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How to solve for x in modular arithmetic when the equation has exponent? Say I have the following equation-->
$x^2=180345$ and modulo is $8$. I am using an online calculator to solve this equation, and the answer comes out to be $x=1$ or $3$ or $5$ or $7$.
I am not getting the maths behind this. It will be very helpful if someone can refer to some materials where these kind of equations are solved or atleast show me the algorithm for solving this equaions. Thank you.
| First thing get that $180345$ to something reasonable. $180345 \equiv 1$ so we have
$x^2\equiv 1 \pmod 8$ so $x^2 -1 = \equiv 1 \pmod 8$. Now we can just test them one after another as there are just $7$ of them....
But let's be clever. $x^2 \equiv 1 \pmod 8$
$x^2 -1 \equiv \pmod 8$
$(x+1)(x-1) \equiv 0 \pmod 8$.
So $x\equiv \pm 1\pmod 8$ are two immediate answers.
Any other solutions must relate to the zero divisors of $8$.
$(x+1)(x-1)= 8k$ for some integer $k$. and $x+1, x-2$ are both even. So we have $x-1, x+1 = 2,4$ or $4,6$ or $6,8\equiv 0$ or $0,2$.
So $x = 3,5, 7\equiv -1, $ or $1$.
=======
Also as player3236 stated in the comments, for any odd number $n = 2k+1$ we have $n^2 \equiv 4k^2 + 4k + 1\equiv 4(k^2+k)+ 1\pmod 8$. And as $k^2 + k$ is always even....
$n^2 \equiv 4(k^2+k) + 1 \equiv 8\frac {k^2 + k}2 + 1\equiv 1\pmod 8$.
So every odd number is a solution. $x\equiv 1,3,5,7$
And clearly $x$ even is not a solution so... that's it.
| {
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Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$ The inequality is
$$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$
for $x,y,z,t>0$.
It originates from a 3-D geometry problem involving volumes of tetrahedra etc. Actually, it is equivalent with that problem (see Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.).
The three variables simpler case
$$\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq \frac{3}{4}$$
can be proved using the Cauchy-Schwarz inequality in "Engel's form".
I have tried variants of Holder type inequalities, until now unsuccessfully.
| We need to prove that:
$$2\sum_{cyc}x^3(y+z)(y+t)(z+t)\geq(x+y)(x+z)(x+t)(y+z)(y+t)(z+t)$$ or
$$2\sum_{sym}\left(x^3y^2z+\frac{1}{3}x^3yzt\right)\geq\sum_{sym}\left(x^3y^2z+\frac{1}{3}x^3yzt+x^2y^2zt+\frac{1}{3}x^2y^2z^2\right)$$ or
$$\sum_{sym}(3x^3y^2z-x^2y^2z^2+x^3yzt-3x^2y^2zt)\geq0,$$ which is true by Muirhead because
$(3,2,1,0)\succ(2,2,2,0),$ $(3,2,1,0)\succ(2,2,1,1)$ and $(3,1,1,1)\succ(2,2,1,1).$
| {
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Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$, $\frac{N^2+N}{2}$ has a remainder of 2
I'd like a correct solution as well. Thanks.
| Your first error is that you you scaled by $\,2,\,$ which is not invertible $\!\bmod 4,\,$ so this will not yield an equivalent congruence. Rather, it yields necessary but not sufficient conditions on roots (so possibly extraneous roots). See here for more on the insufficiency of unidirectional inferences.
To get an equivalent congruence we need to scale the modulus too, since $\,4\mid a/2\iff 8\mid a,\,$ so
$$(n^2+n)/2\equiv 0\!\!\!\pmod{4}\iff n^2+n\equiv 0\!\!\!\pmod{8}\qquad$$
Now we can complete the square as you did, but since this too involves scaling the modulus, this will end up being fruitless, leading back to where we started, namely
$$\begin{align} n^2+n&\equiv 0\pmod{8}\\
\iff\ \ \ \ \ \ \ 4n^2+4n&\equiv 0\pmod{32}\\
\iff 4n^2+4n+1&\equiv 1\pmod{32}\\
\iff \ \ \ \ \ \, \color{#c00}{(2n+1)^2}&\:\color{#c00}{\equiv 1}\pmod{32}\\
\iff (2n)(2n+2)&\equiv 0\pmod{32}\\
\iff\ \ \ \ \ \ \ n(n+1)&\equiv 0\pmod{8}
\end{align}\qquad$$
where we solved $\,\color{#c00}{a^2\equiv 1}\,$ by factoring the difference of squares $\,0\equiv a^2-1\equiv (a-1)(a+1).\,$ You can't simply take square roots as you did, e.g. $\,x^2\equiv 1\pmod{8}\,$ has $4$ roots $\,x\equiv 1,3,5,7$.
Instead, by $\,n,\,n\!+\!1\,$ coprime, $\,8\mid n(n\!+\!1)\iff 8\mid n\,$ or $\,8\mid n\!+\!1,\,$ thus we conclude that $\,n(n+1)/2\equiv 0\pmod{4}\iff n\equiv 0,7\pmod{8}$
Finally, beware that modular fractions are well-defined (uniquely exist) only when they are writable with denominator coprime to the modulus, when $\,a/b := ab^{-1}.\,$ For more on modular fractions see here and here.
| {
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Given an arbitrary quadratic polynomial $ax^2 + bx + c$, how do I find the path the vertex of the parabola travels upon when b is varied? https://www.desmos.com/calculator/ysuklqqwfe
I've made a desmos to illustrate the problem. If you press play on b you'll see what I mean.
I'm pretty sure the arc is parabolic because the y coordinate varies quadratically with respect to x. But I'm not sure how to go about proving it, or to find the equation of the path.
Any insights?
| Let $y = ax^2 + bx + c$. For the time being, suppose $a$, $b$, and $c$ are fixed, with $a \ne 0$.
The vertex of the parabola is located where the parabola reaches an extremum, either a minimum (when $a > 0$) or a maximum (when $a < 0$). To see where this occurs, we complete the square:
$$\begin{align}
y &= a\left(x^2 + \frac{b}{a} x\right) + c \\
&= a \left(x^2 + 2 \cdot \frac{b}{2a} x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c \\
&= a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\
&= a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}.
\end{align}
$$
As a function of $x$, keeping in mind $a, b, c$ are fixed, the only term that changes is the first; moreover, because the square of a real number is never negative, the minimum value that $(x + \frac{b}{2a})^2$ can attain is $0$, when $x = -\frac{b}{2a}$. And as stated previously, this choice leads to a minimum when $a > 0$ and a maximum when $a < 0$. Therefore, this is the $x$-value for the vertex, and the $y$-value is the second term, $-\frac{b^2-4ac}{4a}$; thus the vertex has coordinates
$$(x,y)_{\text{vertex}} = \left(-\frac{b}{2a}, -\frac{b^2-4ac}{4a}\right).$$
Now for a fixed $a$ and $c$, where $b$ is allowed to vary, what is the locus of the vertex--what shape does the vertices of the resulting family of parabolas trace out? This can be realized by expressing the $y$-coordinate of the vertex in terms of the $x$-coordinate, in turn as a function of $b$. That is to say, if $x = -b/(2a)$, then $$y = -\frac{b^2 - 4ac}{4a} = -a \left(-\frac{b}{2a}\right)^2 + c = -ax^2 + c,$$ hence the locus of the vertex is the equation $$y = -ax^2 + c,$$ also a parabola, but with the linear term equal to $0$, and facing the opposite direction to the original family of parabolas. This of course should be obvious in retrospect.
| {
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The $2019$th strictly positive integer $n$ such that $\binom{2n}{n}$ is not divisible by $5$ This is problem $6$ from the $2019$ Pan African Mathematical Olympiad.
Here's my attempt:
We're looking for all integers $n$ such that $5$ doesn't divide $\binom{2n}{n}$.
Which means we need $v_5((2n)!) - 2 v_5(n!) = 0$
By Legendre's formula, this is equivalent to:
$$
\sum_{k = 1}^\infty \left\lfloor \frac{2n}{5^k} \right\rfloor - 2\sum_{k = 1}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor = 0 \\
\iff \sum_{k = 1}^\infty \left(\left\lfloor \frac{2n}{5^k} \right\rfloor - 2\left\lfloor \frac{n}{5^k} \right\rfloor\right) = 0
$$
Which is true if and only if $\left\lfloor \frac{2n}{5^k} \right\rfloor = 2\left\lfloor \frac{n}{5^k} \right\rfloor$ for all strictly positive integers $k$.
Let's find the positive real solutions to the equation $\lfloor2x\rfloor = 2 \lfloor x \rfloor$
$$
\lfloor2x\rfloor = 2 \lfloor x \rfloor \\
\iff 2x - \{2x\} = 2x - 2 \{x\} \\
\iff \{2x\} = 2 \{x\}
$$
Which is true if $\{x\} < 1/2$.
Inversely, if $\{x\} > 1/2$ then $\{2x\} > 1$ which is absurd.
It's also trivially false for $x = 1/2$.
As such, for every strictly positive integer $k$, there exists a positive integer $m_k$ such that:
$$
m_k \le \frac{n}{5^k} < m_k + \frac1{2} \\
\iff 5^k m_k \le n < 5^k m_k + \frac{5^k}{2}
$$
I can see that this can be used to eliminate many possible remainders of $n$ modulo any power of $5$.
For example, we have:
$$
5 m_1 \le n < 5 m_1 + \frac{5}{2} \\
\iff 5 m_1 \le n \le 5m_1 + 2
$$
Which means $n \mod 5 \in \{0 , 1, 2 \}$
And similarly,
$$
25 m_2 \le n \le 25m_2 + 12
$$
So $n \mod 25 \in [0,12]$
But, since $n \mod 5 \in \{0,1,2\}$, we have $n \mod 25 \in \{0,1,2,5,6,7,10,11,12\}$
We can carry on with this for higher powers of $5$.
However, I don't know how exactly to proceed from here.
I don't just want an answer.
I want the reasoning and intuition behind your answer in order to know how I should approach such problems in the future so I would really appreciate it if you include a little bit about how you found the solution if it's not really obvious. I don't mind if you also include failed attempts on the way to finding the solution.
| A possible way to find the answer:
Exploring the numbers for which $\left\lfloor \frac{2n}{5} \right\rfloor - 2 \left\lfloor \frac{n}{5} \right\rfloor =0$ shows the $3$ out of every $5$ integers satisfy this, with the $3$th, $6$th etc. cases being multiples of $5$.
Similarly exploring the numbers for which $\left\lfloor \frac{2n}{5^2} \right\rfloor - 2 \left\lfloor \frac{n}{5^2} \right\rfloor =0$ shows that $13$ out of every $25$ integers satisfy this, with the $13$th, $26$th etc. cases being multiples of $25=5^2$. Overlapping this with the previous result, $9$ out of every $25$ integers satisfy these two.
You can then prove that $\left\lfloor \frac{2n}{5^k} \right\rfloor - 2 \left\lfloor \frac{n}{5^k} \right\rfloor =0$ if and only if $n \in \{0,1,2,\ldots,\frac{5^k-1}{2}\} \mod 5^k$
and by induction that $\sum\limits_{k = 1}^t \left\lfloor \frac{2n}{5^k} \right\rfloor - 2\sum\limits_{k = 1}^t \left\lfloor \frac{n}{5^k} \right\rfloor = 0 $ for $3^t$ out of every $5^t$ integers with the $3^t$th, $2\times 3^t$th etc. being multiples of $5^t$.
To find the solution using this information:
*
*we can find $3^6=729\le 2019$ and $3^7=2187>2019$. So $t=7$ is too big.
*$2019 = 2\times 729+2\times243 + 2\times 27 + 2\times 9 +1\times 3$ which is $2\times 3^6+2\times3^5 + 2\times 3^3 + 2\times 3^2 +1\times 3^1$.
*The desired $n=2\times 5^6+2\times5^5 + 2\times 5^3 + 2\times 5^2 +1\times 5 ^1 = 37805$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$ Find the elliptic integral $$\int\limits^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,\mathrm{d}k,$$
where
$$K(k) = \int\limits_{0}^{1} \frac {\mathrm dt}{\sqrt{(1 - t^2)(1 - k^2t^2)}}=\int\limits_{0}^{\pi/2} \frac{\text{d}\vartheta}{\sqrt{1 - k^2 \sin^2\vartheta}}$$
My approach: Let $k=\dfrac{\sin\theta}{\sin\phi}$
Now after some calculation, I have reached the following: $$\int\limits_{0}^{\pi/2}\int\limits_{0}^{\phi}\dfrac{d\theta d\phi}{\sqrt{\sin^2\phi-\sin^2\theta}}$$
Now how to proceed from this point..please give some idea/hint.
Definition of elliptic integral: https://en.wikipedia.org/wiki/Elliptic_integral
| *
*Replace $K$ with its $_2F_1$ representation (DLMF 19.5.1).
*Evaluate using the Euler integral representation of $_3F_2$ (DLMF 16.5.2).
*Simplify using Dixon's well-poised $_3F_2$ sum (DLMF 16.4.4).
*Simplify using the functional relations for $\Gamma$ (DLMF 5.5).
The result is
$$\frac{\Gamma(1/4)^4}{16\pi}\text{.}$$
In detail:
$\newcommand{\d}{\mathrm{d}}$
$$\begin{split}
\int_0^1 \frac{K(k)}{\sqrt{1-k^2}}\d k
&= \frac{\pi}{2} \int_0^1 \frac{_2F_1(\tfrac{1}{2},\tfrac{1}{2};1;k^2)} {\sqrt{1-k^2}}\d k \\
&= \frac{\pi}{4} \int_0^1 \frac{_2F_1(\tfrac{1}{2},\tfrac{1}{2};1;x)} {\sqrt{x(1-x)}}\d x \\
&= \frac{\pi^2}{4}\,_3F_2(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};1,1;1) \\
&= \frac{\pi^2}{4}\left(\frac{\Gamma(\tfrac{5}{4})\Gamma(1)\Gamma(1)\Gamma(\tfrac{1}{4})}{\Gamma(\tfrac{3}{2})\Gamma(\tfrac{3}{4})\Gamma(\tfrac{3}{4})\Gamma(\tfrac{1}{2})}\right) \\
&= \frac{\Gamma(\tfrac{1}{4})^4}{16\pi}\text{.}
\end{split}$$
| {
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$\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2}.$ I would like to ask why is the sequence convergent to $\dfrac{1}{2}$ when the parameter $|a|<1$ ?
At the same time, I would like to ask if there is a limit for $a<-1$, because the sequence would alternate with signs.
$$\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2}$$
| I have tried to make the answer as simple as possible, so forgive me if it sounds a bit silly to the many experts among you.... :~)
First we simplify $\frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}$
we know that if |a|<1, then, $\lim _{x\rightarrow \infty }a^n=0$ (as when numbers smaller than one are raised to higher powers, they get smaller, eg. (0.01)^2=0.001,etc ...)
Hence, $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}=\frac{0}{2+0}+\frac{1}{2+0}=\frac{1}{2}$
1 raised to any positive power is 1, hence at a=1, we have $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{\left(1^{\infty }+1\right)}{2+1^{\infty }}=\frac{2}{3}$
But if |a|>1, then $\lim _{x\rightarrow \infty }a^n=\infty$ (as when numbers bigger than one are raised to higher powers, they get bigger, eg. 10^2=100,etc ...)
Hence, $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}=\frac{\left(a^n\cdot a^n\right)}{2+a^n}+\frac{1}{2+a^n}=\frac{\left(a^n\right)}{\frac{2}{a^n}+1}+\frac{1}{2+a^n}=\frac{\infty }{\frac{2}{\infty }+1}+\frac{1}{2+\infty }=\infty $
| {
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Let $x \in \mathbb{N}$. Prove that if $x$ is odd, then $\sqrt{2x}$ is not an integer. I need help verifying my proof the following theorem:
Let $x$ be a natural number. Prove that if $x$ is odd, then $\sqrt{2x}$ is not an integer.
The following are the definitions of even and odd numbers:
Definition 3.1. An integer $x$ is odd if there is an integer $n$ such that $x = 2n+1$
Definition 3.2. An integer $x$ is even if there is an integer $n$ such that $x = 2n$
$Proof.$
The contrapositive of the statement is: If $\sqrt{2x}$ is an integer, then $x$ is even.
Because an integer is odd or even and not both and $x$ is a natural number,
\begin{align*}
\sqrt{2x} &= 2n+1 \qquad &\text{for some positive integer n (Definition 3.1)};\\
\qquad &\text{or}\\
\sqrt{2x} &= 2m \qquad &\text{for some positive integer m (Definition 3.2)};\\
\qquad &\text{and not both.}
\end{align*}
In case $\sqrt{2x}$ is odd:
\begin{align*}
(\sqrt{2x})^{2} &= (2n+1)^{2}\\
2x &= 4n^2 + 4n +1\\
x &= 2n^2 + 2n + \frac{1}{2}\\
x &= 2(n^2 + n) + \frac{1}{2}
\end{align*}
Because $n^2 + n$ is some integer, $2(n^2 + n)$ is even (Definition 3.2).
Since no even number added by $\frac{1}{2}$ is an integer, $\sqrt{2x}$ is not odd.
In case $\sqrt{2x}$ is even:
\begin{align*}
(\sqrt{2x})^{2} &= (2m)^{2}\\
2x &= 4m^2\\
x &= 2m^2
\end{align*}
Because $m^2$ is some integer, $2m^2$ is even (Definition 3.2).
Because each implications and its contrapositive are equivalent, given that $x$ is a natural number, if $x$ is odd, then $\sqrt{2x}$ is not an integer.
The question is from the book:
Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 31). New York: Springer.
| Another proof is that if $y=2x$ with $x$ odd, then $2|y$ but $4 \nmid y$.
Hence $\sqrt{y}=k \implies y=k^{2}={p_{1}}^{2}\cdot\cdot\cdot {p_{n}}^{2}$, a contradiction since $p_{i}=2$ for some $i$.
| {
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Evaluate $I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1}$
$$I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1},$$
I used
$$u=\frac{c}{xt}-\frac{mt}{x}$$
and
$$du=-\left(\frac{c}{xt^2}+\frac{m}{x}\right)dt$$
but I'm not able to find the limits at $t=0$ and $t=\infty$. Neither are straight forward to use as
$$\int_0^\infty\frac{1}{1+t^2}dt=\pi/2$$
Any hints on how to proceed?
| Rearrange the integral as follows
\begin{align}
& \int_0^\infty\frac{1}{x^2t^2}\frac{1}{(\frac{c}{xt}-\frac{mt}{x})^2+1}dt \\
=& \frac1{2cx} \int_0^\infty \frac{(\frac{c}{xt^2}+\frac mx)+ (\frac{c}{xt^2}-\frac mx )}{(\frac{c}{xt}-\frac{mt}{x})^2+1}dt\\
=& \frac1{2cx} \int_0^\infty \frac{d(\frac {mt }x-\frac{c}{xt})}{(\frac{c}{xt}-\frac{mt}{x})^2+1}
-\frac1{2cx} \int_0^\infty \frac{d(\frac{c}{xt}+\frac {mt }x )}{(\frac{c}{xt}+\frac{mt}{x})^2+1-\frac{4cm}{x^2}}\\
\end{align}
where the first integral is of the form $\int_0^\infty\frac{du}{1+u^2}$ and the second integral vanishes.
| {
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I want to obtain partition of an integer with an initial value and I want to obtain a partition of an Integer with an initial value and
the value following it is smaller and the value following it is smaller than the previous value and no value repeats itself.
within a limited number of terms
For example
Integer = 41
Intial Value = 11
number of terms = 8
*
*11+9+6+5+4+3+2+1 = 41
*11+8+7+5+4+3+2+1 = 41
let's say I am going to generate such partitions for number 42, how to do that?
| I am not sure whether a super direct approach would exist, these problems with partitions and restrictions are usually complicated to compute since closed formulas are not easy to find and a lot relies on recurrences and bijections between different kind of partitions.
I guess the easiest way to attack what you say is via generating functions: since you don't only want terms to be different, but also have a condition on the total number of terms used, you will need two variables. One, say $x$, will count the actual contribution of the term, while $z$ will count how many terms this contributes. For example, a term $z^2x^4$ would be saying a contribution of $4$ with two terms, i.e. a $2 + 2$ or a $3 + 1$, although we don't know which contributed this term in the final product.
If your biggest term is $N$, then that is a $zx^N$. After that you allow any term to appear, but maybe it won't, so each possible term has a generating function $1 + zx^j$, for $1\le j \le N - 1$. So, the generating function for the partitions whose largest term is $N$, the remaining ones do not repeat and in which you count how much contributions you have is:
$$zx^N(1 + zx^{N - 1})...(1 + zx).$$
To make an example, if $N = 5$ what you have to see is the product
$$\begin{align*}
&\;zx^5(1 + zx^4)(1 + zx^3)(1 + zx^2)(1 + zx)\\
&= zx^5(1 + zx + zx^2 + (z + z^2)x^3 + (z + z^2)x^4 + 2z^2x^5 + (z^2 + z^3)x^6 + (z^2 + z^3)x^7 + z^3x^8 + z^3x^9 + z^4x^{10})\\
&=zx^5 + z^2x^6 + z^2x^7 + (z^2 + z^3)x^8 + (z^2 + z^3)x^9 + 2z^3x^{10} + (z^3 + z^4)x^{11} + (z^3 + z^4)x^{12} + z^4x^{13} + z^4x^{14} + z^5x^{15}
\end{align*}
$$
For example, this would be telling us that for $10$ there are only $2$ ways to make this, each of them uses $2$ terms (that is the $z^2$), and sure enough these are $5 + 4 + 1$ and $5 + 3 + 2$, while for $9$ this would be saying that there is one way with $2$ terms and one with $3$ terms: $5 + 4$ and $5 + 3 + 1$.
These polynomials get complicated to manipulate as your parameters grow, nevertheless if you want to bound the number of terms, then you want to disregard any terms whose power of $z$ is bigger than some number. For example, if you did not want to count partitions with more than $3$ terms you take quotient modulo $z^4$ and the polynomial becomes:
$$
zx^5 + z^2x^6 + z^2x^7 + (z^2 + z^3)x^8 + (z^2 + z^3)x^9 + 2z^3x^{10} + z^3x^{11} + z^3x^{12}
$$
Finally, if what you want at this point is just to see how many there are of a given integer, without knowing how many terms they have since you already took care of your bound then you put $z = 1$ and you get:
$$
x^5 + x^6 + x^7 + 2x^8 + 2x^9 + 2x^{10} + x^{11} + x^{12}
$$
This, for example, would tell you: there are $2$ partitions of $10$ in which:
*
*The biggest term is $5$,
*All terms are different,
*The number of terms is at most $3$.
These would be $5 + 4 + 1$ and $5 + 3 + 2$. For eleven, there would be only one: $5 + 4 + 2$.
For $N = 42$ you can do this, tho you would probably need a computer that makes the calculations for you, and you need to do everything in the correct modulus from the start to make sure computations remain doable.
| {
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Find $\int \frac 1 {(x^2 +a^2)^2} dx $. I am trying to integrate $\int \frac 1 {(x^2+a^2)^2} \ dx$. The only thing that I can think to try is substitution, $u=x^2+a^2$ so that $\frac{du}{dx}=2x \Rightarrow du = 2x\ dx = 2\sqrt{u-a^2}\ du$ and then the integral becomes
$$ \int \frac{1}{u^2} (2\sqrt{u-a^2} \ du) $$
which does not seem productive.
I could try perhaps to separate with integration by parts, setting $u = \frac{1}{x^2 +a^2}$ and $dv = \frac{dx}{x^2+a^2}$. Then we obtain $\frac{du}{dx} = -\frac{1}{(x^2+a^2)^2}(2x)$ and $v = \frac 1 a \tan^{-1}(x/a)$. Then the integral becomes
$$ \begin{align*}
uv - \int v \ du &= \left(\frac{1}{x^2 +a^2}\right)\left(\frac 1 a \tan^{-1}(x/a)\right) - \int \left(\frac 1 a \tan^{-1}(x/a)\right) \left( -\frac{1}{(x^2+a^2)^2}(2x) \right) \ du
\end{align*}$$
but this also looks like it's headed nowhere good. Advice?
| Another thing you can do is to compute first
$$
\int\frac{\text{d}x}{a^2+x^2}
$$
and i'm sure you can compute it.
Then you can use integration by parts on this previous integral, that will make $\displaystyle \frac{1}{\left(a^2+x^2\right)^2}$ appear.
| {
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Simplifying $z^2+i=0$ I need to simplify $z^2+i=0$ and find all solutions for $z$. I have seen that the solutions to $z=\sqrt{i}=\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)$. I was hoping to find a similar solution for $z=\sqrt{-i}\,$ but my attempt gives me $z=\pm i^{\frac{3}{2}}$
$$z=re^{i\theta} \,\,\& \,\, e^{i\pi}=-1 $$
then,
$$(re^{i\theta})^2=-i\\r^2e^{i2\theta}=ie^{i\pi+k(2\pi)}$$ where $k\in\mathbb{Z}$.
So, we have $\begin{cases} r^2=i \,\,\,\therefore r=\sqrt{i}\\ \theta=k\pi
\end{cases}$
Then, $$z_k=\sqrt{i} \, e^{i\left(\frac{\pi}{2}+k\pi\right)}$$
$$z_0=\sqrt{i}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=i^\frac{3}{2}\\
z_1=\sqrt{i}\left(\cos(\frac{\pi}{2}+1)+i(\sin\frac{\pi}{2}+1)\right)=-i^\frac{3}{2}$$
I realized that this is literally the same as just solving $z=\sqrt{-i}=i\sqrt{i}=i^\frac{3}{2}$, however, I was hoping to find a solution of the form $x+iy$. I am not sure how to go about this problem a different way.
| Here's the solution to $z^2 = i$. Give $z^2 + i = 0$ a try after looking through this.
You have the right idea in using $ z = re^{i\theta} $. First, note that we can choose $r = 1$ because $|re^{i\theta}| = |r|$, and $|i| = 1$. Then, substitute:
$$ z^2 = i$$
$$ (e^{i\theta})^2 = i $$
$$ e^{i(2\theta)} = i $$
$$ \cos(2\theta) + i\sin(2\theta) = i . $$
Comparing the real parts of the left and right-hand sides, we have
$$ \cos(2\theta) = 0 $$
$$ \Rightarrow 2\theta = \frac{\pi}{2} + \pi k $$
$$ \Rightarrow \theta = \frac{\pi}{4} + \frac{\pi}{2}k $$
Now comparing the imaginary parts, we have
$$ \sin(2\theta) = 1 $$
$$ \Rightarrow 2\theta = \frac{\pi}{2} + 2\pi k $$
$$ \Rightarrow \theta = \frac{\pi}{4} + \pi k $$
The two solutions have an intersection of $ \theta = \frac{\pi}{4} + \pi k $. In one rotation from $\theta = 0$ to $\theta = 2\pi$, that gives us two unique solutions:
$$ \theta = \frac{\pi}{4} \ \text{or} \ \theta = \frac{5\pi}{4} .$$
Substituting these back into $z = e^{i\theta} = \cos(\theta) + i\sin(\theta) $, we get the solutions you pointed out:
$$ \theta = \frac{\pi}{4} \Rightarrow z = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i , $$
$$ \theta = \frac{5\pi}{4} \Rightarrow z = -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i . $$
Now, for $z^2 = -i$, the solution should be a rotation of $\frac{\pi}{2}$ from the solution we got here, i.e., you should get to $\theta = \frac{3\pi}{4} \ \text{or} \ \frac{7\pi}{4}$.
| {
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How to find the value of $x^3(x^3-18)$? If $x(x-3)=-1$ then the value of $x^3(x^3-18)$ will be ?
$A)1\quad\quad\quad\quad B)-1\quad\quad\quad\quad C)0 \quad\quad\quad\quad D)|-1|$
I tried solving $x(x-3)=-1$:
$x^2-3x+1=0\quad\text{hence}\quad x_{1,2}=\cfrac{3\pm\sqrt{5}}{2}$. But it seems that doesn't help a lot to find the value of $x^6-18x^3$. as an alternative way, from the quadratic we have $x^3=3x^2-x$. if we plug in it $x^3(x^3-18)$ we obtain $(3x^2-x)(3x^2-x-18)$ but I can't continue from here.
Also the options are strange because $|-1|=1$ so it seems the answer should be $B$ or $C$.
| Since $x^2=3x-1$, $x^3=3x^2-x=8x-3$, so$$x^3(x^3-18)=(8x-3)(8x-21)=64x^2-192x+63=192x-64-192x+63=-1,$$so $B$) is correct. Alternatively, use$$x^6-18x^3+1=(x^2-3x+1)(x^4+3x^3+8x^2+3x+1).$$
| {
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The inverse type of Bernhard Leeb's solution for IMO‐1983–inequality
Given three side-lengths $a, b, c$ of a triangle. Prove that
$$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq 3\left ( a+ b- c \right )c\left ( a- b \right )\left ( b- c \right )$$
Source: StackMath/@haidangel ft.@tthnew
I used discriminant to create this inequality, also $constant\!:\!=\!\!3$ is the best here. See_ on.StackMath, that's what I'm doing research on, of course this inequality is a result. Not an answer. I think we use Bernhard Leeb's result to help a real lot
$$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ):= \left ( c+ a- b \right )\left ( c- a \right )^{2}b- \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c$$
| Just a comment but it's too long for a comment.
Given $a, b, c$ be three side-lengths of a triangle. Prove that
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geqslant 3(a+b-c)(a-b)(b-c)c$$
Source: By MSE/@tthnew
Some try to proof: Since $a,b,c$ be three sid-lengths of a triangle, if $(a-b)(b-c)<0$ it's obvious.
So, we only need to consider this inequality when $(a-b)(b-c)\geq 0\to b\equiv \text{mid}\{a,b,c\}.$
But I can't prove this.
By the way if we change $3\to \frac{12}{5}=2.4,$ it's much easy to prove (for the weaker version).
Indeed, let $a=x+y,b=y+z,c=z+x$ and multiply our inequality with $x+y+z.$ The inequality becomes:
$$\frac{1}{20} \left( x-y \right)^2z^2 \left( 15x+31y \right) +\dfrac{1}{20}
\left( 2x-y-z \right)^2y \left( 20x^2+19xy+46zx+37
z^2 \right) $$ $$+{\frac {21}{20}}
\left( zx+y^2-2yz \right)^2x+\frac {17}{10} \left( x-z
\right)^2y^2z+\frac {3}{20} \left( z+x-2y \right) ^2z
^2y$$
$$+\frac{2}{5} \left( x^2-yz \right)^2 \left( 2y+5
z \right) +\frac{1}{5} \left( y-z \right) ^{2}x^3\ge 0.$$
According to Mr. RiverLi's idea,
After let $a=x+y,b=y+z,c=z+x$ then $f(a,b,c)=f(x+y,y+z,z+x)\equiv f(x),$ and we only need to prove
$$\Delta_x=-16 y^2 (y-z)^4 (y+z) (3 y+z) \left(y^2-4 y z+27 z^2\right)\le 0,$$
which is clearly true. I'm still try to find another way.
| {
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Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$
I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$
Am I doing this right? How can I finish this problem?
| It can be proven in a straightforward way, just basic inequalities and definition of the real numbers, we'll have the inequality: $0\leq a\leq b\leq c$.
$$c^2\geq a^2 \Rightarrow \frac{c^2}{a^3}\geq\frac{1}{a}$$
$$\text{and} \quad (\frac{c}{b})^n \geq0> \frac{c^2-b^2}{a^2-b^2} \Rightarrow c^3(a^2-b^2)\geq b^3(c^2-b^2)\Rightarrow c^3a^2-c^3b^2\geq b^3c^2-b^5 \\ a^2-b^2\geq \frac{b^3}{c} - \frac{b^5}{c^3} \Rightarrow \frac{a^2}{b^3}-\frac{1}{b}\geq\frac{1}{c}-\frac{b^2}{c^3} \\ \frac{a^2}{b^3}+\frac{b^2}{c^3}\geq\frac{1}{b}+\frac{1}{c}$$
You can see the same thing happening for any order of inequalities between $a,b,c$, and you combine the two equations from above.
| {
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Divisibility by $12$ of $\overline{abcdef}$ How can one find all six digits numbers formed only with even number that are divisible by $12$?
My attempt was to find all six digits numbers $\overline{abcdef}$ with all digits from $\{0,2,4,6,8\}$ that are divisible with $3$ or $4$. For doing this I used inclusion-exclusion principle. How can I continue?
| A trick is to reduce the restriction "must be divisible by $12$" to a restriction on as few digits as possible and let the other digits "run free".
Must be divisible by $12$ means must be divisible by $3$ and $4$.
Being divisible by $3$ means $a+b+c+d+e+f \equiv 0 \pmod 3$. And that means we can take any one of the digits, say $f$, and note $f \equiv-(a+b+c+d+e)\pmod 3$. So $a,b,c,d,e$ can "run free" and we have reduced the restriction "is divisible by $3$" to a restriction on only one digit.
Being divisible by $4$ means that as $\overline{abcdef} = \overline{abcd}\times 100 + \overline {ef}$ that $\overline {ef} =10e + f$ be divisible by $4$. But if $e$ is already being restricted to being even then $10e +f = 20\cdot \frac e2 + f$ and this reduces to a single restriction on $f$; that $f$ is divisible by $4$. Or that $f = 0,4,8$.
So the restriction that $a,b,c,d,e,f$ are all even and that $\overline{abcdef}$ is divisible by $12$ reduces to a single restriction on $f$ (assuming $a,b,c,d,e,$ are all even); that $f=0,4,8$ and that $f\equiv -(a+b+c+d+e)$.
But $0\equiv 0\pmod 3$ and $4\equiv 1\pmod 3$ and $8\equiv 2\pmod 3$ so if $a,b,c,d,e$ "run free" then our single restriction on $f$ can be expressed precisely as: $f=\begin{cases}0&a+b+c+d+e \equiv 0 \pmod 3\\4&a+b+c+d+e\equiv -1\equiv 2\pmod 3\\8&a+b+c+d+e\equiv -2 \equiv 1 \pmod 3\end{cases}$
So our restrictions are:
*
*$\overline{abcdef}$ is six digits. So $a \ne 0$ and $a$ is even. There are $4$ options.
*$b,c,d,e$ are even. There are $5$ options each or $5^4$ total.
*$f=\begin{cases}0&a+b+c+d+e \equiv 0 \pmod 3\\4&a+b+c+d+e\equiv -1\equiv 2\pmod 3\\8&a+b+c+d+e\equiv -2 \equiv 1 \pmod 3\end{cases}$. There is $1$ option.
So there are $4\cdot 5^4 = 2500$ such numbers.
| {
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What is the domain of k? $2|x-2|-3|x-3|=k$ What is the domain of k?
$2|x-2|-3|x-3|=k$
Using the definition of absolute function, I obtain:
$|x-2|= \begin{cases}
x-2, & x\ge2 \\
2-x, & x<2\\
\end{cases}\\$
$|x-3|= \begin{cases}
x-3, & x\ge3 \\
3-x, & x<3 \\
\end{cases} \\$
For $x<2$
$x-5=k$
For $2\le x<3$
$5x-13=k$
For $x \ge 3$
$-x+5=k$
But I need to find the domain of k.
| *
*If $ -\infty <x≤2$, then you have
$$k=2(2-x)-3(3-x)=x-5 \Longrightarrow - \infty <k=x-5≤-3$$
*
*If $2<x≤3$, then you have
$$k=2(x-2)-3(3-x)=5x-13 \Longrightarrow -3<k=5x-13≤2$$
*
*If $3<x<\infty$, then you have
$$k=2(x-2)-3(x-3)=-x+5 \Longrightarrow 2>k=-x+5> - \infty \Longrightarrow -\infty <k<2$$
Finally, you get
$$k \in (- \infty, -3] ∪ (-3, 2] ∪ (-\infty, 2)= (-\infty,2].$$
| {
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Limit of this sequence $ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$? How to find the limit of this sequence. I tried using the squeeze theorem but no result yet.
$ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$
| It is easy to see that $$\frac{1}{2}(\frac{4^n}{6^n})<\frac{2^n+3^n+4^n}{5^n+6^n}<3(\frac{4^n}{6^n}).$$ Therefore,
$$\frac{4}{6}\sqrt[n]{\frac{1}{2}}<(\frac{2^n+3^n+4^n}{5^n+6^n})^\frac{1}{n}<\sqrt[n]{3}\frac{4}{6}.$$ Since $\sqrt[n]{c}\to 1$ as $n\to\infty$, we have that $$\lim_{n\to\infty}(\frac{2^n+3^n+4^n}{5^n+6^n})^\frac{1}{n}=\frac{4}{6}=\frac{2}{3}.$$
| {
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Trigonometric Manipulation with inequality:$x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.$
Find all angles $\theta,$ $0 \le \theta \le 2 \pi,$ with the following
property: For all real numbers $x,$ $0 \le x \le 1,$
$$x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.$$
I am not exactly sure how to solve this problem. A friend gave me the suggestion to use the substitution $\sin^2(y)=x,\cos^2(y)=1-x$. Then, $\sin^4(y)\cos(\theta)-\sin^2(y)\cos^2(y)+\cos^4(y)\sin(\theta) > 0$. I tried to exploit symmetry by dividing by $\sin^2(y)\cos^2(y)$ to get $\tan^2(y)\cos(\theta)+\cot^2(y)\sin(\theta)>1$. I am not sure how to continue from here. Anything I tried from here was fruitless. What should I do to solve this problem?
| Hint
$$f(x)=x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0\Leftrightarrow$$
$$f(x)=x^2(\sin \theta+\cos \theta+1)-x(1+2\sin \theta)+\sin \theta>0.$$
Let's look at the discriminant,
$$\Delta=(1+2\sin \theta)^2-4(\sin \theta+\cos \theta+1)\sin \theta$$
$$\Delta=1-2\sin (2\theta)\to -1\le \Delta\le 3.$$
First case of solutions,
*
*$\Delta<0$ and $\sin \theta+\cos \theta+1 >0.$
Second case,
*
*$\Delta\ge 0$: $x_1\le x_2$ are the roots of $f(x)$,
a) If $\sin \theta+\cos \theta+1 >0$ then $x_2\le0$ or $x_1\ge 1$;
b) If $\sin \theta+\cos \theta+1 <0$ then $[0,1]\subset[x_1,x_2]$.
| {
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Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$? Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?
I believe it is since $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[b]{a}}^\frac{1}{c}=\frac{1}{\sqrt[bc]{a}}$.
Can someone please confirm my suspicions as I just came across a question that states in the answer that $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\sqrt[bc]{a}$ and I believe that this is a typo.
| ${ \left( {\frac{1}{a}}^{\frac{1}{b}} \right)}^{\frac{1}{c}}= { \frac{1}{a}}^{ \frac{1}{b} \cdot \frac{1}{c}} = { \frac{1}{a} }^{ \frac{1}{bc}} = \frac{1}{a^{\frac{1}{bc}}} = \frac{1}{\sqrt[bc]{a}}$
Thus, yes (assuming $a,b,c$ are all positive reals).
| {
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Extrema of $(1+\sin x)(1+\cos x)$
Find the extrema of $(1+\sin x)(1+\cos x)$ without using calculus.
I was able to figure out the minima by observing that each of the brackets range from $0$ to $2$. Therefore the minima has to be $0$ when either one of the brackets is zero.
However I couldnt figure out the maxima. I tried expanding it to complete the square but it didnt quite work out well.
Any hint is appreciated!
| We can also use the inequality
\begin{gather*}
AM\geq GM\\
So,\ \\
\sqrt{( 1+\sin x)( 1+\cos x)} \leq \frac{1+\sin x+1+\cos x}{2}\\
Now,\ the\ function\ on\ the\ RHS\ is\\
1+\frac{\sin x+\cos x}{2} =1+\frac{1}{\sqrt{2}}\sin\left( x+\frac{\pi }{4}\right)\\
The\ maxima\ of\ this\ function\ is\ 1+\frac{1}{\sqrt{2}} ,\\
because\ maximum\ value\ of\ \sin\left( x+\frac{\pi }{4}\right) =1\\
\sqrt{( 1+\sin x)( 1+\cos x)} \leq 1+\frac{1}{\sqrt{2}}\\
( 1+\sin x)( 1+\cos x) \leq \left( 1+\frac{1}{\sqrt{2}}\right)^{2} =\frac{3+2\sqrt{2}}{2} =\frac{3}{2} +\sqrt{2}
\end{gather*}
| {
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What is the sum of the coefficients? If $h(x) = x^4+ax^3+bx^2+cx+d$ then what is $a+b+c+d$?
I try:
\begin{align}
x=2: 2^4+2^3a+2^2b+2c+d = 3 &\implies 8a+4b+2c+d = -13 \label{I} \tag{I}\\
x=-2:-2^4-2^3a-2^2b-2c+d=3 &\implies -8a+4b-2c+d = -13 \label{II} \tag{II}\\
\eqref{I} + \eqref{II} \colon 8b +2d = -26 &\implies \boxed{4b+d = -13} \ em \tag{I}\\
8a-13+2c=-13 &\implies 8a+2c = 0 \implies\boxed{4a+c = 0}
\end{align}
I stop here..don't find another equation
| Hint
I believe there is a better approach for this problem.
Considering the graph, the polinomial $h(x)-3$ has a double root at $-2$ and $2$. So, it can be writen as
$$h(x)-3=k(x-2)^2(x+2)^2=k(x^2-4)^2.$$
In addition, $h(1)$ give you the sum of the coefficients of $h(x)$.
Can you finish?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Lagrange multipliers to optimize sum of cubes of roots of a quadratic
If $\lambda \geq -2$ such that the sum of cubes of roots of the equation $x^2 + (2-\lambda )x + (10-\lambda) = 0$ is minimum, then the magnitude of difference of root is..?
Let roots be $\alpha $ and $\beta$, we wish to maximize $f(\alpha,\beta) = \alpha^3 + \beta^3$ under the constraint of $ \alpha + \beta = 2- \lambda $ and $\alpha \beta= 10 -\lambda$ , combining the constraints we get:
$$ \alpha + \beta + \alpha \beta = 8$$
Now apply lagrange multiplier equation, we get:
$$ 3 \alpha^2 = \mu ( 1 + \beta)$$
And,
$$ 3 \beta^2 = \mu ( 1 + \alpha)$$
Taking the ratio of the two equations:
$$ \frac{\alpha^2 }{\beta^2} = \frac{1+\beta}{1 + \alpha} \tag{1}$$
Substracting one on both sides:
$$ \frac{\alpha + \beta}{\beta^2} = \frac{-1}{1+\alpha} \tag{2}$$
This leads to:
$$ \alpha + \beta + \alpha^2 + \alpha \beta = - \beta^2$$
Hence,
$$ -12 = \alpha^2 + \beta^2$$
Now.. I am not sure what to do. One way is to ditch the lagrange multiplier approach and directly express the objective function in terms of $\lambda$ and minimize but is there a way to proceed further from the information we have?
| This is a solution without Lagrange Multiplier method. See my edit in the end for Lagrange Multiplier method.
$\alpha + \beta = \lambda-2, \alpha\beta = 10 - \lambda$
$f(\alpha, \beta) = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta(\alpha + \beta)$
So, $f(\alpha, \beta) = (\lambda-2)^3 - 3 (10-\lambda)(\lambda-2) = \lambda^3 - 3 \lambda^2 - 24 \lambda + 52$
Now we take derivative with respect to $\lambda$ and equate to $0$,
$\lambda^2 - 2\lambda-8 = 0 \implies \lambda = -2, 4$
Second derivative test confirms a local minima at $\lambda = 4$. We also observe that the function $f$ is monotonically decreasing for $\lambda \lt - 2$ and is monotonically increasing for $\lambda \gt 4$. So in the given domain $\lambda \geq - 2$, absolute minima occurs at $\lambda = 4$.
And so we have,
$\alpha + \beta = \lambda - 2 = 2, \alpha \beta = 10 - \lambda = 6$
Solving, $\alpha = 1 \pm i \sqrt 5, \beta = 1 \mp i \sqrt5$, which are complex roots of the quadratic for $\lambda = 4$.
For Lagrange Multiplier method,
Say $\alpha + \beta = x, \alpha \beta = y$
You found $\alpha^2 + \beta^2 = - 8$. We also have $y = 8 - x$
$x^2 = - 8 + 2y = - 8 + 16 - 2x \implies x^2 + 2x - 8 = 0 \ $
(where $x = \alpha + \beta$)
Solving $x = 2, y = 6; x = - 4, y = 12$.
We test $\alpha^3 + \beta^3 = x^3 - 3 xy$ for both and obtain $x = 2, y = 6$ to be the minima.
$(\alpha-\beta)^2 = x^2 - 4y = - 20 \implies \alpha-\beta = \pm \sqrt{-20}$
| {
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Expression of $\cos^{-1}\left(4x^3-3x\right)$ The Original question : (translated from French)
Let $f$ be the function
$$
f\left(x\right)=\cos^{-1}\left(4x^3-3x\right)
$$
*
*
*
*Find the definition domain of $f$.
*
*
*Compute $f'\left(x\right)$. Find an expression of $f\left(x\right)$ in terms of $\cos^{-1}\left(x\right)$.
My attempt :
$f$ is defined on $\left[-1;1\right]$ and
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{1-\left(4x^3-3x\right)^2}}
$$
To answer the question I thought i could express it as
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{-1+3x-4x^3}\sqrt{1-3x+4x^3}}
$$
and try a kind of decomposition to find it as
$$
f'\left(x\right)=a\frac{3-12x^2}{\sqrt{-1+3x-4x^3}}+b\frac{3-12x^2}{\sqrt{1-3x+4x^3}}
$$
What am I missing ?
| Since the domain is $[-1,1]$, you can let $t=\cos^{-1} x \implies x=\cos t$. Why? Because $4\cos^3 t -3\cos t = \cos 3t $. $$f(t) = \cos^{-1} (\cos 3t) $$
The range of $\cos^{-1} x$ is $[0,\pi] $, and so you need to be careful and divide $f(t)$ into suitable pieces as follows:
$$f(t) =\begin{cases} 3t, & 0\le t\le \pi/3 \\ 2\pi-3t, & \pi/3 \le t \le 2\pi/3 \\ 3t -2\pi, & 2\pi/3 \le t \le \pi \end{cases} $$
Now just replace $t$ while noting that $0\le \cos^{-1} x \le \pi/3 \iff \frac 12 \le x\le 1$ and $\pi/3 \le \cos^{-1} x \le 2\pi/3 \iff -\frac 12 \le x \le \frac 12$ and $2\pi/3 \le \cos^{-1} x \le \pi \iff -1\le x\le -\frac 12$.
$$f(x) =\begin{cases} 3\cos^{-1} x, & \frac 12 \le x \le 1 \\ 2\pi-3\cos^{-1} x, & -\frac 12 \le x \le \frac 12 \\ 3\cos^{-1} x -2\pi, & -1 \le x \le -\frac 12 \end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The "covering rate" of cosets $7 \Bbb{Z} + a$ by some coset $b \Bbb{Z} +c$. The covering rate of the cosets $7\Bbb{Z} + a$ by the coset $13\Bbb{Z} + 2$:
$$
7(0) -4 = 13(-1) + 2 \\
7(1) -5 = 13(0) + 2 \\
7(2) + 1 = 13(1) + 2 \\
7(4) + 0 = 13(2) + 2 \\
7(6) -1 = 13(3) + 2 \\
7(8) -2 = 13(4) + 2 \\
7(10) -3 = 13(5) + 2 \\
7(12) - 4 = 13(6) + 2 \\
7(14) -5 = 13(7) + 2 \\
$$
It takes minimally an offset of $14$, i.e. $7(x + 14) + a = 13y + 2$ before the cycle starts over again (look at the residue cycling: $-4,-5,1,0,-1,-2,-3,-4,-5, \dots$, the offset separating the two $-5$'s is $14$.
Is $\dfrac{1}{14}$ or for all fixed $x, a \in \Bbb{Z}$, the limit
$$\lim\limits_{y \to \infty} \dfrac{\#(\{7x + a, 7(x+1) + a, \dots, 7(x + y) + a\} \cap (13 \Bbb{Z} + 2))}{y - x} = \dfrac{1}{14}$$.
The covering rate of $7\Bbb{Z} + a$ by $19 \Bbb{Z} + 3$:
$$
7(3) + 1 = 19(1) + 3\\
7(5) + 6 = 19(2) + 3 \\
7(8) + 4 = 19(3) + 3 \\
7(11) + 2 = 19(4) + 3 \\
7(14) + 0 = 19(5) + 3 \\
7(16) + 5 = 19(6) + 3 \\
7(19) + 3 = 19(7) + 3 \\
7(22) + 1 = 19(8) + 3 \\
7(24) + 6 = 19(9) + 3 \\
$$
is therefore $\dfrac{1}{19}$. How can we compute this rate without going through the trouble of enumerating?
| The natural density of a set $S\subseteq\mathbb{N}$ of naturals is defined as
$$ \mu(S)=\lim_{N\to\infty} \frac{|S\cap[1,B]|}{N}. $$
Not all subsets of $S$ have a density - indeed we may construct $S$ so that $|S\cap[1,N]|$ oscillates indefinitely if we want. This density is also not countably additive the way probability measure would be, so for instance every singleton $S=\{n\}$ has density $0$ but their union has density $\mu(\mathbb{N})=1$. It's also worth pointing out it is difficult to prove things about the natural density of certain sets, however it is upgraded by something called the Dirichlet density, defined by a limit with Dirichlet series at $s=1$: whenever the natural density exists, so does the Dirichlet density and they are equal, but the Dirichlet density may exist even when the natural density doesn't and in practice it is easier to work with the Dirichlet density for sets encountered and under scrutiny within analytic number theory.
It should be clear how to extend the definition of natural density to all of $\mathbb{Z}$ if we want a more "symmetric" treatment of cosets. The natural density of a coset is simply $\mu(a+m\mathbb{Z})=\tfrac{1}{m}$. You are interested in the relative density $\mu(A\cap B)/\mu(B)$ for two cosets $A=a+7\mathbb{Z}$ and $B=2+13\mathbb{Z}$.
Fortunately, the intersection of two cosets is itself a coset (or empty). The elements of $A\cap B$ where $A=a+m\mathbb{Z}$ and $B=b+n\mathbb{Z}$ are solutions to the system of congruences
$$ \begin{cases} x \equiv a \mod m \\ x \equiv b \,\mod n \end{cases} $$
The Chinese Remainder Theorem says this is empty if $a\not\equiv b\mod \gcd(m,n)$, or else it is a residue mod $\mathrm{lcm}(m,n)$ (which can be constructed explicitly using a version of Lagrange interpolation and modular inverses, which I can explain if you want). Therefore we have
$$ \mu\big((a+m\mathbb{Z})\cap(b+n\mathbb{Z})\big) =
\begin{cases}
\frac{1}{\mathrm{lcm}(m,n)} & \textrm{ if } a\equiv b\mod\gcd(m,n) \\[5pt]
~~~~~0 & \textrm{ if } a\not\equiv b \mod \gcd(m,n)
\end{cases} $$
| {
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"source": "stackexchange",
"question_score": "1",
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Rewriting function Given function
$$
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
$$
and
$$
g(x) = \frac{2}{1+\sqrt{x+1}}
$$
Prove that $f(x) = g(x)$ where $x\neq 0$.
This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows:
$$
f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2x}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2}{\sqrt{x+1}}
$$
However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.
EDIT:
Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.
| You can start with $f(x)$ by doing the following:
$$
\begin{alignat}{1}
f(x) &= \frac{-2 + 2 \sqrt{x+1}}{x} \cdot \frac{(-2-2\sqrt{x+1})}{(-2-2\sqrt{x+1})} = \frac{4-4(x+1)}{-2x-2x\sqrt{x+1}}=\frac{-4x}{-2x\cdot(1+\sqrt{x+1})}
\\ &= \left(\frac{-2x}{-2x}\right)\cdot \left( \frac{2}{1+\sqrt{x+1}} \right) = \frac{2}{1+\sqrt{x+1}}=g(x)
\end{alignat}
$$
But what's the idea behind it?
When solving this type of problem, look at the structure of the functions you are working with. For the functions in your question:
$$
\newcommand{\altfrac}{\genfrac{}{}{0pt}{}}
\begin{alignat}{1}
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
&\altfrac{\leftarrow \text{Square root here}~~~~~}{\leftarrow \text{No square root here}}
\\[8pt]
g(x) = \frac{2}{1+\sqrt{x+1}}
&\altfrac{\leftarrow \text{No square root here}}{\leftarrow \text{Square root here}~~~~~}
\end{alignat}
$$
Going from $f(x)$ to $g(x)$, you need to make the square root appear in the denominator, so a good start would be to rationalize the numerator of $f(x)$.
Similary, from $g(x)$ to $f(x)$, you need to make the square root appear in the numerator, so a good start would be to rationalize the denominator of $g(x)$, as done in @vitamin d answer.
| {
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"source": "stackexchange",
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Let $a, b, c$ be the three roots of $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$
Let $a, b, c$ be the three roots of the equation $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$.
Here is a problem from a Hong Kong competition that I found a way to solve, but I am interested in knowing other methods since my brute-force way doesn't seem to be the solution intended.
My attempt
Let $y=x-2$, the equation becomes
$$(2+y)^3-(k+1)(2+y)^2+k(2+6)+12=0$$
$$y^3+(5-k)y^2+(8-3k)y-2k+16=0$$
Let $y_1, y_2, y_3$ be the new roots of the equation. By Vieta's formula, we have
\begin{cases}
y_1+y_2+y_3=-5+k \\
y_1y_2+y_2y_3+y_1y_3=8-3k\\
y_1y_2y_3=2k-16
\end{cases}
And by $y_1^3+y_2^3+y_3^3 = (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3)+3y_1y_2y_3$, we have
\begin{align*}
y_1^3+y_2^3+y_3^3 &= (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3) +3y_1y_2y_3\\
&= (-5+k)^3-3(-5+k)(8-3k)+3(2k-16)\\
&=k^3-6k^2+12k-53
\end{align*}
By $y_1^3+y_2^3+y_3^3=-18$ given in the problem, we have
$$k^3-6k^2+12k-35=0$$
And $k=5$ is the only real solution.
| Here is Stinking Bishop's method:
The condition on $a,b,c$ gives $(a^3 + b^3 + c^3) + (-2 \cdot 3)(a^2 + b^2 + c^2) + (4 \cdot 3)(a + b + c) - 3 \cdot 2^3 = -18$ or that $(a^3 + b^3 + c^3) - 6(a^2 + b^2 + c^2) + 12(a + b + c) = 6$.
Then as you said, $(a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc - 6((a+b+c)^2 - 2(ab+bc+ca))$ $+\ 12(a+b+c) = 6$, so by Vieta, $(k+1)^3 - 3(k+1)(k) + 3(-12) - 6((k+1)^2 - 2k) + 12(k+1) = 6$, or $k^3 + 3k^2 + 3k + 1 - 3k^2 -3k - 36 - 6k^2 - 12k - 6 + 12k + 12k + 12 = 6$, which implies $f(k) = k^3 - 6k^2 + 12k - 35 = 0$.
Descartes' rule of signs implies there are either three or one positive roots, and since $f(-k) = -k^3 - 6k^2 - 12k - 35 = 0$, there are no negative roots. Together with the rational root theorem, the only rational candidates (it is unlikely $k$ is irrational in a contest problem) are the factors of $35$: $35, 7, 5, 1$, and $5$ is a root. Dividing by $k = 5$ gives $k^2 - k + 7$ where $\Delta > 0$, hence $k = 5$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrix Multiplication by Peter Freyd I am trying to read “Abelian Categories” of Peter J. Freyd. On page 49, I came across this remark: “The usual rules of matrix multiplication can now be proven”. What does he mean? For instance, I expect to verify
$\left ( \begin{matrix} w&x\\y&z \end{matrix} \right )
\left (\begin{matrix} a\\b \end{matrix} \right ) =
\left ( \begin{matrix} wa+xb\\ya+zb \end{matrix} \right )$
However, if $a$, $b$, $w$, $x$, $y$, and $z$ are all maps $A \to A$, then
$\left ( \begin{matrix} w&x\\y&z \end{matrix} \right )$
is a map $A \oplus A \to A \oplus A$ and
$\left (\begin{matrix} a\\b \end{matrix} \right )$
is a map $ A \oplus A \to A$, so this composition is not defined.
The composition
$\left ( \begin{matrix} w&x\\y&z \end{matrix} \right )
\left (\begin{matrix} a&b \end{matrix} \right ) =
\left ( \begin{matrix} wa+yb\\xa+zb \end{matrix} \right )$
does exist, because $\left (\begin{matrix} a&b \end{matrix} \right ) $ is a map $A \to A \oplus A$. But is this a usual rule of matrix multiplication?
So, my question is, which rules does he mean and how can they be verified, what is the technique? And please, if possible, in the notation of Freyd.
| I managed to verify several equations of the matrix multiplication meant by Freyd. This deals with morphisms in an abelian category. In the text before Freyd’s remark, Freyd introduced addition in abelian categories, so now we can add maps
$A \overset {a} \longrightarrow X$ and
$A \overset {b} \longrightarrow X$ to a map
$A \overset {a+b} \longrightarrow X$
in an abelian category. Below, I verified one of the equations. We have
For maps
$A \overset {x} \longrightarrow X$,
$B \overset {y} \longrightarrow X$,
$X \overset {h} \longrightarrow H$,
$$h \circ
\left ( \begin{matrix} x\\y \end{matrix} \right ) =
\left (\begin{matrix} hx\\hy \end{matrix} \right )
:A \oplus B \longrightarrow H
$$
For maps
$H \overset {h} \longrightarrow X$,
$X \overset {a} \longrightarrow A$,
$X \overset {b} \longrightarrow B$,
$$
\left ( \begin{matrix} a&b \end{matrix} \right )
\circ h =
\left (\begin{matrix} ah&bh \end{matrix} \right )
:H \longrightarrow A \oplus B
$$
For maps
$K \overset {a} \longrightarrow A$,
$K \overset {b} \longrightarrow B$,
$A \overset {x} \longrightarrow Q$,
$B \overset {y} \longrightarrow Q$,
$$
\left ( \begin{matrix} x\\y \end{matrix} \right )
\circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
xa+yb
:K \longrightarrow Q
$$
For maps
$X \overset {x} \longrightarrow Q$,
$Y \overset {y} \longrightarrow Q$,
$Q \overset {a} \longrightarrow A$,
$Q \overset {b} \longrightarrow B$,
$$
\left (\begin{matrix} a&b \end{matrix} \right )
\circ
\left ( \begin{matrix} x\\y \end{matrix} \right ) =
\left ( \begin{matrix} ax&bx\\ay&by \end{matrix} \right )
:X \oplus Y \longrightarrow A \oplus B
$$
For maps
$K \overset {a} \longrightarrow A$,
$K \overset {b} \longrightarrow B$,
$A \overset {w} \longrightarrow X$,
$A\overset {x} \longrightarrow Y$,
$B \overset {y} \longrightarrow X$,
$B \overset {z} \longrightarrow Y$,
$$
\left ( \begin{matrix} w&x\\y&z \end{matrix} \right )
\circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
\left ( \begin{matrix} wa+yb&xa+zb \end{matrix} \right )
:K \longrightarrow X \oplus Y
$$
For maps
$A \overset {w} \longrightarrow X$,
$A\overset {x} \longrightarrow Y$,
$B \overset {y} \longrightarrow X$,
$B \overset {z} \longrightarrow Y$,
$X \overset {c} \longrightarrow Q$,
$Y \overset {d} \longrightarrow Q$,
$$
\left ( \begin{matrix} c\\d \end{matrix} \right )
\circ
\left ( \begin{matrix} w&x\\y&z \end{matrix} \right ) =
\left ( \begin{matrix} cx+dx\\cy+dz \end{matrix} \right )
:A \oplus B \longrightarrow Q
$$
For maps
$K \overset {a} \longrightarrow A$,
$K \overset {b} \longrightarrow B$,
$A \overset {w} \longrightarrow X$,
$A\overset {x} \longrightarrow Y$,
$B \overset {y} \longrightarrow X$,
$B \overset {z} \longrightarrow Y$,
$X \overset {c} \longrightarrow Q$,
$Y \overset {d} \longrightarrow Q$,
$$
\left ( \begin{matrix} c\\d \end{matrix} \right )
\circ
\left ( \begin{matrix} w&x\\y&z \end{matrix} \right )
\circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
cwa+dxa+cyb+dzb
:K \longrightarrow Q
$$
For maps
$G \overset {a} \longrightarrow A$,
$G\overset {b} \longrightarrow B$,
$H \overset {c} \longrightarrow A$,
$H \overset {d} \longrightarrow B$,
$A \overset {w} \longrightarrow X$,
$A\overset {x} \longrightarrow Y$,
$B \overset {y} \longrightarrow X$,
$B \overset {z} \longrightarrow Y$,
$$
\left ( \begin{matrix} w&x\\y&z \end{matrix} \right )
\circ
\left ( \begin{matrix} a&b\\c&d \end{matrix} \right ) =
\left ( \begin{matrix} wa+yb&xa+zb\\wc+yd&xc+zd \end{matrix} \right )
:G \oplus H \longrightarrow X \oplus Y
$$
For maps
$X \overset {x} \longrightarrow H$,
$X\overset {y} \longrightarrow H$,
$H \overset {h} \longrightarrow Q$,
$Q \overset {a} \longrightarrow A$,
$Q \overset {b} \longrightarrow B$,
$$
\left (\begin{matrix} a&b \end{matrix} \right )
\circ h \circ
\left ( \begin{matrix} x\\y \end{matrix} \right ) =
\left ( \begin{matrix} ahx&bhx\\ahy&bhy \end{matrix} \right )
:X \oplus Y \longrightarrow A \oplus B
$$
As an example, we verify for maps
$K \overset {a} \longrightarrow A$,
$K \overset {b} \longrightarrow B$,
$A \overset {x} \longrightarrow Q$,
$B \overset {y} \longrightarrow Q$,
that
$$
\left ( \begin{matrix} x\\y \end{matrix} \right )
\circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
xa+yb
:K \longrightarrow Q
$$
as follows:
There is a unique
$q = \left (\begin{matrix} a&b \end{matrix} \right ) :K \longrightarrow A \oplus B$
such that
$p_1 \circ q = a$ and $p_2 \circ q = b$
There is a unique
$t = \left (\begin{matrix} x\\y \end{matrix} \right ) :A \oplus B \longrightarrow Q$
such that
$t \circ u_1 = x$ and $t \circ u_2 = y$
Then we have
$$t \circ q = $$
$$
= \left (\begin{matrix} x\\y \end{matrix} \right )
\circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
$$
$$
= \left (\begin{matrix} x\\y \end{matrix} \right )
\circ 1_{A \oplus B} \circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
$$
$$
= \left (\begin{matrix} x\\y \end{matrix} \right )
\circ \left ( u_1p_1+u_2p_2 \right ) \circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
$$
$$
= \left ( xp_1+yp_2 \right )
\circ
\left (\begin{matrix} a&b \end{matrix} \right ) =
$$
$$
=xa+yb
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $5\times9^x-8\times15^x+3\times25^x>0$ Question: Solve the inequality:
$5\times9^x-8\times15^x+3\times25^x>0$
So far I have managed to factorise the inequality:
$(5\times3^x-3\times5^x)(3^x-5^x)>0$
From here on, I am stuck.
I would very much appreciate it if you could answer this exam question and lay the working out as if it were an exam. I have an answer from the solutions book so can tell you if your answer is correct using that. Many thanks.
| When both factors are positive, the intersection of $x < 0$ and $x < 1$ is $x < 0$. But when both factors of $(5\times3^x-3\times5^x)(3^x-5^x)$ are negative, their product is still positive, so we must consider this case as well.
$5\times3^x-3\times5^x < 0$ gives $\left( \frac{3}{5}\right)^{x-1} < \left( \frac{3}{5}\right)^{0}$ like before. However, since $3/5 < 1$, the $y$-values are getting smaller as the $x$-values increase. Drawing a sketch of the exponential function and the curve, the inequality sign flips to give $x - 1 > 0, x > 1$.
Similarly, $3^x - 5^x < 0$ leads to $3^x < 5^x$ or $\left( \frac{3}{5}\right)^x < \left( \frac{3}{5}\right)^{0}$, and the inequality sign flips again, thus $x > 0$. Since both factors need to be negative for the product to be positive, we need to take the intersection of $x > 1, x > 0$ which is $x > 1$.
Altogether, we have $x < 0$ and $x > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the gradient of the function How do I find the gradient of a function like this? $F: \mathbb{R}^n-\{0\} \rightarrow \mathbb{R}^n$. $F(x)=\frac{x}{||x||}$ Attempt I guess find the gradient each component $D_1\frac{x_1}{\sqrt{x_1^2+\dots +x_n^2}}=\frac{x_2^2+\dots+x_n^2}{(x_1^2+\dots+x_n^2)^{3/2}}$ $D_2\frac{x_1}{\sqrt{x_1^2+\dots +x_n^2}}=\frac{-x_1x_2}{(x_1^2+\dots+x_n^2)^{3/2}}$ ,$D_3\frac{x_1}{\sqrt{x_1^2+\dots +x_n^2}}=\frac{-x_1x_3}{(x_1^2+\dots+x_n^2)^{3/2}}$,$...$ The answer given to me was $\nabla F(x)=\frac{I}{||x||}-\frac{xx^T}{||x||^3}$
| To simplify, I'll do it with $n=2$. Then$$F(x,y)=\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}}\right).$$Let$$F_1(x,y)=\frac x{\sqrt{x^2+y^2}}\quad\text{and}\quad F_2(x,y)=\frac y{\sqrt{x^2+y^2}}.$$Then\begin{align}\nabla F(x,y)&=\begin{bmatrix}\frac{\partial F_1}{\partial x}(x,y)&\frac{\partial F_1}{\partial y}(x,y)\\\frac{\partial F_2}{\partial x}(x,y)&\frac{\partial F_2}{\partial y}(x,y)\end{bmatrix}\\&=\begin{bmatrix}\frac1{\sqrt{x^2+y^2}}-\frac{x^2}{(x^2+y^2)^{3/2}}&-\frac{xy}{(x^2+y^2)^{3/2}}\\-\frac{xy}{(x^2+y^2)^{3/2}}&\frac1{\sqrt{x^2+y^2}}-\frac{y^2}{(x^2+y^2)^{3/2}}\end{bmatrix}\\&=\frac1{\sqrt{x^2+y^2}}\begin{bmatrix}1&0\\0&1\end{bmatrix}-\frac1{(x^2+y^2)^{3/2}}\begin{bmatrix}x^2&xy\\xy&y^2\end{bmatrix}.\end{align}Besides,$$\begin{bmatrix}x^2&xy\\xy&y^2\end{bmatrix}=\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}x&y\end{bmatrix}^T.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Eigenvalues of correlation matrix $\mathbf{\rho}$ with constants I'm tasked with finding the eigenvalues of $M = \begin{bmatrix} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \end{bmatrix}$. My difficulty here is in finding roots of the characteristic polynomial, since it's a cubic.
Attempt. Let's call the characteristic polynomail by $chp$, and let's use $x$ instead of $\lambda$ to have a more familiar-looking expression. We have
$$
\operatorname{chp}(x) = \det\left(\mathbf{\rho} - xI\right) = \det\begin{bmatrix} 1-x & \rho & \rho \\ \rho & 1-x & \rho \\ \rho & \rho & 1-x \end{bmatrix}
$$
Expanding on the first line, we obtain
$$
chp(x) = (1-x) \det \begin{bmatrix} 1-x & \rho \\ \rho & 1-x \end{bmatrix} - \rho \det \begin{bmatrix} \rho & \rho \\ \rho & 1-x \end{bmatrix} + \rho \det \begin{bmatrix} \rho & 1-x \\ \rho & \rho \end{bmatrix}.
$$
After simplification, you'd get
$$
chp(x) = 1 - 3\rho^2 + 2\rho^3 - 3x + 3\rho^2 x + 3x^2 - x^3.
$$
I couldn't solve $chp(x)=0$ analitically. Using Wolfram though, and coming back to the usual $\lambda$ notation, I got
$$
\lambda_1 = \lambda_2 = 1-\rho \qquad \text{and} \qquad \lambda_3 = 1+2\rho.
$$
Thanks for any help.
| This matrix is
\begin{align}
& (1+2\rho)\left[ \begin{array}{ccc} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{array} \right] + (1-\rho) \left[ \begin{array}{rrr} 2/3 & -1/3 & -1/3 \\ -1/3 & 2/3 & -1/3 \\ -1/3 & -1/3 & 2/3 \end{array} \right] \\[8pt]
={} & (1+2\rho)P+ (1-\rho) Q.
\end{align}
You can check that $P^2=P =P^T$ and $Q^2=Q=Q^T$ and $PQ=QP=0.$
Observe that if all components of a $3\times1$ column vector $\mathbf x$ are equal to each other, then $P\mathbf x=\mathbf x$ and $Q\mathbf x= \mathbf 0,$ and if the sum of the components of a column vector $\mathbf y$ is $0$ (which is the same as saying $\mathbf y\perp\mathbf x$) then $P\mathbf y=\mathbf 0$ and $Q\mathbf y = \mathbf y.$ That means the $2$-dimensional column space of $Q$ is an eigenspace with of $Q$ with eigenvalue $1$ and its $1$-dimensional orthogonal complement is an eigenspace of $Q$ with eigenvalue $0.$ But that same $2$-dimensional column space of $Q$ is an eigenspace of $P$ with eigenvalue $0$ and its orthogonal complement is an eigenspace of $P$ with eigenvalue $1.$
So we have an orthonormal basis with respect to which we have
$$
P = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \text{ and } Q = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]
$$
and therefore the matrix we started with becomes
$$
\left[ \begin{array}{ccc} 1+2\rho & 0 & 0 \\ 0 & 1-\rho & 0 \\ 0 & 0 & 1-\rho \end{array} \right].
$$
So there are the eigenvalues.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Computing $\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$ I'm trying to compute the following limit and would greatly appreciate your feedback:
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$$
Given that $\log(x)$ can be replaced with $\log(x+1)$ because of the identity
$$\log(x) + \log(1) = \log(x) + 0$$
Then using the series:
$$\log(x+1) = x + o(x^n)$$
Then dividing by $x^2+x-2$, I get:
$$\frac{x}{x^2+x-2} = \frac{x + o(x^n)}{(x-1)(x+2)}$$
when $x = 1$
$$\frac{x + o(x^n)}{(x-1)(x+2)} = \frac{1}{0} = 0$$
hence
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} \implies 0$$
| Let $x = e^t$. Then $t\to 0$ as $x\to 1$. The limit then becomes
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} = \lim_{t\to0}\frac{t}{e^{2t}+e^t-2} = \lim_{t\to0}\color{red}{\frac{t}{e^t-1}}\frac{1}{e^t+2} = \color{red}1\cdot \frac{1}{1+2} = \frac{1}{3}$$
$\color{red}{\text{Using}}$ the well-known limit $\lim_{x\to0}\frac{e^x-1}{x}=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Eigenvectors of this matrix I have a matrix A
$$A = \begin{pmatrix}
0 & 1 \\
-1 & 1
\end{pmatrix}
$$
Of which I have calculated the eigenvalues to be $\lambda = e^{\pm i \frac{\pi}{3}}$
Hence we have $\lambda_1 = e^{i \frac{\pi}{3}}$ and $\lambda_2 = e^{-i \frac{\pi}{3}}$
I have then tried to find the eigenvectors as follows:
$$
\begin{pmatrix}
0 & 1 \\
-1 & 1
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix}
=
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix}
\lambda
$$
The way I know how to go now is to 'guess' a value for $x_1$ as the length is irrelevant and the second term $x_2$ will come out as proportional and then I can just normalise it, however I run into some issues. Starting with $\lambda_1$ I guess $x_1$ to be 1:
$$
\begin{pmatrix}
0 & 1 \\
-1 & 1
\end{pmatrix}
\begin{pmatrix}
1 \\
x_2
\end{pmatrix}
=
\begin{pmatrix}
\lambda_1 \\
\lambda_1 x_2
\end{pmatrix}
$$
Which when solving gives me the following equations:
$$
0(1) +1(x_2) = \lambda_1
$$
$$
-1(1) + 1(x_2) = \lambda_1{x_2}
$$
Solving the first gives me:
$$
x_2 = \lambda_1
$$
But the second gives me:
$$
x_2 = \frac{1}{1-\lambda_1}
$$
Which has left me confused because I was under the impression that these should be the same - especially since the way I calculated the eigenvalues should mean that these are linearly coupled. I have seen some stuff online about finding RREF of the matrix however I have not been taught this and I am assuming there is an easier way or I have gone wrong somewhere. Is one of these correct? If so, why is the other not the same? Any help would be appreciated.
| Your two answers for $x_2$ are equivalent: start by converting $\lambda_1$ to Cartesian using Euler's formula. $$\lambda_1 = e^{i\frac{\pi}{3}} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
Now let's simplify your other form for $x_2$.
$$x_2 = \frac{1}{1-\lambda_1} = \frac{1}{1 - {\frac{1}{2} + i\frac{\sqrt{3}}{2}}} = \frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}}$$
Multiplying the numerator and denominator by the conjugate we get:
$$\frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} \cdot \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = \lambda_1$$
Furthermore, notice that the original equation $\lambda_1 = \frac{1}{1 - \lambda_1}$ is equivalent to $\lambda_1(1 - \lambda_1) - 1=0$, and this is likely the same equation you solved to determine your eigenvalues, $\lambda_{1,2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fractional derivative with linear algebra I have this question in linear algebra that asks us to represent $\frac{d}{dx}\sin x$ and $\frac{d}{dx}\cos x$ as a matrix in the basis of $\binom{1}{0}=\sin x, \ \binom{0}{1}=\cos x$ using eigenvectors and eigenvalues. So, I guessed $\left[\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right]_B$, s.t.
$$
\frac{d}{dx}\sin x=\left[\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right]\cdot\binom{1}{0}=\binom{0}{1}=\cos x \\
\frac{d}{dx}\cos x=\left[\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right]\cdot\binom{0}{1}=\binom{-1}{0}=-\sin x
$$
The question then asks me to do it for the square root of a derivative operator, like so $\sqrt{\frac{d}{dx}}$. So, I found the eigenvectors and values.$$
PDP^{-1}=\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} i & 0 \\ 0 & -i \end{matrix}\right]\cdot\frac{1}{2}\left[\begin{matrix} 1 & i \\ 1 & -i \end{matrix}\right] \\
PD^{\frac{1}{2}}P^{-1}=\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} i & 0 \\ 0 & -i \end{matrix}\right]^{\frac{1}{2}}\cdot\frac{1}{2}\left[\begin{matrix} 1 & i \\ 1 & -i \end{matrix}\right] \\
=\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} e^{i\frac{\pi}{4}} & 0 \\ 0 & e^{-i\frac{\pi}{4}} \end{matrix}\right]\cdot\frac{1}{2}\left[\begin{matrix} 1 & i \\ 1 & -i \end{matrix}\right] \\
=\frac{1}{2}\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} e^{i\frac{\pi}{4}} & ie^{i\frac{\pi}{4}} \\ e^{-i\frac{\pi}{4}}& -ie^{-i\frac{\pi}{4}}\end{matrix}\right] \\
=\frac{1}{2}\cdot\left[\begin{matrix}e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}} & i(e^{i\frac{\pi}{4}}-e^{-i\frac{\pi}{4}}) \\ i(e^{i\frac{\pi}{4}}-e^{-i\frac{\pi}{4}}) & i^2(e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}})\end{matrix}\right] \\
=\frac{1}{2}\cdot2\cdot\left[\begin{matrix} \cos{\frac{\pi}{4}} & -\sin{\frac{\pi}{4}} \\ -\sin{\frac{\pi}{4}} & -\cos{\frac{\pi}{4}}\end{matrix}\right] \\
=\frac{1}{\sqrt{2}}\cdot\left[\begin{matrix}1 & -1 \\ -1 & -1\end{matrix}\right]
$$
However, when I tested $\sin x$, I got:$$
\frac{1}{\sqrt{2}}\cdot\left[\begin{matrix}1 & -1 \\ -1 & -1\end{matrix}\right]\cdot\binom{1}{0}=\frac{1}{\sqrt{2}}\cdot\binom{1}{-1} \\
\frac{1}{\sqrt{2}}\cdot\left[\begin{matrix}1 & -1 \\ -1 & -1\end{matrix}\right]\cdot\frac{1}{\sqrt{2}}\cdot\binom{1}{-1}=\binom{1}{0}=\sin x
$$ which is a weird result because I expected $\cos x$. I redid my working again and got the same result. What did I do wrong here? Could someone help me find that error please? Thank you very much for your help.
| The decomposition of $\left[ \begin{array}{cc}
0&-1\\
1&0
\end{array}
\right]$ is wrong. Actually we have
$$\left[ \begin{array}{cc}
0&-1\\
1&0
\end{array}
\right]=(\frac{1}{\sqrt{2}}\left[ \begin{array}{cc}
1&-1\\
1&1
\end{array}
\right])^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Explanation of formula given on Wikipedia (for sign in half-angle formula for sine) This Wikipedia article listing trigonometric identities, states the following identity under half angles:
$$
\sin{\frac{\theta}{2}} = \text{sgn}\bigg(2\pi-\theta+4\pi\bigg\lfloor\frac{\theta}{4\pi}\bigg\rfloor \bigg)\sqrt{\frac{1-\cos{\theta}}{2}}
$$
where $\text{sgn}(x)$ is a function that returns $+1$ if $x>0$ and $-1$ if $x<0$.
I know and understand the part of the formula that says $\sqrt{\frac{1-\cos{\theta}}{2}}$ and understand the fact that the remaining part is to decide the $\pm$ sign in the formula.
From where, however, does the sign-deciding part of the expression come from? And why should it make sense?
| It is from the process of deriving the identity itself.
First, let us use the double-angle formula $\cos2\theta = 2\cos^{2}\theta - 1$. Solving for $\cos^{2}\theta$, we get $$\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}.$$
Replacing $\theta$ by $\frac{\theta}{2}$, we get \begin{align*}\cos^{2}\left(\frac{\theta}{2}\right) &= \frac{1 + \cos\theta}{2} \\ \cos\left(\frac{\theta}{2}\right) &= \pm\sqrt{\frac{1 + \cos\theta}{2}}\end{align*}
Using the identity $\sin^{2}\theta + \cos^{2}\theta = 1$ to get $\sin \frac{\theta}{2}$,
\begin{align*}\sin^{2}\left(\frac{\theta}{2}\right) + \cos^{2}\left(\frac{\theta}{2}\right) &= 1 \\ \sin^{2}\left(\frac{\theta}{2}\right) + \frac{1 + \cos\theta}{2} &= 1 \\ \sin^{2}\left(\frac{\theta}{2}\right) &= 1 - \frac{1 + \cos\theta}{2} \\ \sin^{2}\left(\frac{\theta}{2}\right) &= \frac{2 - 1 - \cos\theta}{2} \\ \sin^{2}\left(\frac{\theta}{2}\right) &= \frac{1 - \cos\theta}{2} \\ \sin\left(\frac{\theta}{2}\right) &= \pm\sqrt{\frac{1 - \cos\theta}{2}}\end{align*}
As you can see, the squares on the equation force it to take the plus-minus sign.
Edit: I don't have an idea about the $\displaystyle \mathrm{sgn}\left(2\pi - \theta + 4\pi\left\lfloor\frac{\theta}{4\pi}\right\rfloor\right)$. I just used the basic ones.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Transform $\frac{-2}{\tan\frac{x}{2}+1}$ to $\tan x-\sec x$? I'm learning to integrate and was asked to integrate $\int\frac{1}{1+\sin x}dx$
I get the answer $-\frac{2}{\tan\frac{x}2+1}+c$ which Symbolab confirms is correct, but the textbook says the answer is $\tan x-\sec x+c$.
I can't seem to transform my answer to that one using trig identities. Any idea how/if one can get that result?
Many thanks,
Andrew
| Let $$f(x) = \frac{-2}{\tan \frac{x}{2} + 1}, \quad g(x) = \tan x - \sec x.$$
Then the first thing to note is that $f(0) = -2$, whereas $g(0) = -1$. So if $f$ and $g$ differ by a constant, this difference must satisfy $g(x) - f(x) = 1$. Consequently,
$$\begin{align}
f(x) + 1 &= 1 - \frac{2}{\tan \frac{x}{2} + 1} \\
&= \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2} + 1} \\
&= \frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} \\
&= \frac{(\sin \frac{x}{2} - \cos \frac{x}{2})^2}{\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}} \\
&= \frac{\sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} + \cos^2 \frac{x}{2}}{-\cos x} \\
&= \frac{\sin x - 1}{\cos x} \\
&= \tan x - \sec x \\
&= g(x).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Second-Order Non-Homogeneous Recurrence Relations I'm trying to solve the following recurrence relation:
$F(n) = F(n -1) + F(n -2) + Θ(n)$ for $n > 1$, and
$Θ(n) = c_1n + c_2$ , where $c_1, c_2 > 0$
and with two initial conditions $F(0) = 0$, $F(1) = 1$
For the homogeneous version of the recurrence, I have:
$F(n) - F(n - 1) - F(n - 2) = 0$
Finding the roots, I get:
$F(n) = α(\frac{1 + √5}{2})^n + β(\frac{1 - √5}{2})^n$
If I take the initial conditions into consideration and incorporate the non-homogeneous part, I have:
$F(0) = α(\frac{1 + √5}{2})^0 + β(\frac{1 - √5}{2})^0 - c_1(0) - c_2 = 0$
$α + β - c_2 = 0$
$α + β = c_2$
$β = c_2 - α$
$F(1) = α(\frac{1 + √5}{2})^1 + β(\frac{1 - √5}{2})^1 - c_1(1) - c_2 = 1$
$α(\frac{1 + √5}{2}) + β(\frac{1 - √5}{2}) - c_1 - c_2 = 1$
$α(\frac{1 + √5}{2}) + β(\frac{1 - √5}{2}) = 1 + c_1 + c_2$
Plugging in value of β from F(0) evaluation
$α(\frac{1 + √5}{2}) + (c_2 - α)(\frac{1 - √5}{2}) = 1 + c_1 + c_2$
$α(\frac{1 + √5}{2}) - α(\frac{1 - √5}{2}) + c_2(\frac{1 - √5}{2}) = 1 + c_1 + c_2$
$α√5 + c_2(\frac{1 - √5}{2}) = 1 + c_1 + c_2$
$α√5 = 1 + c_1 + c_2 - c_2(\frac{1 - √5}{2})$
$α = \frac{1 + c_1 + c_2 - c_2(\frac{1 - √5}{2})}{√5}$
Now that I have the values of α and β, I can plug them into the formula:
$F(n) = α(\frac{1 + √5}{2})^n + β(\frac{1 - √5}{2})^n - c_1n - c_2$
$F(n) = (\frac{1 + c_1 + c_2 - c_2(\frac{1 - √5}{2})}{√5})(\frac{1 + √5}{2})^n + (c_2 - α)(\frac{1 - √5}{2})^n - c_1n - c_2$
$F(n) = (\frac{1 + c_1 + c_2 - c_2(\frac{1 - √5}{2})}{√5})(\frac{1 + √5}{2})^n + (c_2 - \frac{1 + c_1 + c_2 - c_2(\frac{1 - √5}{2})}{√5}))(\frac{1 - √5}{2})^n - c_1n - c_2$
Now, this should be the solution to the recurrence relation, but, for example, when I use $c_1 = 2$ and $c_2 = 3$, I get the following values:
$F(2) = 2$ --> should be 8
$F(3) = 6$ --> should be 18
$F(4) = 13$ --> should be 37
$F(5) = 26$ --> should be 68
And so on ...
Could anybody tell me where I'm going wrong? Thanks
| If I'm following what you've done, you've found that the general solution to the associated homogenous equations is
$$\alpha\left(\frac{1+\sqrt5}2\right)^n+\beta\left(\frac{1-\sqrt5}2\right)^n$$ You have a typo when this formula first appears -- a plus sign instead of a minus sign in the second term.
That looks right. (I haven't checked it, I'm going by memory here.)
The next thing you should do is to find any particular solution to the inhomogeneous equation $$F_n-F_{n-1}-F_{n-2}=c_1n+c_2$$
Because of the form of the right hand side, we guess that there will be a solution of the form $F_n=an+b$ for some constants $a$ and $b$.
Substituting in in the inhomogeneous equation, we get $$an+b-(a(n-1)+b)-(a(n-2)+b)=c_1n +c_2$$ which, if I haven't made a mistake, leads to $$a = -c_1,\ b=-(c_2+3c_1)$$
Now the general solution to the inhomogeneous equation is the general solution to the homogeneous equation plus any particular solution to the inhomogeneous equation:
$$F_n= \alpha\left(\frac{1+\sqrt5}2\right)^n+\beta\left(\frac{1-\sqrt5}2\right)^n-c_1n-c_2-3c_1$$
Now you can substitute the initial values $F_0=0,\ F_1=1$ and determine $\alpha and $\beta$. Check my calculations before you do, though.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\sum_{n\geq 0} \frac{1}{64^n(2n+1)}\left[\frac{{4n\choose 2n}}{{2n\choose n}}\right]$ Since we have encountered several classes of infinite series containing central binomial coefficients which inspired to me introduce following class of series.
$$\sum_{n=0}^{\infty}\frac{1}{64^n(2n+1)}\left[\frac{{4n\choose 2n}}{2n\choose n}\right]$$
My final closed form for this series is
$$\frac{8}{\sqrt{3}}F\left(\frac{\pi}{4},i\sqrt{\frac{2}{3}}\right)+2\sqrt{5}E\left(\frac{\pi}{4},\sqrt{\frac{2}{5}}\right)-2\sqrt{3}E\left(\frac{\pi}{4},i\sqrt{\frac{2}{3}}\right)-\frac{8}{\sqrt{5}}F\left(\frac{\pi}{4},\sqrt{\frac{2}{5}}\right)$$
where $F(\alpha, u)$, $E(\alpha^{\prime},u^{\prime})$ and $i=\sqrt{-1}$ denotes incomplete elliptical integrals of the first kind , second kind and imaginary number respectively.
By Wolfram alpha check here which is further equal to $${}_3F_2\left(\frac{1}{4},\frac{3}{4},1;\frac{1}{2},\frac{3}{2};\frac{1}{16}\right)$$
The closed form in terms of incomplete Elliptical integrals and hypergeometric expression are found to be equal by Wolfram alpha check here and here respectively.
I have no idea if the hypergeometric expression can be deduced to some closed form with elementary functions and the closed form I obtained.
| Not an answer but a lead: Using DLMF 16.5.2 we may write
$$
{_3F}_2\left({1,\frac{1}{4},\frac{3}{4}\atop\frac{3}{2},\frac{1}{2}};\frac{1}{16}\right)=\frac{1}{2}\int_0^1\frac{1}{\sqrt{1-t}}{_2F}_1\left({\frac{1}{4},\frac{3}{4}\atop\frac{1}{2}};\frac{t}{16}\right)\,\mathrm dt.
$$
The reduction formula here then allows us to deduce the integral representation
$$
{_3F}_2\left({1,\frac{1}{4},\frac{3}{4}\atop\frac{3}{2},\frac{1}{2}};\frac{1}{16}\right)=\frac{1}{4}\int_0^1\frac{1}{\sqrt{1-t}}\left(\frac{1}{\sqrt{1+\sqrt{t/16}}}+\frac{1}{\sqrt{1-\sqrt{t/16}}}\right)\,\mathrm dt,
$$
which according to mathematica can be evaluated in terms of elliptic integrals.
| {
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"url": "https://math.stackexchange.com/questions/4093327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show path of the particle is hyperbolic The equation for the path of the particle in cartesian coordinates is $$(1-e^2)x^2+y^2+2exp=p^2$$
where $p$ and $e$ are constant. Under the condition that $0<e<1$, I have shown that this equation describes an elliptical path. Now I'm looking at the situation when $e>1$. This means $0>(1-e^2)$, and dividing with this number gives $$x^2 + \frac{y^2}{(1-e^2)}+\frac{2exp}{(1-e^2)}=\frac{p^2}{(1-e^2)}$$
I complete the square by adding and reducing $(\frac{1}{2}\frac{2exp}{(1-e^2)})^2$. I have $$x^2+\bigg(\frac{exp}{(1-e^2)}\bigg)^2-\bigg(\frac{exp}{(1-e^2)}\bigg)^2 + \frac{y^2}{(1-e^2)}+\frac{2exp}{(1-e^2)}=\frac{p^2}{(1-e^2)}$$
and completing the square gives $$\bigg(x+\bigg(\frac{exp}{(1-e^2)}\bigg)\bigg)^2+\frac{y^2}{(1-e^2)}-\bigg(\frac{exp}{(1-e^2)}\bigg)^2=\frac{p^2}{(1-e^2)}$$
Because I divide $y^2$ with a negative number, I could change the sign in front of $y^2$. Then with a bit of algebra, the RHS can be made equal to one. At that point I would have an equation in ( at least resembling ) the form $$\frac{x^2 -y^2}{a} = 1$$
which describes a hyperbola. Is this correct? The difference between this and the elliptic case is that $(1-e^2)$ is now negative, and it seems natural that this would affect the sign of $y^2$, which in turn changes the path into a hyperbolic one.
| $\small (1-e^2)x^2+y^2 + 2 e p x = p^2$ can be re-written as,
$\displaystyle \small x^2 + \frac{2 e p x}{1-e^2} + \frac{y^2}{1-e^2} = \frac{p^2}{1-e^2}$
$\displaystyle \small (x + \frac{e p}{1-e^2})^2 + \frac{y^2}{1-e^2} = \frac{p^2}{1-e^2} + \frac{e^2p^2}{(1-e^2)^2} = \frac{p^2}{(1-e^2)^2}$
$\displaystyle \small \frac{(x + e p / (1-e^2))^2}{(p / (1-e^2))^2} + \frac{y^2}{(p / \sqrt{1-e^2})^2} = 1$ for $0 \lt e \lt 1$ ...(i)
or $\displaystyle \small \frac{(x - e p / (e^2-1))^2}{(p / (e^2-1))^2} - \frac{y^2}{(p / \sqrt{e^2 -1})^2} = 1$ for $e \gt 1$ ...(ii)
In fact for (ii), we can confirm this is a hyperbola with eccentricity $e$.
Eccentricity of hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ is given by $\sqrt{1+\frac{b^2}{a^2}}$.
Here $a = \frac{p}{e^2-1}, b = \frac{p}{\sqrt{e^2-1}} \ $ so $\sqrt{1+\frac{b^2}{a^2}} = e$.
You can similarly confirm that for $0 \lt e \lt 1$, we get an ellipse, again with eccentricity $e$.
Please note $e = 1$ gives us parabola.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find integral $\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}\,dx\,dy$ Find the integral
$$I=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}\,dx\,dy$$
My try: define
$$I(b):=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}e^{-bxy(x+y)}\,dx\,dy$$
then
$$I'(b)=-\int_0^\infty\int_0^{\infty}\sin x \sin y \sin (x+y) e^{-bxy(x+y)}\,dx\,dy$$
But I can't find $I(b)$.
Thank you
| Note that $I=J(1)$, where
$$J(a) =\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin [a(x+y)]}{xy(x+y)} dx dy$$
and
\begin{align}
J’(a) &=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \cos [a(x+y)]}{xy} dx dy\\
& =\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y [\cos (ax)\cos(ay)- \sin (ax)\sin(ay)]}{xy} dx dy\\
& =\left(\int_0^\infty\frac{\sin x \cos (ax)}xdx \right)^2
- \left(\int_0^\infty\frac{\sin x \sin (ax)}xdx \right)^2\\
& =\frac14\left(\int_0^\infty\frac{\sin (1-a)x + \sin (1+a)x }{x} dx\right)^2\\
& \>\>\>\>\>\>\>-\frac14\left(\int_0^\infty\frac{\cos (1-a)x -\cos (1+a)x }{x} dx\right)^2\\
&=\frac14\left(\pi^2 - \ln^2 \frac{1+a}{1-a} \right)
\end{align}
Thus
$$I=\int_0^1 J’(a)da=\frac14 \int_0^1 \left(\pi^2 - \ln^2 \frac{1+a}{1-a} \right) da=\frac14 \left(\pi^2 - \frac{\pi^2}3 \right)
=\frac{\pi^2}{6}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^2+xy+xz+yz=6+2\sqrt{5}$. Find the minimum of $3x+y+2z$ Let $x,~y,~z>0$ satisfy $x^2+xy+xz+yz=6+2\sqrt{5}$. How to find the minimum of $3x+y+2z$? I can find it by Larange multiplier in calculus, but I wonder if there is a easy way using, say Cauchy-Schwartz inequality?
| You're given $(x+y)(z+x)=6+2\sqrt{5}\iff (x+y)(2z+2x)=12+4\sqrt{5}$. Thus with AM-GM $$3x+y+2z=(x+y)+(2z+2x)\geqslant 2\cdot \sqrt{12+4\sqrt{5}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of an infinite series $\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}$ I want to compute this series, $$\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}.$$ I don't know how to contiune after factoring, which yields $\frac{1}{k(k-1)^2(k+1)^2(k-2)^2(k+2)^2}.$
I know that $\frac{1}{k(k-1)}$ telescopes. However, I can't apply this here, as the product of the sum is not equal to the sum of the product. I only want hints for now.
| Well, you seem to know how to telescope summing:
$$\frac1{k(k-1)} = \frac1{k-1} - \frac1k$$
Similarly, note
$$\frac1{(k-2)^2(k+2)^2} = \frac1{8k}\left(\frac1{(k-2)^2}- \frac1{(k+2)^2}\right)$$
$$\implies \frac1{(k-2)^2(k-1)^2k(k+1)^2(k+2)^2} = \frac1{8(k-1)^2k^2(k+1)^2}\left(\frac1{(k-2)^2}- \frac1{(k+2)^2}\right)$$
which can be noted to telescope, hence the sum is $\dfrac1{8\cdot 2^2\cdot3^2\cdot4^2}\cdot\dfrac1{1^2}=\dfrac1{4608}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two lines passing through $(-2,5)$ making angles $\arctan\frac 34$ & $\arctan\frac 23$ with the line $x-y+5=0$.Find the equations of the lines Two straight lines pass through the point $(-2,5)$ such that one of them makes an angle of $\arctan \frac 34$
with the given line $x-y+5=0$
and the given line makes an angle of $\arctan \frac 23$
with the other line . Find the equations of the two lines.
$\bf{Try} :$
Let the equation of the line be $y=mx+c$.$\cdots (1)$
If $(1)$ makes angle $\theta$ with the given line then $\tan\theta =| \frac {m-1}{1+m}|$
So by the given condition we have
$| \frac {m-1}{1+m}|=\frac 34$
Solving we get $m=7,\frac 17$
Putting these values of $m$ and using the condition that the line passes through $(-2,5)$ we get two lines as
$y=7x+19 \cdots (2)$
$7y=x+37\cdots (3)$
Similarly $| \frac {m-1}{1+m}|=\frac 23$
gives $m=5,\frac 15$ and we get the equations of the lines as
$y=5x+15\cdots (4)$
and $5y=x+27\cdots (5)$
But the answer is given as the lines $(2)$ and $(5)$. Why the lines $(3)$ and $(4)$ are excluded?
| {Red (2) Orange (3) Violet (4) Grey(5)} all for image if I will be able to upload image.
If question strictly wants positive angle for statement then:
(3) makes $-\tan^{-1}(\frac {3}{4})$ with $x-y+5=0$
Similarly (4) makes $-\tan^{-1}(\frac{2}{3})$ with $x-y+5=0$
You may draw graph to check.
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| This solution uses no calculus.
By multiplying constrain with 4 we get $$(2x+y)^2+3y^2=4$$
Let $z= 2x+y$ then we have $z^2+3y^2=4\;\;(*)$ and we are searching for extreme points of $$L:= {1\over 4}(z^2-8zy-y^2)$$
Let $z= 2\cos t$ and $y= {2\sqrt{3}\over 3}\sin t$ be a parametrization of elipse $(*)$. Then we have $$L= {1\over 3}(3\cos ^2t -8\sqrt{3}\sin t\cos t - \sin^2t )$$
By further simplification you can get \begin{align} 3L &= 1+2\cos (2t) -4\sqrt{3} \sin(2t)\\
&= 1+\sqrt{52}\sin (2t+\phi)\\
&\leq 1+\sqrt{52}\end{align}
| {
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Evaluate the sum of series convergent series Compute $\sum\frac{1}{n^2(2n-1)}$ assuming that $\sum\frac{1}{n^2}$ = $\frac{\pi^2}{6}$
If we multiply and divide the fraction by 2n+1 we get $\sum\frac{2n+1}{n^2(4n^2-1)}$
but still I couldn't proceed. How to use $\sum\frac{1}{n^2}$ = $\frac{\pi^2}{6}$
to evaluate the series sum?
| \begin{align*}
\sum\dfrac{1}{n^{2}(2n-1)}&=-\sum\dfrac{2n-1-2n}{n^{2}(2n-1)}\\
&=-\sum\dfrac{1}{n^{2}}+2\dfrac{1}{n(2n-1)}\\
&=-\sum\dfrac{1}{n^{2}}+4\sum\left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the integral ... We need to find the integral
$$\iiiint\limits_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}dxdydudv$$
I was only able to get to this point ...
$$\iiiint\limits_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}dxdydudv=\iint\limits_{x^2+y^2\leq 1}e^{x^2+y^2}\left ( \iint\limits_{u^2+v^2\leq 1-x^2-y^2}\frac{dudv}{e^{u^2+v^2}} \right )dxdy$$
I don’t know how to solve it further ...
| Use polar coordinates (twice):
In the inner integral change variables
$u\rightarrow r\cos{\theta}$,
$v\rightarrow r\sin{\theta}$, to get
$$\iint\limits_{u^2+v^2\leq 1-x^2-y^2}\frac{dudv}{e^{u^2+v^2}}=
\int_{0}^{2\pi}\int_{0}^{\sqrt{1-x^2-y^2}}r e^{-r^2} dr\,d\theta
=\pi \left(1-e^{-1}e^{x^2+y^2}\right)$$
Then use polar coordinates once more to compute
$$\int_{0\leq x^2+y^2\leq 1} e^{x^2+y^2} dx dy=\int_{0}^{2\pi} \int_{0}^{1}
r e^{r^2} dr d\theta=\pi (e-1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many $5$-digit numbers less than $43205$ do not contain any digits greater than $6$? I'm not sure how to solve this question.
The only starting point I've thought of is $3 \times 7^4$ because one case would involve the first digit being either $1, 2$ or $3$, and the remaining $4$ digits would each have $7$ number possibilities.
Would someone be able to explain how to approach this question please?
| Your start is perfect. There are $3\times 7^4$ $5$ digit numbers strictly less than $40000$ that do not contain a digit more than $6$.
So now count how many there are between $40000$ and $43205$. If take the second digit can be $0,1,2$ there are $3\times 7^3$ such numbers between $40000$ and $42999$.
So now count how many there are between $43000$ and $43205$ and ... well, continue. If we take the third digit to be $0$ or $1$ there are $2\times 7^2$ numbers between $4300$ and $43199$.
Now we just need the numbers between $43200$ and $43205$ (exclusive) and there are clearly $5$ of those (as we want numbers strictly less than $43205$... so we want $43200, 43201,43202,43203$ and $43204$).
So there are $3\times 7^4 + 3\times 7^3 + 2\times 7^2 + 5$.
....
But you know.... This is basically counting in base $7$.
Every $5$ digit base $7$ number between $10000_7$ through $43205_7$ will be listed.
And So there will be $43205_7 - 10000_7$ such numbers.
And that is $43205_7 - 10000_7 (4\times 7^4 + 3\times 7^3 + 2\times 7^2 + 0 \times 7 + 5\times 7^0 ) - 7^4 = 3\times 7^4 + 3\times 7^3 + 2\times 7^2 + 5=33205_7$
.....
This basically demonstrates why we can express numbers in a decimal base.
| {
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General deduction of expression $x^n + y^n + z^n $ provided some of its previous powers are known.
If $x^3 -x +1 = 0$ and $\alpha$ , $\beta$ , $\gamma$ are its roots ,find $\alpha^2+\beta^2+\gamma^2$ ; $\alpha^3+\beta^3+\gamma^3$ ; $\alpha^4+\beta^4+\gamma^4$ and $\alpha^5+\beta^5+\gamma^5$
My approach:
1. Find $\alpha^2+\beta^2+\gamma^2$
$$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma + \gamma\alpha)$$
Putting the values we get $$\alpha^2+\beta^2+\gamma^2 = 2$$
2. Find $\alpha^3+\beta^3+\gamma^3$
$$\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma$$
Here we substitute the value of $\alpha^2+\beta^2+\gamma^2$ and get $\alpha^3+\beta^3+\gamma^3=3$
3. Find $\alpha^4+\beta^4+\gamma^4$
$$\alpha^4+\beta^4+\gamma^4 = (\alpha^2+\beta^2+\gamma^2)^2 -2[(\alpha\beta+\beta\gamma + \gamma\alpha)^2 -2\alpha\beta\gamma(\alpha+\beta+\gamma)]$$
Here we substitute the value of $\alpha^2+\beta^2+\gamma^2$ and get $\alpha^4+\beta^4+\gamma^4=2$
But I am stuck at $\alpha^5+\beta^5+\gamma^5$. How to factorize/break it in terms of known expressions?
And is there any general algorithm for breaking down these expressions into known lower level powers since those formualae are pretty hard to remember and also lengthy to derive by trial and error ?
| There is a quite direct way using linear recurrencies.
$x^3-x+1$ is the characteristic polynomial of the linear recurrence
$$a_{n+3} - a_{n+1}+a_n=0$$
This has the general solution
$$a_n = A\alpha^n + B\beta^n + C\gamma^n \text{ for } n \geq 0$$
You are looking for the special solution where
$$A = B = C = 1$$
Now, you have
$$a_0 = A+B+C = 3$$
$$a_1 \stackrel{Vieta}{=} \alpha+\beta+\gamma = 0$$
$$a_2 = a_1^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \stackrel{Vieta}{=} 0-2\cdot(-1) = 2$$
Now, you can start the recursion:
$$a_3 = \alpha^3 + \beta^3 + \gamma^3 = a_1 - a_0 = -3$$
$$a_4 = \alpha^4 + \beta^4 + \gamma^4 = a_2 - a_1 = 2$$
$$ \ldots $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find an exponential equation that passes through the points $(2, 2.25)$ and $(5,60.75)$ I am asked to find an exponential equation that passes through $(2, 2.25)$ and $(5, 60.75)$. My textbook says the solution it's $y=0.25(3)^x$ whereas I got $y=0.028(9)^x$. Here is my working:
Express each coordinate in terms of the exponential function:
$$2.25=ab^2$$
$$60.75=ab^5$$
Solve for $a$ in terms of $b$ in the first equation:
$$a=\frac{2.25}{b^2}$$
$$a=2.25b^{-2}$$
Substitute $a$ in the second equation and solve for $b$:
$$60.75=2.25b^{-2}b^5$$
$$60.75=2.25b^3$$
$$b^3=\frac{60.75}{2.25}$$
$$b=27^{\frac{1}{3}}$$
$$b=9$$
Use the value of b in the first equation to solve for the value of a:
$$a=2.25b^{-2}$$
$$a=2.25(9)^{-2}$$
$$a=\frac{2.25}{9^2}$$
$$a=0.028$$
Thus my solution is:
$$y=0.028(9)^x$$
Where did I go wrong and how can I arrive at $y=0.25(3)^x$?
| Alternatively, taking logarithms on $y = ab^x$, $\ln y = \ln a + x \ln b$, which is a linear equation with $\ln y$ against $x$, as you can rewrite this as $y' = c + mx$.
The line passing through $(2, \ln 2.25)$ and $(5, \ln 60.75)$ has gradient $\frac{\ln 60.75 - \ln 2.25}{5 - 2} = \frac{\ln 27}{3}$, so $\ln b = \frac{\ln 27}{3}, b = e^{(\ln 27) / 3} = 27^{1/3} = 3$. Also, $\ln a = \ln y - x \ln b$, where using the point $(2, \ln 2.25)$ gives $\ln a = \ln 2.25 - (2 \ln 27) / 3 = \ln 2.25 - \ln 27^{2/3} = \ln 0.25$, thus $a = 0.25$. So since $a = 0.25, b = 3$, the equation passing through these two points is $y = 0.25 \cdot 3^x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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strong inquality? Suppose $a,b,c$ are positive real number. Prove that
$$\dfrac{a(b+c)}{(a+b)(a+c)}+\dfrac{b(c+a)}{(b+a)(b+c)}+\dfrac{c(a+b)}{(c+a)(c+b)}\ge \dfrac{(a+b+c)^2}{2(a^2+b^2+c^2)}.$$
I 'd tried to use Cauchy-Schwarz's :
$$LHS=\sum_{\{a,b,c\}}\dfrac{(a(b+c))^2}{a(a+b)(a+c)(b+c)}\ge \dfrac{4(ab+bc+ca)^2}{(a+b+c)(a+b)(b+c)(c+a)}.$$
After that, I used $p,q,r$ tranformation and need to prove
$$\dfrac{4q^2}{p(pq-r)}\ge \dfrac{p^2}{2(p^2-2q)}$$
I wonder if it is a true inequality?
Somebody help me? Or give me some hint?
| We write the inequality as
$$2+\frac{8abc}{(a+b)(b+c)(c+a)} \geqslant \frac{(a+b+c)^{2}}{a^2+b^2+c^2}.$$
But
$$(a+b)(b+c)(c+a) \leqslant \frac{8}{9}(a+b+c)(a^2+b^2+c^2).$$
It's remain to prove that
$$2+\frac{9abc}{(a+b+c)(a^2+b^2+c^2)} \geqslant \frac{(a+b+c)^{2}}{a^2+b^2+c^2},$$
equivalent to
$$a^2+b^2+c^2+\frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
Which is true. The proof is completed.
| {
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Evaluating $\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ I want to evaluate $\displaystyle \int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ but it's quite difficult.
I have tried to rewrite the integral as
$$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{\pi }{2}\int _0^{\frac{\pi }{2}}\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx$$
I've also tried to integrate by parts in multiple ways yet I cant go forth with this integral, I also tried using the substitution $t=\tan{\frac{x}{2}}$ but cant get anything to work, I'll appreciate any sort of help.
I also tried using the classical expansion
$$\ln \left(\cos \left(x\right)\right)=-\ln \left(2\right)-\sum _{n=1}^{\infty }\frac{\left(-1\right)^n\cos \left(2nx\right)}{n},\:-\frac{\pi }{2}<x<\frac{\pi }{2}$$
But it only gets worse.
| A quite elegant solution.
The fact
$$
\Im\left[\log^3\left(\frac{1+e^{2ix}}2\right)\right]=3x\log^2\cos x-x^3
$$
yields
$$
\int_0^{\pi/2}x\cot x\log^2\cos x~d x=\frac16\Im\left[\int_{-\pi/2}^{\pi/2}\cot x\log^3\left(\frac{1+\text e^{2ix}}2\right)d x\right]+\frac13\int_0^{\pi/2}x^3\cot x~d x
$$
The rest two integrals are easy.
$$
\begin{align}
&\int_{-\pi/2}^{\pi/2}\cot x\log^3\left(\frac{1+\text e^{2ix}}2\right)~d x
\\=&\oint\frac{z+1}{z-1}\log^3\left(\frac{1+z}2\right)~\frac{d z}{2iz}\quad z=e^{2ix}
\\=&2\pi~\text{Res}\left[\frac{z+1}{z-1}\log^3\left(\frac{1+z}2\right)~\frac{1}{2iz},0\right]
\\=&i\pi\log^32
\end{align}
$$
The contour is the unit circle with small perturbations near $z=\pm1$ .
$$
\begin{align}
&\int_0^{\pi/2}x^3\cot x~d x
\\=&x^3\log(2\sin x)\Big|_0^{\pi/2}-3\int_0^{\pi/2}x^2\log(2\sin x)d x
\\=&\frac{\pi^3}8\log2+3\int_0^{\pi/2}\sum_{n\ge1}\frac{\cos(2nx)}nx^2~d x
\\=&\frac{\pi^3}8\log2+3\sum_{n\ge1}\frac{\pi(-1)^n}{4n^3}
\\=&\frac{\pi^3}8\log2-\frac{9\pi}{16}\zeta (3)
\end{align}
$$
Where Fourier series of $\log(2\sin x)$ is applied.
Combine them and the result follows.
$$
\int_0^{\pi/2}x\cot x\log^2\cos x~d x=\frac{\pi^3}{24}\log2-\frac{3\pi}{16}\zeta (3)+\frac\pi6\log^32
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
A hypergeometric beast: Simple closed-form for combination of $_4F_3(1)$? We have
$$
\frac{1}{4}
-\frac{1}{4}{_4F}_3\left({-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5}\atop\frac{1}{4},\frac{1}{2},\frac{3}{4}};1\right)
-\frac{1}{\sqrt[4]
{5}}{_4F}_3\left({\frac{1}{20},\frac{9}{20},\frac{13}{20},\frac{17}{20}\atop\frac{1}{2},\frac{3}{4},\frac{5}{4}};1\right)\\
+\frac{1}{5\sqrt{5}}{_4F}_3\left({\frac{3}{10},\frac{7}{10},\frac{9}{10},\frac{11}{10}\atop\frac{3}{4},\frac{5}{4},\frac{3}{2}};1\right)
-\frac{7}{50\,5^{3/4}}{_4F}_3\left({\frac{11}{20},\frac{19}{20},\frac{23}{20},\frac{27}{20}\atop\frac{5}{4},\frac{3}{2},\frac{7}{4}};1\right)\\
=-\frac{1}{4}\left(1+\sqrt[3]{5/9}\left(\sqrt[3]{9+4\sqrt 6}-\sqrt[3]{4\sqrt 6-9}\right)\right).
$$
Through indirect methods, this relationship has been proven true by showing that both sides of the equality are solutions to
$$
4x^3+3x^2+2x+1=0.
$$
How can we prove this relation directly by reducing the hypergeometric functions?
While I have some experience with reducing hypergeometric functions, I don't even know where to begin with this one due to the rational parameters with denominator $20$. My initial thought were to work with the integral representation DLMF 16.5.2 to write the $_4F_3(1)$ functions as double integrals of $_2F_1(\cdot)$. For example, taking the first hypergeometric function above we have
$$
{_4F}_3\left({-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5}\atop\frac{1}{4},\frac{1}{2},\frac{3}{4}};1\right)=(constant)\int_0^1\int_0^1u^{\frac{4}{20}-1}(1-u)^{\frac{1}{20}-1}v^{\frac{8}{20}-1}(1-v)^{\frac{2}{20}-1} {_2F_1}\left({-\frac{4}{20},\frac{12}{20}\atop \frac{15}{20}};uv\right)\,\mathrm du\mathrm dv.
$$
At least in this form there seems to be some hope of reducing the $_2F_1(\cdot)$ function in the integrand, which may then lead to the desired form.
| Remark
There is this paper
Beukers, Frits, Hypergeometric functions, how special are they?, Notices Am. Math. Soc. 61, No. 1, 48-56 (2014). ZBL1323.33007.
Plugging into Theorem 2 of that paper, we find that
$$
{_4F}_3\left({-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5}\atop\frac{1}{4},\frac{1}{2},\frac{3}{4}};x\right),\quad
{_4F}_3\left({\frac{1}{20},\frac{9}{20},\frac{13}{20},\frac{17}{20}\atop\frac{1}{2},\frac{3}{4},\frac{5}{4}};x\right),\quad
{_4F}_3\left({\frac{3}{10},\frac{7}{10},\frac{9}{10},\frac{11}{10}\atop\frac{3}{4},\frac{5}{4},\frac{3}{2}};x\right),\quad
{_4F}_3\left({\frac{11}{20},\frac{19}{20},\frac{23}{20},\frac{27}{20}\atop\frac{5}{4},\frac{3}{2},\frac{7}{4}};x\right)
$$
are all algebraic functions. Probably you would have to consult the references of that paper to find explicit polynomials that they satisfy.
Maple code
> # Beukers, Notices, January 2014, p. 48
> reducemod1:=proc(s)
> {seq(x-floor(x),x in s)};
> end;
>
> isirreducible := proc(l1,l2)
> local s1,s2;
> s1:=reducemod1({op(l1)});
> s2:=reducemod1({op(l2)} union {1});
> if s1 intersect s2 = {} then return(true) else return(false) fi;
> end;
>
> isinterlaced := proc(l1,l2)
> isinterlacedL(reducemod1({op(l2)} union {1}), reducemod1({op(l1)}));
> end;
>
> isinterlacedL := proc(l1,l2)
> local s1,x;
> x:=min(l1);
> if x >= min(l2) then return(false); fi;
> s1 := l1 minus {x};
> if s1 = {} then
> if nops(l2) <= 1 then return(true) else return(false); fi;
> fi;
> isinterlacedL(l2,s1);
> end;
>
> thm2 := proc(l1,l2) # is it algebraic?
> local s1,s2,D,r,x;
> s1:= reducemod1(l1);
> s2:= reducemod1(l2) union {0};
> if s1 intersect s2 <> {} then return("reducible"); fi;
> D:=lcm(seq(map(denom,s1 union s2)));
> for r from 1 to D-1 do
> if igcd(r,D) = 1 then
> if not isinterlaced({seq(r*x,x in s1)},{seq(r*x,x in s2)}) then return("transcendental"); fi;
> fi;
> od;
> return("algebraic");
> end;
>
> thm2([1/4,1/4],[1/2]);
"transcendental"
> thm2([3/10,7/10,9/10,11/10],[3/4,5/4,3/2]);
"algebraic"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113789",
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"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $x^2+\frac{4x^2}{(x+2)^2}=5$ Solve the equation $$x^2+\dfrac{4x^2}{(x+2)^2}=5.$$
For $x+2\ne0 \Rightarrow x\ne -2$ we have $$x^2(x+2)^2+4x^2=5(x+2)^2\\x^2(x+2)^2+4x^2-5(x+2)^2=0.$$
Can I factor the LHS? Thank you in advance!
| Observe what happens when we let $x = u - 1$ to get:
$$\left(u-1\right)^2+\frac{4\left(u-1\right)^2}{\left(u+1\right)^2}-5=0$$
$$\left(u^2-1\right)^2+4\left(u-1\right)^2-5(u+1)^2=0$$
$$u^{4}-2u^{2}+1+4u^{2}-8u+4-5u^{2}-10u-5=0$$
$$u^{4}-3u^{2}-18u=0$$
$$u(u^{3}-3u-18)=0$$
Now by the rational root theorem, the candidate roots for the cubic are the divisors of $18$ up to a sign. We observe that negative candidates all fail because the $-3u$ term is not negative enough to yield negative roots. Checking positive candidates, we have a single root $u = 3$, where there is only 1 positive root by Descartes' rule of signs.
Further factorisation gives $u(u - 3)(u^2 + au + 6)=0$ by comparing coefficients, where $-3 + a = 0 \implies a = 3$. However, the quadratic has discriminant $3^2 - 4 \cdot 1 \cdot 6 < 0$, so the only roots are $x = 0 - 1, 3 - 1$ or $-1, 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.
| Perhaps this will help: $${1\over x}+{1\over x+4} = {2x+4\over x^2+4x}$$
$${1\over x+1}+{1\over x+3} = {2x+4\over x^2+4x+3}$$
$${1\over x+2} = {x+2\over x^2+4x+4}$$
and now we have $$ {x+2\over x^2+4x+4}+ {2x+4\over x^2+4x} + {2x+4\over x^2+4x+3}=0$$
Clearly $x\ne -2$ and let $t=x^2+4x$ then you have:
$${1\over t+4}+{2\over t+3} + {2\over t}=0$$ i.e. $$5t^2 + 25t + 24=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
A die was thrown three times. Find probability that first throw is less than the second and second is less than the third throw. My proposed solution follows. Is this correct? If yes, I'd like to know if there is a better way to solve this problem.
We can see that the value of the first throw could only be $1, 2, 3$ or $4$.
Throws are labeled as $T_1, T_2, T_3$.
If $T_1$ is $1$ then $T_2$, $T_3 \in \{2,3,4,5,6\}$
If $T_1$ is $2$ then $T_2$, $T_3 \in \{3,4,5,6\}$
If $T_1$ is $3$ then $T_2$, $T_3 \in \{4,5,6\}$
If $T_1$ is $4$ then $T_2$, $T_3 \in \{5,6\}$
Hence the number of favorable outcomes is $\dbinom{2}{2}+\dbinom{3}{2}+\dbinom{4}{2}+\dbinom{5}{2} = 20$
And $P(A) = \frac{20}{6\times6\times6}=\frac{5}{54}$
| That method would be hard to scale up to larger numbers.
As an alternative, for an $N$ sided die: First note that the probability that the rolls are distinct is $$1\times \frac {N-1}N\times \frac {N-2}N$$
Given that the rolls are distinct, there are $6$ possible, equally probable, orders for them, only one of which is increasing. Thus the answer to your question is $$\frac 16\times 1\times \frac {N-1}N\times \frac {N-2}N=\boxed {\frac {(N-1)(N-2)}{6N^2}}$$
Note that if $N=6$ this resolves to $$\frac {5\times 4}{6\times 36}=\frac {20}{6^3}$$ as desired.
Should say: it's a numerical coincidence that, in your case, the number of ways to order $3$ distinct items happens to be the same as the number of faces on a standard die.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Convergence of Hypergeometric Distribution to Binomial Assuming that we have a Hypergeometric distribution:
$$f(x | N, M, K) = \frac{{M \choose x}{N - M \choose K- x}}{{N \choose K}}$$
where $x \in \{0, 1, ..., M\}$ and assume $N, M \rightarrow \infty, \frac{M}{N} \rightarrow p$, then
$f(x | N, M, K) \rightarrow {K \choose x}p^{x}(1 - p)^{K - x}$
In short, as the population $N$ and targets $M$ grow, Hypergeometric converges to Binomial(K, p).
The book suggests solving this by utilising Stirling approximation. I have no question about that. By rewriting the binomial coefficients:
$$\frac{{M \choose x}{N - M \choose K- x}}{{N \choose K}} = \frac{M!(N-M)!K!(N-K)!}{x!(M-x)!(N-M-K+x)!(K-x)!N!} \sim \frac{\sqrt{2\pi}M^{M+1/2}e^{-M}...\sqrt{2\pi}(N-K)^{N-K+1/2}e^{K-N}}{\sqrt{2\pi}x^{x + 1/2}e^{-x}...\sqrt{2\pi}N^{N+1/2}e^{-N}}$$
Then using some algebra we take the limit of this, and arrive at the desired result.
BUT
Is it actually justified? I feel like the author (Casella & Berger, exercise 3.11) intends this to be "the way" to solve this task, hence the "hint" about Stirling formula.
But from googling, Stirling formula is a relative approximation! In short, it is $$\lim\limits_{N \rightarrow \infty}\frac{N!}{\sqrt{2\pi}N^{N + 1/2}e^{-N}} = 1$$
In fact, the factorial in the numerator and Stirling function might not "converge" at all in a traditional sense of this word: there is no $\epsilon$, s.t. $|N! - \text{Stirling}(N)| < \epsilon$ for all $N > \text{some number}$.
Thus, even though I establish the limit for the "Stirling"-representation of the Hypergeometric distribution, I actually cannot establish the transitive relation to the original formula in terms of factorials. Thus, I can find $\epsilon > 0$, s.t. "Stirling representation" is as close to Binomial as I want pointwise, I cannot find any $\epsilon_2$, s.t. Hypergeometric is arbitrarily close to its Stirling-representation pointwise.
Is my logic correct?
If it is, is there a way to show this differently?
Because it would mean that the derivation of convergence is incorrect.
| You ask for a way to show this differently.
I leave as an exercise for the reader the fussy correction for $M = [nP-1/2,nP+1/2) \cap \Bbb{Z}$, that is letting $M$ always be the integer closest to $pN$ as $N$ varies. (Alternatively, replace all the factorials with Gammas and let $M$ and $N$ range smoothly through real numbers instead of through integers.) Here, we proceed by approximating $M$ with $nP$ immediately.
You want
\begin{align*}
L &= \lim_{N \rightarrow \infty} \frac{(pN)!(N-pN)!K!(N-K)!}{x! (pN - x)! (K-x)!(N-pN-K+x)!N! } \\
&= \frac{K!}{x! (K-x)!} \cdot \lim_{N \rightarrow \infty} \frac{(N-K)!}{K!} \cdot \frac{(pN)!}{(pN-x)!} \\
&\qquad \cdot \frac{(N-pN)!}{(N-pN-K+x)!} \text{.}
\end{align*}
This really is just an application of "factor out the big part". There are three cases. I show the $x > K$ case. The $x = K$ case is similar, but a little easier since we have $\frac{1}{N^0(1-p)^0}$ in the second step. The $x < K$ case is analogous to the one shown. So, suppose $x > K$ ...
\begin{align*}
L &= \binom{K}{x} \lim_{N \rightarrow \infty} \frac{1}{N(N-1)(N-2) \cdots (N-K+1)} \\
&\qquad \cdot \frac{(pN)(pN-1) \cdots (pN-x+1)}{1} \\
&\qquad \cdot \frac{1}{(N-pN-K+x) \cdots (N-pN+1)} \\
&= \binom{K}{x} \lim_{N \rightarrow \infty} \frac{1}{N^K}\frac{1}{1 \left( 1-\frac{1}{N} \right) \left( 1-\frac{2}{N} \right) \cdots \left( 1- \frac{K-1}{N} \right)} \\
&\qquad \cdot p^xN^x \frac{(1) \left( 1- \frac{1}{pN} \right) \cdots \left( 1-\frac{x-1}{pN} \right)}{1} \\
&\qquad \cdot \frac{1}{N^{x-K}(1-p)^{x-K}}\frac{1}{\left(1+ \frac{-K+x}{N-pN} \right) \cdots \left( 1+ \frac{1}{N-pN} \right)} \\
&= \binom{K}{x} p^x \frac{1}{(1-p)^{x-K}} \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{1}{1 \left( 1-\frac{1}{N} \right) \left( 1-\frac{2}{N} \right) \cdots \left( 1- \frac{K-1}{N} \right)} \\
&\qquad \cdot \frac{(1) \left( 1- \frac{1}{pN} \right) \cdots \left( 1-\frac{x-1}{pN} \right)}{1} \\
&\qquad \cdot \frac{1}{\left(1+ \frac{-K+x}{N-pN} \right) \cdots \left( 1+ \frac{1}{N-pN} \right)} \\
&= \binom{K}{x} p^x (1-p)^{K-x} \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{1}{1 \left( 1-\frac{1}{N} \right) \left( 1-\frac{2}{N} \right) \cdots \left( 1- \frac{K-1}{N} \right)} \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{(1) \left( 1- \frac{1}{pN} \right) \cdots \left( 1-\frac{x-1}{pN} \right)}{1} \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{1}{\left(1+ \frac{-K+x}{N-pN} \right) \cdots \left( 1+ \frac{1}{N-pN} \right)} \\
&= \binom{K}{x} p^x (1-p)^{K-x} \cdot 1 \cdot 1 \cdot 1 \text{.}
\end{align*}
(For evaluating the limits involving products of terms of the form $\displaystyle \left( 1 \pm \frac{A(K,p,x)}{N} \right)$, multiply them out completely, obtaining $1 \pm ($ terms involving $K,p,x$ and various negative powers of $N)$. Of course as $N \rightarrow \infty$, the terms involving negative powers of $N$ go to $0$, leaving only the leading $1$.
| {
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"url": "https://math.stackexchange.com/questions/4123694",
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"source": "stackexchange",
"question_score": "2",
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Does $3 < x+2 < 5$ $\implies$ $\frac{1}{5} < \frac{1}{|x+2|} < \frac{1}{3}$?
$3 < x+2 < 5$ $\implies$ $\frac{1}{5} < \frac{1}{|x+2|} < \frac{1}{3}$
If we bound $x$ by $1<x< 3$, then $x + 2 = |x+2|$. Is this correct?
Why do I ask this? Have a look at the following:
Source: Example 7: Evaluating a limit using the definition
| yes, if $1<x<3$ then this means $x>0$ $\Rightarrow$ $x+2 > 0$
if $x+2 > 0$ then $\mid x+2 \mid = (x+2)$
| {
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"answer_count": 1,
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Number of solutions of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ Find the number of solutions $(a,b,c,d)$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$,$\enspace$$a,b,c,d\in \mathbb{N}$ and $a<b<c<d$.
I have got a solution:
$$\frac{1}{2}\cdot(\frac{1}{2}+\frac{1}{3}+\frac{1}{6})+\frac{1}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1$$
But how to get other solutions.
I received total $6$ solutions from Wolfram Alpha.
| *
*Show $a\le4$.
*Run through various possible values of $a$ and $b$.
*To solve ${1\over c}+{1\over d}={1\over4}$ for example,
$${1\over c}+{1\over d}={1\over4}\\
4d+4c=cd\\
(c-4)(d-4)=16$$
Now just run through the factors of 16.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$ Theorem. $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$.
I'm tasked with proving the given theorem by induction. Here's where I've gotten so far...
Proof. Clearly, the theorem is true for $n=1$, establishing the base case. Moreover, it can easily be shown that the theorem works for $n=2,3,4,5,\ldots, 18$.
Suppose, now, that $k$ is an integer for which the given theorem holds. Consider, then, $k+18$.
$$2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}}$$
By Fermat's Little Theorem, if $n \equiv m \pmod{p-1}$, then $a^m \equiv a^n \pmod{p}$. Clearly, $k+18 \equiv k \pmod{18}$. Then,
$$2^{2^{k+18}} \equiv 2^{2^k} \pmod{19}$$
$$3^{2^{k+18}} \equiv 3^{2^k} \pmod{19}$$
$$5^{2^{k+18}} \equiv 5^{2^k} \pmod{19}$$
Hence, $2^{2^{k+18}}$ may be expressed as $\;2^{2^k} + 19n$ for some integer $n$, $\;3^{2^{k+18}}$ as $\;3^{2^k}+ 19m$ for some integer $m$, and $\;5^{2^{k+18}}$ as $\;5^{2^k} + 19q$ for some integer $q$.
Thus, $2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}} = (2^{2^{k}} + 3^{2^{k}} + 5^{2^{k}}) + 19(m+n+q)$, a sum of integers divisible by $19$, and thus clearly divisible by $19$.
By the principle of induction, we've thus shown that $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$, concluding the proof.
$\blacksquare$
Is this approach valid?
| The direct approach uses $-3\equiv 2^4\pmod{19}$ and $-5\equiv 2^{7}\pmod{19}.$
So when $n>0,$ $(-3)^{2^n}=3^{2^n}$ and $(-5)^{2^n}=5^{2^n},$ and we get:
$$2^{2^n}+3^{2^n}+5^{2^n}\equiv 2^{2^n}+\left(2^{2^{n}}\right)^4+\left(2^{2^{n}}\right)^{7} \pmod{19}$$
So the right hand side is $p(2^{2^n})$ where $$p(x)=x^7+x^4+x=x(x^6+x^3+1).$$
Now, it turns out that $2^{2^n}$ is always a root of $q(x)=x^6+x^3+1,$ modulo $19.$
Specifically, since $$q(x^2)=x^{12}+x^6+1=q(x)(x^6-x^3+1),$$
if $x$ is a root of $q$ then $x^2$ is a root of $q.$
So if $x=2^{2^n}$ is a root of $p(x),$ then so is $\left(2^{2^n}\right)^2=2^{2^{n+1}}$ is a root of $p(x).$
So we only need to deal with the case $n=1.$
At heart, what is happening is that the roots of $x^6+x^3+1$ are the primitive 9th roots of $1.$ That is:
$$\frac{x^9-1}{x^3-1}=x^6+x^3+1$$
Squaring a primitive 9th root of $1$ yields another primitive ninth root of $1.$
We can do examples of the same sort, modulo $37,$ such as:
$$37\mid 3^{2^n}+4^{2^n}+ 7^{2^n}$$
Here, $4^4\equiv -3, 4^7\equiv -7\pmod{37}.$
Other cases:
$$73\mid 2^{2^n}+16^{2^n}+18^{2^n}\\
109\mid 4^{2^n}+34^{2^n}+38^{2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Show a polynomial is reducible over $F=\mathbb{Z}/(p)$ with $p$ prime and $p=a^2+b^2$ I have tried to solve the following question:
Recall that $\mathbb{Z}$ denotes the ring of integers, and $\mathbb{Q}$ is the rationals. We let $F[x]$ denote the polynomial ring in the variable $x$ over a field $F$, and we let $f(x)=x^2+1\in F[x]$
Assume that the integer $p$ is a prime of the form $p=a^2+b^2$. Let $F=\mathbb{Z} /(p)$. Show that $f(x)$ is reducible. (Fact: if $p \equiv 1$ modulo 4, then $p=a^2+b^2)$.
Solution :
If we take $a\in F^*=F\setminus \{0\}$ and $a^2+b^2=0$ in $F$, then $a^{-1}\in F^*$ exist. So :
\begin{align}
a^2+b^2 &= 0\nonumber\\
b^2 &=-a^2 \nonumber\\
b^2 &=-\Big(\frac{1}{a^{-1}} \Big)^2 \nonumber \\
a^{-2}b^2 &= -1 \label{eq:2.3} \\
a^{-2}b^2+1 &= 0 \nonumber
\end{align}
Since $f(x)=x^2+1$ is in $F[x]$, we have to show that $f(x)$ is reducible. If we put $1=-a^{-2}b^2$ in $f(x)$ we get
$$ f(x)=x^2+1=x^2-(a^{-1}b)^2=(x-a^{-1}b)(x+a^{-1}b)$$
As such we have expressed $f(x)$ as factors of degree lower than that of said function, hence $f(x)$ is reducible.
Is this answer correct, and if so. Would this answer be sufficient?
| $a\in F^*$ and $a^2+b^2=0$ in $F$, therefore $b^2a^{-2}+1=0$. Hence $x^2+1=(x-ba^{-1})(x+ba^{-1})$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Almost there?? Help with inequality.(updated) I was recently doing a combinatorics problem from MMO given in book Arthur Engel. I might have solved it (though unlikely) as I arrived at following inequality:-
Both $x$ and $y$ are positive. the first term is not in the pattern, while rest are.
$($$\frac{(x^3+y^3)}{(x+y)^3}$$)$ + $($$\frac{(x^4+y^4+x^3y+y^3x+x^2y^2)}{(x+y)^4}$ + $\frac{(x^5+y^5+x^4y+y^4x+x^3y^2+x^2y^3)}{(x+y)^5}$ ....$)$ $<1$ ?
I don't know if it is true (not good with inequalities). Please verify it.
Any help is appreciated :)
I can post the question if you want to give it a try but please don't tell the solution here.
My method:
$(1)$ All triangles are similar and each triangle has unique a unique area. Two triangles are congruent iff their areas are equal. Then, I constructed an ''area-based representation''
$(2)$ If the bigger RAT(Right angled triangle) into area $x$ and $y$, Then when we divide the RAT with area $x$ we end up with triangles of areas: $x^2/(x+y)$ and $xy/(x+y)$. So if we just keep dividing the RAT (not caring about congruency at this point) We end up with the nth row of pascals triangle/$(x+y)^{(n-2)}$.
1(x+y)
1(x) 1(y)
1(x^2/(x+y)) 2(xy/(x+y)) 1(y^2/(x+y))
1(x^3/(x+y)^2) 3(x^2y/(x+y)^2) 3(xy^2/(x+y)^2) 1(y^3/(x+y)^2)
$(3)$ If we keep the first triangle unchanged, then divide the second triangle into $x$ and $y$, then divide the third into $x^2/(x+y)$,$y^2/(x+y)$,$xy/(x+y)$,$x^2y/(x+y)^2$,$xy^2/(x+y)^2$, then the fourth triangle must be
$($$\frac{(x^3+y^3)}{(x+y)^2}$$)$ + $($$\frac{(x^4+y^4+x^3y+y^3x+x^2y^2)}{(x+y)^3}$ + $\frac{(x^5+y^5+x^4y+y^4x+x^3y^2+x^2y^3)}{(x+y)^4}$ ....$)$
which I wanted to show is less than $x+y$
$Attempt(promising?)$
I noticed that in case of maximum area we can't get $(x^4,y^4,x^5,y^5...)$ , then I mordified the inequality but was only able to give the required proof for $0.276 < x < 0.723 $
Reason: Notice that when we are dividing one single triangle we can get many terms but from the each of the sets $S_x$ and $S_y$ we can only get one term from each ( so in total $2$ distinct terms and think about this yourself )
$S_x = (1,x,x^2,x^3,x^4...) $
$S_y = (1,y,y^2,y^3,y^4...) $
And since there are four triangles we can get (in max area case) ${1,x,y,x^2,y^2,x^3,y^3}$
| Assuming your LHS is correct, you can assume from homogeneity and symmetry, WLOG let $t= x/y > 1$:
$$\frac{x^3+y^3}{(x+y)^3}+\sum_{n=5}^\infty \frac{\sum_{k=0}^n x^{n-k}y^k}{(x+y)^n} = \frac{t^3+1}{(t+1)^3}+ \sum_{n=5}^\infty \frac{\sum_{k=0}^nt^k}{(t+1)^n} = f(t)$$
Then (using GP formula, you can / should do the working yourself)
$$f(t)=\frac{t^2-t+1}{(t+1)^2}+\frac1{t-1}\sum_{n=5}^\infty\left( t\cdot\left(\frac{t}{t+1}\right)^n-\frac1{(t+1)^n}\right) \\= \frac{t^6+2t^5+2t^4+t^3+2t^2+2t+1}{t(t+1)^4}$$
It is clear that $f(t)$ increases without bound (as numerator is a higher degree polynomial), and in fact even$f(2)= \dfrac{181}{162}>1$.
| {
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"url": "https://math.stackexchange.com/questions/4131750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate green area plus red area without using integral calculus? I found this problem on the net, and the truth is I don't know where to start, someone would be so kind to explain it to me, I would be very grateful.
edited
|
There is nothing wrong with the methodology used in kyary’s answer. However, it has generated a surprisingly long thread of comments, mostly exchanges of openions between the asker and the answerer. In view of that, we decided to post the following answer, which gives the complete solution to the problem at hand. We hope that our answer would help OP to clear doubts, if he or she still has some. Please note that certain elements of the method given below are somehow slightly different from that of kyary’s solution.
Let the radius of the circle be $r$. We denote the measure of angles $B\hat{O}A$ and $D\hat{O}C$ by $2\omega$ and $2\phi$ respectively. From this, it follows that
$$ \measuredangle BDA = \omega \quad\text{and}\quad \measuredangle DAC =\phi.$$
The external angle at the vertex $E$ of $\triangle AED$ is equal to $60^o$. Therefore,
$$\phi+\omega=60^o. \tag{1}$$
As shown in the $\mathrm{Fig.\space 1}$, we extend the given configuration by drawing the four radii $OA$, $OB$, $OC$, and $OD$ and the line segment $AD$. Addition of the radii gives us two isosceles triangles $BOA$ and $DOC$, using which we shall write,
$$a=2r\sin \left(\phi\right) \quad\text{and}\quad b=2r\sin \left(\omega\right).$$
When we remove $r$ from these two equations, we get
$$b sin \left(\phi\right) = a\sin \left(\omega\right).$$
Therefore, we have
$$ b sin \left(\phi\right) = a\sin \left(60^o - \phi\right) = \frac{a}{2} \left(\sqrt{3}\cos \left(\phi\right) -\sin \left(\phi\right)\right).$$
By simplifying this trigonometric identity as illustrated below, we can obtain a value for $\phi$ in terms of $a$ and $b$.
$$\phi = \tan^{-1} \left(\frac{\sqrt{3}a}{a+2b}\right)$$
There exists a very similar expression for $\omega$ as well.
$$\omega = \tan^{-1} \left(\frac{\sqrt{3}b}{b+2a}\right)$$
Now, we are in a position to express the radius $r$ in terms of $a$ and $b$.
$$r=\frac{a}{2\sin \left(\phi\right)}=\frac{a}{2\sin \left[\tan^{-1} \left(\dfrac{\sqrt{3}a}{a+2b}\right)\right]} \tag{2}$$
By the way, we need an expression for $\left(\phi - \omega\right)$ as well. We assume that you are versed in manipulating inverse trigonometric functions.
$$\phi - \omega = \tan^{-1} \left(\frac{\sqrt{3}a}{a+2b}\right) - \tan^{-1} \left(\frac{\sqrt{3}b}{b+2a}\right) = \tan^{-1}\left(\frac{\dfrac{\sqrt{3}a}{a+2b}-\dfrac{\sqrt{3}b}{b+2a}}{1+ \dfrac{\sqrt{3}a}{a+2b}\cdot \dfrac{\sqrt{3}b}{b+2a}}\right)$$
Upon simplifying, the above expression yields the following equation.
$$\phi - \omega = \tan^{-1}\left(\frac{\sqrt{3}\left(a^2 – b^2\right)}{ a^2 + b^2+4ab}\right) \tag{3}$$
Now, we can express the two areas $A_{a}$ and $A_{b}$.
$$A_{a}=r^2\phi – \frac{1}{2}r^2\sin \left(2\phi\right)$$
$$A_{b}=r^2\omega– \frac{1}{2}r^2\sin \left(2\omega\right)$$
$$\therefore\quad A_{a} + A_{b}=\left[\phi + \omega - \frac{1}{2}\Big(\sin \left(2\phi\right) + \sin \left(2\omega\right)\Big)\right]r^2$$
$$\qquad\qquad\qquad = \Big[\phi + \omega - \sin\left(\phi + \omega\right) \cos\left(\phi - \omega\right) \Big]r^2. \tag{4}$$
By substituting values from (1), (2), and (3) in (4), we obtain,
$$ A_{a} + A_{b} =\frac{2\pi-3\sqrt{3}\cos\left[\tan^{-1}\left(\dfrac{\sqrt{3}\left(a^2 – b^2\right)}{ a^2 + b^2+4ab}\right) \right]}{24\sin^2\left[\tan^{-1}\left(\dfrac{\sqrt{3}a}{a+2b}\right)\right]}a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all prime numbers a,b,c for which the expression $a^2+b^2+c^2-1$ is a perfect square Determine all prime numbers a,b,c for which the expression $a^2+b^2+c^2-1$ is a perfect square
My approach : we know that $(a+b+c)^2 = a^2+b^2+c^2 +2ab+2bc+2ca$
$\Rightarrow $ the given equation to be a perfect square means $2ab+2bc+2ca =-1$
Is this approach correct, please guide how to take it further.. will be of great help thanks.
| Modulo $4$ all squares are $0$ or $1$. If all of $a,b,c$ are odd primes (not $2$) we have $a^2+b^2+c^2-1\equiv2\bmod4$, which is a contradiction. For similar reasons we cannot have all three of $a,b,c$ being $2$, so there must be one or two $2$'s (which we will assign by default to the earliest variables in alphabetical order). If two, we have $7+c^2=d^2$ with the only solution being $c=3$ (and $d=4$).
It remains to check the one-$2$ case. plop's comment rules out the $b,c>3$ case, so we are left with $a=2$, $b=3$ and the equation $12+c^2=d^2$, whose only solution is $c=2$ – but this is the same as the previous considered case up to variable reordering.
Hence the only solutions to $a^2+b^2+c^2-1=d^2$ with $a,b,c$ primes are $\{a,b,c\}=\{2,2,3\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $1-x^n \le \dfrac{n}{n+1}\cdot \dfrac{1}{x}$ if $0Hy guys. I'm trying to prove this inequality for any $n,k \in \mathbb{N}$
$$1- x^n \le \dfrac{n}{n+1}\cdot \dfrac{1}{x^k}$$
Where $0<x<1$.
My solution attempt was to use induction over $k$
For case $k=1$, we have to prove that
$$1- x^n \le \dfrac{n}{n+1}\cdot \dfrac{1}{x}$$
I thought that maybe an induction on $ n $ would solve it now, but I wondered if there was another way less laborious. Can someone help me?
| WTS $ x ( 1 - x^n) \leq \frac{n}{n+1} $
We will show something stronger, namely $ x ( 1 - x^n ) \leq \frac{ n } { (n+1) \sqrt[n] {n+1}}$.
This follows by applying weighted AM-GM:
$$ \frac{n}{n+1} = \frac{ (nx^n) \times 1 + ( 1 - x^n) \times n } { n+1} \geq \sqrt[n+1] {nx^n ( 1-x^n)^n}.$$
So $ x ( 1 - x^n) \leq \sqrt[n]{\frac{n^{n+1} } { n(n+1)^{n+1}} }= \frac{n}{(n+1) \sqrt[n]{n+1}}.$
Equality holds when $nx^n = 1 - x^n \Rightarrow x = \frac{ 1 } { \sqrt[n] { n+1} } $.
Note: The inequality $ x^k ( 1 - x^n) \leq \frac{n}{n+1}$ can also be done in a similar manner. Try it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.
| When you expand the product you get $3^6$ summands, each summand corresponds to whethere $1,x$ or $x^2$ was picked in each of the factors.
Notice that $6$ is the average "degree", so in order to get $x^6$ you need to select the same number of $1$ as $x^2$, and therefore you need to select an even number of $x$. So if we select $x$ $2i$ times we select $1$ $3-i$ times and $x^2$ also $3-i$ times. The number of ways this can be dome is $\binom{6}{2i,3-i,3-i}$.
We can get that the number of ways to do this is $\sum\limits_{i=0}^3 \binom{6}{2i,3-i,3-i}$.
Hence the answer is $\binom{6}{0,3,3} + \binom{6}{2,2,2} + \binom{6}{4,1,1} + \binom{6}{6,0,0} = 20 + 90 + 30 + 1 = 141$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Equation of a Chord to a Circle If equation of a origin centered $\mathcal circle$ with radius $3$ is $x^2+y^2=9$, what is the $\pmb {equation}$ $\pmb{of ~ the ~chord}$ to this circle which is divided internally by the point $P \equiv (1,2)$ in the ratio $\color{blue} 1: \color{red} 2$?
I've been trying to solve this problem as it appears in my textbook's Circle chapter.
My Attempts
I used section formula and found out if point $A(x_1, y_1)$ and $B(x_2, y_2)$ is the endpoints of the chord $\mathbf {AB}$ (which is also the intersection of the chord and the circle) then using section formula for internal division yields $$\frac{1 \cdot x_2 +2 \cdot x_1}{1+2}=1 \implies 2x_1+x_2=3 \implies x_1+x_2=3- x_1 \qquad (1) $$ and $$ \frac{1 \cdot y_2 +2 \cdot y_1}{1+2} =2 \implies 2y_1+y_2=6 \implies y_1+y_2=6- y_1 \qquad (2) $$ as the point $P(1,2)$ divides the segment (chord) $\mathbf {AB}$ in the ratio $\color{blue} 1: \color{red} 2$ . In the next step, I let the equation of the chord to be $y-2=\frac{y_1-y_2}{x_1 -x_2}(x-1)$ as it passes through the point $P(1,2)$ and $A(x_1, y_1)$, $B(x_2, y_2)$. Then if the center of the circle (which is also the origin) is $O(0, 0)$ the distance $OA$ and $OB$ should be the same as they are the radius of the circle. So, $$ OA^2 = OB^2 \implies x_1^{2} +y_1^{2} = x_2^{2} +y_2^{2} \\ \implies x_1^{2} - x_2^{2} = -(y_1^{2} -y_2^{2}) \\ \implies (x_1 +x_2)(x_1 -x_2)=-(y_1-y_2)(y_1+y_2) \\ \implies \frac{y_1-y_2}{x_1 -x_2} = - \frac{x_1 +x_2}{y_1+y_2} $$
Now, if I plug in the values from $(1)$ and $(2)$ then it becomes- $$\frac{y_1-y_2}{x_1 -x_2} = - \frac{3- x_1}{6- y_1}$$
Now, putting it in the equation of the chord $y-2=\frac{y_1-y_2}{x_1 -x_2}(x-1)$ gives $$y-2=-\frac{3- x_1}{6- y_1} (x-1)$$ Sadly I have $x_1$ and $y_1$ left in my equation.
Was my stride taking me some places or should I change my viewpoint to this particular problem? To be honest, I really can not proceed any further. Please provide
some help.
| From 1) and 2) you got
$$
\left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr} \right.
$$
Now add the condition that A,B are on the circle
$$
\eqalign{
& \left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr
x_{\,2} ^{\,2} + y_{\,2} ^{\,2} = 9 \hfill \cr} \right.\quad \left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr
\left( {3 - 2x_{\,1} } \right)^{\,2} + \left( {6 - 2y_{\,1} } \right)^{\,2} = 9 \hfill \cr} \right. \cr
& \left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr
9 + 4x_{\,1} ^{\,2} - 12x_{\,1} + 36 + 4y_{\,1} ^{\,2} - 24y_{\,1} = 9 \hfill \cr} \right. \cr
& \left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr
36 - 12x_{\,1} - 24y_{\,1} + 36 = 0 \hfill \cr} \right. \cr
& \left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr
x_{\,1} + 2y_{\,1} = 6 \hfill \cr} \right.\quad \quad \left\{ \matrix{
2x_{\,1} + x_{\,2} = 3 \hfill \cr
2y_{\,1} + y_{\,2} = 6 \hfill \cr
2y_{\,1} + x_{\,1} = 6 \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr} \right. \cr
& \left\{ \matrix{
x_{\,2} = 3 - 2x_{\,1} \hfill \cr
y_{\,1} = 3 - {{x_{\,1} } \over 2} \hfill \cr
y_{\,2} = x_{\,1} \hfill \cr
x_{\,1} ^{\,2} + y_{\,1} ^{\,2} = 9 \hfill \cr} \right.\quad \left\{ \matrix{
x_{\,2} = 3 - 2x_{\,1} \hfill \cr
y_{\,1} = 3 - {{x_{\,1} } \over 2} \hfill \cr
y_{\,2} = x_{\,1} \hfill \cr
x_{\,1} ^{\,2} + 9 + {{x_{\,1} ^{\,2} } \over 4} - 3x_{\,1} = 9 \hfill \cr} \right. \cr
& \left\{ \matrix{
x_{\,2} = 3 - 2x_{\,1} \hfill \cr
y_{\,1} = 3 - {{x_{\,1} } \over 2} \hfill \cr
y_{\,2} = x_{\,1} \hfill \cr
\left( {{5 \over 4}x_{\,1} - 3} \right)x_{\,1} = 0 \hfill \cr} \right. \cr}
$$
I suppose you can conclude from here
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$
I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$
| $$
\begin{align}
\sum_{n=1}^\infty\frac1{n(2n+1)}
&=2\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{2n}-\frac1{2n+1}\right)\tag1\\
&=2\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1n-\left(\frac1{2n}+\frac1{2n+1}\right)\right)\tag2\\
&=2\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\sum_{n=2}^{2N+1}\frac1n\right)\tag3\\
&=2\lim_{N\to\infty}\left(1-\sum_{n=N+1}^{2N+1}\frac1n\right)\tag4\\
&=2\lim_{N\to\infty}\left(1-\sum_{n=N+1}^{2N+1}\frac Nn\frac1N\right)\tag5\\
&=2\left(1-\int_1^2\frac{\mathrm{d}x}x\right)\tag6\\[9pt]
&=2(1-\log(2))\tag7
\end{align}
$$
Explanation:
$(1)$: partial fractions
$\phantom{\text{(1):}}$ write infinite sum as a limit of partial sums
$(2)$: arithmetic
$(3)$: combine the inner parenthetical terms into one sum
$(4)$: cancel terms
$(5)$: multiply by $\frac NN$
$(6)$: convert Riemann Sum into integral
$(7)$: evaluate integral
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Non-intersecting lines Given the equations of two lines I have to find the parameter $k$ so the lines don't interesect:
\begin{gather*}
L_1: \left\{(x,y,z)\in \Bbb{R}^3 \,\middle|\, x+2=\frac{3-y}{2}=\frac{z-1}{2} \right\}\\
L_2: \{(x,y,z)\in \Bbb{R}^3 \mid x=a-2k,\ y=2a+k,\ z=4+2k,\ k\in \Bbb{R} \}
\end{gather*}
I am trying to equalize the direction vectors as $(1,-2,2) = k(-2,1,2)$ so the lines are parallels, but I still can't get the value of $k$, any hints?
| First of all, let us re-write the equation of $(L_1)$ under the canonical form:
$$x+2=\frac{y-3}{-2}=\frac{z-1}{2}\color{red}{=p} $$
i.e.,
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-2\\3\\1\end{pmatrix}+p\underbrace{\begin{pmatrix}1\\-2\\2 \end{pmatrix}}_U \tag{1}$$
Then, a major remark is that the second set of equations $(L_2)$ doesn't refer to a single line but a whole (affine) plane,
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\4\end{pmatrix}+a\underbrace{\begin{pmatrix}1\\2\\0\end{pmatrix}}_V+k\underbrace{\begin{pmatrix}-2\\ \ \ 1\\ \ \ 2\end{pmatrix}}_W\tag{2}$$
with normal vector :
$$V \times W = \begin{pmatrix}1\\2\\0\end{pmatrix} \times \begin{pmatrix}-2\\ \ \ 1\\ \ \ 2\end{pmatrix}=\begin{pmatrix}4\\-2\\5 \end{pmatrix} \ \text{not prop. to} \ U=\begin{pmatrix}1\\-2\\2 \end{pmatrix}$$
($U$ has been defined in (1)).
Therefore, $L_1$ intersects the plane, meaning that there is a value $k_0$ of $k$ such that (2) is fulfilled.
Set apart this value, in all other cases, there are no solutions.
Remark: the value $k_0$ of $k$ is obtained by solving the linear system obtained by equating (1) and (2):
$$\begin{pmatrix}-2\\3\\1\end{pmatrix}+p\begin{pmatrix}1\\-2\\2 \end{pmatrix}=\begin{pmatrix}0\\0\\4\end{pmatrix}+a\begin{pmatrix}1\\2\\0\end{pmatrix}+k\begin{pmatrix}-2\\ \ \ 1\\ \ \ 2\end{pmatrix}$$
giving the unique solution:
$k=\dfrac{1}{9}$
for the unique value $a=- \ \dfrac{1}{6}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of an improper integral involving trigonometric functions
Determine the convergence of the following improper integral:
$$\int_{2018}^{\infty}\frac{\sin^3x}{x^{\frac{3}{4}}+x^{\frac{1}{6}}\cos x}\,dx$$
Now I thought about using Dirichlet's criteria, but unfortunately $\frac{1}{x^{\frac{3}{4}}+x^{\frac{1}{6}}\cos x}$ is not monotonically decreasing. No relevant change of variables comes to mind also. Am I on the right track?
Any help would be appreciated.
| Note that $\sin^3 x = \frac{3}{4} \sin x - \frac{1}{4} \sin 3x$ and it is enough to consider the convergence as $c \to \infty$ of the integral
$$\tag{*}\int_{2018}^c\frac{\sin ax}{x^{3/4} + x^{1/6}\cos x}\, dx = \int_{2018}^c\frac{\sin ax}{x^{3/4} }\, dx - \int_{2018}^c\left[\frac{\sin ax}{x^{3/4} + x^{1/6}\cos x} - \frac{\sin ax}{x^{3/4} }\right]\, dx$$
The first integral on the RHS of (*) converges by the Dirichlet test. For the second integral we have
$$\left|\frac{\sin ax}{x^{3/4} + x^{1/6}\cos x} - \frac{\sin ax}{x^{3/4} }\right| = \left|\frac{\sin ax\cdot x^{1/6}\cos x}{(x^{3/4} + x^{1/6}\cos x)x^{3/4}} \right| = \frac{|\sin ax|\, |\cos x|}{(x^{3/4} + x^{1/6}\cos x)x^{7/12}}\\ \leqslant \frac{1}{(x^{3/4} + x^{1/6}\cos x)x^{7/12}} = \frac{1}{x^{4/3}+ x^{3/4}\cos x}$$
Consequently, the second integral on the RHS of (*) is absolutely convergent by the comparison test, since by the limit comparison test (with $x^{-4/3}$) we have
$$\int_{2018}^\infty \frac{dx}{x^{4/3}+ x^{3/4}\cos x} < \infty$$
Therefore, the improper integral on the LHS of (*) is convergent.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$ I am trying to compute the following double sum
$$\boxed{\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)}$$
I proceeded as following
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{\left(m+n \right)\left(m-n \right)}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{2m}\left[\frac{1}{\left(m+n \right)}+\frac{1}{\left(m-n \right)}\right]$$
$$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m+n \right)}}_{2\zeta(3)}+ \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n} \sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m-n \right)}$$
The first sum is given here and I have also evaluated here, therefore we get that
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \sum_{m=1}^{\infty}\frac{1}{m}-\frac{1}{\left(m-n \right)}$$
$$=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \int_{0}^{1}\frac{1-x^{-n}}{1-x}dx$$
$$=\zeta(3)-\frac{1}{2} \int_{0}^{1}\frac{1}{1-x}\left[ \sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{x^{-n}}{n^2}\right]dx$$
$$=\zeta(3)-\frac{1}{2} \int_{0}^{1}\frac{\zeta(2)-Li_{2}\left( \frac{1}{x} \right)}{1-x}dx$$
From this post we can use the relation
$$Li_{2}\left( \frac{1}{x} \right)-Li_{2}\left( x \right)=\zeta(2)-\frac{1}{2}\log^2(-x)$$
To get
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{\zeta(2)}{2} \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2} \int_{0}^{1}\frac{Li_{2}\left( x \right)+\zeta(2)-\frac{1}{2}\log^2(-x)}{1-x}dx$$
$$=\zeta(3)+\frac{\zeta(2)}{2} \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2} \int_{0}^{1}\frac{Li_{2}\left( x \right)}{1-x}dx-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$=\zeta(3)+\frac{\zeta(2)}{2} \log(1-x)\Big|_{0}^{1}+\frac{1}{2}\left\{ -Li_{2}(x)\log(1-x)\Big|_{0}^{1}-\int_{0}^{1}\frac{\log^2(1-x)}{x}dx\right\} -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$=\zeta(3)-\frac{1}{2}2\zeta(3) -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
My questions are the following:
(1)Is the last equality correct?
(2)If so, what is the meaning of $\log^2(-x)=\log^2(e^{i\pi}x)$?
(3) How do we solve the last integral?
| In here, we have that $\displaystyle \sum_{\underset{m \neq n}{m = 1}}^{\infty} \frac{1}{m^2 - n^2} = \frac{3}{4n^2}$, therefore
\begin{aligned} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)} = \frac{3}{4}\sum_{n=1}^{\infty} \frac{1}{n^3} = \frac{3}{4}\zeta(3).\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Multiplicative Order when for integers $m > 2, n > 0$, when $2^m | (3^n - 1)$ It seems to me that for integers $m>2, n>0$, when $2^m | (3^n - 1)$, that the multiplicative order is $2^{m-2}$ so that for $0 < i < 2^{m-2}$, $3^i \not\equiv 1 \pmod {2^m}$ and $3^{2^{m-2}}\equiv 1 \pmod {2^m}$
I am having trouble completing the inductive argument.
Here's what I came up with for the base case. For $m=3$, $2^3 | (3^n-1)$ if and only if $2 | n$ since $3^{2t+1} \equiv 3\times9^t \equiv 3 \pmod 8$ and $3^{2t} \equiv 9^t \equiv 1^t \pmod 8$
Assume that up to some $m \ge 3$, $2^m | (3^n - 1)$ if and only if $2^{m-2} | n$
Assume that $2^{m+1} |(3^n-1)$. It follows that there exists an integer $a$ such that $n = 2^{m-2}a$.
Assume that $a$ is odd. It seems to me that it should be possible to complete this argument through a contradiction. I am not clear how to finish the argument.
| Hint: Note with $n = 2^{m-2}a$ that
$$3^n - 1 = (3^{2^{m-2}} - 1)(\color{blue}{3^{2^{m-2}(a-1)} + 3^{2^{m-2}(a-2)} + \ldots + 3^{2^{m-2}} + 1}) \tag{1}\label{eq1A}$$
There are $a$ terms in $\color{blue}{\text{blue}}$. If $a$ is odd, then since each term is odd, the sum will also be odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Characterizing an integer in terms of associated binary quadratic forms Let’s say I have a positive integer $n$, and I know the following:
*
*$4n+1$ is of the form $a^2+b^2$
*$16n+3$ is of the form $a^2+2b^2$.
*$16n+1$ is of both forms $a^2+b^2$ and $a^2+2b^2$.
QUESTION: Is there a simple/elegant way to characterize all such integers $n$? I’m imagining that quadratic reciprocity will give some constraints… Are there other elementary means one can apply?
EDIT: A brute search on $n ≤ 200$ yields
$$n \in \{1, 3, 6, 15, 18, 21, 28, 36, 39, 42, 48, 55, 58, 60, 61, 78, 81, 88, 105, 106,\\ 111, 120, 126, 130, 133, 139, 150, 151, 175, 189, 190, 193, 196, \dots\}.$$
There are some obvious patterns/members in there (e.g., most [but not all!] triangular numbers), but I’d like a complete characterization, if possible.
| Suppose a is odd and:
$a^2\equiv 1 \bmod 16$
and b is even, then:
$A=4n+1=a^2+b^2\equiv 1\bmod 16$
$B=16n+3=c^2+2d^2$
c must be odd, suppose:
$c^2\equiv 1\bmod 16$
let d be odd and:
$d^2\equiv 1\bmod 16$
therefore:
$c^2+2d^2\equiv 3\bmod 16$
$C=16n+1$
It can be seen that a family of n can be of the form $n=16m+1$ and we have:
$A=4n+1=64m+5$
$B=16n+3=256m+19$
$C=16n+1=256m+17$
For example for $m=5$ we have:
$A=325=15^2+10^2$
$B=1299=7^2+2\times 25^2$
$C=1297=1^2+36^2=35^2+2\times 6^2$
These values of A, B and C are primitive solutions, clearly any multiple such as $k^2$ will give another solution.
Update:
For $m=2^6+1=65$ we obtain:
$A=4165=42^2+49^2$
$B=16659=39^2+2\times 87^2$
$C=16657=67^2+2\times 78^2=129^2+4^2$
I checked more powers up to $m=2^{14}+1$ and could not find more solutions. May be there are some more results with higher powers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $|\frac{x+1}{x-2}|<1$ and $\frac{6w+7}{2w-1}- \frac{6w+1}{2w}=1$. I want to find
the real number such that $|\frac{x+1}{x-2}|<1$ and find the values $w$ such that :
$$\frac{6w+7}{2w-1}- \frac{6w+1}{2w}=1$$.
By playing a little bit with this first inequality I have found out that the real numbers $(-\infty,0]$ satisfy the inequality. But when Im trying to prove this I got te following:
$|\frac{x+1}{x-2}|<1$ means that $-1< \frac{x+1}{x-2}<1$ from the left inequality I got $-x+2<x+1$. Then I got $\frac{1}{2}<x$. From the right inequality I got that $x+1 < x-2$, but this gives me no information for $x$ since its cancelled. So what can I conclude from the right inequality? Does all the real numbers satisfy this inequality are $x \in (- \infty, \frac{1}{2})$?
For the second equality I got that:
$$\frac{6w+7}{2w-1}- \frac{6w+1}{2w}= \frac{6w+7-6w-1}{(2w+1)(2w)}=\frac{6}{4w^{2}+2w}$$
Then I got $0=4w^{2}+2(w)-6$ which can be solved by the quadratic formula but this gives me values $\frac{-6}{4}$ and $1$. But clearly when $w=1$ the equality doesnt hold.
But Im stuck there. Thanks!
| Another way to handle the inequality here is by squaring the absolute value symbol away:
$$\begin{align}
\left|x+1\over x-2\right|\lt1
&\iff\left(x+1\over x-2\right)^2\lt1\\
&\iff(x+1)^2\lt(x-2)^2\\
&\iff x^2+2x+1\lt x^2-4x+4\\
&\iff6x\lt3\\
&\iff x\lt1/2
\end{align}$$
so the inequality is satisfied for all $x\in(-\infty,1/2)$.
The one subtlety here is that we need to check that $x=2$, where the original left hand side is undefined, is not in the interval $(-\infty,1/2)$. If, for example, the inequality sign had pointed the other way, the final answer would be $(1/2,2)\cup(2,+\infty)$ instead of $(1/2,+\infty)$.
For the equation in $w$, you made two errors in the first step. It should be
$${6w+7\over2w-1}-{6w+1\over2w}={(6w+7)(2w)-(6w+1)(2w-1)\over(2w-1)(2w)}$$
That is, you forgot to cross multiply in the numerator, and you changed $2w-1$ to $2w+1$ in the denominator. Can you take things from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the perpendicular distance from a point to a line using vectors (the harder algebraic way) I want to prove that the distance from a point $P\left(x_{0},y_{0}\right)$ to line $l$: $ax+by+c=0$ is given by $\frac{\left|ax_{0}+by_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}$.
So I am stuck on $$\left|\text{perp}_lAP\right|=\left|\binom{x_0}{y_0+\frac{c}{b}}-\frac{-bx_0+ay_0+\frac{ac}{b}}{\sqrt{a^2+b^2}}\:\binom{-b}{a}\right|$$
Where $l$ is the vector parallel to the line,and $AP$ is a vector. See diagram/graph for context.
So I took the projection of $AP$ onto $l$, then found the perpendicular by proj+perp=l
And this should work right? I see no error, but just a very messy simplification.
I know the better way is to do instead, the projection from $PA$ onto $l$, which we know is perpendicular anyway.
But me doing |perp| = |$l$-proj| SHOULD also work right?How do I simplify it (especially turning $bx_0$ into $ax_0$)
Diagram:
https://imgur.com/a/JXfSxko
| The points where $ax+by+c=0$ can be parametrized as
$$
\left(\frac{-bt-ac}{a^2+b^2},\frac{at-bc}{a^2+b^2}\right)\tag1
$$
The square of the distance from $(x_0,y_0)$ is then
$$
\begin{align}
&\left(x_0+\frac{bt+ac}{a^2+b^2}\right)^2+\left(y_0+\frac{-at+bc}{a^2+b^2}\right)^2\\[6pt]
&=\left(x_0^2+y_0^2\right)+\frac{2x_0(bt+ac)+2y_0(-at+bc)}{a^2+b^2}+\overbrace{\frac{(bt+ac)^2+(-at+bc)^2}{\left(a^2+b^2\right)^2}}^{\large\frac{t^2+c^2}{a^2+b^2}}\tag{2a}\\
&=\frac{\left(a^2+b^2\right)\left(x_0^2+y_0^2\right)+2c(ax_0+by_0)+2(bx_0-ay_0)t+c^2+t^2}{a^2+b^2}\tag{2b}\\[3pt]
&=\frac{\left(a^2+b^2\right)\left(x_0^2+y_0^2\right)+2c(ax_0+by_0)+c^2-(bx_0-ay_0)^2+(t+bx_0-ay_0)^2}{a^2+b^2}\tag{2c}\\[3pt]
&=\frac{a^2x_0^2+b^2y_0^2+2abx_0y_0+2c(ax_0+by_0)+c^2+(t+bx_0-ay_0)^2}{a^2+b^2}\tag{2d}\\[3pt]
&=\frac{(ax_0+by_0+c)^2+(t+bx_0-ay_0)^2}{a^2+b^2}\tag{2e}
\end{align}
$$
Explanation:
$\text{(2a)}$: expand the squares collecting the first, middle, and last parts
$\text{(2b)}$: put over a common denominator and collect powers of $t$
$\text{(2c)}$: complete the square in $t$
$\text{(2d)}$: expand $\left(a^2+b^2\right)\left(x_0^2+y_0^2\right)-(bx_0-ay_0)^2$
$\text{(2e)}$: collect $(ax_0+by_0+c)^2$
$\text{(2e)}$ is minimized when $(t+bx_0-ay_0)^2=0$ which gives the minimum distance to be
$$
\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\tag3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?
| By implicit differentiation on $x = t^2 + 2t$, one obtains
$$1 = 2t \cdot \dfrac{\text{d}t}{\text{d}x} + 2\dfrac{\text{d}t}{\text{d}x} = \dfrac{\text{d}t}{\text{d}x}\left(2t+2 \right)$$
hence
$$\dfrac{\text{d}t}{\text{d}x} = \dfrac{1}{2t+2}\text{.}$$
Differentiating a second time on the equation $x = t^2 + 2t$, one obtains by an application of the product rule
$$0 = 2\left(t \cdot \dfrac{\text{d}^2t}{\text{d}x^2} + 1 \cdot \dfrac{\text{d}t}{\text{d}x} \right) + 2\text{.}$$
It follows from the above that
$$\dfrac{\text{d}^2t}{\text{d}x^2} = \dfrac{\dfrac{-2}{2} - \dfrac{\text{d}t}{\text{d}x}}{t} = \dfrac{-1 - \dfrac{\text{d}t}{\text{d}x}}{t} = -\left(\dfrac{1 + \dfrac{\text{d}t}{\text{d}x}}{t} \right)\text{.}$$
Substitute $\dfrac{\text{d}t}{\text{d}x}$ as found earlier, and you're done.
Note that you will have to solve for $t$ to compute this quantity. (Hint: make use of the fact that $x = t^2 = 2t$ and that you have the point $(x, y) = (8, 80)$.)
| {
"language": "en",
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"source": "stackexchange",
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Asymptotic expansion of $\int_0^{2\pi} \mathrm{d}\theta\, \sqrt{k^2\cos^2\theta - \cos\theta + 1}$ as $k\to0$ I have the integral
$$I = \int_0^{2\pi} \mathrm{d}\theta\, \sqrt{k^2\cos^2\theta - \cos\theta + 1}$$
and I would need the asymptotic expansion of the integral for small values of $k$. For $k=0$ we get quite easily $I=4\sqrt{2}$, while for $k\gg 1$, we have $I\sim 4|k|$.
Now, trying to perform the integration using special functions (at first glance, one would think that it could be done in terms of elliptic functions) yields no result. Gradštejn and Ryžik are of no help, nor Mathematica or other softwares.
We can Taylor expand the integrand and integrate each term of the expansion, but the $2n$-th term, for $n>1$, reads
$$ \frac{k^{2n}}{(2n)!}\frac{ (-1)^{n-1} (2n-3)!! }{2^n} \frac{\cos^{2n}\theta}{(1-\cos \theta)^{n-1/2}}$$
having set $k=0$ in the derivative. Of course, integrating each term would mean performing
$$\int_0^{2\pi}\mathrm{d} \theta\,\frac{\cos^{2n}\theta}{(1-\cos \theta)^{n-1/2}}$$
which is divergent for $n>1$.
Any ideas about how to proceed? Are there some kind of 'regularization' techniques that could help?
| Surprisingly, this integral can be written in closed form using a linear combination of complete elliptic functions. This doesn't exactly answer the question but at least the integral disappears.
Step 1: Let $u=\cos\theta$ so that $$I=2\int_{-1}^1\sqrt{\frac{1-u+k^2u^2}{1-u^2}}\,du.$$
Step 2: Notice that an appropriate substitution of the form $u=(Av+B)/(Cv+D)$ allows the linear term to be eliminated in the numerator. Choosing $A=D=1$ and $B=C=K$ where $$K=1+k^2-k\sqrt{2+k^2}$$ yields (after extensive simplification) $$I=\frac{MN}{K^2}\int_{-1}^1\frac1{(v+1/K)^2}\sqrt{\frac{1-v^2/N^2}{1-v^2}}\,dv$$ where $$M=2\sqrt{(1-K^2)(K-K^2-k^2)}\quad\text{and}\quad N=\sqrt{\frac{k^2K^2-K+1}{K-K^2-k^2}}.$$
Step 3: The complete elliptic integral of the first, second and third kinds are, respectively: \begin{align}K\left(\frac1N\right)&=\int_0^1\frac1{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\E\left(\frac1N\right)&=\int_0^1\frac{1-v^2/N^2}{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\\Pi\left(n\mid\frac1N\right)&=\int_0^1\frac1{(1-nv^2)\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv.\end{align} Partial fraction decomposition gives \begin{align}\frac{K^2}{MN}I&=\int_{-1}^1\frac{-\frac1{N^2}+\frac2{KN^2}\cdot\frac1{v+1/K}+\left(1-\frac1{K^2N^2}\right)\cdot\frac1{(v+1/K)^2}}{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\&=-\frac2{N^2}K\left(\frac1N\right)+\frac2{KN^2}J_1+\left(1-\frac1{K^2N^2}\right)J_2.\end{align} Notice that \begin{align}aJ_2&=-bJ_1-c\int_{-1}^1\frac{v^2+d}{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\&=-bJ_1+2cN^2E\left(\frac1N\right)-2c(N^2+d)K\left(\frac1N\right)\end{align} where $$a=-\frac{K^{4}N^{2}+1-K^{2}(N^{2}+1)}{K^{4}N^{2}},\quad b=\frac{2-K^{2}(N^{2}+1)}{K^{3}N^{2}},\quad c=\frac1{N^2},\quad d=-\frac1{K^2}.$$ Thus $$\frac{K^2}{MN}I=fK\left(\frac1N\right)+gE\left(\frac1N\right)+h\int_{-1}^1\frac1{(v+1/K)\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv$$ where \begin{align}f&=-\frac2{N^2}-\frac{2c(N^2+d)}a\left(1-\frac1{K^2N^2}\right)\\g&=\frac{2cN^2}a\left(1-\frac1{K^2N^2}\right)\\h&=\frac2{KN^2}-\frac ba\left(1-\frac1{K^2N^2}\right).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why does the formula for $\tan 3\theta$ work at $\theta=\frac\pi4$ even though its proof involves cancelling $1-\tan^2\theta$? Here's a proof for the $\tan 3\theta$ formula:
According to the procedure for the proof, when we cancel out $1-\tan^2\theta$, the value of $1-\tan^2\theta$ should not be equal to $0$ (in particular, $\theta$ should not be $\frac{\pi}{4}$) as $\frac{ax}{bx}=\frac{a}{b}$ only if $x$ isn't $0$.
But, when we substitute $\theta=\frac{\pi}{4}$ in the final formula, we still get the correct answer:
$$\frac{3\tan\frac{\pi}{4}-\tan^3\frac{\pi}{4}}{1-3\tan^2\frac{\pi}{4}}=\tan\frac{3\pi}{4}=-1$$
How is this possible?
| You could also first compute the triple angle formulas for sine and cosine,
$$
\sin(3x)=\frac1{2i}(e^{i3x}-e^{-i3x})=\sin(x)(e^{i2x}+1+e^{-i2x})=\sin(x)(-4\sin^2(x)+3)\\
\cos(3x)=\frac12(e^{i3x}+e^{-i3x})=\cos(x)(e^{i2x}-1+e^{-i2x})=\cos(x)(4\cos^2(x)-3)\\
$$
so in consequence
$$
\tan(3x)=\tan x\,\frac{3\cos^2x-\sin^2x}{\cos^2x-3\sin^2x}=\tan x\,\frac{3-\tan^2x}{1-3\tan^2x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Behavior of $\sum_i^n (1-\frac{1}{i})^s$ as a function of $s$? I'm interested in behavior of the following sum as a function of $s$
$$\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s$$
For $n=1000$ and $s\in (1,10000)$, this seems almost linear on a log-plot (notebook), any tips how to model this analytically?
| I shall scale $s$ with $n$ and write $s=\alpha n$ with some $\alpha>0$. It will be assumed that $\alpha>0$ is bounded and $n$ is large. First, from the monotonicity properties of the integrand,
\begin{align*}
\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} \le \sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^{\alpha n} } & \le \int_1^{n + 1} {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} \\ & \le \int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} + \left( {1 - \frac{1}{{n + 1}}} \right)^{\alpha n} ,
\end{align*}
i.e.,
$$
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^{\alpha n} } = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} .
$$
Now we write
\begin{align*}
&\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} = \int_{1/n}^1 {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}}
\\ &
= \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}} + \int_{1/n^{2/3} }^1 {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}} .
\end{align*}
Here
$$
\int_{1/n^{2/3} }^1 {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}} \le n^{2/3} \exp \left( {\alpha n\log \left( {1 - \frac{1}{{n^{2/3} }}} \right)} \right) \le n^{2/3} e^{ - \alpha n^{1/3} } ,
$$
and
\begin{align*}
& \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}} = \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha nt - \alpha n\mathcal{O}(t^2 )} \right)\frac{{dt}}{{t^2 }}}
\\ & = \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha nt} \right)\frac{{dt}}{{t^2 }}} \left( {1 +\mathcal{O}\!\left( {\frac{1}{{n^{1/3} }}} \right)} \right) = n\alpha \int_\alpha ^{\alpha n^{1/3} } {e^{ - s} \frac{{ds}}{{s^2 }}} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{n^{1/3} }}} \right)} \right)
\end{align*}
for any fixed $\alpha>0$. Consequently,
$$
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} = \alpha \int_\alpha ^{ + \infty } {e^{ - s} \frac{{ds}}{{s^2 }}} = \alpha \Gamma ( - 1,\alpha )
$$
for any fixed $\alpha>0$. Here $\Gamma(a,z)$ is the upper incomplete gamma function. Finally,
$$\boxed{
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^{\alpha n} } = \alpha \Gamma ( - 1,\alpha )
}$$
for any fixed $\alpha>0$. Note that in terms of the exponential integral, $\alpha \Gamma ( - 1,\alpha ) = e^{ - \alpha } - \alpha E_1 (\alpha )$.
Try to plot, for example,
$$
\frac{1}{{1000}}\sum\limits_{i = 1}^{1000} {\left( {1 - \frac{1}{{i}}} \right)^s } \;\; \text{ and } \;\; \frac{s}{{1000}}\Gamma\! \left( { - 1,\frac{s}{{1000}}} \right)
$$
against each other on the range $1\leq s\leq 10000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Approach to calculating limit via polynomial division Given the following exercise:
$ \lim_{x \to \frac{1}{3}} (\frac{27x^2-6x-1}{6x^2+x-1}) $
The rule of L'Hospital with the first derivative yields:
$\lim_{x \to \frac{1}{3}} \frac{27 x^2 - 6 x - 1}{6 x^2 + x - 1} = \frac{12}{5}$
Another approach would be factoring out the x which also yields:
$\frac{12}{5}$
However, I'm asking myself whether polynomial division, too, could help solving the limit.
Polynomial division yields:
$27 x^2 - 6 x - 1 = \frac{9}{2} \cdot (6 x^2 + x - 1) + \frac{7}{2} - \frac{21x}{2}$
polynomial division on Wolfram Alpha
Thanks.
| Let $$f(x)=\frac{27x^2-6x-1}{6x^2+x-1}$$
let us compute, using division with increasing powers, the limit
$$\lim_{t\to 0}f(t+\frac 13)$$
$$f(t+\frac 13)=\frac{27t^2+12t}{6t^2+5t}=\frac{27t+12}{6t+5}$$
$$=\frac 15\frac{27t+12}{1+\frac 65t}$$
$$=\frac 15(27t+12)(1-\frac 65t+o(t))$$
so the limit is $ \frac{12}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of series $\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2$ Given the series
$$
\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2,
$$
can I conclude from
$$
\begin{align}
\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2
&=\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})(\frac{1+k^2}{1+k^3})\\
&\leq\sum_{k=1}^\infty (\frac{k^2+k^2}{1+k^3})(\frac{k^2+k^2}{1+k^3})\\
&\leq\sum_{k=1}^\infty (\frac{2k^2}{k^3})(\frac{2k^2}{k^3})\\
&=\sum_{k=1}^\infty (\frac{2}{k})(\frac{2}{k})\\
&=\sum_{k=1}^\infty (\frac{2}{k^2}),
\end{align}
$$
that $\sum_{k=1}^\infty (\frac{1+k^2}{1+k^3})^2$ must converge since $\sum_{k=1}^\infty (\frac{2}{k^2})$ converges and the above estimations are valid?
| Yes, it is correct, except that after the final $=$ sign you should get $\sum_{k=1}^\infty\frac4{k^2}$.
Besides, note that the inequality$$\frac{1+k^2}{1+k^3}\leqslant\frac2k\tag1$$is equivalent to $k+k^3\leqslant2+2k^3$, which, in turn, is equivalent to $k\leqslant2+k^3$, which is true, since you actually have $k\leqslant k^3$ for each natural $k$. So, this gives you another way of proving $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Summation problem (not sure of my solution) If $f(x)=3-x$, evaluate $\sum_{k=1}^3f(2k)$.
So this is my solution:
$f[2(3-1)]=4$
$f[2(3-2)]=2$
$f[2(3-3)]=0$
So the required sum is $4+2+0=\boxed{6}$
Is my solution correct?
| With $f(x)=3-x$ we obtain
\begin{align*}
\color{blue}{\sum_{k=1}^3f(2k)}&=f(2\cdot1)+f(2\cdot 2) + f(2\cdot 3)\\
&=f(2)+f(4)+f(6)\\
&=(3-2)+(3-4)+(3-6)\\
&=1-1-3\\
&\,\,\color{blue}{=-3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
If $2^{2x-1}$ = $(\frac{1}{5})^x$ and $\log 2 = a$ prove that: $x=\frac{a}{a+1}$ This is my working out:
$2^{2x} \times 2^{-1} = (5^{-1})^x$
$\log 2^{2x-1} = \log 5^{-x}$
$2x-1 \log 2 = -x\log5$
$2x-1\times a = -x\log5$
At this point, I got stuck.
| $$2^{2x-1}=5^{-x}$$
$$\begin{align}&\implies (2x-1)\log 2=-x \log 5 \\
&\implies\frac{1-2x}{x}=\frac {\log 5}{\log 2} \\
&\implies \frac 1x-2=\frac {\log 5}{\log 2} \\
&\implies \frac 1x=\frac{2\log 2+\log 5}{\log 2}\\
&\implies x=\frac{\log 2}{2\log 2+\log 5}=\frac{\log 2}{\log 20}\end{align}$$
If you mean $\lg$ instead of $\log$, then you have:
$$x=\frac{\lg 2}{\lg 20}=\frac{\lg 2}{1+\lg 2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality, finding Constant Can anyone help me with this Math Olympiad Task from 2007 from Germany? I want to find the smallest $C$, such that for every $x,y \in \mathbb{R}$ the inequality:
$$
1+(x+y)^2 \leq C(1+x^2)(1+y^2)
$$
holds.
I know that I have to maximize the function
$$a(x, y) = \frac{1+(x+y)^2}{(1+x^2)(1+y^2)}$$
which gives me $\frac{4}{3}.$
P.S.: Thanks for the help y'all :)
| Simple Verification
Once we know the constant is $\frac43$, the inequality is fairly simple.
$$
1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag1
$$
is equivalent to
$$
\color{#C00}{1+4x^2y^2}+\color{#090}{x^2+y^2}\ge6xy\tag2
$$
which is true because the AM-GM says $\color{#C00}{1+4x^2y^2}\ge4xy$ and $\color{#090}{x^2+y^2}\ge2xy$. Equality is attained when $x=y=\frac1{\sqrt2}$.
Variational Argument To Get The Constant
Suppose we have have $\left(1+x^2\right)\left(1+y^2\right)$ fixed and we wish to maximize $1+(x+y)^2$. That is, for all $\delta x,\delta y$ so that
$$
2x\left(1+y^2\right)\delta x+2y\left(1+x^2\right)\delta y=0\tag3
$$
we also have
$$
2(x+y)(\delta x+\delta y)=0\tag4
$$
$(3)$, $(4)$, and orthogonality implies that there is a $\lambda$ so that
$$
2(x+y)=\lambda2x\left(1+y^2\right)\qquad\text{and}\qquad2(x+y)=\lambda2y\left(1+x^2\right)\tag5
$$
which implies
$$
x+\frac1x=y+\frac1y\tag6
$$
which means that
$$
y=x\qquad\text{or}\qquad y=\frac1x\tag7
$$
Let $\boldsymbol{y=\frac1x}$
$$
\begin{align}
\frac{1+(x+y)^2}{\left(1+x^2\right)\left(1+y^2\right)}
&=\frac{1+\left(x+\frac1x\right)^2}{\left(1+x^2\right)\left(1+\frac1{x^2}\right)}\\
&=\frac{1+\left(x+\frac1x\right)^2}{\left(x+\frac1x\right)^2}\\
&\le\frac54\tag8
\end{align}
$$
since $x+\frac1x\ge2$ (equality when $x=y=1$).
Let $\boldsymbol{x=y}$
$$
\begin{align}
\frac{1+(x+y)^2}{\left(1+x^2\right)\left(1+y^2\right)}
&=\frac{1+4x^2}{1+2x^2+x^4}\\
&=16\frac{1+4x^2}{16+32x^2+16x^4}\\[6pt]
&=\frac{16}3\frac1{\frac{1+4x^2}3+2+\frac3{1+4x^2}}\\
&\le\frac43\tag9
\end{align}
$$
since $\frac{1+4x^2}3+\frac3{1+4x^2}\ge2$ (equality when $x=y=\frac1{\sqrt2}$).
Therefore, we get that
$$
1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag{10}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the reminder when $1^1+2^2+3^3+\ldots+50^{50}$ is divided by 8 I have a sum that goes as follows $$P=1^1+2^2+3^3+\ldots +49^{49}+50^{50}.$$
The question is to find the reminder when $P$ is divided by 8.
Inorder to find this, I seperated $P$ into $P_1$ and $P_2$ where $$P_1=2^2+4^4+\ldots+50^{50},$$
$$P_2=1^1+3^3+\ldots+49^{49}.$$
And I realised that $P_1 \text{ mod } 8 = 4 $ and $ P_2 \text{ mod } 8 = 1$.
But I want to know if there is an easier, straightforward method for finding the answer.
| This is still kind of brute force, but it's much more optimized, and it's a solution you can easily generalize to $\sum_{k=1}^{n} k^k \pmod{m}$ for higher values of $n$.
By the rules of modular arithmetic, when exponentiating, you can reduce the base modulo 8, and then break the terms into groups of 8. So, instead of writing $1^1 + 2^2 + \cdots + 50^{50}$, you can write $(1^1 + 2^2 + \cdots + 8^8) + (1^9 + 2^{10} + \cdots + 8^{16}) + \cdots + (1^{41} + \cdots + 8^{48}) + 1^{49} + 2^{50}$.
Further rearranging, we can write this as
$$ (1^1 + 1^9 + \cdots + 1^{49}) + \cdots + (8^8 + 8^{16} + \cdots + 8^{48}). $$
Now we can analyze this by casework, into 8 cases.
*
*$1^{n}$ = 1 for all n, so the first subcase sums to 7.
*$2^2 = 4$, and for any $n\geq 3$, $2^n$ is divisible by 8. Hence this second subcase is $4 \pmod{8}$.
*Note that $3^8 = 1 \pmod{8}$. Hence every term is the same mod 8. This contributes $3 \cdot 6 = 18 = 2 \pmod{8}$.
*Writing $4^n$ as $2^{2n}$, we see that all terms are divisible by 8.
*This is the same as the third case; $5^8 = 1 \pmod{8}$. Hence every term is the same mod 8, contributing $5 \cdot 6 = 30 = 6 \pmod{8}$.
*For all $n\geq 3$, $6^n$ is divisible by 8. Hence this case contributes nothing modulo 8.
*$7^8 = 1 \pmod{8}$. Applying the same logic, we have $7\cdot 6=42 = 2 \pmod{8}$.
*This last term contributes nothing because $8 | 8^n$ for all $n\geq 1$.
Adding up all the values, we get $7+4+2+6+2=21 = 5 \pmod{8}$. This is our answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $y''(1+\ln(x)) + \frac{1}{x}y' = 2+\ln(x)$ When $y = \frac{1}{2}$, $y' = 1$, for $x=1$ Solve $$y''(1+\ln(x)) + \frac{1}{x}y' = 2+\ln(x)$$
First I converted it to $$p' +\frac{1}{x(1+\ln(x))}y' = \frac{2+\ln(x)}{1+\ln(x)} $$
Which looks like a nice first order differential equation. However, when $I = 1+ \ln(x)$ you solve for $p$ you get the following DE:
$$y' = \frac{2x+\frac{1}{x}+C}{1+\ln(x)}$$
Which does not produce the right result when I put it through a calculator. Where is the mistake?
| I think your error was the antiderivative of $\ln x$: it is $x\ln x - x$. Here is a complete computation of the result:
Note that
$$
y''(1+\ln x) + \frac{1}{x} y' = \left[y' (1 + \ln x) \right]',
$$
and the equation becomes
$$
\left[y'(1+\ln x)\right]' = 2 + \ln x.
$$
Integrating, it follows that
$$
y' (1+\ln x) = 2x + x\ln x - x + C = x (1 + \ln x) + C,
$$
and
$$
y' = x + \frac{C}{1 + \ln x}.
$$
If $y'(1) = 1$, then $C = 0$ and $y' = x$. Finally, $y(x) = \frac{x^2}{2} + C'$, and $y(1) = \frac{1}{2}$ yields
$$
y(x) = \frac{x^2}{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the solutions of the equation $\frac{a^2(x-b)(x-c)}{(a-b)(a-c)}+\frac{b^2(x-a)(x-c)}{(b-a)(b-c)}+\frac{c^2(x-a)(x-b)}{(c-a)(c-b)}=3x-2$
Find the solutions of the equation $$\frac{a^2(x-b)(x-c)}{(a-b)(a-c)}+\frac{b^2(x-a)(x-c)}{(b-a)(b-c)}+\frac{c^2(x-a)(x-b)}{(c-a)(c-b)}=3x-2$$
If we consider the expression as an identity and substitute $x=a$ we get the value of $a=1,2$ similarly $b=1,2$ and $c=1,2$
If we substitute these values into the expression at least one of the denominator will be equal to $0$ which is not possible therefore this expression is not an identity. How to proceed further?
| Add $-x^2$ on both sides , Let expression on LHS be $f(x)$
Therefore LHS will be a quadratic in $x$ will be become zero three times clearly as
$f(a)=f(b)=f(c)=0$ Now if a quadratic equation has more than three roots, then all complex numbers will be it's roots and therefore it will identically be equal to zero.
$f(x)\equiv 0$ and therefore since $f(x)=3x-2-x^2$
$3x-2-x^2\equiv 0$
which is possible for $x=1,2$ only
| {
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} |
Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$ Let $a,b,c>0$ satisfy $a^2+b^2+c^2=3$ . Prove that: $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$$
My idea is to use a well-known inequality (We can prove by Schur) $$(a^2+2)(b^2+2)(c^2+2)\ge 9(ab+bc+ca)$$ and the problem is prove $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge(a^2+2)(b^2+2)(c^2+2)$$
Anyone have a better idea ? please help me
| Another way: use uvw https://math.stackexchange.com/tags/uvw/info
Since $a^5-2a+4\geq4a^2-5a+4,$ it's enough to prove that:
$$\prod_{cyc}(4a^2-5a+4)\geq9(ab+ac+bc),$$ which is $f(w^3)\geq0,$ where $f$ increases. I got here:
$$f'(w^3)=128w^3+684u+115-240v^2>0.$$
Thus, it's enough to prove the last inequality in two cases:
*
*$w^3\rightarrow0^+$;
*Two variables are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$ I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.
Prove the following inequality without using calculator:
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Can you check my proof? It would be really grateful.
First, we can define function $f(x)$ as following:
$$f(x) = \sqrt[3]{x+1} - \sqrt[3]{x}\space(x > 0)$$
$$f(3) = \sqrt[3]{3+1} - \sqrt[3]{3} = \sqrt[3]{4} - \sqrt[3]{3}$$
$$f(2) = \sqrt[3]{2+1} - \sqrt[3]{2} = \sqrt[3]{3} - \sqrt[3]{2}$$
Then, we will differentiate $f(x)$ to check whether $f(x)$ is a decreasing function or not. $f '(x)$ must be a falling function if $f '(x)$ < 0.
$$f'(x) = \frac{1}{3\sqrt[3]{(x+1)^2}} - \frac{1}{3\sqrt[3]{x^2}}$$
Since the minuend is smaller than the subtrahend (minuend has a bigger denominator than the denominator of subtrahend), we can say $f '(x)$ is less than 0 which makes $f(x)$ a decreasing function. Falling function means that $f(a) > f(a+1)$. Substitute $a=2$ and we get:
$$f(2) > f(3)$$
$$\sqrt[3]{3} - \sqrt[3]{2} > \sqrt[3]{4} - \sqrt[3]{3}$$
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Thank you for reading this text, but it would be more grateful if you check my solution, and comment my solution.
I wish you a beautiful day, and stay safe.
| Alternative approach
No calculators allowed.
Requires knowledge that
$\log_{10} (2) \approx 0.301$ and $\log_{10} (3) \approx 0.477.$
The problem reduces to showing that
$(4)^{(1/3)} + (2)^{(1/3)} < 2(3)^{(1/3)}.$
Cubing both sides, this resolves to showing that
$4 + 2 + 3\left[4^{(2/3)}2^{(1/3)} + 4^{(1/3)}2^{(2/3)}\right] < 24.$
This in turn resolves to showing that
$4^{(2/3)}2^{(1/3)} + 4^{(1/3)}2^{(2/3)} < 6.$
Converting to logarithms, base $(10)$,
this resolves to showing that
$10^{(1.505/3)} + 10^{(1.204/3)} < 6.$
This resolves to showing that
$$10^{(0.5017)} + 10^{(0.4014)} < 6.\tag 1$$
$\log_{10}(3.2) \approx (5 \times 0.301) - 1 = 0.505.$
Therefore, $10^{(0.5017)} < 3.2.$
Further, $\log_{10} (2.7) \approx (0.477 \times 3) - 1 = 0.431.$
Therefore $10^{(0.4014)} < 2.7.$
Therefore, the LHS of expression (1) above is $< 5.9$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $a\in\mathbb{C}$ such that $f=X^4 −(a+4)X^3 +(4a+5)X^2 −(5a+2)X +2a$ has $a$ as a double root. Find all $a\in\mathbb{C}$ such that $f=X^4 −(a+4)X^3 +(4a+5)X^2 −(5a+2)X +2a$ has $a$ as a double root.
Here is what I did:
I know for a fact that all $f\in\mathbb{C}$ of degree $n\geq1$ have exactly $n$ roots. So, I write $f=(X-a)^2(X-b)(X-c)=X^4 - (2a+b+c)X^3 + (bc+2ac+2ab+a^2)X^2 - (2abc+a^2c+a^2b)+a^2bc$. I solved the system of equations ($2a+b+c=a+4$, ...) using Wolfram Alpha and found $\begin{cases} a=1,b=1,c=2 \\ a=1,b=2,c=1 \\ a=2,b=1,c=1\end{cases}$. Hence $a=2$ because when $a=1$, $a$ is a triple root.
I don't reckon this is the solution they were asking for. I mean, I wouldn't try to solve that system of equations without the help of a computer. So what could I do instead?
| I would start with noticing that
\begin{align*}
f(x) & = x^{4} - (a+4)x^{3} + (4a+5)x^{2} - (5a+2)x + 2a\\\\
& = (x^{4} - ax^{3}) - (4x^{3} - 4ax^{2}) + (5x^{2} - 5ax) - (2x - 2a)\\\\
& = x^{3}(x-a) - 4x^{2}(x-a) + 5x(x - a) - 2(x-a)\\\\
& = (x^{3} - 4x^{2} + 5x - 2)(x-a)
\end{align*}
where the first factor from the last expression can be rewritten as
\begin{align*}
x^{3} - 4x^{2} + 5x - 2 & = (x^{3} - x^{2}) - (3x^{2} - 3x) + (2x - 2)\\\\
& = x^{2}(x-1) - 3x(x-1) + 2(x-1)\\\\
& = (x^{2} - 3x + 2)(x-1)\\\\
& = (x-1)^{2}(x-2)
\end{align*}
Consequently, the proposed polynomial has $a$ as a double root iff $a = 2$ (if $a = 1$, then it is a triple root).
Hopefully this helps!
| {
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"url": "https://math.stackexchange.com/questions/4189632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Cubic equation problem $\frac{x^3}{3}-x=k$ The cubic function $\frac{{{x^3}}}{3} - x = k$ has three different roots $\alpha,\beta,\gamma$ about the real number k. Let's call the minimum value of $|\alpha|+|\beta|+|\gamma|$ as $m$. FInd the value of $m^2$.
My approach is as follow
$\frac{{{x^3}}}{3} - x = k \Rightarrow f\left( x \right) = \frac{{{x^3}}}{3} - x - k$
$f'\left( x \right) = {x^2} - 1 = 0$
Hence $x = \pm 1$
$f\left( 1 \right) = \frac{1}{3} - 1 - k = - \left( {k + \frac{2}{3}} \right)\& f\left( { - 1} \right) = - \frac{1}{3} + 1 - k = - \left( {k - \frac{2}{3}} \right)$
For real roots $f\left( 1 \right)f\left( { - 1} \right) < 0$
Therefore $ - \left( {k + \frac{2}{3}} \right) \times \left( { - \left( {k - \frac{2}{3}} \right)} \right) < 0$
$k \in \left( { - \frac{2}{3},\frac{2}{3}} \right)$
We know that $\alpha+\beta+\gamma=0$
But how we will find the minimum values of the sum of the modulus of the roots.
| Note: this is similar to another answer, except that by recognizing only one root needs to be tracked instead of the other two the calculation is simplified.
First observe that if we reverse the sign of $k$ then all roots are also reversed in sign with no effect on the set of absolute values, thus no impact on the sum $|\alpha|+|\beta|+|\gamma|$. So we can cover all values of this sum by considering just the case where $k$ is nonpositive.
Then only one root is negative by Descartes' Rule of Signs. Calling that root $\alpha$ we then have
$|\alpha|+|\beta|+|\gamma|=-\alpha+\beta+\gamma$
$\color{blue}{=-2\alpha\because\alpha+\beta+\gamma=0}$
Thus find the nonpositive value of $k$ that minimizes the absolute value of the negative root $\alpha$. Given that $k=0$ gives $\alpha=-\sqrt3$ and $k<0$ must give $\alpha<-\sqrt3$ (why?), you then get your answer using the blue equation above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If , $\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$ . For , $a_i\in\mathbb Z^+$ & $a_i
$Q.$ If , $$\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$$ For , $a_i\in\mathbb Z^+$ & $a_i<i$ . Find $a_2+a_3+......+a_7$ ?
MY APPROACH :
We have , $$\frac{5}{7}=\frac{3!-1}{3!+1}=\frac{3!+1-2}{3!+1}=1-\frac{2}{3!+1} \Rightarrow 1-\frac{2}{7}$$
Now I've : $$1-\frac{2}{7}=\underbrace{\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}}_{\Lambda}$$ $$\Lambda=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\left(\frac{1}{4!}-\frac{1}{5!}\right)+\left(\frac{1}{5!}-\frac{1}{6!}\right)+\left(\frac{1}{6!}-\frac{2}{7}\right)$$ OR ,$$\frac{a_2}{2!}+\frac{a_3}{3!}+ \frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac{1}{2!}+\frac{2}{3!}+ \frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!}-\frac{1433}{7!}$$
DOUBT : In my solution , all the terms are following the conditions from the Question except $a_7$ not satisfy $a_7\in\mathbb Z^+$ and $a_7\in\{1,2,3,4,5,6\}$ only by this condition $a_i<i$
$\Rightarrow$ Where am I wrong ?
| You did great!
We can absorb the negative sign by using the other terms.
\begin{align}
&\frac1{2!}+\frac2{3!} + \frac3{4!} + \frac4{5!} + \frac5{6!} - \frac{1433}{7!} \\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} + \frac5{6!} \color{blue}{- \frac{1435}{7!} + \frac2{7!}}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} + \frac5{6!}\color{blue}{ - \frac{205}{6!}} + \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} \color{blue}{-\frac{200}{6!}} + \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} \color{blue}{-\frac{204}{6!} + \frac{4}{6!}}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} \color{blue}{-\frac{34}{5!}} + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} \color{blue}{-\frac{30}{5!}} + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} \color{blue}{ -\frac{6}{4!} }+ \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} \color{blue}{- \frac3{4!} } + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} \color{blue}{- \frac{4}{4!} + \frac1{4!} } + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} \color{blue}{- \frac{1}{3!}} + \frac1{4!} + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \color{blue}{\frac{1}{3!} }+ \frac1{4!} + \frac{4}{6!}+ \frac2{7!}\\
\end{align}
I noticed that $1433=1435-2$ and $1435$ is a multiple of $7$, $204$ is a multiple of $6$ and so on.
| {
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"url": "https://math.stackexchange.com/questions/4193132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplify the product $\prod_{r=1}^n\left(\cos{\frac{2 \pi}{n}}+\sin{\frac{2\pi}{n}}\cdot \cot{\frac{(2r-1)\pi}{n}}\right)$, where $n$ is even Question: Simplify the product as far as possible: $$\prod_{r=1}^n\left(\cos{\frac{2 \pi}{n}} + \sin{\frac{2\pi}{n}}\cdot \cot{\frac{(2r-1)\pi}{n}}\right),$$ where $n$ is even.
The question is taken from a University entrance exam from the UK called the STEP, for which I am unable to find the solution to online.
When attempting similar questions my approach was to substitute values in order to spot a pattern however this isn't really feasible in this question.
| Rewrite it as:
$$\frac{\prod_{r=1}^n \left(\cos{\frac{2 \pi}{n}}\sin\frac{(2r-1)\pi}{n}+\sin{\frac{2\pi}{n}}\cos{\frac{(2r-1)\pi}{n}}\right)}{\prod_{r=1}^n \sin{\frac{(2r-1)\pi}{n}}}.$$
The numerator term is $$\cos{\frac{2 \pi}{n}}\sin\frac{(2r-1)\pi}{n}+\sin{\frac{2\pi}{n}}\cos{\frac{(2r-1)\pi}{n}}\\=\sin{\frac{(2r+1)\pi}n}$$
So the numerator and the denominator products are equal, because:
$$\frac{(2r+1)\pi}n=\frac{(2(r+1)-1)\pi}n$$
For $r=1,\dots,n-1$ and:
$$\frac{(2n+1)\pi}n=2\pi+\frac{(2\cdot 1-1)\pi}n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the sum of the series $\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt n}{n}$ have an analytic expression? Just out of curiosity, I'd like to know whether or not the sum of the series
$$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}$$
has a known analytic expression.
I stumbled across this series while trying to evaluate
$$\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx$$
The convergence of this integral can be seen by making use of the inequality $\lfloor x\rfloor > x-1$ and the fact that $\coth^{-1}(t)\to 0$ as $t\to\infty$:
\begin{align*}
\int_1^t\frac{1}{\lfloor x^2\rfloor}dx &= \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{\lfloor x^2\rfloor}dx\\
&< \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{x^2-1}dx\\
&= \int_\sqrt{1}^\sqrt{2}\frac{1}{1}dx-\int_\sqrt{2}^t\frac{1}{1-x^2}dx\\
&= \sqrt{2}-1-\left[\coth^{-1}(t)-\coth^{-1}\left(\sqrt 2\right)\right]\\
&= \sqrt{2}-1-\coth^{-1}(t)+\coth^{-1}\left(\sqrt 2\right)\\
&\to \sqrt{2}-1+\coth^{-1}\left(\sqrt 2\right)\text{ as }t\to\infty\\
\end{align*}
Since this implies that $\int_1^t 1/\lfloor x^2\rfloor dx$ is strictly increasing ($1/\lfloor x^2\rfloor >0$ for every $x\geq 1$) and bounded above, the integral necessarily converges. By breaking up the integral
$$\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx$$
into integrals indexed by the intervals $\left[\sqrt{i},\sqrt{i+1}\right]$ for $i=1,2,3,...,k$ and simplifying the resulting sum, I was able to show that
$$\int_1^{\sqrt{k+1}}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^{k} \frac{\sqrt{n+1}-\sqrt n}{n}$$
is true for every $k\geq 0$, which yields
$$\int_1^\infty \frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}$$
after letting $k\to\infty$. This equality is the main reason why I'm interested in the sum of the aforementioned series.
After some (unsurprisingly) futile attempts to evaluate the integral, I expect there to be no closed-form expression for the sum, which is why I'm open to an analytic expression (gamma function, Bessel functions, Riemann zeta function, etc.). Any help is appreciated.
Edit: after seeing the bounds provided by Markus Scheuer and Jorge, I thought I'd share some of my own.
From the fact that $x-1<\lfloor x\rfloor<x$ is true for every non-integer $x\geq 1$, we can infer that for every integer $k\geq 1$,
$$\int_\sqrt{k+1}^\infty \frac{1}{x^2}dx<\int_\sqrt{k+1}^\infty \frac{1}{\lfloor x^2\rfloor}dx<\int_\sqrt{k+1}^\infty \frac{1}{x^2-1}dx$$
Using
$$\int_{x}^{\infty}\frac{1}{t^2-1}dt=\coth^{-1}(x)$$
and
$$\int_\sqrt{k+1}^\infty\frac{1}{\lfloor x^2\rfloor}dx=\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx-\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}-\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
we deduce that
$$\frac{1}{\sqrt{k+1}}+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}<\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}<\coth^{-1}\left(\sqrt{k+1}\right)+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
| Hint: At least we have nice upper and lower bounds for the series.
We obtain by expanding with $\sqrt{n+1}+\sqrt{n}$:
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}}
&=\sum_{n=1}^{\infty }\frac{1}{n\left(\sqrt{n+1}+\sqrt{n}\right)}\\
&>\sum_{n=1}^{\infty }\frac{1}{(n+1)\left(\sqrt{n+1}+\sqrt{n+1}\right)}\\
&= \frac{1}{2}\sum_{n=1}^{\infty }\frac{1}{(n+1)^{\frac{3}{2}}}\\
&=\frac{1}{2}\sum_{n=2}^\infty \frac{1}{n^{\frac{3}{2}}}\\
&\,\,\color{blue}{=\frac{1}{2}\zeta\left(\frac{3}{2}\right)-\frac{1}{2}}\\
\end{align*}
and on the other hand we have
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}}
&=\sum_{n=1}^{\infty }\frac{1}{n\left(\sqrt{n+1}+\sqrt{n}\right)}\\
&<\sum_{n=1}^{\infty }\frac{1}{n\left(\sqrt{n}+\sqrt{n}\right)}\\
&= \frac{1}{2}\sum_{n=1}^{\infty }\frac{1}{n^{\frac{3}{2}}}\\
&\,\,\color{blue}{=\frac{1}{2}\zeta\left(\frac{3}{2}\right)}\\
\end{align*}
and since the sequence $\left(\sum_{n=1}^{N} \frac{\sqrt{n+1}-\sqrt{n}}{n}\right)_{1\leq N<\infty}$ of partial sums is strictly increasing and bounded above by $\frac{1}{2}\zeta\left(\frac{3}{2}\right)$ we know the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$
I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}-\frac{(7x)^6}{6!}+\dots\}}{3x^2}\\ =\lim\limits_{x\to 0}\frac{x^2(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1}{3}(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)\\ =\frac {49}{6}$
Can this be solved without using the Taylor series?
| Use l'hopital. The derivative of the numerator is $(1-\cos 7x)'=7\sin 7x$ and of the denominator $(3x^2)=6x$. We get
$$\lim_{x\to0}\frac{7\sin(7x)}{6x}=\lim_{x/7\to0}\frac{7\sin(x)}{6(x/7)}=\lim_{x\to0}\frac{49\sin(x)}{6x}=\frac{49}{6}.$$
In the second step we used the substitution $x\mapsto x/7$, see here why it's justified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the following integral $\int\sqrt{x^2+1} dx.$ Please complete my solution. I am reading a calculus book.
There is the following problem with a hint:
Find the following integral $$\int\sqrt{x^2+1} dx.$$
Hint: use the substitution $x+\sqrt{x^2+1}=s$.
I was not able to complete my solution.
My attempt is here:
$s=x+\sqrt{x^2+1}$.
$ds=(1+\frac{x}{\sqrt{x^2+1}})dx=\frac{s}{\sqrt{x^2+1}}dx$.
$dx=\frac{\sqrt{x^2+1}}{s}ds$.
So, $\int\sqrt{x^2+1} dx=\int\frac{x^2+1}{s} ds$.
Please complete my solution.
| $$\int\sqrt{x^2+1} dx=\int\frac{x^2+1}{s} ds\quad (*)$$
Right.
$\ s=x+\sqrt{x^2+1}.\ $ If we expand $\ s^2\ $ we get,
$$s^2 = 2xs+1\implies s^2-1=2xs\implies 2x=\frac{s^2-1}{s}$$
and continue trying to find an expression for $x^2$, then substitute into $(*)$.
| {
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"url": "https://math.stackexchange.com/questions/4198971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closure of a subset in ordered square I want to find the closure of these sets on the ordered square:
$$B=\left\{(1-\frac{1}{n})\times \frac{1}{2}|\ n \in Z_{+})\right\}$$
Consider $I^2 = [0,1] \times [0,1]$ in the dictionary order topology.
Our claim is $\overline{B} = B \cup \{1 \times 0\}$. At first, we will show $1 \times 0$ is a limit point of $B$. Every neighborhood $U$ of $1 \times 0$ is of form $U: = (a \times b, 1 \times c)$ for some $a <1$ and $c > 0$. There exists $n \in \mathbb{N}$ such that $a < 1 - \frac{1}{n} <1$. Now $U$ intersects $B$ at points $1-\frac{1}{n} \times \frac{1}{2}$, hence $1 \times 0 \in \overline{B}$.
Now we will see that no no other points of $I^2$ except $1 \times 0$ is a limit point.
Case 1: The point $0 \times y$ such that $y \in [0, \frac{1}{2})$ is not a limit point of $A$. Clearly $(0 \times 0, 0 \times \frac{1}{2})$ is a neighborhood of the point $0 \times y$, which does not intersect $B$, hence $0 \times y \in \overline{B}$.
Case 2: Let $x \times y \not\in B$, and $x <1$, then there exists $n \in \mathbb{N}$ such that $1-\frac{1}{n} <x < 1-\frac{1}{n+1}$. Then $((1-\frac{1}{n}) \times \frac{1}{2},(1-\frac{1}{n+1}) \times \frac{1}{2}) $ is a neighborhood which does not intersect $B$, hence $x \times y \not\in \overline{B}$.
$\textbf{UPDATE: }$ Let $x \times y \not\in B$, and $x <1$, then there exists $n \in \mathbb{N}$ such that $1-\frac{1}{n} \leq x < 1-\frac{1}{n+1}$. Now we have two cases:
*
*If $x = 1-\frac{1}{n}$ and $y <2$, then $((1-\frac{1}{n}) \times \frac{1}{2},(1-\frac{1}{n+1}) \times \frac{1}{2}) $ is a neighborhood of $x \times y$ which does not intersect $B$, hence $x \times y \not\in \overline{B}$.
*Otherwise, $((1-\frac{1}{n}) \times \frac{1}{2},(1-\frac{1}{n+1}) \times \frac{1}{2}) $ is a neighborhood of $x \times y$ which does not intersect $B$, hence $x \times y \not\in \overline{B}$.
Case 3: The point $1 \times y$ such that $y \in (0,1]$ is not a limit point of $B$. Clearly $(1 \times 1, 1\times \frac{1}{2})$ is a neighborhood which does not intersect $B$, hence $1 \times y \not\in \overline{B}$.
$\textbf{UPDATE:}$ The point $1 \times y$ such that $y \in (0,\frac{1}{2}]$ is not a limit point. Clearly $1 \times 0, 1 \times 1)$ is a neighborhood of $1 \times y$ which does not intersect $B$.
I am not sure about my solution. Is it correct? Please check.
| Just note that $$[0,1]\times [0,1]\setminus \left(B \cup \{1 \times 0\}\right) = [0\times 0, 0\times \frac12\rangle \cup \bigcup_{n=1}^\infty \langle (1-\frac1n) \times \frac12, (1-\frac{1}{n+1}) \times \frac12 \rangle \cup \langle 1 \times 0, 1 \times 1]$$ which is open as a union of basic open sets. So your proposed closure is actually closed and we need to add $1 \times 0$ as a member of $B’$ anyway.
| {
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"url": "https://math.stackexchange.com/questions/4201083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorise $x^{17}-x$ Lets say I have an expression: $$x^{17}-x$$
I have to factorise it.
What I did:
Removed common $x$ outside the expression:$$x(x^{16}-1)$$
What I observed is $x^{16}-1$ can be written as $(x^4)^2-1^2$. So applying the identity of difference of squares $a^2-b^2=(a+b)(a-b)$, I got the answer as $$x(x^4+1)(x^4-1)$$
The last term is a perfect square too. So I again factorised it: $x(x^4+1)(x^2+1)(x^2-1)$
Then Again: $$x(x^4+1)(x^2+1)(x+1)(x-1)$$
But when I used wolfram - alpha to check, I got
$$x(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)$$
Where did I go wrong?
Also, can we write that
*
*$x^4-1$
*$x^2-1$
Are the roots of $x^{17}-x$?
Ideally, if wolfram - alpha gave the result as $(x^8+1)$, so $(x^8-1)$ is also a root if $x^{17}+x$.
| Your mistake is in the third equation. In particular you should have: $$x^{16} - 1=(x^8+1)(x^8 - 1)$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that the limit of $f(x,y)=\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ is $0$ as $(x,y)$ approaches $(0,0)$? I need to show that $\displaystyle\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ exists. I know that it is equal to zero. Until now, all I know how to do is to prove using the $\epsilon-\delta$ definition. I tried to prove it by the following way:
Let $\delta>0$ be a real number such that $0<\sqrt{x^2+y^2}<\delta$. We have $0< x^2+y^2<\delta^2$.
We know that $\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq\left\vert x^2\sin(y)\right\vert+\left\vert y^2\sin(x)\right\vert=x^2\left\vert\sin(y)\right\vert+y^2\left\vert\sin(x)\right\vert\leq x^2+y^2$.
Thus
$$0<\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq x^2+y^2<\delta^2$$
and dividing the inequality by $x^2+y^2>0$ we have
$$ 0<\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}<1<\dfrac{\delta^2}{x^2+y^2}. $$
I thought that this could be useful because $\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}=\left\vert\dfrac{x^2\sin(y)+y^2\sin(x)}{x^2+y^2}\right\vert$, but I don't know how to proceed.
| As suggested in the comments, by polar coordinates
$$\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}=\cos^2\theta \sin (r \sin \theta)+\sin^2\theta \sin (r \cos \theta) \to 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Show that the sum of the squares of 3,4,5 and 6 consecutive numbers can not be a square $\textbf{Edit:}$
Thank you so far for the answers. I still do not understand how to prove that the sum of 6 consecutive squares is not a square.
I've tried ${6d^2+30d+55 \ne n^2 \Rightarrow 6(d^2 + 5d +9)+1 \ne n^2}$ and would somehow like to elaborate this, but (yet again) I get stuck.
Is is correct to say that ${6(d^2 + 5d +9)+1}$ is on the form
${6k+1}$? Because then I could possibly stop there, having
${6k+1 \ne n^2}$
Thanks!
${\rule{8cm}{0.4pt}}$
\begin{equation}
\end{equation}
I've gotten this far:
$\textbf{Sum of 3 consecutive squares}$
Rewrite of problem:
\begin{equation}
\begin{aligned}
(d-1)^2+d^2+(d+1)^2 &\ne n^2\\
3d^2+2 &\ne n^2 \\
\label{p1_1}
\end{aligned}
\end{equation}
Assume ${n=d\cdot3+r}$, where ${r=0,1}$ or $2$.
Then,
\begin{equation}
n^2 \equiv_3 r^2 \equiv_3 0 \vee 1
\end{equation}
Conclusion:
Since ${r=2}$ in ${3d^2+2}$ and ${r=0 \vee 1}$ in ${n^2 \equiv_3 r^2}$ we have shown that
${3d^2+2 \ne n^2}$.
Question: Is my conclusion and reasoning valid and "good enough"?
$\textbf{Sum of 4 consecutive squares}$
Rewrite of problem:
\begin{equation}
\begin{aligned}
d^2+(d+1)^2 + (d+2)^2 + (d+3)^2 &\ne n^2\\
4(d^2+3d+3)+2&\ne n^2\\
\label{p1_b}
\end{aligned}
\end{equation}
${\Rightarrow} {r=2}$.
Assume ${n=d\cdot4+r}$, where ${r=0,1,2}$ or $3$.
Then,
\begin{equation}
n^2 \equiv_4 r^2 \equiv_4 0 \vee 1
\end{equation}
Conclusion:
Same reasoning as in conclusion of sum of 3 consecutive squares.
$\textbf{Sum of 5 consecutive squares}$
\begin{equation}
\begin{aligned}
(d-2)^2+(d-1)^2+d^2+(d+1)^2+(d+2)^2 &\ne n^2\\
5d^2+10 &\ne n^2\\
5(d^2+2) &\ne n^2\\
\label{p1_c}
\end{aligned}
\end{equation}
${\Rightarrow r=2 \vee 10}$ (is this correct?)
${n^2 \equiv_5 r^2 \equiv_5 0 \vee \pm 1}$
Conclusion:
Same reasoning as in conclusion of sum of 3 consecutive squares.
$\textbf{Sum of 6 consecutive squares}$
\begin{equation}
\begin{aligned}
d^2+(d+1)^2+(d+2)^2+(d+3)^2+(d+4)^2+(d+5)^2 &\ne n^2\\
6d^2+30d+55&\ne n^2\\
\label{p1_d}
\end{aligned}
\end{equation}
Here I'm stuck.
I feel pretty unsure about all my reasonings and would therefore appreciate your help. Please, feel free to be verbose in your answers.
Thank you!
| here's an expansion of my comment:
$$0+a\equiv a\pmod r$$ means that since $$a=b^2\equiv 0,1\pmod 4$$ we can ignore the $a\equiv 0\pmod 4$ because they don't change the remainders modulo 4.
By similar logic, since the square they sum to $n^2$ has the same remainder possibilities modulo 4, we get that the number of $a\equiv 1\pmod 4$ that exist, must be $$c\equiv 0,1\pmod 4$$ and since the number of $a\equiv 1\pmod 4$, is also $$\lfloor \frac{k}{2}\rfloor$$ or $$\lceil\frac{k}{2}\rceil$$ we know one of these must be congruent to $c$ .
If we take $k=4,5,6$ we have $2$ for the first form for the first 2 of those, and $3$ for the last, or we get $2,3,3$ and all these values aren't congruent to $c$, so they don't work in $k$. In fact adding 8 to $k$ results in the same remainder for the fractions so all $k\equiv 4,5,6\pmod 8$ are out. The only case spared is $k=3$ but with 1 odd value which was previously ruled out. So we are Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the system $ax + by = 1$, $cx + dy = 2$ From Serge Lang's book:
Let $a, $b, $c, $d be numbers such that $ ad - bc \neq 0 $. Solve the following systems of equations for $x$ and $y$ in terms of $a, $b, $c, $d.
$$
\begin{cases}
ax + by = 1 \\
cx + dy = 2
\end{cases}
$$
What I did so far step by step
\begin{cases}
ax + by = 1 \\
cx + dy = 2
\end{cases}
\begin{cases}
cax + cby = c \\
cax + ady = 2a
\end{cases}
\begin{cases}
cax = c - cby \\
cax + ady = 2a
\end{cases}
\begin{cases}
cax = c - cby \\
c - cby + ady = 2a
\end{cases}
\begin{cases}
cax = c - cby \\
y(ad - cb) = 2a - c
\end{cases}
\begin{cases}
cax = c - cby \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{c - cby}{ca} \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{c(1 - by)}{ca} \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{1 - by}{a} \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
\begin{cases}
x = \dfrac{1 - b * \dfrac{2a - c}{ad-cb}}{a} \\
y = \dfrac{2a - c}{ad-cb}
\end{cases}
Book's solution says that
$x=\dfrac{d -2b}{ad - bc}$
but after multiple tries I can't get my equation to a point where $x$ looks near like the desired answer. Could someone hint me where I did a mistake in my thought process or what I am doing wrong in general?
| What you have so far is fine. You just need to simplify $x$
$x = \dfrac{1 - b * \dfrac{2a - c}{ad-cb}}{a}\\
\dfrac{\left(1 - b * \dfrac{2a - c}{ad-cb}\right)(ad-cb)}{a(ad-cb)}\\
\dfrac{ad-cb - 2ab + cb}{a(ad-cb)}\\
\dfrac{a(d - 2b)}{a(ad-cb)}\\
\dfrac{d - 2b}{ad-cb}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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