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Find $\int_{-1}^{1} \frac{\sqrt{4-x^2}}{3+x}dx$ I came across the integral $$\int_{-1}^{1} \frac{\sqrt{4-x^2}}{3+x}dx$$ in a calculus textbook. At this point in the book, only u-substitutions were covered, which brings me to think that there is a clever substitution that one can use to knock off this integral.
I was able to find the answer using $x= 2 \sin \theta$, doing a bit of polynomial long division and then Weiestrass substitution. However, this solution was rather ugly and I don't think this was what the author had in mind.
What else could I try here? Wolfram gives a closed form $$\pi + \sqrt{5} \left [ \tan ^{-1} \left ( \frac{1}{\sqrt{15}} \right ) - \tan ^{-1} \left ( \frac{7}{\sqrt{15}} \right ) \right ]$$
and the indefinite integral was
$$\sqrt{4-x^2} - \sqrt{5} \tan ^{-1} \left ( \frac{3x+4}{\sqrt{5}\sqrt{4-x^2}}\right ) + 3 \sin ^{-1} \left ( \frac{x}{2} \right )+C $$
| HINT:
Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
$$I+I=6\int_{-1}^1\dfrac{\sqrt{4-x^2}}{9-x^2}dx$$
Putting $x=2\sin y\implies$
$$2I=6\int_{-\pi/6}^{\pi/6}\dfrac{4\cos^2y}{9-4\sin^2y}dy$$
$$=6\int_{-\pi/6}^{\pi/6}\dfrac{5+4(1-\sin^2y)-5}{9-4\sin^2y}dy$$
$$=6\int_{-\pi/6}^{\pi/6}\left(1-\dfrac5{9-\sin^2y}\right)dy$$
Now, $$\int\dfrac{dy}{9-\sin^2y}=\int\dfrac{\csc^2y}{9(1+\cot^2y)-1}dy$$
Set $\cot y=u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
how to change metric variables The metric on unit sphere is given by:
$$g_{ij}=\begin{pmatrix}1 & 0 \\0 & \sin^2{\theta }\end{pmatrix}.$$
The laplacian beltrami operator in $\theta ,\phi$ is
$$\Delta f=\left(\frac{1}{\sqrt{\vert g\vert}} \partial_{i} \ g^{ij}\sqrt{\vert g\vert}\partial_{j}f\right)$$
$$\Delta f=\left( \frac{\partial }{\partial\theta }\sin(\theta)\frac{\partial}{\partial(\theta)}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \phi^2 }\right).$$
How to get these in $x,y$?
| There are probably more sophisticated (and perhaps "better") ways to do this, but I will stick to very elementary concepts:
The metric expresses how to get arc length. If we embed the sphere into 3-space, the metric $(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2$ matches the given metric on the unit sphere. But if we change $x$ and $y$, and stay on the sphere, the change in $z$ is determined:
$$
z = \sqrt{1-x^2-y^2} \\
dz = \frac{-x}{\sqrt{1-x^2-y^2}}dx + \frac{-y}{\sqrt{1-x^2-y^2}}dx
$$
where the two terms come from varying $x$ or $y$ by tiny amounts. Note, by the way, that on the unit sphere equipped with $x,y$ by coordinates which are the projection in the $x,y$ plane of each point on the square, when you are at the equator, a tiny change in $x$ or $y$ forces a huge change in $z$.
We can read the metric from the expression for $(ds)^2$:
$$
(ds)^2 = (dx)^2 + (dy)^2 + \left( \frac{-x\,dx}{\sqrt{1-x^2-y^2}} + \frac{-y\,dy}{\sqrt{1-x^2-y^2}} \right)^2 \\
(ds)^2 = \left(1+ \frac{x^2}{1-x^2-y^2}(dx)^2\right)+ \left(1+ \frac{y^2}{1-x^2-y^2}(dy)^2 \right)
-\frac{2xy}{1-x^2-y^2}(dx)(dy)=
\frac{1}{1-x^2-y^2} \left( (1-y^2) (dx)^2 + (1-x^2) (dy)^2 - 2xy(dx)(dy) \right)\\
g = \frac{1}{1-x^2-y^2} \pmatrix{1-y^2&-xy\\-xy&1-x^2}
$$
Note three points:
*
*The sign of the $-xy$ terms.
*The fact that $||g||=1$ everywhere.
*The (removable) ring of coordinate singularity at $x^2+y^2 = 1$.
In fact, any metric imposed on the unit square will have a coordinate singularity somewhere.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Characteristic polynomials of matrices related How to show that the characteristic polynomials of matrices A and B are $\lambda^{n-1}(\lambda ^2-\lambda -n)=0$ and $\lambda^{n-1}(\lambda^2+\lambda-n)=0$ respectively by applying elementary row or column operations.
$A=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$
$B=\begin{bmatrix}
-1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$
Where $A$ and $B$ are symmetric matrices of order $n+1$.
| Use induction on $n$, we only consider the matrix $A$ because solving $B$ is similar. Before starting, I think that the characteristic polynomial of $A$ and $B$ should be of the form $\color{red}{(-1)^{n+1}}\lambda^{n-1}(\lambda^2-\lambda-n)$ and
$\color{red}{(-1)^{n+1}}\lambda^{n-1}(\lambda^2+\lambda-n)$, respectively.
For $n=1$, then
$$\det(A-tI_2)=\left\vert\begin{matrix}1-t&1\\1&-t\end{matrix}\right\vert=t^2-t-1,$$
which is done.
Now, suppose that the result holds for some $n\in\mathbb{N}$, then for $n+1$,
we have
\begin{align*}
\det(A-tI_{n+2})&=
\left\vert
\begin{matrix}
1-t&1&1&\cdots&1\\
1&-t&0&\cdots&0\\
1&0&-t&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\color{blue} 1&0&0&\cdots&\color{blue}{-t}
\end{matrix}
\right\vert_{n+2}\\
&=(-1)^{n+3}
\left\vert
\begin{matrix}
1&1&1&\cdots&1&1\\
-t&0&0&\cdots&0&0\\
0&-t&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&0\\
0&0&0&\cdots&-t&0
\end{matrix}
\right\vert_{n+1}
-t
\left\vert
\begin{matrix}
1-t&1&1&\cdots&1\\
1&-t&0&\cdots&0\\
1&0&-t&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&\cdots&-t
\end{matrix}
\right\vert_{n+1}\\
&=(-1)^{n+3}
\left\vert
\begin{matrix}
1&1&1&\cdots&1&\color{blue}1\\
-t&0&0&\cdots&0&0\\
0&-t&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&0\\
0&0&0&\cdots&-t&0
\end{matrix}
\right\vert_{n+1}
-t(-1)^{n+1}t^{n-1}(t^2-t-n)\\
&=
(-1)^{n+3}(-1)^{n+2}
\left\vert
\begin{matrix}
-t&0&\cdots&0\\
0&-t&\cdots&0\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&-t
\end{matrix}
\right\vert_{n}
+(-1)^{n+2}t^n(t^2-t-n)\\
&=-(-1)^nt^n+(-1)^{n+2}t^n(t^2-t-n)\\
&=(-1)^{n+2}t^n[(t^2-t-n)-1]\\
&=(-1)^{(n+1)+1}t^{(n+1)+1}[t^2-t-(n+1)].
\end{align*}
This completes the proof for the matrix $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the power series of a complex function So I have the function
$$\frac{z^2}{(z+i)(z-i)^2}.$$
I want to determine the power series around $z=0$ of this function.
I know that the power series is $\sum_{n=0}^\infty a_n(z-a)^n$, where $a_n=\frac{f^{(n)}(a)}{n!}$. But this gives me coefficients, how can I find a series for this?
Edit: maybe we can use partial fractions?
| Note that: $$\frac{z^2}{(z+i)(z-i)^2}\equiv \frac{z^2(z+i)}{(z^2+1)^2}$$
This means, that I would need to only find the power series of $\displaystyle\frac1{(z^2+1)^2}$.
We have: $$\frac1{1-x}\equiv\sum_{n\mathop=0}^\infty x^n$$
Taking derivative of both sides: $$\frac{-1}{(1-x)^2}\equiv\sum_{n\mathop=0}^\infty nx^{n-1}\equiv\sum_{n\mathop=0}^\infty (n+1)x^n$$
Substituting $x=-z^2$: $$\frac{-1}{(1+z^2)^2}\equiv\sum_{n\mathop=0}^\infty (n+1)(-z^2)^n\equiv\sum_{n\mathop=0}^\infty (n+1)(iz)^{2n}$$
Multiplying both sides by $-z^2(z+i)$: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv(z+i)\sum_{n\mathop=0}^\infty (n+1)(iz)^{2n+2}\equiv(z+i)\sum_{n\mathop=1}^\infty n(iz)^{2n}$$
Distributing: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv\sum_{n\mathop=1}^\infty i^{2n}nz^{2n+1}+\sum_{n\mathop=1}^\infty i^{2n+1}nz^{2n}$$
Combining: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv\sum_{n\mathop=1}^\infty (-i)^{2n}nz^{2n+1}+\sum_{n\mathop=1}^\infty (-i)^{2n-1}nz^{2n}$$
$\displaystyle\equiv\sum_{n\mathop=\mbox{odd}}^\infty (-i)^{n+1}\frac{n+1}2z^{n+2}+\sum_{n\mathop=\mbox{even from 0}}^\infty (-i)^{n+1}\frac {n+2}2z^{n+2}$
$\displaystyle\equiv\sum_{n\mathop=\mbox{odd}}^\infty (-i)^{n+1}\left\lceil\frac n2+1\right\rceil z^{n+2}+\sum_{n\mathop=\mbox{even from 0}}^\infty (-i)^{n+1}\frac {n+2}2z^{n+2}$
$\displaystyle\equiv\sum_{n\mathop=0}^\infty (-i)^{n+1}\left\lceil\frac n2+1\right\rceil z^{n+2}$
$\displaystyle\equiv\sum_{n\mathop=2}^\infty (-i)^{n-1}\left\lceil\frac n2\right\rceil z^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the solution $(1+2y)dx+(4-x^2)dy=0$
solve $(1+2y)dx+(4-x^2)dy=0$ using separation of variables
$$(1+2y)dx+(4-x^2)dy=0\implies (4-x^2)dy=-(1+2y)dx\implies \frac{dy}{(-1-2y)}=\frac{dx}{(4-x^2)}$$
Integrate the two sides:
$$\begin{align}
-\frac{1}{2}\ln(-2y-1)&=\int \frac{dx}{4-x^2}\\
\\&\text{now,}\\
\int \frac{dx}{4-x^2}&=\int \frac{1}{2(2-x)}+\int \frac{1}{2(2+x)}\\
&=-\frac{\ln(2-x)}{2}+\frac{\ln(2+x)}{2}\\
\\&\text{now,}\\
-\frac{1}{2}\ln(-2y-1)&=-\frac{\ln(2-x)}{2}+\frac{\ln(2+x)}{2}+C\\
\implies\ln(-2y-1)&={\ln(2-x)}-{\ln(2+x)}-2C\\
\implies e^{\ln(-2y-1)}&=e^{{\ln(2-x)}-{\ln(2+x)}-2C}\\
\implies-2y\cdot e^{-1}&=e^{{\ln(2-x)}-{\ln(2+x)}-2C}\\
\implies y&=-\frac{1}{2}\cdot e^{{\ln(2-x)}-{\ln(2+x)}-2C+1}
\end{align}$$
Is the following correct? are the steps valid?
the answer from wolfram is a bit different
| Not quite. Note that
$$\int \frac{1}{-1-2y}\,dy=-\frac12 \log|2y+1| \tag 1$$
and
$$\begin{align}
\int \frac{1}{4-x^2}\,dx&=\int \left(\frac{1/4}{2-x}+\frac{1/4}{2+x}\right)\,dx\\\\
&=\frac14 \log\left|\frac{x+2}{x-2}\right|+C \tag 2
\end{align}$$
Then, upon equating $(1)$ and $(2)$, we arrive at
$$-\frac12 \log|2y+1|=\frac14 \log\left|\frac{x+2}{x-2}\right|+C $$
from which we obtain
$$|2y+1|=A \sqrt{\frac{|x-2|}{|x+2|}}$$
Note that for $y>-1/2$, we find that
$$y=A' \sqrt{\frac{|x-2|}{|x+2|}} - \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$m$ order determinant related How to find following $m$ order determinant?
$\begin{vmatrix}
1&1&1&1&1&\cdots&1\\
1&-1&0&0&0&\cdots&0\\
1&0&-1&0&0&\cdots&0\\
1&0&0&-1&0&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\cdots&\vdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&0&0&\vdots&-1\\
\end{vmatrix}_m$
| Recall that the determinant does not change if we add one row to another.
By adding all the rows to the first row, we obtain a diagonal matrix whose determinant is easy to evaluate
$\begin{align} \begin{vmatrix}
1&1&1&1&1&\cdots&1\\
1&-1&0&0&0&\cdots&0\\
1&0&-1&0&0&\cdots&0\\
1&0&0&-1&0&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\cdots&\vdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&0&0&\vdots&-1\\
\end{vmatrix}_m
&=
\begin{vmatrix}
m&0&0&0&0&\cdots&0\\
1&-1&0&0&0&\cdots&0\\
1&0&-1&0&0&\cdots&0\\
1&0&0&-1&0&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\cdots&\vdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&0&0&\vdots&-1\end{vmatrix}_m\\\\&=(-1)^{m-1}m\end{align}\\
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can't seem to solve a radical equation? Question is : $\sqrt{x+19} + \sqrt{x-2} = 7$ So there is this equation that I've been trying to solve but keep having trouble with.
The unit is about solving Radical equations and the question says
Solve:
$$\sqrt{x+19} + \sqrt{x-2} = 7$$
I don't want the answer blurted, I want to know how it's done, including steps please.
Thank you!
| $\sqrt{x+19} + \sqrt{x-2} = 7$
Squaring both sides, we have
$x+19+2\sqrt{x+19}\sqrt{x-2}+x-2=49$
Collecting terms, we have
$2x+17+2\sqrt{x^2+17x-38}=49$
$\sqrt{x^2+17x-38}=\dfrac{32-2x}{2}$
Squaring again
$x^2+17x-38=\dfrac{1024-128+4x^2}{4}$
$x^2+17x-38=256-32x+x^2$
$49x=294$
$\therefore x=\dfrac{294}{49}=6$
We can easily verify that this is a correct solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to square both the sides of an equation? Question: $x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}$
My solution: $x^4 (x + 3) = (x + 3)^3$
$=> (x + 3)^2 = x^4$
$=> (x + 3) = x^2$
$=> x^2 -x - 3 = 0$
$=> x = (1 \pm \sqrt{1 + 12})/2$
I understand that you can't really square on both the sides like I did in the first step, however, if this is not the way to do it, then how can you really solve an equation like this one (in which there's a square root on the LHS) without substitution?
| We have
$$x^2\sqrt{x+3} = \sqrt{x+3}\cdot \vert x+3 \vert \implies \sqrt{x+3}\left(x^2-\vert x+3 \vert\right) = 0$$
In the previous statement, we made use of the fact that
$$(x+3)^{3/2} = \sqrt{x+3}\cdot \vert x+3 \vert$$
Hence, we have either
*
*$\sqrt{x+3} = 0 \implies x = -3$
*$x^2-x-3 = 0$ and $x+3 > 0$. This implies $x=\dfrac{1\pm\sqrt{13}}2$
*$x^2+x+3 = 0$ and $x+3<0$. This gives us no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Integrating $\int \frac{\sqrt{x^2-x+1}}{x^2}dx$ Evaluate $$I=\int\frac{\sqrt{x^2-x+1}}{x^2}dx$$ I first Rationalized the numerator and got as
$$I=\int\frac{(x^2-x+1)dx}{x^2\sqrt{x^2-x+1}}$$ and splitting we get
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^2}-\frac{1}{x}}{\sqrt{x^2-x+1}}dx$$ i.e.,
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
First Integral can be evaluated using standard integral. But second one i am not able to do since numerator is not differential of expression inside square root in denominator.
| Using Integration by Parts, We get
Let $$I =\int \sqrt{x^2-x+1}\cdot \frac{1}{x^2}dx = -\frac{\sqrt{x^2-x+1}}{x}+\int\frac{2x-1}{2x\sqrt{x^2-x+1}}dx$$
]
So we get $$I = -\frac{\sqrt{x^2-x+1}}{x}+\int\frac{1}{\sqrt{x^2-x+1}}dx-\frac{1}{2}\int\frac{1}{x\sqrt{x^2-x+1}}dx$$
Now for last Integral, Let $$J = \int \frac{1}{x\sqrt{x^2-x+1}}dx$$
Put $\displaystyle x=\frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get $$J=-\int\frac{1}{\sqrt{t^2-t+1}}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Projectile motion: Proving:$ x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2 = \frac{v^2}{4g^2} $
Question: Projectiles are fired with initial speed $v$ and variable launch angle $0< \alpha < \pi$.
Choose a coordinate system with the firing position at the origin. For each value of $\alpha$ the trajectory will follow a parabolic arc with apex at $(x,y)$ where both $x$ and $y$ depend on $\alpha$. Show that:
$$ x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2 = \frac{v^2}{4g^2} $$
and hence the points of maximum heigh of the trajectories lie on an ellipse.
Requires some basic understanding of physics.
What I have attempted:
As the initial velocity is $v$ , resolving the components
$$ v_x = v\cos(\alpha) $$
$$ v_y = v\sin(\alpha) $$
To find the time to apex (vertex) at $(x,y)$ would be using the formula
$$ v_f = v_i + at $$
$v_f$ at the top is 0 and $a=-g$ and $v_i = v\sin(\alpha)$ hence
$$ t_{top} = \frac{ v\sin(\alpha)}{g} $$
and by using the formula $$ y = v_yt + \frac{1}{2}a_yt^2 $$
$$ y = \frac{1}{2}g t^2 $$
$$ y_{apex} = \frac{v^2\sin^2(\alpha)}{2g} $$
And as horizontal movement is at constant velocity
$$ x = v_x \cdot t $$
$$ x= v\cos(\alpha) \cdot \frac{ v\sin(\alpha)}{g} $$
$$ x = \frac{v^2 \sin(\alpha)\cos(\alpha)}{g} $$
$$ x_{apex} = \frac{v^2 \sin(2\alpha)}{2g} $$
Now I am stuck how do I prove the equation....
| We have:
$$4\left(y-\frac{v^2}{4g}\right)^2=4\left(\frac{v^2\sin^2(\alpha)}{2g}-\frac{v^2}{4g}\right)^2=\frac{v^4\sin^4(\alpha)}{g^2}-\frac{v^4\sin^2(\alpha)}{g^2}+\frac{v^4}{4g^2}$$
$$=\left(\frac{v^4\sin^2\alpha}{g^2}\right)(\sin^2\alpha-1)+\frac{v^4}{4g^2}$$
$$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)(1-\sin^2\alpha)+\frac{v^4}{4g^2}$$
$$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)\cos^2\alpha+\frac{v^4}{4g^2}$$
and
$$x^2= \frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}$$
so
$$x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2=\frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}-\frac{v^4\sin^2\alpha\cos^2\alpha}{g^2}+\frac{v^4}{4g^2}$$
$$=\frac{v^4}{4g^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you prove this without using induction? How do you prove this without using induction
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n-1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}$$
| Let $A(n)=\frac 11 + \frac 12 + \frac 13 + \dots + \frac 1n$. We have
$$
\begin{align}
A(2n)&=\frac 11 + \frac 12 + \frac 13 + \dots + \frac 1n + \dots + \frac 1{2n} \\
&=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)+\left(\frac 12 + \frac 14 + \frac 16 + \dots + \frac 1{2n}\right) \\
&=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)+\frac 12 A(n)
\end{align}
$$
However, we also have
$$
\begin{align}
A(2n)&=\frac 11 + \frac 12 + \frac 13 + \dots + \frac 1n + \dots + \frac 1{2n} \\
&=A(n)+\left(\frac 1{n+1} + \frac 1{n+2} +\dots + \frac 1{2n} + \dots + \frac 1{2n} \right)
\end{align}
$$
Thus
$$
\begin{align}
A(n)+\left(\frac 1{n+1} + \frac 1{n+2} +\dots + \frac 1{2n} + \dots + \frac 1{2n} \right) &=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)+\frac 12 A(n) \\
\frac 1{n+1} + \frac 1{n+2} +\dots + \frac 1{2n} + \dots + \frac 1{2n} &=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)-\frac 12 A(n) \\
&= \frac 11 -\frac 12 + \frac 15-\frac 14 + \dots + \frac 1{2n-1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Symmetric system of equations problem Solve the following simultaneous eqations on the set of real numbers: $$a^2+b^3=a+1$$ $$b^2+a^3=b+1$$
I have found two trivial solutions: $$a=b=1$$ $$a=b=-1$$
but I can't prove that there are no others.
| Subtracting the former from the latter gives
$$b^2-a^2+a^3-b^3=b-a,$$
i.e.
$$(b-a)(b+a)+(a-b)(a^2+ab+b^2)=b-a$$
$$(b-a)(b+a-a^2-ab-b^2-1)=0$$
Case 1 :
If $b=a$, then
$$a^2(a+1)=a+1\iff (a-1)(a+1)^2=0\iff a=\pm 1$$
Case 2 :
If $b+a-a^2-ab-b^2-1=0$, then
$$b^2+(-1+a)b-a+a^2+1=0$$
Since $b$ is real, we have to have
$$(-1+a)^2-4(-a+a^2+1)\ge 0\iff 3\left(a-\frac 13\right)^2+\frac 83\le 0$$
which is impossible since $a$ is real.
So, there are no other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{z}{1-z} = \sum_{j=0}^∞ \frac{2^j}{1 + z^{-2^j}}$ when $z ∈ \mathbb{D}$ The question
Knowing that with $z ∈ \mathbb{D}$:
$$ \prod_{k=0}^∞(1 + z^{2^k}) = \frac{1}{1-z} $$
prove that with $z ∈ \mathbb{D}$:
$$ \sum_{j = 0}^∞ \frac{2^j}{1 + z^{-2^j}} = \frac{z}{1-z} $$
What I've tried
$$ \sum_{j = 0}^∞ \frac{2^j}{1 + z^{-2^j}} = \sum_{j=0}^∞ \frac{2^j z^{2^j}}{1+z^{2^j}} = \frac{1}{1+z} + \frac{2z^2}{1+z^2} + \frac{4z^4}{1+z^4} + … =$$
$$ = \frac{\prod_{k=1}^∞(1+z^{2^k}) + 2z^2 \prod_{k≠1}^∞(1+z^{2^k}) + 4z^4 \prod_{k≠2}^∞(1+z^{2^k})}{\prod_{k=0}^∞ (1 + z^{2^k})} = $$
$$ = (1 - z) \left[ \prod_{k=1}^∞(1+z^{2^k}) + 2z^2 \prod_{k≠1}^∞(1+z^{2^k}) + 4z^4 \prod_{k≠2}^∞(1+z^{2^k}) \right]$$
But I don't know what else I can do right know. Any idea?
| Taking the "logarithmic derivative" $f'/f = (\log f)'$ on both
sides of the equation
$$
\prod_{k=0}^\infty (1 + z^{2^k}) = \frac{1}{1-z}
$$
gives
$$
\sum_{k=0}^\infty \frac{2^k z^{2^k-1}}{1 + z^{2^k}} = \frac{1}{1-z}
$$
which is the desired result.
See $\frac{f'(z)}{f(z)}= \sum_{n=1}^{+ \infty}\frac{f'_{n}(z)}{f_{n}(z)}$ for a justification why this is allowed with an infinite
product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generators for a matrix group Let's denote $\Gamma_0(4)$ the subgroup of $\mathrm{SL}_2(\mathbb Z)$:
$$\Gamma_0(4):=\left\{\begin{pmatrix} a &b\\ 4c&d \end{pmatrix}\in \mathrm{SL}_2(\mathbb Z)\right\}.$$
We also define $A$ and $B$ in $\Gamma_0(4)$ as follow:
$$\gamma_1:=\begin{pmatrix} 1 &1\\ 0&1 \end{pmatrix},$$
$$\gamma_2:=\begin{pmatrix} 1 &0\\ 4&1 \end{pmatrix}.$$
We already know (with sophisticated arguments using group actions) that $\Gamma_0(4)/\{\pm 1\}$ is generated by $\gamma_1$ and $\gamma_2$:
$$\Gamma_0(4)/\{\pm 1\}=\langle \gamma_1,\gamma_2\rangle.$$
I would like to prove it with elementary arguments if possible.
Do you have any leads/solutions to this problem?
| Since I managed to find the answer, I am going to write it here in case it is useful for someone.
Let's defined the group $G=<\gamma_1,\gamma_2>$ generated by:
$$ \displaystyle \gamma_1=\begin{pmatrix} 1 & 0\\ 4 & 1 \end{pmatrix} \text{ and } \displaystyle \gamma_2=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$
First, we compute ${\gamma_1}^n$ and ${\gamma_2}^n$ by induction
$$\forall n \in \mathbb Z\quad {\gamma_1}^n = \begin{pmatrix} 1 & 0\\ 4n & 1 \end{pmatrix} \text{ and } {\gamma_2}^n = \begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix}.$$
Let $\gamma = \begin{pmatrix} a & b\\ 4c & d \end{pmatrix} \in \Gamma_0(4)$.
Everything we will be doing here works if we are dividing by non zero numbers. If it is not the case, there is nothing to prove.
*
*Let $q$ and $r$ be the quotient and the rest of the euclidien division of $a$ by $4c$.
We have $a=4cq+r$, so $r=a-4cq$.
Then we have
$$ {\gamma_2}^{-q}\gamma=\begin{pmatrix} a-4cq & b-dq \\ 4c & d \end{pmatrix} = \begin{pmatrix} r & b'\\ 4c & d \end{pmatrix}. $$
*Let $q'$ and $r'$ be the quotient and the rest of the euclidien division of $c$ by $a$.
We have $c=aq'+r'$, so $4r'=4c-4aq'$.
Then we have
$$ {\gamma_1}^{-q'}\gamma=\begin{pmatrix} a & b \\ 4c-4aq' & d-4bq' \end{pmatrix} = \begin{pmatrix} a & b \\ 4r' & d' \end{pmatrix}.$$
We now apply a slightly modified version of Euclid's algorithm (I can add the details if someone's interested) so the algorithm works on $\mathbb Z$.
We then obtain a sequence of matrixes of the form:
$$ \begin{pmatrix} a & b \\ 4c & d \end{pmatrix} \rightarrow
\begin{pmatrix} r & \star \\ 4c & \star \end{pmatrix} \rightarrow
\begin{pmatrix} r & \star \\ 4r' & \star \end{pmatrix} \rightarrow
\ \cdots \ \rightarrow
\begin{pmatrix} \pm1 & \star \\ 0 & \star \end{pmatrix}. $$
This is beacause, $\gamma\in \text{PSL}_2(\mathbb Z)$, so $ad-4bc=1$.
So with Bezout's theorem, $a\wedge c=1$.
We work in the quotient by $\{\pm 1\}$, which allow us to consider only the positive case.
Then,
$$ \exists \delta \in G,\quad \delta\gamma= \begin{pmatrix} 1 & b' \\ 0 & d' \end{pmatrix}.$$
And $\delta\gamma \in \text{PSL}_2(\mathbb Z)$ so $\det(\delta\gamma)=1$, so $d'=1$.
So,
$$ \delta\gamma= \begin{pmatrix} 1 & b' \\ 0 & 1 \end{pmatrix}.$$
Then,
$${\gamma_2}^{-b'}\delta\gamma= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
So $\gamma\in G$.
So $\Gamma_0(4) \subset G$.
Finally,
$$\Gamma_0(4) / \{\pm 1\}= G.\qquad \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Suppose $\triangle ABC$ is an equilateral triangle inscribed in the unit circle C(0,1). Suppose $\triangle ABC$ is an equilateral triangle inscribed in the unit circle C(0,1). Find the maximum value of $$\overline{PA}\cdot\overline{PB}\cdot\overline{PC}$$
where $P$ is a variable point in $\bar{D}(0,2).$
I am trying to figure a proof for this problem out. I have it all drawn out so I could see conceptually what is going on but I am having trouble finding the answer. I believe the sides of the equilateral triangle are going to be $\sqrt{3}$ but I am not sure about the rest. this question is actually blowing my mind trying to find an answer.
| Supposing
$$\triangle ABC =
\left((0,1) ,\,
\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) ,\,
\left(+\frac{\sqrt{3}}{2},-\frac{1}{2} \right) \right)$$
and $P =(x,y) \in D(0,2) = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 2\}$ and $$\Pi^2 = \overline{PA}\cdot\overline{PB}\cdot\overline{PC},$$
we write
\begin{align}
\Pi^2 &= \sqrt{(x-0)^2+(y-1)^2} \times \sqrt{\left(x-\frac{-\sqrt{3}}{2}\right)^2+\left(y-\frac{-1}{2}\right)^2}\times \sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^2+\left(y-\frac{-1}{2}\right)^2}\\
&=x^6+3 x^4 y^2+3 x^2 y^4+6 x^2 y+y^6-2 y^3+1.
\end{align}
Now, using polar coordinates, such $x\mapsto \rho \cos\theta$ and $x\mapsto \rho \sin\theta$, for $(\rho,\theta) \in [0,2] \times [0,2\pi)$, we have
\begin{align}
\Pi^2 &= 2 \rho ^3 \sin (3 \theta )+\rho ^6+1,
\end{align}
whose maximization is trivial, yielding $\rho=2$ and $\theta = \frac{\pi}{6}+ \frac{2\pi}{3} k, k=\{0,1,2\}$. This result is rather intuitive, being the set of opposite points the vertices of the triangle, where $\Pi^2=81$, implying $\Pi=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Tridiagonal matrix inner product inequality I want to show that there is a $c>0$ such that
$$
\left<Lx,x\right>\ge c\|x\|^2,
$$
for alle $x\in \ell(\mathbb{Z})$ where
$$
L=
\begin{pmatrix}
\ddots & \ddots & & & \\
\ddots & 17 & -4 & 0 & \\
\ddots & -4 & 17 & -4 & \ddots \\
& 0 & -4 & 17 & \ddots \\
& & \ddots & \ddots &\ddots
\end{pmatrix},
$$
is a tridiagonal matrix and
$$
x=
\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix}.
$$
I know that the following holds
\begin{align}
\left<Lx,x\right>&=\left<\begin{pmatrix}
\ddots & \ddots & & & \\
\ddots & 17 & -4 & 0 & \\
\ddots & -4 & 17 & -4 & \ddots \\
& 0 & -4 & 17 & \ddots \\
& & \ddots & \ddots &\ddots
\end{pmatrix}\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix}\right>\\
&=\left< -4\begin{pmatrix}
\vdots \\
x_{2} \\
x_1 \\
x_{0} \\
\vdots
\end{pmatrix} +17\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} -4\begin{pmatrix}
\vdots \\
x_{0} \\
x_{-1} \\
x_{-2} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right>\\
&=-4\left<\begin{pmatrix}
\vdots \\
x_{2} \\
x_1 \\
x_{0} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right> + 17\left<\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right> -4\left<\begin{pmatrix}
\vdots \\
x_{0} \\
x_{-1} \\
x_{-2} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right>.
\end{align}
Hence,
$$
\left<Lx,x\right>=-4k+17\|x\|^2,
$$
where
$$
k=\sum_{j\in \mathbb{Z}}{x_j(x_{j+1}+x_{j-1})}.
$$
Obviously $\|x\|^2\ge 0$, but how can I choose $c$ such that the inequality holds? Here I get stuck, any hints?
| If $V$ is the bilateral shift, we have $$L=17 I - 4(V+V^*).$$
From $\|V\|=1$, we get that $V+V^*$ is a selfadjoint with $\|V+V^*\|\leq2$. Then, for a unit vector $x$,
$$
\langle Lx,x\rangle=17-4\langle(V+V^*)x,x\rangle\geq17-4\|V+V^*\|\geq 17-8=9.
$$
In other words, you can take $c=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
arithmetic mean of smallest numbers of all subsets of r elements formed out of (1,2,..n) Consider all subsets of r elements of the set $\{1,2,3,......,n\}$ where $1 \leq r \leq n$.
Each of these subsets has a smallest member. Let $F(n,r)$ denote the arithmetic mean of these smallest numbers then
$$
F(n,r)=\frac{n+1}{r+1}\;.
$$
I got this solution on a page but couldn't understand it as how he started and what are the different steps one needs to take. Throw light on every single step.
$$\begin{align*}
\sum_{k=1}^nk\binom{n-k}{r-1}&=\sum_{k=1}^{n-(r-1)}k\binom{n-k}{r-1}\\
&=\sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}\binom{n-k}{r-1}\\
&=\sum_{i=1}^{n-(r-1)}\binom{n-i+1}r\\
&=\binom{n+1}{r+1}\;.
\end{align*}$$
And
$$\frac{\binom{n+1}{r+1}}{\binom{n}r}=\frac{n+1}{r+1}\;.$$
| The number of ways that $k$ is the smallest of $r$ numbers from $1\dots n$ is
$$
\binom{n-k}{r-1}\tag{1}
$$
As one would expect, the total number of ways to arrange the $r$ numbers from $1\dots n$ is
$$
\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}=\binom{n}{r}\tag{2}
$$
Thus, the expected smallest of $r$ numbers from $1\dots n$ times $(2)$ is
$$
\begin{align}
\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}k
&=(n+1)\binom{n}{r}-\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}(n-k+1)\\
&=(n+1)\binom{n}{r}-\sum_{k=1}^{n-r+1}\binom{n-k+1}{r}\,r\\
&=(n+1)\binom{n}{r}-\binom{n+1}{r+1}\,r\\
&=(n+1)\binom{n}{r}-\frac{n+1}{r+1}\binom{n}{r}\,r\\
&=\frac{n+1}{r+1}\binom{n}{r}\tag{3}
\end{align}
$$
Therefore, the expected smallest of $r$ numbers from $1\dots n$ is
$$
\frac{n+1}{r+1}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove: $\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$ This is not a homework question, its from sl loney I'm just practicing.
To prove :
$$\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$$
So I changed all the angles to $\arctan$ which gives:
$$\arctan\left(\frac 34\right) - \arctan\left(\frac {12}{5}\right) = \arctan\left(\frac {16}{63}\right)$$
But the problem is after applying formula of $\arctan(X)-\arctan(Y)$ the lhs is negative and not equal to rhs? Is this because I have to add pi somewhere please help.
| If $0<x<1$, then both $\arcsin x$ and $\arccos x$ are in $(0,\pi/2)$. I'll assume $0<x<1$ for the rest of the discussion.
If $\alpha=\arcsin x$, then $\sin\alpha=x$ and $\cos\alpha=\sqrt{1-x^2}$; therefore
$$
\tan\alpha=\frac{x}{\sqrt{1-x^2}}
$$
and
$$
\arcsin x=\alpha=\arctan\frac{x}{\sqrt{1-x^2}}
$$
Similarly, if $\beta=\arccos x$, then $x=\cos\beta$ and
$$
\arccos x=\arctan\frac{\sqrt{1-x^2}}{x}
$$
For $x=3/5$ we have $\sqrt{1-x^2}=4/5$ and so
$$
\arcsin\frac{3}{5}=\arctan\frac{3}{4}
$$
For $x=16/65$ we have $\sqrt{1-x^2}=63/65$, so
$$
\arcsin\frac{16}{65}=\arctan\frac{16}{63}
$$
For $x=12/13$ we have $\sqrt{1-x^2}=5/13$, so
$$
\arccos\frac{12}{13}=\arctan\frac{5}{12}
$$
Now
$$
\tan\left(\arctan\frac{3}{4}-\arctan\frac{5}{12}\right)=
\frac{\dfrac{3}{4}-\dfrac{5}{12}}{1+\dfrac{3}{4}\dfrac{5}{12}}=
\frac{\dfrac{1}{3}}{\;\dfrac{21}{16}\;}=\frac{16}{63}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove: $\frac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$ (for positive $a$, $b$, $c$, $d$) I am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.
Would be grateful for any suggestions!
| Assume that $b,d >0$. Note that $$\frac{a+c}{b+d} = \frac{b}{b+d}\frac{a}{b} +\frac{d}{b+d}\frac{c}{d}.$$ Remark that $0<\frac{b}{b+d}<1$ and the same for $\frac{d}{b+d}$. Hence you have written $\frac{a+b}{c+d}$ as a convex combination of $\frac{a}{b}$ and $\frac{c}{d}$ so you get $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Evaluate $\int_{C[0,3]} \frac{\exp(z)}{(z+2)^2\sin(z)} dz$ Using Residue Theorem $\displaystyle \int_{C[0,3]} \frac{\exp(z)}{(z+2)^2\sin(z)} \, dz$ using Residue Theorem.
I have found singularities within $C[0,3]$, which are $-2$ and $0$. For $z=-2$, it is a pole with degree $2$. However, I do not know what kind of singularities $z=0$ is.
| To answer the specific issue raised in the OP, the pole at $z=0$ is a simple pole since $\sin(z)=z(1+O(z^2))$ and therefore $\frac{1}{\sin(z)}=\frac{1}{z(1+O(z^2))}=\frac{1}{z}+O(1)$
Now, as stated in the comments, there are a number of ways forward to determining the residue at $z=-2$.
METHODOLGY $1$: Apply Standard Limit Formula
First, we can apply the Limit Formula for evaluating the residue at at a pole of order $n$ and write for $n=2$
$$\begin{align}
\text{Res}\left(\frac{e^z}{(z+2)^2\sin(z)}, z=-2\right)&=\lim_{z\to -2}\frac{d}{dz}\left(\frac{e^z}{\sin(z)}\right)\\\\
&\bbox[5px,border:2px solid #C0A000]{=-e^{-2}\frac{1+\cot(2)}{\sin(2)}} \tag 1
\end{align}$$
METHODOLGY $2$: Find the Coefficient on $\frac{1}{z+2}$ of the Laurent Series
Second, we can determine the coefficient of the $(z+2)^{-1}$ term in the Laurent series for the integrand. We proceed by expanding the numerator and denominator in their Taylor series to reveal that
$$\begin{align}
\frac{e^z}{(z+2)^2\sin(2)}&=\frac{e^{-2}\left(1+(z+2)+O(z+2)^2\right)}{(z+2)^2\left(\sin(-2)+\cos(-2)(z+2)+O(z+2)^2\right)}\\\\
&=-\frac{e^{-2}}{\sin(2)}\frac{\left(1+(z+2)+O(z+2)^2\right)\left(1+\cot(2)(z+2)+O(z+2)^2\right)}{(z+2)^2}\\\\
&=-\frac{e^{-2}}{\sin(2)}\left(\frac{1}{(z+2)^2}+\frac{1+\cot(2)}{z+2}+O(1)\right)
\end{align}$$
Inasmuch as the residue is the coefficient on the $\frac{1}{z+2}$ term, we find that
$$\bbox[5px,border:2px solid #C0A000]{\text{Res}\left(\frac{e^z}{(z+2)^2\sin(z)}, z=-2\right)=-e^{-2}\frac{1+\cot(2)}{\sin(2)}}$$
recovering the result reported in $(1)$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Cosine Inequality Show that given three angles $A,B,C\ge0$ with $A+B+C=2\pi$ and any positive numbers $a,b,c$ we have $$bc\cos A + ca \cos B + ab \cos C \ge -\frac {a^2+b^2+c^2}{2}$$
This problem was given in the course notes for a complex analysis course, so I anticipate using $$bc\cos A + ca \cos B + ab \cos C=\mathbf R\mathbf e (bc\text{cis} A + ca \text{cis} B + ab \text{cis} C )\qquad=\mathbf R\mathbf e (c\text{cis}(-C)b\text{cis}(-B)+c\text{cis}(-C)a\text{cis}(-A)+a\text{cis}(-A)b\text{cis}(-B))$$
where $\mathbf R\mathbf e(x)$ denotes the real part of $x$,
is the start, although following this through does not lead me to the inequality I am after.
Also I am unsure of the geometric interpretation of this identity.
This is a similar inequality to $$bc+ca+ab \le a^2+b^2+c^2$$
but the constraints on the angle allow a stronger inequality.
I want some geometric intuition for what the equality is trying to prove, but also I want to see where the inequality comes from.
| Take three vectors $\mathbf a, \mathbf b,$ and $\mathbf c$. We have
$$(\mathbf a + \mathbf b + \mathbf c) \cdot (\mathbf a + \mathbf b + \mathbf c) \ge 0,$$
so
$$\lVert \mathbf a\rVert^2 + \lVert \mathbf b\rVert^2 + \lVert \mathbf c\rVert^2 + 2(\mathbf a \cdot \mathbf b) + 2(\mathbf a \cdot \mathbf c) + 2(\mathbf b \cdot \mathbf c) \ge 0.$$
Now take $a,b,c$ to be the lengths, and $A,B,C$ to be the angles opposite the vectors. Using the fact that $\mathbf a \cdot \mathbf b = \lVert \mathbf a\rVert \cdot \lVert \mathbf b\rVert \cos C$, and similarly for the other two dot products, we get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\frac{1}{a^3}+\frac{1}{b^3}$ Let be a, b solutions for the equation $x^2-2\sqrt{\sqrt{2}+1} (x)+\sqrt{\sqrt{2}+1}=0$
Find the value of $\frac{1}{a^3}+\frac{1}{b^3}$
Using Vieta's formula $a+b=2c$
, $ab=c$ (where $c=\sqrt{\sqrt{2}+1}$) and solving I find that the answer is
$\frac{1}{a^3}+\frac{1}{b^3}=\frac{(a+b)(a^2-ab+b^2)}{(ab)^3}=8-\frac{6}{c}$. It's correct?
| Let $ c = \sqrt{\sqrt{2} + 1} $. Then your given equation is written as :
$ x^2 - 2cx + c = 0 $
Solving it gives you : $ Δ = 4c^2 - 4c > 0 $
$ x_{1,2} = \frac {- 2c \pm \sqrt{Δ}}{2} $
You can then proceed with the calculations by substituting $c$ back in. Just be careful with the square roots and your calculations :).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Isomorphism: $F = \mathbb{Z}_{5}(\alpha)$, $\alpha^2 +2 =0$, and $F'= \mathbb{Z}_{5}(\beta)$, $\beta ^2 + \beta + 1 = 0$. Let $F = \mathbb{Z}_{5}(\alpha)$, where $\alpha^2 +2 =0$, and let $F'= \mathbb{Z}_{5}(\beta)$, where $\beta ^2 + \beta + 1 = 0$.
Exhibit the isomorphism between $F$ and $F'$.
Honestly, I don't know how to solve this problem. I tried to do it by finding roots, which are $\alpha = \pm \sqrt{2}i$ and $\beta = \frac{1}{2} \pm \frac{\sqrt{3}i}{2}$
Please help me to solve this provlem. Thank you!
| Note that $\alpha^2 = 3$, thus $\alpha^4 = 3^2 = 4$, and so $\alpha^8 = 4^2 = 1$. So the order of $\alpha$ is 8.
Now $(\alpha + 1)^2 = \alpha^2 + 2\alpha + 1 = 3 + 2\alpha + 1 = 2\alpha + 4$, and consequently:
$(\alpha + 1)^3 = (\alpha + 1)(2\alpha + 4) = 2\alpha^2 + 4\alpha + 2\alpha + 4 = 2(3) + 4 + \alpha = \alpha$.
This shows that the order of $\alpha + 1$ must be $24$ (do you see why it cannot be a multiple of $3$ less than $24$? Since we already know the order is not $2$, that leaves only $4$ and $8$- but:
$(\alpha + 1)^8 = (\alpha + 1)^6(\alpha + 1)^2 = \alpha^2(\alpha + 1)^2 = 3(2\alpha + 4) = \alpha + 2 \neq 1$).
So $(\Bbb Z_5(\alpha))^{\ast}$ is generated by $(\alpha + 1)$.
On the other hand we have $\beta^3 = \beta(\beta^2) = \beta(4\beta + 4) = 4\beta^2 + 4\beta = 4(4\beta + 4) + 4\beta$
$= \beta + 1 + 4\beta = 1$.
Since $3 = \dfrac{24}{8} = \dfrac{24}{\gcd(8,24)}$, it seems reasonable that we might hope $(\alpha + 1)^8 = \alpha + 2$ is a root of $x^2 + x + 1$.
I leave it you to show this is indeed the case, and then define the sought-after isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find real roots of the equation
Find all real solutions to
$$\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} = x^3-x^2-x+1$$
This question is very similar to one of my previous problem, except I cannot find a monotonic function as has been done in the solution to that problem.
Any help will be appreciated.
Thanks.
| This really is similar in some way to your previous problem.
$$\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} = x^3-x^2-x+1$$
Firstly, for all the terms to be valid, we do domain checking. Solving $x+1\ge0$, $2-x\ge0$, $x^2-x+2\ge0$ and $-x^2+x+1\ge0$, we arrive at $x\in[-0.618, 1.618]$ (actually from the last inequality only)
Putting $x+1:=\alpha$ and $x^2-x+2:=\beta$.
The left hand side is
$$\frac{\sqrt{\alpha}}{2+\sqrt{3-\alpha}} - \frac{\sqrt{\beta}}{2+\sqrt{3-\beta}} $$
Now, consider the function $$f(x)=\frac{\sqrt{x}}{2+\sqrt{3-x}}$$ Indeed it is monotonic. (Check yourself) This implies that $f(\beta)\ge f(\alpha)$ iff $\beta\ge\alpha$. Now see that $\beta-\alpha=x^2-2x+1=(x-1)^2\ge0$ thus $\beta\ge\alpha$ and $f(\beta)\ge f(\alpha)$. Which means,
$$f(\alpha)-f(\beta)=\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} \le 0$$
So the Left hand side is less than or equal to zero.
As for the right hand side, it can be decomposed to $$x^3-x^2-x+1=x^2(x-1)-1(x-1)=(x-1)^2(x+1)$$ Since $(x-1)^2\ge0$, it's behaviour is controlled by $(x+1)$ and it is $0$ at $x=1,-1$ and negative in $x<-1$ which is outside our domain.
So we see that in our domain, $[-0.618, 1.618]$, the Left side is non-positive and the Right side is non-negative. The only possibility for a solution is if they both equal $0$. Hence, solving $\alpha=\beta\implies x+1=x^2-x+2$ and $x^3-x^2-x+1=0$, we get $x=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this inequality with $a+b+c=3$ Let $a,b,c>0$,and $a+b+c=3$,show that
$$\dfrac{a}{2b^3+c}+\dfrac{b}{2c^3+a}+\dfrac{c}{2a^3+b}\ge 1$$
such Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{a}{2b^3+c}+\dfrac{b}{2c^3+a}+\dfrac{c}{2a^3+b}\right)\left(a(2b^3+c)+b(2c^3+a)+c(2a^3+b)\right)\ge (a+b+c)^2=9$$
Therefore,it suffices to prove that
$$(2ab^3+2bc^3+2ca^3)+(ab+bc+ca)\le 9$$
The last inequality doesn't hold for $a=1,b=1.9$,then $2ab^3>9$
I just do it now
Thanks in advance!
| By C-S $\sum\limits_{cyc}\frac{a}{2b^3+c}=\sum\limits_{cyc}\frac{a^2(a+c)^2}{a(a+c)^2(2b^3+c)}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(2b^3+c)}$.
Hence, it remains to prove that $(a+b+c)^2\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq9\sum\limits_{cyc}a(a+c)^2(2b^3+c)$, which is
$\sum\limits_{cyc}(a^6+3a^5b+3a^5c+4a^4b^2+4a^4c^2-14a^3b^3+10a^4bc-a^3b^2c-19a^3c^2b+9a^2b^2c^2)\geq0$,
which is obvious.
For example, $LS\geq\sum\limits_{cyc}(a^6-a^5b-a^5c+a^4bc)+\sum\limits_{cyc}(3a^5b+3a^5c+4a^4b^2+4a^4c^2-14a^3b^3)+$
$+abc\sum\limits_{cyc}(11a^3-a^2b-19a^2c+9abc)\geq0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$, what is the $\max$ value of $|a|+|b|+|c|+|d|$?
My try:
*
*Put $x=0$, we get $p(0)=d$,
*Similarly put $x=1$, we get $p(1)=a+b+c+d$,
*similarly put $x=-1$, we get $p(-1)=-a+b-c+d$,
*similarly put $\displaystyle x=\frac{1}{2}$, we get $\displaystyle p\left(\frac{1}{2}\right)=\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d$
So, given that $|p(x)|\leq 1\forall |x|\leq 1$, we get $|d|\leq 1$.
Similarly $$\displaystyle |b|=\left|\frac{p(1)+p(-1)}{2}-p(0)\right|\leq \left|\frac{p(1)}{2}\right|+\left|\frac{p(1)}{2}\right|+|p(0)|\leq 2$$
Now I do not understand how can I calculate the $\max$ of $|a|$ and $|c|$.
| Let $A=\max(|a|,|c|),C=\min(|a|,|c|),B=\max(|b|,|d|),D=\min(|b|,|d|).$ Then $$|A|+|B|+|C|+|D|=|a|+|b|+|c|+|d|.$$
For $|x|\le1$, $|Ax^2-C|\le|ax^2+c|$ and $|Bx^2-D|\le|bx^2+d|$. Then
$$|(A+B)x^3-(C+D)x|\le |Ax^3-Cx|+|Bx^2-D|\le |ax^3+cx|+|bx^2+d|\le |p(x)|\text{ or } |p(-x)|.$$
Therefore we only need consider $p(x)=ax^3-cx$ with $a\ge c\ge0$.
At $x=1$ we need $a\le c+1$ and at $x=\frac{1}{2}$ we need $|a-4c|\le8$. The required maximum is attained for $a=4,c=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable.
How does one find the exponential of a non-diagonalizable matrix?
My attempt:
Write
$\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix} = M + N$,
with $M = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ and $N = \begin{pmatrix}
0 & 1 & 2 \\
0 & 0 & -4 \\
0 & 0 & 0
\end{pmatrix}$.
We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus:
$$\begin{aligned}
e^{At}
&= e^{(M+N)t} = e^{Mt} e^{Nt} \\
&= \begin{pmatrix}
e^t & 0 & 0 \\
0 & e^t & 0 \\
0 & 0 & e^t
\end{pmatrix} \left(I + \begin{pmatrix}
0 & t & 2t \\
0 & 0 & -4t \\
0 & 0 & 0
\end{pmatrix}+\begin{pmatrix}
0 & 0 & -2t^2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\right) \\
&= e^t \begin{pmatrix}
1 & t & 2t \\
0 & 1 & -4t \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
e^t & te^t & 2t(1-t)e^t \\
0 & e^t & -4te^t \\
0 & 0 & e^t
\end{pmatrix}.
\end{aligned}$$
Is that the right answer?
| (This question was edited a lot, I'm referring to this revision.)
Yes, this is correct. Note however that:
*
*You've used $e^{(M+N)t}=e^{Mt}e^{Nt}$. Note that this is only valid if $M$ and $N$ commute (that is, $MN=NM$). In this case it's ok because $M$ is scalar and commutes with everything, but you should mention it explicitly.
*In general, it may be easier to find the Jordan form of $A$ and use that. You can calculate the exponential in blocks, and there is an elegant expansion for each block.
| {
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"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
integration using change of variables
find $$\iint_{R}x^2-xy+y^2 dA$$ where $R: x^2-xy^+y^2=2$ using $x=\sqrt{2}u-\sqrt{\frac{2}{3}}v$ and $y=\sqrt{2}u+\sqrt{\frac{2}{3}}v$
To calculate the jacobian I take $$\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix}=\begin{vmatrix}
\sqrt{2} &-\sqrt{\frac{2}{3}}\\
\sqrt{2} & \sqrt{\frac{2}{3}}
\end{vmatrix}=\frac{4}{\sqrt{3}}dudv$$
So the integral I have to calculate is now:
$\iint_{R} u^2+v^2\frac{4}{\sqrt{3}}dudv$ or $\iint_{R} u^2+v^2\frac{\sqrt{3}}{4}dudv$
?
| Let be $$I=\iint_{R}(x^2-xy+y^2)\, \mathrm dA$$ where $R: x^2-xy+y^2=2$.
Using the change of variables $x=\sqrt{2}u-\sqrt{\frac{2}{3}}v$ and $y=\sqrt{2}u+\sqrt{\frac{2}{3}}v$ the domain of integration $R$ becomes $S:u^2+v^2=1$ and the integrand function $x^2-xy+y^2$ becomes $2(u^2+v^2)$. The Jacobian determinant is
$$\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\frac{4}{\sqrt 3}$$
Thus we have
$$
I=\iint_{R}(x^2-xy+y^2)\, \mathrm dx\, \mathrm dy=\iint_{S}2(u^2+v^2)\,\frac{4}{\sqrt 3}\, \mathrm du\, \mathrm dv
$$
This integral will be much easier in terms of polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ and then
\begin{align}
I&=\iint_{S}2(u^2+v^2)\,\frac{4}{\sqrt 3}\, \mathrm du\, \mathrm dv\\
&=
\frac{8}{\sqrt 3}\int_0^{2\pi}\int_0^1 (r^2)\cdot r\, \mathrm dr\, \mathrm d\theta=
\frac{8}{\sqrt 3}\int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^1 \, \mathrm d\theta=\frac{8}{\sqrt 3}\int_0^{2\pi} \frac{1}{4}\,\mathrm d\theta=\frac{4\pi}{\sqrt 3}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a$ is not divisible by $7$, then $a^3 - 1$ or $a^3 + 1$ is divisible by $7$ Determine is, in general, true or false. Recall that a universal statement is true if it is true for all possible cases while it is false if there is even one counterexample. Be prepared to prove that your answer is correct by supplying a proof or counterexample, whichever is appropriate.
If $a$ is not divisible by $7$, then $a^3 - 1$ or $a^3 + 1$ is divisible by $7$.
Solution:
According to Fermat's Little Theorem, we know that if $p$ is prime and $a$ is an integer and $p \nmid a$, then $a^{p - 1} \equiv 1 \mbox{ (mod $p$)}$.
Hence, using the above theorem we get that
$$
\begin{array}{rcl}
a^{7-1} & \equiv & 1 \mbox{ (mod 7)}\\
a^6 & \equiv & 1 \mbox{ (mod 7)}\\
a^6 -1 & = & 7k \mbox{ where $k$ is an integer}\\
(a^3-1)(a^3+1) & = & 7k
\end{array}
$$
Since $k$ is integer
$(a^3-1)$ or $(a^3+1)$ is divisible by $7$.
Could you check it for me please is it correct or not?
| It's correct. An alternative can be:
You could realize that
$$x^3 \equiv 0, 1, -1 \pmod 7$$
So, if $7 \nmid x$ then $x^3 \equiv 1, -1 \pmod 7$.
By the definition of congruences we have, either $x^3 + 1 = 7k$ or $x^3 - 1 = 7k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y+z =75$
| If $x=y=z$ then clearly $x=y=z=25$ works.
Otherwise, this is a ternary number with sum of digits $3$. But the unique representation of this number in ternary is
$$1\underbrace{0 \ldots 0}_{26 \text{ times}}$$
which has a sum of digits as $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all integral solutions of the equation $x^n+y^n+z^n=2016$
Find all integral solutions of equation
$$x^n+y^n+z^n=2016,$$
where $x,y,z,n -$ integers and $n\ge 2$
My work so far:
1) $n=2$ $$x^2+y^2+z^2=2016$$
I used wolframalpha n=2 and I received the answer to the problem (Number of integer solutions: 144)
2) $n=3$
I used wolframalpha n=3 and I not received the answer to the problem
How to do without wolframalpha?
| Here an incomplete answer about the possibilities for $2016$. It is clear that the problem would become more difficult for higher numbers.
$$\boxed {n=2}$$
It is known enough that an integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$ so because $2016=4^2(8\cdot15+6)$ we can ensure that
$x^2+y^2+z^2=2016$ has solutions. Calculation gives for integer non negatives
$$(x,y,z)\in\{(4,8,44),(4,20,40),(12,24,36)\}$$
The other solutions are obtained trivially by permutations and change of signs so we get
the $3\cdot8\cdot6=144$ solutions the OP said given by Wolphram.
$$\boxed {n=3}$$ The actual state of the question has been well explained by Dietrich Burde above.
$$\boxed {n=4}$$
We try to see this case in some detail. Note that $6^4=1296<2016<7^4=2401$ so we have to see the $4$-th powers $0,1,16,81,256,625,1296$ and see at the equations
$$\begin{cases}y=\sqrt[4]{2016-z^4}\\y=\sqrt[4]{2015-z^4}\\ y=\sqrt[4]{2000-z^4}\\ y=\sqrt[4]{1935-z^4}\\ y=\sqrt[4]{1760-z^4}\\ y=\sqrt[4]{1391-z^4}\\ y=\sqrt[4]{720-z^4 }\end{cases}$$ where the numbers are $2016-x^4$ corresponding to $x=0,1,2,3,4,5,6$ respectively. There are no solutions (excepting some mistake in calculations).
$$\boxed {n=6,8,…2k}$$
Since $3^6\lt 2016\lt 4^6$ and $2^8\lt 2016\lt 3^8$ and $2^{10}\lt 2016\lt 2^{11}$ it is easy to examine the possibilities (apparently no solutions).
$$\boxed {n=2k+1}$$
It seems that this problem is similar to that in which $n = 3$ due to the appearance of negative integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the probability of selecting five of the winning balls and one of the supplementary balls? So I'm just doing a bit of probability questions and wanted to make sure I got it right.
I have $50$ balls numbered $1-50$, and we pick $6$ winning balls and $2$ supplementary without replacement.
So the chance to get the $6$ winning balls would simply be: $$\frac{6}{50} \cdot \frac{5}{49} \cdot \frac{4}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45} = \frac{1}{15890700}$$
and for $5$ winning balls it would be, same process as above: $$\frac{3}{1059380}$$
Now the part that confuses me is $5$ winning, $1$ supplementary. Would this be given by: $$\frac{6}{50} \cdot \frac{5}{49} \cdot \frac{4}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{44}{45} \cdot \frac{2}{44} = \frac{1}{7945350}?$$
Can someone please check if I did this right, I have a sense that I did not, but not sure where I went wrong.
|
So the chance to get one of the $6$ winning balls would simply be: $\frac{6}{50}*\frac{5}{49}*\frac{4}{48}*\frac{3}{47}*\frac{2}{46}*\frac{1}{45} = \frac{1}{15890700}$
No, that is the chance to pick all of the six winning balls when you pick six numbers. Exactly one winning ball among six picks would be:
$$\frac{6}{50}\times\frac{44}{49}\times\frac{43}{48}\times\frac{42}{47}\times\frac{41}{46}\times\frac{40}{45}\times 6=\dfrac{77572}{189175}$$
(With the last factor accounting for ways to place the non-winning ball in the draw.)
and for $5$ winning balls it would be, same process as above: $\frac{3}{1059380}$
$$\frac{6}{50}\times\frac{5}{49}\times\frac{4}{48}\times\frac{3}{47}\times\frac{2}{46}\times\frac{44}{45}\times 6 = \dfrac{22}{1324225}$$
Now the part that confuses me is 5 winning, 1 supplement. Would this be given by: $\frac{6}{50}*\frac{5}{49}*\frac{4}{48}*\frac{3}{47}*\frac{2}{46}*\frac{44}{45}*\frac{2}{44} = \frac{1}{7945350}?$
$$\frac{6}{50}\times\frac{5}{49}\times\frac{4}{48}\times\frac{3}{47}\times\frac{2}{46}\times\frac{2}{45}\times 6 = \dfrac{1}{1324225}$$
Note: the previous answers assumed you were not concerned about the count of supplemental balls. The probability of $5$ winning and $0$ supplemental balls among six picks would be $\tfrac {21}{1324225}=\tfrac 3{189175}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cyclic Inequality in 3 variables How can I prove the following inequality
$$\frac{2a}{1+b^2}+\frac{2b}{1+c^2}+\frac{2c}{1+a^2}\geq 3, \forall\ a,b,c>0, a+b+c=3.$$
I tried Cauchy inequality, AM-GM, but I don't get anything good...
| By AM-GM $\sum\limits_{cyc}\frac{2a}{1+b^2}=\sum\limits_{cyc}\left(\frac{2a}{1+b^2}-2a\right)+6=\sum\limits_{cyc}\frac{-2ab^2}{1+b^2}+6\geq\sum\limits_{cyc}\frac{-2ab^2}{2b}+6=$
$=6-(ab+ac+bc)\geq6-3=3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\frac{\sqrt{x^2+2x-3}}{x+1}d\,x$ by trig substitution I am preparing for an exam and found this integral in a previous test. Did I do it correctly?
My attempt.
$$
\int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx
$$
Complete the square of $x^2+2x-3$; I changed the integral to
$$
\int\frac{\sqrt{(x-1)^2-4}}{x+1}\,dx
$$
then set $u=x+1$ to get
$$
\int\frac{\sqrt{u^2-4}}{u}\,dx
$$
Using the triangle, $2\sec\theta=u$ and $du=2\sec\theta\tan\theta d\theta$
$$
\int\frac{\sqrt{(2\sec\theta)^2-2^2}}{2\sec\theta}2\sec\theta \tan\theta\, d\theta
$$
This I simplified to
$$
2\int\tan^2\theta\, d\theta = 2\int\sec^2\theta-1\,d\theta
=2[\tan\theta-\theta]+C
$$
Back substitute
$$
\theta=\tan^{-1}\frac{\sqrt{u^2-4}}{2}
$$
and
$$
\tan\theta=\frac{\sqrt{u^2-4}}{2}
$$
Back substitute $u=x+1$
$$
\int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx=
\sqrt{(x-1)^2-4}-\tan^{-1}{\sqrt{(x-1)^2-4}}+C
$$
|
I changed the integral to $\int\frac{\sqrt{(x-1)^2-4}}{x+1}dx$ then $u=x+1$
It should be
$$\int\frac{\sqrt{(x\color{red}{+}1)^2-4}}{x+1}dx$$
I simplified to $2\int tan^2\theta d\theta$ = $2\int\sec^2\theta-1d\theta$
$=2[tan\theta-\theta]+C$
back substitute $\theta=tan^{-1}\frac{\sqrt{u^2-4}}{2}$and $tan\theta=\frac{\sqrt{u^2-4}}{2}$ back substitute $u=x+1$
I think you did nothing wrong here.
$\int\frac{\sqrt{x^2+2x-3}}{x+1}dx={\sqrt{(x-1)^2-4}}-tan^{-1}{\sqrt{(x-1)^2-4}}+C$
This is not correct :
$$\int\frac{\sqrt{x^2+2x-3}}{x+1}dx={\sqrt{(x\color{red}{+}1)^2-4}}-\color{red}{2}\tan^{-1}\frac{\sqrt{(x\color{red}{+}1)^2-4}}{\color{red}{2}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$
Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$
$\bf{My\; Try::}$ I have solved Using Direct formula::
$$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}$$
Now Put $n=7\;,$ We get
$$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}\cdot \sin \frac{4\pi}{7}\cdot \sin \frac{5\pi}{7}\cdot \sin \frac{6\pi}{7}=\frac{7}{2^{7-1}}$$
So $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}$$
Now my question is how can we solve it without using Direct Formula, Help me
Thanks
| $$
\begin{align}
\prod_{k=1}^3\sin\left(\frac{k\pi}7\right)^2
&=\prod_{k=1}^6\sin\left(\frac{k\pi}7\right)\tag{1}\\
&=-\frac1{64}\prod_{k=1}^6\left(e^{ik\pi/7}-e^{-ik\pi/7}\right)\tag{2}\\
&=\frac1{64}\prod_{k=1}^6\left(1-e^{-i2k\pi/7}\right)\tag{3}\\
&=\frac1{64}\lim_{z\to1}\prod_{k=1}^6\left(z-e^{-i2k\pi/7}\right)\tag{4}\\
&=\frac1{64}\lim_{z\to1}\frac{z^7-1}{z-1}\tag{5}\\[6pt]
&=\frac7{64}\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\sin(x)=\sin(\pi-x)$
$(2)$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(3)$: pull $\prod\limits_{k=1}^6e^{ik\pi/7}=-1$ out of the product
$(4)$: $1=\lim\limits_{z\to1}z$
$(5)$: $\prod\limits_{k=1}^6\left(z-e^{-i2k\pi/7}\right)=\frac{z^7-1}{z-1}$
$(6)$: evaluate limit
Therefore, taking the square root of $(6)$, we get
$$
\prod_{k=1}^3\sin\left(\frac{k\pi}7\right)=\frac{\sqrt7}8\tag{7}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Exercise: Evaluate a polynomial function such as $P(x)=2x^3-3x^2+7x-2$ at a surd such as $x=1+2\sqrt{3}$. Exercise: Given polynomial function $P(x)=2x^3-3x^2+7x-2$ evaluate $P(x)$ at the surd $x=1+2\sqrt{3}$.
| Solution: Divide $P(x)$ by the divisor
\begin{equation}
D(x)=[x-(1+2\sqrt{3})]\cdot[x-(1-2\sqrt{3})]=x^2-2x-11
\end{equation}
Then
\begin{equation}
P(x)=(2x+1)(x^2-2x-11)+31x+9
\end{equation}
Therefore
\begin{equation}
P(1+2\sqrt{3})=0+31(1+2\sqrt{3})+9=40+62\sqrt{3}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? $$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$
Find limit of $\dfrac1{f(x)}$ as $x\to0$.
| $$\frac{1}{\sin^2 x}=\frac{1}{(x-x^3/6+O(x^5))^2}=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/6+O(x^4))^2}\\=\frac{1}{x^2}\cdot\frac{1}{(1-x^2/3+O(x^4))}=\frac{1}{x^2}(1+x^2/3+O(x^4))$$
Hence
$$f(x)=\frac{1}{x^2}-\frac{1}{\sin^2 x}=-\frac13+O(x^2)$$
And
$$\frac{1}{f(x)}\underset{x\to 0}{\longrightarrow} -3$$
Without using Taylor expansion, you could apply L'Hôpital's rule to
$$\frac{1}{f(x)}=\frac{x^2\sin^2 x}{\sin^2 x-x^2}$$
However, you will need to differentiate four times, the numerator and the denominator (you get a $0/0$ form for lower derivatives). You will then get after simplification
$$\frac{\mathrm d^4 (x^2\sin^2x)}{\mathrm d x^4}=8(3-x^2)\cos(2x)-32x\sin(2x)$$
$$\frac{\mathrm d^4 (\sin^2x-x^2)}{\mathrm d x^4}=-8\cos(2x)$$
$$\lim_{x\to0}\frac{1}{f(x)}=\lim_{x\to0}\frac{8(3-x^2)\cos(2x)-32x\sin(2x)}{-8\cos(2x)}=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does one parameterize $x^2 + xy + y^2 = \frac{1}{2}$?
Parameterize the curve $C$ that intersects the surface
$x^2+y^2+z^2=1$ and the plane $x+y+z=0$.
I have this replacing equations:
$$ x^2+y^2+(-x-y)^2=1$$
and clearing have the following:
$$ x^2+xy+y^2=1/2$$
which it is the equation of an ellipse but I find it difficult parameterization values
Any advice will be of much help, thanks in advance
| Let
\begin{equation}
x=\sqrt{\frac{2}{3}}\sin(t)
\end{equation}
\begin{equation}
y=\sqrt{\frac{2}{3}}\sin\left(t+\frac{2\pi}{3}\right)
\end{equation}
\begin{equation}
z=\sqrt{\frac{2}{3}}\sin\left(t-\frac{2\pi}{3}\right)
\end{equation}
Then $x+y+z=0$ and $x^2+y^2+z^2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
| Here is a roundabout way.
Let $$I:=\int\frac1{1+x^3}dx, J:=\int\frac x{1+x^3}dx,$$
then
$$I+J=\int\frac {1+x}{1+x^3}dx=\int\frac 1{1-x+x^2}dx=\int\frac1{\left(x-\frac1{2}\right)^2+\frac{3}{4}}dx=\frac2{\sqrt3}\arctan\frac{2x-1}{\sqrt3}+C,$$
$$I-J=\int\frac{1-x}{1+x^3}dx=\int\frac1{1+x}dx-\int\frac{x^2}{1+x^3}dx=\ln\lvert1+x\rvert-\frac1{3}\ln\lvert 1+x^3\rvert+C,$$
solving the above equations yields
$$I=\frac1{\sqrt{3}}\arctan\frac{2x-1}{\sqrt3}+\frac1{2}\ln\lvert1+x\rvert-\frac1{6}\ln\lvert1+x^3\rvert+C,$$
$$J=\frac1{\sqrt{3}}\arctan\frac{2x-1}{\sqrt3}-\frac1{2}\ln\lvert1+x\rvert+\frac1{6}\ln\lvert1+x^3\rvert+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Limit of $\sqrt{\frac{\pi}{1-x}}-\sum\limits_{k=1}^\infty\frac{x^k}{\sqrt{k}}$ when $x\to 1^-$? I am trying to understand if
$$\sqrt{\frac{2\pi}{1-x}}-\sum\limits_{k=1}^\infty\frac{x^k}{\sqrt{k}}$$ is convergent for $x\to 1^-$. Any help?
Update: Given the insightful comments below, it is clear it is not converging, hence the actual question now is to find
$$ \lim_{x\to 1^-}\left(\sqrt{\frac{\pi}{1-x}}-\sum_{n\geq 1}\frac{x^n}{\sqrt{n}}\right)$$
| $f(x):=\sum\limits_{k=1}^\infty\frac{x^k}{\sqrt{k}}$
$g(x):= \sqrt{\frac{2\pi}{1-x}}-f(x)$
$\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{x^k}{\sqrt{k}}=f(x)-\sqrt{2}f(x^2)$ is convergent for $x\uparrow 1$ .
=>
$\sum\limits_{k=1}^\infty (-1)^k \frac{x^k}{\sqrt{k}}+\sqrt{\frac{2\pi}{1-x}}-\sqrt{2}\sqrt{\frac{2\pi}{1-x^2}}= g(x)-\sqrt{2}g(x^2)$.
For $x\uparrow 1$ we get $\sum\limits_{k=1}^\infty (-1)^k \frac{1}{\sqrt{k}}=(1-\sqrt{2})g(1)$.
Therefore is the limit $g(1)=\frac{1}{\sqrt{2}-1}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{1}{\sqrt{k}}$.
EDIT: See above, it's not $\sqrt{2\pi}$, it's $\sqrt{\pi}$
Thank you all for your efforts!
EDIT 2: $g(1)=-\zeta(\frac{1}{2})$
| {
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"url": "https://math.stackexchange.com/questions/1788909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
} |
Characteristic function of Laplace distribution I'm trying to derive the characteristic function for the Laplace distribution with density $$\frac{1}{2}\exp\{-|x|\}$$
My attempt:
$$\frac{1}{2}\int_{\Omega}e^{itx-|x|}\mathrm{d}x$$
$$=\frac{1}{2}\int_0^\infty e^{(-it+1)-x}\mathrm{d}x+\frac{1}{2}\int_{-\infty}^0e^{(it+1)x}\mathrm{d}x$$
$$=\frac{1}{2}e^{(-it+1)}\int_0^\infty e^{-x}\mathrm{d}x+\frac{1}{2}e^{(it+1)}\int_{-\infty}^0e^{x}\mathrm{d}x$$
$$\frac{1}{2}e^{-it+1}+\frac{1}{2}e^{(it+1)}$$
$$\frac{1}{2}e(e^{-it}+e^{it})$$
But this doesn't look like the characteristic function on Wikipedia.
| First I will derive what we will need later using Euler's formula.
\begin{align*}
\int \cos(ux) e^{-x} dx + i \int \sin(ux) e^{-x} dx&= \int (\cos(ux) + i \cdot \sin(ux)) e^{-x} dx \\
&= \int e^{iux} \cdot e^{-x} dx \\
&= \int e^{(iu-1)x} dx \\
&= \frac{1}{iu-1} e^{(iu-1)x} \\
&= \frac{1}{iu-1} \frac{(iu+1)}{(iu+1)} e^{(iu-1)x} \\
&= -\frac{e^{-x}}{u^2+1} (iu+1) \cdot e^{iux} \\
&= -\frac{e^{-x}}{u^2+1} (iu+1) \cdot (\cos(ux) + i \sin(ux)) \\
&= -\frac{e^{-x}}{u^2+1} (iu\cos(ux) - u\sin(ux) + \cos(ux) + i \sin(ux)) \\
&= -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)) + i (-\frac{e^{-x}}{u^2+1} (u\cos(ux) + \sin(ux))).
\end{align*}
Taking the real term,
$\int \cos(ux) e^{-x} dx = -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)). \qquad (1)$
Now, the characteristic function is
\begin{align*}
\varphi_{X} (u) &= E[e^{iuX}] \\
&= \int_{-\infty}^{\infty} e^{iux} \cdot \frac{1}{2} e^{-|x|} dx \\
&= \frac{1}{2} \int_{-\infty}^{0} e^{iux} \cdot e^{-(-x)} dx + \frac{1}{2} \int_{0}^{\infty} e^{iux} \cdot e^{-x} dx \\
&= \frac{1}{2} \int_{-\infty}^{0} (\cos(ux) + i \sin(ux)) \cdot e^{x} dx + \frac{1}{2} \int_{0}^{\infty} (\cos(ux) + i \sin(ux)) \cdot e^{-x} dx \\
&= \frac{1}{2} \int_{0}^{\infty} (\cos(ux) - i \sin(ux)) \cdot e^{-x} dx + \frac{1}{2} \int_{0}^{\infty} (\cos(ux) + i \sin(ux)) \cdot e^{-x} dx \\
&= \int_{0}^{\infty} \cos(ux) \cdot e^{-x} dx \\
&= -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)) \Big|_{0}^{\infty} \qquad (1) \\
&= 0 - (- \frac{1}{u^2+1} (1 - 0)) \\
&= \frac{1}{u^2+1}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Complex numbers, finding solution for z How can you solve this? $z^2+2(1+i)z=2+2(\sqrt{3}-1)i$
I have tried to compare left and right side with real and imaginary part i then get
$ x^2+2x-y^2-2y=2$
$xy+x+y=(\sqrt{3}-1)$
But this equation can not be solved.
What else can i do? Setting in $z=re^{i\theta}$ does not help either
The answer is: $ \sqrt{3}-1$ and $-(1+\sqrt{3})-2i$
| solving
$$
z^2+2(1+i)z-2-2(\sqrt{3}-1)i=0
$$
we have:
$$
z=-1-i\pm\sqrt{(1+i)^2+2+2(\sqrt{3}-1)i}=-1-i\pm\sqrt{2}\sqrt{1+\sqrt{3}i}
$$
now:
$1+\sqrt{3}i=2e^{i\pi/3}$ and
$$
\sqrt{1+\sqrt{3}i}=\sqrt{2}e^{i\pi/6}=\sqrt{2}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)
$$
so:
$$
z=-1-i\pm 2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)=-1-i\pm (\sqrt{3}+i)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Let $A= \{1,2,3,4,5,6,7,8,9,0,20,30,40,50\}$. 1. How many subsets of size 2 are there? 2.How many subsets are there altogether?
Let $A= \{1,2,3,4,5,6,7,8,9,0,20,30,40,50\}$.
1. How many subsets of size $2$ are there?
2.How many subsets are there altogether?
Answer:
1) I think there are $7$ subsets of size two are there, since $14$ elements$/2=7$ and also by grouping them into pairs it shows there are $7$ groups of pairs as follows; $\{\{1,2\}, \{3,4\}, \{5,6\}, \{7,8\}, \{9,0\}, \{20,30\}, \{40,50\}\}$
2) I choose $15$, but I'm not sure.
However, are my answers valid? Hints are much appreciated.
| The comments have already very quickly pointed this out, but perhaps I can give an explanation on why they work.
$1$) André Nicolas' comment has already answered that there are $\binom{14}{2}$ subsets of size $2$. This notation is the binomial coefficient. That is, \begin{align}\binom{14}{2} &= \frac{14!}{(14-2)!2!}\\ &= \frac{14 \times 13 \times \color{red}{12} \times \color{red}{11} \times \color{red}{10} \times \color{red}9 \times \color{red}8 \times \color{red}7 \times \color{red}6 \times \color{red}5 \times \color{red}4 \times \color{red}3 \times \color{red}2 \times \color{red}1}{(\color{red}{12} \times \color{red}{11} \times \color{red}{10} \times \color{red}9 \times \color{red}8 \times \color{red}7 \times \color{red}6 \times \color{red}5 \times \color{red}4 \times \color{red}3 \times \color{red}2 \times \color{red}1)(2 \times 1)}\\ &=\frac{14 \times 13}{2}\\ &= 91. \end{align} This is correct because there are $14$ elements in your set, and we wish to choose $2$ of them each time.
$2$) Ethan Hunt's comment has already answered that there are $2^n$ subsets, including the empty set and the entire set itself. To find all subsets of a set, we take the power set of the set. It's easy to prove that the cardinality of the power set is $2^n$, where $n$ is the number of elements of the set. So in your case, there are $2^{14} = 16384$ subsets.
| {
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"url": "https://math.stackexchange.com/questions/1795982",
"timestamp": "2023-03-29T00:00:00",
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Limit of a Riemann sum: $\lim_{n\to\infty} {n^5 \sum^n_{r=0}\frac1{(n^2+r^2)^3}} $ Required to find
$\lim_{n\to{\infty}} {n^5 \sum^n_{r=0}\frac{1}{(n^2+r^2)^3}} $
$\lim_{n\to{\infty}} \frac{1}{n} \sum^n_{r=0}(\frac{n^2}{n^2+r^2})^3$
$\lim_{n\to{\infty}} \frac{1}{n} \sum^n_{r=0}(\frac{1}{1+\frac{r^2}{n^2}})^3$
$\lim_{n\to{\infty}} \frac{1-0}{n} \sum^n_{r=0}f[0+r(\frac{1-0}{n})]$, where $f(x)=(\frac{1}{1+x^2})^3$.
Since $f(x)$ is continuous on $]0,1[$
$\frac{1}{n} \sum_{r=1}^n f(\frac{r}{n})$ is a Riemann sum.
$\lim_{n\to{\infty}}\frac{1}{n} \sum_{r=1}^n f(\frac{r}{n})=\int_0^1 \frac{1}{(1+x^2)^3}$ $dx$
I am getting problems to compute this integral.
I used a substitution of $x=\tan{u}$
My integral becomes $\int^\frac{\pi}{4}_0 (\cos{u}-\sin^2{u}\cos{u})du=\frac{5\sqrt{2}}{2}$
Hence $\lim_{n\to{\infty}}\frac{1}{n} \sum_{r=1}^n f(\frac{r}{n})=\int_0^1 \frac{1}{(1+x^2)^3}$ $dx=\frac{5\sqrt{2}}{2}$
But the limit is defined to be $\frac{3\pi}{32}+\frac{1}{4}$. I am getting it to be different.
| $$n^5\sum_{k=0}^n\frac1{(n^2+k^2)^3}=\frac1n\sum_{k=0}^n\frac1{\left(1+\left(\frac kn\right)^2\right)^3}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{(1+x^2)^3}$$
Your substitution is fine...yet you must also change the limits accordingly:
$$x=\tan u\implies\begin{cases}x=0=\tan 0\implies u=0\\{}\\x=1=\tan u\implies u=\frac\pi4\end{cases}\;\;,\;\;\;dx=\frac{du}{\cos^2u}$$
and our integral becomes, using $\;1+\tan^2u=\frac1{\cos^2u}\;$ :
$$\int_0^{\pi/4}\frac{du}{\cos^2u}\cdot\frac1{(1+\tan^2u)^3}=\int_0^{\pi/4}\cos^4u\;du=\left.\frac1{32}\left(12u+8\sin2u+\sin4u\right)\right|_0^{\pi/4}=$$
$$=\frac1{32}\left(3\pi+8\right)=\frac{3\pi}{32}+\frac14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Best way to expand $(2+x-x^2)^6$
I've completed part $(a)$ and gotten:
$64+192y+240y^2+160y^3+...$
Using intuition I substituted $x-x^2$ for $y$ and started listing the values for :
$y, y^2 $ and $y^3,$ in terms of $x$.
$y=(x-x^2)\\y^2=(x-x^2)^2 = x^2-2x^3+x^4;\\y^3 = (x-x^2)^3 = (x-x^2)(x^2-2x^3+x^4) = \;...$
Everything became complicated before I had even started to substitute these new-found values back into the polynomial devised from part $(a)$; for a 3 - out of 75 - mark question this seems extremely over-complicated. Am I doing it inefficiently / incorrectly; is there a pre-defined or easier method; or is it simple expansion and substitution that I'm too lazy to complete?
| I think I would say
$((x+2)(x-1))^6 = (x+2)^6(x-1)^6\\
(64 + 192x + 240x^2 + 160x^3 ...)(1 - 6x + 15x^2 - 20x^3 ...)$
I don't care about any powers bigger than 3
$64 + (64(-6) + 192)x + (64*15+192(-6) + 240)x^2 + (64(-20)+ 192*15 + 240*(-6) + 160)x^3...\\
64 - 192 x + 48 x^2 + 320x^3...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the problem. Any help is appreciated.
Find the value of $$\frac{a^2}{a^4+a^2+1}$$ if $$\frac{a}{a^2+a+1}=\frac{1}{6}$$
| You are asked to express $\dfrac1B=\dfrac1{a^2+1+a^{-2}}$ in terms of $\dfrac1A=\dfrac1{a+1+a^{-1}}$.
Squaring "to see",
$$A^2=(a+1+a^{-1})^2=a^2+1+a^{-2}+2a+2+2a^{-1}=B+2A.$$
This gives us
$$B=A^2-2A=6^2-2\cdot6=24.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence
$$(x+y+z)^2-2(xy+yz+xz) < 2(xy + yz + xz)$$
$$ (x+y+z)^2 - 4(xy+yz+xz) < 0 $$
As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 >0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$
Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
| Expanding E.Girgin's answer and putting $x=b+c,y=a+c,z=a+b$ the original inequality turns into:
$$ 2a^2+2b^2+2c^2+2ab+2ac+2bc < 2(a^2+b^2+c^2+3ab+3ac+3bc) $$
that is pretty trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove $(x+y)(y^2+z^2)(z^3+x^3) < \frac92$ for $x+y+z=2$ $x,y,z \geqslant 0$ and $x+y+z=2$, Prove
$$(x+y)(y^2+z^2)(z^3+x^3) < \frac92$$
While numerical method can solve this problem, I am more interested in classical solutions. I tried this problem for the past few months, using all kinds of AM-GM and CS, but still cannot solve it. I hope someone can help me out with this one.
| The proof comes by considering three cases.
Case 1: $z>1$, $y> 0.1$. We have
$$(x+y)(y^2+z^2)(z^3+x^3) - \frac92= (2-z)((2-x-z)^2+z^2)(z^3+(2-z-y)^3) - \frac92\\
< (2-z)((2-z)^2+z^2)(z^3+(2-z-0.1)^3) - \frac92$$
which is less than 0, see this plot:
and in more detail here:
Case 2: $z>1$, $y\leq 0.1$. We have
$$(x+y)(y^2+z^2)(z^3+x^3) - \frac92= (2-z)(y^2+z^2)(z^3+(2-z-y)^3) - \frac92\\
< (2-z)((0.1)^2+z^2)(z^3+(2-z)^3) - \frac92$$
which is less than 0, see this plot:
Case 3:
$z \leq 1$. Here, let $x = \frac{r}{2} (1+q)$ and $y = \frac{r}{2} (1-q)$. Hence, $x+y = r$ and $1 \leq r \leq 2$. Further, $-1\leq q \leq 1$ since $x \geq 0$ and $y \geq 0$. With this notation, we have
$$(x+y)(y^2+z^2)(z^3+x^3) - \frac92= r (\frac{r^2}{4} (1-q)^2+(2-r)^2)((2-r)^3+\frac{r^3}{8} (1+q)^3) - \frac92$$
which is less than 0 in the given ($q,r$)-range, see this contourplot:
Note 1: since no upper limits have been made in rewriting the equation for case 3, it would have been possible to drop the condition $z\leq 1$ and, giving the proof without case discussion, to draw the contourplot for the full range $0\leq r \leq 2$. However, the delicate case where the bound is very tight occurs in case 1 and is easier to see with the given description in case 1. In a contourplot for the full range $0\leq r \leq 2$, this delicate case can only be shown with magnification of the relevant area.
Note 2: The first two cases are in one parameter $z$ only, so showing the negativity over the given $z$-range will be easily possible, instead of plots, by analytic approximations.
Note 3: In the third case, the values are manifestly negative in the whole considered range. Hence, also here, showing negativity over the whole range will be easily possible, instead of plots, by coarse approximations.
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Parametric equations for intersection between plane and circle So I was looking at this question Determine Circle of Intersection of Plane and Sphere
but I need to know how to find a parametric equation for intersections such as these. My particular question is to find a parametric equation for the intersection between the plane
$$x+y+z=1$$ and unit sphere centered at the origin.
I started out my question by substituting
$$
z=-x-y+1$$ into
$$x^2+y^2+z^2=1 $$ deriving
$$x^2+y^2+(-x-y+1)^2=1$$
and getting $$2x^2+2y^2+2xy-2x-2y=0$$ but I am unsure how to proceed from here.
I also tried to use the vector equation of the plane
$$r(u, v)=(0,0,1)+(2,1,-3)u+(1,1,-2)v$$ but I am not sure how that would help.
| First, let's name some points. Let $O=(0,0,0)$, $X=(1,0,0)$, $Y=(0,1,0)$, and $Z=(0,0,1)$. Let $A$ be the center of the circle we are trying to find. Let $K$ be the midpoint of $XY$. Note that $K=\left(\frac{1}{2},\frac{1}{2},0\right)$.
Let's consider right triangle $ZOK$:
We note that A is the base of the altitude to the hypotenuse. Since $OZ=1$ and $OK=\frac{\sqrt{2}}{2}$, $OA=\frac{\sqrt{3}}{3}$. Because $A$ must be of the form $(n,n,n)$, we conclude that $A=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. We can also calculate that $AZ=\frac{\sqrt{6}}{3}$. Note that $AZ$ is the radius of the circle.
Now, let's invent a new 2D coordinate system with new unit vectors. $A$ will be the center of this coordinate system. Rotate $AZ$ 90 degrees clockwise on the plane $x+y+z=1$. This new vector (orange) will be one of our unit vectors, which we will call $u$. The other will be $AZ$ itself (pink), which we will call $v$. Imagine extending vector $u$ into a line and moving it down to the $xy$-plane. It makes a 45-45-90 triangle with the $x$ and $y$ axes. Since $u$ has no $z$-component, we conclude that $u=\left(-\frac{\sqrt{6}}{3}\cdot\frac{\sqrt{2}}{2},\frac{\sqrt{6}}{3}\cdot\frac{\sqrt{2}}{2},0\right)=\left(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},0\right)$.
Drop a perpendicular from point A to the $z$ axis. Call the foot of this perpendicular $H$. Note that triangle $ZOA$ is similar to $ZOK$. Using this fact, we can calculate that $ZH=\frac{2}{3}$. Note that $ZH$ is the $z$-component of $v$. We can also calculate that $AH=\frac{\sqrt{2}}{3}$. Imagine moving $AH$ down to the $xy$-plane. We see that $AH$ forms 45 degrees with the $x$ and $y$ axes. We conclude that $v=\left(-\frac{\sqrt{2}}{3}\cdot\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{3}\cdot\frac{\sqrt{2}}{2},\frac{2}{3}\right)=\left(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)$.
The equation of the circle is $$A+u\cos(t)+v\sin(t)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)+\cos(t)\left(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},0\right)+\sin(t)\left(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)$$
Adding everything together, we get a final answer:
$$\frac{1}{3}\left(1-\sqrt{3}\cdot\cos(t)-\sin(t),1+\sqrt{3}\cdot\cos(t)-\sin(t),1+2\sin(t)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to evaluate $\int_{0}^{1} \frac{\ln x}{x+1} dx$ I want to evaluate:
$$\int_{0}^{1} \frac{\ln x}{x+1} dx$$
If I was asked I would to evaluate:
$$\int_{0}^{1} \frac{\ln x}{x-1} dx$$
That would be easy because if I use the Taylor series for $\ln x$ centered at $1$ then things will cancel out and leave me with a easy integral.
So how about this integral, I'm guessed to use the Taylor series centered around $-1$. But even with that thought in mind, it does not take me anywhere because $\ln (-1)$ is not defined. Can someone help.
| Consider:
$$I=\int_{0}^{1}\frac{\ln(x)}{1+x}dx= \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$$
You can show this result by noticing since $0<x<1$ that $$\frac{1}{1+x}= \sum_{n=0}^{\infty}(-x)^n$$ and performing term by term integration. You will need integration by parts to do that.
Now you can also show that:
$$-I=\int_{0}^{1}\int_{0}^{1}\frac{1}{1+xy}dydx$$ by expanding the integrand into a geometric series and doing term by term integration with respect to $y$ and the with respect to $x$.
Now, in the double integral, let $x=u+v,y=-u+v$
Then $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\begin{vmatrix} 1 && 1\\ -1 && 1 \\ \end{vmatrix}=2$
The change of variables causes the double integral to become:
$$\iint_{T} \frac{2}{1+v^2-u^2}dvdu$$ where $T$ is the square with vertices $(0,0),(\frac{-1}{2},\frac{1}{2}),(0,1),(\frac{1}{2},\frac{1}{2})$
Now, we need to compute two double integrals (over two different subregions of the square) and sum them to get the final answer.
First let $\frac{-1}{2}\leq u \leq 0 , -u \leq v \leq 1+u$.
$$\int_{\frac{-1}{2}}^{0}\int_{-u}^{1+u} \frac{2}{1+v^2-u^2}dvdu=\int_{\frac{-1}{2}}^{0}\frac{2\tan^{-1}\left(\frac{1+u}{\sqrt{1-u^2}}\right)+2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$$
For the first term on the right hand side, let : $z=2\tan^{-1}\left(\frac{1+u}{\sqrt{1-u^2}}\right),dz=\frac{1}{\sqrt{1-u^2}}du$. For the second term, if you draw a right triangle, notice $2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=2\sin^{-1}\left(u\right)$ so let $z=\sin^{-1}\left(u\right),dz=\frac{1}{\sqrt{1-u^2}}du$ for the second term.When evaluating these two terms with these substitutions, the sum of these terms should be $\frac{\pi^2}{24}$.
On the other hand, let $0\leq u \leq \frac{1}{2} , u \leq v \leq 1-u$.
$$\int_{0}^{\frac{1}{2}}\int_{u}^{1-u} \frac{2}{1+v^2-u^2}dvdu=\int_{0}^{\frac{1}{2}}\frac{2\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)-2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$$
For the first term on the right hand side, let $z=2\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right),dz=\frac{-1}{\sqrt{1-u^2}}$ and for the second term, use $2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=2\sin^{-1}\left(u\right)$ and let $z=\sin^{-1}\left(u\right),dz=\frac{1}{\sqrt{1-u^2}}du$ for this.When evaluating these two terms with these substitutions, the sum of these terms should be $\frac{\pi^2}{24}$.
So:
$$\iint_{T} \frac{2}{1+v^2-u^2}dvdu=\frac{2\pi^2}{24}=\frac{\pi^2}{12}$$
thus, $$I=\int_{0}^{1}\frac{\ln(x)}{1+x}dx=\frac{-\pi^2}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A curious approximation to $\cos (\alpha/3)$ The following curious approximation
$\cos\left ( \frac{\alpha}{3} \right ) \approx \frac{1}{2}\sqrt{\frac{2\cos\alpha}{\sqrt{\cos\alpha+3}}+3}$
is accurate for an angle $\alpha$ between $0^\circ$ and $120^\circ$
In fact, for $\alpha = 90^\circ$, the result is exact.
How can we derive it?
| Let $y = \cos \alpha$ and $x = \cos(\alpha/3)$ then we know that
$$y = x(4x^{2} - 3)\tag{1}$$
Your approximation says that
$$x \approx \frac{1}{2}\sqrt{\frac{2y}{\sqrt{y + 3}} + 3}\tag{2}$$
or using $(1)$ we get $$\frac{y}{x} = 4x^{2} - 3 \approx \frac{2y}{\sqrt{y + 3}}\tag{3}$$ Canceling $y$ we get $$\frac{1}{x}\approx \frac{2}{\sqrt{y + 3}}\tag{4}$$ or $$4x^{2} \approx y + 3$$ or $$y \approx 4x^{2} - 3$$ Comparing with $(1)$ we see that the above approximation is true if $x = 1$ and is good enough if $x$ is near $1$. So the overall approximation is good enough if $\alpha$ is near $0$. The approximation is also correct if $y = 0$ because in this case the LHS and RHS of $(3)$ are equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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How to find an upper triangular with $\ U^2 = I $ which gives $\ U $ is its own inverse It's obvious that $\ I $(identity matrix) of any size $\ N $ satisfies$\ I^2 = I$ so that $\ I $ is its own inverse.
However if we consider $\ N = 2$ and attempt to find such a triangular matrix $\ U \neq I $ we have this scenario:
$$\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} .\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} = I$$
$$\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} .\begin{bmatrix} u_1 & u_2 \\ 0 & u_3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
$$\begin{bmatrix} u_1^2 & u_1u_2 + u_2u_3 \\ 0 & u_3^2 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
After solving this we realize that $\ u_1 = 1,u_2, u_3 = -1$ is the form of the solution where $\ U \neq I $ for $\ N = 2 $.
But for $\ N \ge 3 $, what will be the form of such matrices?
| Here's one possible form. Let $a$ be any number. Define an $n \times n$ matrix as given below.
\begin{equation*}
A = \begin{bmatrix}
1 & a & \dfrac{a^2}{2} & \dfrac{a^3}{4}& \cdots & \dfrac{a^{n-1}}{2^{n-2}}\\
0 & -1 & -a & -\dfrac{a^2}{2} & \cdots & -\dfrac{a^{n-2}}{2^{n-3}}\\
0 & 0 & 1 & a & \cdots & \dfrac{a^{n-3}}{2^{n-4}}\\
0 & 0 & 0 & -1 & \cdots & -\dfrac{a^{n-4}}{2^{n-5}}\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \cdots & (-1)^{n-1}
\end{bmatrix}
\end{equation*}
Then $A^2 = I$ (verify).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$
I've tried this far, and I'm stuck
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x (x+12)- \log_x 4^3 &= -1 \\
\log_x(x+12)- \log_x 64 &= -1 \end{align}$$
then I substituted $4^{y+3x} = 64$
$\log_x (x+12) - \log_x 4^{y+3x} = -1$
I don't know what should I do next. any ideas?
| Now you have
\begin{equation}
\log_x\left(\frac{x+12}{64}\right)=-1
\end{equation}
Therefore
\begin{equation}
\frac{x+12}{64}=x^{-1}
\end{equation}
Which leads to
\begin{equation}
x^2-12x-64=0
\end{equation}
Which can be factored
\begin{equation}
(x+16)(x-4)=0
\end{equation}
But of course $x$ cannot equal $-16$ so it must equal $4$.
I believe you can take the problem from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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What to do when the integrating factor is a function of both x and y? I have to solve the following differential equation:
$$(\cos^2x + y \sin 2x) \frac{dy}{dx} + y^2 =0$$
using an integrating factor. An integrating factor that is a function of just $x$ or just $y$ won't work, so we need to find an integrating factor which is a function of both $x$ and $y$.
I have followed the instructions from this other post Finding integrating factor when IF will be a function of x and y which suggests looking for a separable solution:
$$u(x,y)=e^{\int G(x) dx} e^{\int F(y) dy}$$
where $G(x)=\frac{\delta u/\delta x}{u}$ and $F(y)=\frac{\delta u/\delta y}{u}$. However, I get stuck at this point:
$$G(x)(\cos^2 x + y \sin 2x) - F(y) y^2 = 2y + 2 \sin 2x - 2y \cos 2x$$
and I can't find a solution for $G(x)$ and $F(y)$ that would work.
Any ideas? Thanks in advance!
| Let's rewrite the equation in this form:
\begin{equation*}
\underbrace{y^2}_M \,\mathrm dx + \underbrace{(\cos^2 x + y \sin 2x)}_N \,\mathrm dy = 0.
\end{equation*}
Let $u(x,y)$ be an integrating factor. Then after multiplying the above equation by $u$, the resulting equation $uM \,\mathrm dx + uN \,\mathrm dy = 0$ should be exact. Thus, we must have
\begin{align*}
\dfrac{\partial (uM)}{\partial y} = \dfrac{\partial (uN)}{\partial x} \implies\\
u M_y + u_y M = u N_x + u_x N \implies\\
u(M_y - N_x) = u_x N - u_y M.
\end{align*}
Since $M = y^2$ and $N = \cos^2 x + y \sin 2x$, we have $M_y = 2y$ and $N_x = - \sin 2x + 2y \cos2x$, so that $M_y - N_x = \sin 2x + 2y (1 - \cos 2x) = 2\sin x \cos x + 4y \sin^2 x = 2\sin x (\cos x + 2y \sin x)$.
Note that $N = \cos^2 + 2y \sin x \cos x = \cos x (\cos x + 2y \sin x)$.
Thus, we have
\begin{align*}
M_y - N_x & = 2 \sin x (\cos x + 2y \sin x)\\
N & = \cos x (\cos x + 2y \sin x).
\end{align*}
Now, if it happened that $u$ were only a function of $x$, then $u_y = 0$, and our earlier condition for exactness would become
\begin{equation*}
u(M_y - N_x) = u_x N \implies \dfrac{u_x}{u} = \dfrac{M_y - N_x}{N}.
\end{equation*}
Indeed, this would be true if the RHS turned out to be a function of $x$ alone. Observe that this is so, for $\dfrac{M_y - N_x}{N} = 2 \tan x$.
Thus, $$\dfrac{u_x}{u} = 2 \tan x \implies \log u = \int 2 \tan x \,\mathrm dx = 2 \log \sec x.$$ Therefore, $$u = \sec^2 x$$ is the integrating factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating $\lim \limits_{x \to \infty} \frac{x+\frac12\cos x}{x-\frac12\sin x}$ using the sandwich theorem Calculating $\lim \limits_{x \to \infty} \dfrac{x+\frac12\cos x}{x-\frac12\sin x}$
Correct me if I'm wrong:
$\cos x$ and $\sin x$ are bounded so that
$$|\cos x|\le 1,\qquad |\sin x|\le1$$
Therefore I can say:
$$
\frac{x-\frac12}{x+\frac12}\le
\frac{x+\frac12\cos x}{x-\frac12\sin x}\le
\frac{x+\frac12}{x-\frac12}
$$
the limits of the left and right side are equal to 1, therefore the the limit I'm looking for is also equal to 1 . The answer is correct, but what I'm not sure is
$$
\frac{x-\frac12}{x+\frac12}\le\frac{x+\frac12\cos x}{x-\frac12\sin x}
$$
was this step correct?
| Just to make more explicit the steps, we can assume $x>1$, so $x-\frac{1}{2}\sin x>0$ and $x+\frac{1}{2}\cos x>0$.
From $|\sin x|<1$, we get
$$
x-\frac{1}{2}\le x-\frac{1}{2}\sin x\le x+\frac{1}{2}
$$
Similarly,
$$
x-\frac{1}{2}\le x+\frac{1}{2}\cos x\le x+\frac{1}{2}
$$
Since all terms are positive, from
$$
x-\frac{1}{2}\le x-\frac{1}{2}\sin x
\quad\text{and}\quad
x-\frac{1}{2}\le x+\frac{1}{2}\cos x
$$
we obtain
$$
\frac{x-\frac{1}{2}}{x+\frac{1}{2}}\le
\frac{x+\frac{1}{2}\cos x}{x-\frac{1}{2}\sin x}
$$
Similarly,
$$
\frac{x+\frac{1}{2}\cos x}{x-\frac{1}{2}\sin x}\le
\frac{x+\frac{1}{2}}{x-\frac{1}{2}}
$$
and the sandwich theorem allows to conclude.
There is a different and perhaps simpler way to prove the result. It is not restrictive to assume $x>0$.
Since $|\cos x|\le1$, we have
$$
-\frac{1}{2x}\le\frac{\cos x}{2x}\le\frac{1}{2x}
$$
implying $\lim_{x\to\infty}\frac{\cos x}{2x}=0$ by the sandwich theorem. Similarly, $\lim_{x\to\infty}\frac{\sin x}{2x}=0$, so
$$
\lim_{x\to\infty}\frac{x+\frac{1}{2}\cos x}{x-\frac{1}{2}\sin x}
=
\lim_{x\to\infty}\frac{1+\dfrac{\cos x}{2x}}{1-\dfrac{\sin x}{2x}}
=\frac{1+0}{1+0}=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Condition on $a$ for $(x^2+x)^2+a(x^2+x)+4=0$ Find the set of values of $a$ if $$(x^2+x)^2+a(x^2+x)+4=0$$ has
$(i)$ All four real and distinct roots
$(ii)$ Four roots in which only two roots are real and distinct.
$(iii)$ All four imaginary roots
$(iv)$ Four real roots in which only two are equal.
Now if I set $x^2+x=t$ then even if $t^2+at+4=0$ has real roots in is not necessary that $(x^2+x)^2+a(x^2+x)+4=0$ will have real roots too. So how to derive the condition on a?
Could someone give me some direction?
| Suppose we want to find the values of $a$ such that there are four real and distinct roots (values of $x$ such that the equation is true).
This means that if we let $x^2+x=t$ as you did, then we want the two possible values of $t$ to be distinct. Call them $t_1$ and $t_2$. We want that the following two equations both have two roots, and that all of the roots are distinct:
$$x^2+x-t_1=0$$
$$x^2+x-t_2=0$$
The quadratic formula gives
$$x=\frac{-1\pm\sqrt{1+4t_1}}{2}$$
$$x=\frac{-1\pm\sqrt{1+4t_2}}{2}$$
This means we want $t_1\neq t_2$, and $$t_1,t_2 > -\frac{1}{4}$$
Note that this inequality is strict, since a determinant of $0$ would give a double root.
We can now return to the equation involving $a$. We want to find which $a$ allow the two roots of the following equation to satisfy $t_1\neq t_2$ and $t_1,t_2 > -1/4$.
$$t^2+at+4=0$$
We use the quadratic formula again:
$$t_1=\frac{-a + \sqrt{a^2-16}}{2}$$
$$t_2=\frac{-a - \sqrt{a^2-16}}{2}$$
Since we want two distinct real values of $t$, $a$ must have absolute value strictly greater than $4$, to ensure a strictly positive determinant.
We also want $a$ such that both of these inequalities hold:
$$\frac{-a + \sqrt{a^2-16}}{2} > -\frac{1}{4}$$
$$\frac{-a - \sqrt{a^2-16}}{2} > -\frac{1}{4}$$
The second inequality implies the first, so it's sufficient that the second holds. It's equivalent to
$$2a < 1 - 2\sqrt{a^2-16}$$
Clearly all values of $a$ greater than $4$ don't work. We're left with $a < -4$.
At $a=-4$, $$1-2\sqrt{a^2}-16 > 2a$$
Let $f(a)=1-2\sqrt{a^2-16}$ and $g(a)=2a$. Notice that $f(a)$ is greater than $g(a)$ at $a=-4$. The equation $f(a)=g(a)$ has no solutions (to see why, subtract one from both sides and square). This means $f(a)$ never crosses $g(a)$, and since both functions are continuous on $a \in (-\infty, 4]$, we know that $f(a)$ is greater than $g(a)$ for all $a\leq -4$.
The values of $a$ such that there are four real and distinct roots are all $a$ such that $\boxed{a < -4}$.
This same method (using the quadratic formula first on $x^2+x=t$, then on $t^2+at+4=0$, doing casework at each step) can be applied to the other parts of your question. Good luck!
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$$
None of the terms on LHS seem to be divisible by 97!!
Direct calculation shows that: $97\mid 2^{24}+1$ , but how to find it mathematically (of course not using calculator)?
| An elementary method follows:
Let $S$ denote the set $\{1,\cdots,48\}.$ Then every integer $n$ satisfies $n\equiv\pm k\pmod{97}$ for some $k\in S$ and a suitable choice of the sign. Now we consider $2k$ for $k\in S.$
For $k=1,\cdots,24,$ it is clear that $2k\in S$ so that we can take $2k\equiv l\pmod{97}$ for $l=2k.$
For $k=25,\cdots,48,$ clearly $97-2k\in S,$ so that $2k\equiv -l\pmod{97}$ where $l=97-2k.$
Thus $$\begin{align}
2^{48}\prod\limits_{k\in S}k&\equiv\prod\limits_{k\in S}2k\\
&=\prod\limits_{1\le k\le24}2k\prod\limits_{25\le k\le48}2k\\
&\equiv\prod\limits_{1\le k\le24}l\prod\limits_{25\le k\le48}(-l)\\
&\equiv(-1)^{24}\prod\limits_{k\in S}k\pmod{97}
\end{align}$$
This shows that $2^{48}\equiv1\pmod{97}.$
Hope this helps.
Edit:
In fact this argument can also be generalized to prove that $\left(\dfrac{2}{p}\right)=(-1)^{(p^2-1)/8}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I thought maybe it's better to consider odd and even cases for $n$ but there's no unified rule here, because for example: $8^2\equiv{4}\pmod{10}$ and $8^4\equiv{6}\pmod{10}$. Any ideas??
| For $X>0$, and sequence indexed by $n > 0$:
$X^n$ repeats each $2 - 1$ in $\pmod 2$ since $2$ is prime.
$X^n$ repeats each $5 - 1$ in $\pmod 5$ since $5$ is prime.
So the sequence $X^n$ repeats each ${\rm gcd}(2 - 1, 5 - 1)$ in $\pmod {10}$, so you only have to check the value of
$$\sum_{X = 1}^9 X^n \pmod {10}$$
for the values $n \in \{1, 2, 3, 4\}$ and the sequence will repeat, and the case $n = 0$ should be considerred separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$ Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$
Let the circle passes through the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ and the plane $Ax+By+Cz+D=0$
So the equation of the circle is $x^2+y^2+z^2+2ux+2vy+2wz+d+\lambda(Ax+By+Cz+D)=0$
As it passes through $(-1,0,0)$ so
$1-2u+d+\lambda(-A+D)=0$
As it passes through $(0,2,0)$ so
$4+4v+d+\lambda(2B+D)=0$
As it passes through $(0,0,3)$ so
$9+6w+d+\lambda(3C+D)=0$
Here i am stuck.I cannot find the radius of the circle.Is my method correct or not?Is there simpler method possible?Please help.
| You essentially want to find the circumcenter $D$ of the triangle $\triangle_{ABC}$ with vertices
$$A = (-1,0,0), B = (0,2,0), C = (0,0,3)$$
Since $D$ is lying on the plane holding $A, B, C$, there exists $3$ real numbers
$\alpha,\beta,\gamma$ such that
$$D = \alpha A + \beta B + \gamma C\quad\text{ with }\quad \alpha + \beta + \gamma = 1$$
This $3$-tuple is called the baricenteric coordinate of $D$. They can be computed using the sides $a,b,c$ of the triangle alone:
$$\alpha : \beta : \gamma \;=\; a^2(-a^2 + b^2 + c^2) : b^2( a^2 - b^2 + c^2) : c^2(a^2 + b^2 - c^2)$$
For the triangle at hand, we have $(a^2,b^2,c^2) = (13,10,5)$. This leads to
$$\alpha : \beta : \gamma \;=\;
13(-13+10+5) : 10(13-10+5) : 5(13+10-5) = 26 : 80 : 90$$
As a result,
$\displaystyle\;D = \frac{26 A + 80 B + 90C}{26 + 80 + 90} = \left(-\frac{35}{98},\frac{40}{49},\frac{135}{98} \right).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction $\frac{7}{8}+7\left(\frac{1}{8}\right)^2+...+7\left(\frac{1}{8}\right)^n=1-\frac{1}{8^n}$ for every $n \in \mathbb N$
$\frac{7}{8}+7\left(\frac{1}{8}\right)^2+...+7\left(\frac{1}{8}\right)^n=1-\frac{1}{8^n}$ for every $n \in \mathbb N$
I first set $n=k$:
$\frac{7}{8}+7\left(\frac{1}{8}\right)^2+...+7\left(\frac{1}{8}\right)^k=1-\frac{1}{8^k}$
And then $n=k+1$:
$1-\frac{1}{8^k}+7\left(\frac{1}{8}\right)^{k+1}=1-\frac{1}{8^{k+1}}$
And then I tried some Algebra on it:
$1-\frac{1}{8^k}+\left(\frac{7}{8}\right)\left(\frac{7}{8}\right)^k=1-\frac{1}{8^{k+1}}$
But not really sure what to do from here. Hints?
| First comment: you should begin by proving the base case, i.e. for $n=1$.
That being said, for the inductive step:
$$
1-\frac{1}{8^k}+7\frac{1}{8^{k+1}}
=
1-\frac{1}{8^k}+\frac{7}{8}\frac{1}{8^{k}}
=
1-\frac{8}{8}\frac{1}{8^k}+\frac{7}{8}\frac{1}{8^{k}}=
1-\frac{1}{8}\frac{1}{8^k}
=1-\frac{1}{8^{k+1}}
$$
and you can conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 \\
\tan ^2 x & = 3
\end{align}
This is as far as I can get, and when I look at the Mark Scheme no other Trignometric Identities have been used. Thanks
| $$\tan x+\tan2x=\dfrac{\sin(x+2x)}{\cos x\cos2x}$$
So, we need $\sin3x=0\implies3x=n\pi$ where $n$ is any integer
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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evaluating the integral $\int \frac{e^x(x^4+2)dx}{(x^2+1)^{5/2}}$ It accidentally came out like this but I need a fine proof for it.
$$\int e^x\frac{d}{dx}\left(\ln\left(x+\sqrt{x^2+1}\right)+\dfrac{x}{(x^2+1)^{3/2}}\right) dx$$
I am unable to figure out why $$\int\dfrac{1-2x^2}{\left(1+x^2\right)^{5/2}}dx =\dfrac{x}{\left(x^2+1\right)^{3/2}}$$ (It comes in the middle of solving)
Please give hints on how to solve it.
| So you want to find a way to show:
$$\int\frac{1-2x^2}{\left(1+x^2\right)^{5/2}} \,\mbox{d}x =\dfrac{x}{\left(x^2+1\right)^{3/2}}$$
Let $x = \tan t$, then:
$$\int\dfrac{1-2x^2}{\left(1+x^2\right)^{5/2}} \,\mbox{d}x \to \int\dfrac{1-2\tan^2t}{\left(1+\tan^2t\right)^{5/2}} \sec^2 t \,\mbox{d}t$$
Now use $1+\tan^2 = \sec^2t$ and simplify:
$$\int\dfrac{1-2\tan^2t}{\sec^3 t} \,\mbox{d}t
= \int \cos t \left( 1-3\sin^2 t \right) \,\mbox{d}t =\sin t - \sin^3t + C$$
First rewrite (this will make it easier to substitute back and simplify):
$$\sin t - \sin^3t = \sin t \left( 1 - \sin^2t \right) = \sin t\cos^2t$$
Now you can substitute $t = \arctan x$ again and simplify:
$$\sin \left( \arctan x \right) \left( \cos \left( \arctan x \right) \right)^2 = \frac{x}{\sqrt{1+x^2}}\left( \frac{1}{\sqrt{1+x^2}} \right)^2 = \frac{x}{(1+x^2)^{3/2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)$ Wolfram Alpha evaluates this integral numerically as
$$\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=0.379064 \dots$$
Its value is apparently
$$\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)=0.37906401072\dots$$
How would you solve this integral?
Obviously, we can make a substitution $t=x^2$
\begin{align}
\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx&=\frac{1}{2} \int_0^{\infty} \frac{\sqrt{t}}{\cosh^2 (t)} dt\\[10pt]
&=\int_0^{\infty} \frac{\sqrt{t}}{\cosh (2t)+1} dt\\[10pt]
&=\frac{1}{2 \sqrt{2}}\int_0^{\infty} \frac{\sqrt{u}}{\cosh (u)+1} du
\end{align}
We could use geometric series since $\cosh (u) \geq 1$, but I don't know how it will help.
| $$I=\frac{1}{2\sqrt{2}}\int_{0}^{+\infty}\frac{\sqrt{u}\,du}{1+\cosh(u)}=\frac{1}{\sqrt{2}}\int_{1}^{+\infty}\frac{\sqrt{\log v}}{(v+1)^2}\,dv=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{\sqrt{-\log v}}{(1+v)^2}\,dv \tag{1}$$
but since
$$ \int_{0}^{1}v^k \sqrt{-\log v}\,dv = \frac{\sqrt{\pi}}{2(1+k)^{3/2}} \tag{2}$$
by expanding $\frac{1}{(1+v)^2}$ as a Taylor series we get:
$$ I = \frac{1}{\sqrt{2}}\sum_{n\geq 0}(-1)^n (n+1)\frac{\sqrt{\pi}}{2(1+n)^{3/2}} = \color{red}{\frac{\sqrt{\pi}}{2\sqrt{2}}\cdot\eta\left(\frac{1}{2}\right)}\tag{3}$$
and the claim follows from the well-known:
$$ \eta(s) = (1-2^{1-s})\,\zeta(s)\tag{4} $$
that gives an analytic continuation for the $\zeta$ function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
$(\sin^{-1} x)+ (\cos^{-1} x)^3$ How do I find the least and maximum value of $(\sin^{-1} x)+ (\cos^{-1} x)^3$ ?
I have tried the formula $(a+b)^3=a^3 + b^3 +3ab(a+b)$ , but seem to reach nowhere near ?
| If $a+b=k,$
Method $\#1:$
$$a^3+b^3=(a+b)^3-3ab(a+b)=k^3-3kab$$
Now $$(a+b)^2-4ab=(a-b)^2\ge0\iff-4ab\ge-(a+b)^2$$
Method $\#2:$
$$a^3+b^3=a^3+(k-a)^3=k^3-3k^2a+3ka^2=k^3+3k\left(a^2-ka\right)$$
Now $a^2-ka=\dfrac{(2a-k)^2-k^2}4\ge-\dfrac{k^2}4$
For both methods, here $$k=\dfrac\pi2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $U^*DU=D=V^*DV$ for diagonal $D$, is $U^*DV$ diagonal too? All the matrices mentioned are complex $n\times n$ matrices. Let $U, V$ be unitary matrices such that $U^*DU=V^*DV=D$ for a diagonal matrix $D$ with nonnegative diagonal entries. Does this imply that $U^*DV$ is also diagonal? All I understand is that $U^*V$ will commute with $D$.
| In general, no. Consider the permutation matrices
$$
E_1 \;\; =\;\; \left [ \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right ] \;\;\;\;\; E_2 \;\; =\;\; \left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array} \right ].
$$
These will diagonalize any diagonal matrix $D$ using the transformation you have above, but as Niko mentioned in the comments, if we consider $E_1^*IE_2$ we will just obtain
$$
E_1^*E_2 \;\; =\;\; \left [ \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right ] \left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{array} \right ].
$$
Even with some diagonal matrix $D$ we obtain
$$
E_1^*DE_2 \;\; =\;\; \left [ \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right ] \left [ \begin{array}{ccc}
d_1 & 0 & 0 \\
0 & d_2 & 0 \\
0 & 0 & d_3 \\
\end{array} \right ] \left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{ccc}
0 & 0 & d_2 \\
d_1 & 0 & 0 \\
0 & d_3 & 0 \\
\end{array} \right ].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[ \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ?
|
Integration by parts works fine provided one writes the integral $I$ as
$$I=\lim_{\epsilon\to 0^+}\lim_{L\to \infty}\int_{\epsilon}^L\frac{x\log(x)}{(1+x^2)^2}\,dx$$
Then, proceeding with integration by parts, we find
$$\begin{align}
I&=\frac14\lim_{\epsilon\to 0^+}\lim_{L\to \infty}\left.\left(\frac{2x^2\log(x)}{1+x^2}-\log(1+x^2)\right)\right|_{\epsilon}^L\\\\
&=\frac14\lim_{L\to \infty}\left(\frac{2L^2\log(L)}{1+L^2}-\log(1+L^2)\right)\\\\
&-\frac14\lim_{\epsilon\to 0^+}\left(\frac{2\epsilon^2\log(\epsilon)}{1+\epsilon^2}-\log(1+\epsilon^2)\right)\tag 1
\end{align}$$
Now, since $\lim_{\epsilon\to 0^+}\epsilon^2 \log(\epsilon)=0$, the contribution from the evaluation of the lower limit vanishes.
To evaluate the contribution from the upper limit, we write
$$\begin{align}
\frac{2L^2\log(L)}{1+L^2}-\log(1+L^2)&=\frac{2L^2\log(L)}{1+L^2}-2\log(L)-\log(1+1/L^2)\\\\
&=-2\frac{\log(L)}{1+L^2}-\log(1+1/L^2)
\end{align}$$
which clearly approaches zero as $L\to \infty$.
And that is that!
| {
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"source": "stackexchange",
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Why my calculations aren't right? (Maclaurin series) Good evening to everyone!
I tried to calculate $ \cos\left( x- \frac{x^3}{3} + o(x^4)\right) $ using the MacLaurin series but instead of getting the final result equal to $1 - \frac{x^2}{2}+\frac{3x^4}{8} + o(x^4)$ I got this:
$$
\cos\left( x- \frac{x^3}{3} + o(x^4)\right) = 1-\frac{\left(x-\frac{x^3}{3}+o\left(x^4\right)\right)^2}{4}+o\left(x^5\right) = 1-\frac{x^2+\frac{x^6}{9}+o\left(x^8\right)-\frac{2x^4}{3}+2xo\left(x^4\right)-\frac{2x^3o\left(x^4\right)}{3}}{4}+o\left(x^5\right) = 1-\frac{x^2}{4}+\frac{x^6}{36}-\frac{x^4}{6}+o\left(x^4\right)
$$
Where for expanding what's between the parenthesis I'm using the formula $(a-b+c)^2 = a^2+b^2+c^2-2ab+2ac-2bc $.
I would really want to know where I'm doing wrong. Thanks for any possible answers.
| In the first line you should consider the third part of MacLaurin series of $cos(u)$ which is $+\frac{u^4}{4!}$. This way, you would get the right result.
In fact, by now, your first line is not correct because there are some coefficient of $x^4$ in $\frac{(x-\frac{x^3}{3}+o(x^4))^4}{4!}$.
Moreover the second term of MacLaurin series of $cos(u)$ is $-\frac{x^2}{2!}$. You need to correct this as well.
| {
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Finding the second derivative of $f(x) = \frac{4x}{x^2-4}$. What is the second derivative of
$$f(x) = \frac{4x}{x^2-4}?$$ I have tried to use the quotient rule but I can't seem to get the answer.
| To simplify the differentiation, we can first rewrite the function as a sum of partial fractions. To do this, we assert that $\frac{4x}{x^2 - 4}$ can be written as $\frac{A}{x + 2} + \frac{B}{x - 2},$ with $A$ and $B$ constants. If this is the case, then clearly, $(A + B)x + 2(B - A) = 4x,$ from which we conclude that $A = B = 2.$ The decomposition is thus $\frac{2}{x + 2} + \frac{2}{x - 2}.$
Let us differentiate this expression twice. Notice that the first derivative can be computed as
$$\frac{d}{dx}\left[\frac{2}{x + 2} + \frac{2}{x - 2}\right]$$
$$= -2(x + 2)^{-2} - 2(x - 2)^{-2}.$$
The second derivative, likewise, is easy to compute:
$$f''(x) = \boxed{4(x + 2)^{-3} + 4(x - 2)^{-3}}.$$
And we are done! Hope this helped!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $ I need help with this exercise from the book What is mathematics? An Elementary Approach to Ideas and Methods. Basically I need to proove:
$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $$
$i)$ Particular cases
$ Q(1) = \frac{1}{2} ✓ $ ---------- $ P(1) = 2 - \frac{1+2}{2^1} = \frac{1}{2} ✓ $
$+ = 1 ✓ (=Q(1)+Q(2))$
$ Q(2) = \frac{2}{4}(=\frac{1}{2})$----$P(2) = 2 - \frac{2+2}{2^2} = 1 $ ✓
$ii)$ Hypothesis
$$F(k)=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{k}{2^k} = 2 - \frac{k+2}{2^k} $$
$iii)$ Proof
$P1 | F(k+1) = ( 2 -\frac{k+2}{2^k} ) + ( \frac{k+1}{2^{k+1}}) = 2 - \frac{k+3}{2^{k+1}} $
$P2 | \frac{2^{k+1}-k-2}{2^k} + \frac{k+1}{2^{k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
$P3 | \frac{2^{2k+2}-2^{k+1}K-2^{k+2}+2^kk+2k}{2^{2k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
$P4 | \frac{2(2^{k+1}-2^kk-2^{k+1}-2^{k-1}k+k)}{2^{k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
I get stuck here; if you could help me, that would be really kind.
Thanks in advance.
| In $P2$ you have made a misktake. It will be, $$\dfrac{2^{k+1}-k\color{red}{-}2}{2^k}+\dfrac{k+1}{2^{k+1}}=\dfrac{2^{k+2}-k\color{red}{-}3}{2^{k+1}}$$
Now if you simplify LHS then,
$$\dfrac{2^{k+2}-2k-4+k+1}{2^{k+1}}=\dfrac{2^{k+2}-k-3}{2^{k+1}}=2-\dfrac{k+3}{2^{k+1}}$$and you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving integral without partial fraction I would like to expand my pool of integral solving skills and thus try to solve older problems again, however with a different method I had used back then, when I encountered them first. For this problem I omitted the lower and upper bound and just compute the indefinite integral (values are not important to me)
One of them was:
\begin{align}
& \int \frac{x^2-x+2}{x^3-x^2+x-1}dx = \int \frac{x(x-1)+2}{x^2(x-1)+(x-1)}dx \\[10pt]
= {} & \int \frac{x(x-1)}{x^2(x-1)+(x-1)} dx + \int \frac{2}{x^2(x-1)+(x-1)}dx \\[10pt]
= {} & \int \frac{x}{x^2+1}dx + \int \frac{2}{(x^2+1)(x-1)} dx \\[10pt]
= {} & \frac{1}{2} \log(1+x^2) + \int \frac{2(x+1)}{(x^2+1)(x^2-1)} dx \\[10pt]
= {} & \frac{1}{2} \log(1+x^2) + 2 \int \frac{1}{x^4-1} dx + 2 \int \frac{x}{(x^2+1)(x^2-1)} dx \\[10pt]
= {} & \frac{1}{4} (\log(1-x^2)-\log(1+x^2)) + \frac{1}{2} \log(1+x^2) + 2 \int \frac{1}{x^4-1} dx \end{align}
However now I am stuck. I am sure, if I looked up in a formula table I would find something to the last integral, but not being able to solve this, means that I am missing some kind of insight or tool to further proceed. Does someone have an idea what to do here, for the last integral?
Any constructive comment, answer is appreciated. As always thanks in advance.
| Be smart, add zeros!
$$I=\int dx\frac{1}{x^4-1}=\\
\int dx\frac{1+\overbrace{x^2-x^2}^{=0}}{(x^2-1)(x^2+1)}=\\\int dx\frac{1}{x^2-1}-\int dx\frac{x^2}{(x^2-1)(x^2+1)}=\\
\int dx\frac{1}{x^2-1}-\int dx\frac{x^2\overbrace{-1+1}^{=0}}{(x^2-1)(x^2+1)}=\\
\int dx\frac{1}{x^2-1}-\int dx\frac{1}{x^2+1}-\underbrace{\int dx\frac{1}{(x^2-1)(x^2+1)}}_{I}
$$
therefor
$$
2I=\int dx\frac{1}{x^2-1}-\int dx\frac{1}{x^2+1}=\text{arctanh}(x)-\arctan(x)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $\sin x + \csc x =2 \tan x$. Find value of $\cos^9x +\cot^9x +\sin^7x$ Problem:
If $\sin x+\csc x=2\tan x$, Find value of $\cos^9x+\cot^9x+\sin^7x$
Solution:
\begin{align*}&\sin x+\csc x=2\tan x \\ &\sin x+\frac{1}{\sin x}=2\frac{\sin x}{\cos x} \\ &\sin^2x+1=2\frac{\sin^2x}{\cos x} \\ &\sin^2x\cos x+\cos x=2\sin^2x \\ &(1-\cos^2x)\cos x+\cos x=2(1-\cos^2x) \\ &\cos^3x-2\cos^2x-2\cos x+2=0\end{align*}
Am I doing right ?
How to do further ?
| This is not really an answer, but a comment with image indicating that something is wrong with the problem in case the options are $1$, $0$, $-1$ and $2$ as stated in a comment above.
In the picture below I have let Mathematica draw the graphs of $\sin x+\csc x$ (blue), $2\tan x$ (yellow), $\cos^9x+\cot^9x+\sin^7x$ (green), $1$ (red) and $-1$ (purple).
It should be clear that, when the blue and yellow graphs meet each other, the green is not at any of the values stated in the problem.
To make it more evident, here is a zoomed graph around the first of the two points where the graphs meet:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q)^2]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P3| \frac{x\color{red}{q^{x+1}}+[-(x+1)]\color{red}{q^x}+1+[(x+1)(1-q)^2]\color{red}{q^x}}{(1-q)^2} = \frac{x\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2} | $$
Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' $\color{red}{q^x}$ are added next.
$$P4| \frac{xq^{x+1}+[-(x+1)+(x+1)(<1^2q^0+\binom{2}{1}1q-1^0q^2>)]q^x+1}{(1-q)^2}=\frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
$$P5 | \frac{xq^{x+1}+[2xq-xq^2+2q-q^2]q^x+1}{(1-q)^2} = \frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated.
Thanks in advance.
| Here is an alternative approach, motivated by the fact that it's usually wise in such problems to multiply the $(1-q)$ factors through:
\begin{align}
(1-q)^2(1+2q+3q^2+\cdots +nq^{n-1})
&=(1-q)\cdot (1-q)(1+2q+3q^2+\cdots +nq^{n-1})\\
&=(1-q)\cdot (1+q+q^2+\cdots +q^{n-1}-nq^{n})\\
&=1-(n+1)q^{n}+nq^{n+1}.
\end{align}
Induction can be used to verify that multiplying by $(1-q)$ cancels like terms as above.
| {
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"url": "https://math.stackexchange.com/questions/1825825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx $ $$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx$$ where $a>b>0$
What shall I do?
I have diffucty when I meet multi value function.
| If we apply the substitution $x=y^3$ we get:
$$ I(a,b) = 3\int_{0}^{+\infty}\frac{y^3\,dy}{(y^3+a)(y^3+b)}=\frac{3}{a-b}\int_{0}^{+\infty}\frac{a\,dy}{y^3+a}-\frac{3}{a-b}\int_{0}^{+\infty}\frac{b\,dy}{y^3+b}$$
but if we set, for any $c>0$,
$$ J(c) = \int_{0}^{+\infty}\frac{c\,dy}{y^3+c},$$
we simply have $J(c)= c^{1/3}\,J(1)$ through the substitution $y=c^{1/3} z$. At last,
$$ J(1) = \int_{0}^{+\infty}\frac{dy}{y^3+1}=\int_{0}^{1}\frac{dy}{y^3+1}+\int_{0}^{1}\frac{y\,dy}{y^3+1}=\int_{0}^{1}\frac{dy}{1-y+y^2}=\frac{2\pi}{3\sqrt{3}},$$
hence:
$$ \int_{0}^{+\infty}\frac{x^{1/3}\,dx}{(x+a)(x+b)}=\color{red}{\frac{2\pi}{\sqrt{3}}\cdot\frac{a^{1/3}-b^{1/3}}{a-b}}.$$
By considering the limit of the RHS as $b\to a$, we also have:
$$ \int_{0}^{+\infty}\frac{x^{1/3}}{(x+a)^2}\,dx = \frac{2\pi}{3\sqrt{3}\,a^{2/3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find stationary points of the function $f(x,y) = (y^2-x^4)(x^2+y^2-20)$ I have problem in finding some of the stationary points of the function above. I proceeded in this way: the gradient of the function is:
$$ \nabla f = \left( xy^2-3x^5-2x^3y^2+40x^3 ; x^2y+2y^3-x^4y-20y \right) $$
So in order to find the stationary points, I must resolve the system:
$$ \begin{cases}
xy^2-3x^5-2x^3y^2+40x^3 = 0 \\
x^2y+2y^3-x^4y-20y = 0
\end{cases} $$
So far I've found the points:
$$ (0,0) \qquad \left(\pm 2 \sqrt{10 \over 3} , 0 \right) \qquad (0, \pm \sqrt{10}) $$
But, I'm still blocked when I've to found the points deriving by the system:
$$
\begin{cases}
2x^6 + 3x^4 +x^2 -20 = 0 \\
y^2 = \frac{1}{2} \left( x^4 - x^2 + 20 \right) \end{cases}
$$
Which I don't know how to solve. Can someone help me ? Thanks.
| As you implied, you can factor the original system as $x(3x^4-40x^2+2x^2y^2-y^2)$ and $y(x^4-x^2-2y^2+20)=0$.
Taking $y=0$ gives you $x=0,x=\pm2\sqrt{\frac{10}{3}}$. Taking $x=0$ gives you $y=\pm\sqrt{10}$.
So you are left to solve $3x^4-40x^2+2x^2y^2-y^2=0,x^4-x^2-2y^2+20=0$ and $x,y\ne0$. Substituting from the second into the first gives $2x^6+3x^4-39x^2-20=0$ (you dropped the $-40x^2$ here), which factorises as: $$(x-2)(x+2)(x^2+5)(2x^2+1)$$
Can I leave you to finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $5^n$ divides $F_{5^n}$. If $F_n$ denotes the $n$-th Fibonacci number ($F_0 = 0, F_1 = 1, F_{n+2} = F_{n+1} + F_n$), show that $5^n$ divides $F_{5^n}$.
| Note that
$$
\begin{pmatrix}a+b & a \\ a & b\end{pmatrix}^5
=
\begin{pmatrix}* & c \\ c & *\end{pmatrix}
$$
where $c=5 a (a^4 + 3 a^3 b + 4 a^2 b^2 + 2 a b^3 + b^4)$. Therefore, if $5^n$ divides $a$, then $5^{n+1}$ divides $c$.
Apply this to
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n
=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
for a proof by induction that $5^n$ divides $F_{5^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Proving that $3 + 3 \times 5 + 3 \times 5^2 + \cdots+ 3 \times 5^n = [3(5^{n+1} - 1)] / 4$ whenever $n \geq 0$
Use induction to show that $$3 + 3 \times 5 + 3 \times 5^2 + \cdots+ 3 \times 5^n= \frac{3(5^{n+1} - 1)}{4} $$whenever $n$ is a non-negative integer.
I know I need a base-case where $n = 0$:
$$3 \times 5^0 = \frac{3(5^{0+1} - 1)}{4}\\LHS = 3 = \frac{12}{4} = RHS$$
Next I need to show that this is true for the $n + 1$ (next) term through a proof using induction. This is really where I could use a concrete example of a proof; I have yet to find one that I could really understand.
| You are trying to show that
$$\sum_{n=0}^{N}5^n=\frac{5^{N+1}-1}{4}$$
We can leave out the factor of $3$ since it just multiplies both sides. The base case is simple, you just have $1=1$. Now assume it is true for $N$. Then we have
$$\sum_{n=0}^{N+1}5^n=\sum_{n=0}^{N}5^n+5^{N+1}=\frac{5^{N+1}-1}{4}+5^{N+1}=\frac{5^{N+1}-1+4\cdot 5^{N+1}}{4}=\frac{5\cdot 5^{N+1}-1}{4}$$
Which gives
$$\sum_{n=0}^{N+1}5^{n}=\frac{5^{(N+1)+1}-1}{4}$$
Which proves that the statement is true for $N+1$. Thus, the statement is true by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the value of $\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}$ if $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$? If $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$$ then find the values of $$\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}.$$ How can I solve it? Please help me. Thank you in advance.
| You have $$a+b+c = (a+b+c)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right) = \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} + a+b+c.$$
Then you get $$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?
Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
| For $\theta$ an arbitrary constant, we have
\begin{equation}
x^4+2x^2\cos2\theta+1=(x^2-2x\sin\theta+1)(x^2+2x\sin\theta+1)\tag1
\end{equation}
Now, let’s evaluate
\begin{equation}
I=\int_0^\infty\frac{1}{x^4+2x^2\cos2\theta+1}\ dx\tag2
\end{equation}
By making substitution $x\mapsto\frac1x$, then
\begin{equation}
I=\int_{0}^\infty\frac{x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag3
\end{equation}
Adding $(2)$ and $(3)$, we get
\begin{equation}
I=\frac12\int_{0}^\infty\frac{1+x^2}{x^4+2x^2\cos2\theta+1}\ dx=\frac14\int_{-\infty}^\infty\frac{1+x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag4
\end{equation}
Since the integrand is even, an odd function like $2x\sin\theta$ doesn't change the value of the integral, so by using $(1)$ we have
\begin{align}
I&=\frac14\int_{-\infty}^\infty\frac{x^2+2x\sin\theta+1}{x^4+2x^2\cos2\theta+1}\ dx\\[10pt]
&=\frac14\int_{-\infty}^\infty\frac{1}{x^2-2x\sin\theta+1}\ dx\\[10pt]
&=\frac14\int_{-\infty}^\infty\frac{1}{(x-\sin\theta)^2+\cos^2\theta}\ dx\\[10pt]
&=\frac1{4\cos\theta}\int_{-\infty}^\infty\frac{1}{y^2+1}\ dy\quad\longrightarrow\quad y={x-\sin\theta\over\cos\theta}\\[10pt]
&=\frac\pi{4\cos\theta}
\end{align}
For $\theta=\frac\pi3$, then
\begin{equation}
\int_0^\infty\frac{1}{x^4+x^2+1}\ dx=\frac{\pi}{2\sqrt3}
\end{equation}
as announced.
For $J$ we use
\begin{equation}
x^8+2x^4\cos2\theta+1=(x^4-2x^2\sin\theta+1)(x^4-2x^2\sin\theta+1)
\end{equation}
Then apply the same methods for
\begin{equation}
J=\int_0^\infty\frac{1}{x^8+2x^4\cos2\theta+1}\ dx
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 6
} |
$5^{th}$ degree polynomial expression
$p(x)$ is a $5$ degree polynomial such that
$p(1)=1,p(2)=1,p(3)=2,p(4)=3,p(5)=5,p(6)=8,$ then $p(7)$
$\bf{My\; Try::}$ Here We can not write the given polynomial as $p(x)=x$
and $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ for a very complex system of equation,
plz hel me how can i solve that question, Thanks
| Let's do it in the most elementary way. Let $$Q(x)=P(x+1)-P(x)-x+2 \tag{1}$$Observe that $Q$ is of degree $4$ and $Q(3)=Q(4)=Q(5)=0$. Therefore we can write$$Q(x)=a(x-3)(x-4)(x-5)(x-b) \tag{2}$$You have also from $(1)$ that $Q(1)=Q(2)=1$, which after substitution in $(2)$ you get $a=-1/8$ and $b=2/3$. So $$Q(6)=-\frac{1}{8}(6-3)(6-4)(6-5)\left(6-\frac{2}{3} \right)=-4$$And finally$$P(7)=Q(6)+P(6)+6-2=-4+8+6-2=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Find the value of $ [1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$ Assume that [x] is the floor function. I am not able to find any patterns in the numbers obtained. Any suggestions?
$$[1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$$
| Starting from $n=0$, even-$n =2k$ terms are $(4^k - 1)/3$ and the subsequent odd-$n = 2k+1$ terms are $2\times (4^k- 1)/3$. So overall:
\begin{align}
\sum_{k=0}^{49}\left[(1+2)\frac{4^{k} - 1}{3}\right] + \frac{4^{50}-1}{3} &= \sum_{k=0}^{49}(4^{k} - 1) + \frac{4^{50}-1}{3} \\
& = \sum_{k=0}^{49}(4^{k}) + \frac{4^{50}-1}{3}-50 \\
&= \frac{4^{50}-1}{3}+ \frac{4^{50}-1}{3}-50 \\
&= \frac{2}{3}(4^{50}-1)-50 \\
&=845100400152152934331135470200 \tag{W|A}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Find $y(2) $ given $y(x)$ given a separable differential equation Find what $y(2)$ equals if $y$ is a function of $x$ which satisfies:
$x y^5\cdot y'=1$ given $y=6$ when $x=1$
I got $y(2)=\sqrt{6\ln(2)-46656}$
but this answer is wrong can anyone help me figure out the right answer and how I went wrong?
| $$
x \, y^5\, y' = 1 \Rightarrow \\
\int y^5 \, dy = \int \frac{dx}{x} \Rightarrow \\
\frac{1}{6} y^6 = \ln(x) + C \quad (x > 0)
$$
Inserting $y(1) = 6$ gives
$$
\frac{1}{6} 6^6 = \ln(1) + C \Rightarrow \\
C = 6^5
$$
so we got the solution:
$$
y^6 = 6 \ln(x) + 6^6 \quad (x > 0)
$$
which yields
$$
y(2)^6 = 6 \ln(2) + 6^6 \iff \\
y(2) = \left(6 \ln(2) + 6^6\right)^{1/6} = 6.000089135987335\dotsb
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taking Mod on both sides, mathematically correct? When given a equation containing complex numbers such as
$$ \frac{a+ib}{c+id} = x + iy$$
and required to prove
$$ \frac{a^2 +b^2}{c^2+d^2} = x^2 + y^2$$
Is taking the mod of both sides a legal mathematical step? I ask so because my textbook first finds the conjugate of both sides and multiplies the conjugate with the initial equation. This is done everywhere a case like the above one occurs. Hence im a bit confused if the way im thinking of proceeding is mathematically legal.
| Yes, taking the mod of both sides is mathematically valid, but we don't necessarily need to do that.
Convert to polar:
$$\frac{\sqrt{a^2+b^2}\text{cis}(\theta_1)}{\sqrt{c^2+d^2}\text{cis}(\theta_2)}=\sqrt{x^2+y^2}\text{cis}(\theta_3)$$
Separate the moduli and arguments:
$$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}\sqrt{x^2+y^2}}=\frac{\text{cis}(\theta_3)\text{cis}(\theta_2)}{\text{cis}(\theta_1)}=\text{cis}(\theta_3+\theta_2-\theta_1)$$
The left-side is obviously a positive real number, but the only positive real number in the range of $\text{cis}$ is $1$, so we have:
$$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}\sqrt{x^2+y^2}}=1$$
Multiply both sides by $\sqrt{x^2+y^2}$:
$$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}=\sqrt{x^2+y^2}$$
Square both sides:
$$\frac{a^2+b^2}{c^2+d^2}=x^2+y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Q23 from AMC 2012 If $abc+ab+bc+ac+a+b+c=104$, and $a,b,c>0$, what is the value of $ a^2 + b^2 +c^2$?
I tried to make $a,b,c$ the subject of the equation and tried to add them up but it's weird...
I also have lots of more questions which I will ask later from AMC 2012, it's more difficult than what I usually do...
| Given, $abc+ab+bc+ac+a+b+c=104$, and $a,b,c>0$
$abc+ab+bc+ac+a+b+c=104$
$(a+1)(b+1)(c+1)-1=104$
$(a+1)(b+1)(c+1)=105$
$(a+1)(b+1)(c+1)=3\cdot5\cdot7$
Without the loss of generality, let $a\leq b\leq c$
$a+1=3, b+1=5, c+1=7$
$a=2, b=4, c=6$
$\;\therefore\;a^2+b^2+c^2=2^2+4^2+6^2$
Hence, $a^2+b^2+c^2=56$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}^{\infty}{x\over (x^4-x^2+1)^2}dx={1\over 2}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
Substituion of $y=x^3$
$$\int_{-\infty}^{\infty}{x^3\over (x^4-x^2+1)^2}dx={1\over 4}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
As for $\int_{-\infty}^{\infty}{x^2\over (x^4-x^2+1)^2}dx$ and $\int_{-\infty}^{\infty}{x^4\over (x^4-x^2+1)^2}dx$ are difficult to find a suitable substitution. This is the point where I am shrugged with to find a suitable substitution To lead me to a particular standard integral. Need some help, thank.
standard integral of the form
$$\int{1\over (ax^2+bx+c)^2}dx={2ax+b\over (4ac-b^2)(ax^2+bx+c)}+{2a\over 4ac-b^2}\int{1\over ax^2+bx+c}dx$$
And
$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$
| Evaluating the integral of interest can be reduced to evaluating the integral
$$I(a,b)=\int_0^\infty \frac{1}{x^4+bx^2+a} \,dx\tag 1$$
To see this, we exploit first odd symmetry to write the integral of interest as
$$\begin{align}
\int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx&=2\int_0^\infty \frac{x^4+(2\pi +1)x^2+\pi^2}{(x^4-x^2+1)^2}\,dx \tag 2
\end{align}$$
Enforcing the substitution $x\to 1/x$ in $(2)$ reveals
$$\begin{align}
\int_0^\infty \frac{x^4}{(x^4-x^2+1)^2}\,dx&=\int_0^\infty \frac{x^2}{(x^4-x^2+1)^2}\,dx\\\\
&=-\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)} \tag 3
\end{align}$$
Moreover, we have
$$\int_0^\infty \frac{1}{(x^4-x^2+1)^2}\,dx=-\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)} \tag 4$$
Using $(3)$ and $(4)$ in $(2)$ yields
$$\int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx=-4(\pi +1)\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)}-2\pi^2\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)}$$
where $I(a,b)$ as given by $(1)$ can be found using partial fraction expansion or contour integration, for examples.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How to determine which of the following matrices are similar? If we have the following three matrices:
$$
A=\begin{bmatrix}
7 &1 \\
-5 &3
\end{bmatrix},\;\;
B=\begin{bmatrix}
5 &-1 \\
1 &5
\end{bmatrix},\;\;
C=\begin{bmatrix}
5 &1 \\
1 &5
\end{bmatrix}.
$$
What is the right procedure to determine if matrices are similar? I know that if the matrices are similar, then the matrices have the samen eigenvalues. Using the negative of this (determine the eigenvalues of each matrix, if they are not similar they cannot be similar), I got that only $A$ and $B$ potentially could be similar.
I also know that if the 2 matrices $A$ and $B$ are similar then $A=PBP^{−1}$ but I don't know how to work this out further.
So how do I show that matrices are similar, using $A=PBP^{-1}$ for example to show it for $A$ and $B$?
| As similar matrices have similar determinants, matrix $C$ is not similar to $A$ or $B$ as $\det C = 24$ but $\det A = \det B = 26$.
Now we notice that $A$ and $B$ have the similar egeinvalues from characteristic equation
$$
\lambda^2 - 10\lambda + 26 = 0
$$
which gives
$$
\lambda_{1,2} = 5 \pm i.
$$
We may diagonilize bouth matrises $A$ and $B$ and get that $A = PDP^{-1}$ and $B = QDQ^{-1}$ where
$$
D = \begin{pmatrix}
5 - i & 0\\
0 & 5 + i
\end{pmatrix}.
$$
Now one may see that matrix $T = QP^{-1}$ satisfies equation $B = TAT^{-1}$:
$$
T =
\begin{pmatrix}
-i & i \\ 1 & 1
\end{pmatrix}
\begin{pmatrix}
-\frac{5i}{2} & \frac{1}{2}-i \\ \frac{5i}{2} & \frac{1}{2}+i
\end{pmatrix} =
\begin{pmatrix}
-5 & -2 \\ 0 & 1
\end{pmatrix},
$$
$$
T^{-1} =
\frac{1}{5}\begin{pmatrix}
-1 & -2 \\ 0 & 5
\end{pmatrix},
$$
$$
TAT^{-1} =
\frac{1}{5}\begin{pmatrix}
-5 & -2 \\ 0 & 1
\end{pmatrix}\begin{pmatrix}
7 & 1 \\ -5 & 3
\end{pmatrix}\begin{pmatrix}
-1 & -2 \\ 0 & 5
\end{pmatrix} = \begin{pmatrix}
5 & -1 \\ 1 & 5
\end{pmatrix} = B.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$x, y , z$ are respectively the $sines$ and $p, q, r$ are respectively the $cosines$ of the ..... $x, y , z$ are respectively the $sines$ and $p, q, r$ are respectively the $cosines$ of the angles $\alpha, \beta, \gamma$, which are in A.P. with common difference $\frac{2\pi}{3}$.
1. $yz + zx + xy = ?$
2. $x^2 (qy - rz) + y^2 (r - px) + z^2 (px - qy) = ?$
For the first I tried using the formula for $2sinAsinB$ but it is not helping. How do I solve them? Please, help.
| WLOG let $x=\sin\left(A-\dfrac{2\pi}3\right),y=\sin A,z=\sin\left(A+\dfrac{2\pi}3\right)$
$$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)$$
We can prove $x+y+z=0$
Using $\cos2B=1-2\sin^2B,$
$$2(x^2+y^2+z^2)=3-\left\{\cos\left(2A-\dfrac{4\pi}3\right)+\cos2A+\cos\left(2A+\dfrac{4\pi}3\right)\right\}$$
Now $\cos\left(2A-\dfrac{4\pi}3\right)+\cos\left(2A+\dfrac{4\pi}3\right)=2\cos2A\cos\dfrac{4\pi}3=?$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ such that $Q(x)|P(x)$, find $a+b$
Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ be the polynomials where $a$ and $b$ are real numbers. If polynomial $P$ is divisible by $Q$, what is the value of $a+b$.
This is what I have tried so far: Since $Q(x)|P(x)$ we have $P(1)=0$, therefore $a-b+1=0$. Problem is because we cannot obtain system of equations, because polynomial $Q(x)$ has double root at $x=1$. From equation $a-b+1$ we cannot find $a+b$, so how to find out value of $a+b$ or $a^2-b^2$?
| Note that $Q(x) = (x-1)^2$, which means that $(x-1) \mid P(x)$ and $(x-1) \mid P'(x)$. In other words $x=1$ is a zero of both $P(x)$ and $P'(x)$. So after all you're left to solve the system of linear equation:
$$\begin{cases} a-b+1 = 0 \\ 2014a - 2015b = 0 \end{cases}$$
It's easy to conclude that $a=-2015$ and $b=-2014$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?
Thanks
| Write $x=1/a$, $y=1/b$ and $z=1/c$. We get $$xy+yz+zx\leq x^2+y^2+z^2$$ Now use that $$p^2+q^2\geq 2pq$$
for each pair $p,q\in \{x,y,z\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisble by $11$. How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisible by $11$.
My attempt-
A number is divisible by 11 if the alternating sum of its digit is divisible by 11?
The other thing to notice is as it is a 9 digit number formed by digits 1 to 9, exactly once each digit from 1 to 9 will appear in the number.
Basically, the question boils down to how many ways we can arrange 123456789 so that the alternating sum of the digit is divisible by 11.
I am not able to proceed further. Any help would be appreciated.
| $$1+2+...+9=\frac{9\cdot10}{2}=45\\(x_1+x_3+x_5+x_7+x_9)-(x_2+x_4+x_6+x_8)=11m$$
$m$ must be $1$ or $3$ since 45 is odd and $3$ is discarded immediately so we have
$$\begin{cases}X+Y=45\\X-Y=11\end{cases}$$ Hence $X=28$ and $Y=17$ We work with $28$.
1) With $9$ and $8$ one has $x_1+x_2=11$ and $x_1+x_2+x_3=11$ (because $28-(9+8)=11$).
$x_1+x_2=7+4=6+5$ give $2$ cases.
$x_1+x_2+x_3=7+3+1=6+4+1=6+3+2=5+4+2$ give $4$ cases.
2) With $9$ and $7$ without $8$ one has $x_1+x_2=12$ and $x_1+x_2+x_3=12$.
$x_1+x_2=12$ no cases since $6+5=11\lt12$.
$x_1+x_2+x_3=6+5+1=6+4+2=5+4+3$ give $3$ cases.
With $9$ and $6$ without $7$ and $8$ is not possible because $9+6=15$ and $5+4=9\lt 13$.
3) With $8$ and $7$ without $9$ one has $x_1+x_2$ not possible and
$x_1+x_2+x_3=6+5+2=6+4+3$ give $2$ cases.
Hence there are $2+4+3+2=11$ possible cases each of these corresponding to permutations of $5$ and $4$ digits.
Thus there are $11\cdot4!\cdot5!=\color{red}{31680}$ possibilities.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \ x = 0 \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \ x = 0 ;\ x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \ x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x = \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi, \frac{\pi}{3} $$
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.
| There's nothing wrong up to the reduction to
$$
\sin\frac{x}{2}\cos\frac{3x}{2}=0
$$
Then you have either
$$
\sin\frac{x}{2}=0
$$
that is, $x/2=k\pi$ and $x=2k\pi$, or
$$
\cos\frac{3x}{2}=0
$$
so
$$
\frac{3x}{2}=\frac{\pi}{2}+k\pi
$$
and
$$
x=\frac{\pi}{3}+\frac{2k\pi}{3}
$$
Now let's examine the first set of solutions; you want to find the integers $k$ such that
$$
0\le 2k\pi<2\pi
$$
and this is only for $k=0$. For the second set of solutions,
$$
0\le\frac{\pi}{3}+\frac{2k\pi}{3}<2\pi
$$
becomes
$$
0\le 1+2k<6
$$
which gives $k\in\{0,1,2\}$.
Thus you find
$$
x\in\left\{0,\frac{\pi}{3},\pi,\frac{5\pi}{3}\right\}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$
$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\tan x)} = \frac{1+\tan^2 x}{\tan^2(1-\tan x)}$$
Now Put $\tan x= t \in (0,1)\;,$ Then $$h(t) = \frac{1+t^2}{t^2(1-t)}\;\;, 0<t<1 $$ where $h(t)=f(x)$.
Now How can i solve it after that, Help Required ,Thanks
| Since $\displaystyle 0<t<1, \;\frac{1+t^2}{t^2(1-t)}>8\iff 1+t^2>8t^2(1-t)=8t^2-8t^3\iff8t^3-7t^2+1>0$
If $g(t)=8t^3-7t^2+1,\;\; g^{\prime}(t)=24t^2-14t=0\iff t=0 \text{ or }t=\frac{7}{12}$.
Since $g(\frac{7}{12})=\frac{89}{432}$ is the minimum value of $g$ on $(0,1)$, $\;\;g(t)>0$ for $0<t<1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Genereating function of $H_{2n}$ We know the generating function of: $$\sum_{n=1}^{\infty}H_nx^n=\frac{\ln(1-x)}{x-1}$$.
How do we find out the generating function of $$\sum_{n=1}^{\infty}H_{2n}x^n$$
I used the formula: $\displaystyle { H }_{ 2n }=\frac { 1 }{ 2 } \left[ { H }_{ n }+{ H }_{ n-\frac { 1 }{ 2 } } \right] +\ln { 2 } $. But that didn't help.
| Define $$f(x) = \sum_{n=1}^\infty H_{2n} x^{2n}, \quad g(x) = \sum_{n=1}^\infty H_{2n-1} x^{2n-1}.$$ Then since $$H_{2n} = H_{2n-1} + \frac{1}{2n},$$ we have $$f(x) = \sum_{n=1}^\infty H_{2n} x^{2n} = x \sum_{n=1}^\infty H_{2n-1} x^{2n-1} + \sum_{n=1}^\infty \frac{x^{2n}}{2n} = x g(x) - \frac{1}{2}\log(1-x^2).$$ But since $$f(x) + g(x) = \sum_{m=1}^\infty H_m x^m = \frac{\log(1-x)}{x-1} = h(x),$$ we have $$f(x) = x (h(x) - f(x)) - \frac{1}{2} \log(1-x^2),$$ or $$f(x) = \frac{1}{1+x} \left( x h(x) - \frac{\log (1-x^2)}{2} \right).$$ Then the desired sum is simply $$f(x^{1/2}) = \frac{2 x^{1/2} \tanh^{-1} x^{1/2} - \log(1-x)}{2(1-x)} $$ which is a straightforward algebraic exercise.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need a solution to this Integration problem How to evaluate:$\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx$
I have tried substituting $x =y\tan\ A$, but failed.
| Hint
$$I=\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx=\displaystyle\int_{0}^{r}\frac{(x^2+y^2)^2}{(x^2+y^2)^{\frac{3}{2}}}dx-2\displaystyle\int_{0}^{r}\frac{x^2y^2}{(x^2+y^2)^{\frac{3}{2}}}dx-y^2\displaystyle\int_{0}^{r}\frac{1}{(x^2+y^2)^{\frac{3}{2}}}dx$$
$$I=\displaystyle\int_{0}^{r}\sqrt{x^2+y^2}dx-2y^2\displaystyle\int_{0}^{r}\frac{x^2}{\sqrt{x^2+y^2}}dx-y^2\displaystyle\int_{0}^{r}\frac{1}{(x^2+y^2)\sqrt{x^2+y^2}}dx$$
$$I=(1-2y^2)\displaystyle\int_{0}^{r}\sqrt{x^2+y^2}dx+2y^4\displaystyle\int_{0}^{r}\frac{1}{\sqrt{x^2+y^2}}dx-y^2\displaystyle\int_{0}^{r}\frac{1}{(x^2+y^2)\sqrt{x^2+y^2}}dx$$
Now set $x=y\tan\theta$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ Prove that $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ with a,b,c > 0
| Show ${a\over b} +{b\over c} +{c\over a} \ge {c+a\over c+b} + {a+b\over a+c} +{b+c\over b+a}$
But this is the same as (1) a/b+b/c+c/a >= (2) avg(c,a)/avg(c,b)+avg(a,b)/avg(a,c)+avg(b,c)/avg(b,a) . Thus it is sufficient to show a (mean preserving) spread of the inputs increases avg{a/b,b/c,c/a} . This is easily verified.
| {
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What is the range of $λ$? Suppose $a, b, c$ are the sides of a triangle and no two of them are equal.
Let $λ ∈ IR$. If the roots of the equation $x^
2 + 2(a + b + c)x + 3λ(ab +
bc + ca) = 0$ are real, then what is the range of $λ$?
I got that $$λ ≤\frac{
(a + b + c)^
2}
{3(ab + bc + ca)}$$
After that what to do?
| There is a systematic way to get rid of the triangle condition by substituting
$$a = x + y, b = y + z, c = z + x$$
which comes from the tangents to the incircle, explained e.g. here. Now the only condition on $x,y,z$ is that they are positive reals, and additionally they are distinct (because we want $a,b,c$ to be distinct).
The quadratic has real roots iff the discriminant is $\geq 0$, and as you say, if (writing in terms of the new variables)
$$\lambda \leq \frac{4(x+y+z)^2}{3(x^2+y^2+z^2) + 9(xy+yz+zx)}$$
Now we need to find the range of that expression for distinct real $x,y,z$.
Now let $a=x^2+y^2+z^2$, $b=xy+yz+zx$, and observe that $a> b$ from $(x-y)^2+(y-z)^2+(z-x)^2>0$, and moreover $a/b$ can take any value $\alpha> 1$. Hence, since our expression is
$$\frac{4}{3}\frac{a+2b}{a+3b} = \frac{4}{3}\frac{a/b+2}{a/b+3}$$
and we know the range of $a/b$, we get the range of $\lambda$ as well.
| {
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Values of $a$ for Which $|||x-1|-3|-a|=k$ Has 8 Distinct Real Roots Question
$|||x-1|-3|-a|=k$, $a \in \mathbb{N}$ has 8 distinct real roots for some $k$, then find the number of such values of $a$.
My Thought
Sorry I cannot show any work this time because I don't understand how to proceed. Please provide some guidance.
| Of course, we have to assume that $k \ge 0$.
\begin{align}
|||x-1|-3|-a| &= k \\
||x-1|-3|-a &= \pm k \\
||x-1|-3| &= a \pm k \\
|x-1|-3 &= \pm(a \pm k) \\
|x-1| &= 3 \pm(a \pm k) \\
x-1 &= \pm(3 \pm(a \pm k)) \\
x &= 1 \pm(3 \pm(a \pm k)) \\
\end{align}
You get $x \in \left\{ \begin{matrix}
1+3+a+k,\\
1+3+a-k,\\
1+3-a+k,\\
1+3-a-k,\\
1-3+a+k,\\
1-3+a-k,\\
1-3-a+k,\\
1-3-a-k\\
\end{matrix} \right\}$
which simplifies to $x \in \left\{ \begin{matrix}
4+a+k,\\
4+a-k,\\
4-a+k,\\
4-a-k,\\
-2+a+k,\\
-2+a-k,\\
-2-a+k,\\
-2-a-k\\
\end{matrix} \right\}$
For unique roots, you need to find when any pair of solutions is equal and restrict $a$ accordingly.
Being a bit free and easy with the $\pm$s, $4\pm a\pm k = 4\pm a\pm k$ leads to $a \not \in \{k, -k\}$. The same goes for $-2\pm a\pm k = -2\pm a\pm k$. The $16$ equations $4\pm a\pm k = -2\pm a\pm k$ lead to
$a \not \in \{k+3, k-3, -k+3, -k-3\}$. There will also not be eight unique solutions when $k \in \{0, 3, -3\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving $ a^4 \equiv 1 \pmod d$ I need to prove the following statements:
Prove the following statements:
(a) if $a$ is odd then $a^4 ≡ 1 \pmod 4$,
(b) if $5$ does not divide a, then $a^4 \equiv 1 \pmod 5$.
Can I do this inductively? Or should I be adopting another approach? I know for (a), if $a$ is odd, $a^4$ will also be odd, as the product of odd numbers is always odd. This would mean that $4 \mid (a^4 - 1)$, which would always be an even number, but of course not all even numbers are divisible by 4 (or rather $(a^4 - 1)$ would always be an even number). I also know you can obtain all the congruence cases of $1 \pmod 5$ by adding or subtracting $5$, beginning at $1$.
|
If $a$ is odd, then $a^4\equiv1\pmod4$:
*
*$a\equiv\color\red1\pmod4 \implies a^4\equiv\color\red1^4\equiv1\pmod4$
*$a\equiv\color\red3\pmod4 \implies a^4\equiv\color\red3^4\equiv81\equiv1\pmod4$
If $5$ does not divide $a$, then $a^4\equiv1\pmod5$:
*
*$a\equiv\color\red1\pmod5 \implies a^4\equiv\color\red1^4\equiv1\pmod5$
*$a\equiv\color\red2\pmod5 \implies a^4\equiv\color\red2^4\equiv16\equiv1\pmod5$
*$a\equiv\color\red3\pmod5 \implies a^4\equiv\color\red3^4\equiv81\equiv1\pmod5$
*$a\equiv\color\red4\pmod5 \implies a^4\equiv\color\red4^4\equiv256\equiv1\pmod5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Decreasing sequence numbers with first digit $9$
Find the sum of all positive integers whose digits (in base ten) form a strictly decreasing sequence with first digit $9$.
The method I thought of for solving this was very computational and it depended on a lot of casework. Is there a nicer way to solve this question?
Note that there are $\displaystyle\sum_{n=0}^{9} \binom{9}{n} = 2^9$ such numbers.
| If the digit with value $10^j$ is $k$, there are $\binom kj$ options for the digits after that and $2^{8-k}$ options for the digits before that (corresponding to the subsets of the digits between $k$ and $9$), except for $k=9$ there is $1$ option for the digits before (namely none). Thus the sum of the contributions from the $10^j$ digit is
$$
10^j\left(\sum_{k=0}^8k\binom kj2^{8-k}+9\binom9j\right)\;,
$$
and the sum of the contributions from all digits is
\begin{align}
\sum_{j=0}^910^j\sum_{k=0}^8\left(k\binom kj2^{8-k}+9\binom9j\right)
&=
\sum_{k=0}^8k\cdot11^k2^{8-k}+9\cdot11^9
\\
&=
2^8\cdot q\frac{\mathrm d}{\mathrm dq}\left.\frac{q^9-1}{q-1}\right|_{q=\frac{11}2}+9\cdot11^9
\\
&=
2^8\cdot\frac{11}2\cdot\frac{8\left(\frac{11}2\right)^9-9\left(\frac{11}2\right)^8+1}{\left(\frac{11}2-1\right)^2}+9\cdot11^9
\\
&=23259261861\;.
\end{align}
Here's code to check this result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= 16 + 8\cos\theta $
[Adding on both the sides]
=> $64\sin^{2}\theta + 64\cos^{2}\theta= 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $64 = 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $48 = 8\cos\theta + 65\cos^{2}\theta$
=> $48 = \cos\theta(65\cos\theta + 8)$
I can't figure out what to do next !
| Or you can use the half angle formulas:
$\cos(\theta) = \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} $
$\sin(\theta) = \frac{2 \tan(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})}$
So as to get:
$ 8 \frac{2 \tan(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} = 4 + \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} $
$ 3 \tan^2(\frac{\theta}{2}) - 16 \tan(\frac{\theta}{2}) +5 = 0$
Which you can solve for $\tan(\frac{\theta}{2})$ that you can then use to compute $\sin(\theta)$
Edit: I forgot to mention that you can do this only because $\cos(\frac{\theta}{2}) \neq 0$, because $\cos(\frac{\theta}{2}) = 0 \Leftrightarrow \theta = \pi [2\pi] \Rightarrow \sin(\theta) = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
To find the solution of the equation $2\left|z \right|-4az+1+ia=0$ Question:-
For every real number $a \ge 0$, find all the complex numbers $z$, satisfying the equation $2\left|z \right|-4az+1+ia=0$
Attempt at a solution:-
Let $z=x+iy$, then the equation $2\left|z \right|-4az+1+ia=0$ becomes as follows
$$\begin{equation}
2\sqrt{x^2+y^2}-4a(x+iy)+1+ia=0 \\
\left(2\sqrt{x^2+y^2}-4ax+1\right)+i(a-4ay)=0+i\cdot0
\end{equation}$$
Now, equating the imaginary part and real part on both sides, we get
$$\begin{equation}
y=\dfrac{1}{4} \qquad \left(a\neq0\right)
\end{equation}$$
$$\begin{equation}
2\sqrt{x^2+y^2}-4ax+1=0 \tag{1}
\end{equation}$$
Putting $y=\dfrac{1}{4}$ in $(1)$, we get
$$2\sqrt{x^2+\dfrac{1}{16}}-4ax+1=0 \implies 2\sqrt{x^2+\dfrac{1}{16}}=4ax-1$$
On squaring both sides we get the roots of $x$ as $$x=\dfrac{4a\pm\sqrt{4a^2+3}}{4(4a^2-1)}$$
The place where I got stuck:-
From here onward I am not able to think anything
| For $a=0$, there is no solution. In the following, $a\gt 0$.
You have
$$2\sqrt{x^2+\dfrac{1}{16}}=4ax-1\tag2$$
Here, note that you need to have
$$4ax-1\gt 0\iff x\gt \frac{1}{4a}\tag3$$
Squaring the both sides of $(2)$ gives
$$(64a^2-16)x^2-32ax+3=0\tag4$$
Case 1 : For $64a^2-16=0$, i.e. $a=1/2$, $x=3/16$ but this does not satisfy $(3)$.
Case 2 : For $64a^2-16\not=0$,
$$x=\frac{4a\pm \sqrt{4a^2+3}}{16a^2-4}$$
Now we consider the following inequality :
$$x=\frac{4a\pm \sqrt{4a^2+3}}{4(2a+1)(2a-1)}\gt \frac{1}{4a}\tag5$$
Case 2-1 : For $2a-1\gt 0$, i.e. $a\gt 1/2$,
$$(5)\iff 4a^2\pm a\sqrt{4a^2+3}\gt 4a^2-1\iff \pm a\sqrt{4a^2+3}\gt -1$$
Now $a\sqrt{4a^2+3}\gt -1$ holds.
$$- a\sqrt{4a^2+3}\gt -1\iff a\sqrt{4a^2+3}\lt 1$$
This does not hold since $a\gt 1/2$.
So, in this case,
$$x=\frac{4a\color{red}{+} \sqrt{4a^2+3}}{4(2a+1)(2a-1)}$$
Case 2-2 : For $2a-1\lt 0$, i.e. $a\lt 1/2$. Similarly, we get
$$(5)\iff 4a^2\pm a\sqrt{4a^2+3}\lt 4a^2-1\iff \pm a\sqrt{4a^2+3}\lt -1$$
Now $a\sqrt{4a^2+3}\lt -1$ does not hold.
$$-a\sqrt{4a^2+3}\lt -1\iff a\sqrt{4a^2+3}\gt 1$$
This does not hold since $a\lt 1/2$.
So, in this case, there is no solution.
Therefore, the answer is the following :
$$\color{red}{\text{For $\ 0\le a\le\frac 12,\ $ there is no solution}}$$
$$\color{red}{\text{For $\ a\gt \frac 12,\ z=\frac{4a+ \sqrt{4a^2+3}}{4(2a+1)(2a-1)}+\frac 14i$}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits