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Basic Number Theory Question involving quadratic equations and squares Question : Let $m,n\in \mathbb{N}$ such that $2m^2 + m = 2n^2 + n$, then prove that $m - n$ and $2m + 2n + 1$
are perfect squares.
$\begin{align}2m^2-2n^2 &=n-m\\
-2(n-m)(m+n) &= n-m \\\end{align}$
If $m-n \neq 0
$\begin{align}\Rightarrow 2(m+n) &=-1 \\
\therefore 2m+2n+1 &=-1+1=0=0^2 \text{ } \blacksquare\\
\end{align}
If $m-n=0$, then $m-n=0=0^2 \text{ } \blacksquare$
Also, Let $2m^2 + m = 2n^2 + n=k$
Clearly, $m,n$ are solutions to the quadratic equation $2x^2+x-k=0$
This implies that $mn=\frac{-k}{2}$
,Implying either $m$ or $n$ is negative which is incorrect as we are given that $m,n \in N$
Therefore to avoid this contradiction it must be that $m=n$ and there exists another root to the equation $\alpha \in \mathbb{Q}$.
Thus $m-n=0=0^2 \text{ } \blacksquare $
Please just check my solution and tell me if it is correct, if not suggest changes please i wish to learn. Also my main doubt is in the first half of the proof i know it does look very sketchy that the sum of two natural numbers is negative. Thanks a lot for your help and understanding <$3$.
| $$ 2m^2 + m = 2 n^2 + n $$
$$ 2m^2 - 2n^2 + m - n = 0 $$
$$ 2(m^2-n^2) +(m-n) = 0 $$
$$ 2(m+n)(m-n) + (m-n) = 0 $$
$$ ( 2m+2n )(m-n) + 1 (m-n) = 0 $$
$$ (2m+2n+1)(m-n) = 0 $$
With integers, the left factor is odd so nonzero, thus $m=n$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $|a+1|+|2a+5|=5$ How to solve this equation?
$$|a+1|+|2a+5|=5$$
Here is my solution but it seems that my answers are wrong.
$$|a+1|+|2a+5|=5$$
Case 1:
$$a+1+2a+5=5$$
$$a = \frac{1}{3}$$
Case 2:
$$a+1-2a-5=5$$
$$a=-11$$
| Squaring both sides gives
$$(a+1)^2+|(a+1)(2a+5)|+(2a+5)^2=5^2,$$
or
$$|(a+1)(2a+5)|=25-(a^2+2a+1)-(4a^2+20a+25),$$
which gives
$$|(a+1)(2a+5)|=-5a^2-22a-1.$$
Squaring again, we obtain
$$(a+1)^2(2a+5)^2=(5a^2+22a+1)^2,$$
or
$$(2a^2+7a+5)^2=(5a^2+22a+1)^2,$$
or
$$(2a^2+7a+5)^2-(5a^2+22a+1)^2=0.$$
Factorising then gives
$$(-3a^2-15a+4)(7a^2+29a+6)=0,$$ from which you should be able to proceed.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Binomial Expansion $(1+kx)^4 (1+x)^n$
Question
In the expansion of $(1+kx)^4 (1+x)^n$, the coefficient of $x$ is $13$ and the coefficient of $x^2$ is $74$.
Find the possible values of $k$ and $n$.
My solution
$$
\begin{split}
(1+kx)^4 &= 1+4kx+6k^2 x^2+4k^3 x^3+k^4 x^4\\
(1+x)^n &= 1+ \binom{n}{1}x+\binom{n}{2} x^2...
\end{split}
$$
So $\binom{n}{1}x + 4kx=13x$ and
$\binom{n}{2}x^{2}+4k\binom{n}{1}x^{2}+6k^{2}x^{2}=74x^{2}$.
That is, $$\binom{n}{1}+4k=13 \quad (*)$$ and
$$\binom{n}{2}+4k\binom{n}{1}+6k^{2}=74 \quad (**)$$
Since $\binom{n}{1}=n$, we have $n+4k=13$.
From trial and improvement I worked out that $n = 5$ and $k = 2$ satisfies both equations (*) and (**) above.
How do I find the other set of solutions?
| The second equation gives
$$n(n-1) + 8nk + 12k^2 = 148$$
(by multiplying the equation by $2$ and noting $\binom n2 = \frac{n(n-1)}2.$
We have $n+4k=13$. Write $n = 13-4k$, and substitute this into the above equation, to obtain
$$(13-4k)(12-4k) + 8(13-4k)k + 12k^2 = 148,$$
which is a quadratic in $k$. You can now solve for $k$, and then use $n=13-4k$ to find $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate: $\int \frac{dx}{(x^2+x+1)^2},$ without using any kind of substitutions. I know there exists a reduction formula for the integral: $$\int\frac1{(ax^2+bx+c)^n}\,dx.$$ But this uses substitutions. So for the integral in question, let:
\begin{align}
I&=\int\frac{dx}{(x^2+x+1)^2}\\\\
&=\int\frac{dx}{\{(x-\omega)(x-\omega^2)\}^2}\\\\
&=\int\frac{dx}{(x-\omega)^2(x-\omega^2)^2},
\end{align}
Where, $\omega$ is one of the complex cube roots of unity. According to me, resolving $\dfrac1{(x-\omega)^2(x-\omega^2)^2}$ into partial fractions, would serve our purpose. Is this a proper way to approach? Thanks in advance.
| If your condition is integration without substitutions, using partial fractions is a proper approach but not easy in this case because computing the arbitrary constants $A, B, C, D$ is a bit laborious.
$$\frac{1}{(x-\omega)^2(x-\omega^2)^2}=\frac{A}{(x-\omega)}+\frac{B}{(x-\omega)^2}+\frac{C}{(x-\omega^2)}+\frac{D}{(x-\omega^2)^2}$$
Once constant are found, the integration without substitution is quit simple
$$\int \frac{1}{(x-\omega)^2(x-\omega^2)^2}dx$$$$=A\int\frac{dx}{(x-\omega)}+B\int\frac{dx}{(x-\omega)^2}+C\int \frac{dx}{(x-\omega^2)}+D\int \frac{dx}{(x-\omega^2)^2}$$
$$=A\ln|x-\omega|-\frac{B}{x-\omega}+C\ln|x-\omega^2|-\frac{D}{x-\omega^2}$$
Alternatively, one of easiest ways is to use reduction formula:$\int \frac{dx}{(x^2+a)^n}=\frac{x}{(2n-2)(x^2+a)^{n-1}}+\frac{2n-3}{2n-2}\int \frac{dx}{(x^2+1)^{n-1}}$ as follows
$$\int\frac{dx}{(x^2+x+1)^2}$$$$=\int\frac{d\left(x+\frac12\right)}{\left(\left(x+\frac12\right)^2+\frac34\right)^2}$$
Using reduction formula:
$$=\frac{x+\frac12}{(2\cdot 2-2)\left(\left(x+\frac{1}{2}\right)^2+\frac34\right)}+\frac{2\cdot 2-3}{(2\cdot 2-2)}\int \frac{d\left(x+\frac12\right)}{\left(x+\frac{1}{2}\right)^2+\frac34}$$
$$=\frac{2x+1}{4\left(x^2+x+1\right)}+\frac{1}{2}\frac{1}{\sqrt{3}/2}\tan^{-1}\left(\frac{x+\frac12}{\sqrt3/2}\right)+C$$
$$=\frac{2x+1}{4\left(x^2+x+1\right)}+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt3}\right)+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of Squares for $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$ $\color{red}{\textrm{Update}}$
To clarify: I hope to see a (simple) SOS solution similar to the one of the following 3-variable problem, no other assumptions e.g. $a\ge b\ge c\ge d$.
Let us see an example. In the link below, by letting $y = \mathrm{mid}(x, y, z)$,
two SOS solutions are given. They are not what I want. It is more difficult to get a SOS solution without the assumption $y = \mathrm{mid}(x, y, z)$ for that problem.
Prove $2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \le \left(x^2+y^2+z^2+3xyz\right)^2.$
Another example: Mongolian TST 2008, day 2 problem 3, Ji Chen gave a SOS solution with the assumption $x = \min(x, y, z)$.
Actually, I gave a SOS solution without any assumption.
https://artofproblemsolving.com/community/c6h205316p11219067
$\phantom{2}$
Problem 1. Let $a, b, c, d \ge 0$. Prove that $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$.
There are quite a few solutions (including my solution). Here, I am particularly interested in $\color{blue}{\textrm{(simple) Sum of Squares (SOS) solutions}}$.
Denote $f(a, b, c, d) = \mathrm{LHS} - \mathrm{RHS}$.
Then, $f(x^2, y^2, z^2, w^2) \ge 0$ for all real numbers $x, y, z, w$.
However, $f(x^2, y^2, z^2, w^2)$ may not be expressed as SOS (polynomial).
I found that $(x^2+y^2+z^2+w^2)^2f(x^2, y^2, z^2, w^2)$ may be expressed as SOS,
since numerically $(x^2+y^2+z^2+w^2)^2f(x^2, y^2, z^2, w^2) \approx u^\mathsf{T}Q u$ where $Q$ is $185\times 185$ matrix
and $u$ is a vector containing monomials in $x, y, z, w$. But $Q$ is quite large, I have not yet proceeded.
Any comments and solutions are welcome.
Relevant information
For 3-variable problem below, there are quite a few SOS solutions.
Problem 2. Let $a, b, c\ge 0$. Prove that $a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$.
My SOS solution is
\begin{align}
&a^2+b^2+c^2+2abc+1 - 2(ab+bc+ca) \\
=\ & \frac{1}{2(a+b)^2}\Big[(a^2-ac-b^2+bc-a+b)^2 +(a^2-2ab-ac+b^2-bc+a+b)^2\\
&\qquad\qquad\quad + 4ab(a-b)^2 + 4ab(c-1)^2 + 4abc(a+b-2)^2\Big].
\end{align}
FYI, my non-SOS solution for Problem 1:
WLOG, assume that $a\ge b\ge c\ge d$.
If $cd \ge 1$, then $ab \ge 1$ and $(ab-1)(cd-1) \ge 0$, i.e., $ab+cd \le abcd+1$.
Also, $bc + da + ac + bd \le \frac{b^2+c^2}{2} + \frac{d^2+a^2}{2}
+ \frac{a^2+c^2}{2} + \frac{b^2+d^2}{2} = a^2+b^2+c^2+d^2$. Add them up to get the desired result.
If $cd < 1$, let $u = a - \frac{c+d}{cd+1}, \ v = b - \frac{c+d}{cd+1}$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= u^2 + v^2 - (1-cd)uv +\frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\
&\ge 2|uv| - (1-cd) |uv| + \frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\
&\ge \frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\
&\ge \frac{2c^2d^2-c^2d^2-2cd+1}{cd+1}\\
&= \frac{(cd-1)^2}{cd+1}\\
&\ge 0.
\end{align}
(Q.E.D.)
| Since $$1+abcd\geq2\sqrt{abcd}$$ and after replacing $a$ at $a^2$, $b$ at $b^2$, $c$ at $c^2$ and $d$ at $d^2$, we need to prove that
$$a^4+b^4+c^4+d^4+2abcd\geq a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2$$ for still non-negative variables, which is true by SOS!
Indeed, let $a\geq b\geq c\geq d$.
Thus, $$a^4+b^4+c^4+d^4+2abcd-(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)=$$
$$=\frac{1}{12}\sum_{sym}(2a^4-3a^2b^2+abcd)=\frac{1}{12}\sum_{cyc}(2a^4-2a^2b^2-a^2b^2+abcd)=$$
$$=\frac{1}{12}\sum_{cyc}((a^2-b^2)^2-a^2b^2+abcd)=$$
$$=\frac{1}{24}\sum_{cyc}(2(a^2-b^2)^2-2a^2b^2+2a^2bc-2a^2bc+2abcd)=$$
$$=\frac{1}{24}\sum_{cyc}(2(a^2-b^2)^2-d^2(a-b)^2-cd(a^2-ab)-cd(b^2-ab))=$$
$$=\frac{1}{48}\sum_{cyc}(4(a^2-b^2)^2-d^2(a-b)^2-c^2(a-b)^2-cd(2a^2-2ab)-cd(2b^2-2ab))=$$
$$=\frac{1}{48}\sum_{sym}(a-b)^2(4(a+b)^2-(c+d)^2)\geq0,$$
where the last inequality is true by the standard SOS's reasoning.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.
| Characterization of all solutions (non-trivial) to the Diophantine equation $$a^2+ab+b^2=x^2\quad a,b,x\in\mathbb{Z}$$ using Algebraic Number Theory
$\mathrm{WLOG}$ assume that $\gcd(a,b)=1$.We know that $\mathbb{Z}[\omega]$ is a $\mathrm{UFD}$ where $\omega=\frac{1+\sqrt{-3}}{2}$. Now we can factorize $$a^2+ab+b^2=(a-b\omega)(a-b\omega^2)=(a-b\omega)(a+b+b\omega)$$ in $\mathbb{Z}[\omega]$. Let $a^2+ab+b^2=x^2$ for some $x\in\mathbb{N}$. If $\pi\mid(a-b\omega)$ and $\pi\mid(a+b+b\omega)$ then we can show that $\pi\mid1$ and hence $\gcd(a-b\omega,a+b+b\omega)=1$. Then by unique factorization property we conclude $\exists$ $\alpha-\beta\omega,\gamma-\delta\omega\in\mathbb{Z}[\omega]$ such that $a-b\omega=u(\alpha-\beta\omega)^2=u(\alpha^2-2\alpha\beta\omega-\beta^2(1+\omega))=u((\alpha^2-\beta^2)-(2\alpha\beta+\beta^2)\omega)$ and $a+b+b\omega=v(\gamma-\delta\omega)^2$ where $u,v$ are units and hence $u,v\in\{\pm1,\pm\omega,\pm\omega^2\}$. We consider the case $u=1$ only, the other cases gives similar solutions. Comparing we get $a=\alpha^2-\beta^2$ and $b=\beta^2+2\alpha\beta$. We see that in fact $\gamma-\delta\omega=\overline{\alpha-\beta\omega}=\alpha-\beta\omega^2$ and hence $(a^2+ab+b^2)=(\alpha^2+\alpha\beta+\beta^2)^2$. So the set $$\{(\alpha^2-\beta^2,\beta^2+2\alpha\beta,\alpha^2+\alpha\beta+\beta^2):\alpha,\beta\in\mathbb{Z},\gcd(\alpha,\beta)=1\}$$ describes all solutions $(a,b,x)$ to the Diophantine equation $$a^2+ab+b^2=x^2$$ Also we observe that $(a,b,x)$ is a solution if and only if $(a,b,-x),(-a,-b,-x),(-a,-b,x),(b,a,x),(b,a,-x),(-b,-a,-x),(-b,-a,x)$ are solutions. These solutions will be obtained by considering other units except $u=1$. These observations are trivial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=$ is given. Then $\mathrm{det(NM)}$ is?
If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=\pmatrix{8& 2 & -2\\2& 5& 4\\-2& 4&5}$, then $\mathrm{det(NM)}$ is?
($\mathrm{NM}$ is invertible.)
$\mathrm{det(MN)}$ must be (and is) zero. But how to find $\mathrm{det(NM)}$? Any hint?
| One way to proceed, not knowing the relevant theorem( new one for me as well) is to guess that, being symmetric, this is a Gram matrix. Look for integer rectangles...
$$
\left(
\begin{array}{rr}
2 & 2 \\
2 & -1 \\
1 & -2
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 2 & 1 \\
2 & -1 & -2
\end{array}
\right) =
\left(
\begin{array}{rrr}
8 & 2 & -2 \\
2 & 5 & 4 \\
-2 & 4 & 5
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr}
2 & 2 & 1 \\
2 & -1 & -2
\end{array}
\right)
\left(
\begin{array}{rr}
2 & 2 \\
2 & -1 \\
1 & -2
\end{array}
\right) =
\left(
\begin{array}{rr}
9 & 0 \\
0 & 9
\end{array}
\right)
$$
Indeed, the first characteristic polynomial is $x^3 - 18 x^2 + 81x$ and the second is $x^2 - 18 x + 81$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Deriving Law of Cosines from Law of Sines
How to eliminate $\alpha$ from the Law of Sines of plane trigonometry
$$ \dfrac{a}{\sin \alpha}= \dfrac{c}{\sin \gamma} =\dfrac{b}{\sin (\gamma+\alpha)} =2R $$
in order to arrive at the Law of Cosines
$$ c^2= a^2+b^2-2 a b \cos \gamma \;?$$
Starting to isolate $\alpha$
$$ \alpha =\sin^{-1}\big(\dfrac{b-c}{2R} +\sin \gamma \big)-\gamma$$
$$ =\sin^{-1}( b \sin \gamma/c) -\gamma$$
involve $c$ that we are finding and so on, how best to simplify?
| First, let's get the following relationship.
From $$\frac{a}{\sin\alpha}=\frac{c}{\sin\gamma}=\frac{b}{\sin(\alpha+\gamma)}$$
using componendo and dividendo,
$$\frac{a\cos\gamma+c\cos\alpha}{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma}=\frac{b}{\sin(\alpha+\gamma)}$$
$$a\cos\gamma+c\cos\alpha=b$$
However, these are unnecessary stuff since obviously,
Now, let's go back to our problem: deriving law of cosines from law of sines.
From law of sines we have, $$a\sin\gamma=c\sin\alpha$$
Squaring, $$a^2\sin^2\gamma=c^2\sin^2\alpha$$
$$a^2(1-\cos^2\gamma)=c^2(1-\cos^2\alpha)$$
$$a^2=c^2+a^2\cos^2\gamma-c^2\cos^2\alpha$$
$$a^2=c^2+(a\cos\gamma+c\cos\alpha)(a\cos\gamma-c\cos\alpha)$$
$$a^2=c^2+b\cdot(a\cos\gamma-c\cos\alpha)$$
$$a^2=c^2+b\cdot[(b-c\cos\alpha)-c\cos\alpha]$$
$$a^2=b^2+c^2-2bc\cos\alpha$$
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I can not find a way to decompose of combine.
Also, how do you prove that $x^4+y^4+z^2 \ge xyz \sqrt{8}$?
I got $x^4+y^4+z^2\geq3\sqrt[3]{x^4*y^4*z^2}=3xy\sqrt[3]{xyz^2}.$
I do not know what to do from here.
| We have$:$
$$\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1}$$
$$={\sum \frac {2\,b{c}^{2} \left( ab-1 \right) ^{2}+b{c}^{2} \left( a-1
\right) ^{2}}{ 3\left( ab+1 \right) \left( bc+1 \right) \left( ac+1
\right) }} -\frac{f(a,b,c) }{{ 3\left( ab+1
\right) \left( bc+1 \right) \left( ac+1 \right) }}\geq 0$$
where
$$f(a,b,c)= \left( abc-1 \right) \Big[2\,abc \left( a+b+c
\right) -6\,(ab+bc+ca)+3\,a+3\,b+3\,c-9 \Big] =0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When will an ellipse ‘fall’ into a parabola?
Consider the parabola $y=x^2$ and an ellipse which ‘rests’ on it, given by the equation $$\frac{x^2}{a^2} +\frac{(y-h)^2}{b^2}=1$$The goal is to find all ordered pairs $(a,b)$ for which the ellipse doesn’t fall to the origin, namely it touches the parabola at two distinct points.
Replacing $y$ by $x^2$ in the equation of the ellipse, we get $$\frac{x^2}{a^2} +\frac{(x^2-h)^2}{b^2}=1 $$
I could calculate the discriminant of this quadratic in $x^2$ and set it $\gt 0$. Is there a quick and neat way to express $h$ in terms of $a,b$? Or maybe another approach to solve this problem?
| I'd look at the problem in a different way. Suppose we are given $b = h > 0$ such that the ellipse touches the parabola's vertex. What is the largest $a$, say $a^*$, such that the ellipse has no other intersection points with the parabola? For $a > a^*$, the ellipse cannot have $h = b$, thus such an $(a,b)$ pair cannot "fall" all the way down to the vertex.
As such, we require $$\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} = 1, \\ y = x^2,$$ hence $$a^2 y^2 + (b^2 - 2a^2 b) y = 0,$$ for which the unique nonzero root in $y$ is $$y = \frac{(2a^2 - b)b}{a}.$$ Therefore, if this root exists and is positive, the ellipse has another intersection point with the parabola other than its vertex; i.e., if $2a^2 > b$ or $$a > a^* = \sqrt{b/2}.$$ It follows that the set of all ordered pairs for which the ellipse cannot touch the vertex are those pairs for which $a > \sqrt{b/2}$.
By request, there is the question of how to express $h$ as a function of $(a,b)$ for ellipses satisfying the condition $a > \sqrt{b/2}$. This involves only a slight modification of the above calculation, namely we solve the system $$\frac{x^2}{a^2} + \frac{(y-h)^2}{b^2} = 1, \\ y = x^2$$ for $y$, giving $$y = h - \frac{b^2}{2a^2} \pm \frac{b \sqrt{4a^4 + b^2 - 4a^2 h}}{2a^2}.$$ Then we note that under the assumptions $a > \sqrt{b/2}$ and $h \ge b$, we have $h - b^2/(2a^2) > h - b > 0$. When the ellipse is tangent to the parabola, the solution has a unique double root; i.e., the the discriminant $4a^2 + b^2 - 4a^2 h$ must be zero, or $$h = a^2 + \frac{b^2}{4a^2}.$$ This characterizes the location of such an ellipse when $h \ge b$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3721190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$. For $x > 1$, $x^2(x^2+1) > x(x^2+1)$, hence $P>Q$.
The part where I have the problem is I can't prove this for the range $0 < x < 1$ without the help of a graphing calculator. Can anyone help?
What I've done in this region so far is:
*
*Prove that $P$ and $Q$ is always increasing in this region,
*The range for $P$ starts from $1 < P < 3$, and
*The range for $Q$ starts from $0 < Q < 2$.
The only thing I need to prove now is that $P$ and $Q$ will not intersect at $0 < x < 1$, but I can't prove this part.
| Proving that $x^4 + x^2 +1$ is always above $x^3 + x$ si the same as proving that
$$f(x)=x^4-x^3+x^2-x+1$$ is always positive. We have
$$f'(x)=4 x^3-3 x^2+2 x-1$$ has only one real root; using the hyprbolic method, we find that its real root is given by
$$x_*=\frac{1}{12} \left(3+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(3
\sqrt{\frac{3}{5}}\right)\right)\right)\approx 0.605830$$ and the second derivative test reveals that $x_*$ corresponds to a minimum.
$$f(x_*)\approx 0.673553$$ So, it is always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Consider $\dot{x}=4x^{2}-16$. I am solving the ODE above, it is a question from Strogatz Nonlinear dynamics and chaos, chapter 2 question 2.2.1.
Question
\begin{equation}
\dot{x}=4x^{2}-16
\end{equation}
Answer
\begin{equation}
\frac{\dot{x}}{4x^{2}-16} = 1\\
\frac{\dot{x}}{x^{2}-4} = 4\\
{{dx\over dt}\over x^2-4}=4 \\
{{dx\over dt}\over x^2-4}. dt=4. dt \\
{dx\over x^2-4}=4dt \\
\int \frac{1}{x^{2}-4} dx = \int 4 dt \\
\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} \\
x = 2 \frac{1 + C_{2}e^{16t}}{1 - C_{2}e^{16t}}
\end{equation}
\begin{equation}
C_{2}(t=0) = \frac{x-2}{x+2}
\end{equation}
Summary
I am looking to understand the intermediary step in the proof above. How do we get to this step $\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} $ from the previous step. Can we remove the constant $\frac{1}{4} $ then integrate the remaining portion?
| $$I=\int \frac{dx}{x^{2}-4}$$
Partial fraction decomposition gives us:
$$I=\int \left ( \frac A {x-2}-\dfrac B {x+2} \right)dx$$
$$I=\int \frac{x(A-B)+2(A+B)}{x^{2}-4}dx$$
$$\implies A=B=\dfrac 14$$
$$I=\dfrac 14\int \left ( \frac 1 {x-2}-\dfrac 1 {x+2} \right)dx$$
Then integrate with $\ln $ function .
$$I=\dfrac 14 \ln |{x-2}|-\dfrac 14\ln |{x+2}|+C$$
$$I=\dfrac 14 \ln \left | \dfrac {x-2}{x+2} \right |+C$$
| {
"language": "en",
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"answer_count": 2,
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} |
Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2]$$
Since $\sum_{sym} x^3+3xyz-\sum_{sym} xy(x+y)\ge 0$, set $(x,y,z)\rightarrow (ab,bc,ca)$ we obtain $$\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\ge 0$$
Thus $LHS-RHS\ge 0$
I think this's a complicated solution and hard to find it by hand :">
Please give me a simplier solution. Thank you very much.
| For example $$4(a^2+b^2+c^2)^3-(a^3+b^3+c^3+3abc)^2=$$
$$=\sum_{cyc}(a^6+12a^4b^2+12a^4c^2-6a^3b^3-18a^4bc-a^2b^2c^2)\geq0,$$ which is true by Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The quotient of a quotient group by another quotient group Let $G$, $M$, and $N$ be the cyclic groups given by
$$
G \colon= \left\langle a \colon a^{12} = e \right\rangle = \left\{ e = a^0, a, a^2, \ldots, a^{11} \right\},
$$
$$
M \colon= \left\langle a^2 \right\rangle = \left\{ e, a^2, a^4, a^6, a^8, a^{10} \right\},
$$
and
$$
N \colon= \left\langle a^6 \right\rangle = \left\{ e, a^6 \right\}.
$$
Then of course $M$ is a normal subgroup of $G$, and of course $N$ is a normal subgroup of both $G$ and $M$.
Thus the quotient groups $G/M$, $G/N$, and $M/N$ are well-defined.
In fact, we have
$$
\begin{align}
G/M &= \left\{ M, aM, a^3M, a^5M, a^7M, a^9M, a^{11}M \right\} \\
&= \left\{ M, aM \right\} \\
&= \left\{ \, \left\{ e, a^2, a^4, a^6, a^8, a^{10} \right\}, \, \left\{ a, a^3, a^5, a^7, a^9, a^{11} \right\} \, \right\},
\end{align}
$$
$$
\begin{align}
G/N &= \left\{ N, aN, a^2N, a^3N, a^4N, a^5N, a^7N, a^8N, a^9N, a^{10}N, a^{11}N \right\} \\
&= \left\{ N, aN, a^2N, a^3N, a^4N, a^5N \right\} \\
&= \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^4, a^{10} \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\},
\end{align}
$$
and
$$
\begin{align}
M/N &= \left\{ N, a^2N, a^4N, a^8 N, a^{10}N \right\} \\
&= \left\{ N, a^2N, a^4 N \right\} \\
&= \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}.
\end{align}
$$
Furthermore, as $M$ is a normal subgroup of $G$, so the quotient group $M/N$ is a normal subgroup of the quotient group $G/N$.
Therefore we can consider the quotient group $(G/N)/(M/N)$.
Now my question is, is the following construction valid?
We first note that
$$
\begin{align}
\left\{ a, a^7 \right\} (M/N) &= \left\{ a, a^7 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a, a^7 \right\} \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}.
\end{align}
$$
And also
$$
\begin{align}
\left\{ a^3, a^9 \right\} (M/N) &= \left\{ a^3, a^9 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a^3, a^9 \right\} \left\{ e, a^6 \right\}, \, \left\{ a^3, a^9 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a^3, a^9 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\}, \, \left\{ a, a^7 \right\} \, \right\} \\
&= \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}.
\end{align}
$$
Thus we have shown that
$$
\left\{ a, a^7 \right\} (M/N) = \left\{ a^3, a^9 \right\} (M/N).
$$
Similarly we can show that
$$
\left\{ a, a^7 \right\} (M/N) = \left\{ a^5, a^{11} \right\} (M/N).
$$
Using the calculations above, we find that
$$
\begin{align}
(G/N)/(M/N) &= \left\{ \ M/N, \ \left\{ a, a^7 \right\} (M/N) , \ \left\{ a^3, a^9 \right\}(M/N), \ \left\{ a^5, a^{11} \right\} (M/N) \ \right\} \\
&= \left\{ \ M/N, \ \left\{ a, a^7 \right\} (M/N) \ \right\} \\
&= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ a, a^7 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \ \right\} \\
&= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ \, \left\{ a, a^7 \right\} \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \ \right\} \\
&= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\} \ \right\}.
\end{align}
$$
Of course the quotient group $(G/N)/(M/N)$ is isomorphic to the quotient group $G/M$.
Is my construction correct?
Is each and every detail of my calculation above correct? Or, have I made any errors or logical / mathematical mistakes?
Last but not the leat, is my typesetting logically correct and clear enough as well? Or, is there a better way of presenting the work above?
| Theorem: Let N,K be normal subgroups of G with $N\subseteq K\subseteq G$, then (1) N is a normal subgroup of K,(2) $K/N$ is a normal subgroup of $G/N$, and (3) $(G/N)/(K/N)\cong G/K$.
Proof:
(1): Since N is a normal subgroup of G, $N=gNg^{-1}$ for all $g\in G$. In particular, $N=kNk^{-1}$ for all $k\in K$ since $K\subseteq G$, which means that N is a normal subgroup of K.
(2): Define a surjective homomorphism $f: G\rightarrow f(G)$ such that $N=\ker f$. Then $f_K$, the restriction of $f$ onto K, also has kernel N. By the first isomorphism theorem, $G/N\cong f(G), K/N\cong f(K).$ Since f is a homomorphism and $K=gKg^{-1}$ for every $g\in G$, we have $f(K)=f(g)f(K)f(g)^{-1}$ which implies $f(K)$ is normal in $f(G)$. Therefore, $K/N$ is normal in $G/N$.
(3) Define a surjective homomorphism $\pi: G\rightarrow G/K$ canonically by $g \mapsto gK$, then it has kernel K and induces a homomorphism $ \phi: G/N\rightarrow G/K$ with $\phi(gN)=\pi(g)=gK$. Since N is normal to K, the elements in the same coset $gN$ are mapped into the same coset $gK$. Thus, $\phi$ is well-defined. Now, $\phi$ is surjective and $\ker \phi=K/N$, hence by the first isomorphism theorem we have $(G/N)/(K/N)\cong G/K$.
I probably skipped over some details, but my point is that the observation you made is actually something that works for more general objects.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Added angle formula to solve this indefinite integral $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ Starting from this very nice question Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ and the relative answers, I would to understand because this integral $$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx \tag 1$$
must be split thus:
$$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\color{red}{\int A\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +B \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; dx}$$
or it can be splitted in a different way.
Using the added angle formula (for numerator and denominator of the $(1)$) $$a\sin x+b\cos x=\lambda \sin (x+\phi)$$ if $\lambda=\sqrt{a^2+b^2}$ and $\tan \phi=b/a \ $ or $$a\sin x+b\cos x=\lambda \cos (x+\varphi)$$ with $\tan \varphi=-a/b \ $ is it possible to obtain the same result?
| Assuming you have
$$
\int\frac{a\cos x+b\sin x}{c\cos x+d\sin x}\,dx
$$
(with $ad-bc\ne0$, to avoid trivial cases) you can indeed write the denominator as $k\cos(x+\varphi)$ and do the substitution $y=x+\varphi$, so the numerator becomes
$$
a\cos\varphi\cos y-a\sin\varphi\sin y+b\cos\varphi\sin y-b\sin\varphi\cos y
$$
so the integral becomes
$$
\frac{1}{k}\int\Bigl((a\cos\varphi-b\sin\varphi)-(a\sin\varphi-b\cos\varphi)\frac{\sin y}{\cos y}\Bigr)\,dy
$$
which is elementary.
On the other hand, determining $\varphi$ is usually not possible explicitly and, at the end of the day, this is essentially the same as the metod outlined in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding All Solutions For $\sin(x) = x^2$ Hello everyone how can I find the count of the solution for $\sin(x) = x^2$?
I know there is a one solution in $x = 0$ and for the other solutions I tried to find the extreme point of the function: $y = x^2 - \sin(x)$ and $y'$ is:
$y' = 2x -\cos(x)$ but I don't know how to solve this equation.
| Just for the fun of it !
There is no explicit solution for the zero of function $$f(x)=2x -\cos(x)=0$$ If you need it, use Newton method which will converge quite fast as shown in the table
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.000000 \\
1 & 0.500000 \\
2 & 0.450627 \\
3 & 0.450184
\end{array}
\right)$$ Another solution could be a series expansion
$$2x -\cos(x)=1-2 x-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ and use series reversion to get
$$x=t-\frac{t^2}{4}+\frac{t^3}{8}-\frac{11 t^4}{192}+\frac{3 t^5}{128}-\frac{121
t^6}{23040}-\frac{19 t^7}{5120}+O\left(t^8\right)\quad \text{where}\quad t=\frac{1-f(x)}2$$ Making $f(x)=0$ that is to say $t=\frac 12$, you should get, as an approximation,
$$x =\frac{531037}{1179648}\approx 0.450166$$
Amazing would be to use the $\color{red}{1,400}$ years old approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ which would lead to the cubic equation
$$2 x^3+4 x^2+2 \pi ^2 x-\pi ^2=0$$
$$x=-\frac{2}{3} \left(1-\sqrt{3 \pi ^2-4} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{63 \pi ^2-32}{4 \left(3 \pi
^2-4\right)^{3/2}}\right)\right)\right)\approx 0.449785$$
| {
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"url": "https://math.stackexchange.com/questions/3735053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Sum of geometric series when exponent is $2n$, not $n$? I have a probability below which denotes the chance of catching a fish.
$$P = \left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4 + \dots $$
I can find a generalized form of $P$ by assuming the first term is $\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^0$ to come up with an infinite series:
$$P = \sum_{n=0}^\infty \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{2n}$$
I'm learning about series now and I know that if $|r| \geq 1$ the series is divergent. Here I have $|r| = \dfrac{3}{4}$ so there is a sum. I also have $a = \dfrac{1}{4}$. In the examples, the series have an exponent of $n$ but none have $2n$. Were it to be $n$ instead, I could find the sum as:
$$P = \frac{a}{1-r} = \frac{\frac{1}{4}}{1-\frac{3}{4}} = 1$$
But this is with $n$, not $2n$, where the answer I seek is $\dfrac{4}{7}$. How can I approach this number? How do I reconcile the $2n$ and is there a formula for this like there was when it was $n$? I can do partial sums and find an approximation but what is the sum as $n \to \infty$?
I've marked the question calculus as this is a unit in Stewart's Early Transcendentals calculus text.
| Notice, you must take common ratio $r=\left(\frac{3}{4}\right)^2$
$$\therefore \ \ P = \frac{a}{1-r} = \frac{\frac{1}{4}}{1-\left(\frac{3}{4}\right)^2} = \color{blue}{\frac47}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find coordinates of point in $\mathbb{R^4}$. There is a problem here in Q. $5$ on the last page. It states to find coordinates of point $p$.
Taking point $a=(3,2,5,1), \ b=(3,4,7,1), \ c= (5,8,9,3)$.
Also, $b$ has two coordinates in common with $a$, and $p$ lies on the same line as $a,b$.
So, those two coordinates of $p$ are same as $a,b$. Hence, $p= (3,x,y,1)$; where $x,y\in \mathbb{R}$ are unknown.
Given that $\triangle acp, \triangle bcp$ are right-angled; get:
$1. \ \ \triangle acp:\ \ \ \ \ {ac}^2 = {ap}^2 + {cp}^2\implies({(-2)}^2+6^2+4^2+2^2) = ({(x-2)}^2 +{(y-5)}^2) + (2^2+{(8-x)}^2+{(9-y)}^2+{(-2)}^2 )$
$60 = 2x^2+2y^2-20x-28y+182\implies x^2+y^2-10x-14y+61=0$
$2. \ \ \triangle bcp:\ \ \ \ \ {bc}^2 = {bp}^2 + {cp}^2\implies(2^2+4^2+2^2+2^2) = ({(x-2)}^2 +{(y-5)}^2) + (2^2+{(8-x)}^2+{(9-y)}^2+{(-2)}^2 )$
$28 = 2x^2+2y^2-20x-28y+190\implies x^2+y^2-12x-16y+95=0$
From $1,2$, get: $-2x -2y +34 = 0\implies x +y -17=0$.
But, how to proceed it further to find coordinates of $p$ is unclear.
| Using the property of inner product directly is neater.
You can still use the information that $a, b, p$ is collinear.
$$\frac{x-2}{4-2}=\frac{y-5}{7-5}$$
$$x-2=y-5$$
$$y-x=3$$
Along with what you found
$$x+y=17$$
We have $y=10$, $x=7$.
Hence $p=(3,7,10,1).$
Remark: the picture also illustrated that the two circles intersects at two points, hence you still have to rule out one of them say by using collinearity.
| {
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How to simplify $\cos(\frac{1}{3}\cos^{-1}(x))$ and $\sin(\frac{1}{3}\cos^{-1}(x))$? How to simplify this trigonometric expression
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right)
$$
I want to write this equation as
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right) = \cos(\cos^{-1}(B)) = B
$$
Where $B$ is written in terms of $b$.
I couldn't find any identity that would let me do it.
This term appears in the solution of the depressed cubic equation as given in wikipedia.
In the original equation, the term is like
$$
\cos\left(\frac{1}{3}\cos^{-1}(b) - \frac{2\pi n}{3}\right) \\
= \cos\left(\frac{1}{3}\cos^{-1}(b)\right)\cos\left(\frac{2\pi n}{3}\right)+\sin\left(\frac{1}{3}\cos^{-1}(b)\right)\sin\left(\frac{2\pi n}{3}\right)
$$
Here the only term that are bothering me are only those $\cos\left(\frac13\cos^{-1}(b)\right)$ and $\sin\left(\frac13\cos^{-1}(b)\right)$
| Let $y=\frac13\cos^{-1}(b)$ then $\cos(3y)=b$. Which gives $4\cos^3y-3\cos y=b$. Hence your $B$ is a solution of $
4x^3-3x-b=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1})$.
I am stuck at this point and I cannot prove that $\log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}) > 0$.
I know by the definition $\log (1+\frac{1}{n}) =\int_1^{1+\frac{1}{n}}\frac{1}{x}dx$.
But I can only reach to the point where $ \frac{1}{n+1} < \log (1+\frac{1}{n}) < \frac{1}{n}$.
Thanks for your help in advance.
| The derivative of the function being
$$f(n)=\log \left(1+\frac{1}{n}\right)-\frac{n+\frac{1}{2}}{\left(1+\frac{1}{n}\right)
n^2}$$
$f(1)=\log (2)-\frac{3}{4} <0$ and when $n$ is large
$$f(n)=-\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
Morover
$$f'(n)=\frac{1}{2 n^2 (n+1)^2}$$ make that the function never changes concavity.
Then, the function is always decreasing.
| {
"language": "en",
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"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Problem with Summation of series Question: What is the value of $$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3} ...$$ up to infinite terms?
Answer: $\frac{13}{36}$
My Approach:
I first find out the general term ($T_n$)$${T_n}=\frac{1}{(n+2)^2+n}=\frac{1}{n^2+5n+4}=\frac{1}{(n+4)(n+1)}=\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right)$$
Using this, I get,$$T_1=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$$ $$T_2=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}\right)$$ $$T_3=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)$$
I notice right away that the series does not condense into a telescopic series. How do I proceed further?
| You were so close of solving it!
The final step is of the form
$$\frac{1}{3}\sum_{n=1}^\infty{\frac{1}{n+1}-\frac{1}{n+4}}$$
Writing down the series
$$\require{cancel} \frac{1}{3}\left[\left(\frac{1}{2}-\cancel{\frac{1}{5}}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7})\right)+\left(\cancel{\frac{1}{5}}-\frac{1}{8})\right)\right]$$
So we get the first, second and third positive term and the last three, since we are evaluating it up to infinity, these last ones will be zero and we end up with
$$\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{1}{3}\left(\frac{13}{12}\right)=\boxed{\frac{13}{36}}$$
| {
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} |
How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n+2)^2}=-\dfrac{1}{2(n+2)^2}<0$$
then this integral method can't solve (1),so How to prove it?Thanks
| There is a rather elegant solution, which uses the convexity of the function. Since $f(x) = \frac{1}{x^2}$ is convex, let us consider an area element under the curve versus the area of the trapezoid formed by joining $(n+2,f(n+2))$ and $(n+3,f(n+3)$ by a straight line
$$dA < \frac{1}{2}\left(\frac{1}{(n+2)^2} + \frac{1}{(n+3)^2}\right)$$
$$\implies \int_{n+2}^\infty \frac{1}{x^2}dx < \frac{1}{2(n+2)^2} + \sum_{n+3}^\infty \frac{1}{i^2}$$
$$\implies \frac{1}{n+2} + \frac{1}{2(n+2)^2} < \sum_{n+2}^\infty\frac{1}{i^2}$$
EDIT
Added an illustration of the above just to clarify how I got the inequality
| {
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} |
how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x}=\dfrac{1/2}{x-3}+\dfrac{-1/3}{x}=\dfrac{-1/6}{x+3}$$
$$\int \left(\dfrac{1}{2(x-3)}- \dfrac{1}{3x}- \dfrac{1}{6(x+3)}\right)\ dx$$
$$=\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C$$
I am not sure if my answer is correct. my question is can I use substitution to solve above integration? if yes please help me solve it by substitution. thanks
| Your working is correct. Use absolute values for logarithms in final answer
For substitution, let $x=3\sec\theta\implies dx=3\sec\theta\tan\theta d\ \theta $
$$\int \dfrac{2x+3}{x^3-9x}dx=\int \frac{2\sec\theta+3}{3\sec\theta(9\tan^2\theta)}3\sec\theta\tan\theta d\ \theta $$
$$=\frac19\int \frac{2\sec\theta+3}{\tan\theta} d\theta$$
$$=\frac19\int (2\csc\theta+3\cot\theta) d\theta$$
$$=\frac29\int \csc\theta d\theta+\frac13\int \cot\theta d\theta$$
$$=\frac29\ln\left|\csc\theta-\cot\theta\right|+\frac13\ln|\sin\theta|+C$$
Substituting back to $x$,
$$=\frac1{9}\ln\left|\frac{x-3}{x+3}\right|+\frac16\ln\left|\frac{x^2-9}{x^2}\right|+C$$
| {
"language": "en",
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"answer_count": 3,
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} |
Problematic solution related to finding the function range
The Question: Find the range of the following function
$$y=\dfrac{x}{2}+\dfrac{8}{x}$$
Solution $-1.$ (the solution given to me)
By Cauchy inequality,
$$\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$ where $x>0$ $$\dfrac{x}{2}+\dfrac{8}{x}≤-2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=-4$$ where $x<0$ which implies $y \in(-\infty, -4] ∪ [4, +\infty).$
But, as far as I know, we don't define inequality of arithmetic and geometric means for negative numbers.For this reason, I strange this mathematical way.
$$\dfrac{x}{2}+\dfrac{8}{x}≤-2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=-4$$ where $x<0.$
Okay, If our equation were equal to a
$$y=\dfrac{x-2}{2}+\dfrac{8}{2-x}$$ or $$ y=\dfrac{x}{2}+\dfrac{8}{|x|}$$ then we can not apply,
$$y=\dfrac{x-2}{2}+\dfrac{8}{2-x} \leq -2\sqrt{ \dfrac{x-2}{2}× \dfrac{8}{2-x}} \in {\emptyset}.$$
$$ y=\dfrac{x}{2}+\dfrac{8}{|x|} ≤-2\sqrt{ \dfrac{x}{2} × \dfrac{8}{|x|}} \in {\emptyset}.$$ where $x<0$.
What I mean,
For $x<0$ ,the arithmetic meaning of $\dfrac{\dfrac{x}{2}+\dfrac{8}{x}}2$ doesn't exist and the geometric meaning of $\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}$ doesn't exist.
So, for $x<0$ to write the $\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}$, I think, it doesn't make sense.
I would go on like this.
$$y=\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$ where $x>0$
Then, for $x<0$ we have both $\dfrac{x}{2}$ and $\dfrac{8}{x}$ are negative. In this sense we can write
$$\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$
$$-\left(\dfrac{x}{2}+\dfrac{8}{x}\right)\leq-4$$
$$- \dfrac{x}{2}+\left(-\dfrac{8}{x}\right) \leq-4$$ where $x>0$.
It seems more sense to me.
So we get, $y \in(-\infty, -4] ∪ [4, +\infty)$
I don't know how right I am.
My solution:
$$\begin{align} y=\dfrac{x}{2}+\dfrac{8}{x} \Longrightarrow 2yx=x^2+16 \Longrightarrow x^2-2yx+16=0 \Longrightarrow \Delta=y^2-16 \geq0 \Longrightarrow y \in(-\infty, -4] ∪ [4, +\infty). \end{align}$$
*
*Question $-1$ :Do you find the solution $-1$ perfect?
*Question$-2$ :Is my own solution correct?
Remark.
If our function were as follows, we could easily apply the arithmetic-geometric inequality.
$$y=\dfrac{x^2}{2}+\dfrac{8}{x^2}$$
$$y=\dfrac{x^2}{2}+\dfrac{8}{x^2}≥2\sqrt{ \dfrac{x^2}{2}× \dfrac{8}{x^2}}=4$$
$$y \in [4, +\infty)$$
|
*
*Question $-1$ :Do you find the solution $-1$ perfect?
No, I don't.
Firstly, the inequality in solution $-1$ is correct.
If $x\lt 0$, then by the inequality of arithmetic and geometric means, we have
$$\frac x2+\frac 8x=-\bigg(\frac{-x}{2}+\frac{8}{-x}\bigg)\le -2\sqrt{\frac{-x}{2}\times\frac{8}{-x}}=-4$$
So, the inequality in solution $-1$ is correct.
Secondly, however, solution $-1$ is not correct because it does not prove that the range is $y \in(-\infty, -4] ∪ [4, +\infty)$. Solution $-1$ proves that if $x\gt 0$, then $y\ge 4$, and if $x\lt 0$, then $y\le -4$. This does not imply that the range is $y \in(-\infty, -4] ∪ [4, +\infty)$. Solution $-1$ does not prove that $y$ can take every value in $(-\infty, -4] ∪ [4, +\infty)$.
*
*Question$-2$ :Is my own solution correct?
Yes, it is.
(I would add some words as follows : $y=\dfrac{x}{2}+\dfrac{8}{x}$ is equivalent to $x^2-2yx+16=0$. There is at least one $x$ satisfying this quadratic equation on $x$ if and only if the discriminant is non-negative, i.e. $y \in(-\infty, -4] ∪ [4, +\infty)$.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$ Insane integral $$\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln x} \mathop{dx}$$
I know $x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}$ but does it help? I think $u=\ln{x}$ might be necessary some point.
| (edit for a little glitch in writing)
so, basically you want to solve
$$
\int_{0}^{1} \frac{(x^2-1)(x-1)}{(x^5-1)\ln x} \mathrm{d}x
$$
let
$$
I(s) = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{(x^5-1)\ln x} \mathrm{d}x
$$
take derivative of which
$$
\begin{aligned}
I'(s) & = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{x^5-1} \mathrm{d}x = \int_{0}^{1} \frac{x^{s+3}-x^{s+2}-x^{s+1}+x^{s}}{x^5-1} \mathrm{d}x\\
& = \frac1{5} \int_{0}^{1} \frac{x^{\tfrac{s-1}{5}}-x^{\tfrac{s-2}{5}}-x^{\tfrac{s-3}{5}}+x^{\tfrac{s-4}{5}}}{x-1} \mathrm{d}x
\end{aligned}
$$
where you take $x^5\to x$. by recalling the representation of digamma function
$$
\psi(s+1) = -\gamma + \int_{0}^{1} {\frac{x^{s}-1}{x-1} \mathrm{d}x}
$$
we have
$$
I'(s)=\frac1{5}\left(\psi\left(\frac{s+1}{5}\right)+\psi\left(\frac{s+4}{5}\right)-\psi\left(\frac{s+2}{5}\right)-\psi\left(\frac{s+3}{5}\right)\right)
$$
with $\lim_{s\to\infty}I(s)=0$ and asymptotic expansion
$$\ln\Gamma(z) = (z-\tfrac1{2})\ln z - z + \ln2\pi + \tfrac1{12z} + o(z^{-3})$$
to anti-derivative
$$
I(0) = -\int_{0}^{\infty}I'(s)\,\mathrm{d}s = -\ln\left.\left(\frac{\Gamma\left(\frac{s+1}{5}\right)\Gamma\left(\frac{s+4}{5}\right)}{\Gamma\left(\frac{s+2}{5}\right)\Gamma\left(\frac{s+3}{5}\right)}\right)\right|_{s=0}^{\infty} = \ln\left(\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}\right)
$$
where using the reflection formula to obtain final answer
$$
\int_{0}^{1} \frac{x^2-1}{(x^4+x^3+x^2+x+1)\ln x} \mathrm{d}x = \ln\left(\frac{\sqrt5+1}{2}\right)
$$
| {
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} |
Find value of $\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$ My attempt :
$$\dfrac{\left(1+\tan^2\dfrac{5\pi}{12}\right)\left(1-\tan^2\dfrac{\pi}{12}\right)}{\tan\dfrac{\pi}{12}\tan\dfrac{5\pi}{12}}$$
Change into variable form
$$\dfrac{(1+a^2)(1-b^2)}{ab}$$
$$\dfrac{1+a^2-b^2-a^2b^2}{ab}$$
I'm stuck here also I don't think this is the correct way.
| We can write
$$ F=\frac{(1+\tan^2(5\pi/12))(1-\tan^2(\pi/12))}{\tan(5\pi/12) \tan(\pi/12)}$$
$$\implies F=\frac{2}{\frac{2\tan(\pi/12)}{1-\tan^2(\pi/12)}}\frac{\sec^2(5\pi/12)}{\tan(5\pi/12)}.$$
$$F=2 \cot (\pi/6) \frac{\csc^2(\pi/12)}{\cot(\pi/12)}=4 \sqrt{3} \csc(\pi/6)=8\sqrt{3}$$
| {
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A concentric circles related problem. Given a circle of radius $AC= a$ with center in $C(C_x,0)$ and with $C_x>0$. Given an angle $\beta$ (between the points B C F) and a smaller circle of radius $BC =\displaystyle \frac{a+b}{2}$ , also with center in $C$ where the measure $\displaystyle b = \frac{a(1-\sin \beta)}{1+\sin \beta}$. What measure must the angle $\beta$ have for the condition $\displaystyle \left(\frac{a+b}{2}\right)\cos \beta = C_x$ to be true? Which should also mean that the point $\displaystyle F\left(\frac{(a+b)}{2}\sin \beta,\; \frac{(a+b)}{2}\cos \beta\right)$ belongs to the $y$-axis.
My attempt so far was to replace $b$ with
$\frac{a(1-\sin \beta)}{1+\sin \beta}$
in $\displaystyle \left(\frac{a+b}{2}\right)\cos \beta = C_x$
and trying to solve
$\displaystyle \left(\frac{a+\frac{a(1-\sin \beta)}{1+\sin \beta}}{2}\right)\cos \beta = C_x$
then I replaced $cos \beta$ with the sqrt of $1-sin^2\beta$
$\Biggl(\displaystyle \left(\frac{a+\frac{a(1-\sin \beta)}{1+\sin \beta}}{2}\right)\Biggl)^2(1-sin^2\beta)= C^2_x$
$\displaystyle \frac{a^2}{4}\left(1+\frac{1-sin\beta}{1+sin\beta}\right)^2\left(1-sin^2\beta\right)= C^2_x$
but at this point I got lost.
I'm wondering if I am missing a much easier solution or if I am going completely the wrong way.
UPDATE: I made a mistake in setting up the problem this morning when I posted the question, specifically in setting the value of BC. My most sincere apologies, it is now corrected.
| Let's define $\theta:=\frac12(90^\circ-\beta)$, so that $\beta=90^\circ-2\theta$, as this gives
$$b = a\frac{1-\sin\beta}{1+\sin\beta}=a\frac{1-\cos2\theta}{1+\cos2\theta}=a\frac{2\sin^2\theta}{2\cos^2\theta}=a\tan^2\theta \tag{1}$$
Then, defining $c:=C_x$ to reduce visual clutter, the target condition becomes
$$c=\frac12(a+b)\cos\beta=\frac{a}2(1+\tan^2\theta)\cdot\sin2\theta=\frac{a}{2}\cdot\frac{1}{\cos^2\theta}\cdot2\cos\theta\sin\theta=a\tan\theta \tag{2}$$
Therefore,
$$\theta=\arctan\frac{c}{a} \quad\to\quad \beta = 90^\circ-2\arctan\frac{c}{a} \tag{$\star$}$$
To see that this is consistent with @OP's answer, first note that OP's answer reduces to
$$\sin\beta = \frac{-c^2\pm \sqrt{c^4-(c^2+a^2)(c^2-a^2)}}{a^2+c^2}=\frac{-c^2\pm \sqrt{a^4}}{a^2+c^2}\quad\stackrel{\sin\beta\,\geq\,0}{\to}\quad\sin\beta=\frac{a^2-c^2}{a^2+c^2} \tag{3}$$
On the other hand, from $(\star)$,
$$\sin\beta=\cos\left(2\arctan\frac{c}{a}\right)=\cos^2\arctan\frac{c}{a}-\sin^2\arctan\frac{c}{a}=\frac{a^2}{a^2+c^2}-\frac{c^2}{a^2+c^2}\tag{4}$$
which matches $(3)$. $\square$
To address a refinement in the comments, we replace $(1)$ by
$$b=a\frac{1-\sin\beta_n}{1+\sin\beta_n} \qquad \beta_n := \frac{\beta}{n} \tag{1'}$$
for odd integer $n$. We can again define $\theta := \tfrac12(90^\circ-\beta_n)$ so that, as before, $$b = a \tan^2\theta$$
Things get more complicated with $(2)$, which we can start to write as
$$c = \frac12(a+b)\cos(n\beta_n)=\frac12a\sec^2\theta\sin(90^\circ n)\sin 2n\theta \tag{2'}$$
Of course, $\sin(90^\circ n)=\pm 1$ for odd $n$, so that's not a problem; however, expanding $\sin2n\theta$ quickly becomes one, even for $n=3$ and $n=5$. Abbreviating $\tan\theta$ with $t$, we have ...
*
*$n=3$:
$$\begin{align}
2c\cos^2\theta &= -a\sin 6\theta \\[4pt]
\to\quad2c\cos^2\theta &= - 2a\sin\theta\cos\theta\left( 4 \cos^2\theta - 1\right) \left( 4 \cos^2\theta - 3 \right) \\[4pt]
\to\quad\phantom{2\cos^2\theta} c &= - at \left(\frac{4}{1+t^2}-1\right)\left(\frac{4}{1+t^2}-3\right) \\[4pt]
\to\quad\phantom{c\cos^2\theta} 0 &= 3 a t^5 + c t^4 - 10 a t^3 + 2 c t^2 + 3 a t + c \tag{2'.3}
\end{align}$$
*$n=5$:
$$\begin{align}
2c\cos^2\theta &= a\sin 10\theta \\[4pt]
\to\quad\phantom{c\cos^2\theta} 0 &= 5at^9-ct^8-60at^7-4ct^6+126at^5 \\
&\phantom{=\;}- 6ct^4 - 60at^3 - 4ct^2 + 5at - c \tag{2'.5}
\end{align}$$
We almost-never get explicit formulas for $t=\tan\theta$ satisfying $(2'.3)$ and $(2'.5)$. Numerical methods will be required. Once appropriate values of $\theta$ are found, however, we get $\beta$ via
$$\beta = n\beta_n = n (90^\circ-2\theta) \tag{$\star$'}$$
| {
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"answer_count": 2,
"answer_id": 1
} |
Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}
{
x
}
}
$$
Let's begin then.
Using $f(x) = e^{\log{f(x)}} = \exp{\left[\log{f(x)}\right]}$ it gets to:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
x
}
\right]
}
{
x
}
}
$$
Multiplying and dividing by the right terms:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}+\sin^2{x}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Grouping the $\sin{x}$ in the first exponential function:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Applying the know limits, specifically:
$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}+\sin^2{x}\right)}}{\sin{x}+\sin^2{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}\right)}}{\sin{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\sin{x}}{x}=1}$$$$\lim \limits_{x\to0}{\left(1+\sin{x}\right)}=1$$
We should end up with:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
=
\frac
{
\exp{\left[1(1)(1)\right]}
-
\exp{\left[1(1)\right]}
}
{
0
}
= \frac{e-e}{0}
= \frac{0}{0}
$$
Undetermined form, yay. Any hint is really appreciated, and I'm really really sorry for all the spaghetti rendering, it's hard to look at.
| You should note that the expression in numerator is of the form $A-B$ with both $A, B$ tending to $e$. We can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot(\log A - \log B) $$ Since the middle factor tends to $1$ the desired limit is equal to the limit of $$e\cdot\frac{\log A-\log B} {x} =e\cdot\frac{\log(1+\sin x+\sin^2x)-\log(1+\sin x)} {\sin^2x}\cdot\frac{\sin^2x}{x^2}$$ The last factor tends to $1$ and putting $t=\sin x$ we see that desired limit equals the limit of $$e\cdot\frac{\log(1+t+t^2)-\log(1+t)}{t^2}=e\cdot\frac{\log(1+(t^2/(1+t)))}{t^2/(1+t)}\cdot\frac{1}{1+t}$$ and clearly the above tends to $e\cdot 1\cdot 1=e$.
| {
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How find the maximum of the complex number inequality
Let $z_{i}(i=1,2,\cdots,n)$ be complex numbers such that $|z_{i}|\le 1,z_{1}+z_{2}+\cdots+z_{n}=0$. Define
$$f_{n}(z_{1},z_{2},\cdots,z_{n})=|z^3_{1}+z^3_{2}+\cdots+z^3_{n}|$$
*
*If $n=4$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.
*For any positive $n$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.
I have attempted the $n=2,3$ cases, here is my try:
If $n=2$, then $f_{2}(z_1, z_2)=|z_1^3+(-z_1)^{3}|=0$.
If $n=3$, use the identity $a^3+b^3+c^3=3abc$ if $a+b+c=0$, so
$$f_{3}(z_{1},z_{2},z_{3})=3|z_{1}z_{2}z_{3}|\le 3$$
If $z_{1}=1,z_{2}=w,z_{3}=w^2$ is maximum, where $w^3=1,w\neq 1$
For the $n=4$ case, use $$(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$$
then
$$f_{4}(z_{1},z_{2},z_{3},z_{4})=3|(z_{1}+z_{2})(z_{2}+z_{3})(z_{3}+z_{1})|$$but I can't get any farther.
| Here's the asymptotic answer: $n - \max f_n$ converges to zero! Justification will be broken up into mod $3$ cases.
(Sorry, I don't have a more exact answer and this was too long to write as a comment. )
Suppose $n=3k$, then $z_{k} = \exp(k \cdot 2\pi i/3)$ gives $f_n = n$.
Suppose $n=3k+1$, the intuition is to only slightly change the previous and rely on the fact that $\cos'(0) = 0$.
Define $\Delta_n$ to be the smallest positive number such that:
$$-2k \cos(2\pi/3 + \Delta_n) = k+1$$
Choose:
$$z_{1},\ldots,z_{k} = \exp(2\pi/3 + \Delta_n)$$
$$z_{k+1},\ldots,z_{2k} = \exp(4\pi/3 - \Delta_n)$$
$$z_{2k+1},\ldots,z_{n} = 1$$
By construction, $f_n = 2k \cos(3\Delta_n) + k+1$ and the constraint is satisfied.
Clearly $\Delta_n \rightarrow 0$, so a first order Taylor approximation gives:
$$-2k \big(\cos(2\pi/3) + \Delta_n \cos'(2\pi/3) + O(\Delta_{n}^{2})\big) = k+1$$
$$\Rightarrow \Delta_n = (2k\sin(2\pi/3))^{-1} + O(k^{-2})$$
Then,
$$f_n = 2k \big(1 + 3\Delta_n\cos'(0) + O(\Delta_{n}^{2})\big) + k+1$$
$$\Rightarrow f_n = 3k+1 + O(k^{-1})$$
$$\Rightarrow n - f_n = O(k^{-1})$$
Suppose $n=3k+2$, this is the exact same intuition.
Define $\Gamma_n$ to be the smallest positive number such that:
$$-2(k+1) \cos(2\pi/3 + \Gamma_n) = k$$
Choose:
$$z_{1},\ldots,z_{k+1} = \exp(2\pi/3 - \Gamma_n)$$
$$z_{k+2},\ldots,z_{2k+2} = \exp(4\pi/3 + \Gamma_n)$$
$$z_{2k+3},\ldots,z_{n} = 1$$
By construction, $f_n = 2(k+1) \cos(3\Gamma_n) + k$ and the constraint is satisfied.
Clearly $\Gamma_n \rightarrow 0$, so a first order Taylor approximation gives:
$$-2(k+1) \big(\cos(2\pi/3) + \Gamma_n \cos'(2\pi/3) + O(\Gamma_{n}^{2})\big) = k$$
$$\Rightarrow \Gamma_n = (2k\sin(2\pi/3))^{-1} + O(k^{-2})$$
Then,
$$f_n = 2(k+1) \big(1 + 3\Gamma_n\cos'(0) + O(\Gamma_{n}^{2})\big) + k$$
$$\Rightarrow f_n = 3k+2 + O(k^{-1})$$
$$\Rightarrow n - f_n = O(k^{-1})$$
| {
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Let $f(x)$ be a polynomial of degree $8$ such that $f(r)=\frac1r$, for $r=1,2,3,\ldots,9$. Find $\frac1{f(10)}$. Question:
Let $f(x)$ be a polynomial of degree $8$ such that $f(r)=\frac1r$, for $r=1,2,3,\ldots,9$. Find $\frac1{f(10)}$.
My Approach: We know that $f(r)=\frac1r$, which implies that $$rf(r)-1=0$$
Using the information that $f(r)=\frac1r$, for $r=1,2,3,4...8,9$, we get that $1,2,3,4,5...,8,9$ are roots of the equation $rf(r)-1=0$. Which implies that $$rf(r)-1=(r-1)(r-2)(r-3)(r-4)(r-5)(r-6)(r-7)(r-8)(r-9)$$.
Putting $r=10$ in the above equation, we get, $$f(10)=\frac{1+9!}{10}$$ and $\frac{1}{f(10)}$ as $$\frac{1}{f(10)}=\frac{10}{1+9!}$$. But the answer is 5. Please help.
| You are right that $rf(r)-1$ is a polynomial of degree $9$ with zeros at $1, 2, \ldots, 9$. But that determines the polynomial only up to a constant factor, i.e.
$$
rf(r)-1=C(r-1)(r-2)(r-3)(r-4)(r-5)(r-6)(r-7)(r-8)(r-9)
$$
for some constant $C$ (which can be determined by setting $r=0$).
| {
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Calculate the determinant of $n^\text{th}$ order
Calculate the determinant of $n^\text{th}$ order:
$$
\begin{vmatrix}
1 + a_1 & 1 + a_1^2 & \dots & 1 + a_1^n \\
1 + a_2 & 1 + a_2^2 & \dots & 1 + a_2^n \\
\vdots & \vdots & \ddots & \vdots \\
1 + a_n & 1 + a_n^2 & \dots & 1 + a_n^n \\
\end{vmatrix}
$$
So whenever any two of the variables are equal the determinant becomes $0$. Therefore, it has
$$\prod_{1 \le k < i \le n} (a_i - a_k)$$
as a factor. But I haven't been able to find the rest of the factors.
Any help is appreciated.
| When after experimentation we find that the determinant is a product of some factors, it's a good idea to see if the matrix itself can be factored into a product of matrices. If the matrix $A = BC$, where the determinants of $B$ and $C$ are easily calculated, then we can recover $\det(A)$ using the identity $\det (BC) = \det (B) \det (C)$.
Here, as you noted, $\prod_{1 \leq i < j \leq n} (a_j - a_i)$ is a factor of the final determinant. It is well-known that this is the determinant of the Vandermonde matrix
$$V = \begin{bmatrix} 1 & a_1 & a_1^2 & \cdots & a_1^{n-1} \\
1 & a_2 & a_2^2 & \cdots & a_2^{n-1} \\
1 & a_3 & a_3^2 & \cdots & a_3^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & a_n & a_n^2 & \cdots & a_n^{n-1}
\end{bmatrix}$$
so this begs of whether we can factor the given matrix $A$ as $A = V B$ for some matrix $B$. And in fact, we can! In fact, we can show that with
$$B = \begin{bmatrix} 1 & 1 & 1 & \cdots & 1& (-1)^{n-1} e_n + 1 \\ 1 & 0 & 0 & \cdots & 0& (-1)^{n-2} e_{n-1} \\ 0 & 1 & 0 & \cdots & 0 & (-1)^{n-3} e_{n-2} \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
\vdots & \vdots &\vdots & \ddots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 & (-1)^0 e_1\end{bmatrix}$$
the factorization $A = V B$ holds. Here the $e_i$ denotes the degree $i$ elementary symmetric polynomial in the variables $a_1, \cdots a_n$. The correctness of the first $n - 1$ columns of $A$ resulting from this product are easy to verify. What about the pesky last column? We basically want to verify that
$$V \cdot \begin{bmatrix} (-1)^{n-1} e_n \\ (-1)^{n-2} e_{n-1} \\ \vdots \\ e_1 \end{bmatrix} = \begin{bmatrix} a_1^n \\ a_2^n \\ \vdots \\ a_n^n\end{bmatrix}$$
because if this is true, then the last column will also match, since by adding the extra $+1$ in the upper right entry of $B$ the last column will be
$$V \cdot \left(~\begin{bmatrix} (-1)^{n-1} e_n \\ (-1)^{n-2} e_{n-1} \\ \vdots \\ e_1 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}~\right) = V \cdot \begin{bmatrix} (-1)^{n-1} e_n \\ (-1)^{n-2} e_{n-1} \\ \vdots \\ e_1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = \begin{bmatrix} 1 + a_1^n \\ 1+ a_2^n \\ \vdots \\ 1 + a_n^n \end{bmatrix}$$
Note that this is equivalent to asserting
Lemma: For all of the $a_i$, the following holds: $$a_i^n = \sum_{k = 0}^{n-1} (-1)^{n - k - 1} \cdot a_i^k \cdot e_{n - k}$$
Proof of Lemma: This fact follows because the elementary symmetric polynomials are the resulting coefficients of the monic
polynomial $$p(\lambda) = \prod_{i = 1}^n (\lambda - a_i) = \lambda^n - e_1 \lambda^{n-1} + e_2 \lambda^{n-2} - \cdots + (-1)^n e_n$$ Plugging in $\lambda = a_i$, we have $$p(a_i) = \sum_{k = 0}^n (-1)^{n-k} \cdot a_i^k \cdot e_{n-k} = 0 \implies a_i^n = \sum_{k = 0}^{n-1} (-1)^{n-k - 1} \cdot a_i^k \cdot e_{n - k}$$
We've now verified our factorization is correct, so what remains is to calculate the determinant of $B$. Laplace expanding along the last column, we can see that
\begin{align*}
\det B = (-1)^{n+1} [(-1)^{n-1} \cdot e_n + e_0] &+ \sum_{k = 1}^{n-1} \left[(-1)^{n + k - 1} \cdot \det(B_{k, n}) \cdot (-1)^{n - k - 1} \cdot e_{n-k}\right] \\ \\ &= e_n + (-1)^{n-1} \cdot e_0 + \sum_{k = 1}^{n-1} \det(B_{k, n}) \cdot e_{n - k}
\end{align*}
where $B_{k, n}$ is the minor $B$ obtained by removing the $k$th row and $n$th column. It is not too difficult to verify that $\det(B_{k, n}) = (-1)^{k - 1}$ (although if it is not clear, I can post an addendum explaining why). Hence, the determinant of $B$ is
$$e_n + (-1)^{n + 1} e_0 + \sum_{k = 1}^{n-1} \det(B_{k, n}) \cdot e_{n - k} = e_n + e_{n-1} - e_{n-2} + \cdots + (-1)^{n + 1} e_0$$
So it follows that
$$\det A = \det(V) \det (B) = \left[\prod_{1 \leq i < j \leq n} (a_j - a_i)\right] \cdot (e_n + e_{n - 1} - e_{n-2} + \cdots + (-1)^{n+1} e_0)$$
verifying lhf's conjecture. (Note that the above expression also has the correct sign as well). $\square$
| {
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Evaluating $\lim\limits_{x\to \infty}\left(\frac{20^x-1}{19x}\right)^{\frac{1}{x}}$ Evaluate the limit $\lim\limits_{x\to \infty}\left(\dfrac{20^x-1}{19x}\right)^{\frac{1}{x}}$.
My Attempt
$$\lim_{x\to \infty}\left(\frac{20^x-1}{19x}\right)^{\frac{1}{x}}=\lim_{x\to \infty}\left(\frac{(1+19)^x-1}{19x}\right)^{\frac{1}{x}}\\=\lim_{x\to \infty}\left(1+\frac{x-1}{1·2}(19)+\frac{(x-1)(x-2)}{1·2·3}(19)^2+\cdots\right)^{\frac{1}{x}}$$
After this I could not proceed. The answer given is $20$.
| You can calculate the limit directly using the standard limits $\lim_{x\to \infty} x^{\frac 1x}= 1$ and $\lim_{x\to \infty} a^{\frac 1x}= 1$ for any $a>0$ as follows:
Note that
$$\frac{20}{(19x)^{\frac 1x}} =\frac{(20^x)^{\frac 1x}}{(19x)^{\frac 1x}} > \frac{(20^x-1)^{\frac 1x}}{(19x)^{\frac 1x}} >\frac{(20^x-\frac 12\cdot 20^x)^{\frac 1x}}{(19x)^{\frac 1x}} = \frac{20\cdot \left(\frac 12\right)^{\frac 1x}}{(19x)^{\frac 1x}} $$
Now, squeezing gives the desired result.
| {
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Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
$$x^2 -xz+z^2 +y^2-yz+z^2+2\sqrt{(x^2-xz+z^2)(y^2-yz+z^2)} - x^2 - xy - y^2 \geq 0$$
$$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}\geq xz+yz+xy-2z^2$$
Squaring both sides yields
$$4(x^2-xz+z^2)(y^2-yz+z^2)\geq(xy+xz+yz-2z^2)^2$$
After some expanding it becomes
$$(xy-xz-yz)^2 \geq 0$$
which completes the proof.
However, I want to ask whether the squaring in the $3^{rd}$ - $4^{th}$step is problematic. Since it is possible for $xy + xz+ yz -2z^2$ to be negative. Squaring both sides will invert the sign. Do I have to analyse the positive and negative case here seperately?
| Let there be a triangle $ABC$ with vertices as $A(x,0), B(z/2,z\sqrt{3}/2), C(-y/2,y\sqrt{3}/2),$ then $$AB=\sqrt{x^2+z^2-xz},~ BC=\sqrt{y^2+z^2-yz}, ~CA=\sqrt{x^2+y^2+xy}.$$
Finally, the triangular inequality $$AB+BC \ge CA.$$
proves the result. The equality holds when $A,B,C$ are collinear.
| {
"language": "en",
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"source": "stackexchange",
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A tough definite integral using contour integration For some reason, I am guessing that for any fixed $s_1,s_2>0$ and $\varepsilon >0$ being small, we have
\begin{align}&\quad-\int_0^\infty \frac{1}{2\pi} \log\left(\frac{(x-s_1)^2+s^2_2}{(x+s_1)^2+s^2_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1}\,dx\\&= \log\left(\frac{1+s^2_1+s^2_2+2s_2\sin(\frac{\varepsilon}{2})+2s_1\cos(\frac{\varepsilon}{2})}{1+s^2_1+s^2_2+2s_2\sin(\frac{\varepsilon}{2})-2s_1\cos(\frac{\varepsilon}{2})}\right),
\end{align}
and I believe this can be shown by some clever contour integration. However, I really didn't figure out the contour that should be used to evaluate the integral... Any help or suggestions would be greatly appreciated!
| As suggested, a contour integration technique can be used to evaluate this integral. Notice first that the integrand is an even function of $x$, then
\begin{align}
I&=- \frac{1}{2\pi}\int_0^\infty \log\left(\frac{(x-s_1)^2+s^2_2}{(x+s_1)^2+s^2_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1}\,dx\\
&=- \frac{1}{4\pi}\int_{-\infty}^\infty \log\left(\frac{(x-s_1)^2+s^2_2}{(x+s_1)^2+s^2_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1}\,dx
\end{align}
Considering the integral
\begin{equation}
J=- \frac{1}{2\pi}\int_{-\infty}^\infty \log\left(\frac{x-s_1+is_2}{x+s_1+is_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1}\,dx
\end{equation}
where the log function is defined with a branch cut between the points $−s_1−is_2$ and $s_1−is_2$ with $s_2>0$. One can show that it is purely real (see (**)). By expressing the real part (see (*)), we find $J=I$.
The function is holomorphic for $\Im x>0$ except at the poles $x_k$ of the rational fraction with $\Im (x_k)>0$. If the real axis is closed by the upper half-circle $C_R$, the integral can then be evaluated by the residue method. The $C_R$ contribution vanishes as $R\to\infty$.
Assuming $0<\varepsilon<2\pi$, the poles of interest are simple : $x_+=e^{i\varepsilon/2}$ and $x_-=-e^{-i\varepsilon/2}$. The residues are then evaluated as
\begin{align}
R_{\pm}&=\operatorname{Res}\left[ \log\left(\frac{x-s_1+is_2}{x+s_1+is_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1},x_\pm\right]\\
&= \log\left(\frac{x_\pm-s_1+is_2}{x_\pm+s_1+is_2}\right)\frac{4x_\pm\sin\varepsilon}{\left.\frac{d}{dx}\left[x^4-2x^2\cos\varepsilon +1\right]\right|_{x=x_\pm}}\\
&=\log\left(\frac{x_\pm-s_1+is_2}{x_\pm+s_1+is_2}\right)\frac{\sin\varepsilon}{x_\pm^2-\cos\varepsilon}\\
&=\mp i\log\left(\frac{x_\pm-s_1+is_2}{x_\pm+s_1+is_2}\right)
\end{align}
and thus
\begin{align}
I&=-\frac{1}{2\pi}2i\pi \sum_{\pm} R_{\pm}\\
&=-\log\left(\frac{\cos\left(\frac{\varepsilon}{2}\right)-s_1+i(s_2+\sin\left(\frac{\varepsilon}{2}\right))}{\cos\left(\frac{\varepsilon}{2}\right)+s_1+i(s_2+\sin\left(\frac{\varepsilon}{2}\right))}\right)+\log\left(\frac{-\cos\left(\frac{\varepsilon}{2}\right)-s_1+i(s_2+\sin\left(\frac{\varepsilon}{2}\right))}{-\cos\left(\frac{\varepsilon}{2}\right)+s_1+i(s_2+\sin\left(\frac{\varepsilon}{2}\right))}\right)\\
&=-\log\left(\frac{(\cos\left(\frac{\varepsilon}{2}\right)-s_1)^2+(s_2+\sin\left(\frac{\varepsilon}{2}\right))^2}{(\cos\left(\frac{\varepsilon}{2}\right)+s_1)^2+(s_2+\sin\left(\frac{\varepsilon}{2}\right))^2}\right)
\end{align}
Finally,
\begin{equation}
I= \log\left(\frac{1+s^2_1+s^2_2+2s_2\sin(\frac{\varepsilon}{2})+2s_1\cos(\frac{\varepsilon}{2})}{1+s^2_1+s^2_2+2s_2\sin(\frac{\varepsilon}{2})-2s_1\cos(\frac{\varepsilon}{2})}\right)
\end{equation}
as proposed.
(*): using $\log\left( Z \right)=\frac{1}{2}\log\left|Z\right|^2+i\operatorname{Arg}(Z)$
(**): If
\begin{equation}
J=\int_{-\infty}^\infty \log\left(\frac{x-s_1+is_2}{x+s_1+is_2}\right)f(x)\,dx
\end{equation}
where $f(-x)=-f(x)$ and $s_{1,2}$ are real, then the complex conjugate
\begin{align}
J^*&=\int_{-\infty}^\infty \log\left(\frac{x-s_1-is_2}{x+s_1-is_2}\right)f(x)\,dx\\
&=\int_{-\infty}^\infty \log\left(\frac{-x+s_1+is_2}{-x-s_1+is_2}\right)f(x)\,dx\\
&=\int_{-\infty}^\infty \log\left(\frac{y+s_1+is_2}{y-s_1+is_2}\right)f(-y)\,dy\\
&=J
\end{align}
The integral is thus real.
| {
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Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
I tried to add $\frac{1}{(k+1)^2}$ to both sides of this inequality assuming it's true for $n = k$, but eventually it resulted in a complex expression in the second side:
$$
\frac{3k}{2k+1} + \frac{1}{(k+1)^2} = \frac{3k^3+6k^2+5k+1}{(k+1)^2 (2k+1)}
$$
that I'm not sure how to finish the proof using it.
Am I going the right way? If so, how can I finish this proof?
and is there an easier way to prove this by induction without facing such confusing expressions?
Thanks.
| Now, $$\frac{3k}{2k+1}+\frac{1}{(k+1)^2}-\frac{3(k+1)}{2k+3}=\frac{k(k+2)}{(k+1)^2(2k+1)(2k+3)}>0$$ and by induction we are done!
| {
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$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$ I was reading a solution in which author used this identity without giving any hint that how it comes ...{maybe it is obvious}
$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$
by using expansion of $a^n+b^n$ i get
$2^{3^{n}}+1=(2+1)(2^{3^n-1}-2^{3^n-2}+....+1)$
but how we get above ???
| HInt:
Use $$(x+1)(x^2-x+1)=x^3+1$$, successively.
| {
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How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdot 5^3 \cdot 2^4\,.$$ Let $p = 3^a5^b7^c2^d$ and $r = 3^e 5^f 7^g 2^s$. Notice, that given $p$ and $r$, $h$ is determined, so we can count $p$ and $r$. Multiplying $p$ and $r$ we get $$pr = 3^{(a+e)} 5^{(b+f)} 7^{(c+g)} 2^{(d+s)}\,,$$ and so $a+e = 0,1,2$.
For the first case we have $0+1 = 1$ possibility, similarly $2$ and $3$ for the other cases, so the total number is $6$. For $b+f$ we have $b + f = 0,1,2,3$ giving $10$ options. Similarly for $c + g$ we have $3$ choices and for $d + h$ we have $$1+2+3+4+5 = 15$$ choices. Multiplying these together, we get $$15\cdot 3\cdot 6\cdot 10 = 60\cdot 45 = 2700\,,$$ which is not equal to the given answer of $473$.
Edit: sorry for the weird variables. I think I've fixed everything, if not, please do point it out
| I can't really follow what's going on in your attempt - it doesn't help that you use some letters with multiple inconsistent meanings and seem to start off with HCF/LCM swapped.
However, writing $A=2^a3^b5^c7^d$ and $B=2^e3^f5^g7^h$ you have $\max(a,e)=4$, etc. This means there are $2\times 5-1=9$ choices for $(a,e)$: choose one to be $4$; the other is one of $5$ options, and subtract $1$ because you just counted $4,4$ twice.
So the number of ordered pairs is $9\times7\times 5\times 3$. This is $1$ less than twice the number of unordered pairs, since every pair can be ordered in two ways except for $(126000,126000)$.
| {
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"url": "https://math.stackexchange.com/questions/3760396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.
Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.
| The domain gives $0\leq x\leq2$.
Now, $$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{\left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}}\right)^2}=$$
$$=\sqrt{2+2\sqrt{1-2x+x^2}}=\sqrt{2+2|x-1|}.$$
Thus, it's enough to solve $$1+|x-1|=2-x.$$
Can you end it now?
I got $0\leq x\leq1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} + 2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$ How can I compute the sum $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} +
2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$? I think I should expand $(1+ \sqrt{2})^n$ or something like this and then find some kind of linear recurrence, but I'm not sure.
| The expression looks like it has something to do with binomial expansion of $\left(x+1\right)^n=\sum_{k=0}^{n}{\binom{n}{k}x^k}$ evaluated at $x=2$ but with each term being integrated. So we need to integrate both sides with respect to $x$ to get $\frac{\left(x+1\right)^{n+1}}{n+1}+c=\sum_{k=0}^{n}{\binom{n}{k}\frac{x^{k+1}}{k+1}}$
Putting $x=0$ we get $c=\frac{-1}{n+1}$.
Finally we need to evaulate the expression at $x=2$
Which will make the sum equal to $\frac{\left(3\right)^{n+1}}{n+1}-\frac{1}{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $|a + b| = |a| + |b| \iff a\overline{b} \ge 0$ I'm reading a complex analysis book. In this book, the author establishes the following statement
If $a,b \in \mathbb{C}$, then $|a + b| = |a| + |b| \iff \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right)$
This statement didn't seem intuitive to me, so I decided to try to prove it. I denoted the complex numbers $a$ and $b$ as $a = \alpha + i \beta$ and $b = \gamma + i \delta$. Using this, I get that $a\overline{b} = (\alpha \gamma + \beta\delta) + i(\beta \gamma - \alpha\delta)$, which tells us that
$$
\left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right) \iff (\alpha \gamma + \beta \delta \ge 0) \ \ \wedge \ \ (\alpha\delta= \beta \gamma )
$$
From here, I do the following
\begin{align}
&2(\alpha\delta)^2 = 2(\alpha\delta)^2 \iff 2(\alpha\delta)(\alpha\delta) = (\alpha\delta)^2 + (\alpha\delta)^2 \iff 2\alpha\delta\beta \gamma = (\alpha\delta)^2 + (\beta \gamma)^2 \notag \\
\iff& (\alpha\gamma)^2 + 2\alpha\gamma\beta \delta + (\beta \delta)^2 =(\alpha\gamma)^2 + (\alpha\delta)^2 + (\beta \gamma)^2 + (\beta \delta)^2 \iff (\alpha\gamma + \beta \delta)^2 = \left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right) \notag \\
\iff& 2(\alpha\gamma + \beta \delta) = 2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)} \qquad \ \text{(here using the hypothesis that $\alpha \gamma + \beta \delta \ge 0$)} \notag \\
\iff & \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\gamma + \beta \delta) = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 +2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)}\notag\\
\iff& \left(\alpha^2 +2\alpha\gamma + \gamma^2 \right)+ \left(\beta^2 +2 \beta \delta+ \delta^2\right) = \left(\sqrt{\alpha^2 + \beta^2}\right)^2 +2\sqrt{\alpha^2 + \beta^2}\sqrt{\gamma^2 + \delta^2} + \left(\sqrt{\gamma^2 + \delta^2}\right)^2\notag\\
\iff& (\alpha + \gamma)^2 + (\beta + \delta)^2 = \left(\sqrt{\alpha^2 + \beta^2} +\sqrt{\gamma^2 + \delta^2}\right)^2 \iff |a+ b|^2 = \left(|a| + |b|\right)^2 \iff |a+ b| = |a| + |b|
\end{align}
where in the last equivalence I used the fact that $|z|\ge 0, \ \forall z \in \mathbb{C}$.
Is my proof correct? And also, does anyone know a different (possibly shorter) method of proving the above statement? Any and all help would be greatly appreciated. Thank you!
| This could be useful:
$$ |a + b|^2 = |a|^2 +|b|^2 + 2 \mathrm{Re} (a\bar{b}) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\to 0}\dfrac{12^x\ln12-4^x\ln4}{9^x\ln9-3^x\ln3}&=\dfrac{\ln12-\ln4}{\ln9-\ln3}
\\ &=\dfrac{\ln(12/4)}{\ln(9/3)}
\\ &=\dfrac{\ln(3)}{\ln(3)}
\\ &=1
\end{align*}
My answer comes out to be $1$. Can I evaluate this limit without L'H$\hat{\mathrm{o}}$pital's rule? Thanks.
| As an alternative, you can use
$$a^{x} = e^{x \ln(a)} = 1 + x\ln(a) + \frac{x^2\ln^{2}(a)}{2!} + \mathcal{O}(x^{3})$$
therefore
\begin{align}
\frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \frac{x\,(\ln(a) - \ln(b)) + \frac{x^2}{2} \, (\ln^{2}(a) - \ln^{2}(b)) + \mathcal{O}(x^{3})}{x\,(\ln(c) - \ln(d)) + \frac{x^2}{2} \, (\ln^{2}(c) - \ln^{2}(d)) + \mathcal{O}(x^{3})} \\
&= \frac{\ln(\frac{a}{b}) + \frac{x}{2} \, \ln(a b)\,\ln(\frac{a}{b}) + \mathcal{O}(x^{2})}{\ln(\frac{c}{d}) + \frac{x}{2} \, \ln(c d)\,\ln(\frac{c}{d}) + \mathcal{O}(x^{2})}
\end{align}
Taking the limit as $x \to 0$ gives
$$\lim_{x \to 0} \frac{a^{x} - b^{x}}{c^{x} - d^{x}} = \frac{\ln(\frac{a}{b})}{\ln(\frac{c}{d})} $$
hence for your limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\frac{\ln(\frac{12}{4})}{\ln(\frac{9}{3})}=\frac{\ln(3)}{\ln(3)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Use Cauchy formula to solve $\int _0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c} $ given $\sqrt{(a^2+b^2)}=1Use Cauchy formula to solve $$\int _0^{2\pi} \frac{dt}{a \cos t+ b \sin t +c} $$ Given $\sqrt{(a^2+b^2)}=1<c$.
I tried a variable substitution, but nothing elegant. Can anyone solve it using Cauchy formula?
Edit: there are simpler solutions, but the challenge is to solve it with complex analysis.
| This can be done without taking the help of integration limits too.
Evaluate $$\int \frac{dt}{a \cos t+ b \sin t +c} $$
,where $\sqrt{(a^2+b^2)}=1<c$.
I don't know what Cauchy's formula you are talking about. But here is a nice way to do it.
Since $\sin t=\frac{2\tan(\frac x2)}{1+\tan^2(\frac x2)},\cos t=\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}$, we can write it as(You may stop and try it yourself at this point)
\begin{align*}
&\Rightarrow\int \frac{\sec^2(\frac x2)dt}{a(1-\tan^2(\frac x2))+2b\tan(\frac x2)+c(1+\tan^2(\frac x2))}\\
&=\int\frac{2du}{a(1-u^2)+2bu+c(1+u^2)}\\
&=\frac2{c-a}\int\frac{du}{\left(u+\frac{b}{c-a}\right)^2+\frac{c+a}{c-a}-\frac{b^2}{(c-a)^2}}\\
&=\frac2{c-a}\int\frac{du}{\left(u+\frac{b}{c-a}\right)^2+\frac{c^2-(a^2+b^2)}{(c-a)^2}}\\
&=\frac2{c-a}\int\frac{du}{\left(u+\frac{b}{c-a}\right)^2+K^2}\\
&=\frac2{(c-a)K^2}\int\frac{du}{\left(\frac{u+\frac{b}{c-a}}{K}\right)^2+1}\\
\end{align*}
Can you make the substitution and finish it?
| {
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"timestamp": "2023-03-29T00:00:00",
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Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.
I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that satisfy the above conditions. I was also trying to find whether there are additional requirements on $c$ and $d$ so that positive integer solutions for $a$ and $b$ can even exist.
I found that this equation simplifies to $abc - ad - bd = 0$ so that $abc = d(a+b)$.
Also, since the equation is equivalent to $a+b = ab(\frac{c}{d})$, this means $a$ and $b$ are the roots of the quadratic $dx^2-abcx+abd = 0$ since their product is $ab$ and their sum is $a+b = ab(\frac{c}{d})$.
However, after I analyzed the quadratic I just ended up with $a = a$ and $b = b$.
Any ideas on how to solve this further?
Again, I need to find all the conditions on the positive integers $c$ and $d$ (where $\gcd(c,d) = 1$) such that positive integer solutions for $a, b$ can exist. And then also find the positive integer solutions for $a$ and $b$ given that those conditions are satisfied.
| The first equation is equivalent to (ac-d)(bc-d) = d^2. From there, you can find all the ways that two numbers multiply to d^2. If you have two numbers e and f that multiply to be d^2, then you can solve ac-d = e, bc-d = f individually. This stems from Simon's Favorite Factoring Trick that another person mentioned.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)[0,1]
(C)$[-1,\frac{1}{\sqrt{2}}]$
(D)$[\frac{1}{\sqrt{2}},1]$
My approach is as follow
${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) - {\sin ^{ - 1}}x = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)}^2}} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {1 - \left( {\frac{{{x^2} + 1 - {x^2} + 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{2 - 1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
Not able to proceed from here
| The other answers pointed out some very creative ways of solving your problem, but I just wanted to complete your algebraic manipulation for you.
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt
{\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
${x\sqrt {1 - {x^2}} + 1 - {x^2} - x\sqrt
{\left( {1 - 2x\sqrt {1 - {x^2}}} \right)} } = 1$
${x\sqrt {1 - {x^2}} - x\sqrt
{\left( {1 - 2x\sqrt {1 - {x^2}}} \right)} } = x^2$
${\sqrt {1 - {x^2}} - \sqrt
{\left( {1 - 2x\sqrt {1 - {x^2}}} \right)} } = x$
(Since we divide both sides by $x$, you can verify by hand that $x = 0$ is a solution)
${\sqrt {1 - {x^2}} - \sqrt
{\left( {x - \sqrt{1 - {x^2}}} \right)^2} } = x$
${\sqrt {1 - {x^2}} - x + \sqrt{1 - {x^2}}} = x$
$2\sqrt {1 - {x^2}} = 2x$
$\sqrt {1 - {x^2}} = x$
$1 - {x^2} = x^2$
$x^2 = \frac{1}{2}$
$x = \pm \frac{1}{\sqrt{2}}$
Thus, we are left with three solutions: $\{ -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\}$
Only option choice (C) contains all three solutions found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Evaluate trigonometric integral $ \int_{0}^{\pi / 2} \frac{x^{3} \cos x }{3 \sin x-\sin 3 x}dx $ Evaluate: $$
\int_{0}^{\pi / 2} \frac{x^{3} \cos x d x}{3 \sin x-\sin 3 x}
$$
Here I can see that the denominator nicely converts into $4\sin^{3}{x}$ so I basically get $$
\int_{0}^{\pi / 2}\left(\frac{x}{\sin x}\right)^{3} \cos x\>{dx}
$$
After that I tried substituting $\sin{x}$ as $u$ but that only complicates the problem further, leaving me with an inverse function to deal with. Also, the King's property is not useful here.
Can anyone provide an alternate approach to this question?
| \begin{aligned}
& \int_{0}^{\frac{x}{2}} \frac{x^{2} \cos x}{3 \sin x-\sin 3 x} d x \\
=& \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{3 \sin x-3 \sin x+4 \sin ^{3} x} \\
=& \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{\sin ^{3} x} d x \\
=&-\frac{1}{8} \int_{0}^{\frac{\pi}{2}} x^{3} d\left(\frac{1}{\sin ^{2} x}\right) \\
=&-\left[\frac{x^{3}}{8 \sin ^{2} x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{8} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x \\
=&-\frac{\pi^{3}}{64}+\frac{3}{8} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x \\
=& \frac{3 \pi \ln 2}{8}-\frac{\pi^{3}}{64}\quad (\textrm{ Using } \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin ^{2} x} d x=\pi \ln 2)\\
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find $ED$ in the triangle $ABC$
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$
| The radius $r$ of circle inscribed in right $\Delta ABC$ with legs $AB=6$, $BC=8$ and hypotenuse $AC=10$ is given by $$r=\frac{\text{Area}}{\text{semi-perimeter}}=\frac{\frac12(6)(8)}{\frac{1}{2}(6+8+10)}=2\implies OG=OF=GB=r=2$$
$$CG=CB-GB=8-2=6\implies CD=CG=6$$
In similar right triangles $\Delta EGC\sim \Delta ABC$,
$$\frac{CE}{CA}=\frac{CG}{CB}\implies \frac{CD+DE}{10}=\frac{6}{8}$$
$$\frac{6+DE}{10}=\frac68\implies DE=\frac32$$
$$\boxed{\color{blue}{ED=\frac32}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}$$
Using $$\frac{n!}{2^n}=\frac{4!}{2^4}\frac{5}{2}\frac{6}{2}\cdots \cdots >1\Longrightarrow \frac{n!}{2^n}>1$$
So $$\frac{2^n}{n!}\leq 1\Longrightarrow \frac{2^n\cdot n^5}{n!}<n^5$$
How do i solve it Help me please, Thanks
| I don't know "you have to use integral test " or ...? but I suggest ratio test
$$\left|\frac{a_{n+1}}{a_n}\right|=
\\\dfrac{\dfrac{2^{n+1}\cdot (n+1)^5}{(n+1)!}}{\dfrac{2^n\cdot n^5}{n!}}\\=\frac{2(n+1)^5}{(n+1)n^5}\\=\frac{2(n+1)^4}{n^5}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$? I'm confused about a question I posted this morning.
I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$?
I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me closer to understanding my goal.
I noticed that I can remove the fraction inside $-6(x-\frac{1}{2})^2$ by doubling the contents:
$-6(x-\frac{1}{2})^2$ <> $-6(2x-1)^2$ # used <> for does not equal
I don't think I can simply half the factor -6 to get $-3(2x-1)^2$
As you can no doubt see, I am confused.
Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$ ?
If it does could someone show me how to transform from $-6(x-\frac{1}{2})^2$ to $-\frac{3}{2}(2x-1)^2$ in granular baby steps?
| We have
$$-6\left(x-\frac{1}{2}\right)^2=-\frac 6{\color{blue}4}\cdot \color{blue}4\cdot\left(x-\frac{1}{2}\right)^2=-\frac 3{\color{blue}2}\cdot \color{blue}{2^2}\cdot\left(x-\frac{1}{2}\right)^2=\\=-\frac 32\left(\color{blue}2\cdot x-\color{blue}2\cdot\frac{1}{2}\right)^2=-\frac 32(2x-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Conditional Probability Denominator doubt Let X be a rv such that $P(X = 2) = 1/4$ and its CDF is given by
$$F_{X}(x) =
\begin{cases}
0,& x< -3\\
\frac{3}{4}(x+3),& -3\leq x <2 \\
3/4,& 2 \leq x <4\\
\frac{3}{64} x^2,& 4 \leq x < \frac{8}{\sqrt{3}} \\
1,& x \geq \frac{8}{\sqrt{3}}
\end{cases}
$$
$x=2$ in the only discontinuity of $F$.
I have to compute $P(X<3|X \geq 2)$.
My way is: $\frac{P(2 \leq X<3)}{P(X \geq2)}$ = $\frac{F(3)- F(2) - P(X = 3) + P(X=2)}{1 - P(X<2)}$
$P(X= 3) = 0$ because it is continuous there.
How can I compute the denominator.
| Having the CDF no integral is needed.
You can solve the problem in this way
$$\mathbb{P}[Z<3|X\geq 2]=\frac{F_X(3)-F_X(2^-)}{1-F_X(2^-)}=\frac{\frac{3}{4}-[\frac{x+3}{10}]_{x=2}}{1-[\frac{x+3}{10}]_{x=2}}=\frac{1}{2}$$
The current CDF is wrong. It cannot be $\frac{3}{4}(x+3)$ when $-3 \leq x <2$ because it is $F(2^-)>1$
The previous was right...$F_X(x)=\frac{x+3}{10}$, according with $P(X=2)=\frac{1}{4}$
| {
"language": "en",
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"source": "stackexchange",
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} |
Derivatives of $ \frac{1}{r} $ and Dirac delta function I am trying to understand the formula
\begin{equation}
\nabla^2\left(\frac{1}{|{\bf r}-{\bf r}'|}\right) = - 4 \pi \delta(\bf{r}-\bf{r}'), \qquad\qquad {\rm (I)}
\end{equation}
where ${\bf r}=(x,y,z)$. This is something heavily used in electrostatics and the steps to 'show' this is often the following:
The first derivative reads
\begin{equation}
\nabla \frac{1}{| {\bf r} - {\bf r}' |} = - \frac{ {\bf r} - {\bf r}'}{| {\bf r} - {\bf r}'|^3}
\end{equation}
And taking the second derivative gives zero, except for the singularity at ${\bf r} = {\bf r'}$. Then from the divergence theorem we have
\begin{equation}
\int dV \, \nabla^2 \frac{1}{| {\bf r} - {\bf r'}|} = \int dS \,\,{\bf n} \cdot ( \nabla\frac{1}{|{\bf r} - {\bf r}'|}) = -4 \pi
\end{equation}
where the integration is performed over a sphere centered at ${\bf r}'$.
Q1: Is there a more direct proof for equation (I)?
Then my main question is about the separate second-order differentials. For instance, we can obtain, by direct computation
\begin{equation}
\partial_x^2 \, \frac{1}{| {\bf r} - {\bf r}'|} = \frac{ 3 (x-x')^2 }{| {\bf r} - {\bf r'} |^5} - \frac{1}{| {\bf r} - {\bf r}'|^3}
\end{equation}
Q2: Should there be a $\delta$ function on the r.h.s of this equation?
| A1. If you are not familiar with distribution theory, we might consider an alternative approach using the idea of approximate Dirac delta function. Indeed, define
$$ f_{\epsilon}(\mathbf{x}) = \frac{1}{\sqrt{\|\mathbf{x}\|^2+\epsilon^2}}=\frac{1}{\sqrt{x^2+y^2+z^2+\epsilon^2}}. $$
Then its Laplacian is
$$ \Delta f_{\epsilon}(\mathbf{x}) = -\frac{3\epsilon^2}{(x^2+y^2+z^2+\epsilon^2)^{5/2}}. $$
So, if $\varphi$ is any compactly supported smooth function on $\mathbb{R}^3$, then
\begin{align*}
\int_{\mathbb{R}^3} \varphi(\mathbf{x}) \Delta f_{\epsilon}(\mathbf{x}) \, \mathrm{d}\mathbf{x}
&= - \int_{\mathbb{R}^3} \varphi(\mathbf{x}) \frac{3\epsilon^2}{(x^2+y^2+z^2+\epsilon^2)^{5/2}} \, \mathrm{d}\mathbf{x} \\
&= - \int_{0}^{\infty} \int_{\mathbb{S}^2} \varphi(r\omega) \frac{3\epsilon^2 r^2}{(r^2+\epsilon^2)^{5/2}}\, \sigma(\mathrm{d}\omega)\mathrm{d}r \tag{$\mathbf{x}=r\omega$} \\
&= - \int_{0}^{\infty} \int_{\mathbb{S}^2} \varphi(\epsilon s \omega) \frac{3s^2}{(s^2+1)^{5/2}}\, \sigma(\mathrm{d}\omega)\mathrm{d}s, \tag{$r=\epsilon s$}
\end{align*}
where $\mathbb{S}^2$ is the unit sphere centered at the origin and $\sigma$ is the surface measure of $\mathbb{S}^2$. (If this sounds a bit abstract, just think of the spherical coordinates change!) Now letting $\epsilon \to 0^+$, the dominated convergence theorem tells that switching the order of limit and integration is valid in this case, hence the integral converges to
\begin{align*}
\lim_{\epsilon \to 0^+} \int_{\mathbb{R}^3} \varphi(\mathbf{x}) \Delta f_{\epsilon}(\mathbf{x}) \, \mathrm{d}\mathbf{x}
= - \int_{0}^{\infty} \int_{\mathbb{S}^2} \varphi(0) \frac{3s^2}{(s^2+1)^{5/2}}\, \sigma(\mathrm{d}\omega)\mathrm{d}s
= - 4\pi \varphi(0).
\end{align*}
Here, we utilized $\int_{\mathbb{S}^2} \sigma(\mathrm{d}\omega) = 4\pi$ and $\int_{0}^{\infty} \frac{3s^2}{(s^2+1)^{5/2}} \, \mathrm{d}s = 1$.
A2. Still using the above setting, we have
\begin{align*}
\partial^2_x f_{\epsilon}(\mathbf{x})
= \frac{2x^2-y^2-z^2-\epsilon^2}{(\|\mathbf{x}\|+\epsilon^2)^{5/2}}
= \frac{2x^2-y^2-z^2}{(\|\mathbf{x}\|^2+\epsilon^2)^{5/2}} + \frac{1}{3}\Delta f_{\epsilon}(\mathbf{x})
\end{align*}
So it suffices to analyze the contribution of the first term in the last line. To this end, note that if $B_r$ denotes the ball of radius $r$ centered at the origin, then
$$ \int_{B_r} \frac{2x^2-y^2-z^2}{(\|\mathbf{x}\|^2+\epsilon^2)^{5/2}} \, \mathrm{d}\mathbf{x} = 0 $$
by the symmetry, and so, we may write
\begin{align*}
&\int_{\mathbb{R}^3} \varphi(\mathbf{x}) \frac{2x^2-y^2-z^2}{(\|\mathbf{x}\|^2+\epsilon^2)^{5/2}} \, \mathrm{d}\mathbf{x} \\
&= \int_{\mathbb{R}^3} \left( \varphi(\mathbf{x}) - \varphi(0)\mathbf{1}_{B_r}(\mathbf{x}) \right) \frac{2x^2-y^2-z^2}{(\|\mathbf{x}\|^2+\epsilon^2)^{5/2}} \, \mathrm{d}\mathbf{x}
\end{align*}
Introducing the regularizing term $- \varphi(0)\mathbf{1}_{B_r}(\mathbf{x})$ makes the integrand decay fast enough, i.e.,
$$ \left( \varphi(\mathbf{x}) - \varphi(0)\mathbf{1}_{B_r}(\mathbf{x}) \right) (2x^2-y^2-z^2) = \mathcal{O}(\|\mathbf{x}\|^3) $$
as $\|\mathbf{x}\| \to 0$, and so, we can utilize the dominated convergence theorem to conclude that
\begin{align*}
&\lim_{\epsilon \to 0^+} \int_{\mathbb{R}^3} \varphi(\mathbf{x}) \frac{2x^2-y^2-z^2}{(\|\mathbf{x}\|^2+\epsilon^2)^{5/2}} \, \mathrm{d}\mathbf{x} \\
&= \int_{\mathbb{R}^3} \left( \varphi(\mathbf{x}) - \varphi(0)\mathbf{1}_{B_r}(\mathbf{x}) \right) \frac{2x^2-y^2-z^2}{\|\mathbf{x}\|^5} \, \mathrm{d}\mathbf{x}.
\end{align*}
This defines a distribution on $\mathbb{R}^3$ which we may write
$$ \operatorname{p.v.}\left(\frac{2x^2-y^2-z^2}{\|\mathbf{x}\|^5}\right) $$
by analogy with the Cauchy principal value in the one-dimensional setting. In conclusion, we get
$$ \partial_x^2 \frac{1}{\|\mathbf{x}\|} = \operatorname{p.v.}\left(\frac{2x^2-y^2-z^2}{\|\mathbf{x}\|^5}\right) - \frac{4\pi}{3}\delta(\mathbf{x}). $$
| {
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How to transform $z$ into $\hat z$ Assume that we have vector $a = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \in \mathbb R^n$, $b = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} \in \mathbb R^n$, $c = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} \in \mathbb R^n$.
Then, the vector $z = \begin{bmatrix} a \\ b \\ c \end{bmatrix}\in \mathbb R^{3n}$.
How to obtain the vector $\hat z = \begin{bmatrix} a \\ a \\ b \\ b \\ c \\ c \end{bmatrix}\in \mathbb R^{2\times 3n}$ based on $z$?
The Kronecker product is not useful in this case, because it focuses on elements, instead of subvectors.
| I believe the Kronecker product can be pretty much useful. You could take.
Approach 1: Take:
$$z_{1} = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\0 \\ 0\end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \quad z_{3}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\1 \end{pmatrix} $$
so that:
$$\hat{z} = z_{1}\otimes a + z_{2}\otimes b + z_{3}\otimes c$$
Approach 2: Take:
$$z_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \end{pmatrix} \quad z_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$
so that:
$$z'=\begin{pmatrix} a & b & c \\ a & b & c \end{pmatrix} = z_{1}\otimes a + z_{2}\otimes b + z_{3}\otimes c$$
and take $\hat{z}$ to be the vectorization of $z'$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $b^2-4ac$ can not be a perfect square Given $a$,$b$,$c$ are odd integers Prove that $b^2-4ac$ can not be a perfect square.
My try:Let $a=2k_1+1,b=2n+1,c=2k_2+1;n,k_1,k_2 \in I$
$b^2-4ac=(2n+1)^2-4(2k_1+1)(2k_2+1)$
$\implies b^2-4ac=4n^2+4n+1-16k_1k_2-8k_2-8k_1-4 $
| Suppose $b^2 - 4ac$ were a perfect square. Then the quadratic $ax^2 + bx + c$ has a rational root by the quadratic formula. By the rational roots theorem, the root can be written as $\frac{p}{q}$ with $p$ a factor of $c$, $q$ a factor of $a$. In particular, $p$ and $q$ are both odd. Plugging in, we have
$a (\frac{p}{q})^2 + b \frac{p}{q} + c = 0$
$a p^2 + b pq + c q^2 = 0$
All three of $ap^2$, $bpq$, and $cq^2$ are odd. And the sum of three odd numbers is odd. Thus, zero is odd. Contradiction.
Thus, $b^2 - 4ac$ can't be a perfect square.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
| Let $a=\sqrt3\tan\alpha$, $b=\sqrt3\tan\beta$ and $c=\sqrt3\tan\gamma,$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right)$.
Thus, $$\alpha+\beta+\gamma=\frac{\pi}{2}$$ and we need to prove that:
$$\ln\left((1+a^2)(1+b^2)(1+c^2)\right)\geq3\ln2$$ or
$$\ln(1+a^2)-\ln2+\ln(1+b^2)-\ln2+\ln(1+c^2)-\ln2\geq0$$ or
$$\sum_{cyc}(\ln(1+a^2)-\ln2)\geq0$$ or
$$\sum_{cyc}\left(\ln(1+3\tan^2\alpha\right)-\ln2)\geq0$$ or
$$\sum_{cyc}\left(\ln\left(1+3\tan^2\alpha\right)-\ln2-\frac{4}{\sqrt3}\left(\alpha-\frac{\pi}{6}\right)\right)\geq0,$$ which is true because easy to show that for any $x\in\left(0,\frac{\pi}{2}\right)$ we have $f(x)\geq0,$ where
$$f(x)=\ln\left(1+3\tan^2x\right)-\ln2-\frac{4}{\sqrt3}\left(x-\frac{\pi}{6}\right).$$
Also, we see that $$f''(x)=\frac{6(4\cos^4x-6\cos^2x+3)}{\cos^2x(2-\cos2x)^2}>0$$ and our inequality follows from Jensen.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bernoulli's Inequality for $-1 \leq x\leq 0$ My original goal was to prove that
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
using the squeeze theorem as we haven't seen differentiability yet and thus I cannot use arguments such as Taylor series nor Bernoulli's theorem, nor can I use induction. For that I wanted to find a lower and upper bound for $e^x$ in order to apply the squeeze theorem.
For the upper bound I used the fact that $x^n\leq x^2$ for $-1\leq x\leq 1$ and $n\geq 2$ thus one has that
\begin{align*}e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n&=\lim_{n\to\infty}\sum_{k=0}^n{n\choose k}\frac{x^k}{n^k}\\
&=\lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\\
&\leq \lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^2}{n^k}\\
&= \lim_{n\to\infty}1+x+\left(\sum_{k=2}^n{n\choose k}\frac{1}{n^k}\right)\cdot x^2\\
&= \lim_{n\to\infty}1+x+\left(\left(1+\frac{1}{n}\right)^n-2\right)\cdot x^2\\
&= 1+x+\left(e-2\right)\cdot x^2
\end{align*}
I could now potentially bound $x^n\geq -x^2$ in the same interval and obtain the bound
\begin{align*}e^x\geq 1+x-\left(e-2\right)\cdot x^2
\end{align*}
but I am not happy with it as I know that Bernoulli's inequality is stronger and gives
\begin{align*}e^x\geq 1+x.
\end{align*}
For $x\in (0,1)$ it's rather trivial to prove as
$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+\underbrace{{n\choose 2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}}_{\geq 0}\geq 1+x$$
but for $x\in(-1,0)$ the same argument does not apply straightforwardly due to the changing signs. So I modified it as follows:
For $-1\leq x\leq 0$ one has that $x^3\leq x^n$ ($x^3$ is in particular negative)
\begin{align*}
e^x&=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+{n\choose 2}\frac{x^2}{n^2}+\sum_{k=3}^n{n\choose k}\frac{x^k}{n^k}\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\sum_{k=3}^n{n\choose k}\frac{x^3}{n^k}\\
& = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left((1+\frac{1}{n})^n-\frac{n-1}{2n}-2\right)x^3\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left(e-\frac{n-1}{2n}-2\right)x^3\\
&= \lim_{n\to\infty}1+x+x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)
\end{align*}
Now we not that the cubic function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ has a double zero at $x=0$ and the remaining zero is at
$$x=-\frac{\frac{n-1}{2n}}{e-\frac{n-1}{2n}-2}\overset{n\to \infty}{\longrightarrow} -\frac{1/2}{e-1/2-2}\cong -2.29$$
thus for $n$ sufficiently large the last zero is to the left of $-1$ and hence the function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ is positive on $(-1,0)$ thus
\begin{align*}
e^x& \geq \lim_{n\to\infty}1+x+\underbrace{x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)}_{\geq 0,\quad x\in(-1,0)}\\
&\geq 1+x
\end{align*}
Since I wrote up this proof I ask: could please give it a look and tell me if there are any mistakes or if there is a shorter solution which I overlooked?
Many thanks in advance!
| You can use this supplement to Bernoulli's inequality:
If $0<x<1$, we have
$$(1-x)^n<1-nx+\frac{n(n-1)}2x^2.$$
| {
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"url": "https://math.stackexchange.com/questions/3780633",
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Counting the number of integers with given restrictions Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit numbers.
Five digit numbers:
Let $x \in \{2, 3, 4, 5\}$. If the first position in a five digit number is not $x$, then there are $5$ possibilities for this position as there are four values for $x$ and $0$ is inadmissable. The rest of the four positions will have the various permutations of four values of $x$. There are $5 \times 4!$ such numbers. If the non-$x$ value is in the second position, then there are $4$ ways to choose an $x$-value for the first position, $6$ integers for the second position and $3!$ permutations for the rest of the positions. There are $4\times 6 \times 3!$ such numbers. If the non-$x$ value is in the third position, then there are $\binom 42$ ways to choose two $x$-values, $2!$ ways to permute them and $6$ integers for the third position meaning there are $6 \times 2 \times 6$ such numbers. When the non-$x$ value is in the fourth position, there are $4 \times 3! \times 6$ such numbers. Finally, if non-$x$ is in the fifth position, there are $4! \times 6$ such numbers.
Four digit numbers:
We just need to permute the number $2345$. There are $4!$ such permutations.
Thus the number of numbers with the given restrictions is $5\times 4! + 4\times 3!\times 6 + 2\times 6 \times 6 + 4 \times 3! \times 6 + 6 \times 4! + 4! = 648$.
What did I forget to take into account? Thanks.
| To answer the question you asked: In the case where the non-$x$ value is in the third position, you missed permuting the fourth and fifth digits of the number, so that term should be $6\cdot 2\cdot 6\cdot 2$ (rather than $6\cdot 2\cdot 6$).
| {
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Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$ Let $\alpha,\beta \in \mathbb{C}$ such that $\alpha^3+\alpha+1=0$ and $\beta^2+\beta-3=0$ . Find the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
I was trying the usual trick for this kind of problems, that is
Let $\gamma = \alpha+\beta$, and therefore
\begin{align*}
(\gamma - \beta)^3 &= \alpha^3 = -\alpha - 1 \\
\gamma^3 - 3\gamma^2\beta+3\gamma\beta^2-\beta^3 &= -\alpha - 1
\end{align*}
and using $\beta^2+\beta-3=0$, i get $\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1$.
I think the minimal polynomial of $\alpha+\beta$ has degree $6$, then i was trying raise to the $2^{nd}$ power the last equation, but i don't get nothing to obtain the minimal polynomial of $\alpha + \beta$.
| Note that
*
*The polynomial $a^3+a+1$ has three distinct roots, exactly one of which is real.$\\[4pt]$
*The polynomial $b^2+b-3$ has two distinct roots, both of which are real.
By pairing one of the non-real roots of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\gamma$ has at least two distinct non-real roots.
By pairing the real root of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\gamma$ has at least two distinct real roots.
Thus the minimal polynomial of $\gamma$ has degree at least $4$, so $[\mathbb{Q}(\gamma):\mathbb{Q}]\ge 4$.
Continuing the algebra from where you left off . . .
\begin{align*}
&
\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1
\\[4pt]
\implies &
\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -(\gamma-\beta) - 1
\\[4pt]
\implies &
\beta=\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}
\\[4pt]
\implies &
\left(\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}\right)^2
+
\left(\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}\right)
-
3
=
0
\\[4pt]
\implies &
\frac{\gamma^6+3\gamma^5-4\gamma^4-11\gamma^3+25\gamma^2+52\gamma-39}{\left(3\gamma^2+3\gamma+5\right)^2}=0
\\[4pt]
\implies &
\gamma^6+3\gamma^5-4\gamma^4-11\gamma^3+25\gamma^2+52\gamma-39=0
\\[4pt]
\end{align*}
so $\gamma$ is a root of the polynomial
$$p(x)=x^6+3x^5-4x^4-11x^3+25x^2+52x-39$$
hence $4\le [\mathbb{Q}(\gamma):\mathbb{Q}]\le6$.
But from the equation
$$\beta=\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}$$
it follows that $\beta\in\mathbb{Q}(\gamma)$, hence $[\mathbb{Q}(\gamma):\mathbb{Q}]$ must be a multiple of $3$.
Thus we must have $[\mathbb{Q}(\gamma):\mathbb{Q}]=6$, hence $p(x)$ is the minimal polynomial of $\gamma$.
| {
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The value of $\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $ is $$\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $$
My working:
$$\int_0^1\ [(x+1)(x+2)(x+3)...(x+n)]\times[\frac{1}{x+1}\ + \frac{1}{x+2}\ + \frac{1}{x+3}\ +\ ...\ + \frac{1}{x+n}].dx \\$$
$\int_0^1[(x+1)(x+2)(x+3)...(x+n)]\ \times\ \frac{[(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+n-1)]}{(x+1)(x+2)(x+3)...(x+n)}.dx\\
$
Now the numerator of the left expression and the denominator of the right expression will cancel out $\\$
$\int_0^1\require{enclose}\enclose{downdiagonalstrike}{[(x+1)(x+2)(x+3)...(x+n)]}\ \times\ \frac{[(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+n-1)]}{\enclose{downdiagonalstrike}{(x+1)(x+2)(x+3)...(x+n)}}.dx$
Now we are left with
$$\int_0^1 [(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+(n-1))].dx$$
I am unable to think of anything after this step although, I have an intuition that it may have something to do with factorials.
| We have
$$P(x)= \prod\limits_{k=1}^{n}\left(x+k\right)$$
From which
$$P'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x+k\right)$$
then
$$\frac{P'(x)}{P(x)}=\sum\limits_{i=1}^{n}\frac{1}{x+i}$$
As a result
$$\int\limits_0^1 \left(\prod\limits_{r=1}^n (x+r)\right)\cdot \left(\sum_{k=1}^n \frac{1}{x+k}\right)dx =\\
\int\limits_0^1 P(x) \frac{P'(x)}{P(x)}dx=
\int\limits_0^1 P'(x)dx=\\
P(1)-P(0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Proving $\cos a-\cos b-\cos c\geq -\frac{3}{2}$, where $a+b+c=2\pi$ and $a,b,c>0$ I need help proving the following trig inequality:
$$\cos a-\cos b-\cos c\geq -\frac{3}{2}$$ where $a+b+c=2\pi$ and $a,b,c>0$.
I've found that equality occurs when $a=\frac{4\pi}{3},b=c=\frac{\pi}{3},$ but I'm not sure how to prove it in general.
I would also like to only use math covered in precalc and below, if that matters.
| We need to prove that $$\cos{a}-2\cos\frac{b+c}{2}\cos\frac{b-c}{2}+\frac{3}{2}\geq0$$ or
$$2\cos^2\frac{a}{2}-1+2\cos\frac{a}{2}\cos\frac{b-c}{2}+\frac{3}{2}\geq0$$ or
$$4\cos^2\frac{a}{2}+4\cos\frac{a}{2}\cos\frac{b-c}{2}+1\geq0$$ or
$$\left(2\cos\frac{a}{2}+\cos\frac{b-c}{2}\right)^2+\sin^2\frac{b-c}{2}\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Further, the expression $[x_1^2+x_2^2+x_3^2+x_4^2]$ can be written as $[(x_1+x_3)^2+(x_2+x_4)^2-2(x_1x_3+x_2x_4)].$ I'm having trouble eliminating $(x_1x_3+x_2x_4)$ from this expression. Failing to make any sense out of this, I manipulated the existing expressions to deduce $$x_1x_2+x_1x_4+x_2x_3+x_3x_4=\frac{a^2-b^2}{4}$$and $$(x_1^2+x_3^2)-(x_2^2+x_4^2)+2(x_1x_3-x_2x_4)=a\cdot b$$Beyond this, I cannot make sense of the expressions anymore. I have no idea how to proceed with simplifying the expressions further, and would appreciate hints in the same direction.
| Why not use a Lagrangian and find an optimal value for a constrained optimization problem?
That is,
$$
\begin{array}{cl}
\min_{x} & x^T x \\
\text{subject to} & v_1^T x = a, v_2^T x = b
\end{array}
$$
where $x = [\begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \end{array}]^T$, $v_1 = [\begin{array}{cccc} 1 & 1 & 1 & 1 \end{array}]^T$, and $v_2 = [\begin{array}{cccc} 1 & -1 & 1 & -1 \end{array}]^T$.
The Lagrangian is given by
$$
L = x^T x + \lambda_1 (a-v_1^T x) + \lambda_2 (b-v_2^T x).
$$
The gradient of $L$ is $\nabla_x L = 2x - \lambda_1 v_1 - \lambda_2 v_2$, setting it to zero gives the optimal solution
$$
x^* = \frac{\lambda_1 v_1 + \lambda_2 v_2}{2}.
$$
The solution must satisfy the constraints $v_1^T x^* = a$ and $v_2^T x^* = b$, which gives us two equations
$$
\begin{array}{ccl}
\displaystyle \frac{\lambda_1 v_1^T v_1 + \lambda_2 v_1^T v_2}{2} &=& a \\
\displaystyle \frac{\lambda_1 v_2^T v_1 + \lambda_2 v_2^T v_2}{2} &=& b.
\end{array}
$$
By solving these equations, we obtain $\lambda_1 = a/2$ and $\lambda_2 = b/2$. (Notice that $v_1^T v_2 = v_2^T v_1 = 0$ and $v_1^T v_1 = v_2^T v_2 = 4$.)
Finally, the minimum value of $x^T x$ under the constraints $v_1^T x = a$ and $v_2^T x = b$ is given by
$$
\begin{array}{ccl}
x^T x &=& \displaystyle \left(\frac{a v_1 + b v_2}{4}\right)^T \left(\frac{a v_1 + b v_2}{4}\right) \\
&=& \displaystyle \frac{a^2 + b^2}{4}.
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Given $U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$, find $\frac{100U_{10}-1}{U_8}$
If
$$U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$$
Find $\frac{100U_{10}-1}{U_8}$
Answer: $90$
My Attempt:
I tried applying Integration By Parts, and when that failed, I tried the substitution $x\rightarrow \frac{\pi}{2} -x$ , only to establish a (probably useless) relationship between the $U_n$ and its cosine counterpart:
$$U_n=\frac{\pi}{2}\int_0^{\frac{\pi}{2}} \cos^n \ dx-\int_0^{\frac{\pi}{2}}x\cos ^n x\ dx$$
Any help would be appreciated!
| Given,$$u_{n}=\int^{\frac{\pi}{2}}_{0} x \sin^{n}x\ dx=\int^{\frac{\pi}{2}}_{0} (x \cdot \sin x) \sin^{n-1}x\ dx$$
Using Integration by parts:
$$u_{n}=\left[\sin^{n-1}x(-x \cdot \cos x + \sin x)\right]^{\frac{\pi}{2}}_{0}-\int^{\frac{\pi}{2}}_{0} (-x \cdot \cos x + \sin x) (n-1)\sin^{n-2}x \cos x\ dx$$
$$\Rightarrow u_{n}=1+(n-1)\int^{\frac{\pi}{2}}_{0} (x \sin^{n-2}x\cdot \cos^{2}x - \sin^{n-1}x \cos x)\ dx$$
$$\Rightarrow u_{n}=1+(n-1)\int^{\frac{\pi}{2}}_{0} (x \sin^{n-2}x\cdot (1-\sin^{2}x)\ dx - (n-1)\int^{\frac{\pi}{2}}_{0}\sin^{n-1}x \cos x)\ dx$$
$$\Rightarrow u_{n}=1+(n-1)\int^{\frac{\pi}{2}}_{0} x \sin^{n-2}x\ dx- (n-1)\int^{\frac{\pi}{2}}_{0} x\sin^{n}x)\ dx - (n-1) \cdot \frac{1}{n}$$
$$\Rightarrow u_{n}=1+(n-1)u_{n-2}- (n-1)u_{n} - 1+ \frac{1}{n}$$
$$\Rightarrow n \cdot u_{n}=(n-1)u_{n-2}+ \frac{1}{n}$$
$$\Rightarrow \bbox[5px,border:2px solid red]
{u_{n}=\frac{n-1}{n}u_{n-2}+ \frac{1}{n^{2}}}$$
I think you can proceed from here.
| {
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Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The correct answer is : $2, 2 e^{\pi/3}, 2 e^{5\pi/3}$
My calculation is here:
$r=\sqrt{-8^{2}}=8$
Then,
$= 2\ cis\ \frac{2\pi k}{n}$
If k is $0$,
$= 2\ cis\ 0=2$
If k is $1$,
$= 2\ cis\ \frac{\pi }{3}$
If k is $2$,
$= 2\ cis\ \frac{4\pi }{3}$
Therefore, the results are $2$, $= 2\ cis\ \frac{\pi }{3}$, and $= 2\ cis\ \frac{4\pi }{3}$.
So, why the correct answer is $2$, $2\ cis (\frac{\pi}{3})$, $2\ cis (\frac{5\pi}{3})$?
| Since $-8=8e^{i({\pi+2n\pi})}$ we have $(-8)^{\frac{1}{3}}=2e^{i\frac{\pi+2n\pi}{3}}$ for $n\in\mathbb Z$ (and you want the cases when $n=0,1$ and $2$).
| {
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"source": "stackexchange",
"question_score": "1",
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A continuous function $f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\to[-1,1]$ and differentiable on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$.
$\blacksquare~$ Problem: Suppose a continuous function $f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\to[-1,1]$ and differentiable on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$. Then, there exists a point $x_0\in \left(-\frac{\pi}{4},\frac{\pi}{4}\right)$ such that
$$|f'(x_0)|\leqslant 1+f(x_0)^2$$
$\blacksquare~$ My Solution: Let's take $g(x) = \tan^{-1} f(x) $. Then $g : \left[ - \frac{\pi}{4}, \frac{\pi}{4}\right] \to \left[- \frac{\pi}{4}, \frac{\pi}{4} \right] $.
Now, as $f$ is cont in $\left[- \frac{\pi}{4}, \frac{\pi}{4} \right]$ and differentiable in $\left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$, $g$ is also the same.
By LMVT, we have that
$$\frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}} = g'(x_0) \quad \text{for some } x_0 \in \left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$$$$\implies \frac{ \frac{\pi}{4} - \left(- \frac{\pi}{4}\right) }{ \frac{\pi}{2} } \geqslant \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}} = \left(\tan^{-1}f(x_0) \right)' = \frac{f'(x_0)}{1 + f(x_0)^2} $$$$ \implies 1 + f(x_0)^2 \geqslant f'(x_0) $$Again, after the LMVT part, we have that
$$ \implies \frac{ - \frac{\pi}{4} - \left( \frac{\pi}{4}\right) }{ \frac{\pi}{2} } \leqslant \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}} = \left(\tan^{-1}f(x_0) \right)' = \frac{f'(x_0)}{1 + f(x_0)^2} $$$$ \implies - \left( 1 + f(x_0)^2 \right) \leqslant f'(x_0) $$Hence, combining these two we have that$$ \lvert f'(x_0) \rvert \leqslant 1 + f(x_0)^2 \quad \text{for some } x_0 \in \left( - \frac{\pi}{4}, \frac{\pi}{4} \right) $$
Is this fine? Is there any glitch? Another way of a solution will be great!
Regards, Ralph
| Your solution is fine, you may just simpify a bit just one step:
You can write from the begging
$\left\vert \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}}\right\vert = \vert g'(x_0)\vert \quad \text{for some } x_0 \in \left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$ so you can get immediately
$\vert g'(x_0)\vert \le 1$.
| {
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"url": "https://math.stackexchange.com/questions/3793797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$ Question:
The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$
My working:
$D=a^2-8(b+3)\geqq0$
$a^2\geqq8(b+3)=8b+24$
Add $b^2$ to both sides
$a^2+b^2\geqq b^2+8b+24=(b+4)^2+8\geqq 8$
$\therefore {(a^2+b^2)
}_{min} = 8$
However the answer in the book is 9.
What have I done wrong?
| The minimum value for
$ a^2 + b^2 = 8(b+3) + b^2 $, because $ a^2 \geqq 8(b+3) $.
The value of $b$ that minimizes the quadratic $ 8(b+3) + b^2 $ is:
$ 8 + 2b =0 \Rightarrow b = -4$
From $ a^2 \geqq 8(b+3) \Rightarrow b \geqq -3$
Therefore, the minimum value of $ a^2 + b^2 $ is when $ b = -3 $ and $ a = 0 $
$ \Rightarrow a^2 + b^2 \geqq (0)^2 + (-3)^2 $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$
I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
which does not match with the answer on Desmos.
| $\sqrt{9-x^{2}} > x^{2} + 1
\\\textsf{because both the sides are positive, you can square both sides}
\\ 9-x^{2} > (x^2 + 1)^{2}
\\ 9-x^{2} > x^{4} + 2x^{2} + 1
\\ x^{4} + 3x^{2} - 9 < 0
\\ \textsf{assume x^2 = t}
\\ t^2 +3t - 9 < 0
\\ t \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x^{2} \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x \in \left( -\sqrt{\frac{-3+\sqrt{41}}{2}}, +\sqrt{\frac{-3+\sqrt{41}}{2}} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=\frac{1}{9}(a+b+c)^2(a-2b+c)^4\\ &+\frac{2}{3}(a+b+c)^2(a-2b+c)^2(b-c)(a-b)\\ &+\frac{1}{16}(a-b)^2(b-c)^2(a+4b+7c)(7a+4b+c)\\&\geqslant 0\end{align*}
However, this solution is too hard for me to find without computer. Could you help me with figuring out a better soltuion?
Thank you very much
| This is SOS's proof
Since $$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
$$=\sum \Big[{\dfrac {1}{140}}\, \left( 11a+11b-70c \right) ^{2}+{\dfrac {1944
}{35}}\,ab+{\dfrac {999}{140}}\, \left( a-b \right) ^{2}\Big](a-b)^4 \geqslant 0$$
So we are done.
For the best constant, assume $c=\min\{a,b,c\}$$,$ $$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\Big(1+\dfrac{2}{\sqrt{3}}\Big)(a-b)^2(b-c)^2(c-a)^2$$
$$=\dfrac{1}{3} \left( 2\,a+2\,b-c \right) c \left( {a}^{2}-ab-bc+{b}^{2}-ac+{c}
^{2} \right) ^{2}+$$
$$+\frac19 A\cdot \left[ \left( \sqrt {3}-1 \right) {c}^{2
}- \left( \sqrt {3}-1 \right) \left( a+b \right) c-{a}^{2}+ab+\sqrt {
3}ab-{b}^{2} \right] ^{2} \geqslant 0,$$
where $$\text{A}= \left( b-c \right) ^{2}+(2\,\sqrt {3} +2)\left( a-c \right) \left( b-c
\right) + \left( a-c
\right) ^{2} \geqslant 0. $$
Done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
The concept of transformation of roots can be applied here.
So replace
$$x \to \frac{x-1}{x+1}$$
After considerable algebra, a cubic would be obtained, and then the sum can easily be found out
My question: This method is touted around a lot, but I never around how it actually works, especially in this case. If $\frac{x-1}{x+1} =\alpha$, then the transformed equation would be $\alpha^3 -2007 \alpha +2002$, and the sum of all $\alpha$ should still be $2007$. Can I get an explanation for this method?
| If $x$ is any of $a,b,c$ then plugging in $x=\frac{1+y}{1-y}$ gives the new equation $$y^3 - 2y^2 + \frac{999}{1002} \,y+\frac{1}{1002}=0 \, .$$ Finding the objective then amounts to finding the sum $A+B+C$ for the 3 solutions of this equation which is just the negative of the coefficient of the $y^2$ term in the general cubic $$(y-A)(y-B)(y-C)=y^3-(A+B+C)y^2 + (AB+AC+BC)y - ABC=0 \, ,$$ i.e. $A+B+C=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'm not sure.
| We have
$$f'(x) = \frac{(x^4+x^2+2)(x^3+2x^2+2x+2)(x^3-2x^2+2x-2)}{x^2(x^2+1)^2(x^2+2)^2}.$$
Because $x>0,$ so
$$f'(x) = 0 \Rightarrow x = x_0 = \frac{\sqrt[3]{17+3\sqrt{33}}}{3}-\frac{2}{3\sqrt[3]{17+3\sqrt{33}}}+\frac{2}{3} \approx 1.5437$$
Therefore
$$f(x) \geqslant f(x_0) = 5.$$
So $f(x)_{\min} = 5,$ equality occur when $x=x_0.$
Note. The SOS form
$$f(x) = 5 + \frac{(x^3-2x^2+2x-2)^2}{x(x^2+1)(x^2+2)} \geqslant 5.$$
P/s. How do center an image?
| {
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Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity
My attempt:
$(1+\omega^n+\omega^{2n})=0$
$\Rightarrow (1-\omega^n+\omega^{2n})=-2\omega^n$
$\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{1+2+3…2n}$
$\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{n(2n+1)}$
If $n$ is $3k$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$
If $n$ is $3k+1$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$
If $n$ is $3k+2$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{(3k+2)(6k+5)} =2^{2n}\omega^{(3m+1)}=2^{2n}\omega^{3m}\omega=2^{2n}\omega$
What have I done wrong here?
| Since $$a^3+b^3=(a+b)(a^2-ab+b^2),$$ we obtain: $$\prod_{k=1}^{2n}(1-w^{2^{k-1}}+w^{2^{k}})=\prod_{k=1}^{2n}\frac{\left(1+w^{3\cdot2^{k-1}}\right)}{1+w^{2^{k-1}}}=\frac{2^{2n}}{((1+w)(1+w^2))^n}=2^{2n}.$$
| {
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Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle.
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}}$$ are always sides of a triangle.
For all $\triangle MNP$ where $m = MP, n = PM, p = MN$, we have that $$n + p > m, p + m > n, m + n > p$$
We need to obtain that $$\frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}} > \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}$$
First attempt, we have that $$\frac{(a - b)^2}{|a - b|\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{(b - c)^2}{|b - c|\sqrt{b^2 + 1}\sqrt{c^2 + 1}}$$
$$ \ge \frac{(c - a)^2}{\sqrt{b^2 + 1} \cdot \left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)}$$
and $$\left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)^2 \le \left[(b - c)^2 + (a - b)^2\right] \cdot (c^2 + a^2 + 2)$$
It is needed to prove that $$\sqrt{\left[(b - c)^2 + (a - b)^2\right] \cdot (b^2 + 1)(c^2 + a^2 + 2)} < |c - a|\sqrt{c^2 + 1}\sqrt{a^2 + 1}$$
Second attempt, it is to prove that $$|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1} > |c - a|\sqrt{b^2 + 1}$$
According to the Cauchy - Schwarz inequality, we have that $$\left(|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1}\right)^2 \ge 2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)}$$
What needs to be established is $$2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)} > (c - a)^2(b^2 + 1)$$
Third attempt, let $a = \tan\alpha, b = \tan\beta, c = \tan\gamma$ $\left(\alpha, \beta, \gamma \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\right)$, it could be easily deducted that $$\frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}} = \frac{|\tan\gamma - \tan\alpha|}{\sqrt{\tan\gamma^2 + 1}\sqrt{\tan\alpha^2 + 1}} = \frac{\left|\dfrac{\sin(\gamma - \alpha)}{\cos\gamma\cos\alpha}\right|}{\dfrac{1}{\cos\gamma\cos\alpha}} = \pm\sin(\gamma - \alpha)$$
For all of the above attempts, there need to be considered multiple cases of $a, b, c$, whether they're positive and negative, and their arrangements from littlest to greatest.
| Hint.
Given the three sides $l_1,l_2,l_3$ we habe
$$
\cos\theta_1 = \frac{l_2^2+l_3^2-l_1^2}{2l_2l_3}
$$
now making
$$
\cases{
l_1^2 = \frac{(b-c)^2}{\left(b^2+1\right) \left(c^2+1\right)}\\
l_2^2 = \frac{(c-a)^2}{\left(a^2+1\right) \left(c^2+1\right)}\\
l_3^2 = \frac{(a-b)^2}{\left(a^2+1\right) \left(b^2+1\right)}
}
$$
we have
$$
\cos\theta_1 = \frac{b c + 1}{\sqrt{(b^2+1)(c^2+1)}}
$$
and
$$
-1\lt \cos\theta_1 \lt 1
$$
as expected.
NOTE
If $\sin^2\theta_1 = 1-\cos^2\theta_1 = \frac{(b-c)^2}{\left(b^2+1\right) \left(c^2+1\right)}$ then
$$
\frac{l_k^2}{\sin^2\theta_k}=1
$$
so the sinus law is observed as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Limit of an integral $\lim_{r\to 0^+} \int_0^1 \left(\frac{f(x)r}{x^2+r^2}\right )~dx$ Suppose $f$ is a real-valued continuous function on the unit interval $[0,1]$. How can we compute $$\lim_{r\to 0^+} \int_0^1 \left(\dfrac{f(x)r}{x^2+r^2} \right)~dx?$$
If we let $\min f(x)=m$ and $\max f(x)=M$, then we have $$m\tan^{-1}\left(\frac{1}{r}\right)\leq \int_0^1 \left(\dfrac{f(x)r}{x^2+r^2} \right)~dx \leq M\tan^{-1}\left(\frac{1}{r}\right), $$ since $\int_0^1 r/(x^2+r^2)~dx=\tan^{-1}(1/r)$.
| Define $M$ such that $|f(x)|<M$ for $x\in [0,1]$. First, note that for all fixed $\gamma>0$ and all $r>0$ we have
$$r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx\leq rM\int_\gamma^1 \frac{1}{x^2+r^2}dx<rM\int_\gamma^1 \frac{1}{x^2}dx=rM\left(\frac{1}{\gamma}-1\right)$$
$$r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx\geq -rM\int_\gamma^1 \frac{1}{x^2+r^2}dx>-rM\int_\gamma^1 \frac{1}{x^2}dx=-rM\left(\frac{1}{\gamma}-1\right)$$
This implies that for all fixed $\gamma>0$
$$\lim_{r\to 0^{+}}r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx=0$$
Second, since $f(x)$ is continuous at $0$, for all $\epsilon>0$ there exists $\delta>0$ such that $0\leq x\leq\delta$ implies
$$|f(0)-f(x)|<\epsilon$$
$$f(0)-\epsilon<f(x)<f(0)+\epsilon$$
Finally, let $\epsilon>0$ be arbitrary. Split the integral up at $\delta$ (from above):
$$r\int_0^1\frac{f(x)}{x^2+r^2}dx=r\int_0^\delta\frac{f(x)}{x^2+r^2}dx+r\int_\delta^1\frac{f(x)}{x^2+r^2}dx$$
From the first step, we know
$$\lim_{r\to 0^{+}}r\int_0^1\frac{f(x)}{x^2+r^2}dx=\lim_{r\to 0^{+}}\left[r\int_0^\delta\frac{f(x)}{x^2+r^2}dx+r\int_\delta^1\frac{f(x)}{x^2+r^2}dx\right]=\lim_{r\to 0^{+}}r\int_0^\delta\frac{f(x)}{x^2+r^2}dx$$
This integral can be bounded by
$$r\int_0^\delta\frac{f(0)-\epsilon}{x^2+r^2}dx<r\int_0^\delta\frac{f(x)}{x^2+r^2}dx<r\int_0^\delta\frac{f(0)+\epsilon}{x^2+r^2}dx$$
$$r(f(0)-\epsilon)\int_0^\delta\frac{1}{x^2+r^2}dx<r\int_0^\delta\frac{f(x)}{x^2+r^2}dx<r(f(0)+\epsilon)\int_0^\delta\frac{1}{x^2+r^2}dx$$
But we know
$$\int_0^\delta\frac{1}{x^2+r^2}dx=\frac{1}{r}\left[\arctan(\delta/r)-\arctan(0/r)\right]=\frac{1}{r}\arctan(\delta/r)$$
Thus
$$\lim_{r\to 0^{+}}r(f(0)\pm\epsilon)\int_0^\delta\frac{1}{x^2+r^2}dx=\lim_{r\to 0^{+}}(f(0)\pm\epsilon)\arctan(\delta/r)=(f(0)\pm\epsilon)\frac{\pi}{2}$$
This implies
$$f(0)\frac{\pi}{2}-\epsilon\frac{\pi}{2}\leq \lim_{r\to 0^{+}}r\int_0^\delta\frac{f(x)}{x^2+r^2}dx\leq f(0)\frac{\pi}{2}+\epsilon\frac{\pi}{2}$$
However, since $\epsilon$ was arbitrary, this simplifies to
$$f(0)\frac{\pi}{2}\leq \lim_{r\to 0^{+}}r\int_0^\delta\frac{f(x)}{x^2+r^2}dx\leq f(0)\frac{\pi}{2}$$
We conclude
$$\lim_{r\to 0^{+}}\int_0^\delta\frac{f(x)r}{x^2+r^2}dx= f(0)\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.
| If $x\geq1$ so
$$x^8-x^5+x^2-x+1=(x^8-x^5)+(x^2-x)+1>0.$$
If $0\le x<1$ so
$$x^8-x^5+x^2-x+1=1-x+x^2(1-x^3)+x^8>0.$$
If $x<0$ so after replacing $x$ on $-x$ we obtain:
$$x^8+x^5+x^2+x+1$$ has no positive roots by the Descarte's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Generating set for $SL_2(\mathbb{C})$ Let
\begin{equation}
A =
\begin{pmatrix}
\lambda & 0\\
0 & \frac{1}{\lambda}\\
\end{pmatrix}
\end{equation}
be a matrix in $SL_2(\mathbb{C}$). I am trying to prove that $SL_2(\mathbb{C})$ is generated by the matrices
\begin{equation}
\begin{pmatrix}
1 & z\\
0 & 1\\
\end{pmatrix},
\begin{pmatrix}
1 & 0\\
z & 1\\
\end{pmatrix}
\end{equation}
for $z\in\mathbb{C}$. I have already handled all the cases expect for matrices like $A$. How can I write it as a product of matrices of this type? Any hints are appreciated.
| How about this:
$A = \begin{pmatrix}
1 & -\lambda \\
0 & 1
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0\\
\frac{1}{\lambda} & 1\\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0\\
-1 & 1\\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 1\\
0 & 1\\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0\\
\lambda - 1 & 1\\
\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the limit: $\lim_{x\rightarrow\infty}\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac{x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}$ Can someone please evaluate this limit for me, I have been breaking my head for the past 2-3 days..... Any help would be appreciated.
$$\lim_{x\rightarrow\infty}\left \{ \frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac{x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}} \right \}
$$
$a,b,d,e$ are constants; $e\neq 0 $.
$ (a,b,c,d) \in \mathbb{R}$
| Set $p=b^{n+2}e^n-a^{n+2}d^n$
\begin{align}
&\lim_{x\to\infty}\left\{\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac {x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}\right\}=\\
&\qquad\lim_{x\to\infty}\left\{\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e ^n}-\frac{px}{e^{2n}}\right\}=\\
&\qquad\lim_{x\to\infty}\frac{e^{2n}(ax+b)^{n+2}-e^na^{n+2}x^2(ex+d)^n-px( ex+d)^n}{e^{2 n}(ex+d)^n}
\end{align}
now expand the numerator by the binomial theorem
\begin{align}
&\sum_{k=0}^{n+2}\binom{n+2}{k}a^{k}b^{n+2-k}e^{2n}x^k-\sum_{k=0}^n\binom{n}{k}a ^{n+2}e^{n+k}d^{n-k}x^{k+2}-\sum_{k=0}^n\binom{n}{k}e^kd^{n-k}px^{k+1}=\\
&\qquad=\sum_{k=0}^{n+2}\binom{n+2}{k}a^{k}b^{n+2-k}e^{2n}x^k-\sum_{k=2}^{n+2}\binom{n}{k-2}a^{n+2}e^{n+k-2}d^{n-k+2}x^k+{}\\
&\qquad\qquad-\sum_{k=1}^{n+1}\binom{n}{k-1}e^{k-1}d^{n-k+1}px ^k=\\
&\qquad=(a^{n+2}e^{2n}-a^{n+2}e^{2n})x^{n+2}+\sum_{k=2}^{n+1}\left[\binom{n+2}{k}a^{k}b ^{n+2-k}e^{2n}+\right.{}\\
&\qquad\qquad\left.-\binom{n}{k-2}a^{n+2}e ^{n+k-2}d^{n-k+2}-\binom{n}{k-1}e^{k-1}d^{n-k+1}p\right]x^k+{}\\
&\qquad\qquad+[(n+2)ab^{n+1}e^{2 n}-d^np]x+b^{n+2}e^{2n}
=\\
\end{align}
we see that the leading term cancels. The denominator has a non vanishing leading term $e^{3n}x^n,$ while in the numerator remains a term of degree $n+1,$ and if its coefficient does not vanish the limit goes to $\infty.$ The coefficient is given by
$$
(n+2)a^{n+1}be^{2n}+na^{n+2}e^{2n-1}d-e^np
$$
If this coefficient is null, numerator and denominator have the same degree, provided that the next coefficient in the numerator is not null, so the limit is a non vanishing constant, given by
$$
\frac{1}{e^{3n}}\left(\frac{(n+2)(n+1)}{2}a^nb^2e^{2n}-\frac{n(n-1)}{2}a^{n+2}d^2e^{2n-2}-nde^{n-1}p\right)
$$
If also this coefficient vanishes, then the numerator has degree less tha $n$ and the limit is $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
combinatorial identity another solution? I would appreciate if somebody could help me with the following problem:
I need to show that,
$$\sum^{33}_{k=0}\binom{100}{3k}=\sum^{49}_{k=0}4^k$$
My proof is the following:
$$(1+x)^{100} = a_0 + a_1x + a_2x^2 + \cdots + a_{100}x^{100}$$
Let
$A = a_0 + a_3 + a_6 + \cdots + a_{99}, B = a_1 + a_2 + a_4 + a_5 + a_7 + a_8 + \cdots + a_{98} + a_{100}$$
then
$$A+B = 2^{100}$$
$$x^3=1 \to x=1,w,w^2$$
$x$ put $w$, $w^2$ and sum
$$\begin{align*}
(1+w)^{100} + (1+w^2 )^{100}
&=w^{200} + w^{100}\\
&=w^{2} + w\\
&=-1\\
&= 2A + a_1(w+w^2) + a_2(w^2 + w^4) + a_4(w^4 + w^8) + \cdots + a_{100}(w^{100} + w^{200}) \\
&= 2A - B
\end{align*}$$
therfore
$$A+B = 2^{100}, 2A - B=-1$$
$$A = \frac{2^{100} - 1}{3} = \frac{4^{50} - 1}{3} =\sum^{49}_{k=0}4^k $$
But I want to know if exists another proof for this problem.
| My argument uses more or less a discrete Fourier transform in a standard way to pick off every third coefficient. Let $\zeta_3 = \exp(2\pi i/3)$ be a primitive 3rd root of unity. Note that $1^k + \zeta_3^k + \zeta_3^{2k} = 3\delta_{3 \mid k}$. Letting $p(x) := (1+x)^{100}$, we see
$$\begin{align*}
\frac{p(1)+p(\zeta_3)+p(\zeta_3^2)}{3} = \frac{1}{3} \sum_{k=0}^{100} \binom{100}{k} (1^k + \zeta_3^k + \zeta_3^{2k}) = \sum_{k=0}^{33} \binom{100}{3k}.\end{align*}$$
On the other hand, $p(1)=2^{100}$ and $p(\zeta_3) = (1+\zeta_3)^{100}$. As is well-known, $1, \zeta_3, \zeta_3^2$ form the vertices of an equilateral triangle, and $1+\zeta_3, -1, 1+\zeta_3^2$ form the remaining vertices of a regular hexagon. Anyway, $1+\zeta_3$ has a polar angle of $\pi/3$. Hence $(1+\zeta_3)^{100}$ has a polar angle of $100\pi/3 \equiv 4\pi/3$ and a magnitude of $1$. Likewise $(1+\zeta_3^2)^{100}$ has a polar angle of $-4\pi/3$ and a magnitude of $1$. Hence $p(\zeta_3) + p(\zeta_3^2) = 2\cos(4\pi/3) = -1$. So, the left-hand side is
$$\begin{align*}\frac{2^{100} - 1}{3}
&= \frac{4^{50} - 1}{4 - 1} = \sum_{i=0}^{49} 4^i\end{align*}$$
(Note: I wrote this while your argument was unreadable. They seem to be similar in spirit.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $k$ such that $\frac{3x^2+kx-44}{x^2-121}$ simplifies to $\frac{3x-4}{x-11}$
If the rational expression $\frac{3x^2+kx-44}{x^2-121}$, where $k$ is an element of $\mathbb{W}$, simplifies to $\frac{3x-4}{x-11}$, then the value of $k$ must be?
my work:
$$\frac{3x^2+kx-44}{x^2-121}= \frac{3x-4}{x-11}$$
so the answer = value of $k$ is $11$.
| hint
If $$\frac{3x^2+kx-44}{x^2-121}=\frac{3x-4}{x-11}$$
Then, with $ x=1$, we get
$$\frac{3+k-44}{1-121}=\frac{1}{10}$$
and
$$k=-12+41=29$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Can we prove that $\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$? I would like to verify the claim:
$$\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$$
I tried verifying through an example:
Given the following $3$ different matrices:
\begin{align}
A & = \begin{pmatrix}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12 \\
13 & 14 & 15 & 16
\end{pmatrix}\\[2ex]
B & = \begin{pmatrix}
4 & 3 & 2 & 1 \\
3 & 6 & 4 & 2 \\
2 & 4 & 6 & 3 \\
1 & 2 & 3 & 4
\end{pmatrix}\\[2ex]
C & = \begin{pmatrix}
4 & 3 & 2 & 1 \\
5 & 6 & 7 & 8 \\
8 & 7 & 6 & 5 \\
4 & 3 & 2 & 1
\end{pmatrix}
\end{align}
I calculated $\operatorname{Tr}(ABC) = 7930$ and
$\operatorname{Tr}(CBA) = 7510$.
Is there any thing wrong in my calculation, or does this prove that $\operatorname{Tr}(ABC) \neq \operatorname{Tr}(CBA)$?
Many thanks!
| In fact, $\mathrm{Tr}\,(ABC)=\mathrm{Tr}\,(CBA)$ cannot hold for all $A$, $B$, $C$ so long as the ring of scalars is nontrivial. For let
\begin{align}
A &= \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} \\[2ex]
B &= \begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} \\[2ex]
C &= \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix} \\[2ex]
\end{align}
Then $ABC = C$, $CBA = 0$, so $\mathrm{Tr}\,(ABC) = 1$, $\mathrm{Tr}\,(CBA)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Floor function bounding From CMC:
What is the sum of the square of the real numbers $x$ for which $x^2 - 20\lfloor x\rfloor + 19 = 0$?
We use $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$ and eventually get the bounds $1\le x\le19$ and $x\ge 18,x\le 2.$ Of course, it's possible for $x$ not to be an integer, so how do we find the other solutions, other than $19$ and $1$?
Someone wrote this solution:
$x^2 - 20\lfloor x \rfloor + 19 = 0$
Cleary $x\geq \lfloor x \rfloor$ for all real $x$. Thus,
$$x^2-20x+19 \leq x^2 - 20\lfloor x \rfloor + 19=0.$$
Which leads to
$$1 \leq x \leq19.$$Also $x^2=20\lfloor x\rfloor - 19$ which implies $\lfloor x \rfloor=1,17,18,19$.
I'm not sure how we get $\lfloor x\rfloor=17,18$ from this.
| From $\lfloor x\rfloor=(x^2+19)/20\gt0$, we see that we must have $x\gt0$, hence $x=\sqrt{20\lfloor x\rfloor-19}$ (i.e., the positive, not the negative, square root). It follows that $x^2-20\lfloor x\rfloor+19=0$ has a (unique) solution with $\lfloor x\rfloor=n\in\mathbb{Z}^+$ if and only if $n\le\sqrt{20n-19}\lt n+1$. With everything in sight non-negative, we have
$$\begin{align}
n\le\sqrt{20n-19}\lt n+1
&\iff n^2\le20n-19\lt n^2+2n+1\\
&\iff n^2-20n+19\le0\lt n^2-18n+20
\end{align}$$
The first quadratic inequality in the last line tells us $1\le n\le19$; the second tells us either $n\lt9-\sqrt{61}$ or $n\gt9+\sqrt{61}$, which, since $7\lt\sqrt{61}$, tells us either $n\lt2$ or $n\gt16$. We thus have four values for $\lfloor x\rfloor=n$, namely $1$, $17$, $18$, and $19$, with $20n-19$ for the corresponding values of $x^2$. The sum of these squares is
$$(20\cdot1-19)+(20\cdot17-19)+(20\cdot18-19)+(20\cdot19-19)=20(1+54)-76=1024$$
(The fact that the final answer turns out to be a power of $2$ is surely pure coincidence.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse of $f(x)=3x^3+1$ I get $\sqrt[3]{\frac{x-1}{3}}$ whereas textbook shows answer as $\sqrt{\frac{x-1}{3}}3$ I am to find the inverse of $f(x)=3x^3+1$. My working:
$$f(x)=3x^3+1$$
$$x=3y^3+1$$
$$x-1=3y^3$$
$$\frac{x-1}{3}=y^3$$
$$f^{-1}(x)=\sqrt[3]\frac{x-1}{3}$$
The solutions section of my textbook says that the answer is $\sqrt{\frac{x-1}{3}}3$.
How can I arrive at the correct answer? Which step in my working did I go wrong?
| This will be a typo, a misplaced $3$
Your answer $f^{-1}(x)=\sqrt[3]{\dfrac{x-1}{3}}$ is correct, as is $\left(\frac{x-1}{3}\right)^{\frac13}$, though you may need to specify using the real cube root rather the principal cube root when $x<1$
If $\sqrt{\frac{x-1}{3}}3$ had really been intended then I suspect it would have been written more tidily as $\sqrt{3({x-1})}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving for $x$ and $y$ when $(3x + y)(x + 3y)\sqrt{xy} = 14$ Solve for $x$ and $y$ when $$(3x + y)(x + 3y)\sqrt{xy} = 14$$ $$(x+y)(x^2 + 14xy + y^2) = 36.$$
I was thinking of squaring the first equation and moving on from there, but I think it'll be a bit too messy. Is there a better way to start this problem?
| From the first equation we have $xy>0$.
Thus, the second give $x+y>0,$ which says $x>0$ and $y>0$.
Let $y=t^2x,$ where $t>0$.
Thus, $$18(3t^2+1)(t^2+3)t=7(t^2+1)(t^4+14t^2+1)$$ or
$$(t^2-6t+1)(7t^4-12t^3+26t^2-12t+7)=0$$ or
$$t^2-6t+1=0,$$
which gives $y=(17-12\sqrt2)x$ or $y=(17+12\sqrt2)x.$
Let $y=(17-12\sqrt2)x.$
Thus, the first equation gives:
$$(3+10(17-12\sqrt2)+3(17-12\sqrt2)^2)(3-2\sqrt2)x^3=14$$ or
$$x^3=\frac{99+70\sqrt2}{8}$$ or
$$x=\frac{1}{2}(3+2\sqrt2)$$ and
$$y=(3-2\sqrt2)^2\cdot \frac{1}{2}(3+2\sqrt2),$$
which gives $$\left(\frac{3+2\sqrt2}{2},\frac{3-2\sqrt2}{2}\right).$$
The second case by the same way gives:
$$\left(\frac{3-2\sqrt2}{2},\frac{3+2\sqrt2}{2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$
I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?
| Comment:
If you put $x=\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} $ under the third radical you get:
$$\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + }} . . .}} }} = x$$
Suppose there is no negative sign then by squaring both sides you get:
$$X^2=\sqrt 3 +X$$
Which gives:
$$X=\frac{-1 ±\sqrt{1+4\sqrt 3}}{2}≈0.9≈\frac{\sqrt3}{2}$$
So $x<X≈\frac{\sqrt 3}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\left[ \mathbb{Q} \left( \sqrt[3]{4+\sqrt{5}} \right ) : \mathbb{Q} \right] = 6$ Let $\alpha = \sqrt[3]{4+\sqrt{5}}$. I would like to prove that $\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$. We have $\alpha^3 = 4 + \sqrt{5}$, and so $(\alpha^3 - 4)^2 = 5$, hence $\alpha$ is a root of the polynomial $f(x)=x^6 - 8 x^3 + 11$.
I tried to prove with various approaches that $f(x)$ is irreducible over $\mathbb{Q}$ without success, so I devised the following strategy.
Since $x^2 - 5$ is irreducible over $\mathbb{Q}$, we have $\left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] = 2$. Now from $\alpha^3 = 4 + \sqrt{5}$ we get $\sqrt{5} \in \mathbb{Q} \left( \alpha \right )$, so that $\mathbb{Q} \left( \sqrt{5} \right )$ is a subfield of $\mathbb{Q} \left( \alpha \right )$, $\mathbb{Q} \left( \alpha \right )=\mathbb{Q}\left( \sqrt{5}\right) \left( \alpha \right)$, and we have
\begin{equation}
\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{5} \right)\left( \alpha \right ) : \mathbb{Q} \left (\sqrt{5} \right) \right] \left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] .
\end{equation}
Now $\alpha$ is a root of the polynomial $g(x) \in \mathbb{Q} \left (\sqrt{5} \right) [ x ]$ given by $g(x) = x^3 - 4 - \sqrt{5}$. So to prove our thesis it is enough to prove that this polynomial is irreducible in $\mathbb{Q} \left (\sqrt{5} \right) [ x ]$. Being $g(x)$ of third degree, if it were not irreducible, its factorization would have at least one linear factor, so that $g(x)$ would have some root in $\mathbb{Q} \left (\sqrt{5} \right)$. Hence our problem boils down to show that there are no integers $m_0, m_1, n$, with $n \neq 0$, such that
\begin{equation}
\left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5},
\end{equation}
which gives
\begin{equation}
m_0^3 + 5 \sqrt{5} m_1^3 +3 \sqrt{5} m_0^2 m_1 + 15 m_0 m_1^2 = 4 n^3 + \sqrt{5} n^3,
\end{equation}
or
\begin{equation}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 + \sqrt{5} \left( 5 m_1^3 +3 m_0^2 m_1 - n^3 \right)=0,
\end{equation}
which implies, being $\sqrt{5}$ irrational,
\begin{cases}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 = 0, \\ 5 m_1^3 +3 m_0^2 m_1 - n^3 = 0.
\end{cases}
At this point I am stuck, because I do not know how to prove that this system admits the only integer solution $m_0 = m_1 = n = 0$.
Any help is welcome!
| As $\Bbb Q(\sqrt 5)$ is a subfield, its automorphism $\sqrt 5\to-\sqrt 5$ can be extended, which means that $\sqrt[3]{4-\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5})$ and also
$$ \sqrt[3]{11}=\sqrt[3]{4-\sqrt 5}\sqrt[3]{4+\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5}).$$ Now, $[\Bbb Q(\sqrt[3]{11}):\Bbb Q]$ is clearly $3$. With a subfield of degree 2 and another of degree 3 over $\Bbb Q$, ouf field must be at least of degree $6$, hence exactly degree $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$ is non negative Prove that the sequence
$$s_k = \frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$ is non-negative.
I would appreciate an elementary proof. I tried using series / integral comparison without success.
This question is a follow up of this one. There is an answer to that question that uses the integral representation of Hurwitz zeta function, that I'd like to avoid... if possible!
| $f(x) = 1/x^4$ is strictly convex, so that
$$
\frac{1}{(2n)^4} < \frac 12 \left(\frac{1}{(2n-1)^4} + \frac{1}{(2n+1)^4} \right)
$$
or
$$
\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4} < \frac 12 \left(\frac{1}{(2n-1)^4} - \frac{1}{(2n+1)^4} \right)
$$
It follows that
$$
2 \sum_{k=n+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4} \right)
< \sum_{k=n+1}^\infty \left(\frac{1}{(2n-1)^4} - \frac{1}{(2n+1)^4} \right) = \frac{1}{(2k+1)^4} \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below
Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.
All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ and $2^{1/3}\omega^2$, where $\omega = (-1+i\sqrt{3})/2$ is a cube root of unity. Thus $\mathbb{Q}(2^{1/3},\omega)$ or $\mathbb{Q}(2^{1/3},i\sqrt{3})$ is a splitting field of $f(x)$. $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ as irreducble polynomial of $2^{1/3}$ is $x^3-2$(Eisenstien's criteria for $p=2$) and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$, as irreducible polynomial of $\omega$ is $g(x) = x^2+x+1$( $g(x+1)$ is irreducble by eisenstien's criteria for $p=3$). Therefore, if we have a splitting field $\mathbb{Q}(z)$, then degree of irreducible polynomial of $z$ is $6$ by compositum of fields. Now we need to find $z$.
Me and my friend got the idea of taking $z$ as $2^{1/3} + i\sqrt{3}$.
$$\begin{eqnarray*} x & = & 2^{1/3} + i\sqrt{3} \\ \left( x - 2^{1/3} \right)^2 & = & -3
\\ -2^{2/3}x+2^{2/3} & = & -x^2 -3 \\ 4(x-1)^3 & = & (x^3+3)^3 \end{eqnarray*}$$
Thus $h(x) = (x^2+3)^3-4(x-1)^3$ is a degree six polynomial having root $2^{1/3} + i\sqrt{3}$. How do i show from here that it is irreducible in $Q[x]$? It's expanded form is $x^6+9x^4-4x^3+39x^2-12x+31$.
| It is irreducible by Eisenstein shift, which I explain how to find below.
Note$\, \bmod\color{#c00}3\!:\ f(x) := (x^2\!+\!3)^3\!-4(x\!-\!1)^3 \equiv x^6-x^3+1 \equiv (x\!+\!1)^6 \ $ is a prime power.
So Eisenstein works on $\,g(x) = f(x\!-\!1) \equiv x^6\ $ by
$\,g(0) = f(-1) = 96 \not\equiv 0\pmod{\!3^2}$
Key Idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $\,p,\,$ powers of a prime, viz. $\,\equiv x^n,\,$ and products of primes always factor uniquely. The same works for its shift $\,(x-c)^n,\,$ so we seek primes $\,p\,$ such that, mod $\,p,\,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $\,p\,$ that can yield such powers are those dividing the discriminant (here by Alpha = $\,-2^{24}\cdot \color{#c00}3^7\cdot 53),\,$ since if $\,f\equiv a (x-c)^n,\,\ n> 1\,$ then $\,f\,$ and $\,f'\,$ have a common root $\,x\equiv c,\,$ hence their resultant $\, R(f,f')\equiv 0.\,$ But this is, up to sign, the discriminant of $\,f\,$ (presumed monic).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the supremum of $|z|^2+\text{Re}(\overline{z}w)$ subject to $|z|^2+|w|^2=1$. I am trying to find the operator norm of $$A=\begin{pmatrix}3 & 1 \\
1 & 1\end{pmatrix}\in M_n(\mathbb{C}).$$This is what I have so far: If $|z|^2+|w|^2=1$, then $$||A\begin{pmatrix}z\\w\end{pmatrix}||_2=\sqrt2\sqrt{1+4(|z|^2+\text{Re}(\overline{z}w))}$$ Since $|z|^2+\text{Re}(\overline{z}w)\in\mathbb{R}$ and $\sqrt{2}\sqrt{1+4x}$ is an increasing function of $x$, we need to maximise $|z|^2+\text{Re}(\overline{z}w)$. In other words, $$||A||=\sqrt2\sqrt{1+4m},$$ where $m=\text{sup}\lbrace|z|^2+\text{Re}(\overline{z}w): |z|^2+|w|^2=1\rbrace$.
But how do I go about finding $m$? I'm a bit rusty on this stuff, do I need to use Lagrange multipliers?
| You can use Lagrange multipliers. Let$$f(x,y,z,t)=|x+yi|^2+\operatorname{Re}\bigl((x-yi)(z+ti)\bigr)=x^2+y^2+xz+yt$$and let$$g(x,y,z,t)=|x+yi|^2+|z+iy|^2=x^2+y^2+z^2+t^2.$$So, you should solve the system$$\left\{\begin{array}{l}\nabla f(x,y,z,t)=\lambda\nabla g(x,y,z,t)\\g(x,y,z,t)=1\end{array}\right.$$There are several families of solutions:
*
*points of the form $\left(x,-\frac{1}{2} \sqrt{-4 x^2-\sqrt{2}+2},-\left(1+\sqrt{2}\right) x,\frac{1}{2} \left(1+\sqrt{2}\right)\sqrt{-4 x^2-\sqrt{2}+2}\right)$, at which $f$ takes the value $\frac12-\frac1{\sqrt2}$;
*points of the form $\left(x,\frac{1}{2} \sqrt{-4x^2-\sqrt{2}+2},-\left(1+\sqrt{2}\right) x,-\frac{1}{2} \left(1+\sqrt{2}\right) \sqrt{-4 x^2-\sqrt{2}+2}\right)$, at which $f$ also takes the value $\frac12-\frac1{\sqrt2}$;
*points of the form $\left(x,-\frac{1}{2}\sqrt{-4 x^2+\sqrt{2}+2},\left(\sqrt{2}-1\right) x,-\frac{1}{2} \left(\sqrt{2}-1\right) \sqrt{-4 x^2+\sqrt{2}+2}\right)$, at which $f$ takes the value $\frac12+\frac1{\sqrt2}$;
*points of the form $\left(x,\frac{1}{2}\sqrt{-4 x^2+\sqrt{2}+2},\left(\sqrt{2}-1\right) x,\frac{1}{2} \left(\sqrt{2}-1\right) \sqrt{-4 x^2+\sqrt{2}+2}\right)$, at which $f$ also takes the value $\frac12+\frac1{\sqrt2}$.
So, the maximum is $\frac12+\frac1{\sqrt2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $ I have been working on this limit for days, but I am not getting it. The question is
Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$
Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since
$$e^x + \sin tx > e^x-1 > x>0$$
And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.
WHAT I TRIED:
I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens:
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$
Making the change of variables $u = 2n \pi x$ I get
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&=
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$
$$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$
I get the estimate
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align}
Summing all up, I got that
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$
As $n \to \infty$ the series converges to the Riemann integral
$$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$
Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.
Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.
| A slight modification of OP's attempt will lead to a solution. Indeed, write $I(t)$ for the integral and note that
$$ I(t) = \int_{0}^{\infty} \frac{\mathrm{d}x}{e^x + \sin(tx)}
\stackrel{(y=tx)}= \frac{1}{t} \int_{0}^{\infty} \frac{\mathrm{d}y}{e^{y/t} + \sin y}. $$
Also, define $J(t)$ by
$$ J(t)
= \frac{1}{t} \sum_{k=1}^{\infty} \int_{0}^{2\pi} \frac{\mathrm{d}y}{e^{2\pi k/t} + \sin y}
= \sum_{k=1}^{\infty} \frac{2\pi/t}{\sqrt{e^{4\pi k/t} - 1}}, $$
where the second step follows from the integration formula
$$ \int_{0}^{2\pi} \frac{\mathrm{d}y}{c + \sin y} = \frac{2\pi}{\sqrt{c^2 - 1}}, \qquad c > 1. \tag{1} $$
Then similarly as in OP's attempt, we obtain
$$ J(t) \leq I(t) \leq J(t) + \frac{1}{t} \int_{0}^{2\pi} \frac{\mathrm{d}y}{e^{y/t} + \sin y}. \tag{2} $$
Now we observe:
*
*Since the map $ u \mapsto \frac{1}{\sqrt{e^{2u} - 1}} $ is monotone decreasing, we have
$$ \int_{2\pi/t}^{\infty} \frac{\mathrm{d}u}{\sqrt{e^{2u} - 1}} \leq J(t) \leq \int_{0}^{\infty} \frac{\mathrm{d}u}{\sqrt{e^{2u} - 1}}. $$
So by the squeezing theorem, we get
$$ \lim_{t \to \infty} J(t) = \int_{0}^{\infty} \frac{\mathrm{d}u}{\sqrt{e^{2u} - 1}} = \frac{\pi}{2}. $$
*Observe that
$$ \int_{\pi}^{2\pi} \frac{\mathrm{d}y}{c + \sin y}
\stackrel{\text{(1)}}\leq \frac{2\pi}{\sqrt{c^2 - 1}}. $$
From this, we have
\begin{align*}
\frac{1}{t} \int_{0}^{2\pi} \frac{\mathrm{d}y}{e^{y/t} + \sin y}
&\leq \frac{1}{t} \left( \int_{0}^{\pi} \mathrm{d}y + \int_{\pi}^{2\pi} \frac{\mathrm{d}y}{e^{\pi/t} + \sin y} \right) \\
&\leq \frac{1}{t} \left( \pi + \frac{2\pi}{\sqrt{e^{2\pi/t} - 1}} \right).
\end{align*}
It is not hard to check that this bound converges to $0$ as $t \to \infty$.
Combining altogether and applying the squeezing theorem to $\text{(2)}$, we get
$$ \lim_{t\to\infty} I(t) = \frac{\pi}{2}. $$
Further Discussion:
*
*This proof can actually show that $I(t) = \frac{\pi}{2} + \mathcal{O}(t^{-1/2})$ as $t \to \infty$. Can we do better? It is reasonable to suspect that the asymptotic formula takes the form
$$I(t) = \frac{\pi}{2} + \frac{c}{\sqrt{t}} + \cdots \tag{3} $$
Is this true? If so, then what will be the value of $c$? I do not have enough time and energy to pursue in this direction right now (I need to crawl into the bed right now), but it seems an interesting question.
Addendum. Regarding the extra question, the following heuristic approach gives a guess on the value of the constant $c$ in the asymptotic expansion $\text{(3)}$:
Note that, for $x > 0$ and $\theta \in \mathbb{R}$,
\begin{align*}
\frac{1}{e^x + \sin\theta}
&= \frac{1}{\sqrt{e^{2x}-1}} \biggl( 1 + 2 \sum_{k=1}^{\infty} \frac{(-1)^k}{(e^x + \sqrt{e^{2x}-1})^{2k-1}} \sin((2k-1)\theta) \\
&\hspace{7em} + 2 \sum_{k=1}^{\infty} \frac{(-1)^k}{(e^x + \sqrt{e^{2x}-1})^{2k}} \cos (2k\theta) \biggr).
\end{align*}
Using this and substituting $\epsilon = 1/t$, we have
\begin{align*}
I(t)
&= \int_{0}^{\infty} \frac{\mathrm{d}x}{\sqrt{e^{2x}-1}} \\
&\quad + 2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \frac{\sin((2k-1)x/\epsilon)}{\sqrt{e^{2x}-1}(e^x + \sqrt{e^{2x}-1})^{2k-1}} \, \mathrm{d}x \\
&\quad +2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \frac{\cos(2kx/\epsilon)}{\sqrt{e^{2x}-1}(e^x + \sqrt{e^{2x}-1})^{2k}} \, \mathrm{d}x \\
&= \frac{\pi}{2} + 2 \sqrt{\epsilon} \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\sin((2k-1)u)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u \\
&\hspace{3em} + 2 \sqrt{\epsilon} \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\cos(2ku)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u,
\end{align*}
where we utilized the substitution $x = \epsilon u$ in the last step. So it is reasonable to expect that $c$ in $\text{(3)}$ is given by:
\begin{align*}
c
&= 2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \lim_{\epsilon \to 0^+} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\sin((2k-1)u)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u \\
&\quad + 2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \lim_{\epsilon \to 0^+} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\cos(2ku)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u \\
&= 2 \sum_{k=1}^{\infty} (-1)^k \biggl( \int_{0}^{\infty} \frac{\sin((2k-1)u)}{\sqrt{2u}} \, \mathrm{d}u + \int_{0}^{\infty} \frac{\cos(2ku)}{\sqrt{2u}} \, \mathrm{d}u \biggr) \\
&= \sum_{k=1}^{\infty} (-1)^k \biggl( \sqrt{\frac{\pi}{2k-1}} + \sqrt{\frac{\pi}{2k}} \biggr),
\end{align*}
where we utilized the identity
$$ \int_{0}^{\infty} \frac{\sin(a u)}{\sqrt{u}} \, \mathrm{d}u = \int_{0}^{\infty} \frac{\cos(a u)}{\sqrt{u}} \, \mathrm{d}u = \sqrt{\frac{\pi}{2a}}, \qquad a > 0. $$
Of course, interchanging the order of limit operators requires a great deal of care, especially in the situation like this where the absolute convergence fails. So this is not a proof yet, but rather a hand-waving heuristics. (Even the possibility that this guess is not true at all is still open to question!)
| {
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"url": "https://math.stackexchange.com/questions/3836576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Find a mistake in a computation of a determinant. I recently have to solve one exercise in which I need to compute the determinant of the matrix
$$ A= \begin{pmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{pmatrix} ,$$
where $a\in\mathbb{R}$. This is how I proceed using famous properties of the determinant
$$\det A = \begin{vmatrix}
a & 1 & 1 & 1 \\
a-1 & 1-a & 0 & 0 \\
0 & a-1 & 1-a & 0 \\
\textbf{0} & \textbf{0} &\textbf{a-1} & \textbf{1-a} \\
\end{vmatrix} = (1-a) \begin{vmatrix} a & 1 & 1 \\ a-1 & 1-a & 0 \\ 0 & a-1 & 0 \end{vmatrix} + (1-a)\begin{vmatrix} a & 1 & 1 \\ a-1 & 1-a & 0 \\ 0 & a-1 & 1-a \end{vmatrix} ; $$
$$ \det A = (1-a) \begin{vmatrix} a & 1 & \textbf{1} \\ a-1 & 1-a & \textbf{0} \\ 0 & a-1 & \textbf{0} \end{vmatrix} + (1-a)\begin{vmatrix} a & 1 & 2 \\ a-1 & 1-a & 1-a \\ \textbf{0} & \textbf{a-1} & \textbf{0} \end{vmatrix} ; $$
$$ \det A = (1-a) \begin{vmatrix} a-1 & 1-a \\ 0 & a-1 \end{vmatrix} + (1-a)^2\begin{vmatrix} a & 2 \\ a-1 & 1-a \end{vmatrix} ; $$
$$ \det A = (1-a)(a-1)^2 + (1-a)^2[-a^2-a+2] = (1-a)(a-1)^2 - (1-a)^2[a^2 + a -2] ; $$ $$\det A = (1-a)(a-1)^2 - (1-a)^2(a+2)(a-1);$$
$$\det A = -(a-1)^3 - (a-1)^2 (a+2)(a-1) = -(a-1)^3 - (a-1)^3(a+2); $$
$$\det A = -(a-1)^3 [1 + (a+2)] = \boxed{-(a-1)^3(a+3)} $$
I can't see any mistake in my procedure. However, calculating $\det A$ with a computer, I obtained $$\det A= (a-1)^3(a+3)$$
It is the same result except for a minus sign...
EDIT
This is what I am doing in the first steps, basically substracting rows as follows:
$$\det A = \begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{vmatrix} =\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix} =\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
0 & a-1 & 1-a & 0 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix} = \begin{vmatrix}
a & 1 & 1 & 1 \\
a-1 & 1-a & 0 & 0 \\
0 & a-1 & 1-a & 0 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix} $$
So where is exactly my mistake?
| In the first step it seems there is a sign problem indeed
$$\det(A)= \begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{vmatrix}=\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
0 & 0 & 1-a & a-1 \\
\end{vmatrix}=\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
0 & 1-a & a-1 & 0 \\
0 & 0 & 1-a & a-1 \\
\end{vmatrix}=$$
$$=\begin{vmatrix}
a & 1 & 1 & 1 \\
1-a & a-1 & 0 & 0 \\
0 & 1-a & a-1 & 0 \\
0 & 0 & 1-a & a-1 \\
\end{vmatrix}=-\begin{vmatrix}
a & 1 & 1 & 1 \\
a-1 & 1-a & 0 & 0 \\
0 & a-1 & 1-a & 0 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$ How to prove that $$\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$$ with $\varepsilon-\delta$ limit definition?
I know so far that $\forall \varepsilon > 0 \ \exists \delta > 0: 0 < d((x, y), (1, 2)) < \delta \Rightarrow \Big|\frac{x}{x+y} - \frac{1}{3}\Big| < \varepsilon$, so we get
$d = \sqrt{|x-1|^2 + |y-2|^2} \Rightarrow d^2 = |x-1|^2 + |y-2|^2 \Rightarrow \begin{cases}
|x - 1| < d < \delta\\
|y - 2| < d < \delta\\
\end{cases}$
But I don't know how to express $\Big|\frac{x}{x+y} - \frac{1}{3}\Big| = \Big|\frac{3x-(x+y)}{3(x+y)}\Big| = \Big|\frac{3x-x-y}{3x+3y}\Big| = \Big|\frac{2x-y}{3x+3y}\Big|$ using $|x-1|$ and $|y-2|$.
Can anyone help me with that?
| We can choose $\delta$ such that $3x + 3y > c$ for some positive $c$, so we don't have to worry about the denominator.
For the numerator, note that:
$$|2x-y| = |2x - 2 - y + 2| \le |2x-2|+|y-2|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Where did I go wrong in solving this equation?
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$
Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \leftarrow\text{(1)} \\
3a^2 + 3b + 1 &= 4 \leftarrow\text{(2)}
\end{align}
$$ \text{From (1):} $$
\begin{align}
2a^2 + 2b + 1 &= 1 \\
2a^2 &= -2b + 1 - 1 \\
2a^2 &= -2b \\
a^2 &= -b \leftarrow\text{(3)}
\end{align}
$$ \text{Substitute (3) into (2):} $$
\begin{align}
3(-b) + 3b + 1 &= 4 \\
-3b + 3b + 1 &= 4 \\
1 &= 4
\end{align}
Here is my second attempt at solving it:
\begin{align}
ax^2 + bx + 1 &= 1 \\
2a^2 + 2b + 1 &= 1 \\
a^2 + b + \frac{1}{2} &= \frac{1}{2} \\
a^2 + b &= \frac{1}{2} - \frac{1}{2} \\
a^2 + b &= 0 \\
4a^2 + 4b &= 0 \\
4a^2 + 4b + 1 &= 1 \leftarrow\text{(1)}
\end{align}
\begin{align}
3a^2 + 3b + 1 &= 4 \\
3a^2 + 3b &= 4 - 1 \\
3a^2 + 3b &= 3 \\
a^2 + b &= 1 \\
4a^2 + 4b &= 4 \\
4a^2 + 4b + 1 &= 5 \leftarrow\text{(2)}
\end{align}
$$ \text{(1) = (2): } 1 = 5 $$
Where did I go wrong?
| $$ 4a + 2b + 1 = 1 \leftarrow\text{(1)} $$
$$ 9a + 3b + 1 = 4 \leftarrow\text{(2)} $$
$$ 4a + 2b = 0 $$
$$ 9a + 3b = 3 $$
$$a=1, b=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove the inequality $1\le\int_1^4 \frac{1}{1+\sqrt(x)} \,dx$ I want to prove the inequality:
$$1\le\int_1^4 \frac{1}{1+\sqrt{x}} \,dx$$
This is my attempt:
The domain is $(0,\infty)$ and the range is $[0,1]$
So, $\frac{1}{1+\sqrt{x}} \ge \frac{1}{1+\sqrt{9}}=\frac{1}{4}$ (used a value for x=9)
So, $1\ge\int_1^4 \frac{1}{1+\sqrt{x}} \,dx \ge \int_1^4 \frac{1}{4} \,dx = \frac{3}{4}$, but this is not a proof. I don't think this is the way I should calculate this inequality
Can anyone help me where I got it wrong? Am I on the right track? Or am I way wrong?
Many thanks:)
| The idea is correct but you picked $x=9$ instead of $x=4$. We have for $x\in [1,4]$
$$\frac{1}{1+\sqrt{x}}\geq \frac{1}{1+\sqrt{4}}=\frac{1}{1+2}=\frac{1}{3}$$
Then
$$\int_1^4 \frac{1}{1+\sqrt{x}}dx\geq \int_1^4 \frac{1}{3}dx =\frac{3}{3}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.
| $$\frac{1}{\sin{x}}-\frac{1}{\sin\left(\frac{\pi}{3}+x\right)}+\frac{1}{\sin\left(\frac{\pi}{3}-x\right)}=$$
$$=\frac{\sin\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{3}+x\right)+\sin{x}\left(\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)\right)}{\sin{x}\sin\left(\frac{\pi}{3}+x\right)\sin\left(\frac{\pi}{3}-x\right)}=$$
$$=\frac{\frac{1}{2}\left(\cos2x-\cos\frac{2\pi}{3}\right)+\sin{x}\cdot2\sin{x}\cos\frac{\pi}{3}}{\sin{x}\left(\frac{\sqrt3}{2}\cos{x}+\frac{1}{2}\sin{x}\right)\left(\frac{\sqrt3}{2}\cos{x}-\frac{1}{2}\sin{x}\right)}=$$
$$=\frac{\frac{1}{2}-\sin^2x+\frac{1}{4}+\sin^2x}{\sin{x}\left(\frac{3}{4}\cos^2x-\frac{1}{4}\sin^2x\right)}=\frac{3}{\sin{x}(3(1-\sin^2x)-\sin^2x)}=\frac{3}{\sin3x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Limit of multiple absolute values Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.
I broke up each absolute value into parts:
$|1-x^2| = \begin{cases}
1-x^2 & -1 \leq x\leq 1 \\
-(1-x^2) & x>1,x<-1
\end{cases}
$ ,
$|x+1| = \begin{cases}
x+1 & x\geq -1 \\
-(x+1) & x<-1
\end{cases}
$,
$|x^2+x| = \begin{cases}
x^2+x & x\geq0,x\leq -1 \\
-(x+1) & -1<x<0
\end{cases}
$
Thus, when $x\rightarrow -1$, the function will approach the positive value, because by my definitions of the absolute values,each has the positive value at $x=-1$. So then you can take
$\lim_{x\rightarrow -1}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1}\frac{x^2+2x-3(1-x^2)+1}{2(x+1)-(x^2+x)}=\lim_{x\rightarrow -1}\frac{(4x-2)(x+1)}{(x+1)(-x+2)}=-2$.
But looking at the graph, the limit is -6. So I must have messed up in my absolute value declarations, most likely in the third one.
So, how would I correctly declare and solve this limit?
| As an alternative, we have that by $x=y-1$ with $y\to 0$
$$\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\frac{y^2-3|2y-y^2|}{2|y|-|y(y-1)|}=\frac{|y|-3|2-y|}{2-|y-1|}\to \frac{0-6}{2-1}=-6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The Finite Sum $\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$ and failure to Telescope I'm interested in when the partial fraction method of trying to get a series to telescope fails, and have alighted upon the interesting example of,
$$\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$$
The standard method of tackling this would be to use partial fractions to get,
$$\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}=\frac{1}{4}\big(\sum_{r=1}^{n}\frac{1}{3r-2}-\sum_{r=1}^{n}\frac{1}{3r+2}\big)$$
and then look at the fractions generated by this alternative representation of the sum,
$$=\frac{1}{4}\big(\big(\frac{1}{1}-\frac{1}{5}\big)+\big(\frac{1}{4}-\frac{1}{8}\big)+\big(\frac{1}{7}-\frac{1}{8}\big)+\big(\frac{1}{10}-\frac{1}{14}\big)+...+\big(\frac{1}{3n-2}-\frac{1}{3n+2}\big)\big)$$
To me, this does not obviously telescope.
Questions
*
*I am was wondering if anyone can see a way to get this to telescope.
*Failing that, have a nice explanation of why it will not telescope.
*Is there a way of evaluating the sum to $n$ terms via a different approach ?
An Observation
In exploring the series I put it into Wolfram Alpha which reported that the series is convergent and that,
$$\sum_{r=1}^{\infty}\frac{1}{(3r-2)(3r+2)}=\frac{1}{72}\big(2\sqrt{3}\pi+9\big)$$
Given the famous Basel problem for the sum of reciprocals of squares, the appearance of $\pi$ is not, perhaps, a surprise.
| We can write the sum as
\begin{align*}
\sum_{r = 1}^{n}\dfrac{1}{(3r-2)(3r+2)} &= \dfrac{1}{4}\sum_{r = 1}^{n}\left[\dfrac{1}{(3r-2)} - \dfrac{1}{(3r+2)}\right]
\\
&= \dfrac{1}{4}\sum_{r = 1}^{n}\int_{0}^{1}(x^{3r-3}-x^{3r+1})\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\sum_{r = 1}^{n}(x^{3r-3}-x^{3r+1})\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\dfrac{(1-x^4)-(x^{3n}-x^{3n+4})}{1-x^3}\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\dfrac{1-x^4}{1-x^3}\,dx - \dfrac{1}{4}\int_{0}^{1}x^{3n}\dfrac{1-x^4}{1-x^3}\,dx.
\end{align*}
You can easily show that $$0 \le \int_{0}^{1}x^{3n}\dfrac{1-x^4}{1-x^3}\,dx \le \int_{0}^{1}x^{3n}\sup_{x \in [0,1]}\left[\dfrac{1-x^4}{1-x^3}\right]\,dx = \int_{0}^{1}\dfrac{4}{3}x^{3n}\,dx = \dfrac{4}{3(3n+1)}$$ for all integers $n \ge 0$, and thus, $$\int_{0}^{1}x^{3n}\dfrac{1-x^4}{1-x^3}\,dx \to 0 \quad \text{as} \quad n \to \infty.$$
Therefore, $$\sum_{r = 1}^{\infty}\dfrac{1}{(3r-2)(3r+2)} = \dfrac{1}{4}\int_{0}^{1}\dfrac{1-x^4}{1-x^3}\,dx = \dfrac{9+2\pi\sqrt{3}}{72},$$ which can be easily evaluated using partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
$\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no ideas for a further solution.
| $\dfrac{1}{(3n+1)(3n+4)}=\dfrac{1}{3}\bigg(\dfrac{1}{3n+1}-\dfrac{1}{3n+4}\bigg)$
$\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\bigg)$ and so on ...
EDIT
$\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+...+\dfrac{1}{(3n+1)(3n+4)}=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{3n+1}-\dfrac{1}{3n+4}\bigg)=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{3n+4}\bigg)$
So $\lim\limits_{n\to\infty}\bigg(\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+...+\dfrac{1}{(3n+1)(3n+4)}\bigg)=\lim\limits_{n\to\infty}\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{3n+4}\bigg)=\dfrac{1}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$
*
*Using AM-GM Inequality, we have:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$
*
*Therefore, $\min P = -8$
Is this solution correct? If not, then why?
| WLOG $a\ge b\ge c$ and let $x=a-b,y=b-c,z=c-a$
We observe $x+y+z=0$ with $x,y\ge 0$.and as you have found $x^2+y^2+z^2=12$
Elimination of $z$ results in: ${(x+y)}^2=6+xy ...(1)$
since ${(x+y)}^2\ge 4xy$ which means $0\le xy\le 2$
let $xy=t$
since $0\le t\le 2$
Now $x^2y^2z^2=t^2(6+t)\le 6.2^2+2^3=32$....(using (1) and $z=-(x+y)$)
or $|xyz|\le 4\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Using complex numbers, can sine be greater than $1?$ I worked on a brain teaser that started with $\sin{(z)}=10.$ It then solved for $\cos{(z)}=\sqrt{-99}.$
I tried solving for $z,$ using $z=x+iy$ then $z=ae^{bi}$ but got nowhere. So is this even possible?
| Yes, it can.
Let us find such $z$ so that $\sin(z) = 10$.
Firstly, as a consequence of the Euler's formula:
$ e^{iz} = \cos(z) + i \cdot \sin(z)$,
we have the following two identities which hold for all complex numbers $z$:
$$
\sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \\
\cos(z) = \frac{e^{iz} + e^{-iz}}{2}
$$
Then:
$$
\sin(z) = 10 \\
\frac{e^{iz} - e^{-iz}}{2i} = 10 \\
e^{iz} - e^{-iz} - 20i = 0 \\
(e^{iz})^2 - 20i\cdot(e^{iz}) - 1 = 0 \\
$$
Solving the quadratic equation:
$$
\sqrt{D}=\sqrt{(-20i)^2 - 4 \cdot 1 \cdot (-1)} = \sqrt{-396} = 6i\sqrt{11} \\
(e^{iz}) = \frac{-(-20i) \pm \sqrt{D}}{2} = i \cdot (10 \pm 3\sqrt{11}) \\
$$
Complex logarithm:
$$
iz = Ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right)
= \ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) +
i \cdot 2\pi k,
\qquad k \in \mathbb{Z}\\
iz = \ln(i) + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\
iz = \ln\left(e^{i\frac{\pi}{2}}\right) +
\ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\
iz = i\frac{\pi}{2} + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\
$$
The result:
$$
z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(10 \pm 3\sqrt{11}\right),
\qquad k \in \mathbb{Z}
$$
$\cos(z) = \sqrt{-99}$ can be solved in exactly the same way, giving the result:
$$
z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(3\sqrt{11} \pm 10\right),
\qquad k \in \mathbb{Z}
$$
The idea is taken from this video:
https://www.youtube.com/watch?v=3C_XD_cCeeI
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer?
Here how I work, please correct me if I'm wrong
First method :
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\sin x \cdot \sin^2x}\,\mathrm{d}x
\\ &= \int{\sin x (1 - \cos^2x)}\,\mathrm{d}x
\\& = \displaystyle\int{(\sin x - \sin x\cos^2x)}\,\mathrm{d}x
\\& = \dfrac{1}{3}\cos^3x - \cos x + C
\end{align}
Second method :
First, we know that $$\sin 3x = 3\sin x - 4\sin^3x$$
Therefore, $$\sin^3x = \dfrac{3}{4}\sin x - \dfrac{1}{4}\sin 3x$$
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\left(\frac{3}{4}\sin x - \frac{1}{4}\sin 3x\right)}\,\mathrm{d}x\\
& = \frac{1}{12}\cos 3x - \frac{3}{4}\cos x + C
\end{align}
| Since $$\cos^3(x)=\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)$$
Your first integral becomes $$\int \sin^3(x)dx=\dfrac{1}{3}\cos^3x - \cos x + C$$
$$=\frac{1}{3}\big[\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)\big]-\cos(x)+C$$
$$=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)+C$$
Note that the constant of integration are not necessarily the same. For example using $u$-substitutions for the denominator we have $$\int \frac{4x}{4x^2+7}dx=\frac{1}{2}\ln(4x^2+7)+C_{1}$$
$$\int \frac{x}{x^2+\frac{7}{4}}dx=\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}$$
Here we have $C_{2}=C_{1}+\frac{1}{2}\ln(4)$ since they are constants. Indeed we have $$\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}=\frac{1}{2}\ln(x^2+\frac{7}{4})+\frac{1}{2}\ln(4)+C_{1}$$
$$=\frac{1}{2}\big[\ln(x^2+\frac{7}{4})+\ln(4)\big]+C_{1}$$
$$=\frac{1}{2}\ln(4(x^2+\frac{7}{4}))+C_{1}$$
$$=\frac{1}{2}\ln(4x^2+7)+C_{1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
How to calculate the integral $\int_0^{\infty}\frac{x^{1/2}}{1-x^2}\sin(ax)\sin[a(1-x)] dx$ How can I calculate the following integral:
$$\int_0^{\infty}\frac{x^{1/2}}{1-x^2}\sin(ax)\sin[a(1-x)] dx$$
where $a>0$.
It seems like the integrand is well defined without any singularities, but I don't have any clue how to proceed.
Can anyone show me how to do it? Thank you
Edit
Following exactly what @Maxim suggested in the comment
$$\int_0^\infty f(x) dx = \operatorname {Re} \operatorname {v. \! p.} \int_0^\infty g(x) dx = \operatorname {Re} \left(\int_0^{i \infty} g(x) dx + \pi i \operatorname* {Res}_{x = 1} g(x) \right).$$
taking $x=iu$
\begin{align}
\int_0^{i\infty}g(x) dx &=\frac{1}{2}ie^{i\frac{\pi}{4}-a}\int_0^{\infty}\frac{\sqrt{u}e^{-2au}}{1+u^2} du-\frac{1}{2}ie^{i\frac{\pi}{4}}\cos (a) \int_0^{\infty}\frac{\sqrt{u}}{1+u^2} du\\
&=\frac{1}{2}ie^{i\frac{\pi}{4}-a}I-i\frac{\sqrt{2}\pi}{4} \cos (a)e^{i\frac{\pi}{4}}
\end{align}
The integral $I=\int_0^{\infty}\frac{\sqrt{u}e^{-2au}}{1+u^2} du$ can be calculated using the method in the comment of this question.
Let $f_1(u)=\sqrt{u}$ and $g_1(u)=e^{-2au}/(1+u^2)$
\begin{align}
I &=\int_0^{\infty} f_1(u)g_1(u)du\\
&=\int_0^\infty \mathcal L[f](u) \mathcal L^{-1}[g](u) dx\\
&=\frac{\sqrt{\pi}}{2}\int_{2a}^{\infty}\frac{\sin (u-2a)}{u^{3/2}}
\end{align}
then let $u=x^2$
\begin{align}
I &=\sqrt{\pi}\int_{\sqrt{2a}}^{\infty} \frac{\sin (x^2-2a)}{x^2}dx\\
&=\sqrt{\pi}\cos(2a)\int_{\sqrt{2a}}^{\infty}\frac{\sin (x^2)}{x^2}dx-\sqrt{\pi}\sin(2a)\int_{\sqrt{2a}}^{\infty}\frac{\cos (x^2)}{x^2}dx\\
&=\sqrt{\pi}\cos(2a)I_1-\sqrt{\pi}\sin(2a)I_2
\end{align}
Making use of the parameterization technique, we have:
$$I_1=\frac{\sqrt{2\pi}}{2}-2C(\sqrt{2a})+\frac{\sin^2 (\sqrt{2a})}{\sqrt{\pi a}}$$
$$I_2=-\frac{\sqrt{2\pi}}{2}+2S(\sqrt{2a})+\frac{\cos^2 (\sqrt{2a})}{\sqrt{\pi a}}$$
Hence, we can obtain the desired result by substituting $I_1, I_2$ and $I$ into original integral.
My qustion is:
If I am only interested in the asymptotic behavior of the original integral as $a\to +\infty$, is there any simplier way to do it without going through all these steps?
| Following exactly what @Maxim suggested in the comment
$$\int_0^\infty f(x) dx = \operatorname {Re} \operatorname {v. \! p.} \int_0^\infty g(x) dx = \operatorname {Re} \left(\int_0^{i \infty} g(x) dx + \pi i \operatorname* {Res}_{x = 1} g(x) \right).$$
taking $x=iu$
\begin{align}
\int_0^{i\infty}g(x) dx &=\frac{1}{2}ie^{i(\frac{\pi}{4}-a)}\int_0^{\infty}\frac{\sqrt{u}e^{-2au}}{1+u^2} du-\frac{1}{2}ie^{i\frac{\pi}{4}}\cos (a) \int_0^{\infty}\frac{\sqrt{u}}{1+u^2} du\\
&=\frac{1}{2}ie^{i(\frac{\pi}{4}-a)}I-i\frac{\sqrt{2}\pi}{4} \cos (a)e^{i\frac{\pi}{4}}
\end{align}
The integral $I=\int_0^{\infty}\frac{\sqrt{u}e^{-2au}}{1+u^2} du$ can be calculated using the method in the comment of this question.
Let $f_1(u)=\sqrt{u}$ and $g_1(u)=e^{-2au}/(1+u^2)$
\begin{align}
I &=\int_0^{\infty} f_1(u)g_1(u)du\\
&=\int_0^\infty \mathcal L[f_1](u) \mathcal L^{-1}[g_1](u) dx\\
&=\frac{\sqrt{\pi}}{2}\int_{2a}^{\infty}\frac{\sin (u-2a)}{u^{3/2}}
\end{align}
then let $u=x^2$
\begin{align}
I &=\sqrt{\pi}\int_{\sqrt{2a}}^{\infty} \frac{\sin (x^2-2a)}{x^2}dx\\
&=\sqrt{\pi}\cos(2a)\int_{\sqrt{2a}}^{\infty}\frac{\sin (x^2)}{x^2}dx-\sqrt{\pi}\sin(2a)\int_{\sqrt{2a}}^{\infty}\frac{\cos (x^2)}{x^2}dx\\
&=\sqrt{\pi}\cos(2a)I_1-\sqrt{\pi}\sin(2a)I_2
\end{align}
Making use of the parameterization technique, we have:
$$I_1=\frac{\sqrt{2\pi}}{2}-2C(\sqrt{2a})+\frac{\sin (2a)}{\sqrt{2a}}$$
$$I_2=-\frac{\sqrt{2\pi}}{2}+2S(\sqrt{2a})+\frac{\cos (2a)}{\sqrt{2a}}$$
Substituting $I_1$ and $I_2$ into the expression of $I$ yields
$$I=\sqrt{\pi}\cos(2a)\left[\frac{\sqrt{2\pi}}{2}-2C(\sqrt{2a})+\frac{\sin (2a)}{\sqrt{2a}}\right]-\sqrt{\pi}\sin(2a)\left[-\frac{\sqrt{2\pi}}{2}+2S(\sqrt{2a})+\frac{\cos (2a)}{\sqrt{2a}}\right]$$
Finally, the desired integral can be written as:
$$\int_0^{\infty}f(x)dx=\frac{\sqrt{2}}{4}(\sin a -\cos a)I + \frac{\pi}{4}\sin a + \frac{\pi}{4}\cos a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Efficiently enumerate the boolean matrix of sparsity k such that no rows and columns are all 0s. Is there a way to enumerate the boolean matrix of $k$ entries of $1$ with no rows and columns all being $0$s?
e.g.
$k=1$, $\begin{bmatrix}1\end{bmatrix}$
$k=2$, $\begin{bmatrix}1 & 1\end{bmatrix}$
$\begin{bmatrix}1 \\ 1\end{bmatrix}$
$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$
$\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$k=3$, $\begin{bmatrix}1 & 1 & 1\end{bmatrix}$
$\begin{bmatrix}1 & 1 & 0\\ 0&0&1\end{bmatrix}$
$\begin{bmatrix}1 & 1\\0&1\end{bmatrix}$
$\begin{bmatrix}1 & 0&1\\0&1&0\end{bmatrix}$
$\begin{bmatrix}1 & 1\\1&0\end{bmatrix}$
$\begin{bmatrix}0&1 & 1\\1&0&0\end{bmatrix}$
$\begin{bmatrix}1&0&0\\0&1 & 1\end{bmatrix}$
$\begin{bmatrix}1&0\\1 & 1\end{bmatrix}$
$\begin{bmatrix}0&1&0\\1&0 & 1\end{bmatrix}$
$\begin{bmatrix}0&1\\1 & 1\end{bmatrix}$
$\begin{bmatrix}0&0&1\\1 & 1&0\end{bmatrix}$
$\begin{bmatrix}1 \\ 1\\1\end{bmatrix}$
$\begin{bmatrix}0&1 \\0& 1\\1&0\end{bmatrix}$
$\begin{bmatrix}0&1 \\ 1&0\\0&1\end{bmatrix}$
$\begin{bmatrix}1&0 \\ 0&1\\0&1\end{bmatrix}$
$\begin{bmatrix}0&1 \\ 1&0\\1&0\end{bmatrix}$
$\begin{bmatrix}1&0 \\ 0&1\\1&0\end{bmatrix}$
$\begin{bmatrix}1&0 \\ 1&0\\0&1\end{bmatrix}$
$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$
$\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}$
$\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$
$\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$
$\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}$
$k=4$,...
There is a trivial inductive approach that we can just consider where to put the $k$th $1$ based on all of the $(k-1)$th configurations. However, this will make $k!$ duplications for each unique configuration.
Another possible approach is that we can maybe partition $k$ by the rows and columns first. Say $k=4$, partition the rows into $2,1,1$ and cols into $1,1,2$ then enumerate the $3 \times 3$ configurations that satisfy the row col number of $1$s correspondingly. But we still need to consider how to enumerate the balanced matrix efficiently.
Now is there an easy way to enumerate all possibilities? If not, can we find an easy enumerator for just half of all the cases?
| Here's one possible scheme.
We'll fill in the $1$s of the up-to-$n$-by-$n$ matrix one at a time, but going in lexicographic order. The first $1$ can be in any of the positions $(1,1), (1,2), \dots, (1,n)$. If we just placed a $1$ in position $(i,j)$, the next $1$ can be in any of the positions $$(i,j+1), (i,j+2), \dots, (i,n), (i+1,1), (i+1,2), \dots, (i+1,n).$$ This ensures that no zero rows are created, but it doesn't do anything about zero columns. So we must do something about that as well...
Let's keep track of which columns (up to the last column used) are empty so far, and how many of them there are. This has two effects:
*
*If there are $a$ more $1$s to enter after the current one, $b$ empty columns, and the rightmost $1$ already entered appears in column $c$, the current $1$ should be entered no further right than column $c+(a-b)+1$ (creating up to $b-a$ more empty columns).
*If $a$ ever equals $b$, we switch to a mode in which we can only place a $1$ to fill an existing empty column.
How efficient is this scheme? Each possible matrix we want will be listed exactly once. So the only overhead is the overhead of keeping track of the empty columns described above. We can definitely limit this to $O(n)$ processing per $1$ we place, and therefore $O(n^2)$ per matrix, but we could probably be more efficient as well...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use the sum of the first ten terms to approximate the summation $\sum_1^{\infty} \frac{8n}{(n+1)3^n}$ Use the sum of the first ten terms estimate the summation and then to estimate the error
$\sum_1^{\infty} \frac{8n}{(n+1)3^n}$
Okay, so i added up the first ten terms and got approximately 1.101237.
The answer key says:
Now, $ \frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n}$, and so the error is $R_{10} = \sum_{11}^{\infty} \frac{7}{3^n} \cong 7(\frac{\frac{1}{3^{11}}}{1-\frac{1}{3}}) \cong .000059$
Can somebody explain what's going on here? Why are we allowed and how do we know to make the simplification $\frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n}$? Will the infinite summations of these two things be the same since $n$ is going to infinity anyway?? It seemed a bit arbitrary. My textbook doesn't say anything about this. Thanks!!
| This method leads to a conservative estimation for the error since
$$\frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n} \implies R_{10, exact} = \sum_{n=11}^{\infty} \frac{8n}{(n+1)3^n}< R_{10, approx} = 8\sum_{n=11}^{\infty} \frac{1}{3^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
When is $k^4-24k+16$ a perfect square. When is the equation $k^4-24k+16$ perfect square.
(k is an integer.)
I got this equation as discriminant while solving an equation. I tried to solve it but couldn't i tried to write it in a form of a square but couldnt solve it.I bashed a bit and found 0 and 3 as a solution. Any help would be appreciated. Thanks in advance.
| I think this is a correct solution. Suppose that $k^4-24k+16$ is a square.
*
*Suppose that $-24k+16\ge 0$. Then $k^4-24k+16\ge (k^2+1)^2=k^4+2k^2+1$, so $-24k+16\ge 2k^2+1$ or $2k^2+24k-15\le 0$ or $k^2+12k-7.5\le 0$. This means $$-6-\sqrt{36+7.5}\le k\le -6+\sqrt{36+7.5}.$$
Which means $-13\le k\le 0$ (remember that $-24k+16>0$). Checking all these $k$ we get that only $k=-3,0$ are good.
*Suppose that $-24k+16<0$ (in particular $k\ge 0$). Then $k^4-24k+16\le (k^2-1)^2=k^4-2k^2+1$, so $2k^2-24k+15\le 0$, or $k^2-12k+7.5\le 0$. Thus $6-\sqrt{36-7.5} \le k\le 6+\sqrt{36-7.5}$. So $1\le k\le 11$. Checking all these $k$, we get $k=3$.
Thus there are three possibilities: $k=-3,0,3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Subsets and Splits
Fractions in Questions and Answers
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