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Manipulation of Taylor expansion of $e^x$ I have an expression:
$$ f(x) = \sum\limits_{N=0}^\infty (N-x)^2 \frac{x^N}{N!}$$
I want to figure out whose Taylor Expansion this is.
I've found that I can separate out the above expression, and get the following:
$$f(x) = \sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} - 2 \sum\limits_{N=0}^\infty xN \frac{x^N}{N!} + \sum\limits_{N=0}^\infty x^2 \frac{x^N}{N!}$$
I figured out this much:
$$x^2 e^x = \sum\limits_{N=0}^\infty x^2 \frac{x^N}{N!} = \sum\limits_{N=0}^\infty xN \frac{x^N}{N!}$$
$$xe^x = \sum\limits_{N=0}^\infty N \frac{x^N}{N!}$$
But I cannot figure out who the first mysterious sum belongs to!
|
Let's look at the mysterious sum. First, let's just write out some of the terms
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = 0 + x + \frac{4 x^2}{2!} + \cdots$$
so the first term is zero, let's drop it
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} =\sum\limits_{N=1}^\infty N^2 \frac{x^N}{N!}$$
and then shift the index by saying $N=n+1$
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = \sum\limits_{n=0}^\infty (n+1)^2 \frac{x^{n+1}}{(n+1)!}$$
now let's cancel out one of the $n+1$ in the numerator with its match in the denominator
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = \sum\limits_{n=0}^\infty (n+1) \frac{xx^n}{n!}$$
And split the sum
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = \sum\limits_{n=0}^\infty x n \frac{x^n}{n!} + \sum\limits_{n=0}^\infty x \frac{x^n}{n!}$$
The first term is the same as what you identified for $x^2 e^x$ and the second term is $x$ times the series for $e^x$. Therefore
$$\sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} = x^2 e^x + x e^x$$
|
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"url": "https://math.stackexchange.com/questions/3449582",
"timestamp": "2023-03-29T00:00:00",
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|
Value of $L +\frac{153}{L}=$
If $\displaystyle L = \lim_{x\rightarrow 0}\bigg(\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}\bigg).$
Then value of $\displaystyle L +\frac{153}{L}=$
what i try
$$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}$$
from D L Hopital rule
$$L=\lim_{x\rightarrow 0}\frac{(x^2+1)^{-\frac{1}{2}}-(1+x)^{-1}}{\ln(x+\sqrt{x^2+1})(1+x)^{-1}+\ln(1+x)(x^2+1)^{-\frac{1}{2}}}$$
How do i solve it help me please
|
$$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}=\\
\lim_{x\rightarrow 0}\frac{\ln\left(1+\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)\right)}{x\frac{\ln(1+x)}{x}\cdot\frac{\ln(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}-1}\cdot\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{\frac{\ln\left(1+\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)\right)}{\frac{x+\sqrt{x^2+1}}{1+x}-1}\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)}{x\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{\frac{x+\sqrt{x^2+1}}{1+x}-1}{x\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{x+\sqrt{x^2+1}-x-1}{x\left(x+\sqrt{x^2+1}-1\right)(1+x)}=\\
\lim_{x\rightarrow 0}\frac{\frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)}{\sqrt{x^2+1}+1}}{x\left(x+\frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)}{\sqrt{x^2+1}+1}\right)(1+x)}=\\
\lim_{x\rightarrow 0}\frac{\frac{x^2}{\sqrt{x^2+1}+1}}{x\left(x+\frac{x^2}{\sqrt{x^2+1}+1}\right)(1+x)}=\\
\lim_{x\rightarrow 0}\frac{\frac{1}{\sqrt{x^2+1}+1}}{\left(1+\frac{x}{\sqrt{x^2+1}+1}\right)(1+x)}=\frac12$$
|
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|
Solve system of equations $ y + 3x = 4x^3$ and $x + 3y = 4y^3$ Problem: In the set of real numbers find all $x,y$ that satisfy
$ y + 3x = 4x^3,x + 3y = 4y^3$.
What I don't understand, is how do I know that a condition satisfies both equations(I will show later exactly).
The solution goes like this: By addition and substraction of the equations we get the following 2 equations:
$x + y = (x + y)(x^2 β xy + y^2),x β y = 2(x β y)(x^2 + xy + y^2)$.
My question is, how do I know which conditions have to be true? According to the solution either $x+y=0$ or $x-y=0$, which I understand. My problem comes when they mention that both $(x^2 + xy + y^2)=1/2$ and $(x^2 - xy + y^2)=1$ have to be true at the same time. Why both? How do systems of equations work and how do we know that there won't be another solution if we multiply or divide the equations? Any tips for these kinds of problems are greatly appreciated.
Question 2: Why can't we also try to find a solution which satisfies for example $x-y=0$ and $(x^2 β xy + y^2)=1$
|
Hint.
We have with $f(x) = 4x^3-3 x$
$$
y = f(x)\\
x = f(y)\\
$$
or
$$
y = f^2(y)\\
x = f^2(x)
$$
so the solutions are the fixed points
which are
$$
\left\{0,\pm 1, \pm\frac{\sqrt 2}{2},\frac 14\left(-1\pm\sqrt 5\right),\frac 14\left(1\pm\sqrt 5\right)\right\}
$$
NOTE
$$
f^2(x)-x = f(f(x)) -x = 8 (x-1) x (x+1) \left(2 x^2-1\right) \left(4 x^2-2 x-1\right) \left(4 x^2+2 x-1\right)
$$
|
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|
Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc} \sqrt{(a+b)^2-ab}$$
and now I want to apply Cauchy-Schwarz but it is the wrong direction.
|
Observe that, in virtue of the QM-AM inequality, we obtain $$\sqrt{a^2+ab+b^2}\geqslant \frac{a+\sqrt{ab}+b}{\sqrt3}$$ Similarly $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geqslant \frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}$$ We thus want to prove $$\frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}\geqslant\sqrt3\iff 6+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\geqslant 3$$ whicht is obsviously true
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?
$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$
I have already tried to replace $1/x$ with $t \,\,(t\rightarrow 0)$, but the only result I get is infinity, which is incorrect.
|
We can prove by L'Hospital that
$$\lim_{x\to \infty}\frac{\ln\left(1+\frac1x\right)-\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}\right)}{-\frac1{4x^4}}=\lim_{x\to \infty}\frac{\frac1{x^5+x^4}}{\frac1{x^5}}=\lim_{x\to \infty}\frac{x^5}{x^5+x^4}=1 \\\iff \ln\left(1+\frac1x\right)=\frac1x-\frac1{2x^2}+\frac1{3x^3}+O\left(\frac1{x^4}\right)$$
therefore
$$x^2-\frac12 x-(x^3+x+1)\ln\left(1+\frac1x\right)=$$
$$=x^2-\frac12 x-(x^3+x+1)\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}+O\left(\frac1{x^4}\right)\right)=$$
$$=x^2-\frac12 x--x^2+\frac12x-\frac13-1+O\left(\frac1{x}\right)=-\frac43+O\left(\frac1{x}\right) \to -\frac43$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Sketching the graph and finding the function by using the given. A function f that is defined and continuous for $x\neq 0$ satisfies the following conditions:
*
*f($2-{\sqrt 2}$) = $\sqrt[3]{1-1/\sqrt 2}$, f($\frac{2}{3}$) = $0$, f(2) = $\sqrt[3]{2}$, f($2+{\sqrt 2}$) = $\sqrt[3]{1+1/\sqrt 2}$
*$\lim_{x \to 0^-}f(x)$ = $-\infty$, $\lim_{x \to 0^+}f(x)$ = $\infty$, $\lim_{x \to -\infty}f(x)$ = $0$, $\lim_{x \to \infty}f(x)$ = $0$
*$f^\prime$($x$) $<$ $0$ for $x$ $<$ $\frac{2}{3}$ and $x\neq 0$, and for $x$ $>$ $2$; $f^\prime$($x$) $>$ $0$ for $\frac{2}{3}$ $<$ x $<$ $2$
*$\lim_{x \to \frac{2}{3}^-}f^\prime(x)$ = $ -\infty$, $\lim_{x \to \frac{2}{3}^+}f^\prime(x)$ = $\infty$
*$f^{\prime\prime}(x)$ $<$ $0$ for $x$ $<$ $0$ and for $2-{\sqrt 2}$ $<$ $x$ $<$ $2+{\sqrt 2}$ and $x\neq \frac{2}{3}$;$f^{\prime\prime}(x)$ $>$ $0$ for $0$ $<$ $x$ $<$ $2-{\sqrt 2}$ and for $x$ $>$ $2+{\sqrt 2}$
The question asks to come up with function $f$($x$)$ = $ (a$x$ $+$ b)$^{c}$$x$$^{d}$ that satisfies 1-5 conditions. What are a, b, c and d?
I drew the graphand found that d is obviously -1 however I couldn't find a, b, and c. Please help.
|
First, since the left and right limits at zero are unequal (in fact, do not exist), $f$ is discontinuous at $0$, so that $d<0$. Since $f\left(\frac{2}{3}\right)=0$, the factor $(ax+b)^c$ must vanish there, so that $c>0$ and $ax+b = r(3x-2)$ for some constant $r$. So rewriting, we get
$$f(x) = r^c(3x-2)^c x^d.$$
Next, $f(2)>0$ shows that $r^c>0$; assume going forward that in fact $r=1$, and we will try to fix things up later if that proves not to work. This gives
$$f(x) = (3x-2)^c x^d.$$
Then $f(2) = 2^{2c+d} = 2^{1/3}$, so that $2c+d = \frac{1}{3}$. Next,
\begin{align*}
f(2-\sqrt{2})f(2+\sqrt{2}) &= ((4-3\sqrt{2})(4+3\sqrt{2}))^c((2-\sqrt{2})(2+\sqrt{2}))^d \\
&= (-2)^c2^{d} = (-1)^c2^{c+d}\\
\left(1-\frac{1}{\sqrt{2}}\right)^{1/3}\left(1-\frac{1}{\sqrt{2}}\right)^{1/3}
&= \left(\frac{1}{2}\right)^{1/3} = 2^{-1/3}.
\end{align*}
Since these two expressions must be equal, we see that $(-1)^c=1$ and $c+d = -\frac{1}{3}$.
Solving
\begin{align*}
2c+d &= \phantom{-}\frac{1}{3}\\
\phantom{2}c+d &= -\frac{1}{3}
\end{align*}
gives $c=\frac{2}{3}$ and $d=-1$, so that
$$f(x) = (2x-3)^{2/3}x^{-1}.$$
Although we did not use the various conditions on the derivatives, this function is easily seen to satisfy all of them.
A plot is shown below:
|
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|
Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$.
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.
Question from Jalil Hajimir
|
You first fix $y, z$ and let $x > 0$ vary. Taking derivative with respect to $x$, dropping all those nonnegative terms such as $y^2+1$ to simplify notation, leads to
$$ \frac{d (OP\ full\ epxr)}{d x} \approx x - \frac{(x^2+1)}{x+y+z} = \frac{x(y+z) - 1}{x+y+z}, $$
where the $\approx$ means I dropped some positive terms (they do not affect my analysis of the positivity of the derivate).
It is evident that the gradient is negative for small $x$, and once $x > \frac{1}{y+z}$ the gradient becomes positive. Hence the function is minimized at $x = \frac{1}{y+z}$ when $y, z$ are fixed. Similarly, the function is minized at $y = \frac{1}{x+z}$ when $x, z$ are fixed. And the function is minimized at $z = \frac{1}{x+y}$ when $x, y$ are fixed.
Let the global minimizing point be $x_0, y_0, z_0$, given the previous arguments, we must have $x_0= \frac{1}{y_0+z_0}, y_0 = \frac{1}{x_0+z_0}, z_0 = \frac{1}{x_0+y_0} \Rightarrow x_0 = y_0 = z_0 = \frac{\sqrt{2}}{2}$ (otherwise, we can find a point with smaller value).
Hence the global minimum is unique at $x = y = z = \frac{\sqrt{2}}{2}$ if one exists.
For rigorous argument that the a global minimum exists, one can look at a compact set $[\epsilon, N]\times[\epsilon, N]\times [\epsilon, N]$. The function must have a global minimum in the compact set. One can easily argue it does not take minimum at the boundary (contradicts requirements previously stated, or compare function value on the boundary with that of $x = y = z = \frac{\sqrt{2}}{2}$). Hence the minimum MUST be in the interior (with gradients being zero).
Hence $x = y = z = \frac{\sqrt{2}}{2}$ is the unique global minimum.
|
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$\int \frac{dx}{x^3+x^2\sqrt{x^2-1}-x}$ solve:
$$\int \frac{\mathrm dx}{x^3+x^2\sqrt{x^2-1}-x}$$
I tried:
$$\begin{align}\int \frac{\mathrm dx}{x(x^2-1)+x^2\sqrt{x^2-1}}&=\int \frac{\mathrm dx}{x(\sqrt{x^2-1}\sqrt{x^2-1})+x^2\sqrt{x^2-1}}\\&=\int \frac{\mathrm dx}{x\sqrt{x^2-1}(\sqrt{x^2-1}+x)}\end{align}$$
$x=\sin t$
$$\int \frac{\mathrm dt}{\cos t\sin t-\sin^2t}$$
And I can not continue from here.
|
You can proceed from where you are like this, e.g.
\begin{eqnarray}
\mathcal I &=& \int\frac{dx}{x\sqrt{x^2-1}\left(\sqrt{x^2-1}+x\right)} =\\
&=&-\int\frac{\sqrt{x^2-1}-x}{x\sqrt{x^2-1}}dx=\\
&=&-\int\frac1xdx +\int\frac1{\sqrt{x^2-1}}dx=\\
&=&-\log|x|+\log\left(\sqrt{x^2-1}+x\right)+C
\end{eqnarray}
EDIT
For the second integral use $t=\cosh x$, recalling that $\sinh^2 x= \cosh^2-1$, and $\operatorname{arccosh} x=\log(\sqrt{x^2-1}+x)$.
EDIT 2
Alternatively, as in comment, $x = \sec t$, brings the second integral to $\int \sec t dt =\log(\tan t + \sec t) + C= \log(\sqrt{\sec^2t-1}+\sec t)+C\dots$
|
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|
Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$ $\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$
I wanted to use L' Hopital's rule so I wrote the term as:
$$\frac{\sin^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)}{\cos^2{x}}$$
and found the first derivative of the denumaerator & denominator:
$\frac{{\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\sin{x}\cos{x}}{2\cos{x}(-\sin{x})}$
$$\lim_{x\to \frac{\pi}{2}}{\left(-2\left({\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)\sec{x}\csc{x}\right)}$$
But I didn't know how to continue.
I drew a graph in Geogebra:
|
To simplify we have that
$$\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)=$$
$$\sin^2{x}\frac{\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}{\cos^2 x}$$
and since $t=\sin^2 x \to 1$ we can consider the limit
$$\lim_{t\to 1} \frac{\sqrt{2t^2+3t+4}-\sqrt{t^2+6t+2}}{1-t^2}$$
which can be easily solved by rationalization or also by L'Hopital if you prefer to obtain
$$\lim_{t\to 1} \frac{\sqrt{2t^2+3t+4}-\sqrt{t^2+6t+2}}{1-t^2}=\lim_{t\to 1} \frac{\frac{4t+3}{2\sqrt{2t^2+3t+4}}-\frac{2t+6}{2\sqrt{t^2+6t+2}}}{-2t}=\frac1{12}$$
|
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|
Rectangle inscribed on two parabolas but not parallel to the axis Let $f(x)=2x^2$ and $g(x)=36-x^2$. Let $R$ be the closed region between them. Draw a rectangle $ABCD$ such that $A$ and $B$ are points on $f(x)$, and $C$ and $D$ are points on $g(x)$, and the rectangle is not parallel to the axis, or prove that to be impossible.
|
Brute force:
First, we relax the condition to finding any parallelogram with 2 vertices on each equation.
Let $ A = (a,2a^2), B = (a+b, 2a^2 + 4ab + 2b^2), D = (d, 36-d^2)$, which makes $ C = (d+b, 36-(d+b)^2 = 36 - d^2 + 4ab + 2b^2)$. This gives $ b=0$(rejected), $d =-2a- \frac{3b}{2}$.
Now, we to make the parallelogram into a rectangle, we require $ AB \perp AD \Rightarrow (b, 4ab + 2b^2) \cdot (d-a, 36 - d^2 - 2a^2) = 0$. This simplifies to
$$-3/2 b (2 a + b) (8 a^2 + 8 a b + 3 b^2 - 47) = 0$$
Note that $b=0$ is rejected, and $2a+b$ means $AB$ is parallel to the x-axis, hence rejected. So, we're looking for solutions to $8 a^2 + 8 a b + 3 b^2 - 47=0$, which is an ellipse.
As it turns out, $ a = -3, b = 4 - \sqrt{\frac{23}{3}}$ is one such solution, which leads to the rectangle.
$$ A = ( -3, 18), B \approx ( -1.77, 6.26), C\approx (5.38, 7.01),D \approx (4.15, 18.75)$$
Of course, if you don't want $A$ on the intersection point, take $a = -3 + \epsilon$.
|
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If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality when we have two variable.I assumed $y=tx$ and uesd derivative.
Can this be generalized as:
If ${a_i>0}\quad(i=1,2,...,n)$
$$\sum_{i=1}^n a_{i} \sum_{i=1}^n \frac{1}{a_{i}}\geq n^2\sqrt[]\frac{\sum_{i=1}^n a^2_{i} }{\sum_{i=1}^n a_{i}a_{i+1} }$$
$a_{n+1}=a_{1}$
Question from Jalil Hajimir
|
Here I give a proof by using the standard pqr method.
Proof: Let $p = x+y+z$, $q = xy+yz+zx$ and $r = xyz$.
We will use the following facts (see [1], Facts N12 and N6):
(i) $q^3 + 9r^2 \ge 4pqr$.
(ii) $q^3 \ge 27r^2$.
We need to prove that
$$\frac{pq}{r} \ge 9 \sqrt{\frac{p^2-2q}{q}}$$
or
$$\frac{p^2q^2}{r^2} \ge 81 \frac{p^2-2q}{q}$$
or
$$162qr^2 - (81r^2 - q^3)p^2 \ge 0.$$
There are two possible cases:
1) If $81r^2 - q^3 > 0$: From Fact (i), we have $\frac{q^3+9r^2}{4qr} \ge p$. It suffices to prove that
$$162qr^2 - (81r^2 - q^3)\left(\frac{q^3+9r^2}{4qr}\right)^2 \ge 0$$
or
$$\frac{(q^3 - 9r^2)(q^3 - 27r^2)^2}{16q^2r^2} \ge 0.$$
It is true by using Fact (ii).
2) If $81r^2 - q^3 \le 0$, clearly the inequality is true.
We are done.
Reference:
[1] Zdravko Cvetkovski, "Inequalities Theorems, Techniques and Selected Problems", Ch. 14.
https://keoserey.files.wordpress.com/2012/07/zdravko-cvetkovski-inequalities-theorems_-techniques-and-selected-problems.pdf
Remark: In Cvetkovski's book, chapter 14, Page 138, Cvetkovski gave Facts N1 through N13. There is a typo:
Fact N10 should be $2q^3 + 9r^2 \ge 7pqr$ (rather than $2p^3 + 9r^2 \ge 7pqr$ which is not true for $a=4, b=3, c=2$).
|
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|
$(...((x-2)^2-2)^2-...-2)^2\;\;\text{(n parentheses)}$ Find the coefficient next to $x^2$ in the development of a polynomial:
$$(...((x-2)^2-2)^2-...-2)^2\;\;\text{(n parentheses)}$$
I am not sure but I think we get the coefficient in the following way:
We have something like this quasi-recursive formula:
$$...+k_1\cdot x^2+k_2\cdot x+2\;\text{and}\; k_3=k_2^2+4k_1$$
$$k=2^{2(n+1)}+4\cdot k_2$$
$$(...((x-2)^2-2)^2-...-2)^2$$
$$(x^2-4x+4-2)^2=(x^2-4x+2)^2$$
$$((x^2-4x+2)^2-2)^2=(x^4+16x^2+4-8x^3+4x^2-16x-2)^2=(x^4-8x^3+20x^2-16x+2)^2$$
$$((x^4-8x^3+20x^2-16x+2)^2-2)^2$$
Hopefully:
$$20=4^2+4\;\text{next}\;16^2+4\cdot 4=336$$
|
We have that
$$(x-2)^2=x^2-4x+4$$
$$((x-2)^2-2)^2=(x^2-4x+2)^2=\ldots+20x^2-16x+4$$
$$(((x-2)^2-2)^2-2)^2=(\ldots-16x+2)^2=\ldots 336x^2-64x+4$$
therefore we can guess that
*
*coefficient for $x$ are $-4,-16,-64,\ldots \implies k_2(n)=-(4^n)$
*coefficient for $x^2$ are $1,20,336,\ldots \implies k_1(n)=(k_2(n-1))^2+4k_1(n-1)$
which can be proved rigoursly by induction, therefore we have
$$k_1(n)=4k_1(n-1)+4^{2(n-1)}$$
which leads to
$$k_1(n)=\frac13\left(4^{2n-1}-4^{n-1}\right)$$
|
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|
How to prove this $\sum_{i=1}^{n}(x_{i})^{S-x_{i}}>1?$ Question:
Let $x_{i} \in (0,1),i=1,2,\cdots,n$. Show that
$$
x_{1}^{S-x_{1}}+x_{2}^{S-x_{2}}+\cdots+x_{n}^{S-x_{n}}>1
$$
where $S=x_{1}+x_{2}+\cdots+x_{n}$.
I have proved when $n=2$,because it use this Bernoulli's inequality
$$
(1+x)^a\le 1+ax,0<a\le 1,x>-1
$$
so we have
$$
x^y=\dfrac{1}{(1/x)^y}=\dfrac{1}{\left(1+\frac{1-x}{x}\right)^y}\ge\dfrac{1}{1+\frac{1-x}{x}\cdot y}=\dfrac{x}{x+y-xy}>\dfrac{x}{x+y}
$$
and simaler we have
$$
y^x>\dfrac{y}{x+y}
$$
so we have
$$
x^y+y^x>\dfrac{x}{x+y}+\dfrac{y}{x+y}=1
$$
Edit: Now the Mr Michael Rozenberg has prove when $n=3$ and MR Czylabson Asa has prove $n\ge 6$ this inequality can't hold, so how to prove $n=4,5?$
Thanks.
|
Update
An ugly proof for $n=4$:
Due to symmetry, assume that $d = \max(a, b, c, d)$. Note that
$v\mapsto u^v$ is decreasing on $v > 0$ provided $u\in (0,1)$,
and that $x\mapsto x^y$ is convex on $x>0$ provided $y\ge 1$. We have
\begin{align}
a^{S-a} + b^{S-b} + c^{S-c} + d^{S-d} &\ge a^S + b^S + c^S + d^{S-d}\\
& \ge 3\Big(\frac{a+b+c}{3}\Big)^S + d^{a+b+c}.
\end{align}
Let $a+b+c = 3w$. Then $0 < w \le d < 1$.
It suffices to prove that
$$3w^{3w+d} + d^{3w} \ge 1.$$
It is easy to prove that $w^w > \mathrm{e}^{-1/\mathrm{e}}$ for $w\in (0, 1)$.
Also, we have $d^{3w} \ge \frac{d}{d+ 3w - 3dw}$ by using
$u^v \ge \frac{u}{u+v-uv}$ for $u>0, \ v\in (0, 1)$.
Thus, it suffices to prove that
$$3\mathrm{e}^{-3/\mathrm{e}}w^d + \frac{d}{d+ 3w - 3dw} \ge 1$$
or
$$3\mathrm{e}^{-3/\mathrm{e}}w^{d-1} \ge \frac{3(1-d)}{d+ 3w - 3dw}.$$
For each fixed $d\in (0, 1)$, let
$$f(w) = \ln \left(3\mathrm{e}^{-3/\mathrm{e}}w^{d-1}\right) - \ln \frac{3(1-d)}{d+ 3w - 3dw}, \quad w\in (0, d].$$
We have
$$f'(w) = \frac{(1-d)d(3w-1)}{w(d+3w-3dw)}.$$
We split into two cases:
1) $d > \frac{1}{3}$: Note that $f'(w) < 0$ on $w\in (0, \frac{1}{3})$ and $f'(w)>0$ on $w\in (\frac{1}{3}, d]$. It suffices to prove that $f(\frac{1}{3}) \ge 0$ or
$$g(d) = \ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right) + (1-d)\ln 3 - \ln (1-d)\ge 0.$$
We have $g'(d) = -\ln 3 + \frac{1}{1-d} \ge -\ln 3 + \frac{1}{1 - 1/3} > 0$. It suffices to prove that $g(\frac{1}{3}) \ge 0$. It is true.
2) $d \le \frac{1}{3}$: Note that $f'(w) \le 0$ on $w\in (0, d]$. It suffices to prove that $f(d) \ge 0$ or
$$\ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right) - (1-d)\ln d - \ln \frac{3(1-d)}{4d - 3d^2}\ge 0$$
or
$$\ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right) + d\ln d - \ln \frac{3(1-d)}{4 - 3d}\ge 0.$$
It is easy to prove that $d\ln d \ge -\frac{1}{\mathrm{e}} + \frac{1}{2}\mathrm{e}\left(d-\frac{1}{\mathrm{e}}\right)^2$
and $ - \ln \frac{3(1-d)}{4 - 3d} \ge -\ln\tfrac{3}{4} + \frac{1}{4}d + \frac{7}{32}d^2$ for $d\in (0, \frac{1}{3}]$.
Thus, it suffices to prove that
$$\ln \left(3\mathrm{e}^{-3/\mathrm{e}}\right)-\frac{1}{\mathrm{e}} + \frac{1}{2}\mathrm{e}\left(d-\frac{1}{\mathrm{e}}\right)^2
-\ln\tfrac{3}{4} + \frac{1}{4}d + \frac{7}{32}d^2 \ge 0.$$
It is true.
We are done.
|
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|
True/false: $\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$ I have the following proposition about which I have to say whether it is true or false.
$\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$. $I$ is the identity matrix.
I tried brutal ways (taking a generic matrix, evaluating its square and adding $I$), but there are too many calculations and I feel that manner will not lead me to anything of interesting. I also tried to construct a counterexample, but nothing. I am not able to prove nor confute this assertion.
|
You can also check that it is true via the real Jordan normal form.
First if we have a complex eigenvalue $\alpha + i \beta$ and one real eigenvalue $\lambda$, then $A$ is (after conjugation) of the form
$$ A = \begin{pmatrix} \alpha & \beta & 0 \\ -\beta & \alpha & 0 \\ 0 & 0 & \lambda
\end{pmatrix} $$
and thus
$$ det(A^2 + I) = det \begin{pmatrix} \alpha^2-\beta^2+1 & 2\alpha \beta & 0 \\ -2\alpha \beta & \alpha^2 - \beta^2+1 & 0 \\ 0 & 0 & \lambda^2 +1
\end{pmatrix}
= (\lambda^2+1) \left( (\alpha^2 - \beta^2+1)^2 + 4\alpha^2 \beta^2 \right) $$
Now we consider the case when all the eigenvalues $\lambda, \mu, \nu\in \mathbb{R}$.
If they are all simple, then we get $det(A^2 + I) = (\lambda^2 +1) (\mu^2 +1) (\nu^2+1)$.
If one of the eigenvalues has multiplicity 2, then $A$ is (after conjugation) of the form
$$ A = \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu
\end{pmatrix} $$
and thus
$$ det(A^2 + I) = det \begin{pmatrix} \lambda^2+1 & 2\lambda & 0 \\ 0 & \lambda^2+1 & 0 \\ 0 & 0 & \mu^2
\end{pmatrix}
= (\lambda^2+1)^2 (\mu^2+1) $$
Finally, we are left to consider the case when we have one eigenvalue of multiplicity 3, then $A$ is (after conjugation) of the form
$$ A = \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda
\end{pmatrix} $$
and thus
$$ det(A^2 + I) = det \begin{pmatrix} \lambda^2+1 & 2\lambda & 1 \\ 0 & \lambda^2+1 & 2\lambda \\ 0 & 0 & \lambda^2 +1
\end{pmatrix}
= (\lambda^2 + 1)^3 $$
|
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|
particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\
\end{align}
But I stuck at this point because how to solve $\frac{1}{D(D+4i)}x^2?$ The solution provided this
\begin{align}
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{24}[6x^2\cos 2x+x(8x^2-3)\sin 2x]
\end{align}
I still can't figure out how they do this. Any help will be appreciated.
Update: Using @Isham answer I tired
\begin{align}
&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{4i}\left[e^{2ix}\left(2\frac{x^3}{3}+i\frac{x^2}{2}-\frac{x}{4}\right)-e^{-2ix}\left(2\frac{x^3}{3}-i\frac{x^2}{2}-\frac{x}{4}\right)\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}(e^{2ix}-e^{-2ix})+\frac{x^2i}{2}(e^{2ix}-e^{-2ix})-\frac{x}{4}(e^{2ix}-e^{-2ix})\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}2\cos 2x+\frac{x^2i}{2}2\cos 2x-\frac{x}{4}2\cos 2x\right]
\end{align}
But I think I am again Lost.
|
A way to do basically the same thing with less weird operator manipulation is to think of the procedure as using the integrating factor method on two first order ODEs. If you have say
$$y''+4y=x^2 e^{2ix}$$
then you can write it as
$$(D+2i)(D-2i)y=x^2 e^{2ix}.$$
Then you can let $u=(D-2i)y$ and solve $(D+2i)u=x^2 e^{2ix}$ for $u$ by the integrating factor method. So you have
$$e^{2ix} u = \int x^2 e^{4ix} dx$$
which can be evaluated by integrating by parts. Since you just want a particular solution you can set the constant of integration here equal to zero. You wind up with $u$ being some quadratic polynomial times $e^{2ix}$.
Then with $u$ in hand you just solve
$$(D-2i)y=u$$
and get
$$e^{-2ix}y = \int e^{-2ix} u$$
where now the integral winds up being the integral of just a polynomial, because of the "resonance" in the equation. Setting the constant of integration equal to zero again, you wind up with some cubic polynomial with no constant term times $e^{2ix}$.
That said, if you understand how to do this operator shifting trick that you did to pull the $e^{2ix}$ out (I guess $P(D)^{-1} f = e^{ax} P(D+a)^{-1} e^{-ax} f$?), then you can evaluate $\frac{1}{D+4i} x^2$ by just using the geometric series, expanding in powers of $D$. Thus $\frac{1}{D+4i}=\frac{1}{4i} \frac{1}{1+D/4i}=\frac{1}{4i} \sum_{n=0}^\infty (-D/4i)^n$. But now the point is that when you apply this to $x^2$ you get $0$ as soon as $n>2$, so $\frac{1}{D+4i} x^2=\frac{1}{4i} \left ( x^2 - \frac{2x}{4i} + \frac{2}{(4i)^2} \right )$. Then you just apply $D^{-1}$ to that which is just integration.
|
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|
What is the expected length of the hypotenuse formed by bending a unit length randomly at a right angle? This is easy enough to simulate and find the answer is somewhere around .812. However I am not finding it so easy to solve the integral involved which I believe is...
$$\int_0^1 \sqrt{x^2+(1-x)^2} dx$$
I don't seem to get the right answer if I use this derivation. Bonus points for finding the expected area which is a much easier problem!
|
$$\begin{align}I&=\int_0^1 \sqrt{x^2+(1-x)^2} dx\\
&=\int_0^1\sqrt{2x^2-2x+1}dx\\
&=\int_0^1\sqrt2\sqrt{\left(x-\frac12\right)^2+\frac14}dx\end{align}$$
Let $x-\frac12=\frac12\tan\theta$. Then, $dx=\frac12\sec^2\theta d\theta$.
$$\begin{align}I&=\frac{1}{2\sqrt2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}|\sec\theta|\sec^2\theta d\theta\\
&=\frac{1}{2\sqrt2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^3\theta d\theta\\
\end{align}$$
which can be integrated by parts.
Then,
$$\begin{align}I&=\frac{1}{2\sqrt{2}}\left(\frac12\sec\theta\tan\theta+\frac12\ln|\sec\theta +\tan\theta|\right)|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\\
&=\frac{1}{2\sqrt{2}}\left(\sqrt{2}+\frac12\ln\left|\frac{\sqrt2+1}{\sqrt2-1}\right|\right)\end{align}$$
|
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|
evaluation of Trigonometric limit
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$
What i try
Put $\displaystyle \frac{n\pi}{2}=x,$ when $n\rightarrow\infty,$ Then $x\rightarrow \infty$
$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos^2(x)+2x\sin^2(x)}\bigg)$$
How do i solve it help me please
|
Hint:
$$\frac{1}{n^2}=\frac{1}{n\left(\color{red}n\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+\color{red}{\require{cancel}\cancel{n}}\sin^2 \frac{n\pi}{2}\right)}=\frac1n$$
|
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|
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle
Such that : $a+b+c=1$ Then
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$
My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$
Also :
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don't know how to complete it,any help is appreciated !
|
Homogenizing the equation, we WTS
$$\frac{(a+b+c)^3}{2} -(a+b+c)(a^2+b^2+c^2)-4abc > 0$$
Expanding and factoring, we obtain
$$(a+b-c)(a-b+c)(-a+b+c) > 0$$
This is true via the triangle inequality.
|
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|
Does the denominator cancel? I recently asked this question: Simplify expression cross/dot products
The answer was straightforward, yet I can't apply the same to the expression below.
Can the following expression be simplified? In particular, I would like to remove the denominator:
$\frac{( b \times (a \times b)) \cdot (a \times (b\times a))}{(a\times b) \cdot (a\times b)}$
where $a$ and $b$ are 3 dimensional vectors.
Using indexes, the result seems to be:
$- a \cdot b$
but I fail to see how to simplify this using the properties of scalar and cross products.
|
Your numerator is a dot product of two vector triple products. The former is equal to $(b \cdot b)a - (b \cdot a)b = \|b\|^2a - (a \cdot b)b$ and the latter is equal to $(a \cdot a)b - (a \cdot b)a = \|a\|^2b - (a \cdot b)a$, so their dot product is \begin{align*}(\|b\|^2a - (a\cdot b)b)\cdot(\|a\|^2b - (a\cdot b)a) &= \|a\|^2\|b\|^2(a \cdot b) + (a \cdot b)^3 - \|b\|^2(a\cdot a)(a \cdot b) - \|a\|^2(b \cdot b)(a \cdot b) \\&= (a \cdot b)^3 - \|a\|^2\|b\|^2(a \cdot b) \end{align*}
Your denominator is just $\|a \times b\|^2$, so
$$X := \dfrac{(b\times(a\times b))\cdot(a\times(b\times a))}{(a\times b)\cdot(a\times b)} = \dfrac{(a \cdot b)((a\cdot b)^2 - \|a\|^2\|b\|^2)}{\|a\times b\|^2}.$$
Now, if $\theta$ is the angle between $a$ and $b$, the numerator becomes $(a \cdot b)\|a\|^2\|b\|^2(\cos^2\theta - 1)$, while the denominator is $\|a\|^2\|b\|^2(\sin\theta)^2$, so the $\|a\|^2\|b\|^2$ cancel, giving $$X = \dfrac{(a\cdot b)(\cos^2\theta - 1)}{|\sin\theta|^2} = -a\cdot b.$$
|
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|
Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other ways to prove this):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$
|
You can avoid squaring by comparing
$$\eqalign{
\sqrt{12}a&=\sqrt{84}+\sqrt{120}<10+11=21\ ,\cr
\sqrt{12}b&=\sqrt{36}+\sqrt{228}>6+15=21\ .\cr}$$
|
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|
If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the following operation $\biggl|\dfrac{\sqrt{3}n-m}{n}\biggr|\cdot \biggl|\dfrac{\sqrt{3}n+m}{\sqrt{3}n+m}\biggr|$ to get $$\biggl|\dfrac{3n^{2}-m^{2}}{\sqrt{3}n^{2}+mn}\biggr|$$
Since $n,m \ne 0$, we have that $|3n^{2}-m^{2}| \ge 1$. Now for the denominator, we have $$ |\sqrt{3}n^{2}+mn| \le |\sqrt{3n^{2}}| + |mn| $$
Thus it follows that $$\dfrac{1}{|\sqrt{3}n^{2}+mn|} \ge \dfrac{1}{|\sqrt{3}n^{2}| + |mn|}$$
Would I have to work in cases where $m<n$, for example? Then we have $$|\sqrt{3}n^{2}| + |mn| < |\sqrt{3}n^{2}| + n^{2} < 3n^{2} + n^{2} < 5n^{2}$$ which gives us the desired result. Although, the same method doesn't work when $n >m$.
|
We may assume without loss of generality that $m, n$ are both positive as this is equivalent to them both being negative and if only one of them is negative then this inequality is trivial.
We can then consider 2 cases:
Case 1: m< 2n
By your argument we need only show that $|\sqrt{3}n^2|+|mn|<5n^2$. This follows from $|\sqrt{3}n^2|+|mn|<|2n^2|+|2n^2|=4n^2<5n^2$.
Case 2: $m\geq 2n$
In this case we have: $|\sqrt{3} - \frac{m}{n}| > |\frac{9}{5} - 2| = \frac{1}{5} \geq \frac{1}{5n^2}$.
|
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|
Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$
Solve the system:
$$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
x^2-y^2=8\end{array}$$
First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do next?
|
Rewrite the first equation as
$$x^2+y^2 - \dfrac{10}{3}xy = \frac13(x-3y)(3x-y)=0$$
which yields $x=3y$ and $x=\frac y3$. Plug them into $x^2-y^2=8$ to obtain the real solutions $(3,1)$ and $(-3,-1)$.
|
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"url": "https://math.stackexchange.com/questions/3476697",
"timestamp": "2023-03-29T00:00:00",
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|
using absolute function to translate the inequality How to use the absolute value function to translate each of the following statements into a single inequality.
(a) $\ x β (-4,10) $
(b) $\ x β (-\infty,2] \cup[9,\infty) $
I think in the first one the absolute value of $\ x$ should be greater than 4 and less than 10. is that correct?
because the distance from $\ x$ to $\ 0$ should be between $\ 4$ and$\ 10$ in order for $\ x$ to belong in this interval.
|
$x \in (-4,10)\implies$
$-4 < x < 10\implies$
$-4 + k < x + k < 10 + k$
If we have $M=|10+k| = |-4+k|$ (and presumably $-M= -4k < 0 < 10+k=M$) then we would have $-M < x + k < M$ so $|x + k| <M$.
So if $10+k = -(-4+k)=4-k$ then $2k = -6$ and $k -3$ and $M=10 -3=7$ and
$-4 -3 < x - 3< 10-3$
$-7 < x-3 < 7$
$|x-3| < 7$.
b) $x β (-\infty,2] \cup[9,\infty)$ means
$x \le 2$ or $x \ge 9$.
Again $x + k \le 2+k$ or $x \ge 9+k$. If we can get $M=9+k = -(2+k)$ we would have $|x+k | \ge M$.
$9+k = -(2+k) \implies k=-5.5$ so $x-5.5 \le -3.5$ and $x-5.5 \ge 3.5$ so $|x-5.5| \ge 3.5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3477795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\frac{\cos(2x)dx}{1-2\sin^2(2x)}$$
I tried applying the t=tan(y/2) (where y = 2x) but it became a mess so I figure this can be simplified further... help?
|
$\cos^4 x = \frac 18 (\cos 4x + 4\cos 2x + 3)\\
\sin^4 x = \frac 18 (\cos 4x - 4\cos 2x + 3)\\
\cos^4 x+ \sin^4 x = \frac 14(\cos 4x + 3)$
$\frac {4\cos 2x}{\cos 4x + 3}$
That looks a little better.
Lets say $\cos 4x = 1 - 2\sin^2 2x$
$\int \frac {4\cos 2x}{4 - 2\sin^2 2x}\ dx$
Now we can do a u-subtitution
$u = \sin 2x, du = 2\cos 2x$
$\int \frac {2}{4 - 2u^2}\ du$
Separate into partial fractions
$\frac 1{2\sqrt 2} \int \frac {1}{\sqrt 2-u} + \frac {1}{\sqrt 2+u} \ du$
$\frac 1{2\sqrt 2} (\ln (\sqrt 2+u)-\ln (\sqrt 2-u))$
$\frac 1{2\sqrt 2} (\ln (\sqrt 2+\sin 2x)-\ln (\sqrt 2-\sin 2x))$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
How do I solve it? Help me please.
|
For each $n\in\mathbb N$, the first non-null term of the Taylor series of $x^n\sin^n(x)$ centered at $0$ is $x^{2n}$, whereas the first non-null term of the Taylor series of $x^n-\sin^n(x)$ centered at $0$ is $kx^{n+2}$, for some $k\neq0$. So, your limit will be finite and non zero if and only if $2n=n+2$. And this occurs if and only if $n=2$.
|
{
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"url": "https://math.stackexchange.com/questions/3481642",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?
|
We can assume without loss of generality that $x,y\ge 0$ (why?)
So by AM-GM inequality:
$$x^2+y^2=x^2+\frac{y^2}3+ \frac{y^2}3+ \frac{y^2}3\ge4\sqrt[4]{\frac{1}{3^3}x^2y^6}=4\cdot3^{-\frac34}\sqrt{xy^3}.$$
By squaring:
$$(x^2+y^2)^2\geq16\cdot3^{-\frac32}xy^3\approx3.08 xy^3.$$
So we have the sharper bound $$\boxed{(x^2+y^2)^2\geq 3xy^3.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3481757",
"timestamp": "2023-03-29T00:00:00",
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|
solutions of $a+b=c^2 , a^2+c^2=b^2$ ; $a,b,c$ are natural numbers So it all started with a fun observation, $12+13=5^2$ and these are Pythagorean triplets($5,12,13$), so I thought are there more such numbers?
with brute force I was able to get $(24,25,7)$ and $(40,41,9)$.
Then I was able to find 3 families of solutions.
$(50k^2+50k+12 , 50k^2+50k+13 , 10k+5)$
$(10k+4 , 10k+5 , \sqrt{20k+9})$
$(10k, 10k+1 , \sqrt{20k+1})$
ps: I found these by using the property of Pythagorean triplets that they have at least one multiple of 5 in it.
My question is are there more sets of solution and how do I know I haven't missed any?
|
$$a^2+c^2=b^2$$
$$\implies a^2+a+b=b^2$$
$$\implies \Big(a+\frac{1}{2}\Big)^2= \Big(b-\frac{1}{2}\Big)^2$$
$$\implies a+\frac{1}{2}=b-\frac{1}{2}$$
$$\implies b=a+1$$
Hence,
$$2a+1=c^2$$
Therefore, $c$ is odd, let $c=2k+1$. Putting it in above equation, you get, $$a=2k^2+2k$$
$$\implies b=a+1=2k^2+2k+1$$
This is the required general solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3482827",
"timestamp": "2023-03-29T00:00:00",
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|
Finding residues in a complex integral $\int_0^\infty\frac{z^6\,\mathrm{d}z}{(z^4+1)^2}$ using Laurent series expansion I am trying to compute
$$\int_0^\infty\frac{z^6\,\mathrm{d}z}{(z^4+1)^2}.$$
I am getting stuck on computing the residues. I am only considering the residues when $z=\mathrm{e}^{\frac{\mathrm{i}\pi}4}$ and $z=\mathrm{e}^{\frac{3\mathrm{i}\pi}4}$ since we are taking the integral from $0$ to $\infty$. When computing the residues, I am getting a computational nightmare. For instance,
$$\underset{z=\mathrm{e}^{\frac{\mathrm{i}\pi}4}}{\textrm{Res}}\,f(z)=\lim_{z\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}4}}\frac{\mathrm{d}}{\mathrm{d}z}\frac{z^6}{\left(z-\mathrm{e}^{\frac{3\mathrm{i}\pi}4}\right)^2\left(z-\mathrm{e}^{\frac{5\mathrm{i}\pi}4}\right)^2\left(z-\mathrm{e}^{\frac{7\mathrm{i}\pi}4}\right)^2},$$
and the derivative calculation has been quite messy.
I've also tried using $\frac{1+\mathrm{i}}{\sqrt2}$ in place of $\mathrm{e}^{\frac{\mathrm{i}\pi}4}$, and similar representations for the other roots, in the hopes of the calculation maybe "cleaning up" a bit. I was wondering if anyone could see anything that I am missing that would make computing this derivative nicer, or if these residues are just computationally "ugly".
Thank you!
|
Let's compute the residue by finding a few terms of the Laurent series for
$$
p(z) = \frac{z^6}{(z^4+1)^2}
$$
at $\omega = e^{i\pi/4} = \frac{1+i}{\sqrt{2}}$.
To expand in powers of $z-\omega$, I prefer to write $r=z-\omega$ and expand in powers of $r$.
$$
p(r+\omega) = \frac{(r+\omega)^6}{((r+\omega)^4+1)^2}
$$
Do this in small steps. As $r \to 0$:
$$
N:=(r+\omega)^6 = \omega^6+6\omega^5 r + O(r^2)
= -i - \omega r + O(r^2)
\\
(r+\omega)^4 = \omega^4 + 4 \omega^3 r + 6 \omega^2 r^2 + O(r^3)
= -1+4i\omega r + 6 i r^2 + O(r^3)
\\
(r+\omega)^4 + 1 = 4i\omega r + 6 i r^2 + O(r^3)
\\
D = \big((r+\omega)^4 + 1\big)^2 = -16i r^2 - 48 \omega r^3 + O(r^4)
\\
p(r+\omega) = \frac{N}{D} =
\frac{-i - \omega r + O(r^2)}{-16i r^2 - 48 \omega r^3 + O(r^4)}
= \frac{1}{16 r^2}\left(\frac{1 - 6 i\omega r+O(r^2)}{1 - 3 i \omega r+O(r^2)}\right)
\\
=\frac{1}{16 r^2} \big(1 - 3 i \omega r + O(r^2)\big)
= \frac{1}{16} r^{-2} - \frac{3 i \omega}{16} r^{-1} + O(1)
$$
So we see that the residue is
$$
\frac{-3i\omega}{16} = \frac{3-3i}{16\sqrt{2}}
$$
|
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"url": "https://math.stackexchange.com/questions/3484948",
"timestamp": "2023-03-29T00:00:00",
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|
Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$. Exercise 4.4, problem 24, page 144 of the book on differential equations by KKOP [here][1].
Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$, where $a,b,\omega$ are real constants, $a<b$. Consider separately the cases $\omega \ne \sqrt{b^2-a^2}$ and $\omega = \sqrt{b^2-a^2}$ and sketch the solution curve in each case.
I am getting not so nice looking terms. So, I would like to ask if Im proceeding in the right direction, if somebody could verify my steps or help me with the solution, that would be great. My solution is as below.
Solution(My attempt).
The associated homogeneous differential equation is -
$(D^2+2aD+b^2)y=0$
The characteristic equation is:
$\begin{align}
D^2+2aD+b^2&=0\\
(D+a)^2 + (b^2-a^2)&=0\\
(D+a)^2 &= -(b^2-a^2)\\
D &= -a \pm \sqrt{b^2-a^2}i
\end{align}$
The general solution of the homogeneous linear differential equation is :
$\bbox[5px, border: 2px solid blue]{y_h = e^{-ax}(c_1 \cos (\sqrt{b^2-a^2}x) + c_2 \sin (\sqrt{b^2-a^2}x))}$
A particular solution of the nonhomogeneous linear differential equation can be constructed of the form:
$\bbox[5px, border: 2px solid blue]{y_p = c_1(x) e^{-ax} \cos (\sqrt{b^2-a^2}x) + c_2(x) e^{-ax} \sin (\sqrt{b^2-a^2}x)}$
where $c_1(x)$ and $c_2(x)$ are given by the equations:
$\begin{align}
y_1(x)c_1'(x)+y_2(x)c_2'(x)&=0\\
y_1'(x)c_1'(x)+y_2'(x)c_2'(x)&=h(x)\\
\end{align}$
That is,
$\begin{align}
e^{-ax} \cos (\sqrt{b^2-a^2}x) c_1'(x)+e^{-ax} \sin (\sqrt{b^2-a^2}x)c_2'(x)&=0\\
\{-ae^{-ax}\cos (\sqrt{b^2-a^2}x)-\sqrt{b^2-a^2}e^{-ax}\sin (\sqrt{b^2-a^2}x)\}c_1'(x)\\+
\{-ae^{-ax}\sin (\sqrt{b^2-a^2}x)+\sqrt{b^2-a^2}e^{-ax}\cos (\sqrt{b^2-a^2}x)\}c_2'(x)&=\sin (\omega x)\\
\end{align}$
The Wronskian of the functions $e^{-ax}\cos (\sqrt{b^2-a^2}x), e^{-ax}\sin (\sqrt{b^2-a^2}x)$ is:
$\begin{align}
W(x) &=
\scriptsize
e^{-2ax}
\begin{vmatrix}
\cos (\sqrt{b^2-a^2}x) & \sin (\sqrt{b^2-a^2}x)\\
-a\cos (\sqrt{b^2-a^2}x)-\sqrt{b^2-a^2}\sin (\sqrt{b^2-a^2}x) & -a\sin (\sqrt{b^2-a^2}x)+\sqrt{b^2-a^2}\cos (\sqrt{b^2-a^2}x)
\end{vmatrix}\\
&= \sqrt{b^2-a^2}e^{-2ax}
\end{align}$
Consequently,
$\begin{align}
\scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{\sqrt{b^2-a^2}}[\cos (\omega - \sqrt{b^2-a^2}x) - \cos (\omega + \sqrt{b^2-a^2}x)]\\
\scriptsize c_2'(x)&=\scriptsize e^{ax}\frac{\cos \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =\frac{e^{ax}}{\sqrt{b^2-a^2}}[\sin (\omega + \sqrt{b^2-a^2}x) + \sin (\omega - \sqrt{b^2-a^2}x)]
\end{align}$
Integrating these yields -
$\begin{align}
\scriptsize c_1(x)&=\scriptsize \left[\frac{e^{ax}}{a^2+(\omega + \sqrt{b^2-a^2})^2}\{a \sin (\omega + \sqrt{b^2-a^2}) - (\omega + \sqrt{b^2-a^2}) \cos (\omega + \sqrt{b^2-a^2})\}\\
-\frac{e^{ax}}{a^2+(\omega - \sqrt{b^2-a^2})^2}\{a \sin (\omega - \sqrt{b^2-a^2}) - (\omega - \sqrt{b^2-a^2}) \cos (\omega - \sqrt{b^2-a^2})\}\right]\\
\scriptsize c_2(x)&=\scriptsize \left[\frac{e^{ax}}{a^2+(\omega + \sqrt{b^2-a^2})^2}\{a \cos (\omega + \sqrt{b^2-a^2}) + (\omega + \sqrt{b^2-a^2}) \sin (\omega + \sqrt{b^2-a^2})\}\\
+\frac{e^{ax}}{a^2+(\omega - \sqrt{b^2-a^2})^2}\{a \cos (\omega - \sqrt{b^2-a^2}) + (\omega - \sqrt{b^2-a^2}) \sin (\omega - \sqrt{b^2-a^2})\}\right]\\
\end{align}$
|
You first mistake is at the step
$$\begin{align}
\scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{\sqrt{b^2-a^2}}\left[\cos \left(\omega - \sqrt{b^2-a^2}x\right) - \cos \left(\omega + \sqrt{b^2-a^2}x\right)\right]\\
\scriptsize c_2'(x)&=\scriptsize e^{ax}\frac{\cos \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =\frac{e^{ax}}{\sqrt{b^2-a^2}}\left[\sin \left(\omega + \sqrt{b^2-a^2}x\right) + \sin \left(\omega - \sqrt{b^2-a^2}x\right)\right]
\end{align}$$
It should have been
$$\begin{align}
\scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{2\sqrt{b^2-a^2}}\left[\cos \left(\omega - \sqrt{b^2-a^2}\right)x - \cos \left(\omega + \sqrt{b^2-a^2}\right)x\right]\\
\scriptsize c_2'(x)&=\scriptsize e^{ax}\frac{\cos \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =\frac{e^{ax}}{2\sqrt{b^2-a^2}}\left[\sin \left(\omega + \sqrt{b^2-a^2}\right)x + \sin \left(\omega - \sqrt{b^2-a^2}\right)x\right]
\end{align}$$
Part of the error was due to translating your manuscript to MathJax, but you are missing the factors of $2$ in
$$2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta)$$
and
$$2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$$
But you could have corrected the factor of $2$ if you differentiated your solution and substituted into the original differential equation. More serious is the fact that
$$\int e^{ax}\cos\beta x\,dx=\frac{e^{ax}\left(a\cos\beta x+\beta\sin\beta x\right)}{a^2+\beta^2}+C$$
and
$$\int e^{ax}\sin\beta x\,dx=\frac{e^{ax}\left(-\beta\cos\beta x+a\sin\beta x\right)}{a^2+\beta^2}+C$$
which may be checked by differentiation but you seem to have applied switched these formulas around, not to mention the mistake of dropping the factor of $1/\sqrt{b^2-a^2}$ and forgetting about the factor of $x$ in the arguments to the trig functions in transcription to MathJax.
There is a much easier way to solve this kind of problem with a sinusoidally driven damped harmonic oscillator: since $\Im\left(e^{i\omega x}\right)=\sin\omega $ just solve the complex problem
$$z^{\prime\prime}+2az^{\prime}+b^2z=e^{i\omega x}$$
and then take the imaginary part of the solution. Assume a solution $z=Be^{i\omega x}$ for some complex number $B$. Then on differentiation and substitution we get
$$B\left(-\omega^2+2ia\omega+b^2\right)e^{i\omega x}=e^{i\omega x}$$
So
$$\begin{align}z&=Be^{i\omega x}=\frac{e^{i\omega x}}{b^2-\omega^2+2ia\omega}=\frac{\left(b^2-\omega^2-2ia\omega\right)\left(\cos\omega x+i\sin\omega x\right)}{\left(b^2-\omega^2\right)^2+4a^2\omega^2}\\
&=\frac{\left(b^2-\omega^2\right)\cos\omega x+2a\omega\sin\omega x+i\left(\left(b^2-a^2\right)\sin\omega x-2a\omega\cos\omega x\right)}{\left(b^2-\omega^2\right)^2+4a^2\omega^2}\end{align}$$
And our solution is
$$y=\Im z=\frac{\left(b^2-a^2\right)\sin\omega x-2a\omega\cos\omega x}{\left(b^2-\omega^2\right)^2+4a^2\omega^2}$$
When you eventually get your method to work you can expand trig functions like $$\cos\left(\omega+\sqrt{b^2-a^2}\right)x=\cos\omega x\cos\left(\sqrt{b^2-a^2}x\right)-\sin\omega x\sin\left(\sqrt{b^2-a^2}x\right)$$
and then group terms with like powers of trig functions of $\sqrt{b^2-a^2}x$ and $\omega x$ and simplify to compare with the easy answer.
|
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|
How can I calculate the limit $\lim\limits_{n \to \infty} \frac1n\ln \left( \frac{2x^n}{x^n+1} \right)$. I have the following limit to find:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg)$$
Where $n \in \mathbb{N}^*$ and $x \in (0, \infty)$.
I almost got it. For $x > 1$, I observed that:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg) = \lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n(1 + \frac{1}{x^n})} \bigg) = \lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2}{1+\frac{1}{x^n}} \bigg)$$
Because $x>1$, we have that $x^n \rightarrow \infty$ as $n \rightarrow \infty$, so that means that we have:
$$\dfrac{1}{\infty} \cdot \ln \bigg ( \dfrac{2}{1+\frac{1}{\infty}} \bigg ) = 0 \cdot \ln 2 = 0$$
The problem I have is in calculating for $x \in (0, 1]$. If we have that $x \in (0, 1]$ that means $x^n \rightarrow 0$ as $n \to \infty$, so:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n + 1} \bigg ) = \lim\limits_{n \to \infty} \dfrac{\ln \bigg( \dfrac{2x^n}{x^n + 1}\bigg )}{n} $$
And I tried using L'Hospital and after a lot of calculation I ended up with
$$\ln x \lim\limits_{n \to \infty} \dfrac{x^n + 1}{x^n}$$
which is
$$\ln x\cdot \dfrac{1}{0}$$
And this is my problem. Maybe I applied L'Hospital incorrectly or something, I'm not sure. Long story short, I do not know how to calculate the following limit:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n+1} \bigg )$$
when $x \in (0, 1]$.
|
Since $\frac{d}{dn}x^n=x^n\ln x$, $\frac{d}{dn}\frac{2x^n}{x^n+1}=-2\frac{d}{dn}\frac{1}{x^n+1}=\frac{2x^n\ln x}{(x^n+1)^2}$ and $\frac{d}{dn}\ln\frac{2x^n}{x^n+1}=\frac{\ln x}{x^n+1}$. So you want $\lim_{n\to\infty}\frac{\ln x}{x^n+1}$, regardless of $x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron whose vertices are the given points:
$$ ( 0, 0, 0 ), ( 2, 0, 0 ), ( 0, 2, 0 ), ( 0, 0, 2 ) $$
Answer:
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors $A$, $B$ and $C$ then
the area of the object is $A \cdot (B \times C)$. Let $V$ be the volume we are trying to find.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\[4pt]
A &= ( 2, 0, 0) - (0,0,0) = ( 2, 0, 0) \\
B &= ( 0, 2, 0) - (0,0,0) = ( 0, 2, 0) \\
C &= ( 0, 0, 2) - (0,0,0) = ( 0, 0, 2) \\[4pt]
V &= \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix} =
\begin{vmatrix}
2 & 0 &0 \\
0 & 2 & 0 \\
0 & 0 & 2\\
\end{vmatrix} \\
&= 2 \begin{vmatrix}
2 & 0 \\
0 & 2\\
\end{vmatrix} = 2(4 - 0) \\
&= 8
\end{align*}
However, the book gets $\frac{4}{3}$.
|
Yes, because what you are really doing is finding the volume of a volume of a parallelepiped. You can also approach the problem this way:
height= |a|
and the area is given by
|b Γ c|
So, then you have Volume = height * area
Volume= |a||b Γ c|=|a β( b Γ c)|
For b: (0,2,0) β (0,0,0) = (0,2,0)
For c: (0,0,2) β (0,0,0) = (0,0,2)
b Γc=[(2x2)-(0x0),(0x0)-(0x2),(0x0)-(2x0)]=(4,0,0)
For a: (2,0,0) β (0,0,0) = (2,0,0)
β΄ Volume= a β( b Γ c)=(2x4)+(0x0)+(0x0)=8
The volume of tetrahedron is 1/6 that of the parallelepiped
β΄ V_Tetrahedron= (1/6) x 8= 4/3
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3488590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the determinant into that form, but I am not seeing it. And directly computing the determinant by the cofactor expansion looks very nasty. So is there a simpler way to compute this determinant?
|
A determinant of the left matrix is equal to
$$\prod_{cyc}(a^2+b^2)+2a^2b^2c^2-\sum_{cyc}(a^2+b^2)a^2b^2=$$
$$=\sum_{cyc}\left(a^4b^2+a^4c^2+\frac{2}{3}a^2b^2c^2\right)+2a^2b^2c^2-\sum_{cyc}(a^4b^2+a^4c^2)=4a^2b^2c^2.$$
A determinant of the right matrix is equal to
$$0+2abc-0=2abc$$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3489938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluating $\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$ I tried substitution $x=\dfrac{1}{t}$, then $z=t^2$, tried rationalizing the denominator but haven't been able to pull it off. I can't think of any trig substitution either. I prefer an intuitive approach rather than an "plucked-out-of-thin-air" counter-intuitive substitution.
|
Here is an elementary method without using hyperbolic functions:
$$\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$$
$$=\int \frac{1}{x^5(\frac{1}{x^4}+1)\sqrt{\sqrt{\frac{1}{x^4}+1}-1}}dx$$
Now substitute $t=\frac{1}{x^4}+1$, required integral becomes,
$$-\frac{1}{4}\int\frac{1}{t\sqrt{\sqrt {t}-1}}dt$$
Now substitute $y=\sqrt{\sqrt {t}-1}$, to get,
$$t=(y^2+1)^2$$
$$\implies dt=4y(y^2+1)dy$$
$$\implies \text{Required integral}=-\frac{1}{4}\int\frac{4y(y^2+1)}{y(y^2+1)^2}dy$$
$$=-\tan^{-1}y+C$$
$$=-\tan^{-1}\left(\frac{\sqrt{\sqrt{x^4+1}-x^2}}{x}\right)+C$$
$$=\tan^{-1}\left(\frac{x}{\sqrt{\sqrt{x^4+1}-x^2}}\right)+C'$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3490750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Possible values of expression Real numbers $x,y,z$ are selected so, that $\frac{1}{|x^2 +2yz|}$, $\frac{1}{|y^2 +2xz|}$, $\frac{1}{|z^2 +2xy|}$ are sides of a triangle(they satisfy the triangle inequality). Determine the possible values of the expression $xy+xz+yz$.
It is easy to prove that all positive and negative real numbers can be expressed. What about $0$? Any help is appreciated.
|
Suppose that $xy+yz+zx=0$. Then, we have the following identities
$$
\frac{1}{|x^2+2yz|}=\frac{1}{|x^2+2yz-xy-zx-yz|}=\frac{1}{|x-y|\cdot |x-z|},
\\
\frac{1}{|y^2+2zx|}=\frac{1}{|y^2+2zx-xy-zx-yz|}=\frac{1}{|x-y|\cdot |y-z|},
\\
\frac{1}{|x^2+2yz|}=\frac{1}{|z^2+2xy-xy-zx-yz|}=\frac{1}{|z-x|\cdot |y-z|}.
$$
Now, it's easy to prove that sum of two of these number is equal to the third number, so they doesn't satisfy the triangle inequality. Hence, $xy+yz+zx\neq 0$, as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3491231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamma^3})=16$$
Here's what I have tried,
By Vieta's rule
$\alpha+\beta+\gamma=\frac{-1}{2}\text{. ...........}(1)$
$\alpha \beta+\beta \gamma+\alpha \gamma=\frac{1}{2}\text{. ...........}(2)$
$\alpha \beta \gamma=\frac{-1}{2}\text{. ...........}(3)$
Squaring $(1)$,
$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\alpha\gamma)=\frac{1}{4}\text{. ...........}(4)$
From $(2)$,
$\alpha^2+\beta^2+\gamma^2=\frac{-3}{4}\text{. ...........}(5)$
Putting the roots and adding these equations,
$2\alpha^3+\alpha^2+\alpha+1=0$
$2\beta^3+\beta^2+\beta+1=0$
$2\gamma^3+\gamma^2+\gamma+1=0$
We get,
$2(\alpha^3+\beta^3+\gamma^3)+(\alpha^2+\beta^2+\gamma^2)+(\alpha+\beta+\gamma)+3=0$
Putting the values,
$2(\alpha^3+\beta^3+\gamma^3)+\frac{-3}{4}+\frac{-1}{2}+3=0$
$(\alpha^3+\beta^3+\gamma^3)=\frac{-7}{8}$
Then I divided $2x^3+x^2+x+1=0$ by $x$ and I found out $(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma})$ the same way I found out $(\alpha^3+\beta^3+\gamma^3)$. Likewise doing the same, finding $\sum\frac{1}{\alpha^2}$ then $\sum\frac{1}{\alpha^3}$ I found out
$\sum\frac{1}{\alpha^3}=-4$
But still I'm far from the answer, also the $-$ sign is creating problems.
Any help would be highly appreciated
|
Expressing the two complex and conjugate roots as a function of the third root $\gamma$ ,we obtain
$\alpha=-\frac{2\gamma+1}{4}+I\frac{\sqrt{12\gamma^{2}+4\gamma+7}}{4}$
$\beta=\frac{2\gamma+1}{4}-I\frac{\sqrt{12\gamma^{2}+4\gamma+7}}{4}$
therefore
$\Big(\frac{1}{\gamma^{3}}+I\frac{(2\gamma-1)\sqrt{12\gamma^{2}+4\gamma+7}}{(2\gamma^{2}+\gamma+1)^{3}}\Big)$*
$\Big(\frac{1}{\gamma^{3}}-I\frac{(2\gamma-1)\sqrt{12\gamma^{2}+4\gamma+7}}{(2\gamma^{2}+\gamma+1)^{3}}\Big)$*
$\Big(-\frac{1}{\gamma^{3}}+ \frac{(2\gamma+1)(8\gamma^{2}+2\gamma+5)}{(2\gamma^{2}+\gamma+1)^{3}}\Big)=16$
that is
$2\gamma^{3}+\gamma^{2}+\gamma+1=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3492092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Mistake in evaluating $\int_{0}^{1}\frac{x^2\ln x}{\sqrt{1-x^2}}dx$ $$\int_{0}^{1}\dfrac{x^2\ln x}{\sqrt{1-x^2}}dx$$
My attempt is as follows:
$$x=\sin\theta\Rightarrow dx=\cos\theta d\theta$$
$$\int_{0}^{\frac{\pi}{2}}\sin^2\theta\ln (\sin\theta)d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1-\cos2\theta)\ln(\sin\theta)d\theta$$
$$=\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\ln(\sin\theta)d\theta-\int_{0}^{\frac{\pi}{2}}\cos(2\theta)\ln(\sin\theta)d\theta\right)=\frac{1}{2}\left(\dfrac{-\pi\ln(2)}{2}-I'\right)$$
$$I'=\int_{0}^{\frac{\pi}{2}}\cos(2\theta)\ln(\sin\theta)d\theta=\int_{0}^{\frac{\pi}{2}}\cos\left(\pi-2\theta\right)\ln(\cos\theta)d\theta$$
$$\Rightarrow 2I'=\int_{0}^{\frac{\pi}{2}}\cos(2\theta)\ln(\tan\theta)d\theta$$
As $\cos(2\theta)\ln(\tan\theta)d\theta=\cos\left(2\left(\frac{\pi}{2}-\theta\right)\right)\ln\left(\tan\left(\dfrac{\pi}{2}-\theta\right)\right)d\theta$
$$I'=\int_{0}^{\frac{\pi}{4}}\cos(2\theta)\ln(\tan\theta)d\theta$$
$$I'=\int_{0}^{\frac{\pi}{4}}\frac{1-\tan^2\theta}{1+\tan^2\theta}\ln(\tan\theta)d\theta$$
$$\tan\theta=t$$
$$I'=\int_{0}^{1}(1-t^2)\ln(t)dt$$
Applying integration by parts
$$I'=-\lim_{t\to0}\ln(t)\cdot\left(t-\frac{t^3}{3}\right)-\int_{0}^{1}\left(1-\dfrac{t^2}{3}\right)dt$$
$$I'=-\left(1-\frac{1}{9}\right)=-\frac{8}{9}$$
So the answer would be $\dfrac{1}{2}\left(\dfrac{-\pi\ln(2)}{2}-I'\right)=\dfrac{1}{2}\left(\dfrac{-\pi\ln(2)}{2}+\dfrac{8}{9}\right)$
But actual answer is $\dfrac{\pi}{8}\left(1-\ln4\right)$. What mistake am I making here?
|
ActuallyοΌ
\begin{align*}
\int_0^{\frac\pi2}\cos(2\theta)\ln(\sin\theta)d\theta&=
\frac12\int_0^{\frac\pi2}\ln(\sin\theta)d(\sin2\theta)\\
&=\frac12\sin2\theta\ln(\sin\theta)\bigg|_0^{\frac\pi2}-\frac12\int_0^{\frac\pi2}\sin2\theta\cdot\frac{\cos\theta}{\sin\theta}d\theta\\
&=-\frac12\int_0^{\frac\pi2}2\cos^2\theta d\theta=-\frac\pi4.
\end{align*}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $a^{4b}+b^{4a}\geq \frac{1}{2}$ Inspired by a problem of Vasile Cirtoaje I propose this :
Let $a,b>0$ such that $a+b=1$ then we have :
$$a^{4b}+b^{4a}\geq \frac{1}{2}$$
I compute the derivative of $f(x)=x^{4(1-x)}+(1-x)^{4x}$ on $]0,1]$ we get :
$$ f'(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x)) + (1 - x)^{(4 x)} (4 \log(1 - x) - \frac{4 x}{1 - x})$$
If we denote by $g(x)$ the function :
$$g(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x))$$
We can rewrite the derivative as :
$$f'(x)=g(x)-g(1-x)$$
So it's remains to show that $g(x)\geq g(1-x)$ or $g(x)\leq g(1-x)$
So it remains to show that $g(x)$ is increasing or decreasing .
After that I'm stuck...
Any helps are very appreciated !
Thanks a lot for your time .
|
Alternative Proof:
We need to prove that, for all $a$ in $(0, 1/2]$,
$$a^{4 - 4a} + (1 - a)^{4a} \ge 1/2.$$
We split into two cases:
Case I $\,\, a\in (0, 1/4)$:
We have
$$a^{4 - 4a} + (1 - a)^{4a}
\ge (1 - a)^{4a} \ge 1 - a > \frac12.$$
Case II $\,\,a \in [1/4, 1/2]$:
First, using Bernoulli inequality, we have
\begin{align*}
a^{4 - 4a}
&= 2^{-1}2^{-(2 - 4a)}(1 - (1 - 2a))^{3 - 4a} a\\
&\ge 2^{-1}2^{-(2 - 4a)}[1 - (1 - 2a)(3 - 4a)] a\\
&\ge 2^{-1}[1 - (2 - 4a)\ln 2]\,[1 - (1 - 2a)(3 - 4a)]a
\end{align*}
where we have used $2^{-(2 - 4a)} = \mathrm{e}^{-(2 - 4a)\ln 2}
\ge 1 - (2 - 4a)\ln 2$.
Second, using Bernoulli inequality, we have
\begin{align*}
(1 - a)^{4a} &= 2^{-2}2^{2 - 4a}(1 + (1 - 2a))^{4a}\\
&\ge 2^{-2}2^{2 - 4a}[1 + (1 - 2a)\cdot 4a]\\
&\ge 2^{-2}[1 + (2 - 4a)\ln 2]\,[1 + (1 - 2a)\cdot 4a]
\end{align*}
where we have used $2^{2 - 4a} = \mathrm{e}^{(2 - 4a)\ln 2}
\ge 1 + (2 - 4a)\ln 2$.
It suffices to prove that
\begin{align*}
&2^{-1}[1 - (2 - 4a)\ln 2]\,[1 - (1 - 2a)(3 - 4a)]a\\
&\quad + 2^{-2}[1 + (2 - 4a)\ln 2]\,[1 + (1 - 2a)\cdot 4a] \ge \frac12
\end{align*}
or
$$ (1 - 2a)^2[-16(\ln 2)a^2 + (20\ln 2 - 4)a + 2\ln 2 - 1]
\ge 0$$
which is true.
We are done.
|
{
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"url": "https://math.stackexchange.com/questions/3496475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Minimum value of the given function- Find the minimum value of-
$$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$
I tried opening the brackets and trying to cancel the terms and using AM-GM.
|
Hint.
Calling
$$
u = x^3+\frac{1}{x^3}\\
v = \left(x+\frac{1}{x}\right)^3
$$
we have
$$
\frac{v^2-u^2}{v+u}=v-u = 3\left(x+\frac{1}{x}\right)\ge 6
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3501616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
Extrema of $f(x,y) = xy\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$ From Demidovich
Find the extrema of the function of $z = xy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$
What I've done so far:
I've make $z_{x}'= 0$ and $z_{y}' = 0$, wich gives me:
$$\frac{\partial z}{\partial x} = y \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{\partial z}{\partial y} = x \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
And I've tried to simplify, and I got:
$$\frac{y a^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{x b^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
Now the problem is, some cubic terms appear and I don't know how to solve this and find the critical points to proceed with the exercise. Of course the denominator cannot be zero, so my system is composed only by the numerators.
I'm pretty sure I'm doing something wrong, 'cause this seems too much trouble for an introductory exercise (it's one of the firsts in the chapter)
|
I suppose that $a,b>0$ are given parameters. The given function is then defined in the region
$$
R = \left\{\ (x,y)\ :\ \frac {x^2}{a^2}+ \frac {y^2}{b^2}\le 1\ \right\}\ .
$$
The four quadrants $I, II, \dots$ are cutting $R$ in four pieces, $R_I$, $R_{II}$, ... and it is enough to search for the local / global maxima on the piece $R_I$.
On the boundary of $R_I$ (and of $R$) the given function vanishes.
The (local/global) maximal values of the function $f$ on $R_I$ correspond to the (local/global) maximal values of $f^2/(a^2b^2)$, which is a function of $X=x^2/a^2$, $Y=y^2/b^2$, explicitly given by the help function
$$
h(X,Y)=XY\left(1-X-Y\right)=XY-X^2Y-XY^2\ ,
$$
the corresponding region for $(X,Y)$ is a triangle given by $X,Y,1-X-Y\ge 0$. At the boundary the values are zero. Differentiation gives the system of equations for a local extremal point:
$$
\left\{
\begin{aligned}
Y&= 2XY +Y^2\ ,\\
X&= 2XY +X^2\ ,
\end{aligned}
\right.
$$
and subtraction gives $Y-X=Y^2-X^2=(Y-X)(Y+X)$. We have thus either $X=Y$, or $X+Y=1$.
*
*in case of $X=Y$ the common value is $0$ or $1/3$.
*in case of $X+Y=1$ adding the equations we get $1=X+Y=4XY+X^2+Y^2=2XY+1^2$, so $XY=0$, a boundary point.
So we keep only $(X,Y)=(1/3,\;1/3)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$.
find $\alpha^2+\beta^2$
Both equations have a common root
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has equal roots
$$4b^2-20a=
0$$
$$b^2=5a$$
I could substitute the value of a in the above equation, but that gives me a biquadractic equation in b, and I donβt think itβs supposed to go that way. What am I doing wrong?
|
$$a\alpha^2-2b\alpha+5=0\tag{1}$$
$$\alpha^2-2b\alpha-10=0\tag{2}$$
Subtracting $(1)$ and $(2)$
$$\alpha^2(a-1)+15=0$$
$$\alpha^2=\dfrac{15}{1-a}$$
From the first equation $$\alpha^2=\dfrac{5}{a}$$
$$\dfrac{15}{1-a}=\dfrac{5}{a}$$
$$3a=1-a$$
$$a=\dfrac{1}{4}$$
As first equation has equal roots
$$D=0$$
$$b^2-5a=0$$
$$b^2=\dfrac{5}{4}$$
So finally $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$
$$\alpha^2+\beta^2=(2b)^2+20=4b^2+20=25$$
So $25$ is your answer.
|
{
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"url": "https://math.stackexchange.com/questions/3504014",
"timestamp": "2023-03-29T00:00:00",
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|
Find the values for which $A^2 = I_2$, A is a matrix, with $A \neq I_2$ and $A \neq -I_2$ First I tried to find $A^2$ with
$$
A=\begin{bmatrix}
\alpha & \beta\\
\delta & \gamma\\
\end{bmatrix}
$$
I multiplied this by itself and got:
$$
\begin{bmatrix}
\alpha^2+\beta\delta& \beta(\alpha + \gamma)\\
\delta (\alpha + \gamma) & \delta\beta+\gamma^2\\
\end{bmatrix}
$$
I put this in a system:
$$
\left\{
\begin{array}{c}
\alpha^2+\beta\delta = 1 \\
\beta(\alpha + \gamma) = 0 \\
\delta (\alpha + \gamma) = 0 \\
\delta\beta+\gamma^2 = 1 \\
\end{array}
\right.
$$
I tried to solve for $\beta$ first and right away got an issue:
$$\beta = \frac{1-\alpha^2}{\delta}$$
One solution given by my book is:
$$
\begin{bmatrix}
1& 0\\
0 & -1\\
\end{bmatrix}
$$
So $\delta$ can be zero but according to my system it can't. How is this possible?
|
I found a much simpler way to solve this (probably the one intended by the authors).
Since
$$AA = I_2 \Leftrightarrow A = I_2.A^{-1} \Leftrightarrow A = A^{-1}$$
So,
$$A=\begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix} = A^{-1} = \det(A)^{-1}*\begin{bmatrix}
d & -b\\
-c & a\\
\end{bmatrix} $$
$\det(A)^{-1} = \frac{1}{ad-bc} = \delta$
This means that $A=A^{-1}$ equals
$$\begin{bmatrix}
d\delta & -b\delta\\
-c\delta & a\delta\\
\end{bmatrix}$$
Putting this all in a system:
$$
\left\{
\begin{array}{c}
a = d\delta \\
b = -b\delta \\
c = -c\delta \\
d = a\delta \\
\end{array}
\right.
$$
Now there's 2 possibilities: $b=c=0$ or $b,c \neq 0$
Starting with $b=c=0$:
$$
\left\{
\begin{array}{c}
b = c = 0 \\
a = d\delta \\
d = d\delta^2 \Leftrightarrow \pm 1 = \delta \\
\end{array}
\right.
$$
Now for the values of $\delta$ we get:
$\frac{1}{ad} = 1 \lor \frac{1}{ad} = -1 \Leftrightarrow a = 1/d \lor a = -1/d$
$$
\left\{
\begin{array}{c}
b = c = 0 \\
a = 1/d \\
d \in \mathbb{R}\\
\end{array}
\right.
$$
$$
\left\{
\begin{array}{c}
b = c = 0 \\
a = -1/d \\
d \in \mathbb{R}\\
\end{array}
\right.
$$
Now for the case of $b,c \neq 0$ we have:
$$\left\{
\begin{array}{c}
a = -d \\
\delta=-1 \\
\end{array}
\right.$$
Solving for the value of d in this case:
$$\delta = -1 \Leftrightarrow \frac{1}{ad-bc} \Leftrightarrow d^2+bc = 1 \Leftrightarrow \\ \pm d = \sqrt{1-bc}$$
Finally the system becomes:
$$\left\{
\begin{array}{c}
a = -d \\
d = \mp \sqrt{1-bc} \\
b,c \in \mathbb{R} \setminus \{0\}
\end{array}
\right.$$
I think that covers them all... how'd I do?
|
{
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"timestamp": "2023-03-29T00:00:00",
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}
|
What condition the roots of the polynomial meet Coefficients of the polynomial $f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$ meet the condition $a_3^3-4a_2a_3+8a_1=0.$ By substitution $x=y-\frac{a_3}{a_4}$ polynomial $f(x)$ is transformed into a biquadratic polynomial $g(y)=y^4+b_2y^2+b_0,$ where $b_2, b_0$ depend on the polynomial coefficients $f(x).$ What condition the roots of the polynomial $g(y)$ meet. Thanks for your help.
|
It is given $f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$. When we substitute $x=y-\frac{a_3}{a_4}$, we get the following polynomial,
$$f(y)=a_4\left(y-\frac{a_3}{a_4}\right)^4+a_3\left(y-\frac{a_3}{a_4}\right)^3+a_2\left(y-\frac{a_3}{a_4}\right)^2+a_1\left(y-\frac{a_3}{a_4}\right)+a_0$$
Now, we are expanding each term of the polynomial separately.
$$
\begin{matrix}
a_4y^4 & - & 4a_3y^3 & + & 6\left(\frac{a_3^2}{a_4}\right)y^2 & - & 4\left(\frac{a_3^3}{a_4^2}\right)y & + & \frac{a_3^4}{a_4^3}\\
& & a_3y^3 & - & 3\left(\frac{a_3^2}{a_4}\right)y^2 & + & 3\left(\frac{a_3^3}{a_4^2}\right)y & - & \frac{a_3^4}{a_4^3}\\
& & & & a_2y^2 & - & 2\left(\frac{a_3a_2}{a_4}\right)y & + & \frac{a_3^2a_2}{a_4^2} \\
& & & & & &a_1y & - & \frac{a_3a_1}{a_4} \\
& & & & & & & & a_0 \\
\end{matrix}
$$
When we add all these terms together and devide the whole thing by $a_4$, we get,
$$g(y)=y^4-3\frac{a_3}{a_4}y^3+\frac{1}{a_4}\left(3\frac{a_3^2}{a_4}+a_2\right)y^2-\frac{1}{a_4}\left(\frac{a_3^3}{a_4^2}-2\frac{a_3a_2}{a_4}-a_1\right)y+\frac{1}{a_4}\left(\frac{a_3^2a_2}{a_4^2}-\frac{a_3a_1}{a_4}+ a_0\right).$$
Since this transformation is supposed to deliver us the biquadratic polynomial $g(y)=y^4+b_2y^2+b_0,$ we have,
$a_3=0.$
According to the condition satisfied by the coefficients of $f\left(x\right)$, we get,
$a_1=0.$
Therefore, the coefficient of the $2^{nd}$ and $4^{th}$ terms of the polynomial $g\left(y\right)$ vanish giving us the following.
$$g(y)=y^4+\left(\frac{a_2}{a_4}\right)y^2+\frac{a_0}{a_4}$$
If $\beta$ is a root of the equation $g\left(y\right)=0$, then it must meet the following condition.
$$a_4\beta^4+a_2\beta^2+a_0=0.$$
|
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|
Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
|
Since our inequality is homogeneous, we can assume that $a+b+c=3.$
Thus, $$\sum_{cyc}\frac{a}{(b+c)^2}-\frac{9}{4(a+b+c)}=\sum_{cyc}\frac{a}{(3-a)^2}-\frac{3}{4}=\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)=$$
$$=\frac{1}{4}\sum_{cyc}\frac{-a^2+10a-9}{(3-a)^2}=\frac{1}{4}\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}=$$
$$=\frac{1}{4}\sum_{cyc}\left(\frac{(a-1)(9-a)}{(3-a)^2}-2(a-1)\right)=\frac{1}{4}\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\geq0.$$
|
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|
Find the volume of the solid rotating about $x=2$ $y=x^3, y=0, x=1$ rotated about $x=2$
Plugging in $x=1$ into $y=x^3$ we get 1 so our bounds are $[0,1]$
In order to account for rotating about the line $x=2$ subtract 2 from the function
$$\int_0^1 \pi [y^{\frac{1}{3}}-2]^2 dy$$
$$\pi \int_0^1 y^{\frac{2}{3}}-4y^{\frac{1}{3}}+4$$
$$\frac{3}{5}y^{\frac{5}{3}}-3y^{\frac{4}{3}}+4y \Big\vert_0^1$$
pluggin all of this in I get:
$\frac{3}{5}+1=\frac{8}{5} \pi$
The answer is supposed to be $\frac{3\pi}{5}$
|
For any given $y$-value, what is the area of the corresponding annulus ("washer") cross section of our solid of rotation? The outer radius is $2 - y^{1/3}$, while the inner radius is $1$. So the area becomes
$$
\pi(2-y^{1/3})^2 - \pi\cdot 1^2 = \pi\left[(2-y^{1/3})^2 - 1\right]
$$
This is what you should integrate. All in all, it's not difficult to see that the additional $-1$ decreases your answer by $\pi$, which means we get the result we should.
Your calculation didn't take this inner radius into account, and thus calculated the volume of the region between $y= x^3$, $y = 0$, $y = 1$ and $x = 2$, rotated about $x = 2$. This has the extra square bounded by $y = 0, y = 1, x = 1, x = 2$ that shouldn't have been included. Note that the solid of rotation of this region is a cylinder with volume $\pi$.
|
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|
Does $\lim\limits_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1}=-1$? The context: p-adic convergence
I first ran into the formula:
$$
S = \lim_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} = -1
$$
in the context of $p-$adics, where we define a new norm, where for $x = 5^a \cdot b$, $5 \not \lvert b$, the norm is defined as $||x|| \equiv 5^{-a}$. That is, $a$ is the largest power of $5$ in the prime factorization of $x$.
Under this definition, we can see that $\lim_{n\rightarrow \infty}||5^{n+1}|| = 0$, and $\lim_{n\rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^n = \frac{4}{1-5} = -1$. The GP formula is legal since $||5|| = 5^{-1} < 1$.
Hence, the total limit is $-1 - 0 = -1$.
Using the same formula under the usual norm
However, let's now decide to use the $\epsilon-\delta$ definition of convergence of a series, under the usual norm $|x| = abs(x)$ (the absolute value). Now, we claim that the limit of the series $S$ is $-1$. That is, we need to show that
$$
A_n \equiv \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} \\
\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, |A_n - (-1)|< \epsilon
$$
Let's evaluate $|A_n - (-1)|$:
\begin{align*}
|A_n - (-1)| &= |A_n + 1| = |4 + 4\cdot5 + 4 \cdot 5^2 + \dots + 4 \cdot 5^n - 5^{n+1} + 1| \\
&= |5 + 4 \cdot 5 + 4 \cdot5^2 + \cdots 4\cdot 5^n - 5^{n+1}| \\
&= |5\cdot 5+ 4\cdot 5^2 + \cdots + 4\cdot 5^n - 5^{n+1}| \\
&= |5^2 + 4\cdot 5^2 + \cdots + 4\cdot 5^n - 5^{n+1}| \\
&= |5\cdot 5^2 + \cdots + 4\cdot 5^n - 5^{n+1}| \\
&= \cdots \\
&= |5\cdot 5^n - 5^{n+1}| \\
&= |0| = 0
\end{align*}
That seems to imply that the series:
$$
S = \lim_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} = -1
$$
under the usual norm! But this makes no sense, the terms keep getting bigger, there is no reason this should converge? What am I missing?
If it does indeed converge, then why do we need the $p-$adic norm to explain this situation?
|
I think you're conflating a couple of things here. Let $S_n$ be the sequence of partial sums of the geometric series, $S_n=\sum_{i=0}^n4\cdot 5^i$. Then it is true for all $n$ that $S_n=5^{n+1}-1$ by the usual formula, so that if we look at the 'associated' sequence $\mathfrak{S}_n=S_n-5^{n+1}$, then $\mathfrak{S}_n$ is $-1$ for all $n$, so $\lim\limits_{n\to\infty}\mathfrak{S}_n = -1$ because it's the limit of a constant sequence.
But what makes the discussion interesting in the context of the $5$-adics isn't that this limit of $\mathfrak{S}_n$ converges; rather, by the explicit formula we have $|S_n-(-1)|_5$ = $|5^{n+1}|_5$ $=5^{-(n+1)}$. And since the norm of the difference between $S_n$ and $-1$ converges to zero as $n\to\infty$, the limit of the $S_n$ β that is, the limit of the sum itself β exists in the $5$-adic topology, and there $\lim\limits_{n\to\infty}S_n=-1$, so we can assign a meaningful value to the sum $\sum_{i=0}^\infty 4\cdot 5^i$.
|
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|
How to deal with negative area when evaluating a definite integral
Find the area bounded by the curves $y=2-x^2$ and $x+y=0$
$$
-x=2-x^2\implies x^2-x-2=(x-2)(x+1)=0
$$
My Attempt
$A_1:$ Area above the x-axis and $A_2:$ Area below the x-axis
$$
A_1=\int_{-1}^\sqrt{2}(2-x^2)dx-\int_{-1}^0(-x)dx=\Big[2x-\frac{x^3}{3}\Big]_{-1}^{\sqrt{2}}-\Big[-\frac{x^2}{2}\Big]_{-1}^0\\
=2\sqrt{2}-\frac{2\sqrt{2}}{3}+2-\frac{1}{3}-(\frac{1}{2})=\frac{4\sqrt{2}}{3}+\frac{7}{6}=\frac{8\sqrt{2}+7}{6}\\
A_2=\Big|\int_0^{{2}}(-x)dx\Big|-\Big|\int_\sqrt{2}^2(2-x^2)\Big|=\Big|\Big[-\frac{x^2}{2}\Big]_{0}^{2}\Big|-\Big|\Big[2x-\frac{x^3}{3}\Big]_{\sqrt{2}}^{2}\Big|\\
=|-2|-|4-\frac{8}{3}-2\sqrt{2}+\frac{2\sqrt{2}}{3}|=2-\Big|(\frac{4-4\sqrt{2}}{3})\Big|=2-(\frac{4\sqrt{2}-4}{3})=\bigg|\frac{10-4\sqrt{2}}{3}\bigg|\\
=\frac{20-8\sqrt{2}}{6}\\
A=A_1+A_2=\frac{8\sqrt{2}+7+20-8\sqrt{2}}{6}=\frac{27}{6}=\frac{9}{2}
$$
Reference
$$
A=\int_{-1}^2(2-x^2+x)dx=\bigg[2x-\frac{x^3}{3}+\frac{x^2}{2}\bigg]_{-1}^2=4-\frac{8}{3}+2+2-\frac{1}{3}-\frac{1}{2}\\
=5-\frac{1}{2}=9/2
$$
Isn't the attempt in my reference factually incorrect ?
yet why am I getting a same solutions in my attempt, ie. after splitting the areas and subtracting absolute values ?
$\color{red}{\text{Another Example}}$
Area bounded by the curve $y=x^3$, x-axis at $x=-2$ and $x=1$
Method 1
$$
A=|\int_{-2}^{0}(x^3)dx|+|\int_0^1x^3dx|=|\Big[\frac{x^4}{4}\Big]_{-2}^0|+|\Big[\frac{x^4}{4}\Big]_{0}^1|=|-4|+|\frac{1}{4}|=4+\frac{1}{4}=17/4
$$
Method 2
$$
A=|\int_{-2}^{1}(x^3)dx|=|\bigg[\frac{x^4}{4}\bigg]_{-2}^1|=|\frac{1}{4}-4|=|\frac{-15}{4}|=\frac{15}{4}
$$
Here I think we are not getting the correct answer in method 2 because the area is counted negative, right ?
|
The simplest way to do it I see is
$$\int_{-1}^2\left(2-x^2+x\right)dx=\left.2x-\frac {x^3}3+\frac{x^2}2\right|_{-1}^2\\
4-\frac {8}3+\frac 42-(-2)-\frac{1}3-\frac 12=\\
=\frac{9}2$$
This is the approach in your reference, but there is a typo in the upper limit of the second integral. Your $A_1$ is trying to get the area above the $x$ axis, but the $-\frac 13$ should be positive. Your $A_2$ has no term involving the $-x$ integral from $\sqrt 2$ to $2$.
|
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|
Steady state probabilities of divergent Markov matrices I have a problem for which the Markov matrix turns out to be the following:
$$P = \begin{pmatrix}
0 & 0.5 & 0 & 0.5\\
0.5 & 0 & 0.5 & 0 \\
0 & 0.5 & 0 & 0.5\\
0.5 & 0 & 0.5 & 0\\
\end{pmatrix}$$
This matrix has eigenvalues $-1$ and $1$. Hence, it does not converge to some matrix $A$ when raised to some power $n$. However, there exists a solution to $Pv = v$. This is basically, the eigenvector corresponding to the eigenvalue of $1$. Is $v$ the steady state probabilites?
|
This Markov chain is periodic with period $2$, meaning that for any initial state $i$, the limiting probability $\lim_{n\to\infty}\mathbb P(X_n=j\mid X_0=i)$ does not exist. However, $\{X_n\}$ is irreducible, positive recurrent, and has finitely many states, so a unique stationary distribution $\pi$ exists. Since $P$ is doubly stochastic (both the rows and the columns sum to one), it follows that $\pi$ is the uniform distribution over $0,1,2,3$, i.e. $\pi_0=\pi_1=\pi_2=\pi_3=\frac14$.
Now, because $\{X_n\}$ is periodic with period $2$, the limits of $P^{2n}$ and $P^{2n+1}$ as $n\to\infty$ exist. In particular,
$$
P^{2n} = \left(
\begin{array}{cccc}
\frac{1}{2} & 0 & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & 0 & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} \\
\end{array}
\right),\quad P^{2n+1} = P
$$
for all $n$, so
$$
\lim_{n\to\infty} P^{2n} = \left(
\begin{array}{cccc}
\frac{1}{2} & 0 & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & 0 & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} \\
\end{array}
\right),\quad \lim_{n\to\infty}P^{2n+1} = P.
$$
|
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|
$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=b=c$ , $a^3 + b^3 + c^3 = a^3 + a^3 +a^3 = 3a^3 = 3aaa = 3abc$ , hence $a^3 + b^3 + c^3 = 3abc$.
But I was wondering if $a^3 + b^3 + c^3 = 3abc$ can be true even when none of the above two relations are true. Please guide me toward a solution.
|
The equation can also be factorized as follows-
$$\begin{align}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
&=\frac 12(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)\\
&=\frac 12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
\end{align}
$$
Hence the equation can only be true when either $a+b+c=0 $ or $a=b=c$
|
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|
Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\sum_{i = 1}^na_1^2 = 1$. Calculate the maximum value of $\sum_{cyc}|a_1 - a_2|$.
Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$ $(n \in \mathbb N, n \ge 3)$. Calculate the maximum value of $$\large |a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1|$$
There must exist $1 < k < n$ such that $a_{k - 1} \le a_k \le a_{k + 1}$
$ \implies |a_{k + 1} - a_k| + |a_k - a_{k + 1}| = |a_{k + 1} - a_{k + 1}|$.
Repeat the above process for about $n - 1$ times and we have that $$|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1| \le 2 \cdot \min(|a_i - a_j|, 1 \le i < j \le n)$$
Now we just have to find the maximum value of $\min(|a_i - a_j|, 1 \le i < j \le n)$ for $$a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$$, which I don't know how.
|
We have that $|x - y| = 2 \cdot \max(x, y) - (x + y)$
$$\implies \sum_{cyc}|a_1 - a_2| = 2 \cdot \left[\sum_{cyc}\max(a_1, a_2) - \sum_{i = 1}^na_1\right]$$
which could be rewritten as $$\sum_{cyc}|a_1 - a_2| = 2 \cdot \sum_{i = 1}^nx_ia_i$$ where $x_i \in \{-1, 0, 1\}, i = \overline{1, n}$ and $\displaystyle\sum_{i = 1}^nx_i = 0$.
In the case of $n$ being odd-numbered, there must exist $m$ $(1 \le m \le n)$ such that $x_m = 0$, otherwise $\displaystyle\sum_{i = 1}^nx_i$ would be odd.
Let $x_n = 0$, we obtain that $$\sum_{cyc}|a_1 - a_2| \le 2 \cdot \sum_{i = 1}^{n - 1}x_ia_i \le 2 \cdot \sum_{i = 1}^{n - 1}|a_i| \le 2\sqrt{(n - 1) \cdot \sum_{i = 1}^{n - 1}a_i^2} = 2\sqrt{n - 1}$$
The equality sign occurs when $a_i = \pm \sqrt{\dfrac{1}{n - 1}}, i = \overline{1, n - 1}$ and $a_n = 0$ such that $\displaystyle\sum_{i = 1}^{n}a_1 = 0$.
The same progress could be done for even-numbered values of $n$.
|
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|
example for calculation of the exterior derivative I don't understand how to calculate exterior derivatives. For the form
$$
\theta = \frac{x\, dy - y\, dx}{x^2 + y^2}
$$
I arrive at the following solution:
$$
\begin{align*}
d\theta
&= d\left(\frac{x}{x^2 + y^2}\right) \wedge dy
+ d\left(\frac{-y}{x^2 + y^2}\right) \wedge dx \\
&= \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy
- \frac{x^2 - y^2}{(x^2 + y^2)^2}dy \wedge dx \\
&= 0.
\end{align*}
$$
I don't understand the second step. Or also in these examples Exterior Derivatives
I don't see how we get the solutions.
I know that $df_p$ is a function from $T_pM$ to $\mathbb{R}$.
But here we don't have a specific point $p$ and I don't think I can use this definition to calculate the exterior derivative of $x/(x^2 + y^2)$. Is there another definition for $df$?
I was also wondering if there are general rules that could help with this kind of calculations (such as $dx \wedge dx = 0$).
|
If you apply the definition you have:
$$
\begin{align*}
d\left(\frac{x}{x^2 + y^2}\right) \wedge dy
&= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy
+ \partial_y\left(\frac{x}{x^2 + y^2}\right) dy \wedge dy \\
&= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy \\
&= \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy
\end{align*}
$$
I'll let you conclude the exercise.
|
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|
Power Series Representation of $f(x) = \frac{x^3}{(2-x)^3}$ I am trying to represent the function, $f(x) = \frac{x^3}{(2-x)^3}$ as a power series and saw this on a different post.
$
{1\over (1-u)} =\sum_{n=0}^{\infty}u^{n}, \qquad|u|<1, \tag1
$
$-{1 \over (1-u)^2} =\sum_{n=1}^{\infty}nu^{n-1}, \qquad|u|<1,\tag2$
${2 \over (1-u)^3} =\sum_{n=2}^{\infty}n(n-1)u^{n-2}, \qquad|u|<1. \tag3$
I understand up to this part. I began with the general form
$f(x) = \frac{1}{(1-x)}$
and derived it twice to get:
$f''(x) = \frac{2}{(1-x)^3}$
which somewhat resembles the function I need to represent as a power series. However, this is as far as I got and keep hitting a dead-end.
1) $f(x)=\frac{x^3}{(2-x)^3} = [x(\frac{1}{2-x})]^3
= [ \frac{x}{2}(\frac{1}{1-x/2})] ^3,
$
2) $\frac{x^3}{2^3} (\frac{1}{1-x/2})^3
= \frac{x^3}{2^3}\sum_{n=2}^{\infty}n(n-1)(x/2)^{n-2},
$
3)Therefore, $f(x) = \frac{x^3}{(2-x)^3}$
represented as a power series is:
$\sum_{n=2}^{\infty}n(n-1)(x/2)^{n+1}, \qquad|x|<2
$
I have no solution sheet for this but I don't know if I am right or wrong. I also have a different answer from what I saw other people post online so if I am wrong, please tell me why.
|
$$f(x)=\frac{x^3}{(2-x)^3}=\frac{x^3}{8}(1-\frac{x}{2})^{-3}=\frac{x^3}{8}[1-{-3 \choose 1}\frac{x}{2}+{-3 \choose 2} (\frac{x}{2})^2 -{-3 \choose 3}(\frac{x}{2})^3$$ $$+{-3 \choose k} (\frac{x}{2})^4 .....] ~~~~(1)$$
Using $${-p \choose k}=\frac{-p(-p-1)(-p-2)....(-p-k+1)}{k!},$$
we find that $${-3 \choose 1}=-3, {-3 \choose 2}=-3.-4/2=6, {-3 \choose 3}=-3.-4.-5/6 =-10, {-3\choose 4} =-3.-4.-5.-6/24 = 15....$$ So on and so forth.
Inserting these co-efficients in (1) we gwt a serie which is valid for $|x|<2$.
When $|x|>1$, then $$f(x)=-(1-\frac{2}{x})^{-3}=-\left [1-{-3\choose 1}(\frac{2}{x})+{-3 \choose 2} (\frac{2}{x})^2-{-3 \choose 3} (\frac{2}{x})^3+..\right] $$
|
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|
Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagrange multiplier method has not gotten me further either.
|
By AM-GM and C-S we obtain:
$$\frac{\sqrt2(a+b)^2}{4}=\frac{\sqrt{2(a+b)}\sqrt{a+b}(a+b)^2}{4ab}\geq(\sqrt{a}+\sqrt{b})\sqrt{a+b}=$$
$$=\sqrt{a^2+ab}+\sqrt{b^2+ab}\geq\sqrt{a^2+\frac{4a^2b^2}{(a+b)^2}}+\sqrt{b^2+\frac{4a^2b^2}{(a+b)^2}}=$$
$$=\sqrt{a^2+4}+\sqrt{b^2+4}.$$
|
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|
Line integral of vector field when Stokes' theorem cannot be applied directly
Define, on $\mathbb{R}^3 \setminus \{(0,0,z) \in \mathbb{R}^3 \ \mid \ z \in \mathbb{R} \}$, the vector fields $$G(x,y,z) = \left(\frac{x}{x^2+y^2}, \frac{2y}{x^2+y^2}, 2 \right) $$ and $$H(x,y,z) = \left(\frac{-y}{x^2+2y^2}, \frac{x}{x^2+2y^2},3 \right). $$ How do we compute the line integrals of $G$ and $H$ over the closed curve at the intersection of the following surfaces: $$x+y+z = 2 \text{ and } z = x^2+y^2. $$
What is an easy way to solve these types of questions? We can see that $\text{curl}(G) = \text{curl}(H) = 0$, so an idea could be to use the theorem of Stokes (although, I don't think that $H$ is conservative). However, what would be a good surface to choose? The vector fields are not defined on the entire of $\mathbb{R}^3$, so the circle that is enclosed by the curve is not a good choice. Also, a parametrization of the curve would lead to integrals that are hard to solve.
|
If you don't want to use Stoke's theorem then integrate around the path itself.
The cylinder x+ y+ z= 2 and the paraboloid $z= x^2+ y^2$ intersect where $x+ y+ (x^2+ y^2)= 2$. Complete the squares in both variables: $x^2+ x+ \frac{1}{4}+ y^2+ y+ \frac{1}{4}= 2+ \frac{1}{4}+ \frac{1}{4}$, $(x+ \frac{1}{2})^2+ (y+ \frac{1}{2})^2=\frac{5}{2}$. That is a circle with center at (1/2, 1/2, z) and radius $\sqrt{\frac{5}{2}}$. Since $z= 2- x- y$, we can use parametric equations $x= \frac{1}{2}+ \sqrt{\frac{5}{2}}cos(\theta)$, $y= \frac{1}{2}+ \sqrt{\frac{5}{2}}sin(\theta)$, $z= 1- \sqrt{\frac{5}{2}}cos(\theta)- \sqrt{\frac{5}{2}}sin(\theta)$.
|
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|
Minimum without computing the derivative. I am trying to find the minimum of the following function:
$$H(x)=\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})},\hspace{1cm}x>0$$
What I usually do to find extrema of a function is computing the derivative of the function and finding its zeros. However, I think there must be a more efficient approach to this problem, since the derivative of this function is quite long. Any tips?
Thanks.
|
Observe that
$$H(x) = \frac{\left(\left(x+\frac{1}{x}\right)^3\right)^2 - \left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3 + \left(x^3+\frac{1}{x^3}\right)} = \left(x+\frac{1}{x}\right)^3 - \left(x^3+\frac{1}{x^3}\right) = 3\left(x+\frac{1}{x} \right) \geq 6$$
(with equality at $x=1$) where in the last step I have use the inequalities $x>0$ and $(x-1)^2 \geq 0$. On the other hand since $lim_{x \rightarrow 0} H(x) = \infty$, the range of $H$ is $[6, \infty)$.
Edit: Typos corrected, thanks to Rob for pointing them out, apologies for the delay.
|
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|
Calculate $\displaystyle\sum_{k=1}^n \frac{k^2}{2^k}$ Calculate $\sum _{k=1}^n\:\frac{k^2}{2^k}$
I got $\sum _{k=1}^n\:\frac{k^2}{2^k} = (1+2+ \dots +k + \dots +n-1+n)(1+\frac{1}{2} + \dots +\frac{1}{2^{k-1}}+ \dots +\frac{1}{2}^{(n-1)}-\frac{n}{2^n})$. Not sure how to simplify further.
|
Firstly,
$$\sum_{k=1}^n\ \frac k{2^k}=\sum_{k=1}^n\ \left(\frac k{2^{k-1}}-\frac {k+1}{2^k}+\frac 1{2^k}\right)=1-\frac {n+1}{2^n}+\sum_{k=1}^n\ \frac 1{2^k}=2-\frac {n+2}{2^n}$$ .
On the other hand,
$$\sum_{k=1}^n\ \frac {k^2}{2^k}=\sum_{k=1}^n\ \left(\frac {k(k+1)}{2^k}-\frac k{2^k}\right)=\sum_{k=1}^n\ \left(\frac {k^2}{2^{k-1}}-\frac {(k+1)^2}{2^k}+\frac {3k+1}{2^k}\right)-\left(2-\frac {n+2}{2^n}\right)=$$
$$=\left(1-\frac {(n+1)^2}{2^n}\right)-\left(2-\frac {n+2}{2^n}\right)+3\sum_{k=1}^n\ \frac k{2^k}+\sum_{k=1}^n\ \frac 1{2^k}=$$
$$=\frac {-n^2-n+1}{2^n}-1+3\cdot\left(2-\frac {n+2}{2^n}\right)+\left(1-\frac 1{2^n}\right) = 6-\frac{n^2+4n+6}{2^n}$$
|
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|
For real $a$, $b$, $c$ in $(0,1)$, if $a+b+c=2$, then $\frac{a}{1-a}\times\frac{b}{1-b}\times\frac{c}{1-c}\geq 8$
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$
Prove that
$$\frac a{1-a}Γ\frac b{1-b}Γ\frac c {1-c}β₯8$$
I tried by substituting $x$ for$(1-a)$ and similarly for others and used AM-GM inequality, but couldn't get to the desired result.
|
$$a+b-c=2-2c>0,$$ which says that $a$, $b$ and $c$ are sides-lengths of a triangle.
Now, let $a=y+z$, $b=x+z$ and $c=x+y.$
Thus, $x$, $y$ and $z$ are positives, $x+y+z=1$ and by AM-GM
$$\prod_{cyc}\frac{a}{1-a}=\prod_{cyc}\frac{y+z}{x}\geq\prod_{cyc}\frac{2\sqrt{yz}}{x}=8.$$
|
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|
How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0}^{1}\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\left[\int\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b\right]\mathrm{d}b
\end{align}
Based on it, the first summand is given by
\begin{align}\label{eq8}
\int_{0}^{1}\frac{b^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \int_{0}^{1}\frac{(1 - 2b + b^{2}) - (1 - 2b)}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\frac{(2-2b) - 1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - 2\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b + \int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
The first of the last three integrals is given by
\begin{align}
\int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \frac{1}{\theta}\int_{0}^{1}\frac{\theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta}\int_{0}^{1}\frac{1 - \theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta} + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
According to the substitution $u = \sqrt{\theta}(1-b)$, it results that
\begin{align}\label{eq10}
\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\sqrt{\theta}}\int_{0}^{\sqrt{\theta}}\frac{1}{1 - u^{2}}\mathrm{d}u = \frac{1}{2\sqrt{\theta}}\ln\left|\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right|
\end{align}
Finally, based on the same substitution, we have
\begin{align}\label{eq11}
\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\theta}\int_{0}^{\sqrt{\theta}}\frac{u}{1-u^{2}}\mathrm{d}u = -\frac{1}{\theta}\ln|1-\theta^{2}|
\end{align}
Combining all these results, we get the first summand from the integration by parts method. The problem arises when I try to determine the second part.
|
Because of the $b^3$ in numerator, I should rather start with
$$\frac{b^{3}}{1 - \theta(1-b)^{2}}=\frac{b+2}{\theta }+\frac{b (3 \theta +1)-2 \theta +2}{\theta \left(\theta(b-1)^2 -1\right)}$$
$$\frac{b^{3}}{1 - \theta(1-b)^{2}}=\frac{b+2}{\theta }+\frac{b (3 \theta +1)-2 \theta +2}{\theta \left(\theta( b-1)^2 -1\right)}=\frac{b+2}{\theta }+\dfrac{\left(3\theta+1\right)\left(b-1\right)}{\theta\left(b-1\right)^2-1}+\dfrac{\theta+3}{\theta\left(b-1\right)^2-1}$$
Now, we face quite simple antiderivatives at the price of very simple substitutions.
|
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|
Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$ I need to find the number of divisors of $N = 3^55^77^9$ that are of the form $4n+1$, $n\geq 0$.
My try:
I noticed that $5$ itself is a number of the form $4n+1$ so all of its power satisfy the required condition, so number for the exponent of $5$ will be $7+1 = 8$.
Also, even powers of $3$ and $7$ satisfy the required condition.
But, I also noticed that odd powers of $3$ and $7$ multiplied together also satisfy the given condition as $(4m+3)(4k+3) = 4q+1$. How would I take that into account?
|
A generating function approach:
Let $$f(x)=\left(1+x+\cdots+x^5\right)\left(1+x+\cdots+x^9\right)$$
Then the divisors of $3^57^9$ which are of the form $4n+1$ are the number of $3^a7^c$ with $0\leq a\leq 5, 0\leq c\leq 9$ and $a+c$ is even. This is the sum of the even coefficients of $f(x)$ which can be written as:
$$\frac{1}{2}(f(1)+f(-1))$$
But $f(-1)=0,$ so the result is $\frac{60}{2}=30.$
Then multiply by $7+1$ to get the result for $3^55^77^9.$
In general, if we count the divisors of $3^m7^n$ of the form $4n+1,$ you have that the number is $\frac{1}2(m+1)(n+1)$ if either $m,n$ is odd. When both are even, the you get $f(-1)=1$ and the number of divisors is $\frac{1}{2}\left[(m+1)(n+1)+1\right].$
You can take the set $D$ of all divisors $3^a5^b7^c$ where $a+c$ is even, and you get a map $\phi:D\to D$ defined as $\phi(3^a5^b7^c)=3^{5-a}5^b7^c$ then $\phi(\phi(d))=d$ and $d$ is of the form $4n+1$ iff $\phi(d)$ is not. So that means exactly one half of $D$ are of the form $4n+1.$
That only works because $5$ is odd.
|
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|
Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and
$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$
$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$
$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
From this we can conclude that
$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
Is this solution correct?
|
No, it is not, since what you got is not a power series (see what you get if you put $n=0$).
Use the fact that\begin{align}\frac1{x^2}&=\frac14+\left(\frac1{x^2}-\frac14\right)\\&=\frac14+\int_4^x-\frac1x\,\mathrm dx\\&=\frac14-\int_4^x\frac1{-2+(x+2)}\,\mathrm dx\end{align}and you will get that the answer is$$\frac1{x^2}=\sum_{n=0}^\infty\frac{(n+1)}{2^{n+2}}(x+2)^n.$$
|
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|
Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) = \dfrac{\sin^2(2x)}{4}$$
$$\sin^2(2x) - 4 \sin(2x) -4 = 0$$
Here we can use the notation $t = \sin(2x)$ with the condition that $t \in [-1,1]$.
$$t^2-4t-4=0$$
Solving this quadratic equation we get the solutions:
$$t_1 = 2+ 2\sqrt{2} \hspace{3cm} t_2 = 2 - 2\sqrt{2}$$
I managed to prove that $t_1 \notin [-1, 1]$ and that $t_2 \in [-1, 1]$. So the only solution is $t_2 = 2 - \sqrt{2}$. So we have:
$$\sin(2x) = 2 - 2\sqrt{2}$$
From this, we get:
$$2x = \arcsin(2-2\sqrt{2}) + 2 k \pi \hspace{3cm} 2x = \pi - \arcsin(2-2\sqrt{2}) + 2 k \pi$$
$$x = \dfrac{1}{2} \arcsin(2-2\sqrt{2}) + k \pi \hspace{3cm} x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin(2 - 2\sqrt{2}) + k \pi$$
Is this solution correct? It's such an ungly answer, that I kind of feel like it can't be right. Did I do something wrong?
|
As your error has been pointed out, I am providing a different way to tackle the problem without introducing extra solutions.
From the given equation, we have $1=(1-\sin x)(1-\cos x)$, which is equivalent to
$$1=\Biggl(1-\cos\left(\frac{\pi}2-x\right)\Biggr)\left(2\sin^2 \frac{x}{2}\right)=4\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin^2\frac{x}{2}$$
That is
$$2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin\frac{x}{2}=\pm 1.\tag{1}$$
This means
$$\cos\left(\frac{\pi}{4}-x\right)-\cos\frac{\pi}{4}=\pm 1.$$
Therefore
$$\cos\left(\frac{\pi}{4}-x\right)=\frac{1\pm\sqrt{2}}{\sqrt{2}}.$$
But $\frac{1+\sqrt2}{\sqrt2}>1$, so
$$\cos\left(\frac{\pi}{4}-x\right)=\frac{1-\sqrt 2}{\sqrt2}.\tag{2}$$
Therefore
$$2n\pi+\left(\frac{\pi}{4}-x\right) = \pm \arccos \frac{1-\sqrt 2}{\sqrt2}$$
for some integer $n$. This gives us
$$x=\left(2n+\frac14\right)\pi \pm \arccos \frac{1-\sqrt 2}{\sqrt2}.\tag{3}$$
In fact there are also complex solutions to $(1)$, and they are given by
$$x=\left(2n+\frac{1}{4}\right)\pi\pm i\operatorname{arccosh} \frac{1+\sqrt 2}{\sqrt2}.\tag{4}$$
Note that $$\operatorname{arccosh} \frac{1+\sqrt 2}{\sqrt2}=\ln\left(\frac{1+\sqrt2+\sqrt{1+2\sqrt2}}{\sqrt2}\right).$$
All real and complex solutions to the original equation are given by $(3)$ and $(4)$.
Note that
$$\frac\pi4 + \arccos \frac{1-\sqrt 2}{\sqrt2}=\frac12\arcsin(2-2\sqrt2)+\pi$$
and
$$\frac\pi4 -\arccos \frac{1-\sqrt 2}{\sqrt2}=\frac{\pi}{2}-\frac12\arcsin(2-2\sqrt2)-\pi.$$
So your solutions only work for odd $k$. Even values of $k$ do not give solutions.
|
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|
Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'+\frac{4}{x}v=\frac{-10}{x^2}$$
this is now a first order linear ODE where:
$$y(x)=\frac{1}{u(x)}\int u(x)q(x)$$
$u(x)=e^{4\int\frac{1}{x}}=x^4$
$q(x) = \frac{-10}{x^2}$
$$\frac{1}{x^4}\int x^4 \frac{-10}{x^2}=\frac{1}{x^4}\frac{-10x^{2+1}}{2+1}+C$$
which leaves us with :
$$\frac{1}{y^2} = \frac{-10}{3x}+x^{-4}C$$
naturally
$$y^2= \frac{1}{\frac{-10}{3x}+x^{-4}C}$$
The book states the answer as being:
$$y^2= \frac{x}{2+Cx^5}$$
|
You made a sign mistake here:
$$\frac{-1}{2}v'+\frac{2}{x}v = \frac{5}{x^2}$$
And also
$$\left (\frac 1 {y^2} \right )'=-2\frac {y'}{y^3}$$
$$\implies v'=-2\frac {y'}{y^3}$$
Another way:
$$x^2dy+ 2xydx - 5y^3dx=0$$
$$d(x^2y) - 5y^3dx=0$$
$$\frac {dx^2y}{(x^2y)^3} - 5\frac {dx}{x^6}=0$$
$$- \frac {1}{2(x^2y)^2} +\frac {1}{x^5}=C$$
|
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|
An exponential equation over positive real numbers $4^x+14^x+3^x=11^x+10^x$ Solve the following equation over the positive reals:
$$4^x+14^x+3^x=11^x+10^x.$$
By inspecting the graph, the solutions must be $x \in \{1,2\}$
I tried using inequalities like $3^x+4^x<5^x$ for $x>2$ but I couldnβt work it forward. Thanks in advance!
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We'll solve this equation for any real value of $x$.
Indeed, let $x>2$ and $x=1+t$.
Thus, $t>1$ and since $f(x)=x^t$ is a convex function, by Jensen we obtain:
$$4^x+3^x+14^x=(3\cdot3^t+8\cdot14^t)+\left(4\cdot4^t+6\cdot14^t\right)\geq$$
$$\geq11\left(\frac{3\cdot3+8\cdot14}{11}\right)^t+10\left(\frac{4\cdot4+6\cdot14}{10}\right)^t=11^{t+1}+10^{t+1}=11^x+10^x.$$
The equality does not occur, which says that our equation has no roots for $x>2$.
Similarly, for $x<1$ let $x=1+t,$ where $t<0$ and since $g(x)=x^t$ is a convex function,
by the same Jensen we obtain that our equation has no roots for $x<1$.
Now, let $1\leq x\leq 2$ and $$h(x)=\left(\frac{3}{11}\right)^x+\left(\frac{4}{11}\right)^x+\left(\frac{14}{11}\right)^x-\left(\frac{10}{11}\right)^x-1$$ and we'll prove that $h$ is a convex function.
Indeed, $$h''(x)=\left(\frac{3}{11}\right)^x\ln^2\frac{11}{3}+\left(\frac{4}{11}\right)^x\ln^2\frac{11}{4}+\left(\frac{14}{11}\right)^x\ln^2\frac{14}{11}-\left(\frac{10}{11}\right)^x\ln^2\frac{11}{10}\geq$$
$$\geq\left(\tfrac{3}{11}\right)^2\ln^2\tfrac{11}{3}+\left(\tfrac{4}{11}\right)^2\ln^2\tfrac{11}{4}+\tfrac{14}{11}\ln^2\tfrac{14}{11}-\tfrac{10}{11}\ln^2\tfrac{11}{10}>\tfrac{10}{11}\ln^2\tfrac{14}{11}-\tfrac{10}{11}\ln^2\tfrac{11}{10}>0,$$
which says that our equation has maximum two roots on $[1,2].$
But $1$ and $2$ are roots and we are done!
|
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|
Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.
I can't use polar coordinates. I decided to used Lagrange multipliers.
Problem: after using Lagrange multipliers I got this system:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a
\end{array}
\right.
$$
And then I add one more equation to the system: $5=(x-4)^2+(y-2)^2$ and $45=(x-4)^2+(y-2)^2$, so I have to find the $a$ that satisfies both system of equations:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
5=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
and
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
45=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
The thing is that I can't resolve those systems, I already tried a lot of ways but I never get a result. I would really appreciate some help.
|
Hint
Let $a=b^2$ where $b\ge0$
WLOG any point on $$x^2+y^2=b^2$$ be $P(b\cos t,b\sin t)$
Now $$(b\cos t-4)^2+(b\sin t-2)^2=20+b^2-4b(2\cos t+\sin t)$$
Again $$-\sqrt{2^2+1^2}\le-(2\cos t+\sin t)\le\sqrt{2^2+1^2}$$
|
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|
Finding the derivative of $N$ with respect to $t$ of $N=500(1-\frac{3}{(t^2+2)^2})$ Finding the derivative of $N$ with respect to $t$ of $N= 500\left(1-\frac{3}{(t^2+2)^2}\right)$
The first step to simplify things is to note that since $\frac{3}{(t^2+2)^2}=3(t^2+2)^{-2}$, we have
$$N= 500\left(1-\frac{3}{(t^2+2)^2}\right) = 500(1-3(t^2+2)^{-2}) $$
Now let's take the derivative of $N$ with respect to $t$:
\begin{align}
\frac{dN}{dt} &= \frac{d}{dt}500(1-3(t^2+2)^{-2}) \\
&= 500\frac{d}{dt}(1-3(t^2+2)^{-2}) \\
&= 500 \left(\frac{d}{dt}(1) - \frac{d}{dt}3(t^2+2)^{-2} \right) \\
&= 500\left(0 - 3\frac{d}{dt}(t^2+2)^{-2}\right) \\
&= 500\left( -3\frac{d}{dt}(t^2+2)^{-2} \right) \\
&= -1500\frac{d}{dt}(t^2+2)^{-2}
\end{align}
Nowe have to do the chain rule to evaluate
\begin{align}
\frac{d}{dt}(t^2+2)^{-2} &= -1500\left( -2(t^2+2)^{-2-1}\frac{d}{dt}(t^2+2) \right) \\
&= 3000(t^2+2)^{-3}(2t+0) \\
&= \frac{6000t}{(t^2+2)^3} \end{align}
Now, if we want to know what the rate of change of $N$ with respect to $t$ is when t is say 2, then we plug $t=2$ into the formula we just derived:
$$= \frac{6000(2)}{(2^2+2)^3} = 55.555555 \approx 55.6$$
|
Your answer is correct, and it is important initially to write out all the steps, just like you have done. Note that the derivative is positive or negative depending upon whether or not $t$ is, so from the derivative (and from the original function as well) it is found that $N$ has a minimum at $0$, where $N = 125$. Furthermore, we conclude that $N$ is increasing or decreasing in a neighbourhood of $t$, if $t$ is positive or negative respectively.
|
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|
If $5\mid\frac {n^{r-1}-1}{n-1}$ and $r $ is even then $r=10k+6$ Let $5\mid\frac {n^{r-1}-1}{n-1}$ and $r $ is even. I'm trying to show $r=10k+6$.
Since
$ \frac {n^{r-1}-1}{n-1}=n^{r-2}+...+n+1$ and I couldn't find any $n $ for which
$5|n^2+n+1$
$5|n^6+n^5+n^4+n^3+n^2+n+1$
$5|n^8+n^7+n^6+n^5+n^4+n^3+n^2+n+1$
$5|n^{10}+n^9+n^8+n^7+n^6+n^5+n^4+n^3+n^2+n+1$
But for $r=6=10\times0+6$ and $n=6$ we see that
$5|6^4+6^3+6^2+6+1=1555$
|
First of all, let's take $a=r-1$, since it is a bit easier to calculate with it.
Define $l$ like mentioned above $n^a-1=5l(n-1)$. Looking this equation with modulo $n$ we will get:
$$-1\equiv -5l \mod(n) \Leftrightarrow n \mid 5l-1$$
From this we get that 5 can not divide n, since otherwise we would have $5\mid 5l-1$, which is not possible. That means from Fermats Little Theorem we have $5|n^4-1$. If $5|n^a-1, \forall n\in \mathbb{N}$, then $a$ should be of the form $a=2t$ (except for $5\mid n-1$), which contradicts it being odd from your form of $a=r-1=10k+6-1=5(2k+1)$.
For $5\mid n-1$ it follows from the LTE lemma that $5\mid a$, but it can not be derived that $a$ has to be odd. Looking up on wolfram alpha also confirms this ex. $n=6, a=5,10,15,...$ works.
Where is this problem from?
If it really should hold that $a=5(2k+1)$, then we have two forms which need to hold and there can not exist a pair $(n,a)$ for which the above statement holds.
|
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|
Rationalizing the denominator of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$
Simplify
$$\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$$
I think this should be expressed without square roots at the denominator. I tried to multiply by conjugate.
|
$$
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}=\\\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2+\sqrt{2+\sqrt{2}}}}=\\\frac{2 \cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}=\\
\frac{2\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}} \cdot \frac{2-\sqrt{2+\sqrt{2}}}{2-\sqrt{2+\sqrt{2}}}=\\\frac{2\cdot(2-\sqrt{2+\sqrt{2}})\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2}}=\\
\frac{2\cdot(2-\sqrt{2+\sqrt{2}})\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2}} \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}}=\\\frac{2\cdot(2-\sqrt{2+\sqrt{2}})\cdot (2+\sqrt{2})\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2}=\\
(2+\sqrt{2}) \cdot (2-\sqrt{2+\sqrt{2}}) \cdot\sqrt{2+\sqrt{2+\sqrt{2}}}
$$
|
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|
Determine the value of $\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}}{\sqrt{2\sqrt {2\sqrt{2...}}}}$ Determine the value for
$$\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}}{\sqrt{2\sqrt {2\sqrt{2...}}}}$$
I think the formula $S_\infty =\frac {a}{1-r}$ should be used for this question but I donβt know how to find the first term,a and the common ratio, r.
Please show how to solve this question in the simplest way possible.
|
Hint:
$$\sqrt{2\sqrt {2\sqrt{2...}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots}$$
and
$$\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}=4^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots}$$
and now you can use the geometric series formula.
|
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|
Problem regarding unique solution of differential equation
A unique solution to the differential equation $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist
then choose the correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 , y_0)$
My attempt : Here $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$
Now i put $x= e^z$ then $z= \log x$
So $y= Dy- D^2y$
$D^2y-Dy -y=0$
so $(D^2-D-1)y=0$
so auxiliary equation will be $m^2-m-1=0$ ,$m= \frac{1 +_{-}\sqrt - 3}{2}$
so $y= e^{\frac{1}{2}x} (c_1 \cos(\frac{\sqrt - 3}{2} ) + c_2\sin ({\frac{\sqrt - 3}{2}} )x)$
After that im not able to proceed further
|
$$(y')^2-xy'=-y$$
Complete the square:
$$(y')^2-xy'+\frac {x^2} 4 =-y+\frac {x^2} 4 $$
$$(y'-\frac x 2 )^2=-y+\dfrac {x^2} 4 $$
The differential equation reduces to:
$$(z')^2=z$$
Where $z=-y+\dfrac {x^2} 4 $
Now you have to analyse what happens when $z <0,z=0,z>0$
|
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|
Consider the linear function$ f : \mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ defined as follows: Consider the linear function $f : \mathbb{R}^{2\times2} β \mathbb{R}^{3\times2}$ defined as follows:
$$\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix} \in \mathbb{R}^{2\times2}\mapsto f\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix} :=\begin{pmatrix} 3 & 2 \\ -2 & 1 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix}$$
Check whether $f$ is injective and/or surjective. If it is bijective, find its inverse function. Finally, find bases for its Kernel and its Range.
Attempted solution:
This matrix multiplication evaluates to:
$$\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\ -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}$$
$f$ cannot be surjective since $m>n$. However, it can be injective if $n=\text{rank}$. Therefore, it is not bijective.
Now I get stuck. I don't understand the technique to find the bases for its kernel or range nor can I calculate its rank. I know we can row reduce this but it isn't in the proper form so I'm unsure of how to proceed.
|
Let's calculate the kernel first. We are searching for all matrices $R=\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}$ satisfying
$$f(R)=\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\ -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix} =0.
$$
So, from the last line of the above matrix we obtain $r_3=r_4=0$, and from the first line we obtain $r_1=r_2=0$. So the only matrix on the kernel is the zero matrix. Therefore, your map is injective.
Now, let's calculate the range.
$$f(R) = \begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\ -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}.
$$
This matrix can be written as:
$$f(R) = r_1\begin{pmatrix} 3 & 0 \\ -2 & 0 \\ 0 & 0 \end{pmatrix}
+r_2\begin{pmatrix} 0 & 3 \\ 0 & -2 \\ 0 & 0 \end{pmatrix}
+r_3\begin{pmatrix} 2 & 0 \\ 1 & 0 \\ 4 & 0 \end{pmatrix}
+r_4\begin{pmatrix} 0 & 2 \\ 0 & 1 \\ 0 & 4 \end{pmatrix}.
$$
Therefore, the range of $f$ is generated by the matrices
$$\begin{pmatrix} 3 & 0 \\ -2 & 0 \\ 0 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 0 & 3 \\ 0 & -2 \\ 0 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 2 & 0 \\ 1 & 0 \\ 4 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 0 & 2 \\ 0 & 1 \\ 0 & 4 \end{pmatrix}.
$$
Furthermore, from the injectivity of $f$ you actually obtain these matrices are linearly independent, and therefore they form a basis for the range.
|
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|
Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n-1}+z^{n-2}+...+z^2+z+1|}$$
Also $|z|=1$, so if I use triangle inequality, I find:
$$|z-1|\ge \frac{2}{|z^{n-1}|+...+|z|+1}=\frac{2}{n}$$
but this is lower than $\frac{2}{n-1}$. Does this help in any way?
|
The $n$ roots of $z^n=-1$ are $n$ equidistant points on the unit circle. $|z-1|$ is simply the distance between one such root and $1$. Therefore, $|z-1|$ is minimized when $z=e^{\frac{\pi i}{n}}$. Therefore, $|z-1|\geq|e^{\frac{\pi i}{n}}-1|=2\sin{\frac{\pi}{2n}}$(Which can be found using geometry or algebra, as one wishes).
It only remains to show that $\sin{\frac{\pi}{2n}}\geq\frac{1}{n-1}$.
Define $f(x)=(x-1)\sin{\frac{\pi}{2x}}-1$. We will prove that $f(x)\geq0$ for $x\geq3$ and that would imply the claim. Notice that $f(3)=0$. If we now prove $f'(x)\geq 0$ for $x\geq3$, we would be done. This follows as $f'(x)=\sin{\frac{\pi}{2x}}-\frac{\pi(x-1)}{2x^2}\cos{\frac{\pi}{2x}}\geq\sin{\frac{\pi}{2x}}-\frac{\pi}{2x}\cos{\frac{\pi}{2x}}=\cos{\frac{\pi}{2x}}(\tan{\frac{\pi}{2x}}-\frac{\pi}{2x})\geq0$. Note the last inequality is true as $\tan{t}\geq t,\ \forall\ t\in[0,\frac{\pi}{2}]$.
|
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|
Is there non real number of x that sufficient for this $11-\sqrt 7 x = 4x - 10$
Is there non real number of x that sufficient for $11-\sqrt 7 x = 4x - 10$ and $|\sqrt 7 x - 11 | = 4x - 10$
There is a solution which is real number.
|
$|anything|$ is always a positive real number.
So $|anything| = 4x - 10$ will only have real solutions (if any solutions at all).
Also for complex numbers, $x$ we tend not to use the notation $\sqrt{x}$ as it is ambiguous.
I'm going to assume you meant $\sqrt{7x}$ (which is the square root of $7x$) and not $\sqrt{7}x$ (which is $x$ times $\sqrt 7$)
$11 - \sqrt{7x} = 4x -10$
$4x + \sqrt 7\sqrt x -21 = 0$
$\sqrt{x} =\frac {-\sqrt 7 \pm \sqrt{7+4*21*4}}8=$
$\frac {-\sqrt 7\pm \sqrt {342}}8$
Note: $\frac {-\sqrt 7- \sqrt {342}}8$ is negative while $\frac {-\sqrt 7+ \sqrt {342}}8$ is positive. But both are real.
Even if we use $\sqrt{x}$ to allow for other than non-negative real numbers and to apply for any $k$ so that $k^2 = x$ (which is not what we do... at least not for real number-- and if we use complex numbers we avoid the notation $\sqrt{}$ altogether: but for the sake of being thourough, I will pretend we can allow $\sqrt x < 0$.... which we really can not do....) we will nave
$x = (\frac {-\sqrt 7\pm \sqrt {342}}8)^2$ which can only be a positive real number.
Now if we have $|11-\sqrt{7x}|\ne 11- \sqrt{7x}$ and as $11-\sqrt{7x}$ is real mus mean $|11-\sqrt{7x}| = \sqrt{7x} - 11 = 4x-10 = 11-\sqrt{7x}$ which is impossible.
So $x = (\frac {-\sqrt 7\pm \sqrt {342}}8)^2$.
BUT as we DON'T define $\sqrt{x}$ as any $k$ so that $k^2 =x$ but only as the non-negative real $k$ so that $k^2 = x$ we have
$x= (\frac {-\sqrt 7+ \sqrt {342}}8)^2$ is the only real solution.
|
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|
Solving a differential-initial value equation: $y'(x) = \frac{y}{x+y^3}$, $ y(2)=2$
The equation and given values are
$$
\begin{split}
\frac{dy}{dx} &= \frac{y}{x+y^3}\\
y(2) &= 2
\end{split}
$$
At first I thought it was a separation of variables problem, but I then was told later on by a tutor that it could be solved with substitution.
I then subbed-in with the following variables:
$$u+x\frac{du}{dx}=\frac{ux}{x+(ux)^3}$$
Then subtracted the $u$ from both sides and multiplied by the denominator:
$$x\frac{du}{dx}=\frac{ux-u(x+(ux)^3)}{x+(ux)^3}$$
$$x\frac{du}{dx}=\frac{ux-ux-u^4x^3}{x+u^3x^3}$$
$$x\frac{du}{dx}=\frac{-u^4x^3}{x+u^3x^3}$$
Now If I factor out an x from the quotient
$$x\frac{du}{dx}=-\frac{x(u^4x^2)}{x(1+u^3x^3)}$$
$$x\frac{du}{dx}=-\frac{u^4x^2}{1+u^3x^3}$$
Now can I further simplify the quotient till I can integrate or should I go back and do something different to make it easier.
|
The differential equation, $$\frac{dy}{dx}=\frac{y}{x+y^3}$$ is not homogenous so the substitution $y=ux$ is not helpful.
If you change your eqaution into $$\frac{dx}{dy}=\frac{x+y^3}{y}$$
then it is linear and you may solve it using integrating factor method.
$$\frac{dx}{dy}=\frac{x+y^3}{y}=(1/y)x+y^2$$
The integration factor is $1/y$ and after multipluing by $1/y$ and integrating you get the solution $$x=\frac {1}{2} y^3 +cy$$
The initial condition of $y(2)=2$ gives you $$x=\frac {1}{2} y^3-y$$
|
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|
Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
3(3)^2+3(3)+1 = ((3)+1)^3-(3)^3\\
3(4)^2+3(4)+1 = ((4)+1)^3-(4)^3\\
\vdots \\
3(n-1)^2+3(n-1)+1 = ((n-1)+1)^3-(n-1)^3\\
3n^2+3n+1 = (n+1)^3-(n-1)^3\\
$$
The result of adding these formulas is
$$
3[1^2+2^2 + ... + (n-1)^2+n^2] + 3[1 + 2 + ... + (n-1)+n] + n = (n+1)^3-1^3
$$
I am able to follow up-to this point easily. I don't understand how this last equation goes from that to this $ n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$ I know thee sum of arithmetic series to n is $n(n-1)/2$ and that replaces the second expression on the left side.
Can some one please show me the algebra step by step?
|
Go upto $(n+1)$ (not $n$)
You have
\begin{eqnarray*}
3 \sum_{i=1}^{n} i^2 + 3 \sum_{i=1}^{n} i + \sum_{i=1}^{n} 1=(n+1)^3-1.
\end{eqnarray*}
Now use
\begin{eqnarray*}
\sum_{i=1}^{n} 1&=&n \\
\sum_{i=1}^{n} i&=& \frac{n(n+1)}{2}. \\
\end{eqnarray*}
|
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|
minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\displaystyle x+\frac{1}{x}=t,$ then $\displaystyle x^3+\frac{1}{x^3}=t^3-3t$
$$f(t)=\frac{(t-1)^3}{t^3-3t}$$
How do i solve it without derivatives
Help me please
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Hint:
$a=x^2$
$b=-x$
$c=1$ $$Min{\frac{(a+b+c)^3}{a^3+b^3+c^3}}=?$$
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|
Determine the point on the plane $4x-2y+z=1$ that is closest to the point $(-2, -1, 5)$ Determine the point on the plane $4x-2y+z=1$ that is closest to the point $(-2, -1, 5)$. This question is from Pauls's Online Math Notes. He starts by defining a distance function:
$z = 1 - 4x + 2y$
$d(x, y) = \sqrt{(x + 2)^2 + (y + 1)^2 + (-4 -4x + 2y)^2}$
However, at this point, to make the calculus simpler he finds the partial derivatives of $d^2$ instead of $d$. Why does this give you the same answer?
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Here is a simple solution using tools from middle school for the computation:
Denote $x,y,z$ the coordinates of the orthogonal projection of the point $(-2,-1,5)$
It satisfies the equations of proportionality:
$$\frac{x+2}4=\frac{y+1}{-2}=\frac{z-5}1$$
This common ratio is also equal to
$$\frac{4(x+2)-2(y+1)+1(z-5)}{4^2+(-2)^2+1^2}=\frac{(4x-2y+z)+1}{21}=\frac{2}{21}$$
whence the solution
\begin{cases}
x=-2+\dfrac 8{21}=-\dfrac{34}{21},\\[1ex]
y=-1-\dfrac 4{21}=-\dfrac{25}{21}, \\[1ex]
z= 5+\dfrac 2{21}=\dfrac{105}{21}.
\end{cases}
With these elements, the distance squared is
$$d^2=(x+2)^2+(y+1)^2+(z-5)^2=\frac{8^2+(-4)^2+2^2}{21^2}=\frac{84}{21^2}=\frac{4}{21}$$
and finally $\;d=\dfrac 2{\sqrt{21}}.$
As to your exact question, the differential of $d^2$ is $2d D(d)$, hence the critical values are obtained at the same points, and as $d>0$, $d^2$ and $d$ both increase or both decrease.
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|
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
$$a+b=0;a-b+c=1; a-c=0 $$
$$a=c=\frac{1}3 \qquad b=-\frac{1}3$$
Then I wrote:
$$\frac{1}3\int \frac{1}{x-1}\,\mathrm dx-\frac{1}3\int\frac{x-1}{x^2+x+1} \, \mathrm dx$$
so the first one is just $\frac{1}{3}\ln|x-1|$. Which makes my calculations already wrong, most likely.
With the second one I tried a few different things ( involving u-substitution mostly) and got stuck.
I know I'm supposed to get this:
$$\frac{1}6\ln \frac{(x-1)^3}{x^2+x+1}+\frac{1}{\sqrt{3}} \arctan\frac{2x+1}{\sqrt{3}}$$
What have I already done wrong? What am I supposed to do?
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Hint:
$$\int\dfrac{x - 1}{x^2 + x + 1}\,\mathrm dx\equiv\int\dfrac{2x + 1}{2(x^2 + x + 1)} - \dfrac{3}{2(x^2 + x + 1)}\,\mathrm dx$$
Then, let $u = x^2 + x + 1\implies\mathrm du = 2x + 1\,\mathrm dx$. So,
$$\int\dfrac{2x + 1}{x^2 + x+ 1}\,\mathrm dx\equiv\int\dfrac1u\,\mathrm du.$$
Notice that $$\int\dfrac1{x^2 + x + 1}\,\mathrm dx\equiv\int\dfrac1{\left(x + \frac12\right)^2 + \frac34}\,\mathrm dx$$
Let $v = \dfrac{2x + 1}{\sqrt3}\implies\mathrm dx=\dfrac{\sqrt3}2\,\mathrm dv$. So,
$$\int\dfrac1{\left(x + \frac12\right)^2 + \frac34}\,\mathrm dx\equiv\dfrac2{\sqrt3}\int\dfrac1{v^2 + 1}\,\mathrm dv.$$
Edit:
$$\begin{align}\left(x + \dfrac12\right)^2 + \dfrac34 = \left(\dfrac{2x + 1}2\right)^2 + \dfrac34 &= \dfrac14\left(3\left(\dfrac{2x + 1}{\sqrt 3}\right)^2 + 3\right) \\ &= \dfrac34\left(\left(\dfrac{2x + 1}{\sqrt 3}\right)^2 + 1\right)\end{align}$$
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|
Minimum three variable expression $\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$ If $a,b,c \ge 0$ are distinct real numbers, find the minimum value of
$$\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$$
What I did: I used $a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2$, but no success. A friend of mine said the minimum happens when one of the variables is $0$, but I don't really understand why or how to prove this.
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Your friend was right!
Indeed, we can assume that $a<b<c$.
Thus, it's enough to prove that
$$(a+b+c)^3(c-a)(c-b)(b-a)\leq bc(c-b)(b+c)^3$$ or
$$a((b+c)(b^3+c^3)+(b+c)(2b^2+bc+2c^2)a-bca^2-2(b+c)a^3-a^4)\geq0,$$ which is obvious.
Also, $$(a^2+b^2+c^2)^3\geq(b^2+c^2)^3$$ and after assuming $c=xb$ it's enough to find $$\min_{x>1}\frac{(x^2+1)^3}{(x+1)^3(x-1)x},$$ The rest is smooth.
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sum of binomial series with alternate terms
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$
what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displaystyle S_{1}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{0}-4\binom{n}{2}+4^2\binom{n+1}{4}+\cdots +(-4)^n\binom{2n-1}{2n}$
let $\displaystyle S_{2}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{-1}-4\binom{n}{1}+4^2\binom{n+1}{3}+\cdots +(-4)^n\binom{2n-1}{2n-1}$
Help me please
did not know to simplify $S_{1}$ and $S_{2}$
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With the following the purpose was to use slightly different functions
from what we saw in the answer by @MarkusScheuer. Start from
$$\sum_{k=0}^n (-4)^k {n+k\choose 2k}
= \sum_{k=0}^n (-4)^k [z^{n-k}] \frac{1}{(1-z)^{2k+1}}
\\ = [z^n] \frac{1}{1-z}
\sum_{k=0}^n (-4)^k \frac{z^k}{(1-z)^{2k}}.$$
Here the coefficient extractor enforces the range and we get
$$[z^n] \frac{1}{1-z}
\sum_{k\ge 0} (-4)^k \frac{z^k}{(1-z)^{2k}}
= [z^n] \frac{1}{1-z} \frac{1}{1+4z/(1-z)^2}
\\ = [z^n] \frac{1-z}{(1-z)^2+4z}
= [z^n] \frac{1-z}{(1+z)^2}
\\ = [z^n] \frac{1}{(1+z)^2} - [z^{n-1}] \frac{1}{(1+z)^2}
\\ = (-1)^n (n+1) - (-1)^{n-1} n = (-1)^n (2n+1).$$
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|
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63}$. Why is this assumption necessary? Question:
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63} $. Why is this assumption necessary?
Proof:
Since $\gcd(a,3)=1$ $\Leftrightarrow a\equiv 1\pmod 3$ $\Leftrightarrow a^7\equiv 1\pmod3\equiv a\pmod3$
Then using Fermat's Little Theorem:
If $a,p\in\mathbb N$ and $p$ is prime then $a^7\equiv a\pmod7$
$\Rightarrow 3 |a^7-a$ and $7 |a^7-a$
$\Leftrightarrow a^7-a=3k_1$ and $a^7-a=7k_2$
$\Rightarrow (a^7-a)^3=63(k_1)^2k_2$ $\Rightarrow (a^7-a)^3\pmod{63}\equiv 0$
Since $x^m\pmod n\equiv x\pmod n$
$\Rightarrow (a^7-a)^3\pmod{63}\equiv a^7-a\pmod{63}\equiv 0$
$\Leftrightarrow a^7\equiv a\pmod{63}$
The thing I am struggling with is that the question says that the only assumption necessary is that $\gcd(a,3)=1$. Surely there are two assumptions neccessary, since to use Fermat's Little Theorem (in this situation) we need $a\neq 0 \pmod7 \space\space\space(\gcd(a,7)=1)$. I am sure there is something obvious I'm missing -would be great if someone could check over what I have done and point out any mistakes :)
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$\gcd(a,7) = 1$ is not necessary. If $\gcd(a,7)=7$ then $a^7\equiv a \equiv 0 \pmod 7$ and no theorem needed.
FLT says if $\gcd(a,7)=1$ then $a^6\equiv 1\pmod 7$ and from that we conclude that $a^7 \equiv a \pmod 7$ ALWAY; trivialy so if $7|a$--- and by FLT if $\gcd(a,7)=1$.
But we need $\gcd(a,3) = 1$.
Note: $3^7 \equiv 45 \pmod {63}$ so that is a failure.
Fermat's Little theorem applies to primes; and $63= 7*3^2$ is not prime. The prime factor of $7$ is to a single power so we can conlude $a^7\equiv a \pmod 7$.
But for the factor of $9=3^2$ we use Euler's theorem to note $a^6 \equiv 1\pmod 9$ if $\gcd(a,3)=1$ and $a^7\equiv a\pmod 9$ if $\gcd(a,3)=1$.
Now if $3|a$ but $9\not \mid a$ we do not have $a^7\equiv a \equiv 0\pmod 9$ trivially.
In fact we have $a^7\equiv 0 \pmod 9$ while $a^7\equiv a\pmod 7$ so
$a^7 \equiv a + 7m\equiv 0 + 9k\pmod {63}$ where $0 \le m < 9; 0\le k< 7$
So if $3\mid a$ we do not have $a^7\equiv a\pmod {63}$ unless $9\mid a$.
======
BTW, If $9|a$ we do have $a^7\equiv a\pmod {63}$ and we do have $a^7\equiv a\pmod{21}$ always.....
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Simplify $\frac{1}{(p+q)^3}\left(\frac{1}{p^3}+\frac{1}{q^3}\right)+...$
Simplify $\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)$ if $p\ne -q, p\ne0$ and $q\ne 0$.
$\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)=\dfrac{p^3+q^3}{p^3q^3(p+q)^3}+\dfrac{3(p^2+q^2)}{p^2q^2(p+q)^4}+\dfrac{6(p+q)}{pq(p+q)^5}$
I guess now it is when I am supposed to calculate the LCD (least common denominator). I have forgotten the algorithm. Can you give me a hint?
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Big hairy guns:
$\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)=$
$\frac {p^3 + q^3}{p^3q^3(p+q)^3} + 3\frac {p^2 + q^2}{p^2q^2(p+q)^4}+6\frac {p+q}{pq(p+q)^5}=$
$\frac {(p+q)^2(p^3+q^3) + 3(p+q)pq(p^2+q^2) + 6p^2q^2(p+q)}{p^3q^3(p+q)^5}=$
$\frac {(p+q)(p^3+q^3) + 3pq(p^2+q^2) + 6p^2q^2}{p^3q^3(p+q)^4}$
Now $(p+q)^4 = p^4 + 4pq^3 + 6p^2q^2 + 4pq^3 + q^4$ and we seem to have each of those coefficients in the numerator: $(p+q)(p^3+q^3) =p^4 + q^4 + (p^3q + pq^3)$ and $3qp(p^2 + q^2) =3pq^3 + 3p^3q$ and $6p^2q^2$ is $6p^2q^2$ so
$\frac {(p+q)(p^3+q^3) + 3pq(p^2+q^2) + 6p^2q^2}{p^3q^3(p+q)^4}=$
$\frac {p^4 + 4p^3q + 6p^2q^2 + 4pq^3 +q^4}{p^3q^3(p+q)^4}=$
$\frac {(p+q)^4}{p^3q^3(p+q)^4}=$
$\frac {1}{p^3q^3}$
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|
Differentiable function $f(x)=4x^7 β14x^4 +30xβ17$ I am trying to prove that the function $f:\Bbb Rβ\Bbb R$, $f(x)=4x^7 β14x^4 +30xβ17$,is injective. To do this I need to prove it is differentiable from first principles. I can then prove its derivative is strictly increasing to show it is injective. Any help on the proof that it's differentiable would be great, especially in what delta to choose.
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The OP is asking for a first-principles derivation of the derivative of a specific polynomial. Here's one way to do it, using the algebraic identity $(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y^2+\cdots+xy^{n-2}+y^{n-1})$.
If $f(x)=4x^7-14x^4+30x-17$, then
$$\begin{align}
{f(x)-f(a)\over x-a}
&={4(x^7-a^7)-14(x^4-a^4)+30(x-a)\over x-a}\\
&={4(x-a)(x^6+x^5a+x^4a^2+x^3a^3+x^2a^4+xa^5+a^6)-14(x-a)(x^3+x^2a+xa^2+a^3)+30(x-a)\over x-a}\\
&=4(x^6+x^5a+x^4a^2+x^3a^3+x^2a^4+xa^5+a^6)-14(x^3+x^2a+xa^2+a^3)+30
\end{align}$$
hence
$$\begin{align}
f'(a)&=\lim_{x\to a}{f(x)-f(a)\over x-a}\\
&=\lim_{x\to a}(4(x^6+x^5a+x^4a^2+x^3a^3+x^2a^4+xa^5+a^6)-14(x^3+x^2a+xa^2+a^3)+30)\\
&=4(a^6+a^5a+a^4a^2+a^3a^3+a^2a^4+aa^5+a^6)-14(a^3+a^2a+aa^2+a^3)+30\\
&=4\cdot7a^6-14\cdot4a^3+30\\
&=28a^6-56a^3+30
\end{align}$$
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Solve the following trigonometric equation for x
Solve $$ \sin^2x \tan x + \cot x \cos^2 x - \sin2x = 1 + \tan x + \cot x $$
I converted the whole equation in $\sin x$ and $\cos x$ and after rearranging a bit I got $$ (\sin x + \cos x)(1-\sin x \cos x) - 2 \sin^2x \cos ^2x = \sin x \cos x +1 $$
I supposed $\sin x+\cos x $ to be equal to $ t$ and hence, $\sin x\cos x $ will be equal to $ \frac{(t^2-1)}2 $.
Substituting the above values, I got the following equation
$$t^4+t^3-t^2-t+2=0$$
I am unable to find roots of this equation and have got no other way to proceed.
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Let for the sake of writing less stuff $c=\cos x$ and $s = \sin x$.
Case 1: $cs \ne0$; we can multiply your equation by $cs$ without losing roots. We get
$$s^4 + c^4 - 2 s^2c^2 = sc + s^2+c^2$$
after that
$$(s^2+c^2)^2 - 4 s^2c^2 = sc + 1$$
or
$$4s^2c^2+sc=0.$$
Since $sc\ne0$, we get $4sc+1=0$, or, in terms of $x$, $ 2\sin 2x + 1=0$, which is easy to solve.
Cases 2: $cs=0$ is impossible, because otherwise either $\tan x$ or $\cot x$ is undefined.
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|
How do I show this series diverges? The series in question is $$\sum_{n=1}^\infty\frac{n^2+1}{2n^2+5}$$
If $a_n$ is the $n$'th term of this sum, then $a_n \rightarrow \frac{1}{2}$ as $n\rightarrow \infty$. I believe this implies that the series doesn't converge - since this is effectively the negation of the statement "If a series converges, then $a_n$ tends to $0$ as $n$ tends to $\infty$". If I prove that the sequence $x_n=\frac{n^2 + 1}{2n^2 +5}$ converges to $\frac{1}{2}$, have I then proven that the series diverges? Or should I apply a convergence test and show that it doesn't pass one?
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Just for your curiosity.
As already said, the problem is over as soon as you showed that the term tends to anything which is not zero.
If fact, we can even approximate the partial sums
$$\frac{n^2+1}{2n^2+5}=\frac{1}{2}-\frac{3}{4 \left( n^2+\frac52\right)}$$
Now, use partial fraction decomposition
$$\frac{1}{ n^2+\frac52}=\frac 1{(n+i\sqrt{\frac 52})(n-i\sqrt{\frac 52})}=\frac 1 {2a}\left(\frac 1{n-a} -\frac 1{n+a} \right)$$
$$\sum_{n=1}^p\frac 1{n+a} = H_{p+a}-H_a\qquad \text{and}\qquad \sum_{n=1}^p\frac 1{n-a} = H_{p-a}-H_{-a}$$ and using the asymptotics
$$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12
q^2}+O\left(\frac{1}{q^3}\right)$$ After some minor simplifications of the complex numbers, you should end with
$$S_p=\sum_{n=1}^p\frac{n^2+1}{2n^2+5}=\frac{p}{2}+\frac{3}{40} \left(2- \sqrt{10}\, \pi \,\coth \left(\sqrt{\frac{5}{2}} \pi
\right)\right)+\frac{3}{4 p}-\frac{3}{8 p^2}+O\left(\frac{1}{p^3}\right)$$ Try it for $p=10$; the exact value is
$$S_{10}=\frac{255436901298463}{57072223853645}\approx 4.47568 $$ while the truncated expansion given above leads to $4.47608$.
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Inductive proof of inequality involving summation and nested exponents What is the best way to approach this problem?
$$\sum_{k=1}^{n} \frac{1}{2^{k^2}} \leq
1-\frac{1}{2^{n^2}}$$
So far I have proved this works for a base case where $n=1$
$$\sum_{k=1}^{1} \frac{1}{2^{k^2}} \leq
1-\frac{1}{2^{1^2}}$$
$$\frac{1}{2} \leq \frac{1}{2}$$
Now I have the first part of the inductive case where $n=m$
$$\sum_{k=1}^{m} \frac{1}{2^{k^2}} \leq
1-\frac{1}{2^{m^2}}$$
This is the point where I am stuck.
How can the expression above be rearranged to prove that
$$\sum_{k=1}^{m+1} \frac{1}{2^{k^2}} \leq
1-\frac{1}{2^{(m+1)^2}}$$
and therefore complete the inductive case?
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You have the inductive case of $n = m$ giving
$$\sum_{k=1}^{m} \frac{1}{2^{k^2}} \leq 1-\frac{1}{2^{m^2}} \tag{1}\label{eq1A}$$
A common method to use induction with inequalities is to go from the inductive case, given above, to the case you're trying to prove, i.e., for $n = m + 1$, by changing the values so one side is true and then showing the other side is still true as well. In this case, the change is simplest to deal with on the left side as it just involves adding $\dfrac{1}{2^{(m+1)^2}}$. Thus, adding this value to both sides gives
$$\begin{equation}\begin{aligned}
\sum_{k=1}^{m+1} \frac{1}{2^{k^2}} & \leq 1 - \frac{1}{2^{m^2}} + \frac{1}{2^{(m+1)^2}} \\
& = 1 - \frac{1}{2^{(m+1)^2}} + \left(\frac{1}{2^{(m+1)^2}} - \frac{1}{2^{m^2}} + \frac{1}{2^{(m+1)^2}}\right) \\
& = 1 - \frac{1}{2^{(m+1)^2}} + \left(\frac{2}{2^{(m+1)^2}} - \frac{1}{2^{m^2}}\right) \\
& = 1 - \frac{1}{2^{(m+1)^2}} + \left(\frac{1}{2^{(m+1)^2 - 1}} - \frac{1}{2^{m^2}}\right) \\
& \lt 1 - \frac{1}{2^{(m+1)^2}}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Going to the last line can be done since $(m+1)^2 - 1 = m^2 + 2m \gt m^2$, so you have
$$\frac{1}{2^{(m+1)^2 - 1}} \lt \frac{1}{2^{m^2}} \implies \frac{1}{2^{(m+1)^2 - 1}} - \frac{1}{2^{m^2}} \lt 0 \tag{3}\label{eq3A}$$
This shows the problem you're trying to solve for is true for $n = m + 1$, thus completing the inductive process. You can now conclude that
$$\sum_{k=1}^{n} \frac{1}{2^{k^2}} \leq 1-\frac{1}{2^{n^2}} \tag{4}\label{eq4A}$$
is true for all integers $n \ge 1$.
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Find sum of geometric-like series with binomial coefficients using complex analysis Studying analytic number theory, I stumbled across the problem of finding the sum of the series
$\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{1}{5}\right)^n$
A professor gave me the hint of "using basic complex analysis" but honestly, I haven't been able to get anywhere. Appreciate any help/comments :)
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Following the hints we introduce
$${2n\choose n} =
\frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{(1+z)^{2n}}{z^{n+1}} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z}
\sum_{n\ge 0} \frac{(1+z)^{2n}}{5^n\times z^{n}} \; dz.$$
We must now determine $\varepsilon$ for the geometric series to converge.
We need
$$|(1+z)^2| \lt 5 |z|.$$
Now $$|(1+z)^2| \le (1+\varepsilon)^2$$ so we have an admissible
$\varepsilon$ if
$$(1+\varepsilon)^2 \lt 5\varepsilon.$$
which is
$$1-3\varepsilon+\varepsilon^2 \lt 0.$$
The roots are
$$\rho_{0,1} = \frac{3\mp\sqrt{5}}{2}.$$
($\rho_0$ is the smaller of the two)
Then $(\varepsilon-\rho_0) (\varepsilon-\rho_1) \lt 0$ if
$$\rho_0 \lt \varepsilon \lt \rho_1.$$
so we have convergence in an annulus delimited by two circles of radius
$\rho_0 \lt \rho_1.$
We sum the series and get
$$-5 \times \frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z^2-3z+1} \; dz.$$
The only pole inside the circle is the one at $\rho_0$ and we find
$$-5 \times \mathrm{Res}_{z=\rho_0} \frac{1}{z^2-3z+1}
= -5 \frac{1}{2\rho_0-3} = -5 \frac{1}{3-\sqrt{5}-3} = \sqrt{5}.$$
This is our answer.
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Prove that if $a,b,c > 0$ and $a + b + c = 1$, we have: $\frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$ Prove that if $a,b,c > 0$ such that $a + b + c = 1$, then the following inequality holds:
$$S = \frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$$
What I have tried so far is the following:
Firstly I rewrote $S$ as:
$$ S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5\left[ \frac{1}{a(a^3 + 5)} + \frac{1}{b(b^3 + 5)} + \frac{1}{c(c^3 + 5)}\right]$$
Then, in order to upper bound $S$, I used the inequality: $ \frac{x^2}{u} + \frac{y^2}{v} + \frac{z^2}{w} \geq \frac{(x + y + z)^2}{u + v + w} $ for any $u,v,w > 0$.
Therefore, I got:
$$ S \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5 \cdot \frac{(1 + 1 + 1)^2}{a^4 + b^4 + c^4 + 5(a + b + c)} = \frac{45}{a^4 + b^4 + c^4 + 5}$$
Then, given what we want to show about $S$, this would reduce to proving that:
$$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{185 + a^4 + b^4 + c^4}{4(a^4 + b^4 + c^4 + 5)} $$
At which point I got stuck and I am not sure whether I started the right way.
I would be grateful for any suggestions. Many thanks!
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Because by AM-GM we have:
$$\frac{a^2}{a^3+5}\leq\frac{a}{4}$$
$$a^3+2\cdot2.5\geq3\sqrt[3]{a^3\cdot2.5^2}>4a.$$
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|
$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$$
It is given that $x,y,z>1$. Which properties of the logarithm have to be used?
I know that $\log_a b=\log a/\log b\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log ((x^2+4)/(4\sqrt{yz}))/\log xy$
and $\log_a b/c=\log_a b/\log_a c\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log_{yz} (x^2+4)/\log_{yz} (4\sqrt{yz})$
And how to solve this type of equation in general?
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Hint : First use: $a^2 + 4 \ge 4a$ for each term on the left. Then split the log and show next that the left is $\ge 0$. Equality is at $ x = y = z = 2$.
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At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one even number.
I tried to prove this question by contradiction, assuming that each element is odd. There exist the positive integers $k_1, k_2, ..., k_{n+1}$ such that
$$2k_1-1<2^n\sqrt{2}<2k_1$$
$$2k_2-1<2^{n+1}\sqrt{2}<2k_2$$
$$...$$
$$2k_{n+1}-1<2^{2n}\sqrt{2}<2k_{n+1}$$
But i can't find a contradiction among these inequalities.
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Let me continue TonyK's comment. If $\sqrt{2}=1.b_1b_2\ldots b_{n-1}11\ldots11b_{2n+1}\ldots$, then
$$
\sqrt{2}=\frac{k}{2^{n-1}}+\left(\frac{1}{2^n}+\ldots+\frac{1}{2^{2n}}\right)+r,
$$
where $k=(\overline{1b_1b_2\ldots b_{n-1}})_2$ is a positive integer and $r\in(0,\frac{1}{2^{2n}})$ (since $\sqrt{2}$ is not rational number of the form $p/2^{q}$). Hence,
$$
2^{n+1}(k+1)-1<2^{2n}\sqrt{2}<2^{n+1}(k+1).
$$
Denote $m=2^{n+1}(k+1)$, then the last inequality can be rewritten as follows
$$
(m-1)^2<2^{4n+1}<m^2.
$$
However, $m^2$ and $2^{4n+1}$ are divisible by $2^{2n+2}$, so
$$
(m-1)^2\leq 2^{4n+1}\leq m^2-2^{2n+2}.
$$
Thus, $2m-1\geq 2^{2n+2}$, so $m>2^{2n+1}$. Recall that $m=2^{n+1}(k+1)$, so the last inequality means that $k\geq 2^{n}$. But it's impossible because $k=(\overline{1b_1b_2\ldots b_{n-1}})_2<2^n$.
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|
Probability problem; if sum and product are not divisible by number Out of set of numbers {10,11,...,100} three are being chosen randomly. Find these probabilities:
*
*At least one out of three are divisible by 6
*Exactly two are not divisible by 4
*Their sum is not divisible by 4
*Their product is not divisible by 3
So to choose three numbers out of 91 I can do in $\binom{91}{3}$ ways. First probability I solved this way:
$\frac{\binom{15}{1}\binom{76}{2}+\binom{15}{2}\binom{76}{1}+\binom{15}{3}}{\binom{91}{3}}$
Since there are 15 numbers divisible by 6 i have 3 options that 1 is divisible two are not, 2 are divisible 1 is not and all of 3 are divisible.
Second probability:
$\frac{\binom{69}{2}22}{\binom{91}{3}}$
Since there are 22 numbers that are divisible by 4 i subtracted 22 from 91 and got numbers that are not divisible by 4 from which I chose 2 numbers and the third I chose from those 22.
I do not know how to solve third and fourth and also I am not sure if the first two probabilities are correct.
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Well, I took the time to work out the hard part. Here is a table:
$$\begin{array}{cccc|c|c}0&1&2&3&\text{Number}&\text{Total}\\ \hline
23&22&23&23&&\\ \hline
3&0&0&0&\binom{23}{3}&1771\\
1&1&0&1&\binom{23}{1}\binom{22}{1}\binom{23}{1}&11638\\
1&0&2&0&\binom{23}{1}\binom{23}{2}&5819\\
0&2&1&0&\binom{22}{2}\binom{23}{1}&5313\\
0&0&1&2&\binom{23}{1}\binom{23}{2}&5819
\end{array}$$
The fist $4$ columns represent the congruence classes $\pmod4$ and we see how many from each class is in the set to form a number divisible by $4$. The fifth column counts the number of ways such a selection may be made and the sixth column is the numerical result. Adding up the results in the fifth column we see that there are $30360$ ways to sum to $4$, with probability
$$\frac{30360}{\binom{91}{3}}=0.249907$$
But we want the probability that the sum is not divisible by $4$ which is
$$1-0.249907=0.750093$$
In the third problem, since there are $30$ numbers in range divisible by $3$ we need to select $3$ out of the $61$ that are for a probability of
$$\frac{\binom{61}{3}}{\binom{91}{3}}=0.296251$$
Since there are $15$ numbers in range that are divisible by $6$, we get
$$1-\frac{\binom{76}{3}}{\binom{91}{3}}=1-0.578672=0.421328$$
for the first problem. In the second problem, we need $1$ divisible by $4$ and $2$ not, so we get
$$\frac{\binom{23}{1}\binom{68}{2}}{\binom{91}{3}}=0.43128$$
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|
Prove using induction that $1^2 + 3^2 + 5^2 + . . . + (2n + 1)^2 = (n + 1) β (2n + 1) β (2n + 3)/3$ My question reads as follows:
Prove using induction that
$$1^2 + 3^2 + 5^2 + . . . + (2n + 1)^2 = (n + 1) \times (2n + 1) \times (2n + 3)/3.$$
Firstly, I must prove the base case. For $n=0$, my LHS $= 1$ and my RHS $= 1$. Perfect, I've proven the base case.
I must now assume that $n=k$ for some arbitrary $ k\geq 0$.
Secondly, I must assume the induction hypothesis which states that
$S(k) = (k+1)(2k+1)(2k+3)/3$.
We want to prove that for some arbitrary $k$, the predicate of $k+1$ also holds true.
I get to this point and everything is fine until I am unable to return an answer that matches my presumed $S(k+1)$. My answer must be done in this format in order to get full marks.
So far what I have is the following:
In order to get the $(k+1)$th term, we must add the $(k+1)$-th term to the series. Therefore it looks like the following:
$$1^2 + 3^2 + 5^2 + ... + (2k+1)^2 + \mathbf{(2k+3)^2}$$
By substituting in $S(k)$ we get the following:
$$(k+1)(2k+1)(2k+3)/3 + (2k+3)^2$$
It is here that I encounter an issue. No matter which simplification techniques I use, after this step I can never return an answer that is $(k+2)(2k+3)(2k+5)/3$ so I am stuck. And unfortunately, I am not allowed to then assume we can rewrite the series in terms of $(2k-1)^2$ which would make things too easy and I could easily get back the right-hand equivalent to that. In desperate need of help. Thanks.
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$\frac {(k+1)(2k+1)(2k+3)}3 + (2k+3)^2 = \frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 \iff$
$\frac {(k+1)(2k+1)(2k+3)+ 3(2k+3)^2}3= \frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 \iff$
$(k+1)(2k+1)(2k+3)+ 3(2k+3)^2 = (k+2)(2(k+1)+1)(2(k+1)+3)\iff$
$(2k+3)[(k+1)(2k+1) + 3(2k+3)] = (k+2)(2k+3)(2k+5)\iff$
$[(k+1)(2k+1) + 3(2k+3)]=(k+2)(2k+5)$ or $2k+3=0$ which as $k$ is natural never happens$\iff$
$[(2k^2 + 3k + 1) + (6k + 9)= 2k^2 +9k + 10\iff$
$3k^2 + 9k +10 = 2k^3 + 9k + 10$
Which it does
.....
Or just do it:
$\frac {(k+1)(2k+1)(2k+3)}3 + (2k+3)^2=$
$\frac {(k+1)(2k+1)(2k+3)+ 3(2k+3)^2}3=$
$\frac{(2k^2 + 3k + 1)(2k+3) +3(4k^2 + 12k+9)}3=$
$\frac {4k^3 + 24k^2 + 47k + 30}3$.
While $\frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 =$
$\frac {(k+2)(2k+3)(2k+5)}3 =$
$\frac {(2k^2 + 7k + 6)(2k+5)}3=$
$\frac {4k^3 + 24k^2 + 47k + 30}3$.
Those are both equal.
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Differentiate $((x^{2}-1)f_{n-1}(x))^{(n)}$.
Differentiate $((x^{2}-1)f_{n-1}(x))^{(n)}$.
Using Leibniz rule, I obtain the following:
$\frac{x^{2}-1}{2^{n}n!}f_{n-1}^{(n)}(x)+\frac{x}{2^{n-1}(n-1)!}f_{n-1}^{(n-1)}(x)+\frac{1}{2^{n}(n-2)!}f_{n-1}^{(n-2)}(x)$
I am trying to show for Legendre polynomial $\phi_{n}(x)$ of degree $n$ that $x\phi_{n-1}(x)+\frac{x^{2}-1}{n}\phi_{n-1}'(x)=\phi_{n}(x)$, and the first two terms of the derivative above give me this, but how do I show the last term is zero?
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In order to prove that
$$\frac{x^2-1}{n}P_n' = xP_n - P_{n-1}$$
a different route should be followed.
Two other formulas will be used in the derivation below:
\begin{equation}\label{1}
(n+1)P_{n+1} = (2n+1)xP_n-nP_{n-1}
\end{equation}
and
\begin{equation}\label{2}
P_{n+1}'-P_{n-1}' = (2n+1)P_n
\end{equation}
Here we go:
\begin{equation}
\begin{split}
\frac{x^2-1}{n}P_n' &= \frac{x^2-1}{n(2n+1)} \left( P_{n+1}'-P_{n-1}' \right)'\\
&= \frac{x^2-1}{n(2n+1)} \left( P_{n+1}''-P_{n-1}'' \right)\\
&= \frac{1}{n(2n+1)} \left[ (n+1)(n+2)P_{n+1}-2xP_{n+1}' -(n-1)nP_{n-1}+2xP_{n-1}' \right]\\
&= \frac{1}{n(2n+1)} \left[ (n+1)(n+2)P_{n+1} -(n-1)nP_{n-1}-2x(P_{n+1}'-P_{n-1}') \right]\\
&= \frac{1}{n(2n+1)} \left[ (n+2)[(2n+1)xP_n-nP_{n-1}] -n(n-1)P_{n-1}-2x(2n+1)P_n \right]\\
&= \frac{1}{n(2n+1)} \left[ n(2n+1)xP_n-n(2n+1)P_{n-1} \right]\\
&= xP_n-P_{n-1}\\
\end{split}
\end{equation}
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For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
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The trivial solution $n=0$ is seen by inspection.
For $n \ne 0$, $y^2=n^6+n^4+1 \Rightarrow (y+1)(y-1)=n^6+n^4$.
Now $(y+1)-(y-1)=2 \wedge \gcd{(y+1),(y-1)}=1,2$ so we must resolve $n^6+n^4$ into two factors that differ by $2$ and have at most one factor of $2$ in common.
The factors $n^4,(n^2+1)$ are relatively prime, but $n^4-(n^2+1)=2$ has no solutions in the integers.
The factors $n^2,n^2(n^2+1)$ have $n^2$ as their $\gcd$, which is permissible only if $n=1$, but $n=1$ does not solve the original equation.
The factors $n^3,n(n^2+1)$ have $n$ as their $\gcd$, which is permissible if $n=1$ which we have already ruled out, or $n=2$, which solves the original equation. Noting that the original equation has $n$ in even powers, the solution $-2$ is also admitted because $(-2)^{2k}=2^{2k}$.
For completeness, we can look at $n=cd$ to see whether factoring $n^6+n^4$ might be done in any other way. However, $d=1$ changes nothing, and $d>2$ is not permitted, so we are constrained to examine $n=2c$. Also, that factor of $2$ can occur at most once in one of the factors of $n^6+n^4$, so we look at the factorizations $8c^4,2(4c^2+1)$ and $2c^4, 8(4c^2+1)$
After dividing through by $4$, $8c^4-2(4c^2+1)=2 \Rightarrow 2c^4-2c^2=1$ which is impossible because the difference of two even numbers is never $1$.
After dividing through by $2$, $2c^4- 8(4c^2+1)=2 \Rightarrow c^4-16c^2-5=0$. If we treat this as a quadratic equation in $c^2$, we derive $c^2=8\pm \sqrt{69}$ which is not an integer. So alternative factorings of $n$ do not change the outcome.
Solutions are $\{0,\pm2\}$
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For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:
Let $a,b,c$ be non-negative real numbers. Prove that
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
Do you have any hints to solve this inequality? Any hints would be fine.
I tried this:
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
By Cauchy's inequality, $$a^2+1\ge(2a)$$ and I did the same to $b$ and $c$ and applied it to the problem but the results are $$2abc\ge1+a^2b^2c^2$$ and this is wrong.
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First, note that
$$2(1-a+a^2)(1-b+b^2)=1+a^2b^2+(a-b)^2+(1-a)^2(1-b)^2 \geqslant 1+a^2b^2$$
Hence it is enough to show that
$$3(1+a^2b^2)(1-c+c^2) \geqslant 2(1+abc+a^2b^2c^2)$$
Considering the above as a quadratic in $c$, it is clear when $c=0$ this is true, so we just need to confirm the discriminant doesn't get positive, i.e.
$$\Delta = (3+2ab+3a^2b^2)^2-4\cdot (3+a^2b^2)\cdot (1+3a^2b^2) = -3(1-ab)^4 \leqslant 0$$
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Calculus - limit calculate Calculate the limit not using L'Hospitale rule:
$\lim\limits _{x \to 2} \frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$
I tried multiplying for both conjugates but I got nothing. I think I have to make something with $\sqrt{x^3-4}\cdot \sqrt{x^2+12}$, but I don't know what to do.
|
\begin{align}
\lim_{x\to2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}&=\lim_{x\to2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}\color{blue}{\cdot\frac{\sqrt{x^2+12}+4}{\sqrt{x^2+12}+4}\cdot\frac{2+\sqrt{x^3-4}}{2+\sqrt{x^3-4}}}\\
&=\lim_{x\to2}\frac{2+\sqrt{x^3-4}}{\sqrt{x^2+12}+4}\cdot\frac{(x^2+12)-(16)}{(4)-(x^3-4)}\\
&=\frac12\cdot\lim_{x\to2}\frac{x^2-4}{8-x^3}
\end{align}
Can you take it from here?
|
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|
The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this integral see my Answer to: Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$
|
Let $u= \cos x$, $du = - \sin x \ dx$:
$$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
$$= -\int_{1}^{0} \frac{du \ (u^5)}{(1-2(1-u^2)u^2)^2}$$
$$= \int_{0}^{1} \frac{u^5}{(1-2u^2+2u^4)^2} \ du$$
Then substitute again: let $v = u^2, dv = 2u\ du$:
$$= \frac{1}{2} \int_{0}^{1} \frac{v^2 \ dv}{(1-2v+2v^2)^2}$$
and do (not really) partial fractions:
$$ \frac{v^2 }{(1-2v+2v^2)^2} = \frac{a}{(1-2v+2v^2)} + \frac{b}{(1-2v+2v^2)^2}$$
$$ v^2= a(1-2v+2v^2) + b$$
$$a = \frac{1}{2} \Rightarrow v^2 = \frac{1}{2}-v+v^2+b$$
$$b = v - \frac{1}{2}$$
So we have:
$$\frac{1}{4} \int_{0}^{1} \frac{dv}{(1-2v+2v^2)} + \frac{1}{2} \int_{0}^{1} \frac{v \ dv}{(1-2v+2v^2)^2} - \frac{1}{2} \int_{0}^{1} \frac{\ dv}{(1-2v+2v^2)^2}$$
For the first integral, complete the square which resolves to the standard $\arctan$ integral. For the second integral, let $w = 2v - 1$ (AoPS), and for the third integral, complete the square, then substitute $w = \frac{\tan v}{\sqrt 2}$ where you can use $\tan^2 w + 1 = \sec^2 w$.
|
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|
The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$
takes all real values for $x\in R$ are:
My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My attempt is as follows:-
$$y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$$
$$ya+3yx-4yx^2=ax^2+3x-4$$
$$x^2(-4y-a)+x(3y-3)+ya+4=0$$
As $x$ can be any real, so $D\ge0$
$$9y^2+9-18y-4(ya+4)(-4y-a)\ge0$$
$$9y^2+9-18y+4(4y^2a+ya^2+16y+4a)\ge0$$
$$y^2(9+16a)+y(4a^2+64-18)+9+16a\ge0$$
$$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$
As range is $R$, so discriminant of quadratic in $y$ should be less than equal to zero
$$4(2a^2+23)^2-4(9+16a)^2\le0$$
$$(2a^2+23-9-16a)(2a^2+23+9+16a)\le0$$
$$(2a^2-16a+14)(2a^2+16a+32)\le0$$
$$(a^2-8a+7)(a^2+8a+16)\le0$$
$$a\in[1,7]$$
But in such type of questions, we always check at endpoints like here we need to check at $a=1$ and $a=7$. But I don't understand what is so special about endpoints.
From the above calculation I can only say at $a=1,7$ discriminant of quadratic in $y$ is zero, but what is so special about this. Please help me in this.
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The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$
$\dfrac{ax^2+3x-4}{a+3x-4x^2}$ will take all the values for $x \in \mathbb R$ only if the denominator is not equal to zero!
So, For $a + 3x - 4x^2 \neq 0; \iff -3 \pm \sqrt{9 +16a} \neq 0$
We need to check all the points for which the denominator will become equal to zero, and we need to remove those values of $a$ which includes $1, 7$.
|
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"url": "https://math.stackexchange.com/questions/3585702",
"timestamp": "2023-03-29T00:00:00",
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|
$\arctan{x}+\arctan{y}$ from integration I was trying to derive the property
$$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$
for $x,y>0$ and $xy<1$ from the integral representation
$$
\arctan{x}=\int_0^x\frac{dt}{1+t^2}\,.
$$
I am aware of "more trigonometric" proofs, for instance using that $\tan{(\alpha+\beta)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$, but I was willing to see if there is a proof that uses more directly the properties of the integral representation. For instance, if $x>0$, one immediately gets
$$\begin{aligned}
\arctan{x}+\arctan\frac{1}{x}
&=\int_0^x\frac{dt}{1+t^2} + \int_0^{\frac{1}{x}}\frac{dt}{1+t^2}\\
&=
\int_0^x\frac{dt}{1+t^2}+\int_x^\infty\frac{dt}{1+t^2}\\
&=\int_0^\infty \frac{dt}{1+t^2} = \frac{\pi}{2}
\end{aligned}$$
sending $t\to\frac{1}{t}$ in the second integral.
Similarly I tried considering
$$
\int_0^x\frac{dt}{1+t^2} + \int_0^y\frac{dt}{1+t^2}=(x+y)\int_0^1\frac{1+xyt^2}{1+(x^2+y^2)t^2+x^2y^2t^4}\ dt
$$
after rescaling $t\to xt$ and $t\to yt$. On the other hand, via a similar rescaling $t\to \frac{x+y}{1-xy}t$, we have
$$
\int_0^\frac{x+y}{1-xy}\frac{dt}{1+t^2}
=
(x+y)\int_0^1\frac{1-xy}{(1-xy)^2+(x+y)^2t^2}\ dt\,.
$$
By a clever choice of variable it should (must?) be possible to see that these integrals are actually the same, but I can't figure it out...
|
We want show that
\begin{eqnarray*}
\int_x^{ \frac{x+y}{1-xy}} \frac{dt}{1+t^2} = \int_{0}^{y} \frac{du}{1+u^2}
\end{eqnarray*}
that's to say the LHS is actually independent of $x$.
The substitution
\begin{eqnarray*}
t=x+ \frac{u(1+x^2)}{1-ux}
\end{eqnarray*}
will do the trick.
The limits are easily checked and we have
\begin{eqnarray*}
dt= \frac{1+x^2}{(1-ux)^2} du.
\end{eqnarray*}
The rest is a little bit of algebra.
Note the similarity with $ \ln(a)+\ln(b) = \ln(ab)$
\begin{eqnarray*}
\int_{1}^{a} \frac{dt}{t} +\int_{1}^{b} \frac{dt}{t} = \int_{1}^{ab} \frac{dt}{t}.
\end{eqnarray*}
And $ u=at $
\begin{eqnarray*}
\int_{1}^{b} \frac{dt}{t} = \int_{a}^{ab} \frac{du}{u}.
\end{eqnarray*}
|
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|
Inequality for Olympiad students Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that
$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}} \ge 6$
My attempts you can see here: https://scontent-xsp1-2.xx.fbcdn.net/v/t1.0-9/90231854_2782055341915789_3356982430379540480_n.jpg?_nc_cat=101&_nc_sid=8024bb&_nc_ohc=pSSNCiYYb8kAX8lk9Cx&_nc_ht=scontent-xsp1-2.xx&oh=7f6cc512431574e8576250883ef88325&oe=5E9E41A9
|
Another way (L.Hadassy, Y.Ilany).
If $a\geq b\geq c$ we have $$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3b+\frac{1}{b}}$$ because it's $$(a-b)(b-c)\geq0.$$
Thus,
$$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}.$$
If $a\geq c\geq b$ we have $$\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}\geq\sqrt{3b+\frac{1}{a}}+\sqrt{3c+\frac{1}{c}}$$ because it's $$(a-c)(c-b)\geq0,$$ which gives
$$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{a}}+\sqrt{3c+\frac{1}{c}}.$$
Now we see that in any case we need to prove that:
$$\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}\geq6$$ or
$$\sqrt{3ac+1}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)+\sqrt{3b+\frac{1}{b}}\geq6,$$
where $a\geq b\geq c$ or $c\geq b\geq a$.
Now, let $a+c=p=constant,$ $f(a,c)=\sqrt{3ac+1}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)$ and $F(a,c,\lambda)=f(a,c)+\lambda(a+c-p).$
Thus, in the minimum point we have $$\frac{\partial F}{\partial a}=\frac{\partial F}{\partial c}=0,$$ which gives
$$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial c}$$ or
$$(\sqrt{a}-\sqrt{c})(3\sqrt{a^3c^3}-\sqrt{ac}-a-c)=0.$$
1) $a=c$.
Thus, $b=3-2a,$ where $0<a<\frac{3}{2}$ and we need to prove that
$$2\sqrt{\frac{3a^2+1}{a}}+\sqrt{9-6a+\frac{1}{3-2a}}\geq6,$$ which is true.
2) $3\sqrt{a^3c^3}-\sqrt{ac}-a-c=0.$
Let $\sqrt{ac}=x$.
Thus, $a+c=3x^3-x,$ $b=3+x-3x^3$ and since $0<b\leq\frac{3}{2},$ we obtain: $0.932...<x\leq1.11...$ and we need to prove that:
$$\sqrt{3x^2+1}\cdot\sqrt{\frac{a+c+2\sqrt{ac}}{ac}}+\sqrt{3(3+x-3x^3)+\frac{1}{3+x-3x^3}}\geq6$$ or
$$\frac{3x^2+1}{\sqrt{x}}+\sqrt{3(3+x-3x^3)+\frac{1}{3+x-3x^3}}\geq6,$$ which is true and it ends the proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solutions of the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ I have seen the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ in many places and the answer is $x=2$ which is obtained by making the substitution for $x$ inside the radical sign i.e $x=\sqrt{2+x}$ which renders the quadratic $x^2 -x -2=0$
having the solutions $x=-1$ and $x=2$ and we just ignore $x=-1$ since it is negative.
But my question is, we could also have made the substitution for $x$ under the second radical sign i.e. $x=\sqrt{2 + \sqrt{2 +x}}$ and get the bi-quadratic $x^4-4x^2-x+2=0$. Which has the solutions
$$x=-1,2, \frac{-1-\sqrt{5}}{2} , \frac{-1+\sqrt{5}}{2}$$ Or we could also have substituted anywhere else in the infinite radical expression and it would have given us some real roots. So how do we decide that $x=2$ is the correct solution?
|
For $x=\sqrt{2 + \sqrt{2 +x}}$ you got the bi-quadratic $x^4-4x^2-x+2=0$ with solutions
$$x=-1,2, \frac{-1-\sqrt{5}}{2} , \frac{-1+\sqrt{5}}{2}.$$
Now, the first and third values ($x=-1, \frac{-1-\sqrt{5}}{2}$) can not be a solution because they are negative.
The fourth value is approximately $0.61803398875,$ but we know that The value of $x$ has to be greater than $\sqrt 2$. So, this can not be a solution as well.
After eliminating all the other possibilities, $x=2$ is the solution.
If you make the substitution for $x$ under the $n$-th radical sign, you get an equation with a degree $2^n$.
We do not have a method (It cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions) to solve equations of degree $> 5$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $2003$ divides the numerator of $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335}$. I came up with this problem on a book in number theory :
Let $p$ and $q$ be natural numbers such that
$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335} = \frac{p}{q} \,.$$
Show that $2003 \mid p$.
Any ideas of how to start solving it?
Suggestions and helps would be greatly appreciated!
|
Suppose that $k>1$ is a positive integer such that $3k-1$ is a prime number. Then the numerator of the alternating series
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2k-1}$$
is divisible by $3k-1$. For example when $k=2$, we have $2k-1=3$ and $$1-\frac12+\frac13=\frac{5}{6}$$ with numerator divisible by $5=3k-1$. For $k=4$, we have $2k-1=7$ and $3k-1=11$, and $11$ divides the numerator of
$$1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17=\frac{319}{420}.$$ The claim may not hold if $3k-1$ is not prime. For example if $k=3$, we have $3k-1=8$, and $8$ doesn't divide the numerator of
$$1-\frac12+\frac13-\frac14+\frac15=\frac{47}{60}.$$
The OP's problem comes from $k=668$, where $2k-1=1335$ and $3k-1=2003$. Here is a proof of this claim.
Let $S$ denote the series to be evaluated. Then,
$$S=\left(1+\frac12+\frac13+\frac14+\ldots+\frac{1}{2k-1}\right)-2\left(\frac12+\frac14+\ldots+\frac{1}{2k-2}\right).$$
Hence
$$S=\left(1+\frac12+\frac13+\frac14+\ldots+\frac{1}{2k-1}\right)-\left(1+\frac12+\ldots+\frac{1}{k-1}\right).$$
This makes
$$S=\frac1k+\frac1{k+1}+\ldots+\frac1{2k-2}+\frac{1}{2k-1}.$$
Note that $3k-1$ must be odd, so $k$ is even. That is,
$$S=\left(\frac1k+\frac1{2k-1}\right)+\left(\frac{1}{k+1}+\frac{1}{2k-2}\right)+\ldots+\left(\frac{1}{\frac{3}{2}k-1}+\frac{1}{\frac{3}{2}k}\right).$$
That is
$$S=\frac{3k-1}{k(2k-1)}+\frac{3k-1}{(k+1)(2k-2)}+\ldots+\frac{3k-1}{\left(\frac{3}{2}k-1\right)\frac32k}.$$
Thus
$$S=(3k-1)\left(\frac{1}{k(2k-1)}+\frac{1}{(k+1)(2k-2)}+\ldots+\frac{1}{\left(\frac{3}{2}k-1\right)\frac32k}\right).$$
Because $3k-1$ is a prime greater than $k,k+1,\ldots,2k-2,2k-1$, it is not cancelled out from the numerator of $S$.
I think it is an interesting problem to see whether there exists an integer $k>1$ such that $3k-1$ is not prime but $3k-1$ divides the numerator of $S$. I conjecture that such $k$ don't exist.
|
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Fractions in Questions and Answers
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