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Is $\sin^2x$ uniformly continuous on$x\in [0,\infty]$ I have the question that is $sin^2x$ uniformly continuous on $x \in [0,\infty]$ ?
My approach:
Let $\left|x-y\right|<\delta$ we have:-
$$\left|sin^2x-sin^2y\right|=\left|(\sin x+\sin y)(sin x-sin y)\right|$$
$$=4\left|\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\right|\lt4\left|\sin \left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\tag{1}$$
$$\lt \left|(x-y)(x+y)\right|<\left|(x+y)\right|\delta,$$which is dependent on $x$ so $\sin^2\!x$
is not uniformly continuous.
Is this solution correct or not? I have some doubt about validity of inequality $(1)$ also, if it is correct then why?
| Carrying your idea further, we have for $|x-y| < \delta(\epsilon) = \epsilon/2$
$$|\sin^2 x - \sin^2 y|= |\sin x + \sin y||\sin x - \sin y| \\ \leqslant 2 |\sin x - \sin y|\\ = 4\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right| \\ \leqslant 4\left|\sin\left(\frac{x-y}{2}\right)\right|\\ \leqslant 2|x-y| \\< \epsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find the center of circle given two tangent lines (the lines are parallel) and a point. How to find the center of a circle if the circle is passing through $(-1,6)$ and tangent to the lines $x-2y+8=0$ and $2x+y+6=0$?
| Let $(h, k)$ be the center of the circle then the distance of the center $(h, k)$ & $(-1, 6)$ will be equal to the radius ($r$) of circle $$r=\sqrt{(h+1)^2+(k-6)^2}\tag 1$$
Now, the perpendicular distance of the center $(h, k)$ from the line: $x-2y+8=0$
$$r=\left|\frac{h-2k+8}{\sqrt{1^2+(-2)^2}}\right|=\left|\frac{h-2k+8}{\sqrt 5}\right|\tag 2$$
the perpendicular distance of the center $(h, k)$ from the line: $2x+y+6=0$
$$r=\left|\frac{2h+k+6}{\sqrt{2^2+1^2}}\right|=\left|\frac{2h+k+6}{\sqrt 5}\right|\tag 3$$
from (2) & (3), $$|h-2k+8|=|2h+k+6|$$
$$h-2k+8=\pm(2h+k+6)$$ $$h+3k=2\tag 4$$ or $$3h-k+14=0\tag 5$$
from (1) & (2), $$\sqrt{(h+1)^2+(k-6)^2}=\left|\frac{h-2k+8}{\sqrt 5}\right|$$
$$(h+1)^2+(k-6)^2=\frac{(h-2k+8)^2}{5}$$
setting $k=3h+14$ from (5), one should get
$$(h+1)^2+(3h+14-6)^2=\frac{(h-2(3h+14)+8)^2}{5}$$
$$h^2+2h-3=0\implies h=1, -3$$
hence, corresponding values of $k$ are $k=3(1)+14=17$ & $k=3(-3)+14=5$
Hence, the center of the circle is $\color{red}{(1, 17)}$
or $\color{red}{(-3, 5)}$
Note: It can be checked that for $h+3k=2$ from (4), there is no real values of $h$ & $k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given complex $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$ : when and what is $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
If $z_{1},z_{2},z_{3}$ are three complex number Such that $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$
Then $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
$\bf{My\; Try::}$ Let $z_{1}=2\left(\cos \alpha+i\sin \alpha\right)$ and $z_{2}=3\left(\cos \beta+i\sin \beta\right)$ and $z_{3}=4\left(\cos \gamma+i\sin \gamma\right)$
So $$f(\alpha,\beta,\gamma) = 58-\left[12\cos(\alpha-\beta)+24\cos(\beta-\gamma)+16\cos(\gamma-\alpha)\right]$$
Now How can I calculate $\max$ of $f(\alpha,\beta,\gamma)$
Help me
Thanks
| The maximum is $87$.
First we note that by a simple compactness argument, the maximum must be attained for some choice of $z_1$, $z_2$, $z_3$.
Assume first that $z_1$ and $z_2$ are fixed and we let $z_3$ vary. Let $y_3$ be the midpoint of the segment between $z_1$ and $z_2$. The loci on which $Q = |z_3 - z_2|^2 + |z_3 - z_1|^2$ is constant are circles centred at $y_3$. The larger the radius, the larger $Q$ is. Thus $Q$ is maximized when $z_3$ is taken as far away from $y_3$ as possible while keeping $|z_3| = 4$. This occurs when $z_3$ is chosen so that the origin $O$ lies on the median $y_3 z_3$ of triangle $z_1 z_2 z_3$.
Similar arguments can be made for $z_1$ and $z_2$. Therefore, when $z_1$, $z_2$ and $z_3$ are chosen optimally, either $O$ must be the centre of gravity of the triangle $z_1$, $z_2$, $z_3$, or the triangle must be degenerate. Assume first that it is the former.
If we let $a = |z_1 - z_2|$, $b = |z_2 - z_3|$ and $c = |z_3 - z_1|$, and we let $m_a$, $m_b$, $m_c$ be the corresponding medians, then we have $m_a = (3/2)|z_3| = 6$, $m_b = (3/2)|z_1| = 3$, and $m_c = (3/2)|z_2| = 9/2$. Now using $4m_a^2 + a^2 = 2b^2 + 2c^2$ (valid in any triangle, proved using the parallelogram law) and the analogous identities for $m_b$ and $m_c$, we get the system
$$\begin{align*}
144 + a^2 &= 2b^2 + 2c^2, \\
36 + b^2 &= 2c^2 + 2a^2, \\
81 + c^2 &= 2a^2 + 2b^2.
\end{align*}$$
Adding the equalities, we find $a^2 + b^2 + c^2 = 87$. This calculation is also valid in a degenerate triangle, so long as $z_1 + z_2 + z_3 = 0$.
Thus the maximum value is realized either when $z_1$, $z_2$ and $z_3$ are chosen so that $z_1 + z_2 + z_3 = 0$, in which case the value attained is $87$, or when the triangle is degenerate, in which case the maximum value is $86$. (The only degenerate configuration in which $z_i$ and $y_i$ are on opposite sides of $O$ for $i = 1, 2, 3$ is when $z_3$ is on one side of $O$ and $z_1, z_2$ on the other.) Thus the only question is whether a configuration with $z_1 + z_2 + z_3 = 0$ can be realized. But for $|z_1| = 2$ and $|z_2| = 3$, the value of $|z_1 + z_2|$ ranges between $1$ and $5$. Therefore $|z_1 + z_2| = 4$ can indeed be realized.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}{x+4} & \quad -4 < x < 0 \\
\frac{x}{x+4} & \quad x < −4
\end{array}
\right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
| $$\left|\frac{x}{x+4}\right|<4$$
Squaring, you get
$$\frac{x^2}{16(x+4)^2}<1$$
Rearranging we get
$$15x^2 + 128x+256 >0$$
Using quadratic roots
$$(x+\frac{16}{5}) (x+\frac{16}{3}) >0$$
Hence the solution is $$x \in (-\infty,-\frac{16}{3})\cup(-\frac{16}{5},\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Explicit Laplace approximation for tail of gaussian distribution I'm studying some lecture notes by S. R. Srinivasa Varadhan about Large Deviations Theory and I have some trouble understanding a simple equation right on page 2 where it says
$$\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp{\left(-\frac{nx^2}{2}\right)}\mathrm dx=\exp{\left(-\frac{n l^2}{2}+o(n) \right)}.$$
The left side of the equation looks to be the same as $1-\Phi(\sqrt{n}x)=\Phi(-\sqrt{n}x)$ where $\Phi$ is the standard normal CDF. But I don't see how to get there: Is this a simple (approximative) calculation or a fact about the normal distribution?
| Regarding the left hand side,
\begin{align*}
\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp\left\{-\frac{nx^2}{2}\right\}\,dx
&=\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\
&=\frac{\sqrt{n}}{2\pi}\cdot\sqrt{2\pi}(1/\sqrt n) \int_l^\infty \frac{1}{\sqrt{2\pi}(1/\sqrt n)}\exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\
&=\frac{\sqrt{2\pi}}{2\pi}\int_l^\infty \frac{1}{\sqrt{2\pi}(1/\sqrt n)}\exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\
&=\frac{\sqrt{2\pi}}{2\pi}\int_{l\sqrt n}^\infty \frac{1}{\sqrt{2\pi}(1/\sqrt n)}\exp\left\{-\frac{1}{2}u^2\right\}\,\left(\frac{1}{\sqrt n}\,du\right)\\
&=\frac{\sqrt{2\pi}}{2\pi}\int_{l\sqrt n}^\infty \frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}u^2\right\}\,du\\
&=\frac{1}{\sqrt{2\pi}}\left[1-\Phi\left(\sqrt n l\right)\right]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(x)=f\left( \frac{x+1}{2} \right)+f\left( \frac{x-1}{2} \right)$.
But how to find all functions?
| We can prove that if a function $f:\mathbb R\to\mathbb R$ satisfies
$$f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)\tag0\label0$$
for every real numbers $x$ and $y$, then there is a constant real number $k$ such that $f(x)=kx$ for every real number $x$.
First, let $y=x$ in \eqref{0} and get $f(0)=0$. Next, let $x=1$ and $y=-1$ in \eqref{0} and get $f(-1)=-f(1)$. Now, define $k:=f(1)$ and $g(x):=f(x)-kx$. So we have $g(1)=g(-1)=0$ and:
$$g\left(x^2-y^2\right)+k\left(x^2-y^2\right)=(x-y)\big(g(x)+g(y)+k(x+y)\big)$$
$$\therefore\quad g\left(x^2-y^2\right)=(x-y)\big(g(x)+g(y)\big)\tag1\label1$$
Thus letting $y=\pm1$ in \eqref{1} we have:
$$g\left(x^2-1\right)=(x\pm1)g(x)$$
Hence, subtracting these two equations, we conclude that $g(x)=0$ for every real number $x$, which shows that $f(x)=kx$ for every $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| Another way is by induction:
$$
2^1+1 = 3 = 3 \cdot 1
$$
Then, if $2^k+1 = 3j, j \in \mathbb{N}$, then
\begin{align}
2^{k+2}+1 & = 4\cdot2^k+1 \\
& = 4(2^k+1)-3 \\
& = 4(3j)-3 \qquad \leftarrow \text{uses induction hypothesis} \\
& = 3(4j-1)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 5
} |
Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n=2$,
$$
x^2+y^3=z^4
$$
I'm already pretty lost.
After this it gets even more complicated, has anyone encountered this problem before?
I want all the solutions, or at least infinitely many for every equation.
| For $n=2$ and $n=3,4$ we can give infinite families using a Pell equation and elliptic curves, respectively,
$n=2$:
$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3 = (2pq)^4$$
where $p,q$ solve $p^2-d^3q^2=1\tag1$.
$n=3$:
$$(a y)^2 + (ma)^3 + a^4 = a^5$$
and the elliptic curve solvable for an appropriate constant $m$,
$$a^3 - a^2 - m^3 a = y^2\tag2$$
Example: For $m=2$,
$$\begin{aligned}
(4\cdot4)^2 + (2\cdot4)^3 + 4^4 &=4^5\\
(9\cdot24)^2 + (2\cdot9)^3 + 9^4 &=9^5\\
\big(\tfrac{10252a}{125}\big)^2+(2a)^3 + a^4 &= a^5
\end{aligned}$$
and $a=\big(\tfrac{22}{5}\big)^2$. This elliptic curve has an infinite number of rational points and by multiplying by an appropriate factor (such as $5^{60}$), then one can easily clear denominators and get integer solutions. (Note: The initial solution of $(2)$ may have a large height. For example, for $m=13$ we have $a=(\frac{45539}{6612})^2$.)
$n=4$:
Similarly, one can use,
$$(a y)^2 + (2a)^3 + a^4 + a^5 = a^6$$
$$(a y)^2 + (2a)^3 + (3a)^4 + a^5 = a^6$$
both of which entail solving an elliptic curve.
$n>4$:
A nice identity with almost consecutive exponents is one by Enrico Jabara,
$$(7^3m^{11} n^4)^3 + (7\cdot14^4m^{11}n)^4 + (7m^4n^4)^5 + (m n^4)^7 + (14^4 m^6-n^4)^8 = (14^4 m^6+n^4)^8$$
where $m = 28^4$ and arbitrary $n$, and which he probably found by expanding,
$$(an^4)^3 + (bn)^4 + (cn^4)^5 + (dn^4)^7 + (e-n^4)^8 = (e+n^4)^8$$
collecting powers of $n$, and solving for $a,b,c,d,e$.
| {
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"url": "https://math.stackexchange.com/questions/1619089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What does the homogeneous system of equations represent under certain conditions?
Consider the following linear equations
$ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$
1) $a+b+c \neq o$ and $a^2+b^2+c^2=ab+bc+ca$
2) $a+b+c \neq o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
3) $a+b+c = o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
4) $a+b+c = o$ and $a^2+b^2+c^2 = ab+bc+ca$
Now I need to match these with the following options.
a) The equations represent planes meeting only at a single point.
b) The equations represent the line $x=y=z$
c) The equations represent identical planes
d) The equations represent the whole 3D space.
I found out the determinant using Cramer's rule i.e. $-(a^3+b^3+c^3-3abc)$.So using Cramer's rule whenever the determinant is 0 there should be infinite solutions.After that I'm confused as to how to proceed.
| hint
Note that $a^2+b^2+c^2=ab+bc+ca$ is another way of saying $a=b=c$ (just multiply both sides by $2$ and complete the squares).
So if (4) holds, then $a=b=c=0$, in which case the system is actually $0=0$, hence the solution space is all of $\mathbb{R}^3$.
Try to proceed and see if you can complete now. Otherwise I can elaborate.
Further elaboration:
Consider (1), since $a+b+c \neq 0$, then we know even though $a=b=c$ but they are not zero. In which case we are essentially given one plane, namely $x+y+z=0$ That corresponds to (c).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$
Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$ given that:
$a_n=1/{{n}\choose{0}}+1/{{n}\choose{1}}+...+1/{{n}\choose{n}}$
The hint says to consider when $n$ is even and odd. When $n=2k$ I get:
$$a_{n}=1/{{2k}\choose{0}}+1/{{2k}\choose{1}}+...+1/{{2k}\choose{2k}}$$
$$=1+1/{{2k}\choose{1}}+...+1/{{2k}\choose{2k}}$$
$$=1+1/(2k{{2k-1}\choose{0}})+1/(\frac{2k}{2}{{2k-1}\choose{0}})...+1/(\frac{2k}{2k}{{2k-1}\choose{2k-1}})$$
$$=1+\frac{1}{2k}(1/{{2k-1}\choose{0}}+1/{{2k-1}\choose{1}}+1/{{2k-1}\choose{2k-1}})$$
which should be $\frac{2k+1}{4k}a_{n-1}+1$ in the end.
I used:
${{n}\choose{k}}=\frac{n}{k}{{n-1}\choose{k-1}}$ and tried ${{n-1}\choose{k-1}}={{n-1}\choose{n-k}}$
| Let $$f(x) =\sum_{n=1}^{\infty} a_n x^n =\sum_{n=1}^{\infty} \frac{n+1}{2n} a_{n-1} x^n +\sum_{n=1}^{\infty} x^n =1+\sum_{n=1}^{\infty}\frac{n+2}{2(n+1)} a_n x^{n+1} +\frac{x}{1-x}=\frac{1}{1-x} +\frac{1}{2} \int\sum_{n=1}^{\infty}(n+2) a_n x^n dx =\frac{1}{1-x} +\frac{1}{2}\int \frac{1}{x} \left(\sum_{n=1}^{\infty} a_n x^{n+2}\right)' dx=\frac{1}{1-x}+\frac{1}{2}\int \frac{(x^2 f(x))'}{x} dx =\frac{1}{1-x} +\frac{1}{2} \int (2f(x) +xf'(x))dx $$
Hence
$$f'(x) =\frac{1}{(1-x)^2} +f(x) +\frac{xf'(x)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621275",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that $|\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$ I need to show that $ \forall x,y \in \mathbb R, |\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$
I tried using concavity of log function: $\log(1 + x^2) - \log(1 + y^2)=\log(\frac{1 + x^2}{1 + y^2})=\log(\frac{x^2y^2}{(1 + y^2)y^2}+\frac{1}{1 + y^2}) \ge \frac{2(\log(x)-\log(y))}{1+y^2}$
Also the middle value theorem: $0<\frac{\log(1 + x^2) - \log(1 + y^2)}{x^2-y^2} <1$
But both attempts has not led too far.
| The derivative of $f(x)=\log(1+x^2)$ is
$$
f'(x)=\frac{2x}{1+x^2}
$$
and
$$
\left|\frac{2x}{1+x^2}\right|\le1
$$
because
$$
2|x|\le 1+|x|^2
$$
since
$$
(1-|x|)^2\ge0
$$
Thus, by the mean value theorem,
$$
\frac{\log(1+x^2)-\log(1+y^2)}{x-y}=f'(z)
$$
for $z$ between $x$ and $y$ (assuming $x\ne y$ or the inequality is obvious). Then
$$
\left|\frac{\log(1+x^2)-\log(1+y^2)}{x-y}\right|=|f'(z)|\le 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equation with radicals and reciprocals Find all $x\in\mathbb{R}$ satisfying $$x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}.$$
Multiply both sides by $x^{1/2}$ to get $$x^{3/2} = \sqrt{x^2-1} + \sqrt{x-1}.$$Making the substitution $a = \sqrt{x^2 - 1}$, $b=\sqrt{x-1}$, we have $a+b = x^{3/2}$ and $a^2 - b^2 = x^2 - x$, so $a-b = (a^2-b^2)/(a+b) = \frac{x-1}{x^{1/2}}$. Then solving for $b$ gives $b = \frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2}$. Then, we get the equation $$\frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2} = \sqrt{x-1},$$which we can expand/rearrange to get $$x^4 - 2x^3 - x^2 + 2x + 1 = 0,$$which factors as $$\left(x^2 - x - 1\right)^2 = 0 \implies x^2 - x - 1 = 0,$$which has solutions $x = \frac{1\pm\sqrt{5}}{2}$, but note that we must have $x \ge 1$ by the problem, so $\boxed{x = \frac{1+\sqrt{5}}{2}}$.
I would be interested in seeing a slicker solution, though.
| Here is a slicker solution I found.
Use David's substitution $y=\sqrt{x-\frac{1}{x}}, z=\sqrt{1-\frac{1}{x}}$, and note that $x=y^2-z^2+1$. Then multiply $x=y+z\;(\star)$ by $y-z$ to get $$y-z=\frac{y^2-z^2}{x}=z^2.$$ Summing this with $(\star)$ we get $$x-2y+z^2=0\iff(y-1)^2=0.$$ Thus $y=1$, and $x\geq1$ is the root of $x^2-x-1$, so $\boxed{x = \frac{1+\sqrt{5}}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $4 \over {{x^4} + {y^4} + {z^4}}$ from $x + y + z = 1$ and other conditions more? How to calculate $$4 \over {{x^4} + {y^4} + {z^4}}$$
from $$ x + y + z = 1, $$
$$ x^2 + y^2 + z^2 = 9, $$
$$ x^3 + y^3 + z^3 = 1. $$
Alternative answers: A) $1 \over {33}$, B) $2 \over {33}$, C) $4 \over {33}$, D) $16 \over {33}$, E) $64 \over {33}.$
I tried to expanding $(x + y + z)^4 $ or ${(x^2 + y^2 + z^2)^2}$ and so then isolate $x^4 + y^4 + z^4$ but get too long expressions.
What notable product or special product could be applied to this problem?
| \begin{align}
x^2+y^2+z^2&=(x+y+z)^2-2xy-2yz-2zx\\
9&=1-2(xy+yz+zx)\\
xy+yz+zx&=-4\\
(xy+yz+zx)^2&=16\\
x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)&=16\\
x^2y^2+y^2z^2+z^2x^2+2xyz&=16\qquad\text{since }x+y+z=1\\
\frac{(x^2+y^2+z^2)^2-(x^4+y^4+z^4)}{2}+2xyz&=16\tag{1}
\end{align}
On the other hand, from the identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
we have
\begin{align}
1-3xyz&=(1)\left(9-(-4)\right)\\
xyz&=-4
\end{align}
plugging it into $(1)$ we get
\begin{align}
\frac{9^2-(x^4+y^4+z^4)}{2}+2(-4)&=16\\
81-(x^4+y^4+z^4)-16&=32\\
x^4+y^4+z^4&=33
\end{align}
Then
$$\frac{4}{x^4+y^4+z^4}=\boxed{\color{blue}{\frac{4}{33}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
If $\sin x+\sin^{2} x=1$, then the value of
$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
is equal to
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $
$\boxed{\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2\\
=\cos^{6} x\left(\cos^{6} x+3\cos^{4} x+3\cos^{2} x+1\right)+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\cos^{2} x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\sin x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin x[\sin x+1]\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin^{2} x+\sin x\right)^{3}+2\sin^{2} x+\sin x-2\\
=1+\sin^{2} x+\sin^{2} x+\sin x-2\\
=\sin^{2} x} $
I found this solution but considering the time as $1-2$ min to solve this question,
I am looking for a short and simple way.
I have studied maths up to $12$th grade
| Observe that $\sin x = 1-\sin^2 x =\cos^2 x$. Moreover let $\cos^4 x =t$ then the expression given is
\begin{align*}
& = \color{red}{t^3+3t^2\sqrt{t}+3t(\sqrt{t})^2+(\sqrt{t})^3}+2t+\sqrt{t}-2\\
&=\color{red}{(t+\sqrt{t})^3}+2t+\sqrt{t}-2\\
&=(\cos^4x+\cos^2x)^3+2\cos^4x+\cos^2x-2\\
\text{using the fact that $\sin x =\cos^2 x$}\\
&=(\sin^2x+\cos^2x)^3+2\sin^2x+\cos^2x-2\\
&=1+1+\sin^2x-2\\
&=\sin^2x.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Probability of almost correct order Let's say I'm a teacher handing tests back to seven students.
If I do it with my eyes closed, what's the probability I hand exactly 5 of the tests back to the correct students?
There are many possible ways this could happen, and if my understanding of probability is correct, the sum of the individual probabilities of all possibilities where I hand exactly 5 back to the correct student yields the answer I'm looking for.
One such probability (hand out the first two tests incorrectly):
$$\frac{6}{7}\cdot\frac{5}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
Another probability (hand out the last two tests incorrectly):
$$\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
One more:
$$\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
We can simplify these probabilities to:
$$\frac{6\cdot5}{7!}, \frac{1\cdot1}{7!}, \frac{4\cdot2}{7!} $$
More generally, we can think about all possible times when the two incorrect letters might occur, and deduce that the sum of all possible probabilities is:
$$\frac{6(5+4+3+2+1)+5(4+3+2+1)+4(3+2+1)+3(2+1)+2(1)+1(1)}{7!}$$
Is this really the answer? I really feel like I missed something that would simplify this answer.
| There are $$\binom{7}{2}$$ ways to choose the $2$ students who receive the wrong tests. Obviously, the only way for this to happen is if their tests are switched.
There are $7!$ total ways to hand back the tests, so the probability is
$$\frac{\binom{7}{2}}{7!} = \frac{1}{2 \cdot 5!} = \boxed{\frac{1}{240}}$$
A related question would be: what is the probability that you hand back no tests to the correct students? If this interests you, you should read about derangements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$ As stated in the tag, I'm trying to prove by induction the claim $F_{2n} = F^2_{n+1} - F^2_{n-1}$, where $F_{n}$ is the $n^{th}$ Fibonacci number. I've spent hours on the inductive step without substantial progress, and am hoping someone can provide a path to the desired result.
To facilitate the proof, we're given $F_{2n-1} = F^2_{n} + F^2_{n-1}$
(Also, in various past attempts not reflected here I tried to use the following substitutions whenever it seemed prudent: $F_{n} = F_{n-1} + F_{n-2} = F_{n+1} - F_{n-1}$; and $(x^{2} - y^{2}) = (x - y)(x + y)$).
For the inductive step, let $n\geq 1$ and assume $F_{2n} = F^2_{n+1} - F^2_{n-1}$. For $n+1$ we want to show $F_{2(n+1)} = F^2_{n+2} - F^2_{n}$. So
\begin{align}
F_{2(n+1)} &= F_{2n + 2} \\
&= F_{2n+1} + F_{2n} \\
&= (F_{2n} + F_{2n-1}) + F_{2n} \\
&= 2 F_{2n} + F_{2n - 1} \\
&= 2 (F^2_{n+1} - F^2_{n-1})
+ F^2_{n} + F^2_{n-1}
&&\text{by the inductive hypothesis} \tag{1} \\
&=...F^2_{n+2} - F^2_{n}
\end{align}
I have two challenges. The first is squeezing $-2 F^2_{n}$ out of (1). The only reasonable way I can see to do that is to expand the $-F^2_{n-1}$ term in $2 F^2_{n+1} - F^2_{n-1}$ to $-(F_{n} - F_{n-2})^2$. If in instead I combine it with $F^2_{n-1}$ then it seems impossible to generate the $-2 F^2_{n}$ term needed to offset $F^2_{n}$ and produce $-F^2_{n}$.
The second challenge is to find $(F^2_{n+2}$ in (1), and the only way I can see to do that is to expand the $F^2_{n+1}$ term in $2(F^2_{n+1} - F^2_{n-1})$to $(F_{n+2} - F_{n})^2$. Doing that not only generates $2 F^2_{n}$ which totally screws things up, but it also generates a bunch of intermediate terms I can't get rid of, and it's driving me crazy.
| As this is a difficult result to prove, I shall prove another result and then derive the said result through it.
Proposition: $F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$
Base cases: $n = 1$:
$\begin{align}
F_{m+1} &= F_m + F_{m-1} \\
&= F_2 F_m + F_1 F_{m-1}
\end{align}$
Case $n = 2$:
$\begin{align}
F_{m+2}
&= F_{m+1} + F_m \\
&= F_{m-1} + 2 F_m \\
&= F_2 F_{m-1} + F_3 F_m
\end{align}$
Hypothesis: $F_{k+m-1} = F_{k-1} F_{m-1} + F_k F_m$ and
$F_{k + m} = F_k F_{m-1} + F_{k +1} F_m$
Induction step: Adding both the hypothesis statements
$\begin{align}
F_{k + m + 1} &= F_{k+m} + F_{k+m-1} \\
&=F_{k+1} F_{m-1} + F_{k+1+1} F_m
\end{align}$
Hence, proved.
$F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$
Putting $n = m$:
$\begin{align}
F_{2n} &= F_n F_{n-1} + F_{n+1} F_n \\
&= (F_{n+1} - F_{n-1}) F_{n-1} + F_{n+1} F_n \\
&= F_{n+1} F_{n-1} + F_{n+1} F_n - F^2_{n-1} \\
&= F_{n+1}(F_{n-1} + F_n) - F^2_{n-1} \\
&= F^2_{n+1} - F^2_{n-1}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate this series? How do I evaluate this series:
\begin{equation}
\sum_{n=2}^\infty \frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!} = \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} +\ldots
\end{equation}
I wanted to use the Comparison test to show convergence, but I didn't know what to compare it to since my series has a product in the numerator...
I'm lost as to how to evaluate it.
| Defining$$a_n=\frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!}$$ you can get rid of the numerator if you notice that $$\prod_{k=1}^{n-1} (2k-1)= \frac{2^{n-1} \Gamma \left(n-\frac{1}{2}\right)}{\sqrt{\pi }}$$ which makes $$a_n=\frac{\Gamma \left(n-\frac{1}{2}\right)}{2 \sqrt{\pi } n!}$$ which makes $$\frac{a_{n+1}}{a_n}=1-\frac{3}{2 (n+1)}$$ As Brian M. Scott answered, the ratio test doesn’t help much here.
But, for large values of $n$, we can use Stirling approximation and get $$a_n=\frac{\left(\frac{1}{n}\right)^{3/2}}{2 \sqrt{\pi }}+\frac{3
\left(\frac{1}{n}\right)^{5/2}}{16 \sqrt{\pi }}+\frac{25
\left(\frac{1}{n}\right)^{7/2}}{256 \sqrt{\pi
}}+O\left(\frac{1}{n^4}\right)$$ Using the first terms $$\sum_{n=2}^\infty \frac{\left(\frac{1}{n}\right)^{3/2}}{2 \sqrt{\pi }}=\frac{\zeta \left(\frac{3}{2}\right)-1}{2 \sqrt{\pi }}\approx 0.454843$$ $$\sum_{n=2}^\infty\frac{3
\left(\frac{1}{n}\right)^{5/2}}{16 \sqrt{\pi }}=\frac{3 \left(\zeta \left(\frac{5}{2}\right)-1\right)}{16 \sqrt{\pi }}\approx 0.0361244$$ $$\sum_{n=2}^\infty \frac{25
\left(\frac{1}{n}\right)^{7/2}}{256 \sqrt{\pi
}}=\frac{25 \left(\zeta \left(\frac{7}{2}\right)-1\right)}{256 \sqrt{\pi }}\approx 0.00698261$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the straight line from an equation system. I have this problem:
the matrix below is an equation system:
$$
\begin{matrix}
1 & 1 & -1 & 19 \\
5 & 4 & -6 & 43 \\
7 & -1 & 7 & 80 \\
\end{matrix}
$$
It simplifies to
$$
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 35.5 \\
0 & 0 & 1 & 16.5 \\
\end{matrix}
$$
My book claims that the solution set of this problem is a straight line that goes through the point $(11, -3, 0)$ and has the direction vector $(-2, 7, 3)$
How can you get this point and vector from the matrix above?
| The book is wrong, and as one of the commenters said. The solution set is the single point $\begin{bmatrix} 0 \\ 35.5 \\ 16.5 \end{bmatrix}$.
The general rule when row reducing $[A |\mathbf{b}]\rightarrow [RREF(A)|\mathbf{P}]$ where $\mathbf{P}\in\mathbb{R}^n$.
*
*$0$ free variables implies the solution set is only the point $\mathbf{P}$. (Like your example)
*$1$ free variable implies the solution set is a line passing through the coordinates of $\mathbf{P}$ corresponding to fixed variables
*$2$ free varaibles implies the solution set is a 2-dimensional plane passing through coordinates of $\mathbf{P}$ corresponding to fixed variables
*etc.
(It is kind of a weird explanation, because the solution set doesn't necessarily pass through $\mathbf{P}$ but the point constructed from each fixed coordinate(s) of $\mathbf{P}$ assigned to its corresponding variables. I will show by example to make it more clear.)
Example: Let $A\mathbf{x}=\mathbf{b}=\begin{bmatrix} -3 && 0 && -1 \\ 2 && 0 && 2 \\ 2 && 0 && 0\end{bmatrix}\mathbf{x}=\begin{bmatrix} 1 \\ -6 \\ 2 \end{bmatrix}$
and the row equivalent version $RREF(A)\mathbf{x}=\mathbf{P}=\begin{bmatrix} 1 && 0 && 0 \\ 0 && 0 && 1 \\ 0 && 0 && 0\end{bmatrix}\mathbf{x}=\begin{bmatrix} 1 \\ -4 \\ 0 \end{bmatrix}$
The general solution is $\begin{bmatrix} 1 \\ t \\ -4 \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ -4 \end{bmatrix}+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$ ($t\in\mathbb{R}$)
which is clearly a line spanned by $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ that passes through $\begin{bmatrix} 1 \\ 0 \\ -4 \end{bmatrix}$ (at $t=0$)
Also, $\begin{bmatrix} 1 \\ 0 \\ -4 \end{bmatrix}$ is a particular solution to $A\mathbf{x}=\mathbf{b}$ while $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$ is the general solution to $A\mathbf{x}=\mathbf{0}$.
Example: Let $A\mathbf{x}=\mathbf{b}=\begin{bmatrix} 0 && 0 && 1 \\ 0 && 0 && -2 \\ 0 && 0 && 6\end{bmatrix}\mathbf{x}=\begin{bmatrix} 3 \\ -6 \\ 18 \end{bmatrix}$
and the row equivalent version $RREF(A)\mathbf{x}=\mathbf{P}=\begin{bmatrix} 0 && 0 && 1 \\ 0 && 0 && 0 \\ 0 && 0 && 0\end{bmatrix}\mathbf{x}=\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}$
The general solution is $\begin{bmatrix} s \\ t \\ 3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}+\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}s+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$ ($s,t\in\mathbb{R}$)
which is clearly the plane spanned by $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ which passes through $\begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}$ (at $s,t=0$)
Again, $\begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}$ is a particular solution to $A\mathbf{x}=\mathbf{b}$ while $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}s+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$ is the general solution to $A\mathbf{x}=\mathbf{0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$? I have problems to solve this limit
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$
I tried with Taylor:
$$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac{1}{2x}+\frac{1}{24x^2}\right)\right)=\lim _{x\to \infty }\left(x^2-x^2+\frac{x}{2}-\frac{1}{24}\right)=\infty$$
that is the wrong result, in must be $\color{red}{-\frac{1}{6}}$
| Another idea:
$$
\lim_{x\to\infty}\left(x\sqrt{x^2-x}-x^2\cos\left(\frac1{\sqrt{x}}\right)\right) =
\lim_{x\to\infty}\frac{
\left(x\sqrt{x^2-x}-x^2\cos\left(\frac1{\sqrt{x}}\right)\right)\left(x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)\right)}
{x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} =
$$
$$
\lim_{x\to\infty}\frac{
x^2(x^2-x)-x^4\cos^2\left(\frac1{\sqrt{x}}\right)}{x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} =
\lim_{x\to\infty}\frac{
-x^3+x^4\sin^2\left(\frac1{\sqrt{x}}\right)}
{x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} =
$$
$$
\lim_{x\to\infty}\frac{
-x+x^2\sin^2\left(\frac1{\sqrt{x}}\right)}
{\sqrt{1-1/x}+\cos\left(\frac1{\sqrt{x}}\right)} = \cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Real Analysis : Value of $U(f,P)-L(f,P)$ in Darboux Integral? Given $f:[0,4]\rightarrow\mathbb{R}$ with the definition :
$f(x)=\begin{cases}
2x+3, & 0 \leq x<1 \\
3, & x=1 \\
-x+1, & 1<x\leq 3 \\
2, & 3<x\leq 4
\end{cases}$
If we're given the partition $P=\{0,1-h,1+h,3-h,3+h,4\}\subset[0,4]$ for any $h\in(0,\frac{1}{2})$, find the value of $U(f,P)-L(f,P)$.
Can someone please help me? I get stuck every time.
| If $\mathcal P = \{x_0,\ldots,x_n\}$ is a partition then the upper and lower sums are
\begin{align}
U_f(\mathcal P) &= \sum_{j=0}^{n-1}\sup_{x\ \in\ [x_j,x_{j+1}]}f(x)(x_{j+1}-x_j)\\
L_f(\mathcal P) &= \sum_{j=0}^{n-1}\inf_{x\ \in\ [x_j,x_{j+1}]}f(x)(x_{j+1}-x_j).
\end{align}
So we compute
\begin{align}
U_f(\mathcal P) &= \sup_{x\in[0,1-h]}f(x)(1-h-0) + \sup_{x\in[1-h,1+h]}f(x)((1+h)-(1-h)) + \sup_{x\in[1+h,3-h]}f(x)(3-h - (1+h)) + \sup_{x\in[3-h,4]}f(x)(4 - (3-h)) \\
&= (2(1-h)+3)(1-h) + (2+3)(2h) + (-h)(2-2h) + 2(1-h)\\
&= 7 - h + 4h^2
\end{align}
and
\begin{align}
L_f(\mathcal P) &= \inf_{x\in[0,1-h]}f(x)(1-h) + \inf_{x\in[1-h,1+h]}f(x)(2h) + \inf_{x\in[1+h,3-h]}f(x)(2h(h-1)) + \inf_{x\in[3-h,4]}f(x)(1-h) \\
&= 3(1-h) + (-h)(2h) + (-(2-h))(2h(h-1)) + (-2)(1-h)\\
&= 1+3h-8h^2+2h^3.
\end{align}
It follows that
\begin{align}
U_f(\mathcal P) - L_f(\mathcal P) &= 7 - h + 4h^2 - (1+3h-8h^2+2h^3)\\
&= 2(3-2h+6h^2-h^3).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use the partial fraction to evaluate $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$. $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$
My try：
$\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$ ＝
$\int \frac{\left(3y^2+3y+2\right)}{\left(y-1\right)(y+1)\left(y+1\right)}dy$
$\frac{A}{y-1}+\frac{B}{y+1}+\frac{C}{y+1}$ = $\frac{A(y+1)(y+1)+B(y-1)y+1)+C(y-1)(y+1)}{(y-1)(y+1)(y+1)}$ = $\frac{\left(3y^2+3y+2\right)}{\left(y-1\right)(y+1)\left(y+1\right)}$
so that
$A(y+1)(y+1)+B(y-1)y+1)+C(y-1)(y+1)=Ay^2+2Ay+A+By^2-B+Cy^2-C=3y^2+3y+2$
And then I got
A+B+C=3
2A=3
A-B-C=2
=>A=3/2 B=? C=?
And then I found that I cannot figure out the value of BandC. What is wrong with my steps? I am so confusing now.
| Try this format $$\frac{3y^2+3y+2}{\left(y-1\right)\left(y+1\right)^2}=\dfrac{A}{y-1}+\dfrac{B}{y+1}+\dfrac{C}{(y+1)^2}.$$
Actually this happens because of:
$$\begin{align}
\frac{3y^2+3y+2}{\left(y-1\right)\left(y+1\right)^2}
& =\dfrac{A}{y-1}+\dfrac{By+C}{(y+1)^2}\\
& =\dfrac{A}{y-1}+\dfrac{B(y+1)+(C-B)}{(y+1)^2}\\
& =\dfrac{A}{y-1}+\dfrac{B}{y+1}+\dfrac{C-B}{(y+1)^2}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the integral $\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}}$ I successfully evaluated the following integral using partial fraction expansion, but am unsure of a few steps.
$$
\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}} = \int\left( \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1} \right) \\\ \\
\\
x^2-2x-1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2 \\\ \\
\\
x=1 \implies -2 = 2B \implies B = -1 \\\ \\
\begin{align}
x=i \implies & -2-2i = (D+Ci)(i-1^2) = -2i(D+Ci) \\
\implies & -C+Di = 1+i \\
\implies &\ C=-1,\ \ D=1
\end{align} \\\ \\
\\
x^2-2x-1 = A(x-1)(x^2+1) - (x^2+1) - (x-1)^3 \\\ \\
\\
x=0 \implies -1 = -A \implies A = 1 \\\ \\
\\
\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}} = \int\left(\frac{1}{x-1} - \frac{1}{(x-1)^2} + \frac{1-x}{x^2+1}\right) = \int \left( \frac{1}{x-1} - \frac{1}{(x-1)^2} - \frac{x}{x^2+1} + \frac{1}{x^2+1} \right) \\\ \\
\\
= \ln|x-1| + \frac{1}{x-1} - \frac{1}{2}\ln|x^2+1| + \arctan(x) + C\\\ \\
$$
My main points of confusion are:
*
*In general, how do you know which denominators to use in step 1?
*Is it okay to find the variables by plugging in values for $x$? Does this work in general? Is there anything I should be aware of, particularly where $i$ is involved?
| Notice:
*
*Your way of using partial fractions is right, and it works in general.
*Use this to find the partial fractions.
*The integral of $\frac{1}{x}$ is equal to $\ln|x|+\text{C}$.
*For $i$ notice that $i^2=-1$
*When $a,b\in\mathbb{R}$:
$$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|\left(\cos(\arg(a+bi))+\sin(\arg(a+bi))i\right)$$
*
*$|a+bi|=\sqrt{\Re^2(a+bi)+\Im^2(a+bi)}=\sqrt{a^2+b^2}$
$$\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}\space\text{d}x=$$
$$\int\left[\frac{1-x}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}\right]\space\text{d}x=$$
$$\int\left[\frac{1}{x^2+1}-\frac{x}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}\right]\space\text{d}x=$$
$$\int\frac{1}{x^2+1}\space\text{d}x-\int\frac{x}{x^2+1}\space\text{d}x+\int\frac{1}{x-1}\space\text{d}x-\int\frac{1}{(x-1)^2}\space\text{d}x=$$
$$\int\frac{1}{x^2+1}\space\text{d}x-\int\frac{x}{x^2+1}\space\text{d}x+\int\frac{1}{x-1}\space\text{d}x-\int\frac{1}{(x-1)^2}\space\text{d}x=$$
For the intergand $\frac{x}{x^2+1}$, substitute $u=x^2+1$ and $\text{d}u=2x\space\text{d}x$:
$$\int\frac{1}{x^2+1}\space\text{d}x-\frac{1}{2}\int\frac{1}{u}\space\text{d}u+\int\frac{1}{x-1}\space\text{d}x-\int\frac{1}{(x-1)^2}\space\text{d}x=$$
$$\arctan(x)-\frac{1}{2}\int\frac{1}{u}\space\text{d}u+\int\frac{1}{x-1}\space\text{d}x-\int\frac{1}{(x-1)^2}\space\text{d}x=$$
For the intergand $\frac{1}{x-1}$, substitute $s=x-1$ and $\text{d}s=\text{d}x$:
$$\arctan(x)-\frac{1}{2}\int\frac{1}{u}\space\text{d}u+\int\frac{1}{s}\space\text{d}s-\int\frac{1}{(x-1)^2}\space\text{d}x=$$
For the intergand $\frac{1}{(x-1)^2}$, substitute $p=x-1$ and $\text{d}p=\text{d}x$:
$$\arctan(x)-\frac{1}{2}\int\frac{1}{u}\space\text{d}u+\int\frac{1}{s}\space\text{d}s-\int\frac{1}{p^2}\space\text{d}p=$$
$$\arctan(x)-\frac{\ln\left|u\right|}{2}+\ln\left|s\right|+\frac{1}{p}+\text{C}=$$
$$\arctan(x)-\frac{\ln\left|x^2+1\right|}{2}+\ln\left|x-1\right|+\frac{1}{x-1}+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how many $7$ digit numbers can be formed using $1,2,3,4,5,6,7,8,9,0$ How many seven digit numbers can be made if
$(a)$ they must be odd and repetition is not allowed
$(b)$ they must be even and repetition is not allowed
$0532129$ is not a seven digit number
So the question is asking, how many 7 digit numbers can be made if zero can't be the first number but it can be the last one.. am i interpreting this correctly ?
That means there are $5$ evens ${2,4,6,8,0}$ and $5$ odds ${1,3,5,7,9}$.
So for both odds and evens, would it be $6 \times 5 \times 4 \times 5 \times 3 \times 2 \times 1 = 3600$ ?
Any help is appreciated.
| You have 10 numerals. $0,1,2,3,4,5,6,7,8,9$ and want to find the number of orderings of 7 of them so that $0$ is not first, and either $S_O=\{1,3,5,7,9\}$ or $S_E=\{0,2,4,6,8\}$ comes last.
Say we want 0 to come last. We can choose any $6$ of the 9 remaining digits and form a permutation of them, then stick 0 at the end. So there are ${9 \choose 6}\cdot 6!$ valid numbers with 0 last.
Say we want a nonzero numeral to come last. We have $8$ choices for the first number (can't be 0 or the last numeral). To fill in the rest of the number we can choose any 5 of the remaining 8 numerals and form an ordering of them to get: $8\cdot {8 \choose 5}\cdot 5!.$
Thus the number of 7-digit even numbers is ${9 \choose 6}\cdot 6! + 4\cdot 8\cdot {8 \choose 5}\cdot 5!.$
The number of 7-digit odd numbers is $5\cdot 8\cdot {8 \choose 5}\cdot 5!.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of all positive integer ordered pair $(n,r)$ for which $\binom{n}{r} = 2016$
$(1)$ Evaluation of all positive integer ordered pair $(n,r)$ for which $\displaystyle \binom{n}{r} = 120$
$(2)$ Evaluation of all positive integer ordered pair $(n,r)$ for which $\displaystyle \binom{n}{r} = 2016$
$\bf{My\; Try::}$ Here $r=1$ and $n=120\;,$ Then $\displaystyle \binom{120}{1} = 120$
So $\displaystyle (n,r) = (120,1)\;\;,(120,119)$ (Using $\displaystyle \binom{n}{r} = \binom{n}{n-r}$)
Here $\displaystyle \binom{9}{3} = 84$ and $\displaystyle \binom{9}{4} = 126.$
So using Triangular numbers, Here $\displaystyle \binom{n}{r} = 120$ is valid, when $n>9$
Now when $r=2$ and $n=16\;,$ Then $\displaystyle \binom{16}{2} = 120$
So $\displaystyle (n,r) = (16,2)\;\;,(16,14)$
Now we will find when $\displaystyle \binom{n}{r}$ is an Increasing function.
So $\displaystyle \frac{\binom{n}{r+1}}{\binom{n}{r}}\geq 1\Rightarrow \frac{n-r}{r+1}\geq 1\Rightarrow r\leq \frac{n-1}{2}$
So If $r\geq 3\;,$ and $n>9\;,$ Then $\displaystyle 120=\binom{n}{r}>\binom{n}{3} = \frac{n(n-1)(n-2)}{6}>\frac{(n-2)^3}{6}$
So $\displaystyle (n-2)^3<720<(9)^3\Rightarrow n<11$
So we have $n=10$ and $r=3\;,$ So we get $\displaystyle \binom{10}{3} = \binom{10}{7} = 120$
So we get $\displaystyle (n,r) = (10,3) = (10,7)$
So we get Total positive integer ordered pairs $$\displaystyle (m,n) = \left\{(120,1)\;\;,(120,119)\;\;,(16,2)\;\;,(16,8)\;\;,(10,3)\;\;,(10,7)\right\}$$
But i did not understand How can I solve $(2)$ one.
Although we know that $\displaystyle \binom{2016}{1} = \binom{2016}{2015} = 2016$
Help me, Thanks
| I wrote a simple program
to search all
$\binom{n}{k}$
for
$n < 2016$.
The only solutions were the ones you found.
For a more sophisticated search,
I could use the factorization
$2016=2^53^27$,
to limit the possible values
of $n$ and $k$,
but not now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $3x(1-x^2)y^2\frac{dy}{dx}+(2x^2-1)y^3=ax^3$ I am solving this linear Differential equation which can be easily solve by using the formulas for the Bernoulli's Equations
I have solved till
$$\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)y^2}=\frac{ax^3}{3x(1-x^2)y^2}$$
$$y^2\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)}=\frac{ax^3}{3x(1-x^2)}$$
Substituting $y^3=t$
so the equation will be
$$\frac{1}{3}\frac{dt}{dx}+\frac{(2x^2-1)t}{3x(1-x^2)}=\frac{ax^3}{3x(1-x^2)}$$
after this the integrating factor is
$$\frac{1}{x\sqrt{1-x^2}}$$
But I am unable to solve it forward.
| Notice, multiply the integration factor both the sides of the D.E. as follows $$\frac 13\frac{dt}{dx}\frac{1}{x\sqrt{1-x^2}}+\frac{(2x^2-1)t}{3x(1-x^2)}\frac{1}{x\sqrt{1-x^2}}=\frac{ax^3}{3x(1-x^2)}\frac{1}{x\sqrt{1-x^2}}$$
$$\frac{d}{dt}\left(\frac{t}{x\sqrt{1-x^2}}\right)=\frac{ax}{(1-x^2)^{3/2}}$$
$$\int d \left(\frac{t}{x\sqrt{1-x^2}}\right)=\int \frac{ax}{(1-x^2)^{3/2}}\ dx $$
$$\frac{t}{x\sqrt{1-x^2}}=-\frac a2\int \frac{d(1-x^2)}{(1-x^2)^{3/2}}$$
$$\frac{t}{x\sqrt{1-x^2}}=-\frac a2 \frac{(1-x^2)^{-1/2}}{-1/2}+C$$
$$\frac{y^3}{x\sqrt{1-x^2}}= \frac{a}{\sqrt{1-x^2}}+C$$
$$y=(ax+Cx\sqrt{1-x^2})^{1/3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$ by algebraic methods I was trying to solve this equation without using calculus.
Is it possible to be solved by elementary algebraic methods?
$$\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$$
| powering by $3$ we get
$$7x+19+3\sqrt[3]{7x+19}^2\sqrt[3]{7x-19}+3\sqrt{7x+19}(\sqrt[3]{7x-19})^2+7x-19=2$$
and we get
$$14x+3\sqrt[3]{7x+19}\sqrt[3]{7x-19}(\sqrt[3]{7x+19}+\sqrt[3]{7x-19})=2$$
$$14x+3\sqrt[3]{7x+19}\sqrt[3]{7x-19}\sqrt[3]{2}=2$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
derivative of arctan(u) Im trying to find the derivative of $\arctan(x-\sqrt{x^2+1})$ here are my steps if someone could point out where I went wrong.
$$\begin{align}
\frac{\mathrm d~\arctan(u)}{\mathrm d~x} \;& =\; {1\over{1+u^2}}\cdot \frac{\mathrm d~u}{\mathrm d~x}
\\[1ex]
& =\; {1-{x\over{\sqrt{x^2+1}}}\over{1+(x-\sqrt{x^2+1})^2}}
\end{align}$$
Everything after this turns into a huge mess I don't know how to simplify. Is there a trick I missed or something I don't see?
| It seems about right.
$$\begin{align}
\frac{\mathrm d~\arctan(u)}{\mathrm d~x} \;& =\; {1\over{1+u^2}}\cdot \frac{\mathrm d~u}{\mathrm d~x}
\\[1ex]
& =\; {1-{x\over{\sqrt{x^2+1}}}\over{1+(x-\sqrt{x^2+1})^2}}
\\[1ex]
& =\; \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}\;(1+x^2-2x\sqrt{x^2+1}+x^2+1)}
\\[1ex]
& =\; \frac{\sqrt{x^2+1}-x}{2\sqrt{x^2+1}\;(1+x^2-x\sqrt{x^2+1})}
\\[1ex]
& =\; \frac{(\sqrt{x^2+1}-x)\sqrt{x^2+1}\;(1+x^2+x\sqrt{x^2+1})}{2(x^2+1)\;((1+x^2)^2-x^2(x^2+1))}
\\[1ex]
& =\; \frac{(x^2+1-x\sqrt{x^2+1})\;(1+x^2+x\sqrt{x^2+1})}{2(x^2+1)\;((1+x^2)^2-x^2(x^2+1))}
\\[1ex]
& =\; \frac{((1+x^2)^2-x^2(x^2+1))}{2(x^2+1)\;((1+x^2)^2-x^2(x^2+1))}
\\[2ex]
& =\; \frac{1}{2(x^2+1)}
\end{align}$$
[edit]okay, turns out it is a neat derivative[/edit]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
$\sin2(x) - \tan(x) = 0$ , solve for $-180\le x\le 180$ I have been unable to solve the following question,
If $$\sin(2x) - \tan(x) = 0$$
Find $x$ , $-\pi\le x\le \pi$
So far my workings have been
Use following identity:
$$\sin(2x) = 2\sin(x)\cos(x)\\2\sin(x)\cos(x) - \tan(x) = 0\\2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0\\
2\frac{\sin(x)\cos(x)}{1} - \frac{\sin(x)}{\cos(x)} = 0$$
Then cross multiply to give :
$$-\sin x+((2\cos(x)\sin(x))\cos(x))/\cos(x) = 0$$
$$-\sin x+(2\cos^2(x)\sin(x))/ \cos(x) = 0$$
However, I have been unable to get any further.
If someone could help me find a solution to this question it would be very much appreciated.Thank you.
| Once you got to
$$2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0$$
you can pull out a factor of $\sin (x)$ to get
$$\sin(x)\left[2\cos(x)- \frac{1}{\cos(x)}\right]$$
Now, either $\sin(x) = 0$ or $2\cos(x)- \frac{1}{\cos(x)} = 0$. For $\sin(x) = 0$, we have $-180$, $0$, and $180$. For $2\cos(x)- \frac{1}{\cos(x)} = 0$, we can multiply through by $\cos(x)$ to get $2\cos^2(x)-1 = 0$. Solving gives $$\cos(x) = \frac{\sqrt{2}}{2}$$
so x = -45 and 45.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve a linear system of equation involving some recursion $$
\begin{align*}
x_{1} &= 1 + x_{2}\\
x_{2} &= 1 + \frac{1}{2} x_{3} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{i} &= 1 + \frac{1}{2} x_{i+1} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{n-2} &= 1 + \frac{1}{2} x_{n-1} + \frac{1}{2} x_{1}\\
x_{n-1} &= 1 + \frac{1}{2} x_{n} + \frac{1}{2} x_1 \\
x_{n} &= 0 \\
\end{align*}
$$
EDIT: I found one way to solve this. It's simply plugging successive $x_i$ into the first equation. When you reach $x_n$, since $x_n = 0$, you end up with an equation of just $x_1$ which you can solve (using geometric sum) to get $x_1 = 3\times 2^{n-2} - 2$. The rest then is easy.
If you have a more elegant solution, please share.
| I claim that $x_{n-k}=\left(1-\frac{1}{2^k}\right)\left(2+x_1\right)$. To see this, compute the first few terms of the "backwards" sequence $x_{n},x_{n-1},x_{n-2},...$
\begin{align}
x_n&=0 \\
x_{n-1}&=1+\frac{1}{2}\left(1+\frac{1}{2}x_1\right)+\frac{1}{2}x_1 \\
&=\left(1+\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{4}\right)x_1 \\
x_{n-2}&=1+\frac{1}{2}\left(\left(1+\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{4}\right)x_1\right)+\frac{1}{2}x_1 \\
&=\left(1+\frac{1}{2}+\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)x_1 \\
\end{align}
These are very clearly geometric sums in $r=\frac{1}{2}$. Applying the formula $S(n)=\frac{1-r^n}{1-r}$, I get the formula $2\left(1-\frac{1}{2^n}\right)$. To verify, it gives the sequence $0,1,\frac{1}{2},...$, which is what we want.
So, we can write:
\begin{align}
x_{n-k}&=2\left(1-\frac{1}{2^k}\right)+\left(1-\frac{1}{2^k}\right)x_1 \\
&=\left(1-\frac{1}{2^k}\right)(2+x_1)
\end{align}
Now that we have our general formula, we can calculate $x_1$:
\begin{align}
x_1&=x_{n-(n-1)}=\left(1-\frac{1}{2^{n-1}}\right)(2+x_1) \\
0&=2\left(1-\frac{1}{2^{n-1}}\right)+x_1\left(1-1-\frac{1}{2^{n-1}}\right)\\
\frac{x_1}{2^{n-1}}&=2\left(1-\frac{1}{2^{n-1}}\right)\implies x_1=2^n-2 \\
x_{n-k}&=\left(1-\frac{1}{2^k}\right)(2+2^n-2)=2^{n}-2^{n-k}
\end{align}
From this and from our formula for $x_1$ (and the fact that $x_n=0$), it seems clear that $x_i=2^n-2^i$, giving us our solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$
How to start the proof by induction? Is there any way to show this?
| How to start a proof by induction?
What a straight line!
You start at the beginning.
...
Then you show that the middle flows, and conclude that the end is inevitable.
But seriously... You start with $n = 2$.
Prove that $(1 - 1/4) = (2+1)/2*2$.
That's the initial or base step.
Then you do the induction step. You prove that if $(1-1/4)....(1-\frac{1}{k^2}) = \frac{k+1}{2*k}$ for some $k$, then it follows that $(1-1/4) .....(1-\frac{1}{(k+1)^2}) = \frac{k + 2}{2(k+1)}$
And then you are done. It follows that if it is true for $n =2$, and you've proved that if it is true for $n = k$ then it must be true for $n = k + 1$, you have proven it is true for all $n \ge 2$.
The heart is showing that if $(1-1/4)....(1-1/k^2 = (k+1)/2*k$ for some $k$, then it follows that $(1-1/4) .....(1-\frac{1}{(k+1)^2}) = \frac{k + 2}{2(k+1)}$
For that note $(1-1/4) .....(1 - \frac{1}{k^2})(1-\frac{1}{(k+1)^2}) = [![![![![(1-1/4) .....(1 - \frac{1}{k^2})]!]!]!]!](1-\frac{1}{(k+1)^2}) = [\frac{k+1}{2*k}][1-\frac{1}{(k+1)^2})]$
Can you prove that that equals $\frac{k + 2}{2(k+1)}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)( x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$$
$$( x + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)( x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)$$
He then mltiplied both pairs,resulting in:
$$(x^2 - \sqrt{2}x + 1) and (x^2 + \sqrt{2}x + 1)$$
How do I get the first four imaginary points, or the two distinct because I realize that each has a conjugate.
| You can for exemple complete the square $$x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2-2x^2 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1).$$
Now you can use the traditionnal method for each factor.
An other possibility is the used the polar form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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why does matlab give me a negative number? I have the following problem
A steel company has four different types of scrap metal (called Typ-1 to Typ-4) with
the following compositions per unit of volume
They need to determine the volumes to blend from each type of scrap metal so that
the resulting mixture has the following amounts of Al, Si, C, and Fe:
Al - 9 , Si - 6 , C - 8.0 , Fe - 190.0.
Determine volumes of each the 4 scrap metals Typ-1, Typ-2, Typ-3, Typ-4 to blend in
order to reach the desired amounts of Al, Si, C, Fe. Formulate the system and solve it
using matlab (’back-slash’).
However when I go to Matlab, I am using Octave, and enter the following commands
m = [5 3 4 88; 7 6 5 82; 2 1 3 94; 1 2 1 96]
R = [9;6;8;190]
m\R
ans =
22.6980
18.6980
-90.7020
2.2980
How can the amount for type3 be negative?
| Goal
The matrix $\mathbf{A}$ is a map between the $m=4$ material types and $n=4$ metals.
The fraction of the $m=4$ types (I-IV) is $f$. The relative abundance or parts of the $n=4$ elements is $p$.
Blend $m=4$ type to get a material with composition $a$.
Data
$$
%
\begin{align}
%
\mathbf{A} f &= p \\
%
\left(
\begin{array}{rrrr}
5 & 7 & 2 & 1 \\
3 & 6 & 1 & 2 \\
4 & 5 & 3 & 1 \\
88 & 82 & 94 & 96 \\
\end{array}
\right)
%
\left(
\begin{array}{c}
f_{1} \\
f_{2} \\
f_{3} \\
f_{4} \\
\end{array}
\right)
%
&=
%
\left(
\begin{array}{r}
9 \\
6 \\
8 \\
190 \\
\end{array}
\right)
%
=
\left(
\begin{array}{r}
9 \text{ parts Al} \\
6 \text{ parts Si} \\
8 \text{ parts C } \\
190 \text{ parts Fe} \\
\end{array}
\right)
%
\end{align}
%
$$
Solution
$$
%
\begin{align}
%
f &= \mathbf{A}^{-1} p \\
%
&=
\frac{1}{500}
\left(
\begin{array}{rrrr}
607 & -393 & -493 & 7 \\
-255 & 245 & 245 & -5 \\
-403 & 97 & 497 & -3 \\
56 & 56 & -244 & 6 \\
\end{array}
\right)
%
\left(
\begin{array}{r}
9 \\
6 \\
8 \\
190 \\
\end{array}
\right) \\
%
&=
\frac{1}{500}
\left(
\begin{array}{r}
491 \\ 185 \\ 361 \\ 28 \\
\end{array}
\right)
%
\end{align}
%
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show $p(x)$ is a primitive polynomial First the definition:
Polynomial $q(x) \in \mathbb{Z}_p[x]$ of degree $n$ is called primitive, iff:
*
*$q(x) \mid x^{p^n-1}-1$
*$\forall k : 1 \leq k \leq p^{n}-1$ : $q(x) \nmid x^k - 1$
Now the polynomial from my exam, where I should show that it is primitive:
$p(x)=x^6+x^5+x^2+x+1, \quad p(x) \in \mathbb{Z}_2$
So in this case $n=6$ and $p=2$, hence I need to check $1+(2^6-1)=64$ cases. How is this feasible to do on the exam, where I have just few minutes for each task without access to computer? Am I missing some "trick" to show that $p(x)$ is primitive?
| First we show that $p$ is irreducible. If it were not, it would have a root in $F_2$, $F_4$ or $F_8$.
If $a$ were a root in $F_4$, we'd have $a^3 = 1$, hence $0 = a^6 + a^5 + a^2 + a + 1 = a$, which is impossible.
If $a$ were a root in $F_8$, we'd have $a^7 = 1$, hence $0 = a^8 + a^7 + a^4 + a^3 + a^2 = a^4 + a^3 + a^2 + a + 1$. Multiplying by $a - 1$, we find $a^5 = 1$. Combining this with $a^7 = 1$, it follows that $a = 1$, which is absurd.
Since $p$ is irreducible, its roots are distinct and of degree $6$ over $F_2$. They therefore belong to $F_{64}^{*},$ so they satisfy $x^{63} = 1$ by Lagrange's theorem. This proves that $p(x)$ divides $x^{63} - 1$. Moreover, since the roots are conjugate, they all have the same order.
Because the roots have order dividing $63$, the only thing to do now is to check that they do not satisfy $x^9 = 1$ or $x^{21} = 1$. The first part results from the fact that $p(x)$ is irreducible and $x^9 - 1 = (x^3 - 1)(x^6 + x^3 + 1)$. For the second part, we have the factorization $x^{21} - 1 = (x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)(x^{14} + x^7 + 1)$. At this point, I don't see anything better to do than to divide $x^{14} + x^7 + 1$ by $p(x)$ by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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integrate $\int\frac{\sin x}{1+\sin^{2}x}dx$
$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}$$
$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}=\int\frac{\sin x}{2-\cos^{2}x}\mathrm {dx}$$
$u=\cos x$
$du=-\sin x dx$
$$-\int\frac{\mathrm {du}}{2-u^{2}}$$
How could I continue?
| Notice, factorize & use partial fractions as follows $$-\int \frac{du}{2-u^2}=-\int \frac{du}{(\sqrt2-u)(\sqrt2+u)}$$
$$=-\frac{1}{2\sqrt 2}\int \left(\frac{1}{\sqrt 2-u}+\frac{1}{\sqrt 2+u}\right)\ du$$
$$=-\frac{1}{2\sqrt 2} \left(-\ln|\sqrt 2-u|+\ln|\sqrt2+u|\right)+C$$
$$=-\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2+u}{\sqrt 2-u}\right|+C$$
$$=\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2-u}{\sqrt 2+u}\right|+C$$
$$=\color{red}{\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2-\cos x}{\sqrt 2+\cos x}\right|+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Probability of an even number of sixes
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.
I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:
*
*The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.
*Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.
For $n=k+1$, there are two ways the number of sixes are even:
a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n]$
b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
I have two questions
*
*How do I get from $ \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$
*Other than that, is my use of induction correct? Is it rigorous enough to prove the formula?
| Your method is correct but the induction hypothesis should be
$$\text{Probabaility of an even number of sixes when you roll } k \text{ dice} = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^k\right]$$
(you have $n$ instead of $k$) and the thing you want to prove is
$$\text{Probabaility of an even number of sixes when you roll } k+1 \text{ dice} = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^{k+1}\right]$$
and so, using your reasoning, you want to prove that
$$\frac{5}{6}\cdot \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^k\right] +
\frac{1}{6}\left(1- \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^{k}\right]\right) = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^{k+1}\right]$$
[Also, two footnotes: odds isn't the same as probability, so the words shouldn't really be used interchangeably. And $0$ is even! Whoever wrote the question is probably a bit shaky in number theory...]
| {
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"url": "https://math.stackexchange.com/questions/1648601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Prove $ \frac{4cos^{2}(2x)-4cos^2(x)+3sin^2(x)}{4cos^2(\frac{5\pi}{2} - x) -sin^22(x-\pi)} = \frac{8cos(2x)+1}{2(cos(2x)-1)}$
Question: Prove
$$ \frac{4cos^{2}(2x)-4cos^2(x)+3sin^2(x)}{4cos^2(\frac{5\pi}{2} - x) -sin^22(x-\pi)} = \frac{8cos(2x)+1}{2(cos(2x)-1)}$$
My attempt starting with the bottom line on the LHS
$$ {4cos^2(\frac{5\pi}{2} - x) -sin^22(x-\pi)} $$
$$ sin(x) = (\frac{\pi}{2} - x) $$
$$ 4sin^2(x) - sin^2(2x-2\pi)$$
$$ 4sin^2(x) - sin^2(2x)cos^2(2\pi) + cos^2(2x)sin^2(2\pi)$$
$$ 4sin^2(x) - sin^2(2x) $$
$$ 4( \frac{1}{2} - \frac{1}{2}cos(2x)) - (1-cos^2(2x)) $$
$$ 2 - 2cos(2x) - 1 + cos^2(2x) $$
$$ cos^2(2x) - 2cos(2x) + 1 $$
$$ (cos(2x) - 1)(cos(2x) - 1) $$
So I think I have simplified it as much as possible here
Now the top line
$$ {4cos^{2}(2x)-4cos^2(x)+3sin^2(x)} $$
$$ 4cos^2(2x) - 4cos^2(x) + 3(1-cos^2(x))$$
$$ 4cos^2(2x) - 4cos^2(x) + 3-3cos^2(x) $$
$$ 4cos^2(2x) - 7 cos^2(x) + 3 $$
Now I am stuck...
| In the numerator,
$$ 4\cos^2(2x) - 7 \cos^2(x) + 3 $$
As suggested in the comments, use,
$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$
Thus
$$ 4\cos^2(2x) - 7 \cos^2(x) + 3= 4\cos^2(2x)-\frac72-\frac72\cos(2x)+3=\frac12(8\cos^2(2x)-7\cos(2x)-1)=\frac12(8\cos(2x)+1)(\cos(2x)-1)$$
Dividing this with the expression for the denominator you have got gives the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equation of tangent at origin to the curve $y^2=x^2(1+x+x^2)$ How do I find the equation of tangent at $(0,0)$ to the curve $y^2=x^2(1+x+x^2)$ ?
Differentiating and putting the value of $x$ and $y$ gives an indeterminate form.
Can we trace the curve and geometrically make tangents and find their equation ?
| $$y^2=x^2(1+x+x^2)\implies 2yy'=2x(1+x+x^2)+x^2(1+2x)=4x^3+3x^2+2x\implies$$
$$y'(0)=\lim_{x\to 0}\frac{4x^3+3x^2+2x}{2y}=\pm\lim_{x\to0}\frac{4x^2+3x+2}{2\sqrt{1+x+x^2}}=\pm1$$
and thus the tangent doesn't exist since $\;y=|x|\sqrt{1+x+x^2}\;$ and $\;\frac x{|x|}=\pm1\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this indefinite integral using integral substitution? So while working on some physics problem for differential equations, I landed at this weird integral $$ \int \frac 1 {\sqrt{1-\left(\frac 2x\right)}}\,dx $$
So since there is a square root, I thought I could use trig substitution, but I couldn't find anything that works out. How can one solve this integral in a nice simple manner? If you can solve it in a different way, it is still fine. $Thank$ $you!$
This is the answer given to me by symbolab $$ 4\left(-\frac{1}{4\left(\sqrt{1-\frac{2}{x}}-1\right)}-\frac{1}{4\left(\sqrt{1-\frac{2}{x}}+1\right)}-\frac{1}{4}\ln \left|\sqrt{1-\frac{2}{x}}-1\right|+\frac{1}{4}\ln \left|\sqrt{1-\frac{2}{x}}+1\right|\right)+C $$
| Some hints
First of all, simple, thing, just rewrite the integrand this way
$$\frac{1}{\sqrt{\frac{x-2}{x}}}$$
Then substitute
$$y = \frac{x-2}{x} ~~~~~~~ \text{d}y = \frac{1}{x} - \frac{x-2}{x^2}\ \text{d}x$$
You're now with
$$2\int \frac{1}{(1 + y^2)\sqrt{y}}\ \text{d}y$$
And for this one, use
$$z = \sqrt{y} ~~~~~~~ \text{d}z = \frac{1}{2\sqrt{y}}\ \text{d}y$$
and your integral is now:
$$4\int \frac{1}{(1 - z^2)^2}\ \text{d}z$$
which is easily solvable by partial fractions:
$$\frac{1}{(1 - s^2)^2} = \frac{1}{4(z+1)} + \frac{1}{4(z+1)^2} - \frac{1}{4(z-1)} + \frac{1}{4(z-1)^2}$$
Split and do the easy integrals.
Final Result
$$x\sqrt{\frac{x-2}{x}} - \ln\left(\sqrt{\frac{x-2}{x}} - 1\right) + \ln\left(\sqrt{\frac{x-2}{x}} + 1\right) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coordinates of a matrix So on the textbook, it gives an example:
If the basis of B matrix is{$\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \begin{bmatrix}0&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\ 1&1\end{bmatrix}, \begin{bmatrix}1&0\\ 1&0\end{bmatrix}$}
Then the B-coordinates of a matrix $\begin{bmatrix}a&b\\ c&d\end{bmatrix}$ are $\begin{bmatrix}a-c+d\\ b\\d\\c-d\end{bmatrix}$
But can I write the B-coordinates as $\begin{bmatrix}a-d\\ b\\c-d\\d\end{bmatrix}$? I found it much easier to write the column matrix in this way, but is it correct?
| The coordinates of a vector with respect to an ordered basis are unique so there is only one correct answer. In your case,
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} = (a + d - c) \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} + (c - d) \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} $$
so
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}_{B} = \left[ \begin{matrix} a + d - c \\ b \\ d \\ c - d \end{matrix} \right]. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real Analysis proof, show there is a limit and find it. Assume$${ x }_{ 1 }=\sqrt { 3 } ,{ x }_{ 2 }=\sqrt { 3+\sqrt { 3 } } ,..,{ x }_{ n }=\sqrt { 3+\sqrt { 3+..+\sqrt { 3 } } } $$this sequence has a limit and find the limit.
I know I need to show that is is bounded and then show $x_{ n }<{ x }_{ n+1 }$ before finding the limit but have no idea how to go about it. Please help!
| We have the recurrence relationship
$$x_{n+1}=\sqrt{3+x_n}$$
with $x_1=\sqrt{3}$.
We propose that $\lim_{n\to \infty}x_n=\frac{1+\sqrt{13}}{2}$.
First note that $x_n>0$.
Second, we observe that if $x_{n}<\frac{1+\sqrt{13}}{2}$, then $x_{n+1}<\frac{1+\sqrt{13}}{2}$ also.
Therefore, for $x_1=\sqrt3<\frac{1+\sqrt{13}}{2}$, the sequence $x_n$ is bounded above by $\frac{1+\sqrt{13}}{2}$.
Third, note that for $0<x_n<\frac{1+\sqrt{13}}{2}$
$$\begin{align}
x_{n+1}-x_n&=\sqrt{3+x_n}-x_n\\\\
&=-\frac{x_n^2-x_n-3}{\sqrt{3+x_n}+x_n}\\\\
&=-\frac{\left(x_n-\frac{1+\sqrt{13}}{2}\right)\left(x_n-\frac{1-\sqrt{13}}{2}\right)}{\sqrt{3+x_n}+x_n}\\\\
&>0
\end{align}$$
Therefore, the sequence $x_n$ is increasing monotonically and bounded above. Therefore, the limit $\lim_{n\to \infty}x_n$ exists and is finite.
Let $L$ represent the limit. Then,
$$\begin{align}
L&=\lim_{n\to \infty}x_{n+1}\\\\
&=\lim{n\to \infty}\sqrt{3+x_n}\\\\
&=\lim{n\to \infty}\sqrt{3+\lim{n\to \infty}x_n}\\\\
&=\sqrt{3+L} \tag 1
\end{align}$$
whereupon equating the left-hand and right-hand sides of $(1)$ reveal that $L=\frac{1+\sqrt{13}}{2}$ as was proposed!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$ Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
Let $I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
I took $\sqrt{2-x-x^2}$ as first function and $\frac{1}{x^2}$ as the second function and integrated it by parts,
$I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx=\sqrt{2-x-x^2}\int\frac{1}{x^2}dx-\int\frac{-1-2x}{2\sqrt{2-x-x^2}}\times\frac{-1}{x}dx$
$=\sqrt{2-x-x^2}\times\frac{-1}{x}-\frac{1}{2}\int\frac{1+2x}{x\sqrt{2-x-x^2}}dx$
Now i put $x=\frac{1}{t}$ in the integral $\int\frac{1+2x}{x\sqrt{2-x-x^2}}dx$ to get
$I=\sqrt{2-x-x^2}\times\frac{-1}{x}+\frac{1}{2}\int\frac{(t+2)dt}{t\sqrt{2t^2-t-1}}$
I do not know how to solve it further.
| Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int \sqrt{2-x-x^2}\cdot \frac{1}{x^2}dx\;, $$ Now Using Integration by parts
$$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\int\frac{1+2x}{2\sqrt{2-x-x^2}}\cdot \frac{1}{x}dx $$
So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\underbrace{\int\frac{1}{\sqrt{2-x-x^2}}dx}_{J}-\underbrace{\int\frac{1}{x\sqrt{2-x-x^2}}dx}_{K}$$
So for Calculation of $$\displaystyle J = \int\frac{1}{\sqrt{2-x-x^2}}dx = \int\frac{1}{\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{2x+1}{2}\right)^2}}dx$$
Now Let $\displaystyle \left(\frac{2x+1}{2}\right)=\frac{3}{2}\sin \phi\;,$ Then $\displaystyle dx = \frac{3}{2}\cos \phi d\phi$
So we get $$\displaystyle J = \int 1d\phi = \phi+\mathcal{C_{1}} = \sin^{-1}\left(\frac{2x+1}{3}\right)+\mathcal{C}$$
Similarly for calculation of $$\displaystyle K = \int \frac{1}{x\sqrt{2-x-x^2}}dx$$
Put $\displaystyle x=\frac{1}{u}$ and $\displaystyle dx = -\frac{1}{u^2}dt$
So we get $$\displaystyle K = -\int\frac{1}{\sqrt{2u^2-u-1}}du = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\left(u-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}}dx$$
So we get $$\displaystyle J = -\frac{\sqrt{2}}{3}\ln\left|\left(u-\frac{1}{4}\right)+\sqrt{\left(u-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C_{2}}$$
So we get $$\displaystyle J = -\frac{\sqrt{2}}{3}\ln\left|\left(\frac{1}{x}-\frac{1}{4}\right)+\sqrt{\left(\frac{1}{x}-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C_{2}}$$
So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\sin^{-1}\left(\frac{2x+1}{3}\right)+\frac{\sqrt{2}}{3}\ln\left|\left(\frac{1}{x}-\frac{1}{4}\right)+\sqrt{\left(\frac{1}{x}-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The sum of the least common multiple and greatest common divisors of the product Find number of solutions in positive integers the equation
$$[a^2,b^2]+[b^2,c^2]+[c^2,a^2]=(a^2,b^2)(b^2,c^2)(c^2,a^2),$$
where $[m,n] -$ least common multiple, $(m,n) -$ greatest common divisors of positive integers $m$ and $n$.
My work so far:
Let $a=px, b=py, c=pz$, where $p,x,y,z -$ pairwise relatively prime numbers. Then
$$(pxy)^2+(pyz)^2+(pzx)^2=(p^2)^3$$
$$x^2y^2+y^2z^2+z^2x^2=p^4$$
$$x^2(y^2+z^2)=(p^2-yz)(p^2+yz)$$
If $p^2=y^2+yz+z^2$, then
$$x^2(y^2+z^2)=(y^2+2yz+z^2)(y^2+z^2)$$
$$x=y+z$$
Under such conditions, we have a solution
$$(p(y+z);py,pz).$$
Answer: Can we show that there are infinitely many triples of positive integers $(p,y,z)$ for which the equality $p^2=y^2+yz+z^2$?
| The first step isn't possible: $p$ must be the gcd, and you cannot then require that $x$, $y$, and $z$ be coprime, e.g., if $a=6$, $b=10$, $c=15$.
If you write $a$ as $dABC$, where $d$ is $gcd(a,b,c)$, $A$ is $gcd(a/d,b/d)$, and $B$ is $gcd(a/d,c/d)$, and $b$ and $c$ in similar fashion, you should be able to cancel things, and there don't seem to be any solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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integration of $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}*\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{{1+x}+2\sqrt{({1+x})*({1-x})}+1-x}{{1+x}-{1+x}}=\frac{{2}+2\sqrt{1-x^2}}{2x}=\frac{2(1+\sqrt{1-x^2})}{2x}=\frac{(1+\sqrt{1-x^2})}{x}$$
$$\int \frac{(1+\sqrt{1-x^2})}{x}=\int \frac{1}{x}+\int\frac{(\sqrt{1-x^2})}{x}=\ln|x|+\int\frac{(\sqrt{1-x^2})}{x}$$
$x=\sin t$
$dx=\cos t \; dt$
$$\int\frac{(\sqrt{1-\sin^2t})}{\sin t}\cos t\;dt=\int \frac{\cos^2t}{\sin t}\;dt =\int \frac{1-\sin^2t}{\sin t}\; dt=\int \frac{1}{\sin t}\;dt -\int \sin t \; dt$$ $$=\ln\left(\tan\frac{t}{2}\right)+\cos t+c$$
How do I substitute t back to x?
| Since $x=sin(t)$ then $cos(t)= \sqrt{1-x^2}$
To deal with the $tan(t/2)$ term use the half angle formulae to put this in terms of $cos(t)$ and $sin(t)$
| {
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} |
Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$
$$\int\frac{\sin(2x)}{2+\cos x} dx $$
Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 \int 1 - 2\frac{1}{u} du = -2 (u - 2\ln u) + C$$
After substituting for $u$, I get $4 \ln(2+ \cos x) - 2\cos x - 4 + C$. According to Wolfram Alpha, the solution is $4 \ln(2+ \cos x) - 2\cos x + C$. Not sure where my mistake is, but any help is greatly appreciated.
| Nothing's wrong ! Put $u=\cos x +2$
$$-2 (u - 2\ln u) + C=4\ln u -2u +\mathrm C =4 \ln(2+ \cos x) - 2\cos x - 4 + \mathrm C.$$
So $$-4+\textrm C = \rm constant $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Relationships between points lines and planes Develop the Cartesian equation of a plane with $x$-intercept $a$, $y$-intercept $b$ and $z$-intercept $c$.
Show that the distance $d$ from the origin to this plane is given by $$\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$
In the picture below I have included what I have done so far. I was able to set all my variables and begin to define them.
| Another attempt, since the OP finds my first answer confusing ...
Let $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$. The normal to the plane is in the direction $N = (B-A)\times(C-A) = (bc, ca, ab)$. A unit vector in this direction is
$$
U = \frac{N}{\|N\|} = \frac{(bc,ca,ab)}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$
The distance $d$ from the origin to the plane is the length of the projection of $\vec{OA}$ onto the vector $U$. Since $U$ is a unit vector, this projected length is just $A \cdot U$. So, we have
$$
d = A \cdot U = \frac{abc}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$
This formula is correct even if one of $a$, $b$, $c$ is zero and the other two are non-zero, in which case it gives $d=0$, as you would expect.
If $a$, $b$, $c$ are all non-zero, this formula can be rearranged to give
$$
\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$ Does this integral converge to any particular value?
$$\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$$
If the answer is yes, how should I calculate its value?
I tried to use convergence tests but I failed due to the complexity of the integral itself.
| Sure it does.
Collect $e^x$ in the denominator and you will get
$$\int_0^{+\infty}\frac{x}{e^{x}(1 + e^{-x})}\ \text{d}x$$
Since the integral range is from $0$ to infinity, you can see the fraction in this way:
$$\int_0^{+\infty}x e^{-x}\frac{1}{1 + e^{-x}}\ \text{d}x$$
and you can make use of the geometric series for that fraction:
$$\frac{1}{1 + e^{-x}} = \frac{1}{1 - (-e^{-x})} = \sum_{k = 0}^{+\infty} (-e^{-x})^k$$
thence thou have
$$\sum_{k = 0}^{+\infty}(-1)^k \int_0^{+\infty} x e^{-x} (e^{-kx})\ \text{d}x$$
Namely
$$\sum_{k = 0}^{+\infty}(-1)^k \int_0^{+\infty} x e^{-x(1+k)}\ \text{d}x$$
This is trivial, you can do it by parts getting
$$\int_0^{+\infty} x e^{-x(1+k)}\ \text{d} = \frac{1}{(1+k)^2}$$
Thence you have
$$\sum_{k = 0}^{+\infty}(-1)^k \frac{1}{(1+k)^2} = \frac{\pi^2}{12}$$
Which is the result of the integration
If you need more explanations about the sum, just tell me!
HOW TO CALCULATE THAT SERIES
There is a very interesting trick to calculate that series. First of all, let's write it with some terms, explicitly:
$$\sum_{k = 0}^{+\infty}\frac{(-1)^k}{(1+k)^2} = \sum_{k = 1}^{+\infty} \frac{(-1)^{k+1}}{k^2} = -\ \sum_{k = 1}^{+\infty}\frac{(-1)^{k}}{k^2}$$
The first terms of the series are:
$$-\left(-1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} - \frac{1}{25} + \frac{1}{36} - \frac{1}{64} + \frac{1}{128} - \cdots\right)$$
namely
$$\left(1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36} + \frac{1}{64} - \frac{1}{128} + \cdots\right)$$
Now let's call that series $S$, and let's split it into even and odd terms:
$$S = \left(1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \cdots\right) - \left(\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + \cdots\right) ~~~~~ \to ~~~~~ S = A - B$$
Where obviously $A$ and $B$ are respectively the odd and even part.
now the cute trick
take $B$, and factorize out $\frac{1}{4}$:
$$B = \frac{1}{4}\left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \cdots\right)$$
Now the series in the bracket is a well known series, namely the sum of reciprocal squares, which is a particular case of the Zeta Riemann:
$$\zeta(s) = \sum_{k = 1}^{+\infty} \frac{1}{k^s}$$
which is, for $s = 2$
$$\zeta(2) = \sum_{k = 1}^{+\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$$
Thence we have:
$$B = \frac{\zeta(2)}{4} = \frac{\pi^2}{24}$$
Are you seeing where we want to go? But this is not enough since we don't know what $A$ is. To do that, we can again split $B$ into even and odd terms! But doing so, we will find again the initial $A$ and $B$ series:
$$B = \frac{1}{4}\left(\left[1 + \frac{1}{9} + \frac{1}{25} + \cdots\right] + \left[\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots\right]\right) ~~~ \to ~~~ B = \frac{1}{4}\left(A + B\right)$$
This means:
$$4B - B = A ~~~~~ \to ~~~~~ A = 3B$$
So
$$A = 3\cdot \frac{\pi^2}{24} = \frac{\pi^2}{8}$$
Not let's get back to the initial series $S$ we wanted to compute, and substitution this we get:
$$S = A - B = \frac{\pi^2}{8} - \frac{\pi^2}{24} = \frac{\pi^2}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Convergence of Infinite Sum Using Mathematica, I have been able to make the following statement based on numerical evidence:
$$\sum_{i=0}^\infty \frac{2^i}{x^i}=\frac{x}{x-2}$$
for any $x≥3$. How can this be proven?
| This can solved from scratch by taking one term out of the summation, changing the index range and refactoring to obtain
$$
\begin{align}
S=
\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i
&=
1+\sum_{i=1}^{\infty} \left(\frac{2}{x}\right)^i \\
&=
1+\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^{i+1} \\
&=
1+\frac{2}{x}\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^{i} \\
\end{align}$$
Now grouping the series terms yields
$$
\begin{align}
\left(1-\frac{2}{x}\right)\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i
&=
1 \\
\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i
&=\frac{1}{1-\frac{2}{x}}=\frac{x}{x-2} \\
\end{align}$$
where the sum converges for $\vert \frac{2}{x} \vert\lt 1$, not only $x \ge 3$.
Are you assuming positive integer $x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the smallest $a\in\mathbb{N}$, so that $a\equiv3\cdot 5^4\cdot 11\cdot 13^3 \pmod 7$ The problem I am trying to solve is:
Find the smallest $a\in\mathbb{N}$, so that $a\equiv3\cdot 5^4\cdot 11\cdot 13^3 \pmod 7$.
I noticed that $3,5,11$ and $13$ are primes, but I've no idea how that is supposed to help me.
The solution is $4$, but why?
| Note $13^{3} \equiv (-1)^{3} \equiv -1 \pmod 7$
Also, note that $5^{4} \equiv 25^{2} \equiv 4^{2} \equiv 2 \pmod 7$.
Thus $3 \times 5^{4} \times 11 \times 13^{3} \equiv 3 \times 2 \times (-3) \times (-1) \equiv 18 \equiv 4 \pmod 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it
and also plz explain me where i have done wrong in above method
Thanks
| Clearly $$(x + 1)(x + 2)(x + 3)(x + 4)(x + 5) = x^{5} + 15x^{4} + \cdots$$ and as we will show later we don't need to bother about the coefficients of $x^{3}, x^{2}, x, x^{0}$ in order to solve this problem.
Let $P(x)$ be a monic polynomial of degree $n$ and let the coefficient of $x^{n - 1}$ in $P(x)$ be $a$ so that $$P(x) = x^{n} + ax^{n - 1} + bx^{n - 2} + \cdots$$ We prove that $$\lim_{x \to \infty}\left\{\sqrt[n]{P(x)} - x\right\} = \frac{a}{n}$$ and thus for our current problem $n = 5, a = 15$ so that the desired limit is $a/n = 3$. Clearly if we set $g(x) = \sqrt[n]{P(x)}$ then we can see that $$\frac{g(x)}{x} = \sqrt[n]{\frac{P(x)}{x^{n}}} = \sqrt[n]{1 + \frac{a}{x} + \cdots } \to 1\text{ as }x \to \infty$$ and hence
\begin{align}
L &= \lim_{x \to \infty}\left\{\sqrt[n]{P(x)} - x\right\}\notag\\
&= \lim_{x \to \infty}\frac{P(x) - x^{n}}{g(x)^{n - 1} + xg(x)^{n - 2} + \dots + x^{n - 1}}\notag\\
&= \lim_{x \to \infty}\frac{ax^{n - 1} + bx^{n - 2} + \cdots}{g(x)^{n - 1} + xg(x)^{n - 2} + \dots + x^{n - 1}}\notag\\
&= \lim_{x \to \infty}\dfrac{a + \dfrac{b}{x} + \cdots}{\left(\dfrac{g(x)}{x}\right)^{n - 1} + \left(\dfrac{g(x)}{x}\right)^{n - 2} + \dots + \dfrac{g(x)}{x} + 1}\notag\\
&= \frac{a + 0 + \cdots}{1 + 1 + \cdots + n \text{ terms}}\notag\\
&= \frac{a}{n}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 5
} |
L'Hopital's Rule Values for an equation I have the following task:
For what values of $a$ and $b$ is the following equation true?
$$\lim\limits_{x \to 0} \left(\frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}\right) = 0 $$
I want to know the steps I should follow in order to find the solution.
| We have
\begin{align}
L
&= \lim_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2} \\
&= \lim_{x \to 0} \frac{\sin(2x) + a x^3 + b x }{x^3}\\
\end{align}
which is a limit of type $0 / 0$, so we try L'Hôpital's rule:
\begin{align}
L &= \lim_{x \to 0} \frac{2\cos(2x) + 3 a x^2 + b}{3x^2}
\end{align}
The denominator again vanishes for $x \to 0$, the nominator goes to
$2 + b$.
So if $b \ne -2$, the nominator does not vanish and we have
$$
\DeclareMathOperator{sgn}{sgn}
L = \sgn(2 + b) \, \infty
$$
For $b = -2$ we again have a limit of type $0 / 0$ and apply the rule again:
\begin{align}
L
&= \lim_{x\to 0} \frac{-4 \sin(2x) + 6ax}{6x}
\end{align}
The nominator and denominator vanish and we apply the rule once again:
\begin{align}
L
&= \lim_{x\to 0} \frac{-8 \cos(2x) + 6a}{6} = \frac{6a-8}{6}
= a - \frac{4}{3}
\end{align}
This gives the answer that the equation is true, $L$ vanishes,
if $a = 4/3$ and $b = -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Confused about infinite sum $\sum\limits_{a,b,c}\frac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}$ $$ \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}= \ ? \ $$
I calculated its value as $\frac{32}{3}$ but I'm not sure whether I'm right or wrong.
This is what I did :
Considering $a$,$b$ and $c$ as identical and independent,
$$ S= \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})} \\ 3S= \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {(2^{a+b}+2^{b+c}+2^{a+c})}×\left[ \frac{1}{2^a}+\frac{1}{2^b}+\frac{1}{2^c} \right] \\ = \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc}{2^a2^b2^c } \\ = \displaystyle \sum^{\infty}_{a=0}\frac{a}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{1}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{1}{2^c}+ \displaystyle \sum^{\infty}_{a=0}\frac{1}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{b}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{1}{2^c}+ \displaystyle \sum^{\infty}_{a=0}\frac{1}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{1}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{c}{2^c}+ \displaystyle \sum^{\infty}_{a=0}\frac{a}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{b}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{c}{2^c} \\ = 2^3+2^3+2^3+2^3 =32 \\ \Rightarrow S=\frac{32}{3} $$
Please tell me if I'm right.
| Seems OK.
You've used
$$\frac{1}{xy+yz+zx} \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)=
\frac{1}{xyz}$$
and
$$\sum_{a=0}^{\infty} \frac{a}{2^{a}}=
\lim_{n\to \infty} \sum_{k=0}^{n} \frac{k}{2^{k}}=
\lim_{n\to \infty} \left(2-\frac{n}{2^{n}}-\frac{1}{2^{n-1}} \right)=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is this expression strictly positive? $n \left(\cos\frac{k\pi}{n}\right)\left(1-\cos\frac{k\pi}{n}\right)-\sin\frac{k\pi}{n}$ Let us define a function $f(k,n)$ by
\begin{equation}
f(k,n)=n \left (\cos\frac{k\pi}{n}\right) \left(1-\cos\frac{k\pi}{n}\right) - \sin \frac{k\pi}{n}
\end{equation}
where $\frac{k}{n}$ is irreducible with $k,n \in \mathbb{N}$, with $k \leq \left \lfloor{n/2}\right \rfloor$, and $k \geq 2,$ $n \geq 5$.
I suspect that $f(k,n) >0$.
Plugging in a few values of $k$ and $n$ and computing $f(k,n)$ numerically indeed shows that, but how do I prove / disprove this?
| Proof: Since $\frac{k}{n}$ is irreducible and $k\le \lfloor \frac{n}{2}\rfloor$, we have $k \le \frac{n-1}{2}$.
Let $u = \tan \frac{k\pi}{2n}$. Since $2\le k \le \frac{n-1}{2}$ and $n\ge 5$, we have
$$0 < \frac{\pi}{n} \le \tan \frac{\pi}{n} \le u \le \tan \Big(\frac{\pi}{4} - \frac{\pi}{4n}\Big)
= \frac{1 - \tan \frac{\pi}{4n}}{1 + \tan \frac{\pi}{4n}} \le
\frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}} < 1.$$
Using $\cos \frac{k\pi}{n} = \frac{1-u^2}{1+u^2}$
and $\sin \frac{k\pi}{n} = \frac{2u}{1+u^2}$, we have
$$n \Big(\cos \frac{k\pi}{n}\Big) \Big(1 - \cos \frac{k\pi}{n}\Big) - \sin \frac{k\pi}{n} =
\frac{2u(-nu^3-u^2+nu-1)}{(u^2+1)^2}.$$
It suffices to prove that $-nu^3-u^2+nu-1 > 0$. To proceed, we need the following fact.
Fact 1: For each positive integer $n$, $f_n(x) = -nx^3 - x^2 + nx - 1$ is concave on $[0, 1]$
since $f_n''(x) = -6nx - 2 < 0$ for $x > -\frac{1}{3n}$.
As a result, for $0\le a \le x \le b \le 1$, we have $f_n(x) \ge \min(f_n(a), f_n(b))$.
Now, according to Fact 1, letting $a = \frac{\pi}{n}$ and $b = \frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}}$,
since $0 < a \le u \le b < 1$, it suffices to prove that
$f_n(\frac{\pi}{n}) > 0$ and $f_n(\frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}}) > 0$.
We have
$f_n(\frac{\pi}{n}) = -\frac{\pi^3}{n^2}+\pi-\frac{\pi^2}{n^2}-1
\ge -\frac{\pi^3}{5^2}+\pi-\frac{\pi^2}{5^2}-1 > 0$ and
\begin{align}
f_n(\frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}}) &= \frac{2n^3}{(4n+\pi)^3}\Big(-\frac{8\pi^2}{n}+32\pi - \frac{\pi^3}{n^3}-\frac{4\pi^2}{n^2}-\frac{16\pi}{n}-64\Big)\\
&\ge \frac{2n^3}{(4n+\pi)^3}\Big(-\frac{8\pi^2}{5}+32\pi - \frac{\pi^3}{5^3}-\frac{4\pi^2}{5^2}-\frac{16\pi}{5}-64\Big)\\
&> 0.
\end{align}
This completes the proof.
| {
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"url": "https://math.stackexchange.com/questions/1671866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?
My Attempt
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0), (-2,0) are solution of the given equation. Same way solving for $f(0,y)=0$ we get $ y=2 , -1$ are roots and hence (0,2), (0,-1) are solutions. I have tried to factorize $f(x,y)$ or writing it as sum of squares but could not succeed. Is there any other solutions? How do I find them?
| Any $x$ that satisfies the equation must be even. Let $x=2z$. Substituting and dividing by $2$ we get
$$2z^2+2y^2-2zy-2z-2y-4=0.$$
We can rewrite this as
$$(z-1)^2+(y-1)^2+(z-y)^2=6.$$
The only way the sum of three squares is $6$ is if the squares are, in some order, $1$, $1$, and $2$. Now it is a matter of examining a small number of cases. For example $z=0$, $y=2$ gives a solution, as does $z=2$, $y=3$, as does $z=3$, $y=2$.
We could cut down on the arithmetic by letting $s=z-1$ and $t=y-1$. Then we are looking at $s^2+t^2+(s-t)^2=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
The following series with a polynomial as numerator and $e^n$ as denominator converge? The series is the following:
$$\sum_{n=1}^\infty \frac {8n^2-7}{e^n(n+1)^2} $$
| You can also use the ratio test $$u_n=\frac {8n^2-7}{e^n(n+1)^2}$$ $$\frac{u_{n+1}}{u_n}=\frac{(n+1)^2 \left(8 (n+1)^2-7\right)}{e (n+2)^2 \left(8 n^2-7\right)}=\frac 1e \, \frac{8 n^4+32 n^3+41 n^2+18 n+1}{8 n^4+32 n^3+25 n^2-28 n-28}$$ the limit of which being $\frac 1e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic identity involving powers of twin primes Yesterday, I verified that, if $a$,$b$ and $c$ are real numbers such that $a+b+c=0$, then
$$\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2}$$
and
$$\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}$$
Then I asked myself: will it be true that
$$\frac{a^{p_n}+b^{p_n}+c^{p_n}}{p_n}=\frac{a^{p_{n-1}}+b^{p_{n-1}}+c^{p_{n-1}}}{p_{n-1}}\cdot\frac{a^2+b^2+c^2}{2}$$
for any pair of primes satifying $p_n=p_{n-1}+2$, i.e., for any pair of twin primes?
For the first two equalities, I wrote $a=-(b+c)$, expanded the powers $-(b+c)^x$ for $x=2,3,5,7$, multiplied the factors at the right hand side and verified that the left and right hand sides were equal. For the general case, I'm trying to do the same, but I'm having some difficulties. For the left hand side we have
$$-\frac{\sum_{i=1}^{p_n-1}{{p_n}\choose{i}}b^ic^{p_n-i}}{p_n}$$
and for the right hand side we have
$$-\frac{\sum_{i=1}^{p_{n-1}-1}{{p_{n-1}}\choose{i}}[b^{i+2}c^{p_{n-1}-i}+b^{i+1}c^{p_{n-1}+1-i}+b^ic^{p_{n-1}+2-i}]}{p_{n-1}}$$
Note that we can cancel the minus sign. For the term $bc^{p_n-1}$ we have at the left hand side
$$\frac{{p_n}\choose{1}}{p_n}bc^{p_n-1}=\frac{p_n-1}{p_n-1}bc^{p_n-1}$$
and at the right hand side
$$\frac{{p_{n-1}}\choose{1}}{p_{n-1}}bc^{p_{n-1}+2-1}=\frac{p_{n-1}-1}{p_{n-1}-1}bc^{p_n-1}$$
Note that $p_n-1=p_{n-1}+1$. So this term, as well as $b^{p_n-1}c$, is OK. But I can't handle the other terms.
My questions are, is this identity known to be true? If so, how can I manipulate the right hand side?
| $\newcommand{\weird}[4]{\dfrac{{#1}^{#4}+{#2}^{#4}+{#3}^{#4}}{#4}}$
I'd say your suggestion is false, because
$$\weird {(-y-z)}yz{13}-\weird {(-y-z)}yz{11}\cdot\weird {(-y-z)}yz2=\\=-y^3z^3(y+z)^3(y^2+yz+z^2)^2\ne0$$
It's always worth checking the first hard case with a calculator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
arithmetic progession question if $\sqrt{a-x}, \sqrt x, \sqrt{a+x}$ are in AP provided $a>x$ and $a,x$ are positive integers then what is the least possible value of $x$?
| $$\sqrt{a-x},\sqrt{x},\sqrt{a+x}$$ are in an AP. By the definition of an arithmetic progression,
$$\sqrt{a+x}-\sqrt{x}=\sqrt{x}-\sqrt{a-x}$$
Thus
$$2\sqrt{x}=\sqrt{a+x}+\sqrt{a-x}$$
Squaring both sides,
$$4x=a+x+a-x+2\sqrt{a^2-x^2}$$
$$2x-a=\sqrt{a^2-x^2}$$
Squaring both sides,
$$4x^2+a^2-4ax=a^2-x^2$$
$$5x^2-4ax=0$$
$$x(5x-4a)=0$$
Since $x\ne0$,
$$x=\frac{4a}{5}$$
Since $a,x$ are positive integers, for the least value, $a$ must be the smallest positive integer divisible by $5$. Thus, $a=5$.
Correspondingly, the smallest value of $x$ is $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-4\sqrt{3}i}=\sqrt{2^2+3i^2-4\sqrt{3}i}=\sqrt{(2-\sqrt{3}i)^2}$ which made calculation easier.
| First establish a formula for the square root:
$$(a+ib)^2=a^2-b^2+i2ab=x+iy$$ is equivalent to the system
$$a^2-b^2=x,\\2ab=y.$$
If we notice that $$(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2,$$ we have
$$a^2-b^2=x,\\
a^2+b^2=\pm\sqrt{x^2+y^2}.$$
The real solutions are
$$a=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}2},\\b=\pm\sqrt{\frac{\sqrt{x^2+y^2}-x}2},$$
where the signs must be chosen so that $\text{sign}(a)\text{sign}(b)=\text{sign}(y)$.
Applying to the given case, $x=1,y=-4\sqrt3$, we find $a=\pm2,b=\mp\sqrt3$.
Then once again with $x=\pm2,y=\mp\sqrt3$, we get the final answer
$$a=\pm\sqrt{\frac{\sqrt{7}+2}2},b=\mp\sqrt{\frac{\sqrt{7}-2}2},$$ or
$$a=\pm\sqrt{\frac{\sqrt{7}-2}2},b=\pm\sqrt{\frac{\sqrt{7}+2}2}.$$
These four solutions form a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to solve the $\min_x \int_c^d (u-x)^4 e^{-u^2/2} du$ How to solve the following equation
\begin{align}
\min_x \int_c^d (u-x)^4 e^{-u^2/2} du
\end{align}
Observer, that by taking the derivative we get
\begin{align}
\int_c^d -4(u-x)^3 e^{-u^2/2} du
\end{align}
So, if $c=-d$ then the solution is given by
\begin{align}
x=0
\end{align}
However, what if $d$ and $c$ are any numbers?
| Just as Svetoslav answered, you first need to expand $$(u-x)^4=u^4-4 u^3 x+6 u^2 x^2-4 u x^3+x^4$$ and your are first left with the antiderivatives $$I_k=\int u^k e^{-\frac{u^2}{2}} \,du \qquad (k=0,1,2,3,4)$$ The calculations do not present major difficulties (integrations by parts) $$I_0=\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{u}{\sqrt{2}}\right)$$ $$I_1=-e^{-\frac{u^2}{2}}$$ $$I_2=\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{u}{\sqrt{2}}\right)-e^{-\frac{u^2}{2}} u$$ $$I_3=-e^{-\frac{u^2}{2}} \left(u^2+2\right)$$ $$I_4=3 \sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{u}{\sqrt{2}}\right)-e^{-\frac{u^2}{2}}
\left(u^3+3 u\right)$$ Combining all of that $$J=\int (u-x)^4\,e^{-\frac{u^2}{2}} \,du$$ $$J= \sqrt{\frac{\pi }{2}} \left(x^4+6 x^2+3\right)
\text{erf}\left(\frac{u}{\sqrt{2}}\right)+e^{-\frac{u^2}{2}} \left(4
\left(u^2+2\right) x-u \left(u^2+3\right)-6 u x^2+4 x^3\right)$$. Now, use the given bounds, group terms to end with to a fourth degree polynomial in $x$ (say $Ax^4+B x^3 +Cx^2+Dx+E$) that you have to work.
For testing purposes (repeat it), using $c=1$, $d=2$, you should arrive to a minimum for $x\approx 1.42692$ to which corresponds a minimum value of the integral $\approx 0.00363214$.
The other solution to find the value of $x$ corresponding to the minimum is to write (just as you wrote it) $$\frac {dJ}{dx}=-4\int (u-x)^3\,e^{-\frac{u^2}{2}} \,du$$ which uses the same integrals as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Methods for Integrating $\int \frac{\cos(x)}{\sin^2(x) +\sin(x)}dx$ So I've found that there's the Weierstrass Substitution that can be used on this problem but I just want to check I can use a normal substitution method to solve the equation:
$$\int \frac{\cos(x)}{\sin^2(x) +\sin (x)}dx$$
Let $u = \sin(x)$
$du = \cos(x)\, dx$
$dx = \frac {1}{\cos(x)\,} du$
Which becomes:
$$\int \frac{\cos(x)}{u^2 + u} \frac{1}{\cos(x)}du$$
$$\int \frac{1}{u^2 + u}du$$
Factor out u from denominator:
$$\int \frac{1}{u(u + 1)}du$$
Integrate as a partial fraction:
$$\int \frac{1}{u} - \frac{1}{(u + 1)}du$$
Which integrates as:
$$\ln|u| - \ln|(u + 1)| + C$$
Subtitute $u = \sin(x)$ back in and simplifies to:
$$\ln \left|\frac{\sin(x)}{\sin(x)+1} \right| + C$$
Is this correct? From the Weierstrass Substitution, one gets:
$$\ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right| +C $$
| Manipulating the Weierstrass result, we begin with
$$\ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right| +C$$
With $\tan\left(\frac x2\right)={\sin x\over 1+\cos x}$, we then get
$$\ln \left|{\sin x\over 1+\cos x}\right|-2\ln \left|{\sin x\over 1+\cos x}+1\right| +C\\
=\ln \left|(\sin x)(1+\cos x)\right|-2\ln \left|\sin x + \cos x+1\right| +C\\
=\ln \left|\sin x+\sin x\cos x\over(\sin x + \cos x+1)^2\right|+C\\
=\ln \left|\sin x+\sin x\cos x\over \sin^2 x+\cos^2 x +2\sin x\cos x+2\sin x+2\cos x+1\right|+C\\
=\ln \left|\sin x+\sin x\cos x\over 2 +2\sin x\cos x+2\sin x+2\cos x\right|+C_0\\
=\ln \left|\sin x(1+\cos x)\over (1+\sin x)(1+\cos x)\right|+C_1\\
=\ln \left|\sin x\over 1+\sin x\right|+C_1$$
So the two answers differ by a constant $(\ln 2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\int^b_a(x-a)^3(b-x)^4 \,dx$ I came across a question today...
The value of $\int^b_a(x-a)^3(b-x)^4 \,dx$ is
First I tried the property $\int^b_af(x)=\int^b_af(a+b-x)$. I got $\int^b_a(x-a)^4(b-x)^3 \,dx$, which can be simplified to: $\dfrac{b-a}{2}\int^b_a(x-a)^3(b-x)^3 \,dx$. Well now what? Do I have to open these brackets and do the whole thing or is there a short way (as definite integrals always have)?
| How about the following way? (though I'm not sure if you like it)
$$\begin{align}\\&\int_{a}^{b}(\color{red}{x-a})^3(b-x)^4dx\\&=\int_{a}^{b}(\color{red}{x-b+b-a})^3(x-b)^4dx\\&=\int_a^b\left((x-b)^3+3(b-a)(x-b)^2+3(b-a)^2(x-b)+(b-a)^3\right)(x-b)^4dx\\&=\int_a^b\left((x-b)^7+3(b-a)(x-b)^6+3(b-a)^2(x-b)^5+(b-a)^3(x-b)^4\right)dx\\&=\left[\frac{(x-b)^8}{8}+\frac{3(b-a)(x-b)^7}{7}+\frac{3(b-a)^2(x-b)^6}{6}+\frac{(b-a)^3(x-b)^5}{5}\right]_a^b\\&=-\left(\frac{(a-b)^8}{8}+\frac{3(b-a)(a-b)^7}{7}+\frac{3(b-a)^2(a-b)^6}{6}+\frac{(b-a)^3(a-b)^5}{5}\right)\\&=-\left(\frac 18-\frac 37+\frac 36-\frac 15\right)(a-b)^8\\&=\frac{1}{280}(a-b)^8\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt:
$\displaystyle \binom{n}{7}=\binom{n}{8} $
$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$
$$ \frac{6}{7!} = \frac{n-7}{40320} $$
$$ n-7 = 48 $$
$$ n=55 $$
| Reference that $$(a + b)^n = {n \choose 0}a^nb^0 + {n \choose 1}a^{n-1}b^1 + \cdots + {n \choose n-1}ab^{n-1} + {n \choose n}a^0b^n.$$
So we want $a = 2$ and $b = {x \over 3}$. So we are considering the terms $\displaystyle {n \choose 7}a^{n - 7}b^7$ and $\displaystyle {n\choose 8}a^{n-8}b^8.$ So, $${n \choose 7}a^{n - 7}b^7 = {n\choose 8}a^{n-8}b^8$$ $${n! \over 7!(n - 7)!}a^{n-7}b^7 = {n! \over 8!(n-8)!}a^{n-8}b^8$$ $${a^{n-8}a \over 7!(n-7)(n-8)!} = {a^{n-8}b \over 8\times 7!(n-8)!}$$ $${a \over n-7} = {b \over 8}$$ $$n-7 = {8a \over b}$$ $$n = {48 \over x} + 7$$ Can you see where it comes in now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$
| Hint:
\begin{align}
\frac79&=\frac{81^a+9^a+1}{9^a+3^a+1}=\\
&=\frac{{(3^4)}^a+{(3^2)}^a+1}{{(3^2)}^a+3^a+1}=\\
&=\frac{{(3^a)}^4+{(3^a)}^2+1}{{(3^a)}^2+3^a+1};
\end{align}
Let $u=3^a$:
\begin{align}
\frac79&=\frac{u^4+u^2+1}{u^2+u+1}=\\
&=\frac{u^4+2u^2-2u^2+u^2+1}{u^2+u+1}=\\
&=\frac{{(u^2)}^2+2u^2+1+u^2-2u^2}{u^2+u+1}=\\
&=\frac{\left(u^2+1\right)^2-u^2}{u^2+u+1}=\\
&=\frac{\left(u^2-u+1\right)\left(u^2+u+1\right)}{u^2+u+1}=\\
&=u^2-u+1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I simplify this rational expression? I am trying to simplify $$ \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}$$ to get to $$\frac{3(5^{k+2}-1)}{4} $$
I start with $$ \frac{3(5^{k+1} - 1) + 12\;\cdot\;5^{k+1}}{4}$$
$$= \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$
then I am stuck.
| $$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3 \cdot 5^{k+1} - 3 + 12 \cdot 5^{k+1} = \\ 15 \cdot 5^{k+1} - 3 = \\ 3 \cdot 5 \cdot 5^{k+1} - 3 = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$
In going from the first line to the second line, I distribute the $3$ through on the first term in the sum: $3(5^{k+1} - 1) = 3 \cdot 5^{k+1} - 3.$ This is the first two terms in the sum on the second line. Then, I multiply through by $4$ on the second term to complete it: $4(3 \cdot 5^{k+1}) = 12 \cdot 5^{k+1}.$
Here is the same set of equations, substituting $5^{k+1}$ with $X$ until the very end:
$$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3(X - 1) + 4(3 \cdot X) = \\ 3X - 3 + 12X = \\ 15X - 3 = \\ 3 \cdot 5X - 3 = \\ 3 (5X - 1) = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Exponential of a non terminating matric So I understand how to calculate the exponential of matrices that eventually terminate; however, how to approach the cases in which the matrix does not seem to truncate? For example with the matrix $M=$$\quad
\begin{pmatrix}
2 & 1 \\
0 & 2
\end{pmatrix}$
I have calculated the first few terms: for $k=1$ $\quad
\begin{pmatrix}
2 & 2 \\
0 & 2
\end{pmatrix}$
for $k=2$ $\quad
\begin{pmatrix}
4/3 & 2 \\
0 & 4/3
\end{pmatrix}$
for $k=5$ $\quad
\begin{pmatrix}
4/15 & 2/3 \\
0 & 4/15
\end{pmatrix}$
Any advice on summing these together or developing a formula that would help me with the summation?
Similarly for a 3x3 such as $\quad
\begin{pmatrix}
2 & 2 & 0\\
0 & 2 & 1\\
0 & 0 & 2
\end{pmatrix}$
I'm really not sure how to find $e^M$ for this.
| The basic observation that can be used to compute the exponent explicitly is that if $X,Y$ are matrices that commute ($XY = YX$) then $\exp(X + Y) = \exp(X) \exp(Y)$.
Your matrix $M$ can be written as $M = D + N$ where
$$ D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \,\,\, N= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$
Note that $D$ is diagonal and so the exponent of $D$ is easy to compute:
$$ \exp(D) = \begin{pmatrix} \exp(1) & 0 \\ 0 & \exp(1) \end{pmatrix} = \begin{pmatrix} e & 0 \\ 0 & e \end{pmatrix}. $$
The matrix $N$ is nilpotent satisfying $N^2 = 0$ and so the exponent of $N$ "eventually terminates":
$$ \exp(N) = I + N + \frac{N^2}{2} + \dots = I + N = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. $$
Thus,
$$ \exp(M) = \exp(D) \exp(N) = \begin{pmatrix} e & 0 \\ 0 & e \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e & e \\ 0 & e \end{pmatrix}. $$
This also works for your second matrix (and more generally for any upper triangular matrix). For a general matrix, you won't be necessarily able to decompose $M = D + N$ where $D$ is diagonal and $N$ is nilpotent but if you work over the complex numbers, the Jordan decomposition tells you that $M$ is similar to a matrix of the form $D + N$ so you can find an invertible $P$ with $P^{-1}MP = D + N$ and use the property that $\exp(PAP^{-1}) = P\exp(A)P^{-1}$. More explicitly, $M = P(D+N)P^{-1}$ and
$$ \exp(M) = \exp(P(D+N)P^{-1}) = P\exp(D)\exp(N)P^{-1}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction - can not prove the step case I have to prove by induction the following:
$$\sum^{n-1}_{k=2} k\log_{10} k \le \frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2$$
I have proven the base case, but I am stuck at the step case. I have tried different ways but always have stuck somewhere and could not finish it.
Anyone has any idea?
Thank you
| What you want to show is that
$\sum^{n-1}_{k=2} k\log_{10} k
\le \frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2
$
implies
$\sum^{n}_{k=2} k\log_{10} k
\le \frac{1}{2}(n+1)^2\log_{10} (n+1)-\frac{1}{8}(n+1)^2
$.
From your assumption,
$\begin{array}\\
\sum^{n}_{k=2} k\log_{10} k
&=\sum^{n-1}_{k=2} k\log_{10} k+n\log_{10} n\\
&\le\frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2+n\log_{10} n\\
\end{array}
$
Therefore,
you have succeeded
if you can prove
$\frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2+n\log_{10} n
\le \frac{1}{2}(n+1)^2\log_{10} (n+1)-\frac{1}{8}(n+1)^2
$.
I will leave this up to you.
(Since you asked for help.
here it is.
Note that this is just algebra.)
You want to show that
$\begin{array}\\
0
&\le \frac{1}{2}(n+1)^2\log (n+1)-\frac{1}{8}(n+1)^2
-\frac{1}{2}n^2\log n+\frac{1}{8}n^2+n\log n\\
&= \frac{1}{2}(n^2+2n+1)\log (n+1)-\frac{1}{8}(n^2+2n+1)
-\frac{1}{2}n^2\log n-\frac{1}{8}n^2+n\log n\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+\frac{1}{2}(2n+1)\log (n+1)-\frac{1}{8}(2n+1)+n\log n\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+(n+\frac12)\log (n+1)-\frac{1}{8}(2n+1)+n\log n\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+n(\log (n+1)+\log (n))+\frac12\log (n+1)-\frac{1}{8}(2n+1)\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+n(\log (n+1)+\log (n)-\frac14)+\frac12\log (n+1)-\frac{1}{8}\\
\end{array}
$
and all these terms
are positive
for
$n \ge 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An algorithm for creating a circle on a discrete plane and a limit for $\pi$ I know there is a well known algorithm which uses the circle equation to approximate it with pixels.
However, I wanted to approach this problem from the most basic principles.
So we start with a square with the side $a_0$ (all lengths are measured in pixels).
To make it more 'circular' we add several pixel rows on each side. The number of rows $k$ is defined to make the distance between the opposite sides as close as possible to the length of the diagonal.
$$(a_0+2k)^2=2a_0^2$$
$$\left[\frac{\sqrt{2}-1}{2} a_0 \right] \leq k \leq \left[\frac{\sqrt{2}-1}{2} a_0 \right]+1$$
Here $[]$ is the floor function.
Now we need to find the length of each row. Let's consider a rectangle formed by two opposite rows. Its diagonal should be as close as possible to the diagonal of the initial square.
$$a_n^2+(a_0+2n)^2=2a_0^2$$
$$\left[ \sqrt{a_0^2-4na_0-4n^2} \right] \leq a_n \leq \left[ \sqrt{a_0^2-4na_0-4n^2} \right]+1$$
Now if we use the 'exact' value of $k$, we get:
$$k=\frac{\sqrt{2}-1}{2} a_0~~~~\rightarrow~~a_k=0$$
Which means that the correct value for $k$ is:
$$k=\left[\frac{\sqrt{2}-1}{2} a_0 \right]$$
Value of $a_n$ is defined from symmetry considerations:
If $a_0$ is even, all $a_n$ should be even.
If $a_0$ is odd, all $a_n$ should be odd.
If we count the number of pixels in our circle (its discrete area), we get:
$$S_k=a_0^2+4 \sum_{n=1}^k a_n$$
$$\pi=\lim_{k \to \infty} \frac{2 S_k}{a_0^2}$$
And here are two examples:
$$a_0=100~~~~~~~~~k=20~~~~~~~~~S=15496~~~~~~~~~\pi = 3.0992$$
$$a_0=10000~~~~~~~~~k=2071~~~~~~~~~S=157059544~~~~~~~~~\pi = 3.14119$$
This sequence converges to $\pi$ from below. The convergence is stochastic and very slow. See the plot below for first $1000$ terms:
Is this algorithm correct (does it create a 'best' circle approximation for a given $a_0$)? Is there some other way to approximate $\pi$ using this algorithm?
If we just want to compute $\pi$ (without creating the circle) we can simplify the algorithm and get the following:
$$\pi=2+4 \lim_{p \to \infty} \frac{1}{p} \sum_{n=1}^k \sqrt{1-\frac{2n}{p}-\frac{n^2}{p^2}}$$
$$k=\left[(\sqrt{2}-1) p \right]$$
Have you seen this limit before? Maybe it's related to some known formula for $\pi$?
| You state that
$\pi=2+4 \lim_{p \to \infty} \frac{1}{p} \sum_{n=1}^k \sqrt{1-\frac{2n}{p}-\frac{n^2}{p^2}}
$
This is a standard Riemann sum.
$\lim_{p \to \infty} \frac{1}{p} \sum_{n=1}^k \sqrt{1-\frac{2n}{p}-\frac{n^2}{p^2}}
\to \int_0^{\sqrt{2}-1} \sqrt{1-2x-x^2} dx
$
Since
$ \int \sqrt{1-2 x-x^2} dx
= \frac12 \sqrt{-x^2-2 x+1} (x+1)+\sin^{-1}\frac{x+1}{\sqrt{2}}
$,
and
$1-2x-x^2
= 0
$
for $x = \sqrt{2}-1$
($x^2+2x
=3-2\sqrt{2}+2\sqrt{2}-2
= 1
$),
$\begin{array}\\
\int_0^{\sqrt{2}-1} \sqrt{1-2 x-x^2} dx
&= \left(\frac12 \sqrt{-x^2-2 x+1} (x+1)+\sin^{-1}\frac{x+1}{\sqrt{2}}\right)\big|_0^{\sqrt{2}-1}\\
&=(0+ \sin^{-1}\frac{(\sqrt{2}-1)+1}{\sqrt{2}})-(\frac12+\sin^{-1}\frac1{\sqrt{2}})
\\
&=\sin^{-1}(1)-(\frac12+\sin^{-1}\frac1{\sqrt{2}})\\
&=\frac{\pi}{2}-(\frac12+\frac{\pi}{4})\\
&=\frac{\pi}{4}-\frac12\\
\end{array}
$
which verifies your limit.
The standard digital
circle drawing algorithm
is Bresenhams's.
For example,
see here:
https://en.wikipedia.org/wiki/Midpoint_circle_algorithm
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that sum of squares of all the values of $(\sqrt{3}+i)^{3/7}$ is $0$. Show that sum of squares of all the values of $(\sqrt{3}+i)^{3/7}$ is $0$.
Attempt:
Let $z=(\sqrt{3}+i)$ then simplifying I get
$z^{3/7}=2^{3/7}\Big\{\cos{\frac{13k+1}{2\times 7}\pi}+i\sin{{\frac{13k+1}{2\times 7}\pi}}\Big\}$, $k=0,1,2\cdots 6$
Now sum of squares of all values of $z^{3/7}$ is
$S=2^{3/7}\Big[\{\cos{\frac{\pi}{7}}+\cos{\frac{13\pi}{7}}+\cos{\frac{25\pi}{7}}+\cos{\frac{37\pi}{7}}+\cos{\frac{49\pi}{7}}+\cos{\frac{61\pi}{7}}+\cos{\frac{73\pi}{7}}\}+i\{\sin{\frac{\pi}{7}}+\sin{\frac{13\pi}{7}}+\sin{\frac{25\pi}{7}}+\sin{\frac{37\pi}{7}}+\sin{\frac{49\pi}{7}}+\sin{\frac{61\pi}{7}}+\sin{\frac{73\pi}{7}}\}\Big]$
How to show that this is zero?
| Hint: Values of $(\sqrt3+i)^{3/7}$ are roots of $x^7-(\sqrt3+i)^3=0$. Now think Vieta and use $\sum x_i^2 = (\sum x_i)^2-2\sum x_i x_j$.
| {
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"source": "stackexchange",
"question_score": "2",
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On equations $m^2+1=5^n$ I am looking for integer solutions of Diophantine equation $m^2+1=5^n$. I found that $m=0,n=0$ and $m=2,n=1$. I could not find any other solutions. I try to prove this but I could not. Could anyone help me to solve this equation?
| HINT.-If you are interested in another way of indicated in the link given by Yiyuan Lee (a good example of that elementary could be intricate enough and not easy), you could pay attention to the following:
$m^2+1=5^n\Rightarrow n$ is odd ($m^2+1=x^2$ is impossible since the difference between $x^2$ and the square of an integer greater than $x$ is always greater or equal than $2x+1>1$).
So we have the equation $$m^2+1=5x^2\iff m^2-5x^2=-1\quad (*)$$
The Pell-Fermat equation $(*)$ has infinitely many solutions but we have to discriminate these solutions with the restriction $x$ must be a power of $5$.
It is known in Algebraic Number Theory (it is not hard to calculate) that the fundamental unit of $\mathbb Q[\sqrt 5]$ is $\frac{ 1+\sqrt 5}{2}$ so all the solutions $(m_n,x_n)$ of $(*)$ are given by
$$m_k+x_k\sqrt 5=\left(\frac{ 1+\sqrt 5}{2}\right)^k; k\ge 1\qquad (**)$$
Calculation gives
$$2^k(m_k+x_k\sqrt 5)=(1+\sqrt 5)^k=1+\binom k1\sqrt 5+\binom k2(\sqrt 5)^2+\binom k3(\sqrt 5)^4+…..+\binom k2(\sqrt 5)^{k-2}+\binom k1(\sqrt 5)^{k-1}+(\sqrt 5)^k$$
Hence you have according if $k$ is even or odd
$$2^{2k}x_{2k}=\binom {2k}1+5\binom {2k}3+5^2\binom {2k}5+5^3\binom {2k}7+…..5^{k-1}\binom {2k}1$$
$$2^{2k+1}x_{2k+1}=\binom {2k+1}1+5\binom {2k+1}3+5^2\binom{2k+1}5+5^3\binom {2k+1}7+…..+5^{k-1}\binom {2k+1}2+5^k$$
I leave to you the task of verifying that it is not possible that any $x_k$ is a power of $5$.
| {
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"source": "stackexchange",
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} |
Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) &\geq
&\left( 1+1\right) \left( 1+\frac{1}{\cos x}\right) \\
&=&2\left( 1+\frac{1}{\cos x}\right) \\
&\geq &2\left( 1+1\right) =4,
\end{eqnarray*}
but $4\leq 3+\sqrt{2}$.
| Let $\left( 1+\dfrac1{\sin x}\right) \left( 1+\dfrac1{\cos x}\right)=1+y$
$\iff\sec x+\csc x=y-\sec x\csc x$
Squaring both sides, $$\sec^2x+\csc^2x+2\sec x\csc x=y^2-2y\sec x\csc x+\sec^2x\csc^2x$$
But $\sec^2x+\csc^2x=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\sec^2x\csc^2x$
$$\implies y^2-(2y-1)\sec x\csc x=0\iff0=\sin2x y^2-4y-4\le y^2-4y-4$$ as $\sin2x\le1$
As the roots of $y^2-4y-4=0$ are $y=2\pm2\sqrt2$
Either $y\ge2+2\sqrt2$ or $y\le2-2\sqrt2$
But as $x\in \left( 0,\frac{\pi }{2}\right),y>0\implies y\not\le0>2-2\sqrt2$
Can you take it from here?
The equality occurs if $\sin2x=1\iff x=\dfrac\pi4(?)$
| {
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} |
show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
it suffices to
$$4\sqrt{abcd}+64abcd\ge 1$$
| Here is a way without Schur. Let $p = a+b, u = ab$ and $q = c+d, v = cd$. Then we have $p+q = 1$ and wish to show $$F = 3(p^2+q^2-2u-2v)+64uv \ge 1$$
For any fixed $p, q$, the variables $u, v$ can range continuously in $u \in [0, p^2/4]$ and $v \in [0, q^2/4]$. Since $F$, viewed as a function of one variable (either $u$ or $v$) is linear, the minimum has to be when $u, v$ take one of the extreme points.
Now if $u = p^2/4$ and $v = q^2/4$, the variables are equal and we have $F \ge 1 \iff (p-q)^2(p^2+q^2)\ge 0$ so we have equality when $p=q \implies a=b=c=d$ in this case.
Else we have $uv=0$, so WLOG let $d=0$ and the inequality is now the obvious
$$a+b+c=1 \implies 3(a^2+b^2+c^2) \ge 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
| Set $\sin(2x) = u$
$7\cos(2x) - \sin(2x) + 5 = 0$
$\implies 7\sqrt{1-u^2} = u-5$
$\implies 49 - 49u^2 = u^2 - 10u + 25$
$\implies 50u^2 - 10u -24 = 0$
$\implies u_1 = 0.8, u_2 = -0.6$
$ \sin(2x) = 0.8$ or $\sin(2x) = -0.6$
| {
"language": "en",
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"question_score": "2",
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"answer_id": 5
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If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$. If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$.Find $k$
Let $\frac{\pi}{32}=\theta$
$\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\frac{1}{\sin\theta}+\frac{1}{\sin2\theta}+\frac{1}{\sin4\theta}+\frac{1}{\sin8\theta}+\frac{1}{\sin16\theta}$
Now i am stuck.
| Hint: try to work out some smaller cases first, for example, $\csc\frac{\pi}{2}=\cot\frac{\pi}{k}$ or $\csc\frac{\pi}{4}+\csc\frac{\pi}{2}=\cot\frac{\pi}{k}$. Then try to generalize using induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ I want to show that
$$
\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)
= \lim_{n \to \infty} \prod_{k=0}^{n-1}
\left(1-\frac{n}{n^2 - k} \right)
= \mathrm{e}^{-1}
$$
I know that $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ and so my attempt was to write
$$
\left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)\\
= \underbrace{\left(1-\frac{1}{n} \right)^n}_{\to \text{e}^{-1}} \cdot \underbrace{\frac{1-\frac{n}{n^2-1}}{1-\frac{1}{n}}}_{\rightarrow 1} \cdot \ldots \cdot \underbrace{\frac{1-\frac{n}{n^2-n+1}}{1-\frac{1}{n}}}_{\rightarrow 1}
$$
which I thought would solve my problem. But after a second look I see that splitting the limits in the product cannot be allowed. This would be the same nonsense as
$$
\underbrace{\left(1-\frac{n}{n^2} \right)}_{\rightarrow 1} \underbrace{\left(1-\frac{n}{n^2-1} \right)}_{\rightarrow 1} \cdot \ldots \cdot \underbrace{\left(1-\frac{n}{n^2-n+1} \right)}_{\rightarrow 1}.$$
Maybe anybody can make the situation clear to me.
| May be another way to see the limit and how it is approached.
Let $$u_i=1-\frac 1{n^2-i}\implies P_n=\prod_{i=0}^{n-1} u_i$$ Take logarithms $$\log(P_n)=\sum_{i=0}^{n-1} \log(u_i)$$ Now, use Taylor series for large values of $n$ $$\log(u_i)=-\frac{1}{n}-\frac{1}{2
n^2}-\frac{i+\frac{1}{3}}{n^3}-\frac{i+\frac{1}{4}}{n^4}+O\left(\frac{1}{n^5
}\right)$$ Now, compute the sums $$\log(P_n)=-1-\frac{1}{2 n}+\frac{1-3 n}{6 n^2}+\frac{1-2 n}{4 n^3}+\cdots=-1-\frac 1 n-\frac 1 {3n^2}+\frac 1 {4n^3}+\cdots$$ Going back to the exponential $$P_n=\frac 1e\Big(1-\frac{1}{n}+\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right) \Big)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a=b^c$, $b=c^a$, and $c=a^b$, prove that If $$a=b^c, \quad b=c^a, \quad \text{and} \quad c=a^b$$ prove that $$abc=1.$$
My Attempt;
Given,
$$a=b^c$$
$$b=c^a$$
$$c=a^b$$
Now,
$$
\begin{align}
\text{L.H.S.} & =abc \\
& = b^c\cdot b\cdot a^b \\
& =b^c\cdot b\cdot b^{bc} \\
& =b^{c+1+bc} \\
& \,\,\,\vdots
\end{align}
$$
I got struck from here.
Please help to complete.
| In $a=b^c$, substitute $b=c^a$, and you get $a=(c^a)^c=c^{ac}$. Next substitute $c=a^b$ and get $a=(a^b)^{ac}=a^{abc}$. So $a^{1}=a^{abc}$. Either $abc=1$ or $a=1$.
If $a=1$ then $c=a^b=1$ and $b=c^a=1$, so $abc=1$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709483",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Solving $\int{\frac{2x \ dx}{\sqrt{9x^2-6x-8}}}$ I solved the problem but my solution seems too simple to be the right one.
$$\int{\frac{2x \ dx}{\sqrt{9x^2-6x-8}}}$$
$$\frac{1}{3}\int{\frac{2x \ dx}{\sqrt{x^2-2x-\frac{8}{9}}}}$$
$u = x^2 - 2x - \frac{8}{9}$
$du = 2x -2 \ dx$
$2x \ dx = du + 2 = 3 du$
$$\frac{1}{3}\int{\frac{3\ du}{\sqrt{u}}} = \int{\frac{du}{\sqrt{u}}} = 2\sqrt{u} + K = 2\sqrt{x^2-2x-\frac{8}{9}} + K$$
Where did I go wrong?
| Wait, how did you go from $$du = (2x-2) \, dx$$ to $$2x \, dx = 3 \, du?$$ That is not right. You can't separate $(2x-2) \, dx$ into $2x \, dx - 2$ any more than you can write $(a+b)c = ac + b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| $$2(\sin^6x +\cos^6x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^2x +\cos^2x)(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=-2\sin^2x \cos^2 x -\sin^4x-\cos^4x+1=$$
$$-(\sin^2x +\cos^2x)^2+1=-1+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Limit of a sequence ratio Let's assume that $a_{n}$ and $b_{n}$ fulfill the equation
$$
a_{n} + b_{n}\sqrt{3} = (2+\sqrt{3})^n, \hspace{0.5cm} n=1,2,\dots
$$
Compute the limit
$$\lim_{n\to \infty}\frac{a_{n}}{b_{n}}.$$
Being honest, I do not know where to start and would appreciate any help. Thanks in advance!
| A start: Although this is not mentioned explicitly in the post, we assume that the $a_n$, $b_n$ are obtained by taking the natural expansion of $(2+\sqrt{3})^n$, and collecting terms that do not involve $\sqrt{3}$ together, and terms that involve $\sqrt{3}$ together. So for example if $n=2$, then $a_n=7$ and $b_n=4$.
Take the conjugates of both sides. We get
$$a_n-b_n\sqrt{3}=(2-\sqrt{3})^n.$$
Add and subtract to get explicit expressions for $a_n$ and $b_n$.
Added: If $x$ and $y$ are rational, the conjugate of $x+y\sqrt{3}$ is defined to be the number $x-y\sqrt{3}$. To show that $(2-\sqrt{3})^n=a_n-b_n\sqrt{3}$, show that if $s,t,u,v$ are rational then the conjugate of $(s+t\sqrt{3})(u+v\sqrt{3})$ is $(s-t\sqrt{3})(u-v\sqrt{3})$ (the conjugate of a product is the product of the conjugates). This is a computation.
Alternately, one can compare the binomial expansions of $(2+\sqrt{3})^n$ and $(2-\sqrt{3})^n$.
Remark: We sketch another solution that connects the problem more explicitly with number theory. Note that
$$(a_n+b_n\sqrt{3})(a_n-b_n\sqrt{3}=(2+\sqrt{3})^n(2-\sqrt{3})^n=1,$$
since $(2+\sqrt{3})(2-\sqrt{3})=1$.
It follows that $a_n^2-3b_n^2=1$. So the $a_n$, $b_n$ are integer solutions of the Pell equation $x^2-3y^2=1$. It is clear that the $a_n$ and $b_n$ $\to\infty$. But for large $x$ and $y$ such that $x^2-3y^2=1$, we have $x^2\approx 3y^2$, so $\frac{x}{y}\approx \sqrt{3}$.
| {
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How to list all possible dimension of $\ker{T},\ker{T^2},...,\ker{T^{k-1}}$ and the corresponding canonical forms? Let $V$ be $5$-dimension vector space, and $T:\ V\rightarrow V$ a nilpotent linear transformation of order (index) $k$ where $1\le k\le 5$. How to list all possible dimension of $\ker{T},\ker{T^2},...,\ker{T^{k-1}}$ and the corresponding canonical forms?
In my personal opinion, I feel like all dims are possible. For example $T=I$, then $\ker{T}=\ker{T^2}=...=\ker{T^{k-1}}=\{0\}$. So dimension is $1$. But I have no idea for the canonical form.
As for other dimensions, since it's in increasing order, if $\dim(\ker{T^k})=n$, then $\dim(\ker{T^{k-1}})\le n$.
The problem is I have no firm proof of this and I don't know how to provide general canonical form. Could someone give any insight?
| You really should specify what you mean by "canonical form". I will assume that you meant rational canonical form since you did not specify that you were dealing with a complex vector space. In this particular problem, because of the nilpotency, these are also the Jordan forms, but that is just luck, the distinction of which canonical form you are talking about is important generally.
Also, I will leave the computation of the dimensions of the kernels to you. With the canonical forms written down, it will be straightforward.
$k=1:$ $T$ must be the zero transformation. The zero matrix is its canonical form.
$k=2:$ The minimal polynomial is $x^2$. The invariant factors are then $x,x,x,x^2$ or $x,x^2,x^2$. The corresponding rational canonical forms are respectively, $$\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix} \textrm{ or } \begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}_.$$
$k=3:$ The minimal polynomial is $x^3$. The invariant factors are then $x,x,x^3$ or $x^2,x^3$. The corresponding rational canonical forms are respectively, $$ \begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix} \textrm{ or } \begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
\end{pmatrix}_.$$
$k=4:$ The minimal polynomial is $x^4$. The invariant factors are $x,x^4$. The corresponding rational canonical form is, $$ \begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}_.$$
$k=5:$ The minimal polynomial is $x^5$. The invariant factor is $x^5$. The corresponding rational canonical form is, $$ \begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
\end{pmatrix}_.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The Diophantine Equation: $x^3-3=k(x-3)$ I wish to know how to resolve the diophantine equation: $x^3-3=k(x-3)$ ? The problem is:
Find all integers $x\ne3$ such that $x-3\mid x^3-3$.
- From 250 Problem's in Elementary Number Theory, by W. Sierpinski
I have found the solutions to this problem by using rules in divisibility of numbers, as below:
$$x-3\mid x-3 \Rightarrow x-3\mid x^3 - 3x^2$$
$$x-3\mid x^3-3 \quad and \quad x-3 \mid x^3-3x^2$$
Thus, $x-3$ divides their difference:
$$\Rightarrow x-3\mid 3x^2-3 $$
On the other hand:
$$x-3\mid x-3 \Rightarrow x-3 \mid 3x^2-9x$$
Hence their difference is divisible by $x-3$:
$$\Rightarrow x-3\mid 9x-3 \quad and \quad x-3\mid 9x-27$$
$$\Rightarrow x-3\mid 24 \Rightarrow 24=k(x-3)\Rightarrow x=\frac{24}{k}+3$$
So we have the general solution to find the integer values of $x$. But I did not manage to solve the diophantine equation related to the problem: $$x^3-3=k(x-3)$$
Any hints and helps about this equation?
-Thank you in advance.
| Set $x-3=y\iff x=?$
$$\dfrac{x^3-3}{x-3}=\dfrac{(y+3)^3-3}y=y^2+9y^2+27y+\dfrac{3^3-3}y$$
So, $y$ must divide $3^3-3=24$ to keep $k$ an integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question Regarding Integration Within Summation
It expresses an integral within a summation procedure.
| Compute everything from the inside out (one step at a time), don't look at the sum until you've considered the integral.
Observe that
$$
\int_{\frac{1}{k+b}}^{\frac{1}{k+a}}\frac{dx}{1+x}=\ln(1+x)|_{(k+b)^{-1}}^{(k+a)^{-1}}=\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right).
$$
By the properties of the logarithm,
\begin{align*}
\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)&=\ln\left(\frac{k+1+a}{k+a}\right)-\ln\left(\frac{k+1+b}{k+b}\right)\\
&=\ln(k+1+a)-\ln(k+a)-\ln(k+1+b)+\ln(k+b).
\end{align*}
Now, let's consider the sum:
$$
\sum_{k=1}^\infty\ln(k+1+a)-\ln(k+a)-\ln(k+1+b)+\ln(k+b).
$$
This is a telescoping series, so the partial sum when $k=n$ is
\begin{align*}
\sum_{k=1}^n\ln(k+1+a)-\ln(k+a)&-\ln(k+1+b)+\ln(k+b)\\
&=\ln(n+1+a)-\ln(n+1+b)-\ln(1+a)+\ln(1+b).
\end{align*}
Finally, by taking the limit as $n$ approaches infinity, we see that
\begin{align*}
\lim_{n\rightarrow\infty}\ln(n+1+a)-\ln(n+1+b)
&=\lim_{n\rightarrow\infty}\ln\left(\frac{n+1+a}{n+1+b}\right)\\
&=\ln\left(\lim_{n\rightarrow\infty}\frac{n+1+a}{n+1+b}\right)\\
&=\ln(1)=0.
\end{align*}
Therefore, the limit is $\ln\left(\frac{1+b}{1+a}\right)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to factorize a 4th degree polynomial?
I need help to factorise the following polynomial:
$x^4 - 2x^3 + 8x^2 - 14x + 7$
The solution I need to reach is $(x-1)(x^3 - x^2 + 7x - 7)$. I need to factorize to this exactly as it is for a limit question where I cancel out the $(x-1)$ in the numerator and denominator. How do I proceed?
| One may recognize the factorization of the polynomial(#): $$y^2+8y+7=(y+1)(y+7)$$
Also(^)
$$-2x^3-14x=-2x(x^2+7)$$
From (#), when $y=x^2$, we obtain(*)
$$x^4+8x^2+7=(x^2+1)(x^2+7)$$
Combine(*) and (^), we have
\begin{align}
x^4-2x^3+8x^2-14x+7&=(x^2+1)(x^2+7)-2x(x^2+7)\\&=(x^2-2x+1)(x^2+7)\\&=(x-1)^2(x^2+7)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
| You can use the odd-factorial
$$\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)} = \frac{(n!)^2 2^n}{(2n)!}
$$
Taking logarithm,
$$ \log\frac{(n!)^2 2^n}{(2n)!} = 2\log n! + n \log 2 - \log (2n!)
$$
so...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum value of $f(x) = \cos x \left( \sin x + \sqrt {\sin^2x +\sin^2a}\right)$ Can we find maximum value of $$f(x) = \cos x \left( \sin x + \sqrt
{\sin^2x +\sin^2a}\right)$$
where '$a$' is a given constant.
Using derivatives makes calculation too complicated.
| Let $$y=\cos x\left[\sin x+\sqrt{\sin^2 x+\sin^2 a}\right] = \sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
We get $$(\sin^2 x+\cos ^2 x)\cdot \left[\cos^2 x+\sin^2 x+\sin^2 a\right]\geq \left(\sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}\right)^2$$
So we get $$y^2\leq (1+\sin^2 a)\Rightarrow |y| \leq\sqrt{1+\sin^2 a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Finding the matrix of projection Find the matrix A of the orthogonal projection onto the line L in R2 that consists of all scalar multiples of the vector $\begin{pmatrix} 2 \\ 3 \ \end{pmatrix}$. How do I begin to solve this? Any help would be appreciated.
| Use the orthonormal basis for ${\sf R}^2$
$$\gamma=\{u_1,u_2\}=\left\{\frac{1}{\sqrt{13}}\begin{pmatrix}2\\3\end{pmatrix},\frac{1}{\sqrt{13}}\begin{pmatrix}3\\-2\end{pmatrix}\right\},$$
we see that
${\sf T}(u_1)=u_1$, ${\sf T}(u_2)=0$, and $[{\sf T}]_\gamma=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Let $\beta$ be the standard ordered basis for ${\sf R}^2$, and
write $Q=[{\sf I}]_\gamma^\beta=\frac{1}{\sqrt{13}}\begin{pmatrix}2&3\\3&-2\end{pmatrix}$. Then
\begin{align}
A=[{\sf T}]_\beta
&=Q[{\sf T}]_\gamma Q^{-1}
=Q[{\sf T}]_\gamma Q^t
=\frac{1}{13}\begin{pmatrix}2&3\\3&-2\end{pmatrix}
\begin{pmatrix}1&0\\0&0\end{pmatrix}
\begin{pmatrix}2&3\\3&-2\end{pmatrix}
=\frac{1}{13}\begin{pmatrix}4&6\\6&9\end{pmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $ x+iy = \sqrt{\frac{a+ib}{c+id} } ,$Show that$ (x^2+y^2)^2 = \frac {a^2+b^2}{c^2+d^2} $ $ x+iy = \sqrt{\frac{a+ib}{c+id} } , $Show that $ ({x^2+y^2})^2 = \frac {a^2+b^2}{c^2+d^2} $
How do i do this ?I tried squaring both sides but x+iy expansion becomes difficult when squaring the next time .I also tried conjugating the R.H.S
|
Notice, when $q\space\wedge\space s\space\wedge\space z\in\mathbb{C}$, so we set $q=x+yi,s=a+bi,z=c+di$:
*
*$$\left(x^2+y^2\right)^2=\left(\Re^2[q]+\Im^2[q]\right)^2=|q|^4$$
*$$a^2+b^2=\Re^2[s]+\Im^2[s]=|s|^2$$
*$$c^2+d^2=\Re^2[z]+\Im^2[z]=|z|^2$$
So:
$$x+yi=\sqrt{\frac{a+bi}{c+di}}\Longleftrightarrow q=\sqrt{\frac{s}{z}}$$
Now, we can find the absolute value:
$$\left|q\right|=\left|\sqrt{\frac{s}{z}}\right|\Longleftrightarrow$$
$$\left|q\right|=\left|\left(\frac{s}{z}\right)^{\frac{1}{2}}\right|\Longleftrightarrow$$
$$\left|q\right|=\left|\frac{s}{z}\right|^{\frac{1}{2}}\Longleftrightarrow$$
$$\left|q\right|=\left(\frac{\left|s\right|}{\left|z\right|}\right)^{\frac{1}{2}}\Longleftrightarrow$$
$$\left|q\right|^4=\left(\frac{\left|s\right|}{\left|z\right|}\right)^2\Longleftrightarrow$$
$$\left|q\right|^4=\frac{\left|s\right|^2}{\left|z\right|^2}$$
| {
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"url": "https://math.stackexchange.com/questions/1730650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
3D Surface Area Integral find SA of the cone $$z=2\sqrt{(x^2+y^2)}$$ bounded by $$y=x$$ and $$y=x^2$$ in the first quadrant.
This is my integral setup for the surface area of that portion of the cone, what did I do wrong?
$\int_0^1 \int_0^{x^2} \sqrt{\frac{\ 2y+2x}{\sqrt{x^2+y^2}}+1} \, dydx $
Thanks for the help
| Mostly you forgot to square components before adding them under the square root. Along the surface, $\vec r=\langle x,y,2\sqrt{x^2+y^2}\rangle$, so
$$d\vec r=\langle1,0,\frac{2x}{\sqrt{x^2+y^2}}\rangle dx+\langle0,1,\frac{2y}{\sqrt{x^2+y^2}}\rangle dy$$
So
$$\begin{align}d^2\vec A & =\langle1,0,\frac{2x}{\sqrt{x^2+y^2}}\rangle dx\times\langle0,1,\frac{2y}{\sqrt{x^2+y^2}}\rangle dy\\
& =\pm\langle\frac{-2x}{\sqrt{x^2+y^2}},\frac{-2y}{\sqrt{x^2+y^2}},1\rangle dx\,dy\end{align}$$
Taking magnitude,
$$d^2A=||d^2\vec A||=\sqrt{\frac{4x^2+4y^2}{x^2+y^2}+1}\,dx\,dy=\sqrt5\,dx\,dy$$
So you should get
$$A=\int d^2A=\int_0^1\int_{x^2}^x\sqrt5\,dy\,dx=\frac{\sqrt5}6$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume of ice cream cone using cylindrical/spherical coordinates I'm stuck on what the boundaries are for the volume bounded by the cone $z=-\sqrt{(x^2+y^2)}$ and the surface $z=-\sqrt{(9-x^2-y^2)}$ $\,\,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordinates the triple integral boundaries would be
$-\sqrt{(9-r^2)}\le z \le -r$
$0\le r \le 3$
$0\le \theta \le 2\pi$
And for spherical coordinates the triple integral boundaries would be
$0\le r \le 3$
$\pi/2\le \phi \le \pi$
$0\le \theta \le 2\pi$
However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18\pi$.
So somethings wrong with my boundaries, as both integrals should equal the same volume.
This is my working out for cylindrical coordinate integral so far:
$\int_0^{2\pi}\int_0^3\int_{-\sqrt{9-r^2}}^{-r} r\,\, dzdrd\theta$
$\int_0^{2\pi}\int_0^3\ [rz]_{-\sqrt{9-r^2}}^{-r} \,\, dzdrd\theta$
$\int_0^{2\pi}\int_0^3\ -r^2+{r\sqrt{9-r^2}} \,\, dzdrd\theta$
$\int_0^{2\pi}[(\int_0^3\ -r^2 \,dr)+(\int_0^3\ {r\sqrt{9-r^2} \,dr)}]d\theta$
Let $u=9-r^2$
$du=-2r\,dr$
$\int_0^{2\pi}[[\frac{-r^3}{3}]_0^3\,-\frac{1}{2}\int_0^3\ u^{\frac{1}{2}} \,du)]\,d\theta$
$\int_0^{2\pi}[-\frac{27}{3}\,-\frac{1}{2}[\frac{2}{3}u^{\frac{3}{2}}]_{r=0}^{r=3} ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{2}[{\frac{2}{3}}(9-r^2)^{\frac{3}{2}}]_{0}^{3} ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{2}[{\frac{2}{3}}(9-3^2)^{\frac{3}{2}}\,-{\frac{2}{3}}(9-0^2)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{2}{\frac{2}{3}}[(9-9)^{\frac{3}{2}}\,-(9)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{3}[(0)^{\frac{3}{2}}\,-(9)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{3}[-(9)^{\frac{3}{2}}] ]\,d\theta$
$\int_0^{2\pi}[-9\,-\frac{1}{3}[-27] ]\,d\theta$
$\int_0^{2\pi}[-9\,+9]\,d\theta$
$\int_0^{2\pi}0\,d\theta$
$=0$
So as you can see I can't proceed to the third integral since the second integral equals zero
Any help would be greatly appreciated :)
| The surfaces intersect when
$$-r=-\sqrt{9-r^2}$$
which is $r^2=\frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $\phi$ should be from $\frac34\pi$ to $\pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $\rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
I am not able to find an idea on how to proceed with the above questions. I have found only the obvious solution $(1,1,1)$.
Could you please provide some hints and ideas on how to proceed with the above question? Also, can we find the solutions?
Thanks.
|
Let $z=-x$. Then
$$2x^2+y^2=y^3$$
$$2x^2=(y-1)y^2$$
If $\frac{y-1}2=k^2 - $ perfect square, then
$$y=2k^2+1, x=k(2k^2+1)$$
Answer: $$x=k(2k^2+1), y=2k^2+1, z=-k(2k^2+1)$$
Second method:
Let $y=1+a, z=1-a$. Then
$$x^3+2(1+3a^2)=2(1+a^2)+x^2$$
$$x^2-x^3=4a^2.$$
Let $1-x=4p^2$, then
$$x^2(1-x)=(4p^2-1)^24p^2=(2a)^2.$$
Let $a=p(4p^2-1)$. Then
$$(x,y,z)=(1-4p^2, 1+p(4p^2-1), 1-p(4p^2-1))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Roots of quadratic equation are given by $b \pm \sqrt{b^2 - c}$ I was reading slides about the cancellation error in quadratic equations and it's written:
The roots of the quadratic equation:
$$x^2 - 2bx + c = 0$$
with $b^2 > c$ are given by $b \pm \sqrt{b^2 - c}$.
Fact that let me perplexed, since I always thought that the roots can be found using the following formula:
$$\frac{b \pm \sqrt{b^2 - 4ac}}{2a}$$
What the relation between one and the other?
| If we write the quadratic like this $$ax^2+bx+c=0,$$ then the quadratic formula is right as you have written it; the roots are: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But the quadratic $$x^2-2bx+c=0$$ has replaced $b$ with $-2b$ and $a$ with $1$. This makes the quadratic $$\frac{2b\pm\sqrt{(-2b)^2-4(1)c}}{2(1)}=\frac{2b\pm2\sqrt{b^2-c}}{2}=b\pm\sqrt{b^2-c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736021",
"timestamp": "2023-03-29T00:00:00",
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In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}.......(1)$
By Heron's formula,$\Delta=\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}}$
$\Delta^2=\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}$
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=16\Delta^2$
Putting in $(1)$
$16\Delta^2=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$
By formula $R=\frac{abc}{4\Delta}$,we get
$a^2+b^2+c^2=8R^2$
By sine rule,$a=2R\sin A,b=2R\sin B,c=2R\sin C$
$\sin^2A+\sin^2B+\sin^2C=2$
I am stuck here.
| You have
$\sin^2A+\sin^2B+\sin^2C=2
$.
What stands out to me
is that this is true
if the triangle is right:
$\sin C = 1$
and
$\sin^2A+\sin^2B
=\sin^2A+\sin^2(\pi/2-A)
=\sin^2A+\cos^2(A)
=1
$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}} = \frac{1}{4\sqrt[4]{x^3}}$? This is taken from Khan Academy, I don't understand how these equate:
$$\frac{1}{4}x^{-3/4}
= \frac{1}{4x^{3/4}}
= \frac{1}{4\sqrt[4]{x^3}}$$
How come the minus was remove from the original exponent?
| These are properties of exponentiation. In particular,
$$a^{-b} = \frac{1}{a^b}$$
combined with
$$a^{\frac{m}{n}} = \sqrt[\leftroot{-2}\uproot{2}n]{a^m}.$$
In your case,
$$\frac{1}{4}x^\frac{-3}{4} = \frac{1}{4x^\frac{3}{4}} = \frac{1}{4\sqrt[\leftroot{-2}\uproot{2}4]{x^3}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-(4n-2)^2)$$
$$=2(4)+2(6)+2(12)+2(14)+2(20)+2(22)+\cdots+2(8n-4)+2(8n-2)$$
$$=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]$$
How should I proceed further?
| $
=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]
\\ = 4[(2+3)+(6+7)+(10+11)+\cdots+(8n-3)]
\\ = 4[5+13+21+\cdots+(8n-3)]
\\ = 4 \sum_{k=1}^n (8k-3)
\\ = 4 (4n^2+n)
\\ = 16n^2+4n
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
How to solve this question in more time efficient way?
Q) if$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}$$ then find $4z^2(x^2+y^2)$a)$(x^2+y^2)^{3}$b)$(x^2-y^2)^3$c)$(x^2-y^2)^2$d)$(x^2+y^2)^2$
Ans:c
i solved this in a very long way:
$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}=z\tan 2a$$$$\implies x= \frac{z\tan 2a}{\sin a} , y=\frac{z\tan 2a}{\cos a}$$
$$x^2+y^2=\frac{4z^2\tan ^22a}{\sin ^2 2a}=\frac{4z^2}{\cos^2 2a}$$
$$\implies z^2=\cos^2 2a(x^2+y^2)/4\ldots (1)$$
now$$x\sin a= y\cos a\implies -2x^2\sin^2 a=-2y^2\cos^2a$$
adding $x^2$ both sides
$$x^2(1-2\sin^2 a)= x^2-2y^2\cos^2 a$$$$=x^2-2y^2+2y^2\sin^2a$$ $$=x^2-y^2-y^2+2y^2\sin^2 a$$$$=x^2-y^2-y^2(1-2\sin^2 a)=x^2-y^2-y^2\cos 2a$$$$=(x^2+y^2)\cos 2a= x^2-y^2\implies \cos 2a= \frac{x^2-y^2}{x^2+y^2}\ldots (2)$$
hence from (1) and (2)
$$4z^2(x^2+y^2)= \left( \frac{x^2-y^2}{x^2+y^2} \right)^2 (x^2+y^2)^2= (x^2-y^2)^2$$
now you can see that i got the answer but there was a log way to find this. This question has been asked in an exam so there is noway its solution could be soo long there got to be some shorter way.
So how could i solve this question is more time efficient way?
| How about this?
Since $\tan a=y/x$, we have
$$y\cos a=\frac{2z(y/x)}{1-(y/x)^2}\quad\Rightarrow\quad \cos a=\frac{2xz}{x^2-y^2}$$
So, from $1+\tan^2a=1/(\cos^2a)$, we have
$$1+\left(\frac yx\right)^2=\left(\frac{x^2-y^2}{2xz}\right)^2,$$
i.e.
$$4z^2(x^2+y^2)=(x^2-y^2)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that my polynomial has distinct roots? I want to prove that the polynomial
$$
f_p(x) = x^{2p+2} - abx^{2p} - 2x^{p+1} +1
$$
has distinct roots. Here $a$, $b$ are positive real numbers and $p>0$ is an odd integer. How can I prove that this polynomial has distinct roots for any arbitrary $a$,$b$ and $p$.
Thanks in advance.
| Let $c$ denote $ab$. Note that
\begin{equation*}
f(x) = (x^{p+1}-1)^2 - cx^{2p}
\end{equation*}
and
\begin{equation*}
f'(x) = 2(p+1)x^p(x^{p+1}-1) - 2pcx^{2p-1}
\end{equation*}
Then, $f(x) = 0 \iff$
\begin{equation*}
c = \dfrac{(x^{p+1}-1)^2}{x^{2p}} = \varphi(x)\ (\text{say})
\end{equation*}
and $f'(x) = 0 \iff$
\begin{align*}
c & = \dfrac{(p+1)x^p(x^{p+1}-1)}{px^{2p-1}} = \dfrac{(p+1)x}{p}\dfrac{x^{p+1}-1}{x^p} \iff\\
c & = \left(\dfrac{p+1}{p}\right)x \sqrt{\varphi(x)}.
\end{align*}
Thus, $f(x)$ and $f'(x)$ vanish for the same $x$ if and only if for some root $x$ of $f(x)$,
\begin{align*}
c & = \left(\dfrac{p+1}{p}\right)x \sqrt c \iff\\
x & = \dfrac{p\sqrt c}{p+1}.
\end{align*}
Thus, for every $p$ and $c = ab$, if such an $x$ is a root, it is a multiple root.
Now, when does such a root $x$ exist? Let $x = t$ be one such. Then $c = \left(\dfrac{p+1}{p} \right)^2 t^2$. Then, since $f(t) = 0$, we have (from the original form of the equation):
\begin{align*}
t^{2p+2} - \left(\dfrac{p+1}{p}\right)^2 t^{2p + 2} - 2t^{p+1} + 1 = 0\\
-\dfrac{(2p + 1)}{p^2}t^{2p + 2} - 2t^{p+1} + 1 = 0\\
(2p + 1)t^{2(p + 1)} + 2p^2 t^{p + 1} - p^2 = 0.
\end{align*}
When treated as a quadratic equation in $t^{p+1}$, the discriminant is
\begin{equation*}
4p^4 + 4p^2(2p + 1) = 4p^2(p + 1)^2,
\end{equation*}
and therefore, the solutions are
\begin{equation*}
t^{p+1} = -p, \dfrac{p}{2p + 1}.
\end{equation*}
That is,
\begin{equation*}
t = (-p)^{\frac 1 {p + 1}}, \left(\dfrac p {2p + 1} \right)^{\frac 1 {p + 1}}.
\end{equation*}
But substituting the same $c$ in $f'(t) = 0$, we get
\begin{align*}
& 2(p+1)t^p(t^{p+1}-1)-2p\left(\dfrac{p+1}{p}\right)^2t^{2p+1}=0\\
& p(t^{p+1}-1)-(p+1)t^{p+1}=0 \implies\\
& t = (-p)^{\frac{1}{p+1}}
\end{align*}
Thus, only the first of the previous two solutions satisfies both equations.
Then, $c = \left( \dfrac{p+1}{p} \right)^2 t^2$ gives us
\begin{equation*}
\boxed{c= \dfrac{(p+1)^2}{(-p)^{\frac{2p}{p+1}}}}.
\end{equation*}
Thus, the equation has multiple roots exactly when $c$ and $p$ are related as above.
Note that for odd values of $p$, $c$ will be a real number if and only if $p$ is of the form $4k + 1$, and then, $c < 0$. If, as stated in the question, $c = ab$ is a positive real number, the equation will have distinct roots.
Example
For $p = 1$, $f(x) = x^4 - cx^2 - 2x^2 + 1$ and $f'(x) = 4x^3 - 2cx - 4x$.
Then, $f(x) = 0$ and $f'(x) = 0$ imply that $c = \left(\dfrac{x^2 - 1}{x}\right)^2$ and $c = 2x\left(\dfrac{x^2-1}{x}\right)$ respectively. Thus, if $x$ is a multiple root, then $x = \dfrac{\sqrt c}{2}$.
Taking $t$ to be such a root, so that $c = 4t^2$, and substituting in $f(t) = 0$, we get
\begin{align*}
t^4 - 4t^4 - 2t^2 + 1 = 0\\
3t^4 + 2t^2 - 1 = 0.
\end{align*}
Thus, $t^2 = -1, \dfrac 1 3$, of which only the first one satisfies $f'(t)=0$. Thus, $c = -4$.
For $c = 4$, $x^4 + 2x^2 + 1 = 0$ has roots $\pm i, \pm i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Check: Radius of Convergence of the Sum of these Complex Taylor Series I just found the following Taylor series expansions around $z=0$ for the following functions:
*
*$\displaystyle \frac{1}{z^{2}-5z+6} = \frac{1}{(z-2)(z-3)} = \frac{-1}{(z-2)} + \frac{1}{(z-3)} = \sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}} - \sum_{n=0}^{\infty}\frac{z^{n}}{3^{n+1}} = \sum_{n=0}^{\infty}\left( \frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}\right)z^{n}$
*$\displaystyle \frac{1}{1-z-z^{2}} = \frac{1}{\left[z-\left(\frac{1-\sqrt{5}}{-2} \right) \right]\left[z-\left(\frac{1+\sqrt{5}}{-2} \right)\right]} = \frac{2\sqrt{5}}{5(1-\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1-\sqrt{5}} \right)z} - \frac{2\sqrt{5}}{5(1+\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1+\sqrt{5}} \right)z} = \frac{\sqrt{5}}{5} \sum_{n=0}^{\infty}\left(\frac{(-1)^{n}2^{n+1}}{(1-\sqrt{5})^{n+1}} - \frac{(-1)^{n}2^{n+1}}{(1+\sqrt{5})^{n+1}} \right)z^{n}$
I wanted to confirm that the radius of convergence for the first series is $|z|<2$ and that the radius of convergence for the second series is $|z|<\frac{-1+\sqrt{5}}{2}$.
I know that for a series $\sum_{n=1}^{\infty}a_{n}z^{n}$ with radius of convergence $R_{1}$ and $\sum_{n=1}^{\infty}b_{n}z^{n}$ with radius of convergence $R_{2}$, the radius of convergence of the series $\sum_{n=1}^{\infty}(a_{n}+b_{n})z^{n}$ is some $R$ where $R \geq \min(R_{1}, R_{2})$. And it can even be the case that if both of the series have finite $R_{1}$ and $R_{2}$, that $R$ can be infinite!
So, I wanted to make sure this was not the case here, and that I have the correct radii of convergence for both. If not, how do I go about finding them (preferably in the least icky way possible)?
Thank you.
| You can use Hadamard's rule:
$$\frac 1R=\limsup_n a_n^{1/n}=\limsup_n \biggl(\frac1{2^{n+1}}-\frac1{3^{n+1}}\biggr)^{\!1/n}\!.$$
Now rewrite this expression as
$$\frac1{2^{(n+1)/n}}\biggl(1-\frac{2^{n+1}}{3^{n+1}}\biggr)^{\!1/n}=\frac1{2^{1+1/n}}\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{1/n}\!.$$
The first factor tends to $\frac12$. As to the second factor,
$$\frac23<1-\Bigl(\frac23\Bigr)^{n+1}<1,\enspace\text{whence}\quad\Bigl(\frac23\Bigr)^{1/n}<\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{\!1/n}<1, $$
and by the squeeze principle, it tends to $1$. Thus $\limsup_n a_n^{1/n}=\frac12$, and $R=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Study the convergence of the following series $\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $ I have to study the convergence of the following series:
$\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $
Is a positive series, so I should divide for $\frac{1}{\sqrt{n+2}}$
and then use the comparison test ?
Many thanks
| By the Taylor series expansion, as $u \to 0$, one has
$$
\sin u=u+o(u^2)
$$ giving, as $n \to \infty$,
$$
\sin\left(\frac{1}{\sqrt{n+2}}\right)\times \frac{n}{(n^2+3)}=\frac{1}{\sqrt{n+2}}\times \frac{n}{(n^2+3)}+O\left( \frac{1}{n+2}\times \frac{n}{(n^2+3)}\right)
$$ or
$$
\sin\left(\frac{1}{\sqrt{n+2}}\right)\times \frac{n}{(n^2+3)}=\frac{1}{n^{3/2}}+O\left( \frac{1}{n^2}\right)
$$ and the initial series is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proof using deductive reasoning I need to deductively prove that the sum of cubes of $3$ consecutive natural numbers is divisible by $9$. I can prove deductively that they are divisible by $3$ but so far any combination I choose fails to prove the divisibility by $9$. As far as I can see. This is a high school question though, so if someone can explain it to me in a highschool math language, it will be appreciated. Now here is how I try to do it.
Let $X$ stand for any natural number and let $X+1$ and $X+2$ stand for the two consecutive numbers. I will be cubing, expanding and simplifying them
\begin{align*}
& (x)^{3}+(x+1)^{3}+(x+2)^{3}\\
&= x^3+x^3+3 x^2+3 x+1+x^3+6 x^2+12 x+8\\
&=3x^{3}+9x^{2}+15x+9 \\
&= 3\left ( x^{3}+3x^{2}+5x+3 \right )\\
\end{align*}
This can be used to deductively prove that the sum of cube of $3$ consecutive numbers is divisible by $3$ but I can't prove it is divisible by $9$
| $(x-1)^3+x^3+(x+1)^3=3x^3+6x=3(x^3+2x)=3x(x^2+2)$. Now we just have to prove $3|x$ or $3|x^2+2$.
Case $1: x=3k$, then $3|x$.
Case $2: x=3k+1$, then $x^2+2=9k^2+6k+1+2=3(3k^2+2k+1)$.
Case $3: x=3k+2$, then $x^2+2=9k^2+12+4+2=3(3k^2+4k+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the product of $f(x)$ and $g(x)$ given one of them I'm given the final answer which is $$(g \cdot f)(x) = \frac{1}{x^2+4}\;.$$
Also, i'm given $f(x) = x^2+1$.
I've solved this using the composition, however the second part of the question asks me to find the $g(x)$ which would make this multiplication true. How would I do this? Do I divide the final answer by $f(x)$?
| So you are told that $(g \times f)(x) = \dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.
$(g \times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \times f)(x)$ really means $g(x)f(x)$.
So we know what this product is. It is $g(x)f(x) = \dfrac{1}{x^{2} + 4}$. We also know that $f(x) = x^{2} + 1$. So that means:
$$g(x)(x^{2} + 1) = \dfrac{1}{x^{2} + 4} $$
and to solve for $g(x)$, just divide both sides by $x^{2} + 1$ to get:
$$g(x) = \dfrac{\left (\frac{1}{x^{2} + 4} \right )}{x^{2} + 1} $$
Now, how do we simplify this? Well, $x^{2} + 1$ is the same as $\dfrac{x^{2} + 1}{1}$, so the fraction is really $$\dfrac{\left (\frac{1}{x^{2} + 4} \right )}{\left (\frac{x^{2} + 1}{1}\right)} $$
and when we divide two fractions, we invert the bottom one and multiply, so we get:
$$\dfrac{1}{x^{2} + 4} \cdot \dfrac{1}{x^{2} + 1} $$
And this is just $\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is your final answer (unless you want to multiply the denominator out using the FOIL method).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove inequality $n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$
Let $m,n\in\mathbb N$, $n>m$. Prove inequality
$$n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$$
My work so far:
$$\sqrt[n]{n!}=\sqrt[n]{1\cdot2\cdot...\cdot n}\le\frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}2.$$
Then $n\sqrt[n]{n!}\le\frac{n(n+1)}{2}$
$$\sqrt[m]{m!}=\sqrt[m]{1\cdot2\cdot...\cdot m}\le\frac{1+2+...+m}{m}=\frac{m(m+1)}{2m}=\frac{m+1}2.$$
Then $m\sqrt[m]{m!}\le\frac{m(m+1)}{2}$
And $$\frac{n(n+1)}{2}-\frac{m(m+1)}{2}=\frac{(n−m)(n+m+1)}2.$$
But inequalities can be added, without subtract
| The result follows immediately by writing
\begin{align}
n\sqrt[n]{n!}
&=n\sqrt[n]{m!\cdot(m+1)(m+2)\cdots n}\\
&=n\sqrt[n]{(\sqrt[m]{m!})^m\cdot(m+1)(m+2)\cdots n}\\
&\le m\sqrt[m]{m!}+(m+1)+(m+2)+\cdots+n\\
&=m\sqrt[m]{m!}+\frac{(n-m)(n+m+1)}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expand the function $f(z)=\frac{1}{(z-a)(z-b)}$ where $0 < |a| < |b|$ in a Laurent series in different annuli I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)$ and $(c)$, which I'll share below; however, I was not sure what to do about the $0<|z|$ part in (a). I'm kind of teaching these to myself, and the only examples I've seen with a $0$ in them have involved something like $0< |z+2| <|a|$, or something like that where you'd have to make a substitution $w$... Please let me know how to take care of examples like (a).
Also, here are my attempts at parts (b) and (c). Please let me know if they're correct, and if not, let me know what I need to do in order to fix them:
(b) $\mathbf{|a|<|z|<|b|}$. Using partial fractions, I wrote $\frac{1}{(z-a)(z-b)} = \frac{1}{a-b}\left( \frac{1}{z-a}\right) - \frac{1}{a-b}\left( \frac{1}{z-b}\right)$.
Now, if $|z|> |a|$, then $\left\vert \frac{a}{z}\right\vert< 1$, so we write $\frac{1}{(a-b)(z-a)} = \frac{1}{(a-b)z(1-a/z)} = \frac{1}{a-b}\cdot \frac{1}{z} \cdot \sum_{n=0}^{\infty} \left( \frac{a}{z}\right)^{n} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$.
If $|z|<|b|$, then $\left\vert \frac{z}{b}\right\vert < 1$, so we write $\displaystyle \frac{-1}{(a-b)}\cdot \frac{1}{(z-b)} = \frac{-1}{(a-b)}\cdot \frac{1}{b\left( \frac{z}{b} - 1 \right)} = \frac{1}{b(a-b)}\frac{1}{1-\frac{z}{b}} = \frac{1}{b(a-b)}\sum_{n=0}^{\infty}\left( \frac{z}{b}\right)^{n} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{z^{n}}{b^{n+1}}$
So, the whole expansion is $\displaystyle \frac{1}{z-b} \left( \sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}} + \sum_{n=0}^{\infty}\frac{z^{n}}{b^{n+1}}\right)\\ \displaystyle = \frac{1}{a-b} \left( \frac{1}{z} + \frac{a}{z^{2}} + \frac{a^{2}}{z^{3}} + \cdots + \frac{1}{b} + \frac{z}{b^{2}}+\frac{z^{2}}{b^{3}}+\cdots\right) \\ \displaystyle = \frac{1}{z(a-b)} + \frac{a}{z^{2}(a-b)} + \frac{a^{2}}{z^{3}(a-b)} + \cdots + \frac{1}{b(a-b)} + \frac{z}{b^{2}(a-b)}+\frac{z^{2}}{b^{3}(a-b)}+\cdots$
(c) $\mathbf{|b|<|z|}$. If $|a|<|z|$, we have, as in part (b) $\displaystyle \frac{1}{(a-b)}\frac{1}{(z-a)} = \frac{1}{(a-b)}\sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$.
If $|b|<|z|$, $\displaystyle \frac{1}{(a-b)}\frac{1}{(z-b)} = \frac{1}{(a-b)} \frac{1}{z \left( 1 - \frac{b}{z}\right)} \\ \displaystyle = \frac{1}{(a-b)}\frac{1}{z} \sum_{n=0}^{\infty} \left( \frac{b}{z}\right)^{n} \\ \displaystyle = \frac{1}{(a-b)}\frac{1}{z}\left( 1 + \frac{b}{z} + \frac{b^{2}}{z^{2}} + \cdots \right) \\ \displaystyle = \frac{1}{(a-b)} \left( \frac{1}{z} + \frac{b}{z^{2}} + \frac{b^{2}}{z^{3}}+ \cdots\right)$
Then, the required Laurent expansion valid for both $|z|>|a|$ and $|z|>|b|$ is found by subtraction:
$ \displaystyle \frac{1}{(a-b)} \left( \frac{1}{z} - \frac{1}{z} + \frac{a}{z^{2}} - \frac{b}{z^{2}} + \frac{a^{2}}{z^{3}} - \frac{b^{2}}{z^{3}} + \cdots \right) \\ \displaystyle = \frac{1}{(a-b)} \left( \frac{a-b}{z^{2}} + \frac{a^{2}-b^{2}}{z^{3}} + \frac{a^{3}-b^{3}}{z^{4}} + \cdots \right) \\ \displaystyle = \left( \frac{1}{z^{2}} + \frac{a^{2}-b^{2}}{z^{3}(a-b)} + \frac{a^{3}-b^{3}}{z^{4}(a-b)} + \cdots \right)$
Again, please let me know how to do part (a), as well as if my parts (b) and (c) are ok. Thanks.
| For $|z|<|a|<|b|$, there are no singularities of $f(z)=\frac{1}{(z-a)(z-b)}$. Thus, we write
$$\begin{align}
f(z)&=\frac{1}{a-b}\left(\frac{1}{z-a}-\frac{1}{z-b}\right)\\\\
&=\frac{1}{a-b}\left(\frac{1/b}{1-z/b}-\frac{1/a}{1-z/a}\right)\\\\
&=\frac{1}{a-b}\left(\frac{1}{b}\sum_{n=0}^\infty \left(\frac{z}{b}\right)^n-\frac{1}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n\right)\\\\
&=\frac{1}{a-b}\sum_{n=0}^\infty \left(\frac{1}{b^{n+1}}-\frac{1}{a^{n+1}}\right)z^n
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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