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Prove that ${e\over {\pi}}\lt{\sqrt3\over{2}}$ without using a calculator. I have been working on a known question for a long time (this is "Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator") during this time I realized the ${e\over {\pi}}\lt{\sqrt3\over{2}}$.
I have no solution so far. Do you have any idea?
|
Calculation without any computer, only with patience. :-D
$\displaystyle e<\frac{\sqrt{3}}{2}\pi\enspace$ is equivalent to $\enspace\displaystyle \frac{2}{9}\sum\limits_{n=0}^\infty \frac{2^n}{n!} =\frac{2e^2}{9}<\frac{\pi^2}{6}=\zeta(2)$
Case $\,(A)\,$ :
We have $\enspace\displaystyle \prod\limits_{k=0}^n \frac{k+9}{4}>1\enspace$ for all $\,n\geq 0\,$ and therefore $\,4^{n+1}8!<(n+9)!\,$ .
It follows $\enspace\displaystyle \frac{2^{n+1}8!}{(n+9)!}<\frac{1}{2^{n+1}}\enspace$ and with the summation over $\,n=0,1,2,…\,$ we get
$\displaystyle \sum\limits_{n=0}^\infty\frac{2^{n+1}8!}{(n+9)!}< \sum\limits_{n=0}^\infty \frac{1}{2^{n+1}}=1\enspace$ which leads to $\enspace\displaystyle e^2 = \sum\limits_{n=0}^\infty \frac{2^n}{n!} < \sum\limits_{n=0}^7 \frac{2^n}{n!} + \frac{2^9}{8!} \,$ .
Case $\,(B)\,$ :
The Euler–Maclaurin formula (https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula) gives us a good approximation for $\,\zeta(2)\,$ and we get
$\displaystyle \sum\limits_{n=0}^4 \frac{1}{n^2} + \frac{1}{5} + \frac{1}{50} + \frac{1}{6\cdot 125} - \frac{1}{30\cdot 5^5}<\zeta(2)\,$ .
Cases $\,(A)\,$ and $\,(B)\,$ together :
Now we have to prove the middle part of the following inequality.
$\displaystyle \frac{2e^2}{9} = \frac{2}{9}\sum\limits_{n=0}^\infty \frac{2^n}{n!} < \frac{2}{9}\left(\sum\limits_{n=0}^7 \frac{2^n}{n!} + \frac{2^9}{8!}\right)<$
$\displaystyle <\sum\limits_{n=0}^4 \frac{1}{n^2} + \frac{1}{5} + \frac{1}{50} + \frac{1}{6\cdot 125} - \frac{1}{30\cdot 5^5}<\zeta(2)=\frac{\pi^2}{6}\,$
or simply $\enspace\displaystyle \frac{2e^2}{9} < \frac{4658}{2835} < \frac{3701101}{2250000} < \frac{\pi^2}{6}$
And with $\,4658\cdot 2250000 =10480500000 < 10492621335=2835\cdot 3701101$
the claim is proofed.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Compute $\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)$ I note that $\sqrt{x^2-6x+9}=|x-3|$. Splitting upp the limit into cases gives
*
*$x\geq 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(|x-3|+x-1)=2\lim_{x\rightarrow -\infty}(x-2)=-\infty.$$
*
*$x< 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(-x+3+x-1)=\lim_{x\rightarrow -\infty}2=2.$$
I get two different values of the limit, but plotting the function clearly shows that the answer should be $2$.
|
Here is a different approach, setting $x=-t$, the limit becomes:
$$\lim_{t \to \infty} \sqrt{t^2+6t+9} - t - 1$$
Since $t^2+6t+9=(t+3)^2$, we can say for positive t-values that $\sqrt{(t+3)^2}=t+3$, your limit expression becomes $t+3-t-1=2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the limit $\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$, using epsilon-delta definition. $$\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$$
The proof that I have:
Let $\varepsilon > 0$, we must show that
$$\exists \delta >0: 0<x-1<\delta \Rightarrow \left | \frac{1}{\sqrt{x}}-1\right|<\epsilon$$
So usually, when doing these $\varepsilon,\delta$ proofs, I would write $\ldots 0<|x-1|<\delta \ldots$ , but as $x\to 1+$, $x-1$ should always be greater than $0$. Is that correct? Can I just take $|x-1|=x-1$?
We see that
$$ \left | \frac{1}{\sqrt{x}}-1\right|= \left | \frac{1-\sqrt{x}}{\sqrt{x}}\right|=\left | \frac{-(\sqrt{x}-1)}{\sqrt{x}}\right|=\left | \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}\right|=\left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|\left(<\varepsilon\right)$$
We know that $0<x-1<\delta$. But we don't know the estimation for $\frac{1}{\sqrt{x}(\sqrt{x}+1)}$. Let $\delta \leq 1$, so
$$
\begin{align}
0<&x-1<\delta\leq 1\\
0<&x-1<1\\
1<&x<2\\
1<&\sqrt{x}<\sqrt{2}\\
\frac{1}{\sqrt{2}}<&\frac{1}{\sqrt{x}}<1
\end{align}
$$
If I had $|x-1|<\delta$, I would take $\delta\leq\frac{1}{2}$ and get
$$
\begin{align}
&|x-1|<\frac{1}{2}\\
-\frac{1}{2}<&x-1<\frac{1}{2}\\
\frac{1}{2}<&x<\frac{3}{2}\\
\frac{1}{\sqrt{2}}<&\sqrt{x}<\frac{\sqrt{3}}{\sqrt{2}}\\
&\vdots
\end{align}
$$
The estimation for $\frac{1}{\sqrt{x}}$ is known, let's do the same for $\frac{1}{\sqrt{x}+1}$:
$$
\begin{align}
1<&\sqrt{x}<\sqrt{2}\\
2<&\sqrt{x}+1<\sqrt{2}+1\\
\frac{1}{\sqrt{2}+1}<&\frac{1}{\sqrt{x}+1}<\frac{1}{2}
\end{align}
$$
So the estimation for $\frac{1}{\sqrt{x}(\sqrt{x}+1)}$ is
$$
\frac{1}{\sqrt{x}(\sqrt{x}+1)}<\frac{1}{2}
$$
Finally we get
$$
\left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|<\frac{\delta}{2}\leq \varepsilon $$
$$\delta:=\min\{1,2\varepsilon\}$$
How to prove it? Is my approach correct? Any pointers, when doing one-sided limit proofs?
|
Corrections:
We assume that $x$ is chose from $0<x-1<1$:
$$
\begin{align}
0<&x-1<1\\
1<&x<2\\
1<&\sqrt{x}<\sqrt{2}\\
\frac{1}{\sqrt{2}}<&\frac{1}{\sqrt{x}}<1\\
2<&\sqrt{x}+1<\sqrt{2}+1\\
\frac{1}{\sqrt{2}+1}<&\frac{1}{\sqrt{x}+1}<\frac{1}{2}
\end{align}
$$
$$\left | \frac{1}{\sqrt{x}}-1\right|= \left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|<\dfrac{|x-1|}{2}<\dfrac{\delta}{2}<\varepsilon$$
$$\delta:=\min\{1,2\varepsilon\}$$
|
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|
Quadratics and roots Consider the equation (E): $$x^2 - (m+1)x+m+4=0$$ where $m$ is a real parameter
determine $m$ so that $2$ is a root of (E) and calculate the other root.
This is the question.
What I did was basically this:
Let the sum of root 1 and root 2 be $S$ and their product $P$
Let $x_2 = a ; x_1=2$(given)
*
*$S=m+1$
$m+1=2+a$
$m-a=1$
*$P=m+4$
$m+4=2a$
$m-2a=-4$
then these 2 equations form a system whose answers would be $m=6$ and $a=5$.
Is it possible to determine $m$ so that $x^2−(m+1)x+m+4<0$ for all $x \in \mathbb{R}$?
|
If $2$ is a root of $x^2-(m+1)x+(m+4)=0$ then
\begin{align}
&2=\frac{(m+1)\pm\sqrt[\;2]{(m+1)^2-4\cdot (m+4)}}{2}
\\
\Leftrightarrow&
4=(m+1)\pm \sqrt[\;2]{m^2-2m-15}
\\
\Leftrightarrow&
-m+3 =\pm \sqrt[\;2]{m^2-2m-15}
\\
\Leftrightarrow&
(-m+3)^2 =\left(\pm \sqrt[\;2]{m^2-2m-15}\right)^2
\\
\Leftrightarrow&
(-m+3)^2 =\left|m^2-2m-15\right|
\end{align}
Case 1: If $m^2-2m-15=(m+3)(m-5)> 0$ then we have
\begin{align}
(-m+3)^2 =+(m^2-2m-15),& \hspace{1cm} m<-3 \mbox{ or } m>5\\
m^2-6m+9 = m^2-2m-15, & \hspace{1cm} m<-3 \mbox{ or } m>5\\
-4m = -24, & \hspace{1cm} m<-3 \mbox{ or } m>5\\
m = 6, & \hspace{1cm} m<-3 \mbox{ or } m>5\\
\end{align}
Case 2: If $m^2-2m-15=(m+3)(m-5)< 0$ then we have
\begin{align}
(-m+3)^2 =-(m^2-2m-15),&\hspace{1cm} -3<m<5\\
m^2-6m+9 =-m^2+2m+15,&\hspace{1cm} -3<m<5\\
2m^2-8m-6 =0,&\hspace{1cm} -3<m<5\\
m^2-4m-3 =0,&\hspace{1cm} -3<m<5\\
m=\frac{4\pm \sqrt{4^2-4\cdot 1\cdot (-3)}}{2\cdot 1},&\hspace{1cm} -3<m<5\\
m=2\pm \sqrt{7},&\hspace{1cm} -3<m<5\\
\end{align}
In this case there is no integer solution.
Case 3: If $m^2-2m-15=(m+3)(m-5)=0$ then we have
\begin{align}
(-m+3)^2 =0,& \hspace{1cm} m=-3 \mbox{ or } m=5\\
m=3,& \hspace{1cm} m=-3 \mbox{ or } m=5\\
\end{align}
But it's impossible.
|
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|
Probability of a target being hit I don't have the answer of the following question so I wanted to cross check my solution.
$A$ can hit a target 3 times in 5 shots, $B$ 2 times in 5 shots and $C$ 3 times in 4 shots. Find the probability of the target being hit at all when all of them try.
my method
P(target being hit) = $$ \frac{3}{5}*\frac{3}{5}*\frac{1}{4} + \frac{3}{5}*\frac{3}{5}*\frac{3}{4} + \frac{3}{5}*\frac{2}{5}*\frac{1}{4} + \frac{3}{5}*\frac{2}{5}*\frac{3}{4} + \frac{2}{5}*\frac{3}{5}*\frac{3}{4} + \frac{2}{5}*\frac{2}{5}*\frac{1}{4} = 0.82$$
$$ = HMM + HMH + HHM + HHH + MMH + MHM$$
where, H = hit & M = miss
|
Let $a$ indicate that person $A$ hit the target, $b$ indicate that $B$ hit the target, and $c$ that $C$ hit the target.
We expect $a$ to occur $3/5$ of the time, $b$ to occur $2/5$ of the time and $c$ to occur $3/4$ of the time.
The event “at least one hits the target” is the complement of “none of them hits the target,” so
$$\begin{align}
\operatorname{P}(a\cup b\cup c) &= 1-\operatorname{P}\left([a\cup b\cup c]’\right) \\
&= 1-\operatorname{P}\left( a’\cap b’ \cap c’\right)\\
&=1- \operatorname{P}\left( a’ \right) \, \operatorname{P} \left( b’ \right) \, \operatorname{P} \left( c’ \right) \\
&= 1-\left( 1-\frac35 \right) \left( 1-\frac25 \right) \left( 1-\frac34 \right) \\
&= 1-\frac25 · \frac35 · \frac14 \\
&= 1-\frac{3}{50} \\
&= \frac{47}{50} \\
\end{align}$$
|
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|
No. of isosceles triangles possible of integer sides with sides $\leq n$ Prove that the no. of isosceles triangles with integer sides, no sides exceeding $n$ is $\frac{1}{4}(3n^2+1)$ or $\frac{3}{4}(n^2)$ according as n is odd or even, n is any integer.
How to do it? I found that under these conditions no. of triangles possible may be
${n\choose 2}$
|
${n \choose 2}$ is the number of triplets $(k,k, m)$. But not all triplets can be triangles. To be a triangle i) $m < k + k = 2k$ and ii) $k < k + m$. (i) is a essential, ii) is trivially redundant).
So we need to find all possible triplets $(k,k,m)$ where $k,m \le n$ and $m < 2k$. As $m$ is an integer, that means $m \le 2k-1$ and $m \le n$.
That is $\sum_{k=1}^n\sum_{m=1}^{\min (n, 2k - 1)}1=\sum_{k=1}^n{\min (n, 2k - 1)}$
$\sum_{k= 1; k \le \frac n2} (\min (n, 2k - 1)) + \sum_{k> \frac n2}^n(\min(n,2k-1)) =$
$\sum_{k=1; k \le \frac n2} (2k-1) + \sum_{k>\frac n2}^n n$
Case 1: $n$ is even
$= [\sum_{k=1}^{\frac n2}(2k-1)] + [\sum_{k=\frac n2 + 1}^n n]$
$=[(\frac n2)^2] + [\frac n2*n]$
$=\frac 34n^2$.
Case 2: $n$ is odd
$= [\sum_{k=1}^{\frac {n-1}2}(2k-1)] + [\sum_{k=\frac {n+1}2 }^n n]$
$=[(\frac {n-1}2)^2] + [\frac {n+1}2*n]$
$=\frac {(n^2 - 2n + 1)+ (2n^2 + 2n)}4 = \frac 14(3n^2 + 1)$
.....
Example $n = 10$.
Now we need to find all possible $(k,k,m)$ so that $k,m \le 10$ and $m < 2k$.
If $k =1 $ then $m$ may be $1$. Total: $1$ way.
If $k=2$ then $m$ may be $1,2,3$. Total: $3$ + $1$ = $4$ ways.
If $k=3$ then $m$ may be $1...5$. Total: $5+3+1=9$ ways.
If $k=4$ then $m$ may be $1... 7$. Total: $7+5+3+1=16$ ways.
If $k = 5$ then $m$ may be $1....9$. Total: $9 + 7+ ..... + 1 = 25$ ways.
If $k = 6$ then $m$ may be $1....10$. Total: $10 + 25 = 35$ ways.
If $k = 7$ then $m$ may be $1..... 10$. Total: $10 + 10 + 25 = 45$ ways.
If $k = 8 .... 10$ then $m$ may be $1.... 10$. Total $10 + 10 + ..... + 10 + 25$ ways.
Total number of ways: $1 + 3 + 5+ 7 + +9 + 10+10+10 + 10+10 = 75$
For $n$ even we will have
$(1 + 3 + 5 + ..... + (2*(\frac n2)-1) + (n + n + n + .....n) = $
$(\frac n2)^2 + \frac n2 * n = \frac 34 n^2$.
If $n$ is odd we will have.
$(1 + 3 + 5 + ..... + (2 * (\frac {n-1}2) - 1) + (n+n+n .... +n) = $
$(\frac {n-1{2}^2} + \frac {n+1} 2*n = \frac 14(3n^2 + 1))$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the equation $2x^2-[x]-1=0$ where $[x]$ is biggest integer not greater than $x$. I' ve tried with $x^2 = {[x]+1\over 2}$ so $x$ is a square root of half integer. And know? What to do with that?
|
we know $0\leq x-\lfloor x \rfloor <1 $
$$\quad{2x^2-\lfloor x \rfloor-1=0 \to \lfloor x \rfloor=2x^2-1
\\so\\0\leq x-(2x^2-1) <1\to \\
\begin{cases}0\leq x-(2x^2-1) \to & -(x-1)(2x+1)\geq 0 & (*)\\ x-(2x^2-1) <1 \to & x(1-2x)<0 & (**)\end{cases}
\\\begin{cases} (*) \to & x\in[-\frac12,1]\\ (**)\to &x\in (-\infty,0)\cup(\frac12,\infty) \end{cases}\\(*) \cap(*) =[-\frac12,0) \cup (\frac12,1]}$$ now :with respect to $[-\frac12,0) \cup (\frac12,1]$ floor of $x$ can be $\lfloor x \rfloor =-1,0,1$ so
$$\quad{\lfloor x \rfloor =2x^2-1=-1,0,1 \\\lfloor x \rfloor =2x^2-1=-1 \implies 2x^2=0 \implies x=0 \text{ not acceptable}\\
\lfloor x \rfloor =2x^2-1=0 \implies 2x^2=1 \implies x=- \frac{1}{\sqrt 2} \text{ not acceptable} ,x=+ \frac{1}{\sqrt 2}\text{ acceptable}\checkmark\\
\lfloor x \rfloor =2x^2-1=1 \implies 2x^2=2 \implies x=- 1 \text{ not acceptable} ,x=1\text{ acceptable}\checkmark\\} $$ summary :the equation have two solution $\bf{x=1,\frac{1}{\sqrt 2}}$
|
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|
Proof verification: $ \overline{v}_{1} = \overline{v}_{2} \iff v_{1} - v_{2} \in W \iff v_{2} - v_{1} \in W.$
Let W be a subspace of V. Then for $\overline{\textbf{v}}_{i} =
\overline{\textbf{v}}_{i} + W $,
$$ \overline{\textbf{v}}_{1} =
\overline{\textbf{v}}_{2} \iff \textbf{v}_{1} - \textbf{v}_{2} \in W
\iff \textbf{v}_{2} - \textbf{v}_{1} \in W.$$
I need some verification on my proof:
$\Rightarrow$ Let $ x \in \mathbf{v}_{1} + W$ then $x = \mathbf{v}_{1} + \mathbf{w}_{1}$ for some $\mathbf{w}_{1} \in W$.
$x \in \mathbf{v}_{2} + W$ thus $ x = \mathbf{v}_{2} + \mathbf{w}_{2}$
$\mathbf{v}_{1} + \mathbf{w}_{1} = \mathbf{v}_{2} + \mathbf{w}_{2} $ $ \implies $ $\mathbf{v}_{1} - \mathbf{v}_{2} = \mathbf{w}_{2} - \mathbf{w}_{1} \in W.$
$\Leftarrow$ $ \mathbf{v}_{1} - \mathbf{v}_{2} \in W \implies \mathbf{v}_{1} - \mathbf{v}_{2} = \mathbf{w}$ for some $\mathbf{w}$, $\mathbf{v}_{1} = \mathbf{v}_{2} + \mathbf{w}$ and $ \mathbf{v}_{2} = \mathbf{v}_{1} - \mathbf{w}$.
Let $ x \in \mathbf{v}_{1} + W_{1}$, then $ x = \mathbf{v}_{1} + \mathbf{w}_{1} = \mathbf{v}_{2} + (\mathbf{w} + \mathbf{w}_{1}) \in \mathbf{v}_{2} + W$.
Similarly, $\mathbf{v}_{2} + W \subseteq \mathbf{v}_{1} + W$.
|
Yes, it is correct. Note that there is a typo at the statement. You wrote “$\overline{\mathbf{v}}_i=\overline{\mathbf{v}}_i+W$”, but I guess that you meant “$\overline{\mathbf{v}}_i=\mathbf{v}_i+W$”.
|
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|
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, algebraic explanation to this inequality. (I suppose the condition for equality is that $a = b = c$.) I am not familiar with symmetric inequalities in three variables. I would appreciate any references.
|
we have $$\frac{a^4}{ab^2}+\frac{b^4}{bc^2}+\frac{c^4}{ca^2}\geq \frac{a^2+b^2+c^2)^2}{ab^2+bc^2+ca^2}\geq \frac{3(a^2+b^2+c^2)}{a+b+c)}$$
the last is true, since$$a(a-c)^2+b(a-b)^2+c(b-c)^2\geq 0$$
|
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|
Prove that $\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$ $$\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$$
I tried applying a.m. g.m inequality to l.h.s and tried to find upper bound for l.h.s and lower bound for r.h.s but i am not getting answer .
|
By C-S we obtain:
$$\frac{ab}{a+b}+\frac{cd}{c+d}=a+c+\left(\frac{ab}{a+b}-a\right)+\left(\frac{cd}{c+d}-c\right)=$$
$$=a+c-\left(\frac{a^2}{a+b}+\frac{c^2}{c+d}\right)\leq a+c-\frac{(a+c)^2}{a+b+c+d}=\frac{(a+c)(b+d)}{a+b+c+d}.$$
Done!
|
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|
Basic arithmetic with matrices We have matrices.
$$A=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
B^{-1}=\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
C=\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
We need to solve matrix $X$ from equation:
$$A^{-1}XB-C=0$$
$$X=AB^{-1}+C$$
$$X=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
+
\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
$$
X=\begin{bmatrix}
-1 & -18 \\
39 & -102 \\
\end{bmatrix}
$$
This doesn't seem to be correct solutions to this equations ?
|
The equation
$X=AB^{-1}+C \tag 1$
does not follow from
$A^{-1}XB-C= 0; \tag 2$
instead, we have
$A^{-1}XB = C, \tag 3$
$XB = AC,\tag 4$
$X = ACB^{-1}; \tag 5$
if we now perform the indicated matrix arithmetic we arrive at
$X = \begin{bmatrix} -2 & 0 \\ -5 & 6 \end{bmatrix} \begin{bmatrix} -15 & -2\\-8 & -14 \end{bmatrix} \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}$
$= \begin{bmatrix} 30 & 4 \\ 27 & -74 \end{bmatrix}\begin{bmatrix} -7 & 8 \\ 2 & -8 \\ \end{bmatrix} = \begin{bmatrix} -202 & 208\\ -337 & 808 \\ \end{bmatrix}. \tag 6$
As a check of the above, we see that (1) implies
$A^{-1}XB = A^{-1}(AB^{-1}+C)B = A^{-1}AB^{-1}B + A^{-1}CB = I + A^{-1}CB \ne C \tag 7$
in general.
|
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|
a definite integral identity? is it true? $\int_{a}^{b} f(x) dx $ i came across this identity , i don't know if it is true or not
$$\int_{a}^{b} f(x) \ \mathrm{d}x = (b-a) \sum_{n=1}^{\infty} \sum_{k=1}^{2^n - 1} \dfrac{(-1)^{k+1}}{2^{n}} f \left( a+ \left(\frac{b-a}{2^n}\right) k \right)$$
i tried to use the Riemann Sum but i couldn't proceed
Anyone has an idea about it?
|
Note that:
$$
\int_a^bf(x)dx=\lim_{N \to \infty}S_N \\
S_N=\sum_{i=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i) =
\sum_{n=1}^N \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1}f(a+\frac{b-a}{2^n}k)
$$
Prove the last identity by induction:
For $N$=1, trivial.
For $N\to N+1$:
$$
\sum_{n=1}^{N+1} \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1} f(a+\frac{b-a}{2^n}k)\\
=\sum_{n=1}^{N} \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1} f(a+\frac{b-a}{2^n}k)
+\sum_{k=1}^{2^{N+1}-1}\frac{b-a}{2^{N+1}}(-1)^{k+1} f(a+\frac{b-a}{2^{N+1}}k)\\
=\sum_{i'=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i')
+\sum_{k=1}^{2^{N+1}-1}\frac{b-a}{2^{N+1}}(-1)^{k+1} f(a+\frac{b-a}{2^{N+1}}k)\\
=\sum_{i'=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i')
-\sum_{i'=1}^{2^{N}-1}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'))
+\sum_{i'=1}^{2^{N}}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'-1))\\
=\sum_{i'=1}^{2^N-1} \frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'))
+\sum_{i'=1}^{2^{N}}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'-1))\\
=\sum_{i=1}^{2^{N+1}-1} \frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}i)
$$
|
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|
Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$ Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
$$c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$$
$$c{x^2+c(y+1)^2}={x^2+y^2-1}$$
$$c{x^2+cy^2+2cy+c}={x^2+y^2-1}\text{ [expanded]}$$
$$1+c=x^2-cx^2+y^2-cy^2-2cy\text{ [moved to other side]}$$
$$1+c=(1-c)x^2+\color{red}{(1-c)y^2-2cy}\text{ [factor, will complete square of red]}$$
$$1+c=(1-c)x^2+\color{red}{(1-c)(y^2-\dfrac{2cy}{c-1}+(\dfrac{2c}{c-1})^2)-(\dfrac{2c}{c-1})^2}\text{ [completed square]}$$
$$1+c+\color{red}{(\dfrac{2c}{c-1})^2}=(1-c)x^2+\color{red}{(1-c)(y-\dfrac{2c}{c-1})^2}\text{ [done]}$$
So we see that this is an ellipse, stretched in the $y$ direction by a factor of $1-c$, and translated both in $x$ and $y$.
$c\neq 1$.
Consider $c=2$
$19=-x^2-(y+16)^2$
$-19=x^2+(y+16)^2$
But this not a circle nor an ellipse. It's supposed to be a circle? What is the issue here?
|
Here
$1+c=(1-c)x^2+\color{red}{(1-c)(y^2-\dfrac{2cy}{c-1}+(\dfrac{2c}{c-1})^2)-(\dfrac{2c}{c-1})^2}\text{ [completed square]}$
It's not equal to the precedent line..... sign problem and you didnt add the correct square....
By the way put $c=2$ in the first equation, you get :
$x^2+(y+2)^2=1$ which is a circle.
|
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|
Prove using some form of induction that $T(n) = O(n^2)$ given recurrence relation
Here is a recursively defined function where $c \ge 0$.
$T(n) = c$ if $n = 0$
$T(n) = c$ if $n = 1$
$T(n) = 2T(n-1) - T(n-2) + 2$ if $ \ n \geq 2$
Prove using some form of induction that $ \ T(n) = O(n^2)$
My attempt:
We have to show there exist positive constants $ \ c, n_{0}$ such that $ \ T(n) \le cn^2 \ \forall n \ge n_{0}$. Pick $ \ n_{0} = 1$.
Base cases:
$n=1$: $ \ T(1) = c $ by definition and $ \ c \le c(1)^2 = c$ is true
$n=2$. $ \ T(2) = c $ by definition and $ \ c \le c(2)^2 = 4c$ is true
I.H:
Suppose that $ \ n\ge 2$ and that $ \ P(1),P(2), ...., P(n)$ all hold
$ \ T(n+1) = 2T(n-1) -T(n-2) +2 \le 2c(n-1)^2 -c(n-2)^2 + 2$, by I.H.
Then, $ \ 2c(n-1)^2 -c(n-2)^2 + 2 = cn^2 -2c +2 \le cn^2 +2$ since $ -2c \le 0$ and $ \ cn^2 + 2 \le cn^2 + 2cn + c = c(n^2 + 2n +1) = c(n+1)^2$
|
Some corrections are needed. In the induction hypothesis you used the inequality $-T(n-2) \le -c(n-2)^2$ or $T(n-2) \ge c(n-2)^2$ which is not true.
Hint. For $n\geq 1$, let $S(n):=T(n) -T(n-1)$. Then $S(1)=0$ and for $n\geq 2$
$$S(n)=T(n) -T(n-1)=T(n-1) -T(n-2)+2=S(n-1) + 2.$$
Hence (by induction?), for $n\geq 1$, $S(n)=2(n-1)$ that is
$$T(n)=T(n-1) +2(n-1).$$
Now, by using the above recurrence, show that there exist a positive integer $n_0$ and a constant $C$ such that $T(n)\leq Cn^2$ for all $n\geq n_0$ (for example you can take $n_0=1$ and $C=c+1$).
P.S. You can also prove by induction that $T(n)=c+n(n-1)$. Then it is easy to show that $T(n) = O(n^2)$.
|
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|
The sum of series $\frac14+\frac{1\cdot3}{4\cdot6}+\cdots$ The problem I have here today is the following;
$$\frac{1}{4}+\frac{1\cdot3}{4\cdot6}+\frac{1\cdot3\cdot5}{4\cdot6\cdot8}+\cdots$$ the problem is exactly phrased like this (I can't say that the $\infty$ sign is a bit unnecessary at the end),
My Attempts
We can generalize this sum by noticing that each time indices get greater the denominator and numerators are multiplied by $n+2$ for each $n$ either in numerator or denominator, we take $\dfrac{1}{4}$ out of the sum first, so the sum is equal to $$\dfrac{1}{4}+\sum_{k=4} \frac{(k-3)(k-1)}{k(k+3)}$$ then we open up the brackets and we get; ....
Then I was a little stuck here because when I opened up these brackets and try to get the partitions of the sum one of them was logical $\displaystyle\sum\frac{3}{k(k+3)}=\sum\frac{1}{k}-\frac{1}{k+3}$. I couldn't carry it out longer. What do you suggest?
Is this a problem way above elementary solutions?
|
Amusingly, since the sum is $1$, you can actually write this as a question about a random number.
For each $n$, we flip an unfair coin $C_n$ with heads having probability of $\frac{1}{2(n+1)}$.
Let $X$ be the random variable which is $n$ if $C_n$ came up heads and for each $i<n$, $C_i$ came up tails.
Since $\prod_{k=1}^{n}\left(1-\frac{1}{2(k+1)}\right)\to 0$ as $n\to\infty$, you get $P(X<\infty)=1$.
Then it turns out that $$P(X=n)=\prod_{k=1}^{n}\frac{2k-1}{2k+2}$$
Proof: $$\begin{align}P(X=n)&=P(C_n\text{ heads})\prod_{k=1}^{n-1}P(C_k\text{ tails})\\
&=\frac{1}{2n+2}\prod_{k=1}^{n-1}\frac{2k+1}{2k+2}\\
&=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{4\cdot 6\cdots (2n+2)}
\end{align}$$
Since $P(X<\infty)=1$, we get: $\sum_{n=1}^{\infty} P(X=n) = 1$
More generally , give a sequence of real numbers, $a_k$ with $0\leq a_k\leq 1$ and $\sum_{k=1}^{\infty} a_k=+\infty$, then $\prod_{k=1}^{n} (1-a_k)\to 0$, and we get that $b_n=a_n\prod_{k=1}^{n-1}(1-a_k)$ satisfies $$\sum_{n=1}^{\infty} b_n = 1.$$
So if $a_{n}=\frac{1}{an+2}$ then $$b_n = \frac{1}{a+2}\frac{a+1}{2a+2}\cdots\frac{(n-1)a+1}{na+2}=\prod_{k=1}^{n} \frac{a(k-1)+1}{ak+2}$$
satisfies $\sum b_n = 1$.
More generally:
$$\sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{a(k-1)+b-1}{ak+b} = b-1$$
|
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|
Anti-derivative of a function that involves poly-logarithms. Let $n\ge 1$ be an integer and let $0 < z < a$ be real numbers.
Let $Li_n(x):= \sum\limits_{l=1}^\infty z^l/l^n$ by the polylogarithm of order $n$.
The question is to find the following anti-derivative:
\begin{equation}
{\mathfrak J}^{(n)}_a(z):=\int\frac{Li_n(a-z)}{z} dz=?
\end{equation}
By using integration by parts we have found the result for $n<= 3$. We have:
\begin{eqnarray}
{\mathfrak J}^{(1)}_a(z)&=&-\text{Li}_2\left(\frac{-a+z+1}{1-a}\right)-\log \left(\frac{z}{a-1}\right) \log (-a+z+1)\\
{\mathfrak J}^{(2)}_a(z)&=&\frac{1}{6} \pi ^2 \log (z)-\frac{1}{2} \log (a) \log ^2\left(\frac{z}{a}\right)+\text{Li}_2\left(\frac{z}{a}\right) \log \left(\frac{z}{a}\right)-\text{Li}_3\left(\frac{z}{a}\right)+\\
&&\left(\text{Li}_2(-a+z+1)+\text{Li}_2\left(\frac{-a+z+1}{z}\right)-\text{Li}_2\left(\frac{a (-a+z+1)}{z}\right)\right) \log \left(\frac{a (-a+z+1)}{z}\right)+\\
&&-\text{Li}_3(-a+z+1)-\text{Li}_3\left(\frac{-a+z+1}{z}\right)+\text{Li}_3\left(\frac{a (-a+z+1)}{z}\right)\\
{\mathfrak J}^{(3)}_a(z)&=&\text{Li}_2(a-z) \left(-\text{Li}_2\left(\frac{z}{a}\right)+\log \left(\frac{a-z}{a}\right) \log \left(\frac{a}{z}\right)+\frac{\pi ^2}{6}\right)+\\
&&\text{Li}_3(a-z) \log
\left(\frac{z}{a}\right)+\text{Li}_3\left(1-\frac{z}{a}\right) \log (-a+z+1)-\\
&& {\mathfrak J}^{(3)}_{\frac{1}{a}}(\frac{1-a+z}{a})
\end{eqnarray}
What is the result for higher values of $n$?
|
Here we provide an answer for $n=3$. The idea is to expand the function to be sought for in a series about the value $a=1$. Clearly we have:
\begin{equation}
-{\mathfrak J}^{(n)}_a(a-x)=\sum\limits_{m=0}^\infty \int \frac{Li_3(x)}{(1-x)^{m+1}}dx \cdot (1-a)^m
\end{equation}
Now, using integration by parts we can derive the anti-derivative in the sum. There are no essential problems in this process it is just that it is a tedious task and, as usual, requires a lot of double-checking to avoid errors. We just state the result:
\begin{eqnarray}
\int\frac{Li_3(x)}{(1-x)^{m+1}}dx=Li_1(x) Li_3(x)-\frac{1}{2} Li_2(x)^2
\end{eqnarray}
for $m=0$ and
\begin{eqnarray}
&&m \int\frac{Li_3(x)}{(1-x)^{m+1}}dx=\\
&&
(((1-x)^{-m}-1) \text{Li}_3(x)-2 \text{Li}_3(1-x)+
\log(1-x)(-\text{Li}_2(x) -\log(1-x) \log (x)+\frac{1}{3} \pi ^2 ))+\\
&&H_{m-1} \cdot(\frac{1}{2} \log(1-x)^2+\text{Li}_2(x))-\\
&&\sum\limits_{j=2}^m \frac{1}{j-1} \cdot \frac{Li_2(x)}{(1-x)^{j-1}}+
\sum\limits_{j=2}^{m-1} \frac{ \left(H_{m-1}-H_{j-1}\right) }{(j-1)(1-x)^{j-1}}\cdot \left(-\frac{1}{j-1}-\log (1-x)\right)
\end{eqnarray}
for $m\ge 1$.
Now the only thing we need to do is to insert the above expressions into the sum and do the summations. Again, even a glimpse at those equations suffices to see that it is possible to carry out the summations. We state the result:
\begin{eqnarray}
&&\int\limits_0^x \frac{Li_3(t)}{a-t}dt=\\
&&-\frac{\text{Li}_2(x){}^2}{2}-\text{Li}_3(x) \log (1-x)+\\
&&-\log (a) \left(-2 \text{Li}_3(1-x)-\text{Li}_3(x)+
\log(1-x)(-\text{Li}_2(x) -\log (x) \log(1-x)+\frac{1}{3} \pi ^2 )\right)+\\
&&\text{Li}_3(x)
\left(-\log \left(1-\frac{1-a}{1-x}\right)\right)+\\
&&\frac{1}{2} \log ^2(a) \left(\text{Li}_2(x)+\frac{1}{2} \log ^2(1-x)\right)-\\
&&\text{Li}_2(x) \left(\text{Li}_2\left(1-\frac{a}{x}\right)+\log \left(\frac{a-x}{1-x}\right)
\log \left(\frac{a}{x}\right)-\text{Li}_2\left(\frac{x-1}{x}\right)\right)+\\
&&2 Li_3(1) Li_1(1-a)+\\
&&\sum\limits_{j=2}^{\infty}\sum\limits_{m=j+1}^\infty \frac{ \left(H_{m-1}-H_{j-1}\right) }{(j-1)}\cdot \left(\frac{1}{j-1}-\frac{1}{(j-1)(1-x)^{j-1}}-\frac{\log (1-x)}{(1-x)^{j-1}}\right)\cdot \frac{(1-a)^m}{m}
\end{eqnarray}
We will calculate the last double sum and simplify the whole result later on.
|
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Equation of the tangent to a graph where the point is not on the graph. If $f(x)=(x+1)^{3/2}$, provided $x\geq -1$, I am asked to find the equation of all tangent lines to $f(x)$ at the point $(\frac{4}{3},3)$.
Simple enough. I first took the derivative which is:
$$f'(x)= \frac {3\sqrt{x+1}}{2}$$
Since $(\frac{4}{3},3)$ is not on the graph of $f(x)$, I need to find the point of tangency.
So I assumed the point of tangency is at some point $(k,(k+1)^{\frac{3}{2}})$ It also has to pass the point $(\frac{4}{3},3)$ so thus:
$y-y_{1} = m(x-x_{1})$
$3-(k+1)^{\frac{3}{2}} = \frac{3}{2}\sqrt{k+1}(\frac{4}{3}-k)$
$3-\sqrt{(k+1)^2(k+1)} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3-(k+1)\sqrt{k+1} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3-k\sqrt{k+1}-\sqrt{k+1}=2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3+\frac{1}{2}k\sqrt{k+1}-3\sqrt{k+1}=0$
Here is where I am stuck. How do I solve for $k$ (And hence $x$)?
I tried letting $\sqrt{k+1}$ equal $a$ and then try and solve the corresponding cubic but that's not getting me anywhere... Any help on this? Thanks!
|
$$3-\frac{1}{2} k\sqrt{k+1} - 3 \sqrt{k+1} =0$$
$$3=\frac{1}{2}k\sqrt{k+1} + 3 \sqrt{k+1}$$
$$3=\sqrt{k+1}(\frac{1}{2}k+3)$$
$$9=(k+1)(\frac{1}{4}k^{2}+9+3k)$$
$$\frac{1}{4}k^3 + \frac{13}{4}k^2+12k=0$$
$$k(\frac{1}{4}k^2 + \frac{13}{4}k+12)=0$$
from here you now one answer is $k=0$ and others are those who make
$$\frac{1}{4}k^2 + \frac{13}{4}k+12=0$$
and with solving it you got
$$\frac{-13\pm \sqrt{157}}{2}$$
as other answers
but none of these satisfy
$$\sqrt{k+1}>0$$
so your only answer will be k=0.
|
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Prove that if $f(x)= e^{-1/x^2}\sin{\frac{1}{x}}$ for $x\neq0$ and $f(0)=0$, then $f^{(k)}(0)=0$ for all $k$. This question is from Spivak's Calculus (3rd ed) 18-41:
Prove that if $f(x)= e^{-1/x^2}\sin{\frac{1}{x}}$ for $x\neq0$ and $f(0)=0$, then $f^{(k)}(0)=0$ for all $k$.
Solution is:
I need someone to explain how $|\sin(1/x)|\leq1$ and $|\cos(1/x)|\leq1$ is helpful in the proof. I can prove the same result for $f(x)= e^{-1/x^2}$ using the fact $\lim_{x \to \infty}\frac{x^n}{e^x}=0$ but not sure how to deal with the $\sin$ and $\cos$ in this question.
|
I do not know if this is correct, but here is one approach to prove it.
Given that
$$f^{\left( k \right)}\left( x \right) = e^{-\frac{1}{x^2}}\left[ \sum\limits_{i = 1}^{3k} \dfrac{a_i}{x^i} \sin \dfrac{1}{x} + \sum\limits_{i = 1}^{3k} \dfrac{b_i}{x^i} \cos \dfrac{1}{x} \right]$$
We have from here,
$$f^{\left( k \right)}\left( x \right) = e^{-\frac{1}{x^2}} \left[ \dfrac{a_1 \sin \dfrac{1}{x} + b_1 \cos \dfrac{1}{x}}{x} + \dfrac{a_2 \sin \dfrac{1}{x} + b_2 \cos \dfrac{1}{x}}{x^2} + \\ \dfrac{a_3 \sin \dfrac{1}{x} + b_3 \cos \dfrac{1}{x}}{x^3} + \dots + \dfrac{a_{3k} \sin \dfrac{1}{x} + b_{3k} \cos \dfrac{1}{x}}{x^{3k}} \right]$$
$$\therefore f^{\left( k \right)}\left( x \right) = \dfrac{e^{-\frac{1}{x^2}}}{x^{3k}} \left[ x^{3k - 1} \left( a_1 \sin \dfrac{1}{x} + b_1 \cos \dfrac{1}{x}\right) + x^{3k - 2} \left( a_2 \sin \dfrac{1}{x} + b_2 \cos \dfrac{1}{x}\right) + \\ x^{3k - 3} \left( a_2 \sin \dfrac{1}{x} + b_3 \cos \dfrac{1}{x}\right) + \dots + \left( a_{3k} \sin \dfrac{1}{x} + b_{3k} \cos \dfrac{1}{x}\right) \right]$$
From here, we may conclude that $\lim\limits_{x \rightarrow 0} f^{\left( k \right)} \left( x \right) = 0$ which holds true for all positive k
|
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|
Non zero solutions of second order equation by solving for $a$ Im stuck on this problem:
Let a be a real constant. Consider the equation
$y''+5y'+ay=0$ with boundary conditions $y(0)=0$ and $y(3)=0$
For certain discrete values of $a$, this equation can have non-zero solutions. Find the three smallest values of $a$ for which this is the case.
Enter your answers in increasing order.
I assumed the only time the solutions can be non zero are if we have two complex roots and then solving for those.
By solving, I know the roots must be equal to $\frac{-2.5\pm \sqrt{6.25-a}}{2}$
Im stuck after that so any help would be appreciated!
|
The characteristic equation of the homogeneous ODE $y''+5y'+ay=0$ is $r^2+5r+a=0$. The general solution of the equation is $y(x) = \lambda_+e^{r_+ x} + \lambda_-e^{r_- x}$, where
$$
r_\pm = -\frac{5}{2}\pm\sqrt{\frac{25}{4}-a}
$$
and the constants $\lambda_+$, $\lambda_-$ are deduced from the boundary values. Of course, $y=0$ is a solution. So, let us assume that $y\neq 0$. The boundary values yield the linear system
\begin{aligned}
\lambda_+\phantom{e^{3 r_+}} + \lambda_-\phantom{e^{3 r_-}} &= 0 \, ,\\
\lambda_+e^{3 r_+} + \lambda_-e^{3 r_-} &= 0 \, .
\end{aligned}
This system has non-trivial solutions if its determinant is equal to zero, i.e.
$$ \exp\left({3\sqrt{\frac{25}{4}-a}}\right)=\exp\left(-{3\sqrt{\frac{25}{4}-a}}\right) . $$
If $25/4 - a \geq 0$, the only solution for this identity to be true is $a = 25/4$, and $y$ is again trivially equal to zero. However, if $25/4 - a < 0$, then Euler's formula gives
$$
\sin\left({3\sqrt{a-\frac{25}{4}}}\right) = 0 \, ,
$$
i.e.
$$
a \in \left\lbrace \frac{25}{4} + \frac{n^2\pi^2}{9} \, , \quad n \in\Bbb{Z}^* \right\rbrace .
$$
In these cases, a nonzero solution can be found.
The answer is $a_1 = \frac{25}{4} + \frac{\pi^2}{9}$, $a_2 = \frac{25}{4} + \frac{4\pi^2}{9}$, and $a_3 = \frac{25}{4} + \pi^2$.
Note that this post is very similar.
|
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Pauli matrices and the complex number matrix representation The three spin Pauli matrices are:
$
\sigma_1 = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\sigma_2 = \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix},
\sigma_3 = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
$
According to problem 2.2.3 (Mathematical methods for physicists, Arfken Weber & Harris, Seventh Edition) Complex numbers, a + ib, with a and b real, may be represented by (or are isomorphic with) 2 × 2 matrices as follows:
$
a + ib = \begin{pmatrix}
a & b \\
-b & a
\end{pmatrix}
$
This implies that:
$
i = \begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix},
-i = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
$
And therefore according to the above proposition.
$
\sigma_2 = \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix} =
i \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} = i(-i) = 1 =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
$
By exchanging the rows in $\sigma_1$, we can also write it as the identity matrix. Therefore according to this we get $\sigma_1 = \sigma_2$ which to me is quite surprising. They have the property that $\sigma_1^2 = \sigma_2^2 = I$.
One can also prove that $\sigma_1 \sigma_2 = i \sigma_3$. Since $\sigma_1 = \sigma_2 = I$, $\sigma_1 \sigma_2 = I$. The RHS becomes:
$
i\sigma_3 = \begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} =
\begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix} =
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
$
The problem arises when I try to prove the identity $\sigma_i \sigma_j + \sigma_j \sigma_i = 2\delta_{ij}I_2 $ in Problem 2.2.11(c). This clearly does not hold since $\sigma_2 \sigma_3 + \sigma_3 \sigma_2$ $= I \sigma_3 + \sigma_3 I \neq 0$
Where am I making a mistake here ?
|
The mapping $a +ib \Rightarrow \begin{pmatrix}
a & b \\
-b & a
\end{pmatrix}$ where $ a,b \in \mathbb{R}$ does not map any complex number to a Pauli matrix. The Pauli matrices are not of the right form. So saying that $1$ maps to $\sigma_2$ (or even worse, that it is equal to it) is meaningless.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Bag marbles with three colors and probability A bag contains one red, two blue, three green, and four yellow balls. A sample of three balls is taken without replacement. Let $Y$ be the number of yellow balls in the sample. Find the probability of $Y=0$, $Y=1$, $Y=2$ $Y=3$
Attempt 1
all three are yellow would be
$$ \frac{\binom{4}{3}}{\binom{10}{3}}=\frac{4}{120}=\frac{1}{30} $$
two are yellow combinations would be
yy red
$$ \binom{4}{2}\binom{1}{1} =6$$
yy green
$$ \binom{4}{2}\binom{3}{1}=18 $$
yy blue
$$ \binom{4}{2}\binom{2}{1} = 12$$
adding them together and dividing getting $P(Y=2)=.3$
for one yellow
it would be
combinations for Yellow red blue
combinations for Yellow red green
combinations for Yellow b b
combinations for Yellow b g
combinations for Yellow g g
adding them all together and dividing by $120$
for no yellows
just keep going???Better algorithm?
|
We have four yellow balls and $1 + 2 + 3 = 6$ balls that are not yellow. The number of ways we can select exactly $k$ yellow balls and $3 - k$ balls that are not yellow is
$$\binom{4}{k}\binom{6}{3 - k}$$
Since there are $\binom{10}{3}$ possible selections of three of the ten balls, the desired probabilities are
\begin{align*}
P(Y = 0) & = \frac{\dbinom{4}{0}\dbinom{6}{3}}{\dbinom{10}{3}}\\
P(Y = 1) & = \frac{\dbinom{4}{1}\dbinom{6}{2}}{\dbinom{10}{3}}\\
P(Y = 2) & = \frac{\dbinom{4}{2}\dbinom{6}{1}}{\dbinom{10}{3}}\\
P(Y = 3) & = \frac{\dbinom{4}{3}\dbinom{6}{0}}{\dbinom{10}{3}}
\end{align*}
|
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|
How to show that $\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?$ How can we show that $(1)$
$$\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?\tag1$$
$\sin^3(x)={3\over 4}\sin(x)-{1\over 4}\sin(3x)$
$1+\cos^2(x)=2-\sin^2(x)$
$$\int_{0}^{\pi}[{3\over 4}\sin(x)-{1\over 4}\sin(3x)]{\mathrm dx\over (2-\sin^2(x))^2}=1\tag2$$
$2-\sin^2(x)={3\over 2}+{1\over 2}\cos(2x)$
$$\int_{0}^{\pi}{3\sin(x)-\sin(3x)\over 3+\cos(2x)}\mathrm dx\tag3$$
$$\int_{0}^{\pi}{3\sin(x)\over 3+\cos(2x)}\mathrm dx-\int_{0}^{\pi}{\sin(3x)\over 3+\cos(2x)}\mathrm dx\tag4$$
$\cos(2x)={1-\tan^2(x)\over 1+\tan^2(x)}$
$\sin(x)={2\tan(x/2)\over 1+\tan^2(x/2)}$
|
Well, we have:
$$\mathscr{I}_{\space\text{n}}:=\int_0^\text{n}\frac{\sin^3\left(x\right)}{\left(1+\cos^2\left(x\right)\right)^2}\space\text{d}x\tag1$$
Substitute $\text{u}:=\cos\left(x\right)$:
$$\mathscr{I}_{\space\text{n}}:=\int_1^{\cos\left(\text{n}\right)}\frac{\text{u}^2-1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}=\int_1^{\cos\left(\text{n}\right)}\frac{1}{1+\text{u}^2}\space\text{d}\text{u}-2\int_1^{\cos\left(\text{n}\right)}\frac{1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}\tag2$$
Now, we know:
$$\int_1^{\cos\left(\text{n}\right)}\frac{1}{1+\text{u}^2}\space\text{d}\text{u}=\left[\arctan\left(\text{u}\right)\right]_1^{\cos\left(\text{n}\right)}=\arctan\left(\cos\left(\text{n}\right)\right)-\arctan\left(1\right)=$$
$$\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}\tag3$$
We can write:
$$2\int_1^{\cos\left(\text{n}\right)}\frac{1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}=\left[\frac{\text{u}}{1+\text{u}^2}\right]_1^{\cos\left(\text{n}\right)}+\int_1^{\cos\left(\text{n}\right)}\frac{1}{1+\text{u}^2}\space\text{d}\text{u}=$$
$$\left[\frac{\text{u}}{1+\text{u}^2}\right]_1^{\cos\left(\text{n}\right)}+\left[\arctan\left(\text{u}\right)\right]_1^{\cos\left(\text{n}\right)}=$$
$$\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}-\frac{1}{1+1^2}+\arctan\left(\cos\left(\text{n}\right)\right)-\arctan\left(1\right)=$$
$$\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}-\frac{1}{2}+\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}\tag4$$
So, we end up with:
$$\mathscr{I}_{\space\text{n}}=\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}-\left\{\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}-\frac{1}{2}+\arctan\left(\cos\left(\text{n}\right)\right)-\frac{\pi}{4}\right\}=$$
$$\frac{1}{2}-\frac{\cos\left(\text{n}\right)}{1+\cos^2\left(\text{n}\right)}\tag5$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality question: If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. I've tried AM-HM but it gave $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$ which gives $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 2 (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} )\geq 81$
|
$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$ = $\frac{c}{abc} + \frac{a}{abc} + \frac{b}{abc}$ = ${\frac{1}{abc}}$
We know that AM${\ge}$GM
${\frac{a+b+c}{3}}$ ${\ge} \sqrt[3]{ abc}$
${\frac{1}{3}}$ ${\ge} \sqrt[3]{ abc}$
${\frac{1}{27}}$ ${\ge} { abc}$
${\frac{1}{abc}}$ ${\ge} 27$
so minimum value is 27
This minimum value is achieved when a=b=c=${\frac{1}{3}}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that series $(10n+4)^2+1$ contains infinitly many composite numbers Given sequence is
$$4^2+1,\ 14^2+1,\ 24^2+1,\ 34^2+1...$$
How to show that there are infinite amount of composite numbers?
|
If you take any prime $p$ such that $p\mid (10n+4)^2+1$ then $p\mid (10(n+kp)+4)^2+1$ because $$(10(n+kp)+4)^2+1=(10n+4+kp)^2+1=(10n+4)^2+2kp(10n+4)+k^2p^2+1=(10n+4)^2+1+p(2k(10n+4)+k^2p)$$
And trivially $$p\mid p(2k(10n+4)+k^2p)$$
|
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|
Showing that $\frac{\sin(a_{n-1}) + 1}{2}$ is a Cauchy sequence In my homework set, I have the following question:
Show that $$ a_n = \frac{\sin(a_{n-1}) + 1}{2}, \quad a_1=0 $$
satisfies the definition of Cauchy sequence.
As we went over the concept of Cauchy sequences a bit too quickly in class, I'm puzzled about how I should go about showing this. Conceptually, I understand what a Cauchy sequence is, and I know that in $\mathbb{R}$ it is the same as convergence, but I find it hard to apply to the above sequence.
I'd appreciate some hints about how to approach this problem.
Edit: I've been working on this with help from responses below so I thought I'd update my work. It's verbose but I thought that might show my thinking process better.
My work so far:
$$
\begin{align}
|a_{n+1} - a_n| &= \left | \frac{\sin(a_n) + 1}{2} - \frac{\sin(a_{n-1}) + 1}{2} \right | \\
\\
&= \frac{1}{2} \left | \sin(a_n) + 1 - (\sin(a_{n-1}) + 1)\right | \\
\\
&= \frac{1}{2} \left | \sin(a_n) - \sin(a_{n-1}) \right | \\
\\
&= \frac{1}{2} \left | 2 \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\
\\
&\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\
\\
&\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \right | \\
\\
&\leq \frac{\left | a_n - a_{n-1} \right |}{2}
\end{align}
$$
This set of inequalities work because $|sin(x)| \leq |x|, \ \forall x \in \mathbb{R}$.
Hence, we have:
$$
\begin{align}
|a_{n+1} - a_n| &\leq \frac{1}{2} \left |a_n - a_{n-1} \right | \leq \frac{1}{2^2} \left | a_{n-1} - a_{n-2} \right | \leq \dots \leq \frac{1}{2^{n-1}} \left |a_{2} - a_{1} \right |
\end{align}
$$
So we learn that the distance between consecutive terms is becoming smaller, and that:
$$
\begin{align}
\left | a_{n+1} - a_n \right | &\leq \frac{1}{2^{n-1}} \times \left | \frac{1}{2} - 0 \right | = \frac{1}{2} \times \frac{1}{2^{n-1}} = \frac{1}{2^n}
\end{align}
$$
Now, for $n>m$ and $k=m$ we have:
$$
\begin{align}
\left | a_n - a_m \right | &= \sum_{k=m}^{n-1} \left | a_{k+1} - a_k \right | \\
\\
&= \sum_{k=m}^{n-1} \frac{1}{2^k} \\
\\
&= \left ( \frac{1}{2^m} + \frac{1}{2^{m+1}} + \dots + \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}} \right ) \\
\\
&= \frac{1}{2^{m-1}} \left ( \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-m-2}} + \frac{1}{2^{n-m-1}} \right ) \\
\\
&\leq \frac{1}{2^{m-1}}
\end{align}
$$
So, if we have $|a_n - a_{n-1}| < \epsilon$, then $\frac{1}{2^{m-1}} < \epsilon$. If we solve for $m$, we get that $m > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$ and therefore $N > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$.
So we have found $|a_n - a_m| < \epsilon$ for $n, m > N$. Hence, $(a_n)$ is Cauchy.
Resources used:
*
*Answers given below.
*This video was very helpful
*Understanding the definition of Cauchy sequence
*Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy
*Showing a recursive sequence is Cauchy
*How do I find the limit of the sequence $a_n=\frac{n\cos(n)}{n^2+1}$ and prove it is a Cauchy sequence?
|
Hint. Note that the function $f:\mathbb{R}\to \mathbb{R}$,
$$f(x)=\frac{\sin(x)+1}{2}$$
is a contraction: for any $x,y\in \mathbb{R}$ there exists $t$ between $x$ and $y$ such that
$$|f(x)-f(y)|\leq \left|\frac{\cos(t)}{2}\right||x-y|\leq \frac{|x-y|}{2}\tag{1}$$
where we used the Mean value theorem. Hence, for $n\geq 1$, by using (1) $(n-1)$ times, we obtain
$$|a_{n+1}-a_n|\leq \frac{1}{2}|a_{n}-a_{n-1}|\leq \dots\leq \frac{1}{2^{n-1}}|a_{2}-a_{1}|.$$
Can you take it from here?
P.S. If the Mean value theorem is not allowed then we may note that by the sum-to-product identity,
$$\frac{\sin(x)+1}{2}-\frac{\sin(y)+1}{2}=\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right).$$
Since $|\sin(t)|\leq |t|$ and $|\cos(t)|\leq 1$, it follows that (1) holds.
|
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|
Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
|
The equation
$$
x^2 + y^2 + z^2 = 6
$$
gives a sphere around the origin with radius $R=\sqrt{6}$.
$$
0 = x + y + z = (1,1,1) \cdot (x,y,z)
$$
gives a plane $H$ with normal vector $(1,1,1)^\top$ including the origin.
The intersection is a circle of radius $R$ on that plane $H$, with the origin as midpoint.
GeoGebra seems to give:
$$
u(t) = (-\sqrt{3} \cos(t) +\sin(t), \sqrt{3} \cos(t) + \sin(t), -2 \sin(t))^\top
$$
Checking:
$$
(-\sqrt{3} \cos(t) +\sin(t))^2 + (\sqrt{3} \cos(t) + \sin(t))^2 + (-2 \sin(t))^2 = \\
3 \cos^2(t) + \sin^2(t) - 2\sqrt{3} \cos(t)\sin(t) + \\
3 \cos^2(t) + \sin^2(t) + 2\sqrt{3} \cos(t)\sin(t) + \\
4 \sin^2(t) = \\
6 \cos^2(t) + 6 \sin^2(t) = 6
$$
so the curve lies on the sphere.
$$
(1,1,1) \cdot (-\sqrt{3} \cos(t) +\sin(t), \sqrt{3} \cos(t) + \sin(t), -2 \sin(t)) =
0
$$
so that curve is on the plane $H$.
How to get there?
One might try a coordinate transformation of a circle in the $x$-$y$-plane with radius $r$ onto that plane by two rotations:
We would need
$$
T\,(0,0,1)^\top = (1,1,1)/\sqrt{3}
$$
|
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|
Taylor expansion of $\ln(1 + \frac{2^x}{n})$ I have a function $f = \ln(1 + \frac{2^x}{n})$, where $n \to \infty$ and $x \in (0, 1)$.
I want to apply Taylor expansion at $a = 0$ to $f$.
I get $f = \dfrac{2^x}n-\dfrac{4^x}{2n^2}+o(\dfrac{4^x}{2n^2})$
My question is for what $x$ this approximation is true and why?
Wolfram Alpha says that it works for $|x| < 1$.
(I guess the answer for "why" is that the sum that we get in Taylor expansion is only converges for |x| < 1)
And I'm kinda confusing about that we expand it at $0$, then initially I thought it should only work for $x$ that are in $\epsilon$-neighborhood of $0$.
In general, I can expand the function at any point $a$, and this approximation is true $\forall x : \text{expansion converges}$?
|
Let $f(x) = \ln\left( 1 + \frac{2^{x}}{n}\right)$ then
\begin{align}
f'(x) &= \frac{\ln2 \, 2^{x}}{n + 2^{x}} \\
f''(x) &= \frac{n \, \ln^{2}2 \, 2^{x}}{(n+2^{x})^2} \\
f'''(x) &= \frac{n(n-1) \, \ln^{3}2 \, 2^{x} \, (n-2^{x})}{(n+2^{x})^3}
\end{align}
which leads to
$$f_{n}(x) = \ln\left(1 + \frac{1}{n}\right) + \left(\frac{x \, \ln2}{n+1}\right) + \frac{n}{2!} \, \left(\frac{x \, \ln2}{n+1}\right)^2 + \frac{n(n-1)}{3!} \, \left(\frac{x \, \ln2}{n+1}\right)^3 + \cdots$$
For large $n$ values this expansion can be seen as
\begin{align}
f_{n}(x) &= \ln\left(1 + \frac{1}{n}\right) + \left(\frac{x \, \ln2}{n \, \left(1 + \frac{1}{n}\right)}\right) + \frac{1}{2! \, n} \, \left(\frac{x \, \ln2}{n \, \left(1 + \frac{1}{n}\right)}\right)^2 \\
& \hspace{10mm} + \frac{1-\frac{1}{n}}{3!} \, \left(\frac{x \, \ln2}{n \, \left(1+\frac{1}{n}\right)}\right)^3 + \cdots
\end{align}
This leads to
\begin{align}
\lim_{n \to \infty} f_{n}(x) = 0
\end{align}
|
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|
Prove: if |$z|=1$ then $|iz-\overline{z}|\leq 2$
Prove: if |$z|=1$ then $|iz-\overline{z}|\leq 2$
$$|iz-\overline{z}|=|i(a+bi)-(a-bi)|=|-(a+b)+(a+b)i|=\sqrt{(a+b)^2+(a+b)^2}=\sqrt{2(a+b)^2}$$
On the other hand $|z|=1\iff \sqrt{a^2+b^2}=1\iff a^2+b^2=1$
So it seems that $\sqrt{2(a+b)^2}\leq \sqrt{2}$ maybe it is $|iz-\overline{z}|\leq \sqrt{2}$
|
One way to proceed from your point is by using the inequality $2ab\le a^2+b^2$ $$0\le (a+b)^2=1+2ab\le1+a^2+b^2= 2$$ And thus your quantity ends up being in the interval $\left[\sqrt0,\sqrt4\,\right]$.
|
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|
What is $\lim_\limits{x\to 0}\frac{x}{1-\cos{x}}$ equal to? At 29:30 in lecture 8 of UMKC's Calculus I course, the instructor makes the claim that its limit is equal to zero mentioning that he proved this result earlier in the lecture. The thing is that I watched the entire lecture and yet never actually saw him do that. Here's my result:
\begin{align}
\lim_{x\to 0}\frac{x}{1-\cos{x}}
&=\lim_{x\to 0}\frac{x}{1-\cos{x}}\\
&=\lim_{x\to 0}\left(\frac{x}{1-\cos{x}}\cdot\frac{1+\cos{x}}{1+\cos{x}}\right)\\
&=\lim_{x\to 0}\frac{x}{1-\cos^2{x}}\cdot\lim_{x\to 0}(1+\cos{x})\\
&=2\cdot\lim_{x\to 0}\frac{x}{\sin^2{x}}\\
&=2\cdot\lim_{x\to 0}\frac{x}{\sin{x}}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\
&=2\cdot\lim_{x\to 0}\frac{1}{\frac{\sin{x}}{x}}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\
&=2\cdot\frac{\lim_\limits{x\to 0}1}{\lim_\limits{x\to 0}\frac{\sin{x}}{x}}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\
&=2\cdot\frac{1}{1}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\
&=2\cdot\lim_{x\to 0}\frac{1}{\sin{x}}
\end{align}
$$
\lim_{x\to 0^{-}}\frac{1}{\sin{x}}=-\infty \text{ and}
\lim_{x\to 0^{+}}\frac{1}{\sin{x}}=\infty
$$
Thus, the limit does not exit. But the instructor clearly says that it does. What's wrong with him?
|
Write $\dfrac{x}{1-\cos x} = \dfrac{x(1+\cos x)}{1-\cos^2 x}= \dfrac{x^2(1+\cos x)}{x\sin^2 x}= \left(\dfrac{x}{\sin x}\right)^2\cdot \dfrac{1+\cos x}{x}$. The limit does not exist because $\dfrac{1+\cos x}{x} \to +\infty$ and $-\infty$ when $x \to 0^{+}, 0^{-}$ respectively.
|
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|
Prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$ I can show that $5|x^5+y^5$, by considering $(x+y)^5$ and using binomial expansion. But I am not sure how to show that $25|x^5+y^5$.
More generally, if p is a prime and $p>2$, how do I prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$?
|
Notice for all prime $p$ and integer $n$, $p | n^p - n$.
In particular, $p | x^p - x$ and $p | y^p - y$. This means
$$p|x^p + y^p\quad\implies\quad p|x+y$$
Write $x+y$ as $mp$ for some integer $m$, we have
$$\begin{align}
x^p + y^p
&= x^p + (mp-x)^p\\
&= x^p + (-x)^p + \binom{p}{1}(mp)(-x)^{p-1} + (mp)^2\left(\sum_{k=2}^p\binom{p}{k}(mp)^{k-2}(-x)^{p-k}\right)
\end{align}$$
When $p$ is a prime $> 2$, $p$ will be odd and the first two terms cancel out.
The third term is a multiple of $p^2$ because $\binom{p}{1}(mp) = mp^2$.
For the rest, it is a multiple of $p^2$ because of the overall $(mp)^2$ factor.
|
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|
Envelope of a family of curves
Show that the envelope of the family of curves $$\frac{x}{a}+\frac{y}{4-a}=1$$ is the parabola $$\sqrt{y}+\sqrt{x}=2$$
I know how we can get the envelope, but I could not get the required relation. I differentiated the family of curves w.r.t $a$. I got
$$a=\frac{1}{2}(x-y+4)$$
I then substituted by $a$ in the family of curves and got
$$-x^2+2xy-y^2+8y+8x-16=0$$
How to complete to get $\sqrt{x}+\sqrt{y}=2$?
|
Maybe this is a simpler way to approach the problem.
I will assume that ${0\leq a \leq 4}$.
Define $D$ as the union of the family of curves $\begin{align*}\frac{x}{a}+\frac{y}{4-a}=1\end{align*}$. Then there exists a function $~{f(x)}$ defined on the interval $~x\in [0,~4]$ such that the region $~D$ can be written as $\{(x,y)~|0\leq x \leq 4,~0\leq y\leq f(x)\}$.
Rearrange the given equation $\begin{align*}\frac{x}{a}+\frac{y}{4-a}=1\end{align*}$ as $\begin{align*}y=4+x-\frac{4x}{a}-a\end{align*}$. By AM-GM inequality, we have $\begin{align*}4+x-\frac{4x}{a}-a\end{align*}\leq4+x-4\sqrt{x},~$ where the equality holds if and only if $a=2\sqrt{x}\cdots$(1).
Since $0\leq x\leq 4$, by the equation (1) we have $0 \leq a \leq 4$ and the equality condition is satisfied in the given interval of $x$.
Therefore the desired envelope is $y=4+x-4\sqrt{x}$, or $\sqrt{x}+\sqrt{y}=2$
|
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|
Jordan canonical form and reordering
Consider the Jordan canonical form below \begin{equation*} J =
\begin{pmatrix} J_2(\lambda_1)&0& 0& 0\\ 0&J_1(\lambda_2) &0& 0\\ 0&
0& J_3(\lambda_1)& 0\\ 0 &0 &0 &J_2(\lambda_2) \end{pmatrix}
\end{equation*} \begin{equation*}
= \begin{pmatrix} \lambda_1& 1& 0& 0& 0& 0& 0& 0\\ 0 &\lambda_1 &0& 0& 0& 0& 0& 0\\ 0 &0& \lambda_2 &0& 0& 0& 0& 0\\ 0& 0& 0& \lambda_1& 1&
0& 0 &0\\ 0 &0& 0& 0& \lambda_1 &1& 0& 0\\ 0 &0& 0& 0& 0& \lambda_1
&0& 0\\ 0 &0& 0& 0& 0& 0 &\lambda_2 &1\\ 0& 0& 0& 0& 0& 0& 0&
\lambda_2 \end{pmatrix}_{8\times 8} \in M_8(F) \end{equation*} where
$\lambda_1 \neq \lambda_2$ in a field $F$.
Let \begin{equation*} B := (e_1 =
\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix},\cdots e_8 =
\begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix} ) \end{equation*} be the
standard basis of $F_c^8$ . Then we can calculate: \begin{eqnarray*}
p_J(x) &=& (x - \lambda_1)^5(x - \lambda_2)^3,\\ m_J(x) &=& (x -
\lambda_1)^3(x - \lambda_2)^2,\\ V_{\lambda_1} (J)& =&
\text{Span}\{e_1, e_4\},\\ V_{\lambda_2} (J) &= &\text{Span}\{e_3,
e_7\}. \end{eqnarray*}
Then my query is after reordering of $J$, we can also have basis of eigenspace of $\lambda_1$ to be $span\{e_1,e_3\}$ and basis of eigenspace of $\lambda_2$ to be $span\{e_6,e_7\}$ and everytime you rearrange it will be different. Will that cause any problem? The book said the Jordan canonical form matrix $J$ are still similar after reordering, but how come their eigenspace basis is so different for each reordering? Is it normal?
|
Let $A \in \mathbb{C}^{n \times n}$ have the following two Jordan Decompositions:
\begin{align*}
A = X_1 J_1 X_1^{-1}, \quad A = X_2 J_2 X_2^{-1}.
\end{align*}
Here, $J_1, J_2 \in \mathbb{C}^{n\times n}$ is block-diagonal, and $X \in \mathbb{C}^{n\times n}$ is invertible. Then we have
\begin{align*}
X_1 J_1 X_1^{-1} = X_2 J_2 X_2^{-1} \implies J_1 = X_1^{-1} X_2 J_2 X_2^{-1}X_1.
\end{align*}
Now define $X_3 = X_1^{-1} X_2$, so that $X_3^{-1} = X_2^{-1} X_1$, then we have
\begin{align*}
J_1 = X_3 J_2 X_3^{-1}.
\end{align*}
$J_2$ is a "reordering" of $J_1$, but since $J_1$ and $J_2$ are similar matrices, they represent the same linear operator. They will share the same properties that are characteristic of that operator, but they will not share the properties that are dependent on a choice of basis. So, for example, the geometric multiplicies of $J_1$ and $J_2$ will be the same, but the spans of the eigenvectors will probably be different. Wikipedia's discussion on matrix similarity may be enlightening.
|
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|
Integral argument to a polynomial identity One can show that
$$\frac{(X^{15} - 1)(X-1)}{(X^5-1)(X^3-1)} = P^2 + 15 Q^2$$ with $P,Q\in\mathbf{Z}\left[\frac{1}{2}\right][X]$, by simply grouping terms etc. Is it possible to show that such an identity exists by arguments involving $\mathbf{Z}\left[\sqrt{15}\right]$ or other rings over the latter ?
|
Nice question! The LHS clearly is the cyclotomic polynomial $\Phi_{15}(x)$, and by considering its factorization over the ring of integers of $\mathbb{Z}[\sqrt{5}]$ we have
$$ \Phi_{15}(x) = \frac{1}{4}\left(2+\left(-1+\sqrt{5}\right) x+\left(1-\sqrt{5}\right) x^2+\left(-1+\sqrt{5}\right) x^3+2 x^4\right)\cdot\left(2-\left(1+\sqrt{5}\right) x-\left(-1-\sqrt{5}\right) x^2-\left(1+\sqrt{5}\right) x^3+2 x^4\right) $$
so
$$ \Phi_{15}(x)=\frac{1}{4}\left[(1+x+x^2)^2(2x^2-3x+2)^2-5x^2(x^2-x+1)^2\right]$$
and a similar decomposition follows from considering how $\Phi_{15}$ factors over the ring of integers of $\mathbb{Z}[\sqrt{5},\sqrt{-3}]$. One factor is namely given by $$\left(-2+2 \sqrt{-3}+\left(-1-\sqrt{-3}+ \sqrt{5}+\sqrt{-15}\right) x+4 x^2\right)$$
so the claim follows from the Brahmagupta identity.
|
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|
Binomial summation problem
Show that
$$ \frac{\binom{n}{0}}{1} - \frac{\binom{n}{1}}{4} +\dots + (-1)^n \frac{\binom{n}{n}}{3n+1} = \frac{3^n \cdot n!}{ 1\cdot 4\cdot 7\cdots(3n+1)}.$$
I don't know how to proceed in such type of problems. Any help or hint will be much appreciated.
|
Let
$$a_n:=\frac{\binom{n}{0}}{1} - \frac{\binom{n}{1}}{4} +... + (-1)^n \frac{\binom{n}{n}}{3n+1} =\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{3k+1}$$
Then for $n\geq 1$,
\begin{align}
a_n&=1+\sum_{k=1}^n\frac{(-1)^k\frac{n}{k}\binom{n-1}{k-1}}{3k+1}
=1+n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\left(\frac{1}{k}-\frac{3}{3k+1}\right)
\\
&=
1+n\sum_{k=1}^n\frac{(-1)^k\binom{n-1}{k-1}}{k}-3n\sum_{k=1}^n\frac{(-1)^k\binom{n-1}{k-1}}{3k+1}\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}-3n\sum_{k=1}^n\frac{(-1)^k\left(\binom{n}{k}-\binom{n-1}{k}\right)}{3k+1}\\
&=(1-1)^n-3n(a_n-1)+3n(a_{n-1}-1)=-3na_n+3na_{n-1}
\end{align}
which implies that
$$a_n=\frac{3n a_{n-1}}{3n+1}$$
and the desired identity easily follows
$$a_n=\frac{3n a_{n-1}}{3n+1}=\frac{3^2n(n-1) a_{n-2}}{(3n+1)(3(n-1)+1)}=\dots=\frac{3^nn!}{1\cdot 4\cdot 7\cdots(3n+1)}.$$
|
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|
Limit with n roots I have been trying to practice computing limits and this one came across:
$$\lim_{x\to a} \frac{\sqrt[n]{x}-\sqrt[n]{a}}{x-a}$$
I tried L'Hopital and I got this:
$$\lim_{x\to a}{\frac{x^{\frac{1-2n}{n}}\left(1-n\right)}{n}}$$
But I should get as the solution of the limit $\frac{\sqrt[n]{a}}{an}$
Any help?
|
$$\lim _{ x\to a } \frac { \sqrt [ n ]{ x } -\sqrt [ n ]{ a } }{ x-a } =\lim _{ x\to a } \frac { \left( \sqrt [ n ]{ x } -\sqrt [ n ]{ a } \right) \left( \sqrt [ n ]{ { x }^{ n-1 } } +\sqrt [ n ]{ { x }^{ n-2 }a } +\sqrt [ n ]{ { x }^{ n-3 }{ a }^{ 2 } } +...+\sqrt [ n ]{ { a }^{ n-1 } } \right) }{ \left( x-a \right) \left( \sqrt [ n ]{ { x }^{ n-1 } } +\sqrt [ n ]{ { x }^{ n-2 }a } +\sqrt [ n ]{ { x }^{ n-3 }{ a }^{ 2 } } +...+\sqrt [ n ]{ { a }^{ n-1 } } \right) } =\\ =\lim _{ x\to a } \frac { 1 }{ \left( \sqrt [ n ]{ { x }^{ n-1 } } +\sqrt [ n ]{ { x }^{ n-2 }a } +\sqrt [ n ]{ { x }^{ n-3 }{ a }^{ 2 } } +...+\sqrt [ n ]{ { a }^{ n-1 } } \right) } =\frac { 1 }{ n\sqrt [ n ]{ { a }^{ n-1 } } } =\frac { \sqrt [ n ]{ a } }{ na } \\ $$
|
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|
Local extremes of: $f(x,y) = (x^2 + 3y^2)e^{-x^2-y^2}$ I am looking to find the local extremes of the following function:
$$f(x,y) = (x^2 + 3y^2)e^{-x^2-y^2}$$
What have I tried so far?
*
*Calculate the partial derivatives:
$$\frac{\partial f}{\partial x} = 2x(e^{-x^2-y^2}) + (x^2+3y^2)(e^{-x^2-y^2})(-2x)$$
$$=-2(e^{-x^2-y^2})x(-1+x^2+3y^2)$$
$$\frac{\partial f}{\partial y} = 6y(e^{-x^2-y^2}) + (x^2+3y^2)(e^{-x^2-y^2})(-2y)$$
$$=-2(e^{-x^2-y^2})y(-3+x^2+3y^2)$$
*Determine the possible stationary points:
$$\frac{\partial f}{\partial x}=0 \implies -2(e^{-x^2-y^2})x(-1+x^2+3y^2) = 0$$
$$\implies [x=0] \text{ or } -1+x^2+3y^2 = 0 \implies [x=\sqrt{-3y^2+1}]$$
$$\frac{\partial f}{\partial y} = 0 \implies [y=0] \text{ or } -3+x^2+3y^2 = 0 \implies [y=\pm \sqrt{\frac{-x^2+3}{3}}]$$
So now I have:
$$\frac{\partial f}{\partial x}\begin{cases} x = 0 ,\\
\sqrt{-3y^2+1}\end{cases}
\text{ and }
\frac{\partial f}{\partial y}\begin{cases} y = 0 ,\\
\sqrt{\frac{-x^2+3}{3}}\end{cases}$$
From this I determined the following possible points:
$$(0,0) \rightarrow f(0,0) = 0$$
$$(0,1) \rightarrow f(0,1) = 3e^{-1}$$
$$(0,-1) \rightarrow f(0,-1) = -3e^(-1)$$
$$(1,0) \rightarrow f(1,0) = e^{-1}$$
$$(\sqrt{\frac{1}{3}},\sqrt{3}) \rightarrow f(\sqrt{\frac{1}{3}},\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$
$$(-\sqrt{\frac{1}{3}},\sqrt{3}) \rightarrow f(-\sqrt{\frac{1}{3}},\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$
$$(\sqrt{\frac{1}{3}},-\sqrt{3}) \rightarrow f(\sqrt{\frac{1}{3}},-\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$
So far, did I follow the steps correctly? Also, from what I know next is to check the diferencial of the function in each point and those which are zeros are extreme points. Is that correct? And what would follow this?
|
In 1D (single variable Calculus), a stationary-critical point is a point $c$ such that $f'(c)=0$. At $c$ the gradient of $f$ is zero (there is no change).
In 2D, a stationary-critical point is a point $(a, b)$ such that the gradient in $x$ direction and $y$ direction are both zero (at that same point $(a,b)$, simultaneously, as @Moo said). At that point, no change of value of $f$ in both $x$ and $y$ direction. This means :
$$ \frac{\partial f(a,b)}{\partial x } = \frac{\partial f(a,b)}{\partial y } = 0 $$
So, in your case, you must find a point $(a, b)$ such that both
$$ \frac{ \partial f }{\partial x} = - 2 e^{-x^{2} - y^{2}}x \left( -1 + x^{2} + 3y^{2} \right) = 0 $$
$$ \frac{ \partial f }{\partial y} = - 2 e^{-x^{2} - y^{2}}y \left( -3 + x^{2} + 3y^{2} \right) = 0 $$
which means solving both at once. The system to solve is :
*
*$- 2 e^{-x^{2} - y^{2}}x \left( -1 + x^{2} + 3y^{2} \right) = 0$
*$- 2 e^{-x^{2} - y^{2}}y \left( -3 + x^{2} + 3y^{2} \right) = 0$
Since $e^{...}$ is never $0$, you may solve these instead :
*
*$x \left( -1 + x^{2} + 3y^{2} \right) = 0$
*$y \left( -3 + x^{2} + 3y^{2} \right) = 0$
You can continue from here to solve the system.
Hope this helps.
|
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|
$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at solution using strong induction:
Base cases: $n = 1$ and $n = 2 \implies a_1 = a_2 = 1 \implies 1\le 1 \le 2$
Inductive assumption (strong induction): Assume that for all $m\in \mathbb{N}$ such that $1\le m \le k$, where $k\in \mathbb{N}$, the condition $1\le a_m \le 2$ holds True.
Show that $m = k+1$ holds true
$a_{k+1} = \frac{1}{2} \cdot (a_k + \frac{2}{a_{k-1}})$
I know that $a_k$ and $a_{k-1}$ are satisfying $ 1\le a_m\le2$ but I am not sure how to use that to prove that $a_{k+1}$ holds true for the condition.
|
Assume $x,y\in [1,2].$ Then $x\le 2$ and $y\ge 1\implies \dfrac{1}{y}\le 1\implies \dfrac{2}{y}\le 2.$ Thus
$$\dfrac12\left(x+\dfrac2y\right)\le \dfrac12(2+2)=2.$$ On the other hand $x\ge 1$ and $1\le y\le 2\implies \dfrac{1}{y}\ge\dfrac12 \implies \dfrac{2}{y}\ge 1.$ Thus
$$\dfrac12\left(x+\dfrac2y\right)\ge \dfrac12(1+1)=1.$$
|
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|
Finding $S_{n}$ for this geometric series? Here is what was given to me:
$t_{1}$ = $\frac {1}{256}$
$r$ = $-4$
$n$ = $10$
The formula I used is $S_{n}$ = $\frac {t_{1} (r^{n} - 1)}{r - 1}$
Here's what I did:
$S_{10}$ = $\frac {\frac{1}{256} (-4^{10} - 1)}{-4 - 1}$
$S_{10}$ = $\frac {\frac{1}{256} (-1048576 - 1)}{-5}$
$S_{10}$ = $\frac {\frac{1}{256} (-1048577)}{-5}$
$S_{10}$ = $-\frac{1048577}{256}$ $\div$ $-\frac{5}{1}$
$S_{10}$ = $\frac{209715}{256}$
in the textbook, the answer is a negative. how???
I see what I did wrong. Should I always be using the rule mentioned?
|
Given:
$t_{1}$ = $\frac {1}{256}$
$r$ = $-4$
$n$ = $10$
$S_{n} = \frac {t_{1} (r^{n} - 1)}{r - 1}$
then $(-4)^{10} = (-1)^{10} \cdot 1048576 = 1048576$ and
\begin{align}
S_{10} &= \frac{1}{256} \, \frac{(-4)^{10} - 1}{-4 -1} \\
&= \frac{1}{256} \, \frac{1048575}{-5} = - \frac{209715}{256}
\end{align}
|
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Proving $(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3$ for all $n\in\mathbb N$ and all $a\ge -1.$ I was asked to prove the the following without induction. Could someone please verify whether my proof is right? Thank you in advance.
For any real number $a\ge -1$ and every natural number n, the statement, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ holds.
Proof. From the binomial theorem we can see that, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ becomes, $$\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.$$ Also, if $a\lt b$, let us define $\sum^{a}_{i=b}f(i)=0$, by convention. Then for n=1 we have,\begin{align}\sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-\sum^{1}_{k=4}\binom{n}{k}a^{k}\\ \sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-0,\end{align} which is true, since both remaining sums are equal to each other. It is similarly the case for $n=2$ and $n=3$. More generally, \begin{align}\sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0.\end{align}
To determine whether this last statement is true, we first consider $a\ge0$, and we see that it is trivially true. Next we consider the case where $-1\le a\lt0.$ We let $a=-b$, so $0\lt b\le 1$ and notice that, $$\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}$$ then becomes, $$\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^{k}\binom{n}{k}b^{k}+...+(-1)^{n}\binom{n}{n}b^{n}.$$
Also from the binomial theorem we can see that,
\begin{align}
(1+(-1))^{n}-(1+(-1))^{3}+(1+(-1))^{1}\\
=\sum^{n}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}-\sum^{3}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}+\sum^{1}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}\\
0=1-\binom{n}{1}+\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\\
n-1=\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\ge0,
\end{align}
and by noticing that $b^4\ge b^5\ge b^6\ge ...\ge b^k$, we can see that each binomial term $\binom{n}{k}$ is multiplied by a factor of $b^k$, making each term smaller than the term before.
Thus,
\begin{align}
\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^n\binom{n}{n}b^n\ge0\\
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{n}a^n\ge0,\tag{by substitution}
\end{align} as desired.
Working backward, \begin{align}\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ (1+a)^n&\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3.\ \ \ \blacksquare\end{align}
|
Also, we can use the following reasoning.
For $n\in\{1,2,3\}$ it's an identity.
Let $n\geq4$ and $$f(a)= (1+a)^n-1-na-\frac{n(n-1)}{2}a^{2}-\frac{n(n-1)(n-2)}{6}a^3.$$
Thus,
$$f'(a)=n(1+a)^n-n-n(n-1)(n-2)a-\frac{n(n-1)(n-2)}{2}a^2,$$,
$$f''(a)=n(n-1)(1+a)^{n-2}-n(n-1)-n(n-1)(n-2)a$$ and
$$f'''(a)=n(n-1)(n-2)(1+a)^{n-3}-n(n-1)(n-2).$$
We see that $f''$ gets a minimal value for $a=0$.
Thus, $$f''(a)\geq f(0)=0,$$
which gives that $f'($ is an increasing function and $f'(0)=0$.
Thus, since $f'$ has an unique root on $[-1,+\infty)$, we see that $x_{min}=0$,
which gives $f(x)\geq f(0)=0$ and we are done!
|
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Prove that $a_n=(1-\frac{1}{n})^n$ is monotonically increasing sequence I try to solve it Bernoulli inequality but it too complicated, am I missing something easier?
My try-
$$\frac{a_n}{a_{n+1}}=\frac{(1-\frac{1}{n})^n}{(1-\frac{1}{n+1})^{n+1}}\\=(\frac{1}{1-\frac{1}{n+1}})(\frac{\frac{n-1}{n}}{\frac{n}{n+1}})^n\\=(\frac{1}{1-\frac{1}{n+1}})(1-\frac{1}{n^2})^n<1\\\iff (1-\frac{1}{n^2})^n<1-\frac{1}{n+1}$$
here is where that I want to use Bernoulli...
|
$a_n:= (1-1/n)$, $n\in \mathbb{Z+}.$
$a_{n+1} = (1-1/(n+1)).$
$a_n \lt a_{n+1}.$
$\rightarrow:$
$(a_n)^n < (a_{n+1})^n.$
Since $1> a_n >0 :$
$a_n (a_n)^n < (a_{n+1})^n, $
$(a_n)^{n+1} < (a_{n+1})^n.$
Set $s:= \dfrac{1}{n(n+1)} >0$,
Note : $ 0 <s<1$.
Consider $f(x) := x^s,$ $x >0$, real.
$f'(x) = s\dfrac{1}{x^{1-s}}>0.$
Hence $f(x)$ is strictly increasing.
$\Rightarrow:$
$f((a_n)^{n+1}) = (a_n)^{1/n} < $
$f((a_{n+1})^n = (a_{n+1})^{1/(n+1)},$
strictly increasing.
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove the angle is 90 degrees Given the trianle ABC,draw AD, where D is the middle of BC.If the angle BAD is 3 times the angle DAC and the angle BDA is 45 degrees,then prove that the angle BAC is 90 degrees.
I tried to draw a parallel line to BA and compare congruent trianges ,after extending AD to meet the parallel line, but i got no result in proving that BAC is 90 degrees
|
Here is a trigonometric proof.
Let $x=\angle DAC$; then $3x=\angle BAD$. Let $y=BD=DC$. Then some angle chasing gives $\angle ABC=135^{\circ}-3x$ and $\angle ACB=45^{\circ}-x$. By the Law of Sines,
\begin{align*}
\frac{y}{\sin 3x} &= \frac{AD}{\sin(135^{\circ}-3x)} \\
\frac{y}{\sin x} &= \frac{AD}{\sin(45^{\circ}-x)},
\end{align*}
so that
$$\frac{\sin(135^{\circ}-3x)}{\sin 3x} = \frac{\sin(45^{\circ}-x)}{\sin x}.$$
Expanding the numerators and simplifying gives
$$\cot 3x+1 = -1+\cot x-1,\text{ or }\cot 3x = -2+\cot x.$$
Writing $\cot 3x = \frac{\cos 3x}{\sin 3x}$ and using the triple-angle formula, then rewriting in terms of $\cot x$, gives
$$\frac{\cot^3 x-3\cot x}{3\cot^2 x-1} = -2+\cot x.$$
Finally, collecting terms and simplifying, we get
$$2 \cot ^3 x-6 \cot ^2 x+2\cot x +2 = 2(\cot x-1)(\cot^2 x-2\cot x-1)=0.$$
Thus $\cot x=1$ or $\cot x = 1\pm\sqrt{2}$. The only one of these that produces an angle for $\angle BAC$ between $0$ and $180^{\circ}$ is $\cot x = 1+\sqrt{2}$, which gives $x = \frac{\pi}{8}$ so that $\angle BAC = \frac{\pi}{2}$.
|
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|
Solution of a Volterra integral equation Given the Volterra integral equation
$$y(x)=x-\int_0^x xt^2 y(t)\,dt$$
How do I solve it? Predicting $y$ using iteration is seeming difficult. Please help.
|
We start from this equation:
$$y(x) = x\left(1 - \int_0^x t^2 y(t) dt \right)$$
Differentiating both sides, we get:
$$\frac{dy(x)}{dx} = \left(1 - \int_0^x t^2 y(t) dt \right) - x^3y(x).$$
Notice that:
$$y(x) = x\left(1 - \int_0^x t^2 y(t) dt \right) \Rightarrow \left(1 - \int_0^x t^2 y(t) dt \right) = \frac{y(x)}{x}.$$
Then:
$$\frac{dy(x)}{dx} = \frac{y(x)}{x} - x^3y(x) \Rightarrow \\y' = y\left(\frac{1}{x} - x^3\right).$$
Now, we can separate the variables:
$$\int \frac{dy}{y} = \int \left(\frac{1}{x} - x^3\right)dx \Rightarrow \\
\log(y) = \log(x) - \frac{x^4}{4} + C \Rightarrow \\
y = e^{\log(x) - \frac{x^4}{4} + C} \Rightarrow\\
y = Ax e^{-\frac{x^4}{4}}.$$
|
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|
Find the : $\sum_{i=1}^{n}\dfrac{1}{i!}=\text{?}$
Find the :
$$\sum_{i=1}^n \frac{1}{i!}=\text{?}$$
For ex :
$$\sum_{i=1}^{100}\dfrac{1}{i!}=\text{?}$$
My Try :
$$\frac{1}{2\times 1}+\frac{1}{1 \times 2 \times 3 }=\frac{3(1)+1}{1 \times 2 \times 3 }$$
$$\frac{4}{1 \times 2 \times 3 }+\frac{1}{1 \times 2 \times 3 \times 4 } = \frac{4(4)+1}{1 \times 2 \times 3 \times 4}$$
$$\frac{17}{1 \times 2 \times 3 \times 4 }+\frac{1}{1 \times 2 \times 3 \times 4\times 5 }=\frac{5(17)+1}{1 \times 2 \times 3 \times 4 \times 5}$$
Now what ?
|
The "closed form" is $\frac{e \Gamma(n+1,1)}{n!} - 1$ where $\Gamma(\cdot,\cdot)$ is the incomplete Gamma function.
|
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|
How many ways to form 3 unordered partitions from n-element set? Let's say a set $S$ has $n$ elements, and it needs to be partitioned into $3$ different, unordered partitions. How do I obtain a general formula for this?
I think I can calculate it if I know the value of $n$.
For example, if $n=3$ then it's partitioned as $1,1,1$ so there's only $1$ way.
If $n=4$ then it's $2,1,1$ so there's ${4\choose2}=6 $ ways.
If $n=5$ then it's $2,2,1$ or $3,1,1$ and there's $15+10=25$ ways.
I can't figure out $n$ in general though. My latest attempt was summation:
$$\sum_{a=1}^{n-2} \quad \sum_{b=1}^{n-2-a} \quad {n\choose a}{{n-a}\choose b} $$
but the partitions won't be unordered (and the summation is probably wrong too).
I have a hint to regard the first $x$ elements in the first partition, and the $x+1$-th element in the second, but I'm not sure where to proceed from this.
|
The number of partitions of a set of size $n$ into $k$ non-empty sets is counted by the Stirling numbers of the second kind. As explained in the link, they satisfy the recurrence
$$ \left\{ \begin{array}{c} n \\ k \end{array} \right\} = k\left\{ \begin{array}{c} n - 1 \\ k \end{array} \right\}+\left\{ \begin{array}{c} n - 1 \\ k - 1 \end{array} \right\}. $$
In particular, we have
$$ \left\{ \begin{array}{c} n \\ 3 \end{array} \right\} = 3\left\{ \begin{array}{c} n - 1 \\ 3 \end{array} \right\}+\left\{ \begin{array}{c} n - 1 \\ 2 \end{array} \right\} = 3\left\{ \begin{array}{c} n - 1 \\ 3 \end{array} \right\} + 2^{n-1}-1, $$
and therefore,
\begin{align}
\left\{ \begin{array}{c} n \\ 3 \end{array} \right\} &= 3 \left\{ \begin{array}{c} n - 1 \\ 3 \end{array} \right\} + 2^{n-2}-1 \\
&= 3^2 \left\{ \begin{array}{c} n - 2 \\ 3 \end{array} \right\} + 3 \cdot 2^{n-3} + 2^{n-2} -3-1 \\
&= 3^3 \left\{ \begin{array}{c} n - 3 \\ 3 \end{array} \right\} + 3^2 \cdot 2^{n - 4} + 3 \cdot 2^{n - 3} + 2^{n - 2} - 3^2 - 3 - 1 \\
&= 3^r \left\{ \begin{array}{c} n - r \\ 3 \end{array} \right\} + \sum_{i=1}^{r} 3^{i-1} \cdot 2^{n - 1 - i} - \sum_{i=0}^{r-2} 3^i \\
&= 3^{n-3} \left\{ \begin{array}{c} 3 \\ 3 \end{array} \right\} + \sum_{i=1}^{n-3} 3^{i-1} \cdot 2^{n - 1 - i} - \sum_{i=0}^{n-4} 3^i \\
&= 3^{n-3} + 2^{n-2} \frac{(\frac{3}{2})^{n-3} - 1}{\frac{3}{2} - 1} - \frac{3^{n - 2} - 1}{3 - 1} \\
&= \frac{1}{6} \left( 3^n - 3 \cdot 2^n + 3 \right).
\end{align}
In general, one can find a formula for $\left\{ \begin{array}{c} n \\ k \end{array} \right\}$ in two ways. First, one can use the fact that the number of surjective functions $\{1,\dots,n\} \to \{1,\dots,k\}$ is $k!\left\{ \begin{array}{c} n \\ k \end{array} \right\}$ and counting this using inclusion-exclusion:
Let $A_i$, $i = 1,\dots,k$ denote the set of functions $\{1,\dots,n\} \to \{1,\dots,k\}$ whose image does not contain $i$. Let for $S \subseteq \{1,\dots,k\}$, let $A_S = \bigcap_{i \in S} A_i$. Then, by inclusion-exclusion
\begin{align}
\# (A_1 \cup \dots \cup A_k) &= \sum_{\varnothing \ne S \subseteq \{1,\dots,k\}} (-1)^{\#S - 1} \# A_S \\
k^n - k!\left\{ \begin{array}{c} n \\ k \end{array} \right\} &= \sum_{i = 1}^k \sum_{\#S = i} (-1)^{i - 1} (k - i)^n \\
k!\left\{ \begin{array}{c} n \\ k \end{array} \right\} &= k^n + \sum_{i = 1}^k \sum_{\#S = i} (-1)^{i} (k - i)^n \\
&= k^n + \sum_{i = 1}^k \binom{k}{i} (-1)^{i} (k - i)^n \\
&= \sum_{i = 0}^k \binom{k}{i} (-1)^{i} (k - i)^n \\
&= \sum_{i = 0}^k \binom{k}{i} (-1)^{k - i} i^n \\
\left\{ \begin{array}{c} n \\ k \end{array} \right\} &= \frac{1}{k!} \sum_{i = 0}^k \binom{k}{i} (-1)^{k - i} i^n.
\end{align}
Or, one can use a generating function approach:
Let $[x^ny^k]$ denote the coefficient of $x^ny^k$ in a given formal power series. Then
\begin{align}
\left\{ \begin{array}{c} n \\ k \end{array} \right\} &= n![x^ny^k] \exp(y(e^x - 1)) \\
&= n![x^ny^k] \sum_{j = 0}^\infty \frac{y^j(e^x - 1)^j}{j!} \\
&= \frac{n!}{k!}[x^n] (e^x - 1)^k \\
&= \frac{n!}{k!}[x^n] \sum_{i = 0}^k \binom{k}{i} e^{ix}(-1)^{k-i} \\
&= \frac{1}{k!} \sum_{i = 0}^k \binom{k}{i} n![x^n]e^{ix}(-1)^{k-i} \\
&= \frac{1}{k!} \sum_{i = 0}^k \binom{k}{i} i^n (-1)^{k-i}.
\end{align}
Here $e^x - 1$ is the exponential generating function for non-empty sets. That is $n$ is the number of non-empty sets of size $n$. Multiplying by $y$ lets us keep track of the number of non-empty sets we are using and taking $\exp$ gives us sets of non-empty sets (i.e. partitions into non-empty sets).
|
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|
Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
|
Finding the coefficient of $x^{n-2}$ requires picking $2$ terms from the product to multiply the constants. Thus, we get the coefficient to be
$$
\begin{align}
\sum_{k=2}^n\sum_{j=1}^{k-1}jk
&=\sum_{k=2}^n\sum_{j=1}^{k-1}\binom{j}{1}k\\
&=\sum_{k=2}^n\binom{k}{2}k\\
&=\sum_{k=2}^n\binom{k}{2}((k-2)+2)\\
&=\sum_{k=2}^n\left(3\binom{k}{3}+2\binom{k}{2}\right)\\
&=3\binom{n+1}{4}+2\binom{n+1}{3}\\[3pt]
&=\frac{(3n+2)(n^3-n)}{24}
\end{align}
$$
|
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|
Solve $y''+4y=(1+\sin{x})^2$. My attempt: The null-solution is $A\cos{2x}+B\sin{2x}.$ Let's start by rewriting RHS by
$$(1+\sin{x})^2=1+2\sin{x}+\sin^2{x}=\frac{3}{2}+2\sin{x}-\frac{1}{2}\cos{x}.$$ So we can now look at too differential equations:
$$\left\{
\begin{array}{rcr}
y_1''+4y_1 & = & \frac{3}{2}+2\sin{x} \quad \quad \quad \quad (1)\\
y_2''+4y_2 & = & -\frac{1}{2}\cos{x} \quad \quad \quad \quad (2)\\
\end{array}
\right.$$
For (1) we can assume that the particular solution is of the form $y_1=a+b\cos{x}+c\sin{x}$ and then $y_1'=-b\sin{x}+c\cos{x}$ and $y_1''=-b\cos{x}-c\sin{x}$ so
$$y_1''+4y_1= -b\cos{x}-c\sin{x} + 4(a+b\cos{x}+c\sin{x})=4a+3b\cos{x}+3c\sin{x}.$$
Identifying and solving coefficients gives $(a,b,c)=(\frac{3}{8},0,\frac{2}{3}),$ so I have that $y_{1p}=\frac{3}{8}+\frac{2}{3}\sin{x}.$
For (2) we can assume the same form for the particular solution. I get that
\begin{array}{lcl}
y_2 & = & a+b\cos{x}+c\sin{x} \\
y_2' & = & -b\sin{x}+c\cos{x} \\
y_2''& = & -b\cos{x}-c\sin{x}
\end{array}
Substituting in equation (2) gives $3a\cos{x}+3b\sin{x}=-\frac{1}{2}\cos{x}$, thus $(a,b)=(-\frac{1}{6},0)$ so $y_{2p}=-\frac{1}{6}\cos{x}.$
According to the law of super position I have that $$y_p=y_{1p}+y_{2p}=\frac{3}{8}+\frac{2}{3}\sin{x}-\frac{1}{6}\cos{x},$$
which finally gives
$$y(x)=y_h+y_p=A\cos{2x}+B\sin{2x}+\frac{2}{3}\sin{x}-\frac{1}{6}\cos{x}+\frac{3}{8}$$
Correct answer: $$y(x)=A\cos{2x}+B\sin{2x}+\frac{2}{3}\sin{x}-\frac{x}{8}\sin{2x}+\frac{3}{8}.$$
One term difference...why? Please note that I'm not interested in other solutions, I just want to know why my $y_{p2}$ is incorrect.
|
For a particular solution, you can solve for every term independently
$$y''+4y=1\to y=\frac 14.$$
$$y''+4y=\sin x\to y=\frac23\sin x$$ (by indeterminate coefficients).
Then $$y''+4y=\cos2x\to ???$$ (indeterminate coefficients don't seem to work).
It turns out that this RHS can be expressed as an instance of the homogenous solution, so that you need another ansatz.
With $x(a\cos2x+\sin2x)$, you have
$$x(-4a\cos2x-4b\sin2x+4a\cos2x+4b\sin2x)+2(2a\sin2x-2b\cos2x)=\cos2x,$$ giving
$$y''+4y=\cos2x\to \frac{x\sin 2x}4.$$
|
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|
Elementary proof that $4$ never divides $n^2 - 3$ I would like to see a proof that for all integers $n$, $4$ never divides $n^2 - 3$. I have searched around and found some things about quadratic reciprocity, but I don't know anything about that. I am wondering if there is a more elementary proof.
For example, I managed to show that $4$ never divides $x^2 - 2$ by saying that if $4$ does divide $x^2 - 2$, then $x^2 - 2$ is even. And then $x^2$ is even, which means that $x$ is even. So $x = 2m$ for some integer $m$, and so $x^2 - 2 = 4m^2 - 2$ is not divisible by $4$. So I would like to see a similar proof that $4$ doesn't divide $n^2 -3$.
|
It is obviously that if $n$ is even that $n^2-3$ is odd and so it is not divisible even by $2$.
Now suppose $n$ is odd. Among 4 consecutive integer exactly one is divisible by 4. So among $$n^2-3,\;\;\;\;n^2-2,\;\;\;\;n^2-1,\;\;\;\;n^2$$ exactly one is divisible by $4$. Since $n^2-1 = (n-1)(n+1)$ we see that $4|n^2-1$. So $4$ doesn't divide $n^2-3$.
|
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|
Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^4}})}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}\\&
=\lim_{n\rightarrow\infty}\dfrac{n+1-n\sqrt{1+\dfrac{1}{n^4}}}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}
\end{align}$
Given that $\dfrac{1}{n}$ tends to $0$ (so denominator is 2), can I reduce $n$ and $-n\sqrt{1+\dfrac{1}{n^4}}$ and say that the limit is $\dfrac{1}{2}$?
I mean $\dfrac{1}{n^4}$ tends to $0$, so $\sqrt{1+\dfrac{1}{n^4}}$ tends to $1$ and in this case $n-n\sqrt{1+\dfrac{1}{n^4}}$ can be simplified to $n-n$.
Solution given in my book uses conjugate multiplication twice to get rid of all the roots in nominator, but I am curious if my answer is correct or my teacher will tell me that simplifying the way I did it is incorrect.
|
One simple method is to try adding terms to give upper and lower bounds for the given roots. One may confirm (and not hard to find, either)
$$ \left( n + \frac{1}{2} - \frac{1}{8n} \right)^2 < n^2 + n < \left( n + \frac{1}{2} \right)^2 $$
$$ n^4 < n^4 + 1 < \left( n + \frac{1}{4n^3} \right)^4 $$
Together, we get
$$ \frac{1}{2} - \frac{1}{8n} - \frac{1}{4n^3} \; \; < \; \; \sqrt{n^2+n}-\sqrt[4]{n^4+1} \; \; < \; \; \frac{1}{2} $$
Some people call this the Squeeze Theorem, the limit is $\frac{1}{2} $
|
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|
Inverse Trigonometry System of Equations $$2\tan^{-1}\left(\sqrt{x-x^2}\right) = \tan^{-1}\left(x\right)\: +\, \tan^{-1}\left(1-x\right)$$
I have a feeling solution includes drawing triangles but cannot make the leap to get the solution
|
We need $x(1-x)\ge0\iff x(x-1)\le0\iff0\le x\le1$
$\implies\dfrac{x+1-x}2\ge\sqrt{x(1-x)}$
Now use $$\arctan x+\arctan y= \arctan\frac{x+y}{1-xy}$$ if $xy<1$
See Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
OR showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Expected number of games in a best of 7 series Assuming each team has a 0.5 probability of winning. Is there an easier way to do it other than bashing through the probabilities that the series runs 4,5,6,7 games.
|
There are $2 \binom{3}{0}$ ways for the game to finish after $4$ rounds.
There are $2 \binom{4}{1}$ ways for the game to finish after $5$ rounds.
There are $2 \binom{5}{2}$ ways for the game to finish after $6$ rounds.
There are $2 \binom{6}{3}$ ways for the game to finish after $7$ rounds.
Quick sanity check
\begin{eqnarray*}
2 \left( \frac{1}{16} +\frac{4}{32} + \frac{10}{64} +\frac{20}{128} \right) = 1.
\end{eqnarray*}
Right so the expected value is
\begin{eqnarray*}
2 \left( 4\frac{1}{16} +5\frac{4}{32} + 6\frac{10}{64} +7\frac{20}{128} \right) = \color{red}{5 \frac{13}{16}}.
\end{eqnarray*}
|
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|
Improper integral of hyperbolic function I'm looking for an elementary derivation of the following formula:
$$\int_0^\infty \frac{\sinh(ax)}{e^{bx}-1}dx=\frac{1}{2a}-\frac{\pi}{2b}\cot\frac{a\pi}{b}$$
...only where $|a|\lt b$, of course, to ensure convergence. Does anyone know of any elementary ways of proving this identity?
I'm fairly sure that a nice, elementary antiderivative does not exist, and I haven't been able to get anywhere with any of the "definite integral tricks" that I have learned thus far.
|
I give a real method that makes use of series and a bunch of special functions and their associated properties so perhaps this is not quite the elementary method you are after.
Let
$$I = \int^\infty_0 \frac{\sinh (ax)}{e^{bx} - 1} \, dx, \quad |a| < b.$$
Rewriting the hyperbolic sine function term appearing in the numerator in terms of exponentials, since the improper integral converges, we may write
\begin{align*}
I &= \frac{1}{2} \int^\infty_0 \frac{e^{ax}}{e^{bx} - 1} \, dx - \frac{1}{2} \int^\infty_0 \frac{e^{-ax}}{e^{bx} - 1} \, dx\\
&= \frac{1}{2} \int^\infty_0 \frac{e^{-(b - a)x}}{1 - e^{-bx}} - \frac{1}{2} \int^\infty_0 \frac{e^{-(b + a) x}}{1 - e^{-bx}} \, dx.
\end{align*}
Recognising the term $\dfrac{1}{1 - e^{-bx}}$ as the sum of a convergent geometric series, namely
$$\frac{1}{1 - e^{-bx}} = \sum^\infty_{n = 0} e^{-bnx}$$
the integral appearing above, after interchanging the summation with the integration, becomes
$$I = \frac{1}{2} \sum^\infty_{n = 0} \int^\infty_0 e^{-(b + nb - a)x} \, dx - \frac{1}{2} \sum^\infty_{n = 0} \int^\infty_0 e^{-(b + nb + a) x} \, dx.$$
Note that as $|a| < b$ the exponents appearing in each of the exponential terms found in each of the integrals will be negative ensuring convergence of the improper integrals. After performing the integrations we are left with
\begin{align*}
I &= \frac{1}{2} \sum^\infty_{n = 0} \frac{1}{(b - a) + nb} - \frac{1}{2} \sum^\infty_{n = 0} \frac{1}{(a + b) + nb}\\
&= \frac{1}{2b} \sum^\infty_{n = 0} \frac{1}{\left (\frac{b - a}{b} \right ) + n} - \frac{1}{2b} \sum^\infty_{n = 0} \frac{1}{\left (\frac{a + b}{b} \right ) + n}.
\end{align*}
Each of these sums can be expressed in terms of the Hurwitz zeta function $\zeta(s,q)$ which is defined as
$$\zeta(s,q) = \sum^\infty_{n = 0} \frac{1}{(q + n)^s}.$$
Thus
$$I = \frac{1}{2b} \zeta \left (1,1 - \frac{a}{b} \right ) - \frac{1}{2b} \zeta \left (1,1 + \frac{a}{b} \right ).$$
Now the Hurwitz zeta function is related to the polygamma function function of order $m$, which is denoted by $\psi^{(m)} (z)$, by (see here)
$$\psi^{(m)} (z) = (-1)^{m + 1} m! \zeta (m + 1, z).$$
Setting $m = 0$ gives
$$\zeta (1,z) = - \psi (z),$$
where $\psi (z) = \psi^{(0)} (z)$ is the digamma function. So in terms of digamma functions our integral can be written as
$$I = \frac{1}{2b} \left [\psi \left (1 + \frac{a}{b} \right ) - \psi \left (1 - \frac{a}{b} \right ) \right ].$$
Since (see here)
$$\psi (z + 1) = \psi (z) + \frac{1}{z},$$
we can write
$$\psi \left (1 + \frac{a}{b} \right ) = \psi \left (\frac{a}{b} \right ) + \frac{b}{a},$$
and from the reflection formula for the digamma function, namely
$$\psi (1 - z) - \psi (z) = \pi \cot (\pi z),$$
we have
$$\psi \left (1 - \frac{a}{b} \right ) = \psi \left (\frac{a}{b} \right ) + \pi \cot \left (\frac{a \pi}{b} \right ),$$
yielding
$$\int^\infty_0 \frac{\sinh (ax)}{e^{bx} - 1} \, dx = \frac{1}{2a} - \frac{\pi}{2b} \cot \left (\frac{a \pi}{b} \right ),$$
as claimed.
|
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|
How to evaluate the integral $\int_0^{+\infty} \frac{\sin^4{x}}{x^4}dx$? As we know that $\int_0^{+\infty} \frac{\sin{x}}{x}dx=\pi/2$,but how to evaluate the integral $\int_0^{+\infty} \frac{\sin^4{x}}{x^4}dx$?
|
Surely you know that$$\int\limits_0^{\infty}dx\,\frac {\sin^2x}{x^2}=\frac {\pi}2$$
which can be proven using integration by parts. Therefore, we take your integral by using integration by parts twice and a trigonometric identity to deduce$$\begin{align*}I & =-\frac {\sin^4x}{3x^3}\,\Biggr\rvert_{0}^{\infty}+\frac 43\int\limits_0^{\infty}dx\,\frac {\sin^3x\cos x}{x^3}\\ & =-\frac {2\cos x\sin^3x}{3x^2}\,\Biggr\rvert_0^{\infty}+\frac 23\int\limits_0^{\infty}dx\,\left(\frac {\sin^2 2x}{x^2}-\frac {\sin^2x}{x^2}\right)\end{align*}$$Where the second equation is used with the identity$$3\cos^2x\sin^2x-\sin^4=\sin^22x-\sin^2x$$Hence$$\begin{align*}I & =\frac 23\int\limits_0^{\infty}dx\,\frac {\sin^2 2x}{x^2}-\frac 23\int\limits_0^{\infty}dx\,\frac {\sin^2x}{x^2}\\ & =\frac 43\int\limits_0^{\infty}dx\,\frac {\sin^2x}{x^2}-\frac 23\int\limits_{0}^{\infty}dx\,\frac {\sin^2x}{x^2}\\ & =\color{blue}{\frac {\pi}3}\end{align*}$$
|
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|
Product of Trigonometric function
The value of $$\prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)\cdot \prod^{55}_{r=46}\bigg(1+\cot r^\circ\bigg)$$
Attempt: $\displaystyle \prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)=(1+\tan 1^\circ)(1+\tan 9^\circ)\cdots \cdots (1+\tan 4^\circ)(1+\tan 6^\circ)\tan 5^\circ$
from $\tan(A+B) = \tan 10^\circ\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B} = \tan 10^\circ$
could some help me to solve it, thanks
|
$$ \prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)\cdot \prod^{55}_{r=46}\bigg(1+\cot r^\circ\bigg) $$
$$ \prod^{10}_{r=1}\bigg( 1+\tan r^\circ \bigg)\cdot \prod^{10}_{r=1}\bigg(
1+\cot (45^\circ + r^\circ) \bigg)$$
$$ = \prod^{10}_{r=1}\bigg( (1+\tan r^\circ)(1+\cot (45^\circ + r^\circ)) \bigg) $$
$$ = \prod^{10}_{r=1} 2 $$
$$ = 2^{10} $$
The key observation is, both products have the same number of terms.
More details on how to simplify the middle term inside the product:
$$(1+\tan r^\circ)(1+\cot (45^\circ + r^\circ))$$
$$ = (1+\tan r^\circ) \left( 1+{1 \over \tan (45^\circ + r^\circ)} \right) $$
$$ = (1+\tan r^\circ) \left( 1+{1 - \tan 45^\circ \times \tan r^\circ \over \tan 45^\circ + \tan r^\circ} \right) $$
$$ = (1+\tan r^\circ) \left( 1+{1 - \tan r^\circ \over 1 + \tan r^\circ} \right) $$
$$ = (1+\tan r^\circ) \left( {2 \over 1 + \tan r^\circ} \right) $$
$$ = 2 $$
|
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|
Taylor Polynomial $\ln(1-x)$ Given $f(x) = \ln(1-x)$ centered at $0$, with $n = 0,1,2$:
Why is it that the first derivative (for $n=1$): $$\frac{d}{dx}(p(x)) = -x$$ in a Taylor Polynomial? The answer that I got was $$\frac{d}{dx}(p(x)) = 1$$
I got the right answer for the second derivative $$\frac{d^2}{d^2x}(p(x)) = -\frac{x^2}{2}$$
Overall the final Taylor polynomial is:
$$-x-\frac{x^2}{2}$$
I got: $$x-\frac{x^2}{2}$$
Where exactly is that negative coming from?
|
\begin{align}
\int\frac{du}{u-1} = {} & \log\left| u-1 \right| + C \\[10pt]
= {} & \log(1-u) + C \text{ if $ u$ is near $1$} \\
& \text{since in that case, $u-1$ is negative.} \\[10pt]
\text{So } \int_0^x \frac{du}{u-1} & = \log(1-x) - \log(1-0) = \log(1-x).
\end{align}
Thus we have
\begin{align}
\log(1-x) & = \int_0^x \frac{du}{u-1} = \int_0^x \left( -1-u - u^2 -u^3 -u^4 - \cdots \right) \,du \\[10pt]
& = \left[ -u -\frac {u^2} 2 - \frac{u^3} 3 - \frac {u^4} 4 - \cdots \right]_{u\,:=\,0}^{u\,:=\,x} = -x-\frac{x^2} 2 - \frac{x^3} 3 - \frac {x^4} 4 - \cdots
\end{align}
The series on the right above, and the function on the left above, should be equal in the values of their $n$th derivatives for $n=0,1,2,3,\ldots\,.$
|
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|
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
|
Hint:
If $\sqrt{x+2}=a,\sqrt[3]{x+20}=b,\sqrt[4]{x+9}=c$
LCM$(2,3)=6$
$a^6-b^6=(a-b)(\cdots)$
Similarly $c^4-2^4=(c-2)\cdots$
|
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|
Find the real solutions for the system: $ x^3+y^3=1$,$x^2y+2xy^2+y^3=2.$
Find the real solutions for the system:
$$\left\{
\begin{array}{l}
x^3+y^3=1\\
x^2y+2xy^2+y^3=2\\
\end{array}
\right.
$$
From a book with exercises for math contests. The solutions provided are: $(x,y)=(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{\sqrt[3]{2}})$ and $(\dfrac{1}{3^{\frac{2}{3}}},\dfrac{2}{3^{\frac{2}{3}}})$.
Working with the expressions I could find that an equivalent system is
$$\left\{
\begin{array}{l}
(x+y)(x^2-xy+y^2)=1\\
y(x+y)^2=2\\
\end{array}
\right.
$$
Developing these expressions I got stuck.
Hints and answers are appreciated. Sorry if this is a duplicate.
|
Multiply the first equation by $2$ and then set the two left sides equal:
$$2x^3+2y^3 = x^2y+2xy^2+y^3.$$
This is a homogeneous equation of degree $3$, so divide through by $x^3$:
$$2+2\frac{y^3}{x^3} = \frac{y}{x}+2\frac{y^2}{x^2}+\frac{y^3}{x^3}.$$
Substitute $u=y/x$:
$$2+2u^2=u+2u^2+u^3$$
$$u^3-2u^2-u+2=0$$
$$(u^2-1)(u-2)=0$$
$$u=-1, 1 \mbox{ or } 2.$$
Dividing the first equation by $x^3$ gives $1+u^3 = 1/x^3.$ Plug each value of $u$ into this to find values of $x$ and then $y$.
|
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|
Simple solution to this algebraic equation? Consider this system of equations with two variables $x,y$ and all positive parameters
$$\frac{2(x-w_1)}{a_1}=\frac{x}{\sqrt{x^2+y^2}}\,,\quad\frac{2(y-w_2)}{a_2}=\frac{y}{\sqrt{x^2+y^2}}.$$
It at most amounts to an algebraic equation of degree 4, which is always solvable in principle. But is there any possibility to explicitly solve this one? Hopefully some not too complicated form of solution? Thanks.
|
It follows from
\begin{equation}
\frac{(x-w_1)^2}{a_1^2} + \frac{(y-w_2)^2}{a_2^2} = \frac{1}{2^2}
\end{equation}
that
the solutions are lying on the ellipses.
Thus
\begin{equation}
y = w_2 \pm \sqrt{\frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2}}
\end{equation}
Then plugging this into the below
\begin{equation}
\frac{4}{a_1^2}(x-w_1)^2(x^2+y^2) = x^2
\end{equation}
we have
\begin{equation}
-x^2 + \frac{4}{a_1^2}(x-w_1)^2\left(x^2 + \left[w_2 \pm \sqrt{\frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2}} \right]^2\right) = 0
\end{equation}
which is
\begin{equation}
-x^2 + \frac{4}{a_1^2}(x-w_1)^2\left(x^2 + w_2^2 + \frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2} \pm 2w_2\sqrt{\frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2}} \right) = 0
\end{equation}
|
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|
Exercise on elementary number theory Let $a,b,c,d$ integers, $c\not=0$ such that $ad-bc=1$ and $c\equiv 0 \pmod p$ for some prime $p>3$.
Show that if $a+d=\pm1$ then $p\equiv 1\pmod 3$
I don't know how to approach this problem because when I take the expression $ad-bc$ modulo $p$ we have that either $d-d^2\equiv 1\pmod p$ or $d+d^2\equiv 1\pmod p$.
|
Let's consider:
$ad-bc\equiv1 \pmod p\implies ad=1-bc \equiv 1\pmod p$
$(a+d)^2 \equiv a^2+d^2+2ad\equiv a^2+d^2+2\equiv1\pmod p$
$\implies a^2+d^2+1\equiv 0\pmod p$
$\implies (a^2+d^2+1)^2\equiv 0\pmod p$
Since:
$a+d=\pm1 \implies a\cdot d=(k+1)\cdot(-k)=-k^2-k$
Consider the following table $\pmod 3$
$$\begin{array}{ l | c | r }
k & a & d & a^2 & d^2 & p \\
-1 & 0 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 1 \\
1 & -1 & -1 & 1 & 1 & 0 \\
\end{array}$$
$\implies p \equiv1 \pmod 3 \quad \square$
|
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|
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$
My Attempt:
Given
\begin{align}
a(a+7b+49c)+c(a-b+c) &< 0 \\
49a \left( \dfrac {a}{49} + \dfrac {b}{7} + c \right)+c(a-b+c) &< 0 \\
49a \cdot f\left( \dfrac {1}{7} \right) + f(0)\cdot f(-1) &< 0
\end{align}
|
We can assume that $a=1$ and let $f(x) =x^2+bx+c$.
If $c< 0$ then the graph of $f$ cuts the $y-$ axis under the $x-$ axis, so it must have real roots.
If $c=0$ then 0ne root is $0$ and the second is $-b$.
If $c>0$, since we have $$49f({1\over 7})+cf(-1)=0$$
we have 2 possibilities.
a) If $f(-1)=f(1/7)=0$ we are done.
b) If $f(-1)>0>f(1/7)$ or $f(-1)<0<f(1/7)$ then we have one real root in $(-1,{1\over 7})$ and we are done.
|
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|
Closed form of $I(t) = \int_0^{\pi/2}\frac{\cos xdx}{\sqrt{\sin^2 x+ t\cos^2 x}}$ What is the closed form of
$$I(t) = \int_0^{\pi/2}\frac{\cos xdx}{\sqrt{\sin^2 x+ t\cos^2 x}}$$
I tired the change of variables $$\int_a^bf(x) dx= \int_a^bf(a+b-x) dx$$
but is was not fruitful.
|
For $t > 1$
$$I(t) = \int_0^{\pi/2}\frac{\cos xdx}{\sqrt{\sin^2 x+ t\cos^2 x}} = \int_0^{1}\frac{d(\sin x)}{\sqrt{\sin^2 x(1- t)+ t }} = \dfrac{\arcsin\left(\frac{\sqrt{t-1}}{\sqrt{t}}\right)}{\sqrt{t-1}}$$.
For $0 < t < 1$, let $ u = \sin x$
$$I(t) = \int_0^1 \dfrac{du}{\sqrt{u^2(1 - t) + t}} = \dfrac{1}{\sqrt t}\int_0^1 \dfrac{du}{\sqrt {(u((1-t)/t)^{1/2})^2+1} } = \dfrac{1}{\sqrt {1-t}}\int^{\sqrt{\frac{1-t}{t}}}_0 \dfrac{dz}{\sqrt{z^2 +1}}$$
where $z = u \sqrt{\dfrac{1- t}{t}}$.
So $$I(t) = \dfrac{1}{\sqrt{1-t}}\log\left|\sqrt{\dfrac 1t} + \sqrt{\frac{1-t}{t}}\right|$$.
|
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|
Estimate from below of the sine (and from above of cosine) I'm trying to do the following exercise with no success. I'm asked to prove that
$$\sin(x) \ge x-\frac{x^3}{2}\,, \qquad \forall x\in [0,1]$$
By using Taylor's expansion, it's basically immediate that one has the better estimate
$$\sin(x) \ge x-\frac{x^3}{6}\,, \qquad \forall x\in [0,1]$$
as the tail converges absolutely, and one can check that the difference of consecutive terms is positive.
I suppose then, there is a more elementary way to get the first one. Question is: how?
Relatedly, the same exercise asks me to prove that
$$\cos(x) \le \frac{1}{\sqrt{1+x^2}}\,,\qquad \forall x\in [0,1]$$
which again I can prove by using differentiation techniques. But these haven't been explained at that point of the text, so I wonder how to do it "elementarly".
|
A geometric proof is as follows.
Outline:
*
*Show $\cos x > 1-\frac{1}{2}x^2.$
*Show that $\tan x> x.$
From there, you quickly see that $\sin x>x\cos x>x-x^3/2.$
We have that $\sqrt{(1-\cos x)^2+\sin^2 x}$ is the length of segment from $(1,0)$ to $(\cos x,\sin x)$, which is $\leq x$, since the arc along the circle between these two points is length $x$ and the shortest distance between two points is the line.
But $(1-\cos x)^2+\sin^2 x=2-2\cos x$.
So you have $2-2\cos x < x^2$, or $\cos x > 1-\frac12x^2$.
The area of the triangle $(0,0),(1,0),(1,\tan x)$ is $\frac{1}{2}\tan x$ and this triangle contains a region of the unit circle of area $\frac{1}{2}x$. So you get that $\tan x>x.$
This gives the result you want.
The second part uses that $\tan^2 x + 1=\frac{1}{\cos^2 x}$ so $$\cos x = \frac{1}{\sqrt{1+\tan^2 x}}<\frac{1}{\sqrt{1+x^2}}$$ since $x<\tan x$.
Extending to show $\sin x > x-x^3/6$:
Now, if, for all $x\in(0,\pi/2)$, $x>\sin x>x-ax^3$ for $x\in (0,\pi/2)$ then use:
$$\sin x = 2\sin \frac{x}{2}\cos\frac{x}{2}>2\left(\frac{1}{2}x-\frac{a}{8}x^3\right)\left(1-\frac{1}{8}x^2\right)>x-\frac{2a+1}{8}x^3$$
This requires both $1-\frac{a}{4}x^2$ and $1-\frac{1}{8}x^2$ to be positive. Since all values of $a$ in question will be $\leq \frac12$ we want $x<\sqrt{8}$, which is clearly true for $x\in(0,\pi/2).$
If we define $a_0=\frac{1}{2}$ and $a_{n+1}=\frac{2a_n+1}{8}$ then you have that $a_n$ is decreasing and the limit is $\frac{1}{6}.$ Since the $0<a_n\leq \frac{1}{2}$, we get, inductively, for $x\in(0,1)$ that:
$$\sin x > x-a_nx^3$$
In the limit, this means that $\sin x\geq x-\frac{1}{6}x^3.$
|
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|
Find the limit of a series of fractions starting with $\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}$ Problem
Let $a_{0}(n) = \frac{2n-1}{2n}$ and $a_{k+1}(n) = \frac{a_{k}(n)}{a_{k}(n+2^k)}$ for $k \geq 0.$
The first several terms in the series $a_k(1)$ for $k \geq 0$ are:
$$\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}, \, \frac{\frac{1/2}{3/4}/\frac{5/6}{7/8}}{\frac{9/10}{11/12}/\frac{13/14}{15/16}}, \, \ldots$$
What limit do the values of these fractions approach?
My idea
I have calculated the series using recursion in C programming, and it turns out that for $k \geq 8$, the first several digits of $a_k(1)$ are $ 0.7071067811 \ldots,$ so I guess that the limit exists and would be $\frac{1}{\sqrt{2}}$.
|
Let $f_0(z) = z$ and $f_{n+1}(z) = f_n(z) / f_n(z+2^n)$
One can show that when $z \to \infty$, the rational fractions $f_n$ for $n \ge 1$ have asymptotic developments at infinity that converge for $|z| > n$, such that
$f_n(z) = 1 + O(z^{-n})$ and $f_n'(z) = O(z^{-n-1})$
Call $s(k) = +1,-1,-1,+1,\cdots$ the Thue-Morse sequence.
Then the sequence $\prod_{k \ge 0}^n f_2(z+4k)^{s(k)}$ converge pointwise, and the convergence is uniform on half-spaces of the form $\Re(z) \ge A$ for $A > -1$.
The sequence of their derivative also converge uniformly on those opens, so the limit is holomorphic on $\Re(z)> -1$.
Since the sequence $(f_n(z))$ for $n\ge 2$ is a subsequence of this sequence, it also converge to the same limit $f(z)$.
Next, notice that
$f_n(z)/f_n(z+ \frac 12) = f_{n+1}(2z)$.
Taking the limit as $n \to \infty$, this gives the functional equation $f(z)/f(z+ \frac 12) = f(2z)$, or $f(z + \frac 12)f(2z) = f(z)$
(and incidentally, $f$ has a meromorphic continuation to the whole complex plane with zeroes and poles at $0$ and the negative integers).
Let $k$ be the order of $f$ at $0$, so that $f(z) \sim az^k$ for some nonzero $a \in \Bbb C$ as $z \to 0$.
Looking at the functional equation as $z \to 0$, one gets $f(\frac 12) = 2^{-k}$.
And finally, evaluating the functional equation at $z = \frac 12$, $f(1)f(1) = f(\frac 12)$, and so $f(1) = 2^{- \frac 12k}$
Then one simply needs to evaluate $f(1)$ (or $f(\frac 12)$) to enough precision to determine that the order of $f$ at $0$ is $1$ and not anything larger.
So indeed, $f(1) = 2^{- \frac 12}$
|
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|
Simplify expression $\frac{2\cos(x)+1}{4\cos(x/2+π/6)}$ How to simplify the following expression: $$\frac{2\cos(x)+1}{4\cos\left(\frac x2+\fracπ6\right)}$$
I got to: $ \dfrac{2\cos(x)+1}{4\cos\left(\dfrac x2\right)\cdot \dfrac{\sqrt3}2-\sin(x) \cdot \frac 12}$
|
Using half-angle formula: $\cos\dfrac A2=\pm\sqrt{\dfrac{1+\cos A}2}$ with $A=x+\dfrac\pi3$ to give $$\cos\left(\frac x2 + \frac \pi6 \right)=\pm\sqrt{\dfrac{1+\cos \left(x + \frac \pi3 \right)}2}=\pm\sqrt{\dfrac{1+\frac{\sqrt3}2\cos x-\frac12 \sin x }2}$$ so $$\cos\left(\frac x2 + \frac \pi6 \right)=\pm\frac{\sqrt{1+\sqrt3\cos x-\sin x}}2.$$ Hence $$\frac{2 \cos x + 1}{4\cos\left(\frac x2 + \frac \pi6 \right)}=\pm \frac{2(2 \cos x + 1)}{\sqrt{1+\sqrt3\cos x-\sin x}}$$ which is positive if $\dfrac x2 + \dfrac \pi6$ is in quadrant I or IV and negative if $\dfrac x2 + \dfrac \pi6$ is in quadrant II or III.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Non-empty limit set for the dynamical system : $x_1' = x_1 + 2x_2 - 2x_1(x_1^2 + x_2^2)^2, \space x_2' = 4x_1 + 3x_2 - 3x_2(x_1^2 + x_2^2)^2 $
Using the Lyapunov Function $V=\frac{1}{2}x_1^2 + \frac{1}{2}x_2^2$ , prove that the omega(ω)-limit set is non-empty for any initial value given for the dynamical system :
$$x_1' = x_1 + 2x_2 - 2x_1(x_1^2 + x_2^2)^2$$
$$x_2' = 4x_1 + 3x_2 - 3x_2(x_1^2 + x_2^2)^2$$
Attempt :
$$\nabla V(x)F(x)=x_1(x_1 + 2x_2 - 2x_1(x_1^2+x_2^2)) + x_2(4x_1+3x_2-3x_2(x_1^2+x_2^2))$$
$$=$$
$$x_1^2 + 2x_2x_1 - 2x_1^2(x_1^2+x_2^2)^2 + 4x_1x_2 + 3x_2^2 - 3x_2^2(x_1^2 + x_2^2)^2$$
$$=$$
$$x_1^2 + 3x_2^2 + 6x_1x_2 - 2x_1^2(x_1^2 + x_2^2)^2 - 3x_2^2(x_1^2+x_2^2)^2$$
$$\leq$$
$$3x_1^2 + 3x_2^2 + 6x_1x_2 - 2x_1^2(x_1^2 + x_2^2)^2-2x_2^2(x_1^2 + x_2^2)$$
$$=$$
$$3(x_1+x_2)^2 - 2(x_1^2 + x_2^2)^2(x_1^2 + x_2^2)$$
$$\leq$$
$$3(x_1+x_2)^2 - (x_1^2 + x_2^2)^3 = 3(x_1+x_2)^2 - V^3(x)$$
Now, I don't know how to continue from this point forward. I know I have to make the final expression in terms of $V(x)$ but I can't see how I would convert $3(x_1+x_2)^2$ to express it in terms of $V(x)$.
If I have any mistake or if I could go on another approach regarding the inequalities, I would really appreciate some help for these or for what I should do next.
|
The Cauchy's inequality can be rewriten in the form
$$(x_1\cdot 1+x_2\cdot 1)^2\le (x_1^2+x_2^2)\cdot (1^2+1^2)$$
or $$(x_1+x_2)^2\le 2(x_1^2+x_2^2).$$
This implies that
$$
\dot V\leq 3(x_1+x_2)^2 - 2(x_1^2 + x_2^2)^3\le
6(x_1^2+x_2^2)-2(x_1^2+x_2^2)^3
$$
It is easy to check that $\forall (x_1,x_2):\; x_1^2+x_2^2>\sqrt3\;\;$ $\dot V<0$, thus the solution for any initial value enters the bounded set $$
\Omega_C=\left\{ (x_1,x_2):\; x_1^2+x_2^2<C \right\}
$$
in finite time for any $C>\sqrt3$ and stays there forever, hence any solution is bounded and, due to the Bolzano–Weierstrass theorem, has a nonempty $\omega$-limit set.
|
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|
Use Viete's relations to prove the roots of the equation $x^3+ax+b=0$ satisfy $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$ Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.
I know that viete's relations state that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3-px^2+qx-r=0$ have the property $p=x_1+x_2+x_3$, $q=x_1x_2+x_1x_3+x_2x_3$ and $r=x_1x_2x_3$.
My question is whether or not there is a way to do this without multiplying out $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ and showing that it factors into $4(x_1+x_2+x_3)^3+27(x_1x_2x_3)$ because that algebra involved in that looks like it will be nasty.
|
Let
$$f(x)=x^3+ax+b=(x-x_1)(x-x_2)(x-x_3)$$
so that
$$f'(x)=3x^2+a=(x-x_1)(x-x_2)+(x-x_1)(x-x_3)+(x-x_2)(x-x_3)$$
This allows us to get the equation
$$ f'(x_1)f'(x_2)f'(x_3)=(3x_1^2+a)(3x_2^2+a)(3x_3^2+a)=-(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 $$
We try to evaluate
$$
\begin{align}
(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 &=
-(3x_1^2+a)(3x_2^2+a)(3x_3+a) \\
&=-(a^3+3a^2(x_1^2+x_2^2+x_3^2)+9(x_1^2 x_2^2+x_1^2 x_3^2+x_2^2x_3^2)a+27x_1^2x_2^2x_3^2)
\end{align}
$$
Using the Vieta's formulas
$$
\begin{align}
x_1+x_2+x_3 &= 0\\
x_1x_2+x_1x_3+x_2x_3 &= a\\
x_1x_2x_3 &=-b
\end{align}
$$
we obtain
$$
\begin{align}
x_1^2+x_2^2+x_3^2 &= (x_1+x_2+x_3)^2 - 2(x_1x_2+x_1x_3+x_2x_3)\\
&=-2a\\
x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2 &= (x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)\\
&= a^2
\end{align}
$$
Hence we get the final expression
$$
(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = -(a^3-6a^3+9a^3+27b^2)=-(4a^3+27b^2)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all real solutions for the system: $x^3=y+y^5$, $y^5=z+z^7$, $z^7=x+x^3.$
Given: $$\left\{
\begin{array}{l}
x^3=y+y^5\\
y^5=z+z^7\\
z^7=x+x^3
\end{array}
\right.
$$
Find: all real solutions for the system.
From a book on preparation for math contests. The answer states there is just one solution. My problem is showing that this is indeed the case.
My attempt: it is easy to see that one solution is $(x,y,z)=(0,0,0)$. And, adding the equations, we get to $x+y+z=0$. But my problem is showing that this solution is indeed unique, if the answer provided in the book is right.
Hints and answers are appreciated. Sorry if this is a duplicate.
|
Let $x>0$.
Hence, $z^7=x(1+x^2)>0$, which gives $z>0$.
Also, $y^5=z(1+z^6)>0$, which gives $y>0$.
But summing of all equations gives $x+y+z=0$, which is a contradiction.
By the same way we can get a contradiction for $x<0$.
Thus, $x=0$ and from here we obtain $x=y=z=0.$
|
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|
Solve $\sqrt{x^2+1}-\sqrt{4x^4-4x^2+2}=2x^3-x-1$
Could you please help me solve for $x$ in
$$\sqrt{x^2+1}-\sqrt{4x^4-4x^2+2}=2x^3-x-1.$$
I tried this way. But I could not solve further. Please help me.
$$(\sqrt{x^2+1}-\sqrt{4x^4-4x^2+2})^2=(2x^3-x-1)^2$$
|
As $x$ increases:
$\sqrt{4x^4-4x^2+2} = \sqrt{(2x^2-1)^2+1}$ decreases if $x \in [0,\frac12\sqrt2]$ and increases if $x \in [\frac12\sqrt2,-\infty)$.
Symmetrically $\sqrt{4x^4-4x^2+2}$ decreases if $x \in (-\infty,-\frac12\sqrt2]$ and increases if $x \in [-\frac12\sqrt2,0]$.
Therefore $\sqrt{4x^4-4x^2+2} \le \sqrt{2}$ if $|x| \le \frac12\sqrt2$.
If $x \in [0,\frac12\sqrt2]$:
$-(2x^3-x-1) = 1-x·(2x^2-1) \ge 1$.
If $x \in [-\frac12\sqrt2,0]$.
$-(2x^3-x-1) = (1-x)·(\frac12(2x+1)^2+\frac12) \ge \frac12$.
If $|x| \le \frac12\sqrt2$:
$\sqrt{4x^4-4x^2+2} \le \sqrt2 < 1+\frac12 \le 1-(2x^3-x-1)$.
If $x < -\frac12\sqrt2$:
$\sqrt{4x^4-4x^2+2} \le 2x^2 < 1-(2x^3-x-1)$ because $(x+1)·(2x^2-1) < 1$.
Therefore $\sqrt{4x^4-4x^2+2} < \sqrt{x^2+1}-(2x^3-x-1)$ if $x \le \frac12\sqrt2$.
If $x > \frac12\sqrt2$:
$\frac{d}{dx}(\sqrt{x^2+1}-(2x^3-x-1)) = \frac{x}{\sqrt{x^2+1}} - 6x^2+1 < 0$.
Therefore $\sqrt{x^2+1}-(2x^3-x-1) = \sqrt{4x^4-4x^2+2}$ for at most one real $x$, and indeed equality holds when $x = 1$.
|
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|
Get closed form from a complicated closed-form generating function I have a closed form for a generating function:
$A(x)=\frac{x(x-1)(x+1)^3(x^3-x-1)}{(x^3+x^2-1)^2}$
The coefficient of $x^n$ in the above represents $a_n$ (the $n^{th}$ term of a sequence). I want a closed form for the sequence $a_n$. For example, the expansion done in:
http://www.wolframalpha.com/input/?i=x%5E2(1%2Bx%2B(x%5E2(x%2B1)%5E2)%2F(1-x%5E2-x%5E3))%2Bx*(1%2Bx%2B(x%5E2*(x%2B1)%5E2)%2F(1-x%5E2-x%5E3))(x%5E2%2Bx%5E2(1%2Bx%2B(x%5E2*(x%2B1)%5E2)%2F(1-x%5E2-x%5E3)))%2Bx%2B(x%5E2%2Bx%5E2*(1%2Bx%2B(x%5E2*(x%2B1)%5E2)%2F(1-x%5E2-x%5E3)))
shows the first few coefficients: $x+3x^2+4x^3+5x^4+8x^5+O(x^6)$. But I want a closed form for the general $n^{th}$ term $a_n$. What method should I use to get it?
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Hint: Though not a closed form we can use the binomial series expansion to derive an alternate explicit expression for $[x^n]A(x)$, the coefficient of $x^n$ of $A(x)$.
At first we write $A(x)$ with increasing powers of the denominator $1-x^2-x^3$. We obtain using the Euclidian algorithm for polynomials:
\begin{align*}
A(x)&=\frac{x(x-1)(x+1)^3(x^3-x-1)}{(x^3+x^2-1)^2}\\
&=\frac{x^8+2x^7-x^6-5x^5-3x^4+2x^3+3x^2+x}{(x^3+x^2-1)^2}\\
&=\cdots\\
&=x^2-2-\frac{x^2-x}{1-x^2-x^3}+\frac{x^2+x}{\left(1-x^2-x^3\right)^2}\tag{1}
\end{align*}
Next we calculate the coefficient of $x^n$ of $\frac{1}{(1-x^2-x^3)^\alpha}$ which we need with $\alpha\in\{0,1\}$.
We obtain
\begin{align*}
[x^n]\frac{1}{\left(1-x^2-x^3\right)^{\alpha}}&=[x^n]\frac{1}{\left(1-x^2\left(1+x\right)\right)^{\alpha}}\\
&=[x^n]\sum_{j=0}^\infty\binom{-\alpha}{j}(-1)^jx^{2j}(1+x)^j\tag{2}\\
&=\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{\alpha+j-1}{j}[x^{n-2j}](1+x)^j\tag{3}\\
&=\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{\alpha+j-1}{j}\binom{j}{n-2j}\tag{4}
\end{align*}
Comment:
*
*In (2) we apply the binomial series expansion of $(1+z)^{\alpha}$ with $z=-x^2(1+x)$.
*In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (4) we select the coefficient of $x^{2j}$ and set the upper bound of the series to $\left\lfloor\frac{n}{2}\right\rfloor$ since the coefficient of $[x^{n-2j}]$ has to be non-negative.
From (1) and (4) we obtain the coefficient of $A(x)$ as
\begin{align*}
\color{blue}{[x^n]A(x)}&=[x^n]\left(x^2-2\right)-\left([x^{n-2}]-2[x^n]\right)\frac{1}{1-x^2-x^3}\\
&\qquad+\left([x^{n-2}]+[x^n]\right)\frac{1}{\left(1-x^2-x^3\right)^2}\tag{5}\\
&=[[n=2]]-2[[n=0]]-\left([x^{n-2}]-2[x^n]\right)\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{j}{n-2j}x^j\\
&\qquad+\left([x^{n-2}]-[x^{n-1}]\right)\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}(j+1)\binom{j}{n-2j}x^j\tag{6}\\
&\color{blue}{=[[n=2]]-2[[n=0]]+2\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{j}{n-2j}}\\
&\qquad\color{blue}{+\sum_{j=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(j+1)\binom{j}{n-1-2j}
+\sum_{j=0}^{\left\lfloor\frac{n}{2}-1\right\rfloor}j\binom{j}{n-2-2j}}\tag{7}
\end{align*}
Comment:
*
*In (5) we use the linearity of the coefficient of operator.
*In (6) we use the Iverson brackets $[[P]]$ which is $1$ if and only if the proposition $P$ is true and $0$ otherwise. We also use the representation of the coefficients from (2).
*In (7) we select the coefficients accordingly and collect corresponding terms.
Note: The sequence $\left(\sum_{j}\binom{j}{n-2j}\right)_{n\geq 0}=(1,0,1,1,1,2,2,3,4,5,7,9,12,16,\ldots)$ is archived in OEIS as A000931.
|
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|
Evaluating $\lim_{x \to 1^{-}} \prod_{n=0}^{\infty} \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$ If $$\lim_{x \to 1^{-}} \prod_{n=0}^{\infty} \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}=l$$ the question is to find out the integer part of the number $1/l$
I tried bringing the expression into telescopic series but failed to do so.Any ideas?Thanks.
|
Let's consider the logarithm of this product. The partial sum of the resulting series is
\begin{align}
\sum^n_{k=0}x^k\ln\frac{1+x^{k+1}}{1+x^k}&=\sum^n_{k=0}x^k\ln(1+x^{k+1})-\sum^n_{k=0}x^k\ln(1+x^k)
\\&=\sum^n_{k=0}x^k\ln(1+x^{k+1})-\sum^{n+1}_{k=1}x^k\ln(1+x^k)-\ln2+x^{n+1}\ln(1+x^{n+1})
\\&=\sum^n_{k=0}x^k\ln(1+x^{k+1})-\sum^{n}_{k=0}x^{k+1}\ln(1+x^{k+1})-\ln2+x^{n+1}\ln(1+x^{n+1})
\\&=\sum^n_{k=0}(x^k-x^{k+1})\ln(1+x^{k+1})-\ln2+x^{n+1}\ln(1+x^{n+1})
\end{align}
Now letting $n\to\infty$, we see
$$\sum^\infty_{k=0}x^k\ln\frac{1+x^{k+1}}{1+x^k}=\sum^\infty_{k=0}(x^k-x^{k+1})\ln(1+x^{k+1})-\ln2$$ for $x\in[0,1)$. But the series on the RHS is a Riemann sum for $\int^1_0\ln(1+u)\,du$ with a partition $u_k=x^k$, and $\max(u_k-u_{k+1})=1-x\to0$ as $x\to1-$, so the limit of the RHS is
$$\int^1_0\ln(1+u)\,du-\ln2=\ln2-1.$$ So the limit of the infinite product is
$$l=e^{\ln2-1}=\frac2e.$$
|
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|
How to prove $\prod_{k=1}^{\infty}{\frac{p_k^2+1}{p_k^2-1}}=\frac{5}{2}$? I came across the following formula due to Ramanujan:
$$\prod_{k=1}^{\infty}{\frac{p_k^2+1}{p_k^2-1}}=\frac{5}{2}.$$
Can someone show me what the proof of this looks like, or point me to a reference (in English)?
|
By Euler's product, for any $s>1$ we have
$$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \prod_{p}\left(1-\frac{1}{p^s}\right)^{-1} \tag{A}$$
hence
$$ \prod_{p}\frac{p^2+1}{p^2-1} = \prod_p\frac{1-\frac{1}{p^4}}{\left(1-\frac{1}{p^2}\right)^2} = \frac{\zeta(2)^2}{\zeta(4)} = \frac{90}{36} = \frac{5}{2}.\tag{B}$$
Similarly, $\prod_{p}\frac{p^4+1}{p^4-1}=\frac{7}{6}$.
|
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|
Finding Eigenvectors [Confused] I would like to find the eigenvalues for the matrix
$$\begin{pmatrix}
2 & 3-3i \\
3+3i & 5
\end{pmatrix}$$
I find that the eigenvalues are $8$ and $-1$.
For eigenvalue of $8$ I get
$$
\begin{pmatrix}
-2 & 1-1i \\
1+1i & -1
\end{pmatrix}=0
$$
and I get the equation
$$-2x + (1-1i)y = 0\\
(1+1i)x - y = 0
$$
My question is how do I solve this system? I've tried multiple times and have ended up with the wrong eigenvalue when comparing with the back of my textbook.
|
First the eigenvalues
$\begin{align}\det(A-tI)&=\begin{vmatrix}2-t & 3-3i\\ 3+3i & 5-t\end{vmatrix}=(2-t)(5-t)-(3+3i)\overline{(3+3i)}=10-7t+t^2-18\\\\&=t^2-7t-8=(t+1)(t-8)\end{align}$
So you are correct, $\operatorname{Sp}(A)=\{-1,8\}$.
Now for the eigenvectors you want to solve two systems for $v=(x,y)^T$:
*
*$Av=-v$
$\begin{cases}2x+(3-3i)y=-x\\(3+3i)x+5y=-y\end{cases}\iff\begin{cases}(3-3i)y=-3x\\(3+3i)x=-6y\end{cases}\overset{\div 3}{\iff}\begin{cases}(1-i)y=-x\\(1+i)x=-2y\end{cases}$
Since $(1+i)(1-i)=2$ both equations are equivalent, meaning we can choose either $x$ or $y$ and the other is determined.
If you actually report one equation into the other you find
either y=y or x=x, thus either y or x is a free variable.
For instance (1+i)x=(1+i)(1-i)(-y)=-2y <=> -2y=-2y <=> y=y
This is expected since eigenvectors are only defined modulo
proportionality.
We can choose for instance $y=-1$ then $v_{-1}=\begin{pmatrix}1-i\\-1\end{pmatrix}$.
*
*$Av=8v$
$\begin{cases}2x+(3-3i)y=8x\\(3+3i)x+5y=8y\end{cases}\iff\begin{cases}(3-3i)y=6x\\(3+3i)x=3y\end{cases}\overset{\div 3}{\iff}\begin{cases}(1-i)y=2x\\(1+i)x=y\end{cases}$
Since $(1+i)(1-i)=2$ both equations are equivalent, meaning we can choose either $x$ or $y$ and the other is determined.
We can choose for instance $x=1$ then $v_{8}=\begin{pmatrix}1\\1+i\end{pmatrix}$.
The change of basis matrix is then $P=(v_8,v_{-1})=\begin{pmatrix}1 & 1-i\\ 1+i & -1\end{pmatrix}$
And now $P^{-1}AP=\begin{pmatrix}8 & 0\\ 0 & -1\end{pmatrix}$
|
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|
$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
|
$$\lim_{x \to 0}\frac{1}{1-(1-\frac{x^2}{2!}+O(x^4))} - \frac{2}{x^2} = \frac{2}{x^2}-\frac{2}{x^2} = 0$$(higher order terms neglected as the become very small as $x \to0$)
|
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|
find the range of the function : $y=(3\sin 2x-4\cos 2x)^2-5$ find the range of the function :
$$y=(3\sin 2x-4\cos 2x)^2-5$$
My try :
$$y=9\sin^22x+16\cos^22x-24\sin 2x\cos 2x-5\\y=9+7\cos^22x-12\sin4x-5$$
now what do I do؟
|
You can write $$3\sin 2x-4\cos 2x =5\sin(2x+\varphi)$$
for some constant angle $\varphi$,
so $$ y = 25\sin^2(2x+\varphi)-5$$
so $y_{\max} = 20$ and $y_{\min} = -5$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$ If $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?
Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?
|
It's not so bad; after you substitute $y=\sqrt{x-2}$ you will have
$$\sqrt{y^2 - 8y + 16 } + \sqrt{y^2 - 10y+ 25 } = 3$$
and you should be able to take it from here without any squaring of both sides.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3} -\frac{1}{6} +\frac{1}{5}-\dots+\frac{1}{2N-1}$
$= \frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$
We may rewrite the series in pairs as we know it will have an even amount of terms.
$S_{2N+1} = \sum_{n=1}^{2N+1} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3}-\dots+\frac{1}{2N-1} -\frac{1}{2N+2}$
$=\frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} - \frac{1}{2N+2} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}-\frac{1}{2N+2}$
As $n\in\mathbb{N}$, we know that $n\geq1$ so:
$n\geq1 \iff 3n\geq3 \iff 3n^2\geq3n \iff 3n^2-3n\geq0 \iff 4n^2-2n \geq n^2+n$
So: $2n(2n-1)\geq n(n+1) \iff \frac{1}{2n(2n-1)}\leq \frac{1}{n(n+1)}$ for all $n\geq1$.
As the series of the latter sequence converges, we can conclude, by the comparison, test that the series $\sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$ converges.
Suppose it converges to $s$, then we know $^{\lim S_{2N}}_{N\to\infty} = s$ and thus $\lim_{N\to\infty}[S_{2N+1}] = s - (\lim_{N\to\infty}[\frac{1}{2N+2}]) = s-0 = s.$ As the partial sums ending with even and uneven terms both converge to the same limit, the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. $\tag*{$\Box$}$
|
Another way is to observe that
$$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n-(-1)^{n}}=\sum_{n=1}^{\infty}\Bigg[\frac{(-1)^{n}}{n-(-1)^{n}}-\frac{(-1)^{n}}{n}\Bigg]+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}=$$
$$=\sum_{n=1}^{\infty}\frac{1}{n^{2}-(-1)^{n}n}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$$
Now the left series converges by the limit comparison test with $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$, and the right series converges by Leibniz criterion.
|
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|
Proving formally that a set is subspace. I want to prove formally that $U = (x,y,z\in \mathbb{R^3})$
$| \det\begin{bmatrix}a_{11}&a_{12} & x\\
a_{21}&a_{22}&y\\ a_{31}&a_{32}&z \end{bmatrix} = 0$ is a subspace.
I know this is true but the technique I learnt, I think I can't apply it here.
To prove that they are closed under vector addition, suppose for example for some $f =(x,y,z) $ and $g = (x',y',z')$ determinant is $0$ then the I learnt I have to add both vectors $f,g$ and put it in the original expression but then I get $\det(A + B) = \det(A) + \det(B)$ and this way I can't prove that this is a vector space because in general $\det(A+B) \neq \det(A) + \det(B)$
|
Useful observation:
$$\det\begin{bmatrix}a_{11}&a_{12} & x\\
a_{21}&a_{22}&y\\ a_{31}&a_{32}&z \end{bmatrix} = 0 \iff \left\{\begin{pmatrix}a_{11} \\ a_{21} \\a_{31} \end{pmatrix}, \begin{pmatrix}a_{12} \\ a_{22} \\a_{32} \end{pmatrix}, \begin{pmatrix}x \\ y \\z \end{pmatrix}\right\} \text{ is linearly dependent}$$
So if $\begin{pmatrix}a_{11} \\ a_{21} \\a_{31} \end{pmatrix}$ and $\begin{pmatrix}a_{12} \\ a_{22} \\a_{32} \end{pmatrix}$ are linearly dependent, then $U = \mathbb{R}^3$ so it's a subspace.
Otherwise, $\begin{pmatrix}x \\ y \\z \end{pmatrix} \in U$ if and only if $\begin{pmatrix}x \\ y \\z \end{pmatrix} \in \operatorname{span}\left\{\begin{pmatrix}a_{11} \\ a_{21} \\a_{31} \end{pmatrix}, \begin{pmatrix}a_{12} \\ a_{22} \\a_{32} \end{pmatrix}\right\}$.
Thus, $U = \operatorname{span}\left\{\begin{pmatrix}a_{11} \\ a_{21} \\a_{31} \end{pmatrix}, \begin{pmatrix}a_{12} \\ a_{22} \\a_{32} \end{pmatrix}\right\}$ so it is certanly a subspace of $\mathbb{R}^3$.
|
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|
2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation using Rational Root Theorem and got $y = \frac 1 2$
$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$
Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $
$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $
But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$
Why don't I get the second solution correct?
|
All my answers are correct. However, the last answer had to be obtained using a different form. I had used the general solution of $\cos^2 x = \cos ^2 \alpha$ but they expected me to use the general form of $\cos x = 0$
"Correction":
$\cos^2 y = \frac 1 2 \implies \cos 2y = \cos 0 \implies 2y = (2n+1)\frac\pi2 \implies x = (2n+1)\dfrac \pi 8$
Any alternative methods to solve the equation are welcome.
|
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|
Inverse of Laplace operator Let $\ L=\frac{-d^2}{dx^2}$ defined on $\ H^2(]0,1[) ∩ H^1_0(]0,1[)$ \ $\ L$ is the laplacian operator in one dimension \ how can we express the inverse of $\ L $
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The Green's function might need a slight modification.
\begin{align*}
G\left(x,y\right) & =\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\sin\left(n\pi x\right)\sin\left(n\pi y\right)\\
& =\begin{cases}
\left(1-y\right)x & 0\leq x\leq y\\
y\left(1-x\right) & y\leq x\leq1
\end{cases}.
\end{align*}
In details:
Let $L=-\frac{d^{2}}{dx^{2}}$, with domain as specified. Then $L$
is selfadjoint, and it has eigenfunctions (normalized) $\left\{ \varphi_{n}\left(x\right):=\sqrt{2}\sin\left(n\pi x\right)\mathrel{;}n\in\mathbb{N}\right\} $,
so that
\begin{align*}
L^{-1} & =\sum_{n=1}^{\infty}\lambda_{n}^{-1}\varphi_{n}\left(x\right)\varphi_{n}\left(y\right)=\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\sin\left(n\pi x\right)\sin\left(n\pi y\right).
\end{align*}
For the Green's function, assume $LG\left(\cdot,y\right)=\delta\left(\cdot-y\right)$,
$y\in\left(0,1\right)$, subject to the condition $G\left(0,y\right)=G\left(0,1\right)=0$.
So
\begin{align*}
G\left(x,y\right) & =\begin{cases}
ax & x\leq y\\
c\left(x-1\right) & x\geq y
\end{cases}
\end{align*}
where $a=\frac{c\left(y-1\right)}{y}$, since $G\left(\cdot,y\right)$
is continuous at $x=y$. But
\begin{align*}
\lim_{x\rightarrow y^{-}}G_{x}\left(x,y\right)-\lim_{x\rightarrow y^{+}}G_{x}\left(x,y\right) & =1,
\end{align*}
so $c=-y$.
|
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|
Find the Bezout coefficients Find the Bezout coefficients for $a(x)$ and $b(x)$:
$a(x)=3x^4-4x^3-11x^2+4x+9, b(x)=3x^3+5x^2+x-1$
I find the greatest common divisor:
1) $\frac{(3x^4-4x^3-11x^2+4x+9)}{(3x^3+5x^2+x-1)} = x-3$. Remainder of the division: 3x^2+8x+6
2) $\frac{(3x^3+5x^2+x-1)}{(3x^2+8x+6)} = x-1$. Remainder of the division: 3x+5
3) $\frac{(x^2+8x+6)}{(3x+5)} = x+1$. Remainder of the division: $1$
The greatest common divisor: $1$
I find the Bezout coefficients with the help of the extended Euclidean algorithm:
$1\mid 0$
$0\mid 1$
$1\mid -x+3$
$-x+1\mid -x^2+4x-2$
$x^2\mid x^3-4x^2+2x$
I'm doing a check. It should work like this: $(x^2)*a(x) + (x^3-4x^2+2x)*b(x) = 1$
But here an error, the last line $(x^2\mid x^3-4x^2+2x)$ was not calculated correctly. Help me please.
|
I got
$$ \left( 3 x^{4} - 4 x^{3} - 11 x^{2} + 4 x + 9 \right) $$
$$ \left( 3 x^{3} + 5 x^{2} + x - 1 \right) $$
$$ \left( 3 x^{4} - 4 x^{3} - 11 x^{2} + 4 x + 9 \right) = \left( 3 x^{3} + 5 x^{2} + x - 1 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 3 x^{2} + 8 x + 6 \right) $$
$$ \left( 3 x^{3} + 5 x^{2} + x - 1 \right) = \left( 3 x^{2} + 8 x + 6 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 3 x + 5 \right) $$
$$ \left( 3 x^{2} + 8 x + 6 \right) = \left( 3 x + 5 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 1 \right) $$
$$ \left( 3 x + 5 \right) = \left( 1 \right) \cdot \color{magenta}{ \left( 3 x + 5 \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x - 3 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x - 3 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} - 4 x + 4 \right) }{ \left( x - 1 \right) } $$
$$ \color{magenta}{ \left( x + 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{3} - 3 x^{2} + x + 1 \right) }{ \left( x^{2} \right) } $$
$$ \color{magenta}{ \left( 3 x + 5 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 3 x^{4} - 4 x^{3} - 11 x^{2} + 4 x + 9 \right) }{ \left( 3 x^{3} + 5 x^{2} + x - 1 \right) } $$
$$ \left( 3 x^{4} - 4 x^{3} - 11 x^{2} + 4 x + 9 \right) \left( x^{2} \right) - \left( 3 x^{3} + 5 x^{2} + x - 1 \right) \left( x^{3} - 3 x^{2} + x + 1 \right) = \left( 1 \right) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Example of multivalued function (includes pairwise comparisons) that attains maximum when values form evenly spaced vector Suppose, that I have 4 points $x,y,z,w$ (positions on the x-axis), such that $x\le y\le z\le w$.
I find pairwise distances between them: $y-x,z-x,w-x,z-y,w-y,w-z$. Distances should be less or equal to some constant $k$.
Is there a function that contains all pairwise distances and attains the only maximum when $x,y,z,w$ are evenly spaced on the interval $[1, k+1]$ and the only minima when all variables are equal?
The function shouldn't depend on $k$.
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Let $\,g\,$ be a strictly concave and strictly increasing function on $\,[\,0,k\,]\,$ such that $\,g(0)=0\,$.
Define $\,f : \left\{\, (x,y,z,w) \in [\,1,k+1\,]^4 \;\mid\; x \le y \le z \le w \,\right\} \to \mathbb{R}\,$ as:
$$
\begin{align}
f(x,y,z,w) \;&=\; 2g(y-x)+g(z-y)+2g(w-z) \\[3px]
&\quad +\;2 g\left(\frac{z-x}{2}\right) + 2 g\left(\frac{w-y}{2}\right) + \\[3px]
&\quad +\;3 g\left(\frac{w-x}{3}\right)
\end{align}
$$
*
*$\,f\,$ is non-negative since so is $\,g\,$.
*$\,f(x,y,z,w) \gt 0\,$ if at least one of the arguments passed into $\,g\,$ is non-zero, since the respective $\,g(\cdot)\,$ term is strictly positive in that case, so $f(x,y,z,w)=0$ iff all arguments are zero $\;\iff\; x=y=z=w\,$ i.e. the minimum of $\,0\,$ is attained when all variables are equal.
*It follows from Jensen's inequality that the maximum for a fixed total span $\,w-x\,$ is attained when $y-x$ $=z-y$ $=w-z$ $\displaystyle=\frac{z-x}{2}$ $\displaystyle=\frac{w-y}{2}$ $\displaystyle=\frac{w-x}{3}$ which happens iff the values are evenly spaced, and it follows from monotonicity that the absolute maximum is attained when $x=1,w=k+1$ i.e. at $\displaystyle\,\left(1,\frac{k+3}{3},\frac{2k+3}{3},k+1\right)\,$:
$$\require{cancel}
\begin{align}
\displaystyle
f(x,y,z,w) \;&=\; 2g(y-x)+g(z-y)+2g(w-z) + 2 g\left(\frac{z-x}{2}\right) + 2 g\left(\frac{w-y}{2}\right) +\;3 g\left(\frac{w-x}{3}\right) \\[5px]
&\le\; 12 \cdot g\left(\frac{2(y-x)+(z-y)+2(w-z)+2 \cfrac{z-x}{2}+2 \cfrac{w-y}{2}+3 \cfrac{w-x}{3}}{12}\right) \\[5px]
&=\; 12 \cdot g\left(\frac{4(w-x)}{12}\right) \\[5px]
&\le 12 \cdot g\left(\frac{(k+\bcancel{1})-\bcancel{1}}{3}\right) \\[5px]
&= 12 \cdot g\left(\frac{k}{3}\right)
\end{align}
$$
An example could be $\,g(x) =\ln(1+x)\,$, but any $\,g\,$ satisfying the given conditions would work.
[ EDIT ] A more general solution along the same line can be constructed from the identity:
$$
\small a (y - x) + b (z - y) + c (w - z) + (c - b) (z - x) + (a - b) (w - y) + (p - a + b - c) (w - x) \;=\; p(w-x)
$$
Choose $\,0 \le b \le a,c\,$ and $\,p \ge a-b+c\,$ so that all coefficients are positive. Then the function defined below satisfies the posed requirements:
$$
\begin{align}
f(x,y,z,w) \;&=\; a\cdot g(y-x)+ b \cdot g(z-y)+ c \cdot g(w-z) \\[3px]
&\quad + 2(c-b) \cdot g\left(\frac{z-x}{2}\right) + 2(a-b) \cdot g\left(\frac{w-y}{2}\right) \\[3px]
&\quad + 3(p-a+b-c) \cdot g\left(\frac{w-x}{3}\right)
\end{align}
$$
The proof goes just like before, and the maximum is $\displaystyle\,3p \cdot g\left(\frac{k}{3}\right)\,$.
The particular case above corresponds to $b=1, a=c=2, p=4\,$, and the case used in this related answer corresponds to $a=b=c=p=1\,$.
|
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|
$\max a^3+b^3+c^3+4abc$ sub $0\leq a,b,c \le 3/2$ and $a+b+c=3$ Let $S$ be the set of $(a,b,c) \in \mathbb{R}^3$ such that $0\leq a,b,c \leq \frac{3}{2}$ and $a+b+c=3$. Find
$$
\max_{(a,b,c) \in S} a^3+b^3+c^3+4abc.
$$
|
Let us start with a triple $(a, b, c)$. Let us try to "tune" the triple and see if we can come out with a candidate $(a', b', c')$ with a larger $a'^3 + b'^3 + c'^3 + 4a'b'c'$. Heuristic tells us that $(a', b', c') = (a, \frac{b+c}{2}, \frac{b+c}{2})$ worth some consideration.
Let $A = a^3 + b^3 + c^3 + 4abc$ and $B = a'^3 + b'^3 + c'^3 + 4a'b'c'$. By some calculation
$$
\begin{aligned}
A - B &= \left(b^3 + c^3 - \frac{(b+c)^3}{4}\right) + 4a\left(bc - \frac{(b+c)^2}{4}\right)\\
&= (b+c)\left(\frac{3(b-c)^2}{4}\right) - a(b-c)^2 = (b-c)^2\left(\frac{3(b+c)}{4}-a\right)\\
&= (b-c)^2 \left(\frac{3(a+b+c)}{4}- \frac74 a\right) = (b-c)^2\left(\frac{9-7a}{4}\right).
\end{aligned}
$$
This has two implications:
*
*If $a > \frac97$, $(a', b', c')$ is better than $(a, b, c)$.
*If $a < \frac97$, $(a, b, c)$ is better than $(a', b', c')$. Moreover, by looking at the difference, we see that the best we can do to maximize $a^3 + b^3 + c^3+ 4abc$ fixing $a$ is to maximize the difference of $b$ and $c$.
Now, from a set $(a, b, c)$, what we can do is as the follows.
By swapping the variables if necessary, we can assume $a \leq 1$. Let us do the follows:
*
*We know that we can increase $a^3 + b^3 + c^3 +4abc$ by replacing $(a, b, c)$ by $(a, \frac32, \frac32-a)$.
*We also know that we can increase $a^3 + b^3 + c^3 +4abc$ by replacing $(a, \frac32, \frac32-a)$ by $(\frac 32, \frac34, \frac 34)$. From this point we can do no more, hence the maximal value is
$$
\left(\frac32\right)^3 + \left(\frac34\right)^3 + \left(\frac34\right)^3 + 4\times \frac32 \times \frac 34 \times \frac 34 = \frac{243}{32} \approx 7.59.
$$
This is indeed suprising...
|
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|
Integral of rational function - which contour to use? Evaluate : $$\int_{-\infty}^{+\infty} \frac {x}{(x^2+2x+2)(x^2+4)}$$
I found that the integrand can be extended to a function on a complex plane has simple poles at $\pm 2i$ and $-1\pm i$. Now I want to compute the integral by contour integration but I am unable to assume any contour here.
Do excuse me , if my approach is wrong.
|
Hint: Use upper half plane as contour and fraction decomposition
$$\dfrac{z}{(z^2+2z+2)(z^2+4)}=\dfrac{1}{10}\frac{z-2}{z^2+2z+2}-\dfrac{1}{10}\frac{z-4}{z^2+4}$$
then
$$\dfrac{1}{10}\int_C\frac{z-2}{z^2+2z+2}-\frac{z-4}{z^2+4}dz=\dfrac{2\pi i}{10}\left(\operatorname*{Res}_{z=i-1}\frac{z-2}{z^2+2z+2}-\operatorname*{Res}_{z=2i}\frac{z-4}{z^2+4}\right)=\dfrac{2\pi i}{10}\left(\frac{i-3}{2i}-\frac{2i-4}{4i}\right)=\color{blue}{-\dfrac{\pi}{10}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sqrt{s(s-a)(s-b)(s-c)}=A$ In 50 AD, the Heron of Alexandria came up with the well-known formula, that, given the three side lengths of a triangle (or even two and an angle, thanks to trigonometry) you can get the area of said triangle by using this formula:
$$
\text{if } s=\frac{a+b+c}{2},\\
\text{then} A=\sqrt{s(s-a)(s-b)(s-c)}
$$
But the book it came with (mathematics 1001) did not give a mathematical proof to this. So I tried and failed to get a proof to this. This is how my train wreck went…
$$\begin{align}
\text{equation }&\sqrt{s(s-a)(s-b)(s-c)}\\
\text{substitute }&\sqrt{\frac{a+b+c}{2}(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)}\\
\text{eliminate fractions}&\sqrt{a+b+c(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)}\\
\text{comb. Like terms }&\sqrt{a+b+c(b+c-a)(a-b+c)(a+b-c)}\\
\text{associative property }&\sqrt{(a+(b+c)((b+c)-a))((a-(b+c))(a+(b-c)))}\\
\text{difference of 2 squares }&\sqrt{((b+c)^2-a^2)(a^2-b^2-c^2)}\\
\text{square of two numbers }& \sqrt{((b^2+bc+c^2)-a^2)(a^2-b^2-c^2)}\\
\text{polynomial multiplication}& \sqrt{2(a^2b^2-b^2c^2+a^2c^2)+a^2(b-c+a^2)-(b+c)(b^3+c^3)}\\
\end{align}$$
The issue is, I don't know how to get to $A$, the area. Does anyone know how to continue the proof train?? Thanks in advance.
|
Take a look here for the proof: Heron's formula
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Circumcircle and Square Let $P$ be a point on the circumcircle of square $ABCD$. Find all integers
$n > 0$ such that the sum
$$S^n(P) = PA^n + PB^n + PC^n + PD^n$$
is constant with respect to point $P$.
|
Say $x$ is a side of $ABCD$ and $a=PA$, $b=PB$...
If $n=1$ then the sum is not constant:
Say $P$ is on smaller arc $BC$. Then by Ptolomey theorem for $ABPC$ we have $$ax =bx\sqrt{2}+cx$$ and by Ptolomey theorem for $DBPC$ we have $$dx= bx+cx\sqrt{2}$$
so $$a+d = (b+c)(1+\sqrt{2}) \Longrightarrow a+b+c+d = (2+\sqrt{2})(b+c)$$
Since obviously $b+c$ is not constant aslo $a+b+c+d$ is not constant.
If $n=2$ the sum is constant:
Then by theorem of Pythagoras we have $$a^2+c^2 = 2x^2\;\;\;{\rm and}\;\;\;b^2+d^2 = 2x^2\Longrightarrow S^2(P) = 4x^2$$
|
{
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"url": "https://math.stackexchange.com/questions/2582134",
"timestamp": "2023-03-29T00:00:00",
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|
product of terms taken $3$ at a time in polynomial expression
Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$
Try:
$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdots+98\cdot 99\cdot 100)x^{98}+\cdots$$
for $1$ at a time (Coefficient of $x^{99}$) is $\displaystyle \sum^{100}_{i=1}i = 50\cdot 101$
for $2$ at a time (Coefficient of $x^{98}$) is $\displaystyle \sum^{100}_{i=1}\sum^{100}_{j=1\;, (1\leq i<j \leq 100)}i \cdot j= \frac{1}{2}\bigg[\bigg(\sum^{100}_{i=1}i\bigg)^2-\sum^{100}_{i=1}i^2\bigg]$
But could some help me how to find coefficient of $x^{97},$ thanks
|
With some manipulation (using inclusion exclusion principle) you can write the sum as:
$$\left(\sum_{i=1}^{100} i\right)^3 = \binom{3}{1}\left(\sum_{i=1}^{100} i^2\right)\left(\sum_{i=1}^{100} i\right) - 2\left(\sum_{i=1}^{100}i^3\right) + \binom{3}{1} \binom{2}{1}S $$
Where $S$ is the required sum.
|
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"language": "en",
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|
Taylor limits with sine I'm having troubles calculating these two limits (I prefer to write a sigle question including both of them, instead of two different ones).
This one
$$\\ \lim_{x\rightarrow 0} \frac{ x-\sin^2(\sqrt x)-\sin^2(x)} {x^2} $$ I tried expanding with Taylors at different orders but the square root in the sine argument gave problems with the o() grades, leading for example to things like this $$ o(\sqrt x^3) $$ that does not seem easy to manage. I tried rewriting the limit using $$ \sin^2(x) + \cos^2(x) = 1 $$, obtaining
$$\\ \lim_{x\rightarrow 0} \frac{ x-1+\cos^2(\sqrt x)-\sin^2(x)} {x^2} $$
since the McLaurin expasion for the cosine looked a little better to me, but I keep obtaining a wrong result.
The second limit in this one instead
$$\\ \lim_{x\rightarrow 0} \frac{ [\frac{1}{(1-x)} +e^x]^2 -4e^{2x} -2x^2} {x^3} $$
and I happily wrote the McLaurin expansion stopping at the second order, that are
$$(1-x)^{-1} = 1+x+\frac{x^2}{2} + o(x^2)$$
$$e^x = 1+x+\frac{x^2}{2} +o(x^2)$$
$$e^{2x} = 1 + 2x+2x^2+o(x^2)$$
doing the calculation leads to this
$$\\ \lim_{x\rightarrow 0}\frac{\frac{-4}{3}x^3-2x^2}{x^3}$$
$$\\ \lim_{x\rightarrow 0}\frac{-4}{3} - \frac{2}{x}$$ and here I have problems understating what is going on. I don't know if x tends to 0 from right or left, so I can't evaluate the result (that would be, $$ +\infty$$ or $$ -\infty$$ ). Does this mean I did some wrong calculation? Do I need to take some more orders in the McLaurin expansion?
|
Second limit
Note that:
$$(1-x)^{-1}=1+x+x^2+x^3+o(x^3)$$
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)$$
$$({(1-x)^{-1}} +e^x)^2=\left(1+x+x^2+x^3+1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)^2=\left(2+2x+\frac{3x^2}{2}+\frac{7x^3}{6}+o(x^3)\right)^2=4+4x^2+8x+6x^2+6x^3+\frac{14x^3}{3}+o(x^3)=4+8x+10x^2+\frac{32x^3}{3}+o(x^3)$$
$$e^{2x}=1+2x+\frac{4x^2}{2}+\frac{8x^3}{6}+o(x^3)=1+2x+2x^2+\frac{4x^3}{3}+o(x^3)$$
thus
$$\frac{ ({(1-x)^{-1}} +e^x)^2 -4e^{2x} -2x^2} {x^3}=\frac{4+8x+10x^2+\frac{32x^3}{3}-4-8x-8x^2-\frac{16x^3}{3}-2x^2+o(x^3)} {x^3}=\frac{\frac{16x^3}{3}+o(x^3)} {x^3}=\frac{16}{3}+o(1)\to\frac{16}{3}$$
|
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|
How is that $A^n A\neq A A^n$? How is that $A^n A\neq A A^n$? Where $A$ is a $n\times n$ matrix whose elements belong to a set with ring structure(it may not be commutative).
$A^nA$ can be expressed as $(AA\cdots AA)A$, which, by associativity of matrix multiplication it means that it is the same as $A(AA\cdots AA)$
Suppose the $2\times 2$ matrix
$
\begin{pmatrix}a & b \\ c & d \end{pmatrix}
$. Then $A^2=
\begin{pmatrix}
a^2+b c & a b+bd \\
ca+d c & cb+d^2 \\
\end{pmatrix}$
But
\begin{align}
A^3=AA^2&=
\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}
a^2+b c & a b+bd \\
ca+d c & cb+d^2 \\
\end{pmatrix} \\
&= \begin{pmatrix}
a(a^2+b c)+b(ca+d c) & a(a b+bd)+b(cb+d^2) \\
c(a^2+b c)+d(ca+d c) & c(a b+bd)+d(cb+d^2) \\
\end{pmatrix}
\end{align}
and
\begin{align}
A^3=A^2A&=
\begin{pmatrix}
a^2+b c & a b+bd \\
ca+d c & cb+d^2 \\
\end{pmatrix}\begin{pmatrix}a & b \\ c & d \end{pmatrix} \\
&= \begin{pmatrix}
(a^2+b c)a+(a b+bd)c & (a^2+b c)b+(a b+bd)d \\
(ca+d c)a+(cb+d^2)c & (ca+d c)b+(cb+d^2)d \\
\end{pmatrix} \\
A^3=A^2A&\neq AA^2=A^3
\end{align}
knowing the fact that commutativity of its elements is not a given. This means that matrix multiplication over a matrix with elements of rings as its elements is not associative, which means it is not a matrix, right? How is this explained?
|
Do a little more algebra and expand out the terms. For example,
$$ (AA^2)_{1,1} = a(a^2+bc) + b(ca + dc) = a^3 + abc + bca + bdc $$
and
$$ (A^2A)_{1,1} = (a^2+bc)a + (ab + bd)c = a^3 + bca + abc + bdc. $$
Since addition is commutative, these two terms are equal. The remaining entries are similar.
|
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|
How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2}} - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + ..$ How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2} } - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + \frac{\binom{n^2}{4}}{\binom{n+4}{4}} - ......$
I really have no idea how to proceed in this question. Expanding it by the formula isn't helping as much as I can see. And the options are extremely sofisticated as well like 1/n, 1/(n+1), 1. How to proceed?
|
A variation. We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^{n^2}}&\color{blue}{(-1)^j\binom{n^2}{j}\binom{n+j}{j}^{-1}}\\
&=\sum_{j=0}^{n^2}\binom{n^2}{j}\binom{-n-1}{j}^{-1}\tag{1}\\
&=\sum_{j=0}^{n^2}\binom{n^2}{j}(-n)\int_{0}^1t^j(1-t)^{-n-1-j}\,dt\tag{2}\\
&=(-n)\int_{0}^1(1-t)^{-n-1}\sum_{j=0}^{n^2}\binom{n^2}{j}\left(\frac{t}{1-t}\right)^j\,dt\\
&=(-n)\int_{0}^1(1-t)^{-n-1}\left(1+\frac{t}{1-t}\right)^{n^2}\,dt\\
&=(-n)\int_{0}^1(1-t)^{-n^2-n-1}\,dt\\
&=\frac{-n}{-n^2-n}\\
&\color{blue}{=\frac{1}{n+1}}
\end{align*}
Comment:
*
*In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (2) we apply the identity $\binom{n}{r}^{-1}=(n+1)\int_{0}^1t^r(1-t)^{n-r}\,dt$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$
$$\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$
Try: $$y=\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$
$$\log_{e}(y) =n\sum^{n}_{r=1}\log_{e}\bigg(1+\frac{r}{n^2}\bigg)-\frac{n}{2}$$
could some help me to solve it , thanks
|
We have $$\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{r}{n^2}-\frac12\frac{r^2}{n^4}+O\left(\frac{r^3}{n^6}\right),$$ that means
$$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{n(n+1)}2\frac1n-\frac12\frac{n(n+1)(2n+1)}6\frac1{n^3}+O(n^4)\frac1{n^5},$$ so we get
$$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}-\frac{n}2\to\frac12-\frac16=\frac13$$ as $n\to\infty$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Point of intersection when only direction ratios are given I am starting out with 3D Geometry. In one of the test booklets, I found a question for which I have no idea where and how to start from.
If a line with direction ratio $2:2:1$ intersects the line $\frac{x-7}{3}$ = $\frac{y-5}{2}$ = $\frac{z-3}{2}$ and $\frac{x-1}{2}$ = $\frac{y+1}{4}$ = $\frac{z+1}{3}$ at A and B, find AB.
I know the algorithm for finding the points of intersection when a line is in the symmetrical form. I tried with an approach wherein I tried to find A by equating:
$\frac{x}{2}$ = $\frac{y}{2}$ = $\frac{z}{1}$ and $\frac{x-7}{3}$ = $\frac{y-5}{2}$ = $\frac{z-3}{2}$
However, on equating, I found out that these lines never even intersect.
How do I approach this sum? How to start off with this sum?
|
Let $a$ be the common value of $$\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{2}=a$$
We can re-write these equations under the following equivalent parametric form:
$$\tag{1}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}7+3a\\5+2a\\3+2a\end{pmatrix}$$
In the same way, the generic point of the second straight line is : $$\tag{2}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\ \ 1+2b\\-1+4b\\-1+3b\end{pmatrix}$$
Then you just have to express the proportionality of $\vec{AB}$ with the given vector, yielding the following system of 2 equations and 2 unknowns:
$$\dfrac{(1+2b)-(7+3a)}{2}=\dfrac{(-1+4b)-(5+2a)}{2}=\dfrac{(-1+3b)-(3+2a)}{1}$$
giving $a=-2/3$ and $b=1/3$.
It remains to plug these values in (1) and (2) to get the coordinates of
$$A=\begin{pmatrix}5\\11/3\\5/3\end{pmatrix} \ \ \ \text{and} \ \ \ B=\begin{pmatrix}5/3\\1/3\\0\end{pmatrix}.$$
|
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|
How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?
|
I guess you're almost there...
If $a\ge 0$, then $1+a\ge 1$ and so $(1+a)(1+b)\ge 1+b$. This gives you
$$\frac b {(1+a)(1+b)}\le \frac b {1+b}$$
and the result you're looking for follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting for your kindest helps.
Thank you.
|
writing $$\frac{2^x}{8}+\frac{15}{8}2^x=2^8$$ so
$$2^x\left(\frac{1}{8}+\frac{15}{8}\right)=2^8$$
Can you finish?
|
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|
Summation. What does is evaluate to? What is $\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$ if $a_{n+2}=a_{n+1}+a_{n}$ and $a_{1}=a_{2}=1$?
|
$a_{n+2}=a_{n+1}+a_{n}$
with
$a_1=a_2 = 1$.
Let
$f(x)
=\sum_{n=1}^{\infty} a_nx^n
$.
$xf(x)
=\sum_{n=1}^{\infty} a_nx^{n+1}
=\sum_{n=2}^{\infty} a_{n-1}x^{n}
$
and
$x^2f(x)
=\sum_{n=1}^{\infty} a_nx^{n+2}
=\sum_{n=3}^{\infty} a_{n-2}x^{n}
$
so
$\begin{array}\\
xf(x)+x^2f(x)
&=\sum_{n=2}^{\infty} a_{n-1}x^{n}+\sum_{n=3}^{\infty} a_{n-2}x^{n}\\
&=a_1x^2+\sum_{n=3}^{\infty} (a_{n-1}+a_{n-2})x^{n}\\
&=a_1x^2+\sum_{n=3}^{\infty} a_{n}x^{n}\\
&=a_1x^2+f(x)-a_1x-a_2x^2\\
\text{so}\\
f(x)(x^2+x-1)
&=(a_1-a_2)x^2-a_1x\\
\end{array}
$
Put in the initial
$a_1, a_2$
and $x = \frac14$.
Note that
this does not need
the explicit formula
for the $a_n$.
|
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|
Find the maximum of the expression Let $a,b,c$ be real positive numbers so that $abc=1$. Find the maximum value that the following expression can attain:
$$\frac{a}{a^8+1}+\frac{b}{b^8+1}+\frac{c}{c^8+1}$$
My try:
I first though on apply a variable change so that $a=\frac{x}{y}$, $b= \frac{y}{z}$ and $c=\frac{z}{x}$. The problem is that the problem became harder for me:
$$\sum_{cyc} \frac{xy^7}{x^8+x^7}$$
Then I though on applying Holder in the denominator of the first expression, so it would look like:
$$\sum_{cyc} \frac{a}{a^8+1} \leq \sum_{cyc} \frac{2^7a}{(a+1)^8}$$
After that, I tried applying $a+1 \geq 2\sqrt{a}$. But the expression wasn't correct anymore.
|
We'll prove that $$\frac{a}{a^8+1}\leq\frac{3(a^6+1)}{4(a^{12}+a^6+1)}.$$
Indeed, we need to prove that
$$\frac{1}{a^4+\frac{1}{a^4}}\leq\frac{3\left(a^3+\frac{1}{a^3}\right)}{4\left(a^6+\frac{1}{a^6}+1\right)}.$$
Let $a+\frac{1}{a}=2t$.
Thus, by AM-GM $t\geq1$ and we need to prove that:
$$\frac{1}{16t^4-16t^2+2}\leq\frac{3(8t^3-6t)}{4((8t^3-6t)^2-2+1)}$$ or
$$(t-1)(96t^6+32t^5-136t^4-40t^3+44t^2+8t-1)\geq0,$$ which is true because
$$96t^6+32t^5-136t^4-40t^3+44t^2+8t-1\geq$$
$$\geq96t^6+32t^5-136t^4-40t^3+44t^2+4t=$$
$$=4t(t-1)(24t^4+32t^3-2t^2-12t-1)\geq0.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{3(a^6+1)}{4(a^{12}+a^6+1)}\leq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^6+1}{a^{12}+a^6+1}\leq2$$ or
$$\sum_{cyc}\left(\frac{a^6+1}{a^{12}+a^6+1}-1\right)\leq2-3$$ or
$$\sum_{cyc}\frac{a^{12}}{a^{12}+a^6+1}\geq1.$$
Now, let $a^6=\frac{x}{y}$ and $b^6=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.
Thus, $c^6=\frac{z}{x}$ and we need to prove that
$$\sum_{cyc}\frac{x^2}{x^2+xy+y^2}\geq1.$$
Now, by C-S we obtain:
$$\sum_{cyc}\frac{x^2}{x^2+xy+y^2}=\sum_{cyc}\frac{x^2(x+z)^2}{(x^2+xy+y^2)(x+z)^2}\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2)(x+z)^2}.$$
Id est, it's enough to prove that
$$\left(\sum\limits_{cyc}(x^2+xy)\right)^2\geq\sum\limits_{cyc}(x^2+xy+y^2)(x+z)^2$$ or
$$\sum_{cyc}(x^3y-x^2yz)\geq0$$ or
$$\sum_{cyc}(x^3y-2x^2yz+z^2xy)\geq0$$ or
$$\sum_{cyc}xy(x-z)^2\geq0.$$
Done!
|
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|
Definite integral of a rational fraction Can I find the value of $$\int_3^{\infty}\frac{x-1}{(x^2-2x-3)^2}dx$$ by just factoring the fraction?
I tried to wrote:
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{x-1}{(x^2-2x+1-4)^2}=\frac{x-1}{[(x-1)^2-2^2]^2}=\frac{x-1}{(x+1)^2\cdot(x-3)^2}$$ but didn't work out. Any ideas?
|
Use partial fraction decomposition, that is, find $A,B,C,D\in\mathbb{R}$ such that$$\frac{x-1}{(x+1)^2(x-3)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac C{x-3}+\frac D{(x-3)^2}.$$
|
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|
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$
Any point on the ellipse is given by $(a\cos \theta,b\sin \theta)$ where $a=\sqrt 3,b=\frac{\sqrt 3}{\sqrt 2}$.
Now slope of the tangent to the curve at $(a\cos \theta,b\sin \theta)$ is $\dfrac{-a\cos \theta}{2b\sin \theta}$.
Hence we have $\dfrac{b\sin \theta- 2}{a\cos \theta -1}=\dfrac{-a\cos \theta}{2b\sin \theta}$.
On simplifying we get $4b\sin \theta +a\cos\theta =3$
If we can find the value of $\theta $ from above then we can find the two points on the ellipse where the tangents touch them but I am unable to solve them.
Please help to solve it.
Any hints will be helpful
|
If $y=mx+n$ is a tangent then $$n^2=a^2m^2+b^2$$ or
$$n^2=3m^2+\frac{3}{2}.$$
Also, we have $2=m+n$ and we got the following equation on slopes:
$$(2-m)^2=3m^2+\frac{3}{2}.$$
After this use $$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$
I got $$\alpha=\arctan12.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2602561",
"timestamp": "2023-03-29T00:00:00",
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|
Proving a small inequality I am given that $a^3+b^3+c^3=3$ where a, b, c are positive numbers and I need to prove that
$$\frac {3(ab+bc+ac)+a^3c^2+b^3a^2+c^3b^2}{(a+b)(b+c)(a+c)}\ge \frac {3}{2}$$
At first it seems to me that the inequality might be wrong. I have tried using the Cauchy Schwarz , AM GM and some algebraic manipulations to reach the inequality but none of them helped me out.
Can somebody around here help me to prove this small inequality. Thanks in advance.
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We need to prove that
$$\sum_{cyc}\frac{3ab+a^3c^2}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{(a^3+b^3+c^3)ab+a^3c^2}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^4b+a^4c+a^3bc+a^3c^2}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^3(ab+ac+bc+c^2)}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\frac{a^3(b+c)(c+a)}{\prod\limits_{cyc}(a+b)}\geq\frac{3}{2}$$or
$$\sum_{cyc}\frac{a^3}{a+b}\geq\frac{3}{2}$$ and see here:
If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2604465",
"timestamp": "2023-03-29T00:00:00",
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Remainder of $22!$ upon division with $23$? I couldn't solve the problem, but I came to know the answer is $22$.
Then I tried to check the numbers in factorial will be cancelled by their modulo inverses w.r.t $23$. But they didn't.
\begin{array}{|c|c|} \hline \text{Number} & \text{Modulo Inverse w.r.t 23} \\ \hline
2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 2 \\ 6 & 5 \\ 7 & 4 \\ 8 & 7 \\ 9 & 2 \\ 10 & 7 \\ 11 & 1 \\ 12 & 11 \\ 13 & 4 \\ 14 & 11 \\ 15 & 2 \\ 16 & 7 \\ 17 & 3 \\ 18 & 11 \\ 19 & 5 \\ 20 & 7 \\ 21 & 11 \\ 22 & 1 \\ \hline
\end{array}
How to solve this problem?
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Your (multiplicative) inverses should be $$2\times 12=3\times 8=4\times 6=24\equiv 1 \bmod 23$$
$$5\times 14=7\times 10=70\equiv 1\bmod 23$$ $$9\times 18 \equiv 1 \bmod 23 $$
$$11\times21=231\equiv 1\bmod 23$$$$13\times 16=208\equiv 1 \bmod 23$$$$15\times 20=299\equiv 1 \bmod 23$$$$17\times 19=323\equiv 1 \bmod 23$$
You are left with $0$ which isn't part of the product, $1$ and $22\equiv -1$.
So you get a product of $1$s with just one $-1$ to give $-1\bmod 23$
Note: I used some tricks to make this easier eg $4\times 6 =-4\times -6 \equiv 19\times 17$
The fact that $23$ is prime guarantees that the numbers other than $0, \pm 1$ will pair up in this way, and this is one way of proving Wilson's Theorem.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2604626",
"timestamp": "2023-03-29T00:00:00",
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Why the approximate solution of $\frac{1}{{\sqrt {2\pi } }}\frac{n}{x}{e^{ - \frac{{{x^2}}}{2}}} = c$ is $\sqrt {2\log n} $ when $n$ is large I find in a book that when $n$ is large, the approximate solution of $\frac{1}{{\sqrt {2\pi } }}\frac{n}{x}{e^{ - \frac{{{x^2}}}{2}}} = c$, denoted by $x(n,c)$, is about
$x(n,c) \approx\sqrt {2\log n} $
And further the book states the solution of $\frac{1}{{\sqrt {2\pi } }}\frac{n}{x}{e^{ - \frac{{{x^2}}}{2}}} = 1$ is
$x(n,1) = \sqrt {2\log n} (1 - \frac{{\log \log n}}{{4\log n}} - \frac{{\log (2\sqrt \pi )}}{{2\log n}} + o(\frac{1}{{\sqrt {\log n} }}))$
Need help with why this is true. Following is the book contents related to this, and I highlight the part.
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The initial equation can be written
$$e^{-x^2/2-\log x}=\frac{\sqrt{2\pi}c}n,$$
and taking the cologarithm,
$$\frac{x^2}2+\log x=\log n-\log\sqrt{2\pi}c.$$
For very large $n$, this can be approximated by
$$\frac{x^2}2=\log n.$$
Below, a plot of $\dfrac{x^2}2+\log x$ vs. $\dfrac{x^2}2$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2606839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Spectral decomposition of some special matrix Let $A_{n\times n} = aI+bJ$, where $I$ is the identity matrix and $J$ is the matrix of all ones. Is it possible to find the expression of $A^{1/2}$ such that $A^{1/2}A^{1/2} = A$? In particular $A = I_{n\times n} - \frac{(1-\alpha)}{n+\alpha(2-n)}J_{n\times n}$, where $0<\alpha<1$.
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Here is a quick and easy way to get a matrix $A^{1/2}$, provided one is willing to limit oneself to solutions of the form
$C = \alpha I + \beta J; \tag 1$
then
$C^2 = \alpha^2 I + 2 \alpha \beta J + \beta^2 J^2 = \alpha^2 I + 2 \alpha \beta J + \beta^2 n J = \alpha^2 I + (2 \alpha \beta + n \beta^2)J, \tag 2$
where we have used the fact that $J^2 = nJ$, which is easily verified; with
$A = a I + bJ \tag 3$
and
$C^2 = A, \tag 4$
since $I$ and $J$ are linearly independent over $\Bbb C$, we must take
$\alpha^2 = a \tag 5$
and
$2 \alpha \beta + n \beta^2 = b; \tag 6$
then
$\alpha = \pm \sqrt a, \tag 7$
and thus (6) becomes a quadratic equation in $\beta$ whose coefficients are known:
$n\beta^2 + 2 \alpha \beta - b = 0; \tag 8$
enter the quadratic formula:
$\beta = \dfrac{-2\alpha \pm \sqrt{4 \alpha^2 + 4nb}}{2n}; \tag 9$
using (5), we simplify
$\beta = \dfrac{-2\alpha \pm \sqrt{4 a + 4nb}}{2n} = \dfrac{-2\alpha \pm 2\sqrt{a + nb}}{2n} = \dfrac{-\alpha \pm I\sqrt{a + nb}}{n}; \tag{10}$
with $\alpha$ and $\beta$ given by (5) and (10) we take
$A^{1/2} = C. \tag{11}$
Note: It is not yet clear to me whether or not all solutions are of this form. Further investigation required. End of Note.
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{
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"url": "https://math.stackexchange.com/questions/2608354",
"timestamp": "2023-03-29T00:00:00",
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