Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Power series of $f(x)=\frac{x+1}{x^2+x-4}$? I do not understand how I can find the power series with center in x=0 of the function
$$f(x)=\frac{x+1}{x^2+x-4}$$
I tried without succes in the following way:
$$f(x)=\frac{x+1}{x^2+x-4}=\frac{1}{2} \bigg[\frac{2x+1}{x^2+x-4}+\frac{1}{x^2+x-4}\bigg]$$
Now, for $\frac{2x+1}{x^2+x-4}$ I know that
$$\int \frac{2x+1}{x^2+x-4} dx=\mathrm{lg} (x^2+x-4) \implies \frac{2x+1}{x^2+x-4}=\sum_{n \geq 0} \frac{\mathrm{d}}{\mathrm{dx}} ((-1)^n \frac{(x^2+x-4)^{n+1}}{n+1})$$
The last one using the power series of logarithm.
And for $\frac{1}{x^2+x-4}$ I can complete the square and write
$$\frac{1}{x^2+x-4}=\frac{1}{(x-1/2)^2+15/4}=(4/15) \frac{1}{(\frac{x-1/2}{225/16})^2+1}=(4/15) \frac{1}{1-(-(\frac{x-1/2}{225/16})^2)}=(4/15) \sum_{n \geq 0} (-(\frac{x-1/2}{225/16})^2)^{n}$$
But in both cases I don't get a power series with center in $x=0$ (I'm not even sure that these two are really power series), so I'm wrong.
What is the proper way to write the power series of $f(x)$ with center in $x=0$?
| Hint
Just start rewriting $$f(x)=\frac{x+1}{x^2+x-4}=-\frac{1+x}{4-x-x^2}=-\frac 14 \frac {1+x}{1-\frac{x+x^2}4}$$ Now consider the development (by Taylor series) of $\frac 1{1-y}$ and then replace $y$ by $\frac{x+x^2}4$.
When done, multiply by $-\frac{1+x}4$ and you should get something like $$f(x)=-\frac{1}{4}-\frac{5 x}{16}-\frac{9 x^2}{64}-\frac{29
x^3}{256}+O\left(x^4\right)$$ which could have been obtained by division.
Edit
Another way to approach the problem.
From the very first terms obtained by division, you can set $$f(x)=-\sum_{i=0}^\infty \frac {a_i }{4^{i+1}}x^i$$ and write the problem as $$x+1=(x^2+x-4)f(x)\implies (1+x)+(x^2+x-4)\sum_{i=0}^\infty \frac {a_i }{4^{i+1}}x^i=0$$ Looking at $x^0$ and $x^1$, we immediately get $a_0=1$ and $a_1=5$. Now, for any $n\geq 2$, we have, just as B. Goddard commented,$$a_n=a_{n-1}+4a_{n-2}$$ Using the standard method for recurrence relations, we have $$a_n=c_1 \left(\frac{1}{2}-\frac{\sqrt{17}}{2}\right)^n+c_2
\left(\frac{1}{2}+\frac{\sqrt{17}}{2}\right)^n$$ and applying the initial conditions $a_0=1$ and $a_1=5$ $$a_n=\frac{1}{34} \left(\left(17-9 \sqrt{17}\right) \left(\frac{1}{2}
\left(1-\sqrt{17}\right)\right)^n+\left(17+9 \sqrt{17}\right) \left(\frac{1}{2}
\left(1+\sqrt{17}\right)\right)^n\right)$$ which are integer numbers which correspond to sequence $A006131$ in $OEIS$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $a$ for which $f(x)\lt3\;\;\forall x\in\Bbb{R^+}$ and $f(x)\gt3$ for at least one $x\in\Bbb{R}^-$
Problem Statement:-
If $f(x)=\left|\dfrac{x^2+ax+1}{x^2+x+1}\right|$, then find the values of $a$ for which $\forall x\in\Bbb{R^+}, \;f(x)\lt3,\;$ and $f(x)\gt3$ for at least one negative real $x$.
Attempt at a solution:-
As $x^2+x+1\gt0$, $\forall x\in\Bbb{R}$ hence $$f(x)=\dfrac{\left|x^2+ax+1\right|}{x^2+x+1}$$
Now, consider the following cases:-
Case 1:- If $\qquad x^2+ax+1\gt 0$, then $$f(x)=\dfrac{x^2+ax+1}{x^2+x+1}$$
Let $g(x)=2x^2+(3-a)x+2$
For $x^2+ax+1\gt 0$ consider the following subcases:-
Sub Case-1:- $D(x^2+ax+1)\lt0$ which implies that the $\forall x\in\Bbb{R}.\;\;(x^2+ax+1\gt0)$.
So, from the conditions that the question wants to be satisfied, i.e. $\forall x\in\Bbb{R^+}.\;(f(x)\lt3)\;$ and $\exists x\lt0.\;(f(x)\gt3)$, we get
$$\forall x\gt0.\;(g(x)=2x^2+(3-a)x+2\gt0)\\
\exists x\lt0. \;(g(x)=2x^2+(3-a)x+2\lt0)$$
As the situation is same as that in the second sub case, lets save some work here we get
$$a\lt3\\
a\in(-\infty,-1)\cup(7,\infty)$$
So, combining all the inequalities in this sub case we get,
$$\boxed{a\in(-2,-1)}$$
Sub Case-2:- Discriminant of $x^2+ax+1$ is greater than $0$. So, $|a|\gt2$ and
$$x\in\left(-\infty,\dfrac{-a-\sqrt{a^2-4}}{2}\right)\cup\left(\dfrac{-a+\sqrt{a^2-4}}{2},\infty\right)$$
So, $$f(x)=\dfrac{x^2+ax+1}{x^2+x+1}$$
So, as the question asks for $\forall x\gt0.\; (f(x)\gt3)$, which on simplifying gives $$g(x)=2x^2+(3-a)x+2\gt0$$
We see that $g(0)\gt 0$, so either both roots are either positive or negative. As the question also demands for the condition $\exists x\lt0.\; (f(x)\lt3)$, so the both roots of $g(x)$ need to be negative.
So, $g^\prime(0)\gt 0\implies 3-a\gt0 \implies a\lt3$
Also, $D(g(x))\gt0\implies (a+1)(a-7)\gt0\implies a\in(-\infty,-1)\cup(7,\infty)$
And since $|a|\lt2$, the common interval for all the inequalities comes out to be $\boxed{a\in(-\infty, -2)}$
To conclude this case we get that $\boxed{a\in(-\infty,-2)\cup(-2,-1)}$
Case-2:-
If $x^2+ax+1\lt0$, then
$$f(x)=-\dfrac{x^2+ax+1}{x^2+x+1}$$
Also, $D(x^2+ax+1)\gt0\implies a^2-4\gt0\implies |a|\gt2\implies a\in(-\infty,-2)\cup(2,\infty)$
Also, the bounds of $x$ in which the given inequalities are to be examined is
$$x\in\left(\dfrac{-a-\sqrt{a^2-4}}{2},\dfrac{-a+\sqrt{a^2-4}}{2}\right)$$
Let $h(x)=4x^2+(3+a)x+4$ (assumed it here itself to reduce the mess while solving the inequality in this case).
So, from $\forall x\gt0.\;(f(x)\lt3)$ and $\exists x\lt0.\;(f(x)\gt3)$, we get
$$\forall x\gt0.\;(h(x)=4x^2+(3+a)x+4\gt0)\\
\exists x\lt0.\;(h(x)\lt0)$$
As, we can see that $h(0)\gt0$, so same as the last case we conclude that either both the roots are either negative or are either positive. But as $\forall x\gt0.\; (h(x)\gt0)$, so $h^\prime(0)\gt0\implies a\gt-3$.
Lastly as the roots need to be real, so $$D(h(x))\gt0\implies (a-5)(a+11)\gt0\implies a\in(-\infty, -11)\cup(5,\infty)$$
From all the inequalities in this case we get
$$\boxed{a\in(5,\infty)}$$
My deal with the question:-
I am having trouble examining the bounds in which $a$ should be so that interval in which $x$ can be taken also is included in the bounds of $a$. To be more clear how do I incorporate the bounds in which the $x$ has been defined for $x^2+ax+1\lessgtr0$ depending upon the case being considered.
| For $x\gt 0$,
$$\begin{align}f(x)\lt 3&\iff \left|\frac{x^2+ax+1}{x^2+x+1}\right|\lt 3\\\\&\iff \left|1+\frac{(a-1)x}{x^2+x+1}\right|\lt 3\\\\&\iff -3\lt 1+\frac{(a-1)x}{x^2+x+1}\lt 3\\\\&\iff -4(x^2+x+1)\lt (a-1)x\lt 2(x^2+x+1)\\\\&\iff \frac{-4(x^2+x+1)}{x}\lt a-1\lt \frac{2(x^2+x+1)}{x}\\\\&\iff \frac{-4x^2-3x-4}{x}\lt a\lt\frac{2x^2+3x+2}{x}\tag1\end{align}$$
Since we want to find $a$ such that $(1)$ holds for every $x\gt 0$, considering the graphs gives $a\in(-11,7)$.
For $x\lt 0$,
$$\begin{align}f(x)\gt 3&\iff \left|\frac{x^2+ax+1}{x^2+x+1}\right|\gt 3\\\\&\iff \left|1+\frac{(a-1)x}{x^2+x+1}\right|\gt 3\\\\&\iff 1+\frac{(a-1)x}{x^2+x+1}\lt -3\quad\text{or}\quad 1+\frac{(a-1)x}{x^2+x+1}\gt 3\\\\&\iff (a-1)x\lt -4(x^2+x+1)\quad\text{or}\quad (a-1)x\gt 2(x^2+x+1)\\\\&\iff a-1\gt\frac{-4(x^2+x+1)}{x}\quad\text{or}\quad a-1\lt\frac{2(x^2+x+1)}{x}\\\\&\iff a\gt\frac{-4x^2-3x-4}{x}\quad\text{or}\quad a\lt \frac{2x^2+3x+2}{x}\tag2\end{align}$$
Since we want to find $a$ such that there exists at least one $x\lt 0$ satisfying $(2)$, considering the graphs gives $a\in(-\infty,-1)\cup (5,\infty)$.
Therefore, the answer is
$$\color{red}{a\in(-11,-1)\cup (5,7)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has infinitely many solutions in integers $x,y,z$ While solving some old INMO problems I found that one, and I am completely stuck at it. The problem is:
Show that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has infinitely many solutions in integers $x,y,z$.
I shall be thankful if you can provide me any hints or suggestions. Thanks.
| In this type of problems when you have to prove that there are infinitely many solutions it's convenient to look for possible values in P.A. So, suppose that $z-y=y-x=k$, then $z-x=2k$ and our equation becomes $$x^2+(x+k)^2+(x+2k)^2=2k^3.$$
So, after a little calculus we get the equation $3x^2+(6k)x+(5k^2-2k^3)=0$. Using the formula for the quadratic equation gives us $$x=\frac{-6k\pm \sqrt{(6k)^2-12(5k^2-2k^3)}}{6}=-k\pm \frac{k\sqrt{6(k-1)}}{3}.$$
Now, since we want $x\in \mathbb{Z}$ we need that $6(k-1)=u^2$ for some $u\in \mathbb{Z}$. Then $6\mid u^2$, so $6\mid u$ (why?). Set $u=6t$, thus we get $k=6t^2+1$ and hence we get $x=-(6t^2+1)\pm (6t^2+1)(2t)$. If we take the plus sign we get $x=12t^3-6t^2+2t-1$. Now, since $y=x+k$ and $z=x+2k$, replacing $x$ gives us $y=12t^3+2t$ and $z=12t^3+6t^2+2t+1$.
Finally, you can check that $$(12t^3-6t^2+2t-1)^2+(12t^3+2t)^2+(12t^3+6t^2+2t+1)^2=2(6t^2+1)^3.$$
Hence, the equation has infintely many integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
On Reshetnikov's integral $\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}$ V. Reshetnikov gave the remarkable integral,
$$\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})\tag1$$
More generally, given some integer/rational $N$, we are to find an algebraic number $\alpha$ that solves,
$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}\tag2$$
and absolute value $|\alpha|$. (Compare to the similar integral in this post.) Equivalently, to find $\alpha$ such that,
$$\begin{aligned}
\frac{1}{N}
&=I\left(\alpha^2;\ \tfrac12,\tfrac13\right)\\[1.8mm]
&= \frac{B\left(\alpha^2;\ \tfrac12,\tfrac13\right)}{B\left(\tfrac12,\tfrac13\right)}\\
&=B\left(\alpha^2;\ \tfrac12,\tfrac13\right)\frac{\Gamma\left(\frac56\right)}{\sqrt{\pi}\,\Gamma\left(\frac13\right)}\end{aligned} \tag3$$
with beta function $\beta(a,b)$, incomplete beta $\beta(z;a,b)$ and regularized beta $I(z;a,b)$.
Solutions $\alpha$ for $N=2,3,4,5,7$ are known. Let,
$$\alpha=\frac{-3^{1/2}+v^{1/2}}{3^{-1/2}+v^{1/2}}\tag4$$
Then,
$$ - 3 + 6 v + v^2 = 0, \quad N = 2\\
- 3 + 27 v - 33v^2 + v^3 = 0, \quad N = 3\\
3^2 - 150 v^2 + 120 v^3 + 5 v^4 = 0, \quad N = 5\\
- 3^3 - 54 v + 1719 v^2 - 3492v^3 - 957 v^4 + 186 v^5 + v^6 = 0, \quad N = 7$$
and (added later),
$$3^4 - 648 v + 1836 v^2 + 1512 v^3 - 13770 v^4 + 12168 v^5 - 7476 v^6 + 408 v^7 + v^8 = 0,\quad N=4$$
using the largest positive root, respectively. The example was just $N=2$, while $N=4$ leads to,
$$I\left(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{3}{8},\quad\quad I\left(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{5}{8}$$
I found these using Mathematica's FindRoot command, and some hints from Reshetnikov's and other's works, but as much as I tried, I couldn't find prime $N=11$.
Q: Is it true one can find algebraic number $\alpha$ for all $N$? What is it for $N=11$?
|
I. Duplication
Following Nemo's lead in this answer, we find the formula,
$$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{3})=I(1+q^3;\tfrac{1}{2},\tfrac{1}{3})$$
where $p,q$ are related by the $12$-deg,
$$p^2(-2 + 2 q + q^2)^6 = 36(1 + q^3) (4 + 4 q + 6 q^2 - 2 q^3 + q^4)^2$$
This then enables us to find infinitely many $\displaystyle\frac{1}{2^n N}$.
For example, since $I(p^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{3}$ is known, then solving for $I(\alpha^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{6}$ turns out to involve a $36$-deg equation.
II. Triplication
(Courtesy of Nemo.) Starting with,
$$B\left(z;\frac{1}{2},\frac{1}{3}\right)=2 \sqrt{z} \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right).
$$
The transformation
$$
\, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};-\frac{3 z \left(1-\frac{z}{9}\right)^2}{(1-z)^2}\right)=\frac{(1-z) \, }{1-\frac{z}{9}}{}_2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right)
$$
applied two times gives
$$
\frac{1}{3} B\left({\frac{(9-z)^2 z \left(z^3+225 z^2-405 z+243\right)^2}{729 (1-z)^2 (z+3)^6}};\frac{1}{2},\frac{1}{3}\right)=B\left(z;\frac{1}{2},\frac{1}{3}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Can we generalize this formula We have these identities :
$$x^{2n+1} + y^{2n+1} = (x+y)(x^{2n} - x^{2n-1}y + \cdots -y^{2n-1}x + y^{2n})$$
and
$$x^{2n+1} - y^{2n+1} = (x-y)(x^{2n} + x^{2n-1}y + \cdots + y^{2n-1}x + y^{2n})$$
If $n=1$ or $n=2$ , It is obvious that second parenthesis is non-negative .
For example for $n=1$ :
$x^2 - xy + y^2 \ge 0 \iff 2x^2 -2xy + 2y^2 \ge 0 \iff (x-y)^2 +x^2 + y^2 \ge 0 \iff \begin{cases} (x-y)^2 \ge 0 & \\ x^2\ge0 \\ y^2 \ge 0 \end{cases} $
Can we prove that for all $n$ in these identities second parenthesis is non-negative ?
| The left side and the first term on the right will always have the same sign, so the second term on the right can't be negative.
Edit in response to comment asking for more details.
Since the function $x \mapsto x^{\text{odd}}$ is an increasing function the left side of the second equation has the same sign as the first term on the right.
For the first equation the assertion is obvious if $x$ and $y$ have the same sign. If they have opposite signs then the signs are the same as the sign of whichever of $x$ and $y$ is further from $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Diophantine equation in a polynomial ring
I have to calculate the inverse of the polynomial $(2x^2+x+1)$ in the quotient ring $\frac{Z_3[x]}{<x^3+2x+2>}$.
My idea was that I could solve the following equation, but I have no idea where to begin:$$(x^3+2x+2)u+(2x^2+x+1)v = 1.$$
| Hint:
The general method, for any base field, is to use the Extended Euclidean algorithm to obtain a Bézout's relation:
$$u(x)(2x^2+x+1)+v(x)(x^3+2x+2)=1.$$
The the inverse of $2x^2+x+1$ modulo $x^3+2x+2$ is simply $v(x)$.
Here is a layout. Note that $2x^2+x+1=-x^2+x+1$, so, to simplify the computations we'll apply the algorithm to $x^3+2x+2=x^3-x-1$ and $x^2-x-1$:
$$\begin{array}{rccl}
r(x)&u(x)&v(x)&q(x)\\
\hline
x^3-x-1&0&1\\
x^2-x-1&1&0& x+1\\
\hline
x&-x-1&1&x-1\\
-1&x^2&-x+1\\
\hline
\end{array}$$
Thus we have $\;-1=x^2(x^2-x-1)+(-x+1)(x^3-x-1)$, or
$$1=x^2(2x^2+x+1)-+(x-1)(x^3+2x+2),$$
and the inverse of $2x^2+x+1$ is $x^2 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$
Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$
In other words, simplify $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$ into $\log_2x$
I did:
$$\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12) \\
= \log_2(2x+x^2) + \log_{2^2}(36) - \log_2(6x+12) \\
= \log_2(2x+x^2) + \log_2(\sqrt{36}) - \log_2(6x+12) \\
= \log_2{(\frac{(2x+x^2)6}{6x+12})} \\
= \log_2(\frac{12x+6x^2}{6x+12}) \\
= \log_2(\frac{2+x}{12}) \\
= \log_2(\frac{1}{6}+\frac{x}{12}) \\
= \log_2(\frac{12+6x}{72}) \\
= \log_2(12+6x) - \log_2(72)
= ???$$
What do I do next?
| hint: You made a mistake on line #$6$ which should be $\log\left(\dfrac{6x(x+2)}{6(x+2)}\right)= $ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the differential equation $y' = \frac{xy-y^2\ln{y}}{x^2\ln{x}- xy} $ The question is to solve the differential equation given below
$$y' = \frac{xy-y^2\ln{y}}{x^2\ln{x}- xy} $$
First I tried simple substitution but that didn't work , then I tried to cross multiply and arrange things in a manner so that I could (maybe ) solve the question , but that didn't help either.
| \begin{align*}
\frac{dy}{dx} &= \frac{y(x-y\ln y)}{x(x\ln x-y)} \\
&= \frac{x^2y^2 \left( \dfrac{1}{xy}-\dfrac{\ln y}{x^2} \right)}
{x^2y^2 \left( \dfrac{\ln x}{y^2}-\dfrac{1}{xy} \right)} \\
0 &=
\left( \frac{1}{xy}-\frac{\ln y}{x^2} \right)dx+
\left( \frac{1}{xy}-\frac{\ln x}{y^2} \right)dy \\
\frac{\partial}{\partial y}
\left( \frac{1}{xy}-\frac{\ln y}{x^2} \right)
&= -\frac{1}{xy^2}-\frac{1}{x^2y} \tag{$\partial_y M$} \\
\frac{\partial}{\partial x}
\left( \frac{1}{xy}-\frac{\ln x}{y^2} \right)
&= -\frac{1}{x^2y}-\frac{1}{xy^2} \tag{$\partial_x N$} \\
0 &= \left( \frac{\ln y}{x}+\frac{\ln x}{y} \right)' \\
k &= \frac{\ln y}{x}+\frac{\ln x}{y} \\
kxy &= x\ln x+y\ln y \\
e^{kxy} &= x^xy^y
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
proof "$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$" I want to proof:
$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$
I have changed the given eqation $1 = cos^2(x)+sin^2(x)$
$\to$ $cos^2(x) = 1- sin^2(x)$
Then another given eqation:
$cos(2x) = cos^2(x)-sin^2(x)$ $\to$ $-sin^2(x)=cos(2x)-cos^2(x)$
After that I have put those equtions together:
$$cos^2(x)=1-sin^2(x)$$
$$cos^2(x)=1+cos(2x)-cos^2(x)$$
So I have the 1+ {...} structure. How can I go on ?
Or is that the wrong way to proof the equation?
I would really appreciate some hints.
| $$\cos(2y)=2\cos^2(y)-1$$
Let $2y=x$,
$$\cos(x )= 2\cos^2\left(\frac{x}{2}\right)-1$$
Hence
$$\frac{1+\cos(x)}{2}=\frac{1+2\cos^2(\frac{x}{2})-1}{2}=\cos^2\left(\frac{x}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_{\pi/4}^{\pi/2} \frac{3+5cot(x)}{5-3cot(x)}$ I'm asked to evaluate $$\int_{\pi/4}^{\pi/2} \frac{3+5\cot(x)}{5-3\cot(x)}$$.
If this were just an integral for $\displaystyle\frac{5\cot(x)}{-3\cot(x)}$, I would be able to solve it just fine, but I'm a bit stuck on this one. I can't use u-substitution (or if I can, I don't see how) because there's no relation between the numerator and denominator. I also can't split the fraction in two because those constants being added to the cotangent terms keep getting in the way.
Could someone please offer a hint as to how I should get started with this problem?
| Notice that
\begin{align} \int \frac{5 \cot(x) + 3 }{5-3 \cot(x) } ~\text{d}x &= \int \frac{5 \cot(x) + 3 }{5-3 \cot(x) } \cdot \frac{\tan(x) \frac{1}{\cos(x)^2}}{\tan(x) \frac{1}{\cos(x)^2}}~\text{d}x \\
&=\int \frac{(5+3 \tan(x)) \frac{1}{\cos(x)^2}}{(5\tan(x)-3)(\tan(x)^2+1)}~\text{d}x \\
&=\int \frac{5+3y}{(5y+3)(y^2+1)} ~\text{d}y
\end{align}
where we used that $\tan(x)^2+1=\frac{1}{\cos(x)^2}$ and now use partial fraction decomposition
$$\frac{5+3y}{(5y+3)(y^2+1)}=\frac{5}{5y-3}-\frac{y}{y^2+1}$$
to evaluate this integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve $x^6 - 6x + 5 > 0$? Is there any elementary way to solve (i.e. not involving derivatives etc.) $x^6 - 6x + 5 > 0$ ? I can divide it two times by $(x - 1)$ to get $x^6 - 6x + 5 = (x - 1)^2 (x^4 + 2 x^3 + 3 x^2 + 4 x + 5)$, but it is not very helpful, because there is still this $(x^4 + 2 x^3 + 3 x^2 + 4 x + 5)$ part and I don't know how to prove that this is greater than $0$ without using calculus. What I want to say is that $x^6 - 6x + 5$ looks very specific and I am wondering if there exist any simple, yet smart way to solve it.
| It's not possible to solve an inequality, but it can be reduced. The best way to do this is by factoring the left side of this inequality. By finding $x^6-6x+5 = (x-1)^2(x^4+2x^3+3^2+4x+5)$, you're most of the way there. You can first solve $x^6-6x+5 = 0$, in saying that $x = 1$ is a solution, which is correct. Now we have to find the root of $f(x) = x^4 + 2x^3 + 3x^2 + 4x + 5$. Unfortunately, there is not a particularly great way of doing this other than the Binary-Ansatz method if we are not using calculus. What this means is we just want to guess and check pretty much. Let us look at the graph of $x^4 + 2x^3 + 3x^2 + 4x + 5$ and find a range where the function crosses $f(x) = 0$. Looking at the graph we find that this point does not exist, and there is as such no $x$ such that $f(x) = 0$. We can then see that $1$ is the only root for this function. Moreover, we can say that as $f(x) > 0$ and $(x-1)^2 > 0$, the only time when this is not greater than $0$ is when $x = 1$ as $f(x) * (x-1)^2$ at $x = 1$ is $f(1) * 0 = 0$. So we can say that this reduces to $x \neq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
How do I prove this using mathematical induction? \begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}
Using the mathematical induction step I arrive at this :
\begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}
And I don't see any other way to continue except to divide
\begin{align}
n+1(n+3)
\end{align}
into
\begin{align}
n(n+2)+something
\end{align}
and substitute it with the beginning of the fraction.
But that doesn't get me anywhere.
| You still need to prove the base case
Suppose:
$\begin{align}
1*3+2*4+3*5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}$
show that:
$\begin{align}
1*3+2*4+3*5+...+(n+1)(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}$
$\begin{align}
1*3+2*4+3*5+...+n+1(n+3) = \frac{n(n+1)(2n+7)}{6} + (n+1)(n+3)
\end{align}$
By the inductive hypothesis.
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find $\prod_{n=1}^\infty\frac{(4n-3)(4n+1)}{(4n-1)^2}$ Is it true that $$\displaystyle\prod_{n=1}^\infty\frac{(4n-3)(4n+1)}{(4n-1)^2}=\frac{8\pi^2}{\left(\Gamma\left(\frac{1}{4}\right)\right)^4}?$$
How to prove it ? Thank in advances.
| Yes. using the the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ we can prove that $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d
$$ so $$\prod_{n\geq1}\frac{\left(4n-3\right)\left(4n+1\right)}{\left(4n-1\right)\left(4n-1\right)}=\prod_{n\geq0}\frac{\left(n+1/4\right)\left(n+5/4\right)}{\left(n+3/4\right)\left(n+3/4\right)}=\frac{\Gamma\left(\frac{3}{4}\right)^{2}}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}=\color{red}{\frac{8\pi^{2}}{\Gamma\left(\frac{1}{4}\right)^{4}}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the slope of a line that cuts an area in half A region $A$ in the first quadrant is bounded by $y=x^2$, $y=25$ and the $y$-axis. Find the value of $m$ with the property that the line $y = mx$ divides $A$ into two regions with the same area.
| First you can calculate the area of the region A, therefore note that 5^2=25.
Hence
$$
Area_A = \int_0^5 25- x^2 \; dx = \frac{250}3.
$$
Now consider the line $g(x)=5\cdot x$. This line goes through $(0/0)$ as well as $(5/25)$.
The area between $g$ the $y$-axis and $y=25$ is a triangle with area
$$
Area_{Triangle} = \frac{25 \cdot 5}2 =\frac{125}2> \frac{Area_A}2 =\frac{125}3.
$$
This means that the line we are looking for, $h(x)=m\cdot x$ has to satisfy m>5. Taking this into account, the are $B$ bounded by $h$ the $y$-axis and $y=25$ is again a triangle and therefore
$$
Area_B=\frac{25\cdot \frac{25}m}2 =\frac{Area_A}2= \frac{125}3.
$$
We see that $m = \frac{15}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Alternative way to show $\sqrt{2}+ \sqrt{3}$ is algebraic over $\mathbb{Q}$ of degree 4. I want to show that $\sqrt{2}+ \sqrt{3}$ is algebraic over $\mathbb{Q}$ of degree 4. I am aware of the technique of setting $x = \sqrt{2} + \sqrt{3}$ and working backwards to obtain the fourth degree polynomial. I was wondering if there is a different way of doing this, maybe working with field extensions? Particularly, I was thinking that maybe showing that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] =4$ and that $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$ would help me with this. Thanks for any suggestion!
| There's a method I like quite a bit. A spanning set (but not necessairly a basis, it may be too large) for $\mathbb Q(\sqrt{2},\sqrt{3})$ is undoubtedly $\lbrace 1,\sqrt{2},\sqrt{3},\sqrt{2}\sqrt{3}\rbrace$. This is just "every multiplicative combination" of the bases for $\mathbb Q(\sqrt{2})$ and $\mathbb Q(\sqrt{3})$.
What we now want to do is find a matrix that has $\sqrt{2}+\sqrt{3}$ as an eigenvalue.
The way we do this is by first writing:
$$(\sqrt{2}+\sqrt{3})\begin{pmatrix} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{pmatrix}$$
Now, multiply out each row. As an example, the top row is $1\times (\sqrt 2+\sqrt 3) = \sqrt 2+\sqrt 3$. The second row is $\sqrt 2(\sqrt 2+\sqrt 3) = 2+\sqrt 6$. The third row is $\sqrt 3 (\sqrt 2+\sqrt 3) = \sqrt 6+ 3$, and the fourth row is $\sqrt 6(\sqrt 2+\sqrt 3) = 2\sqrt 3+3\sqrt 2$.
We can write this as:
$$(\sqrt 2+\sqrt 3) = \begin{pmatrix} 1 \\ \sqrt 2 \\ \sqrt 3 \\ \sqrt 6\end{pmatrix} = \begin{pmatrix} \sqrt 2+\sqrt 3 \\ 2+\sqrt 6 \\ 3+\sqrt 6 \\ 2\sqrt 3+3\sqrt 2\end{pmatrix}$$
We can "factor" this last vector as the following:
$$\begin{pmatrix} 0 & 1 & 1 & 0 \\ 2 & 0 & 0 & 1 \\3 & 0 & 0 & 1 \\0 & 3 & 2 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ \sqrt 2\\ \sqrt 3\\ \sqrt 6\end{pmatrix}$$
This matrix has $\sqrt 2+\sqrt 3$ as an eigenvalue, so its characteristic polynomial has this as a root.
This will be a degree $4$ polynomial. It's not necessairly the minimal polynomial though, so you still have to check this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Easy way to solve $w^2 = -15 + 8i$
Solve $w^2=−15+8i$, where $w$ is complex.
Normally, I would convert this into polar coordinates, but the problem is that is too slow.
What is another alternative?
| Start out by letting $w = a + bi \implies w^2 = a^2-b^2 + 2abi = -15+8i \implies a^2-b^2 = -15, 2ab = 8 \implies a^2 - \dfrac{16}{a^2} = -15$. Put $x = a^2 \implies x - \dfrac{16}{x} + 15 = 0\implies x^2 + 15x - 16 = 0\implies (x-1)(x+16) = 0\implies x = 1\implies a = \pm 1, b = \pm 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Range of $x$ given $|x^2-a|Exercise:
Using signs of inequality alone (not using signs of absolute value) specify the values of $x$ which satisfy the following relations. Discuss all cases.
$$|x^2-a|<b$$
Solution:
$LHS \geq 0$, so:
$|x^2-a|<b \text{ when } b > 0 \tag{0}$
$-b<x^2-a<b \text{ when } b > 0 \tag{1}$
$a-b<x^2<a+b \text{ when } b > 0 \tag{2}$
$a-b<x^2 \text{ when } b > 0 \tag{2.1}$ $x^2<a+b \text{ when } b > 0 \tag{2.2}$
$0 \leq a-b<x^2 \text{ when } b > 0 \tag{2.1.1}$
$a-b < x^2 \text{ when } 0 < b \leq a \tag{2.1.2}$
$\sqrt{a-b} < |x| \text{ when } 0 <b \leq a \tag{2.1.3}$
$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } 0 < b \leq a \tag{2.1.4}$
$|x|<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.1}$
$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.2}$
Answer:
Formulation 1
$$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } b \leq a$$
$$\text{and}$$
$$-\sqrt{a+b}<x<\sqrt{a+b}$$
$$\text{all when } a > -b, b > 0$$
Formulation 2
$$-\sqrt{a+b}<x<-\sqrt{a-b}, \sqrt{a-b}<x<\sqrt{a+b} \text{, when } b\leq a$$
$$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } b > a$$
$$\text{all when } a > -b, b > 0$$
Request:
Is my answer correct? If not, where in my solution did I go wrong?
Update: I just noticed (graphically) that $b > 0$ for there to be a range of $x$. Where do I prove (algebraically ) that this is so?
Another Update: Ignore the previous update. Thanks to @JeanMarie comment, I realized that this isn't necessarily so.
Yet Another Update: Ignore the previous update, and pay attention to the first. Thanks to @dxiv comment, I realized that this is necessarily so.
An Update Once Again: I've updated my attempt with the input given my @dxiv and @JeanMarie. How is it now?
|
$$|x^2-a|<b$$
The LHS side is non-negative, so there are no solutions if $\,b \le 0 \;\;\;(1)\,$.
For $b \gt 0$ the inequality can be rewritten as:
$$
-b \lt x^2 - a \lt b \quad \iff
\quad \begin{cases}
x^2 \lt a+b \\
x^2 \gt a-b
\end{cases}
$$
*
*$(x^2 \lt a+b)\,$ Since $x^2 \ge 0$ there are no solutions if $a+b \le 0 \;\;\;(2)\,$.
Otherwise for $a+b \gt 0$ the inequality is equivalent to $-\sqrt{a+b} \lt x \lt \sqrt{a+b}\;\;\;(3)$.
*$(x^2 \gt a-b)$ If $a-b \lt 0$ then the inequality holds for $\forall x\;\;\;(4)\,$.
Otherwise for $a-b \ge 0$ the inequality is equivalent to $x \lt -\sqrt{a-b}\,$ or $\,x \gt \sqrt{a-b}\;\;\;(5)\,$.
To combine $(1)\cdots(5)$ into one final answer, note that $\sqrt{a+b} \ge \sqrt{a-b} \ge 0$ when $a \ge b \ge 0$.
*
*From $(1)+(2)\,$: if $b \le 0$ or $a \le -b$ then there are no solutions i.e. $x \in \emptyset$.
*Otherwise ($b \gt 0$ and $a \gt -b$) if $a \lt b$ from $(3)+(4)$: $x \in (-\sqrt{a+b}, \sqrt{a+b})$.
*Otherwise ($a \ge b \gt 0$) from $(3)+(5)$: $x \in (-\sqrt{a+b}, -\sqrt{a-b}) \cup (\sqrt{a-b}, \sqrt{a+b})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find volume of body bounded by $\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right)^2 + \frac{z^4}{c^4} = \frac{z}{k}$
Find the volume of body bounded by
$$\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right)^2 + \frac{z^4}{c^4} = \frac{z}{k}.$$
What substitution is better in this case?
| We change coordinates as follows:
$$
\xi = \frac{x}{a} \qquad \eta = \frac{y}{b} \qquad \zeta = \frac{k^{1/3}}{c^{4/3}} z
$$
In terms of these new coordinates, the bounding surface is given by
$$
\left(\xi^2 + \eta^2 \right)^2 + \frac{c^{4/3}}{k^{4/3}} \zeta^4 = \frac{c^{4/3}}{k^{4/3}} \zeta.
$$
Define $\kappa = c^{2/3}/k^{2/3}$, so we then have
$$
\xi^2 + \eta^2 = \kappa \sqrt{\zeta - \zeta^4}.
$$
This is now a solid of rotation in $(\xi, \eta, \zeta)$ space, with a radial coordinate $\rho^2 = \xi^2 + \eta^2$. The volume of the solid of rotation in these coordinates will be
$$
\tilde{V} = \int \pi \rho(z)^2 \, d\zeta = \pi \kappa \int_0^1 \sqrt{\zeta - \zeta^4} d\zeta.
$$
This last integral can be expressed in terms of the beta function by substituting $u = \zeta^3$:
$$
\int_0^1 \sqrt{\zeta - \zeta^4} d\zeta= \int_0^1 \sqrt{u^{1/3} - u^{4/3}} \frac{du}{3 u^{2/3}} = \frac{1}{3} \int_0^1 u^{-1/2}\sqrt{1 - u} \, du = \frac{1}{3} B\left( \frac{1}{2}, \frac{3}{2} \right) = \frac{\pi}{6}.
$$
Thus, the volume of the solid in $(\xi, \eta, \zeta)$-space is
$$
\tilde{V} = \frac{\pi^2}{6} \frac{c^{2/3}}{k^{2/3}}
$$
and since the volume element in $(x,y,z)$ space is related to that in $(\xi, \eta, \zeta)$ space by
$$
dV = dx \, dy \, dz = (a d\xi)(b d\eta) (\frac{c^{4/3}}{k^{1/3}} d\zeta) = \left(\frac{a b c^{4/3}}{k^{1/3}} \right) d\tilde{V},
$$ we conclude that
$$
\boxed{ V = \frac{\pi^2}{6} \frac{a b c^2}{k}.}
$$
I think. There may be an error or two with my powers in here, but I'm pretty confident in the method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum of the value $\frac{\gcd(n,[n\sqrt{5}])}{\sqrt{n}}$ For any postive integer $n$ find the maximum of the value
$$\dfrac{\gcd(n,\left \lfloor{n\sqrt{5}}\right \rfloor)}{\sqrt{n}}$$
Let $m=[n\sqrt{5}]$,then we have $n\sqrt{5}-1<m\le n\sqrt{5}$,it is said consider pell equation?
| Your inequality can be written as $$\sqrt5-\frac1n < \frac{m}n < \sqrt5 \implies \left| \sqrt5 - \frac{m}n \right| < \frac1n$$
so need good rational approximations to $\sqrt5$. i.e, we need solutions to $5n^2-m^2=k$ for "small" positive $k$. Now, what is "small"? Note $\lfloor n\sqrt5 \rfloor = m \implies m^2 + k < (m+1)^2 \implies k < 2m+1$.
Further note that if $g = \gcd(m, n)$, we may write $m = gx, n = gy, k= g^2z$ for some positive $z$. As we seek maximum of $g^2/n = g/y \approx \sqrt5 g/x$, we need to keep $x, y$ smallest while having largest possible $g$. So write the equation as
$$x^2- 5y^2 = - z$$
where $0< z < 2x/g +1/g^2 \implies gz \le 2x \implies g/x \le 2/z$. Clearly largest value then for $g/x$ is when $z=1$, so we try
$x^2-5y^2=-1$ first, which has the smallest solution $(2, 1)$. Now with $g = 4$, and we get the maximum when $m = gx = 8, n = gy = 4$ to give max $\dfrac{\gcd(4, \lfloor 4\sqrt5\rfloor)}{\sqrt4} = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
If $\log _b a\cdot\log_c a +\log _a b\cdot\log_c b+\log _a c\cdot\log_b c=3$ then find the value of $abc$
If $\log _b a\cdot\log_c a +\log _a b\cdot\log_c b+\log _a c\cdot\log_b c=3$ and $a,b,c$ are different positive real numbers not equal to 1, then find the value of $abc$.
I tried to simplify this by different methods like using the identity $\log_b a=\frac{1}{\log_a b}$, but I couldn't get anywhere. Addition, subraction can't be used. I am not able to figure out a way to simplify the LHS. It would be great if someone could help me to proceed with this problem.
| If you set $X=\log a$, $Y=\log b$ and $Z=\log c$ (any base you like), your condition becomes easily
$$
\frac{X^3+Y^3+Z^3}{XYZ}=3
$$
Now you can use the identity
$$
X^3+Y^3+Z^3=(X+Y+Z)^3-3(X+Y+Z)(XY+YZ+ZX)+3XYZ
$$
to get that
$$
(X+Y+Z)\bigl((X+Y+Z)^2-3(XY+YZ+ZX)\bigr)=0
$$
The second factor can be written
$$
X^2+Y^2+Z^2-XY-YZ-ZX
$$
and we should consider the quadratic form having as matrix
$$
A=\begin{bmatrix}
1 & -1/2 & -1/2 \\
-1/2 & 1 & -1/2 \\
-1/2 & -1/2 & 1
\end{bmatrix}
$$
Since
$$
\det[1]=1>0,
\qquad
\det\begin{bmatrix}1 & -1/2 \\ -1/2 & 1\end{bmatrix}=3/4>0
\qquad
\det A=0
$$
the quadratic form is positive semidefinite. Its null space contains the vector $[1\;1\;1]^T$, so we can conclude that your condition implies
$$
X+Y+Z=0\qquad\text{or}\qquad X=Y=Z
$$
Since by assumption $a$, $b$ and $c$ are pairwise distinct, we end with $X+Y+Z=0$.
Without quadratic forms, you can reason about $X^2+Y^2+Z^2-XY-YZ-ZX=0$ as follows. Since $Z\ne0$ by assumption, we can set $X=uZ$ and $Y=vZ$, so the equation becomes
$$
u^2-uv+v^2-u-v+1=0
$$
and, solving with respect to $v$, $v^2-v(u+1)+u^2-u+1=0$, the discriminant is
$$
(u+1)^2-4(u^2-u+1)=-3(u-1)^2
$$
so the equation has a solution only for $u=1$, which gives $v=1$. Therefore $X=Y=Z$.
Another “elementary” approach. Suppose $(X+Y+Z)^2=3(XY+YZ+ZX)$. Set $s=X+Y+Z$ and $p=XYZ$; then $X$, $Y$ and $Z$ are the roots of the equation
$$
t^3-st^2+\frac{s^2}{3}t-p=0
$$
by Viète's formulas. We can complete the cube getting
$$
\left(t-\frac{s}{3}\right)^3=p-\frac{s^3}{27}
$$
which should have three real roots. This is impossible unless the roots are coincident, so $p=s^3/27$ and $X=Y=Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Can $ x^2+\frac{x^2}{(x+1)^2}=3 $ be solved by 'completing square method'? On using the quartic formula, there are two zeroes of this equation
$$
x^2+\frac{x^2}{(x+1)^2}=3
$$
So, I was wondering if this equation could be solved like a quadratic equation.
Is it possible. If yes, how?
| Write $
x^2+\frac{x^2}{(x+1)^2}=3
$ as
$$x^2(x+1)^2+x^2=3(x+1)^2$$
or
$$x^2(x+1)^2+2x(x+1)+1= 4(x+1)^2$$
which allows the completion of square
$$[x(x+1)+1]^2=[2(x+1)]^2$$
Then, factorize with $a^2-b^2=(a+b)(a-b)$
$$(x^2-x-1)(x^2+3x+3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What's the mistake I'm making in evaluating the below limit? $\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$ Below is my solution for this limit. I get $0$, but the answer, is in fact, $\frac{1}{18}$. What's the mistake I'm making?
{From here: A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$}
$$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$
$$=\lim_{x \to 0}\frac{x\sin(\sin x)}{x^{6}}-\lim_{x \to 0}\frac{\sin^{2}x}{x^{6}}$$
$$=\lim_{x \to 0}\frac{x\sin(\sin x)}{x^{6}}\cdot\frac{\sin x}{\sin x}\cdot\frac{x}{x}-\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{x^4}$$
$$=\lim_{x \to 0}\frac{\sin(\sin x)}{\sin x}\cdot\frac{\sin x}{ x}\cdot\frac{x^2}{x^6}-\lim_{x \to 0}\frac{1}{x^4}$$
$$=\lim_{x \to 0}\frac{1}{x^4}-\lim_{x \to 0}\frac{1}{x^4}
=\lim_{x \to 0}\left(\frac{1}{x^4}-\frac{1}{x^4}\right)=0$$
| $$ \sin t = t - \frac{t^3}{6} + \frac{t^5}{120} \mp ... $$
$$ \sin \sin x = \sin x -\frac{\sin^3 x}{6} + \frac{\sin^5 x}{120} \mp $$
$$ \sin \sin x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} \mp ... \right) - \frac{\left( x - \frac{x^3}{6} \pm \right)^3}{6} + \frac{\left( x - \mp \right)^5}{120} \pm $$
$$ \sin \sin x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} \mp ... \right) - \frac{\left( x^3 - \frac{x^5}{2} \pm \right)}{6} + \frac{ x^5 - \mp }{120} \pm $$
$$ \sin \sin x = x - \frac{x^3}{3} + \frac{x^5}{10} \mp $$
$$ x \sin \sin x = x^2 - \frac{x^4}{3} + \frac{x^6}{10} \mp $$
$$ \sin^2 x = x^2 - \frac{x^4}{3} + \frac{2 x^6}{45} \mp $$
$$ x \sin \sin x - \sin^2 x = \frac{x^6}{10} - \frac{2 x^6}{45} \pm $$
$$ x \sin \sin x - \sin^2 x = \frac{9x^6}{90} - \frac{4 x^6}{90} \pm $$
$$ x \sin \sin x - \sin^2 x = \frac{5x^6}{90} \pm $$
$$ x \sin \sin x - \sin^2 x = \frac{x^6}{18} + O( x^7) $$
Meanwhile, the function depicted is even, all exponents in the final expression must be even, so we can say the slightly stronger
$$ x \sin \sin x - \sin^2 x = \frac{x^6}{18} + O( x^8) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+ab+b^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{1}{\sqrt{a^2+ab+b^2}}+\frac{1}{\sqrt{a^2+ac+c^2}}+\frac{1}{\sqrt{b^2+bc+c^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$$
I tried C-S, Holder and more, but without success.
The equality occurs here also for $(a,b,c)=(1,1,0)$.
| When one of $a, b, c$ is zero, clearly the inequality is true. In the following, assume that $a, b, c > 0$.
We apply Ji Chen's Symmetric Function Theorem for $n=3$:
(see https://artofproblemsolving.com/community/c6h194103p1065812)
Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying
$$x+y+z \ge u+v+w, \quad
xy+yz+zx \ge uv+vw+wu, \quad
xyz \ge uvw.$$
Then $x^d + y^d+z^d \ge u^d + v^d+w^d$.
Now let us prove the inequality. Let
\begin{align}
&X = \frac{1}{a^2+ab+b^2}, \ Y = \frac{1}{b^2+bc+c^2}, \ Z = \frac{1}{c^2+ca+a^2},\\
&U = V = \frac{1}{ab+bc+ca}, \ W = \frac{a+b+c}{3(a^3+b^3+c^3)}.
\end{align}
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge \sqrt{U} + \sqrt{V} + \sqrt{W}$.
Let
\begin{align}
f = X + Y + Z - (U+V+W), \ g = XY+YZ+ZX - (UV+VW+WU), \ h = XYZ - UVW.
\end{align}
We can prove that $f, g, h\ge 0$ using Buffalo Way.
To prove that $f \ge 0$, it suffices to prove that $f_1(a,b,c) \ge 0$ where $f_1(a,b,c)$ is a polynomial.
WLOG, assume that $a\ge b\ge c = 1$. Note that $f_1(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true.
Similarly, we may prove that $g\ge 0$ and $h\ge 0$.
According to Ji Chen's Symmetric Function Theorem, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $S_L= L_n^2L_{n+1} -L_1L_2L_3-L_2L_3L_4-\cdots L_{n-2}L_{n-1}L_n$, $n\ge3$ $2,1,3,4,7,11,...$ for $n=0,1,2,3,4,...$; It is the Lucas numbers
Let the sum of the cube of Lucas series be $S_L$
$n\ge3$
$S_L=2^3+1^3+3^3+4^3+\cdots+L_n^3$,
show that it has a form of $S_L=L_n^2L_{n+1}-L_1L_2L_3-L_2L_3L_4-\cdots-L_{n-2}L_{n-1}L_n$
I try: I can't think of any simple identities to use.
This is the only one might have some sort of link to it,
$L_{n+1}^3+L_{n+2}^3={L_{n+3}\over 2}(L_n^2+L_{n+1}^2+L_{n+2}^2)$
Any further hints?
| In the given form, we can easily prove it via induction. Verify the case $n = 3$ by hand, and for $n \geqslant 4$, use the recurrence to get
\begin{align}
L_n^2L_{n+1} &= L_n^2(L_n + L_{n-1}) \\
&= L_n^3 + L_nL_{n-1}L_n \\
&= L_n^3 + L_nL_{n-1}(L_{n-1} + L_{n-2}) \\
&= L_n^3 + L_{n-1}^2L_n + L_nL_{n-1}L_{n-2},
\end{align}
and thus with the induction hypothesis
\begin{align}
\sum_{k = 0}^n L_k^3 &= \Biggl(\sum_{k = 0}^{n-1} L_k^3\Biggr) + L_n^3 \\
&= \Biggl(L_{n-1}^2 L_n - \sum_{m = 1}^{n-3} L_mL_{m+1}L_{m+2}\Biggr) + L_n^3 \tag{I.H.} \\
&= \Biggl(L_{n-1}^2 L_n - \sum_{m = 1}^{n-3} L_mL_{m+1}L_{m+2}\Biggr) + \bigl(L_n^2L_{n+1} - L_{n-1}^2L_n - L_{n-2}L_{n-1}L_n\bigr) \tag{see above} \\
&= L_n^2L_{n+1} - \sum_{m = 1}^{n-2} L_mL_{m+1}L_{m+2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex Numbers Questions... I am having trouble with these problems:
*
*Find all complex numbers $z$ satisfying the equation
$$\frac{z+1}{z-1} = i.$$
*The value
$$\left(\frac{1+\sqrt 3}{2\sqrt 2}+\frac{\sqrt 3-1}{2\sqrt 2}i\right)^{72}$$
is a positive real number. What real number is it?
On 1) , I was thinking about substituting $z$ for $a+bi$ and then solving. Is this correct?
| For 1 just use simple arithmetic: it's a degree one equation.
For 2, compute first the square:
\begin{align}
\left(\frac{1+\sqrt 3}{2\sqrt 2}+\frac{\sqrt 3-1}{2\sqrt 2}i\right)^{2}
&=
\left(\frac{1+\sqrt 3}{2\sqrt 2}\right)^2-
\left(\frac{\sqrt 3-1}{2\sqrt 2}\right)^2+
2\frac{1+\sqrt 3}{2\sqrt 2}\frac{\sqrt 3-1}{2\sqrt 2}i
\\[6px]
&=\frac{1+3+2\sqrt{3}}{8}-\frac{3+1-2\sqrt{3}}{8}+
2\frac{3-1}{8}i
\\[6px]
&=
\frac{\sqrt{3}}{2}+\frac{1}{2}i
\\[6px]
&=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How do I solve this problem based on induction? Let $a_0 = 0, a_1 = 1$ and $ a_k = a_{k-1} + a_{k-2}$ for $n \ge 3$. Let $b_0 = 2, b_1 = 1$ and $b_n = a_{n+2} - a_{n-2}$. Prove by induction that $a_{2n} = a_nb_n$.
For $n=1$, $a_2 = 1$ and $a_1b_1=1$.
For $n=2$, $a_4 = 3$ and $a_2b_2=3$.
Let's assume it to be true for $n=k-1$ and $n=k$
Then $a_{2k-2} = a_{k-1}b_{k-1}$
and $a_{2k} = a_{k}b_{k}$
$=> a_{2k-2} + a_{2k-1} = a_{k}b_{k}$
$=> a_{k-1}b_{k-1} + a_{2k-1} = a_{k}b_{k}$
$=> a_{2k-1} = a_{k}b_{k}-a_{k-1}b_{k-1} $
$=> 2a_{2k} + a_{2k-1} = 3a_{k}b_{k}-a_{k-1}b_{k-1} $
$=> 2a_{2k} + a_{2k-1} = 3a_{k}b_{k}-a_{k-1}b_{k-1} $
$=> a_{2k} + a_{2k+1} = 3(a_{k+1}-a_{k-1})(a_{k+2}-a_{k-2})-a_{k-1}(a_{k+1}-a_{k-3}) $
$=> a_{2k+2} = 3(a_{k+1}-a_{k-1})(a_{k+3}-a_{k+1}-a_{k-1}+a_{k-3})-a_{k-1}a_{k+1}+a_{k-1}a_{k-3} $
$=> a_{2k+2} =2a_{k+1}a_{k+3} - 3(a_{k+1})^2 +3a_{k+1}a_{k-3} - 3a_{k-1}a_{k+3} + 3(a_{k-1})^2 - 2a_{k-1}a_{k-1} +a_{k+1}a_{k+3} - a_{k+1}a_{k-1}$
$=> a_{2k+2} =2a_{k+1}a_{k+3} - 3(a_{k+1})^2 +3a_{k+1}a_{k-3} - 3a_{k-1}a_{k+3} + 3(a_{k-1})^2 - 2a_{k-1}a_{k-1} +a_{k+1}(a_{k+3} - a_{k-1})$
$=> a_{2k+2} =2a_{k+1}a_{k+3} - 3(a_{k+1})^2 +3a_{k+1}a_{k-3} - 3a_{k-1}a_{k+3} + 3(a_{k-1})^2 - 2a_{k-1}a_{k-1} +a_{k+1}b_{k+1}$
I have to prove here that $$2a_{k+1}a_{k+3} - 3(a_{k+1})^2 +3a_{k+1}a_{k-3} - 3a_{k-1}a_{k+3} + 3(a_{k-1})^2 - 2a_{k-1}a_{k-1} = 0$$
I hope there is an easier method than this.
| Indeed there is a simple method. Denote $c_n=a_{2n} -a_nb_n$ and we want to prove that $c_n=0$ for all $n$. The first thing we want to prove is that $$c_{n+2}=3c_{n+1}-c_n,$$ for which no induction is needed. For example, we only need to show any of $a_{2n}$, $a_{n-2}a_n$ satisfies the recursion. For $a_{2n}$, we want to show $$a_{2n+4} = 3a_{2n+2} - a_{2n},$$ or even simpler $$a_{k+4}=3a_{k+2}-a_k$$ and we can just plug in the recursive relationship for $a_n$ and prove this easily.
Then next we need to show $c_0=c_1=0$ and then prove $c_n=0$ by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
The probability of hitting a bird with The probability of hitting a bird with a stone lying on a tree by A is $\frac {1}{4}$ and by B is $\frac {3}{5}$. Draw a probability tree diagram to show the probability of all possible ways of hitting the bird.
I couldn't draw the tree diagram. What is the sample space here..???
| P(A) = $\frac{1}{4}$
P(A' = Not hitting) = $\frac{3}{4}$
P(B) = $\frac{3}{5}$
P(B' = Not hitting) = $\frac{2}{5}$
There are total three cases -
*
*None of them hit -
P(A') * P(B') = $\frac{3}{4} \times \frac{2}{5} = \frac{6}{20}$
*Either of them hit -
P(A) * P(B') + P(A') * P(B) = $\frac{1}{4} \times \frac{2}{5} + \frac{3}{4} \times \frac{3}{5} = \frac{2}{20} + \frac{9}{20} = \frac{11}{20}$
*Both of them hit -
P(A) * P(B) = $\frac{1}{4} \times \frac{3}{5} = \frac{3}{20}$
Sample space = $ \left[ \frac{6}{20}, \frac{11}{20}, \frac{3}{20} \right]$
And you can draw Probability Tree Diagram with the help of this link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
minimum value of $f(t) = 10t^6-24t^5+15t^4+40t^2+108$ without derivative minimum value of $f(t) = 10t^6-24t^5+15t^4+40t^2+108$ without derivative
for $t\leq 0$ value of function $f(t)\geq 108$
i wan,t be able to proceed after that ,could some help me with this
| Note that
\begin{align}
f(t) &= 10t^6-24t^5+15t^4+40t^2+108
\\
&= 10t^4\left(t^2-\dfrac{12}{5}t+\dfrac{3}{2}\right)+40t^2+108
\\
&= 10t^4\left(t^2-\dfrac{12}{5}t+\dfrac{36}{25}+\dfrac{3}{50}\right)+40t^2+108
\\
&= 10t^4\left(\left(t-\dfrac{6}{5}\right)^2+\dfrac{3}{50}\right)+40t^2+108.
\end{align}
Now, can you show that $f(t) \ge 108$ for all real $t$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that the limit $\lim \limits_{n \to \infty}\left(1 + \frac{1}{\sqrt[3]{n^3 - n^2}}\right)^n$ doesn't exist or find it Given $\lim \limits_{n \to \infty}\left(1 + \frac{1}{\sqrt[3]{n^3 - n^2}}\right)^n$.
My attempt: $\lim \limits_{n \to \infty}\left(1 + \frac{1}{\sqrt[3]{n^3 - n^2}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n$. So as the $n \to \infty \Rightarrow \sqrt[3]{1 - 1/n} \to 1$.
$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.
However, I'm not sure that my proof is strict enough, especially the moment, where I go from $\sqrt[3]{1 - 1/n}$ to $1$.
|
$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n = \lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.
However, I'm not sure that my proof is strict enough, especially the moment, where I go from $\sqrt[3]{1 - 1/n}$ to $1$.
Indeed, this step is not correct, and you need to justify why this is fine (under other circumstances, very similar-looking, it may not give the right result).
As almost always there is a doubt or some non-trivial step to be made, in this situation I would suggest to rewrite the quantity in the exponential form (below, very detailed derivation):
$$
\left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)^n
= \exp\left(n \ln \left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right) \right)
$$
Now,
$$\begin{align}
n \ln \left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)
&= \frac{1}{\sqrt[3]{1 - 1/n}}\cdot n\sqrt[3]{1 - 1/n} \ln \left(1 + \frac{1}{n\sqrt[3]{1 - 1/n}}\right)\\
&= \frac{1}{\sqrt[3]{1 - 1/n}}\cdot \frac{\ln \left(1 + a_n\right)}{a_n}
\end{align}$$
setting $a_n\stackrel{\rm def}{=}\frac{1}{n\sqrt[3]{1 - 1/n}}\xrightarrow[n\to\infty]{}0$. Recalling that $\frac{\ln(1+u)}{u}\xrightarrow[u\to0]{}1$, we get that
$$
\frac{1}{\sqrt[3]{1 - 1/n}}\cdot \frac{\ln \left(1 + a_n\right)}{a_n}
\xrightarrow[n\to\infty]{}1\cdot 1=1
$$
and therefore
$$
\exp\left(\frac{1}{\sqrt[3]{1 - 1/n}}\cdot \frac{\ln \left(1 + a_n\right)}{a_n}\right)
\xrightarrow[n\to\infty]{} e^1 = e
$$
by continuity of the exponential.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Parameterization of the Intersection of Two Parallel Cones I need to parameterize the curve of intersection of two congruent, parallel cones, where $\xi$ is the distance between the axes of symmetry, $\zeta$ is the difference in $z$ position of the vertices, and $\alpha$ is the slope of the cones.
I wrote the equations for the two cones below. $\xi$ and $\zeta$ are both divided by two to keep the problem symmetric about the origin.
$$C_1:(x-\frac{\xi}{2})^2+y^2=(\frac{z-\frac{\zeta}{2}}{\alpha})^2$$
$$C_2:(x+\frac{\xi}{2})^2+y^2=(\frac{z+\frac{\zeta}{2}}{\alpha})^2$$
Normally, to find the intersection of any two equations, I would solve the two equations for the same variable or constant and set the equations equal to eachother. In the case of this problem, it is easy to solve both equations for $y^2$, yielding:
$$C_1:y^2=(x-\frac{\xi}{2})^2-(\frac{z-\frac{\zeta}{2}}{\alpha})^2$$
$$C_2:y^2=(x+\frac{\xi}{2})^2-(\frac{z+\frac{\zeta}{2}}{\alpha})^2$$
Then
$$(x-\frac{\xi}{2})^2-(\frac{z-\frac{\zeta}{2}}{\alpha})^2=(x+\frac{\xi}{2})^2-(\frac{z+\frac{\zeta}{2}}{\alpha})^2$$
I am hesitant to continue, because I just eliminated $y$ from the equation. Intuitively, I know that $y$ should persist throughout the problem, because the curve will occupy the $x$, $y$, and $z$ space.
Am I going about this problem incorrectly? Can someone steer me in the right direction?
| $C_1:(x+\frac{\xi}{2})^2+y^2=(\frac{z-\frac{\zeta}{2}}{\alpha})^2\\
C_2:(x+\frac{\xi}{2})^2+y^2=(\frac{z+\frac{\zeta}{2}}{\alpha})^2$
Note, if $|\frac {\zeta}{\alpha}| > \xi$ then we have the cones nested inside each other. Since they are double cones there will still be a curve of intersection. If they are nested we have an ellipse.
If they are not. We get something hyperbolic.
If the solution is the ellipse we want $x,y,z$ in terms of $\sin\theta,\cos\theta$
If the solution is the hyperbola we want $x,y,z$ in terms of $\sinh\theta,\cosh\theta$ or $\tan\theta, \sec\theta$
Multiply them out:
$x^2 + x\xi + \frac{\xi^2}{4}+y^2=\frac{z^2-z\zeta+\frac {\zeta^2}{4}}{\alpha^2}\\
x^2 - x\xi + \frac{\xi^2}{4}+y^2=\frac{z^2+z\zeta+\frac {\zeta^2}{4}}{\alpha^2}$
Subtract one from the other
$2x\xi = -\frac {2z\zeta}{\alpha^2}\\
x = -\frac {z\zeta}{\xi\alpha^2}$
Substitute into either of the original surfaces to find $y$ in terms of $z.$
$y^2 = $$(\frac {z + \frac {\zeta}{2}}{\alpha})^2 - (\frac {-z\zeta}{\xi\alpha^2}-\frac{\xi}{2})^2\\
\frac {z^2 +z\zeta + \frac {\zeta^2}{4}}{\alpha^2} - (\frac {z^2\zeta^2}{\xi^2\alpha^4} +\frac {z\zeta}{\alpha^2}+\frac {\xi^2}{4})$
$4\xi^2\alpha^4 y^2 = $$(\xi^2\alpha^2-\zeta^2) z^2+\alpha^2(\zeta^2 - \alpha^2\xi^2)\\
(\xi^2\alpha^2-\zeta^2) (z^2-\alpha^2)$
$y^2 = {((\frac {\xi}{2})^2- (\frac {\zeta}{2\alpha})^2)((\frac {z}{\alpha})^2-1)}$
if $(\frac {\xi}{2})^2- (\frac {\zeta}{2\alpha})^2 < 0$ we have the elipse.
$z = \alpha\cos \theta\\
x =-\frac {\zeta}{\xi\alpha} \cos\theta\\
y = \left(\sqrt {(\frac {\zeta}{2\alpha})^2-(\frac {\xi}{2})^2}\right)\sin \theta$
if $(\frac {\xi}{2})^2- (\frac {\zeta}{2\alpha})^2 > 0$ we have the hyperbola
$z = \alpha \sec \theta\\
x =-\frac {\zeta}{\xi\alpha} \sec\theta\\
y = \left(\sqrt {(\frac {\xi}{2})^2-(\frac {\zeta}{2\alpha})^2}\right)\tan \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
finding value of indefinite integration of $\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$ finding value of indefinite integration $\displaystyle \int\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$
$\displaystyle \int\frac{y^5(y^2-1)+y(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2+1-2)}{y^{10}+1}dy$
$\displaystyle =\int\frac{y^5+y}{y^8-y^6+y^4-y^2+1}-2\int\frac{y^5+y}{y^{10}+1}dy$
i wan,t be able to proceed after that, could some help me with this
| Hint: We can write our integral as, $$I = \int \frac{y^7-y^5+y^3-y}{y^{10}+1} dy = \int \frac{y(y^6-y^4 + y^2-1)}{y^{10}+1} dy = \int \frac{y(y^6-y^4+y^2-1)}{(y^2+1)(y^8-y^6+y^4-y^2+1)} dy$$ We can use partial fractions and get $$I = \frac{1}{5}\int \frac{4y^7-3y^5 +2y^3-y}{y^8-y^6+y^4-y^2+1} dy -\frac{4}{5} \int \frac{y}{y^2+1} dy =I_1-I_2$$ I hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Functional equation $(x + y)f(f(x)y) = x^2 f(f(x) + f(y))$ Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that
$$(x + y)f(f(x)y) = x^2 f(f(x) + f(y))
\mbox{, for all } x,y\in\mathbb{R}^+.$$
I tried out various substitutions such as $x=y$, $x=1$, $x+y=x^2$, and nothin' works.
Thanks in advance for your reply.
| $$\frac{f(f(x)y)}{x^2} = \frac{f(f(x) + f(y))}{x + y }\tag{1}$$
Using the symmetry of the right side of $(1)$:
$$\frac{f(f(x)y)}{x^2} = \frac{f(f(x) + f(y))}{x + y }= \frac{f(f(y) + f(x))}{y + x }=\frac{f(f(y)x)}{y^2} \implies\\
$$
$$
\frac{f(f(x)y)}{x^2} =\frac{f(f(y)x)}{y^2} \tag{2} $$
With $ x=y $ in $ (1) $ we have :
$$
\frac{f(xf(x))}{x} = \frac{f(2f(x))}{2} \tag{3}\\
$$
One solution to $ (2) $ is : $ f(x)=x^2 $. A solution to $ (3) $ is: $ f(x) = x $.
Lemma 1 : if $ f(x) $ exists it must be injective.
Let $ f(a)=f(b) $:
$$ | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find angles in a right triangle if $\tan\theta = \frac34$, where $\theta$ is the angle between medians Find angles in right triangle if it is known that $\tan \theta = {3\over4}$, where $\theta$ is the angle between catheti medians. I have tried drawing orthogonal projections and making new right triangles with that same angle but it did not seem to lead anywhere, as well as trying to find something that could be useful for sine theorem, but again got nothing good.
Have tried to make new right triangles from centroid, but did not get anything useful.
Any help would be much appreciated. As it was a problem on a competition, calculators were not allowed
|
The length of one median is $\sqrt{a^2 + 4b^2}$ and the other is $\sqrt{4a^2 + b^2}$
The intersection of the medians split each other with a ratio of $2:1$
Law of sines
$\frac {\sin x}{a} = \frac {\sin y}{\frac 13 \sqrt{a^2 + 4b^2}}$
$\sin x = \frac 35$
$\sin y = \frac {\sqrt{a^2 + 4b^2}}{5a}$
Aslo $\sin y = \frac b{\sqrt{4a^2 + b^2}}$
$\frac {\sqrt{a^2 + 4b^2}}{5a} = \frac b{\sqrt{4a^2 + b^2}}\\
5ab = \sqrt {4a^4 + 17a^2b^2 + 4b^4}\\
25a^2b^2 = 4a^4 + 17a^2b^2 + 4b^4\\
4a^4 - 8a^2b^2 + b^4) = 0\\
4(a^2 - b^2)^2 = 0$
$a = b$
you have an isoceles right triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If two roots of the equation $(a-1)(x^2+x+1)^2-(a+1)(x^4+x^2+1)=0$ are real and distinct, then find the interval in which $a$ lies
If two roots of the equation $(a-1)(x^2+x+1)^2-(a+1)(x^4+x^2+1)=0$ are real and distinct, then find the interval in which $a$ lies.
My Approach:
I have expanded the equation to obtain a quartic equation which I am not able to factorize:
$$ x^4+(1-a)x^3+(2-a)x^2-ax+1=0 $$
If only I could factorize it I would get two quadratic equations, one of which should have real roots. I know how to proceed further in order to find the interval in which a lies.
But as for now I don't know how to proceed further. It would be great if I could get a hint to move forward.
| The term $x^4 + x^2 + 1$ easily gets factorised as follows:
$$
x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)
$$
Applying this in the original equation,
$$
(x^2 + x + 1)[(a - 1)(x^2 + x + 1) + (a + 1)(x^2 - x + 1)] = 0.
$$
Here $x^2 + x + 1 = 0$ is the quadratic with the complex roots. Focusing on the other quadratic, it can be simplified to:
$$
ax^2 - x + a = 0
$$
For this to have real and distinct roots, $1 - 4a^2 > 0 \implies a^2 < 1/4$ which simply gives us
$$
a \in \left (- \frac{1}{2}, \frac{1}{2}\right ).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that: $ m_a + m_b + m_c \le 4R + r $
Let $m_a, m_b, m_c - $ are median of triangle $ABC$.
Prove that:
$$ m_a + m_b + m_c \le 4R + r $$
I used
$$m_a^2=\frac{2b^2+2c^2-a^2}{4}$$
$$R=\frac{abc}{4R}$$
$$r=\frac{2S}{a+b+c}$$
and $R \ge 2r$
| We need to prove that $\sum\limits_{cyc}\left(m_a^2+2m_am_b\right)\leq(4R+r)^2$ and
since by Ptolemy $2m_am_b\leq c^2+\frac{1}{2}ab$, it remains to prove that
$$\sum\limits_{cyc}\left(\frac{1}{4}(2b^2+2c^2-a^2)+c^2+\frac{1}{2}ab\right)\leq\left(\frac{abc}{S}+\frac{2S}{a+b+c}\right)^2$$ or
$$\sum\limits_{cyc}(7a^2+2ab)\leq4\left(\frac{8abc(a+b+c)+16S^2}{8S(a+b+c)}\right)^2$$ or
$$\sum\limits_{cyc}(7a^2+2ab)\leq4\left(\frac{8abc+\prod\limits_{cyc}(a+b-c)}{8S}\right)^2$$ or
$$\sum\limits_{cyc}(7a^2+2ab)\leq4\left(\frac{\sum\limits_{cyc}(-a^3+a^2b+a^2c+2abc)}{8S}\right)^2$$ or
$$\sum\limits_{cyc}(7a^2+2ab)\leq\frac{\left(\sum\limits_{cyc}(-a^3+abc+a^2b+a^2c+abc)\right)^2}{\sum\limits_{cyc}(2a^2b^2-a^4)}$$ or
$$\sum\limits_{cyc}(7a^2+2ab)\leq\frac{\left(\sum\limits_{cyc}(-a^2+ab+ab)\right)^2(a+b+c)}{\prod\limits_{cyc}(a+b-c)}$$ or
$$\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0,$$
which is Schur and it's true even for all positives $a$, $b$ and $c$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
minimum value of $\displaystyle f(x) = \frac{(1+x)^{0.8}}{1+x^{0.8}}$ without derivative minimum value of $\displaystyle f(x) = \frac{(1+x)^{0.8}}{1+x^{0.8}},x\geq 0$ without derivative
Binomial expansion of $\displaystyle (1+x)^{0.8} = 1+0.8 x-\frac{(0.8\cdot (0.8-1)x^2)}{2}+\cdots $
but from above does not get anything
could some help me with this
| Raise your function to the power of five:
$$(f(x))^5=\left(\frac{(1+x)^{0.8}}{1+x^{0.8}}\right)^5$$
$$(f(x))^5=\frac{(1+x)^4}{1+5x^{0.8}+10x^{1.6}+10x^{2.4}+5x^{3.2}+x^4}$$
$$(f(x))^5=\frac{1+4x+6x^2+4x^3+x^4}{1+5x^{0.8}+10x^{1.6}+10x^{2.4}+5x^{3.2}+x^4}$$
$$(f(x))^5=\frac{x^{-2}+4x^{-1}+6+4x+x^2}{x^{-2}+5x^{-1.2}+10x^{-0.4}+10x^{0.4}+5x^{1.2}+x^2}$$
As $(f(x))^5=(f(x^{-1}))^5$ then you will have a minimum (or maximum) when $x=x^{-1}$, i.e. when $x=1$ (ignoring negative values as its unclear which root you would want for $x<0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Similar triangles: the shortest side of triangle $B$ measures $6$. Find the other two sides. Two triangles are similar. Triangle $A$ has sides measuring $3$, $4$, and $5$. The shortest side on triangle $B$ measures $6$. What are the lengths of the other two sides of triangle $B$?
| Let larger side of triangle A is XY = 5, smaller side YZ = 3 and other side XZ = 4.
Now sides of triangle B let larger side PQ = a, smaller QR = 6 and other side PR = b.
Now two triangle are similar so their sides are in proportion.
$\frac{PQ}{XY} = \frac{PR}{XZ} = \frac{QR}{YZ} $
Then you can find other two sides of B.
$\frac{a}{5} = \frac{b}{4} = \frac{6}{3} $
So we have,
$\frac{a}{5} = \frac{6}{3} $
a = 10
Also,
$\frac{b}{4} = \frac{6}{3} $
b = 8
Or you can do like this also,
Smaller side of triangle B is twice than smaller side of triangle A. And similar triangle are in proportion. So you can twice remaining two sides of triangle A to get remaining sides of triangle B.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $(x-2)^6+(x-4)^6=64$ using a substitution. Solve $(x-2)^6+(x-4)^6=64$ using a substitution.
I tired using $t=x-2$.
But again I have to know the expansion of the power $6$.
Can it be transformed to a quadratic equation ?
I know that it can be somehow solved by expanding the powers. But I'm trying to get a good transformation by a sub.
| Let $x-2=a$ and $4-x=b$.
Hence, $a+b=2$ and $a^6+b^6=64$ or
$$(a^2+b^2)(a^4+b^4-a^2b^2)=64$$ or
$$(2-ab)(a^2b^2-16ab+16)=32$$ or
$$ab(a^2b^2-18ab+48)=0.$$
$ab=0$ gives $x=2$ or $x=4$.
$ab=9-\sqrt{33}$ does not give a real roots and $ab=9+\sqrt{33}$ does not give a real roots.
Id est, the answer is $\{2,4\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would want to prove directly that $$f(-x) = -f(x)$$
In other words, I want to solve $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
and to come at the end to this:
$$-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$$
This was my approach:
$$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
$$\ln\left(-x+\sqrt{x^2 + 4}\right) - \ln2$$
$$\ln\left(\frac{-x+\sqrt{x^2 + 4}}{2}\right)$$
$$\ln\left(\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)^{-1}\right)$$
$$-\ln\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)$$
Here I got stuck. I want to get to $-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$ but if I use $\ln\left(\frac ab\right) = \ln a - \ln b$ then I will get back to $f(-x)$ and not to $-f(x)$.
Any help?
| $$\left(x+\sqrt{x^2 + 4}\right)\left(-x+\sqrt{x^2 + 4}\right)=4=2\cdot2$$ so that, taking the logarithm,
$$f(x)+f(-x)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{-2}+u^{-1}) \right|_1^{1+\sqrt{5}}={1\over 2}$$
Prove $(1)$ using an alternative method other than substitution method.
| Using integration by parts : $$\begin{align}\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx &= \int_0^1 \left( - \frac{1}{2\sqrt{5} (1+\sqrt{5}x)^2} \right)'(3-\sqrt{5}x) \, dx \\[8pt] &= \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2}\right]_0^1 - \frac{1}{2} \int_0^1 \frac{dx}{(1+\sqrt{5}x)^2} \\[8pt] &= \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2} + \frac{1}{10x + 2\sqrt{5}}\right]_0^1 \\[8pt] &= \frac{1}{2} \end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve the inequality $\sin(x)\cdot|\tan{x}|\le\frac{3}{2}$ As in the title, solve the inequality $\sin(x)\cdot|\tan{x}|\le\frac{3}{2}$ for $x\in[0;2\pi]$. My concern is that I don't knwo how to get rid of the absolute value sign - should I consider separately cases of the $\tan{x}$ being positive and negative?
Any hints greatly appreciated.
| Let $x\in\left[0,\frac{\pi}{2}\right)\cup\left[\pi,\frac{3\pi}{2}\right)$.
Thus, we need to solve $$\frac{\sin^2x}{\cos{x}}\leq\frac{3}{2}$$ or
$$\frac{(2+\cos x)(2\cos x-1)}{\cos x}\geq0,$$
which is $\cos x\geq\frac{1}{2}$ or $\cos x<0$, which gives $\left[0,\frac{\pi}{3}\right]\cup\left[\pi,\frac{3\pi}{2}\right)$.
Let $x\in\left(\frac{\pi}{2},\pi\right]\cup\left(\frac{3\pi}{2},2\pi\right]$.
Thus, we need to solve $$-\frac{\sin^2x}{\cos{x}}\leq\frac{3}{2}$$ or
$$\frac{(2-\cos x)(2\cos x+1)}{\cos x}\geq0,$$
which is $\cos x\leq-\frac{1}{2}$ or $\cos x>0$, which gives $\left[\frac{2\pi}{3},\pi\right]\cup\left(\frac{3\pi}{2},2\pi\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Divisibility involving the product (ad+bc)(ac+bd) Prove that if $abcd|(ad+bc)(ac+bd)$, then $\frac{ac}{bd}$ is a perfect square of a rational number. $a,b,c,d$ are positive integers.
I am extremely lost on how to do this problem; any help?
| Denote $(ad+bc)(ac+bd)=kabcd.$
Without loss of generality let $(a,b)=1,(c,d)=1,a/b=s,c/d=t,$
then $(s+t)(1+\dfrac{1}{st})=k.$
$$s=\frac{\pm\sqrt{\left(-k t+t^2+1\right)^2-4 t^2}+k t-t^2-1}{2 t}.$$
So $$\left(-k t+t^2+1\right)^2-4 t^2=D^2,\\
(\frac{-k t+t^2+1}{2t})^2-(\frac{D}{2t})^2=1.$$
There exist $m>n\in N,(m,n)=1$, such that
$$\frac{-k t+t^2+1}{2t}=\pm \frac{m^2+n^2}{2mn},\\
\frac{D}{2t}=\frac{m^2-n^2}{2mn}.\\
\frac{-k t+t^2+1}{2t}=\frac{c^2-kcd+d^2}{2cd}=\pm\frac{m^2+n^2}{2mn}$$
Now $(c^2-kcd+d^2,cd)=1,(m^2+n^2,mn)=1,$ hence
$$c^2-kcd+d^2=\pm(m^2+n^2),cd=mn.\\
-k t+t^2+1=\pm (m^2+n^2)/d^2,\\
t=mn/d^2, D=(m^2-n^2)/d^2.$$
$$\frac{ac}{bd}=st=\frac{1}{2}(\pm D-(-k t+t^2+1))=\frac{1}{2d^2}(\pm (m^2-n^2)-\pm (m^2+n^2))\\
=(m/d)^2 \quad or \quad (n/d)^2.\quad (st>0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that the sequence given by its general term $\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}$ is strictly decreasing. Prove that the sequence given by its general term:
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}$$
is strictly decreasing.
$a_n>a_{n+1} \Leftrightarrow \frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}>\frac{\sqrt{(n+1)^2+1}-(n+1)}{\sqrt{(n+1)+3}+2(n+1)}$
Is there some kind of trick to apply in these sorts of problems or do I have to solve this by sheer algebraic manipulation? If I subtract $\frac{\sqrt{(n+1)^2+1}-(n+1)}{\sqrt{(n+1)+3}+2(n+1)}$ from both sides I'll end up with something really ugly so I'm wondering if there's a better way.
| Hint:
$$a_n=\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}=\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{(\sqrt{n+3}+2n)(\sqrt{n^2+1}+n)}=\frac1{(\sqrt{n+3}+2n)(\sqrt{n^2+1}+n)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show $\sum \frac{xy}{xy+x+y} \le \frac{6+x^2+y^2+z^2}{9}$ I've spent some time on this problem but I can't find a method to solve it:
Let $x, y, z>0$, prove that:
$$\frac{xy}{xy+x+y} + \frac{yz}{yz+y+z} + \frac{zx}{zx+z+x} \le \frac{6+x^2+y^2+z^2}{9}$$
That is how I thought :
first of all, if we have $a, b, x>0$ and $a\le b$, then $$\frac{a}{a+x}\le\frac{b}{b+x}$$ (which can be simply proved by making the product on diagonal)
So I try to find a $b$ such as $b\ge xy$.
From means inequality (geometric mean and root mean square), we have $$\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}}$$
or $$xy \le \frac{x^2+y^2}{2}$$
So $$\frac{xy}{xy+x+y} \le \frac{\frac{x^2+y^2}{2}}{\frac{x^2+y^2}{2}+x+y}$$
Now, I need to find something which is $\le x+y$ and that's where I've got stuck.
Some hints would be really apreciated! Thanks!
| Here is an other elementary one.
Since $\displaystyle{\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}}$, we have that:
$\displaystyle{\frac{xy}{xy+x+y}=\frac{1}{1+\frac{1}{x}+\frac{1}{y}}\leq \frac{1}{1+\frac{4}{x+y}}=1-\frac{4}{4+x+y}}$
Similarly we get the other two inequalities. Thus it suffices to prove that:
$\displaystyle{\frac{6+x^2+y^2+z^2}{9}\geq 1-\frac{4}{4+x+y}+1-\frac{4}{4+y+z}+1-\frac{4}{4+z+x}\Leftrightarrow \\ 5+\frac{x^2+y^2}{2}+5+\frac{y^2+z^2}{2}+5+\frac{z^2+x^2}{2}+\frac{36}{x+y+4}+\frac{36}{y+z+4}+\frac{36}{z+x+4}\geq 36}$
But
$\displaystyle{\frac{x^2+y^2}{2}+1=\frac{x^2+1}{2}+\frac{y^2+1}{2}\geq x+y}$ and so:
$\displaystyle{5+\frac{x^2+y^2}{2}+\frac{36}{x+y+4}\geq (5+\frac{x^2+y^2}{2})+\frac{36}{(5+\frac{x^2+y^2}{2})}\geq2\sqrt{(5+\frac{x^2+y^2}{2})\frac{36}{(5+\frac{x^2+y^2}{2})}}=12}$
Simlarly we get the other two inequalities with $y,z$ and $z,x$, we add them and we are done.
Note: No need of AM-GM for $6$ terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For $a+b+c=2$ prove that $2a^ab^bc^c\geq a^2b+b^2c+c^2a$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove that:
$$2a^ab^bc^c\geq a^2b+b^2c+c^2a$$
I tried convexity, but without success:
We need to prove that
$$\ln2+\sum_{cyc}a\ln{a}\geq\ln\sum_{cyc}a^2b$$ and since $f(x)=x\ln{x}$ is a convex function, by Jensen we obtain:
$$\ln2+\sum_{cyc}a\ln{a}\geq\ln2+3\cdot\frac{2}{3}\ln\frac{2}{3}=\ln\frac{8}{9}.$$
Thus, we need to prove that
$$\frac{8}{9}\geq\sum_{cyc}a^2b$$ or
$$(a+b+c)^3\geq9(a^2b+b^2c+c^2a),$$ which is wrong for $c\rightarrow0^+$.
The equality occurs for $a=b=c=\frac{2}{3}$.
| Remark: My first proof does not work for the stronger inequality below.
Let us prove a stronger inequality:
$$2a^a b^b c^c \ge \frac{4}{27}(a + b + c)^3 - abc.$$
(Note: It is well-known that $a^2b + b^2c + c^2a \le \frac{4}{27}(a + b + c)^3 - abc$.)
Letting $x = \frac32 a, y = \frac32 b, z = \frac32 c$, it suffices to prove that, for all $x, y, z > 0$ with $x + y + z = 3$,
$$3 (x^x y^y z^z)^{2/3} \ge \frac{4}{27}(x+y+z)^3 - xyz. $$
(Note: We have $a^a b^b c^c
= (2/3)^{2(x+y+z)/3}(x^xy^yz^x)^{2/3} = \frac49 (x^xy^yz^x)^{2/3} $.)
Using $\mathrm{e}^u \ge 1 + u$, we have
$$(x^x y^y z^z)^{2/3}
= \mathrm{e}^{\frac23(x\ln x + y\ln y + z\ln z)}
\ge 1 + \frac23(x\ln x + y\ln y + z\ln z).$$
It suffices to prove that
$$3 + 2x\ln x + 2y\ln y + 2z\ln z \ge \frac{4}{27}(x+y+z)^3 - xyz.$$
Fact 1: It holds that
$u\ln u \ge u - 1 + \frac12(u-1)^2 - \frac16(u-1)^3
= -\frac13 - \frac12 u + u^2 - \frac16 u^3$
for all $u > 0$.
(The proof is given at the end. Note: The RHS is $3$-th order Taylor approximation of $u\ln u$ around $u = 1$.)
Using Fact 1, it suffices to prove that
\begin{align*}
&3 + \left(-\frac23 - x + 2x^2 - \frac13 x^3\right)
+ \left(-\frac23 - y + 2y^2 - \frac13 y^3\right)\\
&\quad + \left(-\frac23 - z + 2z^2 - \frac13 z^3\right)\\
&\ge \frac{4}{27}(x+y+z)^3 - xyz
\end{align*}
or (using $x + y + z = 3$)
$$3 - xy - yz - zx \ge 0$$
which is true.
We are done.
Proof of Fact 1:
It suffices to prove that
$$\ln u \ge \frac{1}{u}\left(u - 1 + \frac12(u-1)^2 - \frac16(u-1)^3\right).$$
Let $F(u) := \mathrm{LHS} - \mathrm{RHS}$. We have
$$F'(u) = \frac{(u-1)^3}{3u^2}.$$
We have $F'(u) < 0$ on $(0, 1)$, and $F'(u) > 0$ on $(0, \infty)$, and $F'(1) = 0$.
Thus, $F(u) \ge F(1) = 0$ for all $u > 0$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Integration by parts of : $\ln(x^{2} e^{x})$ I need help with this integral,
$$\int \ln(x^2 e^x)\, dx $$
What I have:
I know ,that I can not integer $\ln$ so I rewrite it like this:
$$\int 1 \ln(x^2 e^x)\, dx = \left| \begin{array}{cc} u=\ln(x^2 e^x) & v'=\frac{1}{x^2 e^x} \\ u'=1 & v=x \end{array} \right| = x\ln(x^2 e^x)- \int \frac{1}{x^2 e^x}x = ?$$
Now I am not sure what to do.
The result is:
$$2 (x\ln x -x)+\frac{1}{2}x^2 + C $$
Thank you very much.
| $\int \ln{(x^2e^x)}$ $dx$
$=x \ln {(x^2e^x)}-\int x[\frac{2xe^x+x^2e^x}{x^2e^x}]$ $dx$
$=x\ln {(x^2e^x)}- \int (2+x)$ $dx$
$=x\ln {(x^2e^x)}-2x-\frac{x^2}{2}$
$=x\ln{x^2}+x \ln{e^x}- 2x- \frac{x^2}{2}$
$=2x\ln{x}+x^2-2x-\frac{x^2}{2}+C$
$=2(x\ln{x}-x)+ \frac{x^2}{2} +C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$.
Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.
*Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k}
&=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\
&= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\
\end{align}$$
Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed
$$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\
&\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\
&\small= o\left(\frac{2^n}n \right)
\end{align}$$
Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$
This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).
| $\begin{array}\\
\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}
&=n\sum_{k=1}^n \frac{2^{k-n}}{k}\\
&=n\sum_{k=0}^{n-1} \frac{2^{-k}}{n-k}\\
&=\sum_{k=0}^{n-1} \frac{2^{-k}}{1-k/n}\\
\text{so}\\
\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}-2
&=\sum_{k=0}^{n-1} 2^{-k}(\frac1{1-k/n}-1)-\frac1{2^{n-1}}\\
&=\sum_{k=0}^{n-1} 2^{-k}(\frac{k/n}{1-k/n})-\frac1{2^{n-1}}\\
\end{array}
$
If $k \le cn$,
$2^{-k}(\frac{k/n}{1-k/n})
\le 2^{-k}\frac{c}{1-c}
$
so
$\sum_{k=0}^{\lfloor nc \rfloor} 2^{-k}(\frac{k/n}{1-k/n})
\le 2\frac{c}{1-c}
\lt 4c$
if
$0 < c < \frac12$.
If
$k > cn$,
$\begin{array}\\
\sum_{k=cn}^{n-1} 2^{-k}(\frac{k/n}{1-k/n})
&<\sum_{k=cn}^{n-1} 2^{-cn}(\frac{k/n}{1-k/n})\\
&=2^{-cn}\sum_{k=cn}^{n-1} (\frac{k}{n-k})\\
&<2^{-cn}\sum_{k=cn}^{n-1} n\\
&<n^22^{-cn}\\
&<e^{-cn \ln 2 + 2\ln n}\\
\end{array}
$
For any fixed $c > 0$,
$-cn \ln 2 + 2\ln n
\to -\infty$
as
$n \to \infty$,
so that
$e^{-cn \ln 2 + 2\ln n}
\to 0$.
By first choosing $c$ small
and then $n$ large,
both
$4c$ and
$e^{-cn \ln 2 + 2\ln n}$
can be made as small as we want,
so that
$\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}-2
$
can be made as small as we want,
so that
$\lim_{n \to \infty}\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}
=2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Subtracting two square roots How should one subtract the following square roots? I am aware of the fact that you can only add or subtract two square roots if their 'radical part' is the same.
$\displaystyle y^* = \sqrt{\frac{2KD}{h}} \cdot \sqrt{\frac{s+h}{s}}$
$\displaystyle y_s^* = \sqrt{\frac{2KD}{h}} \cdot \sqrt{\frac{s}{s+h}}$
$\displaystyle y^* - y_s^* = \ ???$
The answer should be $\displaystyle \sqrt{\frac{2KD}{s}} \cdot \sqrt{\frac{h}{s+h}}$
| Let us start by writing
$$A = \frac{2KD}{h}$$
You want to simplify
$$\sqrt{A}\sqrt{\frac{s+h}{s}} - \sqrt{A}\sqrt{\frac{s}{s+h}} = \\
\sqrt{A}\left(\sqrt{\frac{s+h}{s}} - \sqrt{\frac{s}{s+h}} \right)
$$
Now let us rewrite $\sqrt{\frac{s}{s+h}}$:
$$\sqrt{\frac{s}{s+h}} = \sqrt{\frac{1}{\frac{s+h}{s}}} = \frac{1}{\sqrt{\frac{s+h}{s}}}$$
Put that back and reduce to the same denominator:
$$\sqrt{A}\left(\sqrt{\frac{s+h}{s}} - \sqrt{\frac{s}{s+h}} \right) = \\
\sqrt{A}\left(\sqrt{\frac{s+h}{s}} - \frac{1}{\sqrt{\frac{s+h}{s}}}\right) = \\
\sqrt{A}\left(\frac{\frac{s+h}{s}}{\sqrt{\frac{s+h}{s}}} - \frac{\frac{s}{s}}{\sqrt{\frac{s+h}{s}}}\right) = \\
\sqrt{A}\left( \frac{\frac{h}{s}}{\sqrt{\frac{s+h}{s}}} \right)
$$
Can you take it from here? You should put the $\frac{h}{s}$ inside the square root, write $A$ in its original form and manipulate one $h$ and one $s$ from the right factor to the $A$ factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find an non recursive function from recursive function $f(n) = f(n-1) + 3^n$ We have
$f(0) = 1 , f(1) = 4 , f(2) = 13 , f(n) = f(n-1) + 3^n$
(number of all of triangles in Sierpinski triangle)
I want to find a non recursive function.
I know the answer but I want a solution for it.
| You wrote
$$f(n) = f(n-1) + 3^n\tag{1}$$
We note that $f(0) = 3^0 = 1$ and we start rewriting equation $(1)$ by substituting some $f(x)$:
$$f(n) = f(n-1) + 3^n = f(n-2) + 3^{n-1} + 3^n = f(n-3) + 3^{n-2} + 3^{n-1} + 3^n = 3^0 + 3^1 + \cdots + 3^{n-1} + 3^n$$
which is a geometric series of ratio $3$; therefore,
$$f(n) = \frac{3^{n+1} - 1}{2}$$
(assuming you are familiar with geometric series)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is wrong in my method of solving the volume of a solid of uniform cross-section A solid of a uniform cross-section of right-angled isosceles triangles perpendicular to the $x$-axis has a base $x^2 + y^2 = 16$. Solve for the volume.
Let $\theta$ be the angle between the $x$-axis and the radii of a circle for $0 \leq \theta \leq \frac{\pi}{2}$. The hypotenuse of the right-angled triangles is $8\sin \theta$. As the relation between a hypotenuse and area of a right-angled isosceles triangle is $\frac{\text{Hyp}^2}{4}$, the area of a single triangle is $16\sin^2 \theta$. As the solid is symmetric, we integrate over the domain and multiply the answer by 2.
\begin{align*}
&\quad\quad 2\int_0^\frac{\pi}{2}16\sin^2 \theta \quad d\theta\\
&= 32\int_0^\frac{\pi}{2}\frac{1}{2}-\frac{1}{2}\cos 2\theta \quad d\theta\\
&=32\left[\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta \right]_0^\frac{\pi}{2}\\
&= 8\pi
\end{align*}
However, this answer is incorrect as when one uses the more conventional method to solve:
Let the hypotenuse of one of the triangles be $2y$. Then the area is $y^2$.
\begin{align*}
Volume &= \int_{-4}^4y^2dx\\
&= \int_{-4}^416-x^2dx\\
&= \left[16x-\frac{1}{3}x^3\right]_{-4}^4\\
&= \frac{256}{3}
\end{align*}
which appears to be the correct answer. I'm curious as to why my alternative method doesn't work.
| Using a cartesian coordinate system, we have (as you pointed out in your question):
\begin{equation}
V = \int_{a}^{b} A(x) dx = \int_{-4}^{4} \frac{(\sqrt2y)(\sqrt2y)}{2} dx = \int_{-4}^{4} y^2 dx = \int_{-4}^{4} 16-x^2 dx = \frac{256}{3}
\end{equation}
Now if you want to use polar coordinates (where $ \theta $ is the angle between the $ x $ axis and the radius of the circle), you should proceed as follows:
\begin{equation}
V = \int_{a}^{b} A(x) dx = \int_{-4}^{4} y^2 dx
\end{equation}
Since $ y = 4sin(\theta) $ and $ x = 4cos(\theta) \rightarrow dx = -4sin(\theta)d\theta $, we get:
\begin{equation}
\int_{cos^{-1}\big(\frac{-4}{4}\big)}^{cos^{-1}\big(\frac{4}{4}\big)} (4sin(\theta))^2 (-4sin(\theta))d\theta = -4^3 \int_{\pi}^{0} sin^3(\theta) d\theta = 4^3 \int_{0}^{\pi} sin^3(\theta) d\theta = \frac{256}{3}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I calculate this indefinite integral: $\int\sqrt\frac{1}{x^3-1}~dx$ Please help me to find this indefinite integration
$$\int\sqrt\frac{1}{x^3-1}~dx$$
I guess, this is an indefinite integral. Do you have a tip for me to solve this ? Thank you in advance!
| Case $1$: $|x|\leq1$
Then $\int\sqrt{\dfrac{1}{x^3-1}}~dx$
$=\int\dfrac{1}{(x^3-1)^\frac{1}{2}}~dx$
$=\int\dfrac{1}{i(1-x^3)^\frac{1}{2}}~dx$
$=-i\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n}}{4^n(n!)^2}~dx$
$=-\sum\limits_{n=0}^\infty\dfrac{i(2n)!x^{3n+1}}{4^n(n!)^2(3n+1)}+C$
Case $2$: $|x|\geq1$
Then $\int\sqrt{\dfrac{1}{x^3-1}}~dx$
$=\int\dfrac{1}{(x^3-1)^\frac{1}{2}}~dx$
$=\int\dfrac{1}{x^\frac{3}{2}(1-x^{-3})^\frac{1}{2}}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{-3n-\frac{3}{2}}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{-3n-\frac{1}{2}}}{4^n(n!)^2\left(-3n-\dfrac{1}{2}\right)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{(2n)!}{2^{2n-1}(n!)^2(6n+1)x^{3n+\frac{1}{2}}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$
If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$
How do I solve for $\alpha$ and $\beta$ when only one equation is given? By simplification I came up with something like
$$
\sin\frac{\alpha}{2} \sin\frac{\beta}{2} \cos \frac {\alpha +\beta}{2}=\frac{1}{8}.
$$
I don't know if this helps. How to do this?
| HINT:
Use Prosthaphaeresis Formula on $\cos\alpha,\cos\beta$
and Double angle formula on $$\cos 2\cdot\dfrac{\alpha+\beta}2$$ to get
$$2\cos^2\dfrac{\alpha+\beta}2-2\cos\dfrac{\alpha-\beta}2\cos\dfrac{\alpha+\beta}2+\dfrac12=0\ \ \ \ (1)$$ which is a Quadratic Equation in $\cos\dfrac{\alpha+\beta}2$ whose discriminant must be $\not<0$
i.e., $$\left(2\cos\dfrac{\alpha-\beta}2\right)^2-4=-2\sin^2\dfrac{\alpha-\beta}2\ge0$$
But for real $\dfrac{\alpha-\beta}2,$ $$\sin^2\dfrac{\alpha-\beta}2\ge0$$
So, we have $$\sin^2\dfrac{\alpha-\beta}2=0\iff\sin\dfrac{\alpha-\beta}2=0$$
$\implies\dfrac{\alpha-\beta}2=n\pi$ where $n$ is any integer
But as $0<\alpha,\beta<\pi.n=0\implies\alpha-\beta=0$
So, $(1)$ is reduced to $$0=2\cos^2\beta-2\cos\beta+\dfrac12=\dfrac{(2\cos\beta-1)^2}2$$
So, $\cos\beta=\dfrac12=\cos\dfrac\pi6\implies\beta=2m\pi\pm\dfrac\pi6$ where $m$ is any integer
But as $0<\alpha<\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\frac{11x+3y}{3x-1}+\frac{5x+2}{3}=9-\frac{3-y}{x-1}=12$ (multiple equality signs) How to answer an equation with multiple equality signs?
This is the equation:
$$\frac{11x+3y}{3x-1}+\frac{5x+2}{3}=9-\frac{3-y}{x-1}=12$$
I believe it can also be written in this format:
\begin{align}
\frac{11x+3y}{3x-1}+\frac{5x+2}{3}&=9-\frac{3-y}{x-1} \\ \\
\frac{11x+3y}{3x-1}+\frac{5x+2}{3}&= 12\\
9-\frac{3-y}{x-1} &= 12
\end{align}
How can this be solved?
| Hint:
$$9-12=\frac{3-y}{x-1}$$
$$-3(x-1)=3-y$$
$$y=3x$$
Substitute this into
$$\frac{11x+3y}{3x-1}+\frac{5x+2}{3}=12$$
Solve for $x$, this is just a quadratic problem.
After you solve for $x$, compute $y$ by using $y=3x$.
Edit:
From substitution, we have
$$\frac{20x}{3x-1}+\frac{5x+2}{3}=12$$
Multiply $3$ throughout:
$$\frac{60x}{3x-1}+5x+2=36$$
Multiply $(3x-1)$ throughout, we obtain:
$$60x+(5x+2)(3x-1)=36(3x-1)$$
which is a quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $6<3^\sqrt3$ implies $3^\sqrt3 < 7$
Consider the double inequality : $$6<3^\sqrt3 < 7$$
Using $only$ the elementary properties of exponents and inequalities (NO calculator, computer,
table of logarithms, or estimate of √3 may be used), prove that the first inequality implies the
second.
SOURCE : "Inequalities proposed in Crux Mathematicorum" (Page Number 14; Question Number 627)
I have no idea how to solve this problem. I tried taking logarithms, but that did not help. I strongly suspect that some inequality has to be used. I tried AM-GM on a few set of terms, but that just made it messier.
Wolfram Aplha gives the value of ${3^{\sqrt {3}}}$ as $\approx 6.7049918538258$.
Can anyone give me a hint on how to solve this problem ?
| $$6=(6^3)^{\frac{1}{3}}=216^{\frac{1}{3}}<243^{\frac{1}{3}}=(3^5)^{\frac{1}{3}}=3^{\frac{5}{3}}<3^{\sqrt{3}},$$ becouse $$\left(\frac{5}{3}\right)^2=\frac{25}{9}<3=\left(\sqrt{3}\right)^2.$$
Since $$\left(\sqrt{3}\right)^2=3=\frac{48}{16}<\frac{49}{16},$$ then $$\sqrt{3}<\frac{7}{4}.$$
Therefore $$3^{\sqrt{3}}<3^{\frac{7}{4}}=(3^7)^\frac{1}{4}=2187^{\frac{1}{4}}<2401^{\frac{1}{4}}=(7^4)^{\frac{1}{4}}=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^\infty \sqrt{x} \left( \tan^{-1} \left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right)dx$ How to compute the integral
\begin{align}
\int_{0}^{\infty}\sqrt{x}\left[\tan^{-1}\left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right]\mathrm{d}x
\end{align}
for $c,a>0$.
If the it the integration is to difficult can we give a good upper bound?
| With the short hand $b=\sqrt{a^2+c^2}$ and the substitution $x=t^2$, the integral is
\begin{align}
I=& \ 2\int_0^\infty t^2\left( \tan^{-1} \frac{t^2+a}{c} - \tan^{-1} \frac{t^2-a}{c} \right)dt\\
\overset{ibp}=& \ \frac{4c}3 \int_0^\infty
\frac{2at^2-b^2}{t^4-2at^2+b^2}+ \frac{2at^2+b^2}{t^4+2at^2+b^2}\ dt\\
=& \ \frac{4c}3 \bigg(\frac{(2a-b)\pi}{2\sqrt{2(b-a)}}+ \frac{(2a+b)\pi}{2\sqrt{2(b+a)}} \bigg)\\
= &\ \frac{2\pi}3\left(2a\sqrt{b+c}-b \sqrt{b-c} \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a_n=1+n+n^2$ how do you compute the $gcd(a_n+1,a_{n+s})$
Question: If $n$ is a positive number I write $a(n) =1+n+n^2$. Let $s$
be a fixed positive number. Is it true the $\gcd(a_n+1, a_{n+s})$ will be a
divisor of $1+5s^2+s^4$?
For example if $n=1$ and $s=3$ then $\gcd(4,21)=1$ which is a divisor of $1+5*3^2+3^4=127$, since $1 |127$. On the other hand if $n=106$ then $\gcd(10922,11557)=127$ and $127 |127$.
| $g = \gcd(a_n + 1, a_{n+s}) \\= \gcd(n^2 + n + 2, (n+s)^2 + (n+s) + 1) \\= \gcd(n^2 + n + 2, s^2 + 2ns + s - 1)$
So if we have $u \cdot (n^2 + n + 2) + v \cdot (s^2 + 2ns + s - 1) = g$, we can apply the Extended Euclidean Algorithm for polynomials to get:
$(\frac{4s^2}{s^4+5s^2+1}) \cdot (n^2 + n + 2) + (\frac{s^2-2ns-s-1}{s^4+5s^2+1}) \cdot (s^2 + 2ns + s - 1) = 1$
Multiply to flatten the expression:
$(4s^2) \cdot (n^2 + n + 2) + (s^2-2ns-s-1) \cdot (s^2 + 2ns + s - 1) = s^4+5s^2+1$
This means $\gcd(a_n + 1, a_{n+s})$ must divide $s^4+5s^2+1$, so the answer to your question is yes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum of $\frac{1+3a^2}{(a^2+1)^2}$ What is the maximum value of
$\displaystyle{{1 + 3a^{2} \over \left(a^{2} + 1\right)^{2}}}$, given that $a$ is a real number, and for what values of $a$ does it occur ?.
| Writing it as $\cfrac{1+3a^2}{(a^2+1)^2}= \cfrac{3(a^2+1)-2}{(a^2+1)^2}= \cfrac{3}{a^2+1}-\cfrac{2}{(a^2+1)^2}\,$ gives a quadratic in $x=\cfrac{1}{a^2+1}\,$, with a maximum at $x = \cfrac{3}{4}\,$ ($\iff a^2 = 1/3\,$) of value $\cfrac{3 \cdot 3}{4}- \cfrac{2 \cdot 9}{16} = \cfrac{9}{8}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 +1 -\frac{(a^2+b^2)^2}{(2ab)^2} = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 = \frac{(a^2+b^2)^2}{(2ab)^2}-1 \\
z= \frac{a^2 +b^2}{2ab} \pm \sqrt{\frac{(a^2+b^2)^2}{(2ab)^2}-1} \ , \\
$$
which doesn't nearly suggest the simple end result.
I'm sure it will roll out after reorganizing terms, but I'm interested in an alternative approach which more naturally suggests the end result of $(z-a/b)(z-b/a)$, reflecting the word-use of straightforward in the text.
Context.
Using contour integration to evaluate $I = \int_{0}^{2\pi} \frac{\cos(2\theta)}{a^2+b^2-2ab\cos(\theta)}d\theta$
| Indeed, one can see that
$$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$$
Going from the RHS to the LHS is called fraction decomposition.
Your result should be clearer now:
$$z^2 - z\left(\frac{a^2 + b^2}{ab}\right) + 1 = z^2 - z\frac{a}{b} - z\frac{b}{a} + 1 = \left(z - \frac{a}{b}\right)\left(z - \frac{b}{a}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$ How should I determine the following limit?
$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$
| $$\cos\sqrt{x}-\cos\sqrt{x-1}=2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{\sqrt{x-1}-\sqrt{x}}{2}=$$
$$=-2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{1}{2(\sqrt{x-1}+\sqrt{x})}\rightarrow0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find constant that satisfies joint probability mass function for $X$ and $Y$
$f(x,y) = c(1/4)^x(1/3)^y$, $x = 1,.... y= 1,....$
The solution is $6$, but I am having trouble coming to this, and I think my issue has to do with $x$ and $y$ being infinitely countable. I would be grateful for any explanations.
| $$\sum_{x=1}^\infty \sum_{y=1}^\infty c\left(\frac{1}{4}\right)^x\left(\frac{1}{3}\right)^y
= c\sum_{x=1}^\infty \left(\frac{1}{4}\right)^x \times \sum_{y=1}^\infty \left(\frac{1}{3}\right)^y = c\left(\frac{1}{4-1}\right)\left(\frac{1}{3-1}\right) = c\left(\frac{1}{3}\right)\left(\frac{1}{2}\right) = c\left(\frac{1}{6}\right)= 1.$$ And, $$ c = 6 $$
Thanks, my issue was getting formula for geometric series: $$ \sum_{x=1}^\infty \left(\frac{1}{x^n}\right)= \left(\frac{1}{x-1}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$(ax+b)(bx+a)=10x^2+cx+10$ where $a$ and $b$ are positive integers. Find the two possible values of $c$ I know this seems simple but i am really struggling please help me thank you
| $$(ax+b)(bx+a)=10^2+cx+10 \\or\\ (ax+b)(bx+a)=x^2+cx+10 \\or\\(ax+b)(bx+a)=10x^2+cx+10$$I think 3rd is right
$$so\\abx^2+(a^2+b^2)x+ab=10x^2+cx+10\\ab=10\\
ab=10 \to \begin{cases}a=1 & b = 10 \to c=a^2+b^2=1+100=101\\a=10 & b = 1 \to c=a^2+b^2=1+100=101 \\
a=2 & b=5 \to c=a^2+b^2=4+25=29\\
a=5 & b=2 \to c=a^2+b^2=4+25=29\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using the spectral method to solve $Ax=b$ I cannot find where I have made a computational error in applying the spectral method for solving $Ax=b$.
I am given $A = \begin{pmatrix} 7 &-2 & 1 \\ -2 & 10 & -2 \\ 1 & -2 & 7 \end{pmatrix}, b = \begin{pmatrix} 6 \\ 12 \\ 18 \end{pmatrix}$.
I've started by computing the eigenvalues and eigenvectors, I have the eigenvalues $r_1 = 12, r_2 = 6, r_3 = 6$ with eigenvectors $v_1 = (1,-2,1), v_2 = (-1,0,1), v_3 = (2,1,0)$ respectively.
Now, $x$, by the spectral method is given by
$$\displaystyle x= \frac{u_1\cdot b}{r_1}u_1 + \frac{u_2\cdot b}{r_2}u_2+\frac{u_3\cdot b}{r_3}u_3.$$
Normalizing $v_1,v_2,v_3$ I get $u_1=\begin{pmatrix} \frac{1}{\sqrt{6}} \\ \sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{6}} \end{pmatrix} ,u_2 = \begin{pmatrix}\frac{-1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\end{pmatrix}, u_3 = \begin{pmatrix} \frac{2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \\ 0 \end{pmatrix}.$
So $x = 0 + \frac{6\sqrt{2}}{6}\left(-1/\sqrt{2},0,1/\sqrt{2}\right) + \frac{4}{\sqrt{5}}\left(2/\sqrt{5},1/\sqrt{5},0\right)=(7,4,1).$
This does not satisfy $Ax=b$ -- I can't find my error. By normal method of row reduction, I get $x=(1,2,3)$. Wolfram confirms this too! Am I incorrectly applying something here?
| To expand on the answer by user1551: Two eigenvalues of your $A$ are equal, and you have chosen two (correct) eigenvectors: Both $v_2$ and $v_3$ have eigenvalue $6$, but $v_2\cdot v_3\neq 0$, and the "spectral method" formula for $x$ requires orthonormal eigenvectors.
However, any linear combination of $v_2$ and $v_3$ will also be an eigenvector with eigenvalue $6$, so you can choose e.g. $v_2$ and $\tilde{v}_3=v_3+a v_2$ with a suitable $a$ so that $v_2\cdot \tilde{v}_3=0$ and continue from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor-Series at point x = 0 We have
$$ f(x) = (1+x)^{1/3} $$
I have to find the taylor series of f at the point x =0. The problem I am facing is that I do not know how often do I have to derive f
$$ f'(x) = \frac{1}{3}(1+x)^{-2/3} $$
$$ f''(x) = -\frac{2}{9}(1+x)^{-5/3} $$
$$ f'''(x) = \frac{10}{27}(1+x)^{-8/3} $$
$$ f^{4}(x) = -\frac{80}{81}(1+x)^{-11/3}$$
As you can see, I can derive infinity times.
As you also can see, there is a certain pattern when you oberseve the derivatives.
The pattern I recognised is something like that:
$$ f^{(n)}(x) = (-1)^{n+1}\frac{..}{3^n}(1+x)^{-\frac{3n-1}{3}} $$
I don't know what to put in ".." in the first fraction.
I also don't know if I am completely wrong or right.
Thank you in anticipation.
| From
\begin{align*}
f'(x) &= \frac{1}{3}(1+x)^{-\color{blue}{2}/3}\\
f''(x) &= -\frac{2}{9}(1+x)^{-\color{blue}{5}/3} \\
f'''(x) &= \frac{10}{27}(1+x)^{-\color{blue}{8}/3} \\
f^{4}(x) &= -\frac{80}{81}(1+x)^{-11/3}\\
\end{align*}
we conclude the numerator is
\begin{align*}
\color{blue}{2}\cdot \color{blue}{5}\cdot \color{blue}{8}=\prod_{j=1}^3 (3j-1)=8!!!
\end{align*}
with $n!!!=n\cdot (n-3)\cdot (n-6)\cdots $ the triple factorial.
In general $n$ we obtain
\begin{align*}
f^{(n)}(x) &= (-1)^{n+1}\frac{\prod_{j=1}^n (3j-1)}{3^n} (1+x)^{-\frac{3n-1}{3}} \\
&= (-1)^{n+1}\frac{(3n-1)!!!}{3^n} (1+x)^{-\frac{3n-1}{3}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a,b,c \in R$, and $w,w^2$ are non-real cube roots of unity, then find desired value If $w,w^2$ are non-real cube roots of unity and $a,b,c \in R$ such that $$\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}=2 w^2$$ and
$$\frac{1}{a+w^2}+\frac{1}{b+w^2}+\frac{1}{c+w^2}=2 w$$ then find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$$
Could some provide some method to solve this question. I tried adding two equation to get $-2$ on R.H.S. but each term on L.H.S. is not heading towards $a+1$ type in denominator. How to generate $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ ?
| We can say that $$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x} = \frac{2}{x}$$ has $x=\omega\;,\omega^2$ its roots.
Now Simplifying
$$x\bigg((b+x)(c+x)+(c+x)(a+x)+(a+x)(b+x)\bigg) = 2(a+x)(b+x)(c+x)$$
$$\bigg(3x^3+2(a+b+c)x^2+(ab+bc+ca)x\bigg) = 2\bigg(x^3+(a+b+c)x^2+(ab+bc+ca)x+abc\bigg)$$
So $$x^3+0 \cdot x^2-(ab+bc+ca)x-abc = 0$$
So it is a cubic equation whose two roots are $x=\omega\;, \omega^2$
Let third root is $x\;,$ Then $x+\omega+\omega^2 = 0\Rightarrow x = 1$
So $x=1$ is third root of $$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x} = \frac{2}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using mathematical induction prove that $ 11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6 $ is divisible by 8 for all $n \in N$.
Base: $ n = 0 $
$ 8 | 11 + 3 - 6 $ is obvious.
Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$.
So we get $ 11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8.
For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here.
| Suppose $11*3^n + 3*7^n - 6 = 8k$
The $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$
$=3(11*3^n + 3*7^n-2) + 4*3*7^n $
$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$
$= 3(8k) + 4(3*7^n + 3)$; $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.
$= 3(8k) + 8(\frac{3*7^n + 3}2) = 8(3k + \frac{3*7^n + 3}2)$.
======
Actually I like and am inspired by Bill Dubuques answer.
We want to prove $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 0 \mod 8$
And we know $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \mod 8$.
So it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \equiv 6 \mod 8$.
And if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \equiv 3^n + 3(-1)^{n}= f(n) \mod 8$.
So it's now just a matter of showing for $f(0) \equiv f(1) \equiv 6 \mod 8$.
Which is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Given $|x|=|y| = 1$, show that the intersection of two tangents to a unit circle at points $x$ and $y$ is $\frac{2xy}{x+y}$.
Given $|x|= |y| = 1$, show that the intersection of two tangents to a unit circle at points $x$ and $y$ is $\frac{2xy}{x+y}$.
What would be the best place to start to attempt this problem?
What I know:
$x^2= x\overline{x} = 1 = y\overline{y}$
| For an alternative, more geometric proof, consider that the point of intersection $z$ must lie on the bisector of angle $\angle xoy$ by symmetry, so $z = \lambda(x+y)$ for some $\lambda \in \mathbb{R}\,$.
The power of point $z$ with respect to the unit circle is $|z|^2-1$ and, since $zx$ is a tangent, it also equals $|z-x|^2$.
Therefore, using $\,\bar x = 1/x\,$ and $\,\bar y = 1/y\,$ at the last step, since $|x|=|y|=1\,$:
$$
\require{cancel}
\begin{align}
z \bar z -1 & = (z-x)(\bar z - \bar x) \\
\cancel{z \bar z} -1 & = \cancel{z \bar z} - z \bar x - \bar z x + x \bar x \\
z \bar x + \bar z x & = 2 \\
\lambda(x+y)\bar x + \lambda (\bar x + \bar y) x & = 2
\end{align}
$$
$$
\lambda = \frac{2}{x \bar x + \bar x y+\bar x x +x \bar y} = \frac{2}{2 + \frac{x}{y}+\frac{y}{x}} = \frac{2xy}{x^2+y^2+2xy} = \frac{2xy}{(x+y)^2}
$$
Then $z = \lambda (x+y)=\cfrac{2xy}{x+y}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim_{x\to 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ without l'Hôpital's rule. I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$
I did like this:
$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$
$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$
But it's wrong. Where I have gone wrong and how to do it?
| $$\frac{\tan ^3 x - \sin ^3 x}{x^5}=\tan^3x\dfrac{1 - \cos ^3 x}{x^5}=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2}\right)=\\
=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{1-\cos x}{x^2}\right)(1+\cos x+\cos^2x)=\\
=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{\sin x}{x}\right)^2\left(\dfrac{1}{1+\cos x}\right)(1+\cos x+\cos^2x)$$
Now use the fundamental limits:
$$\lim_{x\to 0}\frac{\sin x }{x}=1=\lim_{x\to 0}\frac{\tan x }{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$ If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$.
My Attempt:
$$2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$$
$$2f(x)+3f(\frac {1}{x})=4x + \frac {6}{x}$$
At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method...
| We are trying to find the following:
$$f^{-1}(x)=1\implies x=f(1)$$
Plugging $x=1$ into the original functional equation...
$$2f(1)+3f(1)=\frac{4+6}1$$
$$5f(1)=10\implies x=f(1)=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadratic form. so how do I use the synthetic division
| You can do the division between $x^4-6x^3+16x^2-25x+10$ and $x^2-2x+k$ following the polynomial long division, getting:
$$R=(2k-9)x+(k^2-8k+10)$$
but $R=x+a$, so $2k-9=1\longrightarrow k=5$ and $k^2-8k+10=a\longrightarrow a=-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using the cosine over cosine in square root I got up to
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
| I assume you know (or can figure out) $$\cos(-5\pi/4) = -\frac{\sqrt{2}}2,\ \sin(-5\pi/4) = +\frac{\sqrt{2}}2 $$ Applying the half-angle formulas to that, and noting that $-\pi < -5\pi/8 < -\pi/2$
$$ \eqalign{\cos(-5\pi/8) &= - \sqrt{\frac{1+\cos(-5\pi/4)}{2}} = -\frac{\sqrt{2-\sqrt{2}}}{2}\cr
\sin(-5\pi/8) &= - \sqrt{\frac{1-\cos(-5\pi/4)}{2}} = - \frac{\sqrt{2+\sqrt{2}}}{2}}$$
Similarly, since $-\pi/2 < -5\pi/16 < 0$,
$$ \eqalign{\cos(-5\pi/16) &= +\sqrt{\frac{1+\cos(-5\pi/8)}{2}} = \frac{\sqrt{2-\sqrt{2-\sqrt{2}}}}{2}\cr
\sin(-5\pi/16) &= - \sqrt{\frac{1-\cos(-5\pi/8)}{2}} = - \frac{\sqrt{2+\sqrt{2-\sqrt{2}}}}{2}}$$
so that
$$ \tan(-5\pi/16) = \frac{\sin(-5\pi/16)}{\cos(-5\pi/16)}
= - \frac{\sqrt{2+\sqrt{2-\sqrt{2}}}}{\sqrt{2-\sqrt{2-\sqrt{2}}}}$$
It turns out (but this is somewhat harder) that you can do some simplification here: you can write it as
$$ \tan(-5\pi/16) = 1 - \sqrt{2} - \sqrt{4-2\sqrt{2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Tangents are drawn to the circle $x^2+y^2=a^2$ from a point which always lies on the line $lx+my=1$. Tangents are drawn to the circle $x^2+y^2=a^2$
from a point which always
lies on the line $lx+my=1$. Prove that the locus of
the mid point of
chords of contact is $x^2 + y^2 -a^2(lx+my)=0$
My Attempt::
let mid point of chord of contact be $(h, k)$ &
point $(x1, y1)$ be always
lying on line: $lx+my=1$ then we have
$lx_1+my_1=1$ ...(1)
Now equation of chord of contact to circle:
$x^2+y^2=a^2$ is $xx_1+yy_1=a^2$ but
$(h, k)$ lies on chord of contact hence satisfying
equation of chord gives us
$hx1+ky1=a^2$ .......(2)
Please help me to complete this proof
.
|
You can write the equation of the polar line (green line in the figure):
$$x_1x+y_1y=a^2$$
where $(x_1,y_1)$ is a point of the line $lx+my=1$ (red line).
The equation of the normal (blue line) to the polar line, passing through $(x_1,y_1)$ is:
$$y={y_1\over x_1}x$$
so we can find the middle point $M$ of the chord:
\begin{cases}x_1x+y_1y=a^2\\y={y_1\over x_1}x\end{cases}
Solving this system, we get:
$$M\left(x_M={a^2x_1\over x_1^2+y_1^2},y_M={a^2y_1\over x_1^2+y_1^2}\right)$$
Now we must eliminate $x_1$ and $y_1$, thus:
\begin{cases}x_M={a^2x_1\over x_1^2+y_1^2}\\y_M={a^2y_1\over x_1^2+y_1^2}\\lx_1+my_1=1\end{cases}
Squaring and adding the first and the second equation:
\begin{cases}x_M^2+y_M^2=a^4{x_1^2+y_1^2\over (x_1^2+y_1^2)^2}={a^4\over x_1^2+y_1^2}\\x_M={a^2x_1\over x_1^2+y_1^2}\\y_M={a^2y_1\over x_1^2+y_1^2}\\lx_1+my_1=1\end{cases}
\begin{cases}{x_M^2+y_M^2\over a^2}={a^2\over x_1^2+y_1^2}\\x_M={x_M^2+y_M^2\over a^2}x_1\longrightarrow a^2x_M=(x_M^2+y_M^2)x_1\\y_M={x_M^2+y_M^2\over a^2}y_1\longrightarrow y_M={x_M^2+y_M^2\over a^2}{(1-lx_1)\over m}\longrightarrow a^2my_M-(x_M^2+y_M^2)=-(x_M^2+y_M^2)lx_1\\y_1={1-lx_1\over m}\end{cases}
so $a^2my_M-(x_M^2+y_M^2)=-a^2lx_M\longrightarrow x_M^2+y_M^2-a^2(lx_M+my_M)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$
Is this a correct usage of the chain rule?
| With the chain rule:
$$f'(x)=\frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\left(\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\right)=\frac{1+x^2}{\sqrt{(1+x^2)^2-4x^2}}\left(\frac{2-2x^2}{(1+x^2)^2}\right)$$
$$=\frac{2-2x^2}{1+x^2}\frac{1}{|1-x^2|}=\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
What is the value of $\frac {1}{2\cdot3} + \frac {1}{4\cdot 5} +\frac {1}{6\cdot 7} + \cdots $? What is value of $\dfrac {1}{2\cdot 3} + \dfrac {1}{4\cdot 5} +\dfrac {1}{6\cdot7} + \cdots =?$
1) $\log \left( \dfrac 2e \right)$
2) $\log \left( \dfrac e2 \right)$
3) $\log \left( 2e \right)$
4) $e-1$
I know it converges but I am not getting how to solve
| An alternative solution, just for fun:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{2n(2n+1)} = \int_{0}^{1}\sum_{n\geq 1}\frac{x^{2n}}{2n}\,dx &=& -\frac{1}{2}\int_{0}^{1}\log(1-x^2)\,dx \\(IBP)\quad&=&\int_{0}^{1}x\cdot\frac{1-x}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x}{1+x}\,dx\\&=&1-\log 2=\color{red}{\log\left(\frac{e}{2}\right)}.\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
maximum value of expression $6bc+6abc+2ab+3ac$ If $a,b,c>0$ and $a+2b+3c=15,$ then finding maximum value of $6bc+6abc+2ab+3ac$ is
with the help of AM - GM inequality
$4ab\leq (a+b)^2$ and $4bc \leq (b+c)^2$ and $\displaystyle 4ca \leq (c+a)^2$
and $27(abc)\leq (a+b+c)^3$
could some help me, thanks
| We are nearly done, but we need something other than $a,b,c$ because of the condition of the problem.
By Cauchy-Schwarz we have $$3(x^2+y^2+z^2) \ge (x+y+z)^2 \iff (x+y+z)^2 \ge 3(xy+yz+zx)$$
So put $x=a, y=2b, z=3c$ to get
$$225=(a+2b+3c)^2 \ge 3(2ab+3ac+6bc) \iff 75 \ge 2ab+3ac+6bc $$
Also, note that by AM-GM $$15=a+2b+3c \ge 3 \sqrt[3]{6abc} \iff 125 \ge 6abc $$
Note that both inequalities have equalities when $a=2b=3c$.
So the answer is $200$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove the inequality $\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $a,b,c>0$ As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$.
Thsi inequality can be proved in a pretty straightforward manner utilizing the Muirhead's inequality, yet I ought to prove it using the rearrangement inequality. I can't however figure out any suitable sequences and permutations of them.
| (Added a solution as it's much simpler than the rest.)
Multiplying throughout by $a^3b^3c^3$, we WTS
$$a^8+b^8+c^8\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2.$$
This follows directly by summing up the cyclic inequalities of
$$ 2a^8 + 3b^8 + 3c^8 \geq 8 a^2b^3c^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Repeating decimals
For a certain positive integer $n$ less than $1000$, the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$, a repeating decimal of period $6$, and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$, a repeating decimal of period $4$. In which interval does $n$ lie? $\textbf{(A)}\ [1,200] \qquad
\textbf{(B)}\ [201,400] \qquad
\textbf{(C)}\ [401,600] \qquad
\textbf{(D)}\ [601,800] \qquad
\textbf{(E)}\ [801,999] $
All I have for this is $n \mid 10^6 - 1$ and $n+ 6 \mid 10^4 - 1$. How do I proceed?
| You have $n+6 | 9 \cdot 11 \cdot101$
But if $n+6 | 10^a -1 $ where $a<4 $, then $\frac {1}{n+6} $will have $a$ repeating digits. So checking cases we get
$n+6 = 101, 303 $ or $909$
Now we have $n | 11 \cdot 13 \cdot 7 \cdot27 \cdot37$
So $ n= 297$ is the only solution
EDIT:
$ \frac{1}{297} =0. \overline{003367}$
$ \frac {1}{303} =0. \overline{0033}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Consider $\sum_{k=1}^{\infty}{1\over 3^k}\cdot{\sinh(a/3^k)+\sinh(2a/3^k)\over 2\cosh(a/3^k)+\cosh(2a/3^k)+1.5}=S_a$ Consider
$$\sum_{k=1}^{\infty}{1\over 3^k}\cdot{\sinh(a/3^k)+\sinh(2a/3^k)\over 2\cosh(a/3^k)+\cosh(2a/3^k)+1.5}=S_a\tag1$$
$a\ge1$
How does one show that $$S_a={1\over 2}\cdot{e^a+1\over e^a-1}-{1\over a}?$$
Try to simplify
$x={a\over 3^k}$
$${\sinh(x)+\sinh(2x)\over 2\cosh(x)+\cosh(2x)+1.5}={\sinh(x)(1+2\cosh(x))\over 2\cosh(x)+2\cosh^2(x)+0.5}$$
$$={2\sinh(x)(1+2\cosh(x))\over 3\cosh^2(x)+4\cosh(x)-\sinh^2x}$$
I don't how to simplify down any more.
Edited:
$$\sum_{k=1}^{\infty}{1\over 3^k}\cdot{2\sinh(x)\over 1+2\cosh(x)}$$
$$\sum_{k=1}^{\infty}{1\over 3^k}\cdot{1+2e^{a/3^k}\over 1+2\cosh(a/3^k)}=S_a+{1\over2}$$
| Detailing @JackD'Aurizio Answer,
$$ \begin{align}
\sum_{k=1}^{\infty}\,\frac{1}{3^k}\,\frac{2\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} &= \frac{1}{2}\frac{e^a+1}{e^a-1}-\frac{1}{a} \quad\Rightarrow \\[3mm]
\sum_{k=1}^{\infty}\,\frac{1}{3^k}\,\frac{4\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} &= \coth(a/{\small2})-\frac{1}{a/{\small2}}
\end{align} $$
The idea is that, instead of simplifying the series of LHS to get the expression on RHS,
Start by the expression on RHS and construct the series of LHS.
$$ \begin{align}
\color{red}{\coth(3x)} &= \frac{\cosh(3x)}{\sinh(3x)} = \frac{\cosh(x+2x)}{\sinh(x+2x)} = \frac{\cosh(x)\cosh(2x)+\sinh(x)\sinh(2x)}{\sinh(x)\cosh(2x)+\cosh(x)\sinh(2x)} \\[3mm]
&= \frac{\cosh(x)\left[2\sinh^2(x)+1\right]+\sinh(x)\left[2\sinh(x)\cosh(x)\right]}{\sinh(x)\left[2\cosh^2(x)-1\right]+\cosh(x)\left[2\sinh(x)\cosh(x)\right]} \\[3mm]
&= \frac{4\sinh^2(x)\cosh(x)+\cosh(x)}{4\cosh^2(x)\sinh(x)-\sinh(x)} = \frac{1}{3}\frac{12\sinh^2(x)\cosh(x)+3\cosh(x)}{4\cosh^2(x)\sinh(x)-\sinh(x)} \\[3mm]
&= \frac{1}{3}\frac{8\sinh^2(x)\cosh(x)+2\cosh(x)\left[2\sinh^2(x)+1\right]+\cosh(x)}{2\sinh(x)\left[2\cosh^2(x)-1\right]+\sinh(x)} \\[3mm]
&= \frac{1}{3}\frac{\sinh(x)\left[4\sinh(2x)\right]+\cosh(x)\left[1+2\cosh(2x)\right]}{\sinh(x)\left[1+2\cosh(2x)\right]} \\[3mm]
&= \color{red}{\frac{1}{3}\frac{4\sinh(2x)}{1+2\cosh(2x)} + \frac{1}{3}\coth(x)}
\end{align} $$
Thus,
$$ \begin{align}
\color{red}{\coth(a/{\small2})} &= \frac{1}{3^1}\,\frac{4\sinh\left(a/{\small3^1}\right)}{1+2\cosh\left(a/{\small3^1}\right)} + \frac{1}{3^1}\,\coth\left(\frac{a/{\small2}}{\small3^1}\right) \\[3mm]
&= \frac{1}{3^1}\,\frac{4\sinh\left(a/{\small3^1}\right)}{1+2\cosh\left(a/{\small3^1}\right)} + \frac{1}{3^2}\,\frac{4\sinh\left(a/{\small3^2}\right)}{1+2\cosh\left(a/{\small3^2}\right)} + \frac{1}{3^2}\,\coth\left(\frac{a/{\small2}}{\small3^2}\right) \\[3mm]
&= \,\cdots\, = \sum_{k=1}^{\infty}\,\frac{1}{3^k}\,\frac{4\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} + \color{blue}{\lim_{k\rightarrow\infty}\left[\frac{1}{3^k}\,\coth\left(\frac{a/{\small2}}{\small3^k}\right)\right]} \quad\Rightarrow \\[3mm]
\sum_{k=1}^{\infty}\,\frac{1}{3^k} &\, \frac{4\sinh\left(a/{\small3^k}\right)}{1+2\cosh\left(a/{\small3^k}\right)} = \color{red}{\coth(a/{\small2})} - \color{blue}{\frac{1}{a/{\small2}}}
\end{align} $$
Where,
$$ \lim_{k\rightarrow\infty}\left[\frac{1}{3^k}\,\coth\left(\frac{a/{\small2}}{\small3^k}\right)\right] = \frac{1}{a/{\small2}}\lim_{k\rightarrow\infty}\left[\frac{a/{\small2}}{3^k}\,\coth\left(\frac{a/{\small2}}{\small3^k}\right)\right] = \frac{1}{a/{\small2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}$ Consider
$$\int_{0}^{1}{2n-x-x^3-x^5-\cdots-x^{4n-1}\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}=I\tag1$$
$n\ge1$
How does one show that $I=2n\ln{\Gamma(3/4)\over \Gamma(5/4)}-\ln{[8^n(2n-1)!!]}?$
An attempt:
$J=x+x^3+x^5+x^7+\cdots+$
$J=x(1+x^2+x^4+x^6+\cdots+)$
Geometric series
$1+x+x^2+x^3+\cdots x^{n-1}={x(1-x^n)\over 1-x}$
$1+x^2+x^4+x^6+\cdots+x^{2n-2}={x^2(1-x^{2n})\over 1-x^2}$
$I$ becomes
$$2n\int_{0}^{1}{1\over 1+x^2}\cdot{\mathrm dx\over \ln{x}}-\int_{0}^{1}{1-x^{2n}\over 1+x^2}\cdot{x^3\over \ln{x}}\mathrm dx=I\tag2$$
$x=\tan{y}$ then $dx=\sec^2{y}dy$
$$\int_{0}^{\pi/4}{1\over \ln{\tan{y}}}\mathrm dy-\int_{0}^{\pi/4}{1-\tan^{2n}{y}\over \ln{\tan{y}}}\cdot\tan^3{y}\mathrm dy\tag3$$
$$\int_{0}^{\pi/4}{1-\tan^3{y}\over \ln{\tan{y}}}\mathrm dy-\int_{0}^{\pi/4}{\tan^{2n}{y}\over \ln{\tan{y}}}\cdot\tan^3{y}\mathrm dy\tag4$$
$1-x^3=(1-x)(1+x+x^2)$
I am not sure what to do next
| The first term in the LHS of your $(2)$ is a divergent integral: better to avoid them.
Instead, we may consider that:
$$ I(k)=\int_{0}^{1}\frac{1-x^{2k-1}}{1+x^2}\cdot\frac{dx}{\log(x)}=-\int_{0}^{+\infty}\frac{e^{-x}-e^{-2kx}}{1+e^{-2x}}\cdot\frac{dx}{x} $$
can be easily dealt with through a geometric series expansion and Frullani's theorem, leading to:
$$ I(k) = \log\frac{2k}{1}-\log\frac{2k+2}{3}+\log\frac{2k+4}{5}-\log\frac{2k+6}{7}+\ldots $$
then to:
$$ I(k) = -\log\left(\frac{2k}{1}\cdot\frac{3}{2k+2}\cdot\frac{2k+4}{5}\cdot\frac{7}{2k+6}\cdots\right) $$
and finally to:
$$ I(k) = (-1)^{k+1}\log\sqrt{\pi} +\log\left(\frac{(k-2)!!\,\Gamma\left(\frac{3}{4}\right)}{2(k-1)!!\,\Gamma\left(\frac{5}{4}\right)}\right)$$
that is simple to sum over $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit.
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding this limit?
| Hint: Mulitply by
$$
\frac{\left(\sqrt[3]{n^2+5}\right)^2 + \sqrt[3]{n^2+5}\sqrt[3]{n^2+3} + \left(\sqrt[3]{n^2+3}\right)^2}{\left(\sqrt[3]{n^2+5}\right)^2 + \sqrt[3]{n^2+5}\sqrt[3]{n^2+3} + \left(\sqrt[3]{n^2+3}\right)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$ Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$
My Attempt:
$$L.H.S=\tan^6 20 - 33\tan^4 20 + 27\tan^2 20°$$
$$=\tan^2 20°(\tan^4 20° - 33\tan^2 20°+27)$$
$$=(\sec^2 20° -1)(\tan^4 20° - 33\tan^2 20° + 27)$$
Please help me to continue from here..
| My idea: We know formulas for $\cos(20°)$; we can write it as a relation for $\sec(20°)$, and then use the identity $\sec^2(x)=1+\tan^2(x)$
First of all, we'll compute some formulas related to $\cos(20°)$ (well-known, but I don't renember them xD). We know that
$$ \cos(3x)=\cos(2x)\cos(x)-\sin(x)\sin(2x)=(2\cos^2(x)-1)\cos(x)-2\sin^2(x)\cos(x)$$
So, $\cos(3x)=2\cos^3(x)-\cos(x)-2(1-\cos^2(x))\cos(x)=4\cos^3(x)-3\cos(x)$. Plugging $x=20°$, and calling $a=\cos(20)$, we get
$\frac{1}{2}=\cos(60°)=4\cos^3(20^°)-3\cos(x)=4a^3-3a$
So, $1=8a^3-6a$. Squaring both sides, we get $1=64a^6-96a^4+36a^2$. Now, dividing by $a^6$ both sides, and calling $b=\frac{1}{a}=\frac{1}{\cos(20^°)}=\sec(20°)$, we have
$$ \frac{1}{a^6}=64-\frac{96}{a^2}+\frac{36}{a^4} \Rightarrow b^6=64-96b^2+36b^4 $$
The key step now is try to use the identity $\sec^2(x)=1+\tan^2(x)$. Calling $c=\tan(20^°)$, we can plug $b^2=1+c^2$ on our formula, to get
$$ \begin{align*}
(1+c^2)^3 &= 64-96(1+c^2)+36(1+c^2)^2 \\
1+3c^2+3c^4+c^6 &= 64-96(1+c^2)+36(1+2c^2+c^4) \\
1+3c^3+3c^4+c^6 &= 64-96-96c^2+36+72c^2+36c^4 \\
1+3c^3+3c^4+c^6 &= 4-24c^2+36c^4 \\
c^6-33c^4+27c^2 &= 3 \\
\tan(20°)^6-33\tan(20°)^4+27\tan(20°)^2 &= 3 \\
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Improper integral of $\frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}$ Show that $$\int_{-\infty}^\infty \frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}\,dx = \frac{\pi}{b}\log\left(a+b\right)$$
for $a,b>0\in\mathbb{R}$. I stumbled on this answer empirically, but I'm not sure how to solve it directly.
| Suppose $a \gt b$ for now. Consider the contour integral in the complex plane
$$\oint_C dz \frac{\log{\left ( z^2+a^2 \right )}}{z^2+b^2} $$
where $C$ is a semicircle of radius $R$ in the upper half-plane with a detour down and up the imaginary axis about the branch point $z=i a$. In the limit as $R \to \infty$, the contour integral is equal to
$$\int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} + i \int_{\infty}^a dy \frac{\log{\left ( y^2-a^2 \right )}+i \pi}{b^2-y^2} + i \int_a^{\infty} dy \frac{\log{\left ( y^2-a^2 \right )}-i \pi}{b^2-y^2}$$
Note that the log terms in the latter two integrals vanish. Now, the contour integral is also equal to the residue of the pole of the integrand at $z=i b$. Thus
$$\int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} - 2 \pi \int_a^{\infty} \frac{dy}{y^2-b^2} = i 2 \pi \frac{\log{\left ( a^2-b^2 \right )}}{i 2 b} $$
Accordingly, after doing out that second integral and performing a little algebra, we get...
$$\frac12 \int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} = \frac{\pi}{b} \log{\left ( a+b \right )} $$
ADDENDUM
For $a \lt b$, the answer is the same as above but the contour is altered. This time, the contour $C$ must detour about the pole at $z=i b$ along each side of the branch cut with a semicircle of radius $\epsilon$. The contour integral is this equal to
$$\int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} + i PV \int_{\infty}^a dy \frac{\log{\left ( y^2-a^2 \right )}+i \pi}{b^2-y^2} + i PV \int_a^{\infty} dy \frac{\log{\left ( y^2-a^2 \right )}-i \pi}{b^2-y^2} \\ + i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{\log{\left [- \left (i b + \epsilon e^{i \phi} \right )^2-a^2 \right ]}+i \pi}{\left (i b + \epsilon e^{i \phi} \right )^2+b^2}+ i \epsilon \int_{3 \pi/2}^{\pi/2} d\phi \, e^{i \phi} \frac{\log{\left [ -\left (i b + \epsilon e^{i \phi} \right )^2-a^2 \right ]}-i \pi}{\left (i b + \epsilon e^{i \phi} \right )^2+b^2}$$
Note that the sum of the two final integrals - the pieces that go around the pole - is equal to the residue of the pole at $z=i b$ in the limit as $\epsilon \to 0$. The $\pm i \pi$ pieces cancel. Also note that the principal value integrals are the same as the corresponding integrals above for $a \gt b$. Thus, the result for $a \lt b$ is the same as that for $a \gt b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| There are many good answers already. Here's a very elementary approach:
Let $f(x) = x^4 + 1$.
Over the real numbers, every polynomial factors into a product of linear and quadratic polynomials. We know that $f(x) = 0$ has no solutions in the real numbers. That means $f(x)$ doesn't have any linear factors.
So, it must have only quadratic factors. Now since $f(x)$ is a polynomial of degree $4$, it must be the product of exactly $2$ quadratic polynomials:
$$
x^4 + 1 = (a_1 x^2 + b_1 x + c_1) (a_2 x^2 + b_2 x + c_2)
$$
From above we know that both factors are quadratic; so their $x^2$ terms are nonzero. We might as well assume $a_1=1$ (because if not, we can divide by $a_1$ in the first factor, and multiply the second factor by $a_1$). But then $a_2$ will be $1$ as well, because the $x^4$ term has coefficient $1$.
Similarly, $c_1c_2$ must equal $1$, so $c_2=1/c_1$ (and $c_1\neq 0$).
So let's rewrite the equation, using some new symbols for the coefficients:
$$
x^4 + 1 = (x^2 + ax + b)(x^2 + cx + 1/b)
$$
We do some rearranging:
$$
\begin{eqnarray}
x^4 + 1 & = &(x^2 + ax + b)(x^2 + cx + 1/b)\\
&=& x^4 + ax^3 + bx^2 + cx^3 +acx^2 + bcx + (1/b)x^2 + (a/b)x + 1\\
&=& x^4 + (a+c)(x^3) + (b + ac + 1/b)(x^2) + (bc + a/b)(x) + 1\\
\end{eqnarray}
$$
Since the coefficients on both sides have to be equal, we get three new equations:
$$
\begin{eqnarray}
0& =& a + c\\
0 &=& b + ac + 1/b\\
0& =& bc + a/b
\end{eqnarray}
$$
The first equation says $c = -a$, so let's substitute that into the other two equations:
$$
\begin{eqnarray}
0& =& b - a^2 + 1/b\\
0& = &-ab + a/b\\
\end{eqnarray}
$$
Or, multiplying both sides by $b$ in both equations:
$$
\begin{eqnarray}
a^2b &=& b^2 + 1\\
0& =& -ab^2 + a = a(-b^2 + 1) = a(1+b)(1-b)\\
\end{eqnarray}
$$
The first equation guarantees that $ a\neq 0$, so the second equation now says that $b=\pm 1$. At this point the first equation says that $\pm a^2 = 2$, so in fact $a^2 = 2$ and $a = \pm \sqrt{2}$. Now the first equation says $2b=2$, so $b=1$.
Case I: $a=\sqrt{2}$. Then we get
$$
x^4+1 = (x^2 + x\sqrt{2} + 1)(x^2 - x\sqrt{2} + 1)
$$
Case II: $a=-\sqrt{2}$. Then we get
$$
x^4 + 1 = (x^2 - x\sqrt{2} + 1)(x^2 + x\sqrt{2} + 1)
$$
One case or the other has to be correct, but either way we get the same factorization (just written in a different order). Therefore that factorization must be correct, and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 10
} |
Prove that $\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$ Let $a$, $b$ and $c$ be roots of the equation
$$x^3+15x^2-198x+1=0.$$
Prove that:
$$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$$
I have a solution for this problem, but I want to see another solutions.
My solution is the following.
We'll take the equation $x^3-3x+1=0$,
which has three roots $a=2\cos40^{\circ}$, $b=2\cos80^{\circ}$ and $c=-2\cos20^{\circ}$.
Now, easy to show that
$a^5+b^5+c^5=-15$, $a^5b^5+a^5c^5+b^5c^5=-198$, $a^5b^5c^5=-1$ and since $a+b+c=0$,
we obtain the starting equation.
My question is how we can solve this equation without previous way?
Thank you very much!
| If $\alpha$ is a root of the polynomial $p(x)=x^3+15x^2-198x+1$ then $\alpha^{1/5}$ is a root of the polynomial $p(x^5)=(1-3x+x^3)\cdot q(x)$. If we manage to prove that $(1-3x+x^3)$ is the minimal polynomial of $\alpha^{1/5}$, the claim easily follows from Vieta's theorem. For the last part, it is simpler to go in the opposite direction.
Claim: if $\theta$ is a root of $x^3-3x+1$, then $\theta^5$ is a root of $x^3+15x^2-198x+1$.
Let we work in $\mathbb{Q}[x]/(x^3-3x+1)$. A base of this ring as a vector space over $\mathbb{Q}$ is given by $\{1,x,x^2\}$, and by computing a few polynomial remainders we have:
$$\begin{eqnarray*}1 &=& 1 \\ x^5 &=& -x^2+9x-3\\ x^{10} &=& 90x^2-109x+27\\ x^{15}&=&-1548 x^2+3417 x-1000 \end{eqnarray*} $$
so the previous Claim follows from gaussian elimination.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Factoring irreducible fractions Is there a technique or a formula that factors irreducible fractions into products of polynomials, for example:
$$\frac1{x^8+1}=\frac{1}{(x^4+ \sqrt2x^2+1)(x^4- \sqrt2x^2+1)}$$
Also, is there a way to factor, in the same way as above, the following fraction:
$$\frac1{x^4-x^2+1}$$
I am not interested in the completition of the square. Thank you.
| One has a direct way, writing
$$
x^8+1=(x^8+2x^4+1)-2x^4=(x^4+1)^2-2x^4=(x^4+ \sqrt2x^2+1)(x^4- \sqrt2x^2+1)
$$ similarly
$$
x^4-x^2+1=(x^4+1)-x^2=(x^2+1)^2-3x^2=(x^2+ \sqrt3x+1)(x^2- \sqrt3x+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the $\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}$ Let $a,b,c$ such
$$a\sin^2{x}+b\cos^2{x}=c,~~~\dfrac{a}{\sin^2{x}}+\dfrac{b}{\cos^2{x}}=c$$
find the value
$$\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}$$
| a$\cos^2x$+b$\sin^2x =c$
Divide this equation by $\cos^2x$.
We have thus,a+b $\tan^2x$=c $\sec^2x$.
$\sec^2x$=$\tan^2x$ +1.This gives the value of $\tan^2x$=(a-c)/(c-b).
Thus values of $\cos^2x$ and $\sin^2x$ are obtained. these values are put in equation a/$\cos^2x$ + b/$\sin^2x$=c.
Thus the required value is obtained as c/(a-b) + a/(b-c) + b/(c-a)=0.
Hope u understand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Maximize $P=a^2+b^2+c^2+ab+ac+bc$
For real numbers $a, b, c$ that satisfy $a + b + c = 6$ and $0\leq a,b,c \leq 4$,
maximize $P=a^2+b^2+c^2+ab+ac+bc$.
My try:
$$\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$$ $$(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^2+bc=(6-a)^2-bc \end{cases} \end{align}$$ $$P=a^2+(b^2+c^2+bc)+a(b+c)=a^2+[(6-a)^2-bc]+a(6-a)$$ $$P=(a^2-12a+36)-bc=(a-6)^2-bc (2)\Rightarrow bc\leq 0 \Rightarrow P\geq (a-6)^2$$ When $bc=0 \Rightarrow [{\begin{matrix}b=0\\c=0\end{matrix}}(3)$. From $(1)$ and $(3)$, $\Rightarrow 2\leq a\leq 4(4)$ $P_{max} \implies |a-6|$ max satisfy $(4)$ $\implies a=2$ from $(1)$ and $(3)$ $\implies b=c=4$ $\implies P_{max}(a,b,c)=P(4;2;0)=28$
| From the fact that $a+b+c=6$, we know that $$36=\left( \sum_{cyc} a \right)^2=\sum_{cyc}a^2+2 \sum_{cyc}ab=P+\sum_{cyc}ab$$
So maximizing $P$ becomes equivalent to minimizing $ab+bc+ca$.
WLOG, assume that $a$ is the maximum among $a,b,c$. So $a+b+c =6 \le 3a$, hence $4 \ge a \ge 2$ from the condition.
However, note that $$ab+bc+ca \ge ab+ca= a(b+c)=a(6-a) \ge 8$$
From $4 \ge a \ge 2$.
Hence the minimum of $ab+bc+ca=8$, when $a=4,b=2, c=0$. Thus, the maximum of $P$ is $36-8=28$.
We are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding probability density function of a conditional expectation I´m solving the next exercise of a probability book about conditional expectation:
Let $(X,Y)$ a discrete random vector with probability density function given by
$$f_{x,y}(x,y)=(x+y)/36\quad x,y\in\{1,2,3\}$$ and zero in other case.
a) Find the probability density function of the random variable $E(X|Y).$
b) Check the formula $E(E(X|Y))=E(X)=78/36.$
I've begun computing marginal density for variable $Y.$ My computations lead me to get $$f_{Y}(y)=\dfrac{2+y}{12}\quad\text{if}\quad y\in\{1,2,3\}$$ and zero in other case.
Then I use the definition of $E(X|Y=y),$ so doing the computations I get $$E(X|Y=y)=\dfrac{6y+14}{6+3y}.$$
Due to the above, we conclude that $E(X|Y)=\dfrac{6Y+14}{6+3Y}.$ Then we want to calculate $$P(\dfrac{6Y+14}{6+3Y}=y)$$ which is equivalent to compute $f_{Y}(\frac{6y-14}{6-3y}),$ but here is my problem. I don't know how to calculate that term, because random variable $Y$ is discrete, so the point of evaluation must be a natural number where marginal density of $Y$ takes these values.
How can I compute that value? I would like that $\frac{6y-14}{6-3y}=z$ for every value of $z\in\{1,2,3\},$ but when I do that, the new values of $y$ are rationals.
What am I doing wrong? Is there another way to solve this easily?
Any kind of help is thanked in advanced.
| $\require{cancel}$You've the right calculations for $$f_{Y}(y)=\frac{2+y}{12} \text{ if } y\in\{1,2,3\}$$ and $$E(X \mid Y=y)=\frac{6y+14}{6+3y} \text{ if } y\in\{1,2,3\}.$$
The interpretation of $E(E(X \mid Y))$ should be
\begin{align}
& E(E(X \mid Y)) \\
=& \sum_{y = 1}^3 E(X \mid Y = y) P(Y = y) \\
=& \sum_{y = 1}^3 \frac{6y + 14}{\cancelto{3}{6 + 3y}} \frac{\cancel{2+y}}{12} \\
=& \sum_{y = 1}^3 \frac{6y + 14}{36} \\
=& \frac{6 \times 6 + 3 \times 14}{36} \\
=& \frac{78}{36}.
\end{align}
\begin{align}
& E(X) \\
=& \sum_{x = 1}^3 \sum_{y = 1}^3 x f_{X,Y}(x,y) \\
=& \sum_{x = 1}^3 \sum_{y = 1}^3 x \frac{x+y}{36} \\
=& \sum_{x = 1}^3 \frac{3x^2+6x}{36} \\
=& \sum_{x = 1}^3 \frac{x^2+2x}{12} \\
=& \frac{1+4+9+2\times6}{12} \\
=& \frac{26}{12} \\
=& \frac{78}{36}.
\end{align}
This is not a coincidence. In fact,
\begin{align}
& E(E(X \mid Y)) \\
=& \sum_y E(X \mid Y = y) P(Y = y) \\
=& \sum_y \sum_x x P(X \mid Y = y) P(Y = y) \\
=& \sum_y \sum_x x \frac{f_{X,Y}(x,y)}{\cancel{f_Y(y)}} \cancel{f_Y(y)} \\
=& \sum_y \sum_x x f_{X,Y}(x,y) \\
=& E(X).
\end{align}
Edit in response to OP's comment
\begin{align}
f_Y(y) =& \sum_{x = 1}^3 f_{X,Y}(x,y) \\
=& \sum_{x = 1}^3 \frac{x+y}{36} \\
=& \frac{1+2+3+3y}{36} \\
=& \frac{2+y}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Mathematical Induction: $3^n +1$ is even for all values of positive integers I am not sure how to go about this "proof by induction" problem. below is my attempt.
Base Case: $n = 0$,substituting the value of $n$ to the equation $3^n+1$
$$= 3^0 + 1$$
$$= 1 + 1 = 2 $$
Thus the equation holds true for initial value of $n$ i.e $0$
Induction Hypothesis: Suppose the equation holds true for all the values of $n$: $1,2,3....k$ therefore, $3^k + 1$ results even.
Induction Step: $n = k$ holds true,
to prove: $3^{k+1}+ 1$ for $n = k+1$
LHS:
$$3^{k+1}+ 1 = (3^k \cdot 3) + 1 = (3^k \cdot (2 + 1)) + 1 = 2 \cdot 3^k + 3^k + 1$$
The above solution results even, because since multiplying any integer with $2$ gives even integer, and from the Induction Hypothesis $3^k+1$ is even.
Hence $3^{k+1} + 1$ is even, thus $3^{n+1}$ is even for all values of $n\ge0$
| An easier answer is to note that:
$3^n - 1 \equiv 1^n - 1 \equiv 0 (mod 2)$.Thus, we conclude that $2|(3^n -1)$. Hence, $3^n - 1$ is always even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove that lines $y^3 - x^3 + 3xy(y -x) = 0$ are equally inclined to each other.
Prove that lines $y^3 - x^3 + 3xy(y -x) = 0$ are equally inclined to each other.
I have to find this by using polar form of this equation.
Without using polar form I get,
$$(y - x)(y^2 + 4xy + x^2)= 0$$
$$\implies y = x \ or \ y^2 + 4xy + x^2 = 0$$
Solving second part,
$$y^2 + 4xy + x^2 = 0 \implies x= y(-2 \pm \sqrt{3})$$
Therefore the angles made by these three lines with the x-axis are $45^\circ, 105^\circ$ and $165^\circ$.
Hence proved.
With polar form,
On transforming the equation I get
$$(\sin \theta - \cos \theta)(1 + 4\sin \theta \cos\theta) = 0$$
$$ \implies \tan\theta = 1, \sin 2\theta = -\frac12$$
$$\implies \theta = n\pi + \frac\pi4, {1 \over 2}\left(n\pi + (-1)^n\frac{7\pi}6\right) \tag{*}$$
But I don't know what to do with this now ? how can I show the lines are equally inclined with the help of (*) ?
| Your calculation is correct, so just contrast them. Take $n=1,2,3$
$$\dfrac{7\pi}{12}-\dfrac{\pi}{4}=\dfrac{\pi}3$$
$$\dfrac{11\pi}{12}-\dfrac{7\pi}{12}=\dfrac{\pi}3$$.
note: $0\leq\theta\leq\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find Minima and Maxima of $ y = \frac{x^2-3x+2}{x^2+2x+1}$ $$ y = \frac{x^2-3x+2}{x^2+2x+1}$$
I guess I made some mistakes cause after taking the first derivative and simlifying I have
$$y = \frac{2x^3-4x^2+5}{(x+1)^2}$$
but then numerator has complex roots. which should not be, IMO
| $$y = \frac{x^2 - 3x + 2}{x^2 + 2x + 1} = 1 - \frac{5x-1}{(x+1)^2}$$
\begin{align*}
\frac{dy}{dx} &= \frac{d}{dx} \left[ \frac{1-5x}{(x+1)^2} \right] \\
&= \frac{-5\cdot(x+1)^2-(1-5x)\cdot 2(x+1)}{(x+1)^4} \\
&= \frac{-5\cdot(x+1)-2(1-5x)}{(x+1)^3} \\
&= \frac{-5x-5-2+10x}{(x+1)^3} \\
&= \frac{5x-7}{(x+1)^3}
\end{align*}
$$\frac{dy}{dx} = 0 \Longleftrightarrow x=\frac{7}{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Finding out the solution to the system of equations
The question is to find out the solution of the following system of equations $$\log_{10}(x^3-x^2)=\log_5y^2$$
$$\log_{10}(y^3-y^2)=\log_5z^2$$
$$\log_{10}(z^3-z^2)=\log_5x^2$$
Letting $f(x)=5^{\log_{10}(x^3-x^2)}$.The equations can be expressed as $f(x)=y^2$,$f(y)=z^2$,$f(z)=x^2$.Now $(x^3-x^2)=x^2(x-1)$.So that it is a increasing in nature.Hence if $x>y$ then $f(x)>f(y)\implies y>z\implies z>x$.A contradiction.Hence the only solution is $x=y=z$.Is there another way to solve this problem without going in for contradiction?Thanks.
| $x,y,z>1$
With the equation system we get
$\displaystyle g(x):=( (x^3-x^2)^{1.5\log_{10}5} - (x^3-x^2)^{\log_{10}5} )^ {1.5\log_{10}5} - $
$\hspace{1.2cm}\displaystyle -( (x^3-x^2)^{1.5\log_{10}5} - (x^3-x^2)^ {\log_{10}5})^{ \log_{10}5} - x^{2\log_5 10}=0$
which makes with real $\enspace x\enspace $ only sense for $\enspace x\geq x_0\enspace $ with
$\displaystyle x_0:=\frac{1}{3}\left(1+2^{-\frac{1}{3}}\left( \sqrt[3]{29-3\sqrt{93}}+\sqrt[3]{29+3\sqrt{93}} \right)\right)\approx 1.465571231876768...$ .
It follows $\enspace g(x)<0\enspace $ for $\enspace x<5$, $\enspace g(x)>0\enspace $ for $\enspace x>5$, $\enspace g(5)=0$
and therefore the only solution of the equation system is $\enspace x=y=z=5$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$5^{2^{n-3}}\not \equiv -1 \mod 2^n$ for any $n\geq 3$ I am in the middle of solving an abstract algebra problem and I have narrowed it down to showing that
$$5^{2^{n-3}}\not \equiv -1 \mod 2^n \quad \text{ for any } \quad n\geq 3. $$
I am not very good with congruences. I know this is equivalent to showing that $5^{2^{n-3}}+1\not \equiv 0\mod 2^n$, i.e., $5^{2^{n-3}}$ is not a multiple of $2^n$, but I am still not sure how to prove this.
Thank you for your help.
| prove $5^{2^{n-3}}\not\equiv -1$ mod $2^{n}$, $n\ge 3$
for $n=3$:$\space 2^3=8\nmid 5^{2^0}+1=6$
for $n=4$:$\space 2^4=16\nmid 5^{2^1}+1=26$
for $n\ge 5$: let $m=n-3$:$\space$show that $5^{2^m}\not\equiv -1$ mod $2^{m+3}$, $m\ge 2$
that is, show $2^m2^3\nmid 5^{2^m}+1$
$2^m\nmid 5^{2^m}+1\Longrightarrow 2^m2^3\nmid 5^{2^m}+1$
$5^{2^m}+1=(4+1)^{2^m}+1=\sum_{k=0}^{2^m}{{2^m}\choose k}4^k1^{2^m-k}+1\space$ per binomial theorem.
$=1+{{2^m}\choose 1}4+{{2^m}\choose 2}4^2+\dots +{{2^m}\choose {2^m-1}}4^{2^m-1}+4^{2^m}+1$
$=1+\underbrace{{{2^m}\choose 1}4+{{2^m}\choose 2}4^2+\dots +{{2^m}\choose {2^m-1}}4^{2^m-1}+2^{2^m}2^{2^m}}_{divisible\space by\space 2^m}+1$
$=2^mk+2$ which is not divisible by $2^m$ for $m\gt 1$
therefore $5^{2^m}\not\equiv -1$ mod $2^{m+3}$, $m\ge 2$ and thus $5^{2^{n-3}}\not\equiv -1$ mod $2^{n}$, $n\ge 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Trying to prove that this series of functions do not uniformly converge Consider the series $f_n(x) = \sum_{1}^{n}\frac{nx^2}{n^3+x^3}$ I'm trying to prove that this series does not uniformly converge on $[0,\infty)$
Let $$\lim_{n \to \infty} \sum_{1}^{n}\frac{nx^2}{n^3+x^3} = F(x)$$
While disproving the book has said :
$\lim_{x \to \infty} F(x) = \infty$ while $\lim_{x \to \infty} \frac{nx^2}{n^3+x^3} = 0$ so it can not uniformly converge. My question is how did it deduce that $\lim_{x \to \infty}F(x)= \infty$ I couldn't get that?
| It is easy to verify that $a_k=\frac{kx^2}{k^3+x^3}$ is nonnegative and decreasing sequence, so the integral test tells us:
\begin{align} \int_1^\infty \frac{tx^2\mathrm d t}{t^3+x^3}\leq \sum\limits_{k=1}^{\infty}\frac{kx^2}{k^3+x^3} =F(x) \end{align}
Note that
\begin{align} \int_1^\infty \frac{tx^2\mathrm d t}{(t+x)^3} \leq\int_1^\infty \frac{tx^2\mathrm d t}{t^3+x^3} \end{align}
And integration by parts yields:
\begin{align} \int_1^\infty \frac{tx^2\mathrm d t}{(t+x)^3} =\frac{x^3+2x^2} {2(1+x)^2} \end{align}
The inequality becomes:
\begin{align} \frac{x^3+2x^2} {2(1+x)^2}\leq\sum\limits_{k=1}^{\infty}\frac{kx^2}{k^3+x^3} =F(x) \end{align}
It holds for all $x\in[0, \infty) $, so
\begin{align} \lim\limits_{x \to \infty}\frac{x^3+2x^2} {2(1+x)^2}\leq\lim\limits_{x \to \infty} F(x) \end{align}
Since $\lim\limits_{x \to \infty}\frac{x^3+2x^2} {2(1+x)^2}=\infty$, we have $\lim\limits_{x \to \infty} F(x) =\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that graph of $\tan (\tan^{-1} x) $is a straight line?
How to prove that graph of $\tan (\tan^{-1} x) $is a straight line ?
$y=\tan(\tan^{-1} x) \Rightarrow y'=\sec^2(\tan^{-1} x)\cdot \frac{1}{x^2+1}$
but how can $y'=1$ OR = some constant $k$ be proved?
what I did is:
$y'=\left[\tan^2(\tan^{-1}x)+1\cdot \frac{1} {x^2+1}\right]=\left[(\tan(\tan^{-1}x))^2+1)
\cdot \frac{1} {x^2+1}\right]=\frac{y^2+1}{ x^2+1}$
What next?
| $y = tan(tan^{-1}x)$
$y = x$
$dy/dx = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| x+\sqrt{x^2+a^2}\,\biggr| +C$ I want to show that
$$\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| x+\sqrt{x^2+a^2}\,\biggr| +C$$
and get a slightly incorrect result and I wonder what I am doing wrong.
I let $x = a\tan\theta$ so $dx=a\sec^2\theta\, d\theta$ such that I perform the following operations,
$$\frac aa \int \frac {\sec^2\theta}{\sqrt{\tan^2\theta + 1}}d\theta\;=\; \int \sec\theta\,d\theta$$
Multiplying the integrand by $1 = \frac {\sec\theta+\tan\theta}{\sec\theta+\tan\theta}$ I get
$$\int \sec\theta\, d\theta \;=\; \int \frac {\sec^2\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta} d\theta$$
Then I let $u=\sec\theta+\tan\theta$ and $du = \sec^2\theta+\sec\theta\tan\theta$ such that I get a $\int \frac 1u du$ integral that yields
$$\int \sec\theta \, d\theta \;=\;\ln|\sec\theta+\tan\theta|+C$$
Getting back in $x$ variable terms I get,
$$\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| \frac 1a \left(x+\sqrt{x^2+a^2}\right)\biggr| +C$$
It seems to me that this $\frac 1a$ has to be there according to the substitution $x=a\tan\theta$, the adjacent side to the angle $\theta$ is of length $a$. Then, when substituting back to $x$ I get $\sec\theta= \frac {x^2+a^2}a$ and $\tan\theta = \frac xa$. What am I missing?
| $$\ln \frac{(x+\sqrt{x^2+a^2})}{a} + C$$
$$= \ln (x+\sqrt{x^2+a^2}) - \ln a + C$$
$$= \ln (x+\sqrt{x^2+a^2}) + C'$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The equation $a^x +a^{-x}=b$ Given $a$ is a fixed real number that is greater than one, how many real numbers $b$ are there such that the equation $^{x} + a^{-x} = $ has a unique real solution x?
To solve this problem, I tried to plug in $a = 3$.
So, I want to find how many reals $b$ there are such that $3^{x} + 3^{-x} = b$ has a unique real solution x.
Setting $3^x$ = $t$, I got that $t + \frac{1}{t} = b$. It follows that $t^2 - tb + 1 = 0$. From this, it can be obtained that $t = \frac{b \pm \sqrt{b^2-4}}{2}$.
I set the discriminant equal to zero: $b^{2}-4=0$, so $b = \pm 2$. This means that $t = \pm 1$. Since $t = 3^{x}$ (previously defined), $x$ is real only when $t = 1$.
From the work, doesn't it follow that there is only one real number $b$ (that is when $b$ = 2) such that there is one unique real solution $x$? But the answer I see says there are an infinite values of $b$ such that there is one real solution x.
What am I missing in my analysis?
| If we let $y=a^x$, this becomes
$$y+\frac1y=b,~~y>0$$
$$y^2-by+1=0$$
$$y=\frac{b\pm\sqrt{b^2-4}}2$$
For there to be one solution, $b^2-4=0$, so $b=\pm2$. Clearly, $b>0$, since $y>0$, so the only option is $b=2$.
Alternatively, one could note that for every $x$ that satisfies the equality, $-x$ satisfies the equality, hence one must have $x=0$, and thus $b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the limit without L'hopital rule $\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=$? Find the limit without L'hopital rule
$$\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=?$$
My Try:
$$1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}\\\sin (πx)=\sin (\pi-\pi x)=-\sin \pi(x-1)\\\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=\lim_{ x \to 1}\frac{\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}}{-\sin \pi(x-1)}\\u=x-1⇒x=u+1\\\lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{π}{4}(u+1))}}{-\sin \pi u}$$
now ?
| With your attempt
$$
1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}
$$
you can go further
\begin{align}
\frac{1-\cot( \frac{π}{4}x)}{\sin\pi x}&=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)\sin\pi x}\\
&=\frac{\cos\frac{\pi}{4}x(\tan( \frac{π}{4}x)-1)}{\sin( \frac{π}{4}x)\sin\pi x}\\
&=\frac{\sin\frac{\pi}{4}x-\cos\frac{\pi}{4}x}{\sin( \frac{π}{4}x)\sin\pi x}\tag{1}
\end{align}
use product to sum formula for the denominator
$$\sin( \frac{π}{4}x)\sin\pi x=\frac{1}{2}\left[\cos\left(\frac{3}{4}\pi x\right)-\cos\left(\frac{5}{4}\pi x\right)\right]\tag{2}$$
by substituting $(2)$ in eq. $(1)$ compute the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Perform this integration: $\int\frac{t}{t^3 + 1}$ I am solving the following question
$$\int\frac{\sin x}{\sin^{3}x + \cos^{3}x}dx.$$
I have been able to reduce it to the following form by diving numerator and denominator by $\cos^{3}x$ and then substituting $\tan x$ for $t$ and am getting the following equation. Should Iis there any other way use partial fraction to integrate it further or
$$\int\frac{t}{t^3 + 1}dt.$$
| We have that $-1,\omega,\omega^{-1}$ are the roots of $t^3+1$. In particular, $\frac{t}{t^3+1}$ can be represented as
$$ \frac{t}{t^3+1} = \frac{A}{t+1}+\frac{B}{t-\omega}+\frac{C}{t-\omega^{-1}}$$
with $A+B+C=0$ and
$$ A = \lim_{t\to -1}\frac{t(t+1)}{t^3+1} = \lim_{t\to -1}\frac{t}{t^2-t+1} = -\frac{1}{3} $$
hence:
$$ \frac{t}{t^3+1} = -\frac{1}{3}\cdot \frac{1}{t+1}+\frac{1}{3}\cdot\frac{t+1}{t^2-t+1} $$
and
$$ \int\frac{t}{t^3+1}\,dt = C+\frac{1}{\sqrt{3}}\,\arctan\left(\frac{2t-1}{\sqrt{3}}\right)+\frac{1}{3}\,\log(1+t)-\frac{1}{6}\,\log(1-t+t^2). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is this inequality true? $(a+b)^2\leq 2(a^2+b^2)$ Why is this inequality true? $a,b$ are real numbers.
$$
(a+b)^2=a^2+2ab+b^2\leq 2(a^2+b^2)
$$
I know $(a+b)^2=a^2+2ab+b^2 \geq 0$, but then?
| $$(a-b)^2\geq0\\a^2-2ab+b^2\geq0\\a^2+b^2\geq2ab\\2a^2+2b^2\geq a^2+2ab+b^2\\2(a^2+b^2)\geq(a+b)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Using the Epsilon-Delta definition of a limit Use the definition of a limit to show that:
$$\lim_{(x,y)\to (0,0)} (x+y)\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) = 0$$
Id appreciate the thought process behind the solution, the epsilon-delta definition never makes sense in my head.
| Let $\epsilon>0$. Take $\delta=\frac{\epsilon}{2}$. Let $(x,y)\in\Bbb R^2$ such that $0<|(x,y)-(0,0)|<\delta$. This means that
$$0<\sqrt{(x-0)^2+(y-0)^2}<\delta.$$ We get
$$x^2\leq x^2+y^2<\delta^2\quad\text{and }\quad y^2\leq x^2+y^2<\delta^2.$$ Thus, $$|x|<\delta\quad\text{and}\quad |y|<\delta.$$
Therefore,
$$\begin{align}
\bigg|\bigg[(x+y)\sin\frac{1}{x}\cos\frac{1}{x}\bigg]-0\bigg|&=|x+y|\cdot\bigg|\sin\frac{1}{x}\bigg|\cdot\bigg|\cos\frac{1}{x}\bigg|\\
&\leq |x+y|\\
&\leq |x|+|y|\\
&<\delta+\delta=2\delta=\epsilon.
\end{align}
$$
Hence,
$$\lim_{(x,y)\to (0,0)} (x+y)\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.