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Find all irreducible characters of a matrix group on finite field $\mathbb F_5$ Find all irreducible characters of matrix group $G =\left\{ \left( \begin{array}{cc}
a & b \\0 & a^{-1}\end{array} \right)|\,\,\, a,b \in\mathbb F_5, a\not=0 \right\}$.
The former question is to find irreducible characters of subgroup of $G$
$H =\left\{ \left( \begin{array}{cc}
a & 0 \\0 & a^{-1}\end{array} \right)|\, a\in\mathbb F_5^{\times} \right\}$ and
$N =\left\{ \left( \begin{array}{cc}
1 & b \\0 & 1 \end{array} \right)|\, b\in\mathbb F_5 \right\}$ and I can work it out.
But situation of $G$ is much complicated than I thought.
|
From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$. Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution. This combined with the character table of $C_4$ (below) gives us $\chi_1,\dots\chi_4$.
$$
\begin{array}{c|rrrr}
\rm class&\rm1&\rm2&\rm4A&\rm4B\cr
\rm size&1&1&1&1\cr
\hline
\rho_{1}&1&1&1&1\cr
\rho_{2}&1&1&-1&-1\cr
\rho_{3}&1&-1&-i&i\cr
\rho_{4}&1&-1&i&-i\cr
\end{array}
$$
All that remains now is to find a 2-dimensional representation. We can do this by calculating the quotient group $G/Z(G)$. It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go). Now, we can lift a two dimensional character from $D_5$ (below), say $\eta_3$ and then multiply it by the three linear characters we have already calculated. This will give us $\chi_4,\dots,\chi_8$.
$$
\begin{array}{c|rrrr}
\rm class&\rm1&\rm2&\rm5A&\rm5B\cr
\rm size&1&5&2&2\cr
\hline
\eta_{1}&1&1&1&1\cr
\eta_{2}&1&-1&1&1\cr
\eta_{3}&2&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}\cr
\eta_{4}&2&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}\cr
\end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly. The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
\begin{array}{c|rrrrrrrr}
\rm class&\rm I&\rm -I&\rm4A&\rm4B&\rm5A&\rm5B&\rm10A&\rm10B\cr
\rm size&1&1&5&5&2&2&2&2\cr
\hline
\chi_{1}&1&1&1&1&1&1&1&1\cr
\chi_{2}&1&1&-1&-1&1&1&1&1\cr
\chi_{3}&1&-1&-i&i&1&1&-1&-1\cr
\chi_{4}&1&-1&i&-i&1&1&-1&-1\cr
\chi_{5}&2&2&0&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}\cr
\chi_{6}&2&2&0&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}\cr
\chi_{7}&2&-2&0&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}\cr
\chi_{8}&2&-2&0&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}\cr
\end{array}
$$
|
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|
Solutions of $2^a+5^b=c^2$ I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
\begin{align*}
2^{2n}+5^b &= c^2 \\
(2^n)^2+5^b &= c^2 \\
5^b &= c^2-(2^n)^2 \\
5^b &= (c-2^n)(c+2^n)
\end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided by 5, $c-2^n \neq c+2^n$, so $c-2^n=1$ and $c+2^n=5^b$.
From this equations, I got $1+2^{n+1}=5^b$.
But I don't know how to continue.
|
Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} \equiv -1 \pmod{5^b}$$
implies $n+1$ is divisible by $\varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 \geq 2(5^{b-1})$.
Hence $$5^b \geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.
Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
|
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|
Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\rangle\\ \implies&\begin{cases} -x+6=\lambda(4x-24)\\-y+\frac{13}{2}=\lambda(-4y+26)\\x^2+(12-x)^2-y^2-(13-y)^2=0\end{cases}\end{align*}$
But clearly there is no solution due to the first two equations.
Using Wolfram Alpha, however, yields a maximum at $\displaystyle \left(\frac{17}{2},\,\frac{13}{2}\right)$ being $A=36$ and shows a nice little graph.
|
First, let's re-explain why the first two equations are contradictory. The first equation gives us:
$$\lambda=\frac{-x+6}{4x-24}=-\frac 1 4$$
While the second equation gives us:
$$\lambda=\frac{-y+\frac{13} 2}{-4y+26}=\frac 1 4$$
However, what's important to realize here is that these two equations only work when $4x-24\neq 0$ and when $-4y+26\neq 0$. Therefore, in order to find a solution, we need to consider the other cases: $4x-24=0$ (i.e. $x=6$) and $-4y+26=0$ (i.e. $y=\frac {13} 2$).
Case 1: $x=6$
Let's plug $x=6$ into our constraint equation:
$$6^2+(12-6)^2-y^2-(13-y)^2=0\rightarrow y=\frac{13}{2}\pm \frac{5i}{2}$$
Thus, this equation has no real solutions, and this case can be ignored.
Case 1: $y=\frac {13} 2$
Let's plug $y=\frac {13} 2$ into our constraint equation:
$$x^2+(12-x)^2-\left(\frac{13}2\right)^2-\left(13-\left(\frac{13}2\right)\right)^2=0\rightarrow x=\frac 7 2 \text{ or } x=\frac{17}2$$
Thus, we have two critical points: $(\frac 7 2, \frac{13}2)$ and $(\frac{17}2, \frac{13}2)$. I will leave it to you to show these critical points are maximums.
|
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|
$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$ Reduction Formula I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$\int\frac{x^m}{(ax^2+bx+c)^n}dx=-\frac{x^{m-1}}{a(2n-m-1)(ax^2+bx+c)^{n-1}}-\frac{b(n-m)}{a(2n-m-1)}I_{m-1,n}+\frac{c(m-1)}{a(2n-m-1)}I_{m-2,n}$$
My attempt went as follows:
$$\int\frac{x^m}{(ax^2+bx+c)^n}dx\space\begin{vmatrix}u=\frac{1}{(ax^2+bx+c)^n}\\du=\frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dx\end{vmatrix}\space dv=x^mdx\quad v=\frac{1}{m+1}x^{m+1}\\\int\frac{x^m}{(ax^2+bx+c)^n}dx=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{n}{m+1}\bigg)\int\frac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\\=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{n}{m+1}\bigg)\bigg(2a\int\frac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\\+b\int\frac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\bigg)\\=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{2an}{m+1}\bigg)\int\frac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\\+\bigg(\frac{bn}{m+1}\bigg)\int\frac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\\=\frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+\bigg(\frac{2an}{m+1}\bigg)I_{m+2,n+1}+\bigg(\frac{bn}{m+1}\bigg)I_{m+1,n+1}$$
This is where I'm stuck, any help?
|
Let $f(x)=\frac{x^{m-1}}{(ax^2+bx+c)^{n-1}}$ and differentiate
$$f’(x)
=\frac{a(m-2n-1)x^m+b(m-n)x^{m-1}+c(m-1)x^{m-2}}{(ax^2+bx+c)^{n}}
$$
Integrate both sides
$$f(x) = a(m-2n-1)I_{m,n} +b(m-n) I_{m-1,n} +c(m-1) I_{m-2,n}
$$
and then rearrange to obtain the reduction formula.
|
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|
Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$ Let's look at the following sequence:
$a_n=\left\{1,2,3,1,2,3,1,2,3,1,2,3,...\right\}$
I'm trying to calculate:
$$\sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor$$
The problem is, I'm looking for a closed form for this summation:
$$\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$
Is it possible?
|
If you consider$$b_k=\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}\qquad \text{with}\qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with
$$b_k=2k-\frac{2}{3} \left(1-\cos \left(\frac{2 \pi k}{3}\right)\right)$$
Edit
If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster
$$\frac{1}{2} \left(k^2+k-3 \left\lfloor \frac{k}{3}\right\rfloor ^2-3 \left\lfloor
\frac{k-2}{3}\right\rfloor \left(\left\lfloor \frac{k-2}{3}\right\rfloor
+1\right)-3 \left\lfloor \frac{k-1}{3}\right\rfloor \left(\left\lfloor
\frac{k-1}{3}\right\rfloor +1\right)+3 \left\lfloor \frac{k}{3}\right\rfloor
\right)$$
Which one do you prefer ?
|
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|
show this inequality with $\sum_{i=1}^{n}a_{i}=n$
Let $n\ge 3$ be postive number,$a_{i}>0,i=1,2,\cdots,n$,and $\displaystyle\sum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+\cdots+a^3_{n}a_{1}+n\ge 2(a_{1}a_{2}\cdots a_{n-1}+a_{2}a_{3}\cdots a_{n}+a_{n}a_{1}\cdots a_{n-2})$$
it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3\ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem use three shcur inequaliy
$$a^3+b^3+c^3+3abc\ge \sum ab(a+b)$$
then we have
$$a^2+b^2+c^2+3(abc)^{2/3}\ge 2(ab+bc+ca)$$
|
A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 \ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2)\\
+16a\left(\left(\sum_{cyc} 5u^3-u^2w\right) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w \right)\\
+\left(\left(\sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vw\right) + 68 u^3 v + 4 u w^3 + 68 v^3 w \right)\ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2 \ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 \ge 0,$$
$$\left(\sum_{cyc} 5u^3-u^2w\right) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w\\\ge \left(\sum_{cyc} 4u^3\right) + 4 u^2 v - u w^2 -7vw^2 + 11 v^2 w \\\ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)\ge 0$$
and
$$\left(\sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vw\right) + 68 u^3 v + 4 u w^3 + 68 v^3 w\\\ge\left(\sum_{cyc} 4u^3w-13u^2vw\right) + 68 u^3 v + 4 u w^3 + 68 v^3w\\\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)\ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4\ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 - u v + v^2)\\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)\ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
|
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Rolling $4$ dice and multiplying the results. What is the probability that the product is divisible by $5$ or has $5$ as the least significant digit? Four fair dice are rolled and the four numbers shown are multiplied together. What is the probability that this product
(a) Is divisible by $5$?
(b) Has last digit $5$?
MY ATTEMPT
a) A number is divisible by $5$ if and only if it has at least one factor equal to $5$. Let us denote by $F$ the event "it occurs at least one number five". Hence the sought probability is given by
\begin{align*}
\textbf{P}(F) = 1 - \textbf{P}(F^{c}) = 1 - \frac{5^{4}}{6^{4}} = 1 - \left(\frac{5}{6}\right)^{4}
\end{align*}
b) Here is the problem. I am not able to describe properly the target results.
Am I on the right track? Can someone please help me to solve it? Any help is appreciated.
|
Your answer for part $(a)$ is correct. For part $(b)$, observe that a number $N$ has last digit $5$ iff it's divisible by $5$ and odd. This means that if we denote our rolls by $a,b,c,d$ then $N=abcd$ where at least one of $a,b,c,d$ is a $5$ and none of $a,b,c,d$ is even. This means we have to avoid $2,4,6$ on all rolls, as well as obtain at least one $5$.
*
*Case 1: Exactly one $5$. Choose one of $4$ positions for the $5$ and fill in the remaining three spots with either $1$ or $3$. This can be done in $\binom{4}{1}\cdot 2^{3}$ ways.
*Case 2: Exactly two $5$s. Choose two of $4$ positions for the $5$s and fill in the remaining two spots with either $1$ or $3$. This can be done in $\binom{4}{2}\cdot 2^{2}$ ways.
*Case 3: Exactly three $5$s. Choose three of $4$ positions for the $5$s and fill in the remaining spot with either $1$ or $3$. This can be done in $\binom{4}{3}\cdot 2^{1}$ ways.
*Case 4: All $5$s. There is only one way to do this.
The number of desirable outcomes is therefore
$$\binom{4}{1}\cdot 2^{3}+\binom{4}{2}\cdot 2^{2}+\binom{4}{3}\cdot 2+ 1$$
and since there are $6^{4}$ total outcomes, the desired probability is
$$\frac{\binom{4}{1}\cdot 2^{3}+\binom{4}{2}\cdot 2^{2}+\binom{4}{3}\cdot 2+ 1}{6^{4}}$$
|
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Solving for $x$ in $y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$ I want to confirm my solution of $x$ from
$$y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$$
My answer is:
$$x=\frac{N^{2.6}}{10\cdot y^{2.6}}$$
Is this right? How would you solve?
|
Let $$y = N \cdot (\frac{10}{x})^{-2.6}$$
Then,
$$y \cdot (\frac{10}{x})^{2.6} = N$$
$$ y \cdot 10 ^{2.6} = N \cdot x^{2.6}$$
$$ (y \cdot 10^{2.6})^{1/2.6} = (N \cdot x^{2.6})^{1/2.6}$$
$$ y^{1/2.6} \cdot 10 = N^{1/2.6} \cdot x$$
$$ x = y^{1/2.6} \cdot 10 \cdot N^{-1/2.6} = (\frac{y}{N})^{1/2.6} \cdot 10 $$
|
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Suppose $a, b\in G$ such that $\lvert a\rvert$ is odd and $aba^{-1}=b^{-1}$. Show that $b^2=e$. I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $\lvert a\rvert=2n+1$, $n\in\Bbb N\cup\{0\}$.
We have $$\begin{align}
b&=(b^{-1})^{-1} \\
&=(aba^{-1})^{-1} \\
&=ab^{-1}a^{-1},
\end{align}$$
so, by induction, $b=a(\underbrace{aba^{-1}}_{=b^{-1}})a^{-1}=\dots =a^{2n}ba^{-2n}$, so that
$$\begin{align}
b&=a^{2n}ba^{-2n} \\
&=a^{2n}(ab^{-1}a^{-1})a^{-2n} \\
&=a^{2n+1}b^{-1}a^{-(2n+1)} \\
&=b^{-1},
\end{align}$$
so $b=b^{-1}$, i.e., $b^2=e$.$\square$
Is the above proof okay? I think it is.
However, I'd be interested to see how properties of Baumslag-Solitar groups could be used here, since $aba^{-1}=b^{-1}$ brings them to mind.
|
We have
$$\begin{align}
aba^{-1}=b^{-1}&\implies ab=b^{-1}a=x\\
&\implies (ab) \cdot (b^{-1}a)=x^2 \\
&\implies a^2=x^2.
\end{align}$$
Now:
$$\begin{align} x^{2m+1}&= (x^2)^m \cdot x\\ &= (a^2)^m \cdot (ab)\\ &= a^{2m+1}\cdot b\\ &=b.\end{align}$$
and
$$\begin{align} x^{2m+1}=b &\implies (x^{2m+1})^2=b^2 \\ &\implies (x^2)^{2m+1}=b^2 \\ &\implies (a^2)^{2m+1}=b^2 = (a^{2m+1})^2\\ & \implies e=b^2. \end{align}$$
|
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|
How to apply CRT to a congruence system with moduli not coprime? $x=1 \pmod 8$
$x=5 \pmod{12}$
8 and 12 are not coprime, I could break it to:
$x=1 \pmod 2$
$x=1 \pmod 4$
and
$x=5 \pmod 3$
$x=5 \pmod 4$
But what are the next steps to solve it? By the way, $x$ should be $17$ not sure how to get that number ...
Thanks in advance.
|
No point in figuring it out to any lesser power of $2$ than $2^3$.
Leave it at $x\equiv 1 \pmod 8$ but $x \equiv 5 \pmod {12}$ can be reduced to $x\equiv 5 \equiv 2 \pmod 3$.
So CRT says there is a unique solution $\pmod {28}$; $x \equiv 17 \pmod{24}$.
$x = A \pmod {n_1}$ $(A = 1; n_1 = 8)$
$x = B \pmod {n_2}$ $(B = 2; n_2 = 3)$
$m_1n_1 + m_2n_2 = 1$ (in this case $8m_1 + 3m_2 = 1$ so $m_1 =2; B=-5$ or $A=-1; B=3$ or .....)
Then $x = Am_2n_2 + Bn_1n_1\pmod {n_1n_2}$.
$x = 1*(-5)*3 + 2*2*8= -15+32 = 17$.
There's utterly no point in reducing to $x \equiv 5 \pmod 4$ and $x \equiv 5\pmod 3$ as that will just get you back to $x\equiv 2,5,8,11 \pmod {12}$ and $x \equiv 1,5 \pmod 8$ which is worse than what you started with.
|
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Proof for area of an equilateral triangle with respect to one side? I'm trying to find out the area of an equilateral triangle with respect to one side. Anything wrong with my proof?
An equilateral triangle with sides of length $a$ can be divided in half along the base to create two triangles with right angles. The sides would be $h$ (height), $y$ (hypotenuse) and $b$ (base). I know what some of these sides are:
$$y = a \\
b = 0.5a \\
h = ?$$
Using Pythagorean theorem, I can find the length of $h$:
$$\begin{align}c^2 &= a^2 + b^2 \\
h^2 &= b^2 + y^2 \\
&= 0.5a^2 + a^2 \\
h &= \pm\frac{a\times\sqrt{3}}{2}
\end{align}$$
Since the length can only be positive, we're left with $h =\frac{a\times\sqrt{3}}{2}$.
The area $A$ of a triangle is $A = (b \times h) / 2$. We can plug in our numbers:
$$\begin{align}
A &= \frac{b \times h}{2} \\
&= \frac{\frac{a}{2} \times \frac{a\times\sqrt{3}}{2}}{2} \\
&= \frac{a^2\times\sqrt{3}}{8}
\end{align}$$
But since we halved the equilateral triangle together, the area of the equilateral is twice this number, since we recombine the two triangles:
$$\begin{align}
A &= 2\times\frac{a^2\times\sqrt{3}}{8} \\
&= \frac{a^2\times\sqrt{3}}{4}
\end{align}$$
This is indeed the area of an equilateral triangle with respect to side $a$.
|
Your proof is fine, but in the spirit of visual geometry, let me present a cute alternative: (You might want to break out paper/pencil to follow along with the diagram, but maybe not since the figure is so simple.)
Draw a regular hexagon made from 6 equilateral triangles, then draw the equilateral triangle formed by taking every other vertex of the hexagon. Now, since all equilateral triangles are similar, the area of the large equilateral triangle is $ k^{2}A $, where $ A $ is the area of the smaller equilateral triangles, for some scale factor $k$. On the other hand, the large equilateral triangle is half the area of the hexagon. So, since the hexagon has area $ 6A $, we have that $ k^{2} = 3 $. Now, we can compute the area of the hexagon as 3 times the area of the rhombuses formed by adjacent pairs of smaller equilateral triangles. The area of each rhombus is $ \frac{ka \cdot a}{2} $, if the side length of each small equilateral triangle is $ a $. Thus, we have $ \frac{1}{2}ka^{2} = 6 A $ and $ k = \sqrt{3} $, so $ A = \frac{\sqrt{3}}{4}a^{2} $.
|
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"url": "https://math.stackexchange.com/questions/3082045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integration by substitution to take out square root
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$
In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}{2t^2}dt $ $(3)$
and also $x+1=-\frac{t^2-2t-1}{2t}$ $(4)$
Then
$(1),(3),(4)\Rightarrow$$\int (x+1)\sqrt{x^2+1}dx=$ $\int\frac{t^2-2t-1}{2t} \frac{1-t^2}{2t} \frac{t^2+1}{2t}dt=$
$\frac{1}{8} \int \frac{-t^6+2t^5+t^4+t^2-2t-1}{t^3}dt=$
$\frac{1}{8} [-\int t^3 dt+2\int t^2 dt+\int t dt +\int \frac{1}{t} dt-2 \int \frac{1}{t^2} dt- \int \frac{1}{t^3}dt]=$
$\frac{1}{8}[-\frac{t^4}{4}+2\frac{t^3}{3}+\frac{t^2}{2}+lnt+\frac{2}{t}-\frac{1}{2t^2}]$
We do the substitution from $(2)$ and we get the result
You think that's correct?
|
Before trying the substitution, it is worth to simplify the integrand, by observing that
$$(x^2+1)'=2x$$ so that
$$\int(x+1)\sqrt{x^2+1}\,dx=\frac13(x^2+1)^{3/2}+\int\sqrt{x^2+1}\,dx.$$
For the last integral, substitution with $x=\sinh t$ will allow to get rid of the square root. By parts intégration also works:
$$I:=\int\sqrt{x^2+1}\,dx=x\sqrt{x^2+1}-\int\frac{x^2}{\sqrt{x^2+1}}dx$$
and
$$\int\frac{x^2}{\sqrt{x^2+1}}dx=\int\frac{x^2+1-1}{\sqrt{x^2+1}}dx=I-\text{arshinh } x,$$ from which you draw $I$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3084435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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|
Intuitively, why should I expect a circle in the complex plane from the equation $\left|\frac{z-1}{z+1}\right| = c$? I know how to prove that : ($c \in [0,1[$)
$$C = \{z \in \mathbb{C}: \left|\frac{z-1}{z+1}\right| = c \}$$
is circle in the complex plane. To do so we can for example write $z = x+iy$ and use the brute force approach.
Also, it's worth mentioning that it's intuitive for me that
$$\{z \in \mathbb{C} : \left| z - z_0 \right| = c \}$$
represents a circle.
But I don't see at all why intuitively $C$ is a circle. So is it possible to understand geometrically why $C$ is a circle in the complex plane ?
Thank you very much !
|
Let's explore.
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
\frac{\lvert z - 1 \rvert}{\lvert z + 1 \rvert} = c
} \quad \iff \quad \bbox{
\left\lvert z - 1 \right\rvert = c \left\lvert z + 1 \right\rvert
} \tag{1}\label{NA1}$$
This only makes sense if $0 \lt c \in \mathbb{R}$.
Since $z \in \mathbb{C}$, we can write $z = x + i y$. Since $\lvert z \rvert = \sqrt{x^2 + y^2}$, we have
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
\sqrt{(x-1)^2 + y^2} = c \sqrt{(x+1)^2 + y^2}
} \tag{2}\label{NA2}$$
Since $x, y, z \in \mathbb{R}$ and $c \gt 0$, both sides are positive, and we can square both sides:
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(x-1)^2 + y^2 = c^2 (x+1)^2 + c^2 y^2
} \tag{3}\label{NA3}$$
Expanding and moving all terms to one side, we get
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
x^2 - 2 x + 1 + y^2 - c^2 x^2 - c^2 y^2 - 2 c^2 x - c^2 = 0
} \tag{4}\label{NA4}$$
The case when $c = 1$ is special, because then $\eqref{NA3}$ simplifies to $x = 0$, which is not a circle but a line (unless you say it is an infinite-radius circle centered at real $\pm\infty$). In any case, let's continue the exploration with $0 \lt c \in \mathbb{R}$, $c \ne 1$.
We can collect the terms in $\eqref{NA4}$, getting
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(1 - c^2)\left( (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 \right) = 0
} \tag{5}\label{NA5}$$
Because we already decided $c \ne 1$, this is equivalent to
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 = 0 , \quad c \ne 1
} \tag{6}\label{NA6}$$
This is getting interesting. Compare to the equation of a circle of radius $r$ centered at $x = x_0$,
$$\bbox{ (x - x_0)^2 + y^2 - r^2 = 0 }$$
Now, if we choose
$$\bbox{ x_0 = \frac{2}{1 - c^2} - 1} , \quad \bbox{ r = \sqrt{\left(\frac{1 + c^2}{1 - c^2} \right)^2 - 1} }$$
we find that
$$\bbox{ (x - x_0)^2 + y^2 - r^2 = (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 }$$
Therefore,
$$\bbox[#ffffef]{ \bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
(1 - c^2)\left( (x - x_0)^2 + y^2 - r^2 \right) = 0
} , \quad \bbox{
z = x + i y
} \tag{7a}\label{NA7a} }$$
where
$$\bbox[#ffffef]{ \bbox{
x_0 = \frac{2}{1 - c^2} - 1
} , \quad \bbox{
r = \sqrt{\left(\frac{1 + c^2}{1 - c^2} \right)^2 - 1}
} , \quad \bbox{
c \gt 0
} , \quad \bbox{
c \ne 1
} , \quad \bbox{
c \in \mathbb{R}
} \tag{7b}\label{NA7b} }$$
and describes a circle of radius $r$ centered at $z = x_0$ on the real axis when $c \gt 0$, $c \ne 1$, and a line along the imaginary axis when $c = 1$. No intuition or geometry needed, basic algebra suffices.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3084764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How to evaluate: $\lim\limits_{n\to\infty} \frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$ How to evaluate: $$\lim_{n\to\infty} \dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when
$i)$ $p\in\mathbb R,p\neq0$
$ii)\space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But I'm unsure about it, but for $ii)$ and I don't have a clue where to begin with.
|
We can apply the Stolz-Cesàro theorem for positive real values $p, p\ne 1$
and consider other values of $p$ separately.
We consider for $p\in\mathbb{R}$:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}\tag{1}
\end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{n\geq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
\begin{align*}
a_n&=1^{p-1}+2^{p-1}+\cdots+n^{p-1}\\
b_n&=n^p=\underbrace{n^{p-1}+n^{p-1}+\cdots+n^{p-1}}_{n\ \mathrm{ times}}
\end{align*}
Since
\begin{align*}
\lim_{n\to \infty}\frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=\lim_{n\to\infty}\frac{n^{p-1}}{n^p-(n-1)^{p}}\\
&=\lim_{n\to\infty}\frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+\binom{p}{2}n^{p-2}-\cdots)}\\
&=\lim_{n\to\infty}\frac{n^{p-1}}{pn^{p-1}-\binom{p}{2}n^{p-2}+\cdots}\\
&=\frac{1}{p}
\end{align*}
we have according to the theorem
\begin{align*}
\lim_{n\to \infty}\frac{a_n}{b_n}&=\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}\color{blue}{=\frac{1}{p}}
\end{align*}
Case $p=1$:
We obtain
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}
=\lim_{n\to\infty}\frac{\overbrace{1+1+\cdots+1}^{n\mathrm{\ times}}}{n}=\lim_{n\to\infty}\frac{n}{n}
\color{blue}{=1}
\end{align*}
Case $p=0$:
We obtain
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}
=\lim_{n\to\infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=\sum_{n=1}^\infty \frac{1}{n}\color{blue}{=\infty}
\end{align*}
since the harmonic series is divergent.
Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}
=\lim_{n\to \infty}n^q\left(1+\frac{1}{2^{q+1}}+\cdots+\frac{1}{n^{q+1}}\right)\geq \lim_{n\to\infty} n^q
\color{blue}{=\infty}
\end{align*}
We summarize:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}=
\begin{cases}
\frac{1}{p}&p>0\\
\infty&p\leq 0
\end{cases}
\end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3086947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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|
Finding the closed form of a generating function Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n \choose k}\;\, \forall \;n$, show that: $$f(x)=\frac{x^k}{(1-x)^{k+1}}$$
I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 \choose k}x^{k+1}+{k+2 \choose k}x^{k+2}+...$$$$\;\;\;\;\;\;\;\;\;=x^k\left(1+{k+1 \choose k}x^1+{k+2 \choose k}x^2+...\right)$$$$=x^k\cdot h(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; $$
Where $h(x)$ is defined as the expression in parenthesis, then I tried some manipulation, for example I computed $h(x)-xh(x)$, since: ${k+n+1 \choose k}-{k+n \choose k}={k+n \choose k-1}$ we have:
$h(x)-xh(x)=1+{k \choose k-1}x+{k+1 \choose k-1}x^2+...$
But I'm stuck, not sure how I should procede
|
HINT:
Inductive hypothesis
\begin{eqnarray*}
f_k(x)= \sum_{n=k}^{\infty} \binom{n}{k} x^n = \frac{x^k}{(1-x)^{k+1}}
\end{eqnarray*}
and use
\begin{eqnarray*}
\binom{n}{k} + \binom{n}{k+1}= \binom{n+1}{k+1}.
\end{eqnarray*}
More detail available on request.
Further Hint :
\begin{eqnarray*}
f_{k+1}(x) &=& \sum_{n=k+1}^{\infty} \binom{n}{k+1} x^n = \sum_{n=k}^{\infty} \binom{n+1}{k+1} x^{n+1} \\
&=& \sum_{n=k}^{\infty} \binom{n}{k} x^{n+1} + \sum_{n=k}^{\infty} \binom{n}{k+1} x^{n+1} \\ \cdots
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$ $$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
|
Alternatively:
$$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=\\
\lim_{x \to \frac{\pi}{2}} \left[1+\tan \left(\frac{\pi}{4}\sin x\right)-1\right]^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}\sin x\right)-1}{\tan(\pi \sin x)} \right]=\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}\sin x\right)-\cos \left(\frac{\pi}{4}\sin x\right)}{\sin(\pi \sin x)}\cdot \frac{\cos (\pi \sin x)}{\cos \left(\frac{\pi}{4}\sin x\right)}\right] =\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}\sin x\right)}{\sin(\pi \sin x)}\right] =\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{1}{2\sin(\frac{\pi}{2} \sin x)}\right] =\exp \left[\frac12\right]=\sqrt{e}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3087460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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|
How to prove binomial coefficient $ {2^n \choose k} $ is even number? Prove:
${2^n \choose k}$ (for integers $k$ & $n$ : $0<k<2^n$) is even number.
I have tried induction but was unable to get any useful results.
|
The following solution (copypasted from my old coursework) is purely
elementary number theory:
We set $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $.
Lemma 1. Let $n$ be an integer. Let $m$ be a positive integer. Then,
$\dbinom{n}{m}=\dfrac{n}{m}\cdot\dbinom{n-1}{m-1}$.
Proof of Lemma 1. We have
\begin{align*}
\dbinom{n}{m} & =\dfrac{n\cdot\left( n-1\right) \cdot\cdots\cdot\left(
n-m+1\right) }{m!}\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of
}\dbinom{n}{m}\right) \\
& =\dfrac{n\cdot\left( \left( n-1\right) \cdot\left( n-2\right)
\cdot\cdots\cdot\left( n-m+1\right) \right) }{m\cdot\left( m-1\right)
!}\\
& \qquad\qquad\left(
\begin{array}
[c]{c}
\text{since}\\
n\cdot\left( n-1\right) \cdot\cdots\cdot\left( n-m+1\right) =n\cdot\left(
\left( n-1\right) \cdot\left( n-2\right) \cdot\cdots\cdot\left(
n-m+1\right) \right) \\
\text{and }m!=m\cdot\left( m-1\right) !
\end{array}
\right) \\
& =\dfrac{n}{m}\cdot\dfrac{\left( n-1\right) \cdot\left( n-2\right)
\cdot\cdots\cdot\left( n-m+1\right) }{\left( m-1\right) !}\\
& =\dfrac{n}{m}\cdot\underbrace{\dfrac{\left( n-1\right) \cdot\left(
n-2\right) \cdot\cdots\cdot\left( \left( n-1\right) -\left( m-1\right)
+1\right) }{\left( m-1\right) !}}_{\substack{=\dbinom{n-1}{m-1}
\\\text{(since this is how }\dbinom{n-1}{m-1}\text{ is defined)}}}\\
& \qquad\qquad\left( \text{since }n-m=\left( n-1\right) -\left(
m-1\right) \right) \\
& =\dfrac{n}{m}\cdot\dbinom{n-1}{m-1}.
\end{align*}
Thus, Lemma 1 is proven. $\blacksquare$
Lemma 2. Let $x$, $y$ and $z$ be three integers such that $x\mid yz$ and
$\gcd\left( x,y\right) =1$. Then, $x\mid z$.
Lemma 2 is a classical result in elementary number theory (see, e.g.,
Proposition 1.2.8 in my 18.781 (Spring 2016): Floor and arithmetic
functions).
$\blacksquare$
Lemma 3. Let $p$ be a prime number. Then, every positive divisor of
$p^{\alpha}$ is a power of $p$.
Proof of Lemma 3. Let $d$ be a positive divisor of $p^{\alpha}$. We must
show that $d$ is a power of $p$.
Assume the contrary. Thus, the prime factorization of $d$ must contain at
least one prime $q$ distinct from $p$. Consider this $q$. Now, $q\mid d$
(since the prime factorization of $d$ contains $q$). Hence, $q\mid d\mid
p^{\alpha}$ (since $d$ is a divisor of $p^{\alpha}$). Thus, the prime
factorization of $p^{\alpha}$ contains the prime $q$ (since $q$ is a prime).
Since this prime factorization is clearly $\underbrace{pp\cdots p}
_{\alpha\text{ times}}$, we thus conclude that the prime factorization
$\underbrace{pp\cdots p}_{\alpha\text{ times}}$ contains $q$. Hence, $q=p$.
This contradicts the fact that $q$ is distinct from $p$. This contradiction
proves that our assumption was wrong; hence, Lemma 3 is proven. $\blacksquare$
Lemma 4. Let $p$ be a prime number. Let $\alpha\in\mathbb{N}$. Let $u$ be
an integer such that $u$ is not divisible by $p$. Then, $\gcd\left(
u,p^{\alpha}\right) =1$.
Proof of Lemma 4. The integer $\gcd\left( u,p^{\alpha}\right) $ is a
positive divisor of $p^{\alpha}$, and therefore a power of $p$ (by Lemma 3).
In other words, $\gcd\left( u,p^{\alpha}\right) =p^{\beta}$ for some
$\beta\in\mathbb{N}$. Consider this $\beta$. If $\beta>0$, then $p\mid
p^{\beta}=\gcd\left( u,p^{\alpha}\right) \mid u$, which contradicts the
assumption that $u$ is not divisible by $p$. Hence, we cannot have $\beta>0$,
and thus we must have $\beta=0$. Hence, $p^{\beta}=p^{0}=1$ and $\gcd\left(
u,p^{\alpha}\right) =p^{\beta}=1$. This proves Lemma 4. $\blacksquare$
Theorem 5. Let $p$ be a prime number. Let $\alpha\in\mathbb{N}$ and let
$k$ be an integer such that $0<k<p^{\alpha}$. Then, $\dbinom{p^{\alpha}}{k}$
is divisible by $p$.
Your claim follows from Theorem 5 (applied to $p=2$ and $\alpha=n$), since your $k$ satisfies $0 < k < n \leq 2^n$.
Proof of Theorem 5. Assume the contrary. Thus, $\dbinom{p^{\alpha}}{k}$ is
not divisible by $p$. Hence, Lemma 4 (applied to $u=\dbinom{p^{\alpha}}{k}$)
that $\gcd\left( \dbinom{p^{\alpha}}{k},p^{\alpha}\right) =1$.
Applying Lemma 1 to $n=p^{\alpha}$ and $m=k$, we obtain $\dbinom{p^{\alpha}
}{k}=\dfrac{p^{\alpha}}{k}\cdot\dbinom{p^{\alpha}-1}{k-1}$, so that
$k\dbinom{p^{\alpha}}{k}=p^{\alpha}\dbinom{p^{\alpha}-1}{k-1}$. Thus,
$p^{\alpha}\mid k\dbinom{p^{\alpha}}{k}=\dbinom{p^{\alpha}}{k}k$. Hence, Lemma
2 (applied to $x=p^{\alpha}$, $y=\dbinom{p^{\alpha}}{k}$ and $z=k$) yields
$p^{\alpha}\mid k$ (since $\gcd\left( p^{\alpha},\dbinom{p^{\alpha}}
{k}\right) =\gcd\left( \dbinom{p^{\alpha}}{k},p^{\alpha}\right) =1$). Since
$k$ is positive, this yields $k\geq p^{\alpha}$; but this contradicts
$k<p^{\alpha}$. This contradiction shows that our assumption was wrong. Thus,
Theorem 5 is proven. $\blacksquare$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
What is the dimension of $V$? Let $$A=\begin{pmatrix} 1 & 1& 1\\2 &2&3\\x&y&z\end{pmatrix}$$
and let $V=\{(x,y,z)\in \mathbb{R}^3: \det(A)=0\}$ Then the dimension of $V$ equals,
Efforts:
Applying row transformation $R_2\to R_2-2R_1$, we get $$A=\begin{pmatrix} 1 & 1& 1\\0 &0&1\\x&y&z\end{pmatrix}$$
Now expanding along the second row, I get $\det(A)=y-x$ so informally speaking there are two independent variables.
Hence dimension is 2
Am I right?
Edits: maybe I have received downvote because of poor mathematical language. I meant $x$ is equal to $y$, so we are free to choose $x$ and $z$. Then we can write the basis as well.
|
You should be more precise with the language: “informally speaking” is not enough.
You can go on with the row reduction:
$$
\begin{pmatrix} 1 & 1& 1\\2 &2&3\\x&y&z\end{pmatrix}
\xrightarrow{\begin{aligned}R_2&\gets R_2-2R_1 \\ R_3&\gets R_3-xR_1\end{aligned}}
\begin{pmatrix} 1 & 1& 1\\0 &0&1\\0&y-x&z-x\end{pmatrix}
\xrightarrow{R_3\gets R_3-(z-x)R_2}
\begin{pmatrix} 1 & 1& 1\\0 &0&1\\0&y-x&0\end{pmatrix}
$$
None of these operation changes the determinant. Switching rows 2 and 3 and changing sign to the second row to compensate, we get a triangular form
\begin{pmatrix} 1 & 1& 1\\0&x-y&0\\0 &0&1\end{pmatrix}
and the determinant is $x-y$, so the condition is $x=y$. The variables $y$ and $z$ are “free”. You get two linearly independent solutions by first setting $y=1$ and $z=0$ and then $y=0$ and $z=1$.
Therefore a basis for the subspace you're given is
$$
\left\{
\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\right\}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3088777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove that $\sum_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor (n-k)/2 \rfloor}=\binom{2n+1}{n}$ Where the thing that looks like a floor function is the floor function. This is an interesting result which I hoped to prove by induction, but ran into trouble applying the inductive hypothesis. The base case is trivial. Here's the progress I made:
Inductive hypothesis: $\sum_{k=0}^m 2^k \binom{m}{k} \binom{m-k}{\lfloor{(m-k)/2}\rfloor}=\binom{2m+1}{m}$ for some $m>0$. Then
\begin{align*}
&\sum_{k=0}^{m+1} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\
=&\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}+2^{m+1} \binom{m+1}{m+1} \binom{m+1-m+1}{\lfloor{(m+1-(m+1))/2}\rfloor}\\
=&2^{m+1}+\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}
\end{align*}
Let $A$ be the set of all $k\in[0,m]$ such that $m+1-k$ is even. Let $B$ be the set of $k\in[0,m]$ such that $m+1-k$ is odd. Observe that for each $k\in[0,m]$, $k$ is in exactly one of $A$ or $B$. It follows that
\begin{align*}
&\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\
=&\sum_{k\in A} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}+\sum_{k\in B} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\
=&\sum_{k\in A} 2^k \binom{m+1}{k} \binom{m+1-k}{(m+1-k)/2}+\sum_{k\in B} 2^k \binom{m+1}{k} \binom{m+1-k}{(m-k)/2}
\end{align*}
Splitting the sum into $A$ and $B$ was a last-ditch attempt to form the expression into something to which I could apply the inductive hypothesis. Did I make a mistake somewhere, is this the wrong approach, or am I just missing the next step?
|
Seeking to verify
$$\sum_{k=0}^n 2^k {n\choose k}
{n-k\choose \lfloor (n-k)/2 \rfloor} = {2n+1\choose n}$$
we observe that
$${n\choose k}
{n-k\choose \lfloor (n-k)/2 \rfloor}
= \frac{n!}{k! \times \lfloor (n-k)/2 \rfloor!
\times (n-k-\lfloor (n-k)/2 \rfloor)!}
\\ = {n\choose \lfloor (n-k)/2 \rfloor}
{n-\lfloor (n-k)/2 \rfloor \choose n-k-\lfloor (n-k)/2 \rfloor}.$$
We get
$$2^n \sum_{k=0}^n 2^{-k}
{n\choose \lfloor k/2 \rfloor}
{n- \lfloor k/2 \rfloor \choose k - \lfloor k/2 \rfloor}.$$
This yields two pieces:
$$2^n \sum_{q=0}^{\lfloor n/2 \rfloor} 2^{-2q}
{n\choose q} {n- q \choose q}$$
and
$$2^n \sum_{q=0}^{\lfloor (n-1)/2 \rfloor} 2^{-2q-1}
{n\choose q} {n- q \choose q+1}.$$
We write for the first one
$$2^n \sum_{q=0}^{\lfloor n/2 \rfloor} 2^{-2q}
{n\choose q} {n- q \choose n-2q}
\\ = 2^n \sum_{q=0}^{\lfloor n/2 \rfloor} 2^{-2q}
{n\choose q} [z^{n-2q}] (1+z)^{n-q}
\\ = 2^n [z^n] (1+z)^n \sum_{q=0}^{\lfloor n/2 \rfloor} 2^{-2q}
{n\choose q} z^{2q} (1+z)^{-q}.$$
Now the coefficient extractor enforces the upper limit of the sum
and we find
$$2^n [z^n] (1+z)^n \sum_{q\ge 0} 2^{-2q}
{n\choose q} z^{2q} (1+z)^{-q}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{1}{2^2}\frac{z^2}{1+z}\right)^n
= 2^n [z^n] \left(1 + z + \frac{1}{2^2} z^2\right)^n
\\ = 2^n [z^n] \left(1 + \frac{1}{2} z\right)^{2n}
= {2n\choose n}.$$
The second one is of course very similiar:
$$2^n \sum_{q=0}^{\lfloor (n-1)/2 \rfloor} 2^{-2q-1}
{n\choose q} {n- q \choose n-2q-1}
\\ = 2^n \sum_{q=0}^{\lfloor (n-1)/2 \rfloor} 2^{-2q-1}
{n\choose q} [z^{n-2q-1}] (1+z)^{n-q}
\\ = 2^n [z^{n-1}] (1+z)^n
\sum_{q=0}^{\lfloor (n-1)/2 \rfloor} 2^{-2q-1}
{n\choose q} z^{2q} (1+z)^{-q}.$$
Once more the coefficient extractor enforces the upper limit of the
sum and we find
$$2^{n-1} [z^{n-1}] (1+z)^n \sum_{q\ge 0} 2^{-2q}
{n\choose q} z^{2q} (1+z)^{-q}
\\ = 2^{n-1} [z^{n-1}] (1+z)^n
\left(1+\frac{1}{2^2}\frac{z^2}{1+z}\right)^n
= 2^{n-1} [z^{n-1}] \left(1 + z + \frac{1}{2^2} z^2\right)^n
\\ = 2^{n-1} [z^{n-1}] \left(1 + \frac{1}{2} z\right)^{2n}
= {2n\choose n-1}.$$
Adding the two binomial coefficients we indeed obtain
$$\bbox[5px,border:2px solid #00A000]{
{2n+1\choose n}.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3089150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Find solutions of the given equation in the form of power series Find solutions of the given equation in the form of power series
$$y'' +2xy' = 0$$
My approach:
$$2xy' = 2a_1x + 2\times2a_2x^2 + 2\times3a_3x^3 + ... = \sum 2ia_{i}x^{i}$$
$$y'' = 2a_2 + 3\times2a_3x + 4\times3a_4x^2 + ... = \sum (i+1)(i+2)a_{i+2}x^{i}$$
So to have $y'' +2xy' = 0$ we need to have every coefficient equal to $0$.
So
$x^0$ // $2a_2 + 0 = 0 => a_2=0$
$x^1$ // $3\times2a_3 + 2a_1 = 0 => a_3 = -a_1/3$
$x^2$ // $4\times3a_4 + 2\times2a_2=0 => a_4 = -4a_2/12 = 0$
...
So it was easy to observe that every even coefficient is equal to $0$ and every odd one can be written in terms of $a_0$
Afterwards I wanted to find a recursive formula for odd coefficients and this is a problem that I can't solve.
I tried doing it in that way:
$$2n(2n+1)a_{2n+1} + 2(2n-1)a_{2n-1} = 0$$
$$a_{2n+1} = -\frac{2(2n-1)a_{2n-1}}{2n(2n+1)}$$
Now probably it should be somehow easy to write it as a recursive formula in terms of $a_0$, but I can't really see it.
|
You already did most of the job.
We continue
\begin{align*}
\color{blue}{a_{2n+1}}&= -\frac{2(2n-1)}{(2n+1)2n}a_{2n-1}\\
&=+\frac{2(2n-1)2(2n-3)}{(2n+1)2n(2n-1)(2n-2)}a_{2n-3}\\
&=\cdots\\
&=(-1)^n\frac{2^n(2n-1)!!}{(2n+1)!}a_1\tag{1}\\
&=(-1)^n\frac{2^n}{(2n+1)(2n)!!}a_1\tag{2}\\
&\,\,\color{blue}{=\frac{(-1)^n}{(2n+1)n!}a_1}\tag{3}
\end{align*}
Comment:
*
*In (1) we use the double factorial $(2n-1)!!=(2n-1)(2n-3)\cdots3\cdot 1$.
*In (2) we use $(2n)!=(2n)!!(2n-1)!!$ and $(2n)!!=(2n)(2n-2)\cdots 4\cdot2$.
*In (3) we use $(2n)!!= 2^n n!$.
Since we already know that $a_{2n}=0, n\geq 0$, we get from (3) the generating function $A(x)$
\begin{align*}
A(x)=\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_{2n+1} x^{2n+1}=a_1\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)n!} x^{2n+1}
\end{align*}
and the general solution
\begin{align*}
A(x)+C\color{blue}{=a_1\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)n!} x^{2n+1}+C}\qquad\qquad a_1,C\in\mathbb{R}
\end{align*}
Note that $A(x)=\frac{a_1\sqrt{\pi}}{2}\mathrm{erf}(x)$ where $\mathrm{erf}$ is the error-function.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3089497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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|
What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?
This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=24$. Given that $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3$, $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})= 3(\sqrt {49-x^2} + \sqrt {25-x^2}) =24\implies \sqrt {49-x^2} + \sqrt {25-x^2}=8$
My question is that will this method work for similar problems, and is there a faster method?
Thanks!
|
Another way:
$$1) \ \sqrt{49-x^2}=\sqrt{25-x^2}+3 \Rightarrow \\
49-x^2=25-x^2+9+6\sqrt{25-x^2} \Rightarrow \\
2.5=\sqrt{25-x^2};\\
2) \ \sqrt{25-x^2}=\sqrt{49-x^2}-3 \Rightarrow \\
25-x^2=49-x^2+9-6\sqrt{49-x^2} \Rightarrow \\
\sqrt{49-x^2}=5.5.$$
Hence:
$$\sqrt{49-x^2}+\sqrt{25-x^2}=5.5+2.5=8.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3090437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
Finding an $\epsilon-\delta$ proof for a multivariable limit. Suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is defined as $$(x,y) \longmapsto
\left\{
\begin{array}{cl}
\dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & \mbox {if } (x,y) \neq (0,0) \\
\\
0 & \mbox {if } (x,y) = (0,0)
\end{array}
\right.
$$
I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $\epsilon-\delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)
So for $\epsilon > 0$ I need to find a $\delta$ such that $\|(x,y)\| =\sqrt{x^2 + y^2} < \delta$ implies $|f(x,y)| < \epsilon$
I have tried to work backwards using the inequality $(x^2+y^2)^2 \geq 4x^2y^2$
$$\begin{align*}
\left|\frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}\right|&=\left|\frac{4x^2y^3}{(x^2+y^2)^2} + \frac{x^4 y}{(x^2+y^2)^2} - \frac{y^5}{(x^2+y^2)^2}\right|\\
&\leq\left|\frac{4x^2y^3}{(x^2+y^2)^2}\right| + \left|\frac{x^4 y}{(x^2+y^2)^2}\right| + \left|\frac{y^5}{(x^2+y^2)^2}\right|\\
&\leq\left|\frac{4x^2y^3}{4x^2y^2}\right| + \left|\frac{x^4 y}{4x^2y^2}\right| + \left|\frac{y^5}{4x^2y^2}\right|\\
&= \left|y\right| + \left|\frac{x^2}{4y}\right| + \left|\frac{y^3}{4x^2}\right|
\end{align*}$$
Got stuck here and not sure if my approach is correct.
|
The essential point is that $r:=\sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|\leq \sqrt{x^2+y^2}=r$, and similarly $|y|\leq r$. It follows that
$$\left|{4x^2y^3+x^4y-y^5\over(x^2+y^2)^2}\right|\leq(4+1+1){r^5\over r^4}=6r\ .\tag{1}$$
Given an $\epsilon>0$ choose $\delta:={\epsilon\over6}$. If $|(x,y)|=r<\delta$ then, according to $(1)$, we have
$$\left|{4x^2y^3+x^4y-y^5\over(x^2+y^2)^2}\right|\leq6 r<6\delta=\epsilon\ .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3091176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Reciprocal of $x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))$ where $x_n\searrow0$
Suppose that $\{x_n\}$ is a decreasing sequence having limit $0$. How to find the reciprocal of $x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))$ as $n\to\infty$.
In my textbook it says that
\begin{equation}\frac{1}{x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))}=\frac{1}{x_{n-1}^2}+\frac{1}{3}+o(1) \;(n\to\infty).
\end{equation}
That is,
\begin{align*}
&[x_{n-1}^2-\frac{x_{n-1}^4}{3}+o(x_{n-1}^4)]\cdot[\frac{1}{x_{n-1}^2}+\frac{1}{3}+o(1)] \\
=&1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2)+\frac{x_{n-1}^2}{3}-\frac{x_{n-1}^4}{3}+o(x_{n-1}^4)+o(x_{n-1})+o(x_{n-1}^3)+o(x_{n-1}^4) \\
=&1+o(x_{n-1})\;(n\to0).
\end{align*}
Also,I find that
\begin{align*}
&[x_{n-1}^2-\frac{x_{n-1}^4}{3}+o(x_{n-1}^4)]\cdot\frac{1}{x_{n-1}^2} \\
=&1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2) \\
=&1+o(x_{n-1})\;(n\to 0).&
\end{align*}
So my question is that why the reciprocal should be $\frac{1}{x_{n-1}^2}+\frac{1}{3}+o(1)$.
|
This comes out from the geometric series ($x\rightarrow 0$)
$$
\frac{1}{1-x}=1+x+O(x^2).
$$
In your case, you have to evaluate
$$
\frac{1}{x_n^2\left(1-\frac{x^2_n}{3}+O(x^4_n)\right)}
$$
Using the geometric series you will get
$$
\frac{1}{x_n^2}\left(1+\frac{x_n^2}{3}+O(x_n^4)\right)=\frac{1}{x_n^2}+\frac{1}{3}+O(x_n^2)
$$
that is the result in the textbook.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3093211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\int \frac{x^2}{x-1} \,dx$ Evaluate $\int \frac{x^2}{x-1} \,dx$
(A) $2x^2+x+\ln|x-1|+C$
(B) $\frac{x^2}{2}+x+\ln|x+1|+C$
(C) $\frac{x^2}{2}+x+\ln|x-1|+C$
(D) $x^2+x+\ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$\int \frac{(u+1)^2}{u}\,du\\
=\int u +2+\frac{1}{u}\,du\\
=\frac{u^2}{2}+2u+\ln|u|+C\\
=\frac{(x-1)^2}{2}+2(x-1)+\ln|x-1|+C$
Simplify :
$\frac{x^2-3}{2}+x+\ln|x-1|+C$
It's not on the option.
|
Your answer is right except for the constant I changed.
$$\frac{x^2-3}{2}+x+\ln|x-1|+C_1 = \frac{x^2}{2}+x+\ln|x-1|+C_1-\frac{3}{2} = \frac{x^2}{2}+x+\ln|x-1|+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3095759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$? For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$
I checked in very many cases. Example :$c=1,
a=2,b=\frac{1}{2}...$ then it’s true, but cannot prove that
My attempts:
I consider function $ f(x)=\sqrt{24x^2+25}$
And $f’(x)=\frac{24x}{\sqrt{24x^2+25}}$,
$f’’(x)=\frac{600}{(24x^2+25)(\sqrt{24x^2+25}}>0$.
So $f(x)+f(y)+f(z)\geq 3f(\frac{x+y+z}{3})$.
But cannot prove that $\sqrt{ab}+\sqrt{bc}
+\sqrt{ca}\geq 3$
|
What are the allowed values of $a,b,c$? If they are integers greater than zero then how about this:
Firstly $a = b = c = 1$ meets your constraint, and with these values $3\sqrt{24 +25} = 21$
By my assumption that they are positive integers any valid $a,b$ would satisfy $\sqrt{24ab + 25} \ge \sqrt{24 + 25} = 7$
And since the problem is symmetric in $a,b,c$, we can say that
$\sqrt{24ab+25} + \sqrt{24bc + 25} + \sqrt{24ca + 25} \ge \sqrt{24+25} + \sqrt{24+25} + \sqrt{24+25} = 21$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3098290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2+x^2)(b^2+x^2)}=\frac {Ax+B} {(a^2+x^2)}+\frac {Cx+D} {(b^2+x^2)}$.
From here I get that $A=C=0$, $B=\frac {2a} {(b^2-a^2)}$ and $D=\frac {-2a} {(b^2-a^2)}$
Is this correct? Because when I try to solve $I'(a)$ using these values for $B$ and $D$ I get a different solution from the textbook?
|
$$I(a)=\int_0^\infty\frac{\ln(a^2+x^2)}{(b^2+x^2)}dx$$
$$I'(a)=\int_0^\infty\frac{2a}{(a^2+x^2)(b^2+x^2)}dx$$
now we want to look at simplifying this expression:
$$\frac{2a}{(a^2+x^2)(b^2+x^2)}$$
and we know (due to both factors being quadratic) that it will be of the form:
$$\frac{2a}{(a^2+x^2)(b^2+x^2)}=\frac{Ax+B}{a^2+x^2}+\frac{Cx+D}{b^2+x^2}$$
$$2a=(Ax+B)(b^2+x^2)+(Cx+D)(a^2+x^2)$$
firstly if we look at the cubic terms:
$$0=Ax^3+Cx^3$$
now at the quadratic terms:
$$0=Bx^2+Dx^2$$
now at the linear terms:
$$0=Ab^2x+Ca^2x$$
and finally the constants:
$$2a=Bb^2+Da^2$$
From this we obtain the following simultaneous equations:
$$A+C=0$$
$$B+D=0$$
$$Ab^2+Ca^2=0$$
$$Bb^2+Da^2-2a=0$$
This should give: $A=0$, $C=0$, $B=1/(b^2-a^2)$, $D=1/(a^2-b^2)$
If we put this back in we get:
$$I'(a)=\int_0^\infty\left[\frac{1}{b^2-a^2}\frac{1}{a^2+x^2}+\frac{1}{a^2-b^2}\frac{1}{b^2+x^2}\right]dx$$
we can now split this up into parts to make it easier:
$$I_1=\frac{1}{b^2-a^2}\int_0^\infty\frac{1}{a^2+x^2}dx=\frac{1}{a^2(b^2-a^2)}\int_0^\infty\frac{1}{1+(x/a)^2}dx$$
letting $x=a\tan(u)$ we get $dx=a\sec^2(u)du$ giving:
$$I_1=\frac{1}{a(b^2-a^2)}\int_0^{\pi/2}du=\frac{\pi}{2}\times\frac{1}{a(b^2-a^2)}$$
Similarly we can obtain:
$$I_2=\frac{1}{a^2-b^2}\int_0^\infty\frac{1}{b^2+x^2}dx=\frac{\pi}{2}\times\frac{1}{b(a^2-b^2)}$$
Which gives us:
$$I'(a)=\frac{\pi}{2}\left(\frac{1}{a(b^2-a^2)}+\frac{1}{b(a^2-b^2)}\right)$$
we know now that we must integrate this to obtain an expression for $I(a)$ and work out the constant of integration. To clarify:
$$I(a)=\int\frac{\pi}{2}\left(\frac{1}{a(b^2-a^2)}+\frac{1}{b(a^2-b^2)}\right)da+C$$
To make it easier we will split this up again:
$$I_3=\int\frac{1}{a(b^2-a^2)}da$$
$$I_4=\int\frac{1}{b(a^2-b^2)}da$$
To make this shorter I will just show the answer:
$$I_3=-\frac{\ln|a^2-b^2|-2\ln|a|}{2b^2}$$
$$I_4=-\frac{\ln|a+b|-\ln|a-b|}{2b^2}$$
Giving:
$$I(a)=-\frac{\pi}{4b^2}\left(\ln|a^2-b^2|-2\ln|a|+\ln|a+b|-\ln|a-b|\right)+C$$
Normally this value of $C$ would be calculated where a value of $a$ can be inputted which makes the integral $I(a)$ trivial, its not so easy here but $a=0$ seems the best option
|
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"timestamp": "2023-03-29T00:00:00",
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|
Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$
We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x$ by the condition $-2<x<0$ [neighborhood of $-1$] . For $x$ in this interval, we have $-1<2x+3<3$, so that
$|\frac{x+ 5}{2x+3} -4|= \frac{7}{|2x+3|}|x+1|<...$
Is that true, please? And I don’t know how can I complete.
|
The neighborhood you have chosen is too large, it contains a root of the denominator. Let us try with the bounds
$$-1\pm\frac13$$ (any deviation smaller than $\dfrac12$ can do).
The function is monotonous in that range and the extreme values of $\left|\dfrac7{2x+3}\right|$ are
$$21, \frac{21}{5}$$ so you can write
$$\left|\frac{x+5}{3x+2}-4\right|<21|x+1|.$$
|
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|
remainder in double sum having binomial coefficients find the remainder when
$\displaystyle \sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}$ is divided by $64$
what i try
$$\sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}$$
$$\sum^{2014}_{r=0}(-1)^k\sum^{r}_{k=0}(k+1)(k+2)\binom{2019}{r-k}$$
$$\sum^{0}_{k=0}(k+1)(k+2)\binom{2019}{0-k}-\sum^{1}_{k=0}(k+1)(k+2)\binom{2019}{1-k}+\sum^{2}_{k=0}(k+1)(k+2)\binom{2019}{2-k}+\cdots +\sum^{2014}_{k=0}(k+1)(k+2)\binom{2019}{2014-k}$$
how do i solve it help me please
|
Let us use the followings :
$$\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}=\frac{-2(r-n)(r-n+1)(r-n+2)}{n(n-1)(n-2)}\binom{n}{r}\tag1$$
$$\sum_{r=0}^{n}r^3\binom nr=2^{n-3}n^2(n+3)\tag2$$
$$\sum_{r=0}^{n}r^2\binom nr=2^{n-2}n(n+1)\tag3$$
$$\sum_{r=0}^{n}r\binom nr=2^{n-1}n\tag4$$
$$\sum_{r=0}^{n}\binom nr=(1+1)^n=2^n\tag5$$
See here, here, here, here for proofs.
Using $(1)$, we get
$$\sum^{n-4}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-4-k}=2n-6\tag6$$
$$\sum^{n-3}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-3-k}=2\tag7$$
$$\sum^{n-2}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-2-k}=0\tag8$$
$$\sum^{n-1}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-1-k}=0\tag9$$
$$\sum^{n}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{n-k}=0\tag{10}$$
Let $n=2019$. Then, our sum can be written as
$$\sum^{n-5}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}$$
$$=\underbrace{\sum^{n}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}}_{A}-\underbrace{\sum^{n}_{r=n-4}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}}_{B}$$
$$=\underbrace{\frac{-2}{n(n-1)(n-2)}\sum^{n}_{r=0}(r-n)(r-n+1)(r-n+2)\binom{n}{r}}_A-\underbrace{\bigg((2n-6)+2\bigg)}_B$$
From $(1)(2)(3)(4)(5)$, $$\sum^{n}_{r=0}(r-n)(r-n+1)(r-n+2)\binom{n}{r}$$ is divisible at least by $2^{2019-3}$. Also, $$n(n-1)(n-2)=2019\times 2018\times 2017$$ is not divisible by $2^2$. So, $A$ is divisible by $2^6=64$.
Conclusion :
$$\sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}=A-B\equiv 0-2\equiv \color{red}{62}\pmod{64}$$
|
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|
Prove by induction: $C(2n, 2) = 2C(n, 2) + n^2$
Show that if $n$ is a positive integer, then $C(2n, 2) = 2C(n, 2) + n^2$. Here, $C(a, b)$ means the binomial coefficient $\dbinom{a}{b}$.
Prove this by induction.
Here is my calculation:
$n$ cannot be $1$ because $n$ should be equal or larger than $2$ for $C(n, 2)$.
If $n=2$: $C(4, 2) = 2C(2, 2) + 2^2 = 6$
If $n=m$: $C(2m, 2) = 2C(m, 2) + m^2 = 2m^2 - m$
If $n=m+1$: $C(2m+2, 2) = 2C(m+1, 2) + (m+1)^2 = 2m^2 + 3m + 1$
And I think I have to prove that:
\begin{align*}
C(2m+2, 2) & = [\dots]\\
& = [2C(m, 2) + m^2] + [2C(m, 1) + (2m + 1)]\\
& = 2C(m+1, 2) + (m+1)^2 && \text{(by Pascal's Identity).}
\end{align*}
But I have no idea what to do on $[\dots]$ part.
Thanks.
|
We assume for our inductive hypothesis that $\binom{2n}{2}=2\binom{n}{2}+n^2$ is true for all positive integers $n$ which are less than or equal to some $k$. We then try to prove that this will further imply it is true for $k+1$ as well.
$$\begin{array}{rl}\binom{2(k+1)}{2}&=\binom{2k+2}{2}=\binom{2k+1}{1}+\binom{2k+1}{2}\\
&=\binom{2k+1}{1}+\binom{2k}{1}+\binom{2k}{2}\\
&=2k+1+2k+\binom{2k}{2}\\
~^{\text{I.H.}}&=2k+1+2k+2\binom{k}{2}+k^2\\
&=k+\binom{k}{2}+k+\binom{k}{2}+k^2+2k+1\\
&=\binom{k}{1}+\binom{k}{2}+\binom{k}{1}+\binom{k}{2}+k^2+2k+1\\
&=\binom{k+1}{2}+\binom{k+1}{2}+(k+1)^2\\
&=2\binom{k+1}{2}+(k+1)^2\end{array}$$
which is exactly what we hoped to show. Note how we used (and pointed out our use of) the induction hypothesis going from the third to the fourth line.
The big trick here was to recognize that $n = \binom{n}{1}$ and to recognize that $\binom{n}{r-1}+\binom{n}{r}=\binom{n+1}{r}$ (in particular when $r=2$)
Reiterating what was said in comments above, some times it is preferable to use a combinatorial proof rather than an inductive proof. The combinatorial proof for this situation is relatively straightforward once you know how to spot it.
Suppose we have $2n$ balls, $n$ of which are red, and the other $n$ of which are black, and these balls are all numbered. We count how many ways there are to select two balls from these and do so in two different ways.
The first method, we just count directly. There are $2n$ balls total, and to choose two of them this can be done in $\binom{2n}{2}$ ways.
The second method, we break into cases based on the colors chosen. If both balls chosen were red, there are $\binom{n}{2}$ ways to do this. Similarly there are $\binom{n}{2}$ ways to do so if they are both black. If they are one of each color there are $n^2$ ways to do this, for a total of $2\binom{n}{2}+n^2$ ways to choose two balls.
By the nature of the fact that these expressions both correctly answer the same counting question, despite appearing in different forms they must be equal. Therefore $\binom{2n}{2}=2\binom{n}{2}+n^2$.
|
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|
Is it true that $\int_0^\infty \frac{f(x)}{x} \sin \bigl(\frac{\pi x}{a}\bigr) \,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} f(x) \,\mathrm{d}x$? I dont recall where, but I found this interesting identity a few years ago. It was shown in one of Victor Moll's papers about elliptic integrals.
Corollary 3.1. Let $f$ be an even function with period $a$. Then,
$$\int_0^\infty \frac{f(x)}{x} \sin \bigl(\frac{\pi x}{a}\bigr)
\,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} f(x) \,\mathrm{d}x.$$
To prove this result the following lemma is applied
Lemma 3.1. Let $f$ be an odd periodic function of period $a$. Then
$$ \int_0^{\infty} \frac{f(x)}{x} \,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} f(x) \cot\bigl( \frac{\pi x}{a} \bigr) \,\mathrm{d}x $$
I have no qualms about this lemma. Further, Victor writes that Corollary 3.1 follows from using the lemma on $f(x) = g(x) \sin (\frac{\pi x}{a})$ with the half-angle formula
$$ \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} $$
However, this is where my problems starts. Inserting in our values and using that $\cot x = 1/\tan x$
$$ \cot \bigl( \frac{\pi x}{a} \bigr) = \frac{1 + \cos(2\pi x/a)}{\sin(2\pi x/a)}$$
Inserting this into the lemma with $f = g \cdot \sin(\pi x/a)$ gives
$$
\int_0^{\infty} \frac{g(x)}{x} \sin \bigl( \frac{\pi x}{a} \bigr)\,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} g(x) \sin \bigl( \frac{\pi x}{a} \bigr) \bigl[ 1 + \cos \bigl( \frac{2 \pi x}{a} \bigr) \bigr] \csc \bigl( \frac{2\pi x}{a} \bigr) \,\mathrm{d}x
$$
I would really like the $\csc(x)$ and $\sin(x)$ to cancel out, however the extra factor of $2$ throws the entire calculation of. Any help proving the desired equality would be much obliged.
EDIT: Maybe the lemma is missing a 1/2 factor in the cotan argument?
|
Corollary 3.1
An even function with period $a$ means $f(a-x)=f(x)$.
$$
\begin{align}
\int_0^\infty\frac{f(x)}x\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x
&=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+2ka}-\frac1{x+(2k+1)a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag1\\
&=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+2ka}+\frac1{x-(2k+2)a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag2\\
&=\sum_{k\in\mathbb{Z}}\int_0^af(x)\,\frac1{x+2ka}\,\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag3\\
&=\frac\pi{2a}\int_0^af(x)\cot\left(\frac{\pi x}{2a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag4\\[3pt]
&=\frac\pi{2a}\int_0^af(x)\tan\left(\frac{\pi x}{2a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag5\\[3pt]
&=\frac\pi{2a}\int_0^af(x)\,\mathrm{d}x\tag6\\
&=\frac\pi{a}\int_0^{a/2}f(x)\,\mathrm{d}x\tag7\\
\end{align}
$$
Explanation:
$(1)$: $f(x+a)=f(x)$ and $\sin\left(\frac{\pi(x+a)}a\right)=-\sin\left(\frac{\pi x}a\right)$
$(2)$: $f(a-x)=f(x)$ and $\sin\left(\frac{\pi(a-x)}a\right)=\sin\left(\frac{\pi x}a\right)$ on the second term of the sum
$(3)$: write as a sum over $\mathbb{Z}$
$(4)$: use $(7)$ from this answer
$(5)$: $f(a-x)=f(x)$ and $\sin\left(\frac{\pi(a-x)}a\right)=\sin\left(\frac{\pi x}a\right)$
$(6)$: average $(4)$ and $(5)$
$(7)$: $f(a-x)=f(x)$
Lemma 3.1
An odd function with period $a$ means $f(a-x)=-f(x)$.
$$
\begin{align}
\int_0^\infty\frac{f(x)}x\,\mathrm{d}x
&=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+ka}\right)\,\mathrm{d}x\tag8\\
&=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x-(k+1)a}\right)\,\mathrm{d}x\tag9\\
&=\frac12\sum_{k\in\mathbb{Z}}\int_0^af(x)\left(\frac1{x+ka}\right)\,\mathrm{d}x\tag{10}\\
&=\frac\pi{2a}\int_0^af(x)\cot\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag{11}\\
&=\frac\pi{a}\int_0^{a/2}f(x)\cot\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: $f(x+a)=f(x)$
$\phantom{1}(9)$: $f(a-x)=-f(x)$
$(10)$: average $(8)$ and $(9)$ and write as a sum over $\mathbb{Z}$
$(11)$: use $(7)$ from this answer
$(12)$: $f(a-x)\cot\left(\frac{\pi(a-x)}a\right)=f(x)\cot\left(\frac{\pi x}a\right)$
|
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|
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\
&= x^3 + y^3\\
\end{split}$$
Q.E.D.
Question:
Spivak says there is an easy proof that, if I use this other theorem:
$$
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
$$
then, I will also allow me to find out $x^n+y^n$ whenever $n$ is odd.
How to do this? I fail to see how.
|
Just let $k=-y$ to obtain $$x^3-y^3=(x-y)(x^2+xy+y^2)\iff x^3+k^3=(x+k)(x^2-xk+k^2)$$
Note that the same works whenever $n$ is odd. Given $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots + xy^{n-2}+y^{n-1})$$
Let $k=-y$ and ...
See how it works? Try to explain why it doesn't work whenever $n$ is even!
|
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|
a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc = a^2 b + b^2 c + c^2 a$$
But it seems like there is a way to determine the value.
|
Hint
\begin{align}
a-b&= \frac1c-\frac 1b \\ &=\frac{b-c}{bc}\\ &=\frac{c-a}{abc^2} \\& =\frac{a-b}{a^2b^2c^2}.
\end{align}
|
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|
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \frac{\sqrt{4-0}}{\sqrt[3]{1+0}} \\
& = 2
\end{align}
Despite the steps I've taken seems plausible to me, the answer is given as $-2$.
Is dividing both the numerator and the denominator by $x$ allowed here? Where am I making a mistake?
|
Don't forget that $\sqrt{x^2}=|x|$ and when $x<0$ ($x$ is going to negative infinity), $|x|$ is equivalent to $-x$:
$$
\frac{\sqrt{x^2}\sqrt{4-\frac{2}{x}}}{\sqrt[3]{x^3}\sqrt[3]{1+\frac{1}{x^3}}}=
\frac{|x|\sqrt{4-\frac{2}{x}}}{x\sqrt[3]{1+\frac{1}{x^3}}}=
\frac{-x\sqrt{4-\frac{2}{x}}}{x\sqrt[3]{1+\frac{1}{x^3}}}=\\
-\frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\xrightarrow{x\rightarrow-\infty}-\frac{\sqrt{4-0}}{\sqrt[3]{1+0}}=
-\frac{2}{1}=-2.
$$
|
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|
Evaluating $\int_{} \frac{xe^{2x}}{(1+2x)^2}dx$ via integration by parts $\int_{} \frac{xe^{2x}}{(1+2x)^2}dx$
I am having trouble picking the correct $u/dv$ before integrating by parts. I felt like L.I.A.T.E. did not really help me here...
This is what I tried, but it ended up with integration spiraling into an endless evaluation of an integral...
$$
\begin{align}
u &= (1 + 2x)^2 & dv &= xe^{2x}dx \\
du &= 2(1+2x)dx & v &= \frac{1}{2}x^2 \frac{1}{2}e^{2x} \\
&= 2 + 4xdx & &= \frac{1}{4}x^2e^{2x}
\end{align}
$$
Am I at least correct in choosing the right $u/du$ values? That is all I really want to know, if I am allowed to choose the $u/du$ like how I did
|
Hint:
Let $u=xe^{2x}$, $\text dv=\dfrac{\text dx}{(1+2x)^2}$.
Result:
$$\dfrac{\mathrm{e}^{2x}}{8x+4} + C$$
EDIT: More steps.
\begin{eqnarray*}
\int \frac{xe^{2x}}{(1+2x)^2} \ \text dx &=& \left|\begin{array}{2}
u=xe^{2x} & \text dv=\dfrac{\text d x}{(1+2x)^2} \\
\text du = (1+2x)e^{2x}\ \text dx & v=-\dfrac{1}{2(1+2x)}
\end{array}\right| = \\
&=& -\frac{xe^{2x}}{2(1+2x)} + \frac{1}{2} \int e^{2x}\ \text dx = \\
&=&-\frac{xe^{2x}}{2(1+2x)} + \frac{1}{4}e^{2x} + C = \\
&=& e^{2x}\left( -\frac{x}{2(1+2x)}+\frac{1}{4} \right) + C = \boxed{\frac{e^{2x}}{8x+4} + C}
\end{eqnarray*}
|
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|
Solving $(x^2+4x+3)^x+(2x+4)^x=(x^2+4x+5)^x$ with $x\in(-1,\infty)$ I've been struggling for a few hours on the below pre-calculus olympiad equation to which I still don't have an answer:
$$(x^2+4x+3)^x+(2x+4)^x=(x^2+4x+5)^x$$
where $x \in (-1,\infty)$.
Now, according to WolframAlpha, this has a unique solution, $x=2$, and I need to prove this for one of my student without using derivatives or any more advanced techniques.
Interesting substitutions that I've tried, but with no success:
1) $a=(x+2)-\frac{1}{x+2}, b=(x+2)-\frac{1}{x+2}$ leads to
$$(b^2-a^2)^x=(b^x-a^x)^2$$
2) $\alpha =2\arctan(x+2)$ leads to
$$(-\cos(\alpha ))^{\tan(\alpha /2)-2} + (\sin(\alpha ))^{\tan(\alpha /2)-2} = 1$$
Both of them make the solution $x=2$ easy to see, but its unicity still eludes me.
|
First, the trivial solution is $x=2$.
Then, consider the fact $$(x^2+4x+3)^2+(2x+4)^2=(x^2+4x+5)^2 \tag{1}$$
And the fact $\dfrac{x^2+4x+3}{x^2+4x+5}$ and $\dfrac{2x+4}{x^2+4x+5}$ are both in $(0,1)$ if $x \in (-1,\infty)$. (I believe you can figure this out yourself.)
Consider that $a^x$ is strictly decreasing if $a\in (0,1)$.
We can easily get
$$\left({x^2+4x+3\over x^2+4x+5}\right)^x \gt \left({x^2+4x+3\over x^2+4x+5}\right)^2 \tag{2}$$ if $x\in (-1,2)$, and other cases.
We have:
Case:$x\in (-1,2)$, then
$$\left({x^2+4x+3\over x^2+4x+5}\right)^x+\left({2x+4\over x^2+4x+5}\right)^x \gt \left({x^2+4x+3\over x^2+4x+5}\right)^2+\left({2x+4\over x^2+4x+5}\right)^2=1 \tag{3}$$.
In another case ($x\in (2, \infty)$), we have
$$\left({x^2+4x+3\over x^2+4x+5}\right)^x+\left({2x+4\over x^2+4x+5}\right)^x \lt \left({x^2+4x+3\over x^2+4x+5}\right)^2+\left({2x+4\over x^2+4x+5}\right)^2=1 \tag{4}$$
Hence, the only solution is $x=2$.
|
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|
Integral problem. Unsure of the approach. I have this integral:
$$\int_0^1 \frac{1 + 12t}{1 + 3t}dt$$
I can split this up into:
$$\int_0^1 \frac{1}{1+3t} dt + \int_0^1\frac{12t}{1+3t}dt$$
The left side:
$u = 1+3t$ and $du = 3dt$ and $\frac{du}{3} = dt$
so $$\frac{1}{3} \int_0^1 \frac{1}{u} du = \frac{1}{3} \ln |u| + C$$
But what about the right?
|
My first attempt is to split the integral in another way:
$$\int_0^1{\frac{1+12t}{1+3t}}dt = \int_0^1{\frac{1+3t+9t}{1+3t}} dt = \int_0^1 1 \, dt + \int_0^1 \frac{9t}{1+3t} dt$$
Now observe that
$$\int_0^1 \frac{9t}{1+3t} dt = \int_0^1{\frac{3t}{1+3t}} \cdot 3dt = \int_0^1{\frac{3t}{1+3t}}d(3t) = \int_0^3 \frac{u}{1+u} du$$
, which can be found by another substitution, or
$$\int_0^3 \frac{u}{1+u} du = \int_0^3 \frac{u+1-1}{1+u} du = {\int_0^3 1\,du} - {\int_0^3{\frac{1}{1+u}} du} = 3-\int_1^4 \frac{1}{s} ds = 3-\log(4).$$
Hence $$\int_0^1 \frac{1+12t}{1+3t}dt = 1+3-\log(4)=4-\log(4).$$
|
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"url": "https://math.stackexchange.com/questions/3122196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The intersection of a sphere and the xy-plane is the circle $(x-1)^2+(y-2)^2=5$ and point $(1,2,1)$ is on the sphere The intersection of a sphere and the $xy$-plane is the circle $$(x-1)^2+(y-2)^2=5$$ and point $(1,2,1)$ is on the sphere. Find the center and radius of the sphere.
When the sphere and $xy$-plane intersect than $z=0$.
I tried $(x-1)^2+(y-2)^2+(z+?)^2=5$ but what can I do next?
|
One more :
Equation of the sphere:
$(x-a)^2+(y-b)^2+(z-c)^2=R^2.$
Cut with plane $z=0:$
$(x-a)^2+(y-b)^2 +c^2=R^2.$
1) Compare with the given circle:
$a=1,b=2, R^2-c^2= (√5)^2$
2) $(1,2,1)$ lies on the sphere:
$0+0+(1-c)^2=R^2.$
Hence: $1-2c+c^2=R^2;$
$1-2c= R^2-c^2=(√5)^2.$
$c= (1-5)/2=-2$, and $R^2 =9$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
prove that the triangle is isosceles
In a $\triangle ABC$, If
$\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \;\;\displaystyle\tan \frac{A}{2}+\tan \frac{B}{2}\end{vmatrix}=0$
Then prove that the triangle is isosceles.
Try: Let $\displaystyle \tan \frac{A}{2}=p\;\;,\tan \frac{B}{2}=q\;\;,\tan \frac{C}{2}=r$
Using $$\tan\bigg(\frac{A}{2}+\frac{B}{2}\bigg)=\tan\bigg(\frac{\pi}{2}-C\bigg)$$
So $$\sum \tan\frac{A}{2}\tan\frac{B}{2}=1\Rightarrow pq+qr+rp=1$$
So $$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
p(q+r) & q(r+p)& r(p+q)\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
1-qr & 1-rp& 1-pq\end{vmatrix}=0$$
So $$-\begin{vmatrix}p^2& q^2& r^2\\
\displaystyle p & \displaystyle q & \displaystyle r \\
1 & 1& 1\end{vmatrix}=0$$
So we have $(p-q)(q-r)(r-p)=0.$
So either $p=q$ or $q=r$ and $r=p.$
Could some help me some short way to solve it? Thanks.
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I have got this here for your determinant: $$\tan \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right)-\cot \left(\frac{A}{2}\right)
\tan \left(\frac{B}{2}\right)-\tan \left(\frac{A}{2}\right) \cot
\left(\frac{C}{2}\right)+\cot \left(\frac{A}{2}\right) \tan \left(\frac{C}{2}\right)+\tan
\left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)-\cot \left(\frac{B}{2}\right) \tan
\left(\frac{C}{2}\right)=0$$
With $$C=\pi-A-B$$ we get
$$-\frac{8 (\sin (A-B)+\sin (2 A+B)-\sin (A+2 B))}{-\sin (2 A+2 B)+\sin (2 A)+\sin (2 B)}=0$$
You have to solve
$$\sin(A-B)+\sin(2A+B)=\sin(A+2B)$$
One solution is given by $$A=B$$
|
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|
Prove that $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$ is decreasing Let $\ a>0$ and the sequence $(x_{n})_{n>=0}$ defined by $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$. Prove the sequence is monotonically decreasing and $0<x_n<\frac{4+3a}{8}$, for any $n>0$. I've made little progress towards proving that $x_{n}<x_{n-1}$. First, if a function takes the form of the fraction , $f(x)=\frac{x+a}{x^{3}+2a}$, then it's monotonically decreasing. Also, $x_n$ compared to $x_{n-1}$ boils down to comparing $2\frac{x+2a}{x^{3}+16a}$ with $\frac{x+a}{x^{3}+2a}$. From here I have really no idea how to continue. These simple results are the work of some ruminations and I am unable to bring something out of the blue to complete the proof...
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Let $x_n=\int_{n}^{2n}\frac{x+a}{x^3+2a}\,dx$. Then, the first difference $x_{n+1}-x_n$ is given by
$$\begin{align}
x_{n+1}-x_n&=\int_{n+1}^{2n+2}\frac{x+a}{x^3+2a}\,dx-\int_{n}^{2n}\frac{x+a}{x^3+2a}\,dx\\\\
&=\int_{2n}^{2n+2}\frac{x+a}{x^3+2a}\,dx-\int_n^{n+1}\frac{x+a}{x^3+2a}\,dx\\\\
&=\int_n^{n+1}\left(\frac{4x+2a}{8x^3+2a}-\frac{x+a}{x^3+2a}\right)\,dx\\\\
&=-2\int_n^{n+1} \frac{2x^4+3ax^3-3ax-a^2}{(8x^3+2a)(x^3+2a)}\,dx
\end{align}$$
Now, if the sequence were decreasing, then $\int_n^{n+1} \frac{2x^4+3ax^3-3ax-a^2}{(8x^3+2a)(x^3+2a)}\,dx\ge 0$. However, we see that the sequence is not decreasing for all $n\ge1$ in general.
For example, take $a=100$. Then, the integrand is negative for $n=1$, $n=2$, and $n=3$. But, when $a=100$, the integrand is positive for $n\ge 4$ and hence the sequence $x_n$ is decreasing for $n\ge 4$.
In fact, with $a$ fixed, there exists a number $N$ large enough, so the $2x^4+3ax^3-3ax-a^2\ge 0$ whenever $x\ge N$. So, for an arbitrary $a>0$, the best we can say is there exists a number $N$ such that $x_n$ is decreasing for $n\ge N$.
|
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|
Coefficients of $1,x,x^2$ in $((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Finding coefficients of $x^0,x^1,x^2$ in
$((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
where there are $k$ parenthesis in the left side
Try:
Let $P(x)=((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Let we assume
$P(x)=a_{k}+b_{k}x+c_{k}x^2+d_{k}x^3+\cdots\cdots $
For constant term put $x=0$, we have
$$((\cdots (0-2)^2-2)^2-2)^2\cdots )-2)^2=a_{k}$$
Now i did not know how to find coeff. of $1,x,x^2$ in $P(x)$ . Thanks
|
Lets consider the cases $k=1,2,3,4$:$$P_1(x)=(x-2)^2=x^2-4x+4,\\ P_2(x)=(P_1(x)-2)^2=P_1(x)^2-2P_1(x)+4=x^4-8x^3+20x^2-16x+4,\\ P_3(x)=(P_2(x)-2)^2=x^8+\dots+336x^2-64x+4,\\ P_4(x)=x^{16}+\dots+5440x^2-256x+4.$$
From this it is reasonable to conjecture that the coefficient of $x$ in $P_k(x)$ is $4^k$. Moreover, a quick check using OEIS yields the conjecture that the coefficient of $x^2$ in $P_k(X)$ is given by the sequence $$a_1=1,\\ a_2=20,\\ a_n=20a_{n-1}+64a_{n-2}.$$ This sequence can be written in a compact formula $$a_n=\frac{4\cdot 16^{n+1}-4^{n+1}}{3}.$$
You can prove both conjectures by induction on $k$.
I will show you how to do for the coefficient of $x$. The (harder) coefficient of $x^2$ I leave to you.
We proved the conjecture for $k=1,2,3,4$. This sets up induction. Now assume we know it up to $P_k(x)$. Consider $$P_{k+1}(x)=(P_k(x)-2)^2=P_k(x)^2-4P_k(x)+4.$$ Write $P_k(x)=\sum\limits_{i=0}^{2^k}c_ix^i$. The terms with $x$ in the expression above are $$c_0c_1x+c_1c_0x-4c_1x=(2c_0c_1-4c_1)x.$$ Since $c_0=4$, this yields $8c_1-4c_1=4c_1$. Now use the induction hypothesis to deduce that the coefficient equals $4\cdot 4^k=4^{k+1}$.
|
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|
Prove that the product of the $n$ roots of $e^{-i\cdot \frac{\pi }{2}}$ is given by $e^{i\cdot \frac{\left(2n-3\right)\pi }{2}}$ Let $w=e^{-i\cdot \frac{\pi }{2}}$.
Here's how to tried to solve this:
$\sqrt[n]{w}=e^{i\cdot \left(-\frac{\pi }{2n}+\frac{2\pi }{n}k\right)},\:k\in \left(0;\:1;\:2;\:...;\:n-1\right)$
Factoring out the $1/n$, we get $e^{\frac{i}{n}\cdot \left(-\frac{\pi }{2}+2\pi k\right)}$.
Since for any value of $k$ we get , $e^{\frac{i}{n}\cdot \left(-\frac{\pi }{2}\right)}$, that is, all the roots are the same, the product of the n-th roots would be $\left(e^{\frac{i}{n}\cdot \:\left(-\frac{\pi \:}{2}\right)}\right)^{\left(n-1\right)}$. I get stuck here.
How do I get to $e^{i\cdot \frac{\left(2n-3\right)\pi }{2}}$?
|
Your approach seems correct, except for a small mistake in the "product". I have solved it as follows.
The $n$ roots of $e^{- \iota \frac{\pi}{2}}$ are given by $e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)}$ for $k = 0, 1, 2, \cdots, n - 1$
\begin{align*}
\prod\limits_{k = 0}^{n - 1} e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)} &= e^{- \iota \frac{\pi}{2n}} \prod\limits_{k = 1}^{n - 1} e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)} \\
&= e^{- \iota \frac{\pi}{2n}} \cdot e^{\sum\limits_{k = 1}^{n} \iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)} \\
&= e^{- \iota \frac{\pi}{2}} \cdot e^{\iota \left( - \frac{\pi}{2n} \left( n - 1 \right) + \frac{2\pi}{n} \frac{n \left( n - 1 \right)}{2} \right)} \\
&= e^{- \iota \left( \frac{\pi}{2n} \left( 1 + n - 1 - 2 n \left( n - 1 \right) \right) \right)} \\
&= e^{\iota \frac{\pi}{2} \left( 2n - 3 \right)}\end{align*}
|
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|
Condition when angle between two lines is ${\pi\over 3}$ The whole question is-
Show that if the angle between the lines whose direction cosines are
given $l+m+n=0$ and $fmn+gnl+hlm=0$ is ${\pi\over 3}$ then ${1\over
f}+{1\over g}+{1\over h}=0$.
I'm trying to solve the problem in the following manner-
From the first equation $n=-l-m$, substituting this value of $n$ in the second equation we get-
$fm(-l-m)+g(=l-m)l+hlm=0$
$\implies g\left({l\over m}\right)^2+(f+g+h)\left({l\over m}\right)+f=0$
Now, the roots of this equation are ${l_1\over m_1}$ and ${l_2\over m_2}$. So, product of them ${l_1\over m_1}.{l_2\over m_2}={f\over g}\implies {l_1 l_2\over f}={m_1 m_2\over g}$.
Similarly, we get ${m_1 m_2\over g}={n_1 n_2\over h}$.
Hence, ${l_1 l_2\over f}={m_1 m_2\over g}={n_1 n_2\over h}=K$(say)
Thus, $\cos {\pi\over3}=l_1 l_2+m_1 m_2+n_1n_2=K(f+g+h)\implies K=\frac{\sqrt{3}}{2(f+g+h)}$.
Now, I can't proceed further. I can't prove ${1\over f}+{1\over g}+{1\over h}=0$.
Can anybody solve the problem? Thanks for assistance in advance.
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In general if a, b, c and d, e, f are components of 2 vectors then the direction cosines are for the first vector :
$\frac{a}{\sqrt{a^2 + b^2 + c^2}}$ and $\frac{b}{\sqrt{a^2 + b^2 + c^2}}$ and $\frac{c}{\sqrt{a^2 + b^2 + c^2}}$ similarly for the second vector.
Then the angle $\theta$ between these vectors is
$$cos\theta = \frac{ad + be + cf}{\sqrt{a^2 + b^2 + c^2}\sqrt{d^2 + e^2 + f^2}}$$
Using the definition of dot product. In other words $cos\theta$ equals the sum of product of the directional cosines of corresponding vectors. In this question we have
$$cos\theta = l(fmn) + m(gnl) + n(hlm) = lmn(f + g + h)$$
The next step is to prove this equal 1/2.
From the question we have
$$l + m + n = 0$$
$$l^2 + m^2 + n^2 = 1$$
Since there are three unknowns but only 2 equations, we can only express l and n in terms of m :
$$\implies l^2 + lm + m^2 = \frac{1}{2}$$
$$\implies l = \frac{-m + D}{2}$$
$$\implies n = \frac{-m - D}{2}$$
Where $D = \sqrt{2 - 3m^2}$. It does not matter which root we take for l the other root will be for n.
Also we have other conditions:
$$fmn + gnl + hlm = 0 ... (1)$$
$$(fmn)^2 + (gnl)^2 + (hlm)^2 = 1 ... (2)$$
$$\frac{1}{f} + \frac{1}{g} + \frac{1}{h} = 0 \implies fg + gh + fh = 0... (3)$$
Put l and n into (1) we have
$$-fm(m + D) - g(1 - 2m^2) + hm(-m + D) = 0...(4)$$
Put l and n into (2) we have
$$f^2m^2(2 - 2m^2 + 2mD) + g^2(1 - 2m^2)^2 + h^2m^2(2 - 2m^2 -2mD) = 4...(5)$$
Eliminate f from (3) and (4) we have
$$f^2m(m + D) + fg(2mD - 2m^2 + 1) + g^2(1 - 2m^2) = 0$$
$$\implies f = \frac{2m^2 - 2mD - 1\pm\sqrt{(2mD -2m^2 + 1)^2 - 4(m + D)m(1 - 2m^2)}}{2m(m + D)}g$$
$$\implies f = \frac{2m^2 - 2mD - 1 \pm 1}{2m(m + D)}g$$
We take plus 1 for simplicity since the other root may give a result making $cos\theta$ greater than 1 or less than -1.
$$f = \frac{m - D}{m + D}g$$
Eliminate h from (4) and (5) gives
$$f^2m^2(2 - 2m^2 + 2mD) + g^2(1 - 2m^2)^2 + fgm(m + D)(1 - 2m^2) = 2$$
Put f into this equation we get
$$m^2(m - D)^2g^2 + g^2(1 - 2m^2)^2 + (m - D)mg^2(1 - 2m^2) = 2$$
$$\implies g^2(1 - Dm - m^2) = 2$$
$$\implies g = \sqrt{\frac{2}{1 - Dm - m^2}}$$
$$\implies f = \frac{m - D}{m + D}\sqrt{\frac{2}{1 - Dm - m^2}}$$
$$\implies h = \frac{D - m}{2m}\sqrt{\frac{2}{1 - Dm - m^2}}$$
Hence put l, m, n, f, g, h to the cosine expression we get
$$cos\theta = lmn(f + g +h) = \frac{-(m - D)}{2}m\frac{-(m + D)}{2}\sqrt{\frac{2}{1 - Dm - m^2}}(\frac{m - D}{m + D} + 1 + \frac{D - m}{2m})$$
$$=\frac{m - D}{4}\sqrt{\frac{2}{1 - Dm - m^2}}$$
$$=\frac{1}{4}\sqrt{\frac{2(m^2 + 2 - 3m^2 - 2Dm)}{1 - Dm - m^2}} = \frac{1}{2}$$
|
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|
Finding lengths when circles and squares tangents.
Should one approach by coordinates or by euclidean geometry?
By pure geometry, I am not able to solve.
|
As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $\triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.
Again, denote the side of the sqaure = $x$ and the radius of small semi circle = $r'$. So, the radius of larger semi circle = $r = \frac{x}{2}$.
From $\triangle QEF$, we get
$EQ^2 + QF^2 = EF^2$
$(x-r')^2 + (\frac{x}{2})^2 = (\frac{x}{2} + r')^2$
$x^2 -2xr' +r'^2 + \frac{x^2}{4} = \frac{x^2}{4} + xr' + r'^2 \implies x^2 = 3xr' \implies x = 3r'$
Hence, $r' = \frac{x}{3} \implies r' = \frac{2r}{3}$
$BD$ is the diagonal of the square $ABCD$ and $\angle CBD = \angle LBG = 45^\circ$. So, here we get that $\triangle GLB$ is an isosceles triangle.
Now, by the pythagorian theorem from $\triangle GLB$,
$GL^2 + LB^2 = GB^2 \implies GL^2 + GL^2 = 3^2 \implies 2GL^2 = 9 \implies GL = \frac{3}{\sqrt2}$
Next, $\triangle CFB \sim \triangle CGL$ and from both tne triangle it can be written that
$\frac{BC}{BF} = \frac{x}{y} = \frac{3y}{y} = 3$
and similarly,
$\frac{CL}{GL} = 3 \implies CL = 3 × \frac{3}{\sqrt2} \implies CL = \frac{9}{\sqrt2}$
From that, $CB = CL + LB = CL + GL = \frac{9}{\sqrt2} + \frac{3}{\sqrt2} = \frac{12}{\sqrt2} = 6\sqrt2$. So, the side of the square $ABCD$ = $x$ = $6\sqrt2$
After that,$\triangle CDE \sim \triangle HME$ and \triangle BAD \sim \triangle HMD$. From the similarity of first two triangles, we get
$\frac{CD}{HM} = \frac{DE}{ME}$
Likewise from the next similarity,
$\frac{AB}{HM} = \frac{AD}{MD} \implies \frac{CD}{HM} = \frac{CD}{MD}$.
So, we can write that
$\frac{DE}{ME} = \frac{CD}{MD}$
$\frac{r}{a} = \frac{2r}{r-a}$......(By denoting $ME = a$)
$2ra = r^2-ra \implies 6\sqrt2x = (3\sqrt2)^2 -3\sqrt2x \implies 6\sqrt2x = 18 - 3\sqrt2x = 9\sqrt2x = 18 \implies x = \frac{18}{9\sqrt2} \implies x = \sqrt2$
Then, $DM= 3\sqrt2 - a = 3\sqrt2 - \sqrt2 = 2\sqrt2$
Notice that, $\triangle DMH$ is an isosceles triangle and so,
$DH^2 = 2DM^2 = 2(2\sqrt2)^2 = 2×8 = 16$
And finally, we get $DH = \sqrt16 = 4$.
It could have been solved by any easier effort. But I made it very difficult. I hope that you will understand or if you have any problem, please let me know.
|
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|
$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$ I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$
The provided solution is $21\sqrt{6}$ but I arrive at a different amount.
Here is my working, trying to understand where I went wrong:
First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ * $6\sqrt{6}$ = $6*2*6\sqrt{6}$ = $72\sqrt{6}$
Second expression:
$7\sqrt{54}$ = $7\sqrt{9} * 7\sqrt{6}$ = $147\sqrt{6}$
Third expression is already the remaining common expression $12\sqrt{6}$.
So: $147\sqrt{6} + 72\sqrt{6} - 12\sqrt{6}$ = $207\sqrt{6}$
Where did I go wrong?
|
Notice while simplifying the radicals you are multiplying 6,7 twice...
|
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|
Solving a system of equations with trig How do you solve a system of the following form:
$$a = 2\sin(x) - \sin(y) + \sin(x+y)\tag1$$
$$b = 2\sin(y) - \sin(x) + \sin(x+y)\tag2$$
where $a,b$ are constants, and $x,y$ the variables I'd like to solve for.
Subtracting $(1)-(2)$ gives an expression for $\sin(y)$. However, rewriting
$$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$$ requires an expression for $\cos(y)$. Using $\cos(y)=\sqrt{1-\sin^2(y)}$ results in a complicated equation, which I cannot solve.
Is there an easier way to solve this system?
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You are on the right track. Subtracting $(2)$ from $(1)$ gives $\sin x=c+\sin y$ where $c=\frac{a-b}3$ and putting this into $(1)$ gives $$a=2c+\sin y+\sin x\cos y+\sin y\cos x$$ so $$\cos x=-1-\cos y+\frac{b+c-c\cos y}{\sin y}$$ Thus $$\small \sin^2x+\cos^2x\\=\\\small c^2+2c\sin y+\sin^2y+1+\cos^2y+\left(\frac{b+c-c\cos y}{\sin y}\right)^2+2\cos y-2\frac{b+c-c\cos y}{\sin y}-2\cos y\frac{b+c-c\cos y}{\sin y}$$ so $$\left(\frac{b+c-c\cos y}{\sin y}-1\right)^2-2\cos y\left(\frac{b+c-c\cos y}{\sin y}-1\right)+c^2+2c\sin y=0$$ giving $$\frac{b+c-c\cos y}{\sin y}-1=\frac{2\cos y\pm\sqrt{4\cos^2y-4(c^2+2c\sin y)}}2$$ and the half-angle identities give $$\frac b{\sin y}+c\tan\frac y2=2\cos^2\frac y2\pm\sqrt{1-(c+\sin y)^2}$$ I highly doubt there is an analytical solution to this, but once $y$ is found perhaps numerically, $x=\sin^{-1}(c+\sin y)$.
|
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|
What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder?
If $1000!$ is divided by $4^n$ with a remainder 0, what is the highest
possible value of $n$?
I placed 2, 3, 4, etc value in $n$ but didn't found any possible $4^n$. Moreover I have seen that only $4^1$ can divide 1000! without remainder. Is there any way to prove this or there are some possible $n$?
|
For a solution by hand: $1000!$ has
$500$ multiples of $2^{1}$
$250$ multiples of $2^{2}$
$125$ multiples of $2^{3}$
$62$ multiples of $2^{4}$
$31$ multiples of $2^{5}$
$15$ multiples of $2^{6}$
$7$ multiples of $2^{7}$
$3$ multiples of $2^{8}$
$1$ multiples of $2^{9}$
So $2^{1+3+7+15+...+500}=2^{994}=4^{497}|1000!$
|
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|
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
|
The probability generating function of a Binomiall random variable $X\sim \text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
$$
g_{X}(t)=Et^X=\sum_{k=0}^nP(X=k)t^k=\sum_{k=0}^n\binom{n}{k}\frac{t^k}{2^n}=\frac{1}{2^n}(1+t)^n
$$
In particular the probability that $X$ is even is given by
$$
\sum_{0\leq k\leq n\, k\,{\text{even}}}P(X=k)=\frac{g(1)+g(-1)}{2}=\frac{1+0}{2}=\frac{1}{2}.
$$
|
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|
Finding the sum of first $N$ squares with generating functions I'm trying to prove the well-known result
$$\sum_1^Nn^2=\frac{N(N+1)(2N+1)}6$$
using generating functions. The obvious choice is to set $f(x)=\sum_{n=1}^N n^2x^n$, so that we have $$f(x)=(xD)^2\sum_1^N 1x^n=(xD)^2\left(\frac{x^{N+1}-1}{x-1}\right).$$
It seems like it should be easy to find $f(x)$, then setting $x=1$ to find $f(1)$ should be a slam-dunk. Indeed, the text I'm referring to says that direct evaluation is the way and that "after doing two differentiations and a lot of algebra the answer emerges". Somehow, I can't do this. In particular, I get that
$$f(x)=\frac{xp(x)}{(x-1)^3},$$
where $p$ is a polynomial of degree $n$, and the $(x-1)$s at the denominator does not cancel with anything in the numerator. I'm puzzled how I'm now supposed to substitute $x=1$, since we end up dividing by zero. Can someone help?
|
\begin{align}
f(x)
&=(xD)^2\left(\frac{x^{N+1}-1}{x-1}\right)\\
&=(xD)(xD)\left(\frac{x^{N+1}-1}{x-1}\right)\\
&=x\left(xD\left(\frac{x^{N+1}-1}{x-1}\right)+D^2\left(\frac{x^{N+1}-1}{x-1}\right)\right)\\
&=x^2D\left(\frac{x^{N+1}-1}{x-1}\right)+xD^2\left(\frac{x^{N+1}-1}{x-1}\right)
\end{align}
If $t=x-1$:
\begin{align}
\frac{x^{N+1}-1}{x-1}
&=\frac{(t+1)^{N+1}-1}t\\
&=\sum_{k=1}^{N+1}\binom{N+1}kt^{k-1}
\end{align}
hence
\begin{align}
&\lim_{x\to 1}D\left(\frac{x^{N+1}-1}{x-1}\right)=\binom{N+1}2&
&\lim_{x\to 1}D^2\left(\frac{x^{N+1}-1}{x-1}\right)=2\binom{N+1}3&
\end{align}
thus
\begin{align}
\lim_{x\to 1}f(x)
&=\binom{N+1}2+2\binom{N+1}3\\
&=\frac{(N+1)N}2+\frac{(N+1)N(N-1)}3\\
&=\frac{3(N+1)N+2(N+1)N(N-1)}6\\
&=\frac{(N+1)N(2N+1)}6
\end{align}
|
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|
Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
My idea
I had the idea to check when $$\frac{n}{2^k} < \frac{1}{2}$$
because then $$ \forall_{k_0 \le k} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor=0$$
It should be $$ k_0 = \log_2(2n) $$ but I don't know how it helps me with this task (because I need not only "stop moment" but also sum of considered elements
Book idea
Let $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
then $$ S_n-S_{n-1} = 1$$
and then solve this recursion. But I write $S_n - S_{n-1}$ and I don't see how it can be $1$ , especially that is an infinite sum.
|
Since $S_1=1$ try to prove that $S_n=n$ by induction. Note that if $n=2m$ is even
$$\begin{align*}
S_n=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor
&=\sum_{k=1}^{\infty}\left\lfloor\frac{2m}{2^k}+\frac12\right\rfloor
=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{2^{k-1}}+\frac12\right\rfloor\\
&=\left\lfloor m+\frac12\right\rfloor +\sum_{k=2}^{\infty}\left\lfloor\frac{m}{2^{k-1}}+\frac12\right\rfloor\\
&=\left\lfloor m+\frac12\right\rfloor +\sum_{k=1}^{\infty}\left\lfloor\frac{m}{2^{k}}+\frac12\right\rfloor=m+S_m=m+m=n
\end{align*}$$
On the other hand if $n=2m+1$, then
$$\begin{align*}
S_n=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor
&=\sum_{k=1}^{\infty}\left\lfloor\frac{2m+1}{2^k}+\frac12\right\rfloor
=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{2^{k-1}}+\frac{1}{2^{k}}+\frac12\right\rfloor\\
&=\left\lfloor m+\frac12+\frac12\right\rfloor +\sum_{k=2}^{\infty}\left\lfloor\frac{m}{2^{k-1}}+\frac{1}{2^{k}}+\frac12\right\rfloor\\
&=m+1+\sum_{k=1}^{\infty}\left\lfloor\frac{m}{2^{k}}+\frac{1}{2^{k+1}}+\frac12\right\rfloor\\
&=m+1+S_m=m+1+m=n.
\end{align*}$$
where it remains to show that for all $k\geq 1$,
$$\left\lfloor\frac{m}{2^{k}}+\frac{1}{2^{k+1}}+\frac12\right\rfloor=\left\lfloor\frac{m}{2^{k}}+\frac12\right\rfloor.$$
Can you show this last step?
P.S. Actually $S_n$ is a finite sum. If $n<2^{N}$ then $\frac{n}{2^{N+1}}<\frac12$ and $\left\lfloor \frac{n}{2^{N+1}} + \frac12 \right\rfloor=0$. Hence
$$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac12 \right\rfloor=\sum_{k=1}^{N} \left\lfloor \frac{n}{2^k} + \frac12 \right\rfloor.$$
|
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|
Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$
Find the general formula of $a_{n+1}=\dfrac{a_n^2+4}{a_{n-1}}$ with $a_1=1$, $a_2=5$.
I have tried to write the recursion as a product, make summations, tried to look at patterns but its value grows very fast: $1,5,29,169,985,5741$… So I ran out of ideas.
|
If we start from $a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)$ we will get:
$$a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)\\ \Leftrightarrow\\ a_{n+1}^2+a_{n-1}a_{n+1} = a_n^2+a_{n}a_{n+2}\\ \Leftrightarrow\\ \frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{n}+a_{n+2}}{a_{n+1}}$$
Therefore we have $\frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{0}+a_{2}}{a_1} = 6$. After that, we can follow the solution of JV.Stalker:
$$a_{n+2}-6a_{n+1}+a_n = 0 \Rightarrow r^2-6r+1=0 \Rightarrow r_{1,2} = 3\pm2\sqrt{2}\Rightarrow a_n = c_1(3+2\sqrt{2})^n+c_2(3-2\sqrt{2})^n$$
We can define $a_0$ as $a_0a_2=a_1^2+4\Rightarrow a_0 = 1$. Therefore we have $c_1+c_2=1$ and $3(c_1+c_2)+2\sqrt{2}(c_1-c_2)=1$ and hence $c_1=\frac{2-\sqrt{2}}{4}$ and $c_2=\frac{2+\sqrt{2}}{4}$. If we plug this into the equation for $a_n$ we receive $$a_n = \frac{2-\sqrt{2}}{4}(3+2\sqrt{2})^n+\frac{2+\sqrt{2}}{4}(3-2\sqrt{2})^n\\ \Leftrightarrow\\ a_n = \frac{2-\sqrt{2}}{4}(\sqrt{2}+1)^{2n}+\frac{2+\sqrt{2}}{4}(\sqrt{2}-1)^{2n}\\ \Leftrightarrow\\ a_n = \frac{\sqrt{2}}{4}(\sqrt{2}+1)^{2n-1}+\frac{\sqrt{2}}{4}(\sqrt{2}-1)^{2n-1}$$
|
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|
Finding the least possible value of perimeter of $\triangle ABC$ with given ranges of angle
In $\triangle ABC$,$\angle A >2\angle B$ and $\angle C > 90^\circ$. If the length of all side of triangle $\triangle ABC$ are positive integers, then what is the least possible value of perimeter of $\triangle ABC$?
However, I can't think even of the length of the sides related with the possible values for all angle $\angle A, \angle B$ and $\angle C$. How can I construct the triangle and then get all the side having a length belonging to the positive integers? The problem was very weird for me and all of my effort can be hardly shown or described. And how can I get the minimum possible perimeter?
Thanks in advance.
|
In the standard notation we have:
$$a+b>c,$$
$$c^2>a^2+b^2$$ and
$$\frac{\sin\beta}{b}=\frac{\sin\alpha}{a}>\frac{\sin2\beta}{a}=\frac{2\sin\beta\cos\beta}{a},$$ which gives
$$\frac{a}{b}>\frac{a^2+c^2-b^2}{ac}$$ or
$$a^2(c-b)>b(c^2-b^2)$$ or
$$a^2>bc+b^2.$$
If $b=1$ we obtain $$c<a+1$$ or
$$c-a<1,$$ which is impossible.
Thus, $b\geq2$.
Now, $$a<c<\frac{a^2-b^2}{b},$$ which gives
$$a+1\leq c\leq\frac{a^2}{b}-b-1$$ and
$$a^2-ab-b^2-2b\geq0,$$ which gives
$$a\geq\frac{b+\sqrt{5b^2+8b}}{2}\geq4$$ and since $$c^2>4^2+2^2,$$ we obtain $$c\geq5$$ and $$a+b+c\geq11.$$
Can you end it now?
|
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|
How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
\begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 \Rightarrow f^{\prime}(x) = 3x^2 + 18x + 36 \Rightarrow 3x^2 + 18x + 36 = 9 \Rightarrow\\\\
& 3x^2 + 18x = -27 \Rightarrow 3x ( x + 6 ) = -27 \Rightarrow 3x = -27 x + 6 = -27 \Rightarrow
x = -3 x = -33
\end{align*}
|
You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
|
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|
In how many ways can $4$ boys and $6$ girls be put into $3$ groups of $3$ people so that there is a boy in each group? In how many ways can $4$ boys and $6$ girls be put into $3$ groups of $3$ people so that there is a boy in each group?
I tried by inclusion exclusion principle where: $$ A_{i} = \left\{\text{there is a boy in }i\text{-th group} \right\} $$ and I would look for the complement.
So my answer would be :
$$ \binom{10}{3}\binom{7}{3}\binom{4}{3} - \binom{3}{1}\binom{6}{3}\binom{7}{3}\binom{4}{3} + \binom{3}{2}\binom{6}{3}\binom{3}{3}\binom{4}{3} - 0$$
|
I would condition on whether the one left out is a boy or a girl. If the one left out is a boy, there is one boy in each group. You can choose the first group in $4{6 \choose 2}$ ways, the second in $3{4 \choose 2}$ and the third in $2{2 \choose 2}$ ways, giving $$4{6\choose 2}3{4\choose 2}2{2 \choose 2}=2160$$
If a girl is left out there must be two boys in one group and one in each other. You have $3$ ways to choose the group that gets two boys, $4 \choose 2$ ways to choose the boys, and $6$ ways to choose the girl. You can then choose the boy for the first remaining group in $2$ ways and the girls in $5 \choose 2$ ways. For the third group you have one boy and $3 \choose 2$ ways to choose the girls, giving
$$3{4\choose 2}6\cdot 2{5\choose 2}{3 \choose 2}=6480$$
That gives an overall total of $8640$
|
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|
A logarithmic integral, generalization of a result of Shalev As many of you are already aware, I and Marco Cantarini are currently working on the applications of fractional operators to hypergeometric series, extending the class of $\phantom{}_{p+1} F_p$s whose closed form is provided by FL-expansions (like the ones appearing here, soon in Bollettino UMI). That said, any insightful contribution here might easily result in a collaboration, or at least a citation, in our next work. Fractional operators allow to state that a closed form for any of the following integrals
$$ \int_{0}^{1}\frac{-\log u}{\sqrt{1+6u+u^2}}\,du \tag{A} $$
$$ \int_{0}^{1}\frac{\operatorname{arctanh}(u)}{\sqrt{(1-u^2)(2-u^2)}}\,du\tag{B} $$
$$ \int_{0}^{+\infty}\frac{z}{\sqrt{3+\cosh z}}\,dz\tag{C} $$
$$ \int_{0}^{1}\frac{K(x)}{\sqrt{x}(2-x)}\,dx\tag{D} $$
(where $K(x)=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-x\sin^2\theta}}$) result in a closed form for many $\phantom{}_3 F_2$s with quarter-integer parameters. However, we have not been able to find a closed form for any of the previous integrals. It is relevant to point out that
$$\int_{0}^{1}\frac{-\log x}{\sqrt{x}\sqrt{1-x\sin^2\theta}}\,dx = \frac{4}{\sin\theta}\left[\theta\log(2\sin\theta)+\frac{1}{2}\operatorname{Im}\operatorname{Li}_2(e^{2i\theta})\right]$$
for any $\theta\in\left(0,\frac{\pi}{2}\right)$ thanks to Shalev/nospoon, and the coefficients of the Maclaurin series of $\frac{1}{\sqrt{1+6u+u^2}}$ are given by central Delannoy numbers, i.e. Legendre polynomials evaluated at $3$. I guess this is enough context, so:
How can we express $(A),(B),(C)$ or $(D)$ in terms of standard mathematical constants and values of the $\Gamma$ function and polylogarithms?
Small update/context expansion: if we attack $(D)$ through Taylor series, the problem boils down to finding the twisted hypergeometric series
$$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{\mathscr{H}_n}{2^n},\qquad \mathscr{H}_n=\sum_{k=0}^{n}\frac{1}{2k+1} \tag{E} $$
while in order to tackle $(C)$ through $\int_{0}^{+\infty}\frac{x\,dx}{(\cosh x)^m}$ it might be worth to exploit integral representations for the Riemann $\zeta$ function and Dirichlet L-function $L(\chi_4,s)$, like
$$ \int_{0}^{+\infty}\frac{x^s}{\cosh^2 x}\,dx = \zeta(s)\frac{2(2^s-2)\Gamma(s+1)}{4^s}, $$
$$ \int_{0}^{+\infty}\frac{x^s}{\cosh x}\,dx = 2\,\Gamma(s+1)\,L(\chi_4,s+1)$$
and integration by parts.
March 17th 2019 Update: I realize there was a typo in the original question. $(A)$ should have been
$$ \int_{0}^{1}\frac{-\log u}{\sqrt{\color{red}{u}(1+6u+u^2)}}\,du \tag{A} $$
but I am confident that Shalev's substitution $u=\frac{(1-t)}{t(1+t)}$ simplifies the structure of the integral in this case, too.
|
I'll use the ingenious method of user @FDP from
the IntegralsAndSeries forum.
The idea is to make the following chain of substitutions:
$$ x = \frac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y} $$
$$ z = \frac{y}{2} - \frac{\pi}{8} $$
$$ t = \tan z $$
Together with the following observations:
$$ \frac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y} = \cot \left( \frac{y}{2}+\frac{\pi}{8}\right)\cot \left( \frac{y}{2}-\frac{\pi}{8}\right) \tag1 $$
$$ 1 + \sqrt{2} \cos y = 2 \sqrt{2} \cos \left( \frac{y}{2}+\frac{\pi}{8}\right)\cos \left( \frac{y}{2}-\frac{\pi}{8}\right) \tag2$$
$$ \sqrt{2} \cos z \cos \left( z + \frac{\pi}{4} \right) = \frac{ 1 - \tan z}{ 1 + \tan^2 z} \tag3$$
We find that
$$\begin{align}
I=\int_0^1 \frac{- \ln x}{\sqrt{1+ 6 x +x^2}} \mathrm{d}x
\\&= \int_1^\infty \frac{\ln x}{x\sqrt{1+ 6 x +x^2}} \mathrm{d}x
\\&= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\ln \left( \dfrac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y}\right)}{1+ \sqrt{2} \cos y} \mathrm{d}y
\\&= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln \left( \cot \left( \frac{y}{2}+\frac{\pi}{8}\right)\cot \left( \frac{y}{2}-\frac{\pi}{8}\right)\right)}{2 \sqrt{2} \cos \left( \frac{y}{2}+\frac{\pi}{8}\right)\cos \left( \frac{y}{2}-\frac{\pi}{8}\right)}\mathrm{d}y
\\&= \int_0^{\frac{\pi}{8}} \frac{\ln \left( \cot z \cot \left( z + \frac{\pi}{4} \right) \right)}{ \sqrt{2} \cos z \cos \left( z + \frac{\pi}{4} \right)} \mathrm{d}z
\\&= \int_0^{\frac{\pi}{8}} \frac{\ln \left( \frac{1}{\tan z} \frac{1-\tan z}{1+\tan z} \right)}{1- \tan z} (1+ \tan^2 z) \mathrm{d}z
\\&= \int_0^{\sqrt{2}-1} \frac{ \ln \left( \frac{1}{t} \frac{1-t}{1+t} \right)}{1-t} \mathrm{d}t
\\&= \frac{\pi^2}{12} +\frac12 \ln(2+\sqrt{2}) \ln(2-\sqrt{2}) + \operatorname{Li}_2\left(\frac{1}{\sqrt{2}}\right)-\operatorname{Li}_2 \left( 2-\sqrt{2}\right).
\end{align}$$
Where we just finished off with basic polylog stuff.
Edit.
Now that I think of it, we could have just made the simple Euler substitution $$\sqrt{ x^2 + 6 x +1} = x +t$$ to get
$$ I = \int_1^{2 \sqrt{2}-1} \frac{ \ln \left( \frac{1-t^2}{2 (t-3)} \right)}{t-3} \mathrm{d}t $$
And from here the integral can be reduced to polylogs with little difficulty.
|
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Is $n^2+3n+6$ divisible by 25, where $n$ is a integer? If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.
$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not divisible by $25$.
And so on for $n=5k+2, n=5k+3, n=5k+4.$
*
*Is this a good way to do this?
The other way that came to my mind would be following:
Let's say that $n^2+3n+6$ is divisible by $25$: $$n^2+3n+6=25k$$
$$n^2+3n+(6-25k)=0$$
If we solve this equation, we get $$n_{1,2}=\frac{-3 \pm \sqrt{5} \sqrt{20k-3}}{2}.$$ But $\sqrt{5}$ is a irrational number, so is $n$ irrational number too. This is contradiction, so it's not divisible by $25$.
*Is this a good way to solve this problem?
|
If $25$ divides $n^2+3n+6,$ the later will be divisible by $5$
$n^2+3n+6\equiv n^2-2n+1\pmod5$
So, we need $n\equiv1\pmod5\implies n=1+5m$
$(5m+1)^2+3(5m+1)+6=25m^2+25m+10\not\equiv0\pmod{25}$
|
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|
Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
where $\rho$ is the "Golden Ratio" $\rho = \frac{1+\sqrt{5}}{2}$.
But I can't find a way to prove it.
Prove
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
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Start with $$f(x) = \sum_{n=2}^\infty \frac{x^n}{n(n-1)} = x + (1-x) \log (1-x)$$ If $w$ is a primitive $5$-th root of unity then $$\frac{1}{5}\sum_{k=0}^4 w^{nk} = 1$$ if $5 | n$ and $0$ otherwise. Thus
$$\frac{1}{5}\sum_{k=0}^4 f(w^k) =\sum_{n> 2,\, 5 | n} \frac{1}{n(n-1)} = \sum_{n=1}^\infty \frac{1}{5n(5n-1)}$$
(use the limiting value $2$ for $f(1)$.)
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|
If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach :- Since rank is $2$ , So, the determinant of all the submatrices of order $3 \times 3$ must be zero.
So, $\begin{vmatrix} 1&1 &2 \\ 1&1 &1 \\ a&b &b \end{vmatrix}$ = $0$. After solving it, I am getting $a=b$
and $\begin{vmatrix} 1&2 &2 \\ 1&1 &3 \\ b&b &1 \end{vmatrix}$ = $0$. After Solving it , $b=\frac{1}{3}$
Now, Rank of a matrix is also defined as no. of non-zero rows in row echelon form of that matrix. So, If I convert it into Row Echelon form then
$\begin{bmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{bmatrix}$
Applying $R_{2} \leftarrow R_{2}-R_{1}$
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ a& b &b &1 \end{bmatrix}$
Now, Applying $R_{3}\leftarrow R_{3}-aR_{1}$
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&(b-a) &(b-2a) &(1-2a) \end{bmatrix}$
Now, to make rank of this matrix = $2$ means I should have 2 non-zero rows or all the elements of the last row must be zero simultaneously.
So, $(b-a) =0$ and $(b-2a) =0$ and $(1-2a) =0$
So, $b=a$ and $b=2a$ and $a=\frac{1}{2}$
Now, My doubt is how $b=a$ and $b=2a$ is possible here simultaneously and why this method is giving wrong result. I must be doing some mistake here but I am not getting what mistake I am doing. Please help.
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After getting$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&b-a&b-2a&1-2a\end{bmatrix},$$you can't just say (as you did) that the all the elements of the last row must be zero simultaneously. At this point, you should consider the cases $b=a$ and $b\neq a$. In the first case, you now have the matrix$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&0&-a&1-2a\end{bmatrix}$$and you carry on. If $b\neq a$, you divide the third row by $b-a$. You'll get$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&1&\frac{b-2a}{b-a}&\frac{1-2a}{b-a}\end{bmatrix}$$and, again, you carry on.
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What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:
*
*$x^2-1=(x+1)(x-1)$
*$\sqrt{4x-4}=2\sqrt{x-1}$
*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{ (x+1)(x-1) }+x-1}{ 2\sqrt{x-1} + (x+1)(x-1) }$
So how do I remove what causes the hole function not to become $\frac{0}{0}$ and solve the limit?
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The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.
$$\lim_{x\to 1^+}\dfrac{\sqrt{x^2-1}+x+1}{\sqrt{4x-4}+x^2-1}\to \infty$$
This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf
After the OP's edit: $$\dfrac{\sqrt{x^2-1}+x-1}{2\sqrt{x-1}+x^2-1}=\dfrac{\sqrt{x-1}}{\sqrt{x-1}}\cdot\dfrac{\sqrt{x+1}+\sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \\ \implies \lim_{x\to 1^+}\dfrac{\sqrt{x+1}+\sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}\to \dfrac{\sqrt{2}}{2}$$
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Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$ Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$
My try: Since $y=x^2+ax+a^2+6a$ is an open upward Parabola, the roots $\alpha,\beta$ should be distinct and satisfy $1 \lt \alpha \lt 2$ and $1 \lt \beta \lt 2$ implies both the roots are distinct and lies between $1$ and $2$.
So we have the following conditions:
$1.$ Since it should cut positive Y axis as both the roots are positive we have
$a^2+6a \gt 0$
$2.$ $D \gt 0$ $\implies$ $a^2+8a \lt 0$
$3.$ $f(1) \gt 0$ $\implies$ $a^2+7a+1 \gt 0$
$4.$ $f(2) \gt 0$ $\implies$ $a^2+8a+4 \gt 0$
How to solve all these four inequalities in a simpler way?
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Hint: if you reverse the role of $a$ and $x$ you get $$a^2+a(x+6)+x^2<0$$
Solution of coresponding equation (on $a$) is $$a_{1,2} = {-x-6\pm \sqrt{D}\over 2}$$
where $D= -3(x-6)(x+2)>0$ for $x\in(1,2)$. So $$a_1<a<a_2$$ and now you have to find the maximum and minimum of expression(s):
$${-x-6\pm \sqrt{D}\over 2}$$ while $x\in(1,2)$.
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|
$\sum_{k=0}^\infty\frac{ (-1)^{\left\lfloor\frac kn\right\rfloor}}{k+1}$ While testing a program I came across an interesting equality:
$$
\sum_{k=0}^\infty\frac{ (-1)^{\left\lfloor\frac kn\right\rfloor}}{k+1}
=\frac1n\left[\log2+\frac\pi2\sum_{k=1}^{n-1}\tan\frac{k\pi}{2n}\right],
$$
which generalises the well-known identity for $n=1$. Is there a simple way to prove it?
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For $n \in \mathbb{N}$ write $k = m n + l - 1$ with $m \in \mathbb{N}_0$ and $l \in \{1,\dots,n\}$. Then $\left\lfloor \frac{k}{n} \right\rfloor = m$, so
\begin{align}
S_n &\equiv \sum \limits_{k=0}^\infty \frac{(-1)^{\left\lfloor \frac{k}{n} \right\rfloor}}{k+1} \stackrel{1.}{=} \sum \limits_{l=1}^n \sum \limits_{m=0}^\infty \frac{(-1)^m}{m n +l} = \frac{1}{n} \left[\sum \limits_{m=0}^\infty \frac{(-1)^m}{m+1} + \sum \limits_{l=1}^{n-1} \sum \limits_{m=0}^\infty \frac{(-1)^m}{m + \frac{l}{n}}\right] \\
&\equiv \frac{1}{n} \left[\log (2) + \sum \limits_{l=1}^{n-1} a_l \right] ,
\end{align}
where
$$ a_l = \sum \limits_{m=0}^\infty \frac{(-1)^m}{m + \frac{l}{n}} \stackrel{2.}{=} \sum \limits_{j=0}^\infty \left[\frac{1}{2j + \frac{l}{n}} - \frac{1}{2j+1 + \frac{l}{n}}\right] , \, l \in \{1,\dots,n-1\}.$$
For $l \in \{1,\dots,n-1\}$ consider
\begin{align}
a_l + a_{n-l} &= \sum \limits_{j=0}^\infty \left[\frac{1}{2j + \frac{l}{n}} - \frac{1}{2j+1 + \frac{l}{n}} + \frac{1}{2j+1-\frac{l}{n}} - \frac{1}{2j+2 - \frac{l}{n}}\right] \\
&\stackrel{3.}{=} \frac{\pi}{2} \left[8 \frac{l \pi}{2n} \sum \limits_{j=0}^\infty \frac{1}{(2j+1)^2 \pi^2 - 4 \left(\frac{l \pi}{2n}\right)^2} + \frac{2n}{l\pi} + 2 \frac{l\pi}{2n} \sum \limits_{j=1}^\infty \frac{1}{\left(\frac{l \pi}{2n}\right)^2 - j^2 \pi^2}\right] \\
&= \frac{\pi}{2} \left[\tan \left(\frac{l\pi}{2n}\right) + \cot \left(\frac{l\pi}{2n}\right)\right] = \frac{\pi}{2} \left[\tan \left(\frac{l\pi}{2n}\right) + \tan \left(\frac{(n-l)\pi}{2n}\right)\right] .
\end{align}
The pole expansions of $\tan$ and $\cot$ are discussed here and here, respectively. This also works for $l = \frac{n}{2}$ if $n$ is even, so we obtain
$$ S_n = \frac{1}{n}\left[\log(2) + \frac{\pi}{2} \sum \limits_{l=1}^{n-1} \tan \left(\frac{l \pi}{2n}\right)\right]$$
as conjectured.
Justification of the rearrangements:
*
*For $L \in \{1,\dots,n\}$ and $M \in \mathbb{N}_0$ we have
$$ \sum \limits_{k=0}^{M n + L - 1} \frac{(-1)^{\left\lfloor \frac{k}{n} \right\rfloor}}{k+1} = \sum \limits_{l=1}^L \sum \limits_{m=0}^M \frac{(-1)^m}{m n +l} + \sum \limits_{l=L+1}^n \sum \limits_{m=0}^{M-1} \frac{(-1)^m}{m n +l} \stackrel{M \to \infty}{\longrightarrow} \sum \limits_{l=1}^n \sum \limits_{m=0}^\infty \frac{(-1)^m}{m n +l} \, . $$
Therefore, the sequence $ \left(\sum_{k=0}^{K} \frac{(-1)^{\left\lfloor k/n \right\rfloor}}{k+1}\right)_{K \in \mathbb{N}_0}$ is bounded and has exactly one limit point (namely $\sum_{l=1}^n \sum_{m=0}^\infty \frac{(-1)^m}{m n +l}$), to which it must converge.
*For $M \in \mathbb{N}_0$ we have
\begin{align} \sum \limits_{m=0}^M \frac{(-1)^m}{m + \frac{l}{n}} &= \sum \limits_{j=0}^{\left\lfloor \frac{M}{2}\right\rfloor} \frac{1}{2j + \frac{l}{n}} - \sum \limits_{j=0}^{\left\lfloor \frac{M-1}{2}\right\rfloor} \frac{1}{2j + 1 + \frac{l}{n}} \\
&= \sum \limits_{j=0}^{\left\lfloor \frac{M}{2}\right\rfloor} \left[\frac{1}{2j + \frac{l}{n}} - \frac{1}{2j+1 + \frac{l}{n}}\right] + \frac{\mathbf{1}_{2\mathbb{N}} (M)}{M+1 + \frac{l}{n}}
\end{align}
and letting $M \to \infty$ proves the asserted equality.
*For $J \in \mathbb{N}_0$ we can write
$$ \sum \limits_{j=0}^J \left[\frac{1}{2j + \frac{l}{n}} - \frac{1}{2j+2 - \frac{l}{n}}\right] = \frac{n}{l} - \frac{1}{2J + 2 - \frac{l}{n}} + \sum \limits_{j=1}^{J} \left[\frac{1}{2j + \frac{l}{n}} - \frac{1}{2j - \frac{l}{n}}\right] .$$
Both sides converge as $J \to \infty$ and the limits must be equal.
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Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equivalent to
$$\frac{1}{ab}+\frac{1}{cd} +ab+cd+\left(a+b\right)\left(c+d\right) \geq 8$$
But now I don't know how to change $ab,cd$ into the forms of $a+b$ and $c+d$. Moreover, from the last inequality, we have $\frac{1}{ab}+\frac{1}{cd} \geq 4$, whereas $\left(a+b\right)\left(c+d\right) \leq 4$. I cannot handle this. Please help me.
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Another way.
$$\frac{1}{ab}+\frac{1}{cd}-\frac{a^2+b^2+c^2+d^2}{2}=$$
$$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2+4-\left(ab+cd+\frac{a^2+b^2+c^2+d^2}{2}\right)=$$
$$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2+\frac{(a+b+c+d)^2}{4}-\frac{(a+b)^2+(c+d)^2}{2}=$$
$$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2-\frac{(a+b-c-d)^2}{4}.$$
Now, let $a+b\leq c+d$.
Thus, $0<a+b\leq2$ and by AM-GM $$\frac{1}{\sqrt{ab}}-\sqrt{ab}=\frac{1-ab}{\sqrt{ab}}\geq\frac{1-\left(\frac{a+b}{2}\right)^2}{\sqrt{ab}}\geq0.$$
Id est, it's enough to prove that:
$$\frac{1}{\sqrt{ab}}-\sqrt{ab}\geq\frac{c+d-a-b}{2}$$ or
$$1-ab\geq\sqrt{ab}(2-a-b)$$ or
$$\sqrt{ab}(a+b)-2ab+ab-2\sqrt{ab}+1\geq0$$ or
$$\sqrt{ab}(\sqrt{a}-\sqrt{b})^2+(\sqrt{ab}-1)^2\geq0$$ and we are done!
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Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing from David K's answer:
One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
which is the number of ways to put $3$ or fewer indistinguishable balls in $3$ numbered bins without that constraint, which is the number of ways of putting exactly $3$ indistinguishable balls in $4$ numbered bins.
Multiply by $8$ to take into account all the cases where $x < 0$ or $y < 0$ or $z < 0.$
$x,y,z >0 => x + y + z <= 3, n = Cr(3+4-1,3) * (8) = 160$
Now let $x = 0,$ $y > 0,$ and $z > 0.$
Count the number of ways to put up to $6$ balls in $2$ bins if each bin must contain at least one ball. Multiply by four to account for all the cases were $y < 0$ or $z < 0.$ Multiply that result by $3$ to account for the fact that we could have chosen $y= 0$ or $z=0$ instead of $x = 0.$
$x=0, x,y>0 => y+z <=6, n = Cr(4+3-1, 4) * 3 * 14 = 180$
Now let $x = y = 0$ and $z > 0.$ There are $6$ ways for that to happen. Multiply by $2$ to account for $z < 0,$ then by $3$ to account for the other choices of which variables are zeros.
$x,y = 0, z>0 => z<=6, n = 6 * 2 * 3 = 36$
$x,y,z=0, n = 1$
Sum of all of them is : 160 + 180 + 36 + 1 = 377
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This is $f(6)$ where $f(n)$ for integers $n\ge0$ is the number of lattice
points in the polyhedron $nP$. This is the $n$-fold dilate of $P$, the polyhedron with vertices $(\pm1,0,0)$, $(0,\pm1,0)$ and $(0,0,\pm1)$. By Ehrhart's theorem, $f$ is a degree $3$ polynomial in $n$. Moreover, its leading coefficient is the volume of $P$, namely $\frac43$. We also have $f(0)=1$
and $f(1)=7$. But by Macdonald's reciprocity law $f(-n)$ is the negative
of the number of lattice points in the interior of $nP$, so that $f(-1)=-1$.
The only polynomial satisfying all these conditions is
$$f(n)=\frac43n^3+2n^2+\frac83n+1.$$
Then $f(6)=377$.
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Sum of all real numbers $x$ such that $(\text{A quadratic})^\text{Another quadratic}=1$.
What is the sum of all real numbers $x$ such that
$(x^2-5x+5)^{(x^2-7x+12)}=1$?
So I know that $x^0=1$ and $1^x=1$. So, I can solve for them and find $x$, and add them up.
Solving $x^2-7x+12=0$ for $x^0=1$ gives $x=3, 4$.
Solving $x^2-5x+5=1$ for $1^x=1$ gives $x=1, 4$.
Adding them up gives $1+3+4=8$.
This is wrong. What did I do wrong? Did I miss a case? If so, what case have I missed?
Thanks!
Max0815
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Hint
You have missed $$(-1)^{\text{even number}}=1$$
Now if $x^2-5x+5=-1,x=?$
Which values of $x$ make the exponent even?
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How to prove that $\frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$ when $0$$\frac{a-b}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$$ when $0<b<a$
This might relate to the mean value theorem, but I just can't prove it.
This question was put on hold as off-topic, I couldn't understand the reason why people voted to close it so I add more details and try to re-open it.
After I learned the course of the mean value theorem, my teacher asked us to prove that $$\frac{a-b}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}} < \arctan{a}- \arctan{b} < a-b $$
I found by using the equation $\arctan a - \arctan b = \frac{1}{1+\xi^2}(a-b)$ I could easily prove that $$\frac{a-b}{1+a^2} < \arctan{a}- \arctan{b} < a-b $$
the right inequality related to several questions on StackExchange so I omitted it.
I also tried to use the inequality $\frac{x}{1+x^2} < \arctan x < x$ combined with the equation $\arctan x - \arctan y = \arctan{\frac{x-y}{1+xy}}$ to prove this question but I failed again.
So I went to check if this question is correctly written and my teacher said "Yes, nothing wrong with it".
I found the comment of @YuDing is more useful than the answer of @AdamLatosiński so I didn't tick the answer and I also couldn't tick the comment because it's just a comment.
The comment of @MartinR and the answer of @Matteo provided us a great perspective to solve the question. I ticked the answer for the reason that I couldn't tick a comment. Maybe because the answer is on a purely geometrical perspective so my question is off-topic?
I hope OP could re-open this question because I really appreciate everyone's efforts here. Thank you all.
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You can write right side as
$$ \arctan a - \arctan b = \int_b^a \frac{1}{1+x^2}dx $$
and left side as
$$ \frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} = \int_b^a \frac{\partial}{\partial x}\left(\frac{x-b}{\sqrt{1+x^2}\sqrt{1+b^2}}\right) dx = \int_b^a \frac{xb+1}{(1+x^2)^\frac32\sqrt{1+b^2}} dx$$
Therefore the inequality you want to prove can be written as
\begin{align} 0 &< \int_b^a \left(\frac{1}{1+x^2}-\frac{xb+1}{(1+x^2)^\frac32\sqrt{1+b^2}}\right) dx = \\
&\quad = \int_b^a \frac{\sqrt{1+x^2}\sqrt{1+b^2}-(xb+1)}{(1+x^2)^\frac32\sqrt{1+b^2}} dx = \\
&\quad = \int_b^a \frac{(1+x^2)(1+b^2)-(xb+1)^2}{(1+x^2)^\frac32\sqrt{1+b^2}\big(\sqrt{1+x^2}\sqrt{1+b^2}+(xb+1)\big)} dx = \\
&\quad = \int_b^a \frac{(x-b)^2}{(1+x^2)^\frac32\sqrt{1+b^2}\big(\sqrt{1+x^2}\sqrt{1+b^2}+1+xb\big)} dx\end{align}
which is true because $b<a$ and the integrated function is positive.
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|
A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken? A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?
I don't understand why is this correct.
$$\frac{{20 \choose 6}+{20 \choose 10}}{24 \choose 10} $$
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Because $\tfrac{\binom 44\binom{20}6\binom{14}{14}}{\binom{24}{10}\binom{14}{14}}$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $\tfrac{\binom{20}{10}\binom{4}{4}\binom{10}{10}}{\binom{24}{10}\binom{14}{14}}$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.
$$~\\\dfrac{\binom 44\binom{20}6\binom{14}{14}+\binom{20}{10}\binom{4}{4}\binom{10}{10}}{\binom{24}{10}\binom{14}{14}}=\dfrac{\binom{20}6+\binom{20}{10}}{\binom{24}{10}}$$
|
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series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow \frac{\pi}{2}-x$
$\displaystyle I=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\ln(1-k\sin x)dx$
$\displaystyle I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\bigg[k\sin x-\frac{k^2\sin^2 x}{2}+\frac{k^3\sin^3 x}{3}-\cdots \bigg]dx$
$\displaystyle I =-2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{k^2\sin^2 x}{2}+\frac{k^4\sin^4 x}{4}+\cdots \bigg]dx$
$\displaystyle I =-\pi\bigg[\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots \bigg]$
How can I find sum of that series. Help me please.
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\begin{align}
&\bbox[10px,#ffd]{{1 \over 2}\cdot{1 \over 2}\, k^{2} +
{1 \over 2}\cdot{3 \over 4}\, k^{4} +
{1 \over 2}\cdot{3 \over 4}\cdot{5 \over 6}\, k^{6} + \cdots} =
\sum_{n = 1}^{\infty}{1 \over 2}
\pars{\prod_{j = 1}^{n}{2j - 1 \over 2j}}k^{2n}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{\prod_{j = 1}^{n}\pars{j - 1/2} \over \prod_{m = 1}^{n}m}k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}
{\pars{1/2}^{\overline{n}} \over n!}\,k^{2n}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}
{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!}\, k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}
{\pars{n - 1/2}! \over n!\pars{-1/2}!}\,k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}
{n - 1/2 \choose n}k^{2n}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{{-n + 1/2 + n - 1\choose n}\pars{-1}^{n}}k^{2n} =
{1 \over 2}\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-k^{2}}^{n}
\\[5mm] = &\
{1 \over 2}\braces{\bracks{1 + \pars{-k^{2}}}^{-1/2} - 1} =
\bbx{{1 \over 2}\pars{{1 \over \root{1 - k^{2}}} - 1}}
\end{align}
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"url": "https://math.stackexchange.com/questions/3161755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Evaluate $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$ By accident, I find this summation when I pursue the particular value of $-\operatorname{Li_2}(\tfrac1{2})$, which equals to integral $\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$.
Notice this observation
$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$
And using the Taylor series of $\ln(1-x^{2})$, I found this summation $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$, where $H_{n}$ is the harmonic-numbers.
If the value of $\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$ is given, the result can be easily deduced, which is
$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$
For the original goal is to calculate $\operatorname{Li_2}(\tfrac1{2})$, I expect some other approaches to the summation without using the value of $\operatorname{Li_2}(\tfrac1{2})$. I already knew some famous problem like Euler's Sum, which holds very similar form to this summation, but still in trouble finding the appropriate path.
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Here is an approach that avoids knowing the value of $\operatorname{Li}_2 (\frac{1}{2})$.
Let
$$S = \sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ).$$
Observing that
$$\int_0^1 \frac{x^{2n}}{1 + x} \, dx = H_n - H_{2n} + \ln 2,$$
your sum can be rewritten as
\begin{align}
S &= -\int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{x^{2n}}{n} \, dx\\
&= \int_0^1 \frac{\ln (1 - x^2)}{1 + x} \, dx\\
&= \int_0^1 \frac{\ln (1 + x)}{1 + x} \, dx + \int_0^1 \frac{\ln (1 - x)}{1 + x} \, dx\\
&= I + J.
\end{align}
For the first of the integrals $I$, one has
$$I = \frac{1}{2} \ln^2 2.$$
Now consider $J - I$. Then
$$J - I = \int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{1 + x}.$$
Employing a self-similar substitution of $t = (1-x)/(1+x)$ leads to
\begin{align}
J - I &= \int_0^1 \frac{\ln t}{1 + t} \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 t^n \ln t \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 t^{n + s} \, dt \right ]_{s = 0}\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= -\underbrace{\sum_{n = 0}^\infty \frac{(-1)^n}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\
&= -\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^2}\\
&= -\sum_{n = 1}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= -\frac{1}{2} \zeta (2).
\end{align}
Thus
$$J = I - \frac{1}{2} \zeta (2) = \frac{1}{2} \ln^2 2 - \frac{1}{2} \zeta (2).$$
Since $S = I + J$, we immediately see that
$$\sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ) = \ln^2 2 - \frac{1}{2} \zeta (2),$$
as desired.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3164280",
"timestamp": "2023-03-29T00:00:00",
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Number of solutions to the equation $x_1 + x_2 + \ldots + x_n = k$ when $0 \leq x_i \leq m$ and $m + 1 \leq k \leq 2m + 1$ The problem:
Find the number of solutions to the equation: $x_1+x_2+...+x_n=k$ when $0\leq x_i\leq m$ and $m+1\leq k\leq 2m+1$.
The answer in the book is ${n+k-1\choose k}-{n\choose 1}{n+k-(m+1)-1\choose k-(m+1)}$.
I understand that the idea is taking all solutions, a number equals to ${n+k-1\choose k}$ and then subtracting "bad" solutions, and that ${n\choose 1}$ is there because we can't have more than a single $i$ so $x_i\geq m+1$, what I can't understand is the last part. I think it works only if the "bad" element equals exactly $m+1$, but what if it equals $m+2$? That way it doesn't seem correct anymore.
Thanks in advance.
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A solution of the equation in the nonnegative integers
$$x_1 + x_2 + \ldots + x_n = k \tag{1}$$
corresponds to the placement of $n - 1$ addition signs in a row of $k$ ones. For instance, if $n = 4$ and $k = 10$,
$$1 1 + + 1 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 2$, $x_2 = 0$, $x_3 = 5$, $x_4 = 3$. The number of such solutions is
$$\binom{k + n - 1}{n - 1} = \binom{k + n - 1}{k}$$
since we must choose which $n - 1$ of the $k + n - 1$ positions required for $k$ ones and $n - 1$ addition signs will be filled with addition signs or, equivalently, which $k$ of the $k + n - 1$ positions will be filled with ones.
We wish to solve equation 1 in the nonnegative integers not larger than $m$ when $m + 1 \leq k \leq 2m + 1$. Thus, we must subtract those solutions in which a variable exceeds $m$. At most one variable may exceed $m$ since $2(m + 1) = 2m + 2 > 2m + 1$.
We may choose a variable that exceeds $m$ in $n$ ways. Suppose that variable is $x_1$. Then $x_1' = x_1 - (m + 1)$ is a nonnegative integer. Substituting $x_1' + m + 1$ into equation 1 equation yields
\begin{align*}
x_1' + m + 1 + x_2 + \ldots + x_n & = k\\
x_1' + x_2 + \ldots + x_n & = k - (m + 1) \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{k - (m + 1) + n - 1}{n - 1} = \binom{k - (m + 1) + n - 1}{k - (m + 1)}$$
solutions. Hence, there are
$$\binom{n}{1}\binom{k - (m + 1) + n - 1}{n - 1} = \binom{n}{1}\binom{k - (m + 1) + n - 1}{k - (m + 1)}$$
solutions in which one of the variables exceeds $m$.
Thus, there are
$$\binom{k + n - 1}{k} - \binom{n}{1}\binom{k - (m + 1) + n - 1}{k - (m + 1)}$$
admissible solutions.
Notice that in equation 2, $m + 1 \leq x_1 \leq 2m + 1 \implies 0 \leq x_1' \leq m$. It does not imply that $x_1 = m + 1$.
Let's compare this with what would happen if $x_1 = m + 1$. Then we would have
\begin{align*}
m + 1 + x_2 + \ldots + x_n & = k\\
x_2 + \ldots + x_n & = k - (m + 1)
\end{align*}
which is an equation in the nonnegative integers with
$$\binom{k - (m + 1) + (n - 1) - 1}{(n - 1) - 1} = \binom{k - (m + 1) + (n - 1) - 1}{k - (m + 1)}$$
solutions, which is a smaller number as we would expect.
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"url": "https://math.stackexchange.com/questions/3167115",
"timestamp": "2023-03-29T00:00:00",
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Find the average value of $f$ on the given interval $f(x) = (x-3)^2$, $[2,5]$ After finding the average value of $f$, I need to find $c$ such that $f_\text{avg}= f(c)$
$$f(x) = (x-3)^2 \textrm{, }[2,5]$$
$$\begin{align}
f_\text{avg} &= \frac{1}{5-2} \int_{2}^{5} (x-3)^2dx \\
u = x-3, du = dx & ~~~~~ \frac{1}{3} \int_{2}^{5} u^2du = \frac{1}{3} \bigg[ \frac{1}{3}(x-3)^3 \bigg]_{2}^{5} \\
&= \frac{1}{3} \bigg(\frac{1}{3}(5-3)^3 - (\frac{1}{3}(2-3)^3\bigg) \\
&= \frac{1}{3} \bigg( \frac{8}{3} + \frac{1}{3} \bigg)\\
&= 3
\end{align}$$
The textbook says that $f_{avg} = 1$, not 3.
Did I mess up my u-substitution?
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Just the comments have pointed out, you miscalculated the last line. It should be
$$\frac{1}{3}\cdot\Big(\frac{8}{3}+\frac{1}{3}\Big)=\frac{1}{3}\cdot\frac{9}{3}=1\neq3$$
To solve the second part, it should be an easy substitution
$$(x-3)^2=1 \iff x-3=\pm1 \iff x=4,2$$
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"timestamp": "2023-03-29T00:00:00",
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|
show that if $\frac{2(m^2+n^2+n)+1}{m(2n+1)}=k$, k is the multiple of 3 $$\frac{2(m^2+n^2+n)+1}{(2n+1)m}=k$$
where k, m, and n are all integer.
show that $3|k$
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Write it as
$$
2(m^2+n^2+n)+1=k(m(2n+1))
$$
Compute both sides mod $3$, that is, for $m,n=0,1,2$.
It turns out that there are two cases:
*
*The LHS is not zero and the RHS is zero. This cannot happen.
*The LHS is zero and the RHS is not zero. This implies that $k \equiv 0 \bmod 3$.
$$
\begin{array}{cccc}
m & n & 2(m^2+n^2+n)+1 & m(2n+1) \\
0 & 0 & 1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 2 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 1 & 1 & 0 \\
1 & 2 & 0 & 2 \\
2 & 0 & 0 & 2 \\
2 & 1 & 1 & 0 \\
2 & 2 & 0 & 1 \\
\end{array}
$$
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"url": "https://math.stackexchange.com/questions/3173116",
"timestamp": "2023-03-29T00:00:00",
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|
Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$
The solution proposes that $x^2+y^2+1=k(xy)$ where $k$ is an integer.
It claims that there exists a minimum solution $(x,y)$ that has the minimum value of $x+y$.
So, they use $t$ to replace $x$ to show that $t^2-kty+y^2+1=0$
Then $t_1=x$ is one solution. By vieta formula,$t_1+t_2=ky$
Then $t_2=ky-x=\frac{x^2+y^2+1}{x}-x=\frac{y^2+1}{x}$
which implies $t_2\lt y$ then $t_1+t_2\lt x+y$.
So, the minimum condition only exists when $x=y$
I am ok as far, but after that it says, $x^2$ divided by $2x^2+1$, $x^2$divided by $1$.
So $k=3$.
But, why they can get $k=3$? $k=3$ only when $x$ and $y$ be the minimum solution. Why $k$ cannot be multiple of $3$?
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The proof you've given is incomplete.
You should start assuming $k \neq 3$, so you get $x \neq y$ (otherwise $k=3$ which is absurd).
Then, we have
$x \geq y+1$,
$x^2 \geq (y+1)^2 > y^2+1$,
$x>\frac{y^2+1}{x}=t_2$.
So $(t_2, y)$ is another solution to $\frac{x^2+y^2+1}{xy}=k$ with $k \neq 3$. But $t_2+y<x+y$, which contradicts the minimality condition.
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"url": "https://math.stackexchange.com/questions/3174167",
"timestamp": "2023-03-29T00:00:00",
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|
Proving $\cos^{2}\frac{\pi}{2n} \cos^{2}\frac{3\pi}{2n}\cdots \cos^{2}\frac{\left(n-2\right)\pi}{2n}=n2^{1-n}$ using complex numbers Using $$z^{2n}+1 = \prod_{k=1}^{n}\left(z^2-2z\cos{\left(\frac{\left(2k-1\right)\pi}{2n}\right)}+1\right)$$
and $$z^{2n}+1 = \left(1+z^2\right)\left(1-z^2+z^4-\cdots+z^{2n-2}\right)$$
prove, for $n$ odd, $$\cos^{2}{\left(\frac{\pi}{2n}\right)} \cos^{2}{\left(\frac{3\pi}{2n}\right)} \cos^{2}{\left(\frac{5\pi}{2n}\right)} \cdots \cos^{2}{\left(\frac{\left(n-2\right)\pi}{2n}\right)} = n2^{1-n}$$
The provided solution (from the original source of this problem) involves noting that in the first result, the product will contain a factor of $\left(1+z^2\right)$, where $k=\frac{n+1}{2}$ (which is an integer since $n$ is odd). Thus, substituting in the second result, they cancel the factor of $\left(1+z^2\right)$ on both sides, and then set $z=i$ to deduce the required result.
But my question is, since $z=i$ gives $1+z^2=0$, are we allowed to make this cancellation and then still set $z=i$, since we seem to be dividing by zero? If there is in fact a problem, can their proof be easily repaired?
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There is no problem; you can view the identity as an equality of polynomials in $\Bbb{C}[z]$. This is an integral domain, meaning that you can cancel non-zero factors; the factor $1+z^2$ is a non-zero polynomial, so you can cancel it.
This gives you a simpler equality of polynomials. You can then evaluate these two polynomials in $z=i$ to obtain the given identity.
More formally; combining the given expressions for $z^{2n}+1$ shows that
\begin{eqnarray*}
0&=&\left(1+z^2\right)\left(1-z^2+z^4-\cdots+z^{2n-2}\right)
-\prod_{k=1}^{n}\left(z^2-2z\cos{\left(\frac{\left(2k-1\right)\pi}{2n}\right)}+1\right)\\
&=&(1+z^2)\left(1-z^2+z^4-\cdots+z^{2n-2}
-\prod_{k=1}^{n-1}\left(z^2-2z\cos{\left(\frac{\left(2k-1\right)\pi}{2n}\right)}+1\right)\right),\\
\end{eqnarray*}
as polynomials. This means one of the two factors is the zero polynomial. Clearly this is not $1+z^2$, so
$$1-z^2+z^4-\cdots+z^{2n-2}
-\prod_{k=1}^{n-1}\left(z^2-2z\cos{\left(\frac{\left(2k-1\right)\pi}{2n}\right)}+1\right)=0,$$
which shows that indeed
$$1-z^2+z^4-\cdots+z^{2n-2}
=\prod_{k=1}^{n-1}\left(z^2-2z\cos{\left(\frac{\left(2k-1\right)\pi}{2n}\right)}+1\right).$$
Now plugging in $z=i$ yields the desired identity.
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Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$
Find the degree of splitting field of $f=X^4+2X^2+2$ over $\mathbf{Q}$.
By Eisenstein, $f$ is irreducible. By setting $Y=X^2$, we can solve for the roots: $Y=-1\pm i \iff X=\sqrt[4]{2}e^{a\pi i/8}$, $a\in\{3,5,11,13\}$. Clearly $f$ splits in $\mathbf{Q}(\sqrt[4]{2},\zeta_{16})$. $\zeta_{16}$ is a zero of $X^8+1$, which is irreducible over $\mathbf{Q}$ by Eisenstein applied to $(X+1)^8+1$. I was not able to go further.
Could someone help me to proceed?
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Let $\Omega$ be the OP's splitting field.
We have a chain of 3 field extensions,
$$\tag 1 \mathbb Q \subset \mathbb Q(\sqrt 2) \subset Q(\sqrt 2)\,(\sqrt{1 + \sqrt 2}) \subset Q(\sqrt 2)\,(\sqrt{1 + \sqrt 2})\,(i) = \mathbb F$$
It is easy to explain why the first and last extensions have degree two. For the $2^\text{nd}$ extension we have to show that
$\tag 2 x^2 = {1 + \sqrt 2}$
has no solutions over $\mathbb Q(\sqrt 2)$. If there was a solution then there exists $a,b \in \mathbb Q$ such that
$\tag 3 (a + b\sqrt2)^2 = a^2 +2ab\sqrt2 + 2b^2 = a^2+ 2b^2 + 2ab\sqrt2 = {1 + \sqrt 2}$
Then both $a$ and $b$ are nonzero and $a = \frac{1}{2b}$ and since $a^2+ 2b^2 =1$
$\tag 4 8b^4 - 4b^2 + 1 = 0$
But no rational number $b \in \mathbb Q$ satisfies the equation $\text{(4)}$.
We conclude that the field $\mathbb F$ has degree $8$ over the rationals.
We can write out the four roots $x_1, x_2, x_3 x_4 \in \Omega$ of the quartic $x^4+2x^2+2$ using the $a + bi$ rectangular format,
\begin{align}x_1&=\sqrt{\frac{-1+\sqrt2}{2}}+i\sqrt{\frac{1+\sqrt2}2}\\ x_2&=-\sqrt{\frac{-1+\sqrt2}{2}}-i\sqrt{\frac{1+\sqrt2}2}\\ x_3&=\sqrt{\frac{-1+\sqrt2}{2}}-i\sqrt{\frac{1+\sqrt2}2}\\ x_4&=-\sqrt{\frac{-1+\sqrt2}{2}}+i\sqrt{\frac{1+\sqrt2}2}\end{align}
Since $\sqrt{1 + \sqrt 2} \; \sqrt{\sqrt 2 -1} = 1$, it is now easy to see that all the roots belong to $\mathbb F$, so $\Omega \subset \mathbb F$.
In a chain of logic,
$$ x_1 x_3 = \sqrt 2 = \in \Omega$$
$$ \sqrt 2 \, \frac{x_1 + x_3}{2} = \sqrt{-1+\sqrt2} \in \Omega$$
$$ (\sqrt{-1+\sqrt2})^{-1} = \sqrt{1+\sqrt2} \in \Omega$$
$$ \sqrt 2 \, \frac{x_1 + x_4}{2} \sqrt{-1+\sqrt2} = i \in \Omega$$
Since $\sqrt 2, ,(\sqrt{1 + \sqrt 2}), i \in \Omega$, $\mathbb F \subset \Omega$.
We conclude that $\Omega = \mathbb F$
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"timestamp": "2023-03-29T00:00:00",
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|
Exercise XIX number 15 - Calculus Made Easy $$
\text{Use substitution}\quad\frac{1}{x}=\frac{b}{a}cosh\;u\quad\text{to show that}\quad\\
\int\;\frac{dx}{x\sqrt{a^2-b^2x^2}}=\frac{1}{a}\ln\frac{a-\sqrt{a^2-b^2x^2}}{x}\;+\;C.\\
\text{My way:}\\
\text{Let}\quad\frac{1}{x}=\frac{b}{a}cosh\;u\quad\\
x=\frac{a}{b}\frac{1}{cosh\;u}\\
dx=-\frac{a}{b}\frac{sinh\;u}{cosh^2u}du\\
x^2=\frac{a^2}{b^2}\frac{1}{cosh^2u}\\
u=cosh^-(\frac{a}{bx})\\
\int\frac{\frac{b\;cosh\;u}{a}\cdot-\frac{a\;sinh\;u}{b\;cosh^2u}}{\sqrt{a^2-b^2\cdot\frac{a^2}{b^2\;cosh^2u}}}du=\\
=\int\frac{-\frac{sinh\;u}{cosh\;u}=-tanh\;u}{\sqrt{a^2-\frac{a^2}{cosh^2u}}=\sqrt{a^2(1-\frac{1}{cosh^2u})}=\sqrt{a^2\;tanh^2u}=a\;tanh\;u}du=\\
=\int-\frac{du}{a}=\quad-\frac{1}{a}\int du=\quad -\frac{u}{a}\quad+\quad C
=\quad-\frac{1}{a}cosh^-(\frac{a}{bx})\quad+\quad C=\\
=\quad-\frac{1}{a}\ln\Biggr(\frac{a}{bx}\pm\sqrt{\frac{a^2}{b^2x^2}-1}\Biggr)\quad+\quad C=\\
=\quad-\frac{1}{a}\ln\Biggr(\frac{a\pm\sqrt{a^2-b^2x^2}}{bx}\Biggr)\quad+\quad C=\\
=\quad-\frac{1}{a}\biggr(\ln(a\pm\sqrt{a^2-b^2x^2})-\ln(bx)\biggr)\quad+\quad C=\\
=\quad\frac{1}{a}\biggr(\ln(bx)-\ln(a\pm\sqrt{a^2-b^2x^2})\biggr)\quad+\quad C=\\
=\quad\frac{1}{a}\ln\frac{bx}{a\pm\sqrt{a^2-b^2x^2}}\quad+\quad C\\\text{What went wrong?}
$$
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You only have a minor error: $\cosh^-(\frac{a}{bx})= \ln\left(\frac{a}{bx} + \sqrt{\frac{a^2}{b^2x^2} - 1}\right)$, the minus-sign is not correct. This gives you
$$\int\;\frac{dx}{x\sqrt{a^2-b^2x^2}}\quad = \quad\frac{1}{a}\ln\frac{bx}{a+\sqrt{a^2-b^2x^2}}+ C \\ = \quad \frac{1}{a}\ln\frac{b^2x}{a+\sqrt{a^2-b^2x^2}} - \frac{1}{a}\ln b + C
$$
But
$$\frac{b^2x}{a+\sqrt{a^2-b^2x^2}} \quad = \quad\frac{b^2x(a-\sqrt{a^2-b^2x^2})}{(a+\sqrt{a^2-b^2x^2})(a-\sqrt{a^2-b^2x^2})} \\
= \quad\frac{b^2x(a-\sqrt{a^2-b^2x^2})}{a^2 -(a^2-b^2x^2)} \quad = \quad\frac{a-\sqrt{a^2-b^2x^2}}{x}$$
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|
If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Is it necessarily the case that $m=1$ and that these two divisors are $2^k-1$ and $2^k+1$?
I've tested this up to $y\leq10^{10}$ but I haven't been able to make much progress with standard number theoretic techniques.
If $k=1$ then there are infinitely many solutions of the form $x=y-1$.
Let $(1)$ be the initial version of the problem and let $(2)$ be the supposedly equivalent version of the problem.
$(2)\implies(1)$: Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. We can write $y=m(2^k-1)+1$ for some $m\geq1$. Then
$$2^ky-1=2^k(m(2^k-1)+1)-1=(2^k-1)(2^km+1)$$
so $y-x$ and $y+x$ are two positive divisors of $(2^k-1)(2^km+1)$ which average to $y=m(2^k-1)+1$. By $(2)$, $y-x=2^k-1$ and $y+x=2^k+1$. Then $x=1$ and $y=2^k$.
$(1)\implies(2)$: Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Let $y=m(2^k-1)+1$. We can write the two divisors as $y-x$ and $y+x$ for some $0<x<y$. Thus,
\begin{align*}
y-x&\bigm|2^ky-1,\\
y+x&\bigm|2^ky-1,
\end{align*}
since $2^ky-1=(2^k-1)(2^km+1)$. Manipulating these divisibility relations shows that
\begin{align*}
y-x&\bigm|2^kx-1,\\
y+x&\bigm|2^kx+1,
\end{align*}
where $\gcd(2^kx-1,2^kx+1)=1$. Then $\gcd(y-x,y+x)=1$ so $y^2-x^2\bigm|2^ky-1$. We clearly have $2^k-1\bigm|y-1$. By $(1)$, $x=1$ and $y=2^k$. Then $m=1$ and the two positive divisors were $2^k-1$ and $2^k+1$.
|
I've finally solved it!
My solution to this question proves a generalized version.
To deduce this special case, set $z=2^k$, note that $z$ is divisible by $4$, and use Theorem 9 to conclude that $y=z$ and $yz-1=y^2-x^2$.
Then $y=2^k$ and $x=1$ as desired.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Is The Series $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$ Divergent This is my solution for $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$. First I let the sequence in the series be labeled $A$, then I constructed a new sequence ($B$) that would have the similar behavior as $A$. This gives be the following:
$$A=\frac{n^4-3n+2}{4n^5+7}$$
$$B=\frac{1}{n}$$
Next I compare the value of the two sequances at $n=1$. This gives me $A=0$ and $B=1$, using this information I can concluded that $B \geq A$. Since I know that the $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent, then so must $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$.
Does this solution make sense? I feel like i am missing something
|
So, not quite. Suppose you are considering two series $\sum a_n$ and $\sum b_n$ consisting of non-negative terms. If you know that $\sum a_n$ is divergent and it happens that $ a_n \leq b_n$ for every $n$ then you may conclude that $\sum a_n \leq \sum b_n$. However, since you know the first series is divergent and consists of non-negative terms, you may conclude the same about the second series. But, this only works whenever the divergent series is the "smaller" one in the inequality.
To answer your question, we would likely have to use a limit comparison test.
Observe that $$\lim_{n\to \infty} \frac{\frac{n^4 - 3n +2}{4n^5 + 7}}{\frac{1}{n}} = \lim_{n\to \infty} \frac{n^5 -3n^2 +2n}{4n^5 + 7} = \lim_{n\to \infty} \frac{n^5 -3n^2 +2n}{4n^5 + 7}\cdot \frac{\frac{1}{n^5}}{\frac{1}{n^5}} = \lim_{n\to \infty} \frac{1 - \frac{3}{n^3} + \frac{2}{n^4}}{4 + \frac{7}{n^5}} = \frac{1}{4}$$
Since the limit is a positive, finite number we may conclude by the limit comparison test that since the harmonic series diverges, so must the series you were given.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align*}$$
By using this, we get value of $(3\csc3\theta - 4\sec3\theta) =9.95$ by using calculator.
I want know if there's any way to solve this problem without calculator.
|
Set $3\theta=t$
$\tan3t=\dfrac34\implies\dfrac{\sin3t}3=\dfrac{\cos3t}4=\dfrac1k$
$\implies(k\sin3t)^2+(k\cos3t)^2=3^2+4^2\implies k=5$ as $0<t<\dfrac\pi6$
$$F(t)=\dfrac3{\sin t}-\dfrac4{\cos t}=\dfrac{k\sin3t}{\sin t}-\dfrac{k\cos3t}{\cos t}$$
Method$\#1:$
As $\sin t\cos t\ne0,$ using $\sin3t,\cos3t$ formula,
$$F(t)=k(3-4\sin^2t)-k(4\cos^2t-3)=k(3+3-4)$$
Method$\#2:$
$$F(t)=\dfrac{k\sin(3t-t)}{\dfrac{\sin2t}2}=2k$$ as $\sin t\cos t\ne0$
|
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"url": "https://math.stackexchange.com/questions/3187142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$ Let $x,y,u,v \in \mathbb{R}.$ Prove that
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Proof 1:
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Square both side, we have
$$2\sqrt{(x^2+xy+y^2)(u^2+uv+v^2)} \geq 2xu+xv+yu+2yv$$
If $2xu+xv+yu+2yv<0$ then the inequality is true.
If $2xu+xv+yu+2yv \geq 0$ the square both side, we have
$$x^2v^2-2xvyu+y^2u^2 \geq 0$$
$$ \Leftrightarrow (xv-yu)^2 \geq 0$$
which is true.
Thus the inequality is true for all $x,y,u,v \in \mathbb{R}.$
Proof 2 (need help):
Because
$$x^2+xy+y^2 \geq \frac{3}{4}(x+y)^2$$
for all $x,y \in \mathbb{R},$ we have
$$\begin{matrix}
\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} & \geq & \frac{\sqrt{3}}{2}(|x+y|+|u+v|)\\
& \geq & \frac{\sqrt{3}}{2}|x+u+y+v|\\
& = & \frac{\sqrt{3}}{2}\sqrt{(x+u+y+v)^2}\\
& = & \frac{\sqrt{3}}{2}\sqrt{(x+u)^2+2(x+u)(y+v)+(y+v)^2}
\end{matrix}$$
I'm stuck here. Do you have any idea? Thank you.
|
$$
\overline{AB} = \sqrt{(x+u)^{2} + (x+u)(y+v) + (y+v)^{2}} \\
\overline{BC} = \sqrt{x^{2} + xy + y^{2}} \\
\overline{CA} = \sqrt{u^{2} + uv + v^{2}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$
Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$
So I started by combining the two fractions, which gave me:
$$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{2\sin\theta\cos\theta}{1-\cos^2\theta} = \frac{\sin2\theta}{1-\cos^2\theta}$$
I wasn't sure where to go from here considering I'm aiming for $2\cos\theta / \sin\theta$
|
It's $$\frac{\sin2\theta}{\sin^2\theta}=\frac{2\sin\theta\cos\theta}{\sin^2\theta}=\frac{2\cos\theta}{\sin\theta}=2\cot\theta.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3187825",
"timestamp": "2023-03-29T00:00:00",
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|
Value of $\lim\limits_{n\rightarrow \infty}(a_{1}+a_{2}+\cdots +a_{n})$
If $\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot...\cdot (2n+1)}\bigg)^2.$
Then $\displaystyle \lim_{n\rightarrow \infty}\bigg(a_{1}+a_{2}+...+a_{n}\bigg)$ is
Options:
$(a)$ Does not exists
$(b)$ Greater than $\displaystyle \frac{4}{27}$
$(c)$ Less than $\displaystyle \frac{4}{27}$
$(d)$ None of these
My Try: $$a_{n}=\bigg[\frac{n!\cdot 2\cdot 4 \cdot 6 \cdots (2n)}{1\cdot 2 \cdot 3\cdot 4\cdot \cdots (2n)\cdots (2n+1)}\bigg]^2$$
$$a_{n}=\bigg[\frac{n!\cdot 2 \cdot 4 \cdot 6 \cdots (2n)}{(2n+1)!}\bigg]^2$$
Could some help me to solve it , Thanks
|
Why, why bounty when @gt6989b gave you an awesome hint :) ?
$$a_n=\left(\left(\frac{1}{3}\right)\cdot\left(\frac{2}{5}\right)\cdot...\cdot\left(\frac{n}{2n+1}\right)\right)^2 <
\frac{1}{2^{2n}}=\frac{1}{4^n}$$
and (calculating the first 3 terms and using infinite geometric progression)
$$\sum\limits_{n=1}a_n =
\frac{1457}{11025}+\sum\limits_{n=4} a_n <
\frac{1457}{11025}+\sum\limits_{n=4}\frac{1}{4^{n}}=\frac{1457}{11025}+\frac{1}{4^4}\cdot\frac{1}{1-\frac{1}{4}}=\\
\frac{1457}{11025}+\frac{1}{3\cdot 4^3}=\frac{96923}{705600}<\frac{4}{27}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find : $\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$ I'm try to find this integral
$$\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$$
My try use :
$\ln(\sin x)=-\ln2-\sum\limits_{n=1}^{\infty}\frac{\cos (2nx)}{n}$
But I don't know how to complete summation ...
I will happy if someone help me
Thanks!
|
Your approach works perfectly well:
We can use the Fourier series and integrate by parts to obtain
$$ I \equiv \int \limits_0^{\pi/4} x [- \ln(\sin(x))] \, \mathrm{d} x = \frac{\pi^2}{32} \ln(2) + \frac{1}{4} \sum \limits_{n=1}^\infty \frac{1}{n^2} \left[\frac{\pi}{2} \sin\left(\frac{\pi}{2} n \right) - \frac{1}{n} \left(1 - \cos\left(\frac{\pi}{2} n \right)\right)\right] \, . $$
$\sin\left(\frac{\pi}{2} n \right)$ is non-zero and alternating for odd $n$, while $\cos\left(\frac{\pi}{2} n \right)$ is non-zero and alternating for even $n$. Therefore,
$$ I = \frac{\pi^2}{32} \ln(2) + \frac{\pi}{8} \sum \limits_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} - \frac{1}{4} \sum \limits_{n=1}^\infty \frac{1}{n^3} - \frac{1}{32} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^3} \, .$$
The first series is Catalan's constant $\mathrm{G}$, the second one is $\zeta(3)$ and the third one is $\eta(3) = \frac{3}{4} \zeta(3)$ (with the Riemann zeta function $\zeta$ and the Dirichlet eta function $\eta$), so we obtain
$$ I = \frac{\pi^2}{32} \ln(2) + \frac{\pi}{8} \mathrm{G} - \frac{35}{128} \zeta(3) $$
and your integral is $- I$.
|
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|
Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force. I then found the next pair by using the equation:
$(x + \sqrt{7}y)^i(x-\sqrt{7}y)^i = 1$
$i = 1$ would give me the minimal solution $(8,3)$, so I went to $i = 2$.
$(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2 = 1$
$\left(x^2 + 7y^2 + 2xy\sqrt{7}\right) \left(x^2 + 7y^2 - 2xy\sqrt{7}\right) = 1$
$(x^2 + 7y^2)^2 - 7(2xy)^2 = 1$ (This is of the form $X^2 - 7Y^2 = 1$)
I then let $X = x^2 + 7y^2$ and $Y = 2xy$, so the equation becomes the desired
$X^2 - 7Y^2 = 1$
I then plugged in my minimal solution $(8,3)$ and found $X = (8^3+7(3^3)) = 127$ and $Y = 2(8)(3) = 48$, so my new pair is $(127,48)$. To find the next solution, I let $i = 3$.
$(x + \sqrt{7}y)^3(x-\sqrt{7}y)^3 = 1$
And this is where I got stuck. I tried a few methods to get this new equation into the form of $X^2 - 7Y^2 = 1$, but have been unsuccessful. I tried expanding out the equations completely getting:
$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1$
But I'm fairly sure that's not the right direction. The other method I tried was factoring out a $(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$, so I got:
$[(x + \sqrt{7}y)(x-\sqrt{7}y)](x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$
which simplifying I got
$(x^2-7y^2) \left( (x^2 + 7y^2)^2 - 7(2xy)^2 \right)$
I'm really not sure how to properly factor this cubic to get to the desired function form. Any help would be greatly appreciated!
|
Remember that $x^2-7y^2=1$, so you can write $x^6-21x^4y^2=-14x^4y^2+7x^4$ and keep going
$$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1\\
-14x^4y^2+7x^4+147x^2y^4-343y^6=1\\
7x^4+49x^2y^4-14x^2y^2-343y^6=1\\
7x^4-14x^2y^2+49y^4=1$$
Now you can plug in your fourth degree solution and be done. You can also use the Brahmgupta-Fermat identity to reach the same result
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$ $$\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$$
$$\lim_{n\to \infty} \sum_{r=0}^{n-1} \left[\frac{1}{\sqrt{n^2+(r)n}}\right]$$
How can I convert it to Riemann sum because the higher limit is $(n-1)$ not $n$.
|
Hint
$$\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n(n-1)}}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\sqrt{1+\frac{k}{n}}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3202161",
"timestamp": "2023-03-29T00:00:00",
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|
Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now, divide both sides by three:
$$x^2 - \frac{4}{3}x = \frac{2}{3}$$
Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:
$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$
Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:
$$
\begin{align}
x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\
\left(x - \frac{2}{3}\right)^2 = \frac{10}{9}
\end{align}
$$
Can anyone please explain how they went from the first step to the second step?
|
Try going backward; expand the square $(x-\tfrac{2}{3})^2$ to find that
$$\left(x-\frac{2}{3}\right)^2=\left(x-\frac{2}{3}\right)\left(x-\frac{2}{3}\right)=x^2-\frac43x+\frac49.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3203319",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Showing $\lvert\cos z\rvert^2=a\cos2x+b\cosh2y$, where $z=x+iy$, for numbers $a$ and $b$ to be determined
Show that, for $z=x+iy$,
$$\lvert\cos z\rvert^2=a \cos2x+b\cosh2y$$
where $a$ and $b$ are numbers to be determined.
My incomplete solution:
$$\begin{align}
\lvert z\rvert &= \sqrt{z^*z} \\
\lvert\cos(z)\rvert^2&=\frac{1}{4}e^{2iz}+\frac{1}{4}e^{-2iz}+\frac{1}{2}\\
a\cos(2x)+b\cosh(2y)&= \frac{a}{2} e^{2ix}+\frac{a}{2}e^{-2ix}+\frac{b}{2}\left(e^{2y}+e^{-2y}\right)
\end{align}$$
|
\begin{align}\lvert\cos z\rvert^2&=\cos(z).\overline{\cos(z)}\\&=\cos(z).\cos\left(\overline z\right)\\&=\cos(x+yi)\cos(x-yi)\\&=\bigl(\cos(x)\cosh(y)-\sin(x)\sinh(y)i\bigr)\bigl(\cos(x)\cosh(y)+\sin(x)\sinh(y)i\bigr)\\&=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)\\&=\cos^2(x)\cosh^2(y)+\bigl(1-\cos^2(x)\bigr)\bigl(\cosh^2(y)-1\bigr)\\&=\cos^2(x)+\cosh^2(y)-1\\&=\frac12\bigl(2\cos^2(x)-1+2\cosh^2(y)-1\bigr)\\&=\frac12\cos(2x)+\frac12\cosh(2y).\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3203752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
If $a,b \in \Bbb [0,2]$ and $a+b=2$, maximize $ab(a^2+b^2)$ without differentiation. If $a,b \in \Bbb [0,2]$ and $a+b=2$, maximize $ab(a^2+b^2)$ without differentiation.
My try
$a+b=2 \iff a=2-b \implies \max \{ab(a^2+b^2)\} \iff \max \{[(2-b)b][(2-b)+b^2)\}$
$\iff \max \{-2 b^4 + 8 b^3 - 12 b^2 + 8 b\}$
At this point, i can use differentiaton to find the maximum. It's easy to see that $\max \{-2 b^4 + 8 b^3 - 12 b^2 + 8 b\}=2$ at $b=1$ but this exercise is meant to be done without differentiation.
Any hints?
|
You can use the fact that\begin{align}ab(a^2+b^2)&=ab\bigl((a+b)^2-2ab\bigr)\\&=ab(4-2ab)\\&=2\bigl(2ab-(ab)^2\bigr)\\&=2\left(1-\left(ab-1\right)^2\right)\\&=2-2(ab-1)^2.\end{align}So, the maximum is $2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A modular equation of 23rd degree of Dedekind’s $\eta$ function. Regarding the Post Additional values of Dedekind's $\eta$ function in radical form
I wrote the equation that has as root the value $\frac{\eta(23i)}{\eta(i)}$ that is missing.
Can someone help me solve (in radical form) the following equation, whose solution is the value of Dedekind’s modular $\frac{\eta(23i)}{\eta(i)}$ function?
$$x^{48}+\frac{684}{23^{5}}x^{36}-\frac{2496}{23^{7}}x^{32}+\frac{10944}{23^{9}}x^{28}+\frac{3826738}{23^{11}}x^{24}-\frac{31577472}{23^{13}}x^{20}+$$
$$\frac{785460096}{23^{15}}x^{16}-\frac{2112004548}{23^{17}}x^{12}+\frac{4240221504}{23^{19}}x^{8}+\frac{18998208}{23^{21}}x^{4}-\frac{1}{23^{23}}=0$$
where
$$x=\frac{\eta(23i)}{\eta(i)}.$$
This equation comes from the work of L. Kiepert and specializes for the value reported in the title of the application. My intent is to find the solution in closed form.
|
After substitution $x^4\to x,$ we get a 12-degree equation that factors into cubics over the rationals extended with $\sqrt{23}$ and $\sqrt2\cdot\sqrt[4]{23\,}$. Solving the cubic equation, we get
$$\small\begin{align}
\!\!\frac{\eta(23i)}{\eta(i)}&=\\
&\!\!\!\!\!\!\sqrt[4]{\frac{2\cdot 2^{2/3} \sqrt[3]{\alpha }+\sqrt[3]{\frac{2}{\alpha }} \left(360 \sqrt{23}+9 \sqrt{2} \cdot23^{3/4}-1426-525 \sqrt{2}\, \sqrt[4]{23}\right)-8 \sqrt{23}+6 \sqrt{2}\,\sqrt[4]{23}
\left(3+\sqrt{23}\right)}{6348}},
\end{align}$$
where
$$\small\alpha =32 \sqrt{23} \left(38-9 \sqrt{3}\right)+2484 \left(9 \sqrt{3}-4\right)-9 \sqrt{2}\, \sqrt[4]{23} \,\left(717-454 \sqrt{3}+139 \sqrt{23}+62 \sqrt{69}\right).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove Smaller Distance from Hyperbola to Asymptote There is a canonical hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $$ and the asymptote $$ y= \pm \frac{b}{a}x $$
Let us say that the value of hyperbola at $x$ is given as
$$y=\frac{b}{a}\sqrt{x^2 - a^2}$$ and the value of asymptote at the same $x$ is given as
$$ y = \frac{b}{a}x$$
we will have the difference of hyperbola and asymptote at the same $x$ as
$$ \frac{b}{a}(x-\sqrt{x^2 - a^2})$$
Let us take another point at $x+\delta$ where the difference of hyperbola and asymptote will be
$$ \frac{b}{a}(x+\delta - \sqrt{x^2 - a^2 + 2\delta x+ \delta^2}) $$
Prove that the difference at $x+\delta$ is lower than at $x$.
My attempt is only able to come until this
$$ \sqrt{x^2-a^2+2\delta x+\delta^2}-\delta<\sqrt{x^2-a^2}$$. Then I don't know what to do. Thanks in advance.
|
OK, by using comment from Blue, I will try to answer the question myself. Since the difference is $$\frac{b}{a}(x-\sqrt{x^2 - a^2}) $$ then the derivative will be $$\frac{b}{a}(1-\frac{x}{\sqrt{x^2-a^2}}) $$
Since I am only interested to the value of $x > a$ where $x$ and $a$ is always positive then $$\frac{x}{\sqrt{x^2-a^2}}>1$$ and we always have negative for
$$\frac{b}{a}(1-\frac{x}{\sqrt{x^2-a^2}}) $$
|
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|
How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$?
Prove the identity
\begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
for each nonnegative integer $n$.
To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$
If $n=1$ then I have next expressions
$(1+q)(1+q^2) = \frac{1-q^{2^{1+1}}}{1-q} = $
$ = (1+q)(1+q^2) = \frac{1-q^{4}}{1-q}$
Well played and calculated =(.Speak me where i s mistaked.
|
For the base case $n=1$, prove that $$(1+q)(1+q^2)=\frac{1-q^4}{1-q}\iff \underbrace{(1-q)(1+q)}_{=(1-q^2)}(1+q^2)=1-q^4$$
Which is true since $(a-b)(a+b)=a^2-b^2$.
For the inductive step $\color{blue}{\text{assume that the equation holds for some $n$}}$. Therefore
$$\begin{align*}(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n+1}})&=\color{blue}{(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n}})}\cdot (1+q^{2^{n+1}})\\ &=\color{blue}{\frac{1-q^{2^{n+1}}}{1-q}}\cdot (1+q^{2^{n+1}})\\ &=\frac{1-q^{2^{n+2}}}{1-q}\end{align*}$$
Alternatively
Although induction is usually a nice approach when it comes to proving an equation $\forall n\in\Bbb N$, there is even a nicer way to prove your equation. Observe simply that $$\begin{align*}&(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n}})=\frac{1-q^{2^{n+1}}}{1-q}\\\iff& (1-q)(1+q)(1+q^2)...(1+q^{2^{n}})=1-q^{2^{n+1}}\end{align*}$$ which can be proven using repetidly the fact that $(a-b)(a+b)=a^2-b^2$ Therefore $$\begin{align*}(1-q)(1+q)(1+q^2)...(1+q^{2^{n}})&=(1-q^2)(1+q^2)...(1+q^{2^{n}})\\&=(1-q^4)(1+q^4)...(1+q^{2^{n}})\\&=\;...\\&=(1-q^{2^{n}})(1+q^{2^{n}})\\&=(1-q^{2^{n+1}})\end{align*}$$
Which is what we wanted to prove.
|
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|
$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$.
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \},$$ but I'm stuck proving that it's closed under inverse. If that is true, then the minimal polynomial should be of 3rd degree. True or not, I can't find any such polynomial.
Solution: As @José Carlos Santos mentioned below:
$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha,$$
from which it follows
$$(x - \sqrt[3]3-\sqrt[3]9)(x^2 + x(\sqrt[3]3+\sqrt[3]9) + (\sqrt[3]3+\sqrt[3]9)^2) - 9(x - \sqrt[3]3 - \sqrt[3]9)= x^3 - 12 - 9(\sqrt[3]3+\sqrt[3]9) -9x + 9(\sqrt[3]3+\sqrt[3]9) = x^3 - 9x - 12 = p(x);~~ p(\alpha) = 0.$$
|
Let's have $q=\sqrt[3]{3}\quad$ and $\quad\alpha=q+q^2$.
$$q+q^2+q^3=q(1+q+q^2)\iff3+\alpha=q(1+\alpha)$$
Now since $q^3=3$ then $\alpha$ is solution of $$(3+\alpha)^3=3(1+\alpha)^3\iff 12+9\alpha-\alpha^3=0$$
|
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|
Integration by substitution using $u^2=3-x$ The question says to integrate $\frac x{\sqrt{3-x}}$ using the substitution $u^2=3-x$.
What I did was first rearrange $u^2=3-x$ into $ 3 - u^2 =x$ and then subbed it into the integral .
I then rearranged $u^2=3-x$ to $u=\sqrt{3-x} $ and then also subbed that in.
Next I found the derivative of $u$ which I believe to be $du/dx = -1/(2\sqrt{3-x}$) I then sub in u and then rearrange to get $-2udu = dx $
I then also sub that in then have $y = -2(3-u^2)$ and integrate it.
However when i integrate it i get $-6\sqrt{3-x} + (2(3-x)^3)/3 $
But the answer is $(-2/3)(x+6)\sqrt{3-x} $.
I tried to rearrange my answer to get that, but I was unsuccessful. Is there instead a mistake in my working out?
Side note: I tried to put things to the power of a 1/2 but it did not work out I'm not sure how to do it
thank you
|
Using $u^2 = 3-x$, we have the following
$$\int \dfrac{x}{\sqrt{3-x}} dx = \int \dfrac{3-u^2}{\sqrt{u^2}} dx = \int \dfrac{3-u^2}{u} dx$$
Now, like you did, solve the transformation for $u$ such that $u=(3-x)^\frac{1}{2}$. Then, $du = -\dfrac{1}{2}(3-x)^{-\frac{1}{2}} dx = -\dfrac{1}{2}(u^2)^{-\frac{1}{2}}dx = -\dfrac{1}{2u}dx$ such that $-2u du = dx$. By substitution, we have
$$\int \dfrac{x}{\sqrt{3-x}} dx = -\int \dfrac{(2u)(3-u^2)}{u} du=-\int (6-2u^2) du$$
Thus,
$$\int \dfrac{x}{\sqrt{3-x}} dx = -6u+\dfrac{2}{3}u^3 + C$$
Upon substitution of $u=\sqrt{3-x}$, we have
$$\begin{align}
\int \dfrac{x}{\sqrt{3-x}} dx &= -6\sqrt{3-x}+\dfrac{2}{3}(3-x)^{\frac{3}{2}} + C
\\
&=-6\sqrt{3-x}+\dfrac{2}{3}(3-x)\sqrt{3-x} + C
\\
&=\sqrt{3-x}\Bigg(-6+\dfrac{2}{3}(3-x)\Bigg)+C
\\
&=-\dfrac{2}{3}\sqrt{3-x}\Bigg(9-(3-x)\Bigg)+C
\\
&=-\dfrac{2}{3}\sqrt{3-x}(6+x)+C
\end{align}$$
|
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|
Find $\sum_{k=1}^{2n} k {2n\choose k}^2 (-1)^{k+1}$.
I need to find $$S = {2n\choose 1}^2 -2 {2n\choose 2}^2 + ... - 2n{2n\choose 2n}^2= \sum_{k=1}^{2n} k {2n\choose k}^2 (-1)^{k+1}$$, given $$\sum_{k=1}^{2n} k {2n\choose k} x^{k-1} = 2n(1+ x)^{2n -1}$$
Using $$-(1-x)^{2n} + 1 = \sum_{k=1}^{2n} {2n \choose k} (-1)^{k+1} x^k$$,
$$(1- (1-x)^{2n}) (2n (1+x)^{2n-1}) = \sum_{k=1}^{4n}x^k\sum_{r=0}^k {2n\choose r} (-1)^{r+1} (k - r) {2n \choose k-r}$$
LHS = $2n ( (1+x)^{2n - 1} - (1-x)(1-x^2)^{2n-1}) = 2n((1+x)^{2n - 1} - (1-x^2)^{2n-1} + x(1-x^2)^{2n -1})$
So coefficient of $x^{2n}$ on the LHS is $-2n{2n - 1\choose n} (-1)^n = -n{2n \choose n} (-1)^n$
Coefficient of $x^{2n}$ on the RHS,
$$2n \sum_{r=0}^{2n} {2n\choose r}^2 (-1)^{r+1} - S = - 2n{2n \choose n}(-1)^n - S$$
Equating the coefficient of both sides gives $$S = -n{2n \choose n} (-1)^n$$.
I would like to know different methods for doing this and similar problems as the method I used is cumbersome and prone to miscalculation.
|
$[x^n]:f(x)$ denote the coefficient of $x^n$ in the function $f(x)$
\begin{eqnarray*}
\binom{2n}{k} =\binom{2n}{2n-k}=[x^{2n-k}]: (1+x)^{2n}=[x^{2n}]:x^k(1+x)^{2n}.
\end{eqnarray*}
So your sum can be written as
\begin{eqnarray*}
\sum_{k=1}^{2n} k \binom{2n}{k}^2 (-1)^{k+1} &=& [x^{2n}]: (1+x)^{2n} \sum_{k=1}^{2n} k \binom{2n}{k} (-1)^{k}x^k \\
&=& [x^{2n}]:2n x (1-x)^{2n-1} (1+x)^{2n} \\
&=& [x^{2n}]:2n x(1+x) (1-x^{2})^{2n-1} .
\end{eqnarray*}
We only need to consider the even powers of $x$, so let $y=x^2$ and we have
\begin{eqnarray*}
\sum_{k=1}^{2n} k \binom{2n}{k}^2 (-1)^{k} &=& [y^n]: 2n y (1-y)^{2n-1} =2n \binom{2n-1}{n-1} (-1)^{n-1}.
\end{eqnarray*}
So your answer is correct.
|
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|
The exponential generating function for the central binomial coefficients I am having difficulty proving the following result
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{n}}{n!} = e^{2x} \ I_{0} (2x) \text{,}
\end{align}
where
\begin{align}
I_{0}(y) = \sum_{n=0}^{\infty} \frac{\Big( \frac{y}{2} \Big)^{2n}}{n! \ n!}
\end{align}
is a modified Bessel function of the first kind with zero order.
I approached it using
\begin{align}
e^{2x} \ I_{0} (2x) = \bigg( \sum_{n=0}^{\infty} \frac{(2x)^{n}}{n!}\bigg) \ \bigg( \sum_{k=0}^{\infty} \frac{ x^{2k}}{k! \ k!} \bigg) \text{,}
\end{align}
but I have had no luck.
|
Extracting the coefficient we seek to show
$${2n\choose n} \frac{1}{n!}
= [z^n] \sum_{k\ge 0} \frac{2^k}{k!} z^k
\sum_{k\ge 0} \frac{1}{k! \times k!} z^{2k}.$$
This is
$${2n\choose n} \frac{1}{n!}
= \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{q! \times q!} \frac{2^{n-2q}}{(n-2q)!}.$$
or
$${2n\choose n}
= \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{q! \times q!} \frac{2^{n-2q} \times n!}{(n-2q)!}.$$
or
$${2n\choose n}
= \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{(n-q)!}{q! \times q!}
\frac{2^{n-2q} \times n!}{(n-q)! \times (n-2q)!}.$$
In terms of binomial coefficients
$${2n\choose n}
= \sum_{q=0}^{\lfloor n/2 \rfloor}
{n\choose q} {n-q\choose q} 2^{n-2q}.$$
The RHS is
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
{n\choose q} {n-q\choose n-2q} 2^{n-2q}
\\ = 2^n [z^n] (1+z)^n
\sum_{q=0}^{\lfloor n/2 \rfloor}
{n\choose q} 2^{-2q} (1+z)^{-q} z^{2q}.$$
The coefficient extractor combined with the $z^{2q}$ factor enforces
the upper limit and we may write
$$2^n [z^n] (1+z)^n
\sum_{q\ge 0}
{n\choose q} 2^{-2q} (1+z)^{-q} z^{2q}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{z^2}{4(1+z)}\right)^n
\\ = 2^n [z^n] (1+z)^n
\frac{(4+4z+z^2)^n}{4^n (1+z)^n}
\\ = 2^{-n} [z^n] (z+2)^{2n}
\\ = 2^{-n} {2n\choose n} 2^n = {2n\choose n}.$$
This is the claim.
|
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|
How can I compute this integral in closed form : $\int_0^{\frac{π}{4}}\ln^2(\tan x)dx$ How can I compute this integral in closed form :
$$\displaystyle\int_{0}^{\displaystyle \tfrac{π}{4}}\ln^{2}\left(\tan x\right)dx$$
How can use Fourier series here ?
$$-2\displaystyle \sum_{n=0}^{\infty}\frac{\cos((4n+2)x)}{2n+1}$$
$$=\ln\left(\tan x\right)$$
And what's about if
$\displaystyle\int_0^{\displaystyle \tfrac{π}{4}}\ln^2(\cot x)dx$
Please give me ideas or hints
|
An elementary solution. Utilize
$$ J=\int_0^\infty \frac{(x^2-1)\ln y }{(y+x^2)(y+1)}
\overset{y\to \frac{x^2}y}{dy}=\int_0^\infty \frac{(x^2-1)\ln (x^2) }{(y+x^2)(y+1)}dy- J= {2\ln^2 x}$$
and $x=\tan t $
\begin{align}
&\int_0^{\pi/4}\ln^2(\tan t)\ dt\\
=& \int_0^1\frac{\ln^2x}{1+x^2}\overset{x\to 1/x}{dx}
=\int_1^\infty\frac{\ln^2x}{1+x^2} dx= \frac12\int_0^\infty\frac{\ln^2x}{1+x^2}dx\\
=&\ \frac14\int_0^\infty\int_0^\infty\frac1{1+x^2}\frac{(x^2-1)\ln y}{(y+x^2)(y+1)}\ dy \ dx\\
=& \ \frac14\int_0^\infty\int_0^\infty
\frac{\ln y}{(y+x^2)(y-1)}-\frac{2\ln y}{(1+x^2)(y^2-1)}\ dx \ dy\\
=& \ \frac\pi8 \int_0^\infty \frac{\ln y}{\sqrt y(y-1)} \overset{y\to y^2}{dy}- \frac\pi4 \int_0^\infty \frac{\ln y}{y^2-1}\ dy
=\frac\pi4 \int_0^\infty \frac{\ln y}{y^2-1}\ dy\\
=& \ \frac\pi4 \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy
= \frac{\pi^2}8\int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^3}{16}
\end{align}
|
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|
Is my solution correct? (trigonometry) $$2\cos x\left(\cos x-\sqrt{8}\tan x\right)<5$$
After expanding I get $2\sin^2 x+4\sqrt{2}\sin x-5>0$. I then factorized $(2\sin x+3\sqrt{2})(2\sin x+\sqrt{2})>0$. The first one is always positive so $\sin x>\sin(-\pi/4)$. How to continue?
|
$\sin x>-\sqrt{2}/2$ and you have $\sin(-\pi/4)=\sin(5\pi/4)=-\sqrt{2}/2$. The big arc between $-\pi/4$ and $5\pi/4$ is the solution :
$$\left[\frac{-\pi}{4}, \frac{5\pi}{4}\right]~~~\text{(mod }2\pi\text{)}$$
Another way to denote this solution is
$$\bigcup\limits_{n\in\mathbb{Z}}\left[\frac{-\pi}{4}+2n\pi, \frac{5\pi}{4}+2n\pi\right]$$
|
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|
Find $a-b$ from $(\cos(a)+\cos(b))^{2}+(\sin(a)+\sin(b))^{2}=2\cos^{2}(\frac{a-b}{2})$ $a,b\in \mathbb{R}$ and these numbers satisfy this egality:
$(\cos(a)+\cos(b))^{2}+(\sin(a)+\sin(b))^{2}=2\cos^{2}(\frac{a-b}{2})$
I need to find $a-b$ (The right answer is $a-b=2k\pi+\pi|k\in \mathbb{Z}$)
My try: I know that the expression $(\cos(a)+\cos(b))^{2}+(\sin(a)+\sin(b))^{2}=4\cos^{2}(\frac{a-b}{2})$
So I get $4\cos^{2}(\frac{a-b}{2})=2\cos^{2}(\frac{a-b}{2})$ and I get $\cos^{2}(\frac{a-b}{2})=0$ but my result is $+-\pi+4k\pi$
Where's my mistake?
|
Clayton has the answer for what is wrong with your answer.
An alternative answer is to remember $2\cos^2{x}-1=\cos 2x$ so:
$$2\cos^2\left(\frac{a-b}{2}\right)=\cos(a-b)+1$$
And you have the left side is, after expanding and canceling $\cos^2 a+\sin^2 a = 1=\cos^2b +\sin^2 b$
$$2+2(\cos a \cos b + \sin a \sin b)=2+2\cos(a-b)$$
This gives you the equation $2\cos(a-b)+2=\cos(a-b)+1$ which means $\cos(a-b)=-1.$
|
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|
Series that evaluates to different values I am trying to find a sequence $x_{pq}$ such that $\sum_{p=1}^\infty\sum_{q=1}^\infty x_{pq}$ and $\sum_{q=1}^\infty\sum_{p=1}^\infty x_{pq}$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.
|
The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_{jk} = 1/(k^2-j^2)$ if $j\neq k$ and $a_{jk} = 0$ if $j=k$.
Here we have,
$$\frac{\pi^2}{8}=\tag{*}\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} \neq \sum_{k=1}^\infty \sum_{j=1\\j \neq k}^\infty \frac{1}{k^2-j^2} = -\frac{\pi^2}{8} $$
To obtain the sum, note that
$$\sum_{k=1\\j\neq k}^K \frac{1}{k^2-j^2} = \frac{1}{2j}\sum_{k=1\\j\neq k}^K \left(\frac{1}{k-j} - \frac{1}{k+j} \right) \\= \frac{1}{2j}\left(-\sum_{k=1}^{j-1}\frac{1}{k} + \sum_{k=1}^{K-j}\frac{1}{k} - \sum_{k=j+1}^{K+j}\frac{1}{k} + \frac{1}{2j} \right)\\= \frac{1}{2j}\left(\frac{1}{j} +\sum_{k=j+1}^{K-j}\frac{1}{k} - \sum_{k=j+1}^{K+j}\frac{1}{k} + \frac{1}{2j} \right)\\ = \frac{1}{2j}\left(\frac{3}{2j} - \sum_{k=-j+1}^{j} \frac{1}{K+k}\right) $$
and,
$$\sum_{k=1\\j\neq k}^\infty \frac{1}{k^2-j^2} = \lim_{K \to \infty}\frac{1}{2j}\left(\frac{3}{2j} - \sum_{k=-n+1}^{n} \frac{1}{K+k}\right) = \frac{3}{4j^2}$$
Thus,
$$\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} = \sum_{j=1}^\infty \frac{3}{4j^2} = \frac{3}{4} \frac{\pi^2}{6}$$
|
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|
Find the value of $\lim\limits_{n\rightarrow \infty}\left( \frac{2^{-n^2}}{\sum_{k=n+1}^{\infty}2^{-k^2}}\right)$
Find the value of $$\lim_{n\rightarrow \infty}\left( \frac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)$$
My answer:
Take $r_{n+1} = \sum\limits_{k=n+1}^{\infty}2^{-k^2}$.
If $\lim\limits_{n\rightarrow \infty}\left( \dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)$ exists, then
\begin{align*}
1+\lim\limits_{n\rightarrow \infty}\left( \dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)&=\lim\limits_{n\rightarrow \infty}\left( 1+\dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)\\
&=\lim\limits_{n\rightarrow \infty}\left( \dfrac{\sum\limits_{k=n}^{\infty}2^{-k^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)\\
&=\lim\limits_{n\rightarrow \infty}\dfrac{r_n}{r_{n+1}}=??
\end{align*}
Moreover, I know $r_n$ is a remainder of the series $\sum\limits_{k=1}^{\infty}2^{-k^2}$, which is convergent.$~~~~\left(\mbox{ By ratio test,} ~~\lim\limits_{n\rightarrow\infty} \dfrac{2^{-(n+1)^2}}{2^{-n^2}}=\lim\limits_{n\rightarrow\infty}\dfrac{1}{2.2^{2n}}=0<1 \right)$
|
For $k \geq n+1$ it is easy to see that $n^{2}-k^{2} \leq -k$. Hence $\sum\limits_{k=n+1}^{\infty} 2^{n^{2}-k^{2}} \leq \sum\limits_{k=n+1}^{\infty} 2^{-k} \to 0$. Hence the required limit is $\infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3220564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
Solve $99x^2 \equiv 1 \mod 125$ Solve $$x^{98} \equiv 99 \mod 125$$
Is there any easy way to solve equations like that? My observation is that from Euler's theorem we know that
$$ x^{100} \equiv 1 \mod 125 $$
so
$$x^{98} \equiv 99 \mod 125 \\
x^{100} \equiv 99x^2 \mod 125 \\
99x^2 \equiv 1 \mod 125$$
but what is general method how to deal with equations like that?
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Start mod $5$, and then lift...
$$99 x^2 \equiv 4 x^2 \equiv (2x)^2 \equiv 1 \mod 5$$
so $2 x \equiv \pm 1 \mod 5$, i.e. $x \equiv 2$ or $3 \mod 5$.
If $x \equiv 2 \mod 5$, $x \equiv 2 + 5 y \mod 25$, and then
$$ 99 x^2 - 1 \equiv 5 y + 20 \equiv 0 \mod 25$$
$$ y + 4 \equiv 0 \mod 5$$
$$ y \equiv 1 \mod 5$$
So now $x \equiv 2 + 5 + 25 z \equiv 7 + 25 z \mod 125$, and then
$$ 99 x^2 - 1 \equiv 25 z + 100 \equiv 0 \mod 125$$
$$ z + 4 \equiv 0 \mod 5$$
$$ z \equiv 1 \mod 5$$
Thus one solution is $x \equiv 2 + 5 + 25 \equiv 32 \mod 125$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3221171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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|
Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\infty}$$
$$=\left(\frac{\infty(4\infty+5)}{\infty(4\infty+3)}\right)^{-4\infty}=\left(\frac{4\infty}{4\infty}\right)^{-4\infty} = 1^{-4\infty} = \boxed{1}$$
However, when I try to graph the function, I can't reliably get my answer due to precision limitations, and I feel that this method of calculating limits is less than ideal. How can I confirm that this is indeed the limit?
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Hint:
$$\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)=\frac{(n+2)(4n-3)}{(n+2)(4n-5)}=\frac{4n-3}{4n-5}=\frac{1}{(1-\frac{2}{4n-3})}=(1-\frac{2}{4n-3})^{-1}$$
so
$$\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}=(1-\frac{2}{4n-3})^{4n-3}$$
then use
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3222029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.