4 \\
1\cdot \alpha_1 - 1\cdot \alpha_4 + 3\cdot \alpha_3 + 2\cdot \alpha_2 &=& 5\cdot (\beta - a_4) \\
1\cdot \alpha_1 - 1\cdot \alpha_4 + 2\cdot \alpha_2 - 2\cdot \alpha_3 &=& 5\cdot (\beta - a_4 - a_3)\\
\alpha_1 - \alpha_2 + 2(\alpha_2 - \alpha_3) &=&5\gamma
\end{matrix}$$
Now I am stuck.
Source: Sbírka příkladů z matematiky pro vysoké školy (1987) - F. Jirásek; the assignment was translated from Czech into English
| The key observation is that for every integer $k$ and every choice of $x_0,x_1,x_2,x_3,x_4\in S$ you have
$$k\sum x_i+\sum ix_i\equiv \sum ix_{i-k}\pmod{5},$$
if you take the indices mod $5$. So it suffices to find five elements $x_i\in S$ such that
$$\sum x_i\not\equiv0\pmod{5},$$
as then this sum is coprime to $5$, and hence you can choose $k$ such that
$$\sum ix_{i-k}\equiv k\sum x_i+\sum ix_i\equiv0\pmod{5}.$$
Note that it is not possible to find five such elements if and only if all elements are pairwise congruent mod $5$. Of course in this case any choice of four elements will do, as then
$$\sum ix_i\equiv x_1\sum i=x_1(1+2+3+4)\equiv0\pmod{5}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, why $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$? If we have 4 complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, So, how to prove that $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$？
I try to separate $|a(c+d)|+|b(c-d)|$ to $|a||(c+d)|+|b||(c-d)|$ than I get $|(c+d)|+|(c-d)|$. SO, if $c=d, c+d=2c=2b, c-d=0$
Is my idea good?
| $$|a(c+d)| +|b(c-d)|=|a||c+d| +|b||c-d|=|c+d| +|c-d|=\sqrt{(\cos \alpha +\cos \phi )^2 +(\sin\alpha +\sin\phi )^2 } +\sqrt{(\cos \alpha -\cos \phi )^2 +(\sin\alpha -\sin\phi )^2 }=\sqrt{2+ 2\cos \alpha \cos\phi +2\sin\alpha \sin \phi} +\sqrt{2- 2\cos \alpha \cos\phi -2\sin\alpha \sin \phi} =\sqrt{2+2\cos (\alpha -\phi )}+\sqrt{2-2\cos (\alpha -\phi )}=2\sqrt{\frac{1+\cos (\alpha -\phi )}{2}}+2\sqrt{\frac{1-\cos (\alpha -\phi )}{2}}=2\sqrt{\cos^2 \left(\frac{\alpha -\phi }{2}\right)} +2\sqrt{\sin^2 \left(\frac{\alpha -\phi }{2}\right)}=2\left(\left|\cos \left(\frac{\alpha -\phi }{2}\right)\right|+\left|\sin \left(\frac{\alpha -\phi }{2}\right)\right|\right)\leq 2 \sqrt{1^2 +1^2 }\sqrt{\left|\cos \left(\frac{\alpha -\phi }{2}\right)\right|^2+\left|\sin \left(\frac{\alpha -\phi }{2}\right)\right|^2} =2\sqrt{2}$$
where $c=\cos \alpha +i \sin\alpha , d=\cos\phi +i \sin \phi .$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. let's write some terms of the equation:
\begin{align}
x_0&=1 \\[2ex]
x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex]
x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex]
x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex]
\end{align}
So, we can write:
$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$
Here, I'm not sure how to prove it is bounded above.
| The Limit Exists
Cross-multiplication and comparison shows that
$$
\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}\le1+\frac1{2^k}\le\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}\tag1
$$
Then telescoping products lead to
$$
\underbrace{\prod_{k=1}^\infty\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}}_2
\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)
\le\underbrace{\overset{\substack{k=1\\\downarrow\\[6pt]\,}}{\frac32}\overbrace{\prod_{k=2}^\infty\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}}^2}_3\tag2
$$
Therefore, $x_n$ is an increasing sequence that is bounded above by $3$, so
$$
\lim_{n\to\infty}x_n=\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\tag3
$$
exists.
Bounding The Limit
Using $(1)$ and we get
$$
\left(1+\frac1{2^n}\right)\prod_{k=1}^n\left(1+\frac1{2^k}\right)
\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)
\le\frac1{1-\frac1{2^n}}\prod_{k=1}^n\left(1+\frac1{2^k}\right)\tag4
$$
The greater the $n$ used, the tighter the bounds in $(4)$.
The Limit
Using $(4)$ with $n=30$, we get that
$$
2.38423102903137172\color{#C00}{35}\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\le2.38423102903137172\color{#090}{55}\tag5
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Definite Integral involving logarithm and tangent function
Show that
$$\mathcal{I}:=\int_0^{\frac{\pi}2} \log |1-a^2\tan^2\theta| d\theta= \pi\log\sqrt{a^2+1}.$$
I tried to use the substitution $\tan\theta=z$, to get that
$$\mathcal{I}:=\int_0^{\infty} \frac{\log|1-a^2z^2|}{z^2+1}dz$$
This integral is quite similar to this one:
Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$
However note that the sign inside the logarithm term is different and it seems none of the idea is applicable to this setting. May be there is a way to use the above link result to prove my integral. But I dont know.
Edit: Okay I think I have figured out one way to do it via contour integration. We can take a contour that looks like a large semicircle on the upper half plane so that it encloses only the pole at $z=i$. The contour should also have two holes around $z=\pm \frac1a$ to avoid the singularity coming from $\log$. Then Residue Calculus gives the desired result.
It would be nice to see a proof without residue calculus as well.
| Note that we'll need to take the Cauchy Principal Value. Observe that by DUTIS,
\begin{align*}
I(a) &= \int_0^{\frac{\pi}{2}} \ln\left|1-a^2\tan^2\theta\right|\,\mathrm{d}\theta\\
I'(a) &= \int_0^{\frac{\pi}{2}} \frac{\partial}{\partial a} \ln\left|1-a^2\tan^2\theta\right|\,\mathrm{d}\theta\\
&= \int_0^{\frac{\pi}{2}} \frac{2a \tan^2\theta}{a^2\tan^2\theta-1}\,\mathrm{d}\theta\\
&= \frac{2}{a} \int_0^{\frac{\pi}{2}} 1 + \frac{1}{a^2 \tan^2\theta - 1} \,\mathrm{d}\theta\\\\
&= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \int_0^{\cot^{-1} a - \varepsilon} 1 + \frac{1}{a^2 \tan^2\theta - 1} \, \mathrm{d}\theta + \int_{\cot^{-1} a + \varepsilon}^{\frac{\pi}{2}} 1 + \frac{1}{a^2 \tan^2\theta - 1} \, \mathrm{d}\theta \right ).\\
\end{align*}
so it is sufficient to evaluate the integral of $\dfrac{1}{a^2 \tan^2\theta - 1}$. We have
\begin{align*}
\int \frac{1}{a^2 \tan^2\theta - 1} \, \mathrm{d}\theta &= \int \frac{1}{(u^2 + 1)(a^2 u^2 - 1)} \, \mathrm{d}u && \text{using $u = \tan\theta$}\\
&= -\frac{1}{a^2 + 1} \int \frac{1}{u^2 + 1} \, \mathrm{d}u - \frac{a^2}{2(a^2 + 1)} \int \frac{1}{au+1} \, \mathrm{d}u + \frac{a^2}{2(a^2+1)} \int\frac{1}{au-1} \, \mathrm{d}u\\
&= -\frac{\tan^{-1}(u)}{a^2 + 1} - \frac{a \ln |au + 1|}{2(a^2+1)} + \frac{a \ln |au - 1|}{2(a^2+1)} + C\\
&= \frac{a \ln \left | \frac{a \tan\theta - 1}{a \tan\theta + 1} \right | - 2\theta}{2(a^2 + 1)} + C.
\end{align*}
Applying this to our original equation, we have
\begin{align*}
I'(a) &= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \left [\theta - \frac{a \ln \left | \frac{a \tan\theta - 1}{a \tan\theta + 1} \right | - 2\theta}{2(a^2 + 1)} \right ]_0^{\cot^{-1} a - \varepsilon} + \left [\theta - \frac{a \ln \left | \frac{a \tan\theta - 1}{a \tan\theta + 1} \right | - 2\theta}{2(a^2 + 1)} \right ]_{\cot^{-1} a + \varepsilon}^{\frac{\pi}{2}}\right )\\
&= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \frac{2a^2 \cot^{-1} a - 2a^2 \varepsilon - a \ln \left | \frac{-(a^2+1)\tan\varepsilon}{(1-a^2)\tan\varepsilon+2a}\right |}{2(a^2+1)} + \frac{a^2 \pi}{2(a^2+1)} - \frac{2a^2 \cot^{-1} a + 2a^2 \varepsilon - a \ln \left | \frac{(a^2+1)\tan\varepsilon}{(a^2-1)\tan\varepsilon+2a}\right |}{2(a^2 + 1)} \right )\\
&= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \frac{-4a^2 \varepsilon - a \ln \left | \frac{(1-a^2)\tan\varepsilon+2a}{(a^2-1)\tan\varepsilon+2a}\right |}{2(a^2+1)} + \frac{a^2 \pi}{2(a^2+1)}\right )\\
&= \frac{a \pi}{a^2 + 1}\\
\implies I(a) &= \int \frac{a\pi}{a^2+1} \, \mathrm{d}a\\
&= \pi \ln\sqrt{a^2 + 1} + C.
\end{align*}
Finally, observe $I(0) = 0$, so $C = 0$. Hence, $I(a) = \pi \ln\sqrt{a^2 + 1}$.
| {
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"url": "https://math.stackexchange.com/questions/3839712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx
+ \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\
&= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right)
-2\operatorname{Im} \operatorname{Li}_3(-1-i)
\end{align}
which is equal to $ \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$.
It is perhaps unnecessary, though, to resort to evaluation in complex space. I would like to work out an elementary derivation of this integral result.
| Here is a solution based on real methods. Note that $\frac{1}{(x+1)^2+1}=\frac{x^2-2x+2}{x^4+4}$. Then
\begin{align}
&\int_0^\infty \frac{\ln^2x}{(x+1)^2+1}dx \\
= & \int_0^\infty {\frac{x^2\ln^2x }{x^4+4}}\overset{x^2=1/t^2}{dx}
-2\int_0^\infty {\frac{x\ln^2x }{x^4+4}} \overset{x^2=2t}{dx}
+ 2\int_0^\infty { \frac{\ln^2x }{x^4+4} } \overset{x^2=2t^2}{dx} \\
= &
-\frac18\ln^22\int_0^\infty \frac{1}{t^2+1}dt
-\frac18\int_0^\infty \frac{\ln^2t}{t^2+1}dt \\
& +\frac{1}{2\sqrt2} \ln^22 \int_0^\infty \frac{1}{t^4+1}dt
+ \sqrt2 \int_0^\infty \frac{\ln^2 t}{t^4+1}dt \\
= &
-\frac{\ln^22}8\left(\frac\pi2\right)
-\frac18\left(\frac{\pi^3}8\right) +\frac{\ln^22 }{2\sqrt2}\left(\frac\pi{2\sqrt2}\right)
+ \sqrt2 \left(\frac{3\pi^3\sqrt2}{64}\right) \\
=& \frac{5\pi^3}{64}+\frac\pi{16}\ln^22
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Limit of sequence involving reciprocal of the sum of the first squares Let $(x_n)$ a sequence defined by $x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}.$ What is $\lim_{n\to \infty}x_n$?
My idea: $1^2+\dots+k^2=\frac{k(k+1)(2k+1)}{6}<\frac{k(k+1)(k+2)}{3}$. Thus
$$x_n>3\sum_{k=1}^n\frac{1}{k(k+1)(k+2)}=3\left(\frac 12 -\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{4}\right).$$
I couldn't continue any further. My assumption is that $x_n\to 2$ or something like that. Please help me solve it! Thank you!
| As you properly wrote$$x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}=\sum_{k=1}^n \frac{6}{k (k+1) (2 k+1)}$$ Using partial fraction decomposition, as @Ben Grossmann already did it,
$$x_n=\sum_{k=1}^n \frac{6}{k}+\sum_{k=1}^n \frac{6}{k+1}-\sum_{k=1}^n \frac{24}{2 k+1}$$
$$x_n=(6 H_n)+(6 H_{n+1}-6)-(12 H_{n+\frac{1}{2}}-24+24 \log (2))$$ Using the asymptotics
$$x_n=(18-24 \log (2))-\frac{3}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
expected value rule of Uniform random variable Let be a uniform random variable on the range $\{−1,0,1,2\}$. Let $=^4$. Use the expected value rule to calculate $\mathbb{E}[Y]$.
Why isn’t the answer $1/4 \cdot(-1)^4 +1/4\cdot (0)^4+ 1/4 \cdot(1)^4 + 1/4 \cdot(2)^4$
| The rv X is the following
$$X =
\begin{cases}
\frac{1}{4}, & \text{if $x=-1$} \\
\frac{1}{4}, & \text{if $x=0$} \\
\frac{1}{4}, & \text{if $x=1$} \\
\frac{1}{4}, & \text{if $x=2$}
\end{cases}$$
If you transform $Y=X^4$ when $X=\pm1$ you get that $Y=1$ in both cases, so you have
$$Y =
\begin{cases}
\frac{1}{4}, & \text{if $y=0$} \\
\frac{1}{2}, & \text{if $y=1$} \\
\frac{1}{4}, & \text{if $y=16$}
\end{cases}$$
Thus
$$\mathbb{E}[Y]=\frac{1}{2}+4=\frac{9}{2}$$
Next time, please use MathJax to format your formulas; it is easy and your message is more readeable
In any case, your answer is correct...
$$\mathbb{E}[Y]=(-1)^4\times\frac{1}{4}+(0)^4\times\frac{1}{4}+(1)^4\times\frac{1}{4}+(2)^4\times\frac{1}{4}=\frac{9}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Hint: Start by dividing the whole equation by $a$
At first I have tried solving the equation without using the hint provided in my exercise and directly applying completing square, I get $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-c}{a}}$. So if I am to use the hint, I obtain the appropriate answer. But I wonder if I am asked the same question in my exam where the hint will not be provided then how am I supposed to answer.
I would like to know how one should approach this kind of question and how do I realise when to divide the whole equation with in this case $a$ or is there any other ways so that I can avoid dividing the whole equation by $a$.Thanks in advance for any help you are able to provide!
EDIT: Here's my steps. Please see where have I done wrong.
\begin{align}
ax^2+2bx+c&=0 \\
a\left[\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a^2}\right] + c&=0 \\
a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a} + c&=0 \\
\left(x+\frac{b}{a}\right)^2&=\left(\frac{b^2}{a}-c\right)\left(\frac{1}{a}\right)\\
\left(x+\frac{b}{a}\right)^2&=\frac{a(b^2-c)}{a^2}\\
\left(x+\frac{b}{a}\right)^2&=\frac{b^2-c}{a} \\
x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-c}{a}} \\
\implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-c}{a}}\\
\end{align}
| As you did, complete the square by first taking out a factor of $a$, where $a\neq 0$ since the equation is a quadratic.
\begin{align}
ax^2+2bx+c&=0 \\
a(x^2+2\frac{b}{a}+\frac{c}{a})&=0 \\
a\left[(x+\frac{b}{a})^2-\frac{b^2}{a^2}+\frac{c}{a}\right]&=0 \\
(x+\frac{b}{a})^2&=\frac{b^2-ac}{a^2} \\
x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-ac}{a^2}} \\
\implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-ac}{a^2}}\\
\end{align}
Maybe you made a mistake somewhere?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| We have that
$$\cos^6x + \sin^6x =(\cos^2x +\sin^2x)(\cos^4x -\cos^2x\sin^2x +\sin^4x)=$$
$$=\cos^4x -\cos^2x\sin^2x +\sin^4x$$
and
$$\cos^4x -\cos^2x\sin^2x +\sin^4x=1-3\cos^2x\sin^2x=1-\frac34\sin^2(2x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 4
} |
$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
Find all pairs of integers (m, n) such that the integers $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
I assumed $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ to be $a^3,b^3$ and $c^3$ respectively and deduced that $6(2mn+4)=a^3+6b^3,$ $12(n-m)=a^3-6b^3+36$ but
I can't solve it further. Please help.
| Hint: If I am not mistaken, we have$$m=\frac{24-a^3+6b^3}{12} $$
$$n=\frac{12-a^3+18b^3-12c^3}{12} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.
My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.
| $$2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xy^2z^2$$
$$2xy^2z^2=2yz(x+y+z)=2xyz + 2y^2z+2yz^2$$
$$\implies 2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xyz+2y^2z+2yz^2$$
$$\implies 2y(1-z^2)(1-x^2)=2y-2yz^2-2yx^2+2xyz+2z^2x+2zx^2$$
$$\implies 2z(1-x^2)(1-y^2)=2z-2zx^2-2zy^2+2xyz+2x^2y+2xy^2$$
Summing up gives $2(x+y+z+3xyz)=8xyz$, which reduces to the equality.
I wish to see a simpler proof as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Compute $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$ Problem Statement:
Compute:
$$\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$$
First we note that $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}< \sum_{k\geq0}\frac{2^k}{5^{2^k}}< \sum_{k\geq0}\frac{2^k}{5^{k}}< \infty$, so that this sum converges. We can now do the following manipulations:
\begin{align}
\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}&= \sum_{k\geq0}\frac{\frac{2^k}{5^{2^k}}}{1+ \frac{1}{5^{2^k}}}\\
&= \sum_{k\geq0}\frac{2^k}{5^{2^k}}\left(\sum_{j\geq0}\frac{(-1)^j}{(5^{2^k})^j}\right)\quad \quad \text{geometric series with $r = -\frac{1}{5^{2^k}}$}\\
&= \sum_{k\geq0}2^k\left(\sum_{j\geq0}\frac{(-1)^j}{(5^{2^k})^{j+1}}\right)\\
&=\sum_{j\geq0}(-1)^j\left(\sum_{k\geq0}\frac{2^k}{(5^{j+1})^{2^k}}\right) \quad \quad \text{swap sums due to convergence}
\end{align}
I was hoping that after swapping sums the new summation would be easier to evaluate. However, this doesn't appear to be so. Any hints would be greatly appreciated.
| This is a beautiful sum. Note:
$$\frac 1 {n-1} -\frac1{n+1} =\frac {2}{n^2-1}$$
Thus:
$$\begin{align}\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1} &=\frac {2}{n^2-1}-\frac {2}{n^2+1}\\& = \frac {2^2}{n^{2^2}-1}\\\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1}-\frac {2^2}{n^{2^2}+1} &= \frac {2^2}{n^{2^2}-1}-\frac {2^2}{n^{2^2}+1}\quad\quad\quad\ \ \\&= \frac {2^3}{n^{2^3}-1}\end{align}$$
Now complete the pattern.
EDIT: After a brief discussion with @Integrand, we concluded that this is much easier to explain with a telescoping sum, and by observing:
$$\frac {2^k} {n^{2^k}+1} = \frac{2^k} {n^{2^k}-1} - \frac{2^{k+1}} {n^{2^{k+1}}-1}$$
Now
$$\begin{align}\sum_{k=0}^N\frac {2^k} {n^{2^k}+1} &= \sum_{k=0}^N\left(\frac{2^k} {n^{2^k}-1} - \frac{2^{k+1}} {n^{2^{k+1}}-1}\right)\\&= \sum_{k=0}^N\frac{2^k} {n^{2^k}-1} - \sum_{k=1}^{N+1}\frac{2^{k}} {n^{2^{k}}-1}\\&=\frac{1}{n-1}-\frac{2^{N+1}}{n^{2^{N+1}}-1}\end{align}$$
and we see that $\dfrac{2^{N+1}}{n^{2^{N+1}}-1} \to 0$ as $N\to \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Question about complex integrals I have two questions about integrals.
1.) $$\frac{1}{2\pi i} \int_{\gamma} \frac{dw}{\sin{\frac{1}{w}}}$$ where $\gamma$ is the circle $|w| = \frac{1}{5}$.
2.) $$\int_{\gamma_{a}} \frac{z^{2} + e^{z}}{z^{2}(z-2)}dz$$
where $\gamma_{a}$ is the positively oriented circle $|z| = a, a >0, a \neq2.$
For problem $1$, I concluded that the answer is $\frac{1}{6}$ and I got this answer by computing the residue.
$$\frac{1}{\sin{\frac{1}{w}}} = \frac{1}{\frac{1}{w}} + \frac{\frac{1}{w}}{6} + \frac{7 (\frac{1}{w})^{3}}{360}+...$$
$$ \frac{1}{\sin{\frac{1}{w}}} = w + \frac{1}{6w} + \frac{7}{360w^{3}} +... $$.
But, for the second problem I'm a bit confused about how to go about solving it. If I assume that $a < 2$ then I can do this:
$$ \frac{z^{2} + e^{z}}{z^{2}(z-2)} = \frac{z^{2} +1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+...}{z^{2}(z-2)}$$
$$ \frac{z^{2} + e^{z}}{z^{2}(z-2)} = \frac{3}{2(z-2)}+\frac{1}{z^{2}(z-2)}+ \frac{1}{z(z-2)} + \frac{z}{6(z-2)} + ...$$
hence, the residue is $\frac{3}{2}$. I'm not sure if this is correct so I would appreicate someone looking over this calculation. Also, I don't know what to do when $a > 2$.
EDIT: From the suggestion in the comments, I did partial fractions on some of the terms in problem 2.
$$ \frac{1}{z^{2}(z-2)} = -\frac{1}{4z} -\frac{1}{2z^{2}} + \frac{1}{4(z-2)} $$
$$ \frac{1}{z(z-2)} = -\frac{1}{2z} =\frac{1}{2(z-2)} $$
So, the residue contributions are ($\frac{3}{2} - \frac{1}{4} + \frac{1}{4} -\frac{1}{2} +\frac{1}{2} = \frac{3}{2}.$) Is this correct? Hence, the answer would be the same as it was when $a < 2 .$
| For the first problem, we enforce the transformation $z\mapsto \frac1w$ to obtain
$$\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=\oint_{|z|=5}\frac1{\sin(z)}\frac1{z^2}\,dz\tag1$$
There are 3 poles of $\frac{1}{z^2\sin(z)}$ that are inside the circle $|z|=5$. These include a third order pole at $z=0$ and simple poles at $z=\pm \pi$.
The associated residues are $1/6$, $-1/\pi^2$, and $-1/\pi^2$. Putting it all together, we find that
$$\frac1{2\pi i }\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=\frac16-\frac2{\pi^2}$$
__
Alternatively, we can evaluate the integral on the left-hand side of $(1)$ directly. The poles of $\frac1{\sin(1/w)}$ are located at $w=1/(n\pi)$ for $n\in \mathbb{Z}$. All of these poles for $n>1$ are contained inside the circle $|w|=1/5$.
The residue of the pole at $w=1/(n\pi)$ is $(-1)^{n-1}/(n^2\pi^2)^2$. Hence, we see that
$$\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=2\sum_{n=2}^\infty \frac{(-1)^{n-1}}{n^2\pi^2}=\frac16-\frac2{\pi^2}$$
which agrees with the previous result.
For the second problem, there is a second order pole at $z=0$ and a simple pole at $z=2$.
The residue at $z=0$ is
$$\lim_{z\to0}\frac{d}{dz}\frac{z^2+e^z}{(z-2)}=-3/4$$
The reside at $z=2$ is $\frac{4+e^2}{4}$.
And you can finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| In this case, the rationalisation of the denominator not only simplifies the expression a bit, but reveals a rate of change, which makes the determination of the limit quite fast:
$$\frac{(\sqrt{6-x}-2)\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}}{2-x}=\underbrace{\frac{\sqrt{6-x}-2}{2-x}}_{\textstyle\downarrow\atop \textstyle(-\sqrt{6-x})_{x=2}'}\cdot(\underbrace{\sqrt{3-x}+1}_{\textstyle\downarrow\atop\textstyle2}), $$
hence the limit is $\;-\biggl(\dfrac{1}{2\sqrt{6-x}}\biggr)_{\!x=2}\cdot 2=-\dfrac12.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from Further Mathematics(9231) Paper 1, Question 1, 2009. I tried to solve it but was unable to figure it out, especially how to find the value of $\alpha^6+\beta^6+\gamma^6+\delta^6$. Could anyone try to solve this question and explain how they got the value?
| Making the given substitution yields $y^{4/3}=y+1$, or $y^4=(y+1)^3$ or $y^4-y^3-3y^2-3y-1=0$. The roots of this polynomial are $\alpha^3,\beta^3,\gamma^3,\delta^3$, so we can work out
$$\alpha^6+\beta^6+\gamma^6+\delta^6=(\alpha^3+\beta^3+\gamma^3+\delta^3)^2-2(\alpha^3\beta^3+\cdots)=1^2-2(-3)=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $\frac{\sin(2a-b)}{\sin b}=1+x$ prove that...
if $\frac{\sin(2a-b)}{\sin b}=1+x$ prove that $$\frac{\cos b}{ \cos a}=1+\frac{1}{2}x\tan^2 a$$ approximately .(Take $x$ to be very small)
Attempt by given eq. $$\sin 2a \cos b-\sin b \cos 2a=\sin b +x\sin b$$
or $$2\sin a \cos a \cos b -\sin b (1-2\sin^2 a)=\sin b +x\sin b$$ .
or $$\frac{\cos b}{\cos a}=\frac{1}{2\sin b}\sec^2 a \left(\sin b (1-2\sin^2 a)+\sin b +x\sin b \right)$$ .
How do i proceed?
| $\begin{align}\frac{\cos b}{\cos a} &= \frac{\sec^2a}{2}\left[1-2\sin^2a+(1+x)\right] = \frac{\sec^2a}{2} - \tan^2a +\frac{1+x}{2}\sec^2a
\\&= \left(1+\frac x2\right)\sec^2a -\tan^2a = \left(1+\frac x2\right)(1+\tan^2a)-\tan^2a = \left(1+\frac x2\right)+\frac x2\tan^2a \\&\approx1+\frac x2\tan^2a\end{align} $
As $x$ is very small, you may take $1+\frac x2 \approx 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Subsets and Splits