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Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here. Hints?
| $$f(x,y)=x^2+y^2+(1-x-y)^2$$
$$\frac{\partial f}{\partial x}=2x+2(1-x-y)(-1)=0\tag 1$$
$$\frac{\partial f}{\partial y}=2y+2(1-x-y)(-1)=0\tag 2$$
by subtracting
$$x=y$$
substitute in 1 to get
$$x=\frac{1}{3}$$
so
$$y=\frac{1}{3}$$
and then $$z=1-x-y=\frac{1}{3}$$
to check the critical point, we can reduce the original function, which has three variables, to one varible as follow
$$f(x)=x^2+x^2+(1-x-x)^2=2x^2+(1-2x)^2$$
it is very easy to find the critical point if there is one variable
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 10,
"answer_id": 1
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How would you solve for x in this case using the trig identities?
Solve for all values of $x$:
$$\cos 2x = 2\sin x$$
$$1 - 2\sin^2x - 2\sin x = 0$$
$$-2\sin^2x - 2\sin x + 1 = 0$$
$$\sin^2x - 2\sin x - 2 = 0$$
How would you factor this above to solve for $x$?
| $$\cos 2x = 2\sin x$$
or, $$1-2\sin^2 x = 2\sin x$$
or, $$2\sin^2 x + 2\sin x - 1 = 0$$
So $$\sin x = \frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}=\omega, \omega^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int_{C_N} \frac{dz}{z^2\sin(z)}$ complex integral, problem with residues
Let $C_n$ be the rectangle, positively oriented, which sides are in the lines $$x=\pm(N+\dfrac{1}{2})\pi~~~y=\pm(N+\dfrac{1}{2})\pi$$
with $N\in\mathbb{N}$.
Prove that
$$ \displaystyle\int_{C_N} \dfrac{dz}{z^2\sin(z)}=2\pi i\left[ \frac{1}{6}+2\sum_{n=0}^N \dfrac{(-1)^n}{n^2\pi^2}\right] $$
what I did
If $f(z)=\dfrac{1 }{z^2\sin(z)}$ then $f$ has a pole of order 3 in $z=0$ so the value of the integral is equal to $2\pi i (\text{sum of residues}) $ so, it gives me the factor $\frac{1}{6}$. then every point $z=n\pi, ~~ n=0,\pm1,\pm2,\cdots,\pm N$ are simple poles. my question is:
How i calculate the residues in those points?
What am I missing? Am I doing it wrong?
$$z^2\sin(z)=\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{z^{2n+3}}{(2n+1)!} $$
| The function
$$
f(z) = \frac{1}{z^2 \sin z} = \frac{1/z^2}{\sin z}
$$
has simple poles at $z = n\pi$, where $n \in \mathbb{Z}$, $n \neq 0$, so
$$
\newcommand{\Res}{\operatorname*{Res}\limits} \Res_{z=n\pi} f(z) = \frac{1/z^2}{\cos z} \Big|_{z=n\pi} = \frac{(-1)^n}{n^2\pi^2}.
$$
In particular, the residue at $n\pi$ is equal to the residue at $-n\pi$, so it's enough to sum over the positive integers up to $N$ and multiply by $2$. This leaves us with the residue at $z=0$. Here
\begin{align}
f(z) = \frac{1}{z^2\sin z} &= \frac{1}{z^2 ( z - \frac{z^3}{3!} + \cdots )} \\
&=\frac{1}{z^3} \cdot \frac{1}{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots} \\
&= \frac{1}{z^3} \left(1 + \left( \frac{z^2}{3!} - \frac{z^4}{5!} - \cdots \right) - \left( \frac{z^2}{3!} - \frac{z^4}{5!} - \cdots \right)^2 + \cdots \right) \\
&= \frac{1}{z^3} + \frac{1}{6z} + \text{terms of positive degree}
\end{align}
where the series expansion is valid near $z=0$ (we use the Maclaurin series for $\sin z$ and the geometric series, keeping track of terms of degree $3$ or less). Hence
$$
\Res_{z=0} f(z) = \frac16,
$$
which gives you the final term in the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does a quaternion rotation matrix simplify to this? I'm reading Ken Shoemake's explanation of quaternions in David Eberly's book Game Physics. In it, he defines the rotation matrix for a quaternion $q = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} + w\mathbf{1}$ to be the product of two matrices:
$\begin{align}
\mathbf{Q\overleftarrow{Q}}^T &=
\begin{bmatrix}
w & -z & y & x \\
z & w & -x & y \\
-y & x & w & z \\
-x & -y & -z & w
\end{bmatrix}
\begin{bmatrix}
w & -z & y & -x \\
z & w & -x & -y \\
-y & x & w & -z \\
x & y & z & w
\end{bmatrix} \\
&=
\begin{bmatrix}
\mathbf{R} & 0 \\
0 & xx + yy + zz + ww
\end{bmatrix}
\end{align}
$
where
$
\mathbf{R} = \begin{bmatrix}
ww - zz - yy + xx & -wz -zw + yx + xy & wy + zx + yw + xz \\
zw + wz + xy + yx & -zz + ww - xx + yy & zy - wx - xw + yz \\
-yw + xz - wy + zx & yz + xw + wx + zy & -yy - xx + ww + zz
\end{bmatrix}
$
I can follow this so far. $\mathbf{R}$ is easy to see from the rules of matrix multiplication. But then the book goes on to say that because the bottom right corner of the block matrix (ie, $xx + yy + zz + ww$) is $\textrm{N}(q)$, and because we're concerned with normalized rotations $\textrm{N}(q) = 1$, $\mathbf{R}$ can be simplified to
$
\begin{bmatrix}
1 - 2(z^2 + y^2) & 2(xy - wz) & 2(zx + wy) \\
2(xy + wz) & 1-2(x^2 + z^2) & 2(yz - wx) \\
2(zx - wy) & 2(yz + wx) & 1 - 2(x^2 + y^2)
\end{bmatrix}
$
I don't understand why the elements of the main diagonal can be simplified this way. How does $x^2 + w^2 - y^2 - z^2 = 1 - 2(z^2 + y^2)$? Does this only work because $x^2 + y^2 + z^2 + w^2 = 1$?
| Exactly, you can easily see that
$\begin{align}x^2 + w^2 - y^2 - z^2 &= x^2+w^2+(y^2-2y^2)+(z^2-2z^2) \\
&= \underbrace{x^2+w^2+y^2+z^2}_{=1} - 2(z^2+y^2) \\
&= 1 - 2(z^2 + y^2)\end{align}$
It works exactly the same way for all the entries of the diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Rewrite the equation $(x-a)^2 + (y-b)^2 = r^2$ to make $y$ a function of $x$ I'm trying hard to figure out how
$(x-a)^2 + (y-b)^2 = r^2$
can be written as
$y = b + \sqrt{r^2 - (x-a)^2}$.
My book says that
you’ll want to have $y$ as a function
of $x$.
| $$(x-a)^2+(y-b)^2=r^2$$ implies: $$(y-b)^2=r^2-(x-a)^2$$ Taking the square root of both sides then yields: $$y-b=\sqrt{r^2-(x-a)^2}$$ and adding $b$ to both sides of the equation: $$y=b+\sqrt{r^2-(x-a)^2}$$
Note that when we took the square root, we could have also taken the negative square root, since $x^2=(-x)^2$. So we could also have written: $$y=b-\sqrt{r^2-(x-a)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Complex Numbers (Locus) Find the Cartesian equation of $z$ described by $$\arg\left(\frac{z-2}{z+5}\right)=\frac{\pi}{4}$$
So what I have done is let $z = x+iy$
$$\frac{z-2}{z+5} $$
$$ \frac{x+iy-2}{x+iy+5} $$
$$ \frac{x-2+iy}{x+5+iy} \frac{x+5-iy}{x+5-iy} $$(rationalizing here)
$$ \frac{(x-2)(x+5)+y^2+iy(x+5)-iy(x-2)}{(x+5)^2+y^2}$$
As $\arg\left(\frac{z-2}{z+5}\right)=\frac{\pi}{4}$
$$\arctan(\frac{Im(z)}{Re(z)}) = \frac{\pi}{4}$$
$$\frac{Im(z)}{Re(z)} = \tan(\frac{\pi}{4}) = 1$$
Therefore $$\frac{Im(z)}{Re(z)} = 1$$
Now comparing $im(z)$ and $re(z)$
$$\frac{y(x+5) - y(x-2)}{(x-2)(x+5)+y^2} = 1$$
$$7y = x^2 + 5x - 2x - 10 + y^2$$
$$x^2 + 3x - 10 + y^2 - 7y = 0$$
Now completing the square
$$(x+1.5)^2 + (y - 3.5)^2 = 14.5$$
Therefore locus is a circle with a centre of $(-1.5$ and $3.5i)$ with radius of $\sqrt{14.5}$
Is this correct?
| More generaly you can understand the underlying geometry by considering the general case.
Let's define points $P_{j}(z_j)$ where $j=1,2$ in the Argand plane, then if $Q(z)$ is a variable point then such that the angle suntended by the segment $P_1 P_2$ is $\zeta$ then we've the representation $\left(\frac{z-z_1}{z-z_2} \right) =\rho \,\,e^{i\zeta}$ for some $\rho \in \mathbb{R}$ , therefore it's easy to see that $\arg\,\left(\frac{z-z_1}{z-z_2} \right)=\zeta$,
Thus in the given case we've a circle at such that numbers $-5$ and $2$ in the Argand plane subtend a right angle at the centre, then using geometrical properties of circle, specifically using the result that centre of the circle lies along the perpendicular bisector of the chords and using Pythagorean theorem we end up with the centre corresponding to the number $-\frac{1}{2}+ \frac{3}{2}i$ in the Argand plane, which is it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
finding equation of the normal Find the equation of the normal to the curve $y= 4/x$ at point where $y=1/2$. Find the coordinates of the point where this normal cuts the x axis. I know that the curve cuts the x axis when $y=0$ and I tried doing that for starters.
| Notice, setting $y=\frac{1}{2}$ in the equation $y=\frac{4}{x}$, $$\frac{1}{2}=\frac{4}{x}\implies x=8$$ Thus the normal is drawn at the point $\left(8, \frac{1}{2}\right)$ to the curve.
Now, the slope of tangent to the curve $y=\frac{4}{x}$ $$\frac{dy}{dx}=-\frac{4}{x^2}$$
Now, the slope of the normal at the point $\left(8, \frac{1}{2}\right)$ $$=\frac{-1}{\frac{dy}{dx}}\mathbb|_{\left(8, \frac{1}{2}\right)}=\frac{-1}{-\frac{4}{(8)^2}}=16$$
hence, the equation of the normal at the point $\left(8, \frac{1}{2}\right)$ is given as $$y-\frac{1}{2}=16(x-8)$$
$$\color{red}{32x-2y-255=0}$$
setting $y=0$ in the equation of the normal, one should get
$$32x-2(0)-255=0\implies x=\frac{255}{32}$$
hence the point of intersection of the normal with the x-axis is $\color{blue}{\left(\frac{255}{32}, 0\right)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confusion in evaluating the limit $\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}$ I was solving a question related to functions and i come across a limit which i cannot understand.The question is
If $a$ and $b$ are positive real numbers such that $a-b=2,$ then find the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$
First i found the domain of definition of the function in question $\sqrt{x^2+ax}-\sqrt{x^2+bx}$.The domain is $x\geq0 \cup x\leq -a$.
Then i found the horizontal asymptotes as $x\to \infty$.
$\lim_{x\to\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{x\to\infty}\frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$
$=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{2}{2}=1$
Then i found the horizontal asymptotes as $x\to -\infty$.
$\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{x\to-\infty}\frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\lim_{x\to-\infty}\frac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$
$=\lim_{x\to-\infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{2}{2}=1$
But when i drew the graph using graphing calculator,the horizontal asymptote as $x\to-\infty$ was $-1$I do not understand what mistake i made in calculating the limit as $x\to -\infty$
Then i guessed i should have made the substitution $x=-t$ and
$\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{t\to\infty}\sqrt{t^2-at}-\sqrt{t^2-bt}=\lim_{t\to \infty}\frac{(b-a)t}{\sqrt{t^2-at}+\sqrt{t^2-bt}}=\lim_{t\to\infty}\frac{-2t}{\sqrt{t^2-at}+\sqrt{t^2-bt}}$
$=\lim_{t\to\infty}\frac{-2}{\sqrt{1-\frac{a}{t}}+\sqrt{1-\frac{b}{t}}}=\frac{-2}{2}=-1$
I want to ask why the answer came wrong in the first method and correct in the second method.
Is it always necessary to put $x=-t$ and change the limit to plus
infinity while calculating the limit as $x\to-\infty$
Please help me.Thanks.
| Notice, here is an easier method to solve
let $x=-\large \frac{1}{t}$, $$\lim_{x\to -\infty}(\sqrt{x^2+ax}-\sqrt{x^2+bx})$$
$$=\lim_{t\to 0^+}\left(\sqrt{\left(\frac{-1}{t}\right)^2+a\left(\frac{-1}{t}\right)}-\sqrt{\left(\frac{-1}{t}\right)^2+b\left(\frac{-1}{t}\right)}\right)$$
$$=\lim_{t\to 0^+}\frac{1}{t}\left(\sqrt{1-at}-\sqrt{1-bt}\right)$$
$$=\lim_{t\to 0^+}\frac{\left(\sqrt{1-at}-\sqrt{1-bt}\right)\left(\sqrt{1-at}+\sqrt{1-bt}\right)}{t\left(\sqrt{1-at}+\sqrt{1-bt}\right)}$$
$$=\lim_{t\to 0^+}\frac{1-at-1+bt}{t\left(\sqrt{1-at}+\sqrt{1-bt}\right)}$$
$$=\lim_{t\to 0^+}\frac{b-a}{\sqrt{1-at}+\sqrt{1-bt}}$$
$$=\frac{(b-a)}{1+1}=\color{red}{\frac{b-a}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Find $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$ using the function $f(x)=\frac{1+x}{(1-x)^3}$ Find $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$ using the function $f(x)=\frac{1+x}{(1-x)^3}$
Power series representation of $f(x)$ is $\sum\limits_{n=1}^{\infty}n^2x^{n-1}$.
Question: Why is $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}=\frac{1}{2}f\left(\frac{1}{2}\right)$? Doesn't it has to be $f\left(\frac{1}{2}\right)$?
| It sometimes helps to carefully write out the first few terms:
$$f(x)=\sum_{n=1}^\infty n^2x^{n-1}=1^2x^0+2^2x^1+3^2x^2+\cdots=1+4x+9x^2+\cdots$$
while
$$\sum_{n=1}^\infty{n^2\over2^n}={1\over2}+{4\over2^2}+{9\over2^3}+\cdots={1\over2}\left(1+{4\over2}+{9\over2^2}+\cdots\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577478",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solve the equation: $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$ How do I solve this equation:
$$ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$$
| $\sin x - 3\sin2x + \sin 3x = \cos x -3\cos2x + \cos 3x$
OR
$\sin x + \sin 3x - 3\sin 2x = \cos x + \cos 3x - 3\cos 2x$
$\implies 2\sin 2x \cdot \cos x - 3\sin 2x = 2\cos 2x \cdot \cos x - 3\cos 2x $
$\implies \{\sin 2x - \cos 2x\} \{ 2\cos x - 3\} = 0 $
$\implies \{\sin 2x - \cos 2x\} = 0 $ since $\cos x \neq 3/2 $
$\implies \cos 2x = \cos (\pi/2 - 2x) $
$\implies 4x = 2n\pi + \pi/2 $
$\implies x = \pi/8 \implies n\pi/2 \pm {\pi\over 8} $ in general form .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
show that $a_1+a_2+a_3+a_4=8$ and that $64a_1+27a_2+8a_3+a_4=729$ given the following
Consider the sistem of equations: $$\begin{cases}
a_1+8a_2+27a_3+64a_4=1 \\ 8a_1+27a_2+64a_3+125a_4=27 \\
27a_1+64a_2+125a_3+216a_4=125\\ 64a_1+125a_2+216a_3+343a_4=343\\
\end{cases} $$ These four equations determine $a_1,a_2,a_3,$and $a_4$.
Show that $$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3 \equiv
(2x+1)^3,\tag 1$$i.e.,these two polynomials are identically the same.
Use this to show that $a_1+a_2+a_3+a_4=8$ and that
$64a_1+27a_2+8a_3+a_4=729$.
I was able to prove the first part but I am having trouble showing that $a_1+a_2+a_3+a_4=8$.
My approach was to calculate equation $(1)$ when $x=0,-1,-2,-3$ ,thus yielding:
$a_1=1-(8a_2+27a_3+64a_4) $
$a_2=-1-(8a_3+27a_4)$
$a_3=-27-(-a_1-8a_4)$
$a_4=-125-(-8a_1-a_2)$
So when I add them I get $$a_1+a_2+a_3+a_4=-27-125+(9a_1-7a_2-35a_3-91a_4)$$
The problem is that I don't see how to simplify $9a_1-7a_2-35a_3-91a_4$ to get that it equals $160$.
Can you guys give me some hint ?
| Consider $p(x)=a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3 -
(2x+1)^3$ then $p(0)=p(1)=p(2)=p(3)=0$ that is the polynomial $p(x)$ of degree at most 3 have 4 distinct zeros. Thus $p(x)\equiv0$ i.e., $$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3 \equiv
(2x+1)^3.~~~~~~~~(1)$$
Comparing coefficients of $x^3$ on b/s of $(1)$ we get
$$ a_1+a_2+a_3+a_4=8 $$
Put $x=-5$ in $(1),$ we get
$$ 64a_1+27a_2+8a_3+a_4=729. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ without series expansion or L Hospital rule
If $$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},$$ $x\neq 0$, is continuous at $x=0$, then find $A,B$ and $f(0)$. Do not use series expansion or L Hospital's rule.
As $f(x)$ is continuous at $x=0$,therefore its limit at $x=0$ should equal to its value.
Note that this question is to be solved without series expansion or L Hospital's rule,
I tried to find the limit $\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$
$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3\sin x-4\sin^3x+2A\sin x\cos x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}\times\frac{\sin x}{x}$
$=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}$
As the denominator is zero,so numerator has to be zero,in order the limit to be finite.
So, $3+2A+B=0. (1)$
I tried but I could not get the second equation between $A$ and $B$. I am stuck here. How do I continue?
| Your first step is correct, you need $3+2A+B=0$ for the function to converge. Let's substitute $B=-3-2A$ into the original function. Them you have
$$
f(x)=\frac{\sin(3x)+A\sin(2x)+(-3-2A)\sin(x)}{x^5}
$$
Using the angle sum formulas:
$$
\sin(3x)=\sin(x)\cos(2x)+\cos(x)\sin(2x)=\sin(x)(\cos^2(x)-\sin^2(x))+2\sin(x)\cos(x)^2.
$$
Now, using the standard trig identity, $\sin^2(x)=1-\cos^2(x)$. Therefore,
$$
\sin(3x)=4\sin(x)\cos(x)^2-\sin(x)
$$
Also using the angle sum formulas:
$$
\sin(2x)=2\sin(x)\cos(x)
$$
Substituting these into the original function, we have
$$
f(x)=\frac{4\sin(x)\cos^2(x)-\sin(x)+2A\sin(x)\cos(x)-3\sin(x)-2A\sin(x)}{x^5}.
$$
By combining the terms with $A$ and those without, we have
$$
f(x)=\frac{4\sin(x)(\cos^2(x)-1)+2A\sin(x)(\cos(x)-1)}{x^5}=\frac{\sin(x)}{x}\cdot\frac{4(\cos^2(x)-1)+2A(\cos(x)-1)}{x^4}
$$
Factoring a difference of squares, the numerator becomes
$$
f(x)=\frac{\sin(x)}{x}\cdot\frac{4(\cos(x)-1)(\cos(x)+1)+2A(\cos(x)-1)}{x^4}$$
Factoring again, we have
$$f(x)=\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{x^2}\cdot\frac{4(\cos(x)+1)+2A}{x^2}
$$
Observe that
$$
\lim_{x\rightarrow 0}\frac{\cos(x)-1}{x^2}=\lim_{x\rightarrow 0}\frac{\cos(x)-1}{x^2}\cdot\frac{\cos(x)+1}{\cos(x)+1}=\lim_{x\rightarrow 0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)}
$$
Then, using the standard trig identities, we get that this limit equals
$$
\lim_{x\rightarrow 0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}=\lim_{x\rightarrow 0}\left(\frac{\sin(x)}{x}\right)^2\frac{-1}{\cos(x)+1},
$$
which we can see has a limit of $-\frac{1}{2}$.
Continuing from here, you also need the numerator to vanish for the limit to exist because the denominator vanishes. In other words, $\lim_{x\rightarrow 0}4(\cos(x)+1)+2A=8+2A$ must be $0$, so $A=-4$. After substitution, the entire formula becomes
$$f(x)=\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{x^2}\cdot\frac{4(\cos(x)-1)}{x^2}
$$
Substituting the values that we've calculated the limit as
$$
\lim_{x\rightarrow 0}f(x)=1\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{4}{2}\right)=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
The minimum number of non real roots of the equation $x^4-2x^3+2x^2-x=k$ is?
The minimum number of non real roots of the equation
$x^4-2x^3+2x^2-x=k$ is?
k is any real number.
I plotted this on https://www.desmos.com/calculator/vpfpjwyxz8.It seems that the answer will be 2.But how to solve it manually?
| Method $1$
Let $a_i$ be the roots of the polynomial. We then have
$$\sum_{i=1}^4 a_i = 2$$
$$\sum_{\overset{i,j=1}{i \neq j}}^4 a_ia_j = 2$$
This means we have
$$\sum_{i=1}^4 a_i^2 = \left(\sum_{i=1}^4 a_i\right)^2 - 2\cdot\left(\sum_{\overset{i,j=1}{i \neq j}}^4 a_ia_j\right)= 2^2 -2 \cdot 2 = 0$$
Further all $a_i$'s are not zeros. Hence, there is at-least a pair of complex roots.
Method 2
We have
$$x^4-2x^3+2x^2-x = \left(x^2-x+1/2\right)^2-1/4 = k$$
This means we have
$$x^2-x+1/2 = \pm \sqrt{k+1/4} \implies \left(x-1/2\right)^2 = -\dfrac14 \pm \dfrac{\sqrt{4k+1}}2$$
which clearly shows that there are at-least two complex roots.
In fact the second argument shows that if $k < -3/16$, there are $4$ complex roots and if $k \geq -3/16$, there are $2$ complex roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Last two digits of $2^5+2^{5^2} +\dots +2^{5^{2015}}$
Let $$N=2^5+2^{5^2} +\dots +2^{5^{2015}}.$$ Find the last two digits of $N$.
It suffices to find $N \mod 100$. And I observed that $2^{20+n}=2^n\pmod{100}$ for $n \ge 2$. So is the answer just $(2015)(32) \mod 100 =80$?
| Notice that
$$
32^5=2^{25} \equiv 2^{10}\cdot 2^{10} \cdot 2^5 \equiv 24^2 \cdot 32 \equiv 32 \mod 100
$$
(I think what you wrote there does not make sense as stated)
and it follows that
$$
2^{5^k} \equiv 32 \mod 100
$$
for all $k\ge 1$
So indeed
$$
2^{5}+2^{5^2}+\cdots+2^{5^{2015}} \equiv 32 \cdot 2015 \equiv 64480 \equiv 80 \mod 100
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving Symmetrical Equations Algebraically I'm doing some Cambridge STEP papers and have come across a tricky set of equations.
\begin{align*} 99 &= c^3 + 6 cd^2 \tag{1} \\
70 &= 3c^2d + 2d^3 \tag{2} \end{align*}
From looking around, I've found the easiest way to solve these for the real solutions is to note that $ c^3 < 99 $ and $ 2d^3 < 70 $ and then to work numerically. My question is: How can these be solved algebraically ?
Many thanks!
| \begin{gather}
99=c^3+6cd^2\\
d=\sqrt{\frac{99-c^3}{6c}}
\end{gather}
Substituting into the other equation yields
\begin{gather}
3c^2\sqrt{\frac{99-c^3}{6c}}+2\left(\frac{99-c^3}{6c}\right)^{\frac{3}{2}}=70\\
\sqrt{\frac{99-c^3}{6c}}\left(3c^2+\frac{198-2c^3}{6c}\right)=70\\
\sqrt{\frac{99-c^3}{6c}}\left(\frac{16c^3+198}{6c}\right)=70\\
\left(\frac{99-c^3}{6c}\right)\left(\frac{16c^3+198}{6c}\right)^2= 4900\\
(-c^3+99)(16c^3+198)^2-4900(6c)^3=0\\
-256c^9+19008c^6-470340c^3+3881196=0\\
\end{gather}
which is a rather tedious polynomial to solve. It can be factored using the rational roots theorem, but I will not do that here. Using a calculator gives me the only real root $c=3$. Substituting back into the expression for $d$ gives
$$d=\sqrt{\frac{99-27}{18}}=\sqrt{\frac{72}{18}}=\sqrt{4}=2$$
So a solution of this system of equations is $(c,d)=(3,2)$. $d=-2$ does not work because the second equation will yield $70=-70$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $ \cos(x) \cos(2x) \cos(3x) = \frac{4}{7} $ find $ \frac{1}{\cos^2{x}}+\frac{1}{\cos^2{2x}} + \frac{1}{\cos^2{3x}} $ If $\cos(x) \cos(2x) \cos(3x) = \dfrac{4}{7} $ and $S=\dfrac{1}{\cos^2{x}}+\dfrac{1}{\cos^2{2x}} + \dfrac{1}{\cos^2{3x}} $ when $ S \in \mathbb{R}^{+}$ then $ S = ? $
P.S. I have tried that , but failed many times. Because I suppose that $\cos(x) ,\cos(2x) , \cos(3x)$ be root of $8\cos^{6}(x) -10\cos^{4}(x) + 3\cos^{2}(x) - \frac{4}{7} = 0 $
| Question: $\cos(x) \cos(2x) \cos(3x) = \frac {4}{7}$ and $S=\dfrac{1}{\cos^2{x}}+\dfrac{1}{\cos^2{2x}} + \dfrac{1}{\cos^2{3x}} $ when $ S \in \mathbb{R}^{+}$ then $ S = ?$
Working Out: let $a = \cos(x)$, $b = \cos(2x)$, $c = \cos(3x)$, then:
$$abc = \frac {4}{7}$$ $$S=\dfrac{1}{a^2}+\dfrac{1}{b^2} + \dfrac{1}{c^2} $$
Because $\cos(2x) = 2\cos^2(x) - 1$, $\therefore b = 2a^2 -1$.
Because $\cos(3x) = \cos(x)\cos(2x) - \sin(x)(2\sin(x)\cos(x)) = ab - 2a\sin^2(x)$.
As $2a\sin^2(x) = 2a(1- \cos^2(x)) = 2a(1-a^2)$
$\therefore \cos(3x) = ab(2a^2-1) - 2a(1-a^2) = 4a^3 - 3a$
$\therefore c = 4a^3 - 3a$
So we get the following equation:
$$a(2a^2 - 1)(4a^3 - 3a) = \frac {4}{7}$$
$$56a^6 - 70a^4 + 21a^2 - 4 = 0$$
As I have know special way of solving such polynomials I turned to wolfram alpha and got an approximate root of $a = -0.96392$ and $a = 0.96392$. But $a = \cos(x)$, so when $a = -0.96292$ $x = 164.562(3.dp)$ and when $a = 0.96392$ $x = 15.4378 (4.dp)$. Putting those values into the equation of $S$, we get:
$$S=\dfrac{1}{\cos^2(164.562)}+\dfrac{1}{\cos^2(2 * 164.562)} + \dfrac{1}{\cos^2(3 * 164.562)} $$
$$\therefore S = 10.673$$
Or
$$S=\dfrac{1}{\cos^2(15.4378)}+\dfrac{1}{\cos^2(2 * 15.4378)} + \dfrac{1}{\cos^2(3 * 15.4378)} $$
$$\therefore S = 4.54217$$
Answer: $$S = 4.54217, S = 10.673$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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A valid floor function trick? Given $x\in\mathbb R_+$ and $m,n\in\mathbb Z_+$, is it true that
$$\bigg\lfloor\frac{\lfloor \frac{x}{m}\rfloor}{n}\bigg\rfloor=\bigg\lfloor \frac{x}{mn}\bigg\rfloor?$$
Thanks for at least three different and convincing proofs! I'll use this result trying to code the prime counting formula from Wikipedia $(2)$:
Given $m$, select $y$ so that
$\sqrt[3]{m}\le y\le\sqrt{m}$ and let $n=\pi(y)$. Then
$(1)\;$ $\pi(m)=\phi(m,n)+n-1-P_2(m,n)$, where
$(2)\;$ $\displaystyle \phi(m,n)=\phi(m,n-1)-\phi\Big(\frac{m}{p_n},n-1\Big)$,
$\;\phi(m,0)=\lfloor{m}\rfloor$ and
$(3)\;$ $\displaystyle P_2(m,n)=\sum_{y<p\le\sqrt{m}}
\Big(\pi\Big(\frac{m}{p}\Big)-\pi(p)+1\Big)$, $\;p$ is a prime.
| Let $\left\lfloor\dfrac{x}{mn}\right\rfloor = k \in \mathbb{Z}$. Then $k \le \dfrac{x}{mn} < k+1$. So, $kn \le \dfrac{x}{m} < (k+1)n$.
Since $\dfrac{x}{m} \ge kn$ and $kn$ is an integer, we have $\left\lfloor\dfrac{x}{m}\right\rfloor \ge kn$. Also, $\left\lfloor\dfrac{x}{m}\right\rfloor \le \dfrac{x}{m} < (k+1)n$.
Hence, $kn \le \left\lfloor\dfrac{x}{m}\right\rfloor < (k+1)n$, and thus, $k \le \dfrac{\left\lfloor\tfrac{x}{m}\right\rfloor}{n} < k+1$.
Therefore, $\left\lfloor \dfrac{\left\lfloor\tfrac{x}{m}\right\rfloor}{n} \right\rfloor = k = \left\lfloor\dfrac{x}{mn}\right\rfloor$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the maximum value of $ab+ac+ad+bc+bd+3cd$
If $a,b,c,d>0$ satisfy the condition ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1$, find the maximum value of $ab+ac+ad+bc+bd+3cd$.
I'm not progress in this inequality problem. Please help.
Thank you.
| Hint: consider that the eigenvalues of the symmetric matrix
$$ M = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 3 \\ 1 & 1 & 3 & 0 \end{pmatrix} $$
are $-3,-1,2+\sqrt{5}$ and $2-\sqrt{5}$. In particular, $\left(\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}-1}{2},1,1\right)^T$ in the only eigenvector of $M$ with positive coordinates, and it is associated with the largest eigenvalue $2+\sqrt{5}$. It follows that:
$$ \max_{\substack{a^2+b^2+c^2+d^2=1 \\ a,b,c,d>0}} \frac{1}{2}(a\, b\, c\, d)\, M\, (a\, b\, c\, d)^T = \frac{2+\sqrt{5}}{2} = \color{red}{1+\frac{\sqrt{5}}{2}}. $$
Such a maximum is attained by $(a,b,c,d)=(x,x,y,y)$ with $x=\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{5}}}$ and $y=\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{5}}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that $\int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} = \frac{\pi^2}{8}$ $\displaystyle \int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} = \frac{\pi^2}{8}$
I had been told that $x =\displaystyle \frac{1-y}{1+y}$ solves it, but it takes me back to where I started.
| In fact, the correct answer is 0. As @JimmyK4542 computed, we know from the substitution $x \mapsto (1-x)/(1+x)$ that
$$ I = \int_{0}^{1} \arctan\left( \frac{3(1+x)}{1-2x-x^2} \right) \, \frac{dx}{1+x^2} = \int_{0}^{1} \arctan\left( - \frac{3(1+x)}{1-2x-x^2} \right) \, \frac{dx}{1+x^2}. $$
But the identity $\arctan(-x) = -\arctan x$ is true for any $x \in \Bbb{R}$. Consequently
$$ I = -I $$
and hence $I = 0$. So why we have vanishing quantity here? This is because the function satisfies
$$ x \mapsto \arctan\left(\frac{3(1+x)}{1-2x-x^2}\right) \quad \text{is} \quad \begin{cases}
\text{positive when } 0 < x < \sqrt{2}-1 \\
\text{negative when } \sqrt{2}-1 < x < 1.
\end{cases} $$
So the graph of this function looks like this:
Somehow magically, the contribution from the positive part and that from the negative part cancel out each other. This observation is also supported by numerical calculation:
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Solving logarithm problem, Found the value This is my first question in this place, I don't know how to solve my problem.
I have this equation, I need to find the central value based on this equation:
$k = 0.2$
$$k = 2^{(\frac{1}{24} + \frac{1}{24\cdot2^x})}-2^{(\frac{1}{24} - \frac{1}{24\cdot2^x})}$$
I need to figure out the x value, the known value is k.
$$x=?$$
if I define y like: $y = \frac{1}{24\cdot2^{x}}= 2^{-x}/24$, and l like: $l = k/2^{\frac{1}{24}}$
$$k = 2^{\frac{1}{24}}\cdot2^y-2^{\frac{1}{24}}\cdot2^{-y}$$
$$l = 2^y-2^{-y}$$
$$\log_{2}(l)=\text{ ?}$$
$$y=\text{ ?}$$
Please help me.
| First of all, set $a=1/24$ and $t=2^{-x}$, so the equation becomes
$$
k=2^{a+at}-2^{a-at}
$$
We now see a further simplification: set $b=2^a$, so the equation becomes
$$
k=b^{1+t}-b^{1-t}
$$
or else
$$
k=bb^t-\frac{b}{b^t}
$$
that can be rewritten as
$$
bb^{2t}-kb^t-b=0
$$
This is a quadratic in $b^t$ and we can immediately discard the negative root, so
$$
b^t=\frac{k+\sqrt{k^2+4b^2}}{2b}
$$
Therefore
$$
t=\frac{\log(k+\sqrt{k^2+4b^2})-\log b-\log 2}{\log b}
$$
Recalling that $t=2^{-x}$ we get
$$
-x\log 2=\log\frac{\log(k+\sqrt{k^2+4b^2})-\log b-\log 2}{\log b}
$$
The logarithm can be with respect to any basis, the formula is simpler if we use base $2$, so $\log_2b=\log_22^a=a$ and $\log_22=1$, so
$$
-x=\log_2\frac{\log_2(k+\sqrt{k^2+4b^2})-a-1}{a}
$$
Since $b=2^a$ we have $b^2=2^{2a}$ and $4b^2=2^{2+2a}$ and we have
$$
x=\log_2\frac{a}{\log_2(k+\sqrt{k^2+2^{2+2a}})-a-1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a trig. equation Solve for x $$\tan 2x = 3 \tan x$$
Well, I'm stuck with this one. This is how far I got.
$$\tan 2x = 3 \tan x$$
$$\frac{2 \tan x}{1 - \tan^2 x} = 3 \tan x$$
$$2 \tan x = 3 \tan x - 3 \tan^3 x$$
$$3 \tan^3 x = \tan x$$
$$3 \tan^3 x - \tan x = 0$$
$$\tan x (3 \tan^2 x - 1) = 0$$
So, then a part of the solution is $\tan x = 0$, so $x = k\pi$
This leaves $\tan^2x = 1$. I thought to take the root of both sides, leaving me with $\tan x = 1$. So I figured $x = \frac{1}{4}\pi + k\pi$ so we the first and third quadrants, where the tangent is positive.
But, to my deep deep sadness, this last part turned out to be wrong.
What am I missing here?
| The equation $3\tan^3 x - \tan x = 0$ factors as $\tan x(\color{red}{3}\tan^2 x - 1) = 0$ (you forgot the $3$).
Hence, $\tan x = 0$ or $\tan x = \pm \dfrac{1}{\sqrt{3}}$. Both of these should be easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to prove $\left(1+\frac{1}{\sin a}\right)\left(1+\frac{1}{\cos a}\right)\ge 3+2\sqrt{2}$? Prove the inequality:
$$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$
| Differentiate to find the minimum of the LHS :
$$\frac{\mathrm{d}}{\mathrm{d}a}\left(\,\left(1+\frac{1}{\sin\,a}\right)\left(1+\frac{1}{\cos\,a}\right)\,\right)=0$$
$$\left(-\frac{1}{\sin^2 a}\right)\cos a\left(1+\frac{1}{\cos a}\right)+\left(1+\frac{1}{\sin a}\right)(-\sin a)\left(-\frac{1}{\cos^2 a}\right)=0$$
$$\left(\sin\,a+1\right)\left(-\frac{1}{\cos^2 a}\right)=\left(\cos\,a+1\right)\left(\frac{1}{\sin^2 a}\right)$$
$$(1+\sin a)(\sin^2a)=(1+\cos a)(\cos^2a)=(1+\cos a)(1-\sin^2a)$$
$$(1+\sin\,a)(\sin^2a)=(1+\cos\,a)(1+\sin\,a)(1-\sin\,a)$$
$$\sin^2a=1-\cos^2a=(1-\cos\,a)(1+\cos\,a)=(1+\cos\,a)(1-\sin\,a)$$
$$1-\cos\,a=1-\sin\,a$$
$$\cos\,a=\sin\,a$$
This only happens at $a=\frac{\pi}{4}$ in the given range.
And here $\sin\,a=\cos\,a=\frac{1}{\sqrt{2}}$.
So the minimum of the LHS is $(1+\sqrt{2})^2=1+2+2\sqrt{2}=3+2\sqrt{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
| In fact we can even prove a general form of this inequality. $$\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)\leqslant(1+n)^n(*)$$
Where: $x_1, x_2, ..., x_n\gt0$ and $x_1+x_2+...+x_n=1$.
Proof:
Firstly let's consider this matrix
\begin{pmatrix}
a_{11} & a_{12} & ... & a_{1n}\\
a_{21} & a_{22} & ... & a_{2n}\\
. & . & . & .\\
. & . & . & .\\
a_{m1} & a_{m2} & ... & a_{mn}
\end{pmatrix}
And prove the following lemma:
$$A(G_1, G_2, ..., G_m)\leqslant G(A_1, A_2, ..., A_n)(**)$$
where $A(x_1, x_2, ..., x_n)=\frac{x_1+x_2+...+x_n}{n}$,
$G(x_1, x_2, ..., x_n)=\sqrt[n]{x_1x_2...x_n}$,
$A_i$ is an arithmetic mean of i-th column of the matrix, $i={1,2,...,n}$.
And $G_j$ is a geometric mean of j-th row of the matrix, $j={1,2,...,m}$.
If any of $A_i=0$ then $(**)$ is obviously true, so let $A_i \gt 0$ then by
the Cauchy–Schwarz inequality we have:$$\sum_{j=1}^n \frac{a_{ij}}{A_j}\geqslant\frac{nG_i}{G(A_1, A_2, ..., A_n)}$$
from ths we get $$\frac{mA_1}{A_2}+\frac{mA_2}{A_2}+...+\frac{mA_n}{A_n}\geqslant\frac{nmA(G_1, G_2, ..., G_m)}{G(A_1, A_2, ..., A_n)}$$ so finally $$A(G_1, G_2, ..., G_m)\leqslant G(A_1, A_2, ..., A_n)$$ which proves our lemma.
Now let's construct following matrix
\begin{pmatrix}
1 & 1 & ... & 1\\
\frac{1}{x_1} & \frac{1}{x_2} & ... & \frac{1}{x_n}\\
\end{pmatrix}
Now
$G(A_1, A_2, ..., A_n)=G(\frac{1+\frac{1}{x_1}}{2}, \frac{1+\frac{1}{x_2}}{2}, ..., \frac{1+\frac{1}{x_n}}{2})=\frac12\sqrt[n]{(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})}$,
$A(G_1, G_2)=A(1, \sqrt[n]{\frac{1}{x_1x_2...x_n}})=\frac12(1+\frac{1}{\sqrt[n]{x_1x_2...x_n}})$
Now using our lemma we have: $$(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})\geqslant \left(1+\frac{1}{\sqrt[n]{x_1x_2...x_n}}\right)^n$$
Then we only need to show that $\frac{1}{\sqrt[n]{x_1x_2...x_n}}\geqslant n\iff \frac{1}{n}\geqslant \sqrt[n]{x_1x_2...x_n}\iff\frac{x_1+x_2+...+x_n}{n}\geqslant\sqrt[n]{x_1x_2...x_n}$ because of our assumption $\sum_{k=1}^n x_k=1$
And the last inequality is true by the AM-GM.
Thus $(*)$ is proven.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. Prove that if c ϵ C and $x = \frac{(c+\sqrt{c^2+4})}{2}$ so $x-\frac{1}{x} $ ϵ C. I have no idea how do this. Please help me.
| We have
$x-\frac{1}{x}$ = $\frac{x^2 - 1}{x}$ = $\frac{(x-1)(x+1)}{x}$
x-1 = $\frac{c + \sqrt{ c ^ {2}+4 }}{2}$ -1 = $\frac{c + \sqrt{ c ^ {2}+4 }-2}{2}$
x+1 = $\frac{c + \sqrt{ c ^ {2}+4 }+2}{2}$
So : $x-\frac{1}{x}$ = $\frac{\frac{c + \sqrt{ c ^ {2}+4 }-2}{2}*\frac{c + \sqrt{ c ^ {2}+4 }+2}2}{\frac{c + \sqrt{ c ^ {2}+4 }}{2}}$ = $\frac{(c + \sqrt{ c ^ {2}+4 })^2 - 4 }{4}$*$\frac{2}{c + \sqrt{ c ^ {2}+4 }}$ = $\frac{2c^2 + 2c \sqrt{ c ^ {2}+4 }}{2(c + \sqrt{ c ^ {2}+4} )}$=c
We have proved $x-\frac{1}{x}$ ϵ C
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evalute $\int\frac{3x-2}{x^2-4x+5} \, dx$ Evalute $$\int\frac{3x-2}{x^2-4x+5} \, dx$$
So, I tried using $x^2-4x+5 = t$, which unfortunately didn't help much.
I also tried making the denominator $(x-2)^2 +1$ and then using $x-2 = t$, which made everything messy. Any suggestions? It looks so simple, and I bet my first approach is right.
| Using your second idea, you get
$\displaystyle\int\frac{3(t+2)-2}{t^2+1}\;dt=\int\frac{3t+4}{t^2+1}\;dt=\frac{3}{2}\int\frac{2t}{t^2+1}\;dt+4\int\frac{1}{t^2+1}\;dt$
$\displaystyle=\frac{3}{2}\ln(t^2+1)+4\arctan t+C=\frac{3}{2}\ln(x^2-4x+5)+4\arctan(x-2)+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
The probability that in a game of bridge each of the four players is dealt one ace The question is to show that the probability that each of the four players in a game of bridge receives one ace is $$ \frac{24 \cdot 48! \cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$. I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?
| Since the order doesn't matter, let's make hands one by one.
*
*For the first, there are 4 Aces and I need to choose 1 to give to this person, $\binom{4}{1}$. Then I need to choose 12 cards from the 48 non-Aces to complete this hand, $\binom{48}{12}$. Thus the number of ways to make this hand is
$$\binom{4}{1}\binom{48}{12}.$$
*I already gave the first person an Ace, so I have three left and I need to choose one to give to this person. I also used up 12 non Aces on the last person, so there are 36 non Aces left and I need to choose 12. The number of ways to make this hand is thus,
$$\binom{3}{1}\binom{36}{12}.$$
*and 4. follow the same logic and so I have
$$\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}$$
ways to make hands 3 and 4.
Finally, all the possible ways to make 4 13-card hands are
$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} = \binom{52}{13,13,13,13} = \frac{52!}{13!\,13!\,13!\,13!}.$$
Lastly, putting it all together, the probability of each person getting an Ace
is thus
\begin{align*}\frac{\binom{4}{1}\binom{48}{12}\binom{3}{1}\binom{36}{12}\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}}{\binom{52}{13,13,13,13}} &= \frac{4!}{1!3!}\cdot\frac{48!}{12!36!}\cdot\frac{3!}{1!2!}\cdot\frac{36!}{12!24!}\cdot\frac{2!}{1!1!}\\
&\qquad\times\frac{24!}{12!12!}\frac{1!}{1!0!}\cdot\frac{12!}{12!0!}\cdot\frac{13!\,13!\,13!\,13!}{52!}\\
&=\frac{24 \cdot 48! \cdot13^4}{52!}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
| $ \int{\frac{(7x^2 + 1)dx}{(x+1)(x-1)(x+3}} $ = $ \int{\frac{A dx}{(x+1)}} + \int{\frac{B dx}{(x-1)}} + \int{\frac{C dx}{(x+3)}}$
You need to solve the equation:
$ 7x^2 + 1 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1) $
To find A, B and C you need to solve an simultaneous equation:
$A + B + C = 7$
$2A + 4B = 0$
$-3A+3B-C = 1$
After solving it you can find out that $ A = 1$, $B=-2$ and $ C = 8$
So our integral can be written as:
$ \int{\frac{(7x^2 + 1)dx}{(x+1)(x-1)(x+3}} $ = $ \int{\frac{dx}{(x+1)}} - \int{\frac{2 dx}{(x-1)}} + \int{\frac{8 dx}{(x+3)}}$
Each of this integrals can be solved by substitution, you just need to use $t$ as $x+1$, $x-1$ and $x+3$
For example:
$x+1 = t$, so $dx = dt$
$\int{\frac{dx}{(x+1)}} $= $\int{\frac{dt}{t}}$ = $\ln{|t|} + C $ = $\ln{|x+1|} + C$
When you solve all of these 3 internals you will see that the answer is $\ln{|x+1|} -2\ln{|x-1|} + 8\ln{|x+3|} + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
number of ternary trees: finding a recurrent relationship If $t_n$ is the number of ternary trees with n nodes, with $t_0=0$, what would be the convenient manner for finding a recurrent relationship for $t_n$? It is given that $t_1=1, t_2=3, t_3=12$.
A sketch for $t_1, t_2, t_3$
https://drive.google.com/file/d/0Bzsj_3-U1ZOhS2dFaFNyWFlGOHFnLXdtdEFxQkYxbl9JRzNR/view?usp=sharing
I had found this link on the stackexchange: https://oeis.org/A001764
my guesswork is that $t_n$ of the form $C_1t_{n-1}+C_2t_{n-2}+C_3t_{n-3}$
| Suppose we are interested in computing a recurrence for the number of
ternary trees where there is one leaf and one ternary tree on one
node. This gives the species
$$\mathcal{T} =
\epsilon + \mathcal{Z}\mathfrak{S}_{=3}(\mathcal{T})$$
This yields the functional equation
$$T(z) = 1 + z T(z)^3$$
which immediately produces the recurrence $T_0 = 1$ and
$$T_n = \sum_{a=0}^{n-1}\sum_{b=0}^{n-1-a} T_a T_b T_{n-1-a-b}.$$
For the closed form we get the integral
$$T_n =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z) \; dz.$$
Use the functional equation and put $T(z) = w$ to get
$$z = \frac{w-1}{w^3} = \frac{1}{w^2}-\frac{1}{w^3}$$
and $$dz = -\frac{2}{w^3}+\frac{3}{w^4} \; dw$$
to get for the integral (by Lagrange inversion)
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon} \frac{w^{3n+3}}{(w-1)^{n+1}}
w \left(-\frac{2}{w^3}+\frac{3}{w^4}\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}}
\left(-2w^{3n+1}+3 w^{3n}\right) \; dw .$$
This evaluates by inspection to
$$-2{3n+1\choose n} + 3{3n\choose n}
=\left( -2\frac{3n+1}{2n+1} + 3\frac{2n+1}{2n+1} \right)
{3n\choose n}
\\ = \frac{1}{2n+1}{3n\choose n}.$$
Here we have used the fact that
$$w^{3n+1} = \sum_{q=0}^{3n+1} {3n+1\choose q} (w-1)^q
\quad\text{and}\quad
w^{3n} = \sum_{q=0}^{3n} {3n\choose q} (w-1)^q.$$
Addendum. For an alternate recurrence observe that
$$(2n+3) T_{n+1} =
{3n+3\choose n+1}
= \frac{(3n+3)(3n+2)(3n+1)}{(2n+2)(2n+1)(n+1)}
{3n\choose n}
\\ = \frac{(3n+3)(3n+2)(3n+1)}{(2n+2)(n+1)} T_n.$$
This yields
$$T_{n+1} =
\frac{(3n+3)(3n+2)(3n+1)}{(2n+3)(2n+2)(n+1)} T_n
\\ = 3\frac{(3n+2)(3n+1)}{(2n+3)(2n+2)} T_n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\mathrm dx$ What substitution is useful for this integral?
$$\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\mathrm dx$$
Substitutions $u=x^{\frac{2}{3}},u=(x+1)^{\frac{2}{3}},u=(x-2)^{\frac{2}{3}}$ are not working.
Can't find useful trigonometric substitution.
| Change variable to
$$x - \frac12 = \frac32 u \iff u = \frac{2x-1}{3}
\quad\text{ followed by }\quad
(u^2-1)^{1/3} = v \iff u = \sqrt{1+v^3}$$
We have
$$\int\frac{dx}{\sqrt[3]{(x+1)^2(x-2)^2}}
= \int\frac{dx}{(x^2-x-2)^{2/3}}
= \int\frac{dx}{\left(\left(x-\frac12\right)^2-\frac{9}{4}\right)^{2/3}}\\
= \left(\frac{2}{3}\right)^{1/3}\int\frac{du}{(u^2-1)^{2/3}}
= \left(\frac{2}{3}\right)^{1/3}\int\frac{d\sqrt{1+v^3}}{v^2}
= \left(\frac{3}{2}\right)^{2/3}\int \frac{dv}{\sqrt{1+v^3}}
$$
Using result from this answer,
the integral becomes
$$\frac{3^{5/12}}{2^{2/3}}
\left.F\left(\cos^{-1}\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt[3]{\frac49(x+1)(x-2)}}-1\right)\right|\frac{2+\sqrt{3}}{4}\right)
+ \text{constant}$$
where $\displaystyle\;F(\phi|m) = \int_0^\phi \frac{d\theta}{\sqrt{1-m(\sin\theta)^2}}\;$
is the incomplete elliptic integral of the first kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\frac1{a^3(b+c)}+\frac1{b^3(a+c)}+\frac1{c^3(a+b)} \geq \frac32$
Let $a,b,c$ be positive real numbers with $abc = 1$. Prove that $$\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)} \geq \dfrac{3}{2}.$$
Attempt
We can rewrite it as $\dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}$. Then, using AM-GM we have $$\dfrac{bc}{2a^2\sqrt{bc}}+\dfrac{ac}{2b^2\sqrt{ac}} \dfrac{ab}{2c^2\sqrt{ab}} \geq \dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}.$$
I get stuck here.
| Change notation to $a=\frac{1}{x}$ ,$b=\frac{1}{y}$ , $c=\frac{1}{z}$ with $xyz=1$ . Then the inequality is :
$$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geq \frac{3}{2}$$
But using Nesbit's inequality and then AM-GM gives:
$$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \geq \frac{x+y+z}{2} \geq \frac{3}{2} \sqrt[3]{xyz}=\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\dfrac{1}{2}\sqrt{4\sin^2(36^{\circ})-1} = \cos(72^{\circ})$
Prove that $\dfrac{1}{2}\sqrt{4\sin^2(36^{\circ})-1} = \cos(72^{\circ})$
This question seemed pretty simple so I first started out by turning the left-hand side into terms of $\cos(x)$. We have $\dfrac{1}{2}\sqrt{4(1-\cos^2(36^{\circ}))-1} = \dfrac{1}{2}\sqrt{3-4\cos^2(36^{\circ})}$. Then I would use $\cos(72^{\circ}) = 2\cos^2(36^{\circ})-1$. Thus, $\dfrac{1}{2}\sqrt{3-4\cos^2(36^{\circ})} = \dfrac{1}{2}\sqrt{3-2(\cos(72^{\circ})+1)} = \dfrac{1}{2}\sqrt{1-2\cos(72^{\circ})}$. I get stuck here.
| We need $$4\sin^236^\circ-4\cos^272^\circ=1$$
Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$
$$\iff2(1-\cos72^\circ)-2(1+\cos144^\circ)=1$$
As $\cos(180^\circ-B)=-\cos B,$ $$\iff\cos36^\circ-\cos72^\circ=\dfrac12$$
Utilize Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding Stationary Points of Natural Log Function $$
f(x) = x - 2\ln(x^2 + 3)
$$
I started by using the chain rule on $x^2 + 3$ which gives me
$\frac{2x}{x^2} + 3$.
At this point I tried to multiple $\frac{2x}{x^2} + 3$ by $x - 2$ - is this correct?
How can I then find the stationary point?
I understand that you may not wish to give me the answer on a silver platter. Perhaps somebody could kindly provide me with an example in a similar format and the steps involved in solving it.
| The first derivative of your function can be found like this:
$$
\begin{eqnarray}
f'(x) &=& d(x - 2 \ln(x^2 + 3))/dx \\
&=& d(x)/dx - d(2 \ln(x^2 + 3))/dx \\
&=& 1 - \frac{2 d(x^2 + 3)/dx}{x^2 + 3} \\
&=& 1 - \frac{4x}{x^2 + 3} \\
\end{eqnarray}
$$
And the stationary points can be found by solving $f'(x) = 0$:
$$
\begin{eqnarray}
1 - \frac{4x}{x^2 + 3} &=& 0 \\
\frac{4x}{x^2 + 3} &=& 1 \\
x^2 + 3 &=& 4x \\
x^2 - 4x + 3 &=& 0 \\
(x - 1)(x - 3) &=& 0 \\
\end{eqnarray}
$$
So you have $x = 1$ and $x = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Cosine rule problem Question:
In triangle ABC, BC = a, AC= b, AB = c, and BD is perpendicular to AC.
If $\angle ABC = 60^{\circ}$, prove that $c = \dfrac{1}{2}a \pm \sqrt{b^2-\dfrac{3}{4}a^2}$
My approach: $b^2 = a^2 + c^2 - 2ac\cos B$
$b^2 = a^2 + c^2 - 2ac \cos{60}^\circ$
$b^2 = a^2 + c^2 - ac$
$b^2 = (a-c)^2+ac$
After this I don't know how to isolate the $c$ term.
| From $b^2 = a^2 + c^2 -ac$
$b^2 - a^2 = c^2 - ac$
$b^2 - \dfrac{3}{4}a^2 = c^2- ac + \dfrac{1}{4}a^2$
$b^2 - \dfrac{3}{4}a^2 = (c-\dfrac{1}{2}a)^2$
$c-\dfrac{1}{2}a = \pm \sqrt{b^2 - \dfrac{3}{4}a^2}$
$c = \dfrac{1}{2}a \pm \sqrt{b^2 - \dfrac{3}{4}a^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of the infinite product $\sqrt\frac12\cdot\sqrt{\frac12+\sqrt\frac12}\cdot\sqrt{\frac12+\sqrt{\frac12+\sqrt\frac12}}\cdots$ Find $\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
Let $x=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
$\log x=\frac{1}{2}\log(\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}})+....$
I do not know how to solve it further.
| Hint: Let $a =\dfrac{1}{2}$ and $b$ be the expression you are trying to calculate, then $b^2 = a + b$. You can use quadratic equation to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
$\varepsilon-\delta$ definition I am trying to prove the continuity of $f \colon [0, \infty) \to \mathbb{R}$
$$f(x) = \frac{x^2}{x+1}.$$
I tried to use $|f(x) - f(x_0)|$ So:
$$\left|\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}\right|$$
How to get to $|x - x_0|$ so that I can use $|f(x) - f(x_0)| < \delta$?
| Consider
$$
\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}=
\frac{x^2x_0+x^2-xx_0^2-x_0^2}{(x+1)(x_0+1)}=
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}(x-x_0)
$$
Thus, if $|x-x_0|<\delta$, you have
$$
\left|\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}\right|<
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}\delta
$$
If you can find an upper bound for the factor to $\delta$, say
$$
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}<M
$$
you can take $\delta=\varepsilon/M$ and then
$$
\left|\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}\right|<
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}\delta<M\frac{\varepsilon}{M}=\varepsilon
$$
whenever $|x-x_0|<\delta$.
Now
$$
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}=
\frac{xx_0+x+x_0+1-1}{(x+1)(x_0+1)}=
1-\frac{1}{(x+1)(x_0+1)}<1
$$
so you can use $M=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
integrate $\int \frac{x^4}{(1-x)^3}dx$
$$\int \frac{x^4}{(1-x)^3}$$
Because the degree of the numerator is bigger than the denominator I need to do a long division, should I open $(1-x)^3$ or is there a shorter way?
I got to $-\int x-\int 3-\int \frac{6x^2-10x-3}{(x-1)^3}$
So $6x^2-10x-3=A(x-1)^2+B(x-1)+C$
So $A=6$ And $B=2$ AND $C=-9$ which is wrong
| HINT:
$$\int\frac{x^4}{(1-x)^3}\space\text{d}x=$$
$$-\int\frac{x^4}{(x-1)^3}\space\text{d}x=$$
Do long division:
$$-\int\left[x+3+\frac{6}{x-1}+\frac{4}{(x-1)^2}+\frac{1}{(x-1)^3}\right]\space\text{d}x=$$
$$-\left[\int x\space\text{d}x+\int3\space\text{d}x+\int\frac{6}{x-1}\space\text{d}x+\int\frac{4}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\int x\space\text{d}x+3\int1\space\text{d}x+6\int\frac{1}{x-1}\space\text{d}x+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\frac{x^2}{2}+3x+6\int\frac{1}{x-1}\space\text{d}x+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
For the integrand $\frac{1}{x-1}$, substitute $u=x-1$ and $\text{d}u=\text{d}x$:
$$-\left[\frac{x^2}{2}+3x+6\int\frac{1}{u}\space\text{d}u+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\frac{x^2}{2}+3x+6\ln\left|u\right|+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\frac{x^2}{2}+3x+6\ln\left|x-1\right|+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Would the order of Taylor Polynomial change after substitution? I found the order of Taylor Polynomial is kind of confusing.
For example, we know:
$$T_4e^x = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!}$$
After substitute $x$ as $t^2$, we would have:
$$T_4e^{t^2} = \underbrace{1 + t^2 + \frac {t^4} {2!} + \frac {t^6} {3!} + \frac {t^8} {4!}}_\text{It seems to be 8th order?}$$
or:
$$T_4e^{t^2} = 1 + t^2 + \frac {t^4} {2!}$$
Could anyone tell me the relationship between the order of Taylor Polynomials and the degree of variables inside the polynomial?
In this case, which $T_4e^{t^2}$ is correct?
| You mistake is a quite common one to make and stems from thinking $T_4 e^x$ can be though of as a function of $x$. When writing
$$T_4 e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$$
we see that we have $x$ on both sides so it looks like we could just substitute $x=t^2$ to get
$$T_4 e^{t^2} = 1 + t^2 + \frac{t^4}{2!} + \frac{t^6}{3!} + \frac{t^8}{4!}$$
This is not correct and the reason is that $T_4$ is not a function, but rather a functional that takes in a function $e^x$ and outputs a polynomial of degree $4$.
What is true however, and which you can show with some work, is that $[T_n f(x)]_{x = t^2} = T_{2n} f(x^2)$ so substituing $t=x^2$ in a Taylor polynomial of $f(x)$ (centered at zero) gives the Taylor polynomial of $f(x^2)$ to double degree.
| {
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Locus of point of intersection of tangents at $A$ and $B$ From a Point $P$ on $C_1 \equiv x^2+y^2=9$ two tangents are drawn to $C_2 \equiv x^2+y^2=1$ which meets $C_1$ at $A$ and $B$. Find the Locus of point of intersection of tangents at $A$ and $B$ on $C_1$.
My Try: Let $P(3\cos\alpha, 3\sin\alpha)$, $A(3\cos\beta, 3\sin\beta)$ and $B(3cos\gamma, 3sin\gamma)$. Now $PA$ is chord of $C_1$ and tangent to $C_2$ say at point $M$. Since center of two circles is same, let it be $O$.
Now $OM$ is perpendicular to $PA$ at $M$ and since $PA$ is chord of $C_1$, $M$ is Mid Point of $PA$. So coordinates of $M$ is $M(\frac{3\cos\alpha+3\cos\beta}{2},\frac{3\sin\alpha+\sin\beta}{2})$
Now since $M$ lies on $C_2$ we have
$$\left(\frac{3\cos\alpha+3\cos\beta}{2}\right)^2+\left(\frac{3\sin\alpha+3\sin\beta}{2}\right)^2=1$$ Simplifying we get
$$\cos(\alpha-\beta)=\frac{-7}{9} $$ By similar analysis
$$\cos(\alpha-\gamma)=\frac{-7}{9} $$ which means
$$(\alpha-\beta)=2\pi+(\alpha-\gamma) \implies \gamma-\beta=2\pi \tag{1}$$
Now Equations of tangent to $C_1$ at $A$ and $B$ are
$$x\cos\beta+y\sin\beta=3$$ and $$x\cos\gamma+y\sin\gamma=3$$ Their Point of Intersection is
$$(h,k) =\left(\frac{3\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos\left(\frac{\gamma-\beta}{2}\right)},\frac{-3\sin\left(\frac{\gamma+\beta}{2}\right)}{\cos\left(\frac{\gamma-\beta}{2}\right)}\right)$$
Now $$h^2+k^2=\frac{18}{1+cos(\gamma-\beta)}$$ Now using $(1)$ i will get the locus as $$h^2+k^2=9$$ which is the same circle $C_1$. Can i know where i went wrong and any better approaches?
| A syntetic aproach:
Let $I,J$ the tangent points in $C_2$ from $P$. Then $PI=PJ$. Moreover, by Pitagoras applyed to $\triangle PIA$, we have that $PI$ is the same value for all $P\in C_1$. On the other hand, as $OI\perp PA$ and $PA$ is a chord of $C_1$, then $I$ is the midpoint of $PA$. Thus $PA=2PI=2PJ=PB$. So, $PA=PB$ and this common value is the same for all $P$. As $PO=OA$ for all values independent from $P$, all triangles $OAP$ are congruents. In particular $\angle APO$ is independent from $P$.
Next, note that $\angle HAO=90$, cause $HA$ is tangent to $C_1$. Then, $\angle AOH=90-\angle AHO$. On the other hand, $\angle POA=180-2\angle APO=180-\angle APB=180-\frac{1}{2}\angle AOB=180-\angle AOH$. Then $\angle AOH+\angle POA=90-\angle AHO+180-\angle AOH=270-(\angle AHO+\angle AOH)=270-90=180$. It proves that $H$, $O$ and $P$ are collinear.
Therefore $\angle AOH=\angle OAP+\angle OPA=2 \angle OPA$. But the last is independent of $P$. Thus, $\angle AOH$ not depends on $P$. So, all triangules $\triangle AOH$ are congruents. Therefore $OH$ is independent of $P$. That is, $OH$ has the same value for all $P$. Then, $H$ is in a circunferecence centered in $O$.
Finally, we have $\sin\alpha=1/3$. But $3/OH=\cos\beta=\cos 2\alpha=1-2\sin^2\alpha=1-2/9=7/9$. Then $OH=27/7$.
Conclusion: The focus searched is $x^2+y^2=(27/7)^2$
| {
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Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$ Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$
$a.)\ 1 \\
\color{green}{b.)\ 13} \\
c.)\ 15 \\
d.)\ 16 $
$ 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)\\
=3(1-\sin (2x))^{2}+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=3(1-2\sin (2x)+\sin^{2} (2x))+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=9+3\sin^{2} 2x+4(\sin^{6} x+\cos^{6} x)\\
=9+12\sin^{2}x\times \cos^{2}x+4(\sin^{6} x+\cos^{6} x)\\$
I am stucked.
I look for a short and simple way.
I have studied maths upto $12$th grade.
| $$3\left(\sin(x)-\cos(x)\right)^4+6\left(\sin(x)+\cos(x)\right)^2+4\left(\sin^6(x)+\cos^6(x)\right)=$$
$$3\left(-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)^4+6\left(\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)^2+4\left(\frac{3\cos(4x)+5}{8}\right)=$$
$$3\left(-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)^4+6\left(\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)^2+\frac{12\cos(4x)+20}{8}=$$
$$3\left(-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)^4+6\left(\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)^2+\frac{3\cos(4x)+5}{2}=$$
$$3\left(4\sin^4\left(\frac{\pi}{4}-x\right)\right)+6\left(2\sin^2\left(\frac{\pi}{4}+x\right)\right)+\frac{3\cos(4x)+5}{2}=$$
$$12\sin^4\left(\frac{\pi}{4}-x\right)+12\sin^2\left(\frac{\pi}{4}+x\right)+\frac{3\cos(4x)+5}{2}=$$
$$\frac{12\sin(2x)+3\cos(4x)+17}{2}-\frac{12\sin(2x)+3\cos(4x)-9}{2}=$$
$$\frac{12\sin(2x)+3\cos(4x)+17-12\sin(2x)-3\cos(4x)+9}{2}=$$
$$\frac{0+3\cos(4x)+17-0-3\cos(4x)+9}{2}=$$
$$\frac{3\cos(4x)+17-3\cos(4x)+9}{2}=$$
$$\frac{0+17-0+9}{2}=$$
$$\frac{17+9}{2}=$$
$$\frac{26}{2}=13$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the indefinite integral $\int \frac{dx}{({x^2-2x+5})^\frac{3}{2}} $ I have this.
$$\int \frac{dx}{({x^2-2x+5})^\frac{3}{2}} $$
I tried to define -
$ u = x-1 $
$\int \frac{du}{8(\frac{u^2}{4}+1)^{\frac{3}{2}}} = \frac{1}{8}\int \frac{du}{(\frac{u^2}{4}+1)^{\frac{3}{2}}}$
Now, im pretty stuck.
Thanks.
| Hint...substitute $x=1+2\tan\theta$
| {
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Compute the limit of $n\cdot\left(\frac{\arccos\left(3/n^3\right)}{\arccos\left(3/(n+1)^3\right)}-1\right)$ when $n\to\infty$ How to solve limit like this?
$$\lim_{n\to\infty}{n\left(\frac{\arccos\frac3{n^3}}{\arccos\frac3{(n+1)^3}}-1\right)}$$
$$=\lim_{n\to\infty}{n\left(\frac{\frac\pi2}{\frac\pi2}-1\right)}$$
$$=\lim_{n\to\infty}{n(1-1)} = 0$$
but $\infty*0$ is not define, how to solve it?
| This is just a naive thought, but have you tried the first order approximations $\arccos\left(\frac{3}{n^3}\right)\approx\frac{\pi}{2}-\frac{3}{n^3}$ and $\arccos\left(\frac{3}{(n+1)^3}\right)\approx\frac{\pi}{2}-\frac{3}{(n+1)^3}$ for $n\rightarrow\infty$? In this way the whole expression becomes
$$-\frac{18n^2+18n+6}{\pi n^5+3\pi n^4+3\pi n^3+(\pi+3)n^2},$$
whose limit is $0$ (as it should be).
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin\left(x-3\right)}}$ without L'Hôspital I have to calculate this limit $$\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin (x-3)}}$$
without L'Hôpital's rule.
These are my steps
$$\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\:=\:e^{\ln\left(\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\right)}=e\:^{\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}}$$
now i will caculate only $\frac{x\left(3x-8\right)}{\sin\left(x-3\right)}$
$$ t = x-3 \Rightarrow x = t+3$$
$\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\:=\lim \:_{t\to \:0}\:\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}$
And here im stack.
Help someone ?
Thanks.
| Rewrite
$$
\frac{x \ln(3x-8)}{\sin(x-3)} = x \frac{\ln(1+(3x-9))}{3x-9}\frac{x-3} {\sin(x-3)}\frac{3x-9}{x-3}\to3\cdot1\cdot1\cdot 3=9
$$
as $x\to3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$
Is there an easy way to factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$?
I have tried various methods such as completing the square, but I wasn't able to figure it out as it became very messy.
| View the expression as a polynomial in $a$. Plug in $a=b+c$. If the polynomial vanishes $(b+c-a)$ is a factor of the polynomial. Repeat for other factors. As the degree of the polynomial is $4$, it can't have more than $4$ linear factors and you're done.
Note: This method is often used to factor multivariate polynomials and is only as good as your guess.
| {
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"timestamp": "2023-03-29T00:00:00",
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Lim of $a_{n} = (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$ $a_{n} = (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$
I first find that $a_{n}$ is descending
$$\frac{a_{n+1}}{a_{n}}=\cfrac{(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})(1-\frac{1}{(n+1)^2})}{(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})} = 1-\frac{1}{(n+1)^2}<1$$
So $a_{n}$ is descending and also it is $$0< 1-\frac{1}{(n+1)^2}<1$$
So I can say $$\exists\lim\limits_{n \to \infty} a_{n} =l\in\Bbb{R} $$
How do i continue?
| We have
$$a_n = \dfrac{1 \cdot 3}{2\cdot2} \cdot \dfrac{2 \cdot 4}{3\cdot3} \cdot \dfrac{3 \cdot 5}{4\cdot4} \cdot \dfrac{4 \cdot 6}{5\cdot5} \cdots \dfrac{(n-1)\cdot (n+1)}{n \cdot n}$$
Telescopic cancellation shows that
$$a_n = \dfrac12 \cdot \dfrac{(n+1)}{n},\,\, \forall n \geq 2$$
You can also check this by induction. Now it is easy to see what $\lim_{n \to \infty} a_n$ is.
| {
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prove that $-1 \le \frac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$
For the real numbers $a, b, c, d$ prove that $$-1 \le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$$
Actually if we let $\vec{u} = (a, b)$ and $\vec{v} = (c, d)$ then by dot product we got $$\begin{split} cosθ & = \dfrac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|} \\ & =\dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \end{split}$$
But $-1 \le cosθ \le 1$ hence $$-1 \le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$$
Is it possible to prove the above inequality using only standard inequalities like AM-GM, Cauchy-Swarz, Rearrangement inequality or Jensen inequality?
| This is equivalent to $$\sqrt{(ac+bd)^2}=|ac+bd|\leqslant\sqrt{a^2+b^2}\sqrt{c^2+d^2},$$ which is an immediate consequence of the Cauchy–Schwarz inequality.
| {
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Inequality in triangle.
If $a,b,c$ are sides of a triangle prove that-
$$\frac a{c+a-b}+\frac b{a+b-c}+\frac c{b+c-a}\geq3$$
I am having problem in approaching the problem as the sides are not mentioned as integers.How do I approach the problem?
Thanks for any help!!..
| By using the Ravi transformation let $a=x+y$, $b=y+z$ and $c=z+x$, then
\begin{align}
\frac a{c+a-b}+\frac b{a+b-c}+\frac c{b+c-a}&=\frac{x+y}{2x}+\frac{y+z}{2y}+\frac{z+x}{2z}\\
&=\frac{3}{2}+\frac{1}{2}\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)
\end{align}
Now use AM-GM inequality.
| {
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Is my choice of $\delta$ under the given bounds for M correct in this $\epsilon-\delta$ proof? Are the choices of < or = correct?
Prove that
$$\lim_{x \to 9} \sqrt{x} = 3$$
Let $\epsilon > 0$.
$\color{red}{\text{Choose} \ \delta = \min\{\epsilon(\sqrt{9-M} + 3), M\} \ \text{where} \ M \in (0,9)}$.
Case 1: $\epsilon \le \frac{M}{\sqrt{9-M} + 3}$
Then $\delta = \epsilon(\sqrt{9-M} + 3)$
$$|\sqrt{x} - 3| = \frac{|x - 9|}{|\sqrt{x} + 3|}$$
$$= \frac{|x - 9|}{\sqrt{x} + 3}$$
$$< \frac{\delta}{\sqrt{x} + 3}$$
$$= \frac{\epsilon(\sqrt{9-M} + 3)}{\sqrt{x} + 3}$$
$$< \frac{\epsilon(\sqrt{9-M} + 3)}{\sqrt{9-M} + 3} = \epsilon$$
because $|x-9| < \epsilon(\sqrt{9-M} + 3) \to |x-9| < M \to \frac{1}{\sqrt{x} + 3} < \frac{1}{\sqrt{9-M} + 3}$
Case 2: $\epsilon \ge \frac{M}{\sqrt{9-M} + 3}$
Then $\delta = M$
$$|\sqrt{x} - 3| = \frac{|x - 9|}{|\sqrt{x} + 3|}$$
$$ = \frac{|x - 9|}{\sqrt{x} + 3}$$
$$< \frac{\delta}{\sqrt{x} + 3}$$
$$ = \frac{M}{\sqrt{x} + 3}$$
$$< \frac{M}{\sqrt{9-M} + 3} \le \epsilon$$
because $|x-9| < M \to \frac{1}{\sqrt{x} + 3} < \frac{1}{\sqrt{9-M} + 3}$
QED
Is that right?
| To be honest, I do not quite understand what you wrote. What I can tell you is that $\lim\limits_{x\to a^2}\sqrt x = a$, because we can choose $\delta = \epsilon$ and write $$\vert \sqrt x-a\vert =\frac{\vert x-a^2\vert}{\vert \sqrt x+a\vert}< \vert x-a^2\vert <\delta=\epsilon.$$ $\require{enclose} \enclose{horizontalstrike}{\textrm{This will always work.}}$
Edit: As Jyrki Lahtonen pointed out, this will of course only work if we know that $\vert \sqrt x+a\vert>1$. If this is not the case, then we have to resort to other (more cumbersome, for as far as I can see) methods.
| {
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How to integrate $\frac{1}{(1 + x^5)(1 + x^7)}$ My cousin who is in high school asked me if it is possible to integrate
$$
\int \frac{1}{(1 + x^5)(1 + x^7)} \, dx
$$
I checked the list of integrals of rational functions on Wikipedia link
and it doesn't seem to be here.
Is this not possible to do using elementary functions?
Wolfram Alpha can do this but the answer is extremely complicated.
| Solve:
$$\begin{align}\frac{1}{(x^5+1)(x^7+1)} & =\frac{a_0}{(x+1)^2}+\frac{a_1}{x+1}\\
&\quad+\frac{a_2x+b_2}{x^2-2\cos(\pi/5)x +1}\\&+\frac{a_3x+b_3}{x^2-2\cos(3\pi/5)x+1}\\
&\quad+\frac{a_4x+b_4}{x^2-2\cos(\pi/7)x +1}\\
&\quad+\frac{a_5x+b_5}{x^2-2\cos(3\pi/7)x +1}\\
&\quad+\frac{a_6x+b_6}{x^2-2\cos(5\pi/7)x +1}\\
\end{align}$$
Then you can use basic integrals to solve each:
$$\int \frac{ax+b}{x^2-2cx+1}\,dx$$
You'll get lots of logarithms and arctans, and it will be just horrible. And that doesn't even count how horrible the $a_i$ and $b_i$ are going to be.
| {
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If there are nonzero elements $a$ and $b$ in $A$ such that $(a+b)^2 = a^2 + b^2$, then $A$ has characteristic 2. Let $A$ be a finite integral domain. If there are nonzero elements $a$ and $b$ in $A$ such that $(a+b)^2 = a^2 + b^2$, then $A$ has characteristic 2.
I was thinking, if $(a+b)^2 = a^2 + 2ab + b^2 = a^2 + b^2$, wouldn't that mean that $2ab = 0$? But I thought that would be a contradiction since $A$ is a finite integral domain and has no zero divisors. I am not sure where I am confused.
However, I do think I know how to prove the statement. We know that in any integral domain of characteristic $p$, then $(a+b)^p = a^p + b^p$. Thus 2 is either the characteristic or 2 divides the characteristic. Since the characteristic must be prime, then 2 is the characteristic for $A$.
| It is an if and only if, prove both ways.
*
*If the characteristic is $2$, then $2 = 0$, so $(a + b)^2 = a^2 + 2 a b + b^2 = a^2 + b^2$
*If $a, b \ne 0$ are such that $(a + b)^2 = a^2 + b^2$, then $2 a b = (a + b)^2 - a^2 - b^2 = 0$. As $a b \ne 0$, and it is an integral domain, it has no zero divisors, so $2 = 0$.
| {
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Evaluating $\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$ without L'Hospital's Theorem I've been trying to evaluate$$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$$
I tried:
(a) Rationizing the numerator -> Error
(b) Rationizing the denominator -> Error
(c) Factoring out $x$ -> Error
Finally, I used L'Hospital's Rule and got the answer $3/2$, but that is not what I am supposed to do, for not a single word about this Theorem was mentioned during my lectures.
Is there any other way to solve this limit without this Theorem?
| \begin{align*}
\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}
&=\lim_{x\to1}\frac{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}\cdot\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}\\
&=\lim_{x\to1}\frac{x^2-1}{x^2-1}\cdot\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}\\
&=\frac32
\end{align*}
using $(a-b)(a+b)=a^2-b^2$.
| {
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In need of assistance with evaluation of a tricky limit I am trying to evaluate a limit, but without much luck.
I keep getting infinity as the answer, but both Wolfram and my textbook state otherwise.
The limit problem goes as follows:
$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}}) $
/* EDIT */
This is my original work, I wonder why it has failed to yield the result of zero as expected.
Please solve it step by step, in order for me to understand my error.
Thank you!
| $$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}})=$$
$$\lim_{n \to \infty} n^{2/3}({\sqrt{n+1}}+{\sqrt{n-1}}-2{\sqrt{n}})\frac{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{({\sqrt{n+1}}+{\sqrt{n-1}})^2-4n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{2\sqrt{(n+1)(n-1)}-2n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{2\sqrt{(n+1)(n-1)}-2n}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}\frac{2\sqrt{(n+1)(n-1)}+2n}{2\sqrt{(n+1)(n-1)}+2n}=$$
$$\lim_{n \to \infty}n^{2/3}\frac{4(n+1)(n-1)-4n^2}{{\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}}}\frac{1}{2\sqrt{(n+1)(n-1)}+2n}=$$
$$\lim_{n \to \infty} n^{2/3}\frac{-4}{(\sqrt{(n+1)(n-1)}+2n)({\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}})}=$$
$$\lim_{n \to \infty} \frac{-4}{(\sqrt{(n^{1/3}+n^{-2/3})(n^{1/3}-n^{-2/3})}+2n^{1/3})({\sqrt{n+1}}+{\sqrt{n-1}}+2{\sqrt{n}})}=0$$
Misread $\frac{2}{3}$ as $\frac{3}{2}$ in the original post
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Compound angle formula I understand how to use the compound angle formula when solving $\sin(\pi/12)$.
However I dont understand how I can use a compound angle formula to show
$$\arctan(3)-\arctan(1/2)=\pi/4$$
Thankyou
Any references would be highly appreciated, I looked online but couldn't find any that targeted the arc-tan/sin/cos.
| Hint:
I assume you know the formula for $\tan(A\pm B)$ (otherwise, your teacher is evil :) ).
Plug $A=\tan^{-1}a$, $B=\tan^{-1}b$ into that formula and rearrange it to get a formula for $tan^{-1}a - tan^{-1}b = tan^{-1}$(some expression involving $a$ and $b$).
We start with the known $\tan$ compound angle formula:
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
Substituting $A=\tan^{-1}a$, $B=\tan^{-1}b$ : $a = \tan A$, $b = \tan B$
$$\tan(\tan^{-1}a + \tan^{-1}b) = \frac{a + b}{1 - ab}$$
$$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$$
Now let $a = 3$, $b = -\frac{1}{2}$
Note than $\tan(-x) \equiv -\tan x$ for all $x$
$$\tan^{-1}3 - \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{3 - \frac{1}{2}}{1 + 3\cdot\frac{1}{2}}\right)$$
$$\tan^{-1}3 - \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{6 - 1}{2 + 3}\right) = \tan^{-1}1 = \frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine if $T$ is a Linear Transformation or Not Suppose that
$$
T\begin{pmatrix}
x \\
y \\
\end{pmatrix} =
\begin{pmatrix}
x-y\\
x+y \\
2x
\end{pmatrix}
$$
I tried multiplying with a scalar as follows
$$
T \left\{\alpha\begin{pmatrix}
x\\
y \\
\end{pmatrix}\right\}
=T\begin{pmatrix}
\alpha x\\
\alpha y \\
\end{pmatrix}
= \begin{pmatrix}
\alpha x- \alpha y\\
\alpha x+\alpha y \\
2\alpha x
\end{pmatrix}
= \alpha \begin{pmatrix}
x-y\\
x+y \\
2x
\end{pmatrix}
=\alpha T\begin{pmatrix}
x\\
y \\
\end{pmatrix}
$$
Is the above correct?
And I don't know how to show the addition part.
| The first part for the multiplication is correct.
For the addition, consider two different vectors
$V_1=\begin{pmatrix}
x\\
y \\
\end{pmatrix}$
and
$V_2=\begin{pmatrix}
u\\
v \\
\end{pmatrix}$
and show that
$$T(V_1+V_2)=T(V_1)+T(V_2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Emden‐Fowler differential equation Good day.
I am trying to solve the following equation:
$$\ddot{y}(x)-\frac{A}{x}\dot{y}(x)+\frac{Bx^2}{2}y(x)=0.$$
WolframAlpha says it is an Emden‐Fowler equation, but I have no idea how to solve this. Can you give me some tips?
In case of $$A=B/2=1$$ WolframALpha gives an analytical solution $$y(x)=c_1 sin\frac{x^2}{2}+c_1 cos\frac{x^2}{2}.$$
If there is no analytical way to solve, can I do it numerical?
Thank you.
| $$\frac{d^2y}{dx^2}-\frac{A}{x}\frac{dy}{dx}+\frac{Bx^2}{2}y(x)=0.$$
Changing $x$ into $-x$ doesn't change the equation. This draw us to a change of variable
$t=x^2 \quad;\quad dt=2x\:dx \quad;\quad \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=2x\frac{dy}{dt}\quad;\quad \frac{d^2y}{dx^2}=2\frac{dy}{dt}+2x\frac{d^2y}{dt^2}\frac{dt}{dx}=2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}$
$$2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}-2A\frac{dy}{dt} +\frac{B}{2}ty=0.$$
$$4t\frac{d^2y}{dt^2}+2(1-A)\frac{dy}{dt} +\frac{B}{2}ty=0.$$
This is an ODE of the Bessel kind.
In the case of $A=1$ the ODE reduces to :
$$4\frac{d^2y}{dt^2} +\frac{B}{2}y=0.$$
The solution is wellknown :
$$y=c_1 \cos\left(\sqrt{\frac{B}{8}}t\right) + c_2 \sin\left(\sqrt{\frac{B}{8}}t\right)$$
$$y=c_1 \cos\left(\sqrt{\frac{B}{8}}x^2\right) + c_2 \sin\left(\sqrt{\frac{B}{8}}x^2\right)$$
In the case of $A\neq 1$ the solutions are expressed thanks to the Bessel functions :
$$y=C_1 x^{\frac{A+1}{2}}J_{\frac{A+1}{4}}\left(\sqrt{\frac{B}{8}}x^2 \right) + C_2 x^{\frac{A+1}{2}}Y_{\frac{A+1}{4}}\left(\sqrt{\frac{B}{8}}x^2 \right)$$
The functions $J$ and $Y$ are the Bessel functions of the first and second kind respectively.
| {
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"url": "https://math.stackexchange.com/questions/1640471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Probability of an array having all distinct numbers Suppose you have an array of size $2n$. There are two times $2n^2$ distinct numbers that can be put into the array without replacement, i.e. for each choice of number, there are two copies, so a maximum of two cells in the array can have the same value. What is the probability that all $2n$ numbers in the array are distinct?
This is what I came up with: The probability that the second number is distinct from the first is $(4n^2 - 2)/(4n^2 - 1)$, the probability that the third number is distinct from the first and the second, is $(4n^2 - 4)/(4n^2-2)$, and so on...Conditioning gives
$$
\frac{4n^2 - 2}{4n^2 - 1}\frac{4n^2 - 4}{4n^2-2}... \frac{4n^2 - 2(2n-1)}{4n^2 - 2n+1}
$$
I'm not sure if this is correct though.
| Your solution is correct, though it may be incomplete. I'll start with what you have, including the $\frac{4n^2}{4n^2}$ term:
$$\frac{(4n^2)(4n^2-2)(4n^2-4)\cdots(4n^2-2(2n-1))}{(4n^2)(4n^2-1)(4n^2-2)\cdots(4n^2-(2n-1))}$$
Removing a factor of $2$ from each term in the numerator gives:
$$2^{2n}\times\frac{(2n^2)(2n^2-1)(2n^2-2)\cdots(2n^2-(2n-1))}{(4n^2)(4n^2-1)(4n^2-2)\cdots(4n^2-(2n-1))}\tag{1}$$
Notice now that the numerator is $\frac{(2n^2)!}{(2n^2-2n)!}$ and the denominator is $\frac{(4n^2)!}{(4n^2-2n)!}$
Let's multiply each by $\frac{(2n)!}{(2n)!}$ and see what happens:
$$\frac{(2n)!}{(2n)!}\frac{(2n^2)!}{(2n^2-2n)!}=(2n)!\frac{(2n^2)!}{(2n)!(2n^2-2n)!}=(2n)!\binom{2n^2}{2n}$$and:$$
\frac{(2n)!}{(2n)!}\frac{(4n^2)!}{(4n^2-2n)!}=(2n)!\frac{(4n^2)!}{(2n)!(4n^2-2n)!}=(2n)!\binom{4n^2}{2n}$$
Substituting these into $(1)$ gives:
$$2^{2n}\times\frac{(2n)!\binom{2n^2}{2n}}{(2n)!\binom{4n^2}{2n}}=2^{2n}\times\frac{\binom{2n^2}{2n}}{\binom{4n^2}{2n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ Any ideas on evaluating the definite integral
$$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx$$
The best numerical approximation I could get is $0.2796245358$.
Is there even a closed form solution?
| Here is an elementary approach. Denote $ K= \int_0^\infty \frac{\ln t\ln(1+t^2) \tan^{-1}t}{1+t^2}dt $
\begin{align}
I= &\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx
\overset{t=\tan x}= \frac14 \int_0^\infty \frac{\tan^{-1}t \ln^2\frac{t^2}{1+t^2}}{1+t^2}dt\\
=& \int_0^\infty \frac{\tan^{-1}t\ln^2t}{1+t^2}\overset{t\to 1/t}{dt} +\frac14
\int_0^\infty \frac{\tan^{-1}t\ln^2(1+t^2)}{1+t^2}\overset{t\to 1/t}{dt}
-K\\
=& \frac\pi8 \int_0^\infty \frac{\ln^2t}{1+t^2}dt
+ \frac\pi{16}\int_0^\infty \frac{\ln^2(1+t^2)}{1+t^2}dt -I -K\\
=& \frac\pi8 \cdot\frac{\pi^3}8
+\frac\pi{16}\left(\frac{\pi^3}6+2\pi\ln^22\right) -I-K
=\frac{5\pi^4}{192}+\frac{\pi^2}8\ln^22-\frac12K\tag1
\end{align}
Note that, with the substitution $x=\frac{(1+t^2)y}{1-y}$
\begin{align}
\int_0^1 \frac{2t\ln\frac{1-y}y}{1+t^2y^2}dy
= \int_0^\infty \frac{2t\ln\frac{1+t^2}x}{(1+x)^2+t^2}dx=\ln(1+t^2)\tan^{-1}t
\end{align}
Then, integrate $K$ as follows
\begin{align}
K=& \int_0^\infty \frac{\ln t}{1+t^2}\int_0^1 \frac{2t\ln\frac{1-y}y}{1+t^2y^2}dy \>\overset{t^2\to t}{dt}\\
= &\frac12 \int_0^1 \ln\frac{1-y}y\int_0^\infty
\frac{\ln t}{(1+t)(1+y^2t)}dt\>dy
=\int_0^1 \ln\frac{1-y}y
\frac{\ln^2y}{1-y^2}dy\\
=&\frac{\pi^4}{16}+\frac12\int_0^1\frac{\ln^2y\ln(1-y)}{1-y}dy
+ \frac12\int_0^1\frac{\ln^2y\ln(1-y)}{1+y}dy
\end{align}
The pair of integrals above are known and given below
\begin{align}
&\int_0^1\frac{\ln^2y\ln(1-y)}{1-y}dy= -\frac{\pi^4}{180}\\
&\int_0^1\frac{\ln^2y\ln(1-y)}{1+y}dy=
-4Li_4\left(\frac12\right) +\frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\
\end{align}
Plug them into $K$ and then into (1) to obtain
$$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx
= Li_4\left(\frac12\right)-\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Calculate $A^8$ using Cayley-Hamilton Therorem Find $A^8$ using Cayley Hamilton Therorem, when
$$A = \begin{pmatrix}
0 & 0 & 0 & -1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 0
\end{pmatrix}$$
I found that $P(t) =t^4-2x^2+1 = (t-1)^2(t+1)^2$.
But how can I use Cayley-Hamilton to find $A^8$?
| Given that $P(t) = t^4-2t^2+1$, the Cayley-Hamilton Theorem yields that
$$P(A) = O,$$
where $O$ is $4$ by $4$ zero matrix. Then
$$O = A^4 - 2A^2 + I \iff A^4 = 2A^2 - I \implies A^8 = (2A^2 - I)^2.$$
EDIT: As suggested, this can be further simplified such that
$$ A^8 = 4A^4 - 4A^2+ I = 4(2A^2 - I) - 4A^2 + I = 4A^2 -3I .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a > 0$,$b>0$, and $\frac{1}{a} + \frac{1}{b}$ is an integer, prove that $a=b$. And show that $a = 1$ or $2$ If $a$ and $b$ are positive integers, and $\frac{1}{a} + \frac{1}{b}$ is an integer, prove that $a=b$. And show that $a = 1$ or $2$
-I played around with numbers and the conditions and it seems that it's common sense that if $a = b$ then $a = b = 1$ or $2$, For example:
If $a = b = 4$, then $\frac{1}{a} + \frac{1}{b}$ would not be an integer.
-Any help is appreciated
| Suppose $a>b$, then $\frac{1}{a}+\frac{1}{b}<\frac{2}{b}$, and since $\frac{1}{a}+\frac{1}{b}$ is integer with $a>0, b>0$, $$\frac{2}{b}>\frac{1}{a}+\frac{1}{b}\geq1$$or, $2>b$, hence $2>b>0$, and $b=1$.
Hence, $\frac{1}{a}+\frac{1}{b}=\frac{1}{a}+1$ is integer, and so $a=1$ (because $\frac{1}{a}$ is positive integer).
But this is a contradiction, since we supposed that $a>b$. Therefore $a \le b$.
Similarly, if we start by suposing that $a<b$ we find again that $a=b$ : a contradiction. So $a \ge b$.
Combining these two statements gives $a=b$ as required.
In addition, $\frac{1}{a}+\frac{1}{b}=\frac{2}{a}$ is integer, so, since $a>0$, we have $a=1$ or $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Showing Trigonometric Identity Prove that:
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Attempt:
\begin{align*}
L.H.S & = \cos^2\theta\sin^4\theta\\
& = \cos^2\theta\sin^2\theta\sin^2\theta\\
& = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\
& = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta)
\end{align*}
Now, what should I do?
|
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Use $\sin^2 \theta + \cos^2 \theta = 1$ to put everything in terms of cosine:
$$\begin{align}
\cos^2\theta (1 - \cos^2 \theta )^2 &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \\
\cos^2\theta (1 - 2\cos^2\theta + \cos^4 \theta ) &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \\
\cos^2\theta - 2\cos^4\theta + \cos^6 \theta &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)
\end{align}$$
Use $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ :
$$
\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^2
- 2\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^4
+ \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^6
=
\frac{1}{32}\left(
\left(\frac{e^{6i\theta} + e^{-6i\theta}}{2}\right) -
\left(\frac{e^{2i\theta} + e^{-2i\theta}}{2}\right)
+ 2
- 2\left(\frac{e^{4i\theta} + e^{-4i\theta}}{2}\right)
\right)$$
Multiply it out:
$$
\begin{align}
\frac{1}{4} & \left( e^{2i\theta} + 2 + e^{-2i\theta} \right) \\
- \frac{1}{8} & \left( e^{4i\theta} + 4 e^{2i\theta} + 6 + 4 e^{-2i \theta} + e^{-4i\theta} \right) \\
+ \frac{1}{64} & \left( e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \right) \\
=
\frac{1}{32} & \left(
\frac{
\left( e^{6i\theta} + e^{-6i\theta} \right) -
\left( e^{2i\theta} + e^{-2i\theta} \right)
+ 4
- \left(2e^{4i\theta} + 2e^{-4i\theta}\right)
}{2}
\right)
\end{align}$$
Combine like terms:
$$\begin{array} {c}
16 e^{2i\theta} + 32 + 16 e^{-2i\theta} \\
-~ 8 e^{4i\theta} - 32 e^{2i\theta} - 48 - 32 e^{-2i \theta} - 8 e^{-4i\theta} \\
+~ e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \\
=~
e^{6i\theta}
- 2e^{4i\theta}
- e^{2i\theta}
+ 4
- e^{-2i\theta}
- 2e^{-4i\theta}
+ e^{-6i\theta}
\end{array}$$
So
$$\begin{array} {c}
e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \\
= \\
e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta}
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Proof of limit of $x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)$ correct? I want to show that $$\lim_{x\to\infty} x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= 1 $$
But I'm not sure if it's correct:
Because $\arcsin:[-1;1]\to \mathbb{R}$ is continuous and differentiable on $(-1,1)$, the mean value theorem says that there is a $c$ in $\left(\frac{1}{x+1} , \frac{1}{x}\right)$ for $f(t)=\arcsin(t)$ such that
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= f'(c_x) {\frac{x}{1+x}} $$
then
$f'(c_x)=\frac{1}{(\sqrt{1-c_x})}(\frac{x}{x+1}))\leqslant \frac{1}{(\sqrt{1-c_x})}\times\frac{x}{x} = \frac{1}{(\sqrt{1-c_x})} $
with Sandwichlemma for $x\to\infty, c_x\to0$
so the limit of $(x^2(\arcsin{\frac{1}{x}} -\arcsin{\frac{x}{1+x}} )= 1$
| You need to bound the function on both sides, one side is not enough. I'd proceed in a slightly different way, for applying the mean value theorem.
Consider the function
$$
f(x)=\arcsin\frac{1}{x}
$$
defined for $x>1$ (it is defined on a larger set, but $(1,\infty)$ is sufficient for our purposes).
By the mean value theorem, there exists $c_x\in(x,x+1)$ such that
$$
\frac{f(x)-f(x+1)}{x-(x+1)}=f'(c_x)
$$
so
$$
x^2(f(x)-f(x+1))=-x^2f'(c_x)
$$
Now
$$
f'(x)=\frac{1}{\sqrt{1-(1/x)^2}}\left(-\frac{1}{x^2}\right)=
-\frac{1}{x\sqrt{x^2-1}}
$$
so we end up with
$$
x^2(f(x)-f(x+1))=\frac{x^2}{c_x\sqrt{c_x^2-1}}
$$
Let's consider
$$
g(t)=t\sqrt{t^2-1}
$$
with
$$
g'(t)=\sqrt{t^2-1}+\frac{t^2}{\sqrt{t^2-1}}>0
$$
for $t>1$. Thus we can say, from $x<c_x<x+1$, that
$$
g(x)<g(c_x)<g(x+1)
$$
so
$$
\frac{x^2}{g(x+1)}\le x^2(f(x)-f(x+1))\le\frac{x^2}{g(x)}
$$
which is the same as
$$
\frac{x^2}{(x+1)\sqrt{(x+1)^2-1}}\le
x^2(f(x)-f(x+1))\le
\frac{x^2}{x\sqrt{x^2-1}}
$$
The squeeze theorem allows you to conclude.
| {
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"url": "https://math.stackexchange.com/questions/1650854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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integrate $\int \frac{dx}{x\sqrt{1-x}}$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int \frac{\sqrt{u}}{1-u}du-2{\sqrt{u}}$$
$$\int \frac{\sqrt{u}}{1-u}du=\int \frac{1+\sqrt{u}-1}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1+\sqrt{u}}{(1+\sqrt{u})(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1}{(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du$$
$$\int \frac{1}{(1-\sqrt{u})}du$$
How do I continue from here? it seems that I have made it harder
| Set $t=\sqrt{1-x}$, so $x=1-t^2$ and $dx=-2t\,dt$, so the integral becomes
$$
\int\frac{-2t}{(1-t^2)t}\,dt=\int\frac{2}{t^2-1}\,dt=
\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)\,dt
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Principal roots in the derivation of the quadratic formula The following is an abbreviation of a common derivation of the quadratic formula:
$$
ax^2+bx+c=0\\
\vdots\\
\sqrt{(x+\frac{b}{2a})^2}=\sqrt{\frac{b^2-4ac}{4a^2}}\\
x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
$$
Following the square root extraction, the equation should have three plus-minus signs:
$$
\pm(x+\frac{b}{2a})=\frac{\pm\sqrt{b^2-4ac}}{\pm2a}\\
$$
But instead there's only one. Is it because the signs cancel each other out, or are moved from one expression to the other (through multiplication in $-1$ or $\frac{-1}{-1}$) in six of the eight permutations that the three plus-minus signs allow; or is it a matter of choosing a principal root for some arcane reason beyond the scope of elementary algebra (eg. branch points)?
| The solutions of $Y^2=(b^2-4 a c)/4 a^2\;$ are $\;Y=+\sqrt {b^2-4 a c}/2 a\;$ and $\;Y=-\sqrt {b^2-4 a c}/2 a.$
If $(x+b/ 2 a )^2=Y^2$ then $x+b/2 a$ must be equal to one of these solutions or to the other.
So $x+b/2 a=\pm \sqrt {b^2-4 a c}/2 a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a formula for $\sin(xy)$ Can we express a trigonometric function for the product of two angles as a function of trigonometric functions of its factors?
For example: Is there a formula for $\sin(xy)$ as a function of $\sin x$ and $\sin y$ or other trigonometric functions of $x$ and $y$.
| Unlikely to be explicitly/purely in terms of $\sin x$, $\sin y$, etc.
But
$\, \sin xy=\sin \left[ \frac{(x+y)^{2}}{4}-\frac{(x-y)^{2}}{4} \right]
=\sin \frac{(x+y)^{2}}{4} \cos \frac{(x-y)^{2}}{4}-
\cos \frac{(x+y)^{2}}{4} \sin \frac{(x-y)^{2}}{4}$
or approximately for small $\, x$ and $y$,
$\sin xy \approx \sin x \sin y
\left( 1+\frac{\sin^{2} x+\sin^{2} y}{6}+\frac{3}{40}\sin^{4} x-\frac{5}{36}\sin^{2} x \sin^{2} y+\frac{3}{40} \sin^{4} y \right)$
| {
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"url": "https://math.stackexchange.com/questions/1656407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation of this huge series The value of $\dfrac{2^2+1}{2^2-1}+\dfrac{3^2+1}{3^2-1}...+\dfrac{2011^2+1}{2011^2-1}$ is:
*
*In the interval $(2010,2010\frac{1}{2})$
*In the interval $(2011-1/2011,2011-1/2012)$
*In the interval $(2011,2011\frac{1}{2})$
*In the interval $(2012,2012\frac{1}{2})$
I'm staring at it but can't see any trick to solve it.
I think there is some trick as its of the form $a^2+b^2/(a-b)(a+b)$.
| Note that $$\frac{k^2+1}{k^2-1}=\frac{k^2-1}{k^2-1}+\frac{2}{k^2-1}=1+\frac{2}{k^2-1}=1+\frac{1}{k-1}-\frac{1}{k+1}$$
So that \begin{align}
\sum_{k=2}^{2011}\frac{k^2+1}{k^2-1}&=\sum_{k=2}^{2011}\left(1+\frac{2}{k^2-1}\right)\\
&=2010+\sum_{k=2}^{2011}\frac{1}{k-1}-\sum_{k=2}^{2011}\frac{1}{k+1}\\
&=2010+\frac{1}{1}+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011}+\sum_{k=3}^{2010}\frac{1}{k}-\sum_{k=3}^{2010}\frac{1}{k}\\
&=2010+\frac{1}{1}+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011}
\end{align}
so that is just a little less than $2011\frac12$, so it's in the interval $(2011,2011\frac12)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find Eigen values and eigen vectors $\displaystyle F=\begin{pmatrix}1 & 1 & 2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}$
Expanding along $R_3$, I got the equation $(1-\lambda)(\lambda)(\lambda-2)=0$
Eigenvector corresponds to $\lambda=1$: $\displaystyle \begin{pmatrix}1\\-2\\0\end{pmatrix}$.
Got stuck for $\lambda=2$.
$$\begin{pmatrix} -1 & 1 & 2\\0 & -1 & 0\\0 & 0 & -1\end{pmatrix}$$
Cannot obtain the eigenvector $\displaystyle \begin{pmatrix}1\\0\\0\end{pmatrix}$.
| The matrix is triangular, hence its eigenvalues are on the diagonal. In this case we have $\lambda=1$ with algebraic multiplicity of $3$.
For $\lambda=1$ we get $(F-\lambda I)v=0\Rightarrow \begin{pmatrix}0&1&2\\0&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}=0$ And we get $v_2=-2v_3$, thus the solution is of the form $\displaystyle \begin{pmatrix}v_1\\-2v_3\\v_3\end{pmatrix}=v_1\begin{pmatrix}1\\0\\0\end{pmatrix}+v_3\begin{pmatrix}0\\-2\\1\end{pmatrix}$, hence the eigenspace corresponds to $\lambda=1$ is $\displaystyle \left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\-2\\1\end{pmatrix}\right\}$.
As we can see, the geometric multiplicity of $\lambda=1$ is $2$, so this matrix is not diagonalizable.
| {
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"url": "https://math.stackexchange.com/questions/1657040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$ by mathematical induction
Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$
My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.
(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6
We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$
I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?
| I'm utterly confused. Three hours before asking this, you asked (and had answered) this: Proof of for all integer $n \ge 2$, $n^3-n$ is divisible by 6 by mathematical induction.
In that one you asked, and got answer for, how to show $6| 3k^2 + 3k$. In this one you are asking for how to show $6| 3k^2 + 3k + 6$.
How can you know the answer to one but not the answer to the other?
Answer to both: $3|3*h$ for any integer h, so $3|3(k^2 + k)$. If $k$ is odd so is $k^2$ so $k^2 + k$ is the sum of two odd numbers and is even. If $k$ is even so is $k^2$ and so is $k^2 + k$. So $k^2 + k$ is even. So $2|(k^2 +k)$ so $6|3(k^2 + k)$ so $6|3(k^2 + k) + 6$.
BTW: $n(n^2 + 5) = n^3 + 5n = (n^3 -1) + 6n$ so one is divisible by 6 if and only if the other one is.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$
My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+\cos^2\theta)\cos^2\theta}{\sin^5\theta}d\theta$
I could not solve it further.
| Take the partial fractions of the integrand which are $\frac{1}{2(x-1)^3} -\frac{1}{2(x+1)^3}$ which can be easily integrate using substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) =$
If $\alpha,\beta,\gamma$ be three distinct real values such that $$\displaystyle \frac{\sin \alpha+\sin \beta+\sin \gamma}{\sin(\alpha+\beta+\gamma)} = \frac{\cos \alpha+\cos \beta+\cos \gamma}{\cos(\alpha+\beta+\gamma)}=2$$
Then value of $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) = $
$\bf{My\; Try::}$ Using
$\sin (\alpha+\beta+\gamma) = \sin \alpha \cos \beta \cos \gamma+\cos \alpha \sin \beta \cos \gamma+\cos \alpha \cos \beta \sin \gamma-\sin \alpha \sin \beta \sin\gamma$
$\cos (\alpha+\beta+\gamma) = \cos \alpha \cos \beta \cos \gamma-\sin \alpha \sin \beta \cos \gamma-\cos \alpha \sin \beta \sin \gamma-\sin \alpha \cos \beta \sin \gamma$
But putting that values in above expression form very lengthy expression,
How can we solve it in less complex way,
Thanks
| The three expressions are symmetric wrt $\alpha, \, \beta, \, \gamma$, which means that,
if they are valid for a given triple, they shall be valid for any permutation of that triple.
Thus if $(\alpha, \, \beta, \, \gamma)$ is a solution that lies on the plane $\alpha + \beta + \gamma=s$, the permutations will also lie on that plane.
Therefore
$$
\eqalign{
& \cos \left( {\alpha + \beta } \right) + \cos \left( {\beta + \gamma } \right) + \cos \left( {\alpha + \gamma } \right) = \cr
& = \cos \left( {s - \alpha } \right) + \cos \left( {s - \beta } \right) + \cos \left( {s - \gamma } \right) = \cr
& = \cos s\;\left( {\cos \alpha + \cos \beta + \cos \gamma } \right) + \sin s\left( {\sin \alpha + \sin \beta + \sin \gamma } \right) \cr
& = \cos ^{\,2} s\;{{\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)} \over {\cos s}}
+ \sin ^{\,2} s{{\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)} \over {\sin s}} = \cr
& = 2\left( {\cos ^{\,2} s + \sin ^{\,2} s} \right) = 2 \cr}
$$
which is independent of $s$, and is therefore the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this equation if $x\in \mathbb{R}$, $x ^3+1 =2\sqrt [ 3 ]{ 2x-1 }$, find $x$
If $x\in \mathbb{R}$, and $$x ^3+1 =2\sqrt [ 3 ]{ 2x-1 }$$ find $x$.
I asked my teacher to help me in solving it, he responded I can't! So, I hope you help me in solving it or even giving me hints that will help me in solving.
| By factoring we see that
$$
x^9+3x^6+3x^3-16x+9=(x^6 + 2x^4 + 2x^3 + 4x^2 + 2x + 9)(x^2 + x - 1)(x - 1),
$$
so that we have solutions $x=1$, $x=\frac{\pm \sqrt{5}-1}{2}$ and $6$ further complex solutions. So we have exactly three real solutions, easy to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$
Integrate $$\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$$
$$\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}=\int \frac{\mathrm dx}{(x-2)(\sqrt{(x-2)^2-(1)^2})}=\operatorname{arcsec}(x-2)+\rm C$$
Wolfram writes that the answer is $-\arctan\left(\frac{1}{\sqrt{x^2-4x+3}}\right)+\rm C\;.$
| if you draw a right angled triangle you can see that $\operatorname{arcsec}(x-2)$ is equivalent to $\arctan\sqrt{x^2-4x+3}$ which in turn is equivalent to $\frac{\pi}{2}-\arctan\frac{1}{\sqrt{x^2-4x+3}}$, so the answers are the same give or take a constant
| {
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"url": "https://math.stackexchange.com/questions/1661343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\int \frac{\sin(3x)}{\sin(5x)} \, dx$ How to find integral of
$$\int \frac{\sin(3x)}{\sin(5x)}dx$$
I wrote $\sin(3x)=\sin(8x-5x)$ but it generated $\frac{\sin(8x) \cos(5x)}{\sin(5x)}$.
How should I proceed?
| Using Chebyshev polynomials (that can be used also for sine when you have odd multipliers for the angle) we have: $$
\sin3x = 3\sin x-4\sin^3x \\
\sin5x = 16\sin^5x-20\sin^3x+5\sin x
$$
So: $$
\frac{\sin3x}{\sin5x}=\frac{3-4\sin^2x}{16\sin^4x-20\sin^2x+5}=\frac{-\left (4\sin^2x-\frac{5}{2} \right )+\frac{1}{2}}{\left (4\sin^2x-\frac{5}{2} \right )^2-\frac{5}{4}}
$$ Let's convert sines to cosines (they look better when you integrate): $$
\frac{\sin3x}{\sin5x}=\frac{-\left (4(1-\cos^2x)-\frac{5}{2} \right )+\frac{1}{2}}{\left (4(1-\cos^2x)-\frac{5}{2} \right )^2-\frac{5}{4}}=\frac{\left (4\cos^2x-\frac{3}{2} \right )+\frac{1}{2}}{\left (4\cos^2x-\frac{3}{2} \right )^2-\frac{5}{4}}
$$ Now we can decompose the denominator: $$
\frac{u+\frac{1}{2}}{u^2-\frac{5}{4}}=\frac{a}{u-\sqrt{\frac{5}{4}}}+\frac{b}{u+\sqrt{\frac{5}{4}}} \\ a=\frac{\sqrt{5}+1}{2\sqrt{5}}, b=\frac{\sqrt{5}-1}{2\sqrt{5}}
$$ Then: $$
\frac{\sin3x}{\sin5x}=\frac{\frac{\sqrt{5}+1}{2\sqrt{5}}}{4\cos^2x-\frac{3}{2}-\sqrt{\frac{5}{4}}}+\frac{\frac{\sqrt{5}-1}{2\sqrt{5}}}{4\cos^2x-\frac{3}{2}+\sqrt{\frac{5}{4}}}
$$ For both the denominators we have a negative constant term, so let's try to find a solution for: $$
I(a^2, b^2)=\int {\frac{1}{a^2cos^2x-b^2}}dx=\int {\frac{1+\tan^2x}{a^2-b^2-b^2\tan^2x}}dx$$ using $u=\tan x$, $du=(1+\tan^2x)dx$: $$
I(a^2, b^2)=\int {\frac{1}{a^2-b^2-b^2u^2}}du=\frac{1}{b^2}\int {\frac{1}{\frac{a^2-b^2}{b^2}-u^2}}du=\frac{1}{b^2}\frac{b}{\sqrt{a^2-b^2}}\tanh^{-1}\frac{bu}{\sqrt{a^2-b^2}}=\frac{1}{b\sqrt{a^2-b^2}}\tanh^{-1}\frac{b\tan x}{\sqrt{a^2-b^2}}
$$ Carefully doing the substitutions I found: $$
\int {\frac{\sin3x}{\sin5x}}dx= I\left (4, \frac{3+\sqrt{5}}{2}\right ) + I\left (4, \frac{3-\sqrt{5}}{2}\right )= \frac{\sqrt{5}+1}{4\sqrt{5}}\frac{1}{\sqrt{3+\sqrt{5}}\sqrt{5-\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3+\sqrt{5}}}{\sqrt{5-\sqrt{5}}}\tan x+\frac{\sqrt{5}-1}{4\sqrt{5}}\frac{1}{\sqrt{3-\sqrt{5}}\sqrt{5+\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3-\sqrt{5}}}{\sqrt{5+\sqrt{5}}}\tan x + C
$$
| {
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"url": "https://math.stackexchange.com/questions/1667307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ conditionally convergent $$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$$
I want to show first that $S_{3n}$, $S_{3n+1}$ and $S_{3n+2}$ converges to the same limit, I show
$$S_{3n} = (1 + {1 \over 3} - {1 \over 2}) + ({1 \over 5} + {1 \over 7} - {1 \over 4}) + ({1 \over 9} + {1 \over 11} - {1 \over 6}) + ... + ({1 \over 4n-3} + {1 \over 4n-1} - {1 \over 2n})$$
but how to proceed from here?
| $$|S_{3n+1}-S_{3n}| \leq \frac{1}{n}$$
$$|S_{3n+2}-S_{3n}| \leq \frac{1}{n}$$
So if $(S_{3n})$ converges, all the three converges
No to prove that $(S_{3n})$ converges, all you need to do is to write $\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}$ over the same denominator and use a comparaison.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integrate $\int \frac{dx}{e^{2x}+e^{x}+6}$
$$\int \frac{dx}{e^{2x}+e^{x}+6}$$
I have tried to start with substitution.
Is $u=e^x$ than $du=e^xdx\Rightarrow \frac{du}{e^x}=dx$ therefore we have
$$\int \frac{du}{u(u^{2}+u+6)}$$
Is applicable?
| You're correct. To continue, use Partial Fractions: $$\frac{1}{u\left(u^2+u+6\right)}=-\frac{1}{6}\cdot \frac{u+1}{u^2+u+6}+\frac{1}{6}\cdot \frac{1}{u}$$
$$\int \frac{du}{u\left(u^2+u+6\right)}=-\frac{1}{12}\left(\int \frac{d\left(u^2+u+6\right)}{u^2+u+6}+\int \frac{du}{u^2+u+6}\right)+\frac{1}{6}\int \frac{du}{u}$$
$$=-\frac{1}{12}\left(\ln\left|u^2+u+6\right|+2\int \frac{d(2u+1)}{(2u+1)^2+23}\right)+\frac{1}{6}\ln|u|$$
Let $2u+1=t\sqrt{23}$.
$$\int\frac{d(2u+1)}{(2u+1)^2+23}=\frac{1}{\sqrt{23}}\int \frac{dt}{t^2+1}=\frac{1}{\sqrt{23}}\arctan t+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Laurent series for $f(z)=\frac{2}{(z-4)}-\frac{3}{(z+1)}$ The question is 2 parts - I'm to find the Laurent series valid for $$1 < |z| <4$$ and $$|z| > 5$$ I've already solved the first part, but I have a conceptual question about the second part.
My result for the first part:
I found that after putting the $f(z)$ into the standard geometric series form, it resulted in
$$-\sum_{n=0}^\infty \frac{z^n}{2^{2n+1}}; |z|<4$$
$$-\sum_{n=0}^\infty \frac{2^{2n-1}}{z^n}; |z|>4$$
$$-\sum_{n=0}^\infty (-3)^nz^n; |z|<1$$
$$-\sum_{n=0}^\infty \frac{(-3)^n}{z^n}; |z|>1$$
Thus,
$$f(z)=-\sum_{n=0}^\infty \frac{z^n}{2^{2n+1}}-\sum_{n=0}^\infty \frac{(-3)^n}{z^n} $$ for $1<|z|<4$.
The way I understand this is that a Laurent series is simply another way to represent a function, $f(z)$, but in a way that puts into focus what happens around the singularities of the function.
My attempt for part 2:
Given my answer above and my reasoning for what it means, I'd say that this sum is the answer. It explains what happens at all radius greater than $4$.
$$f(z)=-\sum_{n=0}^\infty \frac{2^{2n-1}}{z^n}; |z|>4$$
Is this the correct way to understand this?
| For $|z| < 4$,
$$\begin{align}\frac 2{z - 4} &= \frac 2{-4}\frac 1{1 - \frac z 4}\\
&=\frac {-1}2\sum_{n=0}^\infty \left(\frac z4\right)^n\\
&=-\sum_{n=0}^\infty\frac {z^n}{2^{2n+1}}\end{align}$$
For $|z| > 4$,
$$\begin{align}\frac 2{z - 4} &= \frac 2{z}\frac 1{1 - \frac 4 z}\\
&=\frac 2z\sum_{n=0}^\infty \left(\frac 4z\right)^n\\
&=\sum_{n=0}^\infty\frac {2^{2n+1}}{z^{n+1}}\\
&= \sum_{n=1}^\infty\frac {2^{2n-1}}{z^{n}}\end{align}$$
for $|z| < 1$,$$\begin{align}\frac {-3}{z + 1} &= -3\frac 1{1 - (-z)}\\
&=-3\sum_{n=0}^\infty (-z)^n\\
&=-3\sum_{n=0}^\infty(-1)^nz^n\end{align}$$
for $|z| > 1$ $$\begin{align}\frac {-3}{z + 1} &= \frac {-3}z\frac 1{1 - (-\frac 1z)}\\
&=\frac {-3}z\sum_{n=0}^\infty \left(-\frac 1z\right)^n\\
&=3\sum_{n=0}^\infty\frac {(-1)^{n+1}}{z^{n+1}}\\
&=3\sum_{n=1}^\infty\frac {(-1)^{n}}{z^{n}}\end{align}$$
Hence for $|z| < 1$,
$$f(z) = -\sum_{n=0}^\infty\frac {z^n}{2^{2n+1}} - 3\sum_{n=0}^\infty(-1)^nz^n = -\sum_{n=0}^\infty \frac{1 + 3(-1)^n}{2^{2n+1}}z^n$$
For $1 < |z| < 4$,
$$f(z) = -\sum_{n=0}^\infty\frac {z^n}{2^{2n+1}} + \sum_{n=1}^\infty\frac {3(-1)^{n}}{z^{n}}$$
For $4 < |z|$,
$$f(z) = \sum_{n=1}^\infty\frac {2^{2n-1}}{z^{n}} + 3\sum_{n=1}^\infty\frac {(-1)^{n}}{z^{n}} = \sum_{n=1}^\infty \frac{2^{2n-1} + 3(-1)^n}{z^n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining whether there are solutions to the cubic polynomial equation $x^3 - x = k - k^3$ other than $x = -k$ for a given parameter $k$ Let $k$ be a real parameter, and consider the equation
$$x^3 - x = k - k^3 .$$
Obviously, $x=-k$ is a solution. Is it the only one? How to prove it?
| Rearranging gives $$x + k = x^3 + k^3 = (x + k) (x^2 - x k + k^2),$$ and for any solution other than $x = -k$, we can divide both sides by $x + k$, leaving
$$x^2 - x k + k^2 = 1.$$
As a function of $(x, k)$ the left-hand side is a positive-definite quadratic form, so the solution set $(x, k)$ of this latter equation over $\Bbb R$ is a (nondegenerate) ellipse in the $xk$-plane. So, whether there are solutions other than $x = -k$ depends on whether the line corresponding to the value of the parameter $k$ intersects this ellipse or not.
We now use some elementary calculus to find for which values it does: If we regard $k$ as an implicit function of $x$, differentiating gives
$$2 x - \left(x \frac{dk}{dx} + k\right) + 2 k \frac{dk}{dx} = 0 ,$$ and where $k$ takes on an extremal value we have $\frac{dk}{dx} = 0$. Substituting and rearranging gives $x = \frac{k}{2} ,$ and substituting in the ellipse equation gives
$$\left(\frac{k}{2}\right)^2 - \left(\frac{k}{2}\right) k + k^2 = 1.$$ Simplifying gives $$\frac{3 k^2}{4} = 1,$$ and solving gives the extremal values of $k$ on the ellipse, that is, the extremal values of $k$ for which there is a solution other than $x = -k$, namely, $x = \pm \frac{2}{\sqrt{3}}$. (Strictly we must check that such solutions are not all on the line $x = -k$, but this is immediately clear from a graph, and is anyway not hard to see algebraically.) We conclude that there is a solution to the equation other than $x = -k$ if the parameter $k$ satisfies
$$-\frac{2}{\sqrt{3}} \leq k \leq \frac{2}{\sqrt{3}} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Riemann integral of $ f(x) = \begin{cases} \frac{1}{n}, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\ 0, & \text{other where} \end{cases}$ Let $f:[0, 1]\to\mathbb{R}$ and $ f(x) =
\begin{cases}
\frac{1}{n}, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\
0, & \text{other where}
\end{cases}$
I want to find $\int_{0}^{1} f\, dx.$ I think the answer is $0$.
Any ideas or insight would be greatly appreciated
| You have $0 \le f \le \bar{f}$ where
$$ \bar{f}(x) =
\begin{cases}
1, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\
0, & \text{else}
\end{cases}$$
The Riemann integral of $\bar{f}$ is equal to $0$, hence $f$ Riemann integral is also equal to $0$.
To prove it consider the step functions
$$ \bar{f_p}(x) =
\begin{cases}
1, & \text{if $x \in (\frac{1}{n}-\frac{1}{2^p n(n+1)},\frac{1}{n}+\frac{1}{2^p n(n+1)}) \cap [0,1] \ \ n=1, 2, \dots ,p$ } \\
1, & \text{if $x \in [0,\frac{1}{p+1})$ } \\
0, & \text{else}
\end{cases}$$
For all $p \ge 1$ you have $$0 \le f \le \bar{f} \le \bar{f_p}$$ and $$\int_0^1 \bar{f_p(x)} \ dx = \frac{1}{p+1} + \frac{1}{2^{p+1}}\sum_{k=1}^p \frac{1}{k(k+1)}$$
As the RHS is converging to $0$ as $p \to \infty$, you are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
For which integers $a,b$ does $ab-1$ divide $a^3+1$? A problem I wasn't able to solve:
For which values of $a,b\in\mathbb{Z}$ does $ab-1$ divide $a^3+1$?
I am looking for every possible solution. Some of them are trivial, like $a=0,b=0$ or $(a,b)\in\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,5),(5,3)\}.$
You may notice it is very similar to the famous 1988 IMO problem #6, and I bet Vieta jumping is the key, but an elliptic curve seems to be involved. You may also notice that $(ab-1)\mid (a^3+1)$ implies $(ab-1)\mid (a^3 b^3+b^3)$, hence $(ab-1)\mid (a^3+1)$ is equivalent to $(ab-1)\mid (b^3+1)$.
In order to avoid trivial cases, we may assume $|a|>1$ and $|b|>1$ without loss of generality. Given
$$ a^3+1 = (ab-1)\cdot k \tag{1}$$
we must have $k\equiv -1\pmod{a}$, i.e. $k=(ac-1)$. The previous identity then becomes:
$$ a^2-(bc)a+(b+c)=0 \tag{2}$$
hence every solution $(a,b)$ is associated with other solutions $\left(a,\frac{a^2+b}{ab-1}\right),\left(a,\frac{b^2+a}{ab-1}\right)$.
$b^2 c^2-4(b+c)$ has to be a square: that obviously cannot happen if $b^2 c^2-4(b+c)<b^2 c^2$ and $b^2 c^2-4(b+c)>(bc-1)^2$: that observation leads to the fact that the only solutions in $\mathbb{N}$ are the ones listed above, but what about the other solutions in $\mathbb{Z}$?
| A solution can be obtained in a more direct manner than Vieta jumping.
First, if $ab-1\mid a^3+1$, then $ab-1\mid a^2+b$ since $b\cdot(a^3+1)=a^2\cdot(ab-1)+a^2+b$. Similarly, $ab-1\mid b^2+a$ and $ab-1\mid b^3+1$. So, $ab-1$ divides either all or none of $a^3+1$, $a^2+b$, $b^2+a$, and $b^3+1$.
In particular, as has already been noted, if $(a,b)$ is a solution, then so is $(b,a)$, so we may restrict ourselves to solutions with $|a|\le|b|$. For small, fixed $a$, we may simply check all $b=(k+1)/a$ for integers $k\mid a^3+1$. This gives us the following solutions with $|b|\ge|a|$, assuming I haven't missed any:
*
*$a=0$ and any $b$
*$a=1$ and $b=-1,2,3$
*$a=-1$ and any $b$
*$a=2$ and $b=-4,2,5$
*$a=-2$ and $b=-4,3$
*$a=3$ and $b=-9,5$
*$a=-3$ and $b=-9,4$
In the following, we may therefore assume $4\le|a|\le|b|$.
If $ab-1\mid a^2+b$, either $b=-a^2$, which is always a solution, or $|a^2+b|\ge|ab-1|$. The latter, we split into two cases based on the sign of $a^2+b$.
If $a^2+b>0$, we get $a^2+b\ge|ab-1|$ from which follows that $|a|^2+|b|\ge|a|\cdot|b|-1$. Still assuming $|b|\ge|a|$, this gives
$$
\bigl(|b|-|a|-1\bigr)\bigl(|a|-1\bigr)=|a|\cdot|b|-|a|^2-|b|+1\le2
$$
which implies that $|b|$ is either $|a|$ or $|a|+1$ when $|a|>3$. We can then check the alternatives $b=a$, $b=-a$, $b=a\pm 1$, $-b=a\pm 1$ (with $\pm$ depending on sign of $a$) to verify these give no additional solutions.
Similarly, if $a^2+b<0$, we have $-b>a^2$, which with $|b|\ge|a|\ge4$ makes
$$|ab-1|\ge|a|\cdot|b|-1>|b|=-b>-b-a^2>0$$
for which there is no solution.
Although it is quite likely that I've made some mistake along the way, the approach should work.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How do you integrate $(x+2)\ln(x-3)$? I got $$\left(\frac {x^2}{2} +2x\right)\ln(x-3)-\left(\frac {x^2}{4}-\frac {7x}{2} -\frac {21}{2}\right)\ln(2x-6)$$ as my answer... Not sure If I got it right. Please correct me, thank you!
| With $u=\ln(x-3)$, $du = \frac{1}{x-3} dx$, $dv = (x-3) dx$, $v = \left(\frac{x^2}{2}+2x\right)$,
$$ \int (x+2) \ln(x-3) dx = \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{x-3} \left(\frac{x^2}{2}+2x\right) dx \\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{w} \left(\frac{(w+3)^2}{2}+2(w+3)\right) dw \\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{w} \left(\frac{w^2+6w+9}{2}+2w+6)\right) dw \\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \left( \frac{1}{2}w + 5 +\frac{21}{2w} \right)dw \\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \frac{w^2}{4} -5 w - \frac{21}{2} \ln w + C\\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \frac{(x-3)^2}{4} -5 (x-3) - \frac{21}{2} \ln (x-3) + C\\
= \left(\frac{x^2}{2}+2x - \frac{21}{2}\right) \ln(x-3) - \frac{(x-3)^2}{4} -5x + C$$
EDIT: Here's how you can check if your answer is right.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 4
} |
Find the limit of the cosine sequence. The question is to find $\lim\limits_{n\to\infty} \cos \frac{x}{2} \cdot \cos \frac{x}{2^2} \cdot \cos \frac{x}{2^3}\cdot\cdot\cdot \cos \frac{x}{2^n}$.
I first thought of several facts:
1). $-1 \leq \cos(a) \leq 1$ (From this I know the sequence is also bounded by -1 and 1)
2). $\cos(0) = 1$ (This is what the items in the sequence tend to, as n becomes bigger)
3). When $x=0$, the limit is 1.
Then I failed to think more and the above do not get me anywhere close to finding the limit...
Can anybody teach me how to find the limit of this sequence? Thanks a lot.
| If $x \ne 0$, then
\begin{align}
\lim_{n\to\infty}\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^n}&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^n}\sin\frac{x}{2^n}}{\sin\frac{x}{2^n}}\\
&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^{n-1}}\sin\frac{x}{2^{n-1}}}{2\sin\frac{x}{2^n}}\\
&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^{n-2}}\sin\frac{x}{2^{n-2}}}{2^2\sin\frac{x}{2^n}}\\
&=\cdots\\
&=\lim_{n\to\infty}\frac{\sin x}{2^n \sin\frac{x}{2^n}}\\
&=\frac{\sin x}{x}.
\end{align}
If $x=0$, then given limit goes to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \frac{2x}{x^2 +2x+1}$ Here is my attempt:
$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$
Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.
| Your answer is almost correct, but you should put modulus sign in your log's argument as it must be positive i.e your answer should be;
$$2\ln|x+1|+\frac{2}{x+1}+C$$
You can try one more method too.
$$\int \frac{2x}{x^2+2x+1}dx$$
Try to create derivative of denominator in numerator
$$\int \frac{2x+2-2}{x^2+2x+1}dx$$
$$\int \frac{2x+2}{x^2+2x+1}dx-2\int \frac{1}{x^2+2x+1}dx$$
$$\int \frac{2x+2}{x^2+2x+1}dx-2\int \frac{1}{(x+1)^2}dx$$
For first integral simply take $x^2+2x+1=t$, then $(2x+2)dx=dt$, hence you will get:
$$\ln(x^2+2x+1)+\frac{2}{x+1}+C$$
$$\ln(x+1)^2+\frac{2}{x+1}+C$$
$$2\ln|x+1|+\frac{2}{x+1}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Finding max perimeter of triangle of three circulating points I'm thinking a plane geometry problem, and it seems quite puzzling.
Here it is.
Question:
Consider three concentric circles with radius 3, 5 and 7 each.
and construct a triangle by picking one point from each circle.
Question is, in which condition that triangle has maximum perimeter?
Maybe we can try to use law of cosine and calculus? ... but I want more geometric argument.
How can we find the condition by using some geometry?
| Summary
*
*the condition is: $O$ is the in-center for $\triangle ABC$.
*the maximal perimeter $p \approx 26.47085983465459$
*
*Let $a = 3$, $b = 5$ and $c = 7$.
*For any $r > 0$, let $\mathcal{C}_r$ be the circle centered at $O$ with radius $r$.
*Let $A, B, C$ be three points lie on the circle $\mathcal{C}_a$,$\mathcal{C}_b$ and $\mathcal{C}_c$ respectively.
*Furthermore, assume $A, B, C$ are positioned such that the perimeter $p \stackrel{def}{=} AB + BC + CA$ of $\triangle ABC$ , is maximized.
Let $\mathcal{E}$ be the locus of point $P$ such that $PB + PC = AB + AC$.
$\mathcal{E}$ an ellipse with foci $B$, $C$ intersecting $\mathcal{C}_a$ at $A$. $\mathcal{E}$ cannot intersect $\mathcal{C}_a$ transversely. Otherwise, we can a find a point $P$ on $\mathcal{C}_a$ near $A$ outside $\mathcal{E}$. This contradict with the condition that the perimeter of $\triangle ABC$ is maximized.
As a result, $\mathcal{E}$ and $\mathcal{C}_a$ are tangent to each other at $A$.
As an ellipse, the normal line of $\mathcal{E}$ at $A$ is a angular bisector of the $\angle CAB$. Since $\mathcal{E}$ and $\mathcal{C}_a$ share the same normal line at $A$ and $O$ lies on this normal line. We find $O$ lies on the angular bisector for $\angle CAB$.
Apply a similar argument to $B$, $C$, we find $O$ lies on the three angular bisectors of $\triangle ABC$ and hence is the in-center.
*
*Let $r$ be the in-radius for $\triangle ABC$.
*Let $\alpha = \frac12 \angle CAB$, $\beta = \frac12 \angle ABC$ and $\gamma = \frac12 \angle BCA$.
By elementary geometry, we know
$a\sin\alpha = b\sin\beta = c\sin\gamma = r$
and
$$
p = 2(a\cos\alpha+b\cos\beta+c\cos\gamma)
= 2\left(\sqrt{a^2-r^2}+\sqrt{b^2-r^2}+\sqrt{c^2-r^2}\right)
$$
Since $\alpha + \beta + \gamma = \frac{\pi}{2}$, it is easy to show they satisfy
following trigonometric identity.
$$1 - \sin\alpha^2 - \sin\beta^2 - \sin\gamma^2 - 2\sin\alpha\sin\beta\sin\gamma = 0$$
This implies $r$ is a positive root of the cubic equation:
$$1 - \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)r^2 - \frac{2}{abc} r^3 = 0
$$
Define $\rho$ and $t$ such that
$$\frac{1}{\rho^2} = \frac13 \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)\quad\text{ and }\quad r = \frac{\rho}{2t}$$
We can simplify above to
$$4t^3 - 3t = \frac{\rho^3}{abc}$$
Compare this with the triple angle formula for cosine: $\cos(3\theta) = 4\cos\theta^3 - 3\cos\theta$, we get
$$r = \frac{\rho}{2\cos\phi}\quad\text{ where }\quad
\phi = \frac13\left[ \cos^{-1}\left(\frac{\rho^3}{abc}\right) + 2\pi N\right]
\quad\text{ for some }\quad N \in \{0,1,2\}
$$
For the case at hand, $(a,b,c) = (3,5,7)$ and $\rho = 105\sqrt{\frac{3}{1891}}$.
One can check that only $N = 0$ gives us a positive $r$. As a result, the inradius
$r$ for the optimal configuration is
$$r =
\frac{
105\sqrt{\frac{3}{1891}}
}{
2\cos\left\{
\frac13\left[
\cos^{-1}\left(105^2\left(\frac{3}{1891}\right)^{3/2}\right)
\right]
\right\}} \approx 2.167727449130353
$$
and the maximal perimeter equals to
$$p = 2\left(\sqrt{3^2-r^2} + \sqrt{5^2-r^2}+\sqrt{7^2-r^2}\right)
\approx 26.47085983465459
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $=$ $32$ and $\log_3(x+y)+\log_3(x-y)=1$
Question:
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $= 32$ and $\log_3(x+y)+\log_3(x-y)=1$
My attempt: With the first equation
$$4^{\frac{x}{y} + \frac{y}{x}} = 32$$
$$2^{2(\frac{x}{y} + \frac{y}{x})} = 2^5$$
$$ 2(\frac{x}{y} + \frac{y}{x}) = 5 $$
$$ \frac{x}{y} + \frac{y}{x} = \frac{5}{2} $$
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
Now with the second equation
$$\log_3(x+y)+\log_3(x-y)=1$$
$$\log_3((x+y)(x-y)) = 1 $$
$$ \log_3(x^2-y^2) = 1 $$
$$ x^2-y^2 = 3$$
Now I have 2 equations:
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
$$ x^2-y^2 = 3$$
Now I am stuck..
| You have the two equations $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\tag 1$$ $$x^2-y^2 = 3\tag 2$$ From $(2)$, $y=\pm \sqrt{x^2-3}$. So, plugging in $(1)$ $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\implies 2(x^2+y^2)=5 x y \implies 2(2x^2-3)=\pm 5x\sqrt{x^2-3}\tag 3$$ Square both sides $$4(2x^2-3)^2=25x^2(x^2-3)$$ Expand and group terms to get $$-9 x^4+27 x^2+36=0$$ which is a quadratic in $t$ if $t=x^2$; its roots are $-1$ and $4$.
I am sure that you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to solve this irrational integral I want to solve the integral
$$\large\int\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx$$
I've gone with the varialbe change method using
$$
2x-3=u\
$$
and got
$$
\frac{1}{2}\int \frac{2u^\frac{1}{5}-1}{uu^\frac{1}{5}+u^\frac{1}{5}} du\
$$
then I used shift
$$
u=t^5
$$
and got
$$
\frac{5}{2}\int\frac{2t^4-t^3}{t^5+1} dt\
$$
This is the part that I'm stuck on.
Any kind of advise would be helpfull.
| HINT:
$$\int\frac{2\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\space\text{d}x=2\int\frac{\sqrt[5]{2x-3}-1}{\left(2x-3\right)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\space\text{d}x=$$
Substitute $u=2x-3$ and $\text{d}u=2\space\text{d}x$:
$$\int\frac{\sqrt[5]{u}-1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\int\left[\frac{\sqrt[5]{u}}{u\sqrt[5]{u}+\sqrt[5]{u}}-\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\right]\space\text{d}u=$$
$$\int\frac{\sqrt[5]{u}}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\int\frac{1}{1+u}\space\text{d}u-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$
Substitute $s=1+u$ and $\text{d}s=\text{d}u$:
$$\int\frac{1}{s}\space\text{d}s-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\ln\left|s\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$
$$\ln\left|1+u\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=\ln\left|1+2x-3\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$
$$\ln\left|2x-2\right|-\int\frac{1}{u\sqrt[5]{u}+\sqrt[5]{u}}\space\text{d}u=$$
Substitute $p=\sqrt[5]{u}$ and $\text{d}p=\frac{1}{5\sqrt[5]{u^4}}\space\text{d}u$:
$$\ln\left|2x-2\right|-5\int\frac{p^4}{p^6+p}\space\text{d}p=\ln\left|2x-2\right|-5\int\frac{p^3}{p^5+1}\space\text{d}p$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding an angle in a triangle, when the length of one side is unknown and the distances from each vertex to an arbitrary point is known Given the triangle $\triangle ABN$ I would like to find the angle $\angle ANB$. The lengths of $AN$ and $BN$ are known, but the length $AB$ is unknown.
There exists another point $S$ that forms two more triangles $\triangle ANS$ and $\triangle BNS$. The lengths of $NS, AS$ and $BS$ are known. The location of $S$ is arbitrary, it may be inside or outside $\triangle ABN$.
Using these triangles $\angle ANS$ and $\angle BNS$ can be calculated. These can then be used to find $\angle ANB$. However, There are multiple ways to calculate ANB depending on the position of $S$, as shown by these two figures:
I've also come up with other equations that are required when $S$ is in different positions.
Is there a better way to find $\angle ANB$?
If there is not, is there a way to determine which of the equations to use to calculate $\angle ANB$ from $\angle ANS$ and $\angle BNS$?
The coordinates of the four points $A$, $B$, $N$ and $S$ are all unknown.
|
As stated, the problem does not necessarily have a single solution. Example:
AN = 1.0, BN = 0.65, NS = 0.92, AS = 0.81, BS = 0.35
B's coordinates can be either {0.45509,0.354363} or
{0.733145,0.592696}, and the values of angle ANB is different for
these two sets of coordinates.
Note that I am using approximate numbers here, but the roundoff errors
are small enough to be ignored.
I hope to expand on my answer by providing a closed formula for the
two possible values of ANB (when they exist).
EDIT: General answer:
Create a Cartesian grid such that $N$ is at the origin, $A$ lies on
the positive x axis and $S$ lies in the upper half plane. You can
always do this using translation, rotation, and reflecting without
affecting any lengths:
(note that $A$ and $N$ above are flipped from my counterexample
earlier, as it turns out the calculations are easier with $N$ at the
origin; I've also replaced the side lengths with lowercase letters
since I'm using Mathematica to help solve this problem)
Although I've draw B and S in quadrant I, B can be in any quadrant,
and S can be in quadrants I or II (since we explicitly chose our axis
system so that sy > 0).
We can now find sx and sy by solving the simultaneous equations:
$\text{sx}^2+\text{sy}^2=d^2$
$(\text{sx}-c)^2+\text{sy}^2=e^2$
The are two solutions, but only one with sy > 0:
$
\left\{\text{sx}\to \frac{c^2+d^2-e^2}{2 c},\text{sy}\to
\sqrt{d^2-\frac{\left(c^2+d^2-e^2\right)^2}{4 c^2}}\right\}
$
We can now solve for bx and by using:
$(\text{bx}-\text{sx})^2+(\text{by}-\text{sy})^2=f^2$
$\text{bx}^2+\text{by}^2=g^2$
and plugging in the values of sx and sy we found earlier, yielding two
solutions, both valid:
$
\left\{\text{bx}\to \frac{\left(c^2+d^2-e^2\right)
\left(d^2-f^2+g^2\right)+\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g)
(d+f-g) (d-f+g) (d+f+g)}}{4 c d^2},\text{by}\to -\frac{c^4
\left(d^2-f^2+g^2\right)+c^2 \left(\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e)
(d-f-g) (d+f-g) (d-f+g) (d+f+g)}-2 \left(d^2+e^2\right)
\left(d^2-f^2+g^2\right)\right)+(d-e) (d+e) \left(\sqrt{(c-d-e) (c+d-e)
(c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g) (d+f+g)}+(d-e) (d+e)
\left(d^2-f^2+g^2\right)\right)}{4 c d^2 \sqrt{-(c-d-e) (c+d-e) (c-d+e)
(c+d+e)}}\right\}
$
$
\left\{\text{bx}\to \frac{\left(c^2+d^2-e^2\right)
\left(d^2-f^2+g^2\right)-\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g)
(d+f-g) (d-f+g) (d+f+g)}}{4 c d^2},\text{by}\to \frac{c^4
\left(-\left(d^2-f^2+g^2\right)\right)+c^2 \left(\sqrt{(c-d-e) (c+d-e)
(c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g) (d+f+g)}+2 \left(d^2+e^2\right)
\left(d^2-f^2+g^2\right)\right)+(d-e) (d+e) \left(\sqrt{(c-d-e) (c+d-e)
(c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g) (d+f+g)}-(d-e) (d+e)
\left(d^2-f^2+g^2\right)\right)}{4 c d^2 \sqrt{-(c-d-e) (c+d-e) (c-d+e)
(c+d+e)}}\right\}
$
To find ABN, we simply take the arctangent of
$\frac{\text{by}}{\text{bx}}$. Since the angle will always be between
0 and 180 degrees (if it's larger than 180 degrees, we measure the
angle clockwise), we can use the single argument form of arctangent
(I'd stated incorrectly earlier than we needed the two argument
form. This yields:
$
-\tan ^{-1}\left(\frac{c^4 \left(d^2-f^2+g^2\right)+c^2 \left(\sqrt{(c-d-e)
(c+d-e) (c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g) (d+f+g)}-2
\left(d^2+e^2\right) \left(d^2-f^2+g^2\right)\right)+(d-e) (d+e)
\left(\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g)
(d+f+g)}+(d-e) (d+e) \left(d^2-f^2+g^2\right)\right)}{\sqrt{-(c-d-e) (c+d-e)
(c-d+e) (c+d+e)} \left(\left(c^2+d^2-e^2\right)
\left(d^2-f^2+g^2\right)+\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g)
(d+f-g) (d-f+g) (d+f+g)}\right)}\right)
$
$
\tan ^{-1}\left(\frac{c^4 \left(-\left(d^2-f^2+g^2\right)\right)+c^2
\left(\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g)
(d+f+g)}+2 \left(d^2+e^2\right) \left(d^2-f^2+g^2\right)\right)+(d-e) (d+e)
\left(\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g) (d+f-g) (d-f+g)
(d+f+g)}-(d-e) (d+e) \left(d^2-f^2+g^2\right)\right)}{\sqrt{-(c-d-e) (c+d-e)
(c-d+e) (c+d+e)} \left(\left(c^2+d^2-e^2\right)
\left(d^2-f^2+g^2\right)-\sqrt{(c-d-e) (c+d-e) (c-d+e) (c+d+e) (d-f-g)
(d+f-g) (d-f+g) (d+f+g)}\right)}\right)
$
Mathematica can't find a simpler form for the answer, and I'm not sure
how useful the above is, but there you have it.
Note that I assume the triangle inequality throughout. When the
triangle inequality doesn't hold, some of the square roots above are
of negative values, and thus have no real number solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$
Find $$\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$$
What do you guys think about this approach, and are there any faster and easier approaches?
My approach:
*
*$\frac{\tan(x^3+y^3)(x^2+y^2)}{\sin(x^2+y^2)(x^2+y^2)}$
2.$\frac{\tan(x^3+y^3)}{(x^2+y^2)}$
*$\frac{\sin(x^3+y^3)}{\cos(x^3+y^3)\rightarrow 1(x^2+y^2)}$
*$\frac{\sin(x^3+y^3)}{(x^2+y^2)}$
*Polar: $rcos^3\phi + rsin^3\phi = 0 + 0 = 0$
Thus,
$$\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)} = 0$$
| I would say for $\vert z \vert$ small enough $$\vert \tan z \vert \le 2 \vert z \vert$$
Hence for $\vert x \vert + \vert y \vert$ small enough
$$\left\vert \frac{\tan(x^3+ y^3)}{\sin (x^2 +y^2)} \right\vert \le \frac{2(\vert x \vert^3+ \vert y \vert^3)}{\sin (x^2 +y^2)} \le 2 \frac{x^2+y^2}{\sin (x^2+y^2)}(\vert x \vert + \vert y \vert)$$ and you are done as $\lim\limits_{z \to 0} \frac{\sin z}{z} = 1$ and $\lim\limits_{(x,y) \to 0} \vert x \vert + \vert y \vert =0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that $7 | (3^{2n + 1} + 2^{n +2})$ Prove that $7 | (3^{2n + 1} + 2^{n +2})$
So far I have:
Base case: n = 1
$ = (3^{2(1) + 1} + 2^{(1) +2})$
$ = (3^{3} + 2^{3})$
$ = (35)$ which divides 7
Inductive Step:
$ = (3^{2(n +1) + 1} + 2^{(n +1) +2})$
$ = (3^{2n + 3} + 2^{n + 3})$
$ 7j = (3^{2n + 3} + 2^{n + 3})$
-I'm a bit stuck as to how to proceed, any help or correcitons are appreciated.
- I believe I understand where we need to go now but i'm not sure how to get there. It seems obvious that any n plugged into this formula would be divisible by 7 given that $3^1 + 2^2 = 7$ Since you have that $3^1+2^2$ nested inside the given formula it seems easily understood that any sum with any positive n would also be divisible by 7. I just don't see the steps in between to actually prove it.
| Hint: if you want to use induction:
if $7\mid 3^{2k+1}+2^{k+2}=3\cdot 9^k+4\cdot 2^k$ for some $k\in\mathbb Z_{\ge 0}$, then:
$$3\cdot 9^{k+1}+4\cdot 2^{k+1}=2\left(3\cdot 9^k+4\cdot 2^k\right)+7\left(3\cdot 9^k\right)$$
Otherwise, if you know Modular Arithmetic:
$$3\cdot 9^n+4\cdot 2^n\equiv 3\cdot 2^n+4\cdot 2^n\equiv 7\cdot 2^n\equiv 0\pmod{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\lim_{x\to 0}\frac{\cos^2x-\cos x-e^x\cos x+e^x-\frac{x^3}{2}}{x^n}$ is a non zero finite number number, find $n$ where $n\in\mathbb{N}$
If the following limit $$\lim_{x\to 0}\frac{\cos^2x-\cos x-e^x\cos x+e^x-\frac{x^3}{2}}{x^n}$$
is a non zero finite number number, find $n$ where $n\in\mathbb{N}$
$$\lim_{x\to 0}\frac{\cos^2x-\cos x-e^x\cos x+e^x-\frac{x^3}{2}}{x^n}=\lim_{x\to 0}\frac{(\cos x-e^x)(\cos x-1)-\frac{x^3}{2}}{x^n}=\lim_{x\to 0}\frac{x^2(\cos x-e^x)\frac{(\cos x-1)}{x^2}-\frac{x^3}{2}}{x^n}$$
Since $\lim_{x\to 0}\frac{(\cos x-1)}{x^2}=-\frac{1}{2}$
$$\implies \lim_{x\to 0}\frac{x^2(\cos x-e^x)\frac{(\cos x-1)}{x^2}-\frac{x^3}{2}}{x^n}=\lim_{x\to 0}\frac{-\frac{x^2}{2}(\cos x-e^x)-\frac{x^3}{2}}{x^n}$$
Since $\cos x$ is $1$ when $x$ tends to zero,
$$\lim_{x\to 0}\frac{-\frac{x^2}{2}(\cos x-e^x)-\frac{x^3}{2}}{x^n}=\lim_{x\to 0}\frac{-\frac{x^2}{2}(1-e^x)-\frac{x^3}{2}}{x^n}$$
Since $\frac{1-e^x}{x}$ tends to $-1$ when $x\to 0$,
$$\lim_{x\to 0}\frac{-\frac{x^2}{2}(1-e^x)-\frac{x^3}{2}}{x^n}=\lim_{x\to 0}\frac{\frac{x^2}{2}x-\frac{x^3}{2}}{x^n}$$
Isnt the answer $n=3$? Answer given is $n=4$
| Taylor series of $\cos^2 x - \cos x - e^x \cos x + e^x$ is
$$\frac{x^3}{2}+\frac{x^4}{2}+O(x^5)$$
Therefore,
$$\lim_{x \to 0} \frac{\cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}}{x^n} = \lim_{x \to 0} \frac{\frac{x^4}{2} + O(x^5)}{x^n}$$
If $n = 3$, the limit goes to 0, but it is given that the limit is non-zero finite. Therefore $n = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $z=18+26i$ where $z_0=x_0+iy_0, \ x_0, y_0\in \Bbb R$ is the cube root of $z$ having least positive argument. Find the value of $x_0y_0(x_0+y_0)$. Let $z=18+26i$ where $z_0=x_0+iy_0, \quad x_0,y_0\in R$ is the cube root of $z$ having least positive argument.Find the value of $x_0y_0(x_0+y_0)$.
Let $\sqrt[3]{18+26i}=x_0+iy_0$
$18+26i=x_0^3+3ix_0^2y_0-3x_0y_0^2-iy_0^3$
$18=x_0^3-3x_0y_0^2,26=-y_0^3+3x_0^2y_0$.
I am stuck here. What should i do to find $x_0y_0(x_0+y_0)$.
| I recommend another approach. Let's use polar form to find $z_0$.
We see that $|z|=\sqrt{18^2+26^2}=\sqrt{1000}$. Thus $|z_0|$ is the cube root of that, namely $|z_0|=\sqrt{10}$.
To find the argument of $z_0$ (let's call it $\theta$), note that the argument of $z_0^3=z=18+26i$ is $3\theta$ and is given by $\tan3\theta=\frac{26}{18}=\frac{13}9$. Expanding,
$$\begin{align}
\frac{13}9 &= \tan 3\theta \\[2ex]
&= \tan(2\theta+\theta) \\[2ex]
&= \frac{\tan 2\theta+\tan\theta}{1-(\tan 2\theta)(\tan\theta)} \\[2ex]
&= \frac{\frac{2\tan\theta}{1-\tan^2\theta}+\tan\theta}
{1-\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)(\tan\theta)} \\[2ex]
&=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}
\end{align}$$
Letting $t=\tan\theta$ and rearranging that equation, we get
$$9t^3-39t^2-27t+13=0$$
Using the Rational Root Theorem we can factor that into
$$(3t-1)(3t^2-12t-13)=0$$
Which gives us the solutions
$$\tan\theta=t=\frac 13, \ \frac{6\pm5\sqrt 3}{3}$$
The smallest positive argument is given by the first solution, $\tan\theta=\frac 13$. We can quickly combine that with $|z_0|=\sqrt{10}$ to get the answer
$$z_0=3+i$$
This could have been seen by inspection much earlier, but I wanted to show a fuller exposition of getting that answer. Checking quickly shows us this is the desired cube root of $z$. In any case, we get
$$x_0=3, \ y_0=1$$
and the final answer
$$x_0y_0(x_0+y_0)=3\cdot 1(3+1)=12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{cyc}^{}\sqrt{a^{2}+1}\leq\sqrt{2}\sum_{cyc}^{} a$ $a$,$b$,$c$,$d>0$
$abc+bcd+cda+dab=a+b+c+d$
Prove the following inequality
$\sqrt{a^{2}+1}+\sqrt{b^{2}+1}+\sqrt{c^{2}+1}+\sqrt{d^{2}+1}\leq\sqrt{2}(a+b+c+d)$
| Easy to see that $\sum\limits_{cyc}\frac{a^2+1}{a+b}=a+b+c+d$.
Hence, by C-S $a+b+c+d=\sum\limits_{cyc}\frac{a^2+1}{a+b}\geq\frac{\left(\sum\limits_{cyc}\sqrt{a^2+1}\right)^2}{2(a+b+c+d)}$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^{2\pi}\sqrt{1+\sin x}dx$ Problem Evaluate the following integral
$$\int_0^{2*\pi}\sqrt{1+\sin x}dx$$
Attempted solution
Note that
\begin{align*}
& \int_{0}^{2\pi}\sqrt{1+\sin x}dx\\
= & \int_{0}^{\pi}\sqrt{1+\sin x}dx+\int_{\pi}^{2\pi}\sqrt{1+\sin x}dx\\
= & \int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}dx+\int_{0}^{\pi}\sqrt{1+\sin\left(x-\pi\right)}d\left(x-\pi\right)\\
= & \int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}dx+\int_{0}^{\pi}\sqrt{1-\sin x}dx\\
= & \int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}dx+\int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}}dx\\
= & \int_{0}^{\pi}\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|dx+\int_{0}^{\pi}\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|dx\\
= & \int_{0}^{\frac{\pi}{2}}\cos\frac{x}{2}+\sin\frac{x}{2}dx+\int_{\frac{\pi}{2}}^{\pi}\cos\frac{x}{2}+\sin\frac{x}{2}+\int_{0}^{\frac{\pi}{2}}\cos\frac{x}{2}-\sin\frac{x}{2}dx+\int_{\frac{\pi}{2}}^{\pi}\sin\frac{x}{2}-\cos\frac{x}{2}dx\\
= & \int_{0}^{\frac{\pi}{2}}2\cos\frac{x}{2}dx+\int_{\frac{\pi}{2}}^{\pi}\sin\frac{x}{2}dx\\
= & \left.4\sin\frac{x}{2}\right|_{0}^{\frac{\pi}{2}}-\left.4\cos\frac{x}{2}\right|_{\frac{\pi}{2}}^{\pi}=\boxed{4\sqrt{2}}.
\end{align*}
Question I don't know if I solved this integral in the best way possible. I really appreciate if someone can offer some alternative solution.
| If you multiply and divide by $\sqrt{1-\sin x}$, the integrand becomes
$$
\frac{|\cos x|}{\sqrt{1-\sin x}}.
$$
Now, you can split the integrand according to the sign of the cosine, and use the subtitution $u=1-\sin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find $\int\frac{x}{(1+x)(x^4+4)}dx$ Find $$\int\frac{x}{(1+x)(x^4+4)}dx$$
I tried to do partial fractions.
$$\int\frac{x}{(1+x)(x^4+4)}dx=\int\frac{x+1-1}{(1+x)(x^4+4)}dx=\int\frac{dx}{x^4+4}-\int\frac{dx}{(1+x)(x^4+4)}$$
I am stuck now.I do not know how to integrate now.
| HINT:
$$x^4+4=(x^2+2)^2-(2x)^2=\cdots$$
Now use Partial Fraction Decomposition,
$$\dfrac x{(x+1)(x^4+4)}=\dfrac A{x+1}+\dfrac{Bx+C}{x^2+2x+2}+\dfrac{Dx+E}{x^ 2-2x+2}$$
Write $$\dfrac{Dx+E}{x^ 2-2x+2}=\dfrac{D(x-1)}{(x-1)^2+1^2}+\dfrac{D+E}{(x-1)^2+1^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{1}{n} + \frac{1}{n+1} + .... + \frac{1}{2n} > 0.63 $ I need to prove this:
$\frac{1}{n} + \frac{1}{n+1} + .... + \frac{1}{2n} > 0.63 $.
I know its $\log{(2)} $ and it's something like $\int_{0}^{1} \ \frac{dx}{1+x}\ = \log{(2)}$, but I don't know the steps between those.
| Let $(a_n)$ be the sequence
$$a_n= \frac{1}{n}+\frac{1}{n+1} + \dots + \frac{1}{2n},$$
then $a_1=1.5$ and $a_{n+1}-a_n= \frac{1}{2n+2} +\frac{1}{2n+1} -\frac{1}{n}<0$, then the sequence is decreasing, as u have noticed that $a_n \to \log 2$, it won't be hard to figure out that $a_n>\log 2 \approx 0.693$ for all integers $n>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Factorisation and hence ... Factorise $2x^2 + 8x + 6$ completely . Hence express 286 as the product of three prime factors .
Workings -
$2x^2 + 8x + 6
= 2(x^2 + 4x + 3)
=2(x+3)(x+1)$
How do I use my answer above to express 286 as the product of three prime factors ?
Thanks in advance ...
| You are right- $2x^2+8x+6=2(x+3)(x+1)$
Now, $2x^2+8x+6=286$ will give you $x=10$
So, $286=2(x+3)(x+1)=2\times 13 \times 11$
The prime factors are $2,13,11$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\lim\limits_{x\to 4}\frac{2x-5}{6-3x} = -\frac{1}{2} $
Proving that $$\lim_{x\to 4}\frac{2x-5}{6-3x} = -\frac{1}{2} $$
Now of course when you plug in $4$ you get $-\frac{1}{2}$, but I need to prove it using:
$\forall \varepsilon \gt 0,\;\exists\delta \gt 0\;\forall x \in D $ (where $D$ is the domain of the function)
$$0\lt |x-x_0|\lt \delta \implies |f(x)-L| \lt \varepsilon $$
Where $L$ is the limit we "guessed" and $x_0$ is where $x$ approaches)
So far I have:
$$\begin{align}
0\lt |x-4|\lt \delta &\implies \left|f(x)-\left(-\frac{1}{2}\right)\right| \lt \varepsilon\\
\iff 0\lt |x-4|\lt \delta &\implies \left|\frac{2x-5}{6-3x}+\left(\frac{1}{2}\right)\right| \lt \varepsilon\\
\iff 0\lt |x-4|\lt \delta &\implies \left|\frac{4-x}{6(x-2)}\right| \lt \varepsilon
\end{align}$$
This is where I'm stuck. What exactly do I need to prove after this last line?
| Given any $\varepsilon > 0$, let $\delta = \min\{1, \varepsilon/M\} > 0$, where $M$ is some fixed magical positive constant. While doing this proof, we will eventually figure out what $M$ should be, then go back and erase $M$ and replace it with this constant. Now if $0 < |x - 4| < \delta$, then observe that:
\begin{align*}
\left| \frac{2x - 5}{6 - 3x} - \frac{-1}{2} \right|
&= \left| \frac{4 - x}{6(x - 2)} \right| \\
&= \frac{1}{6|x - 2|}|x - 4| \\
&< \frac{1}{6|x - 2|} \cdot \frac{\varepsilon}{M} &\text{since } |x - 4| < \delta \leq \frac{\varepsilon}{M} \\
\end{align*}
Now we take a break from our proof and try to figure out a way to bound $\frac{1}{6|x - 2|}$ by a constant $M$ given that $x$ is at most $1$ away from $4$, implying that $|x - 2|$ won't be too close to zero. Indeed, if $|x - 4| < 1$, then observe that:
\begin{align*}
|x - 2|
&= |(x - 4) - (-2)| \\
&\geq ||x - 4| - \left| -2 \right|| &\text{by the reverse triangle inequality} \\
&= 2 - |x - 4| &\text{since } |x - 4| < 1 \leq 2 \\
&> 2 - 1 &\text{since } |x - 4| < 1 \\
&= 1
\end{align*}
so that:
$$
\frac{1}{6|x - 2|} < \frac{1}{6}
$$
Thus, by taking $M = \frac{1}{6}$, we can finish up our original proof:
Given any $\varepsilon > 0$, let $\delta = \min\{1, 6\varepsilon\} > 0$. Then if $0 < |x - 4| < \delta$, observe that:
\begin{align*}
\left| \frac{2x - 5}{6 - 3x} - \frac{-1}{2} \right|
&= \left| \frac{4 - x}{6(x - 2)} \right| \\
&= \frac{1}{6|x - 2|}|x - 4| \\
&< \frac{1}{6|x - 2|} \cdot 6\varepsilon &\text{since } |x - 4| < \delta \leq 6\varepsilon \\
&< \frac{1}{6} \cdot 6\varepsilon &\text{since } |x - 4| < \delta \leq 1 \\
&= \varepsilon
\end{align*}
as desired. $~\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find derivative? How can I get from step $3$ to step $4$ in the problem?
I've tried this:
$6x^2 - 9x + 8x -12 + 2\Delta x - 6x^2 -8x +9x + 12 + 3\Delta x$
It doesn't cancel out to $17\Delta x$, though.
I've also tried this:
$(3x+4)(2x-3) - (2x-3)(3x+4) = 0$, then the only things left are $2\Delta x$ and $3\Delta x$. That adds up to $5\Delta x$, which isn't the answer.
How can I solve this problem?
Thanks.
| Step 3: $$\frac{(3x+4)[(2x-3)+2\Delta x]-(2x-3)[(3x+4)+3\Delta x])}{(3x+4)(3x+3\Delta x+4)}$$
$$=\frac{(3x+4)(2x-3)+2(3x+4)\Delta x-(2x-3)(3x+4)-3(2x-3)\Delta x)}{(3x+4)(3x+3\Delta x+4)}$$
Now first and third term cancel out and you will get
$$=\frac{2(3x+4)\Delta x-3(2x-3)\Delta x)}{(3x+4)(3x+3\Delta x+4)}$$
$$=\frac{17\Delta x}{(3x+4)(3x+3\Delta x+4)}$$
which is your step 4
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\gcd(n-1,n+10)$, for $n$ a natural number? Currently I have,
let $d=\gcd(n-1,n+10)$
then, $d\mid n-1$ and $d\mid n+10$
and hence $d$ must also divide $11$
not really sure what to do now, any help would be greatly appreciated.
| As you have already observed, $d$ must divide $11$, thus $d$ is either $1$ or $11$.
However, one thing to note that is both are possible-for example, for $n=2$, $d=1$, and for $n=12$, $d=11$.
A similar idea would be using the Euclidean Algorithim, which yields that $$\gcd(n-1,n+10)=\gcd(n-1,n+10-n+1)=\gcd(n-1,11)$$ which is $1$ or $11$.
EDIT
An important observation would be noting when it is $1$, and when it is $11$.
This matter can be resolved using our upper equation, which says that $\gcd(n-1,n+10)=\gcd(n-1,11)=11$. This implies that $11|n-1$, further showing that $\gcd(n-1,11)=11 \Rightarrow 11|n-1 \Rightarrow n \equiv 1 \pmod {11}$.
Thus, $\gcd(n-1,11)=11$ implies that $n \equiv 1 \pmod {11}$. Further note that the converse is true-if $n \equiv 1 \pmod {11}$, then $n=11k+1$.
This gives us $$\gcd(n-1,11)=\gcd(11k,11)=11$$ Thus we have $$\gcd(n-1,n+10)=11 \Leftrightarrow n \equiv 1 \pmod {11}$$
Since $\gcd(n-1,n+10)$ is either $1$ or $11$, this gives us that $$\gcd(n-1,n+10)=1 \Leftrightarrow n \not \equiv 1 \pmod {11}$$
This can be demonstrated in my upper example-if $n=2$, then $n \not \equiv 1 \pmod {11}$ thus implying that $\gcd(n-1,n+10)=1$.
If $n=12$, then $n \equiv 1 \pmod {11}$, thus implying $\gcd(n-1,n+10)=11$.
| {
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"url": "https://math.stackexchange.com/questions/1707242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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integral $\int \frac{2x}{(3x-2)^2+2^2} \, \mathrm{d}x$
$$\int \frac{2x}{(3x-2)^2+2^2}\, \mathrm{d}x$$
$\int \frac{1}{(3x-2)^2+2^2}\, \mathrm{d}x=\frac{1}{2}\arctan(\frac{3x-2}{2})+C$
But what can I do with the $2x$ in the numerator?
| This can be written as:
$$\frac{1}{9} \int \frac{(18x-12+12)dx}{9x^2-12x+8}=$$
$$\frac{1}{9} \int \frac{(18x-12)dx}{9x^2-12x+8}+\frac{4}{3} \int \frac{dx}{(3x-2)^2+2^2}=$$
$$\frac{1}{9} \ln(9x^2-12x+8)+\frac{2}{9}\tan^{-1}{\frac{3x-2}{2}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality $abdc$ $\leq$ $3$ $a+b+c+d=6$
and
$a^2+b^2+c^2+d^2=12$.
and $a,b,c,d$ are reals.
Prove: $abcd$ $\leq$ $3$ without Lagrange multipliers, complex numbers or convexity help.
Using Cauchy–Schwarz inequality I found: $a,b,c,d \in [0,3]$.
How solve inequality?
| If $abcd\le0$ there is nothing to prove, so either (1) $a,b>0$ and $c,d<0$ or (2) $a,b,c,d>0$.
Case (1). We have $a+b>6$, so $a^2+b^2\ge 3^2+3^2>12$. Contradiction.
Case (2) There is a standard result (well, standard amongst those who mess around with Olympiad inequalities) that for positive $a_1,\dots,a_n$ if we fix $\sum a_i$ and $\sum_{i\ne J}a_ia_j$, then the maximum of $a_1a_2\dots a_n$ is achieved when the $a_i$ assume at most two distinct values between them. So in this case we must have either (A) $a=b,c=d=3-a$ or (B) $b=c=d=6-a$.
Using $a^2+b^2+c^2+d^2$, (A) gives $ac=\frac{3}{2}$ and hence $abcd=\frac{9}{4}<3$, whilst (B) gives $a=b=c=1,d=3$ and hence $abcd=3$.
[see http://artofproblemsolving.com/community/c6h1187374p5777294 for the "generalised uvw" theorem]
| {
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"timestamp": "2023-03-29T00:00:00",
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How to obtain $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$ without L'Hospital's rule?
$$\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$$
How would I find the limit with out using conjugate nor L'Hospital's rule.
| You may observe that, by the Taylor expansion, you have, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}6+O(x^5) \tag1
\\ \tan x&=x+\frac{x^3}3+O(x^5) \tag2
\end{align}
$$ and, as $u \to 0$, $$
\begin{align}
\sqrt{1+u}&=1+\frac{u}2-\frac{u^2}8+\frac{u^3}{16}+O(u^4). \tag3
\end{align}
$$ Thus
$$
\sqrt{1+\tan x}-\sqrt{1+\sin x}=\frac{x^3}4+O(x^4) \tag4
$$ giving, as $x \to 0$,
$$
\lim_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}} = \frac14.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using Hensel's Lemma to find the number of solutions. I am trying to determine the number of solutions of the congruence
$x^2 ≡ 1 \mod 2^k$ when $k \ge 3$
The statement of Hensel's Lemma that I use is the following:
Attempt: $x \equiv 1,3,5,7 \mod 8$, and for all of these $\tau = 1$ so by (i) since we have 2 distinct solutions $\mod 4$ we have two distinct solutions $\mod 8$? This is where I get lost.
| The following is not really an answer to the question, since after the first sentence it does not mention Hensel's lemma.
Let $k\ge 3$. We are trying to solve $x^2\equiv 1\pmod{2^k}$. For concreteness let $2^k=1024$. We want to solve the congruence
$$(x-1)(x+1)\equiv 0\pmod{1024}.$$
So $x-1$ and $x+1$ will need to be consecutive even integers.
Of any two consecutive even integers, one is congruent to $2$ modulo $4$, and the other is congruent to $0$ modulo $4$. The one congruent to $2$ modulo $4$ has only one $2$ to contribute to the product $(x-1)(x+1)$. So the other one must supply the rest of the $2$'s.
Thus $(x-1)(x+1)$ is divisible by $1024$ if and only if one of $x-1$ and $x+1$ is divisible by $512$.
This gives the four solutions $x\equiv 1\pmod{1024}$, $x\equiv 513\pmod{1024}$, $x\equiv 1023\pmod{1024}$ and $x\equiv 511\pmod{1024}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.