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Sum of fourth powers in terms of sum of squares The sum of the fourth powers of the first $n$ integers can be expressed as a multiple of the sum of squares of the first $n$ integers, i.e.
$$\begin{align}
\sum_{r=1}^n r^4&=\frac {n(n+1)(2n+1)(3n^3+3n-1)}{30}\\
&=\frac{3n^2+3n-1}5\cdot \frac {n(n+1)(2n+1)}6 \\
&=\frac{3n^2+3n-1}5\sum_{r=1}^nr^2 \end{align}$$
Question: Is it possible to show this, purely by manipulating the summand, and without first expressing the summation in closed form and then factoring the sum of squares?
From the question here, we see that
$$\sum_{r=1}^n r^4=\left(\sum_{r=1}^n r^2\right)^2-2\sum_{r=1}^n r^2\sum_{j=1}^{r-1}j^2\\
=\sum_{r=1}^n r^2 \left(\sum_{i=1}^n i^2-2\sum_{j=1}^{r-1}j^2\right)$$
but this does not appear to lead anywhere closer to answering the question.
Another approach might be to use Abel's summation formula
$$\sum_{r=1}^n f_r (g_{r+1}-g_r)=\left[f_{n+1}g_{n+1}-f_1g_1\right]-\sum_{r=1}^n g_{r+1}(f_{r+1}-f_r)$$
Putting $f_r=r^2$ and $g_r=\frac {r(r-1)(2r-1)}6$ gives
$$\sum_{r=1}^n r^4=(n+1)^2\cdot \frac {n(n+1)(2n+1)}6-\sum_{r=1}^n \frac {r(r+1)(2r+1)}6\cdot (2r+1)$$
but again this does not seem to get us any further.
1st Edit
Putting $T_m=\sum_{r=1}^n r^m$, the original problem can be restated as an attempt to prove that
$$5T_4=(6T_1-1)T_2$$
2nd Edit
This paper might be useful.
In section 4 (p$206$), it is stated that
$$\frac{\sigma_4}{\sigma_2}=\frac {6\sigma_1-1}5$$
which is derived from the Faulhabner polynomials.
$\sigma_m$ has the same definition as our $T_m$ as defined above.
|
Statement we'd like to show:
$$
\sum^{n}_{r=1}r^4=\frac{3n^2+3n-1}{5}\sum^{n}_{r=1}r^2
$$
Or equivalently:
$$
5\sum^{n}_{r=0}r^4=(3n^2+3n-1)\sum^{n}_{r=0}r^2
$$
We may recall one of approaches how to get closed form of this expression and utilize it in a slightly different way.
$$
\sum^{n}_{r=0}r^5 + (n+1)^5=\sum^{n}_{r=0}(r+1)^5\\
(r+1)^5=r^5+5r^4+10r^3+10r^2+5r+1
$$
Which leads to:
$$
(n+1)^5=5\sum^{n}_{r=0}r^4+10\sum^{n}_{r=0}r^3+10\sum^{n}_{r=0}r^2+5\sum^{n}_{r=0}r+\sum^{n}_{r=0}1
$$
Similar actions give us:
$$
(n+1)^4=4\sum^{n}_{r=0}r^3+6\sum^{n}_{r=0}r^2+4\sum^{n}_{r=0}r+\sum^{n}_{r=0}1\\
(n+1)^3=3\sum^{n}_{r=0}r^2+3\sum^{n}_{r=0}r+\sum^{n}_{r=0}1
$$
So let us combine it:
$$
5\sum^{n}_{r=0}r^4=(n+1)^5-\left(10\sum^{n}_{r=0}r^3+10\sum^{n}_{r=0}r^2+5\sum^{n}_{r=0}r+\sum^{n}_{r=0}1\right)\to\\
5\sum^{n}_{r=0}r^4=(n+1)^5-\left(\frac{5}{2}(n+1)^4-5\sum^{n}_{r=0}r^2-5\sum^{n}_{r=0}r-\frac{3}{2}\sum^{n}_{r=0}1\right)\to\\
5\sum^{n}_{r=0}r^4=(n+1)^5-\left(\frac{5}{2}(n+1)^4-\frac{3}{2}(n+1)^3-\frac{1}{2}\sum^{n}_{r=0}r^2-\frac{1}{2}\sum^{n}_{r=0}r\right)\to\\
5\sum^{n}_{r=0}r^4=n(n+1)^3\left(n-\frac{1}{2}\right)+\frac{1}{2}\sum^{n}_{r=0}r+\frac{1}{2}\sum^{n}_{r=0}r^2
$$
At this point we have to express $\sum^{n}_{r=0} r=\frac{n(n+1)}{2}$, which after some simplifications leads us to:
$$
5\sum^{n}_{r=0}r^4=3\left(n^2+n-\frac{1}{2}\right)\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\sum^{n}_{r=0}r^2
$$
We recognize $\sum^{n}_{r=0}r^2=\frac{n(n+1)(2n+1)}{6}$, and finish with:
$$
5\sum^{n}_{r=0}r^4=(3n^2+3n-1)\sum^{n}_{r=0}r^2
$$
|
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|
Calculus Integral $\int \frac{dt}{2^{t} + 4}$
I try resolve this integral, with $u = 2^t + 4 $, but can´t ... plz
$$\int \frac{dt}{2^t + 4} = \frac{1}{\ln 2}\int\frac{du}{u(u-4)} = \text{??} $$
|
$\displaystyle \int\dfrac{dt}{2^t+4}$
on u substitution, i.e., $2^t+4=u,\;$ we get-
$\dfrac{1}{\ell n2}\displaystyle \int\dfrac{du}{u(u-4)}\;....(1),\;$ as you already done.
Now, we need partial fraction to solve it further.
For this, $\dfrac{1}{u(u-4)}=\dfrac{A}{u}+\dfrac{B}{u-4}$
$=\dfrac{Au-4A+Bu}{u(u-4)}$
$\dfrac{1}{u(u-4)}=\dfrac{(A+B)u-4A}{u(u-4)}$
On comparing L.H.S. and R.H.S.
$(A+B)u=0,\; -4A=1$
$\implies A+B=0,\;A=-\dfrac{1}{4}$
$thus, A=-\dfrac{1}{4}, \;B=\dfrac{1}{4}$
Now, expression (1) becomes, $\dfrac{1}{\ell n2}\displaystyle \int\dfrac{-\dfrac{1}{4}}{u}+\dfrac{\dfrac{1}{4}}{u-4}$
$\dfrac{1}{\ell n2}\displaystyle \int\left(-\dfrac{1}{4u}+\dfrac{1}{4(u-4)}\right)du$
$\dfrac{1}{4\ell n2} \big(\ell n(u-4)-\ell nu\big)$
put the value of u
$\dfrac{1}{4\ell n2} \big(\ell n(2^t)-\ell n(2^t+4)\big)$
$\dfrac{1}{4\ell n2} \big(t\ell n2-\ell n(2^t+4)\big)$
$\dfrac{t}{4}-\dfrac{\ell n(2^t+4)}{4\ell n2}$
|
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|
Find the $n$th derivative of $f(x)=x\sin(x)\cos(2x)$ If it helps, it ask the value for $n=100$ and $x=\pi/2$.
I can't do it by induction because it has too many factors and trying to use an equality for $\cos(2x)$ didn't helped.
I don't see the relation in the derivatives.
|
Observe
\begin{align}
x\sin x \cos 2x =&\ x\frac{e^{ix}-e^{-ix}}{2i} \frac{e^{2ix}+e^{-2ix}}{2} = \frac{x}{4i}\left(e^{3ix}-e^{ix}+e^{-ix}-e^{-3ix} \right)\\
=&\ \frac{1}{2}x\sin 3x-\frac{1}{2}x\sin x \\
=&\ \frac{1}{2}\left(x-\frac{\pi}{2}\right)\sin 3x -\frac{1}{2}\left(x-\frac{\pi}{2}\right)\sin x + \frac{\pi}{4}\sin 3x -\frac{\pi}{4}\sin x\\
=&\ -\frac{1}{2}\left(x-\frac{\pi}{2}\right)\cos\left(3 \left(x-\frac{\pi}{2} \right) \right)- \frac{1}{2}\left(x-\frac{\pi}{2}\right)\cos \left(x-\frac{\pi}{2}\right) \\
&- \frac{\pi}{4}\cos\left(3 \left(x-\frac{\pi}{2} \right) \right)-\frac{\pi}{4}\cos\left(x-\frac{\pi}{2}\right).
\end{align}
Since
\begin{align}
\cos\left(3 \left(x-\frac{\pi}{2} \right) \right)=\sum^\infty_{k=0} (-1)^k\frac{3^{2k}(x-\pi/2)^{2k}}{(2k)!}
\end{align}
and
\begin{align}
\cos \left(x-\frac{\pi}{2} \right)=\sum^\infty_{k=0} (-1)^k\frac{(x-\pi/2)^{2k}}{(2k)!}
\end{align}
then it follows the
\begin{align}
f^{(100)}(\pi/2) = -\frac{\pi}{4}100! \left(\frac{1}{100!}+\frac{3^{100}}{100!} \right) = -\frac{\pi}{4}(3^{100}+1).
\end{align}
|
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|
Find the limit of the sequence $x_{n+1}=\sqrt{3x_n}$
Let $x_{n+1}=\sqrt{3x_n}$ and $x_1=1$. Prove $x_n=3^{1-(\frac{1}{2^{n-1}})}$ for all $n$ and find the limit of $\{x_n\}$.
Notes: The first few terms of the sequence are $1,\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}} ...$ I do not know from this how to find the equation for the $x_n$ term in terms of $3$. The second part can be answered as follows, please tell me if my thinking is correct.
\begin{align*}
x_n&=3^{1-(\frac{1}{2^{n-1}})} \\
\lim_{n\rightarrow \infty}x_n&=\lim_{n\rightarrow \infty}3^{1-(\frac{1}{2^{n-1}})} \\
&=3^{1-\lim_{n\rightarrow \infty}(\frac{1}{2^{n-1}})} \\
&=3^{1-(\frac{1}{\lim_{n\rightarrow \infty}2^{n-1}})} \\
&= 3
\end{align*}
Also I have looked at Show that the sequence $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$ converges and find its limit. But there is no discussion on how to find something similar to $x_n=3^{1-(\frac{1}{2^{n-1}})}$ .
|
If you want to see where the formula comes from, notice the pattern:
\begin{eqnarray}
x_1 &=& 1\\
x_2 &=& \sqrt{3} &=& 3^{\frac12}\\
x_3 &=& \sqrt{3\sqrt{3}} &=& \bigl(3\cdot3^{\frac12}\bigr)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 &=& 3^{\frac12 + \frac14}\\
x_4 &=& \sqrt{3\sqrt{3\sqrt{3}}} &=& \left(3\bigl(3\cdot3^{\frac12}\bigr)^\frac12\right)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 \cdot 3^\frac18 &=& 3^{\frac12 + \frac14 + \frac18}\\
\end{eqnarray}
In general, the sequence can be written
$$x_n = 3^{\sum_{k=1}^{n-1}\frac{1}{2^k}}$$
Inductively,
\begin{eqnarray}
x_{n+1} &=& \sqrt{3x_n}\\
&=& \sqrt{3 \cdot 3^{\sum_{k=1}^{n-1}\frac{1}{2^k}}}\\
&=& \sqrt{3^{1+\sum_{k=1}^{n-1}\frac{1}{2^k}}}\\
&=& 3^{\frac12+\frac12\sum_{k=1}^{n-1}\frac{1}{2^k}}\\
&=& 3^{\sum_{k=1}^{(n-1)+1}\frac{1}{2^k}}\\
\end{eqnarray}
The sum for the geometric series is $\displaystyle\sum_{k=0}^{n-1} a r^k = a \dfrac{1-r^n}{1-r}$, so
$$1+\sum\limits_{k=1}^{n-1}\frac{1}{2^k} = \sum\limits_{k=0}^{n-1}\frac{1}{2^k} = \frac{1-(\frac12)^n}{1-\frac12} = \frac{1-\frac{1}{2^n}}{\frac12} = 2 - \frac{1}{2^{n-1}}$$
Subtracting $1$ from both sides of the equation, we can write
$$x_n = 3^{\left(1 - \frac{1}{2^{n-1}}\right)}= 3^{1-2^{1-n}}$$
|
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|
Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$ Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$
My Attempt:
$LHS=\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{(\sin\frac{x}{2}-\cos\frac{x}{2})(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\sin\frac{x}{2}\cos\frac{x}{2})}{2+2\sin \frac{x}{2}\cos\frac{x}{2}}=\frac{\sin\frac{x}{2}-\cos\frac{x}{2}}{2}$
$RHS=\frac{\cos x}{3}=\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{3}$
But my LHS and RHS are not equal.Please help me.
|
You have correctly proved they are not equal.
WolframAlpha agrees with you.
|
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|
Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4k^2 + 4k + 1 - (4l^2 + 4l + 1) \\
&= 4k^2 + 4k - 4l^2 - 4l
\end{align}
My two concerns:
1) Is this correct so far?
2) How would I deal with the “$8\;\mid$” part?
|
$8 \mid x$ if and only if $x\equiv 0\mod 8$. Then you have $4k^2 + 4k - 4l^2 - 4l$ If $k$ is odd, $k^2$ is too. Then $4k^2 + 4k \equiv 0 \mod 8$ if $k$ is even then so is $k^2$ then $4k^2 + 4k \equiv 0\mod 8$. A similar argument for $l$ will finish your proof.
|
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|
Necessary conditions for a Sudoku puzzle to have no repetitions Is it true that if a Sudoku puzzle has the following features there will be no repetitions in rows, columns and $3 \times 3$ subsquares?
*
*The sum of each row must be $45$
*The sum of each column must be $45$
*The sum of each $3 \times 3$ subsquare must be $45$
If so, why? Is there a mathematical proof? If not, why? Is there a case where these conditions are satisfied, but is there at least one repetition?
Thanks!
|
If all the cells are distinct 1 through 9 then the sum is 1+2+....+9 =45. But there is utterly no reason earth to assume the converse, that is $a+b+.... +i = 45$ then they are all distinct.
For any $b,...,h =N$ we can have $a$ be any $1 \le a \le 45-N $ and $i = 45-N-a$. And we can determine values for the other rows and columns. Yes, it takes a bit of thought to actually work this out but there is no reason that that keeping them distinct will be a requirement.
Let's suppose for instance we have a grid labeled A1....A9..... I1.... I9 where every row, column and quadrant add up to 45. Then lets say we replace mk (where $A \le m \le I$ and $1\le k \le 9$) with mk + 1. Then we replace mj in the same column and quadrant with mk - 1$, replace nk in the same column and quadrant with nk-1 and nj with nj + 1. Then all the quadrants, columns and rows still add to 45 but one or the other or both grids are no longer distinct.
e.g suppose we have:
$\begin{array}{|ccc|ccc|ccc|}
\hline 1&2&3&4&5&6&7&8&9\\
4&5&6&7&8&9&1&2&3\\
7&8&9&1&2&3&4&5&6\\
\hline 2&3&4&5&6&7&8&9&1\\
5&6&7&8&9&1&2&3&4\\
8&9&1&2&3&4&5&6&7\\
\hline {\color{red}3}&4&{\color{red}5}&6&7 &8&9&1&2\\
6&7&8&9&1&2&3&4&5\\
{\color{red}9}&1&{\color{red}2}&3&4&5&6&7&8\\
\hline
\end{array}$
and we replace it with
$\begin{array}{|ccc|ccc|ccc|}
\hline 1&2&3&4&5&6&7&8&9\\
4&5&6&7&8&9&1&2&3\\
7&8&9&1&2&3&4&5&6\\
\hline 2&3&4&5&6&7&8&9&1\\
5&6&7&8&9&1&2&3&4\\
8&9&1&2&3&4&5&6&7\\
\hline {\color{blue}4}&4&{\color{blue}4}&6&7 &8&9&1&2\\
6&7&8&9&1&2&3&4&5\\
{\color{blue}8}&1&{\color{blue}3}&3&4&5&6&7&8\\
\hline
\end{array}$
Note, the sums must be the same but values need not be distinct.
|
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|
Prove that for $n\ge 2$, $(n + 1)^n n! < (2n)! < 4^n (n!)^2 .$ Prove that for all real $n\ge 2$, the following inequalities hold: $$(n + 1)^n n! < (2n)! < 4^n (n!)^2 .$$
I have posted my proof below, if anyone has a shorter, or more interesting proof, please share :-)
|
First note we can re-write the left hand side as:
$$(n+1)^n n!=1\cdot2\cdot3\cdots n\cdot \underbrace{(n+1)\cdot(n+1)\cdots(n+1)}_{n\text{ times}},$$
and the middle term as
$$(2n)!=1\cdot2\cdot3\cdots n\cdot(n+1)\cdot (n+2)\cdots (n+(n-1))\cdot 2n.$$
The first $n+1$ terms are equal, and the remaining terms can be compared as follows:
$$\begin{align}
n+2 &> n+1,\\
n+3 &> n+1,\\
&\vdots\\
2n-1&> n+1,\\
2n &> n+1.
\end{align}$$
It follows that $(2n)!>(n+1)^n n!$.
It remains to prove that $4^n(n!)^2>(2n)!$, we proceed by induction. For the base case it is clear that $4^2(2!)^2=64>24=(2\cdot 2)!$. Assume the statement is true for $n=k$, and consider $n=k+1$. The middle term becomes
$$(2(k+1))!=(2k)! \cdot (2k+1) (2k+2),$$
and the right hand side is
$$4^{k+1}
((k+1)!)^2=4^k(k!)^2\cdot 4(k+1)(k+1).$$
From the inductive hypothesis, we know that $4^k(k!)^2>(2k)!$, and clearly $$4(k+1)(k+1)=2(k+1)\cdot 2(k+1)=(2k+2)(2k+2)>(2k+1)(2k+2),$$
and the result follows.
|
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|
Question on Partial Fraction Decomposition I'm supposed to evaluate the following integral by decomposing the integrand into a sum of partial fractions. Note, this question isn't about evaluating.
$$\int\frac{6x+6}{(x^2+1)(x-1)^3}dx$$
From my understanding the numerator should be equated to the following form:
$$\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}$$
I know how to evaluate for constants $A$ through $E$, but I'm still unsure as to how the above form was equated to the integrand. If someone could help clarify that'd be great.
|
By setting $f(x)=\frac{6x+6}{(x^2+1)(x-1)^3}$ we have:
$$ E = \lim_{x\to 1}(x-1)^3 f(x) = \lim_{x\to 1}\frac{6x+6}{x^2+1} = 6 \tag{E} $$
$$ D = \lim_{x\to 1}\frac{d}{dx}(x-1)^3 f(x) = -\lim_{x\to 1}\frac{6(x^2+2x-1)}{(x^2+1)^2}=-3\tag{D} $$
$$ C = \frac{1}{2}\lim_{x\to 1}\frac{d^2}{dx^2}(x-1)^3 f(x) = 6\lim_{x\to 1}\frac{(x-1)(1+4x+x^2)}{(1+x^2)^3}=0 \tag{C} $$
and
$$ f(x)-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3} = \frac{3}{1+x^2}\tag{A,B}$$
so $(A,B,C,D,E)=\color{red}{(0,3,0,-3,6)}$. This method clearly shows the strong relation between partial fraction decomposition and the residue theorem. The reason for such a decomposition to hold is that $f(x)$ is a meromorphic function with a triple pole at $x=1$ and simple poles at $x=\pm i$. If we remove from $f(x)$ the "polar part" given by the triple pole at $x=1$, i.e. if we consider
$$g(x)=f(x)-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3}$$
we know in advance that such a meromorphic function is regular in a neighbourhood of $x=1$ and has two simple poles at $x=\pm i$. By removing from $g(x)$ the "polar part" associated with such poles, i.e. by subtracting from $g(x)$ something of the form $\frac{Ax+B}{x^2+1}$, we get that
$$p(x)=f(x)-\frac{Ax+B}{x^2+1}-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3}$$
is an entire function and a rational function, i.e. a polynomial. By inspecting the removed part and the degrees of the polynomials whose ratio defines $f(x)$, we may also easily get what the degree of $p(x)$ has to be, and deduce
$$f(x)=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}.$$
$p(x)\equiv 0$ also follows from computing $\lim_{x\to +\infty}f(x)$ and $\lim_{x\to +\infty}$ of the removed part.
|
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|
Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula Let $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real.
It can be shown using a distance formula that the minimal value of $f(x)$ is
$D = \sqrt{(a-c)^2+(|b|+|d|)^2}$.
Show that result without derivatives and without a distance formula. At what value of $x$ does the minimum of $f(x)$ occur?
Hint: this is a generalization of this question.
|
Let, w.l.o.g., $b > 0$ and $d > 0$.
Define
$$
x_0 = \frac{ad+bc}{b+d}
$$
Further, as already stated in the question, the minimum value of $f$ is
$D = \sqrt{(a-c)^2+(b+d)^2}$.
The following equality holds
$$
f(x) = \sqrt {(g(x))^2 + (h(x))^2}+\sqrt {(g(x))^2 + (D - h(x))^2}
$$
where
$$g(x) = \frac{b+d}{D} \; (x-x_0) $$
and
$$h(x) =\frac{1}{D} \; \Big[(c-a)\, x + a^2 - c a + b^2 + d b\Big] $$
Now from $(g(x))^2 \ge 0$ follows
$$
f(x) \ge \sqrt {(h(x))^2}+\sqrt { (D - h(x))^2} = D
$$
with equality for $g(x) =0$ which corresponds to $x=x_0$. So indeed $D$ is the smallest value that $f(x)$ can attain, and this happens at $x=x_0$. This completes the proof without distance measures and without derivatives. $\quad \Box$
If required, the formulae can be re-interpreted geometrically. Note that $f(x)$ is the sum of the distances from point $N = (x,0)$ to the points $P_1 = (a,b)$ and $ P_2 = (c,-d)$. Consider the line $P_1 P_2$. From a distance argument, the minimum of $f(x)$ is the length of $P_1 P_2$ which is $D$.
Define a point $R$ on that line $P_1 P_2$ such that $NR$ is perpendicular to $P_1 P_2$. Then $g(x)$ is the length of $NR$, $h(x)$ is the length of $P_1R$, and $D-h(x)$ is the length of $P_2R$. Since $NR$ is perpendicular to $P_1 P_2$, the two square roots then calculate the distances $NP_1$ and $NP_2$ in a Pythagorean way. The minimum of $f(x)$ is obtained for $N = R$, i.e. $g(x) = 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $d$ must be a perfect square
Prove that if $a,b,c,d$ are integers such that $$(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2 = d,$$ then $d$ is a perfect square.
In order for $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2$ to be an integer, either $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ has to be the square root of a positive integer or $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ must be a perfect square. How do we deal with those conditions?
|
Theorem: Suppose that $a+2^{1/3}b+2^{2/3}c=0$ where $a,b,c\in\mathbb{Z}$. Then $a=b=c=0$.
Proof: Note that
$$
\begin{align}
&\left(a+2^{1/3}b+2^{2/3}c\right)\left(\left(a^2-2bc\right)+2^{1/3}\left(2c^2-ab\right)+2^{2/3}\left(b^2-ac\right)\right)\\
&=a^3+2b^3+4c^3-6abc\tag{1}
\end{align}
$$
Thus, if $a+2^{1/3}b+2^{2/3}c=0$, then $(1)$ says that
$$
a^3+2b^3+4c^3-6abc=0\tag{2}
$$
If it is not the case that $a=b=c=0$, we can assume that $\gcd(a,b,c)=1$.
Looking at $(2)$ mod $2$, we see that $\phantom{2}a^3\equiv0\pmod2$; therefore $a\equiv0\pmod2$.
Looking at $(2)$ mod $4$, we see that $2b^3\equiv0\pmod4$; therefore $b\equiv0\pmod2$.
Looking at $(2)$ mod $8$, we see that $4c^3\equiv0\pmod8$; therefore $c\equiv0\pmod2$.
Therefore, $2\mid\gcd(a,b,c)$. Contradiction. Thus, $a=b=c=0$.
QED
Answer to the Question
Suppose that
$$
\left(a+2^{1/3}b+2^{2/3}c\right)^2=d\tag{3}
$$
Expanding $(3)$ gives
$$
\left(a^2+4bc-d\right)+2^{1/3}\left(2ab+2c^2\right)+2^{2/3}\left(b^2+2ac\right)=0\tag{4}
$$
The Theorem guarantees that $2abc+2c^3=0$ and $b^3+2abc=0$. That is, $2c^3=b^3$. Since $2^{1/3}\not\in\mathbb{Q}$, we must have that $b=c=0$.
The Theorem also guarantees that $a^2+4bc-d=0$. Since $b=c=0$, this says that $d=a^2$, which is a perfect square.
|
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|
How do I solve differentil equation if the begining values are given? How do I solve differentil equation if the begining values are given?
$\frac{dx}{dt}=x^2+5x$ with x(0)=-3. I need to find x(t).
$\int \frac{dx}{x^2+5x}=\int dt$
So when I put it on Symbollab I get that left side is $-\frac{2}{5}arctanh(\frac{2}{5}(x+\frac{2}{5}))$, I already see that is going to be hard to extract x from here.
What have I done?
$\int \frac{dx}{x^2+5x}=\int \frac{dx}{x^2(1+\frac{x}{5})}$ than I use substitution that $1+\frac{x}{5}$ is u, than $-\frac{x^2}{5}du$.
Than I get $\frac{-1}{5}ln(1+\frac{x}{5})=t+C$.
$ln(1+\frac{x}{5})=-5t+C$
$1+\frac{x}{5}=e^{-5t}C$
$\frac{x}{5}=e^{-5t}C-1$
$x=\frac{5}{e^{-5t}C-1}$
$-3=\frac{5}{e^{0}C-1}$
$-3=\frac{5}{C-1}$
Than I calculate C. This all seems nice to me.
But when I put than integral in WolframAlpha I get that $\int \frac{dx}{x^2+5x}=\frac{1}{5}(log(x)-log(x+5))+constant$
Why I don't get when I integrate those results from WolframAlpha or Symbollab?
|
If $y = \mbox{arc}\tanh x$, then $$x = \tanh y =\frac{e^y - e^{-y}}{e^y+e^{-y}}= \frac{e^{2y} - 1}{e^{2y}+1}.$$ So
$$ x(e^{2y}+1) = e^{2y}-1$$
or $$e^{2y}(x-1) = -x-1$$
So $$e^{2y} = \frac{1+x}{1-x}.$$ That is
$$2y = \log\left( \frac{1+x}{1-x}\right)$$
and
$$y= \frac{1}{2}(\log(1+x)-\log(1-x)).$$
So we have $\mbox{arc}\tanh x = \frac{1}{2}(\log(1+x)-\log(1-x).$$
Partial fractions is giving the $\log$ version of the answer and WA is giving the $\mbox{arc}\tanh$ version.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trouble with integral expressing $\mathbb E[X^2]$ where $X \sim N(0,1)$ Let $X \sim N(0,1)$
Then to find $E[X^2]$, we can do:
$$
\int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx
$$
Let's say we didn't know this is an odd function, and decided to take the integral by splitting it into 2 pieces:
$$
\frac{1}{\sqrt{2\pi}} \left( \int_{-\infty}^0 x^2e^{-\frac{x^2}{2}} \, dx + \int_0^\infty x^2 e^{-\frac{x^2}{2}} \, dx \right)
$$
Then substitute $t = \frac{x^2}{2}$, we get:
$$
\frac{1}{\sqrt{2\pi}} \left( \int_\infty^0 \sqrt{2t}e^{-t} \, dt + \int_0^\infty \sqrt{2t}e^{-t} \, dt \right) \\
= \frac{1}{\sqrt{2\pi}} \left( - \int_0^\infty \sqrt{2t}e^{-t} \, dt + \int_0^\infty \sqrt{2t}e^{-t} \, dt \right)
= 0
$$
Which is $= 0 $, but obviously it shouldn't be $0$. Those two integrals should be adding together instead of canceling each other out. Where did I make a mistake in my math? Thanks!
|
Let
\begin{align}
I(a) &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \mathrm{d}x \\
&= \frac{1}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\
&= \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{a}} \mathrm{erf}(y) \Big|_{0}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{a}}
\end{align}
using the substitution $y^{2} = ax^{2}$.
Then
\begin{align}
I &= \frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{\infty} x^{2} \mathrm{e}^{-x^{2}/2} \mathrm{d}x
= \frac{2}{\sqrt{2 \pi}} \int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}/2} \mathrm{d}x \\
&= -\frac{2}{\sqrt{2 \pi}} \lim_{a \to 1/2} \frac{\partial I(a)}{\partial a}
= \frac{2}{\sqrt{2 \pi}} \lim_{a \to 1/2} \int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-ax^{2}} \mathrm{d}x
= \frac{2}{\sqrt{2 \pi}} \int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}/2} \mathrm{d}x \\
&= -\frac{2}{\sqrt{2 \pi}} \frac{\sqrt{\pi}}{2} \lim_{a \to 1/2} \frac{\partial}{\partial a} a^{-1/2}
= -\frac{1}{\sqrt{2}} \left(-\frac{1}{2}\right) \lim_{a \to 1/2} a^{-3/2} \\
&= 1
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the 2016th power of a complex number Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac{\sqrt 3 - 1}{2} + i\frac{\sqrt 3 + 1}{2}$.
$$z = r(\cos \phi + i \sin \phi)$$
$$r = |z| = \sqrt 2$$
$$\phi = \arctan {\sqrt 3 + 1}$$
Now, I don't know what to do with that $\sqrt 3 + 1$. How do I calculate $\phi$ ?
Thank you in advance!
|
$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$
Lets simplify $\frac{-1 + i\sqrt 3}{1 + i}$
$\frac {1}{1+i}(-1 + i\sqrt 3)\\
\frac {1-i}{2}(-1 + i\sqrt 3)\\
(1-i)(-\frac12 + i\frac{\sqrt 3}2)$
Convert to polar:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)$
Now we have a choice...we could raise to the 2016 power right now, or we could mulitiply those two complex numbers first then raise to the 2016 power.
$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} =
\sqrt 2^{2016} (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016}(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)^{2016}$
Applying deMoivres theorem:
$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = (\cos \frac {-2016\pi}{4} +\sin \frac {-2016\pi}{4} )$
$8$ divides $2016$
$\frac {-2016\pi}{4} = 2n\pi$
$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = 1$
$6$ divides $2016$, too.
$2^{1008}$
alternatively:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3) = \sqrt 2 (\cos \frac {5\pi}{12} +\sin \frac {5\pi}{12})$
and we get to the same place.
|
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|
Continuity of this function in the origin Does anyone have any idea on how can I prove if this partial derivative is continuous at the origin:
$f(x,y) = \frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}$
Of course I know the definition: for every $\epsilon$ there is a $\delta$ if $ ||(x,y)||$ is less that $\delta$ than $|f(x,y)|$ is less than $\epsilon$.77
I'm finding difficult do work with $|f(x,y)|$
I already tried $|f(x,y| = |\frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}| \leq |\frac{x(x^2 + 2y^2)}{(x^2+y^2)^{1/2}}|$ because $|\frac{y}{(x^2+y^2)}|\leq1$
But I'm know kind of stucked in there. Can someone help me or give a hint on how to proceed? Thanks!
|
HINT:
Transforming to polar coordinates $(\rho,\phi)$, we have
$$\begin{align}
f(x,y)&=\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\\\\
&=\frac{\rho^4\sin(\phi)\cos(\phi)(\cos^2(\phi)+2\sin^2(\phi))}{\rho^3}
\end{align}$$
Then, the limit as $\rho\to0$ is trivially $0$. If $f(0,0)=0$, then $f$ is continuous at the origin.
If one wishes to proceed without a coordinate transformation, simply note that
$$\begin{align}
\left|\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\right|&\le \left|\frac{2xy(x^2+y^2)}{(x^2+y^2)^{3/2}}\right|\\\\
&=\left|\frac{2xy}{\sqrt{x^2+y^2}}\right| \\\\
&\le \frac{x^2+y^2}{\sqrt{x^2+y^2}}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Combinatorial proof of $\sum_{1\le i\le n,\ 1\le j\le n}\min(i,j)=\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$ $$\sum_{1\le i\le n,\ 1\le j\le n}\min(i,j)=\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$$
Is there a link, if any, between these two identical sums?
|
$$\begin{array}{rr|rrrrrrr}
\hline
\min(i,j)&&i\\
&&1&2&3&4&\cdots &(n-2)&(n-1)&n\\
\hline
j&1&1&1&1&1&\cdots &1&1&1\\
&2&1&2&2&2&\cdots &2&2&2\\
&3&1&2&3&3&\cdots &3&3&3\\
&4&1&2&3&4&\cdots &4&4&4\\
&\vdots&&&&&\ddots \\
&\vdots \\
&\vdots \\
&(n-2)&1&2&3&4&&\color{green}{(n-2)}&\color{green}{(n-2)}&\color{green}{(n-2)}\\
&(n-1)&1&2&3&4&&\color{green}{(n-2)}&\color{blue}{(n-1)}&\color{blue}{(n-1)}\\
&n&1&2&3&4&&\color{green}{(n-2)}&\color{blue}{(n-1)}&\color{red}n\\
\hline
\end{array}$$
From the table above it is clear that
$$\begin{align}
\sum_{1\le i,j\le n}\min(i,j)
&=\color{red}{1\cdot n}+\color{blue}{3\cdot (n-1)}+\color{green}{5\cdot (n-2)}+\cdots+(2n-1)\cdot 1 \\
&=\sum_{r=1}^n (2r-1)\color{orange}{(n+1-r)}\\
&=\sum_{r=1}^n \color{orange}{\sum_{i=r}^n}(2r-1)\\
&=\sum_{i=1}^n\sum_{r=1}^i (2r-1)
\color{lightgrey}{=\sum_{i=1}^n\sum_{r=1}^i r^2-(r-1)^2}\\
&=\sum_{i=1}^n i^2\quad\blacksquare\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the roots of any cubic with trigonometric roots.
Question: How would you find the roots of a cubic polynomial whose roots are only expressive in trigonometric forms?
I'm really confused on how you would solve it. Some examples:$$x^3+x^2-2x-1=0\\x_1=2\cos\frac {2\pi}7,x_2=2\cos\frac {4\pi}7,x_3=2\cos\frac {8\pi}7\tag{1}$$$$x^3-x^2-9x+1=0\\x_1=4\cos\frac {2\pi}7,x_2=4\cos\frac {4\pi}7+1,x_3=4\cos\frac {8\pi}7+1\tag{2}$$$$x^3+x^2-6x-7=0\\x_1=2\left(\cos\frac {4\pi}{19}+\cos\frac {6\pi}{19}+\cos\frac {10\pi}{19}\right)\\x_2=2\left(\cos\frac {2\pi}{19}+\cos\frac {14\pi}{19}+\cos\frac {16\pi}{19}\right)\\x_3=2\left(\cos\frac {8\pi}{19}+\cos\frac {12\pi}{19}+\cos\frac {20\pi}{19}\right)\tag{3}$$
With $(3)$ being a very famous relation with this problem.
Now that I know it's possible, I'm wondering if there is a simple way to find the roots of any cubic with trigonometric roots. And is it possible to use the method to find the roots of cubics such as $x^3+x^2-10x-8=0$?
|
For $x^3+x^2-10x-8 = 0$.
$(\Bbb Z/31 \Bbb Z)^*/\{\pm 1\}$ is cyclic of order $15$, so it has only one subgroup $H$ of index $3$.
$H = \{\pm 1; \pm 2; \pm 4; \pm 8; \pm 15\}$ (generated by $\pm 2$)
So, the roots should be integer combinations of the quantities
$\sum_{n \in kH} 2\cos(\frac {2n\pi}{31})$ where $kH \in G/H$ are the three cosets of $H$
In fact, the roots are exactly those sums
$\sum_{n \in H} 2\cos(\frac {2n\pi}{31}) = 3.083872... \\
\sum_{n \in 3H} 2\cos(\frac {2n\pi}{31}) = -0.786802... \\
\sum_{n \in 9H} 2\cos(\frac {2n\pi}{31}) = -3.297071...$
However, it can't be related to an equation like $x = \sqrt {2a + \sqrt {2a +\sqrt {2a + x}}}$ because modulo $2$, the action of the Frobenius automorphism would permute the roots cyclically, while here, the Frobenius acts trivially because $\pm 2 \in H$ (or equivalently, because $2$ is a cube mod $31$).
|
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|
Elementary number theory proof without contradiction Question: suppose $a, b, c$ are integers and $\gcd(a,b) = 1$. Prove that if $c|(a^2+b^2)$ then $\gcd(a,c) = gcd(b,c) = 1$.
Prove by contradiction (naive approach) : assume $p \in PRIMES$ and $\ne1$ such that: $p|a$ and $p|c$. Since $c|(a^2+b^2)$ then $p|(a^2+b^2) \Rightarrow p|b^2$ (because $p|a$ which means $p|a^2$). Thus, as $p|b^2$ then $p|b$. It contradicts with the fact that: $\gcd(a, b) = 1$
My question is that is there another way to prove it? (preferably without contradiction).
Any help would be appreciated.
|
Suppose $a^2+b^2=kc$ and, as implied by $\text{gcd}(a,b)=1$, we have $ma+nb=1$ for some integers $m$ and $n$. Then
$$
1=m^2a^2+n^2b^2+2mnab.
$$
Consider the RHS above:
\begin{aligned}
1=\text{RHS}&=m^2(a^2+b^2)+(n^2-m^2)b^2+2mnab=c(km^2)+b[(n^2-m^2)b+2mna];\\
1=\text{RHS}&=n^2(a^2+b^2)+(m^2-n^2)a^2+2mnab=c(kn^2)+a[(m^2-n^2)a+2mnb].
\end{aligned}
Then, respectively, the two lines above prove $\text{gcd}(c,b)=1$ and $\text{gcd}(c,a)=1$.
|
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|
Discuss the inequality different methods inequality $\sum_{k=1}^{n}k^2y_{k}\ge (n+1)\sum_{k=1}^{n}ky_{k}$
Nice Problem: Let $n\ge 3$,and $y_{1},y_{2},\cdots,y_{n}$ be real numbers, and such that $$2y_{k+1}\le y_{k}+y_{k+2}.1<k\le n-2$$
and $\displaystyle\sum_{k=1}^{n}y_{k}=0$. Show that
$$\sum_{k=1}^{n}k^2y_{k}\ge (n+1)\sum_{k=1}^{n}ky_{k}.$$
I have understand this following nice Proof from AMM11866:
Note \begin{align*}
&0\le\sum_{k=1}^{n-2}\binom{k+1}{2}\binom{n-k}{2}(y_{k+2}-2y_{k+1}+y_{k})\\
&=\sum_{k=1}^{n}\left(\binom{k+1}{2}\binom{n-k}{2}-2\binom{k}{2}\binom{n-k+1}{2}+\binom{k-1}{2}\binom{n-k+2}{2}\right)y_{k}\\
&=\sum_{k=1}^{n}\left(3k^2-3(n+1)k+\binom{n+2}{2}\right)y_{k}\\
&=3\left(\sum_{k=1}^{n}k^2y_{k}-(n+1)\sum_{k=1}^{n}ky_{k}\right)
\end{align*}
Question 1: I want to know this proof with the coefficient $\binom{k+1}{2}\binom{n-k}{2}$. How to get it?
Question 2: This problem have other methods? (without integral)
|
The inequality to prove is obviously equivalent to $\sum_{k=1}^{n} k(n-k+1)y_k \le 0$.
Question 1: Idea in the given proof is to use summation by parts to transform the above sum in $y_k$ into a sum in $\Delta_k^{2} \;y_k = y_{k+2}-2 y_{k+1} + y_k$ where the terms are known to be positive by convexity of the sequence. The chosen coefficients "happen" to satisfy the $2^{nd}$ order finite difference equality $$\nabla_{k}^2 \;\binom{k+1}{2} \binom{n-k}{2} = \frac{1}{2}\Big(-6 k(n-k+1) + (n+1)(n+2)\Big)$$
which "conveniently" leads to the sought inequality. Working backwards to actually derive the coefficients from $k(n-k+1)$ is trickier, and requires the appropriate choices of constants in the indefinite sums $\nabla_k^{-1}$, based on the boundary conditions, particular expressions, and general flair.
Question 2: The following will use Chebyshev's sum inequality to prove the stronger statement:
For $n \ge 3$ and $y_k \mid k=1,\cdots,n$ a convex sequence i.e. $y_k \le \frac{1}{2}(y_{k-1}+y_{k+1})$ for $1 \lt k \lt n$:
$$\sum_{k=1}^{n} k(n-k+1)y_k \le \frac{(n+1)(n+2)}{6} \sum_{k=1}^{n} y_k$$
In brief: note first that the coefficients $k(n-k+1)$ are increasing with $k$ for $2k \le n$ $\iff k \le \lfloor \frac{n+1}{2}\rfloor$, and symmetric WRT $k \mapsto n-k+1$. Then group the symmetric terms in $\sum_{k=1}^{n} k(n-k+1)y_k \le 0$ and reduce it to a sum of products of what turns out ot be inversely monotonic sequences, where Chebyshev's inequality applies.
In more detail: $y_k$ is a convex sequence, and it follows from the premise $y_k\le \frac{1}{2}(y_{k-1}+y_{k+1})$ that the finite differences $y_{k+1}-y_{k} \ge y_{k}-y_{k-1}$ form an increasing sequence.
Let $S = \sum_{k=1}^{n} y_k$ and $z_k=y_k+y_{n-k+1}$. Then:
*
*$\quad z_k = z_{n-k+1}$
*$\quad \sum_{k=1}^{n}z_k = 2 \sum_{k=1}^{n}y_k = 2 S$
*$\quad z_{k+1}-z_k = (y_{k+1}-y_k)-(y_{n-k+1}-y_{n-k}) \le 0$ $\iff k \le n-k$ $\iff 2k \le n$ $\iff k \le \lfloor \frac{n+1}{2} \rfloor$ so the sequence $z_k$ is decreasing for $k=1, 2, \;\cdots\;, \lfloor \frac{n+1}{2} \rfloor$
Case 1: Assume $n=2m$ even (so that $\lfloor \frac{n+1}{2} \rfloor = m$) then $\sum_{k=1}^{m}z_k = \sum_{k=1}^{n}y_k = S$ and, by Chebyshev's inequality:
$$
\begin{align}
\sum_{k=1}^{n} k(n-k+1)y_k & = \sum_{k=1}^{m} k(2 m-k+1)z_k \\
& \le \frac{1}{m}\left(\sum_{k=1}^{m} k(2m-k+1)\right)\left(\sum_{k=1}^{m} z_k\right) \\
& = \frac{1}{m} \frac{m(m+1)(2m + 1)}{3} S \\
& = \frac{(n+1)(n + 2)}{6} S
\end{align}
$$
Case 2: Assume $n=2m+1$ odd (so that $\lfloor \frac{n+1}{2} \rfloor = m+1$ and $z_{m+1}=2 y_{m+1}$) then:
*
*$\quad \sum_{k=1}^{m}z_k + y_{m+1} = S\;\;$ so $\;\;\sum_{k=1}^{m}z_k = S - y_{m+1}$
*$\quad y_{m+1} \le \frac{S}{2m+1}\;\;$ by Jensen's inequality for abscissae $1,2,\cdots,2m+1$
Again by Chebyshev's inequality:
$$
\begin{align}
\sum_{k=1}^{n} k(n-k+1)y_k & = \sum_{k=1}^{m} k(2 m-k+2)z_k + (m+1)^2 y_{m+1} \\
& \le \frac{1}{m}\left(\sum_{k=1}^{m} k(2m-k+2)\right)\left(\sum_{k=1}^{m} z_k\right) + (m+1)^2 y_{m+1} \\
& = \frac{1}{m}\frac{m(m+1)(4m+5)}{6}(S - y_{m+1}) + (m+1)^2 y_{m+1} \\
& = \frac{m+1}{6}\Big( (4m+5)S + (2m+1)y_{m+1}\Big) \\
& \le \frac{m+1}{6}\Big( (4m+5)S + S\Big) \\
& = \frac{(m+1)(2m+3)}{3} S \\
& = \frac{(n+1)(n+2)}{6} S
\end{align}
$$
This concludes the proof, and the case $S=0$ gives AMM 11886.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is $\lim_{n \rightarrow \infty} \bigg( \bigg| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} \bigg| \bigg)=e$? I know that the following is the correct limit, but I have difficulties in seeing just why this is.
$$\lim_{n\to\infty}\left| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}}\right|=e$$
|
Hint. By using the Taylor series expansion, as $x \to 0$, one has
$$
\log(1+x)=x-\frac{x}2+\frac{x^3}3+o(x^3)
$$ giving, as $n \to \infty$,
$$
\begin{align}
-n^2\log\left(1+\frac1n\right)&=-n+\frac12-\frac1{3n}+o\left(\frac1{n}\right)
\\\\
(n+1)^2\log\left(1+\frac1{n+1}\right)&=n+\frac12+\frac1{3n}+o\left(\frac1{n}\right)
\end{align}
$$ then one may write, as $n \to \infty$,
$$
\begin{align}
\frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}}&=\frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{n^2}(n+1)^{(n+1)^2}}
\\\\&=\left(1+\frac1n \right)^{-n^2}\left(1+\frac1{n+1} \right)^{(n+1)^2}
\\\\&=e^{-n^2\log(1+1/n)}\cdot e^{(n+1)^2\log(1+1/(n+1))}
\\\\&=e^{1/2+1/2+o(1/n)}
\\\\&=e^{1+o(1/n)}
\end{align}
$$ which yields the announced result.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Limit involving a trigonometric function as $x\rightarrow +\infty$ I am probably supposed to use this identity: $$\lim_{x\to0} \frac{\sin(x)}{x} = 1$$
I have this mathematical problem:
$$\lim_{x\to+∞} (x^2-1) \sin(\frac{1}{x-1}) = \lim_{x\to+∞} (x+1)(x-1) \sin(\frac{1}{x-1}) = ???$$
Can I do this to the sinus limit?
$$\lim_{x\to+∞} \frac{1}{\sin(\frac{x-1}{1})}$$
|
Let $X=\frac{1}{x-1}$ so that $x-1=\frac{1}{X}$ and $x+1=2+\frac{1}{X}$
We have $\lim_{x \to +\infty} (x^2-1) \sin(\frac{1}{x-1})=\lim_{X \to 0^+} (2+\frac{1}{X}) \frac{\sin(X)}{X}=+\infty$
|
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|
How was this calculated $(1-1/z)^{(\frac{-1}{3})}=(1+\frac{1}{3z}+\dots)$ *Background: a simplification step from an interesting longer problem here where the residue is needed so only the first two numbers are needed.
$$(1-\tfrac1z)^{-\frac13}=(1+\tfrac1{3z}+\cdots)$$
Was the likely step to compute the cube root: $(1+\frac{1}{z}+(\frac{1}{z})^2\dots)^{\frac{1}{3}}$?
If so, how do I do that?
|
Considering $$A=\frac{1}{\sqrt[3]{1-\frac{1}{z}}}$$ let for simplicity $z=\frac 1x$ to make $$A=(1-x)^{-1/3}$$ and use either the generalized binomial theorem or Taylor series around $x=0$. This will give $$A=1+\frac{x}{3}+\frac{2 x^2}{9}+\frac{14 x^3}{81}+\frac{35 x^4}{243}+\frac{91
x^5}{729}+O\left(x^6\right)$$ Replace back $x$ by $\frac 1z$ to get $$A=1+\frac{1}{3 z}+\frac{2}{9 z^2}+\frac{14}{81 z^3}+\frac{35}{243 z^4}+\frac{91}{729
z^5}+O\left(\frac{1}{z^6}\right)$$
For illustration purposes, using $z=2$, the above expansion gives $\frac{29317}{23328}\approx 1.25673$ while the exact value $\sqrt[3]{2}\approx 1.25992$. For sure, it will be better and better when $z$ will increase.
For example using $z=10$, the above expansion gives $\frac{75505741}{72900000}\approx 1.035744047$ while the exact value $\approx 1.035744169$.
|
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|
Is it 'not mathematical' to compare the L.H.S. and R.H.S in such type of equations? $$x+\frac{1}{x}=25 + \frac{1}{25}$$
The solution is very simple. But the problem is whether my solution is correct or not. I did it by simply comparing the LHS and the RHS. Thus, I got $x=25$ or $\frac{1}{25}$. But my book does it in this way
$x-25=\frac{1}{25}-\frac{1}{x}=\frac{x-25}{25x}$. So, $x=25$ or $1=\frac {1}{25x}\implies x=\frac{1}{25}$.
I asked my teacher whether my method was correct or not. She told me that the method in the book is correct and that my method of comparing will not be accepted during the exam as it is 'not mathematical' and is 'some sort of hit and trial'.
Now, I am not worried about whether I'll be awarded marks for my method or not. But is it 'not mathematical' to compare the L.H.S. and R.H.S in such type of equations?
|
The "longer" method is:
\begin{align}
x + \frac{1}{x} &= 25 + \frac{1}{25} \\
x^{2} + 1 &= \left(25 + \frac{1}{25} \right) x \\
x^{2} - \left(25 + \frac{1}{25} \right) x + 1 &= 0 \\
x &= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \sqrt{\left(25 + \frac{1}{25} \right)^2 - 4} \\
&= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \sqrt{\left(25 - \frac{1}{25} \right)^2} \\
&= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \left(25 - \frac{1}{25} \right) \\
x &= 25 \hspace{5mm} \mbox{or} \hspace{5mm} \frac{1}{25}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding if sentence is true using induction: $\sqrt1 + \sqrt2 +\dots+\sqrt{n} \le \frac23(n+1)\sqrt{n+1}$ I have to prove
$\sqrt{1} + \sqrt{2} +...+\sqrt{n} \le \frac{2}{3}*(n+1)\sqrt{n+1}$
by using math induction.
First step is to prove that it works for n = 1 , which is true.
Next step is to prove it for n + 1. We can rewrite the formula using
$\sum_{i=1}^{n+1} \sqrt{i}= \sum_{i=1}^{n}\sqrt{i} + \sqrt{i+1}$
and we can substitute sum
$\frac{2}{3}(n+1)\sqrt{n+1} +\sqrt{n+1} \le \frac{2}{3}(n+2)\sqrt{n+2}$
we can transform the left side into
$\sqrt{n+1}(\frac{2}{3}(n+1)+1)$
but how to I further transform the formula in order to find if the sentence is true?
Thanks for all help!
|
To finish your proof by induction we could use the fact the following inequalities are equivalent to each other
\begin{align*}
\frac23(n+1)^{3/2}+\sqrt{n+1} &\le \frac23(n+2)^{3/2}\\
\frac32\sqrt{n+1} &\le (n+2)^{3/2} - (n+1)^{3/2}\\
\frac32\sqrt{n+1} &\le (\sqrt{n+2} - \sqrt{n+1}) (n+2 + \sqrt{(n+2)(n+1)} + n+1)\\
\frac32\sqrt{n+1}(\sqrt{n+2} + \sqrt{n+1}) &\le n+2 + \sqrt{(n+2)(n+1)} + n+1\\
\frac32(n+1 + \sqrt{(n+2)(n+1)}) &\le 2n+3 + \sqrt{(n+2)(n+1)}\\
\frac12\sqrt{(n+2)(n+1)} &\le \frac{n+3}2
\end{align*}
I have:
Used $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=\sqrt{n+2}$ $b=\sqrt{n+1}$
Multiplied both sides by $\sqrt{n+2}+\sqrt{n+1}$ to use that $(\sqrt{n+2} - \sqrt{n+1})(\sqrt{n+2} + \sqrt{n+1}) = (n+2)-(n+1)=1$.
If you already know how to integrate, you can simply use that
$$\newcommand{\dx}{\; \mathrm{d}x}\sqrt k \le \int_{k}^{k+1} \sqrt x \dx.$$
Simply by adding these inequalities together for $k=0$ to $n$ you get
$$ \sum_{k=0}^n \sqrt n \le \int_0^{n+1} \sqrt x \dx = \frac{(n+1)^{3/2}}{3/2} = \frac23 (n+1)\sqrt{n+1}.$$
This can be also visualized by noticing that one side is are of rectangle under the curve $y=\sqrt x$ and the other is the whole area under the curve.
I will add the following - somewhat similar - picture which might help (it was taken from this post):
|
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|
How to split 59 in $\mathbb{Q}(\sqrt{13})$ How to split 59 in $\mathbb{Q}(\sqrt{13})$
as i know that 59 is prime and we can write $(x+y\sqrt{13})(x-y\sqrt{13})=59$
$x^2-13y^2=59$ but i cant find the x,y
|
Agree with Parcly Taxel.
In fact for squarefree $-50 \leq d \leq 50$, $59$ splits in $\Bbb{Q}[\sqrt{d}]$ only for \begin{align}
d &=-43 & 59 &= 4^2 + 43 \cdot 1^2 \\
d &=-34 & 59 &= 5^2 + 34 \cdot 1^2 \\
d &=-23 & 59 &= 6^2 + 23 \cdot 1^2 \\
d &=-11 & 59 &= \left(\frac{15}{2}\right)^2 + 11 \cdot \left(\frac{1}{2}\right)^2 & [11 \cong 3 \pmod{4}] \\
d &=-10 & 59 &= 7^2 + 10 \cdot 1^2 \\
d &= -2 & 59 &= 3^2 + 2 \cdot 5^2 \\
d &= 1 & 59 &= 30^2 - 1 \cdot 29^2 & [\text{"splits" as $59 \cdot 1$}] \\
d &= 5 & 59 &= 8^2 - 5 \cdot 1^2 \\
d &= 17 & 59 &= 22^2 - 17 \cdot 5^2 \\
d &= 22 & 59 &= 9^2 - 22 \cdot 1^2 \\
d &= 29 & 59 &= \left(\frac{31}{2}\right)^2 - 29 \cdot \left(\frac{5}{2}\right)^2 \\
d &= 41 & 59 &= 10^2 - 41 \cdot 1^2 \\
d &= 46 & 59 &= 75^2 - 46 \cdot 11^2
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve Recurrence Relation for Maximum and Minimum in an Array. I know that recursive equation of this algo is
$T\left ( n \right )=2T\left ( \frac{n}{2} \right )+2 $
Given that $T\left ( 1 \right )=0,T\left ( 2 \right )=1 $
and its solution is also given here, ijust want to clear my doubt where i am stuck at..
I solved this eqtn as--:
$\Rightarrow T\left ( n \right )=2\left ( 2*T\left ( \frac{n}{4} \right )+2 \right )+2 $
$\Rightarrow T\left ( n \right )=2^{2}+T\left ( \frac{n}{2^{2}} \right )+2^{2}+2^{1}$
$\vdots$
$\Rightarrow T\left ( n \right )=2^{k}*T\left ( \frac{n}{2^{k}} \right )+2^{k}+\cdots 2^{2}+2^{1}$
Taking $T\left ( 1 \right )=0 $
$\frac{n}{2^{k}}=1 \Rightarrow k=\log_{2}n $
And our equation becomes
$2^{k}+2^{k-1}+\cdots 2^{2}+2^{1} = \frac{2*\left ( 1-2^{k} \right )}{1-2}=2n-2 ,\left \{ 2^{k}=2^{log_{2}n}=n\right \} $
But answer is $\frac{3*n}{2}-2$..please correct me in this equation.
|
Changing my base case from $T\left ( 1 \right )=0$ to $T\left ( 2 \right )=1 $ .my equation looks like
$\Rightarrow T\left ( n \right )=2^{k}*T\left ( \frac{n}{2^{k}} \right )+2^{k}+\cdots 2^{2}+2^{1}$
Taking $T\left ( 2 \right )=1 $
$\frac{n}{2^{k}}=2 \Rightarrow k=\log_{2}\frac{n}{2} $
And our equation becomes
$T\left(n \right)=2^{k}+2^{k}+2^{k-1}+\cdots 2^{2}+2^{1}$
$2^{k}=\frac{n}{2}$
$T\left(n \right)=2^{k}+\sum_{j=1}^{j=k}2^{j}$
$T\left(n \right)=2^{k}+2*\frac{2^{k}-1}{2-1}$
$T\left(n \right)=\frac{n}{2} +2*\left(\frac{n}{2} -1 \right) $
$T\left(n \right)=\frac{3n}{2}-2$
|
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|
Evaluating $ \int_{-\infty}^{\infty} \frac{e^{-ix}}{\sqrt{x+2i} + \sqrt{x+5i}}dx$ I want to compute following improper integral via residue theorem.
$ \int_{-\infty}^{\infty} \frac{e^{-ix}}{\sqrt{x+2i} + \sqrt{x+5i}}dx$
Note this is a exercise 4.57 in Krantz, complex analysis textbook. (Function theory of one complex variable)
First my trial is using $(A+B)(A-B)= A^2-B^2$, change above integral into the form of
\begin{align}
\frac{1}{-3i}\int_{-\infty}^{\infty} e^{-ix} ( \sqrt{x+2i} - \sqrt{x+5i})dx
\end{align}
which seems no pole.
I know the form of integral
\begin{align}
\int_{-\infty}^{\infty} \frac{e^{-im z}}{\sqrt{z+ia}} = 2 e^{-m a} e^{-\frac{\pi i}{4}} \sqrt{\frac{\pi}{m}}
\end{align}
where $a$ and $m$ is positive. I obtain above results by taking keyhole countour around branch cut $z=-i a$. $i.e$, take three parts $z=-ia + x e^{-\frac{i\pi}{2}}$, $z=-ia + \epsilon e^{i\theta}$, $z=-ia + x e^{\frac{i3\pi}{2}}$ where $\epsilon \leq x < \infty$ and $-\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$ and take limt $\epsilon \rightarrow 0$.
But to apply this to above form to make
\begin{align}
\sqrt{x+2i} + \sqrt{x+5i} =0 \qquad \Rightarrow \qquad \sqrt{x+2i} = -\sqrt{x+5i} = \sqrt{ix-5} \qquad x = \frac{-5-2i}{1-i}
\end{align}
|
We take a branch of $\sqrt{z}$ as follows:$$
\sqrt{z}=\sqrt{r}e^{i\frac{\theta }{2}},\quad \quad z=re^{i\theta} \,\,\left(-\frac{\pi}{2}<\theta <\frac{3\pi}{2}\right)
.$$
Then $\sqrt{z}$ is analytic and single-valued in $\mathbb{C}\setminus [0,-i\infty)$.
We consider a contour $\Gamma$ consisting of the horizontal line segment $A=[-R, R]$, the quadrant $B$ of radius $R$ traced clockwise from $R$ to $-iR$, the line segment $C$ from $-iR$ to $-5i$, the line segment $D$ from $-5i$ to $-2i$, the line segment $E$ from $-2i$ to $-5i$, the line segment $F$ from $-5i$ to $-iR$ and the quadrant $G$ of radius $R$ traced clockwise from $-iR$ to $-R$. Ofcourse, for instance, $C$ is the limit of $C_\varepsilon =[\varepsilon -5i, \varepsilon -iR]$ as $\varepsilon \to 0.$
See the diagram.
Since $\sqrt{z+2i}+\sqrt{z+5i}$ is single-valued inside $\Gamma$ (by the definition above) and has no zeros, we have $$
\int_\Gamma \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz=0$$
by Cauchy's theorem.
Since $\frac{1}{|\sqrt{z+2i}+\sqrt{z+5i}|} \to 0$ $(|z|\to \infty)$ uniformly, we have $$
\lim_{R\to \infty } \int_{B+G} \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz=0
$$
by Jordan's lemma.
We evaluate integrals on $C, F$.
\begin{align}
\int_C \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz&=\frac{i}{3}\int_C e^{-iz}(\sqrt{z+2i}-\sqrt{z+5i})\,dz\\
&=\frac{1}{3}\int_R^5 e^{-t}(\sqrt{-it+2i}-\sqrt{-it+5i})\,dt\quad (z=-it)\\
&=-\frac{1}{3}\int_5^R e^{-t}(e^{-\frac{\pi i}{4}}\sqrt{t-2}-e^{-\frac{\pi i}{4}}\sqrt{t-5})\,dt\\
&=\frac{-1+i}{3\sqrt{2}}\int_5^R e^{-t}(\sqrt{t-2}-\sqrt{t-5})\,dt.
\end{align}
Simirarly we have
\begin{align}
\int_F \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz&=\frac{1}{3}\int_5^R e^{-t}(\sqrt{-it+2i}-\sqrt{-it+5i})\,dt\\
&=\frac{1}{3}\int_5^R e^{-t}(e^{\frac{3\pi i}{4}}\sqrt{t-2}-e^{\frac{3\pi i}{4}}\sqrt{t-5})\,dt\\
&=\frac{-1+i}{3\sqrt{2}}\int_5^R e^{-t}(\sqrt{t-2}-\sqrt{t-5})\,dt.
\end{align}
Therefore we have$$
\int_{C+F} \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz=\frac{(-1+i)\sqrt{2}}{3}\int_5^R e^{-t}(\sqrt{t-2}-\sqrt{t-5})\,dt.
$$
Next we evaluate integrals on $D, E$.
\begin{align}
\int_D \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz&=\frac{1}{3}\int_5^2 e^{-t}(\sqrt{-it+2i}-\sqrt{-it+5i})\,dt\\
&=-\frac{1}{3}\int_2^5 e^{-t}(e^{-\frac{\pi i}{4}}\sqrt{t-2}-e^{\frac{\pi i}{4}}\sqrt{5-t})\,dt\\
&=\frac{-1+i}{3\sqrt{2}}\int_5^R e^{-t}(\sqrt{t-2}-i\sqrt{5-t})\,dt.
\end{align}
Similarly we have
\begin{align}
\int_E \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz&=\frac{1}{3}\int_2^5 e^{-t}(\sqrt{-it+2i}-\sqrt{-it+5i})\,dt\\
&=\frac{1}{3}\int_2^5 e^{-t}(e^{\frac{3\pi i}{4}}\sqrt{t-2}-e^{\frac{\pi i}{4}}\sqrt{5-t})\,dt\\
&=\frac{-1+i}{3\sqrt{2}}\int_2^5 e^{-t}(\sqrt{t-2}+i\sqrt{5-t})\,dt.
\end{align}
Therefore we have$$
\int_{D+E} \frac{e^{-iz}}{\sqrt{z+2i}+\sqrt{z+5i}}\,dz=\frac{(-1+i)\sqrt{2}}{3}\int_2^5 e^{-t}\sqrt{t-2}\,dt.
$$
Thus,tending $R$ to $\infty$, we get
\begin{align}
\int_{-\infty}^\infty \frac{e^{-ix}}{\sqrt{x+2i}+\sqrt{x+5i}}\,dx&=-\frac{(-1+i)\sqrt{2}}{3}\left(\int_5^\infty e^{-t}(\sqrt{t-2}-\sqrt{t-5})\,dt+\int_2^5 e^{-t}\sqrt{t-2}\,dt\right)\\
&=\frac{(1-i)\sqrt{2}}{3}\left(\int_2^\infty e^{-t}\sqrt{t-2}\,dt-\int_5^\infty e^{-t}\sqrt{t-5}\,dt\right)\\
&=\frac{(1-i)\sqrt{2}}{3}(e^{-2}-e^{-5})\int_0^\infty e^{-u}\sqrt{u}\,du\\
&=\frac{(1-i)(e^{-2}-e^{-5})}{3}\sqrt{\frac{\pi}{2}}.
\end{align}
|
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|
How prove this $BC$ always passes through a fixed point with $\frac{x^2}{4}+y^2=1$
if the point $A(0,1)$ on the ellipse $\Gamma:$ $\dfrac{x^2}{4}+y^2=1$ and the circle $\tau:$ $(x+1)^2+y^2=r^2(0<r<1)$,if $AB,AC$ tangent the circle $\tau$ ,$B,C\in \Gamma$,show that the line $BC$ always passes through a fixed point
I try Let $AB:y=kx+1$ then
$$\begin{cases}
\dfrac{x^2}{4}+y^2=1\\
y=kx+1
\end{cases}
$$
so we have
$$x^2+4(kx+1)^2=4\Longrightarrow (4k^2+1)x^2+8kx=0$$
so we have
$$B(-\dfrac{8k}{4k^2+1},\dfrac{1-4k^2}{1+4k^2})$$
But for $C$ it hard to find it.
|
When $r\rightarrow 1$, it's easy to see that $y$-axis will be a tangent line, which means that the fixed point should be on the $y$-axis. So later, we need to put $x=0$ in the equation of tangent line to find the fixed $y$-coordinate. Use the tangent conditions, get $\frac{|k-1|}{\sqrt{k^2+1}}=r$ (here, we get $k_1$ and $k_2$), that is, $(1-r^2)k^2-2k+1-r^2=0$, or
\begin{eqnarray*}
k^2=\frac{r^2+2k-1}{1-r^2}.
\end{eqnarray*}
Use Vita's law ($k_1k_2=1$ and $k_1+k_2=\frac{2}{1-r^2}$) to simply the equations of $AB$, you can get
\begin{eqnarray*}
y=-\frac{2}{3(1-r^2)}(x+\frac{8k_1}{4k_1^2+1})-1+\frac{2}{4k_1^2+1}.
\end{eqnarray*}
Set $x=0$, use the first equation to substitute $k^2$ in the above equation and simply, you will get a constant $-\frac{2}{3}-1=-\frac{5}{3}$. So $AB$ must pass through the point $(0,-\frac{5}{3})$.
|
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}
|
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
|
Induction hypothesis:
$$9|5\times10^k+10^{k-1}+3$$
Thus,
$$9|10(5\times10^k+10^{k-1}+3)$$
$$9|5\times10^{k+1}+10^k+30$$
$$9|5\times10^{k+1}+10^k+3+27$$
But $9|27$. Hence,
$$9|5\times10^{k+1}+10^k+3$$
which is what had to be shown.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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|
Symmetric ratio of binomial probabilities around the midpoint $\frac{m+1}{2}$ greater $\frac{p}{1-p}$ for $p>0.5$ I am trying to show the following:
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)}\geq \frac{p}{1-p}$,
with $n>m\geq \frac{n+1}{2}$, $1>p>0.5$ and $n,m$ are both odd Natural Numbers with $n\geq3$.
The probabilities I am considering come from a binomial distribution, so that I get
$\frac{\displaystyle \sum_{i=\frac{m+1}{2}}^{\frac{n-1}{2}}\binom{m}{i}p^i (1-p)^{m-i}}{\displaystyle \sum_{i=\frac{2m-n+1}{2}}^{\frac{m-1}{2}} \binom{m}{i}p^i(1-p)^{m-i}}\geq\frac{p}{1-p}$
It is worth noting that the concerned intervals lie symmetrically around $\frac{m}{2}$ on the line $0,\ldots,m$:
$m-\frac{m+1}{2}=\frac{2m-n+1}{2}=\frac{2m-n+1}{2}-0$ and
$\frac{m+1}{2}-\frac{m}{2}=\frac{1}{2}=\frac{m}{2}-\frac{m-1}{2}$.
Now a as $p>0.5$, it seems intuitive that the probability above the midpoint is higher than the one below it. But I just don't know how to show that it igretaer than $\frac {p}{1-p}$. All my other results indicate that it must be true, but I am unable to show it directly.
Since I have absolutely no idea how to tackle this problem, I would appreciate any hints, ideas or help.
Thanks in advance.
Update:
I have played around a bit with the above expression and reformulated it. Maybe this helps?
$P_r(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2})=P_r(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2})-P_r(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2})$
Therefore,
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}= \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)-P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}=\frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}-1 $
Thus
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)}\geq \frac{p}{1-p} \iff \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}-1 \geq \frac{p}{1-p}$
$\iff \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)} \geq \frac{1}{1-p} \iff (1-p) \geq \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}$
$\iff
1-\frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)} \geq p
\iff \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)-P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)} \geq p$
$\iff \frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)} \geq p$
From here on, I don't know how to show that:
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}\geq p$
|
Note that
\begin{align}
\sum_{i=(2m-n+1)/2}^{(m-1)/2} \binom{m}{i}p^i(1-p)^{m-i} &=~\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{m-i}p^{m-i}(1-p)^i \\
&=~\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{m-i}(1-p)^i
\end{align}
Therefore,
\begin{align}
&\frac{\Pr\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{\Pr\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)} =~\frac{\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{i}(1-p)^{m-i}}{\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{m-i}(1-p)^i} \\
\geq~&\min\left\{\frac{\binom{m}{(m+1)/2}p^{(m+1)/2}(1-p)^{(m-1)/2}}{\binom{m}{(m+1)/2}p^{(m-1)/2}(1-p)^{(m+1)/2}},\ \cdots,\ \frac{\binom{m}{(n-1)/2}p^{(n-1)/2}(1-p)^{(2m-n+1)/2}}{\binom{m}{(n-1)/2}p^{(2m-n+1)/2}(1-p)^{(n-1)/2}}\right\} \\
=~&\min\left\{\frac{p^{(m+1)/2}(1-p)^{(m-1)/2}}{p^{(m-1)/2}(1-p)^{(m+1)/2}},\ \cdots,\ \frac{p^{(n-1)/2}(1-p)^{(2m-n+1)/2}}{p^{(2m-n+1)/2}(1-p)^{(n-1)/2}}\right\} \\
=~&\min\left\{\left(\frac{p}{1-p}\right)^{\frac{m+1}{2} -\frac{m-1}{2}},\ \cdots,\ \left(\frac{p}{1-p}\right)^{\frac{n-1}{2}-\frac{2m-n+1}{2}}\right\} \\
=~&\min\left\{\left(\frac{p}{1-p}\right)^1,\ \cdots,\ \left(\frac{p}{1-p}\right)^{n-m-1}\right\} \\
=~& \frac{p}{1-p}
\end{align}
where the first inequality is because $\frac{a_1 + \cdots + a_n}{b_1 + \cdots + b_n} \geq \min\{\frac{a_1}{b_1}, \cdots, \frac{a_n}{b_n}\}$ for positive $a_i$s and $b_i$s.
|
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"url": "https://math.stackexchange.com/questions/2000363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$f(x) = x^2$. For each positive integer $n$, let $P_n$ be the partition $P_n = \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n},1 \}$ of $[0,1]$. Let $f(x) = x^2$. For each positive integer $n$, let $P_n$ be the partition $P_n = \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n},1 \}$ of $[0,1]$.
Show that $S(P)=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}$ .Find a similar expression for $s(P)$.
I know that $S(P)=\sum_{i=1}^{n}M_j(x_j-x_{j-1})$ where $M_j=\sup_{[x_{j-1},x_j]}f(x)$. I'm not sure how to go from there.
|
For $j=1,2...n\;$, we have
$$x_j=j\frac{1}{n},$$
$$x_{j}-x_{j-1}=\frac{1}{n},$$
$$\displaystyle{M_j=sup_{x\in[x_{j-1},x_j]}f(x)=f(x_j)=x_j^2=\frac{j^2}{n^2}},$$
since $f$ is increasing.
and
$$S(P)=\sum_{j=1}^n(x_j-x_{j-1})M_j=\frac{1}{n^3}\sum_{j=1}^n j^2$$
$$=\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}$$
$$=\frac{1}{6n^2}(2n^2+3n+1)$$
$$=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}.$$
if $n\to+\infty$, $\; S(P)$ goes to
$$\int_0^1 x^2dx=\frac{1}{3}.$$
|
{
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"url": "https://math.stackexchange.com/questions/2001079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determine whether $\sum\limits_{n=1}^{\infty} \frac{ \sqrt{n} }{n^2+3}$ converges or diverges. - Please check my reasoning. Determine whether $\sum_{n=1}^{\infty} \dfrac{ \sqrt{n} }{n^2+3}$ converges or diverges.
I would appreciate it if someone could check my reasoning for this problem and indicate if I have made any mistakes, why they are mistakes, and how to properly reason for the answer.
My reasoning is as follows.
We know that $ 0 > \dfrac{ \sqrt{n} }{n^2+3} > \dfrac{1}{n^2} $.
We also know that $\dfrac{1}{n^2}$ diverges by the p-series test, since $p=2>1$.
Therefore, since $\dfrac{1}{n^2}$ diverges, $\sum_{n=1}^{\infty} \dfrac{ \sqrt{n} }{n^2+3}$ also must diverge, since $0 >\dfrac{ \sqrt{n} }{n^2+3} > \dfrac{1}{n^2}$.
Thank you.
|
One may observe that, for $n>1$,
$$
0<\frac{\sqrt{n}}{n^2+3} < \frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}
$$ since $\displaystyle \sum\frac{1}{n^{3/2}}$ is convergent then the initial series $\displaystyle \sum\frac{\sqrt{n}}{n^2+3} $ is convergent by the comparison test.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2003145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Generalization of Leibniz formula for $\pi$ The well-known Leibniz formula for $\pi$ is
$$
1-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{\pi}{4}.
$$
Looking at some nonstandard techniques, I've happened upon the following formula:
\begin{equation} \label{armdark}
\begin{split}
\Big( \frac{1}{1} + \frac{1}{3} &+ \ldots + \frac{1}{2q-1}\Big) - \Big( \frac{1}{2q+1} + \ldots + \frac{1}{4q-1}\Big) \\
&+ \Big( \frac{1}{4q+1} + \ldots + \frac{1}{6q-1}\Big) - \Big( \frac{1}{6q+1} + \ldots + \frac{1}{8q-1}\Big) + \ldots \\
& \qquad \qquad = \frac{\pi}{4q} \sum_{k=0}^{q-1} \Big(\frac{1+\tan(\frac{\pi}{4}(\frac{q-1-2k}{q}))}{1-\tan(\frac{\pi}{4}(\frac{q-1-2k}{q}))} \Big).
\end{split}
\end{equation}
In other words, we are summing the reciprocals of the odd integers, but rather than signs which alternate at each successive term we are adding $q$ positive terms in a row, then $q$ negative ones, then $q$ positive ones, etc. Clearly, this reduces to the Leibniz formula if $q=1$, but if $q=2$ we get
$$
\frac{1}{1} + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{13} - \frac{1}{15} + \ldots = \frac{\pi \sqrt{2}}{4},
$$
and for $q=3$,
$$
\frac{1}{1} + \frac{1}{3} + \frac{1}{5} - \frac{1}{7} - \frac{1}{9} - \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \frac{1}{17} - \ldots = \frac{5 \pi}{12},
$$
and so forth. If we want to do this sort of thing for all positive integers, rather than just the odd ones, then the formula is
\begin{equation} \label{embeth}
\begin{split}
\Big( \frac{1}{1} + \frac{1}{2} &+ \ldots + \frac{1}{q}\Big) - \Big( \frac{1}{q+1} + \ldots + \frac{1}{2q}\Big) \\
& + \Big( \frac{1}{2q+1} + \ldots + \frac{1}{3q}\Big) - \Big( \frac{1}{3q+1} + \ldots + \frac{1}{4q}\Big) + \ldots \\
& \qquad \qquad = \frac{\ln 2}{q} + \frac{\pi}{2q} \sum_{k=1}^{q-1} \Big(\frac{1+\tan(\frac{\pi}{4}(\frac{q-2k}{q}))}{1-\tan(\frac{\pi}{4}(\frac{q-2k}{q}))} \Big).
\end{split}
\end{equation}
This reduces to known results for $q=1,2$ (e.g. for $q=1$ the sum on the right is empty, and we get $1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = \ln 2$), and taking for instance $q=3$ we get
$$
\frac{1}{1} + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} - \ldots = \frac{2\pi}{3\sqrt{3}} + \frac{\ln 2}{3},
$$
and $q=4$ gives
$$
\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} - \frac{1}{8} + \ldots = \frac{\pi(1+2\sqrt{2})}{8} + \frac{\ln 2}{4}.
$$
I don't know too much about these types of sums, so my question is whether anyone has seen these sums before, and/or how someone might go about proving them. I mean, I have proofs of them, but I'm curious to see whether and how more standard techniques can produce them.
Thanks in advance,
Greg
|
The idea is to use $\sum_{n=1}^\infty \exp(2\pi i nm/\ell)/n=-\log(1-\exp(2\pi i m/\ell))$ for any $m$ not divisible by $\ell$. From here you can find $\sum_{n=1}^\infty c_n/n$ for any $\ell$-periodic sequence $c_n$ such that $\sum_{n=1}^\ell c_n=0$, by finding $d_m$'s such that $c_n=\sum_{m=1}^\ell d_m\exp(2\pi i nm/\ell)$, and hence
$$\sum_{n=1}^\infty c_n/n=-\sum_{m=1}^\ell d_m\log(1-\exp(2\pi i m/\ell))\qquad(*)$$
The numbers $d_m$ can be found as (inverse finite Fourier transform if you wish)
$$d_m=\frac{1}{\ell}\sum_{n=1}^\ell c_n \exp(-2\pi i nm/\ell)$$
(notice that $d_\ell=0$). You now need to compute $d_m$'s for your favorite sequence $c_n$ and simplify the finite sum $(*)$.
|
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|
Inequality based on AM/GM Inequality Find the greatest value of $x^3y^4$
If $2x+3y=7 $ and $x≥0, y≥0$.
(Probably based on weighted arithmetic and geometric mean)
|
My usual attempt to generalize.
To find the max of
$x^m y^n$
subject to
$ax+by = c$
where
$m$ and $n$
are integers.
Following
Macavity's solution,
use the
AGM inequality
in the form
$\sum_{i=1}^p a_i
\ge p (\prod_{i=1}^p a_i)^{1/p}
$.
Then
$\begin{array}\\
c
&=m\frac{ax}{m}
+n\frac{by}{n}\\
&\ge (m+n)((\frac{ax}{m})^m(\frac{by}{n})^n)^{1/(m+n)}
\qquad (m \text{ copies of }\frac{ax}{m}
\text{ and }n\text{ copies of }\frac{by}{n})\\
\end{array}
$
with equality,
by the AGM inequality,
if and only if
$\frac{ax}{m}
=\frac{by}{n}
$.
If this is true,
then
$y
=\frac{anx}{bm}
$
so
$\begin{array}\\
c
&=m\frac{ax}{m}
+n\frac{by}{n}\\
&=m\frac{ax}{m}
+n\frac{ax}{m}\\
&=ax(1+\frac{n}{m})\\
&=ax\frac{m+n}{m}\\
\end{array}
$
or
$x
=\frac{mc}{a(m+n)}
$
and
$y
=\frac{mc}{a(m+n)}\frac{an}{bm}
=\frac{nc}{b(m+n)}
$.
Finally,
$x^m y^n
=(\frac{mc}{a(m+n)})^m(\frac{nc}{b(m+n)})^n
=(\frac{m}{a})^m(\frac{n}{b})^n(\frac{c}{m+n})^{m+n}
$.
If $m+n = c$,
as in OP's case,
$x
=\frac{m}{a}
$
and
$y
=\frac{n}{b}
$
and the max is
$x^m y^n
=(\frac{m}{a})^m(\frac{n}{b})^n
$.
In this case,
$m=3, n=4, a=2, b=3, c=7$.
The result is
$x
= \frac{3}{2}
$,
$y
= \frac{4}{3}
$,
and the max is
$(\frac{3}{2})^3(\frac{4}{3})^4
=\frac{2^5}{3}
$.
Note that
this generalizes easily to
maximize
$\prod_{i=1}^m x_i^{n_i}
$
subject to
$c
=\sum_{i=1}^m a_ix_i
$
with
$c, n_i,$
and $a_i$
being specified.
|
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"url": "https://math.stackexchange.com/questions/2006086",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Prove that $\left(0.5^x-0.25^x\right)^{\frac{1}{x}}$ is strictly increasing The problem is to prove that
$$\left(0.5^x-0.25^x\right)^{\frac{1}{x}}$$
is strictly increasing on $\mathbb{R}^+$. The problem I'm having is that once I found the derivative to be
$$\left(\dfrac{1}{2^x}-\dfrac{1}{4^x}\right)^\frac{1}{x}\left(\dfrac{\frac{\ln\left(4\right)}{4^x}-\frac{\ln\left(2\right)}{2^x}}{x\left(\frac{1}{2^x}-\frac{1}{4^x}\right)}-\dfrac{\ln\left(\frac{1}{2^x}-\frac{1}{4^x}\right)}{x^2}\right),$$
I don't know how to show that this expression is always positive. Any help?
|
Assume $a$ is a real number such that $a>1$. Then let
$$
f_a(x):=\left(1-\frac1{a^x}\right)^{1/x}, \qquad x>0.
$$ We have
$$
\begin{align}
f_a'(x)&=\left( -\frac1{x^2}\cdot \ln \left(1-\frac1{a^x}\right)+\frac1x \cdot \frac{\ln a \cdot \frac1{a^x}}{1-\frac1{a^x}} \right)\cdot\left(1-\frac1{a^x}\right)^{1/x}
\\\\&=\frac1x \left( -\frac1{x}\cdot \ln \left(1-\frac1{a^x}\right)+ \frac{\ln a }{a^x-1} \right)\cdot\left(1-\frac1{a^x}\right)^{1/x}
\\\\&>0 \quad (\text{a sum of positive terms})
\end{align}
$$ thus $f_a$ is increasing over $(0,\infty)$. Now write
$$
\left(0.5^x-0.25^x\right)^{\frac{1}{x}}=\frac12 \cdot\left(1-\frac1{2^x} \right)^{1/x}=\frac12 f_2(x)
$$ to conclude.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a_{ij}=\max(i,j)$, calculate the determinant of $A$
If $A$ is an $n \times n$ real matrix and
$$a_{ij}=\max(i,j)$$
for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.
So, we know that
$$A=\begin{pmatrix}
1 & 2 & 3 & \dots & n\\
2 & 2 & 3 & \dots & n\\
3 & 3 & 3 & \dots & n\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
n& n & n & \dots & n
\end{pmatrix}$$
but what do I do after?
|
Well you can prove this using induction too. The n = $1$ case is trivial. Let, $\det(A_{n}) = n(-1)^{n-1}$. We'll need to prove $$\det(A_{n+1}) = (-1)^{n}(n+1)$$Now $A_{n+1}$ looks like:-
$$A_{n+1}=\begin{pmatrix}A_{n}&P\\
P^{T}&n+1
\end{pmatrix}
$$
where $P^{T} = (n+1)(1,1,...1)^{T}$ $$\det(A_{n+1}) = \det(A_{n} - P\frac{1}{(n+1)}P^{T})(n+1)$$
$$\det(A_{n+1}) = \det(A_{n} - (n+1)J)(n+1)\cdots\cdots\cdots\cdots [1]$$
where J is a square matrix of only ones. Now,
$$-(A_{n} - (n+1)J)=
\begin{pmatrix}
n & n-1 & n-2 & \cdots &1 \\
n-1 & n-1 & n-2 & \cdots & 1 \\
n-2&n-2&n-2&\cdots &1\\
\vdots & \vdots &\vdots &\ddots&\vdots\\
1 & 1 & 1&\cdots& 1 \\
\end{pmatrix}
$$
Inverse of such matrix is of the form:-
$$D = -(A_{n} - (n+1)J)^{-1}=
\begin{pmatrix}
1 & -1 & 0 & \cdots &0 \\
-1 & 2 & -1 & \cdots & 0 \\
0&-1&2&\cdots &0\\
\vdots & \vdots &\vdots &\ddots&\vdots\\
0 & 0 & 0&\cdots& 2 \\
\end{pmatrix}
$$
By expanding along row 1 we could break det(D) in terms of determinant of a simpler matrix B i.e.$$\det(D_{n}) = \det(B_{n-1})-\det(B_{n-2})$$ where, $$B =
\begin{pmatrix}
2 & -1 & 0 & \cdots &0 \\
-1 & 2 & -1 & \cdots & 0 \\
0&-1&2&\cdots &0\\
\vdots & \vdots &\vdots &\ddots&\vdots\\
0 & 0 & 0&\cdots& 2 \\
\end{pmatrix}
$$
B follows the recursion formulla $$\det(B_{n}) = 2\det(B_{n-1})-\det(B_{n-2})$$ Base case $\det(B_{1}) = 2$, which gives $\det(B_{n}) = n+1. $
Hence, For such matrices D we have $\det(D_{n}) = 1$ $\space$ $\forall n\geq2$
Now from [1] we have $$\det(A_{n+1}) = (-1)^{n}(1)(n+1)$$
|
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|
Taylor's series problems Can you help please me solve this by using Taylor series?
I would be grateful if you can explain how did you solve it
$$\lim_{x\to 0}{\cosh{2x\over 2+x^4}+\cos{2x\over 2+x^4}-2e^{x^4\over 2}\over \tan\sqrt{1+x^4}-\tan\sqrt{1-x^4}}$$
https://i.stack.imgur.com/xUZA1.png
|
Let's develop every term to the fifth order:
$$\cosh \frac{2x}{2+x^4} \approx 1+ \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^5)$$
$$\cos \frac{2x}{2 + x^4} \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^5)$$
$$2\ e^{x^4/2} \approx 2 + x^4 + \mathcal{O}(x^5)$$
$$\sqrt{1 + x^4} \approx 1 + \frac{x^4}{2} + \mathcal{O}(x^5)$$
$$\tan\sqrt{1+x^4} \approx \tan(1) + \frac{1}{2}\left(1 + \tan(1)^2\right)x^4 + \mathcal{O}(x^5)$$
$$\sqrt{1 - x^4} \approx 1 - \frac{x^4}{2} + \mathcal{O}(x^5)$$
$$\tan\sqrt{1+x^4} \approx \tan(1) +\left(- \frac{1}{2} - \frac{\tan(1)^2}{2}\right)x^4 + \mathcal{O}(x^5)$$
Accodino to that, we rewrite the function as
$$\lim_{x\to 0} \frac{1+ \frac{x^2}{2} + \frac{x^4}{24} + 1 - \frac{x^2}{2} + \frac{x^4}{24} - 2 - x^4}{\tan(1) + \frac{1}{2}\left(1 + \tan(1)^2\right)x^4 -\tan(1) -\left(- \frac{1}{2} - \frac{\tan(1)^2}{2}\right)x^4 }$$
$$\lim_{x\to 0} \frac{-\frac{22}{24}}{\frac{2 + 2\tan(1)^2}{2}}$$
$$\lim_{x\to 0} \frac{-\frac{11}{12}}{\tan(1)^2 + 1}$$
Now, generally speaking
$$\tan^2(x) + 1 = \frac{\sin^2(x)}{\cos^2(x)} + 1 = \frac{\sin^2(x) + \cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}$$
In your case
$$1 + \tan^2(1) = \frac{1}{\cos^2(1)}$$
hence the final result is
$$\boxed{-\frac{11}{12}\cos^2(1)}$$
|
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|
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-25\tag{2}$$
From $(1)$ and $(2)$, we have
$$\alpha+\beta=\dfrac{(x-5)(x+5)}{x-5}=x+5\qquad(\therefore x\neq5)\tag{3}$$
From $(1)$ and $(3)$, we get
$$\alpha=\sqrt{3x^2-7x-30}=x\qquad\qquad\beta=\sqrt{2x^2-7x-5}=5$$
On solving any one of $\alpha=x$ or $\beta=5$, we get $x=6,-\dfrac{5}{2}$.
On putting $x=-\dfrac{5}{2}$, $\alpha=x$ is not satisfied. Hence, $x=6$ is the only solution.
But as in $(3)$, we have ruled out $x=5$, as a solution then we can't put $x=5$ in the original equation. But, if we do put $x=5$ in the original equation we see that it is indeed satisfied. So, I tried a solution which also gives $x=5$ as a solution. So I picked the last solution from the $(2)$ without cancelling $(x-5)$.
$$(x-5)(\alpha+\beta)=x^2-25\implies (x-5)(\alpha+\beta-(x+5))=0$$.
Now, I am pretty much stuck here.
Edit-1:- I just saw now and feel pretty stupid about it.
If we proceed from the last step i.e. $(x-5)(\alpha+\beta-(x+5))=0$, we get
$$(x-5)(\alpha+\beta-(x+5))=0\implies (x-5)=0\;\;\;\text{ or }\;\;\;\alpha+\beta=x+5$$
So we have the following equations to be solved:-
$$\alpha-\beta=x-5\\
\alpha+\beta=x+5\\
x-5=0$$
Which results in $x=5,6$.
Now, since I have put so much effort in posting this question, I might as well ask for better solutions if you can come up with one. And, please don't post a solution which includes a lot of squaring to result into a quartic equation.
|
Your $(2)$ is incorrect. Instead, we should square both sides of $(1)$ to get
$$(\alpha-\beta)^2=(x-5)^2$$
$$\alpha^2-2\alpha\beta+\beta^2=(x-5)^2$$
$$2\alpha\beta=(x-5)^2-\alpha^2-\beta^2$$
Square both sides again
$$4\alpha^2\beta^2=\left[(x-5)^2-\alpha^2-\beta^2\right]^2$$
Now all radicals are removed and you should get something like two quartic polynomials. Using the rational roots theorem, we get
$$x=5,6\tag{Click here}$$
And keeping in mind that we must have
$$3x^2-7x-30>0,\ 2x^2-7x-5>0$$
These are the only solutions.
|
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|
Derivative with quotient rule Trying to get $\dfrac{d}{dx}\left[\frac{20e^x}{(e^x+4)^2}\right]$.
After quotient rule: $$f'(x)=20\dfrac{e^x(e^{2x}+8e^x+16)-e^x(2e^x+8e^x)}{(e^x+4)^4}\\\\\\= 20\dfrac{e^{3x}+8e^{2x}+16e^x-2e^{2x}-8e^x}{(e^x+4)^4}\\\\=20\dfrac{e^{3x}+6e^{2x}+8e^x}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+4)(e^x+2)}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+2)}{(e^x+4)^3}$$
But Wolframalpha says it should be: $-\dfrac{20e^x(e^x-4)}{(e^x+4)^3}$
Where is my mistake?
|
$$
f'(x)=20\frac{e^x(e^{2x}+8e^x+16)-e^x(2e^{\color{red}{2x}}+8e^x)}{(e^x+4)^2}
$$
You also do $e^x\cdot 8e^{x}=8e^x$, which is wrong.
On the other hand, if you consider
$$
g(x)=\frac{x}{(x+4)^2}
$$
you have
$$
g'(x)=\frac{(x+4)^2-x\cdot 2(x+4)}{(x+4)^2}=
\frac{x+4-2x}{(x+4)^3}=-\frac{x-4}{(x+4)^3}
$$
Since
$$
f(x)=20g(e^x)
$$
you have
$$
f'(x)=20g'(e^x)\cdot e^x=-\frac{20e^x(e^x-4)}{(e^x+4)^3}
$$
|
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|
Limit of $\mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}}$ How can I calculate the below limit?
$$
\lim\limits_{x\to \infty} \left( \mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}} \right)
$$
In fact I know should use the L’Hospital’s Rule, but I do not how to use it.
|
Using the fact that $\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =1\\ \\ $ we can write
$$\lim _{ x\rightarrow \infty }{ \left( e^{ \sqrt { x+1 } }-e^{ \sqrt { x } } \right) } =\lim _{ x\rightarrow \infty }{ { e }^{ \sqrt { x } }\left( e^{ \sqrt { x+1 } -\sqrt { x } }-1 \right) } =\lim _{ x\rightarrow \infty }{ { e }^{ \sqrt { x } }\left( \frac { e^{ \frac { 1 }{ \sqrt { x+1 } +\sqrt { x } } }-1 }{ \frac { 1 }{ \sqrt { x+1 } +\sqrt { x } } } \right) } \cdot \frac { 1 }{ \sqrt { x+1 } +\sqrt { x } } =\\ =\lim _{ x\rightarrow \infty }{ \frac { { e }^{ \sqrt { x } } }{ \sqrt { x+1 } +\sqrt { x } } } =+\infty $$
|
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|
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
|
Hint:
Apply $a^3+b^3 = (a+b)(a^2-ab+b^2)$ where you take $a = \cos^2 x$ and $b=\sin^2 x$.
|
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|
Prove that this is a valid formula using axioms of propositional logic The question is how to prove using basic axioms this expression:
$$(A \to B) \to ((\lnot C \to A) \to (\lnot C \to B))$$
I have the list of axioms, one of them looks like this: $A \to (B \to A)...$ But I don't understand how to apply this to my expression...
|
Start with a premised proof:
$$\begin{array} {rl}
\text{Premise} :& A \Rightarrow B
\\ \text{Premise} :& \lnot C \Rightarrow A
\\ \text{Premise} :& \lnot C
\\ \text{MP} :& A
\\ \text{MP} :& B
\end{array}$$
Then apply deduction theorem:
$$\begin{array} {rl}
\text{Ax 1} :& (A \Rightarrow B) \Rightarrow \lnot C \Rightarrow A \Rightarrow B
\\ \text{Ax 2} :& (\lnot C \Rightarrow A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{Premise} :& A \Rightarrow B
\\ \text{Premise} :& \lnot C \Rightarrow A
\\ \text{MP} :& \lnot C \Rightarrow A \Rightarrow B
\\ \text{MP} :& (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{MP} :& \lnot C \Rightarrow B
\end{array}$$
Then apply deduction theorem again:
$$\begin{array} {rl}
\text{Ax 1} :& (A \Rightarrow B) \Rightarrow \lnot C \Rightarrow A \Rightarrow B
\\ \text{Ax 2} :& (\lnot C \Rightarrow A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{Premise} :& A \Rightarrow B
\\ \text{MP} :& \lnot C \Rightarrow A \Rightarrow B
\\ \text{MP} :& (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\end{array}$$
Then apply deduction theorem again:
$$\begin{array} {rl}
\\ \text{Ax 2} :& (\lnot C \Rightarrow A \Rightarrow B) \Rightarrow
\\ & (\lnot C \Rightarrow A) \Rightarrow
\\ & \lnot C \Rightarrow B
\\ \text{Ax 1} :& ((\lnot C \Rightarrow A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B) \Rightarrow
\\ & (A \Rightarrow B) \Rightarrow
\\ & (\lnot C \Rightarrow A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{MP} :& (A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{Ax 2} :& ((A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B) \Rightarrow
\\ & ((A \Rightarrow B) \Rightarrow \lnot C \Rightarrow A \Rightarrow B) \Rightarrow
\\ & (A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{MP} :& ((A \Rightarrow B) \Rightarrow \lnot C \Rightarrow A \Rightarrow B) \Rightarrow (A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\\ \text{Ax 1} :& (A \Rightarrow B) \Rightarrow \lnot C \Rightarrow A \Rightarrow B
\\ \text{MP} :& (A \Rightarrow B) \Rightarrow (\lnot C \Rightarrow A) \Rightarrow \lnot C \Rightarrow B
\end{array}$$
|
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|
BMO1 2011 Problem 1: Find $n$ such that $n^2+20n+11$ is a perfect square
*
*Find all (positive or negative) integers
$n$ for which
$$n^2+20n+11$$
is a perfect square. Remember that you must justify that you have found
them all.
For the first part I did so:
$$n^2+20n+11=M^2 $$ where M is an integer.
That factors down to:
$$ (n+10)^2-M^2=89$$
$$ (n+10-M)(n+10+M)=89$$
From there it is easy to find solutions as 89 is prime and it is a diophantine equation.
So the first part of the question is quite simple, however I am completely stumped on how to do the second part, how might we show that those are the only solutions?
|
Note that
$$n^2+20n+11=(n+10)^2-89$$
so we have to find all the pairs of squares $a^2$, $b^2$ such that $a^2-b^2=89$. Of course, $a$ and $b$ are integers. We see that $|a|>|b|$.
Since $a^2-b^2=(a+b)(a-b)$ and $89$ is a prime number, then we have these possibilities:
*
*$a+b=89$ and $a-b=1$, that is, $a=45$, $b=44$.
*$a+b=1$ and $a-b=89$, that is, $a=45$, $b=-44$.
*$a+b=-89$ and $a-b=-1$, that is, $a=-45$, $b=-44$.
*$a+b=-1$ and $a-b=-89$, that is, $a=-45$, $b=44$.
So clearly, $a^2=2025$ and $b^2=1936$.
Then $n+10=\pm45$ which gives two values for $n$, namely $n=35$ and $n=-55$.
EDIT: How can be proved that there are no more solutions?
Indeed, this is already done. Our reasoning is like this:
*
*Proposition $A$: "$n$ is such that $n^2-20n+11$ is a perfect square".
*Proposition $B$: "$n$ is $35$ or $-55$".
We have shown, assuming that $n$ is integer, that $A$ implies $B$. So if $n$ is another integer number, $B$ is false. Hence, $A$ is false. That is, if $n$ is not $35$ or $-55$ then $n^2+20n+11$ is not a perfect square.
|
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|
Find all $n\in\mathbb N$ such that $\sqrt{n^2+8n-5}$ is an integer. $n^2+8n-5$ has to be a perfect square.
How to find all $n$?
|
Note that $n^2 + 8x - 5 = (n + 4)^2 - 21$. Now this number will certainly be no square if it is bigger than $(n + 3)^2$ (since there are no squares strictly between $(n + 3)^2$ and $(n + 4)^2$), i.e. if
$$\begin{align*}
&n^2 + 8n - 5 > (n + 3)^2 \\
\iff& n^2 + 8n - 5 > n^2 + 6n + 9 \\
\iff& 2n > 14 \\
\iff& n > 7
\end{align*}$$
So we know that $n^2 + 8n - 5$ is no square if $n$ is greater than $7$. But this leaves only $n = 1, 2, \ldots, 7$ to check manually. This shows that $n = 1$ and $n = 7$ are the only numbers for which $n^2 + 8n - 5$ is a square.
|
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|
determine the roots of the equation $7\sec^2{x}+2\tan{x}-6=2\sec^2{x}+2$ Determine the roots of the equation $7\sec^2{x}+2\tan{x}-6=2\sec^2{x}+2$, when $0≤x≤2π$. I don't know how to solve this equation algebraically. Please help?
|
Using $1+tan^2=sec^2$,
$$7\sec^2(x)+2\tan(x)-6=2\sec^2(x)+2\iff $$
$$7+7\tan^2(x)+2\tan(x)-6=2+2\tan^2(x)+2\iff $$
$$5\tan^2(x)+2\tan(x)-3=0 \iff$$
$$(\tan(x)+1)(5\tan(x)-3)=0 \iff$$
$$\tan(x)=-1 \hbox{ or } \tan(x)=3/5.$$
Thus, in $[0,2\pi)$, $x=3\pi/4$ or $x=7\pi/4$ or $x=\tan^{-1}(3/5)$ or $x=\tan^{-1}(3/5)+\pi.$
|
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|
A nice infinite series: ${\sum\limits_{n=1}^\infty}\frac{1}{n!(n^4+n^2+1)}=\frac{e}{2}-1$ - looking for a more general method while clearing out some stuff I found an old proof I wrote for an recreational style problem a while back, here's the sum:
${\sum\limits_{n=1}^\infty}\frac{1}{n!(n^4+n^2+1)}=\frac{e}{2}-1$
I'll show how I got to the limit in a moment, but because my argument was a bit long winded / ad-hoc (first thing that came to mind at the time), just wondering if this kind of inverse polynomial/factorial sum can be attacked using other techniques, is it part of a well known / classical family of these kind of sums / any useful references? Thanks in advance!
Proof:
\begin{align}
\displaystyle{\sum\limits_{n=1}^\infty}\frac{1}{n!(n^4+n^2+1)} &= \displaystyle{\sum\limits_{n=1}^\infty}\frac{1}{n!{(n^2+1)^2-n^2}}=
\displaystyle{\sum\limits_{n=1}^\infty}\frac{1}{n!(n^2+1-n)(n^2+1+n)}\nonumber \\
&= \displaystyle{\sum\limits_{n=1}^\infty \frac{n}{n \cdot n!(n^2+1-n)(n^2+1+n)} }\nonumber\\
&= \displaystyle{\frac{1}{2}\sum\limits_{n=1}^\infty\frac{1}{n \cdot n!} \left(\frac{1}{n^2+1-n}-\frac{1}{n^2+1+n}\right)}\end{align}
Now this almost telescopes, the factorial term being the issue, but if we slightly change the groupings of the summands (essentially starting the summation having moved one index along) we get something that we can further simplify by taking the telescoping part out of the parentheses and putting the factorial part in:
$
= \displaystyle{\frac{1}{2}\left[\frac{1}{1\cdot1!}\cdot\frac{1}{1}+\sum\limits_{n=2}^{\infty}\frac{1}{n^2+1-n}\left\{\frac{1}{n \cdot n!}-\frac{1}{(n-1) (n-1)!}\right\}\right]}$
We have taken out the 1/quadratic that was common to successive telescoping summands in the original summation, and taken the difference of the factorials instead since as we will see this simplifies nicely:
$ \begin{align}&= \displaystyle{\frac{1}{2}\left[1+\sum\limits_{n=2}^{\infty}\frac{1}{n^2+1-n}\left\{\frac{n-1-n^2}{n\cdot(n-1)\cdot n!}\right\}\right]}\nonumber \\
&= \displaystyle{\frac{1}{2}\left[1+\sum\limits_{n=2}^{\infty}\frac{1}{n\cdot(n-1)\cdot n!}\right]}\end{align}$
Using the same partial fraction trick we used in (1) to pseudo-telescope and regroup terms to get to (2) helps again:
\begin{align} &= \displaystyle{\frac{1}{2}\left[1+\sum\limits_{n=2}^{\infty}\frac{1}{n!}\left(\frac{1}{n-1}-\frac{1}{n}\right)\right]} =\displaystyle{\frac{1}{2}\left[1-\left\{\frac{1}{2!}+\sum\limits_{n=3}^{\infty}\frac{1}{(n-1)}\left(\frac{1}{n!}-\frac{1}{(n-1)!}\right)\right\}\right]}\nonumber \\ &=\displaystyle{\frac{1}{2}\left[1-\left\{\frac{1}{2}-\sum\limits_{n=3}^{\infty}\frac{1}{n!}\right\}\right]}
=\displaystyle{\frac{1}{2}\left[\frac{1}{2}+\left\{\sum\limits_{n=0}^{\infty}\frac{1}{n!}\right\}-2 \frac{1}{2}\right]}\\
&=\displaystyle{\frac{1}{2}\left\{\sum\limits_{n=0}^{\infty}\frac{1}{n!}\right\}-1=\frac{e}{2}-1}\,\,\,\square
\end{align}
|
Here is my heuristics understanding on why the summation is so special. Consider the function
$$ f(z) = \sum_{n=0}^{\infty} \frac{1}{z(z-1)\cdots(z-n)}. $$
Let $a \in \Bbb{C} \setminus \{0,1,2,\cdots\}$ and $\epsilon > 0$ be sufficiently small. Putting the matter of rigor aside, computing the following contour integral
$$ \frac{1}{2\pi i} \oint_{|z| = \epsilon} \frac{f(z)}{z-a} \, dz $$
in two ways yields
$$ \sum_{n=0}^{\infty} \frac{1}{n!}\frac{1}{n-a} = -e f(a). \tag{1}$$
Moreover, by the direct computation we check that
$$ f(z-1) = zf(z) - 1. \tag{2}$$
Now what is so special about our sum is that, if we write $\omega = \frac{1}{2} + i\frac{\sqrt{3}}{2}$, then applying $\text{(1)}$ to the partial fraction decomposition
$$ \frac{1}{z^4 + z^2 + 1} = \frac{1}{2i\sqrt{3}} \left( \frac{\bar{\omega}}{z-\omega} - \frac{\bar{\omega}^2}{z - (\omega-1)} - \frac{\omega}{z-\bar{\omega}} + \frac{\omega^2}{z - (\bar{\omega}-1)} \right), \tag{3} $$
we find that
$$\sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{n^4+n^2+1}
= -\frac{e}{2i\sqrt{3}} \left[ \bar{\omega}^2 (\omega f(\omega) - f(\omega-1)) - \omega^2 (\bar{\omega} f(\bar{\omega}) - f(\bar{\omega} - 1)) \right]. $$
Now applying $\text{(2)}$ gets rid of the occurrence of the nebulous function $f$ and simply we get
$$\sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{n^4+n^2+1} = \frac{e}{2}. $$
As we see, all this argument relies on the special structure of the partial fraction decomposition $\text{(3)}$.
EDIT. By translating $\text{(2)}$ in terms of the left-hand side of $\text{(1)}$, we can give a more elementary proof.
Let $a \in \Bbb{C}\setminus\{0,1,2,\cdots\}$. Then
\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{n!} \left( \frac{1}{n-(a-1)} - \frac{a}{n-a} \right)
&= \sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{n-(a-1)} - \sum_{n=0}^{\infty} \frac{1}{n!} \frac{a}{n-a} \\
&= \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \frac{1}{n-a} - \bigg( -1 + \sum_{n=1}^{\infty} \frac{1}{n!} \frac{a}{n-a} \bigg) \\
&= 1 + \sum_{n=1}^{\infty} \frac{1}{n!} \bigg( \frac{n}{n-a} - \frac{a}{n-a} \bigg) \\
&= e.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2029003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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|
How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$$
I suppose in the second I may not take into account arctan and sin as sinx approximately equals x
|
Hint: Multiply the top and bottom of the first limit in order to rewrite it as
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3 (\sqrt{1+\tan(x)} + \sqrt{1 + \sin(x)})} =
\frac 12 \lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}
$$
Then we have
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3} =
\lim_{x \to 0} \frac 1{\cos(x)}\frac{\sin(x)}{x} \frac{(1 - \cos(x))}{x^2} = \\
\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}
$$
now apply a similar trick.
For the second, we can write
$$
\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} =
\lim_{x \to 0} \frac{\arctan 2x}{\sin(2 \pi x)} = \\
\lim_{x \to 0} \frac{1}{2\pi}\frac{\arctan 2x}{x}
\cdot \frac{2 \pi x}{\sin (2 \pi x)} =\\
\frac{1}{2\pi}\lim_{x \to 0} \frac{\arctan (2(0+x)) - \arctan(2(0))}{x}
$$
you might find it easier to find that second limit if you replace $x$ with an $h$ or a $\Delta x$. Alternatively, with $\theta = \arctan(2x)$, we have
$$
\lim_{x \to 0} \frac{\arctan(2x)}{x} = \lim_{\theta \to 0} \frac{\theta}{\tan(\theta/2)} =
2\lim_{\theta \to 0}
\cos (\theta/2)\frac{\theta/2}{\sin(\theta/2)}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2031082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
Prove:$x^2+2xy+3y^2-6x-2y\ge-11\;\;\forall(x,y)\in\Bbb{R}$
Problem Statement:-
Prove that for all real values of $x$ and $y$
$$x^2+2xy+3y^2-6x-2y\ge-11$$
I have no idea how to approach this question all I could think on seeing it was tryin to find the linear factors, turns out that the determinant
$$\begin{vmatrix}
a & h & g\\
h & b & f\\
g & f & c\\
\end{vmatrix}\neq0$$
So finding linear factor just flew out of the window, so I plotted the equation $x^2+2xy+3y^2-6x-2y=0$ turns out it is an ellipse. I could conclude no further as to how to approach this problem.
|
The matrix for the quadratic form is
$$
A=\begin{bmatrix}
1 & 1 & -3 \\
1 & 3 & -1 \\
-3 & -1 & 11
\end{bmatrix}
$$
A sufficient condition for the quadratic form to be positive semidefinite is that the leading principal minors but the last (that ought to be nonnegative) are positive. If also the full determinant is positive, the matrix is positive definite.
The leading principal minors are
$$
\det[1]=1,
\qquad
\det\begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix}=2,
\qquad
\det A=0
$$
Thus $A$ is a positive semidefinite symmetric matrix and thus $v^TAv\ge0$ for all $v\in\mathbb{R}^3$, in particular for $v=[x\;y\;1]^T$, which proves the thesis.
The null space can be determined by Gaussian elimination:
\begin{align}
\begin{bmatrix}
1 & 1 & -3 \\
1 & 3 & -1 \\
-3 & -1 & 11
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & -3 \\
0 & 2 & 2 \\
0 & 2 & 2
\end{bmatrix}&&R_2\gets R_2-R_1, R_3\gets R_3+3R_1
\\&\to
\begin{bmatrix}
1 & 1 & -3 \\
0 & 1 & 1 \\
0 & 2 & 2
\end{bmatrix}&&R_2\gets \tfrac{1}{2}R_2
\\&\to
\begin{bmatrix}
1 & 1 & -3 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}&&R_3\gets R_3-2R_2
\\&\to
\begin{bmatrix}
1 & 0 & -4 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}&&R_1\gets R_1-R_2
\end{align}
So an eigenvector for the zero eigenvalue is
$$
\begin{bmatrix}4 \\ -1 \\ 1\end{bmatrix}
$$
and, indeed, the point with coordinates $(4,-1)$ is the only one where the form vanishes.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2031706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
How to prove that $\int_{-1}^{+1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$ I'am trying to prove that
$$\int_{-1}^{+1} (1-x^2)^n\:\mathrm{d}x = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$$
Here what I have done so far. We know that:
\begin{equation}
\sin^{2}x + \cos^{2}x = 1
\end{equation}
Let $x = \sin\alpha$ so $\mathrm{d}x = \cos\alpha\:\mathrm{d}\alpha$
\begin{equation}
x^{2} = \sin^{2}\alpha
\end{equation}
\begin{equation}
\cos^{2}x = 1 -\sin^{2}x
\end{equation}
We have $x = 1 \Rightarrow \alpha = \frac{\pi}{2}$ and $x = -1 \Rightarrow \alpha = - \frac{\pi}{2}$, so we replace the original integral by:
$$\int_{-1}^{1} (x-1^2)^n\:\mathrm{d}x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha$$
Now integrating by parts,
$$u = \cos^{2n}\alpha \quad \Rightarrow\:\mathrm{d}v = \cos\alpha\:\mathrm{d}\alpha \\
\mathrm{d}u = -2n\cos^{2n-1}\alpha\sin\alpha\:\mathrm{d}\alpha \quad \Rightarrow v = \sin\alpha \\
\int udv = uv - \int vdu$$
Then,
\begin{align}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha &= \underbrace{\cos^{2n}\alpha \sin\alpha \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}_{\text{= 0}} + 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha \sin^{2} \alpha\:\mathrm{d}\alpha \\[10pt]
&= 2n \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha (1-\cos^2\alpha)\:\mathrm{d}\alpha \\
&= 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha\:\mathrm{d}\alpha - 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha
\end{align}
Now we see that
$$(2n+1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha = 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos^{2n-1}\alpha\:\mathrm{d}\alpha$$
I am stuck here, it's almost the same integral as the original one. Any suggestions or corrections will be very welcome. Thanks.
|
Consider the integral $\int_{0}^{+1} (1-x^2)^n dx$
Let $I(n)$ be the above integral do the substitution $x=\sin t$. Then $dx=\cos t dt$ and
$1-x^2=cos^2 t$.
Also $t=0$ for $x=0$ and $t=\pi/2$ for $x=1$ so
$I(n) = \int_0^{\pi/2} \cos^{2n+1} t \ dt = \int_0^{\pi/2} \cos^{2n} t \ \cos t \ dt$.
To integrate by parts do the substitution
$$u = \cos^{2n} t$$ and $$dv=\cos t dt$$ From where we get
$$du = 2n \cos^{2n-1} t (-\sin t) dt $$ and
$$v=\sin t$$
Now, $uv$ is zero at both limits so
$$I(n) = 2n \int \cos^{2n-1} t \sin^2 t dt = 2n \int \cos^{2n-1} t (1-\cos^2 t) dt = 2n [ I(n-1) - I(n) ]$$ from where we get
$$I(n) = \frac{2n}{2n+1}I(n-1)$$
It is clear that $I(0)=1$ so
$$I(n) = \frac{(2n) 2(n-1) 2(n-2) ... 2}{ [ (2n+1) (2n-1) ... 3]}$$
Multiplying both the numerator and the denominator by $(2n) 2(n-1) 2(n-2) ... 2$ the denominator becomes $(2n+1)!$ where as the numerator is the square of $2^n n!$ So, finally
$I(n)= \frac{2^{2n} (n!^2)} { (2n+1)!}$
The asked integral is just the double of the computed integral.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2033851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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|
power series solution for $xy''+y'+xy=0$ I need to solve
$$xy''+y'+xy=0$$
at $x_0 =1$
so I supposed:
$$y = \sum_{n=0}^{\infty}a_n(x-1)^n\implies$$
$$y'= \sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}\implies$$
$$y'' = \sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}$$
Then I did:
$$(x-1+1)\sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}+\sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}+(x-1+1) \sum_{n=0}^{\infty}a_n(x-1)^n = $$
$$\sum_{n=2}^{\infty}n\cdot (n-1)\cdot (n-2)(x-1)^{n-1}+\sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}+\sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}+\sum_{n=0}^{\infty}a_n(x-1)^{n+1}+\sum_{n=0}^{\infty}a_n(x-1)^n$$
as you can see, I get this $(x-1)^{n+1}$ term. When it happens, I normally shift the index to get $(x-1)^n$, but in this case I'm already in the index $0$. What should I do?
|
Rewrite as $$(x-1)y''+(x-1)y+y+y'+y''=0 $$
The coefficient of $(x-1)^n$
*
*in $y$ is $a_n$
*in $(x-1)y$ is $a_{n-1} $ (with $a_{-1}=0$ understood)
*in $y'$ is $(n+1)a_{n+1}$
*in $y''$ is $(n+2)(n+1)a_{n+2}$
*in $(x-1)y''$ is $(n+1)na_{n+1}$
Thus you get the equation
$$(n+1)na_{n+1} + a_{n-1}+a_n+(n+1)a_{n+1}+(n+2)(n+1)a_{n+2}=0$$
or the recursion
$$a_{n+2}=-\frac{(n+1)^2a_{n+1} + a_{n-1}+a_n}{(n+2)(n+1)} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2034549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
What's wrong with the way I solved the complex equation $z^5=-16\overline{z} (z \in \mathbb{C})$
Solve the fallowing complex equation($z \in \mathbb{C})$: $z^5=-16\overline{z}$.
Here's how I tried to solve the equation:
$$ z^5=-16\overline{z} \rightarrow z^6=-16(x^2+y^2)$$
$$z= \varepsilon(\cos{\theta} +i\sin{\theta}) \rightarrow z^6=\varepsilon^6(\cos{6\theta} +i\sin{6\theta})$$
$$w=r(\cos{\varphi} +i\sin{\varphi})=-16(x^2+y^2) $$
$$r=\sqrt{(-16)^2(x^2+y^2)^2} = 16(x^2+y^2)$$
$$ \varphi = \arctan{0} = \pi$$
$$ 6\theta = \varphi + 2k\pi, k\in \mathbb{Z}\rightarrow \theta_0 = \frac{\pi}{6}; \theta_1 = \frac{\pi}{2}; \theta_2 = \frac{5\pi}{6} \theta_3 = \frac{7\pi}{6}; \theta_4 = \frac{3\pi}{2}; \theta_5 = \frac{11\pi}{6}$$
$$ z_0=16(x^2+y^2)e^{i\frac{\pi}{6}};z_1=16(x^2+y^2)e^{i\frac{\pi}{2}};z_2=16(x^2+y^2)e^{i\frac{5\pi}{6}};z_3=16(x^2+y^2)e^{i\frac{7\pi}{6}};z_4=16(x^2+y^2)e^{i\frac{3\pi}{2}};z_5=16(x^2+y^2)e^{i\frac{11\pi}{6}} $$
But my teacher told me that I didn't solve it correctly. Where am I wrong?
|
$x$ and $y$ are unknown, so it makes no sense saying $z_0=16(x^2+y^2)e^{i\pi/6}$. Recall that $z=x+iy$, according to what you wrote.
A proper way would be first noting that $z=0$ is a solution, so we can henceforth assume $z\ne0$.
Write $z=re^{i\varphi}$; then the equation becomes
$$
r^5e^{5i\varphi}=-16re^{-i\varphi}
$$
and therefore, being $r\ne0$,
$$
r^4e^{6i\varphi}=-16
$$
Then, as $r>0$, we get $r=2$ and $e^{6i\varphi}=-1=e^{i\pi}$, so
$$
6i\varphi=\pi+2k\pi
$$
You get six distinct solutions for $k=0,1,\dots,5$.
Alternatively, take the modulus of both sides:
$$
|z|^5=16|\bar{z}|=16|z|
$$
so $|z|^4=16$ and $|z|=2$ (we're still assuming $z\ne0$). Write $z=2u$; then the equation becomes $u^5=-\bar{u}=-u^{-1}$, since $|u|=1$. Thus we get $u^6=-1$ and so $u$ is one of the sixth roots of $-1$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2035046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Different sites give different answers for this series sum For the series $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}$ wolframalpha gives answer $\displaystyle\frac{\pi^2}{64}$, but another site gives $\displaystyle\frac{\pi^2}{64}-\frac{1}{12}$: http://www.emathhelp.net/calculators/calculus-2/series-calculator/?f=1%2F%28%282n%2B1%29%5E2-4%29%5E2&var=&a=0&b=inf.
What is the correct answer?
|
$$\begin{align}
S&=\sum_{n=0}^\infty\frac{1}{((2n+1)^2-4)^2}=\sum_{n=0}^\infty\frac{1}{(4n^2+4n-3)^2}=\sum_{n=0}^\infty\frac{1}{((2n-1)(2n+3))^2}\\
&=\frac{1}{4}\sum_{n=0}^\infty\frac{2n+3-(2n-1)}{((2n-1)(2n+3))^2}=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)^2(2n+3)}-\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)^2}\\
&=\frac{1}{16}\sum_{n=0}^\infty\frac{1}{(2n-1)^2}-\frac{2}{16}\sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)}+\frac{1}{16}\sum_{n=0}^\infty\frac{1}{(2n+3)^2}
\end{align}$$
Using this we see
$$\begin{align}
&\sum_{n=0}^\infty\frac{1}{(2n-1)^2}=\frac{3}{4}\zeta(2)+1\\
&\sum_{n=0}^\infty\frac{1}{(2n+3)^2}=\frac{3}{4}\zeta(2)-1
\end{align}$$
And we can use telescoping on the middle-term as follows
$$\begin{align}
\sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)}&=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)}-\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n+3)}\\
&=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)}-\frac{1}{4}\sum_{n=2}^\infty\frac{1}{(2n-1)}\\
&=\frac{1}{4}(-1+1)=0
\end{align}$$
And so
$$S=\frac{1}{16}\times\frac{3}{2}\zeta(2)=\frac{\pi^2}{64}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2038557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.
For $x^2\equiv 1\pmod {7}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$.
For $x^2\equiv 1\pmod {13}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$ (\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$.
Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?
|
You have $x\equiv\pm1\mod7$ and $x\equiv\pm1\mod13$. For all the solutions, you have to consider the systems:
$$x\equiv1\mod7$$
$$x\equiv1\mod13$$
and
$$x\equiv-1\mod7$$
$$x\equiv1\mod13$$
and
$$x\equiv1\mod7$$
$$x\equiv-1\mod13$$
and
$$x\equiv-1\mod7$$
$$x\equiv-1\mod13$$
as each system will get you a valid answer. I think you only had the first and the last systems and that you only considered the cases where the signs were similar.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2041254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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|
Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan \theta = \frac{5}{12}$$
Thus $$\sin\theta=\frac{5}{13}$$
But If I do like this , $$\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$$
$$ 2\sin\theta +2=3\cos \theta$$
$$ (2\sin\theta +2)^2=9\cos^2 \theta$$
$$ 4\sin^2\theta+8\sin\theta +4=9-9\sin^2\theta$$
$$13\sin^2\theta+8\sin\theta-5=0$$
Therefore I get two answers $$\sin\theta=\frac{5}{13} , \sin\theta =-1$$
What is the reason behind this ? Why am I getting two answers in one method and one in another ?
|
The second solution $\sin(\theta) = -1$ doesn't work with the original equation because then $\tan$ and $\sec$ are both undefined.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2042019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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|
Closed form of recursively defined sequence Let $a_0, a_1 \in \mathbb R$. I am given the sequence $(a_n)_{n\in \mathbb N}$ recursively defined by $$a_n = \frac{2}{5} a_{n-2} + \frac{3}{5} a_{n-1}$$
for $n \geq 2$. I want to find an explicit term describing $a_n$, I tried to figure it out by writing out the first 3-4 terms but I can't really find a pattern since the coefficients are not monotone. Or is there another, easier way to show that the sequence converges?
|
Given the recurrence
$$
a_{\,n} = \frac{3}
{5}a_{\,n - 1} + \frac{2}
{5}a_{\,n - 2} \quad \left| {\;a_{\,n\, < \,0} = 0} \right.
$$
in my opinion, the best way to solve it is through the generating function.
So let's write it as:
$$
a_{\,n} - \frac{3}
{5}a_{\,n - 1} - \frac{2}
{5}a_{\,n - 2} - \left[ {1 = n} \right]a_{\,1} - \left[ {0 = n} \right]a_{\,0} = 0\quad \left| {\;a_{\,n\, < \,0} = 0} \right.
$$
where $[P]$ is the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
Then, multiplying by $z^n$ and summing
$$
\begin{gathered}
0 = \sum\limits_{0\, \leqslant \,n} {a_{\,n} \;z^{\,n} } - \frac{3}
{5}\sum\limits_{0\, \leqslant \,n} {a_{\,n - 1} z^{\,n} } - \frac{2}
{5}\sum\limits_{0\, \leqslant \,n} {a_{\,n - 2} z^{\,n} } - a_{\,1} \sum\limits_{0\, \leqslant \,n} {\left[ {1 = n} \right]z^{\,n} } - a_{\,0} \sum\limits_{0\, \leqslant \,n} {\left[ {0 = n} \right]z^{\,n} } = \hfill \\
= \sum\limits_{0\, \leqslant \,n} {a_{\,n} \;z^{\,n} } - \frac{3}
{5}z\sum\limits_{0\, \leqslant \,n} {a_{\,n - 1} z^{\,n - 1} } - \frac{2}
{5}z^{\,2} \sum\limits_{0\, \leqslant \,n} {a_{\,n - 2} z^{\,n} } - a_{\,1} \;z - a_{\,0} = \hfill \\
= A(z) - \frac{3}
{5}z\,A(z) - \frac{2}
{5}z^{\,2} \,A(z) - a_{\,1} \;z - a_{\,0} \hfill \\
\end{gathered}
$$
which gives:
$$
\begin{gathered}
A(z) = \frac{{a_{\,1} \;z + a_{\,0} }}
{{1 - \frac{3}
{5}\,z - \frac{2}
{5}z^{\,2} }} = 5\frac{{a_{\,1} \;z + a_{\,0} }}
{{5 - 3\,z - 2z^{\,2} }} = \hfill \\
= \frac{5}
{7}\left( {\frac{{a_{\,1} \; + a_{\,0} }}
{{1 - z}} + \frac{{2a_{\,0} - 5a_{\,1} }}
{{5 + 2z}}} \right) = \hfill \\
= \frac{1}
{7}\left( {\frac{{5\left( {a_{\,0} + a_{\,1} } \right)}}
{{1 - z}} + \frac{{\left( {2a_{\,0} - 5a_{\,1} } \right)}}
{{\left( {1 - \left( { - \frac{2}
{5}z} \right)} \right)}}} \right) = \hfill \\
= \frac{1}
{7}\left( {\frac{{5\left( {a_{\,0} + a_{\,1} } \right)}}
{{1 - z}} + \frac{{\left( {2a_{\,0} - 5a_{\,1} } \right)}}
{{\left( {1 - \left( { - \frac{2}
{5}z} \right)} \right)}}} \right) = \hfill \\
= \frac{1}
{7}\left( {\sum\limits_{0\, \leqslant \,n} {\left( {5\left( {a_{\,0} + a_{\,1} } \right) + \left( {2a_{\,0} - 5a_{\,1} } \right)\left( { - 1} \right)^{\,n} \frac{{2^{\,n} }}
{{5^{\,n} }}} \right)\;z^{\,n} } } \right) = \hfill \\
= \frac{{5\left( {a_{\,0} + a_{\,1} } \right)}}
{7}\left( {\sum\limits_{0\, \leqslant \,n} {\left( {1 + \left( {\frac{{2a_{\,0} - 5a_{\,1} }}
{{a_{\,0} + a_{\,1} }}} \right)\left( { - 1} \right)^{\,n} \frac{{2^{\,n} }}
{{5^{\,n + 1} }}} \right)\;z^{\,n} } } \right) \hfill \\
\end{gathered}
$$
That tells you that $a(n)$ is converging, oscillating, to $5/7(a_0+a_1)$.
|
{
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"url": "https://math.stackexchange.com/questions/2043490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Trouble with a substitution I'm struggling to show that
$$ \int_{-1}^{1} \frac{(1-x^2)^{1/2}}{1+x^2} dx $$
to
$$ -\pi + \int_{-\pi}^{\pi} (1+\cos^2(\theta))^{-1}d\theta$$
with $x=\cos(\theta)$
I'm aware I'm missing something obvious but I end up with a stray $\sin(\theta)$
|
You can use $u=\tan(\frac{\theta}{2})$ in the calculation. In fact,
\begin{eqnarray}
\int_{-\pi}^{\pi}\frac{1}{1+\cos^2\theta}d\theta&=&\int_{-\pi}^{\pi}\frac{1}{1+\frac{1+\cos(2\theta)}{2}}d\theta\\
&=&\int_{-\pi}^{\pi}\frac{2}{3+\cos(2\theta)}d\theta\\
&=&\int_{-2\pi}^{2\pi}\frac{1}{3+\cos(\theta)}d\theta\\
&=&2\int_{-\pi}^{\pi}\frac{1}{3+\cos(\theta)}d\theta\\
&=&4\int_{0}^{\pi}\frac{1}{3+\cos(\theta)}d\theta\\
&=&4\int_{0}^{\pi}\frac{1}{3+\frac{1-\tan^2(\frac\theta2)}{1+\tan^2(\frac\theta2)}}d\theta\\
&=&4\int_{0}^\infty\frac{1}{3+\frac{1-u^2}{1+u^2}}\frac{2}{1+u^2}du\\
&=&8\int_{0}^\infty\frac{1}{4+2u^2}du\\
&=&4\int_{0}^\infty\frac{1}{2+u^2}du\\
&=&4\cdot\frac{1}{\sqrt2}\arctan\frac{u}{\sqrt2}\bigg|_0^\infty\\
&=&\sqrt2\pi.
\end{eqnarray}
|
{
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"url": "https://math.stackexchange.com/questions/2045406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Finding particular solution to 1D - wave equation given general solution. Given the general solution of the wave equation:
$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$
Find the particular solution given that:
$\displaystyle \frac{\partial}{\partial t}u(x,0) = x$ for $0\leq x \leq 1$
First Attempt:
$\displaystyle u_t(x,0) = \sum_{n=0}^\infty b_nc\left(n+\frac{1}{2}\right)\pi\cdot \sin\left[\pi\left(n+\frac{1}{2}\right) x\right] = x $
Then we find $b_n$ as shown below:
$b_n = \frac{2}{\pi} \int_0^\pi \pi x c(n+\frac{1}{2})) \cdot sin\big(\pi(n+\frac{1}{2}))\big)dx $
which simplifies to:
$2c(n+\frac{1}{2})\Big(\Big[-x\pi (n+\frac{1}{2}) \cdot cos\big(\pi x(n+\frac{1}{2})\big)\Big]_0^\pi + \pi (n+\frac{1}{2}) \int_0^\pi cos\big( x\pi (n+\frac{1}{2})\big)\Big)$
But is this the right way to go?
Second Attempt:
$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$
let $\alpha = (n+\frac{1}{2})$
$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \cdot sin\big(ct\alpha \pi\big) \cdot sin\big(\pi\alpha x\big)$
$\displaystyle u_t(x,t) = \sum_{n=0}^\infty b_n \cdot c\pi \alpha \cdot cos\big(ct\alpha \pi\big) \cdot sin\big(\pi\alpha x\big)$
How do I calculate $b_n$ from here?
Me attempting to find $b_n$:
$\displaystyle b_n = \frac{2}{\pi} \int_0^1 \pi x c\alpha \cdot sin\big(\pi\alpha\big)dx $
goes to:
$b_{n} = \frac{2}{(c \pi^{2} \alpha)} \int_{0}^{1} x \sin (\alpha \pi x) dx$
|
$$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$$
$$\displaystyle u(x,t) = \sum_{n=0}^\infty \frac{b_n}{2}\left[ \cos\left[\pi\left(n+\frac{1}{2}\right)(x-ct) \right]-\cos\left[\pi\left(n+\frac{1}{2}\right) (x+ct)\right] \right]$$
$$u(x,t) =f(x-ct)-f(x+ct)$$
Where $$f(y)=\sum_{n=0}^\infty \frac{b_n}{2} \cos\left[\pi\left(n+\frac{1}{2}\right)(y) \right]$$
Now the boundary condition($\displaystyle \frac{\partial}{\partial t}u(x,0) = x$) gives $f(x)= -\frac{x^2}{4c}+d$ which gives $$u(x,t) =\frac{(x+ct)^2}{4c} -\frac{(x-ct)^2}{4c}$$
|
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"url": "https://math.stackexchange.com/questions/2048161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Zeta Infinite Summation $\sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n)$ Let $Re\{s\}\gt0$ :
$$
\sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} = \sum_{n=1}^{\infty} \left( \frac{n}{n+1}\right) n^{-(s+1)} = \sum_{n=1}^{\infty} \left( 1 - \frac{1}{n+1}\right) n^{-(s+1)} = \zeta(s+1) - \sum_{n=1}^{\infty} \frac{n^{-(s+1)}}{n+1} = \\[6mm]
\zeta(s+1) - \zeta(s+2) + \sum_{n=1}^{\infty} \frac{n^{-(s+2)}}{n+1} = \zeta(s+1) - \zeta(s+2) + \zeta(s+3) - \sum_{n=1}^{\infty} \frac{n^{-(s+3)}}{n+1} = \text{...} \Rightarrow \\[6mm]
\sum_{n=1}^{N} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - (-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1} \Rightarrow \\[6mm]
\boxed{ \quad \sum_{n=1}^{\infty} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - \lim_{N\rightarrow\infty}\left[(-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1}\right] \quad } \\[6mm]
$$
Does the limit exist? and What does it equal?
$$ L = \lim_{N\rightarrow\infty}\left[(-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1}\right] \qquad\qquad\colon\space Re\{s\} \gt 0 \tag{1}$$
$$\sum_{n=1}^{\infty} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - L \qquad\colon\space Re\{s\} \gt 0 \tag{2}$$
Without the outer sign, the limit is:
$$
\small \sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n^2}\lt\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1}\lt\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n} \Rightarrow \zeta(s+N+2)-1 \lt \sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1} \lt \zeta(s+N+1)-1 \\
\small \text{Let}\space\left\{N\rightarrow\infty\right\}\space\text{and use the limit}\space\left\{\lim_{z\rightarrow\infty}\zeta(z)=1\right\} \Rightarrow \lim_{N\rightarrow\infty}\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1}=0 \Rightarrow \color{red}{\lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1}=\frac{1}{2}} \\
$$
NB: appreciating your explanations on a similar previous question. Many Thanks.
conclusion:
As of the correct answer(s):
$$
\sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n) = \sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}\left[\color{red}{\zeta(s+2n-1)-\zeta(s+2n)}\right] \\[6mm]
\quad = \lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1} = \color{red}{\frac{1}{2}} \quad\colon\space Re\{s\}\ge0 \quad\{\small\text{holds for s=0 too}\normalsize\} \\[6mm]
$$
|
Let
$S(N)
=\sum_{n=1}^{N} \frac{n^{-s}}{n+1}
=\sum_{n=1}^{N} \frac{1}{n^s(n+1)}
$.
If $Re(s)> 0$,
$\lim_{N \to \infty} S(N)$
exists by comparison with
$\sum_{n=1}^{N} \frac{1}{n^{1+Re(s)}}
$.
Then,
since
$\frac1{1+x}
=\sum_{k=0}^{2m} (-1)^k x^k
-\frac{x^{2m+1}}{1+x}
$
$\begin{array}\\
S(N)
&=\sum_{n=1}^{N} \frac{1}{n^s(n+1)}\\
&=\sum_{n=1}^{N} \frac{1}{n^{s+1}(1+1/n)}\\
&=\sum_{n=1}^{N} \frac{1}{n^{s+1}}(\sum_{k=0}^{2m} (-1)^k n^{-k}
-\frac{n^{-2m-1}}{1+1/n})\\
&=\sum_{n=1}^{N} \frac{1}{n^{s+1}}\sum_{k=0}^{2m} (-1)^k n^{-k}
-\sum_{n=1}^{N} \frac{1}{n^{s+1}}\frac{n^{-2m-1}}{1+1/n}\\
&=\sum_{n=1}^{N} \sum_{k=0}^{2m} (-1)^k \frac1{n^{k+s+1}}
-\sum_{n=1}^{N} \frac{1}{(1+1/n)n^{s+2m+2}}\\
&= \sum_{k=0}^{2m}\sum_{n=1}^{N} (-1)^k \frac1{n^{k+s+1}}
-\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&= \sum_{k=0}^{2m}(-1)^k\sum_{n=1}^{N} \frac1{n^{k+s+1}}
-\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&= S_1(N)-S_2(N)\\
\end{array}
$
As $N \to \infty$,
$S_1(N)
= \sum_{k=0}^{2m}(-1)^k\sum_{n=1}^{N} \frac1{n^{k+s+1}}
\to \sum_{k=0}^{2m}(-1)^k \zeta(k+s+1)
$
and
$\begin{array}\\
S_2(N)
&=\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&\gt \frac12\\
\text{and}\\
S_2(N)
&=\frac12+\sum_{n=2}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\
&\le\frac12+\sum_{n=2}^{N} \frac{1}{(n+1)2^{2m}n^{s+1}}\\
&=\frac12+\frac1{4^m}\sum_{n=2}^{N} \frac{1}{(n+1)n^{s+1}}\\
&\lt\frac12+\frac1{4^m}\sum_{n=2}^{N} \frac{1}{(n+1)n}\\
&=\frac12+\frac1{4^m}\sum_{n=2}^{N} (\frac1{n}-\frac{1}{n+1})\\
&<\frac12+\frac1{2\cdot 4^m}\\
\end{array}
$
Therefore
$\lim_{m \to \infty} S_2(N)
= \frac12
$,
so that
$\lim_{m \to \infty, N \to \infty} S(N)
=-\frac12+\sum_{k=0}^{2m}(-1)^k \zeta(k+s+1)
$.
Note that,
if we write
$\sum_{k=0}^{2m}(-1)^k \zeta(k+s+1)
=\sum_{k=0}^{m-1}(\zeta(2k+s+1)-\zeta(2k+s+2))
+\zeta(2m+s+1)
$,
the sum converges properly.
If we do this grouping
earlier on
(even and odd terms together
with opposite signs),
this all becomes rigorous.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of digits after the decimal point of the order of $10$ in $(\frac{\Bbb{Z}}{p\Bbb{Z}})^*$ Let $p$ be a prime number and $a$ an integer such that $0<a<p.$
If I look at $(\frac{\Bbb{Z}}{p\Bbb{Z}})^*$, and the order of $10.$
I discover numerically that if the order of $10$ is even $2n$ then if we look at the list $(q_1,q_2,\cdots,q_{2n})$ the digits after the decimal point of $a/p$ then we have always $q(k)+q(k+N)=9$ for $k=1,\cdots,n.$
Can anyone enlighten me on how to prove this ?
|
If the order of $10$ modulo $p$ is $r$, then we have
$$\frac{1}{p} = \frac{Q}{10^{r}-1} = \sum_{m = 1}^\infty \frac{Q}{(10^{r})^m},$$
where
$$Q = \sum_{k = 1}^r q_k \cdot 10^{r-k}.$$
If $r$ is even, $r = 2n$, we can write $Q = 10^n\cdot a + b$, where $0 \leqslant a,b < 10^n$, and then
$$\frac{1}{p} = \sum_{m = 1}^\infty \frac{c_m}{(10^n)^m}$$
with $c_m = a$ if $m$ is odd, and $c_m = b$ if $m$ is even. Thus we find
$$10^n\cdot \frac{1}{p} = a + \sum_{m = 1}^\infty \frac{c_{m+1}}{(10^n)^m}$$
and therefore
$$\frac{10^n+1}{p} = a + \sum_{m = 1}^\infty \frac{c_m + c_{m+1}}{(10^n)^m} = a + \sum_{m = 1}^\infty \frac{a+b}{(10^n)^m}.$$
Since $p$ divides $10^{2n} - 1 = (10^n-1)(10^n+1)$ but not $10^n - 1$ (the order of $10$ modulo $p$ is $2n$), it follows that $p$ divides $10^n+1$. It follows that
$$g := \sum_{m = 1}^\infty \frac{a+b}{(10^n)^m} = \frac{a+b}{10^n - 1}$$
is an integer. Since $0 \leqslant a+b \leqslant 2\cdot (10^n-1)$, $g \in \{0,1,2\}$. $g = 0$ would imply $a = b = 0$, hence $Q = 0$ and so $\frac{1}{p} = 0$, which is absurd. $g = 2$ would imply $a = b = 10^n - 1$, hence $Q = 10^{2n}-1$, and thus $\frac{1}{p} = 1$, which is also absurd. It follows that $g = 1$, i.e. $a + b = 10^n - 1$. This is just the assertion $q_k + q_{n+k} = 9$ for $1 \leqslant k \leqslant n$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ Divergent? Did I do this right? $$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$$
$$a_n = \frac{1}{\sqrt{n^2+1}} <\frac{1}{\sqrt{n^2}} \le \frac{1}{n}, \forall n\ge1$$
Then by The Comparison Test with $b_n$, where $b_n$ is a divergent p-series = $\frac{1}{n}$ and $a_n \le b_n$,
Then
$\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ is Divergent.
|
This conclusion is not quite correct. Note that $n^2+1>n^2 \implies \frac{1}{\sqrt{n^2+1}}<\frac1n$ is correct. But $\frac{1}{n^2}\le \frac1n$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
Instead, we can assert that for $n\ge 1$, $n^2+1\le 2n^2$. Therefore, $\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{2}\,n}$. And since $\sum_{n=1}^\infty\frac1n$ diverges, so does $\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2052618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
repeating decimals equation problem The repeating decimals $0.abab\overline{ab}$ and $.abcabc\overline{abc}$ satisfy
$0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37}$
where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.
|
The numbers can be expressed as $0.\overline {ab}$ and $0.\overline {abc}$ and written as fractions fractions as ${ 0.\overline {ab} }= {ab\over99} $ and $ 0.\overline {abc} = { abc\over999} $:
$$ {ab\over99} + { abc\over999} = {33\over 37} $$
Let $h = ab $ and $k = c \implies abc = 10h + k$
where $0\leq h<100 $ and $0 \leq k<10 $.
Rewriting the same expression using $h$ and $k$ :
$$ {h\over99} + {10 h+k\over999} = {33\over 37} $$
We notice that $999 = 37 \times 3^3 $ and $99 = 3^2\times11$
$$ {h\over3^2\times11} + {10 h+k\over37 \times 3^3} = {33 \times 3^3 \over 37 \times 3^3 } $$
multiply by $3^2$:
$$ {h\over11} + {10 h+k\over37 \times 3} = {11 \times 3^3 \over 37 } $$
Since $h$ and $k$ are natural numbers, the only way this expression can be satisfied is by having $h$ to be divisible by 11 and $10h + k$ to be divisible by 3.
let $h = 11n$
where $n \in \{0,1,2,3,4,5,6,7,8,9\} $
so
$$ {n} + {10 \times 11 n +k\over37 \times 3} = {11 \times 3^3 \over 37 } $$
some algebra and we get
$$ 221 n + k = 891 $$
since $k<10$ and $221 \times 4 = 884$ we see that $n=4$ gets us close.
$$ 884 + k = 891 \implies k = 7$$
putting it all together:
$$c = 7$$
$$ab = h = 11\times 4 = 44 \implies a=4,b=4$$
|
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"url": "https://math.stackexchange.com/questions/2052959",
"timestamp": "2023-03-29T00:00:00",
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|
Binomial coefficients are odd if and only if $n = 2^k-1$
Let $n$ be an integer, $n > 0$. Prove that all the coefficients of the expansion of the Newtonian binomial $(a+b)^n$ are odd if and only if $n$ is of the form $2^k-1$.
Why in the solution below do they take $n > 8$? Also, I don't get the middle paragraph argument about $n_1$. What do those values represent using $n_1$?
Solution:
For $n \leq 8$ the theorem is immediately verifiable. It is therefore sufficient to assume that the theorem is true for the binomials $a+b,(a+b)^2,\ldots,(a+b)^{n-1}$ for $n > 8$, and prove that the theorem holds for $(a+b)^n$. The coefficients of the binomial expansion (except for the extreme ones both equal to $1$) are $$\dfrac{n}{1}, \dfrac{n(n-1)}{1 \cdot 2},\ldots,\dfrac{n(n-1) \cdots 1}{1 \cdot 2 \cdots (n-1)}.$$ A necessary and sufficient condition in order that all these numbers be odd is that $n$ is odd and the numbers obtained by deleting the odd factors from the numerators and denominators of the remaining numbers be odd.
But setting $n = 2n_1+1$ these numbers can be represented by the terms of the sequence $$\dfrac{n_1}{1},\dfrac{n_1(n_1-1)}{1 \cdot 2}, \ldots, \dfrac{n_1(n_1-1) \cdots 1}{1 \cdot 2 \cdots (n_1-1)}.$$ Since $n_1 < n$, the latter are all odd if and only if $n_1$ is of the form $2^k-1$, i.e. if and only if $n$ is of the form $2(2^k-1)+1 = 2^{k+1}-1$.
|
If $n=2n_1+1$, then the coefficients are:
$$\frac{2n_1+1}{1}, \frac{(2n_1+1)\cdot (2n_1)}{1\cdot 2}, \cdots , \frac{(2n_1+1)\cdot (2n_1)\cdots 1}{1\cdots (2n_1)}. $$
Removing all the odd factors from the numerator and denomiator we get the sequence:
$$ 1, \frac{2n_1}{2}, \frac{2n_1}{2},\frac{(2n_1)\cdot(2n_1-2)}{2\cdot 4},\frac{(2n_1)\cdot(2n_1-2)}{2\cdot 4},\cdots ,\frac{(2n_1)\cdot(2n_1-2)\cdots 2}{2\cdot 4\cdots 2n_1}. $$Dividing through we get that these are equal to
$$ 1, \frac{n_1}{1}, \frac{n_1}{1},\frac{(n_1)\cdot(n_1-1)}{1\cdot 2},\frac{(n_1)\cdot(n_1-1)}{1\cdot 2},\cdots ,\frac{(n_1)\cdot(n_1-1)\cdots 1}{1\cdot 2\cdots n_1}. $$These however are just the coefficients in $(a+b)^{n_1}$ repeated twice. Hence by an induction argument we can conclude that these are odd and hence the coefficients in $(a+b)^n$ are odd.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
I don't have any idea as to how to proceed with this question. I have figured out that $x^3+2x^2-x-2=(x-1)(x+1)(x+2)$.
But I don't know how to use this information to find the remainder $x^3+2x^2-x-2$ would give. A small hint as to how to proceed would be enough.
|
Write $\ f = q\,g + r\ $ with $\,\deg r \le 2 = \deg g,\ $ by the Division Algorithm.
$\color{#0a0}5 \ = f(1)\ =\ r(1)\ \ $ by $\ g(1)\ =0\ $
$\color{#0a0}5\! =\! f(-2)\!=\! r(-2)\ $ by $\ g(-2)\!=\!0,\ $ so $\ r = \color{#0a0}5 + \color{#c00}c\,(x\!-\!1)(x\!+\!2)$
$\!\begin{align} 3\! =\! f(-1)\!\! &=\! r(-1)\ \, {\rm by}\ \ g(-1)\! =\! 0\\
&= 5+\color{#c00}c(-2)\ \Rightarrow\, \color{#c00}{c= 1}\ \Rightarrow\ r = 5\ +\ (x\!-\!1)(x\!+\!2)\end{align}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding an unknown length in a triangle. I have incurred a question and I am having a hard time with it. Please refer the image below
Here it is given that $AP:PC = 3:4$, $QM:MP = 3:2$ and $QB=12cm$. We have to find the length of $AB$. I have tried as many theorems I knew but was not able to conclude to the answer. Please help me to find the answer. Any help is appreciated. $$Thank you$$
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In following link, you may see the triangle drawn.
http://i.imgur.com/AUskpyR.png
Solution:
$\frac{AP}{PC} = \frac{3}{4}, \frac{QM}{MP} = \frac{3}{2} $ are given in the question. Lets draw a line from A to T through M.
$\frac{AP}{AC} \times \frac{CT}{TQ} \times \frac{QM}{MP} = 1$ by the menelaus theorem. When we plug the values into this:$
\frac{3}{7} \times \frac{CT}{TQ} \times \frac{3}{2} = 1,
\frac{CT}{TQ} = \frac{14}{9}\tag{1} $$
\frac{QT}{QC} \times \frac{CP}{PA} \times \frac{AM}{MT} = 1$ by the menelaus theorem. When we plug the values into this:$
\frac{9}{23} \times \frac{4}{3} \times \frac{AM}{MT} = 1,
\frac{AM}{MT} = \frac{23}{12}\tag{2} $$
\frac{CT}{CQ} \times \frac{QB}{BA} \times \frac{AM}{MT} = 1$ by the menelaus theorem. When we plug the values into this:$
\frac{14}{23} \times \frac{12cm}{BA} \times \frac{23}{12} = 1,
\frac{12cm}{BA} = \frac{12}{14}\tag{3}$$Thus, BA = 14 cm.$
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|
Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$
I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know how simplify it to 2.
I try:
Let $a=F_n$, $b=F_{n+1}$ and $c=F_{n+2}$
$a^4+b^4+c^4+2(ab)^2+2(ac)^2+2(bc)^2=2a^4+2b^4+2c^4$
$2(ab)^2+2(ac)^2+2(bc)^2=a^4+b^4+c^4$
I am not sure, what to do next. Can anyone help by completing the prove?
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As $F_{n+2}=F_{n+1}+F_n$. One first solves the characteristic polynomial.
$$
X^2=X+1
$$
on gets two roots $r=\frac{1+\sqrt{5}}{2}$ and $\bar{r}=\frac{1-\sqrt{5}}{2}$.
One now solves $\alpha.r^n+\beta.\bar{r}^n=F_n$ for the two first indices.
The OP did not indicate his initialization so I suppose that $F_0=0;\ F_1=1$, one gets then $F_n=\frac{1}{\sqrt{5}}(r^n-\bar{r}^n)$. This, with the relations $r+\bar{r}=1;\ r.\bar{r}=-1;\ r-\bar{r}=\sqrt{5}$ will, I think, be sufficient to get the desired result.
Coda In fact, it is true in general that
$$
\Big(X^2+Y^2+(X+Y)^2\Big)^2=2\Big(X^4+Y^4+(X+Y)^4\Big)
$$
thus, the fraction holds true for any sequence s.t. $s_{n+2}=s_n+s_{n+1}$.
|
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|
$\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{\sin{4x}}{4}$ - solving a trigonometric equation The question is to solve the following equation:
$$ \sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{\sin{4x}}{4} $$
There is a tedious and mistake-prone way to do this, that is using trigonometric identities to write the equation in terms of $\sin{x}$ exclusively. But what I'm concerned about, is if there's another, perhaps somewhat clever way to deal with such problems?
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$$\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{2\cdot\sin{2x}\cdot \cos 2x}{4}$$
$$2\cdot\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\sin{2x}\cdot \cos 2x $$
$$\sin 2x(2\cdot\sin{x}\cdot\sin{3x}-\cos 2x) =0$$
$$\sin 2x(-\cos 4x+\cos 2x-\cos 2x)=0 \Rightarrow \sin2x \cdot\cos4x=0$$
$$\sin 2x =0 \Rightarrow x=\frac{k\pi}{2}$$
$$\cos 4x =0 \Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}$$
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|
Finding $f(x)$ from a equation $f(x)+f(\frac{x-1}{x}) = x+1$ is given.
find $f(x)$.
I did tried to change it into another form buy substituting $x$ with $\frac{x}{x-1}$
The result were $f(\frac{1}{x})+f(\frac{x}{x-1})=\frac{2x-1}{x-1}$
I don't know what to do next.
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All you have to do is make the substitution again.
$$f(x)+f\bigg(\frac{x-1}{x}\bigg)=x+1$$
$$\implies f\bigg(\frac{x-1}{x}\bigg)+f\bigg(\frac{1}{1-x}\bigg)=\frac{x-1}{x}+1$$
$$\implies f\bigg(\frac{1}{1-x}\bigg)+f(x)=\frac{1}{1-x}+1$$
Now subtract the first and second equations to get
$$f(x)-f\bigg(\frac{1}{1-x}\bigg)=x-\frac{x-1}{x}$$
Then add this to the third equation to get
$$f(x)+f(x)=x-\frac{x-1}{x}+\frac{1}{1-x}+1$$
$$2f(x)=x-\frac{x-1}{x}+\frac{1}{1-x}+1$$
$$2f(x)=x-1+\frac{1}{x}+\frac{1}{1-x}+1$$
$$2f(x)=x+\frac{1}{x}+\frac{1}{1-x}$$
$$2f(x)=x+\frac{1}{x(1-x)}$$
$$\color{red}{f(x)=\frac{x^2(1-x)+1}{2x(1-x)}}$$
|
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|
Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.
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More generally, the coefficient of $x^{n}$ is given by
$$[x^{n}](1+x^2(1-x))^8=[x^{n}]\sum_{k=0}^8\binom{8}{k}x^{2k}(1-x)^k
=(-1)^n\sum_{n/3\leq k\leq \min(n/2,8)}\binom{8}{k}\binom{k}{n-2k}.$$
For $n=10$ we get
$$\sum_{4\leq k\leq 5}\binom{8}{k}\binom{k}{10-2k}=\binom{8}{4}\binom{4}{2}+\binom{8}{5}\binom{5}{0}=476.$$
|
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|
Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two different ways. One example I got is $50$ which is $49+1$ and $25+25$. But here second pair contains same numbers. Does any one have other examples ?
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Note that $a^2 + b^2 = c^2 + d^2$ is equivalent to $a^2 - c^2 = d^2 - b^2$, i.e. $(a-c)(a+c) = (d-b)(d+b)$. If we factor any odd number $m$ as $m = uv$, where $u$ and $v$ are both odd and $u < v$, we can write this as $m = (a-c)(a+c)$ where $a = (u+v)/2$ and $c = (v-u)/2$. So any odd number with more than one factorization of this type gives an example.
Thus from $m = 15 = 1 \cdot 15 = 3 \cdot 5$, we get $8^2 - 7^2 = 4^2 - 1^2$, or $1^2 + 8^2 = 4^2 + 7^2$.
From $m = 21 = 1 \cdot 21 = 3 \cdot 7$ we get $11^2 - 10^2 = 5^2 - 2^2$, or $2^2 + 11^2 = 5^2 + 10^2$.
|
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|
If $x:y = 7:3$ , then find the value of $\frac{y}{x-y}$ If $x:y = 7:3$ , then can I in order to find the value of $\frac{y}{x-y}$ replace $x$ and $y$ with $7$ and $3$ respectively ?
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$ \frac{x-y}{y}=\frac{x}{y}-1=\frac{7}{3}-1=\frac{4}{3}$. Hence, what you want $=\frac{3}{4}$
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|
Why is $2^b-1=2^{b-1}+2^{b-2}+...+1$? Can I get an intuitive explanation why the formula in the title holds?
I know that it works but I am not sure why
$2^b-1=2^{b-1}+2^{b-2}+...+1$
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Consider the number $N=2^{b-1}+2^{b-2}+\dots+2^2+2^1+2^0$. Now let's look at $N+1$:
$$N+1 = 2^{b-1}+2^{b-2}+\dots+2^2+2^1+2^0+2^0\\
=2^{b-1}+2^{b-2}+\dots+2^2+2^1+2^1\\
=2^{b-1}+2^{b-2}+\dots+2^2+2^2\\
\vdots\\
=2^{b-1}+2^{b-2}+2^{b-2}\\
=2^{b-1}+2^{b-1}\\
=2^b$$
(since $2^k+2^k=2^{k+1}$ for any $k$), so $N=2^b-1$.
|
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|
Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&=2\sin x (1 - \sin^2 x) + \sin x - 2\sin^3 x\\
&= " 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \\
&=3\sin x - 4\sin^3 x"
\end{align}
$$
The part of the problem I'm having trouble with is in quotations.
My question:
is how does $\sin x - 2\sin^3 x = 4\sin^3 x$?
I see it as this $\sin x - 2\sin^3 x = 2\sin^{3-1} x = 4\sin x$.
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Observe it's rather:
$$
\begin{align}
\color{blue}{2\sin x} - 2\sin^3 x + \color{blue}{\sin x} - 2\sin^3 x &=\color{blue}{2\sin x} + \color{blue}{\sin x}- 2\sin^3 x - 2\sin^3 x
\\\\&=3\sin x - 4\sin^3 x.
\end{align}
$$
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|
Compute $ \lim_{n\rightarrow \infty }\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k} $ Calculate $ \lim_{n\rightarrow \infty }\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k} $.
So far I found that $ \frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=\frac{(k-5)(k-4)(k-3)}{(k-2)(k-1)k(k+1)(k+2)}=\frac{((k-3)!)^{2}}{(k-6)!\cdot (k+2)!}. $
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You did a big part of the job.
$k^{5}-5k^{3}+4k=(k-2)(k-1)k(k+1)(k+2)$
Now you can rewrite $$\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=\frac{A}{k-1}+\frac{B}{k}+\frac{C}{k+1}+\frac{D}{k-2}+\frac{E}{k+2}$$.
By multiplying the numerators and denominators of the RHS, and identifying the new numerator in terms of $A,B,C...$ to the numerator of the lHS.
You will get
$$\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=4\sum_{k=5}^{n-1}{\frac{1}{k}}-15\sum_{k=6}^{n}{\frac{1}{k}}+20\sum_{k=7}^{n+1}{\frac{1}{k}}-\frac{35}{4}\sum_{k=8}^{n+2}{\frac{1}{k}}-\frac{1}{4}\sum_{k=4}^{n-2}{\frac{1}{k}}$$
or equivalently
$$4\frac{1}{5}-20\frac{1}{6}+\frac{35}{4}(\frac{1}{6}+\frac{1}{7})-\frac{1}{4}(\frac{1}{4}+\frac{1}{5})+(4-15+20-\frac{35}{4}-\frac{1}{4})\sum_{k=6}^{n}{\frac{1}{k}}-4*\frac{1}{n}+20\frac{1}{n+1}-\frac{35}{4}(\frac{1}{n+2}+\frac{1}{n+1})+\frac{1}{4}(\frac{1}{n-1}+\frac{1}{n})$$
We have $$(4-15+20-\frac{35}{4}-\frac{1}{4})=0$$, therfore the limit is $$4\frac{1}{5}-20\frac{1}{7}+\frac{35}{4}(\frac{1}{8}+\frac{1}{7})-\frac{1}{4}=\frac{1}{16}$$
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|
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2-1}}{2^{2n}}
\end{align}
I could not move on. Please help me to continue.
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Try taking out $2^n$ and $2^{n-1}$ from the denominator and the numerator. I got $$\frac{1}{2^{(n-1)^2}}.$$
If you can't get it let me know. I'll show the work.
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|
Factorize $x^4+16x-12$ over reals Factorize $x^4+16x-12$ over reals.
The factor is $x^4+16x-12=(x^2-2x+6)(x^2+2x-2)$
It can be factorized again but I am stuck in this step.If we want to add and then subtract we have a lot of thing to add and subtract.Another idea that I saw in books is writing as this:
$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$
and then find $a,b,b',a'$ but there are two problems I can't find these here and we can say maybe it can factorized into one degree $3$ and one degree $1$ polynomial.
Isn't there a nice way to factor this?
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@JohnHughes explains how to find that this is not a cubic times linear factor. Once you figure that out, we know it's a quadratic times quadratic. To solve this, I'm going to go off the equation you had:
$$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$$
Expand the right side:
$$x^4+16x-12=x^4+(a+a')x^3+(b+b'+aa')x^2+(ab'+ba')x+bb'$$
Now, we have $0x^3$ on the left side and $(a+a')x^3$ on the right side, so we have $a+a'=0 \implies a'=-a$. Furthermore, we have $-12$ as our constant on the left side and $bb'$ as our constant on the right side, so we have $bb'=-12 \implies b'=-\frac{12}b$. Susbtitute:
$$x^4+16x-12=x^4+\left(b-\frac{12}{b}-a^2\right)x^2+\left(-\frac{12a}{b}-ab\right)x-12$$
We have $0x^2$ and $+16x$ on the left side, so by comparing such with the $x^2$ and $x$ coefficients on the right side, we get the following equations:
$$b-\frac{12}{b}-a^2=0$$
$$-\frac{12a}{b}-ab=16$$
Multiply both sides by both equations $b$:
$$b^2-12-ba^2=0$$
$$-12a-ab^2=16b$$
However, instead of guess and check, we can solve for $a$ in terms of $b$ in the second equation (I choose the second equation because it is easier):
$$a=\frac{-16b}{12+b^2}$$
Now, remember that $b$ is a factor of $12$ since $bb'=-12$, so $b \in \{-12,-6,-4,-3,-2,-1,1,2,3,4,6,12\}$. Guess and check values of $b$ and solve for $a$. When you get an integer value of $a$, you know you've solved the problem, so substitute back into $a+a'=0$ to find $a'$ and $bb'=-12$ to find $b'$.
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|
Find integral $\int\frac{\arcsin(x)}{x^{2}}dt$ Find integral $$\int\frac{\arcsin(x)}{x^{2}}dx$$
what I've done: $$\int\frac{\arcsin(x)}{x^{2}}dx=-\int\arcsin(x)d(\frac{1}{x})=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$ I got stuck with that
|
Perform integration by parts to get that $\displaystyle \int \frac{\arcsin(x)}{x^2} dx$ = $\displaystyle \frac{-\arcsin(x)}{x} + \int \frac{1}{x\sqrt{1 -x^2}}dx$.
To solve this latter integral, let $x = \sin(\theta)$. Then $dx = \cos(\theta)d\theta,$ so the integral becomes $\displaystyle \int\frac{\cos(\theta)}{\sin(\theta)\cos(\theta)}d\theta = \int \csc(\theta) d\theta = -\ln|\csc(\theta) + \cot(\theta)| + C.$
Now since we have $x = \sin(\theta)$, then $\frac{1}{x} = \csc(\theta)$ and $\cot(\theta) = \frac{\sqrt{1 - x^2}}{x}$.
Putting it all together, we have $\displaystyle \int \frac{\arcsin(x)}{x^2} dx =\displaystyle \frac{-\arcsin(x)}{x} - \ln|\frac{1+ \sqrt{1 - x^2}}{x}| + C. $
|
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|
Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$.
So, I just want to understand which convention is correct.
This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.
Here is the answer sheet from the book (answer 6, 3rd element):
*
*$3,8,15,24,35$
*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
*$2, 4, 8, 16 \text{ and } 32$
*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
*$25,-125,625,-3125,15625$
*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
*$65, 93$
*$\dfrac{49}{128}$
*$729$
*$\dfrac{360}{23}$
*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
*$1,2,\dfrac{3}{5},\dfrac{8}{5}$
|
I don't think I've ever seen $x \frac{y}{z}$ used to mean $x + \frac{y}{z}$ except when $x$, $y$ and $z$ are literal integers (e.g. $2 \frac{3}{4}$). That's not to say it never happens, but it would be terribly confusing.
|
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|
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx $ Problem :
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx$.
My approach :
\begin{align}
&\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx \\
=& \ln(1^x +2^x +3^x +6^x) \frac{x^2}{2} - \int^1_{-1} \frac{1}{1^x+2^x+3^x+6^x}(2^x\log2+3^x \log3+6^x \log 6 )\frac{x^2}{2} \,dx
\end{align}
[By using by parts]
Is it the correct method of solving this, please suggest, will be of great help, thanks.
|
$$I=\int_{-1}^0f(x)dx+\int_0^1f(x)dx$$
$$=\int_0^1(f(-x)+f(x))dx$$
$$=\int_0^1(-f(x)+x\ln(6^x)+f(x))dx$$
$$=\ln(6)\int_0^1x^2dx$$
$$=\frac{\ln(6)}{3}.$$
|
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|
Finding linear asymptote We have
$$ f(x) = (2x^2-x^3)^{1/3} $$
A linear asymptote is like: $$y = px+q$$ $$p \not= 0$$
We had this definition in our lecture:$$ p = \lim\limits_{x \rightarrow \infty}{\frac{f(x)}{x}}
$$
so
$$ p = \lim\limits_{x \rightarrow \infty}{\frac{(2x^2-x^3)^\frac{1}{3}}{x}} $$
$$ =\lim\limits_{x \rightarrow \infty}{(\frac{2x^2}{x^3} - \frac{x^3}{x^3} )^\frac{1}{3}} $$
$$ = (-1)^\frac{1}{3}$$
So, $$p = (-1)^\frac{1}{3}$$
And
$$q = \lim\limits_{x \rightarrow \infty}{f(x)-px} $$
This would be $$ \lim\limits_{x \rightarrow \infty}{(2x^2-x^3)^\frac{1}{3}} - (-1)^\frac{1}{3}x$$
In Wolframalpha this lim has a complex solution. Does this function have a linear asymptote ? Can a real number function (from $$R -> R$$ ) can have a complex asymptote? I guess not, because then it wouldn't be linear anymore.
|
I would use the generalized
binomial theorem:
$\begin{array}\\
f(x)
&= (2x^2-x^3)^{1/3}\\
&= (-x)(1-\frac{2}{x})^{1/3}\\
&= (-x)(1-\frac{2}{x}\frac13+(-\frac{2}{x})^2\frac13 (-\frac23)+O(\frac1{x^3}))\\
&= (-x)(1-\frac{2}{3x}-\frac{8}{9x^2}+O(\frac1{x^3}))\\
&= -x+\frac{2}{3}+\frac{8}{9x}+O(\frac1{x^2}))\\
\end{array}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2076637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find Delta for given limit of function I have : $$\lim_{x \to-2}\frac{3x+6}{x^3+8} = \frac{1}{4}$$
This is what i've come up with so far :
$$|f(x) - L| = \frac{3x+6}{x^3+8} -\frac{1}{4}| =|\frac{12x+24}{4x^3+32} -\frac{x^3+8}{4x^3+32}| = |\frac{-x^3+12x+16}{4x^3+32}|$$
I am looking for a $\delta(\epsilon)$ so that $|x - x_0| < \delta \rightarrow |f(x) - L| < \epsilon$
My problem is that I don't know how to work with that fraction .
|
$$\left|\frac{3(x+2)}{x^3+8}-{1\over4}\right|<\epsilon\Longleftrightarrow\left|\frac{3(x+2)}{(x+2)(x^2-2x+4)}-{1\over4}\right|<\epsilon$$
$$x\ne-2\hspace{1cm}\left|\frac{3}{x^2-2x+4}-{1\over4}\right|<\epsilon$$ $$\left|\frac{-x^2+2x+8}{4(x^2-2x+4)}\right|<\epsilon\Longleftrightarrow-\epsilon<\frac{-x^2+2x+8}{4(x^2-2x+4)}<\epsilon\Longleftrightarrow \begin{cases}{x^2(-1-4\epsilon)+x(2+8\epsilon)+8-16\epsilon\over4(x^2-2x+4)}<0\\{x^2(-1+4\epsilon)+x(2-8\epsilon)+8+16\epsilon\over4(x^2-2x+4)}>0\end{cases}$$
but $4(x^2-2x+4)>0\ \forall x\in\mathbb R$, so:
\begin{cases}x^2(-1-4\epsilon)+x(2+8\epsilon)+8-16\epsilon<0\\x^2(-1+4\epsilon)+x(2-8\epsilon)+8+16\epsilon>0\end{cases}
If you solve this system, you get:
$${-1+4\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}<x<{-1-4\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}$$
that is a neighborhood of $-2$, hence:
$$2+{-1+4\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}<x+2<2+{-1-4\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}$$
$${-3+12\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}<x+2<{-3-12\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}$$
$$\left|x+2\right|<\min\left\{\left|{-3+12\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}\right|;{-3-12\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}\right\}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alternative proof, or at least a hint to it. My proof as follows:
Bernoulli inequality states that for $-1<x, x\neq 0, n\in \mathbb{N},n>1$ the following is true:$(1+x)^n>1+nx$.
Thus, $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ is also true.
Then I need to show that $(1-x)^n<\frac{1}{(1+x)^n}$, which is equivalent to $\frac{1}{(1-x)^n}>(1+x)^n$, which I prove by induction:
Basecase: $n=1$ $1=\frac{1-x}{1-x}\Leftrightarrow 1=\frac{1}{1-x}-\frac{x}{1-x}\Leftrightarrow 1+\frac{x}{1-x}=\frac{1}{1-x}$
Let $a,b\in \mathbb{R_{>0}}$ and $0>b>1$ Then $\left[ a>ab \right]\Leftrightarrow \left[a<\frac{a}{b}\right]$. Thus $\left[ 0<x<1\right] \Rightarrow \left[ x<\frac{x}{1-x}\right]$
Thus $1+\frac{x}{1-x}=\frac{1}{1-x} \Rightarrow 1+x<\frac{1}{1-x} \square$
Inductive step: Assume $(1+x)^n<\frac{1}{(1-x)^n}$. Need to show $(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}$.
$(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}\Leftrightarrow (1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x}$
Let $a,b,c,d \in \mathbb{R_{>0}}$. Then $[a>c]\wedge[b>d] \Rightarrow [ab>cd]$. $(1+x)^n<\frac{1}{(1-x)^n}$ was the assumption and $1+x>\frac{1}{1-x}$ was the basecase, therefore $(1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x} \Leftrightarrow (1+x)^{n+1}<\frac{1}{(1-x)^{n+1}} \square$
Thus $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ and $(1-x)^n<\frac{1}{(1+x)^n}$ are both true, which implies the original statement $(1-x)^n<\frac{1}{1+nx} \square$ If I were to count the proof of the Bernoulli inequality by induction, it would mean that I used induction twice in order to prove something that basic, which to me doesn't seem to be a sensible thing to do.
|
Let $f(x)=(1-x)^{-n}$. Then by applying mean value theorem for $f$ we have,
$$\frac{f(x)-f(0)}{x-0}=f'(\zeta)$$(for some $0<\zeta<x<1$)
$$\frac{(1-x)^{-n}-(1-0)^n}{x-0}=(-n)(1-\zeta)^{-n-1}(-1)=\frac{n}{(1-\zeta)^{n+1}}>n$$
The above line is true as $1-\zeta<1\Rightarrow\frac{1}{1-\zeta}>1$
$$\Rightarrow (1-x)^{-n}-1>nx$$
$$\Rightarrow (1-x)^{-n}>1+nx$$
Now $1-x>0, 1+nx>0$ implies,
$$\Rightarrow \frac{1}{1+nx}>(1-x)^n$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
What fractions can fill a $N$ by $N$ matrix given that their sum is always 2? My problem is the following: I have a matrix $N$ by $N$ in size. I want to fill it with fractions of $1$ of increasing denominator in relation to their distance from the center of the matrix. The central value is always $1$.
$N$ is always odd.
Example for $N = 3$:
1/12, 1/ 6, 1/12
1/ 6, 1 , 1/ 6
1/12, 1/ 6, 1/12
Proof:
$1/12 * 4 + 1/6 * 4 + 1 = 2$
But what would be the formulaic approach for a, say, $5*5$ matrix?
|
Let's say we have a $2n+1 \times 2n+1$ matrix $A$ with a $1$ in the center, which is $A_{n+1,n+1}$, and $\frac{1}{d\cdot \text{taxicab}(x,y,n+1,n+1)}$ for $A_{x,y}$ for any cell that is not the center. We want all of this to sum to $2$, so without the $1$ in the center, we want the elements to sum to $1$.
In Python, this means:
from fractions import Fraction
def taxicab(a,b,c,d): return abs(a-c)+abs(b-d)
# Set this to any non-negative integer you want:
n = 2
# We need to sum x,y from 1 to 2n+1, inclusive on both ends, which is range(1, 2n+2) in Python
sum([Fraction(1, d*taxicab(x,y,n+1,n+1)) for x in range(1, 2*n+2) for y in range(1, 2*n+2) if (x, y) != (n+1, n+1)]) == 1
All of the terms in the sum have a $\frac{1}{d}$ in them, so we can factor out that $\frac 1 d$ and then multiply both sides by $d$ to get $d$ on the right side. Now, switch both sides of the equation to get:
d = sum([Fraction(1, taxicab(x,y,n+1,n+1)) for x in range(1, 2*n+2) for y in range(1, 2*n+2) if (x, y) != (n+1, n+1)])
And then, we can print out the matrix:
# This prints out d:
print("d =", d)
# This prints out the matrix:
for x in range(1, 2*n+2):
for y in range(1, 2*n+2):
# Assume that it's the center and our element is 1:
this_element = "1"
# If it's not the center, then change the element accordingly:
if (x, y) != (n+1, n+1):
this_element = str(Fraction(1, d*taxicab(x,y,n+1,n+1)))
print(this_element, end="")
# For formatting:
for i in range(10-len(this_element)): print(end=" ")
# For formatting:
print("")
For example, for $n=5$, we get $d=\frac{35}{3}$, giving us:
$$\left[\begin{matrix}\frac{3}{140} \ \frac{1}{35} \ \frac{3}{70} \ \frac{1}{35} \ \frac{3}{140} \\ \frac{1}{35} \ \frac{3}{70} \frac{3}{35} \ \frac{3}{70} \ \frac{1}{35} \\ \frac{3}{70} \ \frac{3}{35} \ 1 \ \frac{3}{35} \ \frac{3}{70} \\ \frac{1}{35} \ \frac{3}{70} \frac{3}{35} \ \frac{3}{70} \ \frac{1}{35} \\ \frac{3}{140} \ \frac{1}{35} \ \frac{3}{70} \ \frac{1}{35} \ \frac{3}{140}\end{matrix}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2078547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Trignometric integration We have to solve the following integration.
$$\int\frac{\tan 2\theta}{\sqrt{\cos^6 \theta+\sin^6\theta}}\ d\theta$$
In this, I write the denominator as $\sqrt 1-3\sin^2\theta \cos^2\theta$
But now how to proceed?
|
as you write $\displaystyle \sin^6 \theta+\sin^6 \theta = 1-3\sin^2 \theta \cos^2 \theta=\frac{1}{4}\left(4-3\sin^2 2 \theta\right)$
let $\displaystyle \mathcal{I} = \int\frac{\tan 2 \theta }{\sqrt{\sin^6 \theta+\sin^6 \theta}}d\theta = 2\int\frac{\sin 2 \theta }{\cos 2 \theta \sqrt{4-3\sin^2 2 \theta}}d\theta = 2\int\frac{\sin 2 \theta\cos 2 \theta }{\cos^2 2 \theta \sqrt{4-3\sin^2 2 \theta}}d\theta$
substitute $\sin 2 \theta = t$ and $2\cos 2 \theta d \theta = dt$
$\displaystyle \mathcal{I} = \int\frac{t}{(1-t^2)\sqrt{4-3t^2}}dt$
substitute $4-3t^2 = u^2$ and $\displaystyle tdt = -\frac{1}{3}udu$
$\displaystyle \mathcal{I} = -\int\frac{1}{3-3u^2}du = -\int\frac{1}{u^2-1}du$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
Solve a matrix equation I need to find $X$ from
$$\begin{pmatrix}
1 & 2\\
-3 &-6
\end{pmatrix} X \begin{pmatrix}
1 &2 \\
-1 &-2
\end{pmatrix}=\begin{pmatrix}
2 &4 \\
-6 & -12
\end{pmatrix}$$
I wrote $X$ as
$$X=\begin{pmatrix}
a & b\\
c &d
\end{pmatrix}$$
and I got $a+2c-b-2d=2$ but I do not know what to do next. Please help.
|
You are on good way. When you multiply matrices on left side, you finally get $$a-b+2c-2d = 2$$ which is a linear equation. Solutions are given with three parameters, and thus
$$ X = \begin{pmatrix}
a & b\\
c & \frac{a-b+2c-2}{2}
\end{pmatrix}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2081979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal.
\begin{equation*}
A =
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
\end{equation*}
I know that a $n \times n$ matrix is called orthogonal if $A^TA$ $=$ $id$ which means:
\begin{equation*}
\begin{pmatrix}
1&a \\
2&b
\end{pmatrix}
\cdot
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
=
\begin{pmatrix}
1+a^2&2+ab \\
2+ab&4+b^2
\end{pmatrix}
\stackrel{?}{=}
\begin{pmatrix}
1&0 \\
0& 1
\end{pmatrix}
\end{equation*}
|
You are right. Now solve the system of equation in terms of $a$ and $b$. However, if $b$ is restricted to be an integer, the equation $$4 + b^2 = 1$$ does not have a solution.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
|
Since $$(1 - \sqrt{5})^3 = 16 - 8 \sqrt{5}$$ and similarly $$(1 + \sqrt{5})^3 = 16 + 8 \sqrt{5}$$ it follows that $$(2 + \sqrt{5})^{1/3} + (2 - \sqrt{5})^{1/3} = \left(\frac {16+8\sqrt{5}} 8 \right)^{1/3} +\left(\frac {16-8\sqrt{5}} 8 \right)^{1/3} = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1.$$
|
{
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"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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|
Probability event with $70$% success rate occurs three consecutive times for sample size $n$ It has been a long time since I've done probability so I am not sure which to do (if either are correct). Thank you for taking the time to look at my work.
Probability an event occurs is $70$%.
I'm looking for the probability our event occurs three times in a row for sample size $n$.
$(.7)^3=.343$ is the probability to occur three consecutive times
$1-.343=.657$ would be the chance to fail.
First idea:
For $n=3$ our success rate is $.343$
$n=4$ we have two opportunities for success, thus $1-(.657)^2=.568351$
$n=5$, three opportunities for success, thus $1-(.657)^3=.71640$...
Generalization: Probability for success: $$1 - (.657)^{n-2}$$
Second idea:
Probability when $n=3$ would be $(.7)^3$
At $n=4$ we'd have $(.7)^3+(.3)(.7)^3$
For $n=5$ we'd have $(.7)^3+(.3)(.7)^3+(.3)^2(.7)^3+(.3)(.7)^4$
I'm leaning towards the second idea...but I'm failing to see a generalization for it.
Please excuse my LaTeX it has been a long time since I've asked/answered any questions. Thank you.
|
Let's denote the event under consideration with $a$
\begin{align*}
P(X=a)=0.7
\end{align*}
and we denote the complementary event with $b$.
We are looking for the words of length $n$ and their probability of occurrence which do not contain three or more consecutive $a$'s. The result is $1$ minus this probability.
We can describe the set of these invalid words as the words built from the alphabet $V=\{a,b\}$ which contain at most one or two consecutive $a$'s.
These are words
*
*starting with zero or more $b$'s:$$b^*$$
*followed by zero or more occurrences of $a$ or $aa$ each followed by one or more $b$'s: $$(ab^+|aab^+)^*$$
*and terminating with zero, one or two $a$'s
$$(\varepsilon|a|aa)$$
We obtain
\begin{align*}
b^*(ab^+|aab^+)^*(\varepsilon|a|aa)\tag{1}
\end{align*}
The regular expression (1) generates all invalid words in a unique manner. In such cases we can use it to derive a generating function $$\sum_{n=0}^\infty a_n z^n$$ with
$a_n$ giving the number of invalid words of length $n$.
In order to do so all we need to know is the geometric series expansion since the $star$ operator
\begin{align*}
a^*=\left(\varepsilon|a|a^2|a^3|\cdots\right)\qquad\text{ translates to }\qquad 1+z+z^2+z^3+\cdots=\frac{1}{1-z}
\end{align*}
Accordingly $a^+=aa^*$ translates to $\frac{z}{1-z}$ and alternatives like $(\varepsilon|a|aa)$ can be written as $1+z+z^2$.
We translate the regular expression (1) into a generating function (by mixing up somewhat the symbolic to provide some intermediate steps).
Since we want to calculate the probabilities of occurrence of $P(X=a)$ we keep track of $a$ and $b$ by respecting them as corresponding factors in the generating function.
\begin{align*}
b^*\left(ab^+|aab^+\right)^*(\varepsilon|a|aa)
&\longrightarrow \quad \frac{1}{1-bz}\left(\left.\frac{abz^2}{1-bz}\right|\frac{a^2bz^3}{1-bz}\right)^*\left(1+az+a^2z^2\right)\\
&\longrightarrow \quad \frac{1}{1-bz}\left(\frac{abz^2+a^2bz^3}{1-bz}\right)^*(1+az+a^2z^2)\\
&\longrightarrow \quad \frac{1}{1-bz}\cdot\frac{1}{1-\frac{abz^2+a^2bz^3}{1-bz}}(1+az+a^2z^2)\\
&\quad\quad=\frac{1+az+a^2z^2}{1-bz-abz^2-a^2bz^3}\tag{1}
\end{align*}
$$ $$
We conclude: The number of invalid words is given by (1). So, the number of valid words is the number of all words minus the number of invalid words. We obtain the generating function $A(z)$
\begin{align*}
A(z)&=\sum_{n=0}^\infty(a+b)^nz^n-\frac{1+az+a^2z^2}{1-bz-az^2-a^2z^3}\\
&=\frac{1}{1-(a+b)z}-\frac{1+az+a^2z^2}{1-bz-az^2-a^2z^3}\\
&=\frac{a^3z^3}{(1-(a+b)z)(1-bz-abz^2-a^2bz^3)}\\
&=a^3z^3+a^3(a+2b)z^4+a^3(\color{blue}{1}a^2+\color{blue}{4}ab+\color{blue}{3}b^2)z^5\\
&\qquad a^3(a+b)^2(a+4b)z^6+a^3(a^4+7a^3b+18a^2b^2+16ab^3+5b^4)z^7+\cdots
\end{align*}
The expansion was done with the help of Wolfram Alpha. We see that e.g. the number of valid words of length $5$ is $\color{blue}{1}+\color{blue}{4}+\color{blue}{3}=8$.
$$ $$
Out of $2^5=32$ binary words of length $5$ there are $8$ valid words which are marked $\color{blue}{\text{blue}}$ in the table below.
We obtain
\begin{array}{cccc}
\color{blue}{aaa}aa\qquad&ab\color{blue}{aaa}\qquad&b\color{blue}{aaa}a\qquad&bb\color{blue}{aaa}\\
\color{blue}{aaa}ab\qquad&abaab\qquad&b\color{blue}{aaa}b\qquad&bbaab\\
\color{blue}{aaa}ba\qquad&ababa\qquad&baaba\qquad&bbaba\\
\color{blue}{aaa}bb\qquad&ababb\qquad&baabb\qquad&bbabb\\
aabaa\qquad&abbaa\qquad&babaa\qquad&bbbaa\\
aabab\qquad&abbab\qquad&babab\qquad&bbbab\\
aabba\qquad&abbba\qquad&babba\qquad&bbbba\\
aabbb\qquad&abbbb\qquad&babbb\qquad&bbbbb\\
\end{array}
We denote with $[z^n]$ the coefficient of $z^n$ in a series.
We conclude
*
*The number of occurrences of wanted words of size $n$ is the coefficient of $z^n$ of $A(z)$ evaluated at $a=b=1$ and given as OEIS sequence 050231.
\begin{align*}
\left.[z^n]A(z)\right|_{a=b=1}
\end{align*}
*
*The wanted probability of valid words of size $n$ is the coefficient of $z^n$ of $A(z)$ evaluated at $a=0.7,b=0.3$.
\begin{align*}
\left.[z^n]A(z)\right|_{a=0.7,b=0.3}
\end{align*}
*We obtain for $n=0$ up to $n=7$
\begin{array}{c|cl}
n&\left.[z^n]A(z)\right|_{a=b=1}&\left.[z^n]A(z)\right|_{a=0.7,b=0.3}\\
\hline
0&0&0\\
1&0&0\\
2&0&0\\
3&1&0.343\\
4&3&0.4459\\
5&8&0.5488\\
6&20&0.6517\\
7&47&0.7193\\
\end{array}
|
{
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"url": "https://math.stackexchange.com/questions/2082969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]
But, in my book the integration of $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.
We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?
|
WLOG, $a>0$. As there are singularities at $x=-a$ and $x=a$, the expressions can differ in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$, depending on the signs of $x+a$ and $x-a$.
Correct expressions are
$$\int\frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|.$$
(But you are not allowed to evaluate the definite integrals across the singularities.)
Then the effect of a change of sign is
$$\int\frac{dx}{a^2-x^2}=-\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{x-a}{x+a}\right|=\frac1{2a}\log\left|\frac{a-x}{a+x}\right|.$$
|
{
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"url": "https://math.stackexchange.com/questions/2083201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Expected value and probability
The distribution function of a discrete random variable X is given
$$F_X(x)=\begin{cases} 0, &x<3\\ \frac{6}{11},& 3\leq x< 4 \\ \frac{7}{11}, & 4\leq x<5 \\ \frac{8}{11}, & 5\leq x<6 \\ 1, & 6\leq x \end{cases} $$
Let $A=(X=3)\cup (X=5)$. Calculate: $P(A)$ and $E(X)$
How I did it:
$P(1)=0, P(2)=\frac{6}{11}, P(3)=\frac{1}{11}, P(4)=\frac{1}{11}, P(5)=\frac{3}{11}$
$P(A)=\frac{1}{11}+\frac{3}{11}=\frac{4}{11}$ (Correct solution: $\frac{7}{11}$)
$E(X)=2\cdot\frac{6}{11}+3 \cdot \frac{1}{11} + 4 \cdot \frac{1}{11} +5\cdot \frac{3}{11}=\frac{34}{11}$ (Correct solution: $ \frac{45}{11}$)
Where is my mistake?
Should I start with $P(0)$ or $P(1)$?
|
The values that you got for the probabilities $P(X=x)$ are wrong. $P(2)=0$ and $P(3)=\frac6{11}$ and so on. If you correct this, then you get for the first one:\begin{align}P(X=3)&=P(X\le 3)-P(X<3)=F_X(3)=\frac6{11}\\[0.2cm]P(X=5)&=P(X\le 5)-P(X<5)=F_X(5)-F_X(5-)=\frac8{11}-\frac7{11}=\frac1{11}\end{align} Hence $P(A)=P(X=3)+P(X=5)=\frac6{11}+\frac1{11}=\frac7{11}$.
For the expected value, note that from the given $F$ you can conlude that $$P(X=3)=\frac6{11},\;P(X=4)=\frac1{11},\;P(X=5)=\frac1{11},\;P(X=6)=\frac3{11}$$ Hence $$E[X]=3\cdot\frac6{11}+4\cdot\frac1{11}+5\cdot\frac1{11}+6\cdot\frac3{11}=\frac{45}{11}$$
|
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|
Find all $n$ such that $n \mid x_n$ where $x_n = x_{n-1} + \lfloor n^2 / 4 \rfloor$, $x_0 = 0$ $X_n$ is sequence such that $x_n=x_{n-1}+[\frac{n^2}{4}]$ and $x_0=0$. Find all positve integers $n$ for which $x_n$ is divisible by $n$.
[X] means integer part.
|
$$x_{2k}=x_{2k-1}+k^2=x_{2k-2}+2k^2-k=2k^2-k+(2(k-1)^2-(k-1))+x_{2k-4}=\cdots=\sum_{n=1}^{k} 2n^2-\sum_{k=1}^{k} n=\frac{k(k+1)(2k+1)}{3}-\frac{k(k+1)}{2}=\frac{k(k+1)(4k+2-3)}{6}=\frac{k(k+1)(4k-1)}{6}$$
Also
$$a_{2k+1}=a_{2k}+k^2+k\\a_{2k+1}=\frac{k(k+1)(4k-1)}{6}+\frac{6k(k+1)}{6}=\frac{k(k+1)(4k+5)}{6}$$
Now splitting into two cases we can see that
$$2k+1\mid \frac{k(k+1)(4k+5)}{6}$$ Works only for $k=1$ since $\gcd(2k+1,k)=\gcd(2k+1,k+1)=1$ and $\gcd(4k+5,2k+1)\mid 3$ that implies that only possible solution is $2k+1=1$ or $2k+1=3$
$$2k\mid\frac{k(k+1)(4k-1)}{6}$$
Works for $k=12t+7$ and $k=12t+11$,this is because $$\frac{(k+1)(4k-1)}{12}\in\mathbb{N}$$
Putting it all together for $n=3$ and $n=24t+14,n=24t+22$ we have that $n\mid a_n$
|
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|
Integrability of $f(t) =\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{\Gamma \left( \frac{it+1}{1.5} \right) }{\Gamma \frac{ it+1}{2} }$ Can we show that the following function is integrable
\begin{align}
f(t) =\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{\Gamma \left( \frac{it+1}{1.5} \right) }{\Gamma \left(\frac{ it+1}{2} \right)},
\end{align}
where $t \in \mathbb{R}$ and $i=\sqrt{-1}$.
That is can we show that
\begin{align}
\int_{-\infty}^{\infty} |f(t)| dt<\infty.
\end{align}
I was wondering if Stirling's approximation can be used, since this is a complex case?
Note if Stirling's approximation can be used than
\begin{align}
f(t) \approx \sqrt{\frac{1.5}{2}}\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}} \frac{ \left( \frac{it+1}{1.5 e} \right)^{\frac{1+it}{1.5}} }{ \left(\frac{ it+1}{2 e} \right)^{\frac{1+it}{2}}}.
\end{align}
Note that
\begin{align}
|f(t)| &\approx \left| \frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}}\right| \left| \frac{ \left( \frac{it+1}{1.5 e} \right)^{\frac{1+it}{ 1.5}} }{ \left(\frac{ it+1}{2 e} \right)^{\frac{1+it}{2}}}\right|\\
&=\left| \frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}}\right| \left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right| \left|\frac{ \left(2 e\right)^{\frac{1+it}{2 }}} {(1.5 e)^{\frac{1+it}{ 1.5}}} \right|
\end{align}
Also we have that
\begin{align}
&\left|\frac{2^{\frac{it+1}{1.5}}}{2^{\frac{it+1}{2}}}\right|= 2^{\frac{2}{3}-\frac{1}{2}} \\
& \left|\frac{ \left(2 e\right)^{\frac{1+it}{2 }}} {(1.5 e)^{\frac{1+it}{ 1.5}}} \right| =\frac{ \left(2 e\right)^{\frac{1}{2 }}} {(1.5 e)^{\frac{1}{ 1.5}}}
\end{align}
So, in the end we if everthing is corect we have to show that
\begin{align}
g(t)=\left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right|
\end{align}
is integrable, but I am not sure how to show if the above equation is integrable or not?
Any ideas would be greatly appreciated. Thank you.
Edit
The integrability of $g(t)$ has been shown in one of the answers.
My question now is the following: Since we have that
\begin{align}
|f(t)| =|g(t)| +e
\end{align}
and $g(t)$ is integrable does this mean that $f(t)$ is inegrable? Can we prove that the error term $e$ is also integrable?
|
We have $$\left| \frac{ \left( it+1\right)^\frac{1+it}{1.5} }{ \left( it+1 \right)^{\frac{1+it}{2}}}\right| = |(1+it)^{\frac{1+it}{6}}|= \left|\exp\left(\left(\frac{1+it}{6}\right)\ln(1+it)\right)\right|.$$ Remember that $|e^z|=e^{\Re(z)}$ and $$\Re\left(\left(\frac{1+it}{6}\right)\ln(1+it)\right)=\frac{\ln(\sqrt{1+t^2})}{6}-\frac{t\arg(1+it)}{6}.$$ (Here we have used the definition of the complex logarithm $\ln(z) = \ln|z|+i\arg(z)$.) Hence $$g(t) = (1+t^2)^{1/12} e^{-\arg(1+it)t/6}.$$ If $t>0$ is big enough, we can assume that $\arg(1+it)>\pi/4$, hence $$g(t) < (1+t^2)^{1/12} e^{-\frac{\pi t}{24}}$$ is clearly integrable at $+\infty.$ You can procede in a same way for $-\infty.$
|
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|
Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)}.$$
Let's prove $\frac{n+2}{3n+1}$ is a decreasing sequence.
$$a_n>a_{n+1} \Leftrightarrow \frac{n+2}{3n+1}>\frac{n+3}{3n+4}\Leftrightarrow(n+2)(3n+4)>(n+3)(3n+1)\Leftrightarrow3n^2+10n+8>3n^2+10n+3\Leftrightarrow 8>3$$
So $\frac{n+2}{3n+1}$ is a decreasing sequence and we know that $\frac{1}{n(3n+1)}$ is also decreasing so our given sequence is a decreasing sequence as a sum of $2$ decreasing sequences.
|
Your solution looks good.
Another approach could be:
$$a_n=\frac{3n^2+n}{(n+1)^2}=\frac{3(n+1)^2-5n-3}{(n+1)^2}=3-\left[\frac{5n}{(n+1)^2}+\frac{3}{(n+1)^2}\right]=3-\left[\frac{5}{n+2+\frac{1}{n}}+\frac{3}{(n+1)^2}\right]$$
$a_n$ is increasing so what can we conclude about $\frac{1}{a_n}$?
|
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|
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
I don't know how to start...
|
Let's start by applying the quadratic formula.
We have: $(x^{2}+4x+8)^{2}+3x\hspace{1 mm}(x^{2}+4x+8)+2x^{2}=0$
$\Rightarrow x^{2}+4x+8=\dfrac{-3x\pm\sqrt{(3x)^{2}-4(1)(2x^{2})}}{2(1)}$
$\hspace{27.75 mm} =\dfrac{-3x\pm{x}}{2}$
$\hspace{27.75 mm} =-2x,-x$
Then,
$\Rightarrow x^{2}+4x+8=-2x$
$\Rightarrow x^{2}+6x+8=0$
$\Rightarrow x^{2}+2x+4x+8=0$
$\Rightarrow x\hspace{1 mm}(x+2)+4\hspace{1 mm}(x+2)=0$
$\Rightarrow (x+2)(x+4)=0$
$\Rightarrow x=-2,-4$
or
$\Rightarrow x^{2}+4x+8=-x$
$\Rightarrow x^{2}+5x+8=0$
$\Rightarrow x=\dfrac{-5\pm\sqrt{5^{2}-4(1)(8)}}{2(1)}$
$\hspace{9 mm} =\dfrac{-5\pm\sqrt{-7}}{2}$
$\hspace{9 mm} =\dfrac{-5\pm\sqrt{7}i}{2}$
$\hspace{9 mm} =\dfrac{-5-\sqrt{7}i}{2},\dfrac{-5+\sqrt{7}i}{2}$
However, $x\in\mathbb{Q}$.
Therefore, the solutions to the equation are $x=-2$ and $x=-4$.
|
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|
Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\dfrac{1}{2}-\sqrt{x^2-x+1} \\
&=\dfrac{(x-\dfrac{1}{2})^2-|x^2-x+1|}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}\\
&= \dfrac{-3/4}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}
\end{aligned}
i don't know if $(x-\dfrac{1}{2})+\sqrt{x^2-x+1} $ positive or negative for all x in R
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Note that this proof is a bit longer than expected, but it shows some line of thoughts....We have: If $x \le 0 \implies f(x) \le 0 < \dfrac{1}{2} \implies f(x) < \dfrac{1}{2}$. Thus if $x > 0 \implies f(x) = \dfrac{x-1}{x+\sqrt{x^2-x+1}}$. Here we can reason a little bit. If $0 < x \le 1 \implies f(x) \le 0 \implies f(x) < \dfrac{1}{2}$. Thus $x > 1$. If again $1 < x < 2\implies x^2 - x + 1 > x^2-4x+4 \implies x^2-x+1 > (x-2)^2\implies \sqrt{x^2-x+1} > |x-2| = 2-x\implies f(x) < \dfrac{x-1}{x+2-x}= \dfrac{x-1}{2} < \dfrac{1}{2}$ since $x < 2$. Finally, if $x \ge 2\implies |x-2| = x-2\implies f(x) < \dfrac{x-1}{x+x-2}= \dfrac{x-1}{2x-2}= \dfrac{x-1}{2(x-1)} = \dfrac{1}{2}$. Thus for all $x \in \mathbb{R}$, $f(x) < \dfrac{1}{2}$.
|
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|
Prove that $\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$ Let $x_1,x_2,x_3,\cdots ,x_n (n\ge2)$ be real numbers greater than $1.$ Suppose that $|x_i-x_{i+1}|<1$ for $i=1,2,3,\cdots,(n-1)$.
Prove that $$\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$$
Please help!!!
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Set $d_i = x_{i+1} - x_i$ for $i = 1, \ldots, n-1$. Then
$$
\frac{x_n}{x_1} = 1 + \sum_{i=1}^{n-1} \frac{d_i}{x_1} \quad ,
\quad \frac{x_i}{x_{i+1}} = 1 - \frac{d_i}{x_{i+1}}
$$
and therefore
$$
\frac{x_n}{x_1} + \sum_{i=1}^{n-1} \frac{x_i}{x_{i+1}} =
n + \sum_{i=1}^{n-1} d_i \left( \frac{1}{x_1} - \frac{1}{x_{i+1}} \right) \\
\le n + \sum_{i=1}^{n-1} \lvert d_i \lvert \left\lvert \frac{1}{x_1} - \frac{1}{x_{i+1}} \right \rvert < n + (n-1) = 2n-1
$$
because
*
*$\lvert d_i \lvert < 1$ and
*all numbers $1/x_i$ are in the
interval $(0, 1]$ so that the (absolute value of) any difference
$1/{x_1} - 1/{x_{i+1}}$ is less than one.
The inequality can actually be improved a bit.
For fixed $i$ set $m = \min(x_1, x_{i+1})$ and $M = \max(x_1, x_{i+1})$.
Then $M < m + i$ and therefore
$$
\left\lvert \frac{1}{x_1} - \frac{1}{x_{i+1}} \right \rvert
= \frac 1m - \frac 1M < \frac 1m - \frac{1}{m+i}
= \frac{i}{m(m+i)} \le \frac{i}{i+1}
$$
and the above method gives
$$
\frac{x_n}{x_1} + \sum_{i=1}^{n-1} \frac{x_i}{x_{i+1}}
< n + \sum_{i=1}^{n-1}\frac{i}{i+1}
= (2n-1) - \sum_{i=2}^{n} \frac 1i \, .
$$
Choosing $x_i = i + (i-1) \varepsilon$ shows that this bound is sharp.
|
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|
Using AM-GM inequality Let $a,\ b,\ c$ be positive real numbers. Prove that:
$$\left(a+\dfrac{bc}{a}\right)\left(b+\dfrac{ca}{b}\right)\left(c+\dfrac{ab}{c}\right)\geq 4\sqrt[3]{\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)}\tag1$$
|
By AM-GM $$(a^2+bc)(b^2+ac)=c(a^3+b^3)+ab(c^2+ab)\geq2\sqrt{abc(a^3+b^3)(c^2+ab)}.$$
Thus, $$\prod_{cyc}\left((a^2+bc)(b^2+ac)\right)\geq8\sqrt{a^3b^3c^3\prod_{cyc}(a^3+b^3)\prod_{cyc}(c^2+ab)}$$ or
$$\prod_{cyc}(a^2+bc)^3\geq64a^3b^3c^3\prod_{cyc}(a^3+b^3)$$ or
$$\prod_{cyc}\left(a+\frac{bc}{a}\right)\geq4\sqrt[3]{\prod_{cyc}(a^3+b^3)}.$$
Done!
|
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|
Show that $\mathbb Q(\sqrt[3]{2})[Y]/\langle Y^2+Y+1\rangle$ is a splitting field of $X^3-2$ Show that $\mathbb Q(\sqrt[3]{2})[Y]/\langle Y^2+Y+1\rangle$ is a splitting field of $f(X) = X^3-2$ where $Y$ is an indeterminate over $\mathbb Q$
|
Since $Y^2 + Y + 1$ is irreducible, the elements of $\mathbb{Q}(\sqrt[3]{2})(Y)/ \langle Y^2 + Y + 1 \rangle$
are of the form $ a+b \theta$, where $\theta$ is a root of $Y^2 + Y + 1$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$. The roots of $Y^2 + Y + 1$ are
$$ -\frac{1}{2}\pm i\frac{\sqrt{3}}{2}.$$
Choose the one with the plus sign, hence the elements of $\mathbb{Q}(\sqrt[3]{2})(Y)/ \langle Y^2 + Y + 1 \rangle$
are of the form $ a+b \theta$, where $\theta=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$
Now, the roots of $X^3-2=0$ are $\sqrt[3]{2}$, $\zeta \sqrt[3]{2}$ and $\zeta^2 \sqrt[3]{2}$ where $\zeta$ is a the primitive third root of $1$, hence its splitting field is $\mathbb{Q}(\sqrt[3]{2}, \zeta)$. Since
$$ \zeta = e^{\frac{2\pi i}{3}}=\cos \frac{2\pi i}{3} + i \sin \frac{2\pi i}{3} = -\frac{1}{2}+i \frac{\sqrt{3}}{2},$$
then $\mathbb{Q}(\sqrt[3]{2}, \zeta) =\mathbb{Q}(\sqrt[3]{2},-\frac{1}{2}+i \frac{\sqrt{3}}{2})$, that is, the elements of the form $a+b \theta$, where $\theta=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$.
|
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|
Compute ${11 \choose 1} + {11 \choose 3} + ... + {11 \choose 11}$? I did this with brute force, and got 1024.
What is the faster method of solving this?
|
Since $\binom{11}{1}=\binom{11}{10},\binom{11}{3}=\binom{11}{8}$ and so on,
$$ \binom{11}{1}+\binom{11}{3}+\ldots+\binom{11}{11} = \binom{11}{10}+\binom{11}{8}+\ldots+\binom{11}{0} $$
so both the terms are half the sum $\binom{11}{0}+\binom{11}{1}+\binom{11}{2}+\ldots+\binom{11}{11}=2^{11}$, i.e. $2^{10}=\color{red}{1024}$.
|
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|
Help to Prove that $\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}$ I need help on proving $(1)$.
$$I=\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}\tag1$$
This is what I have attempted;
Enforcing a sub: $u=\cot^2{x}$ then $du=-2\cot{x}\csc^2{x}dx$
Recall $1+\cot^2{x}=\csc^2{x}$
$$I={1\over2}\int_{1}^{\infty}\arctan{u}\cdot{\mathrm dx\over u^{1/2}+u^{3/2}}$$
Recall $u^3+1=(u+1)(u^2-u+1)$
$$I={1\over2}\int_{1}^{\infty}\arctan{u}\left({A\over u^{1/2}}+{B\over u+1}+{Cu+D\over u^2-u+1}\right)\mathrm du$$
I am stuck at this point.
Can anyone help to prove $(1)$?
|
We first write $I$ as
$$ I= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \left( \arctan(\cot^2 x) - \arctan(1) \right) \, dx. $$
Now using addition formulas for $\arctan$ and $\cos$, we have
$$ \arctan(\cot^2 x) - \arctan(1)
= \arctan\left(\frac{\cot^2 x - 1}{\cot^2 x + 1} \right)
= \arctan(\cos 2x). $$
Consequently we have
\begin{align*}
I
&= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \arctan(\cos 2x) \, dx \\
&= \frac{\pi^2}{16} + \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \arctan(\sin \theta) \, d\theta,
\end{align*}
where the last line follows from the substitution $\theta = \frac{\pi}{2} - 2x$. The last integral can be computed in terms of the Legendre chi function $\chi_2$:
$$ \int_{0}^{\frac{\pi}{2}} \arctan(\sin \theta) \, d\theta = 2\chi_2(\sqrt{2}-1). \tag{1} $$
For a proof of $\text{(1)}$, see my previous answer for instance. There are only a handful of known special values of $\chi_2$, but thankfully
$$\chi_2(\sqrt{2}-1) = \frac{\pi^2}{16} - \frac{1}{4}\log^2(\sqrt{2}+1) \tag{2} $$
is one of them. Summarizing, we have
$$ I = \frac{\pi^2}{8} - \frac{1}{4}\log^2(\sqrt{2}+1), $$
which coincides with the proposed answer.
Addendum. The identity $\text{(2)}$ follows by plugging $x = \sqrt{2}-1$ to the identity
$$ \chi_2\left(\frac{1-x}{1+x}\right) + \chi_2(x) = \frac{\pi^2}{8} - \frac{1}{2}\log x \log\left(\frac{1-x}{1+x}\right), $$
which can be easily checked by differentiating both sides.
|
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|
Probability with two boxes and balls There are $10$ white and $10$ black balls in the first box, and $8$ white and $10$ black balls in the second box. Two balls are put from the first to the second box, then we choose one ball from the second box. What is the probability that the chosen ball from the second box is white?
From the first box, we can choose two balls in ${20\choose 2}=190$ ways. Possible combinations are $\{BB,BW,WB,WW\}$.
After putting two balls from the first to the second box, there are total $18$ balls in the first, and $20$ in the second.
From the second box, we can choose one ball in ${20\choose 1}=20$ ways.
How can we find the probability that the chosen ball from the second box is white?
|
Case $1$: $2$ white balls are put from first box to second and one ball chosen from the second box is white
probability $=\dfrac{\dbinom{10}{2}}{\dbinom{20}{2}} \times
\dfrac{10}{20}$
Case $2$: $2$ black balls are put from first box to second and one ball chosen from the second box is white
probability $=\dfrac{\dbinom{10}{2}}{\dbinom{20}{2}} \times
\dfrac{8}{20}$
Case $3$: $1$ white ball and $1$ black balls are put from first box to second and one ball chosen from the second box is white
probability $=\dfrac{\left[\dfrac{\dbinom{10}{1}\dbinom{10}{1}}{2!}\right]}{\dbinom{20}{2}} \times \dfrac{9}{20}$
Sum of these three cases will give required probability
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2093792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.