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How to find all values of $z$ such that $z^3=-8i$ If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?
I have the following formula:
$$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)\right]$$
for $k=0,\pm1, \pm2,...$
Applying this formula, I find the cubed root of $8$, which is $2$. And then when I apply it to the formula, I get the following:
$$z = 2\left[\cos\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$
I am confused, because the given solution is as follows:
$$z = 2\left[\cos\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$
Where did I go wrong? How would my approach changed if I was asked to find all values for $-8$, or $8i$?
|
Short way write it as $$z^{ 3 }=-8i\\ z^{ 3 }=8{ e }^{ i\frac { 3\pi }{ 2 } }\\ z=\sqrt [ 3 ]{ 8 } { e }^{ i\left( \frac { \frac { 3\pi }{ 2 } +2\pi k }{ 3 } \right) }=2{ e }^{ i\left( \frac { 3\pi +4\pi k }{ 6 } \right) },k=0,1,2\quad $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2203770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Functions satsify $f'=f^{-1}$ with $f^{-1}$ is compostional inverse of$ f$ let $f$ be a function such that :$f:\mathbb{C}\to \mathbb{C}$ and $f^{-1}$ is the compositional inverse of $f$, I seek for the analyticity of $f$ at $0$, then my question here is :
Question:
Are there functions satisfy:$f'=f^{-1}$ with $f^{-1}$ is compostional inverse of $f$ ?
|
If
$f'=f^{-1}$
and $f(x) = ax^b$,
then
$f'(x) = ab x^{b-1}$
and
$f^{(-1)}(x)
=(x/a)^{1/b}
$
so
$(x/a)^{1/b}
=ab x^{b-1}
$.
If $b \ne 0$,
then,
raising to the $b$ power,
$x/a
=(ab)^b x^{b(b-1)}
$
or
$a^{-b-1}b^{-b}
= x^{b(b-1)-1}
$.
For the right side
to be constant,
we must have
$0
=b(b-1)-1
=b^2-b-1
$
so
$b
=\dfrac{1\pm \sqrt{1+4}}{2}
=\dfrac{1\pm \sqrt{5}}{2}
$
and
$a^{-b-1}b^{-b}
= 1
$
or
$a^{-b-1}
= b^b
$
or,
since
$b+1 = b^2$,
$a
=b^{\frac{b}{-b-1}}
=b^{\frac{b}{-b^2}}
=b^{\frac{-1}{b}}
$.
|
{
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"url": "https://math.stackexchange.com/questions/2204476",
"timestamp": "2023-03-29T00:00:00",
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|
prove that the following function is convex at least I want a hint Let we have the following function
$$ψ:R^{*+} \to R$$
$$x\to ψ(x)=x^3$$
How can I prove that $ψ$ is a convex function by using the definition ?
I meant that I have to prove that
$$ψ(λx+(1-λ)y) \le λψ(x)+ (1-λ)ψ(y)$$ where $λ \in [0,1] $ $x$ and $y$ belong to $R^{*+}$
|
Take $a,b$ positive reals. Then for $a \ge b$, we have $a^2+ab \ge 2b^2$, that is, $a^2 + ab+b^2 \ge 3b^2$. Multiplying by $(a-b)$, we get $a^3 - b^3 \ge 3b^2(a-b)$.
For $a\le b$, we have $a^2+ab \le 2b^2$, that is, $a^2+ab+b^2 \le 3b^2$. Here $(a-b) \le 0$, so multiplying by $(a-b)$, we have $a^3-b^3 \ge 3b^2(a-b)$.
Thus for any positive reals $a,b$, we have $a^3-b^3 \ge 3b^2(a-b)$, that is, $a^3 \ge b^3 + 3b^2(a-b). \tag{1}$
Taking $a = x, b = \lambda x + (1- \lambda)y$ in the above inequality, we get
$$ x^3 \ge (\lambda x + (1-\lambda)y)^3 + 3b^2(1-\lambda)(x-y) \tag{2}$$
Taking $a = y, b = \lambda x + (1- \lambda)y$ in $(1)$, we get
$$ y^3 \ge (\lambda x + (1-\lambda)y)^3 + 3b^2(-\lambda)(x-y) \tag{3}$$
Now, $\lambda \times (2) + (1-\lambda) \times (3)$ gives
$$\lambda x^3 + (1-\lambda)y^3 \ge (\lambda x + (1-\lambda)y)^3 $$
That proves convexity the way you want.
Note though that everything we did above follows from Taylor's theorem and non-negativity of the second derivative and is in fact a proof in disguise of the fact that a non-negative second derivative implies convexity. You are better off just using that.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$
Find $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$
Since $\frac{k}{n^2+k^2}\leq \frac{k}{k^2+k^2}=\frac{1}{2k}$, then $\sum_{k=1}^n \frac{k}{n^2+k^2}\leq \sum_{k=1}^n \frac{1}{2k}=\frac{1}{2} \sum_{k=1}^n\frac{1}{k}$.
Now we send $n$ to infinity, then since $\sum_{k=1}^\infty \frac{1}{k}$ is harmonic, $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$ doesn't exist.
I wonder if my thinking is right.
|
Just for the fun.
Consider $$S_n= \sum_{k=1}^n \frac{k}{n^2+k^2}=\sum_{k=1}^n \frac{k}{(n^2+ik)(n-ik)}=\frac i2 \sum_{k=1}^n\left(\frac{1}{n+i k}-\frac{1}{n-i k} \right)$$ Using generalized harmonic numbers $$S_n=\frac{1}{2} \left(-H_{-i n}-H_{i n}+H_{(1-i) n}+H_{(1+i) n}\right)$$ Using asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^4}\right)$$ and applying to each term, after simplifications, you should end with $$S_n=\frac{\log (2)}{2}+\frac{1}{4 n}-\frac{1}{12
n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.
Use it with $n=10$ which gives $$S_{10}=\frac{2892380100711541}{7801656832544900}\approx 0.370739$$ while the expansion gives $$S_{10}\approx\frac{\log (2)}{2}+\frac{193}{8400}\approx 0.369550$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help with a Binomial Coefficient Identity The following (conjectured) identity has come up in a research problem from representation theory that I am working on:
$$\binom{i}{n-1}
=\frac{(n+k+i)!}{(n-1)!(k+i)!} \sum_{j=0}^{n-1} \frac{(-1)^j }{(n+k+i-j)}\binom{n-1}{j} \binom{2n+k-j-2}{n-1}$$
where $k, i, n$ are positive integers and I interpret $\binom{i}{n-1}=i*(i-1)\cdots (i-n+1)/(n-1)!$ even if $i<n-1$. (That is, $\binom{i}{n-1}=0$ if $i<n-1$.)
I've verified the identity holds for small values of $n$, but all of my naive attempts to prove it have failed. Any suggestions for how to prove it would be most appreciated!
|
Let us re-write this as follows:
$${q\choose n} =
(n+1) {n+1+k+q\choose n+1}
\sum_{j=0}^n \frac{(-1)^j}{n+1+k+q-j}
{n\choose j} {2n+k-j\choose n}
\\ = (n+1+k+q) {n+k+q\choose n}
\sum_{j=0}^n \frac{(-1)^j}{n+1+k+q-j}
{n\choose j} {2n+k-j\choose n}.$$
where clearly $q\ge n.$ We have
$${n+k+q\choose n} {n\choose j}
= \frac{(n+k+q)!}{(k+q)! j! (n-j)!}
= {n+k+q\choose j} {n+k+q-j\choose n-j}.$$
Substititute into the identity to get for the RHS
$$\sum_{j=0}^n (-1)^j {n+k+q+1\choose j}
{n+k+q-j\choose n-j} {2n+k-j\choose n}.$$
We put
$${n+k+q-j\choose n-j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+k+q-j}
\; dz$$
and
$${2n+k-j\choose n} =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} (1+w)^{2n+k-j}
\; dw.$$
Observe that the first of these two vanishes when $j\gt n$ so it
provides range control and we may raise the upper limit of the sum to
$n+k+q+1.$ We get
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} (1+w)^{2n+k}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k+q}
\\ \times \sum_{j=0}^{n+k+q+1} {n+k+q+1\choose j}
(-1)^j \frac{z^j}{(1+z)^j (1+w)^j}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} (1+w)^{2n+k}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k+q}
\left(1-\frac{z}{(1+z)(1+w)}\right)^{n+k+q+1}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} \frac{1}{(1+w)^{q-n+1}}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{1}{1+z}
\left(1+w+wz\right)^{n+k+q+1}
\; dz\; dw
.$$
We evaluate this using the fact that residues sum to zero. We get for
the residue at $z=-1$ (flip sign)
$$-(-1)^{n+1}\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} \frac{1}{(1+w)^{q-n+1}}
\; dw
\\ = (-1)^{n}
\times
(-1)^n {q-n+n\choose q-n}
\; dw = {q\choose q-n} = {q\choose n}.$$
This is the target result, so we just need to show that the
contribution from the residue at infinity is zero. We find
$$-\mathrm{Res}_{z=\infty} \frac{1}{z^{n+1}} \frac{1}{1+z}
\left(1+w+wz\right)^{n+k+q+1}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1} \frac{1}{1+1/z}
\left(1+w+w/z\right)^{n+k+q+1}
\\ = \mathrm{Res}_{z=0} \frac{1}{z} z^{n+1} \frac{1}{1+z}
\frac{1}{z^{n+k+q+1}}
\left(z(1+w)+w\right)^{n+k+q+1}
\\ = \mathrm{Res}_{z=0}
\frac{1}{z^{k+q+1}} \frac{1}{1+z}
\left(z(1+w)+w\right)^{n+k+q+1}.$$
Extracting coefficients yields
$$\sum_{p=0}^{k+q} {n+k+q+1\choose k+q-p}
(1+w)^{k+q-p} w^{n+p+1} (-1)^{p}.$$
On substituting this into the integral in $w$ we obtain
$$[w^n] \frac{1}{(1+w)^{q-n+1}}
\sum_{p=0}^{k+q} {n+k+q+1\choose k+q-p}
(1+w)^{k+q-p} w^{n+p+1} (-1)^{p}
\\ = \sum_{p=0}^{k+q} {n+k+q+1\choose k+q-p}
[w^n] w^{n+p+1} (1+w)^{k+n-p-1} (-1)^{p}.$$
This is zero since
$$[w^n] w^{n+p+1} (1+w)^{k+n-p-1} =
[w^0] w^{p+1} (1+w)^{k+n-p-1} = 0$$
(no constant coefficient in $w$ because $p\ge 0$, we can have
$k+n-p-1$ positive or negative here since there is never a pole at
zero). This concludes the argument. We have shown that the sum indeed
evaluates to ${q\choose n}$ as claimed.
|
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|
Solve $2\cos(x)^2 - 5\sin(x) - 4 = 0$ on the interval $0 \leq x \leq 2\pi$ This is homework.
I got it down to either $\sin(x) = -2$ or $\sin(x) = -1/2$ but I have absolutely no idea where to go from here.
|
$2\cos(x)^2 - 5\sin(x) - 4 = 0$ on the interval $0 \leq x \leq 2\pi$
$2(1-\sin^2x)-5\sin x-4=0$
$2\sin^2x +5\sin x+2=0$
$(2\sin x+1)(\sin x+2)=0$
Then either $\sin x = -2$ which has no solutions since $|\sin x |\leq1$
or $\sin x = -\frac{1}{2}$
Now, let's consider the graph of $\sin x$:
So we have $ x = -\frac{\pi}{6} + 2k\pi,k\in\mathbb{Z}$ - this is because $y=\sin x $ has period $2\pi$, which means $\sin(x+2k\pi) = \sin x$
Now also note the following property:
$\sin(\pi - x) = \sin \pi \cos x - \sin x \cos \pi = \sin x$
Therefore, $x = (\pi - -\frac{\pi}{6}) + 2k\pi = \frac{7\pi}{6} +2k\pi,k\in\mathbb{Z}$
Now since the interval we are required to give solutions in is $0\leq x\leq 2\pi$, the solutions are $x= \frac{11\pi}{6}$ or $x= \frac{7\pi}{6}$
|
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|
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\beta=\frac{c}{a}$$
Therefore the roots of $9x^2-2x+7=0$ are $(\alpha+2)$ and $(\beta+2)$. So, $$\alpha+\beta+4=\frac{2}{9}\implies4-\frac{b}{a}=\frac{2}{9}\implies\frac{b}{a}=\frac{34}{9}$$ and
\begin{align*}
(\alpha+2)(\beta+2)=\frac{7}{9}\\
\Rightarrow\alpha\beta+2(\alpha+\beta)+4=\frac{7}{9}\\
\Rightarrow\frac{c}{a}-2\frac{b}{a}+4=\frac{7}{9}\\
\Rightarrow\frac{4a-2b+c}{a}=\frac{7}{9}\\
\end{align*}
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.
There is another similar question: If the roots of $px^2+qx+r=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then what will be the expression of $r$ in terms of $a$, $b$, and $c$?
|
You can check the other answers for (possible faster) alternatives, but you were doing fine.
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.
Notice that the system
$$\left\{ \begin{array}{l}\dfrac{b}{a}=\dfrac{34}{9} \\[8pt] \dfrac{4a-2b+c}{a}=\dfrac{7}{9}
\end{array}\right. \iff \left\{ \begin{array}{l}34a-9b=0 \\[8pt] 29 a - 18 b + 9 c = 0
\end{array}\right.$$
has an infinite number of solutions, given by:
$$\left\{ \begin{array}{l} a = 9t \\ b = 34t \\ c = 39t
\end{array}\right. \quad t \in \mathbb{R_0}$$
This is logical because we can simply divide $ax^2+bx+c=0$ (for $a$ non-zero) by $a$ to obtain a quadratic equation with the same roots.
|
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|
Right Triangle Problem with Angle Bisector Theorem Right triangle $ABC$ has $AC = 8$ and $CB = 6$. $M$ is the midpoint of $AB$. Pick point $N$ on line $CM$ with $M$ between $C$ and $N$ such that $∠CAB = ∠BAN$. Compute $MN$. Express your answer as a common fraction.
I figured out that $CM$, $BM$, and $AM$ were $5$, but I can't figure out how to continue.
|
I think it is better to leave the trigonometry out of the solution.
Since $\angle BCA = 90^{\circ}$ one can conclude that $AM = BM = CM$. By Pythagoras one gets $AB = 10$ so $AM = BM = CM = 5$. Triangle $ACM$ is isosceles and $$\angle \, ACN = \angle \, ACM = \angle\, CAM = \angle \, CAB = \angle \, BAN = \angle \, MAN$$ Let's look at triangles $ACN$ and $AMN$. We have concluded that $\angle\, ACN = \angle \, MAN$. Since $\angle \, ANM = \angle \, CNA$ as a common angle at vertex $N$, triangles $ACN$ and $AMN$ are similar and therefore
$$\frac{AM}{CA} = \frac{MN}{AN} = \frac{AN}{CN}$$ However, since $AM = 5, \,\,$ $AC=8$ and $CN = CM + MN = 5 + MN$, we get the relations $$\frac{5}{8} = \frac{MN}{AN} = \frac{AN}{5+MN}$$ Thus, one can express $AN$ in two different ways
$$AN = \frac{8}{5}\, MN \,\,\, \text{ and } \,\,\, AN = \frac{5}{8}\, (MN + 5)$$ Equate the two, obtaining the equation
$$\frac{8}{5}\, MN = \frac{5}{8} (MN + 5)$$ Solve the equation for $MN$ in order to obtain $$MN = \frac{125}{39}= 3 + \frac{8}{39}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Coordinate system $(a,\tau)$ for butterfly circles? Consider this picture
which shows a coordinate system $(a,\tau)$ on the Cartesian coordinate frame $(x,y)$, which is very similar to the bipolar coordinate system, with isosurfaces (i.e. circles) $\tau$ and their respective foci $(-a,0)$ and $(a,0)$.
Can anyone help me derive formulae on how I might be able to go from Cartesian coordinates $(x,y)\mapsto (a,\tau)$ or polar coordinates $(\rho,\theta)\mapsto (a,\tau)$, similar to what you might find on https://en.wikipedia.org/wiki/Bipolar_coordinates?
Essentially, this problem arose because I'd like to sample a function of the form $f(\rho,\theta)=g(C\rho(\cos\theta,\sin\theta-1))$.
|
We have one family of curves $(x-a)^2+y^2=a^2$. Calculating the partial derivatives
\begin{eqnarray*}
\left( \frac{\partial a}{\partial x},\frac{\partial a}{\partial y} \right) = \left( \frac{x^2-y^2 }{2 x^2},\frac{y}{x} \right)
\end{eqnarray*}
We need a vector that is orthogonal to this
\begin{eqnarray*}
\left( \frac{\partial a}{\partial x},\frac{\partial a}{\partial y} \right) \cdot \left( \frac{\partial b}{\partial x},\frac{\partial b}{\partial y} \right) =0
\end{eqnarray*}
So we need to solve the differential equation
\begin{eqnarray*}
(x^2-y^2) \frac{\partial b}{\partial x}+ 2xy \frac{\partial b}{\partial y} =0
\end{eqnarray*}
Which is equivalent to solving
\begin{eqnarray*}
\frac{dy}{dx} =\frac{ (x^2-y^2)}{ 2xy }
\end{eqnarray*}
Let $y=ux$ and after a little calculus & algebra (& neat choice of arbitary constant)
\begin{eqnarray*}
x =\frac{2 bu}{ (1+u^2) }
\end{eqnarray*}
So
\begin{eqnarray*}
x^2 +(y-b)^2=b^2
\end{eqnarray*}
So the other family of curves comes out to be circles as well ! ... but this time their centers run up the y-axis.
\begin{eqnarray*}
x^2 +(y-b)^2=b^2
\end{eqnarray*}
To summarise ...
\begin{eqnarray*}
x & =& \frac{a b^2}{a^2+b^2} \ a & =& \frac{x^2+y^2}{2x} \\
y & =& \frac{a^2 b}{a^2+b^2} \ b & =& \frac{x^2+y^2}{2y} .
\end{eqnarray*}
|
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|
Show that if $a\neq b$, then we have for the $n\times n$-matrix $\textrm{det}=\frac{a^{n+1}-b^{n+1}}{a-b}$. The Problem
Show that if $a\neq b$, then we have for the $n\times n$-matrix
$$\textrm{det}\begin{pmatrix} a+b & ab & 0 & \ldots & 0 & 0 \\
1 & a+b & ab & \ldots & 0 & 0 \\
0 & 1 & a+b & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 & \ldots & a+b & ab \\
0 & 0 & 0 & \ldots & 1 & a+b\end{pmatrix} =\frac{a^{n+1}-b^{n+1}}{a-b}$$
What if $a=b$?
My Questions
I am not entirely sure how to begin this proof. I suspect I am missing something that makes it quite simple. I tried looking for similar problems, and I noticed a common theme being the use of row operations to rewrite the matrix, from which the determinant was found. However, I still didn't understand many of the intermediate computations when it came to actually finding the determinant. My questions are as follows.
*
*Should I use row operations to rewrite this matrix? If so, what would be an example of how that would look computationally?
*In either case, how should I go about actually computing the determinant to show the statement is true? Is there an algorithm that is helpful here?
*I noticed for the follow up question that, if $a=b$, then $$\textrm{det}\begin{pmatrix} a+a & aa & 0 & \ldots & 0 & 0 \\
1 & a+a & aa & \ldots & 0 & 0 \\
0 & 1 & a+a & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 & \ldots & a+a & aa \\
0 & 0 & 0 & \ldots & 1 & a+a\end{pmatrix}.$$ Am I right in my thinking there?
Other Details
The book used in the course is Abstract Linear Algebra by Curtis. It has been of little help to me here...
|
Let's call $A_n$ the matrix for which you want to compute the determinant and let's see how it can be constructed iteratively:
$$ A_{n+1} = \begin{pmatrix} a+b & ab & 0 & \ldots & 0 & 0 \\
1 & a+b & ab & \ldots & 0 & 0 \\
0 & 1 & a+b & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 & \ldots & a+b & ab \\
0 & 0 & 0 & \ldots & 1 & a+b\end{pmatrix}
= \begin{pmatrix}
A_n & \begin{matrix} 0 \\ \cdots \\ 0 \\ ab \end{matrix} \\
\begin{matrix} 0 & \cdots & 0 & 1 \end{matrix} & a+b
\end{pmatrix} $$
Such layout cries out for induction to compute the determinant $D_n = \det(A_n)$ . For $n=1$, we have $A_1 = (a+b)$ so $D_1 = a+b$ and $D_1 \cdot (a-b) = a^2 - b^2 $.
Now, if we assume the formula is true up to an integer $n$, can we prove that it is also true for $n+1$ ?
We can use Laplace expansion applied to the last column to find out that:
$$ D_{n+1} = (a+b)\cdot D_n - ab \cdot D_{n-1}$$
Therefore:
$$ \begin{align} (a-b)\cdot D_{n+1} & = (a+b)\cdot (a-b) D_n - ab \cdot(a-b) D_{n-1} \\
& = (a+b)\cdot( a^{n+1} - b^{n+1} ) - ab \cdot( a^{n} - b^{n} ) \\
& = a^{n+2}- ab^{n+1} + ba^{n+1} - b^{n+2} - ba^{n+1}+ ab^{n+1} \\
& = a^{n+2}- b^{n+2}\\
\end{align} $$
And we are done for the first question.
The second question asks what is happening when $a=b$. Obviously we can't apply the formula above since it is only valid for $a \ne b $. But we can take $b$ as close as we want to $a$, let's say $b_k = a + \frac{1}{k}$. Now we can apply the previous result, meaning that for all positive integer $k$:
$$ \det \begin{pmatrix} a+(a + \frac{1}{k}) & a(a + \frac{1}{k}) & \ldots & 0 & 0 \\
1 & a+(a + \frac{1}{k}) & \ldots & 0 & 0 \\
0 & 1 & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & \ldots & a+(a + \frac{1}{k}) & a(a + \frac{1}{k}) \\
0 & 0 & \ldots & 1 & a+(a + \frac{1}{k})\end{pmatrix} = \frac{\left(a + \frac{1}{k}\right)^{n+1} - a^{n+1}}{ \frac{1}{k} } $$
To conclude, you can use the continuity of the determinant and take the limit on both sides as $k \to \infty $. On the left you have the determinant of the matrix $A_n$ with $a=b$ and on the right you have the derivative of the function $ x \mapsto x^{n+1} $ at $x=a$, which is $(n+1)a^n$.
|
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|
Equation $2n^2=m^2+1$ Given the equation
$$2n^2=m^2+1$$
Is there a general solution?
since $m$ is odd for example $$2\cdot5^2=7^2+1$$ $$2\cdot29^2=41^2+1$$
Any hint would be appreciation.
|
If
$2n^2=m^2+1$
then
$2(3n+2m)^2 = 18n^2 + 14nm +8m^2$
$=9(m^2 + 1) + 14nm + 8m^2$
$=9(m^2+1) + 14nm + 8(2n^2-1)$
$=16n^2 + 14nm + 9m^2 + 1$
$=(4n+3m)^2 + 1$
so we have (1,1) -> (5,7) -> (29,41) etc.
|
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|
Sign of determinant when using $det A^\top A$ We have been given matrix:
$$A =
\begin{pmatrix}
a& b& c &d \\
b &−a& d& −c\\
c& −d &−a& b \\
d &c& −b& −a\\
\end{pmatrix}
$$
...and have been asked to calculate $\det(A)$ using $AA^T$.
We see that:
$$AA^T=\begin{pmatrix}
a^2+ b^2+ c^2+ d^2& 0& 0&0\\
0 &a^2+ b^2+ c^2+ d^2& 0& 0\\
0& 0 &a^2+ b^2+ c^2+ d^2& b \\
0 &0& 0& a^2+ b^2+ c^2+ d^2\\
\end{pmatrix}
$$
So, $\det(AA^T)= (a^2+ b^2+ c^2+ d^2)^4$
Now which should I choose: $(a^2+ b^2+ c^2+ d^2)^2 $ or $ -(a^2+ b^2+ c^2+ d^2)^2$ ?
Please explain me which one and why.
|
This might be an overkill but it is a useful observation nonetheless. You have a matrix $A = A(a,b,c,d)$ which depends on four real parameters and you computed $\det(A)^2$. From your computation, it is clear that $\det(A) = 0$ iff $a = b = c = d = 0$. Set $U = \mathbb{R}^4 \setminus \{ (0,0,0,0) \}$. If $(a,b,c,d) \in U$ then $\det(A) \neq 0$ so either $\det(A) > 0$ or $\det(A) < 0$. However, the set $U$ is connected and the function $U \rightarrow \mathbb{R}$ given by $(a,b,c,d) \mapsto \det(A(a,b,c,d))$ is continuous so we must have $\det(A) > 0$ for all $(a,b,c,d) \in U$ or $\det(A) < 0$ for all $(a,b,c,d) \in U$ and we can check which by plugging in specific values for $a,b,c,d$. For example, when $b = c = d = 0$ we have
$$ \det(A) = a(-a)^3 = -a^4 $$
so $\det(A) < 0$ if $a \neq 0$ and hence
$$ \det(A(a,b,c,d)) = -(a^2 + b^2 + c^2 + d^2)^2 $$
for all $(a,b,c,d) \in U$.
|
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|
if $\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$ then $6(\alpha+\beta)=?$ Let $\alpha, \beta \in \mathbb{R}$ be such that $$\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$$ then what is the value of $6(\alpha+\beta)$
|
$$
\begin{aligned}
\lim _{x\to 0}\left(\frac{x^2\sin \left(\beta \:x\right)}{\alpha \:x-\sin \:x}\right)
& = \lim _{x\to 0}\left(\frac{x^2\left(\beta x+o\left(x\right)\right)}{\alpha \:x-\left(x-\frac{x^3}{3!}+o(x^3)\right)}\right)
\\& = \beta \:\cdot \lim _{x\to 0}\left(\frac{6x^2}{6α+x^2-6}\right) (\text{so } \color{red}{\beta \ne 0})
\\& = \beta \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = 6\beta \text{ (with } \color{red}{\alpha = 1})
\\& = \color{red}{\frac{1}{6}} \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = \color{blue}{1} \text{ (with } \color{red}{\beta = \frac{1}{6}})
\end{aligned}
$$
So, with
$$\alpha = 1, \beta = \frac{1}{6}$$
$$6(\alpha + \beta)=6\left(1+\frac{1}{6}\right) = \color{gold}{7}$$
|
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|
$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?
|
We know
$$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots$$
Fixing a positive integer $n$, let
\begin{align*}
x &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!}\\[4pt]
r &= \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots\\[4pt]
\end{align*}
Thus, we have
$$e = x + r$$
and we want to show
$$r < \frac{1}{n{(n!)}}$$
Then
\begin{align*}
r\, &= \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots\\[4pt]
\implies\; (n+1)r\, &= \frac{n+1}{(n+1)!} + \frac{n+1}{(n+2)!} + \frac{n+1}{(n+3)!} + \cdots\\[4pt]
&< \frac{n+1}{(n+1)!} + \frac{n+2}{(n+2)!} + \frac{n+3}{(n+3)!} + \cdots\\[4pt]
&= \frac{1}{n!} + \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots\\[4pt]
&= \frac{1}{n!} + r\\[16pt]
\text{Then}\;\,(n+1)r\, &< \frac{1}{n!} + r\\[4pt]
\implies\; nr\, &< \frac{1}{n!}\\[4pt]
\implies\;\;\;\, r\, &< \frac{1}{n{(n!)}}\\[4pt]
\end{align*}
as required.
|
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|
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} dt + B\int {1 \over t+w_2}dt \\= {w - 1\over w - w_2} \ln|t + w| + {1- w_2\over w - w_2}\ln|t + w_2| + C $$
After finding the value of $w, w_2$ final answer I got is
$${\sqrt{5} + 1\over 2\sqrt{5}}\ln|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\ln|t + 1/2 + \sqrt{5}/2| + C$$
But the given answer is :
$$\bbox[7px,Border:2px solid black]{ \frac{\ln\left(\left|t^2+t-1\right|\right)}{2}+\frac{\ln\left(\left|2t-\sqrt{5}+1\right|\right)-\ln\left(\left|2t+\sqrt{5}+1\right|\right)}{2\cdot\sqrt{5}}+C}$$
Where did I go wrong ? especially that first term of the answer is a mystery to me.
|
Another approach would be to rewrite the integrand as
$$\int\frac{2 t+1}{2 \left(t^2+t-1\right)}\,dt+\int\frac{1}{2 \left(t^2+t-1\right)}\,dt$$
Then let $u=t^2+t-1$ such that $du=2t+1\, dt$ so the integral becomes
$$\frac 12 \int \frac 1u\,du+\frac 12\int\frac{1}{ \left(t^2+t-1\right)}\,dt \;=\; \frac {\ln(|t^2+t-1|)}2+\frac 12\int\frac{1}{ \left(t^2+t-1\right)}\,dt$$
Completing the square we get $t^2+t-1 \,=\, (t+\frac 12)^2-\frac 54$ so we let $v=t+\frac 12$ such that $dv=dt$ and we get
$$\begin{align}
\frac {\ln(|t^2+t-1|)}2+\frac 12 \int \frac 1{v^2-\frac 54}\,dv &\;=\;\frac {\ln(|t^2+t-1|)}2+\frac{\ln \left(|\sqrt{5}-2 v|\right)-\ln \left(|2 v+\sqrt{5}|\right)}{2\sqrt{5}}\\
& \;=\;\frac {\ln(|t^2+t-1|)}2+\frac{\ln \left(|\sqrt{5}-2 t+ 1|\right)-\ln \left(|2 t+ 1+\sqrt{5}\right|)}{2\sqrt{5}}+C\\
\end{align}$$
|
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|
Show the value of determinant is 0 I dont have any idea how to show the value of the following determinant is 0 without expanding the determinants
$$
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}
$$
Any ideas?
|
Subtract the first row from each of the other rows:
$$
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}
=\begin{vmatrix}
1 & a & a^2-bc \\
0 & b-a & (b-a)(b+a+c) \\
0 & c-a & (c-a)(c+a+b) \\
\end{vmatrix}\\
$$
Note that the second and third rows are both multiples of $\begin{bmatrix}0&1&a+b+c\end{bmatrix}$; therefore, they are linearly dependent.
|
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|
Knowing that $\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3}$ and $\tan \alpha > 0$, find the exact value of $\tan \alpha$ First I tried to solve for $\alpha$:
$$\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3} \Leftrightarrow \alpha = \arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2} \lor \alpha = -\arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2}$$
Then I put $\tan(\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ and $\tan(-\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ in the calculator and $\tan(\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ was positive while the other was negative, so I assumed the positive one was the correct expression.
But I don't know how to simplify it to get the exact value. My book says the solution is $\frac{\sqrt{2}}{4}$. How do I solve this?
|
$\cos\left(\frac{3\pi}{2}+a\right)=\cos\left(\frac{3\pi}{2}\right)\cos(a)-\sin\left(\frac{3\pi}{2}\right)\sin(a)=-\sin(a)=-\frac{1}{3}=>\sin(a)=\frac{1}{3}$
$\cos(a)=\sqrt{1-\sin^2{a}}=\sqrt{1-\frac19}=\sqrt{\frac89}$
$\tan(a)=\frac{\frac{1}{3}}{\sqrt{\frac89}}=\frac{1}{\sqrt{8}}=\boxed{\frac{\sqrt{2}}{4}}$
|
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|
Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, an explicit expression for $f(x)$ for all positive rational numbers $x$.
Every number I have evaluated is of the form $ f ( x ) = \frac x { x + 1 } $ and this clearly fits the functional equations, but I can't prove that it's the only solution. Can anyone help me? I have put down the start of my workings which led me to the conjecture of $ f ( x ) = \frac x { x + 1 } $.
Plugging in $ x = 1 $ clearly gives $ f ( 1 ) = \frac 1 2 $ and $ f ( 2 ) = 2 f \big( f ( 1 ) \big) = 2 f \left( \frac 1 2 \right) $ which we can plug back into the first equation to get that $ f ( 2 ) = \frac 2 3 $. Working in this vein I have been able to show that $ f ( x ) = \frac x { x + 1 } $ for particular values of $ x $, but not in general.
The most difficult part appears to be proving it for the even integers. To prove $ x = 8 $, we have
$$ f ( 12 ) = 2 f \left( \frac 6 7 \right) = 4 f \left( \frac 3 { 10 } \right) = 4 - 4 f \left( \frac { 10 } 3 \right) \\
= 4 - 8 f \left( \frac 5 8 \right) = 8 f \left( \frac 8 5 \right) - 4 = 16 f \left( \frac 4 9 \right) - 4 \\
= 32 f \left( \frac 2 { 11 } \right) - 4 = 64 f \left( \frac 1 { 12 } \right) - 4 = 60 - 64 f ( 12 ) \text , $$
giving us $ f ( 12 ) = \frac { 12 } { 13 } $. This will probably be the main area of difficulty in the proof.
|
For each coprime $n, m \in \mathbb N_+$, $f\left( \dfrac nm\right)=\dfrac{n}{n+m}$ is a solution.
To prove this, we induct on $n+m$. The base case is clearly true. Suppose it's true for $n+m\leq k$ for some natural number $k$ and consider $n+m=k+1$.
We consider two cases:
Case 1, $n>m$:
Using the first condition and Case 2, we get that
$$f\left(\dfrac{n}{m}\right) = 1 - f\left(\dfrac{m}{n} \right)=1-\frac{m}{m+n}=\frac{n}{n+m}\,\,\,\,\,\,\,\,\,\blacksquare$$
Case 2, $m>n$:
By the condition, we can assume $m=n+l$ for some natural $l$. Then,
\begin{align*}f\left( \dfrac{n}{m}\right)=f\left( \frac{n}{n+l}\right) = f\left( f\left( \frac{n}{l}\right)\right) = \frac 12f\left( \frac{2n}{l}\right) = \frac{n}{2n+l} = \frac{n}{n+m}\,\,\,\,\,\,\,\,\,\,\blacksquare
\end{align*}
Finally, we plug the function in and see that it indeed satisfies the conditions.
This was problem B4 of Ireland MO 1991.
|
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|
Area enclosed by the curve $\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$
What is the area enclosed by the curve $$\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$$ $$x\ge y, \forall x, y \ge 0$$
$\lfloor x\rfloor$ stands for the Greatest Integer Function.
I think that the curve will start from $\left(\cfrac{5}{2},\cfrac{5}{2}\right)$ and then be a straight line parallel to the $y$ axis till $\left(\cfrac{7}{2},\cfrac{5}{2}\right)$ (not including this last point). But then I am finding it difficult to manipulate it further.
|
Blue tiles:
$$
\lfloor{x + y}\rfloor + \lfloor{x - y}\rfloor = 5
$$
Gray triangle
$$
y > 0, \quad x \le y
$$
Find the area of the blue tiles within the gray triangle.
The black dots are the vertices of the first square:
$$
\left( \begin{array}{c}
3 \\ 0
\end{array} \right), \quad
%
\left( \begin{array}{c}
3.5 \\ 0.5
\end{array} \right), \quad
%
\left( \begin{array}{c}
3 \\ 1
\end{array} \right), \quad
%
\left( \begin{array}{c}
2.5 \\ 0.5
\end{array} \right)
%
$$
The side length for each square is $s =\frac{1}{\sqrt{2}}$. Therefore each square has area $A=s^{2} = \frac{1}{2}.$ There are three squares within the second region. Therefore the total area is
$$A_{total} = 3 A = \frac{3}{2}$$
|
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|
Showing that roots of a quadratic polynomial are of opposite signs.
Prove that, for all values of $k$, the roots of the quadratic polynomial $x^2 - (2 + k) x - 3$ are real. Show further that the roots are of opposite signs.
For the first part I was able to demonstrate such by using the discriminant of the quadratic, then using the discriminant of the discriminant.
For the second part I was not able to demonstrate such.
|
Well, the solutions to the equation are
$$x=\frac{2+k}{2}\pm\frac12\sqrt{(-2-k)^2-4\cdot1\cdot-3}$$
or, simplified,
$$x=\frac{2+k}2\pm\frac12\sqrt{(2+k)^2+12}$$
and this is equal to
$$x=\frac{2+k}2\pm\sqrt{\left(\frac{2+k}2\right)^2+3}$$
The important step here is
$$\sqrt{\left(\frac{2+k}2\right)^2+3}>\sqrt{\left(\frac{2+k}2\right)^2}=\left|\frac{2+k}2\right|$$
and so if we choose $+$, then we see the root is positive:
$$\frac{2+k}2+\sqrt{\left(\frac{2+k}2\right)^2+3}>\frac{2+k}2+\left|\frac{2+k}2\right|\geq0$$
When we choose $-$, we get something negative:
$$\frac{2+k}2-\sqrt{\left(\frac{2+k}2\right)^2+3}<\frac{2+k}2-\left|\frac{2+k}2\right|\leq0$$
thus, the roots must be of different signs.
|
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|
Problem on number of ways of writing $n$ as a sum of distinct integers vs a sum of odd integers I found this problem while doing a few problems on partitions of an integer:
Prove that the number of ways of writing $n$ as a sum of distinct positive integers is equal to the number of ways of writing $n$ as a sum of odd positive integers.
Attempt at solution:
I tried to perform some manipulations on the generating function for the partitions of $n$ to get the latter, but could not do so.
|
I think this was proved by Euler.
The number of partitions of $n$ as a sum of odd positive integers is the coefficient of the $x^n$ in the expansion of :
$$\frac{1}{(1-x)(1-x^3)(1-x^5)(1-x^7)\dots} $$
and the number of ways of writing $n$ as a sum of distinct positive integers is the coefficient of $x^n$ in
$$(1+x)(1+x^2)(1+x^3)\dots$$
Trivially we have $\frac{1}{(1-x)(1-x^3)(1-x^5)(1-x^7)\dots} = \frac{1-x^2}{1-x} \cdot \frac{1-x^4}{1-x^2} \cdot \frac{1-x^6}{1-x^3} \cdot \frac{1-x^8}{1-x^4}\dots$
$$ = (1+x)(1+x^2)(1+x^3)\dots$$
|
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|
Algebraic simplification problem
I must simplify this expression:
$$\frac{a^2+ac}{a^2c-c^3}-\frac{a^2-c^2}{a^2c+2ac^2+c}+\frac{2c}{c^2-a^2}-\frac{3}{a+c}$$
I managed to simplify it to:
$$\frac{( a+1)^2-(a+c)^2}{c(a+c)(a+1)^2}$$
However, I am stuck now. Therefore, I would like some help to simplify it further.
|
Hint
$$\frac{a(a+c)}{c(a-c)(a+c)}-\frac{(a-c)(a+c)}{ac(a+c)+c(a+c)}+\frac{2c}{(c-a)(a+c)}-\frac{3}{a+c}$$
|
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|
Help to use change of variables to solve the double integral I'm trying to evaluate:
$$\iint_D \left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy$$
where $D_{xy}$ is the disk $x^2+y^2\le a^2$.
The exercise is to use change of variables to solve this integral.
My solution
I chose $\varphi (r,\theta)=(ra\cos\theta,ra\sin\theta)$, where $0\le r\le 1$ and $0\le \theta\le 2\pi$ to be the change of variables.
The determinant of the Jacobian is $ra^2$ and
\begin{align*}
&\iint_{D_{xy}}\left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy \\&=\int_0^{2\pi}\int^1_0\left(\sqrt{a^2-r^2a^2}-ra\right)ra^2 drd\theta\\
&=2\pi a^3\int^1_0 \left(r\sqrt{1-r^2}-r^2 \right)dr\\
&=2\pi a^3\left(\int^1_0r\sqrt{1-r^2}dr-\int^1_0r^2dr\right)\\
&=2\pi a^3\left( \frac{1}{3}-\frac{1}{3} \right)\\
&=0
\end{align*}
I would like to know where I'm mistaken. The answer in the end of the book shows $\pi a^3/3$.
|
Just for fun, I figured I'd show that the answer is indeed zero by a direct substitution without any evaluation of an integral.
Let
$$
(x,y)=\frac{\sqrt{a^2-\hat{x}^2-\hat{y}^2}}{\sqrt{\hat{x}^2+\hat{y}^2}} (\hat x,\hat y)
$$
Now, a quick calculation will confirm that the determinant of the Jacobian is $-1$ (and so the multiplying factor is 1), and that it maps the disc of radius $a$ onto itself.
Applying our transformation, we get
$$\begin{align}
I&=\iint_D\left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}\right)dxdy\\
&=\iint_D\left(\sqrt{\hat x^2+\hat y^2}-\sqrt{a^2-\hat x^2-\hat y^2}\right)d\hat xd\hat y\\
&=-I
\end{align}$$
Therefore, $I=0$.
|
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|
mathematical induction with exponent $$\frac 13 + \frac 1{3^2} + \frac 1{3^3} + \dots + \frac 1{3^n} + = \frac 12 \times \left( 1 - \frac{1}{3^n} \right)$$
Step 1 - $n=1$
$$\begin{align}
\frac 1 {3^1} & = \frac 1 2 \times \left( 1 - \frac 1 {3^1} \right) \\
\frac 1 3 & = \frac 1 2 \times \left( 1 - \frac 1 3 \right) \\
\frac 1 3 & = \frac 1 2 \times \frac 2 3 \\
\frac 1 3 & = \frac 1 3 \\
\end{align}$$
Step 2 - n$=k$
$$ \frac 13 + \frac 1{3^2} + \frac1 {3^3} + \dots + \frac 1 {3^k} = \frac 12 \times \left( 1 - \frac 1 {3^k} \right)$$
Step 3 - $n=k+1$
Having problems solving step 3.
|
You must prove that $$\sum_1^k\frac{1}{3^i} + \frac{1}{3^{k+1}} = \frac{1}{2}\left(1-\frac{1}{3^{k+1}}\right).$$
Applying the assumption that $$\sum_1^k\frac{1}{3^i} = \frac{1}{2}\left(1-\frac{1}{3^k}\right)$$, you can get the result.
|
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|
$\int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$
$$I = \int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$$
Substituting for $e^x$,
$$I = \int {u^2 - 1 \over u^4 + u^2 + 1} du = \int { u^4 + u^2 + 1 + - 2 - u^4 \over u^4 + u^2 + 1} du = u - \int {u^4 + 2 \over u^4 + u^2 + 1} du $$
Now I don't know anything I can do to last integral except partial fraction decomposition but I am pretty sure that $u^4 + u^2 + 1$ does not have any factors in real numbers.
Is this integral computable on real numbers ? How do I compute it ?
|
Observe that $$u^4 + u^2 + 1 = (u^4 + 2u^2 + 1) - u^2 = (u^2 + 1)^2 - u^2 = (u^2 + u + 1)(u^2 - u + 1).$$ From this, we try a partial fraction decomposition of the form $$\frac{Au + B}{u^2 - u + 1} + \frac{Cu + D}{u^2 + u + 1} = \frac{u^2 - 1}{u^4 + u^2 + 1},$$ and after multiplying out and comparing like coefficients of $u$, we have $$\begin{align*} A + C &= 0 \\ A+B-C+D &= 1 \\
A+B+C-D &= 0 \\
B+D &= -1. \end{align*}$$ From here, we get $$B = D = -1/2,$$ and $$A = 1, C = -1.$$ Consequently, the integrand becomes $$\frac{1}{2}\left( \frac{2u-1}{u^2-u+1} - \frac{2u+1}{u^2+u+1}\right),$$ and the rest is straightforward. As a bonus, you even find that the numerator of each term is the derivative of the respective denominator. You can't get much nicer than that.
|
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|
Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.
|
We have the following:
$$(log_ab+1)(log_bc+1)(log_ca+1)=(\frac {logb}{loga}+1)(\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$
Expanding yields:
$$(\frac {logc}{loga}+\frac {logb}{loga}+\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$
$$=(\frac {loga}{loga}+\frac {logc}{loga}+\frac {logb}{logc}+\frac {logb}{loga} +\frac {loga}{logb}+\frac {logc}{logb}+\frac {loga}{logc}+1)$$
$$=(log_aa+log_ac+log_cb+log_ab+log_ba+log_bc+log_ca+1)$$
$$=(log_ab+log_bc+log_ca)+(log_ba+log_cb+log_ac)+2$$
Thus finally we see, via your initial condition:
$$(log_ab+1)(log_bc+1)(log_ca+1)=8+13+2=23$$
|
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|
Prove that $\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$ is a closed and bounded set.
Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.
The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\textbf{x}\in\mathbb R^n: f(\textbf{x})=0\}$ is closed. I would need to have a hint for the boundedness of $V$. Thanks!
|
The equation $x^2 - x y + y^2 = 1$ can be rewritten as follows
$$\begin{bmatrix} x\\ y\end{bmatrix}^{\top} \begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$
Computing the spectral decomposition,
$$\begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} = \frac 12 \begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 0.5 & 0\\ 0 & 1.5\end{bmatrix} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix}$$
we then obtain the following
$$\frac{1}{2} \left( \frac{x + y}{\sqrt 2} \right)^2 + \frac{3}{2} \left( \frac{x - y}{\sqrt 2} \right)^2 = 1$$
Thus, the set is an ellipse, which is closed and bounded. The ellipse's semimajor and semiminor axes are $\sqrt 2 \approx 1.414$ and $\sqrt{\frac 23} \approx 0.816$, respectively. Plotting the ellipse,
|
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|
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
|
Hint:
For $x\ne2$,
$$|3x+1|<|x-2|.$$
Then we need to distiguish three cases:
*
*$x\le-\frac13\to-3x-1<-x+2$, or $x>-\frac32$,
*$-\frac13\le x\le 2\to3x+1<-x+2$, or $x<\frac14$,
*$2\le x\to 3x+1<x-2$ or $x<-\frac32$, which is impossible.
|
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|
On Chinese remainder theorem Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as solution.
My query is say instead of $3$ we choose $3+11k$ and instead of $7$ we choose $7+13\ell$ for some $k,\ell\in\Bbb Z$ can we still recover $124$?
Instead of $696$ is there a direct way to get $124$?
We have $696=124+4(143)$. In general what is the quantity that goes in instead of $4$?
|
Some people do CRT theorem problems like this. If $x\equiv 3\pmod{11},$ then $x = 3+11k$. Plug that into $x \equiv 7 \pmod{13}$ to get $3+11k \equiv 7 \pmod{13}$ which reduces to
$$11k \equiv 4 \pmod{13}$$
$$-2k \equiv 4 \pmod{13}$$
$$k\equiv -2 \equiv 11 \pmod{13}$$
So $x = 3+11\cdot 11 =124.$
|
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|
Multiples of six in an interval How many multiples of $6$ are there between the following pairs of numbers?
$0$ and $100$ and $-6$ and $34$.
*
*$16$ and $6$
*$17$ and $6$
*$17$ and $7$
*$16$ and $7$
My attempt:
Total number of integer between $0$ to $100 = 101$
So, total number of multiplier between $0$ to $100 = \left\lfloor\frac{101}{6}\right\rfloor = 16$
And, Total number of integer between $-6$ to $34 = 41$
So, total number of multiplier between $-6$ to $34 = \left\lfloor\frac{34}{6}\right\rfloor = 6$
But, somewhere it explained as:
Number of multiples of $6$ between $1$ and $100 = 100/6 = 16$
Since the range starts from zero, we need to take zero too. [zero is a multiple of every integer (except zero itself)].
So, answer $= 16+1 = 17$
Number of multiples of $6$ between $1$ and $34 = 34/6 = 5$
Since the range is $-6$ to $34$, we need to take $-6$ and zero.
So, answer $= 5+2 = 7$.
However, this small question and we can count on fingers.
But, what is wrong with my approach, can you explain it, please?
|
There are no multiples of $m$ bewteen $km+1$ and $km + (m-1)$ and the only multiple between $km$ and $km + (m-1)$ is $km$.
So....
The multiples of $m$ between $km + 1$ and $jm$ will be the $lm$ for $k < l \le j$ or $0 < l-j \le j-k$ and there will be $j-k$ of them.
The multiples of $m$ between $km$ and $jm$ will be the $lm$ for $k \le l \le j$ or $0 \le l-j \le j-k$ and there will be $(j-k) + 1$.
The multiples of $m$ between $km + i;0\le i < m$ and $jm + n; n\le 0 < m$ will be $j-k$ if $i > 0$ and $(j-k)+1$ if $i= 0$.
So....
Between:
$0=6*0$ and $100=96+4$ will have $\frac {96}6 + 1$.
$−6=-6$ and $34=30+4$ will be $\frac {30-(-6)}6 + 1$.
$16= 12 + 4$ and $6$ will be $\frac{12-6}{6}+1$
$ 6= 6$ and $17=12+5$ will be $\frac{12-6}{6} + 1$
Those all had the lower term itself being a multiple of $6$.
$7 = 6+1$ and $17= 12 + 5$ will be $\frac{12-6}{6}$. This was the only one were the lowest term was not itself a multiple of $6$.
|
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Number Theory: Find possible $n$ values that $n(n+1)(5n+2)$ is divisible by $2017^{2018}$ Assume $m= 2017^{2018}.$ How can I find possible value(s) of $n$ that $n < m$ and $n(n+1)(5n+2)$ is divisible by $m$.
Any hints?
|
$2017$ is prime. $\gcd(n,n+1) = 1$, $\gcd(5n+2,n) = \gcd(2,n) = \{1,2\}$, $\gcd(5n+2, n+1) = \gcd(-3, n+1) = \{3,1\}$. So if $2017^{2018}|n(n+1)(5n+2)$ then either
$2017^{2018}|n$ and $n = k*m$.
or $2017^{2018}|n+1$ and $n= k*m - 1$.
or $2017^{2018}|5n + 2$ and $n = \frac{km- 2}5$. This requires $km \equiv 2 \mod 5$ As $2017 \equiv 2 \mod 5$ and $2^4 \equiv 1 \mod 5$, $2017^{2018} \equiv 2^2 \equiv 4 \mod 5$ so we need $4k \equiv 2 \mod 5$ so $k \equiv 3 \mod 5$. $n = \frac{(5j+3)m - 2}5$
..... But (dopeslap!) $n < m$ (didn't see that at first)
So $n = m -1$ or $n = \frac {3m - 2}5$
Oh, or $n = 0$... or $n - 1$ or any of the negative multiples above. I assume that $n$ is natural number....
|
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|
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
|
Cauchy- Schwarz works:
$$x^2+y^2+z^2=\frac{1}{3}(1^2+1^2+1^2)(x^2+y^2+z^2)\geq\frac{1}{3}(x+y+z)^2=\frac{1}{3}$$
|
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How do I prove using the definition of limit that $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$? I want to use the definition of limit in order to prove $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$, but I almost crack my head open trying to do this, it seems to be easy way but y cant find a $\delta$ such that $|\frac {x^2+y^3}{y^2 - x + xy}| \leq \epsilon$, any idea of how I might be able to find an inequality to transform the denominator in a nicer thing like $\frac{1}{\left\lVert (x,y)\right\rVert}$?
|
The limit does not exist. You can write $$
y^{2}+xy-x=\left( y+\frac{1}{2}x+\frac{1}{2}\sqrt{x\left( x+4\right)
}\right) \left( y-\frac{1}{2}\sqrt{x\left( x+4\right) }+\frac{1}
{2}x\right).
$$
Take $y=x^{2}+\frac{1}{2}\sqrt{x\left( x+4\right) }-\frac{1}{2}x$ with $x>0$, to get
\begin{align*}
& \frac{x^{2}+\left( x^{2}+\frac{1}{2}\sqrt{x\left( x+4\right) }-\frac
{1}{2}x\right) ^{3}}{\left( x^{2}+\sqrt{x\left( x+4\right) }\right)
x^{2}}\\
& =\frac{x^{2}+x^{3/2}\left( x^{3/2}+\frac{1}{2}\sqrt{x+4}-\frac{1}{2}%
x^{1/2}\right) ^{3}}{\left( x^{3/2}+\sqrt{x+4}\right) x^{5/2}}\\
& =\frac{1}{x}\frac{\left[ x^{1/2}+\left( x^{3/2}+\frac{1}{2}\sqrt
{x+4}-\frac{1}{2}x^{1/2}\right) ^{3}\right] }{x^{3/2}+\sqrt{x+4}}%
\rightarrow\infty
\end{align*}
On the other hand if $y=0$ then $-\frac{x^2}{x}=-x\to 0$.
EDIT: robjohn yes there are simpler ways to show that the limit does not exist, but I was trying to show a general method. If you are computing a limit
$$\lim_{(x,y)\to (0,0)}\frac{x^my^n}{Q(x,y)(y-f(x))^l}$$
where $Q(x,y)$ is a polynomial with $Q(x,y)>0$ for $(x,y)\ne (0,0)$ near $(0,0)$ and $f$ is a regular nonzero function with $f(0)=0$, or similar limits where the denominator vanishes along a regular curve (or union of curves) passing through $(0,0)$ then usually you can show that the limit does not exist. For example
$$\lim_{(x,y)\to (0,0)}\frac{x^{100}y^{2000}}{x(y-x)}$$
or$$\lim_{(x,y)\to (0,0)}\frac{x^{100}y^{2000}}{x(y-\sin x)}$$
|
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|
If $A= 2 \cdot \pi/7$ then show that $ \sec A+ \sec 2A+ \sec 4A=-4$ If $A= 2 \times \pi/7$ then how to show,
$$\sec A+ \sec 2A+ \sec 4A=-4$$ I have tried using formula for $\cos 2A$ but I failed.
|
Let us consider $\Phi_7(x)=\frac{x^7-1}{x-1}$. It is a palyndromic polynomial, hence $\frac{\Phi_7(x)}{x^3}$ can be written as a polynomial of $x+\frac{1}{x}$. Given that $\Phi_7(x)$ vanishes at the primitive seventh roots of unity, we get a polynomial that vanishes at $\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}\right\}=\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{8\pi}{7}\right\}=A$:
$$ f(x)=8x^3+4x^2-4x-1 = 8\prod_{\zeta\in A}(x-\zeta) \tag{1}$$
from which:
$$ \frac{f'(x)}{f(x)} = \frac{d}{dx}\log f(x) =\sum_{\zeta\in A}\frac{1}{x-\zeta}\tag{2}$$
and:
$$ \sum_{\zeta\in A}\frac{1}{\zeta} = -\frac{f'(0)}{f(0)} = -\frac{-4}{-1} = \color{red}{-4} \tag{3}$$
as wanted.
|
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|
Given: $\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$, find $f(x)$.
$$\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$$
On differentiating, I get,
$$f(x)\sin x\cos x = {f^\prime(x)\over f(x)}{1\over 2( b^2 - a^2)}$$
$$\sin 2x (b^2 - a^2) = {f^\prime( x)\over (f(x))^2} $$
On integrating,
$${-1\over f(x)} = {-(b^2 - a^2)\cos 2x\over 2} \implies f(x) = { 2\over(b^2 - a^2)\cos 2x}$$
The answer given is $\displaystyle f(x) = {1\over a^2 \sin^2 x + b^2 \cos^2 x}$.
I am unable to get the given result, the closest I got is, $$f(x) = {2\over b^2 \cos^2x -b^2\sin^2 x- a^2\cos^2x+ a^2\sin^2 x}$$.
How to simplify further to get the given answer ?
Related but not duplicate.
|
$f(x) = {2\over b^2 \cos^2x -b^2\sin^2 x- a^2\cos^2x+ a^2\sin^2 x}$
$= {2\over b^2 \cos^2x -b^2(1-\cos^2 x)- a^2(1-\sin^2x)+ a^2\sin^2 x}$
$= {2\over b^2 \cos^2x -b^2 + b^2\cos^2 x- a^2 +a^2\sin^2x+ a^2\sin^2 x}$
$= {2\over -(a^2 + b^2 ) + 2b^2\cos^2x +2a^2\sin^2x}$
As you can see one extra term $-(a^2+b^2)$ in denominator and its not possible to eliminate it.
Edit -
From @mickep comment as you have constant term after integration. If we consider $= {2\over -(a^2 + b^2 )}$ as constant term which is possible we get expected result.
|
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Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}-\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
But by symmetry $$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
so
$$2S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}$$ and by Cauchy Scwartz inequality
$$2S \le \sqrt{4-x+4-y+4-z}\times \sqrt{3}$$ so
$$2S \le \sqrt{24}$$
so
$$S \le \sqrt{6}$$
Is this approach correct?
|
$$\mathbf{\color{green}{New\ version\ of\ 10.02.2018}}$$
Let us find the greatest value of $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$ under the condition $x+y+z=4.$
Substitutions
$$a=\sqrt{x+y},\quad b=\sqrt{y+z},\quad c=\sqrt{z+x}$$
lead to the task on the greatest value of $a+b+c$ under the condition $a^2+b^2+c^2=8.$
Stationary points can be found using Lagrange multipliers method with the function
$$f(a,b,c,\lambda)=a+b+c+\lambda(a^2+b^2+c^2-8)$$
via solution of the system $f'_a=f'_b=f'_c=f'_\lambda=0,$ or
\begin{cases}
1+2\lambda a = 0\\
1+2\lambda b = 0\\
1+2\lambda c=0\\
a^2+b^2+c^2=8\\
(a,b,c)\in \mathbb R_+.
\end{cases}
Summation of $(1.1)-(1.4)$ with factors $a,b,c,-\lambda$ gives
$$a+b+c=-16\lambda,$$ and after substitution of this to $(1)$ one can get the system
$$a(a+b+c)=b(a+b+c)=c(a+b+c),\quad a^2+b^2+c^2=8,$$
with the evident solution
$$a=b=c=\sqrt{\dfrac83},\quad x=y=z=\sqrt{\dfrac43},\quad f=\sqrt{24},$$
so
$$\boxed{S=\sqrt6\approx2.449490}$$
is the least value.
For example, in the point $(x,y,z)=(0,1,3)$
$$S_1=\dfrac{\sqrt1+\sqrt4+\sqrt3}2=\dfrac{3+\sqrt3}2\approx2.366025 < S.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2250109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Showing that $\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$ for any fixed $x \in \mathbb{R}_{>0}$ How can I show that
$$\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$ when $x$ is a positive real number ?
|
$$ \prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$
Proof:
$$\frac{1+kx}{2+kx}=1+\frac{-1}{2+kx} \leqslant\exp \left( \frac{-1}{2+kx} \right) \leqslant \exp \left( \frac{-x^{-1}}{k} \right)$$
So,
$$ \prod_{k = 1}^{N} \frac{1+kx}{2+kx} \leqslant \prod_{k = 1}^{N} \exp \left( \frac{-x^{-1}}{k} \right) = \exp(-x^{-1}H_N)$$
where $H_N$ is the $N^{th}$ harmonic number.
Using the simple estimate $H_N > \log N$, we have that
$$\prod_{k = 0}^{N} \frac{1+kx}{2+kx} \le \frac{1}{2} \exp(-x^{-1}\log N)=\frac{1}{2N^{x^{-1}}}\to 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2250474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Number Theory: Prove that $7^{7^n}+1$ can be shown as product of at least $2n+3$ prime numbers Prove that for each natural number $n$, $7^{7^n}+1$ can be represented as product of at least $2n+3$ prime numbers. (Prime numbers are not necessarily different)
Any hints how to start the proof?
|
Aurifeuillean factorization gives
$$7^{14k+7} + 1 = (7^{2k+1}+1)\cdot(C-D)(C+D)$$
with $$C,D=(7^{6k+3} + 3\cdot 7^{4k+2} + 3\cdot 7^{2k+1} + 1),(7^{5k + 3} + 7^{3k+2} + 7^{k+1})$$
so for $14k+7=7^{n+1}$:
$$\begin{align*}7^{7^{n+1}} + 1 \;= &\;\;(7^{7^n}+1)\\
&\cdot\left[(7^{3\cdot7^n} + 3\cdot 7^{2\cdot7^n} + 3\cdot 7^{7^n} + 1)+(7^{(5\cdot7^n+1)/2} + 7^{(3\cdot7^n+1)/2} + 7^{(7^n+1)/2})\right]\\
&\cdot\left[(7^{3\cdot7^n} + 3\cdot 7^{2\cdot7^n} + 3\cdot 7^{7^n} + 1)-(7^{(5\cdot7^n+1)/2} + 7^{(3\cdot7^n+1)/2} + 7^{(7^n+1)/2})\right]\end{align*}$$
For $n=0$ this gives $$7^{7^1}+1=2^3\cdot \text{ two factors}$$
So by induction $7^{7^n}+1$ is a product at least $2n+3$ factors $>1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2256547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find $m\in\mathbb N$, $n\in\mathbb N$, and $f(0)$ where $f(x)=ax^3+bx^2+cx+d$ $(a,b,c,d\in\mathbb Z)$, $f(mn)=1$, $f(m)=n^2$, $f(n)=m^2$, ...
Question: Find $m\in\mathbb N$, $n\in\mathbb N$, and $f(0)$ where ($m,n\gt 1$)
*
*$f(x)=ax^3+bx^2+cx+d$ $(a,b,c,d\in\mathbb Z)$
*$f(mn)=1$
*$f(m)=n^2$
*$f(n)=m^2$
*$f(1)=m^2n^2$
What I tried so far was:
\begin{align}
&f(mn)-f(1)=(mn-1)\left(a(m^2n^2+mn+1)+b(mn+1)+c\right)=(mn-1)(mn+1)\\
&\therefore a(m^2n^2+mn+1)+(b-1)(mn+1)+c=0\tag1\\
&f(m)-f(n)=(m-n)\left(a(m^2+mn+n^2)+b(m+n)+c\right)=(m-n)(m+n)\\
&\therefore m=n,\quad\text{or}\quad a(m^2+mn+n^2)+(b-1)(m+n)+c=0\tag2\\
&\\
&m\ne n:\\
&(1)-(2)\Rightarrow (m-1)(n-1)\left(a(m+1)(n+1)+(b-1)\right)=0\\
&b-1=-a(m+1)(n+1)\\
&(2)\rightarrow c=-a(m^2+mn+n^2)+a(m+1)(n+1)(m+n)\\
&d=m^2n^2-a-b-c=mn(mn-a(m+n))-1\\
&f(x)=ax^3-a(m+1)(n+1)x^2+x^2-a(m^2+mn+n^2)x+a(m+1)(n+1)(m+n)x+mn(mn-a(m+n))-1\\
&
\end{align}
And then I gave up...
|
By 2., we can use the Factor Theorem to write
$$
f(x)-1=q(x)(x-mn)
$$
for polynomial $q$ with integer coefficients (and degree at most $2$, though that doesn't directly figure into this proof).
From 3. it follows that
$$
n^2-1=q(m)(m-mn)
$$
and so (since $n>1$)
$$
q(m)=\frac{n^2-1}{m-mn}=-\frac{n+1}{m}
$$
Similarly,
$$
q(n)=-\frac{m+1}{n}
$$
But $q$ has integer coefficients, which means it takes on integer values at integer arguments. So $n+1$ is a multiple of $m$ and $m+1$ is a multiple of $n$.
Without loss of generality, suppose $n \geq m$. Since $n > 1$ and $m+1$ is a multiple of $n$, we must have $n=m+1$. But then $m+2$ is a multiple of $m$. As $m > 1$, this implies that $m=2$ (and hence $n=3$).
So the conditions reduce to
$$
f(6)=1\\
f(3)=4\\
f(2)=9\\
f(1)=36
$$
from which it follows by Lagrange interpolation that
$$
f(x)
=−2x^3+23x^2−82x+97
$$
So $f(0)=97$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2256985",
"timestamp": "2023-03-29T00:00:00",
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|
Find the radius of convergence of $\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$ Find the radius of convergence of the following:
$$\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$$
My attempt:
I used Ratio Test and managed to get until
$$\lim_{n\to\infty} \bigg|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \left(\frac{n}{n+1}\right)^{3} (x-2)\bigg|$$
I need to use L'Hopital Rule to get the answer.
|
$$\sum_{n=1}^\infty\dfrac{5^n+(-1)^n}{n^3}(x-2)^n=\sum_{n=1}^\infty\dfrac{\{5(x-2)\}^n}{n^3}+\sum_{n=1}^\infty\dfrac{\{(-1)(x-2)\}^n}{n^3}$$
Can you apply Ratio Test separately?
OR
$$\lim_{n\to\infty} \bigg|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \left(\frac{n}{n+1}\right)^{3} (x-2)\bigg|$$
$$=\lim_{n\to\infty}\left|(x-2)\left(\dfrac1{1+\dfrac1n}\right)^3\cdot\dfrac{5-\left(\dfrac15\right)^n}{1+\left(\dfrac15\right)^n}\right|=|5(x-2)|$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2257178",
"timestamp": "2023-03-29T00:00:00",
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|
Two-row notation and disjoint cycles $$\pi = (12)(13)(826)(58)$$
I am asked to express $\pi$ as a product of disjoint cycles.
I am given the answer $(513268)$, why is it not $(513)(268)$
|
$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline (58)x & 1 & 2 & 3 & 4 & \color{blue}8 & 6 & 7 & \color{blue}5 \\
\hline (826)(58)x & 1 & \color{blue}6 & 3 & 4 & \color{blue}2 & \color{blue}8 & 7 & 5 \\
\hline (13)(826)(58)x & \color{blue}3 & 6 & \color{blue}1 & 4 & 2 & 8 & 7 & 5 \\
\hline (12)(13)(826)(58)x & 3 & 6 & \color{blue}2 & 4 & \color{blue}1 & 8 & 7 & 5 \\
\hline
\end{array}$
Result : $1\to 3\to 2\to 6\to 8\to 5\to 1$ and $4$ and $7$ do not move.
$(12)(13)(826)(58)=(132685)$ which is the same as the stated $(513268)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2258354",
"timestamp": "2023-03-29T00:00:00",
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|
Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my soliution:
$$S(x)=1+\frac{x^{4}}{4\cdot2^{4}}+\frac{x^{7}}{7\cdot2^{7}}+\frac{x^{10}}{10\cdot2^{10}}+\cdots=1+\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}=1+S_1(x)$$
$$(S_1(x))_x'=\left(\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}\right)_x'=\sum_{n=1}^\infty \frac{x^{3n}}{2^{3n+1}}=\frac{1}{x}\sum_{n=1}^\infty \left(\frac{x}{2}\right)^{3n+1}$$
Now let's take $\frac{x}{2}=y$, then
$$S_2(y)=\sum_{n=1}^\infty y^{3n+1}=y^4+y^7+y^{10}+\cdots=\frac{y^4}{1+y^3},|y|\le1$$
$$\left(S_1(y)\right)'=\frac{1}{2y}\cdot\frac{y^4}{1-y^3}=\frac{1}{2}\cdot\frac{y^3}{1-y^3}$$
$$S_1(y)=\frac{1}{2}\int\frac{y^3}{1-y^3}dy=\frac{1}{2}\int\left(-1+\frac{1}{1-y^3}\right)dy=-\frac{1}{2}y+\frac{\sqrt{3}}{6}\arctan\left(\frac{2\left(y+\frac{1}{2}\right)}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\lvert y-1\rvert+C$$
$$S_1(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|+C$$
$$S_1(0)=0, C=-\frac{\sqrt{3}\pi}{36}$$
$$S(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|-\frac{\sqrt{3}\pi}{36}+1$$
$$S(1)=\frac{3}{4}+\frac{\sqrt{3}}{3}\arctan\left(\frac{2\sqrt{3}}{3}\right)+\frac{1}{6}\ln(7)-\frac{\sqrt{3}\pi}{36}$$
By writing this for about 2 hours I deserve extra 50 points or at least good answers... Ha ha, thanks!
|
Similar to the answer by Jack D'Aurizio and yours, but let $$f(x) = \sum_{n=0}^\infty \frac{1}{3n+1}\cdot x^{3n+1}\implies f'(x) = \sum_{n=0}^\infty x^{3n} =\frac{1}{1-x^3}$$
Hence,
$$f(x) = \int\frac{1}{1-x^3}dx = \frac{1}{6}\left(\ln(x^2+x+1)-2\ln(1-x)+2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)+C$$ Because $f(0)=0$ we have $C = -\frac{\sqrt{3}}{18}\pi$. We get our result by noticing that our result is $\frac{1}{2}+f\left(\frac{1}{2}\right)$, which is approximately $1.016849...$.
|
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|
If $a,b,c$ are positive, prove that $\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \geq \frac{9}{a+b+c}$ If $a,b,c$ are positive real numbers, prove that
$$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$
|
By Cauchy-Schwartz. $$\sum_{cyc}\frac{1}{a+b}\sum_{cyc}(a+b)\geq (1+1+1)^2=9\implies 2\cdot\sum_{cyc}\frac{1}{a+b}\geq2\cdot \frac{9}{\sum_{cyc}(a+b)}=\frac{9}{a+b+c}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving a Line to bisect a line in a Triangle From point $A$ tangents $AB$ and $AC$ to a circle are drawn ($B$ and $C$ tangent points); $PQ$ is a diameter of the circle; line $L$ is tangent to the circle at point $Q$. Lines $PA$, $PB$, and $PC$ intersect line $L$ at points $A_1, B_1, C_1$. Prove that $A_1B_1 = A_1C_1$.
|
Not a complete solution yet, but this is the beginning of pure "bashing" with coordinates.
Choose a coordinate system so that $B_1 = (1,0)$ and $C_1 = (0, -1).$ Then, it suffices to prove that $A_1 = (0,0).$
Let $P = (a,2b)$ and $Q = (a,0)$, so the equation of the circle is $(x-a)^2+(y-b)^2 = b^2$. This is simplified to $x^2+y^2-2ax-2by+a^2 = 0.$ The equation of the line passing through the points $P$ and $B_1$ is simply $y=\dfrac{2b}{a-1}x-\dfrac{2b}{a-1}$. Thus, coordinate of the point $B$ will satisfy the above two equations, since it is the intersection of the circle and that line.
Substituting $y$ in terms of $x$ and denote $p = \dfrac{2b}{a-1}$, we get: $$0 = (1+p^2)x^2-2x(a+bp+p^2)+(p^2+2bp+a^2).$$
Now when you solve this equation, you will get two solutions $x_1 = a$ and $x_2 = \dfrac{p^2+a}{p^2+1}$, but we have to take the latter because $P\neq B.$ Thus, $B = \Big(\dfrac{p^2+a}{p^2+1}, \dfrac{2b}{p^2+1}\Big)$.
Similarly, the equation of the line passing through $P$ and $C$ is $y = \dfrac{2b+1}{a}x - 1$ and when you plug this into the equation of the circle you will find $C = \Big(\dfrac{q+a}{q^2+1}, \dfrac{aq-1}{q^2+1} \Big)$, where $q = \dfrac{2b+1}{a}.$
Now, it is straightforward to find the coordinate of point $A$ because it is the intersection of two tangents to the circle at $B$ and $C$. After this, find the equation of the line $l = PA$, then the coordinate of $A_1$ will be found by intersecting $l$ with the $x$ axis.
I will not have time to complete the remaining computations, but it is completely straightforward at this point. But our goal is to prove that $A_1 = (0,0)$; therefore, if the problem is true, then it the equation of the line $l$ must have been found in the form $y = cx, c\in\mathbb{R}.$
|
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|
2D Geometry. Find equation of a circle
The line $y=x$ touches a circle at $P$ so that $OP=4\sqrt{2}$ where $O$ is the origin. The point $(-10,2)$ lies inside the circle and the length of the chord $x+y=0$ is $6\sqrt{2}$.
Find the equation of the circle.
How to approach this sum?
My approach to the sum: I found that the lines $y=x$ and $x+y=0$ are perpendicular and meet at the origin . Now how to use the given information about the lengths and find the radius? Since the lines are perpendicular, I know I have to use Pythagorean theorem to find the radius.
|
The only points on $y=x$ which are $4 \sqrt{2} $ away from the origin are $(4,4)$ and $(-4,-4)$.
The perpendicular distance from the centre of the circle to the chord $x+y=0$ is $4 \sqrt{2} $.
So the radius of the circle is $\sqrt{(4\sqrt{2}) ^2 +(6\sqrt{2} \div 2)^2 } =5\sqrt{2} $.
The centre of the circle is $(4,4)\pm(5,-5)$ or $(-4,-4)\pm(5,-5)$.
The only possible case that $(-10,2)$ is inside the circle is when its centre is $(-9,1)$.
The equation of the circle is $(x+9)^2+(y-1)^2=50$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a, b, c$ are positive and $a+b+c=1$, prove that $8abc\le\ (1-a)(1-b)(1-c)\le\frac{8}{27}$ If $a, b, c$ are positive and $a+b+c=1$, prove that $$8abc\le\ (1-a)(1-b)(1-c)\le\dfrac{8}{27}$$
I have solved $8abc\le\ (1-a)(1-b)(1-c)$ (by expanding $(1-a)(1-b)(1-c)$)
but do not get how to show that $(1-a)(1-b)(1-c)\le\frac{8}{27}$
|
For the 1st one: $(1-a)(1-b)(1-c) = (b+c)(a+c)(b+a) \ge 2\sqrt{bc}\cdot 2\sqrt{ac}\cdot 2\sqrt{ba} = 8abc$, and the 2nd one is by AM-GM as shown by the other answer.
|
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|
Continued Cosine Product. Is there a way to evaluate,
$$
\large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)}
$$
I asked this to one of my teachers and what he told is something like this,
Multiply and divide the last term of $(1)$ with $\boxed{\sin \frac{x}{2^{n-1}}}$
So,
$$
\large \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}} \\ \tag*{(2)}
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \sin \frac{x}{2^{n-1}}} \\
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \frac{\sin \frac{x}{2^{n-1}}}{\frac{x}{2^{n-1}}} \cdot \frac{x}{2^{n-1}}} \tag*{(3)}
$$
Now, as $n \to \infty$ , we have $x \to 0$,
Using this, $\lim$ we have,
$$
\boxed{ \lim_{x \to 0} \frac{\sin x}{x} = 1}
$$
Using this in $(3)$, we have,
$$
\large \boxed{\frac{\sin (2x)}{2x}} \tag*{(4)}
$$
All the steps sort of make sense. My doubts are,
*
*How do I do this for other trigonometric ratios?
*How does the step 2 happen?
I need help looking into it more intuitionally.
Please provide necessary reading suggestions.
Regards.
|
How does step 2 happen:
multiplying the last factor by $\frac {\sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$
gives us
$\cos x \cdot \cos \frac x2\cdots \cos \frac x{2^{n-2}}\cdot \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$
Double angle formula.
$\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}} = \frac 12 \sin \frac{x}{2^{n-2}}$
Applying this we get:
$\cos x \cdot \cos \frac x2\cdots \cdot \frac{\cos \frac x{2^{n-2}}\cdot\sin \frac{x}{2^{n-2}}}{2\sin \frac{x}{2^{n-1}}}$
And we can apply the double angle formula again. And do it repeatedly until all of the $\cos$ factors have been devoured.
|
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|
Equality involving greatest integer I am trying to find the integral solutions to the equation
$$[x][y]=x+y$$
and show that all non-integral solutions lie on exactly two lines. Here $[x]$ denotes greatest integer function.
For integer solutions I solved the equation $xy=x+y.$
Upon solving it I got $y=\frac{x}{x-1}$ and was not able to proceed further. For non integral values, I got two inequalities $y>1$, $x>1$ but these aren't lines. These I got by breaking greatest integer of $x$ and $y$ into their subsequent fractional and integer parts and using the concept of basic inequalities but my answer is still incomplete. How do I proceed?
|
First suppose $x,y$ are integers.
\begin{align*}
\text{Then}\;\;&\lfloor x \rfloor \lfloor y \rfloor = x + y\\[4pt]
\iff\; &xy = x + y\\[4pt]
\iff\; &xy - x - y = 0\\[4pt]
\iff\; &xy - x - y + 1 = 1\\[4pt]
\iff\; &(x-1)(y-1) = 1\\[4pt]
\iff\; &x-1 = 1\;\;\text{and}\;\;y-1=1\\[0pt]
&\;\;\;\;\text{or}\\[0pt]
&x-1 = -1\;\;\text{and}\;\;y-1= -1\\[4pt]
\iff\; &(x,y)=(2,2)\;\;\text{or}\;\;(x,y)=(0,0)\\[4pt]
\end{align*}
Next suppose at least one of $x,y$ is not an integer.
\begin{align*}
\text{Let}\;\; &a=\lfloor x \rfloor,\;\;r=x-\lfloor x \rfloor\\[4pt]
&b=\lfloor y \rfloor,\;\;s=y-\lfloor y \rfloor\\[4pt]
\end{align*}
Thus we have $x = a + r$ and $y = b + s$, where $a,b$ are integers, and
$0 \le r,s < 1$.
By assumption, $r,s$ are not both equal to $0$, hence $0 < r + s < 2$.
\begin{align*}
\text{Then}\;\;&\lfloor x \rfloor \lfloor y \rfloor = x + y\\[4pt]
\implies\; &ab = (a + r) + (b + s)\\[4pt]
\implies\; &ab - a - b = r + s\\[4pt]
\implies\; &r + s \in \mathbb{Z}\\[4pt]
\implies\; &r + s = 1\;\;\text{and}\;\;r,s>0\\[4pt]
\implies\; &ab - a - b = 1\\[4pt]
\implies\; &ab - a - b + 1 = 2\\[4pt]
\implies\; &(a-1)(b-1) = 2\\[4pt]
\end{align*}
From the equation$\;(a-1)(b-1) = 2,\;$you get $4$ cases for the pair $(a,b)$, one for each of the factorizations
$$(1)(2)=2
,\;\;\;
(2)(1)=2
,\;\;\;
(-1)(-2)=2
,\;\;\;
(-2)(-1)=2
$$
Noting that $r+s=1$ and $r,s>0$, for each valid pair $(a,b)$, the pair $(x,y)$ must satisfy
\begin{align*}
&\begin{cases}
x=a+r\\[2pt]
y=b+(1-r)\\
\end{cases}
\\[2pt]
&\;\;\;\,0 < r < 1\\[2pt]
\end{align*}
which is the segment of the line $x+y = a + b + 1$, strictly between the points $(a,b+1),\;(a+1,b)$.
Conversely, if $a,b$ are integers such that $ab - a - b = 1$, then for any point $(x,y) \in \mathbb{R}^2$ satisfying
\begin{align*}
&\begin{cases}
x=a+r\\[2pt]
y=b+(1-r)\\
\end{cases}
\\[2pt]
&\;\;\;\,0 < r < 1\\[2pt]
\end{align*}
\begin{align*}
\qquad\qquad\text{we have}\;
\lfloor x \rfloor \lfloor y \rfloor
&= ab\\[4pt]
&= a + b + 1\\[4pt]
&=(a+r)+\left(b+(1-r)\right)\\[4pt]
&=x + y\\[4pt]
\end{align*}
At this point, I'll let you finish it.
If you work it out correctly, the set $S$ of all points $(x,y) \in \mathbb{R}^2$ satisfying
$$\lfloor x \rfloor \lfloor y \rfloor = x + y$$
has $4$ components$\,-\,$an isolated point, and $3$ open line segments.
More precisely, $S = \{P\} \cup L_0 \cup L_1 \cup L_2$, where
\begin{align*}
{\small{\bullet}}\;\,&P\;\text{is the point}\;(2,2)\text{.}\\[5pt]
{\small{\bullet}}\;\,&L_0\;\text{is the segment of the line}\;x+y=0\;\text{strictly between}\\[-0.5pt]
&\text{the points}\;(-1,1),\;(1,-1)\text{.}\\[5pt]
{\small{\bullet}}\;\,&L_1\;\text{is the segment of the line}\;x+y=6\;\text{strictly between}\\[-0.5pt]
&\text{the points}\;(2,4),\;(3,3)\text{.}\\[5pt]
{\small{\bullet}}\;\,&L_2\;\text{is the segment of the line}\;x+y=6\;\text{strictly between}\\[-0.5pt]
&\text{the points}\;(3,3),\;(4,2)\text{.}\\[5pt]
\end{align*}
|
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|
$a+b+c+d+e=79$ with constraints How many non-negative integer solutions are there to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$?
I get that for $a\ge7$ you do $79-7=72$, $\binom{72+5-1}{5-1}=\binom{76}4$. For $b\ge35$ I think it's $\binom{47}4$ and I'm not too sure what it is for $3\le c\le41$ and I also have no clue as to how to do them all at the same time.
|
So we are looking for non-negative solutions for to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$
I always like, in this kind of problem, to work with generating functions. Everything variable gets a polynomial such that its powers correspond to the restraints, and such that he requested solution would be the coefficient of $x^{79}$ in their product:
The restriction on $a$ translates to the function $$(x^7 + x^8 + x^9 + \ldots) = x^7\left(\frac{1}{1-x}\right)$$ For $b$ we have $$(1+ x + x^2 + \ldots x^{34}) = \frac{1-x^{35}}{1-x}$$ all using standard geometric series. For $c$ we have $$\left(x^3 + x^4 + \ldots + x^{41}\right) = x^3\left(1+x+ \ldots + x^{38}\right) = x^3\left(\frac{1-x^{39}}{1-x}\right)$$ while $d$ and $e$ have no restrictions so we use $$(1+x+x^2 + \ldots) = \frac{1}{1-x}$$
So the answer to your question is the coefficient of $x^{79}$ in:
$$x^7 \frac{1}{1-x} (1-x^{35})\frac{1}{1-x} x^3 (1-x^{39})\frac{1}{1-x}\left(\frac{1}{1-x}\right)^2 $$ which comes down to the coefficient of $x^{69}$ (removing the always present $x^{10}$) in:
$$(1-x^{35})(1-x^{39})(1-x)^{-5} = (1 - x^{35} - x^{39} + x^{74})\sum_{k=0}^\infty {k+4 \choose k} x^k$$ using the generalised binomial formula.
And this coefficient equals $${73 \choose 69} - {38 \choose 34}- {34 \choose 30}$$
|
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|
Differentiation - Power rule Find derivative of this function
$$g(t)=-3t(6t^4-1)^{4/3}$$
I have tried it till the answer:
$ -3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) $
$-3t x \frac{4}{3} (6t^4-1)^\frac{1}{3} ( 6X4 t^3 -0 ) $
$ -4t (6t^4 -1)^\frac{4}{3} (24t^3) $
$-96t(6t^4-1)^\frac{1}{3} $
However , I checked and saw that the answer is
$$-3(6t^4-1)^{1/3}(38t^4-1)$$
I just started learning differentiation, and I don't understand my mistake on why I can't achieve the answer. Thanks!
|
As there are two terms. You need to use product rule.
Product rule -
$(UV)' = UV' + U'V$
So it should be -
$g(t) = -3t(6t^4-1)^\frac{4}{3}$
$-3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) - 3.\frac{d}{dt} t.(6t^4-1)^\frac{4}{3}$
Take common terms and simplify.
|
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|
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$.
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}{\sqrt[a+1]{b}}+\frac{c}{\sqrt[b+1]{c}}+\frac{a}{\sqrt[c+1]{a}}$$
but I do not yet know how to finish my proof from there (if this is helpful at all?!).
|
Now I see. Using Especially Lime's comment, we can see that $$\left(\frac{1+ab}{1+a}\right)\left(\frac{1+bc}{1+b}\right)\left(\frac{1+ca}{1+c}\right)=1.$$
Hence, we can set $$x:=\frac{1+ab}{1+a}, y:=\frac{1+bc}{1+b}$$ and our inequality becomes $$x+y+\frac{1}{xy}\geq 3$$ which is equivalent to $$\frac{x+y+\frac{1}{xy}}{3}\geq 1.$$ But using the AM-GM inequality for three variables, we can see that $$\frac{x+y+\frac{1}{xy}}{3}\geq \sqrt[3]{xy\cdot\frac{1}{xy}}=1,$$and our claim follows.
|
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|
Express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$
If $\cos^2\theta+\cos\theta = 1$, express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$.
We have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta &= \sin^6\theta(\sin^2\theta+1)+\sin^2\theta(\sin^2\theta+1)-2\\&=\sin^6\theta(2-\cos^2\theta)+\sin^2\theta(2-\cos^2\theta)-2,\end{align*} but I didn't see how to use that $\cos^2\theta+\cos\theta = 1$.
|
$S=\sin^8x+\sin^6x+\sin^4x+\sin^2x-2=\cos^4x+\cos^3x+\cos^2x+\cos x-2$
Now $S=(\cos^2x+\cos x-1)\cos^2x+2(\cos^2x+\cos x-1)-\cos x=?$ as $\cos^2x+\cos x-1=0$
|
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|
Simpson's Rule in Numerical Integration I have a problem in proving that the simple Simpson’s rule $$\int_a^b f(x) dx \approx \frac{(b − a)}6[ f (a) + 4 f \left(\frac{a + b}{2}\right) + f (b)]$$
is exact for all cubic polynomials.
I am a little bit confused about using the change of variable $$x = a + t (b − a)$$ for a general interval [a, b].
|
If the change of variables bothers you, you don't really need it: it just makes some of the formulas somewhat simpler, because it means you can take $a=0$.
If $f(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3$,
$$ \eqalign{\int_a^b f(x)\; dx &= \left. c_0 x + c_1 \frac{x^2}{2} + c_2 \frac{x^3}{3} + c_3 \frac{x^4}{4} \right|_{a}^b \cr &= c_0 (b-a) + c_1 \frac{b^2-a^2}{2} + c_2 \frac{b^3 - a^3}{3} + c_3 \frac{b^4 - a^4}{4}}$$
$$ \eqalign{\frac{b-a}{6} &\left(f(a) + 4 f\left(\frac{a+b}{2}\right) + f(b)\right)\cr
&= \frac{b-a}{6}\left(6 c_0 + c_1 \left(a + 4\left(\frac{a+b}{2}\right)+b\right) + c_2 \left(a^2 + 4\left(\frac{a+b}{2}\right)^2+b^2\right) + c_3 \left(a^3 + 4\left(\frac{a+b}{2}\right)^3+b^3\right) \right)}$$
Expand this all out and see that this is the same expression you got for
the integral.
|
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|
If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$ THE PROBLEM: If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$
MY THOUGHT PROCESS: We have to prove that $ab+bc+ca=0$.
One method using which we can do this is, if we can somehow obtain the equation $k(ab+bc+ca)=0$ we can deduce that $ab+bc+ca=0$ using the zero product rule.
Another method to do this would be to obtain an expression which has $(ab+bc+ca)$ in it. The identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ has $2(ab+bc+ca)$ in it. Therefore if we can somehow show that $(a+b+c)^2=a^2+b^2+c^2$ our task would be over.
MY ATTEMPT: I have proved the result using the first approach. I tried to do it using the second one but could not proceed far. I was facing problems showing that $(a+b+c)^2=a^2+b^2+c^2$. If i get any help i shall be very grateful.
|
I don't know whether I am just repeating your first solution.
Let $a\cos\theta=k$. Then
\begin{align*}
k(ab+bc+ca)&=abc\cos\left(\theta+\frac{4\pi}{3}\right)+abc\cos\theta+abc\cos\left(\theta+\frac{2\pi}{3}\right)\\
&=abc\left[\cos\left(\theta+\frac{2\pi}{3}\right)+2\cos\left(\theta+\frac{2\pi}{3}\right)\cos\frac{2\pi}{3}\right]\\&=0
\end{align*}
|
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|
showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$, where $f$ is a periodic arithmetical function of period $4$, with the values $f(1)=f(3)=f(4)=1$ and $f(2)=-3$. Since $\sum_{1 \le i \le 4} f(i)=0$, it is assured that this series is convergent.
|
sum of all odd terms: $X = (1 + 1/3 + 1/5 + ... )$
sum of all terms (4n+2): $Y = -3/2(1 + 1/3 + 1/5 + ...) = -3X/2$
Sum of remaining terms = $Z = 1/4 + 1/8 + 1/12 + ... = 1/4(1 + 1/2 + 1/3 + 1/4 + 1/5 + ...)$
splitting into odd and even terms: $Z = 1/4(1+ 1/3 + 1/5 +...) + 1/8(1 + 1/2 + 1/3 + ...)$
So $Z = X/4 + Z/2 \implies Z = X/2$
So $X + Y + Z = X - 3X/2 + X/2 = 0$
|
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|
Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
...
Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
|
So your final expression is: $\cos\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)$
Now we know that:
$\tan^{-1}p=\cos^{-1}\left(\dfrac{1}{\sqrt{1+p^2}}\right)\\
\implies\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt{1+{\left(\dfrac{1}{\sqrt{1+x^2}}\right)}^2}}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt{1+{\dfrac{1}{1+x^2}}}}\right)\\
\implies\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos^{-1}\sqrt{\dfrac{1+x^2}{2+x^2}}$
Therefore, $\cos\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos\cos^{-1}\sqrt{\dfrac{1+x^2}{2+x^2}}=\sqrt{\dfrac{1+x^2}{2+x^2}}$
|
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|
Evaluate the following triple summation
Evaluate $$\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \sum_{i=0}^{\infty} \frac{1}{3^i3^j3^k}$$ for $i\neq j\neq k$
My Attempt
Evaluate the summation with no restriction on $i,j,k$. Let this be $a_0$
When $i= j=k$. Let this be $a_1$.
When $i\neq j=k$, let this be $a_2$.
The required answer would be $a_0-a_1-3a_2$.
I got \begin{align}a_0&=\frac{27}{8}\\\\
a_1&=\frac{27}{26}\\ \\
a_2&=\frac{135}{208}\end{align}
And therefore, my answer is
\begin{align}a_0-a_1-3a_2&=\frac{27}{8}-\frac{27}{26}-3\times\frac{135}{208}\\\\
&=\frac{81}{208}\end{align}
Is everything correct? The method and the calculation? Also if anyone has a better method, please tell.
|
I have the same answer.
\begin{align*}
\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \sum_{i\in\mathbb{N}\setminus\{j,k\}} \frac{1}{3^i3^j3^k}&=\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \left(\frac{1}{3^j3^k}\sum_{i\in\mathbb{N}\setminus\{j,k\}} \frac{1}{3^i}\right)\\
&=\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \left[\frac{1}{3^j3^k}\left(\frac{1}{1-\frac{1}{3}}-\frac{1}{3^j}-\frac{1}{3^k}\right)\right]\\
&=\sum_{k\in\mathbb{N}}\left[\frac{1}{3^k}\left(\frac{3}{2}-\frac{1}{3^k}\right)\sum_{j\in\mathbb{N}\setminus\{k\}} \frac{1}{3^j}-\frac{1}{3^k}\sum_{j\in\mathbb{N}\setminus\{k\}} \frac{1}{3^{2j}}\right]\\
&=\sum_{k\in\mathbb{N}}\left[\frac{1}{3^k}\left(\frac{3}{2}-\frac{1}{3^k}\right) \left(\frac{3}{2}-\frac{1}{3^k}\right)-\frac{1}{3^k}\left(\frac{1}{1-\frac{1}{9}}-\frac{1}{3^{2k}}\right)\right]\\
&=\sum_{k\in\mathbb{N}}\left(\frac{9}{8}\cdot\frac{1}{3^k}-3\cdot\frac{1}{3^{2k}}+2\cdot\frac{1}{3^{3k}}\right)\\
&=\frac{9}{8}\cdot\frac{3}{2}-3\cdot\frac{9}{8}+2\cdot\frac{1}{1-\frac{1}{27}}\\
&=\frac{9}{8}\cdot\frac{3}{2}-3\cdot\frac{9}{8}+2\cdot\frac{27}{26}\\
&=\frac{81}{208}
\end{align*}
|
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|
Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \text{if} \ x<0 \\
\end{array}
\right.$$
For the first equation, we have
$$|x^2-2x|\Rightarrow\left\{\begin{array}{rcr}
x^2-2x & = & 1, \ \text{if} \ x^2\geq 2x \\
x^2-2x & = & -1, \ \text{if} \ x^2<2x \\
\end{array}
\right.$$
and for the second equation, we have
$$|x^2+2x|\Rightarrow\left\{\begin{array}{rcr}
x^2+2x & = & 1, \ \text{if} \ x^2+2x\geq 0 \\
x^2+2x & = & -1, \ \text{if} \ x^2+2x<0 \\
\end{array}
\right.$$
Solving for all of these equations, we get
$$\left\{\begin{array}{rcr}
x^2-2x & = 1 \Rightarrow& x_1=1+\sqrt{2} \ \ \text{and} \ \ x_2=1-\sqrt{2}\\
x^2-2x & =-1 \Rightarrow& x_3=1 \ \ \text{and} \ \ x_4=1\\
x^2+2x & = 1 \Rightarrow& x_5=-1-\sqrt{2} \ \ \text{and} \ \ x_6=-1+\sqrt{2}\\
x^2+2x & =-1 \Rightarrow& x_7=-1 \ \ \text{and} \ \ x_8=-1
\end{array}
\right.$$
So we have the roots $$\begin{array}{lcl}
x_1 = & 1+\sqrt{2} \\
x_2 = & 1-\sqrt{2} \\
x_3 = & -1+\sqrt{2} \\
x_4 = & -1-\sqrt{2} \\
x_5 = & 1 \\
x_6 = & -1
\end{array}$$
But according to the book, the answer is
\begin{array}{lcl}
x_1 & = & 1+\sqrt{2} \\
x_4 & = & -1-\sqrt{2} \\
x_5 & = & 1 \\
x_6 & = & -1
\end{array}
What happened to $x_2$ and $x_3$? Any other way to solve this equation quicker?
|
\begin{align*}
|x^2-2|x||&=1\\
x^2-2|x|&=1\quad\text{or}\quad -1\\
|x|^2-2|x|-1&=0\quad\text{or}\quad |x|^2-2|x|+1=0\\
|x|&=1+\sqrt{2} \quad\text{or}\quad 1\qquad(|x|=1-\sqrt{2}<0\text{ is rejected})\\
x&=\pm(1+\sqrt{2}) \quad\text{or}\quad \pm1
\end{align*}
|
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|
Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$ Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$.
My work : I noticed that A is symmetric . Hence all of its eigen vectors are real .. That is all i have got .
Your help will be highly appreciated .Thank you .
|
If $n=1$, $r(A) = 1$.
Otherwise if $n>1$, then $r(A) = 2$. Notice that
$$A =
\begin{pmatrix}
2 & 3 & \cdots & n+1 \\
3 & 4 & \cdots & n+2 \\
&&\cdots&\\
n+1&n+2&\cdots &2n
\end{pmatrix}
$$
Use elementary row operations to subtract the 1st row from the $i$th row for $2 \leq i \leq n$. Then we get
$$ A' = \begin{pmatrix}
2 & 3 & \cdots & n+1 \\
1 & 1 & \cdots & 1 \\
2 & 2 & \cdots & 2 \\
&& \cdots & \\
n-1 & n-1 & \cdots & n-1
\end{pmatrix} $$
And this matrix has rank $2$. Since we only used elementary row operations, $A$ also has rank $2$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2283993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
How to find the inverse Laplace transform of the following fraction? Please help me to find the inverse Laplace transform of the following:
$$\frac{s^2+6s+9}{(s-1)(s-2)(s-3)}$$
Do I have to divide the fractions and use partial fractions decomposition?
Like this:
$$\frac{s^2}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$
$$\frac{6s}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$
$$\frac{9}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$
|
The partial expansion is:
$$\frac{s^2 + 6 s + 9}{(s - 3) (s - 2) (s - 1)}=\frac{18}{s-3}-\frac{25}{s-2}+\frac{8}{s-1}$$
Inverting term-by-term we get:
$$\mathcal{L}_s^{-1}\left(\frac{18}{s-3}\right)=18 e^{3 n}$$
$$-\mathcal{L}_s^{-1}\left(\frac{25}{s-2}\right)=-25 e^{2 n}$$
$$\mathcal{L}_s^{-1}\left(\frac{8}{s-1}\right)=8 e^n$$
And, putting it all together...
$$8 e^n-25 e^{2 n}+18 e^{3 n}$$
EDIT:
To get the quantities, $A$, $B$, $C$, we need to crate the partial fraction template using the denominator $(s-3)(s-2)(s-1)$
$$\frac{s^2 + 6 s + 9}{(s-3)(s-2)(s-1)}=\frac{A}{s-3}-\frac{B}{s-2}+\frac{C}{s-1}$$
Next, multiply the equation by the denominator:
$$\frac{(s^2 + 6 s + 9)(s-3)(s-2)(s-1)}{(s - 3) (s - 2) (s - 1)}=\frac{A(s-3)(s-2)(s-1)}{s-3}-\frac{B(s-3)(s-2)(s-1)}{s-2}+\frac{C(s-3)(s-2)(s-1)}{s-1}$$
And, after we simplify, it turns out that:
$$s^2 + 6 s + 9=A (s - 2) (s - 1)-B(s - 3) (s - 1)+C(s - 3) (s - 2)$$
Solve for the unknown parameters by plugging in the roots of the denominators
For example, for root 3, plug in $s=3$ into the equation
$$3^2+3\dot6+9 = A(3-2))(3-1)+B(3-3)(3-1)+C(3-3)(3-2)$$
$$36=2A$$
$$A=18$$
Can you do the rest?
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2284323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$ How do you evaluate $$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$$ using the identity $$\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)?$$
I assume I'll have to express $1+\frac{1}{n^2}+\frac{1}{n^4}$ as $\left(1+\frac{a}{n^2}\right)\left(1+\frac{b}{n^2}\right)$ for some $a,b\in\mathbb{C}$ so that I could use the identity given, but I can't seem to factor the above appropriately.
|
We have $1+z+z^2=\Phi_3(z)=(z-\omega)(z-\bar{\omega})=(1-\omega z)(1-\bar{\omega}z)$, with $\omega=\exp{\frac{2\pi i}{3}}$.
By considering $z=\frac{1}{n^2}$ and $\eta=\exp\frac{\pi i}{3}$ we have that
$$ \frac{\sin(\pi w)}{\pi w}=\prod_{n\geq 1}\left(1-\frac{w^2}{n^2}\right)\tag{1} $$
implies:
$$ \prod_{n\geq 1}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right) = \frac{\sin(\pi\eta)\sin(\pi\bar{\eta})}{\pi^2}=\frac{\cos(\pi(\eta-\bar{\eta}))-\cos(\pi(\eta+\bar{\eta}))}{2\pi^2}\tag{2}$$
so that:
$$ \prod_{n\geq 1}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right) = \color{red}{\frac{1+\cosh(\pi\sqrt{3})}{2\pi^2}}=\left(\frac{1}{\pi}\,\cosh\frac{\pi\sqrt{3}}{2}\right)^2.\tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2284630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Tail Probability for Student's t-Distribution Let $n$ be a positive integer, $t < 0$, and let $F_n$ the cumulative distribution function of Student's t-distribution with $n$ degrees of freedom. Then it is a well known fact (which I always found stated without proof) that the sequence $(F_n(t))_{n=1}^{\infty}$ is strictly decreasing.
Do you know some proof of the fact that $(F_n(t))_{n=1}^{\infty}$ is a strictly decreasing sequence?
NOTE. Note that since we have $F_n(t) \rightarrow \Phi(t)$ as $n \rightarrow \infty$, where $\Phi$ is the cumulative distribution function of the standard normal distribution, from the proof of the monotonicity property above we would get the well known inequality $F_n(t) > \Phi(t)$ for all $n$ (which I found always quoted without proof, too).
|
I finally found the proof, by following Henry's suggestion.
For every positive integer $m$, let $f_m(t)$ be the density function of Student's t-distribution with $m$ degrees of freedom, that is
\begin{equation}
f_m(t)=\frac{\Gamma \left(\frac{m+1}{2} \right)}{\sqrt{m \pi} \Gamma \left( \frac{m}{2} \right)} \left( 1 + \frac{t^2}{m} \right)^{-\frac{m+1}{2}} \quad (t \in \mathbb{R}).
\end{equation}
Fix a positive integer $n$ and define the ratio
\begin{equation}
R(t)=\frac{f_{n+1}(t)}{f_{n}(t)} \quad (t \in \mathbb{R}).
\end{equation}
First let us note that
\begin{equation}
\lim_{t \rightarrow - \infty} R(t) =0.
\end{equation}
Moreover, we have
\begin{equation}
R'(t)=R(t) \frac{t(1-t^2)}{n(n+1) \left(1+\frac{t^2}{n} \right) \left(1+\frac{t^2}{n+1} \right)} \quad (t \in \mathbb{R}),
\end{equation}
so that $R'(t) > 0$ for $t < -1$ and $R'(t) <0$ for $-1 < t < 0$.
Now, let us note that since $F_n(0)=F_{n+1}(0)=1/2$, there must exist at least one value $\bar{t} < 0$ such that $R(\bar{t})=1$. Clearly we must have $\bar{t} < -1$, and since $R(t)$ is strictly increasing on $(-\infty,-1]$, there exist only one $\bar{t} \in (-\infty,-1]$ such that $R(\bar{t})=1$. Now note that $R(t)$ is strictly decreasing on $(-1,0]$, so in order to prove that there is no $s \in (-1,0]$ such that $R(s)=1$, it is enough to prove that $R(0) > 1$.
In order to prove that $R(0)>1$, we shall us two remarkable inequalities proved by Wallis in his great work "Arithmetica Infinitorum": for any positive integer $k$, we have
\begin{equation}
\left[\frac{(2k-1)!!}{(2k-2)!!}\right]^2 \frac{1}{2k} \sqrt{\frac{2k+1}{2k}} < \frac{2}{\pi} < \left[\frac{(2k-1)!!}{(2k-2)!!} \right]^2 \frac{1}{2k}\sqrt{\frac{2k}{2k-1}}. \tag{I}
\end{equation}
For a short proof and very elegant proof, see the work by Dutka On Some Gamma Function Inequalities.
Now, assume that $n$ is even. Then we have
\begin{equation}
R(0)=\left(\frac{2 \cdot 4 \cdot 6 \cdot \dots \cdot (n-2)}{3 \cdot 5 \cdot \dots \cdot (n-1)} \right)^2 \frac{2 n \sqrt{n}}{\pi \sqrt{n+1}}.
\end{equation}
The inequality $R(0) > 1$ can be so written:
\begin{equation}
\frac{\pi}{2} < \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \dots \cdot \frac{n-2}{n-3} \cdot \frac{n-2}{n-1} \cdot \frac{n}{n-1} \cdot \sqrt{\frac{n}{n+1}},
\end{equation}
which is the first inequality in (I).
Analogously, if $n$ is odd, we get
\begin{equation}
R(0)=\left(\frac{3 \cdot 5 \cdot 7 \cdot \dots \cdot (n-2)}{2 \cdot 4 \cdot \dots \cdot (n-1)} \right)^2 \frac{\pi n \sqrt{n}}{2 \sqrt{n+1}}.
\end{equation}
The inequality $R(0) > 1$ now reads
\begin{equation}
\frac{\pi}{2} > \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \dots \cdot \frac{n-1}{n-2} \cdot \frac{n-1}{n} \cdot \frac{n+1}{n} \cdot \sqrt{\frac{n}{n+1}},
\end{equation}
which is the second inequality in (I).
So, we have that $R(t) < 1$ for $t < \bar{t}$ and $R(t) > 1$ for $\bar{t} < t < 0$. This result, when combined to the fact that $F_n(0)=F_{n+1}(0)=1/2$, allows us to conclude that $F_{n+1}(t) < F_{n}(t)$ for each $t < 0$.
QED
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$
Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$
My Try: By Cauchy Schwarz Inequality we have
$$\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)\le \sqrt{3}\sqrt{S}$$
$\implies$
$$\sqrt{3S} \ge 12+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
Now by $AM \ge HM $ inequality we have
$$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ $\implies$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{4}$$ hence
$$\sqrt{3S} \ge 12+\frac{3}{4}=\frac{51}{4}$$
hence
$$3S \ge \frac{2601}{16}$$
so $$S \ge \frac{867}{16}$$
is this approach correct and any better approach please share.
|
Just another way: By QM-AM inequality,
$$S=\sum \left(a+\frac1b\right)^2 \geqslant \frac13 \left(\sum a + \sum \frac1b\right)^2 \geqslant \frac13\left(12 + \frac34\right)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2287621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find the locus of $\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3}$ I tried simplifying this first:
$$\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} \Leftrightarrow \\
\frac{\pi}{3}\le \arg(-(z+(3-i)))\le \frac{5\pi}{3} \Leftrightarrow \\
\frac{\pi}{3}\le \arg(z+(3-i))+\pi \le \frac{5\pi}{3} \Leftrightarrow \\
???$$
My book does the following:
$$\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} \Leftrightarrow \\
-\frac{2\pi}{3}\le \arg[z-(3+i)]\le \frac{2\pi}{3}$$
With the expression in this form I can easily find the locus, what i don't understand is how my book does this simplification. Can anyone explain it to me?
|
Here is how I was taught how to do this (I think it might be a slightly longer method):
$$ \frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} $$
Split this inequality into two inequalities:
$$ \frac{\pi}{3} \le \arg(-z+3+i) ~~\text{and}~~ \arg(-z+3+i) \le \frac{5\pi}{3}$$
Let's solve the first inequality first: Let $z=x+iy$
$$\begin{align*}
\frac{\pi}{3} &\le \arg(-z+3+i)\\
\frac{\pi}{3} &\le \arg(-x-iy+3+i) \\
\frac{\pi}{3} &\le \arg((3-x)+i(1-y))\\
\frac{\pi}{3} &\le \arctan \left( \frac{1-y}{3-x}\right)\\
\sqrt{3} &\le \frac{1-y}{3-x} \\
\sqrt{3} (3-x) &\le 1-y \\
3\sqrt{3} - \sqrt{3}x &\le 1-y\\
y &\le \sqrt{3}x + (1-3\sqrt{3})\\
\end{align*}$$
Second inequality:
$$\begin{align*}
\arg(-z+3+i) &\le \frac{5\pi}{3}\\
\arg(-x-iy+3+i) &\le \frac{5\pi}{3} \\
\arg((3-x)+i(1-y)) &\le \frac{5\pi}{3} \\
\arctan \left( \frac{1-y}{3-x} \right) &\le \frac{5\pi}{3}\\
\frac{1-y}{3-x} &\le -\sqrt{3} \\
y &\ge -\sqrt{3}x+(3\sqrt{3}+1) \\
\end{align*}$$
So you have both of those regions with a hole at $x=3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
equation involving absolute value function has $4$ solutions
If the equation $|x^2-5|x|+k|-\lambda x+7 \lambda = 0$ has exactly $4$ solution, Then $(\lambda,k)$ is
Attempt:$x^2-5|x|+k = \lambda x-7 \lambda$
$\star$ For $x>0,$ then $x^2-5x+k=\lambda x- 7 \lambda.$
$\star$ For $x\leq 0,$ then $x^2+5x+k=\lambda x- 7 \lambda.$
could some help me how to go further,thanks
|
No harm in doing it in a case by case basis.
You have
$x^2\pm 5x +k = \pm(\lambda x-7 \lambda)$
That's four equations to solve:
1) $x^2+ 5x +k = \lambda x-7 \lambda\implies x^2 + x(5 - \lambda) + (k+7\lambda) = 0$
2) $x^2- 5x +k = \lambda x-7 \lambda \implies x^2 + x(-5 - \lambda) + (k+7\lambda) = 0$
3) $x^2+ 5x +k = -\lambda x+7 \lambda\implies x^2 + x(5 + \lambda) + (k-7\lambda) = 0$
4) $x^2- 5x +k = -\lambda x+7 \lambda\implies x^2 + x(-5 + \lambda) + (k-7\lambda) = 0$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find $xyz$, given that the value of $x^2+y^2+z^2$, $x+y+z=x^3+y^3+z^3=7$
Given that $$x^2+y^2+z^2=49$$ $$x+y+z=x^3+y^3+z^3=7$$
Find $xyz$.
My attempt,
I've used a old school way to try to solve it, but I guess it doesn't work.
I expanded $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+yz^2)+6xyz$
Since I know substitute the given information into the equation and it becomes $112=x^2y+xy^2+xz^2+yz^2+2xyz$
In another hand, I also expanded $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$7^2=49+2(xy+xz+yz)$,
So from here, I know that $xy+xz+yz=0$.
It seems that I stuck here and don't know how to proceed anymore.
How to continue from my steps? And is there another trick to solve this question? Thanks a lot.
|
If we identify $x,y,z$ as the roots of a cubic polynomial $av^3+bv^2+cv+d$ with $a\neq0$ then there are recursions in terms of $a,b,c,d$ for the power sums
$$P_i:=x^i+y^i+z^i.$$
These are the newton identities. As seen in this link
\begin{align}
P_0=&+3\\
P_1=&-\dfrac{b}{a}\\
P_2=&-\dfrac{b}{a}P_1-\dfrac{c}{a}2\\
P_3=&-\dfrac{b}{a}P_2-\dfrac{c}{a}P_1-\dfrac{d}{a}P_0\\
\end{align}
or
\begin{align}
P_1a&+b&=0\\
P_1b&+P_2a+2c&=0\\
P_0d&+P_1c+P_2b+P_3a&=0
\end{align}
We see that these equations are linear in $a,b,c,d$. Inserting $P_0=3,P_1=7,P_2=49,P_3=7$ and rearranging:
\begin{align} \tag1
7a&&+b&&&&&=0\\\tag2
49a&&+7b&&+2c&&&=0\\\tag3
7a&&+49b&&+7c&&+3d&=0
\end{align}
Multiply $(1)$ by $7$ and subtract the result by $(2)$. The result is $-2c=0$ so $c=0$. Multiply $(2)$ by $7$ and subtract the result by $(3)$. The result is $336a+7c-3d=0$. Substituting $c=0$ and dividing by $3$ yields $112a-d=0$. Finally according to the link $xyz=-\dfrac{d}{a}=-112$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Use integration to find which function has Fourier series $\frac{4}{\pi} \sum\limits^{\infty}_{n = 1, n\text{ odd}} \frac{\sin(nt)}{n^3}$ Problem
It can be shown that the function
$$
f(t) = \begin{cases}
\dfrac{\pi}{2} + t & , -\pi < t < 0,\\
\dfrac{\pi}{2} - t & , 0 < t < \pi,
\end{cases}
$$
has Fourier series $FS_f(t) = \dfrac{4}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\cos(nt)}{n^2}$.
Use integration to find which function has Fourier series $\dfrac{4}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\sin(nt)}{n^3}$.
I understand that term-by-term integration of a $2\pi$-periodic Fourier series is the formula $\int^t_{-\pi} f(\alpha) \ d\alpha = \sum_{n = 1}^{\infty} \dfrac{a_n}{n}\sin(nt) - \sum_{n = 1}^{\infty} \dfrac{b_n}{n}(\cos(nt) - \cos(n\pi))$. However, I do not understand how to solve this problem.
I would greatly appreciate it if people could please take the time to explain the reasoning involved in solving this problem.
|
Using the property that the function evaluated at $t=0$ is the average of the left and right piece-wise function evaluated at zero and
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 \, n + 1)^3} = \frac{\pi^3}{32}$$
then the following is obtained:
Integrating
$$
f_{c}(t) = \begin{cases}
\frac{\pi}{2} + t & , -\pi < t < 0,\\
\frac{\pi}{2} - t & , 0 < t < \pi,
\end{cases}
$$
becomes
$$
f_{s}(t) = \begin{cases}
c_{0} + \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
c_{1} + \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
$$
Now, using $f_{s}(0) = (f_{s}(0-) + f_{s}(0+))/2$, then $c_{1} = - c_{0}$ and
$$
f_{s}(t) = \begin{cases}
c_{0} + \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
-c_{0} + \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
$$
In order to determine $c_{0}$ evaluate the series at the point $t = \pi/2$ leading to
$$f_{s}\left(\frac{\pi}{2}\right) = - c_{0} + \frac{\pi^2}{4} - \frac{\pi^2}{8} = - c_{0} + \frac{\pi^2}{8} $$
From the series it is determined that
$$f_{s}(t) = \frac{4}{\pi} \, \sum_{n=0}^{\infty} \frac{\sin(2\,n +1)t}{(2\, n +1)^3}$$
and
$$f_{s}\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}.$$
This yields $c_{0}=0$ and
$$
f_{s}(t) = \begin{cases}
\frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
\frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
$$
One can verify this by using
\begin{align}
f_{s}(t) &= \sum_{n=0}^{\infty} B_{n} \, \sin((2\, n +1) \, t) \\
B_{n} &= \frac{1}{\pi} \, \int_{- \pi}^{\pi} g(t) \, \sin((2 \, n +1) t) \, dt \\
g(t) &= \begin{cases}
\frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
\frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
\end{align}
|
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|
how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f,g,h,i\in \{0,1,2\}$
$$M^{T}M= \begin{pmatrix}
a & d & g\\
b & e & h\\
c & f & i
\end{pmatrix}\begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix} $$.
sum of diagonal entries $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2 = 5$$
How can I form different cases?
|
Positioning the $0$'s, we get $\dbinom94=126$ for the $'000011111'$ case.
For the $'000000012'$ case, we have $\dbinom97=36$ ways to position the $0$'s, we need to multiply this by $2!$ to account for the two cases $12, 21$.
Total: $126+72=198$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given that $a\cos\theta+b\sin\theta+A\cos 2\theta+B\sin 2\theta \leq 1$ for all $\theta$, prove that $a^2+b^2\leq 2$ and $A^2+B^2 \leq 1$ I tried to solve this question but things got too much complicated and hence my efforts were completely futile.
Let $a,A,b,B \in \mathbb{R}$ and $$F(\theta)= 1- a\cos \theta - b\sin \theta- A\cos 2\theta- B\sin 2\theta$$
It is given that $$F(\theta) \ge 0 \;\forall\; \theta $$
and we have to prove that $\color{red}{a^2+b^2 \le 2}$ and $\color{green}{A^2+B^2 \le 1}$.
MY ATTEMPT
We need to prove that $$a\cos \theta + b\sin \theta+A\cos 2\theta+ B\sin 2\theta \le 1$$
$$\begin{align}
& = a\cos \theta + b\sin \theta+A( \cos^2 \theta- \sin^2 \theta)+ B \sin \theta \cdot \cos \theta + B \sin \theta \cdot \cos \theta \le 1 \\
& =\ cos \theta (a+A \cos \theta+ B \sin \theta)+\sin \theta(b-A \sin \theta+B \ \cos \theta) \le1 \\
\end{align}$$
We know that $-\sqrt{x^2 + y^2} \le x \cos \theta + y \sin \theta \le \sqrt{x^2 + y^2}$
$$\Rightarrow (a+A \cos \theta+ B \sin \theta)^2 + (b-A \sin \theta+B \ \cos \theta)^2 \le 1$$
After solving this equation we get
$$a^2 + b^2 + 2(A^2 +B^2)+ \cos \theta (2aA+2bB) + \sin \theta (2aB-2bA) \le1$$
Now If I again apply the same property, certainly the things are going to become more complicated and hence I think my approach is not at all right. Kindly Help me with this question.
|
Here is another (?) approach: define $c=a-bi$, $d=A-Bi$. Then we are given that
$$
\operatorname{Re}(cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{1}
$$
Replacing $z$ with $-z$ and with $iz$ in (1) we get two more inequalities
\begin{eqnarray}
z\mapsto -z:\quad\operatorname{Re}(-cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1,\tag{2}\\
z\mapsto iz:\quad\ \operatorname{Re}(icz-dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{3}
\end{eqnarray}
Adding (1) and (2) gives
$$
\operatorname{Re}(dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1\quad\Leftrightarrow\quad |d|\le 1.
$$
Similarly adding (1) and (3) gives
$$
\operatorname{Re}((1+i)cz)\le 2,\quad\forall z\in\Bbb C\colon |z|=1\quad\Leftrightarrow\quad |(1+i)c|\le 2.
$$
|
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"url": "https://math.stackexchange.com/questions/2291584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
|
If you suspect there exists a factorization over $\mathbb{Q}[x]$ into polynomials of degree 2 and 3, but you just don't know their coefficients, one way is to write it down as
$x^5 + x + 1 = (x^2 + ax + b)(x^3 + cx^2 + dx + e)$ where the coefficients are integers (by Gauss' lemma). And then expand and solve the system.
Then $a + c = 0, b + ac + d = 0, bc + ad + e = 0, ae + bd = 1, be = 1$.
So we get $c = -a$ and $b = e = 1$ or $b = e = -1$.
In the first case we reduce to $1 - a^2 + d = 0, -a + ad + 1 = 0, a + d = 1$ which gives $d = 1 - a, 1 - a^2 + 1 - a = 0, 1 - a^2 = 0$ so $a = 1, b = 1, c = -1, d = 0, e = 1$.
Substituting gives us $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$
If the factorization was not over $\mathbb{Q}[x]$ then things would get more complicated because I could not assume $b = e = +/- 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Manipulating a Summation Series
Summation: How does$$\begin{align*} & \frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\tag1\\ & =\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k\tag2\\ & =\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k=\sum\limits_{k=1}^n\frac 1{k+n}\tag3\end{align*}$$
Given that$$\sum\limits_{k=1}^n\frac 1{(2k)^3-2k}=\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k$$
I'm not sure how they got from the first step to the second. The $\tfrac 12\sum\limits_{k=1}^n\tfrac 1k$ didn't change, so that must mean that$$\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac n{2n+1}=\sum\limits_{k=1}^n\frac 1{2k-1}$$But I don't see how. I tried expanding the LHS, and rearranging to get the RHS, but that didn't go far.$$\begin{align*}\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}\right)+\frac n{2n+1} & =\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 1{2k+1}\right)+\frac n{2n+1}\\ & =\frac 12\sum\limits_{k=1}^n\frac {4k}{4k^2-1}+\frac n{2n+1}\\ & =2\sum\limits_{k=1}^n\frac k{(2k)^2-1}+\frac n{2n+1}\end{align*}$$I feel like I'm close, but I just can't seem to finish it. I believe that you'll need to use a summation identity.
Questions:
*
*How do you get from step $(1)$ to step $(2)$?
*How do you get from $(2)$ to $(3)$
*Is there a PDF that lists all the summation rules and identities?
|
\begin{align}
&\sum\limits_{k=1}^n\frac 1{2k+1}\\
= & \ \sum\limits_{k=1}^{n-1}\frac 1{2k+1} + \frac{1}{2n+1} \\
= & \ 1 + \sum\limits_{k=1}^{n-1}\frac 1{2k+1} + \frac{1}{2n+1} -1\\
= & \ \sum\limits_{k=0}^{n-1}\frac 1{2k+1} + \frac{1 -2n-1}{2n+1}\\
= & \ \sum\limits_{l=1}^{n}\frac 1{2l-1} - \frac{2n}{n+1} \tag{where $l = k+1$}\\
= & \ \sum\limits_{k=1}^{n}\frac 1{2k-1} - \frac{2n}{n+1} \tag{$\ast$}
\end{align}
Therefore,
\begin{align}
&\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\\
= & \ \frac 12\sum\limits_{k=1}^n\frac 1{2k-1} + \frac{1}{2}\left( \sum\limits_{k=1}^{n}\frac 1{2k-1} - \frac{2n}{2n+1} \right) -\frac 12\sum\limits_{k=1}^n\frac 1k + \frac n{2n+1}\tag{using $(\ast)$}\\
= & \ \sum\limits_{k=1}^n\frac 1{2k-1} -\frac 12\sum\limits_{k=1}^n\frac 1k
\end{align}
This gives $(1)$ to $(2)$.
Now,
\begin{align}
&\sum_{k=1}^{n}\frac{1}{2k-1} \\
= & \ \sum_{k=1}^n\frac{1}{2k-1} + \sum_{k=1}^n\frac{1}{2k} - \sum_{k=1}^n\frac{1}{2k}\\
= & \ \sum_{k=1}^{2n} \frac{1}{k}- \frac{1}{2}\sum_{k=1}^n\frac{1}{k}\\
\end{align}
And then,
\begin{align}
&\sum\limits_{k=1}^n\frac 1{2k-1} -\frac 12\sum\limits_{k=1}^n\frac 1k \\
= & \ \sum_{k=1}^{2n} \frac{1}{k}- \frac{1}{2}\sum_{k=1}^n\frac{1}{k} -\frac 12\sum\limits_{k=1}^n\frac 1k\\
= & \ \sum_{k=1}^{2n} \frac{1}{k}- \sum_{k=1}^n\frac{1}{k} \\
= & \ \sum_{k=n+1}^{2n}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k}
\end{align}
This gives $(2)$ to $(3)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solutions of $\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$
:Solutions of:
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$$
My try :
$$\lfloor 2x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac12 \rfloor $$
$$\lfloor 3x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac13 \rfloor +\lfloor x+\dfrac23 \rfloor $$
$$\lfloor 4x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor $$
Now :
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0 $$
$$(\lfloor x \rfloor)^2+(\lfloor x \rfloor)(\lfloor x+\dfrac13 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac23 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac12 \rfloor)+(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac13 \rfloor)\\(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac23 \rfloor) +\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor=0$$
Now what ?
|
HINT:
If $0<b < 1$ then $\lfloor x + b \rfloor = \lfloor x \rfloor$ if $x-\lfloor
x\rfloor + b < 1$ or $\lfloor x \rfloor + 1$ otherwise
you might have to look at the different cases that arise
|
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|
Is the $n^{th}$ derivative of $\sin(x)$ just a translation of $\sin(x)$? I noticed that
$$\frac d{dx}\sin x=\cos x=\sin\left(x+\frac\pi2\right)$$
$$\frac{d^n}{dx^n}\sin x=\sin\left(x+\frac{\pi n}2\right)$$
Does this hold for any positive real value of $n$?
If so, does anybody have any reasoning behind why it's just a simple translation?
|
Indeed, we know that
$$
\frac{d}{dx} \sin{x} = \cos{x} = \sin{\left( x + \frac{\pi}{2} \right) }
$$
and then
$$
\frac{d^2}{dx^2} \sin{x} = \cos{\left(x + \frac{\pi}{2} \right)}= \sin{\left(x + \frac{2\pi}{2} \right)}
$$
Proof by induction:
Let's assume that
$$
\frac{d^k}{dx^k} \sin{x} = \sin{\left( x+ \frac{k\pi}{2} \right)}
$$
for some integer $k$. We have already seen that this is true for $k=0,1,2$.
Then we take $k$ to $k+1$:
$$
\frac{d^{k+1}}{dx^{k+1}} \sin{\left(x \right)}= \frac{d}{dx} \sin{\left( x+ \frac{k\pi}{2} \right)} = \cos{\left(x + \frac{k\pi}{2} \right)}
$$
$$
= \sin{\left(x + \frac{\pi}{2} + \frac{k\pi}{2} \right)}= \sin{\left( x + \frac{(k+1)\pi}{2}\right)}
$$
which shows that the assumption is correct for whole numbers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability for 5 dice (high d6 result +1 per additional die 4 or higher) I think I am calculating these odds properly. Hopefully someone can confirm that I am, or tell me where I am going wrong if I'm not.
When rolling 5 regular six-sided dice, the total result I am looking for is the value of the highest die +1 for every additional die that comes up 4+.
I have a 59.81% chance of one of the 5 dice coming up a 6. Of the remaining 4 dice there is a 93.75% chance that one of them is 4+, so $59.81\% \times 93.75\%$ gives me a 55.89% chance of getting a 7, or 6+1.
Of the remaining 3 dice there is a 87.5% chance of getting a 4+. So the odds of getting a 6 on 1 of 5 dice (0.5981) times the odds of getting a 4+ on 4 dice (0.9375) times the odds of getting a 4+ on 3 dice (0.875) gives me a 49.06% chance of getting 6+2.
Is this correct?
|
You get $6 + 1$ if you throw one six, one $4$ or $5$ and three dice lower than $4$, or if you throw two sixes and three dice lower than $4$. The probability of the former equals:
$$\frac{{5 \choose 1}1^4{4 \choose 1}2^1{3 \choose 3}3^3}{6^5} = \frac{1080}{7776}$$
The probability of the latter equals:
$$\frac{{5 \choose 2}1^2{3 \choose 3}3^3}{6^5} = \frac{270}{7776}$$
As such, the probability of getting $6 + 1$ equals:
$$\frac{1080+270}{7776} = \frac{1350}{7776} \approx 0.174$$
A similar approach can be used for $5 + 2$, resulting in a probability of:
$$\frac{{5 \choose 1}1^1{4 \choose 2}1^2{2 \choose 2}3^2 + {5 \choose 2}1^2{3 \choose 1}1^1{2 \choose 2}3^2 + {5 \choose 3}1^3{2 \choose 2}3^2}{6^5} = \frac{270 + 270 + 90}{7776} = \frac{630}{7776}$$
Again, the same approach can be used for $4 + 3$, resulting in a probability of:
$$\frac{{5 \choose 4}1^4{1 \choose 1}3^1}{6^5} = \frac{15}{7776}$$
As such, the probability of getting $7$ equals:
$$\frac{1350 + 630 + 15}{7776} \approx 0.257$$
|
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|
Circle equation - diametric form - polar coordinates. A line segment joining $(a,\alpha)$, $(b,\beta)$ in polar coordinates is the diameter of a circle. I want to find the equation of this circle. It can be done by converting into the Cartesian system but I want to find the equation without moving out of polar.
|
General equation of circle:
$$r^2-2rr_0\cos (\theta-\phi)+r_0^2=R^2$$
Now
\begin{align}
(r_0 \cos \phi,r_0 \sin \phi) &=
\left(
\frac{a\cos \alpha+b\cos \beta}{2},
\frac{a\sin \alpha+b\sin \beta}{2}
\right) \\
r_0^2 &=
\left( \frac{a\cos \alpha+b\cos \beta}{2} \right)^2+
\left( \frac{a\sin \alpha+b\sin \beta}{2} \right)^2 \\
R^2 &=
\left( \frac{a\cos \alpha-b\cos \beta}{2} \right)^2+
\left( \frac{a\sin \alpha-b\sin \beta}{2} \right)^2
\end{align}
On simplifying,
\begin{align}
r^2-r[a\cos (\theta-\alpha)+b\cos (\theta-\beta)]+
ab\cos (\alpha-\beta) &=0 \\[5pt]
[r-a\cos (\theta-\alpha)][r-b\cos (\theta-\beta)]+
ab\sin (\theta-\alpha) \sin (\theta-\beta) &= 0
\end{align}
|
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|
Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing
Since there are 20 presents to be distributed among 4 children,
$$C_1 + C_2 + C_3 + C_4 = 20$$
By bars and stars,
$$\binom{ 20 + 4-1 }{ 4-1 } = \binom{ 23 }{ 3 } = 1771$$
$C_2 + C_3 + C_4 = 14 $ (after giving 6 presents to $C_1$)
Assuming $C_1$ gets 6 present = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_2$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_3$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_4$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
No one gets exactly 6 presents = Total - Everyone gets 6 presents
= $\binom{ 23 }{ 3 } - \binom{ 4 }{ 1 }\binom{ 16 }{ 2 } = 1771 - 480 = 1291$
I am new to combinatorics and not sure if I'm on the right path any help would be appreciated.
|
The presents are different so the stars and bars formula doesn't work here. For example, the number of ways to distribute $20$ different presents to $4$ children is $4^{20}$ since each present can go to $4$ possible children (think of stamping the name of the child on each present).
Let $U$ be the set of all distributions of the $20$ different presents without constraints and let $A_i$ be the set of distributions in which child $i$ has exactly 6 presents. Then by PIE, the desired quantity equals
$$
|\bar{A}_1\cap\bar{ A}_2\cap \bar{A}_3\cap \bar{A}_4|=|U|-S_1+S_2-S_3+S_4\tag{1}
$$
where
$$
S_k=\sum_{1\leq i_1<\dotsb<i_k\leq 4}|A_{i_1}\cap\dotsb A_{i_k}|.\tag{2}
$$
From the above discussion $|U|=4^{20}$ and clearly, $$S_4=|A_1\cap\dotsb \cap A_4|=0.\tag{3}$$
To find $S_1$ note that $|A_i|=\binom{20}{6}3^{14}$ for all $i$ since we choose which six of the $20$ presents child $i$ gets and we distribute the remaining $14$ presents among the remaining $3$ children. Hence
$$
S_1=\binom{4}{1}\times\binom{20}{6}3^{14}\tag{4}.
$$
To find $S_2$ note that for $i\neq j$, we have that $|A_i\cap A_j|=\binom{20}{6}\binom{14}{6}2^{8}$ since we distribute $6$ presents to each of $i$ and $j$ and then the remaining $8$ are distributed without constraints. Hence
$$
S_2=\binom{4}{2}\times\binom{20}{6}\binom{14}{6}2^{8}.\tag{5}
$$
Finally, using similar reasoning as above
$$
S_3=\binom{4}{3}\times\binom{20}{6}\binom{14}{6}\binom{8}{6}1^{2}.\tag{6}
$$
Thus, by (1)
$$
\text{number of ways}=4^{20}-\binom{4}{1}\binom{20}{6}3^{14}+\binom{4}{2}\binom{20}{6}\binom{14}{6}2^{8}-\binom{4}{3}\binom{20}{6}\binom{14}{6}\binom{8}{6}1^{2}+0.
$$
|
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|
Mass Point Geometry Question In the triangle $ABC$, let $E$ be a point on $BC$ such that $BE : EC =
3: 2$. Pick points $D$ and $F$ on the sides $AB$ and $AC$ , correspondingly, so that
$3AD = 2AF$ . Let $G$ be the point of intersection of $AE$ and $DF$ . Given that $AB = 7$
and $AC = 9$, find the ratio $DG: GF$.
I have been working on trying to solve this problem. I am having difficulty relating the length of $AB =7$ and the ratios given to find $AD:BD$. Similarly I am having trouble finding ratio of $FC:AF$. I am sure that I can solve this problem if someone can give me a hint on how to find those ratios. Any help would be appreciated.
|
Let $D'\in AB$ be such that $CD'\parallel FD$ and let $G'=AE\cap CD'$.
By similarity, $\frac{DG}{GF}=\frac{DG'}{G'C}$. Let $H=BG'\cap AC$. By Ceva's theorem
$$ 1=\frac{BE}{EC}\cdot \frac{CH}{HA}\cdot \frac{AD'}{D'B} = \frac{3}{2}\cdot\frac{CH}{HA}\cdot\frac{\frac{2}{3}b}{c-\frac{2}{3}b} $$
hence $\frac{CH}{HA}=\frac{c-\frac{2}{3}b}{b}$. By Van Obel's theorem
$$ \frac{CG'}{G'D}=\frac{CE}{EB}+\frac{CH}{HA} = \frac{2}{3}+\frac{c}{b}-\frac{2}{3}=\frac{c}{b} $$
hence if $c=AB=7$ and $b=AC=9$ it follows that $\frac{DG}{GF}=\color{red}{\large \frac{9}{7}}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, taking into account the time zone). Here's one question that I've been stuck with:
With $a, b, c$ $\in \mathbb{R}^{+}$ such that $ab + bc + ca + abc =2$, find the max of
$ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$
I'm just curious what kind(s) of techniques these junior high students can use, since as far as I remember when I was at that age, calculus and derivatives aren't taught until high school.
|
The functional form of $M$ suggests us to change variable to $u = a+1, v = b+1, w = c + 1$.
In terms of them, the expression of $M$ simplifies to
$$M = \frac{u}{1+u^2} + \frac{v}{1+v^2} + \frac{w}{1+w^2}$$
The constraint $abc + ab + bc + ca = 2$ can be rewritten as
$$(a+1)(b+1)(c+1) - (a+b+c) - 1 = 2\quad\iff\quad u + v + w - uvw = 0$$
Since $a,b,c > 0$, $u,v,w \in (1,\infty)$. Pick $A, B, C \in (\frac{\pi}{4},\frac{\pi}{2})$ such that $u = \tan A$,$v = \tan B$ and $w = \tan C$.
The constraint tell us
$$\tan(A+B+C) = \frac{u+v+w - uvw}{1 - uv - vw - wu} = 0
\quad\implies\quad A+B+C = k\pi, \quad k \in \mathbb{Z}
$$
Since $A + B + C \in (\frac{3\pi}{4}, \frac{3\pi}{2})$, $k$ equals to $1$
and $A + B + C = \pi$.
In terms of $A,B,C$, the expression we want to maximize becomes
$$M = \frac{\tan A}{1+\tan^2 A} + \frac{\tan B}{1+\tan^2 B} + \frac{\tan C}{1+\tan^2 C} = \frac12 \left[ \sin(2A) + \sin(2B) + \sin(2C) \right]$$
This is nothing but the area of triangle inscribed inside unit circle with
angles $2A$, $2B$ and $2C$ subtended at center. It is well known for such triangles, the one which maximize the area is an equilateral triangle. This implies $M$ is maximized when
$$A = B = C = \frac{\pi}{3} \implies a = b = c = \tan\frac{\pi}{3} - 1 = \sqrt{3} - 1$$
and the maximum value of $M$ is $\displaystyle\;\frac{3}{2}\sin\frac{2\pi}{3} = \frac{3\sqrt{3}}{4}$.
|
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|
Infinite zeros in infinite series The problem:
Given that
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$
My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} -\frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
\frac{\pi}{12} & = \frac{1}{3} - \frac{1}{9} + \frac{1}{15} - \frac{1}{21} + \frac{1}{27} -\frac{1}{33} + \frac{1}{39} - \frac{1}{45} + \ldots\\
\\
& = 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + 0 + 0 - \frac{1}{21}
\end{align}
$$
now add them together:
$$
\begin{align}
\frac{\pi}{4} + \frac{\pi}{12} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
& + 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + \ldots \\
\end{align}
$$
and we will get:
$$
\begin{align}
\frac{\pi}{3} & = 1 + 0 + \frac{1}{5} - \frac{1}{7} + 0 -\frac{1}{11} + \frac{1}{13} + 0 + \ldots \\
& = 1 + \frac{1}{5} - \frac{1}{7} -\frac{1}{11} + \frac{1}{13} + \ldots
\end{align}
$$
My questions:
*
*I inserted/removed infinite zeros into/from the series, is that OK?
*My solution relies on the fact that $\Sigma a_n + \Sigma b_n = \Sigma (a_n + b_n)$ and $k \Sigma a_n = \Sigma k a_n$. Is this always true for convergent infinite series? If so, why is it? (yeah I know this is a stupid question, but since I'm adding infinite terms up, I'd better pay some attention.)
*Bouns question: Can I arbitrarily (arbitrariness isn't infinity, you know) insert/remove zeros into/from a convergent infinite series, without changing its convergence value?
|
Both the given identities follow from the fact that $\frac{\pi}{\text{something}}$ is related with the integral over $(0,1)$ of a rational function. The first identity is a consequence of
$$\frac{\pi}{4} = \arctan(1)=\int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{1}\left(1-x^2+x^4-x^6+\ldots\right)\,dx $$
and for the second series we may perform the same manipulation in the opposite direction, leading to:
$$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)&=&\int_{0}^{1}(1+x^4-x^6-x^{10})\sum_{n\geq 0}x^{12n}\,dx\\&=&\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx\\&=&\int_{0}^{1}\left(\frac{1}{1+x^2}+\frac{x^2}{1+x^6}\right)\,dx\\(x\mapsto z^{1/3})\qquad &=&\frac{\pi}{4}+\frac{1}{3}\int_{0}^{1}\frac{dz}{1+z^2}=\color{red}{\frac{\pi}{3}}.\end{eqnarray*} $$
Indeed, we are just multipling the first series by $\frac{4}{3}$ :D
|
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|
Polynomial remainder problem. A polynomial $f(x)$ is such that upon division by $(x-3)$ it leaves a remainder of $15$ and upon division by $(x-1)^2$, it leaves a remainder of $2x+1$.
What is the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$?
|
Hint. We have that
$$f(x)=A(x)(x-3)+15=B(x)(x-1)^2+2x+1=C(x)(x-3)(x-1)^2+r(x)$$
where $r(x)=ax^2+bx+c$ (because the degree of $(x-3)(x-1)^2$ is $3$) and $A,B,C$ are polynomials.
In order to find $a$, $b$, and $c$, let $x=1$ and $x=3$.
Consider also the derivative of $f$ at $x=1$.
Therefore we obtain the equations:
$$r(3)=15,\quad r(1)=3,\quad r'(1)=2.$$
Are you able to find $a,b,c$ now?
|
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|
How do you integrate $\frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)}$?
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)} \,dx$$
I tried by dividing the terms in both the numerator and denominator by $\cos^5x$ but still cant find my way.
|
The problem was changed and my following work is not necessary.
Let $\tan{x}=t$.
Hence, $dt=\frac{1}{\cos^2x}dx=(1+t^2)dx$.
Thus,
$$\int\frac{\cos^5x}{\sin^5x+\cos^5x}dx=\int\frac{1}{1+\tan^5x}dx=\int\frac{1}{(1+t^2)(1+t^5)}dt.$$
Let $t+\frac{1}{t}=u$.
Hence, $$1+t^5=(1+t)(1-t+t^2-t^3+t^4)=t^2(1+t)\left(t^2+\frac{1}{t^2}-t-\frac{1}{t}+1\right)=$$
$$=t^2(1+t)(u^2-2-u-1)=t^2(1+t)(u^2-u-1)=t^2(1+t)\left(u-\frac{1-\sqrt5}{2}\right)\left(u-\frac{1+\sqrt5}{2}\right)=$$
$$=(1+t)\left(t^2-\frac{1-\sqrt5}{2}t+1\right)\left(t^2-\frac{1+\sqrt5}{2}t+1\right).$$
The rest is smooth:
Let $$\frac{1}{(1+t)(1+t^2)(\left(t^2-\frac{1-\sqrt5}{2}t+1\right)\left(t^2-\frac{1+\sqrt5}{2}t+1\right)}=$$
$$=\frac{A}{1+t}+\frac{Bt+C}{1+t^2}+\frac{Dt+E}{t^2-\frac{1-\sqrt5}{2}t+1}+\frac{Ft+G}{t^2-\frac{1+\sqrt5}{2}t+1}$$
and solve the system.
|
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|
Solve $\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$ for $x$ I have been scratching my head for solving this equation but I am unable to do this. Even I am unable to get how to use the hint. A way to solve this would be of great help
Solve for $x$ :-
$$\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$$
(Hint: $-3-4=-7$, if $\frac 1a+\frac 1b +\frac 1c \ne 0$)
|
\begin{align}
\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} &= -7\\
\frac{a+b-x}c+\frac{c}{c} + \frac{a+c-x}b+\frac{b}{b} + \frac{b+c-x}a+\frac{a}{a} -\frac{4abc}{a+b+c} &= -7+3\\
\frac{a+b+c-x}c + \frac{a+b+c-x}b + \frac{a+b+c-x}a -\frac{4abc}{a+b+c} &= -4\\
(a+b+c-x)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&=\frac{4abc}{a+b+c}-4\\
a+b+c-x&=\frac{4abc(abc-a-b-c)}{(a+b+c)(ab+bc+ca)}
\end{align}
So, $\displaystyle x=a+b+c-\frac{4abc(abc-a-b-c)}{(a+b+c)(ab+bc+ca)}$.
|
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|
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.
LHS $= 2\cos^{2}3x-1$
$=2{(4\cos^{3}x-3\cos{x})}^2-1$
$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$
$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$
Still thank you for the answers!
|
Simpler way!
Use the fact $\displaystyle (\cos(x)+i\sin(x))^6 = \cos(6x)+i\sin(6x)$.
Expand, use $\sin^2(x)=1-\cos^2(x)$ to simplify higher powers of sine in the real terms, and then organizing by real terms, we get
$\cos(6x)+i\sin(6x)=$
$=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)+32i\sin(x)\cos(x)^5-32i\sin(x)\cos(x)^3+6i\sin(x)\cos(x)-1$
Consider the real part.
This gives you $\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$.
If you consider the imaginary part, you can isolate $i\sin(6x)$.
This also gives you $\sin(6x)=2\sin(x)\cos(x)^5-32\sin(x)\cos(x)^3+6\sin(x)\cos(x)$
|
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|
$u$-substitution always evaluates to $0$ Consider the integral
$$ \int_a^b f(x) \,dx $$
now make the $u$-substitution $u \mapsto c + (x-a)(x-b)$. The resulting integral is
$$ \int_c^c h(u) \,du $$
where $h(u)$ is the integrand $f$ after the substitution, however, regardless of $f$ the integral $\int_c^c du = 0$. Looking at the definition Wikipedia provides I believe the substitution meets every condition. It's differentiable and has a integrable derivative, because it's a polynomial. So what's wrong with this substitution such that it always results in $0$?
|
Assume $a<b$.
If $u=c+(x-a)(x-b)$. Then
\begin{align}
u&=x^2-(a+b)x+ab+c\\
&=\left(x-\frac{a+b}{2}\right)^2+ab-\left(\frac{a+b}{2}\right)^2+c\\
&=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2+c\\
x&=\frac{a+b}{2}\pm\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}
\end{align}
When $\displaystyle x<\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and
$$\frac{du}{dx}=2x-(a+b)=-\sqrt{(a-b)^2-4c-4u}$$
When $\displaystyle x\ge\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and
$$\frac{du}{dx}=2x-(a+b)=\sqrt{(a-b)^2-4c-4u}$$
When $\displaystyle x=\frac{a+b}{2}$, $\displaystyle u=c-\left(\frac{a-b}{2}\right)^2$
Therefore,
\begin{align}
\int_a^bf(x)dx&=\int_a^{\frac{a+b}{2}}f(x)dx+\int_{\frac{a+b}{2}}^bf(x)dx\\
&=\int_0^{c-\left(\frac{a-b}{2}\right)^2}\left[\frac{f\left(\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{-\sqrt{(a-b)^2-4c-4u}}\right]du\\
&\qquad +\int_{c-\left(\frac{a-b}{2}\right)^2}^0\left[\frac{f\left(\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{\sqrt{(a-b)^2-4c-4u}}\right]du
\end{align}
It looks quite complicated. But my point is that the definite integral should be broken into two parts with different integrands.
|
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|
Why multiply a matrix with its transpose? This might be a very stupid question, but I do not seem to understand why I would multiple a matrix with its transpose. I am not a mathematician, but I am very interested in understanding the practical usage of equations:
Imagine I have three products sales Apple, Orange and Pear for the last 3 days in a matrix form called A:
$$ A=
\begin{bmatrix}
Apple & Orange & Pear \\
10 & 2 & 5 \\
5 & 3 & 10 \\
4 & 3 & 2 \\
5 & 10 & 5 \\
\end{bmatrix}$$
What will $AA^{\rm T}$ tell me?
I have seen this long answer link: Is a matrix multiplied with its transpose something special?, but I did not get it at all.
I see that a lot of equations use the product $AA^{\rm T}$ and I really hope that someone will give a very simple answer.
|
It can be a part of a bigger question (context needed, for example, linear regression as mentioned by lhf in the linked post).
Assume that the vector (Q) of quantity apple sales and the vector (R) of revenues on four days are:
$$Q=\begin{pmatrix}10\\ 5\\ 4 \\ 8\end{pmatrix}; \ \ R=\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pmatrix}$$
And now we want to find the linear revenue function $R=aQ+b$. Obviously, it is $R=2Q$:
$$R=\begin{pmatrix}R_1\\ R_2\\ R_3\\ R_4\end{pmatrix}=2\begin{pmatrix}10\\ 5\\ 4\\ 8\end{pmatrix}=\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pmatrix}$$
Keeping in mind that the predicted function is not always perfectly linear fit and for demonstration purpose, assume that we want to find the linear revenue function $y=b_0+b_1x$ using linear regression, where $y=R,x=Q$ and $b_0,b_1$ are the parameters to be found. Then the linear function can be written in the matrix form as:
$$Y=\begin{pmatrix}y_1\\ y_2\\ y_3\\ y_4\end{pmatrix}=\begin{pmatrix}1&x_1\\ 1&x_2\\ 1&x_3\\ 1&x_4\end{pmatrix}\begin{pmatrix}b_0\\ b_1\end{pmatrix}=Xb$$
Now we need to solve the matrix equation:
$$Y=Xb \Rightarrow \\
X^TY=X^TXb \Rightarrow \\
b=(X^TX)^{-1}X^TY=\\
\left[\begin{pmatrix}1&1&1&1\\x_1&x_2&x_3&x_4\end{pmatrix}\begin{pmatrix}1&x_1\\ 1&x_2\\ 1&x_3\\ 1&x_4\end{pmatrix}\right]^{-1}\begin{pmatrix}1&1&1&1\\x_1&x_2&x_3&x_4\end{pmatrix}\begin{pmatrix}y_1\\ y_2\\ y_3\\ y_4\end{pmatrix}=\\
\left[\begin{pmatrix}1&1&1&1\\10&5&4&8\end{pmatrix}\begin{pmatrix}1&10\\ 1&5\\ 1&4\\ 1&8\end{pmatrix}\right]^{-1}\begin{pmatrix}1&1&1&1\\10&5&4&8\end{pmatrix}\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pmatrix}=\\
\left[\begin{pmatrix}4&27\\27&205\end{pmatrix}\right]^{-1}\begin{pmatrix}54\\410\end{pmatrix}=\\
\frac{1}{91}\begin{pmatrix}205&-27\\ -27&4\end{pmatrix}\begin{pmatrix}54\\410\end{pmatrix}=\begin{pmatrix}0\\2\end{pmatrix}=\begin{pmatrix}b_0\\b_1\end{pmatrix}$$
as expected: $y=0+2x \iff R=2Q$.
|
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|
Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$ The binomial expression given as follows:
$$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$
results nicely into the following closed form:
$$(-1)^{n}\binom{2n}{n}$$
I wish to know how exactly is it being done? I haven't been able to make much progress in solving it.
My approach:
\begin{align}
\sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{2} = \binom{2n}{0}^{2} - \binom{2n}{1}^{2} + \binom{2n}{2}^{2} - ... -\binom{2n}{2n-1}^{2} + \binom{2n}{2n}^{2} \\
= \binom{2n}{0}.\binom{2n}{0} - \binom{2n}{1}.\binom{2n}{1} + \binom{2n}{2}.\binom{2n}{2} - ... -\binom{2n}{2n-1}.\binom{2n}{2n-1} + \binom{2n}{2n}.\binom{2n}{2n} \\
\text{By Symmetry of binomial coefficients} \\
= \binom{2n}{2n}.\binom{2n}{0} - \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} - ... -\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} \\
= \binom{2n}{2n}.\binom{2n}{0} + \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} + ... +\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} - 2.\left(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1}\right) \\
\text{By Vandermond's identity, first component, i.e.not enclosed within -2.(...) evaluates to C(4n, 2n)} \\
\binom{4n}{2n} - 2.(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1})
\end{align}
I'm lost beyond this point. It will be extremely helpful if someone can direct me in the right direction or provide the answer to this perplexing and challenging problem. Thank you.
|
I will use the notation $[x^k]\,f(x)$ for denoting the coefficient of $x^k$ in the Taylor/Laurent expansion of $f(x)$ around the origin. We have:
$$ S(n)=\sum_{k=0}^{2n}(-1)^k \binom{2n}{k}^2 = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}\binom{2n}{2n-k}=\sum_{\substack{a,b\geq 0 \\ a+b=2n}}(-1)^a\binom{2n}{a}\binom{2n}{b} $$
and since
$$ \sum_{c\geq 0}(-1)^c \binom{2n}{c} x^c = (1-x)^{2n}, \qquad \sum_{d\geq 0}\binom{2n}{d} x^d = (1+x)^{2n} $$
it follows that:
$$ S(n) = [x^{2n}] (1-x)^{2n}(1+x)^{2n} = [x^{2n}](1-x^2)^{2n} \stackrel{x^2\mapsto z}{=} [z^n](1-z)^{2n}$$
so $S(n) =\color{red}{ (-1)^n \binom{2n}{n}}$ just follows from the binomial theorem.
|
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|
How to find a degree 2 homogeneous linear recurrence relation given a formula Given the formula $a_n = (-1)^n + 2^{n+1}$, how do you find a degree 2 homogeneous linear recurrence relation?
|
Let $a_{n+2} = a_{n+1}x + a_ny$. This gives
$$
(-1)^{n+2} + 2^{n+3} = ((-1)^{n+1} + 2^{n+2})x + ((-1)^n + 2^{n+1})y
$$
Separating this into two equations, one for the $-1$ term and one for the $2$ term, we get
$$
\cases{(-1)^{n+2} = (-1)^{n+1}x + (-1)^{n}y\\2^{n+3} = 2^{n+2}x + 2^{n+1}y}
$$
Dividing the first equation by $(-1)^n$ and the second by $2^{n+1}$, we get
$$
\cases{1 = -x+y\\4 = 2x + y}
$$
which may be solved for $x$ and $y$.
|
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|
Find the number of integers 1≤x≤210 which are divided by two of the numbers 3,5,7 I went with;
Numbers that can be divided by $3,5$ can also be divided with $15$. These are $15,30,45,60,75,90,105,120,135,150,165,180,195,210$
Numbers that can be divided by $3,7$ can also be divided with $21$. These are $21,42,63,84,105,126,147,168,189$
Numbers that can be divided by $5,7$ can also be divided with $35$. These are
$35,70,105,140,175,210$
So total number of integers are $14+9+6=$$29$
But I'm not sure about my answer.
|
$$210= 3 \times 5 \times 7 \times 2$$
$$\frac{210}{3 \times 5} = 2 \times 7$$
$$\frac{210}{3 \times 7} = 2 \times 5$$
$$\frac{210}{5 \times 7} = 3 \times 2$$
If a number is divisible by all $3$ numbers, it will appear in each list, hence we need to substract $2 \times$ number of multiples of $3 \times 5 \times 7$.
Hence the desired number is
$$2 \times 7+ 2 \times 5 + 3 \times 2 - 2 \times \frac{210}{3 \times 5 \times 7}=2(7+5+3-2)=26$$
|
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|
Inequality in three variables $a,b,c$ given $abc\geq 1$
Let $a,b,c$ be three real positive numbers such that $abc\geq 1$. Prove that $$\frac{a}{b} + \frac{b}{c} +\frac{c}{a} ≥ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$
Here's what I have done. The expression is equivalent to
$$a^2c+b^2a+c^2b \geq ab+bc+ac .$$
Denote $f(a,b,c)=a^2c+b^2a+c^2b-(ab+bc+ac)$. We need to show that $f(a,b,c)\geq 0$.
By AM-GM, $a^2c+b^2a+c^2b \geq 3$ (Given, $abc\geq 1$). By AM-HM, $(a+b+c)(1/a +1/b +1/c)\geq 9$ by AM-GM, min value of $a+b+c=3$ so $1/a + 1/b + 1/c \geq 9/3=3$. Thus, The minimal value of $f(a,b,c)$ is equal to the min value of $a^2c+b^2a+c^2b$ minus the min value of $ab+bc+ca=3-3=0$.
Please let me know if there is fault in argument.
|
By AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b}=\frac{1}{3}\sum_{cyc}\left(\frac{a}{b}+\frac{2b}{c}\right)\geq\sum_{cyc}\sqrt[3]{\frac{ab}{c^2}}=\sum_{cyc}\sqrt[3]{\frac{abc}{c^3}}\geq\sum_{cyc}\frac{1}{a}.$$
Done!
|
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|
Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{(x+4)}{\sqrt{6}})+C$
However the result is $\frac{arctan(\frac{(x+4)}{\sqrt{6}})}{\sqrt{6}}+C$
Why is my result wrong? I can't see any mistake.
|
Anyway everyone should know that
$$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\arctan\frac xa.$$
|
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|
Finding $\int_0^7 g(r)~dr$ from $ g(r)=\begin{cases} \sqrt{9-(r-3)^2}, & 0\le r<3 \\ r+2, & 3\le r<5 \\ \sqrt{4-(r-5)^2}, & 5 \le r \le 7\end{cases} $ I have some piece-wise integration that I for some reason am not getting.
$$
g(r) =
\begin{cases}
\sqrt{9-(r-3)^2} && 0 \le r \lt 3 \\
r+2 && 3 \le r \lt 5 \\
\sqrt{4-(r-5)^2} && 5 \le r \le 7
\end{cases}
$$
I must find the area of $\int_0^7g(r) \ dr$.
I know that this must be equivalent to:
$$
\int_0^2\sqrt{9-(r-3)^2} \ dr \ + \ \int_3^4(r+2) \ dr \ + \ \int_5^7\sqrt{4-(r-5)^2} \ dr
$$
But my solution of $4.125 + 5.5 + \pi$ yields an incorrect answer?
|
I have no idea why you separated it like how you did it, this is incorrect:
$$\int_0^7 g(r)~dr= \int_0^2 \sqrt{9-(r-3)^2}~dr+\int_3^4 (r+2)~dr+\int_5^7 \sqrt{4-(r-5)^2}~dr$$
Doing your method will give this area instead:
Splitting definite integrals in two does not work like when you split summations!
$$\sum_{k=0}^n f(k)\equiv \sum_{k=0}^m f(k)+\sum_{k=m+1}^n f(k)$$
$$\int_0^n f(x)~dx\not\equiv \int_0^m f(x)~dx+\int_{m+1}^n f(x)~dx$$
In general, one can separate a definite integral into two definite ones as follows:
$$\int_a^c f(x)~dx=\int_a^b f(x)~dx+\int_b^c f(x)~dx$$
Hence, you want to separate it as follows:
$$\begin{align} \int_0^7 g(r)~dr&=\int_0^3 \sqrt{9-(r-3)^2}~dr+\int_3^5 (r+2)~dr+\int_5^7 \sqrt{4-(r-5)^2}~dr \\&=\frac{9\pi}{4}+12+\pi\\&=\frac{13\pi}{4}+12\end{align}$$
Since you managed to evaluate your other integrals, I suppose you know how to evaluate these ones to $\frac{9\pi}{4}$ and $12$, so I left it as an exercise for the reader.
|
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|
convergence of $\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $ For which value $\alpha$ does the following integral converges:
$\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $
I tried to use: $ \sqrt{x^2 + 2x + 2} - 1 = \frac{x^2 + 2x + 1}{\sqrt{x^2 + 2x + 2} + 1} $
I wanted to simplify it more but that didn't lead to anything.
Thank you for your help.
|
$I_a =\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^a dx
$
Use the standard method of
$a-b
=\dfrac{a^2-b^2}{a+b}
$.
$\begin{array}\\
\sqrt{x^2 + 2x + 2} -( x + 1)
&=\dfrac{(x^2 + 2x + 2)-(x+1)^2}{\sqrt{x^2 + 2x + 2} + x + 1}\\
&=\dfrac{(x^2 + 2x + 2)-(x^2+2x+1)}{\sqrt{x^2 + 2x + 2} + x + 1}\\
&=\dfrac{1}{\sqrt{x^2 + 2x + 2} + x + 1}\\
\end{array}
$
Therefore
$\dfrac1{2(x+1)}
\gt \sqrt{x^2 + 2x + 2} -( x + 1)
\gt \dfrac1{2(x+2)}
$.
Therefore
$I_a$ converges
if and only if
$\int_0^{\infty} \dfrac{dx}{(x+1)^a}
$
converges
and this is when
$a > 1$.
|
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|
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me anymore:
$$\begin{align*}
&= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\
&= \frac{(k+1)^2(k+2)^2}{4}.\\
\end{align*}$$
I understand that first $(k+1)^3$ was changed to have the same denominator as the main term (which is $4$). Can someone help me break down the steps how the polynomials are added then after that, the powers confuse me a bit.
|
Just looking at the numerators,
$$k^2(k+1)^2+4(k+1)^3 =k^2(k+1)^2+4(k+1)(k+1)^2$$
$$ = (k+1)^2(k^2+4(k+1)) = (k+1)^2(k^2+4k+4)$$
$$=(k+1)^2(k+2)^2$$
For the second equality, I factored out the $(k+1)^2$ that appears in both terms; something of the form $ba+ca$ can be rewritten as $a(b+c)$. For the third equality, I just distributed the $4$, noting that $4(k+1)=4k+4$. Finally, $k^2+4k+4$ is of the form $a^2+2ab+b^2$, which can be rewritten as $(a+b)^2$; here, $a=k$ and $b=2$, giving us $(k+2)^2$.
|
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|
Solve the following quadratic equation: $4\tan^2(\theta)x^4-4x^2+1=0$ $$4\tan^2(\theta)x^4-4x^2+1=0$$
It can be seen that as $\theta$ goes to $0$, the leftmost term disappears and $x=\pm \frac{1}{2}$.
But when I try to solve this using the quadratic formula I get:
$$x=\pm \sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}}$$
In this expression as $\theta$ goes to $0$ the denominator goes to zero.
I guess there may be a way of manipulating this expression such that as $\theta$ goes to zero $x$ goes to $\pm 0.5$.
How could this expression be manipulated?
|
$$\lim_{\theta\to0}\frac{1-\sqrt{1-\tan^2\theta}}{2\tan^2\theta}
= \lim_{\theta\to0}\frac{(1-\sqrt{1-\tan^2\theta})(1+\sqrt{1-\tan^2\theta})}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})} = \lim_{\theta\to0}\frac{\tan^2\theta}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})}=\lim_{\theta\to0}\frac{1}{2(1+\sqrt{1-\tan^2\theta})}=\frac{1}{2}$$
|
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|
Finding minima and maxima to $f(x,y) = x^2 + x(y^2 - 1)$ in the area $x^2 + y^2 \leq 1$ I'm asked to find minima and maxima on the function
$$f(x,y) = x^2 + x(y^2 - 1)$$
in the area
$$x^2 + y^2 \leq 1.$$
My solution:
$$\nabla (f) = (2x + y^2 - 1, 2xy)$$
Finding stationary points
$$2xy = 0$$
$$2x + y^2 - 1 = 0$$
gets me $(0,1),(0,-1),(\frac{1}{2},0)$.
Stuyding the boundary:
$$x^2 + y^2 = 1$$
$$y^2 = 1 - x^2 $$
$$f(x,y) = f(x, 1-x^2) = x^2 -x^3$$
Finding stationary points on the boundary:
$$f'(x,y) = 2x - 3x^2 = 0$$
gives $(0, 1), (0,-1), (\frac{2}{3}, \sqrt{\frac{5}{9}}),(\frac{2}{3}, -\sqrt{\frac{5}{9}})$.
So in total i've got the stationary points
$$(\frac{1}{2} , 0) | (0,1)| (0,-1)| (\frac{2}{3},\sqrt{\frac{5}{9}})|(\frac{2}{3}, -\sqrt{\frac{5}{9}}).$$
which give the function values of:
$$-\frac{1}{4}, 0,0,0.15,0.15$$.
which gives the minima: $-\frac{1}{4}$ and maxima $0.15$.
The minima is correct but the maxima should be 2. Why? What stationary point am I missing?
|
Just as you check the boundary of the 2_D region, you also have to check the endpoints of your interval. $f(x)=x^2-x^3$ on the interval $[-1,1]$. You find that $x=-1, y=0$ is your max.
|
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|
Prove the sequence of general term $\sum\limits_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k}$ converges
Given the sequence $$a_n :=\sum_{k=1}^n \frac{(-1)^{k+1}}{k}$$
Prove the sequence $(a_{2n+1})_{n \geq 0}$ converges.
My thoughts
I've proven that the sequence $(a_{2n+1})_{n \geq 0}$ is monotone decreasing. So now I want to prove that the sequence is bounded below, since then I can prove the sequence converges.
I wanted to prove the sequence is bounded below by induction, however I'm not quite sure how.
Since $$a_{2n+1} :=\sum_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k}= 1 -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$$
I think it is bounded by $\frac{1}{2}$.
Using induction:
*
*For $n=0$, we know $a_{2n+1} = a_1 = 1 \geq\frac{1}{2}$
*Suppose that $a_{2n+1} \geq \frac{1}{2}$. Prove that $a_{2(n+1)+1)} = a_{2n+3} \geq \frac{1}{2}$.
I know that $$a_{2n+3} :=\sum_{k=1}^{2n+3} \frac{(-1)^{k+1}}{k} = a_{2n+1} -\frac{1}{2n+2} + \frac{1}{2n+3} = a_{2n+1} -\frac{1}{(2n+2)(2n+3)}$$ But I don't know how to proceed. Any help?
|
To show that $a_{2n+1} > \frac12$, note that
$$
a_{2n+1} = \sum_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k} = \underbrace{1 - \frac12}_{=\frac12} + \underbrace{\frac13 - \frac14}_{>0} + \underbrace{\frac15 - \frac16}_{>0} + \ldots + \underbrace{\frac{1}{2n-1} - \frac{1}{2n}}_{>0} + \underbrace{\frac{1}{2n+1}}_{>0} > \frac12.
$$
|
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|
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
|
You have proved that the LHS is
$$
LHS=\frac{1-\cos x}{1+\cos x}
$$
And if you know the formula for $\tan \frac{x}{2}$ you have done (bisection).
If not use:
$$
\tan^2 \frac{x}{2}=\frac{\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}}=\frac{\frac{1-\cos x}{2}}{\frac{1+\cos x}{2}}
$$
|
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|
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
|
Work out $a=z+1/z$. Note that $z^3+1/z^3=a^3-3a$.
|
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|
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians.
Find the area of $ ABC $
This problem is too difficult for me , the teacher said it is a challenge problem.
Any help will be appreciated , thank you.
|
Let $AB=c$, $AC=b$ and $BC=a$.
Thus, $a^2=b^2+c^2-bc$, $\frac{1}{2}\sqrt{2a^2+2c^2-b^2}=9$ and $\frac{1}{2}\sqrt{2a^2+2b^2-c^2}=6$, which gives
$$\frac{2(b^2+c^2-bc)+2c^2-b^2}{2(b^2+c^2-bc)+2b^2-c^2}=\frac{9}{4}$$
|
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|
$a-b \nmid a^2+b^2$ if $a>b+2$ and $\gcd(a,b)=1$ Any two co-prime number $a,b$ with $a>b+2$ we have $a^2+b^2$ is not divisible by $a-b$, $a,b \in \mathbb{N}$.
But how to prove this?
|
Since $a-b$ divides $a^2-b^2$, if it also divides $a^2+b^2$ then $a-b$ divides $2b^2$.
Now use that $a$ and $b$ are coprime to see that $a-b$ and $b^2$ are coprime. So by Euclides' lemma $a-b$ divides $2$. This contradicts the condition $a-b>2$.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.