Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\int \int \sqrt{(x^2 + y^2)}$ $dx dy$ over a triangle with corners (0, 0), (4, 4), and (4, 0) using $x = u$, $y = uv$. Help?
So far I have:
$$a = x = 4\\ b = y = 0\\ c = y = x$$
$$y=x \rightarrow uv = u\rightarrow v=u/u \rightarrow v=1$$
$$x = 4 \rightarrow u = 4$$
$$y = 0\rightarrow uv = 0 \rightarrow u = 0 \rightarrow v = 0$$
$$\sqrt {(x^2 + y^2)} = (u^2 +uv^2)^{(1/2)}$$
\begin{align} \|J\| & =det\left[ {\begin{array}{cc}
\frac {∂x}{∂u} & \frac{∂x}{∂v} \\
\frac{∂y}{∂u} & \frac{∂y}{∂v}\\
\end{array} } \right] = \left[ {\begin{array}{cc}
1 & v \\
0 & u\\
\end{array} } \right] = u = 4
\end{align}
\begin{align*}
\int_{0}^{4}\int_{0}^{1} \sqrt{(u^2 + uv^2)} + 4 ~dvdu = \int_{0}^{4}\int_{0}^{1} (u^2 + uv^2)^{1/2} + 4 ~dvdu
\end{align*}
I can already see this is incorrect due to the given answer.
This gives me J = 0, however J is supposed to = 1? and the answer to the problem is supposed to be:
$$\frac{32(\sqrt2+ln(1+\sqrt2))}3$$
according to the website: https://www.whitman.edu/mathematics/calculus_online/section15.07.html
| By switching to polar coordinates, the wanted integral equals
$$ \int_{0}^{\pi/4} \int_{0}^{\frac{4}{\cos\theta}}\rho^2\,d\rho\,d\theta = \int_{0}^{\pi/4}\frac{1}{3}\left(\frac{4}{\cos\theta}\right)^3\,d\theta $$
and by using Weierstrass substitution $\theta=2\arctan\frac{t}{2}$ that boils down to
$$ \frac{32}{3}\left(\sqrt{2}+2\,\text{arctanh}\tan\frac{\pi}{8}\right)=\color{red}{\frac{32}{3}\left[\sqrt{2}+\log(1+\sqrt{2})\right]}$$
as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
| Let's see. We have
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$$
now set $x=y+\frac12$. Then
$$\sqrt[3]{5y+\tfrac{19}2}-\sqrt[3]{5y-\tfrac{19}2}=1$$
Now cubing yields
$$19-3\sqrt[3]{(5y)^2-(\tfrac{19}2)^2}\left(\sqrt[3]{5y+\tfrac{19}2}-\sqrt[3]{5y-\tfrac{19}2}\right)=1$$
which rewrites to
$$\sqrt[3]{(5y)^2-(\tfrac{19}2)^2}=6$$
Which is easy to solve, $25y^2=6^3+(\frac{19}2)^2$ or $y=\pm \frac 72$. Now we conclude $x=4$ or $x=-3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trouble with replacing a Cartesian equation with polar equation. This is the last question on my assignment and I can't figure out how to solve it. Replace the Cartesian equation with an equivalent polar equation:
$$
\frac{x^2}{4} + \frac{y^2}{49}=1
$$
I know that $x^2+y^2=r^2$ but when I try to work this out...
$$
\frac{4y^2+49x^2}{196}=1
$$
$$
\frac{4(r\sin\theta)^2+49(r\cos\theta)^2}{196}=1
$$
$$
\frac{4r^2\sin^2\theta+49r^2\cos^2\theta}{196}=1
$$
$$
r^2(4\sin^2\theta+49\cos^2\theta)=196
$$
$$
r^2=\frac{196}{49\cos^2\theta+4\sin^2\theta}
$$
$$
r=\sqrt{\frac{196}{49\cos^2\theta+4\sin^2\theta}}
$$
But I know this is very wrong.
| You are totally correct : You have done what is done to find polar form of any equation, i.e.
Assume the polar coordinates of a curve to $\big(r(\theta)\cos \theta,r(\theta)\sin \theta \big)$ and put it into the cartesian equation of the curve, and then solve for $r(\theta)$.
For an ellipse :
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
The polar coordinates of point $\text{P}$ are given by :
$$\text{P} \equiv\Big(r(\theta)\cos \theta,r(\theta)\sin \theta\Big) ; ~\text{where}~r(\theta)=\frac{ab}{\sqrt{a^2 \sin^2 \theta+b^2 \cos^2 \theta}} $$
You can confirm yourself here on Wikipedia.
| {
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Finding the laurent series for $e^{\sin(1/z)}$ I began with the expansion of $e^z$, which is \begin{eqnarray} e^z\ = \sum_{n=0}^\infty \frac{z^n}{(n)!} \end{eqnarray}
Then I found the expansion for sin(1/z), which is \begin{eqnarray} \sin\left(\frac{1}{z}\right) = \sum_{n=0}^\infty (-1)^n \frac{z^{-2n-1}}{(2n+1)!} \end{eqnarray}
But now I don't know how to go further?
Should I just substitute the first four values for $\sin\left(\frac 1 z \right)$ into the formula for $e^z$?
| Substitute $\sin( \frac{1}{z})=\frac{1}{z}-\frac{1}{6z^3}+\cdots$ into the series for $e^z$ ... we have
\begin{eqnarray*}
e^{\sin( \frac{1}{z})}=1+(\frac{1}{z}-\frac{1}{6z^3}+\cdots)+\frac{1}{2}(\frac{1}{z}-\frac{1}{6z^3}+\cdots)^2+\frac{1}{6}(\frac{1}{z}-\frac{1}{6z^3}+\cdots)^3 +\frac{1}{4!}(\frac{1}{z}+\cdots)^3 \cdots
\end{eqnarray*}
Expanding & collecting terms ... we have
\begin{eqnarray*}
e^{\sin( \frac{1}{z})}=1+\frac{1}{z}+\frac{1}{2}\frac{1}{z^2}+\frac{1}{z^3}(-\frac{1}{6}+\frac{1}{6})+\frac{1}{z^4}(\frac{1}{4!}-\frac{2}{2 \times 6} )+\cdots
\end{eqnarray*}
So the first $4$ non zero terms are
\begin{eqnarray*}
e^{\sin( \frac{1}{z})}=1+\frac{1}{z}+\frac{1}{2}\frac{1}{z^2}-\frac{1}{8}\frac{1}{z^4}+\cdots
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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A Problem From My Exam In the image, a semicircle with diameter $AD$ has smaller semicircles with diameters $AB$, $BC$, $CD$, all next to each other inside it. The area between them is shaded.
The perimeter of the painted area is $24\pi$, what is the area?
| Ratio of diameters
$$
\begin{align}
AB : BC : CD = 1 : 2 : 3 =
r_{1} : r_{2} : r_{3}
\end{align}
$$
Radius of perimeter segment: $\pi r = 24 \pi$.
$$
r = 24
$$
Add up the sgements:
$$
\begin{align}
r &= r_{1} + r_{2} + r_{3} \\
&= r_{1} + 2 r_{1} + 3 r_{1} = 6 r_{1} \\
24 &= 6 r_{1} \\
4 &= r_{1}
\end{align}
$$
Shaded area
$$
\begin{align}
%
A &= \frac{\pi}{2} \left( r^{2} -
\left(
r^{2}_{1} + r^{2}_{2} + r^{2}_{3}
\right)
\right) \\
%
&=\frac{\pi}{2} \left( 24^{2} -
\left(
4^{2} + 8^{2} + 12^{2}
\right)
\right) \\
&= 352 \frac{\pi}{2} = \boxed{176 \pi}
%
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Differentiation to tangent and normals The equation of the curve is given by
$ y = \frac{6}{2x^2+1} $
Find the equation of the tangent and the normal at the point when
$ x = -1$
My workings
When $x= -1, y= 2 $
$\frac{d}{dx} (Y) = \frac{-24}{(2x^2+1)^2} $
Sub $x=-1$ to the above expression to find the gradient ...
$\frac{d}{dx} (Y) = \frac{-8}{3} $
Now , to find the equation of the tangent , what can I do ?
Is it $y=mx+c$ ?
Sub in $x=-1 , y=2$ to find the y intercept and then continue to find the equation of the tangent ?
Which is
$y= \frac{-8}{3} x - \frac{2}{3} $
But that don't feel right to me .. so im here to ask .
| The derivative should be
$\frac{d}{dx} (Y) = \frac{-24*x}{(2x^2+1)^2} $
when $x=-1 $, it gives $24/9 $
So the tangent equation is $ y-2=8/3 *(x+1) $ and the normal is $ y-2 =-3/8 *(x+1)$
Does it sound good ?
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$? The function
$$
f(z) = \frac{z}{z^2 - 1}
$$
has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$,
\begin{align*}
f(x) =& \frac{z}{z^2 - 1} \\[3mm]
=& \frac{z}{(z+1)(z-1)} \\[3mm]
=& \frac{w-1}{w(w-2)} \\[3mm]
=& \frac{1-w}{2w} \cdot \frac{1}{1 - \frac{w}{2}} \\[3mm]
=& \frac{1-w}{2w} \sum_{n=0}^\infty \frac{w^n}{2^n} \\[3mm]
=& \sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{w^n}{2^{n+1}} \\[3mm]
=& \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}
\end{align*}
This is almost in the form of a Laurent expansion, but not quite. How can I continue this to get the correct final result?
| You did the right thing, and can conclude as the others suggested, but notice that this way you are expanding around $z=-1$ and not around $z=1$.
| {
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Prove by induction: inequality Use induction to prove the inequality:
$$\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}\leq \frac{1}{\sqrt{n+1}}$$
I tried by multiplying $\frac{(2n+1)}{(2n+2)}$ to both sides but I don't know how to get $\frac{1}{\sqrt{n + 2}}$ on the RHS.
| Hint: For positive integers $n$ one has
$\frac{1}{\sqrt{n+1}}\cdot\frac{(2n+1)}{(2n+2)}\leq \frac{1}{\sqrt{n+2}}$
is true if and only if
$\frac{1}{n+1}\cdot \frac{4n^2+4n+1}{4n^2+8n+4}\leq \frac{1}{n+2}~~~~~$ (seen by squaring both sides)
is true if and only if
$(n+2)(4n^2+4n+1)\leq (n+1)(4n^2+8n+4)~~~~~$ (seen by cross multiplying)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x)\cos(x)-\sin(x)+\cos(x)2\sin(x)\cos(x) = \\
\sin(x)(2\cos^2(x)-1+2\cos^2(x)) = \\
\sin(x)(4\cos^2(x)-1) = \\
4\cos^2(x)\sin(x)-\sin(x) = ???$$
What do I do next?
| Also it can be proved using complex numbers.
$(\cos x+i\sin x)^3=\cos 3x+i\sin 3x $ (Moivre's formula)
$(\cos x+i\sin x)^3=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2 x\sin x-\sin^3x)$ (binom)
Comparing these two formulas, we obtain
$\cos 3x=\cos^3x-3\cos x\sin^2x=\cos^3x-3\cos x(1-\cos^2x)=\ldots$
$\sin 3x =3\cos^2 x\sin x-\sin^3x=\ldots$
| {
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Is there a straight forward way to prove that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{31} > 3$ Using brute force, it is straight forward to calculate that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{31} > 3$
Is there a more elegant way to demonstrate this?
What if I want to find the smallest $n$ such that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} > 4$
Is there a standard way to solve for $n$ without using brute force?
| BBP-type formulas for $\log(3)$ and $\log(7)$ in base $3$
$$\log(3)=\frac{1}{9}\sum_{k=0}^\infty \left(\frac{9}{2k+1}+\frac{1}{2k+2}\right)\frac{1}{9^{k}}$$
$$\log(7)=\frac{1}{3^5}\sum_{k=0}^\infty \left(\frac{405}{6k+1}+\frac{81}{6k+2}+\frac{72}{6k+3}+\frac{9}{6k+4}+\frac{5}{6k+5}\right)\frac{1}{3^{6k}}$$
give lower bounds
$$\log(3)>\left(9+\frac{1}{2}\right)\frac{1}{9}+\left(\frac{9}{3}+\frac{1}{4}\right)\frac{1}{81}=\frac{355}{324}$$
and
$$\log(7)>\left(405+\frac{81}{2}+\frac{72}{3}+\frac{9}{4}+\frac{5}{5}\right)\frac{1}{3^5}=\frac{1891}{972}$$
From Series and integrals for inequalities and approximations to $\log(n)$
$$\log(2)=\frac{7}{10}-\int_0^1 \frac{x^4(1-x)^2}{1+x^2} dx$$
the upper bound
$$\log(2)<\frac{7}{10}$$
is obtained, and also the convergent approximation
$$\gamma>\frac{15}{26}$$
will be useful.
For your particular case $n=31$, the bound
$$ H_n \ge \log(n+\frac12)+\gamma$$
given by @leonbloy becomes
$$\begin{align}
H_{31} &\ge \log(2·31+1)-\log(2)+\gamma\\
&=2\log(3) + \log(7) - \log(2) + \gamma\\
&>2\frac{355}{324}+\frac{1891}{972}-\frac{7}{10}+\frac{15}{26}=\frac{253589}{63180}\\
&=4+\frac{869}{63180} \gt 4
\end{align}$$
which proves $H_{31}>4$.
To compute the smallest $n$ such that $H_n$ exceeds an integer $N$,
$$log\left(n+\frac{1}{2}\right)+\gamma>N$$
$$log\left(n+\frac{1}{2}\right)>N-\gamma$$
$$n+\frac{1}{2}>e^{N-\gamma}$$
$$n>e^{N-\gamma}-\frac{1}{2}$$
so
$$n=\lceil e^{N-\gamma}-\frac{1}{2}\rceil$$
The PARI code
for (N=0,28,print(N," ",ceil(exp(N-Euler)-1/2)))
shows that this formula produces the values published in http://oeis.org/A002387, although this does not imply that it agrees forever. The inequality and the formula derived here guarantees exceeding $N$, not doing so with the lowest $n$ possible, as in the OEIS sequence.
| {
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Factoring $x^5 - 5x^4 + 1$ The original problem was stated like this:
Prove that the polynomial $x^5 - 5x^4 + 1$ does not have roots of multiplicity 4.
So, factoring the polynomial would answer the question, but I don't know how to do it.
Using the root test, ${1, -1}$ should be roots, but If I divide by $x-1$ or $x+1$ $\frac {x^5 - 5x^4 + 1}{x-1}$, or $\frac {x^5 - 5x^4 + 1}{x-1}$ I get a remainder, so those are not roots?!
| Let $f(x) = x^5 + 5x^4 + 1.$
Rather than perform division of polynomials, a simple way to test possible rational roots is just to plug them into the equation. For $x=-1$ and $x=1$ this gives us
\begin{align}
f(-1) &= (-1)^5-5(-1)^4 + 1 = -5, \\
f(1) &= (1)^5-5(1)^4 + 1 = -3.
\end{align}
But of course $f(0) = 1.$ So there is one root between $-1$ and $0$ and another between $0$ and $1.$ There must also be a root greater than $1$ since we know $f(x)$ eventually grows without bound.
That's three distinct roots. If any of these had multiplicity $4,$
then $f(x)$ would have a total of at least six roots (including multiple roots).
But $f(x)$ is only fifth degree and can have at most five roots.
Therefore $f(x)$ has no roots of multiplicity $4.$
| {
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How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms? How to find the sum of this series?
$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$
Answer is given as:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
| Let $T_n$ be the $n$th triangle number, which are defined by
$$T_n := \sum_{k=1}^n k = 1 + 2 +...+n.$$
Then we can see that the sum you are looking for (which wee will call $S_n$) equates to
$$S_n = T_{T_n}$$
since $S_n$ sums the first $T_n$ natural numbers.
Thus, using the fact that $T_n = \frac{n(n+1)}{2}$, we have
$$S_n = T_{T_n} \\ = \frac{\frac{n(n+1)}{2}\cdot\left(\frac{n(n+1)}{2}+1\right)}{2} \\ = \frac{n(n+1)(n^2+n+2)}{8}$$
| {
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How to prove that the product of eight consecutive numbers can't be a number raised to exponent 4? How to prove this? I tried something like $$P(n,8)=\frac{n!}{(n-8)!} = b^4$$ but I can't proceed to a solution.
| Let
$$f(x) = (x)\cdots(x+7)$$
$$g(x) = x^2 + 3x + 1$$
$$h(x) = x^2 + 11x + 29$$
Then for $x > 0$,
$$f(x) < (g(x)h(x) - 1)^2$$
and for $x \ge 4$,
$$(g(x)h(x)-2)^2 < f(x)$$
hence, if $n$ is a positive integer with $n \ge 4$,
$$(g(n)h(n)-2)^2 < f(n) < (g(n)h(n)-1)^2$$
so $f(n)$ can't be a perfect square.
It's easily verified that $f(1),f(2),f(3)$ are not perfect squares (since for example, they're multiples of $7$, but not multiples of $7^2$).
Therefore the product of $8$ consecutive positive integers can't be a perfect square.
Some explanation of where $g,h$ came from . . .
Identically,
\begin{align*}
x(x+1)(x+2)(x+3) &= \bigl((x)(x+3)\bigr)\bigl((x+1)(x+2)\bigr)\\[4pt]
&=(x^2+3x)(x^2+3x+2)\\[4pt]
&=\bigl((x^2+3x+1)-1)\bigr)\bigl((x^2+3x+1)+1)\bigr)\\[4pt]
&=(x^2+3x+1)^2-1\\[4pt]
\end{align*}
Then, replacing $x$ by $x+4$,
$$(x+4)(x+5)(x+6)(x+7) = (x^2+11x+29)^2-1
\qquad\qquad\qquad\qquad\qquad\;\;\;$$
Thus, $f(x)$ is algebraically "close" to
$$(g(x)h(x))^2$$
which motivates the idea of trying to "trap" $f(x)$ between two nearby perfect squares.
| {
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An integral for $\frac{9801}{2206\sqrt{2}}-\pi$ From integrals
$$\pi=\frac{24\sqrt{2}}{11} + \frac{8}{11} \int_0^1 \frac{x (1 - x)^2(1 + 2 \sqrt{2} x^4)}{1 + x^2 + x^4 + x^6} dx$$
and
$$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6} dx$$
the following linear combination for Ramanujan's approximation $\pi\approx\frac{9801}{2206\sqrt{2}}$ is obtained:
$$\pi=\frac{9801}{2206\sqrt{2}} - \frac{1}{8824}\int_0^1 \frac{x (1 - x)^2 (124 (1 + 2 \sqrt{2} x^4) - 5769 \sqrt{2} x^3 (1 - x)^2)}{1 + x^2 + x^4 + x^6} dx$$
This integrand is small but has sign changes in $\left(0,1\right)$, so it does not provide a direct proof that $\pi<\frac{9801}{2206\sqrt{2}}$ such as Dalzell integral for $\pi<\frac{22}{7}$.
$$\pi = \frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$
Is there an integral for $\frac{9801}{2206\sqrt{2}}-\pi$ with positive integrand?
Related questions
Why some curious almost-identities
Is there an integral that proves $\pi > 333/106$?
Is there an integral or series for $\frac{\pi}{3}-1-\frac{1}{15\sqrt{2}}$?
An integral for $2\pi+e-9$
| HINT: write your Integrand in the form
$$2\,\sqrt {2}x-4\,\sqrt {2}+{\frac {2\,\sqrt {2}+1}{{x}^{2}+1}}+{\frac
{2\,{x}^{2}\sqrt {2}-2\,\sqrt {2}x-{x}^{2}+2\,\sqrt {2}+x-1}{{x}^{4}+1
}}
$$
| {
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"answer_id": 0
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Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$ Let $x,y,z>0$ and $x+y+z=xyz$. What is the minimum value of $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}?$$
In the case when $x=y=z$, the equation $x+y+z=xyz$ translates to $3x=x^3$, or $x=\sqrt{3}$. If $A$ denotes the quantity that we want to minimize, then $A=\sqrt{3}$ as well.
If we use the inequality of arithmetic and geometric means, we get
$$A\geq \frac{3}{\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{x+y+z}}.$$
| For $x=y=z=\sqrt3$ we get a value $\sqrt3$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\geq\sqrt{\frac{3(x+y+z)}{xyz}}$$ or
$$\sum_{cyc}\left(\frac{x}{y^2}-\frac{2}{y}+\frac{1}{x}\right)+\sum_{cyc}\frac{1}{x}-\sqrt{\frac{3(x+y+z)}{xyz}}\geq0$$ or
$$\sum_{cyc}\frac{(x-y)^2}{y^2x}+\frac{xy+xz+yz-\sqrt{3xyz(x+y+z)}}{xyz}\geq0$$ or
$$\sum_{cyc}\frac{(x-y)^2}{y^2x}+\frac{\sum\limits_{cyc}z^2(x-y)^2}{2xyz\left(xy+xz+yz+\sqrt{3xyz(x+y+z)}\right)}\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show $\lim_{x \to \sqrt{3}^+}\sqrt{x^{2}-3}=0$ I understand that given any $\epsilon>0$, I need to find a $\delta > 0$ such that $|f(x) - \sqrt{3}|<\epsilon$ whenever $0 < x - \sqrt{3} < \delta$.
Starting with $|\sqrt{x^{2}-3}|$, how do I simplify this to show this is $< \epsilon$ for an appropriately choosen $\delta$ using $0 < x - \sqrt{3} < \delta$?
| We want to show that for all $\epsilon > 0$, there exists a $\delta > 0$ such that $x- \sqrt{3} < \delta$ (this is where the $\sqrt{3}^+$ comes in!) implies $\sqrt{x^2 - 3} < \epsilon$.
Let $\epsilon > 0$. Pick $\delta = \min(1, \frac{\epsilon^2}{1+ 2\sqrt{3}})$.
Assume that $x - \sqrt{3} < \delta$. Then we have $x+ \sqrt{3} < \delta + 2\sqrt{3}$, so taking their product,
$$ x^2 - 3 < \delta(\delta + 2\sqrt{3}) .$$
Sinc $\delta = \min(1, \frac{\epsilon^2}{1+ 2\sqrt{3}})$, in particular $\delta \leq 1$. So $\delta + 2\sqrt{3} \leq 1 + 2\sqrt{3}$, which means that
$$ x^2 - 3 < \delta (1 + 2\sqrt{3}) $$
Now use that $\delta \leq \frac{\epsilon^2}{1+ 2\sqrt{3}}$ to get
$$ x^2 - 3< \epsilon^2$$
and thus $\sqrt{x^2 - 3} < \epsilon.$
Edit: Here's some motivation behind this real analysis black magic.
You're given $x - \sqrt{3} < \delta$ and want $\sqrt{x^2 - 3} < \epsilon$. First of all, notice that $\sqrt{x^2 - 3} = \sqrt{x - \sqrt{3}}\sqrt{x + \sqrt{3}}$. The $\sqrt{x - \sqrt{3}}$ goes to $0$, but the $\sqrt{x + \sqrt{3}}$ term doesn't. It sort of hovers around $\sqrt[4]{3}$. Ideally, we'd like to do something like the following:
$$ x - \sqrt{3} < \delta \implies (\sqrt{x - \sqrt{3}})\sqrt{x + \sqrt{3}} < \epsilon .$$
So it seems natural to "solve" for the part we know, $x - \sqrt{3}$, to get
$$ x - \sqrt{3} < \frac{\epsilon^2}{x + \sqrt{3}} $$
Now this should start to look familiar. We're almost done! If we would be able to set
$$ \delta := \frac{\epsilon^2}{x + \sqrt{3}} $$
we'd have a proof. Unfortunately we need to get rid of the $x$ in the denominator.
Specifically, we need some lower bound for $\frac{1}{x + \sqrt{3}}$ that's a number independent of $x$. Luckily, since $x$ is "close" to $\sqrt{3}$, $\frac{1}{x + \sqrt{3}}$ won't grow without bound anywhere. To bound this denominator, we'll additionally require that $\delta \leq 1$. This gives us
$$ \frac{1}{x + \sqrt{3}} \geq \frac{1}{1 + 2\sqrt{3}},$$
so by requiring
$$ \delta = \frac{\epsilon^2}{1 + 2\sqrt{3}} $$
our proof will stay correct.
The way to formally write this is
$$ \delta := \min(1, \frac{\epsilon^2}{1 + 2\sqrt{3}}) .$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is the mistake of this derivation? It is asked to find which on is greater:
$$A) \frac{x^2}{y+\frac{1}{y}}$$
$$B) \frac{y^2}{x+\frac{1}{x}}$$
It is given that, $xy\neq0 $ & $x>y$.
So I solve this like this
$$x>y$$
$$\implies x^3+x>y^3+y$$
$$\implies \frac{1}{x^3+x}<\frac{1}{y^3+y}$$
$$\implies \frac{x^2 y^2}{x^3+x}<\frac{x^2 y^2}{y^3+y}$$
$$\implies \frac{x^2 y^2}{x^2(x+1/x)}<\frac{x^2 y^2}{y^2(y+1/y)}$$
$$\implies \frac{y^2}{(x+1/x)}<\frac{x^2}{(y+1/y)}$$
But if I use $x=2 $ & $y=-2$ then
$$\frac{y^2}{(x+1/x)}=\frac{(-2)^2}{(2+1/2)}=\frac{8}{5}$$
$$\frac{x^2}{(y+1/y)}=\frac{2^2}{(-2+1/(-2))}=-\frac{8}{5}$$ which gives total opposite result.
Could you please show me where I did wrong?
| $$a>b$$
Doesn't imply
$$\frac1a <\frac 1b$$
This is true only if both $a$ and $b$ are positive.
For example $$3>2 \implies \frac 13<\frac 12$$
So far so good, but $$-2>-3 \not \implies -\frac{1}{2}<-\frac 13$$
And
$$2>-3 \not \implies \frac{1}{2}<-\frac 13$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve $\sqrt{3+\sqrt{3+x}}=x$
Solve
$$\sqrt{3+\sqrt{3+x}}=x$$
My try:
$$\sqrt{3+\sqrt{3+x}}=x \\
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
| $$\sqrt{3+x}=x^2-3$$
$$3+x=(x^2-3)^2$$
$$x^4 -6x^2 +9 -x - 3=x^4 -6x^2-x+6=0$$
$$(x-1)(x+2)(x^2-x-3)=0$$
None of $1, -2, \frac{1-\sqrt{13}}{2}$ are not answers, due to our double-squaring. Basically, we need the following inequalities to be true:
$$x+3 \ge 0$$
$$x\ge 0$$
$$\sqrt{3+x} = x^2 - 3 \ge 0$$
Thus $x \ge \sqrt{3}$, and $x=\frac{1+\sqrt{13}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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PreCalc Roots of Unity Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$$
So far, I have simplified that down to $\frac{-2}{(1+\omega)^2}$, but I don't know how to continue. Thank you!
| Let $f(x) = x^3 / (1 + x)$. Then, you're looking for
$$ f(\omega) + f(\omega^2) + f(\omega^3) + f(\omega^4) $$
But we can simplify the formula by multiplying the top and bottom by an appropriate formula:
$$ f(x) = \frac{x^3 (1 - x + x^2 - x^3 + x^4)}{1 + x^5} $$
$$ f(x) \equiv \frac{x^3 - x^4 + 1 - x + x^2}{2} \pmod{x^5 \equiv 1} $$
Since $\omega + \omega^2 + \omega^3 + \omega^4 = -1$, we can add up the five terms of this formula separately to get
$$ f(\omega) + f(\omega^2) + f(\omega^3) + f(\omega^4)
= \frac{(-1) - (-1) + 4 - (-1) + (-1)}{2}
\\= 2
$$
This works because whenever $\gcd(n, 5) = 1$, the four possible values of $(\omega^i)^n$ are all four of the values $\omega, \omega^2, \omega^3, \omega^4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove an inequality involving powers of two Let $a,b,c$ be mutually distinct positive integers.
Prove
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
Any hints on how to start proving it?
| It can also be proven by induction.
Assume it's true for $(a,b,c)$. Then:
\begin{align}
2^{a+b+(c+1)}+1&=2\cdot 2^{a+b+c}+2-1\\
&=2\cdot 2^{a+b+c}+2-1\\
&=2(2^{a+b+c}+1)-1\\
&>2(2^{a+b}+2^{b+c}+2^{c+a})-1\\
&>2^{a+b}+2^{b+(c+1)}+2^{(c+1)+a}+2^{a+b}-1\\
&>2^{a+b}+2^{b+(c+1)}+2^{(c+1)+a}\\
\end{align}
Note that we didn't assume $a,b,c$ are distinct, or anything about the order. Thus, from $(a,b,c)$ we can not only derive $(a,b,c+1)$, but also $(a,b+1,c)$ and $(a+1,b,c)$. Now check the base case:
$$2^{1+2}+2^{2+3}+2^{3+1}<2^{1+2+3}+1$$
so that it's true for all $a,b,c\in\Bbb N$ (such that they are larger than $1$, $2$ and $3$; they don't have to be distinct, but we still needed that for the base case).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Root of the equation $x^2+ix+2=0$
What are the roots of the equation $x^2+ix+2=0$, where $i=\sqrt{-1}$?
*
*$(-1, 1)$
*$(-2i, i)$
*$( i, 1)$
*no root exists
I don't know the method for finding roots of this type of equation.
I have tried by the method of $b^2-4ac$ but it doesn't work
| Use the quadratic formula:
\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
&=\frac{-i\pm\sqrt{i^2-4\times 2}}{2}\\
&=\frac{-i\pm\sqrt{(-1)-8}}{2}\\
&=\frac{-i\pm\sqrt{-9}}{2}\\
&=\frac{-i\pm i\sqrt{9}}{2}\\
&=\frac{-i\pm3i}{2}\end{align}
So we have $$x=\frac{-i+3i}{2}=\frac{2i}{2}=i$$ and $$x=\frac{-i-3i}{2}=\frac{-4i}{2}=-2i$$
So our roots are $(-2i,i)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the derivative of $y^3-xy^2+\cos xy=2$ Find the derivative of $y^3-xy^2+\cos xy=2$
My Attempt:
$$y^3-xy^2+\cos xy=2$$
$$\dfrac {d}{dx} [y^3-xy^2+\cos xy]=\dfrac {d}{dx} [2]$$
$$3y^2.\dfrac {dy}{dx} -[1.y^2+2xy\dfrac {dy}{dx}] + (-\sin xy) \dfrac {dy}{dx}(xy)=0$$
$$3y^2 \dfrac {dy}{dx} - y^2 - 2xy\dfrac {dy}{dx} - \sin xy (y+\dfrac {dy}{dx} x)=0$$
How do I proceed further?
| \begin{align}
y^3 - xy^2 + \cos xy=& 2\\
3y^2 \frac{dy}{dx} - y^2 - 2yx\frac{dy}{dx}-\sin xy(y+x\frac{dy}{dx}) =& 0\\
3y^2 \frac{dy}{dx} - y^2 - 2yx\frac{dy}{dx}-y\sin xy-x\sin xy\frac{dy}{dx} =& 0\\
\frac{dy}{dx}(3y^2 - 2xy - x\sin xy) =&y^2+y\sin xy
\end{align}
Hence
$$\frac{dy}{dx} = \frac{y^2+y\sin xy}{3y^2 - 2xy - x\sin xy}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$$
how should I approach this?
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)}=0$$ but I still have $$\lim_{(x,y)\to(0,0)}\frac{1}{x^2y^2}$$
The answer says there is no limit
| Hint :
$$\lim_{(x,y)\to (0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)}=\lim_{(x,y)\to (0,0)}\frac{2\sin^2\left (\frac{x^2+y^2}{2}\right)}{x^2+y^2} = \lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof the area of a given triangle with coordinates is half determinant I was given a problem, tried to solved it but couldn't get to a solution.
It goes like that:
There's a triangle ABC with area S.
$$ \vec{AB} = (a,b) $$
$$ \vec{AC} = (c,d) $$
Prove that
$$ S = \frac{\lvert ad - bc \rvert}{2} $$
I tried to solve it that way:
Express $AB \cdot AC$ as $ac+bd$ and as $\sqrt{a^2+b^2}\sqrt{c^2+d^2}\cos(\alpha)$
then I expressed with $a,b,c,d \sin(\alpha)$ (by squaring both sides and using $\cos^2\alpha=1-\sin^2\alpha$)
then I expressed S as $|AC||AB|\sin(\alpha)/2$, put $\sin(\alpha)$, $|AC|$ and $|AB|$ expressed with $a,b,c,d$ and got a disgusting expression which is probably not equal $ad-bc$...
Would be happy to get your help,
Thanks
| Your approach:
From
$$\cos\alpha=\frac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}},$$
you draw
$$1-\cos^2\alpha=\frac{(a^2+b^2)(c^2+d^2)-(ac+bd)^2}{(a^2+b^2)(c^2+d^2)}.$$
You transform the numerator with
$$a^2c^2+a^2d^2+b^2c^2+b^2d^2-a^2c^2-2abcd-b^2d^2=a^2d^2+b^2c^2-2abcd=(ad-bc)^2$$
hence the claim.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Different Representations of GF(8) Can anyone point me in the right direction with the following problem?
Given that
$$GF(8)=\frac{Z_{2}}{x^3+x^2+1}= \frac{Z_{2}}{x^3+x+1}$$
Find $\beta$ as a function of $\alpha$ , where $\alpha$ is a root of $x^3+x+1$ and $\beta$ is a root of $x^3+x^2+1$
| Let $\alpha$ be a root of $x^3+x+1$. Then a general element of $GF(8)$ has the form $a+b\alpha + c\alpha^2$ for some $a,b,c \in GF(2)$. Need to solve for for $a,b,c$ in $(a+b\alpha + c\alpha^2)^3+(a+b\alpha + c\alpha^2)^2+1 = 0$ Recall that $u^2=u$ for any $u \in GF(2)$ and $2u=0$ for any $u \in GF(8)$. After simplifying you'll find
$$ (1+b+c+bc) + (b+c+ab+ac)\alpha + (b+ab)\alpha^2 = 0$$
from which you can determine that $\beta$ can be taken to be $1+\alpha, 1+\alpha^2, $ or $1+\alpha+\alpha^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\frac{x^\alpha}{(1 + x^2)^2}$ from 0 to $\infty$ using contour integration for $-1 < \alpha <3$ I have tried to evaluate $\int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx$ for $-1 < \alpha < 3$ using the keyhole contour, but am not sure about my working, especially in defining the analytic branch of log $z$ when we consider the function $f(z) = \dfrac{z^\alpha}{(z^2 + 1)^2}$ where $z^\alpha = e^{\alpha\text{log}\ z}$. Below is my working:
Using the keyhole contour $K_{\varepsilon,M}$ and considering the complex function $\dfrac{z^\alpha}{(z^2 + 1)^2}$, we inspect the integral of the function over each part of the contour (namely $C_{\varepsilon},C_M, I_1$ and $I_2$) in the limit $\varepsilon \to 0,\ M \to \infty$. For $C_{\varepsilon}$, we have $|z^2 +1|$ to be the modulus of points lying on a circle of radius $\varepsilon^2$ centered at $z = 1$ and hence $|z^2 + 1| > 1/2$ for $\varepsilon$ small enough (e.g. $\varepsilon^2 < 1/2$). We thus have:
\begin{align*}
\int_{C_{\varepsilon}} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz \ll \pi\varepsilon\ \text{max}_{C_\varepsilon}\bigg|\dfrac{z^\alpha}{(z^2 + 1)^2}\bigg| \ll 4\pi\varepsilon^{\alpha + 1} \Rightarrow \lim_{\varepsilon \to 0}\int_{C_{\varepsilon}} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = 0
\end{align*} For any $z \in C_M$, we have $|z^2 + 1| \geq |z^2| = M^2$ for $M$ big enough. We thus have:
\begin{align*}
\int_{C_M} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz \ll 2\pi M \ \text{max}_{C_M}\bigg|\dfrac{z^\alpha}{(z^2 + 1)^2}\bigg| \ll 2\pi M^{\alpha - 3} \Rightarrow \lim_{M \to \infty}\int_{C_M} \dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = 0
\end{align*}
For $I_1$, we have $\begin{aligned} = \lim_{\varepsilon \to 0,\ M \to \infty}\int_{I_1}\dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = \int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx \end{aligned}$. As for $I_2$, we have $z^\alpha = e^{\alpha(\text{ln}x + 2\pi i)} = x^{\alpha}e^{2\pi i\alpha}$, and hence we have $\begin{aligned} = \lim_{\varepsilon \to 0,\ M \to \infty}\int_{I_2}\dfrac{z^\alpha}{(z^2 + 1)^2}\ dz = -e^{2\pi i\alpha}\int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx \end{aligned}$. Then, by the Residue Theorem, we have:
\begin{align*}
\int_0^\infty \dfrac{x^\alpha}{(x^2 + 1)^2}\ dx &= \frac{2\pi i}{1 - e^{2\pi i\alpha}}\sum_k \text{Res}\bigg(\dfrac{z^\alpha}{(z^2 + 1)^2}; z_k\bigg) \\
&= \frac{2\pi i}{1 - e^{2\pi i\alpha}}\bigg(\lim_{z \to i} \frac{d}{dz}\bigg(\dfrac{z^\alpha}{(z + i)^2}\bigg) + \lim_{z \to -i} \frac{d}{dz}\bigg(\dfrac{z^\alpha}{(z - i)^2}\bigg)\bigg) \\
&= \frac{2\pi i}{1 - e^{2\pi i\alpha}}\bigg(\frac{\alpha -1}{4}i^{\alpha -3} + \frac{\alpha -1}{4}(-i)^{\alpha -3}\bigg) \\
&= \frac{\pi i(\alpha -1)\text{cos}(\pi(\alpha - 3)/2)}{1 - e^{2\pi i\alpha}}
\end{align*}
where the keyhole contour is given as !Keyhole Contour 1
Any help in correcting my working or giving hints or posting your solution is appreciated.
| Complex Analysis Approach
Consider the function
$$f(z) = \frac{z^{\alpha}}{(z^2+1)^2}$$
Note that $z^{\alpha} = e^{\alpha \log(z)}$ and choose the branch cut on the imaginary axis where $$\log(z) = \log|z|+ i\theta , \,\,\,\theta \in (-3\pi/2,3\pi/2)$$
Consider the following contour
$$\int^{-r}_{-R}f(z) \,dz +\int^{R}_{r}f(z) \,dz+ \int_{C_R}f(z)\,dz+\int_{c_r}f(z)\,dz =2\pi i \mathrm{Res}(f,i)$$
As you showed the integrals on the semi-circles go to zero as $r \to 0, R \to \infty$ hence
$$\int^{0}_{-\infty}\frac{e^{\alpha(\log|x|+i\pi)}}{(x^2+1)^2}\,dx+\int_{0}^{\infty}\frac{e^{\alpha(\log|x|)}}{(x^2+1)^2}\,dx = 2\pi i \mathrm{Res}(f,i)$$
$$(1+e^{\alpha \pi i})\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx = 2\pi i \mathrm{Res}(f,i)$$
Note that
$$\mathrm{Res}(f,i) = \lim_{z \to i}\frac{d}{dz} \frac{e^{\alpha \log(z)}}{(z+i)^2} = \lim_{z \to i} \frac{\alpha z^{\alpha-1}(z+i)-2z^{\alpha}}{(z+i)^3} = \frac{2e^{i\alpha \pi/2}(\alpha-1)}{-8 i }$$
$$\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx=\frac{\pi e^{i\alpha \pi/2}(1-\alpha)}{2(1+e^{\alpha \pi i}) } = \frac{\pi (1-\alpha)}{4\cos\left( \frac{\pi}{2}\alpha\right)}$$
Real Analysis Approach
$$\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx$$
Let $x^2 = t$
\begin{align}\frac{1}{2}\int_{0}^{\infty}\frac{t^{(\alpha-1)/2}}{(t+1)^2}\,dt &= \frac{1}{2}\Gamma \left(\frac{\alpha +1}{2} \right)\Gamma \left(2-\frac{\alpha +1}{2}\right)
\tag{1}\\ & = \frac{(1-\alpha)}{4}\Gamma \left(\frac{\alpha +1}{2} \right)\Gamma \left(1-\frac{\alpha +1}{2}\right)
\tag{2}\\ & = \frac{\pi (1-\alpha)}{4 \sin\left((1-\alpha)\pi/2 \right)}
\tag{3}\\ & = \frac{\pi (1-\alpha)}{4\cos\left( \frac{\pi}{2}\alpha\right) }\tag{4}\end{align}
(1). $$\int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}}\,dt = B(x,y)$$
(2). $$\Gamma(x+1) = x \Gamma(x)$$
(3). $$\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which occur at $x = \frac{(4n+1)\pi}{2} $ and $\frac{-(4n+3) \pi}{2}$
Equating this to its Taylor series, we get
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x)(1 + \frac{2}{3\pi}x)(1 - \frac{2}{5\pi}x)(1 + \frac{2}{7\pi}x)... $$
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x(1-\frac13)+O(x^2))(1 - \frac{2}{\pi}x(\frac15-\frac17)+O(x^2))... $$
$$1 - x + O(x^3) = 1 - \frac{2}{\pi}x(1-\frac13+\frac15-\frac17+...)+O(x^2)$$
$$x - O(x^3) = \frac{2}{\pi}x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}-O(x^2)$$
Equating $x$ terms,
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{2}$$
However, I seem to be off by a factor of $2$ (actual answer is $\frac{\pi}{4} $). Does anyone see where this went wrong?
| The Taylor series expansion for the arctangent function (see this) is
$$\begin{align}
\arctan(x)
&=\sum_{n=0}^\infty \frac {(-1)^n}{2n+1}x^{2n+1}=x-\frac 13 x^3+\frac 15x^5-\frac 17 x^7+\cdots
\end{align}$$
Putting $x=1$ gives
$$\arctan(1)=\color{red}{\frac {\pi}4}=\sum_{n=0}^\infty \frac {(-1)^n}{2n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Finding maximum area of a pentagon Out of all the pentagons in the shape of a rectangle overlapped by an isosceles triangle, with perimeter P fixed, determine the dimensions of the one with maximum area.
| The object is to maximize the area $A(x,y,z)$ of the pictured pentagon while keeping the perimeter $P(x,y,z)$ constant.
\begin{eqnarray}
A(x,y,z)&=&xy+\frac{1}{2}x\sqrt{z^2-\left(\frac{x}{2}\right)^2}\\
&=&xy+\frac{x}{4}\sqrt{4z^2-x^2}
\end{eqnarray}
\begin{equation}
P(x,y,z)=x+2y+2z
\end{equation}
This may be accomplished using the method of LaGrange multipliers, solving
\begin{equation}
\nabla A=\lambda\nabla P
\end{equation}
\begin{eqnarray}
\nabla A(x,y,z) &=&\begin{pmatrix}
{y+\dfrac{2z^2-x^2}{2\sqrt{4z^2-x^2}}}\\\\
x\\\\
\dfrac{xz}{\sqrt{4z^2-x^2}}
\end{pmatrix}\\
\nabla P(x,y,z)&=&\begin{pmatrix}
1\\2\\2
\end{pmatrix}
\end{eqnarray}
So we have
*
*$$ y+\dfrac{2z^2-x^2}{2\sqrt{4z^2-x^2}}=\lambda$$
*$$ x=2\lambda $$
*$$ \dfrac{xz}{\sqrt{4z^2-x^2}}=2\lambda $$
*$$ P=x+2y+2z $$
Since $P$ is the only fixed quantity it would be best to solve for $x,\,y$ and $z$ in terms of $P$. One way to do that is to first solve for $x,\,y$ and $z$ in terms of $\lambda$ then use the fourth equation to solve for $\lambda$ in terms of $P$.
This strategy results in
*
*$$x=2\lambda$$
*$$y=\left(\dfrac{3+\sqrt{3}}{3}\right)\lambda$$
*$$ z=\left(\dfrac{2\sqrt{3}}{3}\right)\lambda $$
*$$ \lambda=\left(\dfrac{2-\sqrt{3}}{2}\right)P $$
Solving for the values of $x,\,y$ and $z$ in terms of $P$ gives
*
*$$x=\left(2-\sqrt{3}\right)P $$
*$$ y=\left(\dfrac{3-\sqrt{3}}{6}\right)P$$
*$$ z=\left(\dfrac{2\sqrt{3}-3}{3}\right)P$$
I have omitted algebraic details of the last six equations because of the length of time required to render them into MathJax. If OP finds an error in any of these, let me know and I will double-check my steps.
Addendum: I have checked my results twice and found no errors.
There is also a curious feature of the solution.
The following are easily verified:
*
*$xy=\dfrac{9-5\sqrt{3}}{6}P^2$
*$x-y=\dfrac{9-5\sqrt{3}}{6}P$
Therefore
$$ \frac{xy}{x-y}=P $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An nth term for a Fibonacci series? Say the first two terms of a sequence are $a_0,a_1$, then the remaining terms meet the formula $$a_{n+2}=a_{n+1}+a_n$$
What is the $n_{th}$ term formula?
I figured that $$\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}=\Phi = \frac{1+\sqrt 5}{2}\approx 1.618$$
Using this fact, find the $n_{th}$ term formula for the Fibonacci Series.
| Let $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$, where $\alpha, \beta$ are roots of $x^2 = x + 1$, then given $a_{n+2} = a_{n+1} + a_{n}$ it is seen that:
$$a_{n} = A \, \alpha^n + B \, \beta^n. $$
Now
\begin{align}
a_{0} &= A + B \\
a_{1} &= A \, \alpha + B \, \beta
\end{align}
which yields $(\alpha - \beta) \, A = a_{1} - a_{0} \beta$, $(\alpha - \beta) \, B = a_{0} \alpha - a_{1}$, and
$$a_{n} = \frac{1}{\alpha - \beta} \, (a_{1} \, (\alpha^n - \beta^n) + a_{0} \, (\alpha^{n-1} - \beta^{n-1})) = a_{1} \, F_{n} + a_{0} \, F_{n-1}. $$
Here
$$F_{n} = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$
are the Fibonacci numbers.
Given: $(a_{0}, a_{1}) = (0,1)$ then $a_{n} = F_{n}$ and $(a_{0}, a_{1}) = (2,1)$ then $a_{n} = F_{n} + 2 \, F_{n-1} = F_{n+1} + F_{n-1} = L_{n}$, where $L_{n}$ are the Lucas numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?
| We have $$\begin{align}\frac{\frac{1}{x+1} - \frac{1}{9}}{(x-8)((x+1)^{-1/2} + 1/3)} &= \frac{\frac{8-x}{9(x+1)}}{(x-8)((x+1)^{-1/2} + 1/3)} \\&= -\frac{1}{9(x+1)((x+1)^{-1/2} + 1/3))} \\ &\longrightarrow-\frac{1}{9(9)(2/3)} = \color{blue}{-\frac{1}{54}}.\end{align}$$
In particular you forgot to take into account the $9(x+1)$ term you get when simplifying $\frac{1}{x+1} - \frac{1}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$
let :
$$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$
My try :
$$\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=\sqrt3+\sqrt 3\times\sqrt2+4+3\sqrt2 \\ \sqrt3(1+\sqrt2+3)+4$$
Now what ?
| Achille hui's comment expanded:
$$x-1\:=\:\frac{\left(\sqrt 6 +4+3\sqrt 2\right)-\left(\sqrt 2 +2\right)}{\sqrt{3}+\sqrt{2}+2}\:
=\:\frac{\sqrt{2}\sqrt{3}+\sqrt 2\sqrt 2+\sqrt 2\cdot2}{\sqrt{3}+\sqrt{2}+2}\:=\:\sqrt 2 \\[4ex]
\implies\; x=\sqrt 2 +1\quad\implies\;\frac 1x=\sqrt 2-1
$$
which yields
$$x+\frac 1x\;=\;2\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor=
\lim_{x\to 0}\left\lfloor\dfrac{2\cos x}{\cos 2x}\right\rfloor $$
Now what?
| The function in question is even and hence it is sufficient to deal with $x\to 0^{+}$. And then we have the famous inequality $$\sin x<x<\tan x\tag{1}$$ so that $$\frac{\tan 2x}{\sin x} =\frac{\tan 2x}{2x}\cdot 2\cdot\frac{x}{\sin x} > 1\cdot 2\cdot 1=2$$ And clearly from the above equation it is evident that $(\tan 2x)/\sin x\to 2$ so that as $x\to 0^{+}$ we have $(\tan 2x)/\sin x<5/2$. It follows that the function $[(\tan 2x)/\sin x] $ is constant with value $2$ as $x\to 0^{+}$ and hence the desired limit is $2$.
Your approach is also fine and you may do without inequality $(1)$. You just have to note that as $x\to 0^{+}$ we have $x<2x$ and $\cos$ is decreasing so that $\cos x>\cos 2x$ and then $$\frac{2\cos x} {\cos 2x}>2$$ and the limit of the fraction is $2$ so that it is less than $5/2$ as $x\to 0^{+}$. And as before the desired limit is $2$. The question is much simpler than it appears.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integration Using Trig Sub and Partial Fractions $\displaystyle\int\frac{1}{x^4 + x^2 + 1}\mathrm{d}x$ using the trig substitution.
My attempt: I got $\left(x^2 + \frac{1}{2}\right)^2+\frac{3}{4}$ and did $x= \frac{1}{\sqrt2}\tan \theta$ but did not know how to proceed because I got $\displaystyle\int\frac{\sec^2\theta}{\sec^4\theta+ 3}\mathrm{d}\theta$ after factoring out the constant
| HINT:
$$\dfrac2{x^4+x+1}=\dfrac{x^2+x+1-(x^2-x+1)}{x(x^2+x+1)(x^2-x+1)}=\dfrac1{x(x^2-x+1)}-\dfrac1{x(x^2+x+1)}$$
Set $2x+1=\sqrt3\tan y$
$$\int\dfrac1{x(x^2+x+1)}dx=\int\dfrac4{x\{(2x+1)^2+3\}}dx=\int\dfrac{4dy}{(\sqrt3\tan y-1)\sqrt3}$$
$$=\dfrac4{\sqrt3}\int\dfrac{\cos y}{\sqrt3\sin y-\cos y}dy$$
$$=-\dfrac2{\sqrt3}\int\dfrac{\cos\left(y+\dfrac\pi3-\dfrac\pi3\right)}{\cos\left(y+\dfrac\pi3\right)}dy$$
$$=-\dfrac1{\sqrt3}\int\dfrac{\cos\left(y+\dfrac\pi3\right)-\sqrt3\sin\left(y+\dfrac\pi3\right)}{\cos\left(y+\dfrac\pi3\right)}dy=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the sum of the series $\frac{1}{2^a 3^b 5^c}$ I'm trying to compute this sum:
$$\sum_{1 \le a \lt b \lt c; a,b,c \in\mathbb N}^n \frac{1}{2^a 3^b 5^c}$$
I've tried to try to compute $\sum\frac{1}{2^{a+b+c}}$ and $\sum \frac{1}{5^{a+b+c}}$ (perhaps they will end up being equal) but the three indexes are giving me the trouble.
I've tried dropping the $a<b<c$ restriction (perhaps I'll be able to compute special cases ($a = 1$, $b = 2$) separately, but I'm still stuck.
| \begin{align}
\sum_{1 \le a \lt b \lt c; a,b,c \in\mathbb N} \frac{1}{2^a 3^b 5^c}&=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\sum_{c=b+1}^{\infty}\frac{1}{2^a3^b5^c}\\
&=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\frac{1}{2^a3^b}\sum_{c=b+1}^{\infty}\frac{1}{5^c}\\
&=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\frac{1}{2^a3^b}\cdot\frac{5^{-b-1}}{1-\frac{1}{5}}\\
&=\sum_{a=1}^{\infty}\frac{1}{4\cdot2^a}\sum_{b=a+1}^{\infty}\frac{1}{15^b}\\
&=\sum_{a=1}^{\infty}\frac{1}{4\cdot2^a}\cdot\frac{15^{-a-1}}{1-\frac{1}{15}}\\
&=\frac{1}{56}\sum_{a=1}^{\infty}\frac{1}{30^a}\\
&=\frac{1}{56}\cdot\frac{30^{-1}}{1-\frac{1}{30}}\\
&=\frac{1}{1624}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find if the function is Continuous and whether the partial derivatives exist
$$f(x,y)=\begin{Bmatrix}
0 & (x,y)=(0,0)\\
\frac{xy}{x^{2}+y^{2}}&(x,y)\neq (0,0)
\end{Bmatrix}$$
How to find out if this is continuous?
And do the partial derivatives exist at $(0,0)$?
| Note that it is not differential at $(0,0)$ $$\left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) \overset { n\rightarrow \infty }{ \longrightarrow } \left( 0,0 \right) ,\left( \frac { 2 }{ n } ,\frac { 1 }{ n } \right) \overset { n\rightarrow \infty }{ \longrightarrow } \left( 0,0 \right) \\ f\left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) =\frac { \frac { 1 }{ { n }^{ 2 } } }{ \frac { 2 }{ { n }^{ 2 } } } \longrightarrow \frac { 1 }{ 2 } \\ f\left( \frac { 2 }{ n } ,\frac { 1 }{ n } \right) =\frac { \frac { 2 }{ { n }^{ 2 } } }{ \frac { 5 }{ { n }^{ 2 } } } \longrightarrow \frac { 2 }{ 5 } \\ \quad $$
And partial derivatives are $${ f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { x\cdot 0 }{ { x }^{ 2 }+{ 0 }^{ 2 } } -0 }{ x } } =0\\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( 0,y \right) -f\left( 0,0 \right) }{ y } } =\lim _{ y\rightarrow 0 }{ \frac { \frac { 0\cdot y }{ { 0 }^{ 2 }+{ y }^{ 2 } } -0 }{ y } } =0\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that the square of any given odd number gives a remainder of 1 when divided by 4 I'm supposed to prove this by induction, and I get that if I assume $n$ is odd such that $n^2$ gives a remainder of 1, then $n^2=4m+1$ for a number $m$ where $m \in \mathbb{Z}$, but I honestly can't get any further than that. Could someone complete the proof for me so that I can get a proper idea of how I can do this?
| Here is a proof that doesn't require induction:
We want to show that $n^2 - 1 \equiv 0 \mod{4}$ for $n$ odd.
Assume n is odd. Note that $n^2 = (n-1)^2 + 2(n-1) + 1$, or, visually for n =5:
Since $n$ is odd, $n-1$ is even and can be written as $2k$ for some $k$. So we have that:
$$\begin{align}
n^2 - 1 &= (n-1)^2 + 2(n-1) \\
&= (2k)^2 + 2(2k) \\
&= 4k^2 + 4k \\
&= 4(k^2 + k)
\end{align}$$
So we have shown that $n^2 -1$ is a multiple of 4, so by definition $n^2 - 1 \equiv 0 \mod{4}$ or $n^2 \equiv 1 \mod{4}$
$\blacksquare$
Unfortunately, some professors require a certain type of proof even when many are valid. We can try to turn this reasoning into an induction proof. First, we prove the claim for the base case:
Let $n = 1$. Then $1^2 = 1 \equiv 1 \mod{4}$, so the claim holds for the base case. Now, we move to prove the inductive step. Assume the claim holds for some odd n. We want to show that it holds for $n+2$ (the next odd n). See the figure below to visualize this.
What we are saying is that $\color{red} n^2 \color{black} \equiv 1 \mod{4}$, so we want anything we add in to get to $(n+2)^2$ to be equivalent to 0 mod 4. That is, we need to check that
$$
2\color{blue}n + 2\color{orange}{(n+1)}\color{black} + \color{purple} 2 \color{black} \equiv 0 \mod{4}
$$
Well, we see this must be the case because
$$\begin{align}
2n + 2(n+1) + 2 &= 2n + 2n + 2 + 2 \\
&= 4n + 4 \\
&= 4(n+1) \\
\end{align}$$
is a multiple of four, so $2n + 2(n+1) + 2 \equiv 0 \mod{4}$. So putting this together we get that:
$$
(n+2)^2 = n^2 + 2n + 2(n+1) + 2 \equiv 1 \mod{4}
$$
as desired. $\blacksquare$.
The other proofs here are perfectly valid, I was hoping only to give a little perspective.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Integrate with residues $I=\int_0^\infty \frac {\ln^2(x)\ln(1+x)}{1+x^2}dx$ I was trying next: $I=\int_0^\infty \frac {\ln^2(x)\ln(1+ax)}{1+x^2}$;
$\frac{dI}{da}=\int_0^\infty \frac{ x\ln^2(x)}{(1+x^2)(1+a*x)}\ dx$;
$f(z)= \frac {z\ln^3(z)}{(1+az)(1+z^2)}$ ; $f(\overline{z})= \frac {z(\ln(z+2\pi*i))^3}{(1+az)(1+z^2)}$;
$\textrm{Res}_{z=i}f(z)=-\frac{\pi^3(i+a)}{16(a^2+1)}$;
$\textrm{Res}_{z=-i}f(z)=\frac{\pi^3(i-a)}{16(a^2+1)}$;
$\textrm{Res}_{z=-\frac{1}{a}}f(z)=\frac{(\ln(a)-\pi i)^3}{a^2+1}$;
sum of integrals around the contour: $\int_0^\infty\frac {z\ln^3(z)}{(1+az)(1+z^2)}dz-(\int_0^\infty\frac {z\ln^3(z)}{(1+az)(1+z^2)}dz+6\pi i \int_0^\infty\frac {z\ln^2(z)}{(1+az)(1+z^2)}dz-12\pi^2\int_0^\infty\frac {z\ln(z)}{(1+az)(1+z^2)}dz-8\pi^3 i\int_0^\infty\frac {z}{(1+az)(1+z^2)}dz)$
$\int_0^\infty\frac {z}{(1+az)(1+z^2)}dz=\frac{\pi a-2\ln((a)}{2(a^2+1)}$,
so,
$-6\pi i\frac{dI}{da}+8\pi^3 i\frac{\pi a-2\ln((a)}{2(a^2+1)}=2\pi i\sum{Res}=\frac{8\pi i \ln^3(a)-24\pi i\ln^2(a)-24\pi^2\ln(a)-8\pi^3 i-\pi^3 a}{4(1+a^2)} $;
taking imaginary parts:
$\frac{dI}{da}=\frac{17\pi^3 a-8\pi^2\ln(a)-8\ln^3(a)}{24(a^2+1)}$, and $I=\int_{0}^{1}\frac{17\pi^3 a-8\pi^2\ln(a)-8\ln^3(a)}{24(a^2+1)}da$,
but final answer doesn't connect with mathematica. Where is my mistake?
| To apply Feynman's trick is a nice idea:
$$ I=\int_{0}^{+\infty}\frac{\log^2(x)\log(1+x)}{1+x^2}\,dx = \int_{0}^{1}\int_{0}^{+\infty}\frac{x\log^2(x)}{(1+x^2)(1+a x)}\,dx\,da \tag{1}$$
leads to:
$$ I = \frac{\pi^3}{8}\int_{0}^{1}\frac{a\,da}{1+a^2}-\frac{\pi^2}{3}\int_{0}^{1}\frac{\log(a)}{1+a^2}\,da-\frac{1}{3}\int_{0}^{1}\frac{\log^3(a)}{1+a^2}\,da \tag{2} $$
and since $\int_{0}^{1}a^k\log(a)\,da = -\frac{1}{(k+1)^2}$, $\int_{0}^{1}a^k\log^3(a)\,da = -\frac{6}{(k+1)^4}$ it turns out that
$$\boxed{ I = \color{red}{\frac{\pi^3\log 2}{16}+\frac{\pi^2}{3}\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^2}+2\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^4}}}\tag{3} $$
depending on quite non-elementary constants, as already pointed out by Zaid Alyafeai.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.
Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.
| Hint: $$\frac{a}{b+c}=\frac{a+b+c}{b+c}-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 1
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Parametrising a given surface to verify Stokes' Theorem
Question
Let $M := \{ (x,y,z) \in \mathbb{R}^{3}: 4z^{2} - x^{2} = 1 \text{ and } x^{2} + y^{2} <1 \text{ and } z > 0 \}$ be a surface in $\mathbb{R}^{3}$. Verify Stokes' Theorem for $\mathbf{F} = (xy, -2z^{2}, 1)$ on $M$.
I am having trouble finding the correct surface parametrisation for $M$, because of the $4z^{2} - x^{2} = 1$ condition I tried $z = \frac{1}{2} \cosh(t)$ and $x = \sinh(t)$ but I cannot figure out then how to find the values that $t$ must take or to incorporate the unit circle constraint.
How can I work this out?
| Since you have the condition $z>0$, you can use $x$ and $y$ as parameters and parametrize the surface $M$ as follows:
\begin{cases}
x=x \\
y=y \\
z= \sqrt{\frac{1+x^2}{4}}
\end{cases}
with $(x,y)\mid x^2+y^2 \le 1$.
For any curve $C$ that bounds this surface, e.g., the circle $x^2+y^2 =1$, you have to verify that the following holds
$$
\oint_C \vec{F}\cdot d\vec{r} = \iint_M \nabla \times\vec{F}\cdot d\vec{S}
$$
For the left hand term, since you are on a unit circle at $z=0$, compute
$$
\oint_C \vec{F}\cdot d\vec{r} = \oint_0^{2\pi} \pmatrix{\cos t \sin t \\ 0 \\ 1}\cdot \pmatrix{- \sin t \\ \cos t \\0}dt=0
$$
For the right hand term, use the above parametrization:
\begin{align}
\iint_M \nabla \times\vec{F}\cdot d\vec{S} &= \iint_{x^2+y^2\le 1}\pmatrix{4\sqrt{\frac{1+x^2}{4}} \\0 \\-x}\cdot \pmatrix{1\\0\\\frac{x}{2\sqrt{1+x^2}}} \times \pmatrix{0\\1\\0} dxdy
\\&= \iint_{x^2+y^2\le 1}\pmatrix{4\sqrt{\frac{1+x^2}{4}} \\0 \\-x}\cdot
\pmatrix{-\frac{x}{2\sqrt{1+x^2}}\\0\\1}dx dy
\\&= \iint_{x^2+y^2\le 1}-x \;dx dy=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
| As an alternative approach, since by computing residues we have
$$ \frac{3n+7}{n(n+1)(n+2)} = \frac{7}{2}\cdot\frac{1}{n}-4\cdot\frac{1}{n+1}+\frac{1}{2}\cdot\frac{1}{n+2} \tag{1}$$
it follows that
$$ S=\sum_{n\geq 1}\frac{3n+7}{n(n+1)(n+2)} = \int_{0}^{1}\left(\frac{7}{2}-4x+\frac{x^2}{2}\right)\sum_{n\geq 1}x^{n-1}\,dx \tag{2}$$
hence:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{7-8x+x^2}{1-x}\,dx = \frac{1}{2}\int_{0}^{1}(7-x)\,dx = \frac{1}{2}\left(7-\frac{1}{2}\right)=\color{red}{\frac{13}{4}}.\tag{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $p, q, r$ are the roots of $x^3 -x+1=0$, what is $p^5 + q^5 + r^5$? Can someone please help us solve this equation. It was on our college entry exam, but no one managed to solve it. The problem is:
Let $p,q,r$ be the roots of $x^3 -x+1=0$. Then $p^5+q^5+r^5 = ?$
Correct answer was $-5$, but no one managed to solve this problem. We tried different methods, but none came up with a solution. Any help is appreciated.
| For any three numbers $p$, $q$, and $r$, $p^5+q^5+r^5$ is equal to\begin{multline*}(p+q+r)^5-5 (p q+p r+q r) (p+q+r)^3+5 p q r (p+q+r)^2+\\+5 (p q+p r+q r)^2(p+q+r)-5 p q r (p q+p r+q r).\end{multline*}Since $p+q+r=0$, $pq+pr+qr=-1$, and $pqr=-1$,$$p^5+q^5+r^5=-5\times(-1)\times(-1)=-5.$$
Added note: The way I used to express $p^5+q^5+r^5$ in function of $p+q+r$, $pq+pr+qr$, and $pqr$ was this: as a first step, I saw that the greatest power of $p$ in this expression was $p^5$ and that there was no $q$ and no $r$ here. So, I subtracted $(p+q+r)^5$ from my expression, obtaining\begin{multline*}-5 p^4 q-5 p^4 r-10 p^3 q^2-20 p^3 q r-10 p^3 r^2-10 p^2 q^3-30 p^2 q^2 r-30 p^2 q r^2+\\-10 p^2 r^3-5 p q^4-20 p q^3 r-30 p q^2 r^2-20 p q r^3-5 p r^4-5 q^4 r-10 q^3 r^2-10 q^2 r^3-5 q r^4.\end{multline*}Now, the greatest power of $p$ is $p^4$, and among those monomials with $p^4$, the one with the greatest power of $q$ is $-5p^4q=-5p^3(pq)$. So, now I add $5(p+q+r)^3(pq+pr+qr)$ to my expression, getting\begin{multline*}5 p^3 q^2+15 p^3 q r+5 p^3 r^2+5 p^2 q^3+30 p^2 q^2 r+30 p^2 q r^2+5 p^2 r^3+\\+15 p q^3 r+30 p q^2 r^2+15 p q r^3+5 q^3 r^2+5 q^2 r^3\end{multline*}
I suppose that by now you got the pattern. At this point, the greatest power of $p$ is $p^3$, and among those monomials with $p^3$, the one with the greatest power of $q$ is $5p^3q^2=5p(pq)^2$. So, now I subtract $5(p+q+r)(pq+pr+qr)^2$ from my expression, getting$$5 p^3 q r+5 p^2 q^2 r+5 p^2 q r^2+5 p q^3 r+5 p q^2 r^2+5 p q r^3\text,$$and so on.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x,y$ are distinct real numbers If $x,y$ are distinct real numbers, prove that $x+y=-2$ if and only if $(x+1)^2=(y+1)^2$
How would I prove this? do I use contrapositive, contradiction or direct proof?
| $(x+1)^2=(y+1)^2 \iff (x+1)^2-(y+1)^2=0\iff (x+1-(y+1))(x+1+y+1)=0$
$ \iff x+1=y+1$ or $x+y+2=0$
Since we are excluding $x+1=y+1$ we have that the first is equivalent to $x+y=-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find all polynomials satisfying $P(x^2+x-4)=P^2(x)+P(x)$?
Find all polynomials with real coefficients $P(x)$ such that $$P(x^2+x-4)=P^2(x)+P(x).$$
$P=0$ is a solution. For non-constant polynomial $P$ comparing both sides gives the leading coefficient as $1$.
We have $$P(x^2+x-4)=P(x) \big(P(x)+1 \big)$$ If $z$ is a complex root of $P$ then $w=z^2+z-4$ is also a root of $P$. Now if we let $x=w$ then $w^2+w-4$ is a root too and etc. But then $P$ has infinitely many roots. And this is a contradiction. Thus we must have $z=z^2+z-4$ and $z= \pm 2$ and $P$ is in the form $$P(x)=(x-2)^n(x+2)^m$$
I know that this solution is not complete and above arguments have some flaws! I just wrote it to let you know how I thought about it! I would love to see an elementary solution to this problem.
| For $x=2$ the relation gives $\require{cancel}\,\bcancel{P(2)}=P^2(2)+\bcancel{P(2)}\,$, therefore $P(2)=0\,$. Assuming $P \not \equiv 0$, there must exist a multiplicity $n \ge 1$ and some polynomial $Q(x)$ such that $P(x)=(x-2)^n\cdot Q(x)$ and $Q(2) \ne 0$. Substituting back into the relation:
$$
\left(x^2+x-6\right)^n \cdot Q(x^2+x-4)=(x-2)^{2n} \cdot Q^2(x)+(x-2)^n \cdot Q(x)
$$
Since $x^2+x-6=(x-2)(x+3)\,$:
$$
\bcancel{(x-2)^n} \cdot (x+3)^n \cdot Q(x^2+x-4)=\bcancel{(x-2)^n} \cdot (x-2)^n \cdot Q^2(x)+\bcancel{(x-2)^n} \cdot Q(x) \\[5px]
\iff (x+3)^n \cdot Q(x^2+x-4)=(x-2)^n \cdot Q^2(x)+Q(x)
$$
For $x=2\,$, the above reduces to $5^n \cdot Q(2) = Q(2)\,$, but the equality cannot hold for $n \ge 1$ and $Q(2) \ne 0\,$. Therefore, the only solution is the zero polynomial $P \equiv 0\,$.
| {
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"url": "https://math.stackexchange.com/questions/2319986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $p+q+r=0$, find the value of the determinant
If $p+q+r=0$, prove that the value of the determinant
$$ \Delta= \begin{vmatrix}
pa & qb &rc \\
qc & ra &pb\\
rb& pc & qa \\
\end{vmatrix} =-pqr \begin{vmatrix}
a & b &c \\
b & c &a\\
c& a & b \\
\end{vmatrix}$$
My Try:Since $p+q+r=0$ we have
$$a(p+q+r)+b(p+q+r)+c(p+q+r)=0$$ $\implies$
$$(ap+qc+rb)+(qb+ra+pc)+(rc+pb+qa)=0 \tag{1}$$
Now applying $C_1 \to C_1+C_2+C_3$ and then applying $R_1 \to R_1+R_2+R_3$ for $\Delta$ we get
$$\Delta= \begin{vmatrix}
0 & qb &rc \\
qc+ra+pb & ra &pb\\
rb+pc+qa& pc & qa \\
\end{vmatrix}$$
Any clue here?
| By Sarrus' rule, you have$$\begin{vmatrix}pa&qb&rc\\qc&ra&pb\\rb&pc&qa\end{vmatrix}=prq(a^3+b^3+c^3)-abc(p^3+q^3+r^3).$$But\begin{align*}p^3+q^3+r^3&=\overbrace{(p+q+r)^3}^{\phantom{0}=0}-3(p q+r q+p r) (\overbrace{p+q+r}^{\phantom{0}=0})+3 p q r\\&=3pqr\end{align*} and therefore your determinant is equal to \begin{align*}pqr(a^3+b^3+c^3-3abc)&=-pqr(3abc-a^3-b^3-c^3)\\&=-pqr\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix},\end{align*}again by Sarrus' rule.
| {
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"source": "stackexchange",
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Computation of $\int_0^\infty xe^{-\frac{x^2}{2}}\sin(x \xi) \, dx$ How to compute $$\int_0^\infty xe^{-\frac{x^2}{2}}\sin(x \xi) \, dx$$
I tried to integrate by part but it was bad idea.
| Fourier transform, contour shifting, differential equation and other nice techniques work nicely here. Since they are already explained by other users, let me show a brutal-force computation.
Using the substitution $u = x^2/2$,
\begin{align*}
\int_{0}^{\infty} x \sin(\xi x) e^{-x^2/2} \, dx
&= \int_{0}^{\infty} \sin(\xi\sqrt{2u}) e^{-u} \, du \\
&= \int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{2n+1} (2u)^{n+\frac{1}{2}} \right) e^{-u} \, du \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\xi^{2n+1} 2^{n+\frac{1}{2}} \Gamma\left(n + \frac{3}{2}\right) \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\xi^{2n+1} 2^{n+\frac{1}{2}} \cdot \frac{\sqrt{\pi}}{2}\prod_{k=1}^{n} \left(k + \frac{1}{2}\right) \\
&= \sqrt{\frac{\pi}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^n}\xi^{2n+1} \\
&= \sqrt{\frac{\pi}{2}} \xi e^{-\xi^2/2}.
\end{align*}
Here, we utilized the following identities
$$ \Gamma(s) = \int_{0}^{\infty} u^{s-1}e^{-u} \, du, \qquad \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}. $$
Also, interchanging the summation and integral is justified by Fubini's theorem together with the estimation
$$\int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{|\xi|^{2n+1}}{(2n+1)!} (2u)^{n+\frac{1}{2}} \right) e^{-u} \, du
= \int_{0}^{\infty} \sinh(|\xi|\sqrt{2u}) e^{-u} \, du < \infty. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the radius of the circle $B$? If one of the diameters of the circle $A$ with equation $x^2+y^2-2x-6y+6=0$ is a chord to the circle $B$ with centre $(2, 1)$ then the radius of the circle $B$ is
$(1)\ \sqrt 3$
$(2)\ \sqrt 2$
$(3)\ 3$
$(4)\ 2$
If I draw a perpendicular from $(2,1)$ to that diameter of $A$ then since it is the chord of $B$ the foot of the perpendicular should be $(1,3)$ since the centre of $A$ is the midpoint of that diameter.So the perpendicular distance of that chord of $B$ (equivalently the diameter of $A$) from the centre of $B$ is $\sqrt 5$ and hence the radius of $B$ is $\sqrt {2^2 + (\sqrt 5)^2} =\sqrt 9 = 3$. So $(3)$ is the correct answer.
Is it correct at all? Please verify it.
Thank you in advance.
| Yes, it is correct. Note that the equation $x^2+y^2-2x-6y+6=0$ can be rewritten as
$$(x-1)^2+(y-3)^2=1+9-6=2^2$$
so the circle $A$ has centre $c_A=(1,3)$ and radius $r_A=2$. Therefore the radius of $B$ is given by
$$r_B=\sqrt{r_A^2+d(c_A,c_B)^2}=\sqrt {2^2 + (1-2)^2+(3-1)^2} =\sqrt 9 = 3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:
This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
| Not a completely geometric solution and I am assuming that only integers are allowed.
Since area of yellow triangle is twice the red one, so $a=2d$. Now divide the square into four rectangles by drawing horizontal line from point between $a$ and $b$ and a vertical line from point between $c$ and $d$. Then $$4+8+6-a.\frac{a}{2}=y^2$$ which gives $$a^2+2y^2=36.$$ This has solution $a=2, y=4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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What is my mistake in this integral: $\int \tan^5(x) dx$? I'm doing an exercise from Stewart's Calculus textbook in which I have to evaluate the following integral:
$\int \tan^5(x) dx$
I start by rewriting the integral this way:
$\int \tan^5(x) dx = \int \frac{\sin^5(x)}{\cos^5(x)} dx = \int \frac{\sin^4(x)}{\cos^5(x)} \sin(x) dx$
And here I make a substitution:
$u = \cos(x)$
$-du = \sin(x)dx$
So the integral with the substitution becomes:
$\int \frac{(1-u^2)^2}{u^5} (-1) du$
$\int \frac{(1-2u^2+u^4)}{u^5} (-1) du$
$\int \frac{(2u^2-u^4-1)}{u^5} du$
$\int \frac{(2u^2)}{u^5} - \frac{(u^4)}{u^5} - \frac{1}{u^5} du = 2 \frac{u^{-2}}{(-2)} - \ln\vert u \vert - \frac{u^{-4}}{(-4)} + constant$
$= \frac{-1}{u^2} - \ln\vert u \vert + \frac{1}{4u^4} + constant $
$= \frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant $
This antiderivative seems to be correct, because if I differentiate it I get:
$\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant] $
$-(-2)\cos^{-3}(x)(-\sin(x))-\frac{1}{\cos(x)}(-\sin(x))+\frac{1}{4}(-4)\cos^{-5}(x)(-\sin(x))$
$ \frac{-2\sin(x)}{\cos^3(x)} +\tan(x)+\frac{\sin(x)}{\cos^5(x)}$
$\tan(x)[1-\frac{2}{\cos^2(x)} + \frac{1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-2\cos^2(x)+1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+1-\cos^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(\cos^2(x)-1)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(-\sin^2(x))+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(1-\cos^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(\sin^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^4(x)}{\cos^4(x)}] = \tan^5(x)$
However, when I graph $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $ and $\tan^5(x)$ at a first glance they look like they are the same, but when I start to zoom in I notice that these curves are actually not equal:
The magenta curve is $\tan^5(x)$ and the black curve is $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $
The answer that Stewart provides at the end of the book is the following:
$ \int \tan^5(x) dx = \frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C$
Which is confusing me even more, because if I try to differentiate his answer I get:
$\frac{d}{dx}[\frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln\vert1/\cos(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln(1) - \ln\vert\cos(x)\vert+ C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)- \ln\vert\cos(x)\vert+ C]$
$= \frac{1}{4}(-4)\cos^{-5}(x)-2\tan(x)\sec^2(x)-\frac{1}{\cos(x)}(-\sin(x))$
$= \frac{-1}{\cos^5(x)}-2\tan(x)\sec^2(x)+\tan(x)$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-2\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-\sec^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(-\tan^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}-\tan^3(x)-\tan(x)\sec^2(x)$
$= \frac{-1}{\cos^5(x)}-\frac{\sin^3(x)}{\cos^3(x)}-\frac{\sin(x)}{\cos^3(x)}$
$=\frac{-1-\cos^2(x)\sin^3(x)-\cos^2(x)\sin(x)}{\cos^5(x)}$
$=\frac{-1-(1-\sin^2(x))\sin^3(x)-(1-\sin^2(x))\sin(x)}{\cos^5(x)}$
$=\frac{-1-(\sin^3(x)-\sin^5(x))-(\sin(x)-\sin^3(x))}{\cos^5(x)}$
$=\frac{-1-\sin^3(x)+\sin^5(x)-\sin(x)+\sin^3(x)}{\cos^5(x)}$
$=\frac{-1-\sin(x)+\sin^5(x)}{\cos^5(x)}$
$=\tan^5(x)-\frac{(1+\sin(x))}{\cos^5(x)}$
If I plot the derivative of Stewart's answer along the derivative of my answer and the original function $\tan^5(x)$ this is how the the graph looks like:
*
*Magenta curve: $\tan^5(x)$
*Black curve: $\frac{d}{dx}[\text{my answer}]$
*Blue curve: $\frac{d}{dx}[\text{Stewart's answer}]$
What's going on? Is my answer correct? If it isn't, what am I doing wrong? And what about Stewart's answer? Am I making a mistake when I differentiate it or is it correct?
Thanks for your help!
| First of all, congrats on going to all the extra effort to check your work, comparing your answer to the book's, and questioning the discrepancy among the graphs. That's truly commendable.
As others have noted, your answer is correct and equivalent to the book's. The remaining, unanswered, question concerns the fact that the graphs begin to separate when you zoom in on them. I was able to duplicate this on my Mac in its built-in "Grapher" program (which the OP identified in comments below Jaideep Khare's answer as the graphing program used in the question).
If you go into Grapher's preferences and select "Advanced," you'll see a numerical accuracy slider for derivatives. As you slide it to the right, you need to zoom in more and more to get the graphs to separate. I have no idea how the program works, really, but I suspect it's computing derivatives with a numerical approximation, e.g., $(f(x+h)-f(x))/h$ with some very small $h$, rather than doing any symbolic algebra. If you slide the derivative accuracy slider all the way to the right and continue to zoom in, you'll see some rather spectacular effects.
In short, what you've encountered is a limitation of Grapher's accuracy. There is interesting mathematics in the things that can go wrong when working numerically, so you'll do well to remember what you observed in this problem as you continue to study.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding all pairs $(a,b)$ of positive integers such that $a^2+nab+b^2$ is a perfect square. When $n=2$, the question is trivial. Is there a general method to find all such pairs for $n\ge{3}$ and $n\in{\mathbb{N}}$?
| Let $a^2+nab+b^2 = c^2$
Let $x = \dfrac ac$ and $y=\dfrac bc$.
Then we need to find rational soloutions $(x,y)$ to $x^2 + nxy + y^2 = 1$
We start with the particular solution $(x,y) = (-1,0)$ and "draw" the line
$y = \dfrac uv(x+1)$, where $u,v$ are integers, through that point. It should intersect the hyperbola $x^2 + nxy + y^2 = 1$ at a rational point.
\begin{align}
x^2 + nxy + y^2 &= 1 \\
x^2 + \dfrac uvnx(x+1) + \dfrac{u^2}{v^2}(x+1)^2 &= 1 \\
x &= \dfrac{v^2-u^2}{u^2+uvn+v^2} &\text{(We discarded $x=-1$.)} \\
y &= u\dfrac{nu+2v}{u^2+nuv+v^2}
\end{align}
We get $a^2+nab+b^2 = c^2$ where
\begin{align}
a &= v^2-u^2 \\
b &= nu^2+2uv \\
c &= u^2+nuv+v^2
\end{align}
If $(a,b)$ is a solution, then so too is $(-a,-b)$, $(b,a)$, and $(-b,-a)$. So all solutions can be characterized as
$$\{a,b\} \in \{v^2-u^2, nu^2+2uv\}, \quad c = u^2+nuv+v^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you evaluate high powers in modulo expressions? How would you evaluate $ 2^{2143} \pmod{17}?$
For instance, I was looking through my notes and the example I have is :
$5^{131} \pmod{14}$
We look at the powers $ 5^i$ modulo $14$ and notice that they repeat at $i = 6$
Now since $131 = 5 \pmod{6}$ then $5^{131} = 5^5 = 3 \pmod{14}$
The answer is $3$. However, I do not understand this logic at all.
| Fermat's Little Theorem:
If $p$ is a prime and $gcd(a,p) = 1$, then $a^{p-1} \equiv 1 (mod \; p)$.
Generalization of Fermat's Little Theorem (Euler's Theorem):
Let $a,n \in \mathbb{N}$ and $gcd(a,n) = 1$, then $a^{\phi(n)} \equiv 1 (mod \; n)$, where $\phi(n)$ = number of positive integers less than $n$ and coprime to $n$.
In the first example, 17 is a prime and $gcd(2, 17) = 1$. Hence $2^{16} \equiv 1 (mod \; 17)$. Hence $2^{2143} = 2^{16 \times 133 + 15} = (2^{16})^{133} \cdot 2^{15} \equiv 1^{133} \cdot 2^{15} \equiv 2^{15} (mod \; 17)$. Also note that $2^4 \equiv -1 (mod \; 17) \implies 2^{15} = (2^4)^3 \cdot 2^3 \equiv (-1)^3 \cdot 8 \equiv -8 \equiv 9 (mod \; 17)$.
In the second example, we have to use the generalization because 14 is not a prime and $gcd(5,14) = 1$. Also, $\phi(14) = \phi(2 \times 7) = \phi(2) \times \phi(7) = 1 \times 6 = 6$. (This comes from the fact that $\phi$ is a multiplicative function. A number theoretic function $f$ is multiplicative if $f(mn) = f(m) \cdot f(n)$ whenever $gcd(m,n) = 1$).
Hence $5^6 \equiv 1 (mod \; 14)$. Also, $131 = 6 \times 21 + 5$. Thus, $5^{131} = (5^6)^{21} \cdot 5^5 \equiv 1^{21} \cdot 5^5 \equiv 5^5 (mod \; 14)$.
If you want more transparency in the logic, write $131 = 6q + r$ where $q$ is the quotient and $r$ is the remainder when 131 is divided by 6. Repeat the calculation above using $q$ and $r$ to note that the value of $q$ has no effect on the answer. All that matters is the value of $r$ which is nothing but $131 (mod \; 6)$.
| {
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"source": "stackexchange",
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Find sum of series $\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$ How to find sum of above series
$$\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$$
How to find sum of series I can find its convergence but not sum of series.
Can anyone explain?
| Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
as $$\dfrac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}=\dfrac32\cdot\dfrac{-2/3\cdot-5/3\cdot-8/3\cdot-11/3}{4!}\left(-\dfrac36\right)^4$$
$$\dfrac16=\dfrac32\cdot\dfrac{-2/3}{1!}\left(-\dfrac36\right)^1$$
So, the sum $$=-1+\dfrac32\cdot\left(1-\dfrac36\right)^{-2/3}$$
| {
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Number of ways of splitting $2310$ as product of three factors Find Number of ways of splitting $2310$ as product of three factors.
My Try:
$$N=2310=3 \times 7 \times 2 \times 5 \times 11$$
$1.$ if two of the factors are ones ten trivially it is one way.
$2.$ if exactly one of the factors in one then number of ways is $\frac{(1+1)(1+1)(1+1)(1+1)(1+1)-2}{2}=15$ ways
Can i have any clue if None of the factors is one
| We can choose prime factors for each of three factors. I am assuming the order of the factors does not matter.
*
*All five primes are in one factor, leaving $0$ primes for each of the other two factors. Note that this corresponds to the factorization $1 \cdot 1 \cdot 2310$.
$$\binom{5}{5}$$
*Four primes are in one factor and the remaining prime is in another factor. In this case, one of the factors is $1$, one is prime, and the other is composite.
$$\binom{5}{4}\binom{1}{1}$$
*Three primes are in one factor and both of the remaining primes are in another factor. In this case, one of the factors is $1$ and the other two are composite.
$$\binom{5}{3}\binom{2}{2}$$
*Three primes are in one factor and each of the remaining factors is prime.
$$\binom{5}{3}$$
Note that we do not need to choose the second factor since the order of the two prime factors does not matter.
*Two primes are in one factor, two of the remaining primes are in another factor, and the remaining factor is prime.
$$\frac{1}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}$$
where we multiply by $1/2$ since choosing $2$ and $3$ for the first factor and $5$ and $7$ for the second factor yields the same factorization as choosing $5$ and $7$ for the first factor and $2$ and $3$ for the second factor. Alternatively, we choose the prime factor, then choose which of the three primes will be paired with the smallest remaining prime.
$$\binom{5}{1}\binom{3}{1}$$
Observe that the cases are disjoint, so the number of ways of factoring $2310$ into three factors is found by summing the above results.
| {
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Can anyone explain this solution (contractive function, interval, first derivative related)
We are trying to find a positive null $x \geq 0$ of $$f(x) =
\sqrt{x+1}+\frac{\text{sin}(x)}{10}-2$$
Choose $X=\left[0,8\right]$. For $x \in X$ we have that
$$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqrt{x+1}}+\frac{\text{cos}(x)}{10}\leq
\frac{1}{2}+\frac{1}{10} = \frac{3}{5}$$
(YOU COULD STOP READING HERE BECAUSE I'M RATHER INTERESTING IN JUST UNDERSTANDING THE STEPS TILL HERE.)
Now try to find a $\beta$ such that $\text{sup} \left|1+\beta
f'(x)\right|<1$, i.e.
$-2 < \beta f'(x) <0 \text{ }\forall x \in M$. Because of the
inequality above, we have that $$-\frac{10}{3} < \beta <0$$
We can thus set $\beta = -3$ and $\phi(x) = x-3f(x) =
6+x-3\sqrt{x+1}-\frac{3}{10}\text{sin}(x)$. Thus
$\left|\phi'(x)\right| \leq \frac{4}{5}= 0.8$
So, we have a contractive mapping.
[... further things that need to be shown so we satisfy all conditions (that there even exists a null). I leave them out to not make this question longer.]
I'm mainly interesting in understanding the beginning where we have $$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqrt{x+1}}+\frac{\text{cos}(x)}{10}\leq
\frac{1}{2}+\frac{1}{10} = \frac{3}{5}$$
Can you tell me how you get that? I understand that in the middle we have the first derivative of $f$. On the right side, we took the beginning of interval zero and put it into the first fraction, so $\frac{1}{2\sqrt{0+1}}= \frac{1}{2}$. Then put zero in the second fraction: $\frac{\text{cos}(0)}{10}=\frac{1}{10}$. This explains why we have $\frac{3}{5}$ on the right side.
But how do you get the left side, especially how you get $-\frac{1}{10}?$ We insert the end of interval on the left, $8$, I think. But why we have $-\frac{1}{10}$ ?
| As, you rightly calculated $f^{'}(x) = \frac{1}{2\sqrt{x+1}} + \frac{cos(x)}{10}$.
Now at the boundary point $x=8$, $$f^{'}(8) = \frac{1}{6} + \frac{cos(8)}{10}$$
Finally as $cos(8)>-1$,(i.e. the step you were looking for) we get
$$f^{'}(8) \geq \frac{1}{6} - \frac {1}{10} $$
| {
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Area under the graph - integration
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .
Firstly , on the first step , in already stuck ...
I used definite integral to find the area of $P$ -
$$\int^a_0\ (3x-x^2)dx=\frac{9a^2-2a^3}{6}$$
However when I calculate area of $Q$ , it's the same as Area of $P$ - $$\frac{9a^2-2a^3}{6}$$
Then since
$Q= 2P$
$9a^2 - 2a^3 = 18a^2 - 4a^3 $
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?
| $$\int _{ a }^{ 2a }{ \left( 3x-{ x }^{ 2 } \right) dx } =2\int _{ 0 }^{ a }{ \left( 3x-{ x }^{ 2 } \right) dx } \\ \frac { 36{ a }^{ 2 }-16{ a }^{ 3 } }{ 6 } -\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 6 } =\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 3 } \\ \frac { 27{ a }^{ 2 }-14{ a }^{ 3 } }{ 6 } =\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 3 } \\ 27{ a }^{ 2 }-14{ a }^{ 3 }=18{ a }^{ 2 }-4{ a }^{ 3 }\\ 9{ a }^{ 2 }-10{ a }^{ 3 }=0\\$$ clearly $ a\neq 0$ so the answer is $$ \color{red}{a=\frac { 9 }{ 10 }} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate this : $\int \frac{\cos5x+\cos 4x}{1-2\cos 3x}\,dx$ How to integrate this :
$\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx$
My approach :
We know that $\cos A+cosB = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$
But it is not working here, please suggest, thanks.
| $$\begin{equation}\begin{split}\int\dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx&=\int\dfrac{2\cos \left(\frac{5x+4x}{2}\right)\cos \left(\frac{5x-4x}{2}\right)}{1-2\left[2\cos^2 \left(\frac{3x}{2}\right)-1\right]}\,dx\\&=\int\dfrac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)}{3-4\cos^2 \left(\frac{3x}{2}\right)}\,dx\\&=\int\dfrac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)}{3\cos \left(\frac{3x}{2}\right)-4\cos^3 \left(\frac{3x}{2}\right)}\,dx\\&=-\int\dfrac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)}{\cos \left(\frac{9x}{2}\right)}\,dx\\&=-\int 2\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)\,dx\\&=-\int\left(\cos 2x + \cos x\right)\,dx\\&=-\int\cos 2x\,dx - \int\cos x\,dx\\&=-\dfrac{\sin 2x}{2} - \sin x + C\\&=-\left(\dfrac{\sin 2x}{2} + \sin x\right) + C
\end{split}\end{equation}$$
Remember the fact that
$$\cos 3x = 4\cos^3x - 3\cos x$$
| {
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Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$
$$
\begin{array}{l}
\color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2 \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2) \\
\Leftrightarrow \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2)-\log _{2} x
\end{array}
$$
| This is a property of logarithms -
$$\color{blue}{\log_{m^a}{n}}=\frac {1}{\log_n m^a}=\frac{1}{a\log_n m}=\color{blue}{\frac {1}{a} \log_{m} n}$$
Therefore, $$\frac 12 \log_{\sqrt 2} (x-2)=\frac 12 \log_{2^{1/2}} (x-2)=\frac 12 \frac{1}{\frac 12} \log_{ 2} (x-2)= \log_{2} (x-2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of $1/(4k+3),1/(4k-1),1/(2k-1)$ and $1/(2k+1)$
Question: How do I show that
$$\varphi(2,n)-\varphi(4,n)=2\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}$$
Where$$\varphi(2,n)=1+2\sum\limits_{k=1}^n\frac 1{(2k)^3-2k}$$$$\varphi(4,n)=1+2\sum\limits_{k=1}^n\frac 1{(4k)^3-4k}$$
I started with the LHS, and tried to manipulate it to the RHS.$$\begin{align*}\varphi(2,n)-\varphi(4,n) & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}-\sum\limits_{k=1}^n\frac 1{2k}\\ & \\ & =\left(1+\cdots+\frac 1{2n-1}\right)+\left(\frac 13+\cdots+\frac 1{2n+1}\right)-\left(\frac 13+\cdots+\frac 1{4n-1}\right)-\left(\frac 15+\cdots+\frac 1{4n+1}\right)-\sum\limits_{k=1}^n\frac 1{2k}\end{align*}$$
But, that's as far as I got to. I'm not sure what to do nexy to get the summation. Breaking it apart, we get$$2\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}=\sum\limits_{k=1}^n\frac 1{4k-3}+\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{2k-1}$$
However, I am not aware as to how the $4k-3$ and $4k-1$ arrived.
| This is the correct version:
$(1/2)(\varphi(2) - \varphi(4)) = \sum\limits_{k=1}^\infty\frac 1{\{2(2k-1)\}^3-2(2k-1)}$
where $\varphi(a) = \lim_{n \to \infty} \varphi(a,n)$
We can show that $\varphi(2) = 2 \log 2$ and $\varphi(4) = (3/2) \log 2$ and the value of the big sum on the right as $(1/4) \log 2$.
Proof:
Everything starts with the identity: $$\frac{1}{x^3 -x} = \frac{1}{2(x-1)} + \frac{1}{2(x+1)} -\frac{1}{x}$$
I am going to keep the upper limit of the sum as $n$ for now.
$$\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}=\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-3}+\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-1}-\frac{1}{2}\sum\limits_{k=1}^n\frac 1{2k-1}$$
Let RHS = $\frac{y}{2}$
The first two terms of $y$ above give the first two terms below and last term of $y$ above gives the last term below.
$$y= \sum\limits_{k=1}^{4n}\frac{1}{k} -
\sum\limits_{k=1}^{2n}\frac{1}{2k} -\left(\sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n}\frac{1}{2k}\right)$$
$$y= \left(\sum\limits_{k=1}^{4n}\frac{1}{k} - \log 4n\right) +\log 4n - \frac{1}{2}\left(\sum\limits_{k=1}^{2n}\frac{1}{k} -\log2n\right) - (1/2)\log2n -\left(\sum\limits_{k=1}^{2n}\frac{1}{k}-\log 2n\right) - \log 2n + \frac{1}{2}\left(\sum\limits_{k=1}^{n}\frac{1}{k} - \log n\right) + (1/2)\log n$$
As $n \to \infty$, each of the expressions in parentheses reduces to Euler's constant and cancel out. The remaining terms are:
$$ \log 4n - \log 2n - (1/2)\left(\log 2n - \log n\right) = \log 2 - (1/2) \log 2 = (1/2) \log 2$$
Hence original sum is $(1/4) \log 2$
You can do $\varphi(2)$ and $\varphi(4)$ similarly.
| {
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Convergence/divergence of $\int_{0}^{\infty}{x^{3}\,\mathrm{d}x \over 1 + x^{a} \sin^2{x}}$ I am trying to conclude about the convergence/divergence of
$$
\int_{0}^{\infty}{x^3 \over 1+x^a \sin^{2}\left(\,x\,\right)}\,\mathrm{d}x
\qquad\mbox{for}\quad a \in \mathbb{R}.
$$
First, we notice
$${x^3 \over 1+x^a} \leq{x^3 \over 1+x^a \sin^2{x}}$$
$${x^3 \over 1+x^a} \approx {x^3 \over x^a}={1 \over x^{a-3}}$$
So for $a \leq 4$ the integral diverges by the comparison test.
How do I approach this for $a > 4$ ?. We do seem to run into trouble for
$x = k\pi$.
| First note that, for $m>0$ we have
$$\eqalign{\int_{0}^\pi\frac{dx}{1+ m \sin^2x}&=2\int_{0}^{\pi/2}\frac{1}{1+m+\cot^2x}\frac{dx}{\sin^2x}\cr
&=2\int_0^\infty\frac{dt}{1+m+t^2}=\frac{\pi}{\sqrt{1+m}}
}$$
Thus, if
$$a_n=\int_{n\pi}^{(n+1)\pi}\frac{x^3}{1+x^a\sin^2x}dx
=\int_{0}^{\pi}\frac{(n\pi + u)^3}{1+(n\pi +u)^a\sin^2x}dx
$$
Then
$$ \frac{\pi^4n^3}{\sqrt{1+\pi^a\min(n^a,(n+1)^a)}}\le a_n\le \frac{\pi^4(n+1)^3}{\sqrt{1+\pi^a\max(n^a,(n+1)^a)}}$$
It follows that
$$\lim_{n\to\infty} n^{\frac{a}{2}-3} a_n =\pi^{4-a/2}>0$$
So, the series $\sum a_n$, and consequently the considered integral, does converge if and only if $\frac{a}{2}-3>1$ or equivalently $a>8$.$\square$
| {
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how many sequences of the length 10 and elements $\{a,b,c\}$? How many sequences of the length $10$, with elements from $\{a, b, c\}$ have exactly four $“a”$ and at least three $“b”$?
Can anyone tell me if I am right?
$$\binom{10}{4}\binom{6}{3} +\binom{10}{4}\binom{6}{4}+ \binom{10}{4}\binom{6}{5} +\binom{10}{4}\binom{6}{6} $$
| You're given that there are exactly four $a$ and at least three $b$. Thus, you need only choose three more elements from the set $\left\{b,c\right\}$ to get the total number of multisets satisfying these constraints, which is these four:
*
*$\left\{a,a,a,a,b,b,b,b,b,b\right\}$
*$\left\{a,a,a,a,b,b,b,b,b,c\right\}$
*$\left\{a,a,a,a,b,b,b,b,c,c\right\}$
*$\left\{a,a,a,a,b,b,b,c,c,c\right\}$
Adding the number of ways to permute each of these gives the total number of sequences possible, given these constraints. In each case, we can start with $10!$, and then divide out the $4!$ ways to permute the $a$'s that make no difference. This gives us a starting point of $\frac{10!}{4!} = \binom{10}4\cdot6!$ for each case.
Now, to finish the job, we note that in (1) we have to divide out the $6!$ ways to permute the $b$'s that make no difference, giving only $\binom{10}4$ here (or, equivalently, $\binom{10}4\binom66$, as you wrote). For case (2), we divide out $5!$, giving $\binom{10}4 \cdot 6 = \binom{10}4\binom65$. Similarly, we will have to divide out $2!$ and $4!$ for (3), giving $\binom{10}4\binom64$, and we will divide out $\left(3!\right)^2$ for (4), giving $\binom{10}4\binom63$.
Adding up these possibile permutations for all the multisets gives $\binom{10}4\binom63 + \binom{10}4\binom64 + \binom{10}4\binom65 + \binom{10}4\binom66$, so you are indeed correct. However, I felt that a (hopefully) intuitive explanation of why that is so might be of use to you.
| {
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Integration of secant $$\begin{align}
\int \sec x \, dx
&= \int \cos x \left( \frac{1}{\cos^2x} \right) \, dx \\
&= \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\
& = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\
&= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx
\end{align}$$
I am stuck in here. Any help to integrate secant?
| After $\int \cos x \left(\frac{1}{1-\sin^2x}\right)dx$ use the transformation $z = \sin x$ and $dz = \cos x \, dx$.
Edit:
$$\int\frac{1}{1-u^2}\,du = \frac{1}{2}\int\frac{(1+u)+(1-u)}{(1+u)(1-u)} = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u}\,du$$
And use, $\int \frac{1}{u}\,du = \ln|u|$
| {
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If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated!
Please suggest a simpler process.
| Although three years old, this is a good question with some terrific answers. I thought I'd add mine to the collection...
Let$$x=\left(\frac{1-a}{1+a}\right), y=\left(\frac{1-b}{1+b}\right)$$
Then,
$$x+y=\left(\frac{1-a}{1+a}\right)+\left(\frac{1-b}{1+b}\right)=\frac{2(1-ab)}{1+(a+b)+ab}=\frac{2\left(1-\left(\frac{1}{3}\right)\right)}{1+\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}=2$$
and $$xy=\left(\frac{1-a}{1+a}\right)\left(\frac{1-b}{1+b}\right)=\frac{1-(a+b)+ab}{1+(a+b)+ab}=\frac{1-\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}{1+\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}=3$$
From the Binomial Theorem,
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
we get,
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
$$\left(\frac{1-a}{1+a}\right)^3+\left(\frac{1-b}{1+b}\right)^3=2^3-3\times 2\times3=8-18=-10$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to solve in integers the equation, $(x^2-y^2)^2=1+16y$? How to solve in integers the equation,
$$(x^2-y^2)^2=1+16y$$
By observation we can take $y=3,5$
Is there another method to solve the equation?
| Notice that $y\ge 0$, $y\neq 1$ and $x \neq 0$. If $(x, y)$ is a solution then $(-x, y)$ is a solution too. Hence W.L.O.G $x>0$.
It seems that $x$ and $y$ cannot be too far from each other, otherwise LHS would become larger than RHS. We'll find out this. Suppose that $z=|x-y|$. Notice that $z\neq 0$. Now, we have
$$z^2(x+y)^2=|x-y|^2 (x+y)^2=(x^2-y^2)^2=16y+1$$
And since $y\ge 0$, $y\neq 1$ and $x > 0$, then
$$z^2=\frac{16y+1}{(x+y)^2}<\frac{4(1+y)^2}{(1+y)^2}=4 \ \ \Longrightarrow \ \ z<2$$Therefore $z=1$. It follows that $x=y \pm 1$ and equation turns into a simple quadratic for $y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
Prove $\lim_{x \to 1} \frac{x^b - 1}{x - 1} = b$ Prove that
$$
\lim_{x \to 1} \frac{x^b - 1}{x - 1} = b
$$
(No L'Hospital's rule, or series)
I'm not sure how to go about this.
I have that $x^b = e^{b \ln(x)}$, which gives
$$
\lim_{x \to 1} \frac{x^b - 1}{x - 1} =
\lim_{x \to 1} \frac{e^{b \ln(x)}- 1}{x - 1}
$$
But this doesn't (seem to) do much.
I also have that $1 - \frac{1}{x} < \ln(x) < x - 1$.
| Another approach, as indicated by the proposer:
\begin{align}
x^{b} &= e^{b \, \ln(x)} = e^{b \, \ln(1 - (1-x))} = Exp\left[-b \, \sum_{k=1}^{\infty} \frac{(1-x)^{k}}{k} \right] \\
&= 1 - b \, \left((1-x) + \frac{(1-x)^2}{2} + \mathcal{O}((1-x)^{3}) \right) + \frac{(-b)^{2}}{2!} \, \left( (1-x)^{2} + \mathcal{O}((1-x)^{3}) \right) \\
& \hspace{10mm} + \frac{(-b)^{3}}{3!} \, \mathcal{O}((1-x)^{3}) \\
&= 1 + b \, (x-1) + \frac{b \, (b-1)}{2!} \, (x-1)^{2} + \mathcal{O}((1-x)^{3})
\end{align}
which leads to
\begin{align}
x^{b} - 1 &= b \, (x-1) + \frac{b \, (b-1)}{2!} \, (x-1)^{2} + \mathcal{O}((1-x)^{3}) \\
\frac{x^{b} - 1}{x-1} &= b + \frac{b(b-1)}{2} \, (x-1) + \mathcal{O}((1-x)^{2}) \\
\lim_{x \to 1} \, \frac{x^{b} - 1}{x-1} &= b.
\end{align}
| {
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} |
How to calculate $\iint_D{} (x^2 + y^2) \, dxdy$ with $D = \{(\frac{x}{2})^2 + (\frac{y}{3})^2 \leq 1\}$?
I'm asked to calculate
$$\iint_D{} (x^2 + y^2)\, dxdy,\quad D = \left\{\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 \leq 1\right\}.$$
My attempt:
Set
$$\frac{x}{2} = r\cos{\theta},\quad\frac{y}{3} = r\sin{\theta}$$
which nets the functional matrix i $6r$.
$$\iint_D{(4r^2\cos^2{\theta} + 9r^2\sin^2{\theta})6r \, drd\theta}$$
$$\iint_D{24r^3 + 30r^3\sin^2{\theta} \, drd\theta}$$
$$\int^{2\pi}_0\int^1_0{24r^3 + 30r^3\sin^2{\theta} \,drd\theta}$$
$$\int^{2\pi}_0{6 + 10\sin^2{\theta} \, d\theta}$$
$$\left[6\theta + \frac{10}{2}(\theta - \sin{\theta}\cos{\theta})\right]^{2\pi}_0$$
which equals to $22\pi$.
However, the answer is supposed to be $\frac{39\pi}{2}$. What am I doing wrong / missing?
| In going from the third-to-last displayed expression to the second-to-last, you have $\int_0^1 30r^3 \: dr = 10$. This integral is $30/4$, so the second-to-last displayed expression should be
$$ \int_0^{2\pi} 6 + {30 \over 4} \sin^2 \theta \: d\theta $$
which after integration will give you $39\pi/2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\int \cos^2x \ dx$ I am new to integration and I want to evaluate
$ \int \cos^2x \, dx $
What I done
Using simple chain rule,
$ \int(\cos x)^2 \\
= \int t^2 \,dt \ (t = \cos x) \\
= \frac{t^3}{3} \ + C $
By substitution of $t = \cos(x)$ I got with answer.
| Another way is to use integration by parts:
$$ \int \cos^2x \, dx $$
$$=\int \cos x \cdot \cos x \, dx $$
$$=\int \cos x \,d(\sin x)$$
$$=\cos x \cdot \sin x -\int \sin x \, d(\cos x) $$
$$=\cos x \cdot \sin x +\int \sin^2 x \, dx $$
So
$$\int \cos^2x \, dx =\cos x \cdot \sin x +\int (1- \cos^2 x) \, dx $$
$$=\cos x \cdot \sin x + x + C-\int \cos^2 x \, dx $$
Thus
$$\int \cos^2 x \, dx = \frac{1}{2} (\cos x \cdot \sin x + x + C)$$
$$=\frac{1}{4}\sin 2x + \frac{1}{2} x + C'$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\begin{cases} z_1z_2=10\operatorname{cis}(\frac{4\pi}5)\\\frac{z_1}{\overline{z_2}^2}=\frac2{25}\operatorname{cis}(\frac{\pi}5)\end{cases}$
Solve
$$\begin{cases}z_1z_2=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\\frac{z_1}{\overline{z_2}^2}=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}$$
over $\mathbb C$. The answer should be in trigonometric form. $z=a+ib$.
Let:
$$
z_1=r_1(\cos\theta_1+i\sin\theta_1)\\
z_2=r_2(\cos\theta_2+i\sin\theta_2)\\
\frac{z_1}{\overline{z_2}^2}=\frac{z_1z_2^2}{\overline{z_2}^2z_2^2}=\frac{r_1r_2^2}{|z_2|^4}\stackrel{(\ast)}{=}\frac{r_1r_2^2}{r_2^4}=\frac{r_1}{r_2^2}
$$
then:
$$
\begin{cases}r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\[10pt]
\dfrac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2+i\sin2\theta_2)}\stackrel{(\ast\ast)}{=}\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}
$$
According to De Moivre's rule we know that $z^n=r^n \operatorname{cis}(n\theta)$ but in the transition $(\ast)$ why does $|z_2|^4=r_2^4$?
Lastly in the transition $(\ast\ast)$ is there some trig identity which leads to that?
Please explain the points $(\ast)$ and $(\ast\ast)$.
| $r_1r_2=10$ and $\frac{r_1}{r_2^2}=\frac{2}{25}$, which gives $r_1=2$ and $r_2=5$.
Let $z_1=2cis\theta_1$ and $z_2=5cis\theta_2$.
Thus, $\theta_1+\theta_2=\frac{4\pi}{5}+2\pi k$ and $\theta_1+2\theta_2=\frac{\pi}{5}+2\pi m$, where $\{k,m\}\subset\mathbb Z$.
Finally we obtain: $\theta_1=\theta_2=\frac{7\pi}{5}$
Your $*$ is true by the De Moivre's rule.
Your $**$ is true because $\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ and $\overline{cis\theta}=cis(-\theta)$.
$\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ because
$$\frac{cis\theta}{cis\phi}=\frac{cis\theta\overline{cis\phi}}{cis\phi\overline{cis\phi}}=cis(\theta-\phi)=$$
| {
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"source": "stackexchange",
"question_score": "1",
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Implicit function theorem on 3 equations Check for $x,y,z$ and $w$ if the equation system
$$3x + y-z+w^2 = 0$$
$$x-y+2z+w=0$$
$$2x+2y-3z+2w=0$$
can be solved respectively by the other three variables.
I tried to simplify the equation system $(III+II-I) \Longrightarrow w=0 \land w=3$ for example but i am not really sure if thats the right way.
I also had in mind to maybe calculate the roots of the equation system and then calculate the Jacobian Matrix and then check if $a_{i,j} \neq 0$ for the roots to use the implicit function theorem. But i am really unsure so any tips are appreciated
| The system can be written as
$$\begin{pmatrix}3&1&-1\\1&-1&2\\2&2&-3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-w^2\\-w\\-2w\end{pmatrix}$$
As you've shown, $w=0\lor w=3$. So let us consider each of the cases:
$w\neq0\land w\neq3 \implies$ There are no solutions for $x,y,z$.
$w=3\implies$ Reducing $\left(\begin{array}{ccc|c}3&1&-1&-9\\1&-1&2&-3\\2&2&-3&-6\end{array}\right)$ using row reduction gives $\left(\begin{array}{ccc|c}1&0&\frac{1}{4}&-3\\0&1&\frac{-7}{4}&0\\0&0&0&0\end{array}\right)$
$\implies \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-3\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}\frac{1}{4}\\\frac{-7}{4}\\-1\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb R$
$w=0 \implies$ Reducing $\left(\begin{array}{ccc|c}3&1&-1&0\\1&-1&2&0\\2&2&-3&0\end{array}\right)$ using row reduction gives $\left(\begin{array}{ccc|c}1&0&\frac{1}{4}&0\\0&1&\frac{-7}{4}&0\\0&0&0&0\end{array}\right)$
$\implies \begin{pmatrix}x\\y\\z\end{pmatrix}=\lambda\begin{pmatrix}\frac{1}{4}\\\frac{-7}{4}\\-1\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb R$
Note that I was lazy and used a calculator to reduce the augmented matrix. If done algebraically, you'll get as the solution
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}f_1(w)\\f_2(w)\\f_3(w)\end{pmatrix}+\lambda\begin{pmatrix}\frac{1}{4}\\\frac{-7}{4}\\-1\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb R$$
in each case for some $f_1,f_2,f_3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Z transform of a polynomial signal $$x(k)=\begin{cases} k^n \qquad k\ge 0 \\ 0 \ \ \qquad k<0 \end{cases}$$
I have this formula to calculate the Z transform:
$$\mathscr{Z} \{x(k) \}=\Big(-z \ \frac{d}{dz}\Big)^n \Big[ \frac{z}{z-1} \Big]$$
I have considered this particular case: $$x(k)=\begin{cases} k^2 \qquad k\ge 0 \\ 0 \ \ \qquad k<0 \end{cases}$$
$$f(z)=\frac{z}{z-1}$$
$$f'(z)=\frac{-1}{(z-1)^2}$$
$$-z f'(z)=\frac{z}{(z-1)^2}$$
$$g(z)=\frac{z}{(z-1)^2}$$
$$g'(z)=\frac{(z-1)^2-2z(z-1)}{(z-1)^4}=\frac{-z-1}{(z-1)^3}$$
$$-z g'(z)=\frac{z^2+z}{(z-1)^3}$$
So $$\mathscr{Z} \{ x(t) \}=\frac{z (z+1)}{(z-1)^3}$$
But, I have tried to calculate the inverse Z transform:
$$\mathscr{Z}^{-1} \Big\{\frac{z (z+1)}{(z-1)^3} \Big\}=\mathscr{Z}^{-1} \Big\{H(z) \Big\}$$
Partial fraction decomposition of $\frac{H(z)}{z}$:
$$\frac{z+1}{(z-1)^3}=\frac{1}{(z-1)^2}+\frac{2}{(z-1)^3}$$
$$\mathscr{Z}^{-1} \Big\{\frac{z}{(z-1)^2} \Big\}=k \ u(k)$$
$$\mathscr{Z}^{-1} \Big\{\frac{2z}{(z-1)^3} \Big\}=k^2 \ u(k)$$
$$\mathscr{Z}^{-1} \Big\{H(z) \Big\}=(k+k^2) \ u(k)$$
Where is the mistake?
I know this formula to calculate the inverse Z transform:
$$\mathscr{Z^{-1}}\Big[ \frac{C_{i,j} \ z}{(z-p_i)^j}\Big]=C_{i,j} \ \frac{k^{(j-1)}}{(j-1)!} \ p_i^{k-j+1} \ u(k)$$
Is it not applicable to $\frac{z(z+1)}{(z-1)^3}$?
Thanks
| Consider:
$$z(z+1) = z^{2} + z = (z^{2} - 2 z + 1) + 3(z-1) + 2$$
which leads to
\begin{align}
\frac{z(z+1)}{(z-1)^{3}} &= \frac{(z-1)^{2} + 3 \, (z-1) + 2}{(z-1)^{3}} \\
&= \frac{1}{z-1} + \frac{3}{(z-1)^{2}} + \frac{2}{(z-1)^{3}}.
\end{align}
Using
\begin{align}
\frac{1}{1-x} &= \sum_{k=0}^{\infty} x^{k} \\
\frac{1}{(1-x)^{2}} &= \sum_{k=0}^{\infty} (k+1) \, x^{k} \\
\frac{1}{(1-x)^{3}} &= \frac{1}{2} \, \sum_{k=0}^{\infty} (k+1)(k+2) \, x^{k}
\end{align}
provides
\begin{align}
\frac{z(z+1)}{(z-1)^{3}} &= \sum_{k=0}^{\infty} \left[ 1 - 3 \, (k+1) + 2 \, \frac{(k+1)(k+2)}{2} \right] \, z^{-k} \\
&\Doteq \left[ 1 - 3 \, (k+1) + (k^{2} + 3 k + 2) \right] \\
&\Doteq k^2.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2351704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating the expression $\cos(\frac{1}{2}\tan^{-1}(-\frac{4}{3}))$ I can prove that $\cos(\tan^{-1}(x)) = +\dfrac{1}{\sqrt{1+x^2}}$, set $y=\tan^{-1}(x)$ and therefore $y=\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
Dividing both sides by $2$:
$\dfrac{y}{2}=\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
$\cos(\dfrac{1}{2}\tan^{-1}(x))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}}))$
$\cos(\dfrac{1}{2}\tan^{-1}(-\frac{4}{3}))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{3}{5}))$
This is my approach to the problem, are there any better ways to simplify the initial expression and perhaps find an approximation even without using a calculator?
| Inverse trig functions should always be thought of as angles. Let $\theta = \tan^{-1}\left(\frac{-4}{3}\right)$. Then your expression is $\cos \frac{\theta}{2},$ which might suggest the half-angle identity $\cos \frac{\theta}{2} = \pm\sqrt{\frac{1+\cos \theta}{2}}.$ Since you've already evaluated $\cos \theta,$ you're done.
| {
"language": "en",
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3abc$$
It would e helpful if someone could tell me how should I proceed.
| Another approach we can use is AM-GM inequality .
$$\frac{a^2+b^2}{2} \ge \sqrt{a^2b^2}$$
$$\sum_{a,b,c}\frac{a^2+b^2}{2} \ge \sum{ab}$$
Now equality will hold only when $a=b=c$
| {
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Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram. Problem
Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.
My Solution
B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogram.
A (Hypothesis): Let $A$, $B$, $C$, $D$ be four points such that they form a space quadrilateral.
B1: $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ where $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B}$ and $\dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ are congruent sides. The same can be said for the other two sides.
A1: $\mathbf{A} + \mathbf{B} = \mathbf{C} + \mathbf{D}$ by the definition of quadrilaterals.
$\implies \dfrac{1}{2} \left( \mathbf{A} + \mathbf{B} \right) = \dfrac{1}{2} \left( \mathbf{C} + \mathbf{D} \right)$
$\implies \dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$
$Q.E.D.$
I would greatly appreciate it if people could please review my proof for correctness.
| In general, the midpoints of any convex quadrilateral form a parallelogram, and you can prove that quite easily by drawing diagonals of the initial quadrilateral, but I'm not exactly sure what a space parallelogram is either, nor do I know how to prove this using vectors or check your proof as I have close to none understanding of them.
| {
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"timestamp": "2023-03-29T00:00:00",
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Telescoping(?) an infinite series Find the value of the sum $\displaystyle \sum\limits_{n=1}^{\infty} \frac{(7n+32) \cdot3^n}{n(n+2) \cdot 4^n}.$
Using partial fraction decomposition, I found the above expression is equivalent to $\displaystyle \sum\limits_{n=1}^{\infty} \frac{25}{n} \cdot \left(\frac{3}{4}\right)^n - \sum\limits_{n=1}^{\infty} \frac{18}{n+2} \cdot \left(\frac{3}{4}\right)^n,$ where I got couldn't find a closed form of either expression because of the $\left(\dfrac{3}{4}\right)^n.$
Similarly, trying to telescope one of the the sums with $\displaystyle \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+2}\right) \cdot \left(\frac{3}{4}\right)^n$ fails for the same reason. How can I further simplify the above expression? Thanks.
| Note that
$$\sum_{n=1}^\infty \frac {r^n}n=\sum_{n=1}^\infty \int_0^x r^{n-1}dr=\int_0^x\sum_{n=1}^\infty r^n dr=\int_0^x \frac 1{1-r}dr=\bigg[-\ln(1-r)\bigg]_0^x=\ln\left(\frac 1{1-x}\right)$$
and
$$\sum_{n=1}^\infty\frac {r^n}{n+2}=\sum_{n=3}^\infty \frac {r^{n-2}}{n}=\frac 1{r^2}\sum_{n=3}^\infty\frac {r^n}n=\frac 1{r^2}\left[\left(\sum_{n=1}^\infty\frac {r^n}n\right)-r-\frac {r^2}2\right]=\frac 1{r^2}\left[\ln\left(\frac 1{1+r}\right)\right]-\frac 1r-\frac 12$$
Putting $r=\dfrac 34$ gives
$$\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}n=\ln 4$$
and $$\sum_{n=1}^\infty \frac {\left(\frac 34\right)^n}{n+2}=\frac {16}9\ln 4-\frac {11}6$$
Hence
$$\begin{align}
\sum_{n=1}^\infty \frac {7n+32}{n(n+2)}\cdot \left(\frac 34\right)^n
&=16\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n}
-9\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n+2} \\
&=16\;\;\ln4\;\;\;\;-\;\;9\left(\frac {16}9 \ln4 -\frac {11}6\right)\\
&=\color{red}{\frac{33}2}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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integration of $\int_{0}^{\infty}\frac{2\sin x-\sin2x}{x^n}dx$ How to find $$\int_{0}^{\infty}\cfrac{2\sin x-\sin2x}{x^n}dx$$? Any suggestion will be very helpful.
| For $n=3$ we have
$$\int\limits_{0}^{\infty}\cfrac{2\sin x-\sin2x}{x^n}dx=4\int\limits_{0}^{\infty}\left(\frac{\sin{x}}{x}\cdot\frac{\sin\frac{x}{2}}{x}\cdot\frac{\sin\frac{x}{2}}{x}\right)dx=4\cdot\frac{\pi}{2}\cdot1\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{\pi}{2}.$$
There is the following result.
If $a_k>0$ and $a_0>\sum\limits_{k=1}^na_k$ then
$$\int\limits_0^{+\infty}\prod_{k=0}^n\frac{\sin{a_kx}}{x}dx=\frac{\pi}{2}\prod_{k=0}^na_k$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| i would write
$$-1\le \frac{2x}{x^2+1}\le 1$$ this is equivalent to
$$-x^2-1\le 2x$$ or $$(x+1)^2\geq 0$$ and $$2x\le x^2+1$$ this is equivalent to $$(x-1)^2\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 3
} |
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.
My Attempt:
\begin{align}
\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}}
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}}
\\
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}.
\end{align}
How do I proceed?
| $$\lim_{\theta \to \frac {\pi}{4}} \frac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}} $$
Let $x:=\theta-\pi /4$, the limit you want to find equals to
$$\lim_{x \to 0} \frac {\cos (x+\pi /4) - \sin (x+\pi /4)}{x} $$
$$=\lim_{x \to 0} -\frac {\sqrt 2(\sin (x+\pi /4)\cos (\pi /4)-\sin (\pi /4)\cos (x+\pi /4))}{x} $$
As something we (intentionally) got in the bracket is the compound angle formula for $\sin$, we have the above limit to equal
$$\lim_{x \to 0} -\frac {\sqrt 2(\sin (x+\pi /4-\pi /4))}{x} $$
$$=\lim_{x \to 0} -\frac {\sqrt 2\sin{x}}{x} $$
$$=-\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Integral $\int \frac{x^2}{x^4+x^2+1}\ dx$ I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get
$$\int \frac{x}{2u(2x^2+1)}\ du$$
| By decomposition:
$$\frac{4x^2}{x^4+x^2+1}=\frac{2x}{x^2-x+1}-\frac{2x}{x^2+x+1}
\\=\frac{2x-1}{x^2-x+1}+\frac1{(x-\frac12)^2+\frac34}-\frac{2x+1}{x^2+x+1}+\frac1{(x+\frac12)^2+\frac34}.$$
The corresponding antiderivatives are
$$\log(x^2\pm x+1)$$ and $$\sqrt{\tfrac43}\arctan\left(\sqrt{\tfrac43}(x\pm\tfrac12)\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Logical suite and inequalities This is a related problem (see here)
We have the following inequalities :
For $n=3$ with $a,b,c$ real numbers the following inequality holds.
$$\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(c-b)^2+(c-a)^2+(a-b)^2\geq \sqrt{54}$$
For $n=4$ with $a,b,c,d$ real numbers
$$\frac{1}{(a-d)^2}+\frac{1}{(d-b)^2}+\frac{1}{(d-c)^2}+\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(d-b)^2+(c-d)^2+(a-d)^2+(a-b)^2+(b-c)^2+(a-c)^2\geq \sqrt{288}$$
If we continue with $n=5,6,7\cdots$ there is a logical suite but I don't know how to prove this and what is the following numbers .
Thanks.
| Your first inequality it's just AM-GM.
Indeed, let $a=\min\{a,b,c\}$, $b=a+x$ and $c=a+y$.
Thus, $x>0$ and $y>0$ and
$$\sum_{cyc}\left((a-b)^2+\frac{1}{(a-b)^2}\right)=x^2+y^2+(x-y)^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{(x-y)^2}=$$
$$=2(x^2-xy+y^2)+\frac{(x^2-xy+y^2)^2}{x^2y^2(x-y)^2}\geq2\sqrt{\frac{2(x^2-xy+y^2)^3}{x^2y^2(x-y)^2}}=$$
$$=2\sqrt{\frac{2\left((x-y)^2+2\frac{xy}{2}\right)^3}{x^2y^2(x-y)^2}}\geq2\sqrt{\frac{2\left(3\sqrt[3]{(x-y)^2\cdot\left(\frac{xy}{2}\right)^2}\right)^3}{x^2y^2(x-y)^2}}=\sqrt{54}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $s(n)=2n+[\log_2n]-1$ what is $s(n+1)-s(n)$?
Question: If $s(n)=2n+[\log_2n]-1$ what is $s(n+1)-s(n)$ ?
Let $n$ be positive nonzero integer. I write $[a]$ to denote the greatest integer less than or equal to $a$. If $s(n)=2n+[\log_2n]-1$ then what is $s(n+1)-s(n)$ equal to? I have by numerical inspection
$$s(n+1)-s(n) =
\begin{cases}
3, & \text{if $\color{blue}{n+1=2^t}$ for some integer $t>0$} \\[2ex]
2, & \text{otherwise}
\end{cases}$$
Surely I only need to calculate ? $$\left(2(n+1)+[\log_2(n+1)]-1\right)-\left(2n+[\log_2n]-1\right) $$
But when I reduce the above expression I get $2$. Here are two working examples:
Example 1: Suppose $n=5$. Then $2*5+[\log_25]-1=10+2-1=11$. On the other hand $n+1=6$ and $2*6+[\log_26]-1=12+2-1=13$. So
$$s(6)-s(5)=13-11=2$$
also
Example 2: Suppose $n=15$. Then $2*15+[\log_215]-1=30+3-1=32$. On the other hand $n+1=16=2^4$ and $2*16+[\log_216]-1=32+4-1=35$ So
$$s(16)-s(15)=35-32=3$$
| Consider $$s\left(2^n\right)-s\left(2^n-1\right)=\left\lfloor \log _2(2^n)\right\rfloor +2 \cdot 2^n-1-\left(\left\lfloor \log _2(2^n-1)\right\rfloor +2 \cdot(2^n-1)-1\right)$$
we have that $\log_2 (2^n)=n$
while $n-1<\log_2(2^n-1)<n$ thus $\lfloor\log_2(2^n-1)\rfloor=n-1$
Therefore the previous calculation gives
$$s\left(2^n\right)-s\left(2^n-1\right)=n+2^{n+1}-1-(n-1)-2^{n+1}+2+1=3$$
If the argument of $s(n)$ is not a power of $2$ we have
$$s(n+1)-s(n)=\left\lfloor \log _2(n)\right\rfloor +2 n-1-(\left\lfloor \log _2(n-1)\right\rfloor +2 (n-1)-1)=2$$
because $\left\lfloor \log _2(n)\right\rfloor=\left\lfloor \log _2(n-1)\right\rfloor$ if $n\ne 2^k$ for some $k$
Hope this helps
| {
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"url": "https://math.stackexchange.com/questions/2371292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can the estimate $ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1} $be made? Let's define the average value of binomial coefficients by $$
A (n) = \frac {1}{n+1} \sum_{i=0}^n \binom{n}{i}
$$
Clearly, $$A= \frac {2^n}{n+1} \qquad .$$
Now, this gives us the trivial estimate for $$ \binom{n}{\frac {n}{2}} $$ as
$$ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1}
$$
I know that the largest possible value of $\binom{n}{k}$ is at $k= \frac {n}{2}$
The estimate holds just because this largest value is greater than the average value of binomial coefficients?
Is it enough to say that this estimation holds since the maximum number in a set is always at least the average value?
| Just induction!
Firstly, we'll replace $n$ at $2n$.
Thus, we need to prove that $\binom{2n}{n}\geq\frac{2^{2n}}{2n+1}.$
For $n=1$ it's obvious.
Let $$\binom{2n}{n}\geq\frac{2^{2n}}{2n+1}.$$
Hence,$$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}\geq\frac{2^{2n}}{2n+1}\cdot\frac{2(2n+1)}{n+1}=$$
$$=\frac{2^{2n+2}}{2n+3}\cdot\frac{2n+3}{2(n+1)}>\frac{2^{2n+2}}{2n+3}$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, can you guys help me out because I can't understand which formula to use, Do we have to use the Discriminant formula ($b^2-4ac$)?
| Hint: Calculate
$$(\sqrt2+\sqrt3)^2.$$
Then do the same with a sign change.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Minimize $x^4+ \frac{1}{x^4} +y^4+ \frac{1}{y^4}$
Minimize $${x^4+ \frac{1}{x^4} +y^4+ \frac{1}{y^4}} $$
Subject to $$x^2+y^2=4$$
I've managed to solve this by substituting $x=(2+t)^\frac{1}{2}$ and $y=(2-t)^\frac{1}{2}$ and I arrived at a minimum of $\frac{17}{2}$ when $t=0$, but it's kind of tedious.
So I'm asking if there are straightforward ways of finding the minimum using inequalities and not using symmetry.
| Let $f(x)=x^2+\frac{1}{x^2}$.
Thus, $f''(x)>0$.
Thus, by Jensen
$$x^4+\frac{1}{x^4}+y^4+\frac{1}{y^4}\geq2\left(\left(\frac{x^2+y^2}{2}\right)^2+\frac{1}{\left(\frac{x^2+y^2}{2}\right)^2}\right)=\frac{17}{2}.$$
The equality occurs for $x=y=\sqrt2$, which says that we got a minimal value.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many solutions does the equation $a + b + c + d + e = 21$ have in the nonnegative integers if $a \leq 3$, $0 < b < 4$, and $c \geq 15$? How many solutions are there to the equation in the nonnegative integers:
$$a+b+c+d+e = 21$$
Conditions:
a) $ a \le 10$
I understood the solution which is total number of possibilities - the number of possibilities with $a \gt 10$ which is $\binom{5+21-1}{21} - \binom{5+10-1}{10}$.
b) $a \le 3$, $0 \lt b \lt 4$, $c \ge 15$.
How do I solve this part? Thanks in advance.
| If you don't mind, I'll solve this with generating functions. We represent the different choices for each letter as a power series, where the coefficient represents the allowed numbers. In your case, $a$ will be represented by $(1+x+x^2+x^3)$ since it can be $0,1,2,3$, similarly $b$ will be $(x+x^2+x^3)$ , $c$ will be $\sum_{k=15}^\infty x^k = x^{15}\sum_{k=0}^\infty x^k$, and $d$ and $e$ will be $\sum_{k=0}^\infty x^k$ since they can be anything. Now, when we multiply these power series together, for every way that we can choose a term from the letters and make the power $21$, we will add $1$ to the coefficient of $x^{21}$ in the product power series.
To convince yourself that this works, you could look at the generating function $(1+x)^n = \sum_{k=0}^n\binom{n}kx^k$. This is like having $n$ letters that can be $0$ or $1$, and the coefficient of $x^k$ in the product power series is the number of ways which you can write $k$ as sum of $0$s and $1$s, which is just the number of ways which you can choose which $k$ of the $n$ variables to take on the value $1$.
Now, our product series is
\begin{align*}
(1+x+x^2+x^3)x(1+x+x^2)x^{15}\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=0}^\infty x^k\right)^2 &= x^{16}\frac{1-x^4}{1-x}\frac{1-x^3}{1-x}\frac1{(1-x)^3} \\
&= \frac{x^{16}(1-x^4)(1-x^3)}{(1-x)^5}.
\end{align*}
So, we want the coefficient of $x^{21}$ of the power series expansion of the above, which is just the coefficient of $x^5$ in the expansion of
$$
\frac{(1-x^4)(1-x^3)}{(1-x)^5} = (1-x^3-x^4+x^7)\sum_{k=0}^\infty\binom{k+4}4 x^k.
$$
The relevant terms come from $1\cdot\binom{5+4}4 x^5$, $-x^3\cdot\binom{2+4}4 x^2$, $-x^4\cdot\binom{1+4}4 x$, so your answer is
$$
\boxed{\binom{5+4}4-\binom{2+4}4-\binom{1+4}4.}
$$
Please let me know if I have made any mistakes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\tan(\alpha)=2-\sqrt{3},$ find the value of the acute angle $\alpha$ I rewrote it as $$\tan\left(2\cdot\frac{\alpha}{2}\right)=\frac{2\tan(\frac{\alpha}{2})}{1-\tan^2(\frac{\alpha}{2})}=2-\sqrt{3} \ \Rightarrow \ 2(1-\tan^2(\frac{\alpha}{2})-\sqrt{3}(2\tan(\frac{\alpha}{2}).$$ Obviously, it didn't bring the answer. My other attempts also were untrue. I would like somebody to hint at other method(s).
| You should pay more attention to your hints.
$\tan \alpha = 2-\sqrt{3}$
Since we are given that $\alpha$ is acute and $0 < \tan \alpha < 1$, then
$0 < \alpha < \frac{\pi}{4}$ and $0 < 2\alpha < \frac{\pi}{2}$.
Since \begin{align}
\tan 2\alpha
&= \dfrac{2\tan \alpha}{1 - \tan^2 \alpha} \\
&= \dfrac{4-2\sqrt 3}{1 - (7-4 \sqrt 3)} \\
&= \dfrac{4-2\sqrt 3}{-6 + 4 \sqrt 3} \\
&= \dfrac{2(2 - \sqrt 3)}{2\sqrt 3(2 - \sqrt 3)} \\
&= \dfrac{1}{\sqrt 3} \\
&= \tan \frac{\pi}{6}
\end{align}
Then $\alpha = \dfrac{\pi}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find the number of rearrangements of the string 123456 in which none of the sequences 123, 321, 456, and 654 occur. My attempt:
Let $A_{1}$ denotes where 123 occurs, $A_{2}$ denotes where 321 occurs, $A_{3}$ denotes where 456 occurs, and $A_{4}$ denotes where 654 occurs. Also $|U|$=6!
$$\begin{aligned}|A_{1} &\cup A_{2} \cup A_{3} \cup A_{4}|\\
&= |A_{1}|+|A_{2}|+|A_{3}|+|A_{4}|\\
&\quad -(|A_{1}A_{2}|+|A_{1}A_{3}|+|A_{1}A_{4}|+|A_{2}A_{3}|+|A_{2}A_{4}|+|A_{3}A_{4}|)\\
&\quad +|A_{1}A_{2}A_{3}|+|A_{1}A_{2}A_{4}|+|A_{1}A_{3}A_{4}|+|A_{2}A_{3}A_{4}|\\
&\quad-|A_{1}A_{2}A_{3}A_{4}|\\
=& \ ? \end{aligned}$$
Then,
$$\begin{aligned}|A_{1}'A_{2}'A_{3}'A_{4}'|& =|U|-|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}|\\
&= 6!-? \end{aligned}$$
I am pretty sure this is ALL wrong. Please help me with this problem. Thank you!
| Your approach is quite ok.
We can enhance readability a bit by denoting the sets $A_1,A_2,A_3,A_4$ with
\begin{align*}
A_{123},A_{321},A_{456},A_{654}
\end{align*}
instead. We apply the IEP
and obtain
\begin{align*}
|A|&=6!-\left(|A_{123}|+|A_{321}|+|A_{456}|+|A_{654}|\right)\\
&\qquad+\left(|A_{123}\cap A_{456}|+|A_{123}\cap A_{654}|+|A_{321}\cap A_{456}|+|A_{321}\cap A_{654}|\right)
\end{align*}
We do not have to respect more terms, since $$A_{123}\cap A_{321}=A_{456}\cap A_{654}=\emptyset$$
Since $|A_{123}|=4!$ and $|A_{123}\cap A_{456}|=2!$ we get due to symmetry
\begin{align*}
\color{blue}{|A|}&=6!-4\cdot4!+4\cdot 2!\\
&=720-96+8\\
&=\color{blue}{632}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit if it exists of $S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})$
Suppose that $S_0$ and A are positive numbers, let $$S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right)$$ with $n \geq 0 $.
(a)Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$
(b)Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists
(d) find s
(a) Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$
Given $$P_n: S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \geq \sqrt{A}$$
$$P_0: S_{1} = \frac{1}{2}\left(S_0 +\frac{A}{S_0}\right) \geq \sqrt{A} $$
We assume that $P_n$ is true
$$P_{n+1}: S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right)$$
by assumption
$$S_{n+2}= \frac{1}{2}\left(S_{n+1}\left(1 +\frac{A}{(S_{n+1})^2}\right)\right) \geq \frac{1}{2}\left(\sqrt{A}\left(1 +\frac{A}{(\sqrt{A})^2}\right)\right)$$
$$ S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right) \geq \sqrt{A} $$
It follows that $S_{n+1} \geq \sqrt{A} $
(b) Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
$$S_{n+1} \leq S_n$$
$$\frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \leq S_n $$
Dividing by $S_n$
$$\frac{1}{2}\left(1 +\frac{A}{S_n^2}\right) \leq 1 $$
$$\frac{A}{2S_n^2} \leq \frac{1}{2}$$
$$A \leq S_n^2$$
$$S_n \geq \sqrt{A}$$
As $S_{n+1} \leq S_n$ yields a true statement, it follows $S_{n+1} \leq S_n$ is true.
(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists
Since $S_{n+1} \leq S_n$, the sequence is non-increasing,
using the non-increasing theorem stating that
if $\{S_n\}$ is non-increasing then $$\lim\limits_{n \rightarrow
> \infty} S_n = \inf\{S_n\} $$
(d) find s
Is the argumentation in (a) and (b) appropriate? Also, I have to admit I m getting less confident in my argumentation (c) and (d). How to proceed in (c) and (d)?
Much appreciated for your input or help.
| a) $\frac{1}{2}(S_{n+1}(1 +\frac{A}{(S_{n+1})^2}) \geq \frac{1}{2}(\sqrt{A}(1 +\frac{A}{(\sqrt{A})^2})$ is not necessarily true for $S_{n+1} \ge \sqrt{A}$ because $\frac{A}{(S_{n+1})^2} \le \frac{A}{(\sqrt{A})^2}$. You can prove $\frac{1}{2}(S_{n+1} +\frac{A}{S_{n+1}}) \ge \sqrt{A}$ simply transforming it into $(S_{n+1} - \sqrt{A})^2 \ge 0$
d) To find $s$ just replace $S_{n+1}$ and $S_{n}$ in the recurrence formula with $s$ and solve the resulting equation
| {
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"url": "https://math.stackexchange.com/questions/2384298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
| Your last line is $$L=-\frac{1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}$$ So, as other answers used, by Taylor around $x=0$, $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(\cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$ $$x^2+2\ln\left(\cos{x}\right)=-\frac{x^4}{6}+O\left(x^6\right)$$ $$\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}=-\frac{1}{6}+O\left(x^2\right)$$ making $$L=-\frac{1}{3}-\frac{1}{6}=-\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
On a determinant identity
Show that
$$\det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = (-1)^n(1+x-n)(1+x)^{n-1}$$
for all $n \in \mathbb{N}$.
I tried using induction but I didn't manage (using row/column swaps). Does anyone have any idea?
| $$\begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = 1_n 1_n^\top - (x+1) \mathrm I_n = - (x+1) \left( \mathrm I_n - \frac{1}{x+1} 1_n 1_n^\top \right)$$
Using the Weinstein-Aronszajn determinant identity,
$$\begin{array}{rl} \det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} &= (-1)^n (x+1)^n \det \left( \mathrm I_n - \frac{1}{x+1} 1_n 1_n^\top \right)\\ &= (-1)^n (x+1)^n \left( 1 - \frac{1}{x+1} 1_n^\top 1_n \right)\\\\ &= (-1)^n (x+1)^n \left( 1 - \frac{n}{x+1} \right)\\\\ &= (-1)^n (x+1)^{n-1} \left( x+1 - n \right)\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
General solution of $\cos{4x}=\cos{2x}$ My attempt:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
\cos^2{2x}-\sin^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-1+\cos^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-\cos{2x}-1 &= 0 \\
(2\cos{2x}+1)(\cos{2x}-1) &= 0,
\end{align}
which leads to
$$\cos2x= -\frac{1}{2} \hspace{5mm} \text{or} \hspace{5mm} \cos2x=1,$$
$$2x=2n\pi\pm{2\pi\over3} \hspace{5mm} \text{or} \hspace{5mm} 2x=2n\pi$$
and
$$x=n\pi\pm{\pi\over3} \hspace{5mm} \text{ or } \hspace{5mm} x=n\pi.$$
Book Solution:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
-2\sin{3x}\sin{x} &=0 \\
\sin{3x}\sin{x} &=0.
\end{align}
$$\sin{3x}=0 \hspace{5mm} \text{or} \hspace{5mm} \sin{x}=0$$
$$3x=n\pi \hspace{5mm} \text{or} \hspace{5mm} x=n\pi$$
$$x=\frac{n\pi}{3} \hspace{5mm} \text{or} \hspace{5mm} x=n\pi$$
My question is: How can there be two general solutions of a same equation Or did I made a mistake somewhere?
| The solution is much more direct if you think in terms of congruences:
\begin{align}
\cos 4x =\cos 2x&\iff 4x\equiv \pm2x \mod 2\pi\iff\begin{cases}2x\equiv 0\pmod{2\pi},\\ 6x\equiv 0\pmod{2\pi},\end{cases}\\
&\iff\begin{cases}x\equiv 0\pmod{\pi},\\[1ex] x\equiv 0\mod{\Bigl(\dfrac\pi3\Bigr)}.\end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int{\frac{1}{x^2-x+1}dx}$ Original question was to integrate $$\int{\frac{(x+1)}{x^2-x+1}dx}$$
But I was able to break it into 2 parts: $$\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}+\frac{3}{2}\int{\frac{1}{x^2-x+1}dx} $$
The first part $\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}$ could be easily integrated by substituting $u={x^2-x+1}$ and thus, getting $\frac{\ln(x^2-x+1)}{2}$ as the answer. But I have no idea on how to integrate part 2 of the equation.
Any help will be appreciated.
| I hope this explains why you need to make a perfect square.
Starting with your original problem ....
$$\begin{align}
\int{\frac{x+1}{x^2-x+1}dx} &= \int{\frac{x+1}{x^2-\frac{1}{2}x-\frac{1}{2}x+1}dx}\\
&= \int{\frac{x+1}{x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{4} + \frac{3}{4}}dx}\\
&= \int{\frac{x+1}{x(x-\frac{1}{2})-\frac{1}{2}(x-\frac{1}{2}) + \frac{3}{4}}dx}\\
&= \int{\frac{x+1}{(x-\frac{1}{2})^2 + \frac{3}{4}}dx}\\
&= \int{\frac{x+1}{\frac{3}{4}\cdot \frac{4}{3}(x-\frac{1}{2})^2 + \frac{3}{4}}dx}\\
&= \frac{1}{\frac{3}{4}}\int{\frac{x+1}{\frac{4}{3}(x-\frac{1}{2})^2 + 1}dx}\\
&= \frac{1}{\frac{3}{4}}\cdot\frac{\frac{4}{3}}{\frac{4}{3}}\int{\frac{x+1}{\frac{2^2}{\sqrt{3}^2}(x-\frac{1}{2})^2 + 1}dx}\\
&= \frac{4}{3}\int{\frac{x+1}{\bigg(\frac{2}{\sqrt{3}}(x-\frac{1}{2})\bigg)^2 + 1}dx}\\
&= \bigg(\frac{2}{\sqrt{3}}\bigg)^2\int{\frac{x-\frac{1}{2}+\frac{1}{2}+1}{\bigg(\frac{2}{\sqrt{3}}(x-\frac{1}{2})\bigg)^2 + 1}dx}\\
&= \frac{\bigg(\frac{2}{\sqrt{3}}\bigg)^2}{\color{red}{\frac{2}{\sqrt{3}}}}\int{\frac{\color{red}{\frac{2}{\sqrt{3}}}(x-\frac{1}{2}+\frac{3}{2})}{\bigg(\frac{2}{\sqrt{3}}(x-\frac{1}{2})\bigg)^2 + 1}dx}\\
&= \frac{2}{\sqrt{3}}\int{\frac{\color{green}{\frac{2}{\sqrt{3}}(x-\frac{1}{2})}+\sqrt{3}}{\bigg(\color{green}{\frac{2}{\sqrt{3}}(x-\frac{1}{2})}\bigg)^2 + 1}dx}\\
\end{align}$$
Now, just make the substitution $\displaystyle u = \frac{2}{\sqrt{3}}(x-\frac{1}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Sum of the recursive series Let $\langle a_n \rangle$ be a recursive sequence given by $a_1>2$ and,
$a_{n+1}=a_n^2-2$ for $n \in \mathbb N$
Show that
$\sum_{n=1}^\infty \frac{1}{a_1a_2\cdots a_n} = \frac{a_1-\sqrt{a_1^2-4}}{2}$
I have reached this step:
$\frac{1}{a_1a_2a_3\cdots a_n}=\frac{1}{2} (\frac{a_n}{a_1a_2\cdots a_n-1} -\frac{a_n-1}{a_1a_2\cdots a_n})$
But I am not able to obtain the final expression .Please help me to obtain it. Thanks for help in advance.
| Notice the function $x \mapsto x^2 - 2$ send $(2,\infty)$ to $(2,\infty)$.
Start from $a_1 > 2$, it is easy to see $a_n > 2$ for all $n$.
For each $n$, pick a $x_n > 1$ such that $a_n = x_n + \frac{1}{x_n}$, the recurrence relation becomes
$$x_{n+1} + \frac{1}{x_{n+1}} = \left(x_n + \frac{1}{x_n}\right)^2 - 2
= x_{n}^2 + \frac{1}{x_{n}^2}\quad\implies\quad x_{n+1} = x_n^2$$
Let $\lambda = x_1$. Solving above recurrence relation gives us
$$x_n = \lambda^{2^{n-1}}\quad \implies\quad a_n = \lambda^{2^{n-1}} + \lambda^{-2^{n-1}}
= \frac{\lambda^{2^n} - \lambda^{-2^n}}{\lambda^{2^{n-1}} - \lambda^{-2^{n-1}}}
$$
The summands in the sum are telescoping products.
$$\begin{align}\prod_{k=1}^n \frac{1}{a_k} &= \prod_{k=1}^n
\frac{\lambda^{2^{k-1}} - \lambda^{-2^{k-1}}}{\lambda^{2^{k}} - \lambda^{-2^k}}
= \frac{\lambda - \lambda^{-1}}{\lambda^{2^n} - \lambda^{-2^n}}\\
&= \frac{\lambda^2-1}{\lambda}\frac{\lambda^{2^n}}{(\lambda^{2^n})^2 - 1}
= \frac{\lambda^2-1}{\lambda}\left[\frac{1}{\lambda^{2^n}-1} - \frac{1}{\lambda^{2^{n+1}} - 1}\right]
\end{align}
$$
The sum itself is also a telescoping one. At the end, we have
$$\begin{align}\sum_{n=1}^\infty \prod_{k=1}^n \frac{1}{a_k}
&= \frac{\lambda^2-1}{\lambda} \sum_{n=1}^\infty
\left[\frac{1}{\lambda^{2^n}-1} - \frac{1}{\lambda^{2^{n+1}} - 1}\right]
= \frac{1}{\lambda}\\
&= \frac12\left(
(\lambda + \lambda^{-1}) - (\lambda - \lambda^{-1})
\right)
=
\frac12\left(
(\lambda + \lambda^{-1}) - \sqrt{(\lambda + \lambda^{-1})^2 - 4}
\right)\\
&= \frac{a_1 - \sqrt{a_1^2-4}}{2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$ The question is to evaluate $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.
| $$
\cos(2x) = 1 - 2\sin^2 x.
$$
\begin{align}
\int_0^{\pi/4} (\cos(2x))^{3/2} \cos x\,dx & = \int_0^{\pi/4} \left( 1-2\sin^2 x \right)^{3/2} \overbrace{\Bigg( \cos x\,dx\Bigg)}^{\large\text{This will be } du.} \\[10pt]
& = \int_0^{1/\sqrt 2} (1-2u^2)^{3/2} \, du \\[10pt]
& = \int_0^1 (1-w^2)^{3/2} \, \frac{dw}{\sqrt 2} \\[10pt]
& = \frac 1 {2\sqrt 2} \int_{-1}^1 (1-w^2)^{3/2} \, dw \text{ (since this is an even function)} \\[10pt]
& = \frac 1 {2\sqrt 2} \int_{-1}^1 (1+w)^{3/2}(1-w)^{3/2} \, dw \\[10pt]
& = \frac 1 {2\sqrt 2} \int_0^1 (2v)^{3/2}(2(1-v))^{3/2} (2 \, dv) \\[10pt]
& = \frac 8 {\sqrt 2} \int_0^1 v^{3/2} (1 - v)^{3/2} \, dv \\[10pt]
& = \frac 8 {\sqrt 2} \operatorname{B}\left( \frac 5 2, \frac 5 2 \right) = \frac 8 {\sqrt 2} \cdot\frac{\Gamma(5/2)\Gamma(5/2)}{\Gamma(5)} = \frac {3\pi} {16\sqrt 2}.
\end{align}
APPENDIX:
\begin{align}
\int_0^1 x^{\alpha-1}(1-x)^{\beta-1} \, dx & = \operatorname{B}(\alpha,\beta) \quad \text{(the Beta function)} \\[10pt]
& = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \\[10pt]
\text{and } \Gamma(\alpha+1) & = \alpha\Gamma(\alpha), \\[10pt]
\text{so that } \Gamma\left( \frac 5 2 \right) & = \frac 3 2 \Gamma\left( \frac 3 2 \right) = \frac 3 2 \cdot \frac 1 2 \cdot \Gamma\left( \frac 1 2 \right) = \frac 3 2 \cdot \frac 1 2 \cdot \sqrt \pi.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the area of the shaded region of this figure
Find the area of the shaded region. (Each arcs of circles in the figure are assumed to be $\frac{1}{4}$ of a full circle)
| $FC=2x\sin 15°$
$\sin 15°=\sqrt{\dfrac{1-\cos 30°}{2}}=\sqrt{\dfrac{1-\frac{\sqrt 3}{2}}{2}}=\dfrac{1}{2}\,\dfrac{\sqrt{3}-1}{\sqrt{2}}$
$FC=2x\dfrac{\sqrt{3}-1}{2 \sqrt{2}}=x\dfrac{\sqrt{3}-1}{ \sqrt{2}}\\
Area_{FHGC}=FC^2=\left(x\dfrac{\sqrt{3}-1}{ \sqrt{2}}\right)^2=x^2(2-\sqrt 3)$
$area_{red}=\dfrac{1}{2} x^2 (t-\sin t)\\
area_{red}=\dfrac{1}{2}x^2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right)
$
$Area=x^2\left[2-\sqrt 3+2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right)\right]\\
Area=x^2\left(1+\dfrac{\pi}{3}-\sqrt{3}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving this inequality $\left(1+\frac{1}{n}\right)^n < 3$ How would I prove this, particularly using a method using a geometric series and binomial coefficient.
I'm having trouble giving a reason for $(\star)$ especially.
This is what I did
For $n=1$, LHS = $(1+1)^1 = 2 < 3$ = RHS.
Thus, for $n=2$,
$$\left(1+\frac{1}{n}\right)^n \equiv {\large{\sum_{k=0}^n}} \begin{pmatrix}n \\ k \end{pmatrix}\frac{1}{n^k} \\ = {\large{\sum_{k=0}^n}}\frac{n!}{k!(n-k)!n^k} \\ < \sum_{k=0}^n \frac{1}{k!} \\ < \sum_{k=0}^\infty \frac{1}{k!} \\ < \sum_{k=0}^\infty \frac{1}{(\frac{3}{2})^k} \qquad (\star)\\ = 3. $$
$(\star)$ since $k>2 > \frac{3}{2} \implies k! > 2 > \frac{3}{2}$ and $k! > (\frac{3}{2})^k$ for $k>2$.
I just know this to be true (the last step of raising 1.5 to the power $k$) by checking my calculator, but I don't see a reason why the factorial grows faster or becomes larger than the exponential.
| Have a look at page 99 of these notes, where I proved the sharper inequality $\left(1+\frac{1}{n}\right)^n\leq \frac{20}{7}$ by combining the AM-GM inequality with a creative telescoping idea.
Another possible approach, once it is proved that the sequence $\left(1+\frac{1}{n}\right)^n$ is increasing and convergent to $e$, it is to notice that $x(1-x)$ is non-negative and bounded by $\frac{1}{4}$ on $(0,1)$, hence
$$ 0\leq -1+\frac{3}{e} = \int_{0}^{1}x(1-x) e^{-x}\,dx \leq \frac{1}{4} $$
immediately leads to $e<3$.
The layman's way is just to recall that $e^{-x}$ is an entire function, hence $\frac{1}{e}$ is represented by a fast-convergent series:
$$ \frac{1}{e} = \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\ldots $$
and due to the alternating signs:
$$ \frac{1}{3}=\frac{1}{2!}-\frac{1}{3!}<\frac{1}{e}< \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} = \frac{3}{8}$$
hence $e\in\left(\tfrac{8}{3},3\right).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?
| Any number is congruent to $0$, $1$ or $2$ modulo $3$, and upon squaring you get any square is congruent to $0$, $1$ or $1$ modulo $3$. Now take an arbitrary number $a$. If $3$ divides $a$ then you're done, if not $a^2+2$ is, by the above, congruent to $0$ modulo $3$ since $1+2=3$, and so you're good, too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Trignometric Equality Proof
Prove that $$\frac{\cos 3A + Cos 3B}{2\cos(A-B) -1} =(\cos A + \cos B)\cos(A+B)-(\sin A+\sin B)\sin(A+B)$$
I used $\cos 3A=4{\cos ^3A}-3\cos A$, but it is getting more and more complicated.
| $$(\cos\alpha+\cos\beta)\cos(\alpha+\beta)-(\sin\alpha+\sin\beta)\sin(\alpha+\beta)=$$
$$\cos\alpha\cos(\alpha+\beta)-\sin\alpha\sin(\alpha+\beta)+\cos\beta\cos(\alpha+\beta)-\sin\beta\sin(\alpha+\beta)=$$
$$=\cos(2\alpha+\beta)+\cos(2\beta+\alpha)=$$
$$=2\cos\frac{3\alpha+3\beta}{2}\cos\frac{\alpha-\beta}{2}$$ and
$$\frac{\cos3\alpha+\cos3\beta}{2\cos(\alpha-\beta)-1}=\frac{2\cos\frac{3\alpha+3\beta}{2}\cos\frac{3\alpha-3\beta}{2}}{2\cos(\alpha-\beta)-1}.$$
Thus, it remains to prove that
$$\cos\frac{3\alpha-3\beta}{2}=\cos\frac{\alpha-\beta}{2}\left(2\cos(\alpha-\beta)-1\right),$$
which is true because
$$\cos\frac{3\alpha-3\beta}{2}=4\cos^3\frac{\alpha-\beta}{2}-3\cos\frac{\alpha-\beta}{2}=$$
$$=\cos\frac{\alpha-\beta}{2}\left(2(1+\cos(\alpha-\beta))-3\right)=$$
$$=\cos\frac{\alpha-\beta}{2}\left(2\cos(\alpha-\beta)-1\right).$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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how to solve systems of congruence with more than 2 equations $x \equiv 1 \bmod 7$
$x \equiv 2 \bmod 3$
I can solve this with the formula
$x = m_1a_2x+m_2a_1y\ $ mod $(m_1m_2)$
By doing euclid's algorithm I get, $7(1) + 3(-2) = 1$, then
$x = 7(2)(1) + 3(1)(-2)\ $ mod(21)
$x = 8$
When I have more than 2 congruences, how do I solve using this same formula ?
| One option is to solve two of them, then solve the result with the next one, etc. Note that your final line there is $x\equiv 8 \bmod 21$.
For systems of more than two congruences you could try the Chinese Remainder Theorem for multiple coprime moduli, although the amount of work is similar (but the intermediate numbers are usually smaller):
Let $m_1,m_2,\ldots,m_k$ be pairwise relatively prime moduli. Then the
system of congruences:
$x \equiv c_1\pmod {m_1} \\
x \equiv c_2\pmod { m_2} \\
\vdots \\
x \equiv c_r \pmod {m_r}$
has a unique solution modulo the product $m = m_1m_2\cdots m_r$ as follows:
Take $s_i := \dfrac{m}{m_i}$. Since the $m_i$ share no factors, $\gcd(s_i,m_i ) = 1$. Therefore, for each $i$
we can compute $t_i$, the inverse of $s_i \pmod{mi}$
Then
$x ≡ c_1 s_1 t_1 + c_2 s_2 t_2 +\cdots + c_r s_r t_r \pmod m $ solves
the system since
$c_1 s_1 t_1 + c_2 s_2 t_2 +\cdots + c_r s_r t_r \equiv c_i s_i t_i\equiv c_i \pmod {m_i}$
Note that for two congruences, it reduces to your formula, since then $s_1 = m_2$ etc.
Example:
$x \equiv \color{blue}1 \bmod 7$
$x \equiv \color{blue}2 \bmod 3$
$x \equiv \color{blue}4 \bmod 5$
$m=7\cdot 3\cdot 5 = 105$
$(s_1,s_2,s_3) = (15,35,21)$
$t_1:\quad 15^{-1}\equiv 1^{-1} \equiv \color{red}1 \bmod 7$
$t_2:\quad 35^{-1}\equiv 2^{-1} \equiv \color{red}2 \bmod 3$
$t_3:\quad 21^{-1}\equiv 1^{-1} \equiv \color{red}1 \bmod 5$
$x= \color{blue}1\cdot 15\cdot \color{red}1
+ \color{blue}2\cdot 35\cdot \color{red}2
+ \color{blue}4\cdot 21\cdot \color{red}1
= 15+140+84 \equiv 29\bmod 105$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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partitions of positive integer $n$ with respect to a multiset Recently, I think on a new problem related to partitions.
Let $n$ be a non-negative integer and $\mathbb{A}=\{a_1,\ldots,a_k\}$ be a multiset with $k$
not necessarily distinct positive integers. We denote by $D(n\mid\mathbb{A})$, the number
of ways to partition $n$ in the form $a_1x_1+\cdots+a_kx_k$, where $x_i$'s are positive integers
and $x_i\leqslant x_{i+1}$ whenever $a_i=a_{i+1}$.
I would like to calculate generation function of $D(n|A)$.
| Let's try again. (The original text is gone. If one would like to waste one's time, it is available through the "edited" link just below the examples.)
Instead of starting with a structureless multiset, let's go with: $I$ is the cardinality of the set of distinct elements in $A$, and $B = \{(b_i, m_i), i \in I\}$ where the $b_i$ are the distinct elements of $A$ and $m_i$ is the multiplicity of $b_i$ in $A$. So, for example, $\sum_{i \in I} m_i = |A|$.
For each $i$ in $I$, we want a monotonically nondecreasing sequence $n_{i,1} \leq n_{i,2} \leq \cdots \leq n_{i,m_i}$. We make the change of variables \begin{align*}
d_{i,1} &= n_{i,1} \text{,} \\
d_{i,2} &= n_{i,2} - n_{i,1} \text{,} \\
&\vdots \\
d_{i,j} &= n_{i,j} - n_{i,j-1} & (2 \leq j \leq m_i) \text{,} \\
&\vdots \\
d_{i,m_i} &= n_{i,m_i} - n_{i,m_i-1} \text{.}
\end{align*}
Then the monotonically nondecreasing condition on the $(n_{i,j})_j$ becomes $d_{i,1} \geq 1$ and $d_{i,j} \geq 0$ for $1 \leq j \leq m_i$. Observe that
\begin{align*}
\sum_{j=1}^{m_i} n_{i,j} &= (d_{i,1}) + (d_{i,1} + d_{i,2}) + \cdots + (d_{i,1} + d_{i,2} + \cdots + d_{i,m_i}) \\
&= \sum_{j=1}^{m_i} (m_i - j +1) d_{i,j}
\end{align*}
Then $D(n|A)$ is the number of ways of choosing all these $d_{i,j}$ such that \begin{align*}
n &= \sum_{i \in I} b_i \sum_{j =1}^{m_i} n_{i,j} \\
&= \sum_{i \in I} b_i \sum_{j=1}^{m_i} (m_i - j +1) d_{i,j} \\
&= \sum_{i \in I} \left( b_i m_i d_{i,1} + \sum_{j=2}^{m_i} b_i (m_i - j +1) d_{i,j} \right) \text{,} \\
d_{i,1} &\geq 1 \qquad (i \in I) & \text{, and } \\
d_{i,j} &\geq 0 \qquad (i \in I, 2 \leq j \leq m_i) \text{.}
\end{align*}
The generating function for $D$ is then
$$ \prod_{i \in I} \left( \frac{x^{b_i m_i}}{1-x^{b_i m_i}} \prod_{j = 2}^{m_i} \frac{1}{1-x^{b_i(m_i - j +1)}} \right) \text{.} $$
Examples:
\begin{align*}
\mathrm{gf}(D(n|\{1,2\})) &= x^3 + x^4 + 2 x^5 + 2 x^6 + 3 x^7 + 3 x^8 + 4 x^9 + 4 x^{10} + \cdots \text{,} \\
\mathrm{gf}(D(n|\{1,2,2\})) &= x^5 + x^6 + 2 x^7 + 2 x^8 + 4 x^9 + 4 x^{10} + 6 x^{11} + 6 x^{12} + 9 x^{13} \\
&\qquad + 9 x^{14} + 12 x^{15} + 12 x^{16} + 16 x^{17} + 16 x^{18} + 20 x^{19} + 20 x^{20} + \cdots \text{, and} \\
\mathrm{gf}(D(n|\{1,1,1\})) &= x^3 + x^4 + 2 x^5 + 3 x^6 + 4 x^7 + 5 x^8 + 7 x^9 + 8 x^{10} + \cdots \text{.}
\end{align*}
Continuing:
We can simplify the gf a bit. \begin{align*}
\prod_{i \in I} \left( \frac{x^{b_i m_i}}{1-x^{b_i m_i}} \prod_{j = 2}^{m_i} \frac{1}{1-x^{b_i(m_i - j +1)}} \right)
&= \prod_{i \in I} \frac{x^{b_i m_i}}{1-x^{b_i m_i}} \frac{1}{\prod_{j = 2}^{m_i}1-x^{b_i(m_i - j +1)}} \\
&= \prod_{i \in I} \frac{x^{b_i m_i}}{\prod_{j = 1}^{m_i}1-x^{b_i(m_i - j +1)}} \\
&= \prod_{i \in I} \prod_{j = 1}^{m_i} \frac{x^{b_i}}{1-x^{b_i(m_i - j +1)}} \text{.}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Review. Prove: if $a$ and $b$ are odd integers then $8\mid (a^2-b^2)$ Please notice that this post is not a duplicate from the ones posted by @Lil or @LizW since I want to use another result.
Proof:
Lets prove the next statement first:
If $n$ is an odd integer, then $n$ is equal to $4k+1$ or $4k+3$
Any integer $n$ divided by $4$ has remainder 0, 1, 2 or 3 using the quotient remainder theorem. Therefore any integer can be written as $4k, 4k+1, 4k+2$ and $4k+3$ for some integer $k$. Clearly, $4k$ and $4k+2$ are even numbers. Therefore, all odd integer numbers are $4k+1$ or $4k+3$.
Using this result, if $a$ and $b$ are odd integers then, $a=4m+1$ and $b=4n+1$ for some integers $m$ and $n$.
By algebra, $a^2=(4m+1)^2$ and $b^2=(4n+1)^2$
Then,
$\begin{align*}a^2-b^2=&(4m+1)^2-(4n+1)^2\\ =&16m^2+8m+1-16n^2-8n-1\\ =&8(2m^2+m-2n^2-n)\end{align*}$
This shows that $8\mid (a^2-b^2)$
Should I prove it for $4k+3$?
| Hint: if $a$ is an odd integer, then $a=4m\pm1\,$, so $a^2=16m^2\pm 8m+1=8(2m^2\pm m)+1\,$. Therefore $\,a^2-b^2\,$ is the difference of two numbers that each are a multiple of $\,8\,$ plus $\,1\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral involving Dilogarithm $\int_{1/2}^{1} \frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$ I need your help in evaluating the following integral in closed form. $$\displaystyle\int\limits_{0.5}^{1}
\frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$$
Since the function is singular at $x = 0.5$, we are looking for Principal Value. The integral is finite and was evaluated numerically.
I expect the closed form result to contain $\,\mathrm{Li}_{3}$ and $\,\mathrm{Li}_{2}$.
Thanks
| Note that
\begin{align}
I_1=&\int_0^1 \frac{\ln^2(1-x)\ln(1+x)}{1+x}dx =\int_0^1 \frac{\ln^2x \ln(2-x)}{2-x}dx\\
=& \>\ln2 \int_0^1 \frac{\ln^2x }{2-x}dx +\int_0^1 \frac{\ln^2x }{2-x} \left( -\int_0^1 \frac x{2-xy}dy\right) dx\\
=&\>2\ln2 Li_3(\frac12)+\int_0^1 \int_0^1 \frac{\ln^2x }{1-y}\left(\frac1{2-xy}-\frac1{2-x}\right)dy\>dx\\
= &\>2\ln2 Li_3(\frac12)+ 2\int_0^1\frac{Li_3(\frac y2)}ydy +2 \int_0^1 \frac{Li_3(\frac y2)-Li_3(\frac12)}{1-y}\>\overset{ibp}{dy}\\
= & \> 2\ln2 Li_3(\frac12)+2Li_4(\frac12)+
2\int_0^{\frac12}\underset{=J}{\frac{\ln(1-2y)Li_2(y)}y}dy\tag1
\end{align}
A similar procedure establishes
\begin{align}
I_2=\int_0^1 \frac{\ln^2(1-x)\ln x}{1+x}dx
= \>2Li_4(1)+ 2J
+ 2\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy\tag2
\end{align}
Combine (1) and (2) to express the original integral as
\begin{align}
\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy
=\frac12 (I_2-I_1)+\ln2 Li_3(\frac12)+Li_4(\frac12)-Li_4(1)\tag3
\end{align}
where the integrals $I_1$ and $I_2$ are known, given by
\begin{align}
&I_1 =-\frac{\pi^4}{360} +2\ln2 \zeta(3)-\frac{\pi^2}6\ln^22+\frac14 \ln^42\\
&I_2 = -6Li_4(\frac12)+\frac{11\pi^4}{360}-\frac14\ln^42
\end{align}
Substitute into (3) to obtain the close-form
\begin{align}\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy
=& -2{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln2\zeta(3) + \frac{\pi^4}{180}- \frac{1}{12}\ln^42
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area of rhombus given circumradii of its contained triangles Here is a question I came across
Find the area of rhombus ABCD given that the radii of the circles
circumscribed around triangles ABD and ACD are 12.5 and 25,
respectively.
With the diagonals of the rhombus perpendicularly bisecting each other, let $a$ be half of diagonal BD and $b$ half of diagonal AC. Thus, the sides of the rhombus are $\sqrt{a^2 + b^2}$ long.
The area of any triangle can be expressed as $\frac{\text{product of side lengths}}{\text{4 multiplied by the circumradius}}$. Thus, the area $\Delta ABC$ is $\frac{2b (a^2 + b^2)}{4(12.5)}$, while the area of $\Delta ABD$ is $\frac{2a (a^2 + b^2)}{4(25)}$. These two areas are equal since both of the said triangles are half of the rhombus. Setting these two expressions equal to each other and simplifying gives $b = 2a$.
I've been stuck here for the past few hours. I'm blatantly missing something here. What's the next step?
|
Let $\angle DAB=\alpha$, $R_1=12.5$, $R_2=2R_1=25$.
Using the general formula for the area of triangle $ABC$
in terms of its curcumradius $R$ and sinuses of its angles $A,B,C$
\begin{align}
S_{ABC}&=2R^2\sin A\sin B\sin C
\tag{1}\label{1}
,
\end{align}
we have
\begin{align}
S_{ABCD}&=2S_{ABD}=2S_{ACD}
\tag{2}\label{2}
,\\
S_{ABD}&=2R_1^2\sin\alpha\sin^2(90^\circ-\tfrac\alpha2)
=2R_1^2\sin\alpha\cos^2\tfrac\alpha2
\tag{3}\label{3}
,\\
S_{ACD}&=2R_2^2\sin(180^\circ-\alpha)\sin^2\tfrac\alpha2
=2R_2^2\sin\alpha\sin^2\tfrac\alpha2
\tag{4}\label{4}
,
\end{align}
hence we must have
\begin{align}
\tan\tfrac\alpha2&=\frac{R_1}{R_2}=\frac{12.5}{25}=\frac12
,
\end{align}
and the rest is straightforward:
\begin{align}
\sin\alpha&=\frac{2\tan\tfrac\alpha2}{1+\tan^2\tfrac\alpha2}=\frac45
.
\end{align}
Since given $R_2=2R_1$, we can rewrite \eqref{4} as
\begin{align}
S_{ACD}
&=
2(2R_1)^2\sin\alpha\sin^2\tfrac\alpha2
=
8R_1^2\sin\alpha\sin^2\tfrac\alpha2
\tag{6}\label{6}
,
\end{align}
and avoid calculation of
$\sin\tfrac\alpha2$
and
$\cos\tfrac\alpha2$
\begin{align}
4S_{ABD}+S_{ACD}&=5S_{ABD}
\\
&=8R_1^2\sin\alpha(\sin^2\tfrac\alpha2+\cos^2\tfrac\alpha2)
=8R_1^2\cdot\frac45
,\\
S_{ABD}&=\frac{32R_1^2}{25}=200
,
\end{align}
and the answer is
\begin{align}
S_{ABCD}&=400
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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