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Conjugacy classes in a matrix group Consider the matrix group $PGL_{2}(\mathbb{F}_{q})$ for $q$ odd. Why is it that $\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}$ has $q(q + 1)/2$ elements in its conjugacy class while $\begin{pmatrix} 2 & 0\\ 0 & 1\end{pmatrix}$ has $q(q + 1)$ elements?
|
The centralizer of $\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}$ in $PGL(2,q)$ is given by all matrices $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$, such that
$$
\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} ka & kb\\ kc & kd\end{pmatrix}
$$
for some constant $k\in\mathbb{F}_q$.
We get that
$$
\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} a & -b\\ -c & d\end{pmatrix}
,
$$ which fits our pattern, if either $b=c=0$ or $a=d=0$. Thus we get $2(q-1)$ total matrices, for a conjugacy class of size $\dfrac{q(q-1)(q+1)}{2(q-1)}=\dfrac{q(q+1)}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
|
Use that $z=-(x+y)$, so you have that the numerator turns out to be $$x^7+y^7-(x+y)^7=-7x^6y-21x^5y^2-35x^4y^3-35x^3y^4-21x^2y^5-7xy^6$$Also, the denominator turns out to be $$xy(-x-y)(x^4+y^4+(-x-y))^4=-2x^6y-6x^5y^2-10x^4y^3-10x^3y^4-6x^2t^5-2xy^6$$You can factor a $7$ from the numerator and a $2$ from the denominator, and the answer turns out to be $$\frac{7}{2}$$
|
{
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"url": "https://math.stackexchange.com/questions/115520",
"timestamp": "2023-03-29T00:00:00",
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|
Find the minimum value.
Find the minimum value of $4^x + 4^{1-x}$ , $x\in\mathbb{R}$.
In this I used the property that $a + \frac{1}{a}\geq 2$.
So I begin with
$$ 4^x + \left(\frac{1}{4}\right)^x + 3\left(\frac{1}{4}\right)^x \geq 2 + 3\left(\frac{1}{4}\right)^x$$
So I think the minimum value be between 2 to 3.
But the answer is 4.
Thanks in advance.
|
Your approach was a good one, except for the lack of "$x$, $-x$" symmetry.
But that is not hard to fix. Since $1/2$ is the midway point between $0$ and $1$, it seems natural to look instead at $4^{x-1/2}+4^{1/2-x}$. This is a close relative of our expression, since
$$4^x+4^{1-x}=4^{1/2}(4^{x-1/2}+4^{1/2-x}).$$
Now it's over. By the result you quoted, the expression $4^{x-1/2}+4^{1/2-x}$ reaches a minimum of $2$ when $x=1/2$. So the minimum value of our original expression is $(4^{1/2})(2)$, which is $4$.
An alternate way of viewing the matter is that we made the symmetrizing change of variable $u=x-1/2$. Then $4^x=4^{1/2}4^u$ and $4^{1-x}=4^{1/2}4^{-u}$. The procedure is quite analogous to what happens when we complete the square in $x^2-x$. There, we can let $u=x-1/2$, and the parabola $y=x^2-x$ becomes the nicer parabola $y=u^2-\frac{1}{4}$, from which the minimum can be read off.
|
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|
Partial Fraction Decomposition of $\frac{x^4+2}{x^5+6x^3}$ I tried answering the following question but I'm getting it wrong for some reason. I would appreciate any help.
$$\frac{x^4+2}{x^5+6x^3}$$
My answer:
$$\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^3}+\frac{Fx+G}{x^2+6}$$
What am I doing wrong?
|
The typical way to deal with $(x+b)^n$ in the denominator is to have the terms
$$ \frac{A_1}{x+b} + \frac{A_2}{(x+b)^2} + \dots + \frac{A_n}{(x+b)^n}$$
Note that $A_2$, $A_3$ etc are constant terms.
In your case, try expressing it in the form
$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+6}$$
|
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|
Eigenvalues of a matrix Let $A$ be a square matrix of order, say, $4$.
Consider the matrix $$B=\left( \begin{array}{ccc}A &I \\ I & A\end{array} \right)$$ where $I$ is the identity matrix of order $4$.
Let $\lambda$ be an eigenvalue of $A$. Then there exists a nonzero vector $\bf{x}=$ $\left( \begin{array}{c} x_1 &x_2 &x_3 &x_4\end{array} \right)^T$ such that $A\bf{x}=\lambda \bf{x}$
Now,
$$\begin{eqnarray*}\left( \begin{array}{ccc}A &I \\ I & A\end{array} \right) \left( \begin{array}{rrrrrrrr}x_1 \\x_2 \\x_3 \\x_4 \\x_1 \\x_2 \\x_3 \\x_4\end{array} \right)&=&\left( \begin{array}{rr}A\bf{x}+I\bf{x} \\ I\bf{x}+A\bf{x}\end{array} \right)\\&=&\left( \begin{array}{rr}\lambda \bf{x}+I\bf{x} \\ I\bf{x}+\lambda \bf{x}\end{array} \right)\\&=& (\lambda+1) \left( \begin{array}{rrrrrrrr}x_1 \\x_2 \\x_3 \\x_4 \\x_1 \\x_2 \\x_3 \\x_4 \end{array} \right)\end{eqnarray*}$$
So $\lambda+1$ is an eigenvalue of $B$. So, in this way we are able to say $4$ eigenvalues of $B$. Can we say anything about the other $4$ eigenvalues of $B$?
Is there anything special about this kind of matrix $B$?
|
Here is a useful observation. For any $k$, let $I_k$ denote the $k \times k$ identity matrix.
Fixing a positive integer $n$, you can check that the $(2n) \times (2n)$ square matrix
$$
C = \frac{1}{\sqrt{2}} \begin{pmatrix} I_n & -I_n \\ I_n & I_n \end{pmatrix}
$$
satisfies $C^T C = CC^T = I_{2n}$ (here $C^T$ denotes the transpose of $C$). In other words, $C^T = C^{-1}$, or: $C$ is a unitary matrix.
Moreover, you can check that that for any square $n \times n$ matrices $A$ and $B$, you have
$$
C^T \begin{pmatrix} A & B \\ B & A \end{pmatrix} C = \begin{pmatrix} A + B & 0 \\ 0 & A - B \end{pmatrix}
$$
(where the $0$s on the extreme right hand side denote the $n \times n$ zero matrices).
This shows that the block matrices $\begin{pmatrix} A & B \\ B & A \end{pmatrix}$ and $\begin{pmatrix} A + B & 0 \\ 0 & A - B \end{pmatrix}$ are similar. Similar matrices have the same characteristic polynomials, and in particular, the same eigenvalues, with the same algebraic multiplicities. (The matrix $C$ implementing the similarity even allows you to pass back and forth between eigenvectors of one matrix and eigenvectors of the other, so even the geometric multiplicities of the eigenvalues coincide.)
It should be clear that the eigenvalue list of the block matrix $\begin{pmatrix} A + B & 0 \\ 0 & A - B \end{pmatrix}$ consists of the eigenvalue list of $A + B$, together with the eigenvalue list of $A - B$. This explains why in the special case $B = I$ the eigenvalues of your matrix are the eigenvalues of $A \pm I$, which are, of course, the eigenvalues of $A$, plus or minus $1$.
For more general $A$ and $B$, of course, the eigenvalues of $A \pm B$ are not obtained simply by adding (or subtracting) eigenvalues of $B$ to (or from) the eigenvalues of $A$. (As an example, if $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then the eigenvalues of $A$ are $0,2$, and $0$ is the only eigenvalue of $B$, but the eigenvalues of $A+B$ are $1 \pm \sqrt{2}$, and $1$ is the only eigenvalue of $A - B$.) But the computation above still shows that the problem of determining the eigenstuff of any $(2n) \times (2n)$ matrix of the form $\begin{pmatrix} A & B \\ B & A \end{pmatrix}$ reduces to determining the eigenstuff of the two $n \times n$ matrices $A + B$ and $A-B$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many times do these two graphs intersect for values x >0? When the curves
$$y = x^2 + 4x -5$$
and
$$y = \frac{1}{1+x^2} $$
are drawn in the $xy$-plane, how many times do the two graphs intersect for values of $x > 0$ ?
I equate the value of $y$, then the equation comes in the fourth power of $x$ . How I can solve this?
Thanks in advance.
|
While polynomials of degree 4 can be solved by radicals, that is not needed here.
The first graph, $y=x^2+4x-5 = (x+5)(x-1)$ is positive if $x\lt -5$ or if $x\gt 1$. It is increasing on $x\gt 1$. The graph of $y=\frac{1}{1+x^2}$ is always positive, and is decreasing on $x\gt 0$.
When we look at the portions of the graphs that are on the first quadrant, the graph of $y=x^2+4x-5$ is going up, the graph of $y=\frac{1}{1+x^2}$ is going down. And $y=x^2+4x-5$ is smaller than $y=\frac{1}{1+x^2}$ when $x=1$, but is larger when $x=2$.
So...
|
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|
Maclaurin expansion $\log\left( \frac{1+x}{1-x}\right)$, show equality of two sums I am supposed to find the Maclaurin expantion of
$ \log\left( \frac{1+x}{1-x} \right) $
So I noticed the obvious that $\log(1+x) - \log(1-x)$
Then Maclaurin polynomial of $\log (1+x)$ equals $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$
So by doing a quick quick change of variables (-x) I obtained the maclaurin expansion of $\log(1-x)$
In total, I obtained that the Maclaurin expansion of $ \log\left( \frac{1+x}{1-x} \right) $ equals.
$ \displaystyle P(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n}$
But in my book the answer says $2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$, I can see that this is correct, by writing out a few terms. But how do I show this by algebra?
My question is therefore, how do I show that
$\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n} = 2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$
?
|
The expansion for $\log (1+x)$ is not $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$, but
$$\begin{equation*}
\log (1+x)=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}%
x^{n+1}=\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n+1}}{n}x^{n+1}
\end{equation*}.$$
Consequently
$$
\begin{eqnarray*}
\log (1-x) &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}\left(
-x\right) ^{n+1}=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left(
-1\right) ^{n+1}}{n+1}x^{n+1} \\
&=&-\sum_{n=0}^{\infty }\frac{1}{n+1}x^{n+1},
\end{eqnarray*}
$$
and
$$
\begin{eqnarray*}
\log \left( \frac{1+x}{1-x}\right) &=&\log (1+x)-\log (1-x) \\
&=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}x^{n+1}+\sum_{n=0}^{
\infty }\frac{1}{n+1}x^{n+1} \\
&=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}+1}{n+1}x^{n+1} \\
&=&\sum_{n=0}^{\infty }\frac{2}{2n+1}x^{2n+1},
\end{eqnarray*}$$
because
$$
\begin{equation*}
\left( -1\right) ^{n}+1=\left\{
\begin{array}{c}
2\quad \text{if }n\text{ even} \\
0\quad \text{if }n\text{ odd}.
\end{array}
\right.
\end{equation*}$$
|
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|
Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$. I have been running into this type of problem a lot:
Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$.
First of all, by Euler's theorem, we have that
$$a^{\phi(42)}\equiv a^{12}\equiv1\pmod{42}.$$
Notice that
$$a^6a^6\equiv1\pmod{42}\text{ and }168=4\cdot42.$$
It follows that
$$a^{12}\equiv1\pmod{168},$$
$$a^{12}-a^6\equiv1-a^6\pmod{168},$$
$$a^6(a^6-1)\equiv1-a^6\pmod{168},$$
$$a^6-1\equiv a^6(1-a^6)\pmod{168},$$
$$a^6-1\equiv a^6-a^{12}\pmod{168}.$$
I get stuck here: How can I show that the RHS is congruent to zero modulo $168$?
|
By Fermat's Theorem, $a^6\equiv 1 \pmod 7$. Also by Fermat's Theorem, or otherwise, $a^2\equiv 1 \pmod 3$. Thus $a^6\equiv 1 \pmod 3$. So far, we have that
$$a^6\equiv 1 \pmod {21}.$$
But $a$ is odd, so $a^2\equiv 1 \pmod 8$. It follows that
$$a^6 \equiv 1 \pmod {8}.$$
Now it's over.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of integer solutions to $3i^2 + 2j^2 = 77 \cdot 6^{2012}$ here is another problem I did not manage to solve in the contest I mentioned in my previous question.
Determine the number of integer solutions $(i, j)$ of the equation:
$3i^2 + 2j^2 = 77 \cdot 6^{2012}$.
Applying logarithms is not useful, since on the left hand side we have a sum; I also tried some algebraic manipulations that led me to nothing useful. Is there a simple solution to the problem?
Thank you,
rubik
|
$A)~ i^2=25x^2 ~\text {and}~ j^2=x^2$
$3 \cdot 25x^2+2x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :
$i= \pm 5x \Rightarrow i = \pm 5 \cdot 6^{1006}$
$j= \pm x \Rightarrow j = \pm 6^{1006}$
$R_A :$
$(i,j) \in \{(-5 \cdot 6^{1006},-6^{1006}),(-5 \cdot 6^{1006},6^{1006}),(5 \cdot 6^{1006},-6^{1006}),(5 \cdot 6^{1006},6^{1006})\}$
$B)~ i^2=9x^2 ~\text {and}~ j^2=25x^2$
$3 \cdot 9x^2+2 \cdot 25x^2=77 \cdot 6^{2012} \Rightarrow x^2=6^{2012}$ , hence :
$i= \pm 3x \Rightarrow i = \pm 3 \cdot 6^{1006}$
$j= \pm 5x \Rightarrow j = \pm 5 \cdot 6^{1006}$
$R_B :$
$(i,j) \in \{(-3 \cdot 6^{1006},-5 \cdot 6^{1006}),(-3 \cdot 6^{1006},5 \cdot 6^{1006}),(3 \cdot 6^{1006},-5 \cdot 6^{1006}),(3 \cdot 6^{1006},5 \cdot 6^{1006})\}$
|
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|
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.
|
This is the same as proving $(10^n +3⋅4^n +5)$ is divisible by 9. The rule for divisibility by $9$ is if it's digital sum is $9$, then it's divisible by $9$
The sum of the three parts should give a digital sum of $9$.
$10^n$ digital sum $=1$
$4^n$ digital sum repeats in the cycle $(4,7,1)$
multiplying these by 3 gives the cycle$(12,21,3)$
The digital sum of $12,21,3$ are each $3$
$3\times4^n$ digital sum $=3$
$5$ is last part
$1 +3 +5 =9$ and is therefore divisible by $9$
|
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|
Minimum of integral Let $S$ be the set of all integrable on $[0,1]$ such that $$\int\limits_0^1f(x)dx=\int\limits_0^1xf(x)dx+1=3.$$
Prove that $S$ is infinite and evaluate $$\min\limits_{f\in S}\int\limits_0^1f^2(x)dx.$$
|
We have $3 = \int_0^1 f(x) \cdot \left(x +\frac{1}{3} \right) \, dx$. Let's use Cauchy–Schwarz inequality:
$$3 = \int_0^1 f(x) \cdot \left(x +\frac{1}{3} \right) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 \left(x +\frac{1}{3} \right)^2 dx} = \sqrt{\frac{7}{9}} \sqrt{ \int_0^1 f^2(x) \, dx}$$
Hence:
$$\int_0^1 f^2(x) \, dx \ge \left( \frac{9 \cdot 3}{\sqrt{7}} \right)^2 = \frac{81}{7} \approx 11.57$$
Edit
$$2 = \int_0^1 x f(x) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 x^2 dx} = \frac{1}{\sqrt{3}} \sqrt{ \int_0^1 f^2(x) \, dx}$$
So:
$$\int_0^1 f^2(x) \, dx \ge 12 $$
And equality holds if $f(x) = 6x$
|
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|
Gaussian proof for the sum of squares? There is a famous proof of the Sum of integers, supposedly put forward by Gauss.
$$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$
$$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$
$$S=\frac{n(1+n)}{2}$$
I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$
I've tried the same approach of adding the summation to itself in reverse, and I've found this:
$$2S=(1^2+n^2)+(2^2+n^2+1^2-2n)+(3^2+n^2+2^2-4n)+\cdots+(n^2+n^2+(n-1)^2-2(n-1)n$$
From which I noted I could extract the original sum;
$$2S-S=(1^2+n^2)+(2^2+n^2-2n)+(3^2+n^2-4n)+\cdots+(n^2+n^2-2(n-1)n-n^2$$
Then if I collect all the $n$ terms;
$$2S-S=n\cdot (n-1)^2 +(1^2)+(2^2-2n)+(3^2-4n)+\cdots+(n^2-2(n-1)n$$
But then I realised I still had the original sum in there, and taking that out mean I no longer had a sum term to extract.
Have I made a mistake here? How can I arrive at the answer of $\dfrac{n (n + 1) (2 n + 1)}{6}$ using a method similar to the one I expound on above? I.e following Gauss' line of reasoning?
|
Since I think the solution Tyler proposes is very useful and accesible, I'll spell it out for you:
We know that
$$(k+1)^3-k^3=3k^2+3k+1$$
If we give the equation values from $1$ to $n$ we get the following:
$$(\color{red}{1}+1)^3-\color{red}{1}^3=3\cdot \color{red}{1}^2+3\cdot \color{red}{1}+1$$
$$(\color{red}{2}+1)^3-\color{red}{2}^3=3 \cdot \color{red}{2}^2+3 \cdot \color{red}{2}+1$$
$$(\color{red}{3}+1)^3-\color{red}{3}^3=3 \cdot \color{red}{3}^2+3 \cdot \color{red}{3}+1$$
$$\cdots=\cdots$$
$$(\color{red}{n-1}+1)^3-(\color{red}{n-1})^3=3(\color{red}{n-1})^2+3(\color{red}{n-1})+1$$
$$(\color{red}{n}+1)^3-\color{red}{n}^3=3\color{red}{n}^2+3\color{red}{n}+1$$
We sum this orderly in columns.
Note that in the LHS the numbers cancel out with each other, except for the $(n+1)^3$ and the starting $-1$ ($2^3-1^3+3^3-2^3+4^3-3^3+\cdots+n^3-(n-1)^3+(n+1)^3-n^3$). We get:
$$(n+1)^3-1 = 3(1+2^2+3^2+\cdots +(n-1)^2+n^2)+ 3(1+2+3+\cdots +(n-1)+n)+(\underbrace{1+1+\cdots+1}_{n})$$
We can write this in sigma notation as:
$$(n+1)^3-1=\sum\limits_{k=1}^n(3k^2+3k+1)$$
Naming our sum $S$ we have that:
$$(n+1)^3-1=3S+\sum\limits_{k=1}^n(3k+1)$$
We know how to compute the sum in the RHS, because
$$\sum\limits_{k=1}^n 3k =3\frac{n(n+1)}{2}$$
$$\sum\limits_{k=1}^n 1 =n$$
(We're summing $n$ ones in the last sum.)
$$(n+1)^3-1=3S+3 \frac{n(n+1)}{2}+n$$
$$n^3+3n^2+3n=3S+\frac{3}{2}n^2+\frac{3}{2}n+n$$
$$n^3+\frac{3n^2}{2}+\frac{n}{2}=3S$$
$$\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=S$$
This factors to
$$\frac{n(2n+1)(n+1)}{6}=S$$
which is what you wanted.
|
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Prove for any positive real numbers $a,b,c$ $\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$ Since the problem sheets says I should use Cauchy-Schwarz inequality, I used
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3}$
$\geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$
I first multiplied each term by $a,b,c$ to get a perfect square on top like
$\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$
$=\frac{a^4}{a(a^2+ab+b^2)}+\frac{b^4}{b(b^2+bc+c^2)}+\frac{c^4}{c(c^2+ca+a^2)}$
$\hspace{120pt}\geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} $
But I am still stuck for few hours. This is not a homework, but is a set of problems to prepare for AMC/USAMO.
(Note: I started off with a general question and got closed as off topic. I have another genuine Math Question now, and I am just trying to see if math.se is going to be useful)
|
We need to prove that
$$\sum_{cyc}\left(\frac{a^3}{a^2+ab+b^2}-\frac{a}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{2a^3-a^2b-ab^2}{a^2+ab+b^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(a-b)(2a+b)}{a^2+ab+b^2}-(a-b)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(a+b)}{a^2+ab+b^2}\geq0.$$
Done!
|
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|
Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
How do you find the sum of these and prove it by induction? Can someone help me get through this?
|
If you don't have to use induction, you can do this:
1)
$$
\sum_k k\binom{n}{k} = \sum_k k \frac{n(n-1)\cdots(n-k+1)}{k!} = n\sum_k \binom{n-1}{k-1} = n \sum_k \binom{n-1}{k} = n2^{n-1}
$$
Here $k$ runs over all integers and by convention $\binom{n}{k}$ is defined as zero if $k<0$ or $k > n$. The last equality follows from the fact that $\binom{n-1}{k}$ counts the $k$-element subsets of $\{1,\ldots,n-1\}$, so $\sum_k \binom{n-1}{k}$ counts the number of all subsets which is $2^{n-1}$.
2) Note that $\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}$ so the sum is a telescope sum:
$$\sum_{k=1}^{n-1} \frac{1}{k(k+1)} = \sum_{k=1}^{n-1}\left( \frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
can't figure out what I been doing wrong on simple integration question $$\int_0^1 \sqrt{(\sqrt{5})^2+(2t)^2}\;dt$$
Based on the formula
$\int \sqrt{a^2+x^2}\;dx=\frac{1}{2}[x\sqrt{a^2+x^2}+a^2\log(x+\sqrt{a^2+x^2})]$
I just plug in above input into the formula above
However I can only find $3+\frac{5}{2}\log(5)$
but answers that I get from Mathematica is $\frac{3}{2}+\frac{5}{8}\log(5)$
i been trying to figuring out what I been doing wrong for days but I still can't find out what I been doing wrong.
Appreciate if someone can show what I'm been doing wrong
|
we have
\begin{align*}
\int_0^1 \sqrt{5 + (2t)^2}\, dt &= \frac 12 \int_0^2 \sqrt{5 + x^2}\, dx\\
&= \frac 14\left[x\sqrt{5 + x^2} + 5\log\bigl(x + \sqrt{5 + x^2}\bigr)\right]_0^2\\
&= \frac 14\left(6 + 5\log 5 - 0 - 5 \log \sqrt 5\right)\\
&= \frac 32 + \frac 58 \, \log 5.
\end{align*}
I don't know what you did wrong ;-), because I don't know what you did at all,
|
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|
How to Derive a Double Angle Identity. How does one derive the following two identities:
$$\begin{align*}
\cos 2\theta &= 1-2\sin^2\theta\\
\sin 2\theta &= 2\sin\theta\cos\theta
\end{align*}$$
|
Hints:
For the $\cos 2\theta$ formula, use the sum identity (with $x=y=\theta$)
$$
\cos(x+y)=\cos x\cos y - \sin x\sin y,
$$
followed by the Pythagorean identity $\cos^2 x=1-\sin^2 x $.
For the $\sin 2\theta$ formula , use the sum identity (with $x=y=\theta$)
$$
\sin(x+y)=\sin x\cos y +\sin y\cos x.
$$
Or, for the $\cos2\theta$ formula and $0<\theta<\pi/2$, consider the diagram:
We have
$$
\cos\theta ={ {1+\cos2\theta}\over\sqrt{2+2\cos 2\theta}};
$$
whence
$$
\sqrt 2\cos\theta=\sqrt{1+1\cos\theta},
$$
or,
$$
2\cos^2\theta= 1+1\cos2\theta
$$
From this, we have
$$
2-2\sin^2\theta=1+\cos2\theta,
$$
or
$$
\cos2\theta =1-2\sin^2\theta.
$$
(Having this in hand, we could also use the diagram to derive the formula for $\sin2\theta$.)
|
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|
Real VS Complex for integrals: $\int_0^\infty \frac{dx}{1 + x^3}$ The integral
$$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt2}$$
can be evaluated both by a complex method (residues) and by a
real method (partial fraction decomposition).
The complex method works also for the integral
$$\int_0^\infty \frac{dx}{1 + x^3} = \frac{2\pi}{3\sqrt3}$$
but partial fraction decomposition does not give convergent integrals. I would like to know if there is some real method for evaluating this last integral.
|
Note that for $a > 0$, $$\int_0^N \frac{1}{x+a}\ dx = \ln(N+a) - \ln(a) = \ln(N) - \ln(a) + o(1)\ \text{as} \ N \to \infty$$ while
$$\eqalign{\int_0^N \frac{x+a}{(x+a)^2 + b^2}\ dx &= \frac{1}{2} \left(\ln((N+a)^2+b^2) - \ln(a^2+b^2)\right)\cr &= \ln(N) - \ln(a^2+b^2) + o(1) \ \text{as} \ N \to \infty\cr}$$
and (if $b > 0$)
$$ \eqalign{\int_0^N \frac{1}{(x+a)^2+b^2}\ dx = \frac{\arctan\left(\frac{N+a}{b}\right) - \arctan\left(\frac{a}{b}\right)}{b} = \frac{\pi}{2b} - \frac{\arctan\left(\frac{a}{b}\right)}{b} + o(1)
\ \text{as} \ N \to \infty\cr}$$
In particular, from the partial fraction decomposition
$$ \frac{1}{1+x^3} = \frac{1/3}{x+1} + \frac{(2-x)/3}{x^2 - x + 1} = \frac{1/3}{x+1} +
\frac{1/2}{(x-1/2)^2+3/4} - \frac{(x-1/2)/3}{(x-1/2)^2 + 3/4}$$
you get
$$ \int_0^N \frac{1}{1+x^3} \ dx = \frac{\ln(N) - \ln(1))}{3} + \frac{\pi/2 + \arctan(1/\sqrt{3})}{\sqrt{3}} - \frac{\ln(N) - \ln((1/2)^2 + 3/4)}{3} + o(1)$$
i.e.
$$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$
|
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|
Graphing simple fractional function I am trying to graph
$$f(x) = \frac{x^2-4}{x^2+4}$$
It seems pretty simple to me but I can't finish it correctly.
I know that there is a horizontal asymptote at $1$ beacuse the degrees on the variable at the same and $\frac{x}{x}$ is 1.
I know that it is a negative function until zero, and then positive because the only critical number of the derivative
$$\frac{16x}{(x^2+4)^2}$$
is going to be zero since the denominator can't be zero and the only root of the top is $0$.
This then tells me that there is no local max but only a minimum which is at $0$ which gives me $-1$.
Trying to find concavity
$$\frac{16(x^2+4)^2 - 16x (4x(x^2+4))}{(x^2+4)^4}=
\frac {-64x^5 + 16x^4 + 256x^3 + 128x^2 + 256}{(x^2+4)^4}$$
Which I have no idea how to really work with I can't see to get anything workable out of that.
|
Correct second derivative:
$$\begin{align*}
f(x) &= \frac{x^2-4}{x^2+4}\\
f'(x) &= \frac{(x^2+4)(x^2-4)' - (x^2-4)(x^2+4)'}{(x^2+4)^2}\\
&= \frac{(x^2+4)(2x) - (x^2-4)(2x)}{(x^2+4)^2}\\
&= \frac{2x(x^2+4-x^2+4)}{(x^2+4)^2}\\
&= \frac{16x}{(x^2+4)^2}.\\
f''(x) &= \frac{(x^2+4)^2(16x)' - 16x\Bigl( (x^2+4)^2\Bigr)'}{\Bigl((x^2+4)^2\Bigr)^2}\\
&= \frac{16(x^2+4)^2 - 16x(2(x^2+4)(x^2+4)')}{(x^2+4)^4}\\
&= \frac{16(x^2+4)^2 - 16x(2)(x^2+4)(2x)}{(x^2+4)^4}\\
&= \frac{16(x^2+4)\Bigl( x^2+4 - 4x^2\Bigr)}{(x^2+4)^4}\\
&= \frac{16(4-3x^2)}{(x^2+4)^3}.
\end{align*}$$
The denominator is never zero; the numerator is zero exactly when $4-3x^2=0$. This is a parabola that opens down, so it will be positive between the roots and negative before and after both roots.
|
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|
Maximize volume of box in ellipsoid I need to find the dimensions of the box with maximum volume (with faces parallel to the coordinate planes) that can be inscribed in ellipsoid
$$\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$$
A hint given was:
If vertex of box in first octants is (x,y,z) then volume is 8xyz.
So I first find the partial derivatives
$V_x = 8yz$
$V_y = 8xz$
$V_z = 8xy$
Now the partial derivatives for constraint:
$g_x = \lambda\frac{x}{2}$
$g_y = \lambda\frac{2y}{9}$
$g_z = \lambda\frac{z}{8}$
Then $f_\alpha = \lambda g_\alpha$ so:
$8yz = \lambda\frac{x}{2} \rightarrow \lambda = \frac{16yz}{x}$
$8xz = \lambda\frac{2y}{9} \rightarrow \lambda = \frac{36xz}{y}$
$8xy = \lambda\frac{z}{8} \rightarrow \lambda = \frac{64xy}{z}$
$\frac{16yz}{x} = \frac{36xz}{y} \rightarrow y = \frac{3x}{2}$
$\frac{36xz}{y} = \frac{64xy}{z} \rightarrow y = \frac{3z}{4}$
$\frac{3x}{2} = \frac{3z}{4} \rightarrow x = \frac{z}{2}$
Now how do I continue? I seems to be missing something?
|
You can cheat here, because the property of being a maximum-volume axis-parallel box is preserved by "stretching" transformations of the form $(x,y,z) \mapsto (ax,by,cz)$ (because such a transformation preserves the property of being axis-parallel, and multiplies all volumes by the constant $abc$).
So start with a sphere of radius 1: then the biggest box has corners
$$\left(\pm\sqrt \frac{1}{3},\pm\sqrt \frac{1}{3},\pm\sqrt \frac{1}{3}\right)$$
and a volume of $\left(2\sqrt \frac{1}{3}\right)^3 = \frac{8}{3}\sqrt \frac{1}{3}$ (see the comments for more about this). Now stretch it by a factor of $2$ along the $x$-axis, $3$ along the $y$-axis, and $4$ along the $z$-axis. The corners go to
$$\left(\pm2\sqrt \frac{1}{3},\pm3\sqrt \frac{1}{3},\pm4\sqrt \frac{1}{3}\right)$$
and the volume is $64\sqrt \frac{1}{3}$.
|
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|
How to compute $\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$? $$\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$$
So ... how do I start? Numerator cant be factorized it seems, and this looks like a complicated expression ...
I tried expanding the denominator to see if integration by substitution will work, but it didn't give any ideas
$$\int{\frac{5x^3+8x^2+x+2}{2x^4+x^2}} dx$$
I don't see a clear way to integrate by parts either. Can anyone enlighten me?
|
If you wish to avoid partial fractions, it is possible to clean up that denominator with a simple substitution. To make that $2x^2+1$ factor more manageable, make the substitution
$x=\frac{\sqrt2}2\tan\theta,dx=\frac{\sqrt2}2\sec^2\theta d\theta$
$\int\dfrac{5x^3+8x^2+x+2}{x^2(2x^2+1)}dx=\int\dfrac{\frac{5\sqrt2}4\tan^3\theta+4\tan^2\theta+\frac{\sqrt2}2\tan\theta+2}{\frac12\tan^2\theta(\tan^2\theta+1)}\times\frac{\sqrt2}2\sec^2\theta d\theta$
$\dfrac{\frac{\sqrt2}2\sec^2\theta}{\frac12(\tan^2+1)}$ reduces to $\sqrt2$. Then dividing through gets
$\int\frac52\tan\theta+4\sqrt2+\cot\theta+2\sqrt2\cot^2\theta d\theta$
Can you take it from there?
|
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|
$3^k$ not congruent to $-1 \pmod {2^e}, e > 2$. $3^k \not\equiv -1 \pmod {2^e}$ for $e > 2, k > 0$. Is this true? I have tried to prove it by expanding $(1 + 2)^k$. [Notation: $(n; m) := n! / (m! (n - m)!)$] E.g., for $e = 3$ I get: $(1+2)^k + 1 = 2 + (k; 1) 2 + (k; 2) 2^2 + (k; e) 2^e + ...$ So, here it's enough to prove that $2^3$ does not divide $2 + (k; 1) 2 + (k; 2) 2^2$. The validity for general e seems very hard to prove.
|
Based on pedja,we only prove that $8 \nmid 3^k+1$,
1.when $k=2m+1$,$3^k+1=3^{2m+1}+1=9^m \times 3+1 \equiv 4 \pmod{8} $
2.when $k=2m$,$3^k+1=3^{2m}+1=9^m+1 \equiv 2 \pmod{8} $
so we have $8 \nmid 3^k+1$,this is
$3^k \not\equiv -1 \pmod {2^e}$
|
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|
Differentiation of $y = \tan^{-1} \Bigl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\Bigr\}$ How do i differentiate the following: $$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}$$
I know that $\text{derivative}$ of $\tan^{-1}{x}$ is $\frac{1}{1+x^{2}}$ but not sure as to how to do this.
|
Putting $x^2=\cos2z,1+x^2=1+\cos2z=2\cos^2z,1-x^2=1-\cos2z=2\sin^2z$
$$\frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}=\frac{\cos z-\sin z}{\cos z+\sin z}=\frac{1-\tan z}{1+\tan z}=\tan\left(\frac\pi4-z\right)$$
$$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}=\tan^{-1}\tan\left(\frac\pi4-z\right)=\frac\pi4-z=\frac\pi4-\frac12\cos^{-1}x^2$$
$$\implies \frac{dy}{dx}=\frac{\frac\pi4-\frac12\cos^{-1}x^2}{dx}=-\frac12\left(-\frac{2x}{\sqrt{1-(x^2)^2}}\right)=\frac x{1-x^4}$$
|
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|
How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
|
Hint $\ $ The LHS is a difference of squares $\rm\:y^2\!-\!1,\:$ hence so too is $\rm\:(y^2\!-\!1)-3\: =\: y^2\!-\!2^2,\:$ viz.
$\rm\qquad\ \:\! (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\ =\ (x^2\!-\!5x+4)(x^2\!-\!5x+6)\ =\ (x^2\!-\!5x+5)^2 \!-\! 1^2 $
$\rm\ \ \Rightarrow\ (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\!-\!3\ =\ (x^2\!-\!5x+5)^2 \!-\! 2^2\ =\ (x^2\!-\!5x+3)(x^2\!-\!5x+7)$
|
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|
Another way to go about proving Binet's Formula As I showed in another question of mine, it is easy to prove that
$$\tag{1}\phi^{n+1} =F_{n+1} \phi+F_{n }$$
given $F_1=1$ , $F_2=1$ , $F_{n+1}=F_n+F_{n-1}\text{ ; }n\geq2$.
Now, extending $(1)$ to negative and zero indices naturally yields:
$$\eqalign{
& {F_{ 0}} = 0 \cr
& {F_{ - 1}} = 1 \cr
& {F_{ - 2}} = - 1 \cr
& {F_{ - 3}} = 2 \cr
& {F_{ - 4}} = - 3 \cr
& {F_{ - n}} = {\left( { - 1} \right)^{n+1}}{F_{n}} \cr} $$
From this one has
$${\phi ^{ - n}} = {\left( { - 1} \right)^{n + 1}}\left( {{F_{n }}\phi - {F_{n + 1}}} \right)$$
A few examples:
$$\eqalign{
& {\phi ^{ - 1}} = \phi - 1 \cr
& {\phi ^{ - 2}} = 2 - \phi \cr
& {\phi ^{ - 3}} = 2\phi - 3 \cr} $$
Then one has that
$$\eqalign{
& {\left( { - \frac{1}{\phi }} \right)^n} = {F_{n + 1}} - {F_n}\phi \cr
& {\phi ^n} = {F_n}\phi + {F_{n - 1}} \cr} $$
and consequently
$$\eqalign{
& {\phi ^n} + {\left( { - \frac{1}{\phi }} \right)^n} = {F_n}\phi + {F_{n - 1}} + {F_{n + 1}} - {F_n}\phi \cr
& {\phi ^n} + {\left( { - \frac{1}{\phi }} \right)^n} = {F_{n + 1}} + {F_{n - 1}} \cr
& {\phi ^n} + {\left( { - \frac{1}{\phi }} \right)^n} = {L_n} \cr} $$
How can this be put into an acceptable proof? (Note that given the relation between Lucas and Fibonacci numers, this straightforwardly gives Binet's Formula:
$${F_n} = \frac{1}{{\sqrt 5 }}\left[ {{\phi ^n} - {{\left( { - \frac{1}{\phi }} \right)}^n}} \right]$$
|
The set of sequences $x_n$ that have the property that $x_{n+2} = x_{n+1} + x_n,$ with, say, real number values, makes a vector space over the reals of dimension 2. Taking the two roots of $\lambda^2 = \lambda + 1,$ the larger being
$$ \phi = \frac{1 + \sqrt 5}{2} $$ and the smaller being $-1 / \phi,$ any such sequence, including index shifts, whatever you like, is a linear combination of two basis sequences,
$$ \ldots, 1/\phi, 1, \phi, \phi^2, \ldots $$ and
$$ \ldots, - \phi, 1, -1 / \phi, 1/ \phi^2, \ldots $$
where I am demanding that the element with value 1 have index 0 in both sequences. If I have any favorite sequence with $x_0, x_1$ specified, the equation to be solved for coefficients $A,B$ are
$$ A + B = x_0, \; \; \; A \phi - B / \phi = x_1. $$
So
$$
\left( \begin{array}{cc}
1 & 1 \\
\phi & -1/\phi
\end{array}
\right)
\left( \begin{array}{c}
A \\
B
\end{array}
\right) \; \; = \; \;
\left( \begin{array}{c}
x_0 \\
x_1
\end{array}
\right).
$$
still composing...
$$
\left( \begin{array}{c}
A \\
B
\end{array}
\right) \; \; = \; \; \frac{-1}{\sqrt 5}
\left( \begin{array}{cc}
-1/\phi & -1 \\
-\phi & 1
\end{array}
\right)
\left( \begin{array}{c}
x_0 \\
x_1
\end{array}
\right).
$$
getting there...
Evidently you are taking $x_0 = 0, x_1 = 1,$ so we get
$$
\left( \begin{array}{c}
A \\
B
\end{array}
\right) \; \; = \; \; \frac{-1}{\sqrt 5}
\left( \begin{array}{cc}
-1/\phi & -1 \\
-\phi & 1
\end{array}
\right)
\left( \begin{array}{c}
0 \\
1
\end{array}
\right) \; \; = \; \;
\left( \begin{array}{c}
1 / \sqrt 5 \\
-1/ \sqrt 5
\end{array}
\right)
.
$$
My first basis sequence has elements labelled $y_n = \phi^n,$ and the second has $z_n = (-1/\phi)^n,$ so the Fibonacci sequence obeys $$ x_n = \frac{1}{\sqrt 5} \; \; \phi^n \; \; \; - \; \; \; \frac{1}{\sqrt 5} \; \; \left( \frac{-1}{\phi} \right)^n. $$
|
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|
Complex Series question Would you please give hint for these two(not homework)~~ $$\Pi_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$$ and $$\Pi_{1}^{\infty}(1+\frac{}{})e^{\frac{-z}{n}} \text{ converges absolutely and uniformly on every compact set }$$
|
For $(1)$, note that
$$1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{n-1}{n}\frac{n+1}{n}.$$
Now write down the first few terms of the product, in factored form. (So each term of the original product is represented as the product of two terms.) We get
$$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\frac{8}{7}\cdots.$$
Note that second and third term cancel, as do fourth and fifth, and so on. Almost everything dies, and we are left with $\frac{1}{2}$.
This does not quite complete things, we should be more formal about the limiting process. But we can use the cancellation to produce an explicit formula for
$$\prod_{k=2}^n \left(1-\frac{1}{k^2}\right).$$
Then it is easy to show that for large $n$ the above product is not far from $\frac{1}{2}$.
|
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|
More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$ So the question I am trying to work through is:
Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$
for convergence.
The solution (using D'Alembert's ratio test) is:
$$u_n=\frac{n^3}{3^n}\;,$$
so
$$\begin{align*}
\frac{|u_{n+1}|}{|u_n|} &=\frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3}\;.
\end{align*}$$
How do we get from there to...
$$=\frac{n^3+3n^2+3n+1}{3n^3}$$
What happens with $3^n$ in the numerator and power of $n+1$ in the denominator? How do they cancel out?
Also, in the very next step that all goes to being equal to
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=\frac{1}{3}<1\;,$$
which means the series is convergent.
But how do we get to $\dfrac{1}{3}$?
|
What happens with the $3^n$ in the numerator and the $3^{n+1}$ in the denominator?
Recall the following laws of exponents:
$$a^{b}a^c = a^{b+c}, \quad \text{ and } \quad \frac{a^b}{a^c} = a^{b-c} = \frac{1}{a^{c-b}},$$
for any $a>0$, and any real numbers $b,c$. In particular, if $a=3$, $b=n$, and $c=n+1$, then:
$$\frac{3^n}{3^{n+1}}=\frac{1}{3^{(n+1)-n}} = \frac{1}{3^1}=\frac{1}{3}.$$
|
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|
Determine convergence of $\sum_{n=1}^{\infty} (\cos{\frac{2}{n}}-\cos{\frac{4}{n}})$ Determine convergence of $$\sum_{n=1}^{\infty} \left(\cos{\frac{2}{n}}-\cos{\frac{4}{n}}\right)$$
In the answer, it says
$$\cos{\frac{2}{n}}-\cos{\frac{4}{n}} = 2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 2\cdot \frac{3}{n} \cdot \frac{1}{n} = \frac{6}{n^2}$$
But how do I get the above trig substitution? I guess removing the fractions, I will get $\cos{x}-\cos{2x}=2\sin{(2x-1)}\sin{(x-1)}$ ... probably this is wrong, but how do I get that?
|
I suggest another way:
$\cos\left(\frac{2}{n}\right)-\cos\left(\frac{4}{n}\right)=-\left(1-\cos\left(\frac{2}{n}\right)\right)+1-\cos\left(\frac{4}{n}\right)\sim -\frac{4}{2n^2}+\frac{16}{2n^2}=\frac{6}{n^2}$ and $\sum \frac{6}{n^2}$ converges.
|
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|
Polynomial regression interpolation?
Possible Duplicate:
Writing a function $f$ when $x$ and $f(x)$ are known
I'm not versed in mathematics, so you'll have to speak slowly...
If I want to fit a curve to the points,
X Y
1 0.5
2 5.0
3 0.5
4 2.5
5 5.0
6 0.5
Where would I begin? For my purposes, this needs to be a sixth-order fit...
|
Here is a step by step way of finding a polynomial $f(x)$ such that for $1 \leq i \leq 6$, $f(i)$ has the values you stated. We will "build up" $f(x)$ slowly, calling the intermediate stages $f^{(1)}(x), f^{(2)}(x), \ldots $ and so on.
*
*Set $y = f^{(1)}(x) = 0.5$ where the right side doesn't really depend on $x$ and the $0.5$ was chosen from the given data: when we set $x=1$ in $f^{(1)}(x)$, this (zero-degree) polynomial evaluates to $0.5$, and thus fits the point $(1,0.5)$ that was given to you.
*If we set $x=2$ in $f^{(1)}(x)$, we will get $0.5$ but we really need
to get $5.0$. So, let us test $$y = f^{(2)}(x) = f^{(1)}(x) + c(x-1)
= 0.5 + c(x-1)$$
where $c$ is just some number (for now) to see what happens. If we set $x=1$
in $f^{(2)}(x)$, we get $0.5$ as before (yay!) but
if we set $x=2$, we get $0.5 + c$ and so by choosing $c = 4.5$, we can get
$f^{(2)}(x)$ to give the desired values for both $x = 1$ and $x = 2$.
$f^{(2)}(x) = 0.5 + 4.5(x-1)$ is a polynomial such that $f^{(2)}(1) = 0.5$ and
$f^{(2)}(2) = 5.0$.\
Notice that $4.5 = 5.0 - f^{(1)}(2)$ which I will write for future reference
as $\displaystyle\frac{5.0 - f^{(1)}(2)}{2-1}$.
*$f^{(2)}(3) = 0.5 + 4.5(3-1) = 9.5$ which is not what we want. So,
let us try
$$y = f^{(3)}(x) = f^{(2)}(x) + c(x-1)(x-2)$$
where, as before, we will choose the value of $c$ in just a bit. Note
that if we set $x$ to either $1$ or $2$ in $f^{(3)}(x)$, that last term
evaluates to $0$ and so $f^{(3)}(1) = f^{(2)}(1) = 0.5$,
$f^{(3)}(2) = f^{(2)}(2) = 5.0$ and so we are OK there. But,
$$f^{(3)}(3) = f^{(2)}(3) + c(2)(1) = 9.5 + 2c$$
and so choosing $c = -4.5$ will make $f^{(3)}(3)$ evaluate to
the desired $0.5$.
$f^{(3)}(x) = 0.5 + 4.5(x-1) - 4.5(x-1)(x-2)$ is a polynomial such that
$f^{(3)}(1) = 0.5$, $f^{(3)}(2) = 5.0$, and $f^{(3)}(3) = 0.5$.
Note that
$-4.5 = \displaystyle\frac{0.5 - f^{(2)}(3)}{(3-1)(3-2)}$.
Since I am getting tired of typing, let me briefly explain how the
rest of the thing will work. We set
$$f^{(4)}(x) = f^{(3)}(x)
+ \frac{2.5 - f^{(3)}(4)}{(4-1)(4-2)(4-3)}(x-1)(x-2)(x-3)$$
where the second term vanishes when we set $x = 1, 2, 3$ and so
we get the desired values of $y$, and when we set $x = 4$,
we get
$$f^{(4)}(4) = f^{(3)}(4) + \frac{2.5 - f^{(3)}(4)}{(4-1)(4-2)(4-3)}(4-1)(4-2)(4-3) = 2.5$$
which is exactly what we want. Lather, rinse, and repeat for
$f^{(5)}(x)$ and $f^{(6)}(x)$.
If you feel inclined to learn more about this idea, look for
Newtonian interpolation on the Internet.
|
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|
How to find limit of function: $\lim\limits_{n \to \infty} \frac{\sqrt{n}}{2} \arccos(\frac{n-2}{22+n})$ How would I find this limit?
$\lim_{n \to \infty} \frac{\sqrt{n}}{2} \bigl(\arccos(\frac{n-2}{22+n}))$
|
As the argument of $\arccos$ approaches 1 for large $n$, the inverse cosine function approaches zero. Rewriting the expression as quotient of two expression approaching zero for
large $n$
$$
\frac{ \arccos\left( \frac{1-2/n}{1+22/n} \right)}{2 n^{-1/2}}
$$
we can apply the L'Hospital's rule:
$$ \begin{eqnarray}
\lim_{n \to \infty} \frac{ \arccos\left( \frac{1-2/n}{1+22/n} \right)}{2 n^{-1/2}} &=&
\lim_{n \to \infty} \left(-n^{3/2} \right) \left(-\frac{2 \sqrt{3}}{\sqrt{n+10} (n+22)} \right) \\ &=& 2 \sqrt{3} \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{10}{n}} \left( 1 + \frac{22}{n} \right)} = 2 \sqrt{3}
\end{eqnarray}$$
|
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|
Minimal polynomial of the root of algebraic number I have obtained the minimal polynomial of $9-4\sqrt{2}$ over $\mathbb{Q}$ by algebraic operations:
$$ (x-9)^2-32 = x^2-18x+49.$$
I wonder how to calculate the minimal polynomial of $\sqrt{9-4\sqrt{2}}$ with the help of this sub-result? Or is there a smarter way to do this (not necessarily algorithmic)?
|
$\rm x^4-18\,x^2+49\:$ will be the minimal polynomial, unless $\rm\:\sqrt{9-4\sqrt{2}}\:$ denests to $\rm\:a + b\sqrt{2}.\:$
This can be tested by a radical denesting formula that I discovered as a teenager.
Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $
and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$
Here $\:9-4\sqrt{2}\:$ has norm $= 49.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 7\ $ yields $\ 2-4\sqrt{2}\:$
and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ so,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1-2\sqrt{2}.$
Checking we have
$\ \smash[t]{\displaystyle \left(1-2\sqrt{2}\right)^2 =\ 1+8 -4\sqrt{-3}\ =\ 9-4 \sqrt{2}}.$
See this answer for general radical denesting algorithms.
|
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|
Calculating the shortest possible distance between points Question:
Given the points $A(3,3)$, $B(0,1)$ and $C(x,0)$ where $0 < x < 3$, $AC$ is the distance between $A$ and $C$ and $BC$ is the distance between $B$ and $C$. What is x for the distance $AC + BC$ to be minimal?
What have I done?
I defined the function $AC + BC$ as:
$\mathrm{f}\left( x\right) =\sqrt{{1}^{2}+{x}^{2}}+\sqrt{{3}^{2}+{\left( 3-x\right) }^{2}}$
And the first derivative:
$\mathrm{f'}\left( x\right) =\frac{x}{\sqrt{{x}^{2}+1}}+\frac{3-x}{\sqrt{{x}^{2}-6\,x+18}}$
We need to find the values for $\mathrm{f'}\left( x\right) = 0$, so by summing and multiplying both sides I got to the equation:
$2x^4 - 12x^3 + 19x^2 -6x+18 = 0$
But I don't think the purpose should be to solve a 4th grade equation, there should be another way I'm missing..
|
The following little trick (method?) is useful in many places. Let $y=3-x$. We want to minimize $\sqrt{1+x^2}+\sqrt{3^2+y^2}$, where $x+y=3$.
Differentiate with respect to $x$. Using the fact that $\frac{dy}{dx}=-1$, we arrive at the equation
$$\frac{x}{\sqrt{1+x^2}}=\frac{y}{\sqrt{9+y^2}}.$$
Note that this is (apart from a little minus sign problem) the equation you arrived at, with $3-x$ replaced by $y$.
Square both sides, cross-multiply. We get
$$x^2(9+y^2)=y^2(1+x^2),$$
which simplifies to $9x^2=y^2$.
No fourth degree equation here! Since $x$ and $y$ are non-negative, we get $y=3x$, a linear equation. Now from $x+y=3$ we obtain $x=3/4$.
|
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|
Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$
Now I am not sure what to do.
|
But $$\frac{5x+1}{(2x+1)(x-1)}\ne\frac{5x+1}{2x+1}+\frac{5x+1}{x-1}\;,$$ as you’ll see if you combine the fractions on the righthand side over a common denominator: you get
$$\begin{align*}
\frac{5x+1}{2x+1}+\frac{5x+1}{x-1}&=\frac{5x+1}{2x+1}\cdot\frac{x-1}{x-1}+\frac{5x+1}{x-1}\cdot\frac{2x+1}{2x+1}\\\\
&=\frac{(5x+1)(x-1)+(5x+1)(2x+1)}{(2x+1)(x-1)}\\\\
&=\frac{(5x+1)\big((x-1)+(2x+1)\big)}{(2x+1)(x-1)}\\\\
&=\frac{(5x+1)(3x)}{(2x+1)(x-1)}\;,
\end{align*}$$
clearly not the same as $$\frac{5x+1}{(2x+1)(x-1)}\;.\tag{1}$$ You have to do a bit of algebra to split into partial fractions. Set it up in the usual way:
$$\frac{5x+1}{(2x+1)(x-1)}=\frac{A}{2x+1}+\frac{B}{x-1}=\frac{A(x-1)+B(2x+1)}{(2x+1)(x-1)}\tag{2}\;.$$
We want to choose $A$ and $B$ so that the fractions on the two ends of $(2)$ really are equal; they have the same denominator, so they must have the same numerator, and therefore $$A(x-1)+B(2x+1)=5x+1$$ or, after multiplying out and collecting terms on the lefthand side, $$(A+2B)x+(-A+B)=5x+1\;.\tag{3}$$ The only way that the two sides of $(3)$ can be the same polynomial is to have $A+2B=5$ and $-A+B=1$. Solve this little system for $A$ and $B$, which turn out to be nice numbers, and plug those values back into the middle expression in $(2)$. You’ll end up with the integration $$\int\frac{A}{2x+1}dx+\int\frac{B}{x-1}dx$$ (with specific numbers for $A$ and $B$), and this is pretty straightforward.
|
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|
Arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$ Arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$ over $1 \leq x \leq 2$
I know that the first thing I need to do is take the derivative.
$$y' = x^2 - 4x^{-2}$$
Then I take the integral on that range using the arc length formula.
$$\int_1^2 \sqrt{1 + (x^2-4x^{-2})^2}$$
$$(x^2-4x^{-2})^2 = -16x^{-4} - 8 + x^4$$
$$\int_1^2 \sqrt{-16x^{-4} - 7 + x^4 }$$
From here I have no idea how to factor this but I am pretty sure I must have messed up something before that.
|
Your derivative is wrong, and your squaring is wrong.
*
*Your derivative is wrong:
$$\left(\frac{x^3}{3} +\frac{1}{4x}\right)' = \left(\frac{1}{3}x^3 + \frac{1}{4}x^{-1}\right)' = x^2 - \frac{1}{4}x^{-2} = x^2 - \frac{x^{-2}}{4}.$$
*You squared incorrectly: if you square $x^2-4x^{-2}$, you get:
$$(x^2-4x^{-2})^2 = x^4 - 8x^2x^{-2} + 16x^{-4} = x^4 - 8+16x^{-4}.$$
Note the plus sign on $16x^{-4}$; you have a minus sign.
If you square the correct function, you get
$$\left( x^2 - \frac{1}{4}x^{-2}\right)^2 = x^4 - \frac{1}{2} + \frac{1}{16}x^{-4}.$$
So the integral would be
$$\int_{1}^2 \sqrt{ x^4 + \frac{1}{2} + \frac{1}{16}x^{-4}}\,dx.$$
As for solving it, note that:
$$x^4 + \frac{1}{2} + \frac {1}{16}x^{-4} = (x^2)^2 + 2x^2\left(\frac{1}{4}x^{-2}\right) + \left(\frac{1}{4}x^{-2}\right)^2 = \left( x^2 + \frac{1}{4}x^{-2}\right)^2.$$
|
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|
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
|
The calculation is correct. There are many alternate forms of the integral, because of the endlessly many trigonometric identities.
If you differentiate the expression you got and simplify, you will see that you are right.
The answer you saw is likely also right. What was it?
Added: Since $\sec x=\frac{1}{\cos x}$, we have $\ln(|\sec x|)=-\ln(|\cos x|)$.
Also, since $\cos 2x=2\cos^2 x -1$, we have $\frac{\cos 2x}{4}=\frac{\cos^2 x}{2}-\frac{1}{4}$.
So your answer and the book answer differ by a constant. That's taken care of by the arbitrary constant of integration. As a simpler example,
$\int 2x\, dx=x^2+C$ and $\int 2x\, dx=(x^2+17)+C$ are both right.
|
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|
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Compute
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$
|
Consider:
$$I(a) = \int_0^1 \frac{\ln (1+ax)}{1+x^2} \, dx$$
than, the derivative $I'$ is equal:
$$I'(a) = \int_0^1 \frac{x}{(1+ax)(1+x^2)} \, dx = \frac{2 a \arctan x - 2\ln (1+a x) + \ln (1+x^2)}{2(1+a^2)} \Big|_0^1\\
= \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2}$$
Hence:
$$I(1) = \int_0^1 \left( \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2} \right) \, da \\
2 I(1) = \int_0^1 \frac{\pi a + 2 \ln 2}{4(1+a^2)} \, da = \frac{\pi}{4} \ln 2$$
Divide both sides by $2$ and you're done.
|
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|
When does $\sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$ converge? For what values of $z \in \mathbb{C}$ does the following series converge:
$$\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n\quad ?$$
|
You're given
$$f(z)=\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$$
A sensible solution would be using Cauchy's Root test. We want to find
$$\lim\limits_{n\to\infty}\left(\frac{2^n+n^2}{3^n+n^3}\right)^{1/n} =$$
$$=\lim\limits_{n\to\infty}\frac 2 3\left(\frac{1+n^2/2^n}{1+n^3/3^n}\right)^{1/n} =$$
$$=\frac 2 3\left(\frac{1+0}{1+0}\right)^{0}=\frac 2 3 $$
Then the sum converges for $|z|<\dfrac 3 2 $
|
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|
Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$ I would like to show that
$$ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx \sim_{n\rightarrow \infty} \frac{1}{n}$$
Using the change of variable $u=x^n$:
$$ I_{n}=\frac{1}{n^2} \int_0^1 \frac{u^{1/n} \ln u}{u^{1/n}-1} \mathrm du=\frac{1}{n^2}\left(\int_0^1 \ln x \mathrm dx+\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \right)=\frac{-1}{n^2}+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx=o(1/n)+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx$$
So I have to show that
$$ \int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \sim_{n \rightarrow \infty} n$$
Could you help me?
|
$$\frac{x^n}{x-1} = \frac{x^n-1}{x-1} + \frac{1}{x-1}= \left(1+ x+ \cdots + x^{n-1}\right) +\frac{1}{x-1}.$$
So $$I_n = \int^1_0 (1+x+\cdots + x^{n-1}) \log x + \int^1_0 \frac{\log x}{x-1} dx.$$
Integration by parts shows $ \int^1_0 x^k \log x = -1/(k+1)^2$ and expanding $\log$ by Taylor series will show $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ so $$I_n = \sum_{k=n+1}^{\infty} \frac{1}{k^2}.$$
Thus $$nI_n = \frac{1}{n} \sum_{k=n+1}^{\infty} \frac{1}{(k/n)^2} \to \int_1^{\infty} \frac{1}{x^2} dx=1$$
so $I_n \sim 1/n.$
You can skip showing $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ if you expand $x^n/(x-1)$ as a geometric series from the start.
Instead of using Riemann sums we could also have noted that $1/x^2$ is monotonically decreasing and use the well known theorem that if $f$ is monotone then $\displaystyle \int^n_1 f(x) dx \sim \sum_{k=1}^n f(n).$
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|
Another Congruence Proof I've been asked to attempt a proof of the following congruence. It is found in a section of my textbook with Wilson's theorem and Fermat's Little theorem. I've pondered the problem for a while and nothing interesting has occurred to me.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2} (\text{mod} p)$
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An idea that, perhaps, will appeal to you besides the ones already given above:$$1^23^2\cdot\ldots\cdot (p-1)^2=\frac{\left(1\cdot 2\cdot\ldots\cdot (p-1)\right)^2}{\left(2\cdot 4\cdot\ldots\cdot (p-1)\right)^2}\,\,(**)$$Now we use Wilson's theorem, Fermat's Little Theorem and arithmetic modulo $p$: $$(**)\,\,=\frac{(-1)^2}{\left(2^{\frac{p-1}{2}}\right)^2\left(1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\right)^2}=\frac{1}{1\cdot\left(1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\right)^2} \,\,\,(***)$$
Let us now check more closely Wilson's theorem (again, arithmetic modulo $p$):$$-1=1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\ldots\cdot (p-1)=$$$$=1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\left(-\frac{p-1}{2}\right)\cdot\ldots\cdots (-2)(-1)=$$$$=\left(-1\right)^{\frac{p-1}{2}}\left(1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\right)^2$$So: $$(1)\,\,p=3\pmod 4\Longrightarrow \frac{p-1}{2}\text{ is odd}\Longrightarrow (-1)^\frac{p-1}{2}=-1\Longrightarrow $$$$\Longrightarrow\,\,(***)=1$$$$(2)\,\,p=1\pmod 4\Longrightarrow \frac{p-1}{2}\text {is even}\,\Longrightarrow (-1)^{\frac{p-1}{2}}=1\Longrightarrow$$$$\Longrightarrow (***) =-1$$
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|
Limit involving $(\sin x) /x -\cos x $ and $(e^{2x}-1)/(2x)$, without l'Hôpital Find:
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
I have factorized it in this manner in an attempt to use the formulae.
I have tried to use that for $x$ tending to $0$, $\dfrac{\sin x}{x} = 1$ and that $\dfrac{e^x - 1}x$ is also $1$.
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You are given
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
I guess you know
$$\lim_{x\to 0}\dfrac{\sin x}{x}=1$$
$$\lim_{x\to 0} \dfrac{e^{2x} - 1}{2x}=1$$
The most healthy way of solving this is using
$$\frac{\sin x}{x} = 1-\frac {x^2}{6}+o(x^2)$$
$$\frac{e^x-1}{x}=1+\frac x 2 +o(x^2)$$
$$\cos x = 1-\frac {x^2}{2}+o(x^2)$$
This gives
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \;\frac{{1 - \dfrac{{{x^2}}}{6} + o({x^2}) - 1 + \dfrac{{{x^2}}}{2} - o({x^2})}}{{2x\left( {1 + x + o({x^2}) - 1} \right)}} \cr
& \mathop {\lim }\limits_{x \to 0} \;\frac{{\dfrac{{{x^2}}}{3} + o\left( {{x^2}} \right)}}{{2{x^2} + 2xo\left( {{x^2}} \right)}} \cr
& \mathop {\lim }\limits_{x \to 0} \;\frac{{\dfrac{1}{3} + \dfrac{{o\left( {{x^2}} \right)}}{{{x^2}}}}}{{2 + 2\dfrac{{o\left( {{x^2}} \right)}}{{{x^2}}}}} = \dfrac{1}{6} \cr} $$
Note that
$$\eqalign{
& \frac{{o\left( {{x^2}} \right)}}{{{x^2}}} \to 0 \cr
& \frac{{2o\left( {{x^2}} \right)}}{x} \to 0 \cr} $$
|
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|
Solving a differential equation I am trying to find the solution of the equation
t $y''-(\cos x) y'+(\sin x )y = 0$.
I need help urgently.Thanks
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This is a linear ODE of trigonometric function coefficients. The current approach of solving it is to transform it to a linear ODE of polynomial function coefficients first.
Let $u=\sin x$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=(\cos x)\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left((\cos x)\dfrac{dy}{du}\right)=(\cos x)\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)-(\sin x)\dfrac{dy}{du}=(\cos x)\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}-(\sin x)\dfrac{dy}{du}=(\cos x)\dfrac{d^2y}{du^2}\cos x-(\sin x)\dfrac{dy}{du}=(\cos^2 x)\dfrac{d^2y}{du^2}-(\sin x)\dfrac{dy}{du}$
$\therefore(\cos^2 x)\dfrac{d^2y}{du^2}-(\sin x)\dfrac{dy}{du}-(\cos^2 x)\dfrac{dy}{du}+(\sin x)y=0$
$(1-\sin^2 x)\dfrac{d^2y}{du^2}+(\sin^2 x-1-\sin x)\dfrac{dy}{du}+(\sin x)y=0$
$(1-u^2)\dfrac{d^2y}{du^2}+(u^2-u-1)\dfrac{dy}{du}+uy=0$
This belongs to a Heun’s confluent equation (http://dlmf.nist.gov/31.12) (http://www.maplesoft.com/support/help/Maple/view.aspx?path=HeunC), however in this case the properties are even simpler, since in this case is lucky that sum of the coefficients is equal to zero.
$\therefore y=e^u$ is a particular solution.
Let $y=e^uv$ ,
Then $\dfrac{dy}{du}=e^u\dfrac{dv}{du}+e^uv$
$\dfrac{d^2y}{du^2}=e^u\dfrac{d^2v}{du^2}+e^u\dfrac{dv}{du}+e^u\dfrac{dv}{du}+e^uv=e^u\dfrac{d^2v}{du^2}+2e^u\dfrac{dv}{du}+e^uv$
$\therefore(1-u^2)\left(e^u\dfrac{d^2v}{du^2}+2e^u\dfrac{dv}{du}+e^uv\right)+(u^2-u-1)\left(e^u\dfrac{dv}{du}+e^uv\right)+ue^uv=0$
$(1-u^2)\left(\dfrac{d^2v}{du^2}+2\dfrac{dv}{du}+v\right)+(u^2-u-1)\left(\dfrac{dv}{du}+v\right)+uv=0$
$(1-u^2)\dfrac{d^2v}{du^2}+2(1-u^2)\dfrac{dv}{du}+(1-u^2)v+(u^2-u-1)\dfrac{dv}{du}+(u^2-u-1)v+uv=0$
$(1-u^2)\dfrac{d^2v}{du^2}-(u^2+u-1)\dfrac{dv}{du}=0$
$(u^2-1)\dfrac{d^2v}{du^2}=-(u^2+u-1)\dfrac{dv}{du}$
$\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}=-1-\dfrac{u}{u^2-1}$
$\int\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}~du=\int\left(-1-\dfrac{u}{u^2-1}\right)~du$
$\ln\dfrac{dv}{du}=-u-\dfrac{1}{2}\ln(u^2-1)+c_1$
$\dfrac{dv}{du}=\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}$
$v=\int\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}~du$
$ye^{-u}=\int\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}~du$
$y=e^u\int\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}~du$
$y=e^{\sin x}\int\dfrac{c_2e^{-\sin x}}{\sqrt{\sin^2 x-1}}~d(\sin x)$
$y=e^{\sin x}\int\dfrac{c_2e^{-\sin x}}{i\cos x}\cos x~dx$
$y=e^{\sin x}\int C_2e^{-\sin x}~dx$
$y=e^{\sin x}\left(C_2\int_0^xe^{-\sin x}~dx+C_1\right)$
$y=C_1e^{\sin x}+C_2e^{\sin x}\int_0^xe^{-\sin x}~dx$
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|
The rank of a multiplication map $L\colon M_{2 \times 3} \to M_{3 \times 3}$
Let $L\colon M_{2 \times 3} \to M_{3 \times 3}$ be the linear transformation defined by $L(A) = \left[\begin{array}{rr}2 & -1 \\ 1 & 2 \\ 3 & 1\end{array}\right]\,A$. Find the dimension of the range of $L$.
Answer: $6$
How is the answer $6$? Isn't it $2$?
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If we identify $M_{2 \times 3}$ with $\Bbb{R}^6$ and similarly with the other vector space, we see that $L$ is a linear transformation from $\Bbb{R}^6$ to $\Bbb{R}^9$. Now I claim that the kernel of $L$ is trivial. Indeed, suppose there is a matrix
$$A = \left[\begin{array}{ccc} a & b & c & \\ d & e & f \end{array}\right]$$
such that $L(A) = 0$. Then we get the following system of equations:
$$\begin{eqnarray*} 2a - d &=& 0 \\ 2b - e &=& 0 \\ 2c - f &=& 0 \\ a + 2d &=& 0 \\ b + 2e &=& 0 \\ c + 2f&=& 0 \\ 3a + d &=& 0\\ 3b + e &=& 0 \\ 3c + f &=& 0. \end{eqnarray*}$$
In other words we are trying to find the null space of the matrix
$$\left[\begin{array}{cccccc} 2 & 0 & 0 & -1 & 0 & 0 \\ 0& 2 & 0 & 0 & -1 & 0 \\ 0& 0& 2 & 0 & 0 & -1 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0& 1 & 0 & 0 & 2 & 0 \\ 0& 0& 1 & 0 & 0 & 2\\ 3 & 0 & 0 & 1 & 0 & 0 \\ 0& 3 & 0 & 0 & 1 & 0 \\ 0& 0& 3 & 0 & 0 & 1 \end{array}\right].$$
Upon row reduction, this matrix in reduced row echelon form is
$$\left[\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0& 1 & 0 & 0 & 0 & 0 \\ 0& 0& 1 & 0 & 0 & \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 0 & 1 & 0 \\ 0& 0& 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right].$$
You can see that there are no free variables, so that the dimension of the kernel if zero. By rank nullity, we have
$$\begin{eqnarray*} \dim \textrm{ran} L &=& \dim \Bbb{R}^6 - \dim \textrm{ker} L \\
&=& 6 - 0 \\
&=& 0 \end{eqnarray*}$$
from which it follows that the range of $L$ has dimension 6.
|
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|
Doubt on displacement of a parabola(Again) In another exercise is given:
Find the parabola which is a displacement of $y = 2x^2 - 3x + 4$ which passes though the point $(2, -1)$ and has $x = 1$ as its symmetry axis.
I've reduced the based equation to the form: $y = 2(x - \frac{3}{4})^2 + \frac{23}{8}$, so the vertex is $(\frac{3}{4}, \frac{23}{8})$.
As the displaced equation has its symmetry axes on $x = 1$ it's known that its vertex is $(1, x)$ what can I do more to find the $y$ of the displaced equation ?
Thanks in advance.
EDIT: Corrected the typo
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Your equation for the original parabola should read $y = 2\left(x-\dfrac{3}{4}\right)^2+\dfrac{23}{8}$. The displaced parabola will have the equation $y = 2(x-x_v)^2 + y_v$, where $(x_v, y_v)$ is the vertex. Since the axis is $x=1$, $x_v$ is $1$. The point $(2,-1)$ is on the parabola, so we have $-1 = 2(2-1)^2 + y_v \Rightarrow y_v = -1 -2 = -3$. Then, the equation of the displaced parabola is $y = 2(x-1)^2 -3$.
|
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|
Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?
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Note that $5^3 = 125 = 2(61)+3$. So $5^{20} = (5^3)^6\times 5^2\equiv 3^6\times 5^2 \pmod{61}$.
Now, $3^3\equiv 27$, $3^3\times 5 = 135 \equiv 13\pmod{61}$. So
$$5^{20}\equiv (5^3)^6\times 5^2\equiv 3^6\times 5^2 =(3^3\times 5)^2 \equiv 13^2 \equiv 169\pmod{61}$$
and since $169 = 2(61) + 47$, we finally have that $5^{20}\equiv 47\pmod{61}$.
If you want to be systematic, then repeated squaring works well: $20 = 16 + 4$.
So first compute $5^2=25$. Then square that and reduce modulo $61$ to get $5^4$; then square that and reduce to get $5^8$; square again to get $5^{16}$. Then multiply the results of $5^4$ and $5^{16}$ to get $5^{20}$. This amounts five products and reductions.
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|
$|2^x-3^y|=1$ has only three natural pairs as solutions Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$?
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Case 1: If $2^x=3^y+1$ then $x \ge 1$. If $y=0$ then $x=1$.
If $y \ge 1$ then $3^y+1 \equiv 1 \pmod{3}$. Therefore $2^x \equiv 1 \pmod{3}$. Hence $x=2x_1$ with $x_1 \in \mathbb{N}^*$. The equation is equivalent to $$3^y= (2^{x_1}-1)(2^{x_1}+1)$$
Since $2^{x_1}+1>2^{x_1}-1$ and $\gcd (2^{x_1}-1,2^{x_1}+1) \ne 3$ then $2= 3^{m} (3^{n-m}-1)$ with $2^{x_1}-1=3^m,2^{x_1}+1=3^n$. Thus, $m=0,n=1$. We obtain $x_1=1$ or $x=2$. Thus, $y=1$.
Case 2. If $3^y=2^x+1$. If $x=0$ then there is no natural number $y$. If $x=1$ then $y=1$. If $x \ge 2$ then $3^y \equiv 1 \pmod{4}$. Thus, $y$ is even, let $y=2y_1$ with $y_1 \in \mathbb{N}^*$. The equation is equivalent to $$2^y=(3^{y_1}-1)(3^{y_1}+1)$$
Thus, $2=2^n(2^{m-n}-1)$ with $2^m=3^{y_1}+1,2^n=3^{y_1}-1$. It follows that $n=1,m=2$. Thus, $y_1=1$ or $y=2$. We get $x=3$.
The answer is $\boxed{ (x,y)=(1,0),(1,1),(2,1),(3,2)}$.
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Show $\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$ I have some difficulty to prove the following limit:
$$\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$$
Can someone help me? Thanks.
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$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$
You are looking for $f(1)$
$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$
$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3} (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$
$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation
After solving limits. We get:
$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$
$$f(x)=\ln(x+1)$$
$$f(1)=\ln(2)$$
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Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. First approach.
$\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$
From this relationship, I get:
$2\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$ Then:
$\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{1}{2}\left[\frac{x}{1+x^2}+\arctan x\right]+C$ This is a recursive solution.
Second approach.
$x=\tan t$ in $t\in (- \pi/2, \pi/2)$, i.e. $t=\arctan x$, then $dx=(1+x^2) dt$.
$\int \frac{1}{\left(1+x^2\right)^2}dx=\int \frac{1}{1+x^2}dt=\int \frac{\cos^2t}{\sin^2t+\cos^2t}dt=\int \cos^2t dt=\frac{1}{2}\int \left(1+\cos 2t \right) dt=\frac{t}{2}+\frac{1}{4}\sin 2t$
This result can be rewritten (using trigonometric formulas):
$\frac{t}{2}+\frac{1}{4}\sin 2t=\frac{t}{2}+\frac{1}{2}\sin t \cos t$
From $\cos^2 t=\frac{1}{1+x^2}$, I have:
$|\cos t|=\sqrt{\frac{1}{1+x^2}}$ but in $t\in (- \pi/2, \pi/2)$, $|\cos t|=\cos t$. So:
$\cos t=\sqrt{\frac{1}{1+x^2}}$. Now I have a problem: $|\sin t|=\sqrt{\frac{1}{1+x^2}}$, but $|\sin t|\neq \sin t$ for $t\in (- \pi/2, \pi/2)$. Any suggestions, please? This integral can be solved in other ways?
Thanks.
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This integral can be evaluated in that way too as shown below :
$$\int \frac{1}{a^2+x^2} dx=\frac{1}{a} \arctan \frac{x}{a} +c(a) $$
$$\frac {d}{da} (\int \frac{1}{a^2+x^2} dx)= \frac {d}{da}(\frac{1}{a} \arctan \frac{x}{a} +c(a) ) $$
$$\int \frac{-2a}{(a^2+x^2)^2} dx= \frac{-1}{a^2} \arctan \frac{x}{a} + \frac{1}{a} \frac{-x/a^2}{(1+x^2/a^2)}+c_1(a) ) $$
for $a=1$
$$-2\int \frac{1}{(1+x^2)^2} dx= - \arctan x + \frac{-x}{(1+x^2)}+c_1(1) $$
$$\int \frac{1}{(1+x^2)^2} dx= \frac{1}{2} \arctan x + \frac{x}{2(1+x^2)}+k $$
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|
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
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Case 1. If $a,b,c$ are lengths of triangle.
Since
$$
2\sqrt{xy}\leq x+y\qquad
2\sqrt{yz}\leq y+z\qquad
2\sqrt{zx}\leq z+x
$$
for $x,y,z\geq 0$, then multiplying this inequalities we get
$$
8xyz\leq(x+y)(y+z)(z+x)
$$
Now substitute
$$
x=\frac{a+b-c}{2}\qquad
y=\frac{a-b+c}{2}\qquad
z=\frac{-a+b+c}{2}\qquad
$$
Since $a,b,c$ are lengths of triangle, then $x,y,z\geq 0$ and our substitution is valid. Then we will obtain
$$
(-a+b+c)(a-b+c)(a+b-c)\leq abc\tag{1}
$$
Case 2. If $a,b,c$ are not lengths of triangle.
Then at least one factor in left hand side of inequality $(1)$ is negative. In fact the only one factor is negative. Indeed, without loss of generality assume that $a+b-c<0$ and $a-b+c<0$, then $a=0.5((a+b-c)+(a-b+c))<0$. Contradiction, hence the only one factor is negative. As the consequence left hand side of inequality $(1)$ is negative and right hand side is positive, so $(1)$ obviously holds.
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"url": "https://math.stackexchange.com/questions/170813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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|
Trigonometric Identities To Prove
*
*$\tan\theta+\cot\theta=\dfrac{2}{\sin2\theta}$
Left Side:
$$\begin{align*}
\tan\theta+\cot\theta={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={\sin^2\theta+\cos^2\theta\over\cos\theta\sin\theta}
= \dfrac{1}{1\sin\theta\cos\theta}
\end{align*}$$
Right Side:
$$\begin{align*}
\dfrac{2}{\sin2\theta}=\dfrac{2}{2\sin\theta\cos\theta}=\dfrac{1}{1\cos\theta\sin\theta}
\end{align*}$$
I got it now. Thanks!
|
$$\tan(\theta) + \cot(\theta) = {\sin(\theta)\over \cos(\theta)} + {\cos(\theta)\over \sin(\theta)} = {\sin^2(\theta) + \cos^2(\theta) \over\cos(\theta)\sin(\theta)}
= {1\over\sin(\theta)\cos(\theta)}.$$
Now avail yourself of the fact that $$\sin(2\theta) = 2\cos(
\theta)\sin(\theta).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/170951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to evaluate $\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx$ by hand How can I evaluate$$\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx\text{ and }\int_0^\infty\frac{\sin x}{e^x-1}\,\mathrm dx.$$
Thanks in advance.
|
Let us start from the Weierstrass product for the sine function:
$$ \sin(x) = x \prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)\tag{1} $$
This identity over the real line can be proved by exploiting Chebyshev polynomials, but it holds over $\mathbb{C}$ and it is equivalent to
$$ \sinh(x) = x \prod_{n\geq 1}\left(1+\frac{x^2}{n^2\pi^2}\right).\tag{1'} $$
If we apply $\frac{d}{dx}\log(\cdot)$ (the logarithmic derivative) to both sides of $(1)$ we end up with
$$ \cot(x)-\frac{1}{x} = \sum_{n\geq 1}\frac{2x}{x^2-n^2 \pi^2} \tag{2}$$
$$ \coth(x)-\frac{1}{x} = \sum_{n\geq 1}\frac{2x}{x^2 + n^2 \pi^2} \tag{2'}$$
and by expanding the main term of the RHS of $(2)$ as a geometric series we have
$$ \frac{1-x\cot x}{2} = \sum_{n\geq 1}\frac{\frac{x^2}{n^2 \pi^2}}{1-\frac{x^2}{n^2\pi^2}} =\sum_{n\geq 1}\sum_{m\geq 1}\left(\frac{x^2}{n^2 \pi^2}\right)^m = \sum_{m\geq 1}x^{2m}\frac{\zeta(2m)}{\pi^{2m}} \tag{3}$$
$$ \frac{1-x\coth x}{2} = \sum_{m\geq 1}(-1)^m x^{2m}\frac{\zeta(2m)}{\pi^{2m}}\tag{3'} $$
$(3')$ alone is enough to prove that $\zeta(2m)$ is always a rational multiple of $\pi^{2m}$, related to a coefficient of the Maclaurin series of $\frac{z}{e^z-1}=-\frac{z}{2}+\frac{z}{2}\coth\frac{z}{2}$, i.e. to a Bernoulli number, but that is not the point here.
Let us deal with the second integral:
$$ \int_{0}^{+\infty}\frac{\sin z}{e^z-1}\,dz = \sum_{m\geq 1}\int_{0}^{+\infty}\sin(z) e^{-mz}\,dz = \sum_{m\geq 1}\frac{1}{m^2+1} $$
can be evaluated through $(2')$ by picking $x=\pi$. In a similar fashion
$$ \int_{0}^{+\infty}\frac{\cos z}{e^z+e^{-z}}\,dz = \sum_{m\geq 0}(-1)^m\int_{0}^{+\infty}\cos(z)e^{-(2m+1)z}\,dz = \sum_{m\geq 0}(-1)^m \frac{(2m+1)}{(2m+1)^2+1} $$
can be computed from the Weierstrass product of the cosine function,
$$ \cos(x) = \prod_{m\geq 0}\left(1-\frac{4x^2}{(2m+1)^2\pi^2}\right)\tag{4} $$
$$ \cosh(x) = \prod_{m\geq 0}\left(1+\frac{4x^2}{(2m+1)^2\pi^2}\right)\tag{4'} $$
leading by logarithmic differentiation to
$$ \sum_{m\geq 0}\frac{x}{x^2+(2m+1)^2} = \frac{\pi}{4}\tanh\left(\frac{\pi x}{2}\right)\tag{5} $$
and also to
$$ \sum_{m\geq 0}\frac{(-1)^m(2m+1)}{x^2+(2m+1)^2} = \frac{\pi}{4}\operatorname{sech}\left(\frac{\pi x}{2}\right).\tag{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$, $P(1)=10$, $P(2)=20$, $P(3)=30$
What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$
provided that $P(1)=10$, $P(2)=20$, $P(3)=30$?
I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems?
|
I'm not sure this is the smartest way, but here's one way.
Define $Q(x) = P(x) - x^4$ which satisfies the conditions $$Q(1) = 9, Q(2) = 4, Q(3) = -51.$$
So $a,b,c,d$ satisfy the matrix identity
$$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \end{pmatrix} \begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix} = \begin{pmatrix} 9 \\ 4 \\ -51 \end{pmatrix}.$$ To solve this, find the kernel of the matrix (a bit of linear algebra), which turns out to be $$<\begin{pmatrix} -6 \\ 11 \\ -6 \\ 1 \end{pmatrix}>$$ and add a particular solution, for example where $a = 0$: $b = -25, c = 70, d = -36$.
So your polynomial is given by $$P(x) = x^4 + ax^3 + (-25-6a)x^2 + (11 + 70a)x + (-36 - 6a)$$ for some $a$ which we can't determine.
You then get $$P(12) + P(-8) = 17940 + 990a + 1900 - 990a = 19840.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/172366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$ $$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.
|
$$\text{Let us check the value of }\frac a{\sin\theta}+\frac b{\cos\theta}$$
$$\frac a{\sin\theta}+\frac b{\cos\theta}=\frac{a\cos\theta+b\sin\theta}{\cos\theta\sin\theta}$$
Putting $a=r\sin\alpha,b=r\cos\alpha$ where $r>0$
Squaring & adding we get $r^2=a^2+b^2\implies r=+\sqrt{a^2+b^2}$
$$\implies \frac a{\sin\theta}+\frac b{\cos\theta}=\frac{2\sqrt{a^2+b^2}(\sin\theta\cos\alpha+\cos\theta\sin\alpha)}{\sin2\theta}$$
$$=2\sqrt{a^2+b^2}\cdot\frac{\sin(\theta+\alpha)}{\sin2\theta}\text{ as }\sin2\theta=2\sin\theta\cos\theta$$
Now, the solution of $P\sin x= Q\sin A $ is general intractable unless $P=0$ or $Q=0$ or $P=\pm Q\ne0$
Here the coefficients of $\sin(\theta+\alpha),\sin2\theta$ can not be $0$
So, either $\sin2\theta=\sin(\theta+\alpha)$ or $\sin2\theta=-\sin(\theta+\alpha)$
$$\begin{array}{|c|c|c|}
\hline \text{ Case } & \sin2\theta=\sin(\theta+\alpha) & \sin2\theta=-\sin(\theta+\alpha)=\sin(-\theta-\alpha) \text{ as }\sin(-x)=-\sin x \\
\hline \text{General Solution} & 2\theta=n180^\circ+(-1)^n(\theta+\alpha)\text{ where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(-\alpha-\theta)\text{ where }n\text{ is any integer } \\
\hline n=2m & \alpha=\theta-m360^\circ\equiv\theta\pmod{360^\circ} & \alpha=m360^\circ-3\theta\equiv-3\theta \\
\hline n=2m+1 & \alpha=(2m+1)180^\circ-3\theta\equiv 180^\circ-3\theta & \alpha=\theta-(2m+1)180^\circ\equiv\theta+180^\circ \\
\hline
\end{array} $$
Here $a=1,b=-\sqrt3$ and $\theta=10^\circ$
Taking $\sin2\theta=\sin(\theta+\alpha), \alpha=\theta=10^\circ$ or $=180^\circ-3\theta=150^\circ$
$\implies r=+\sqrt{a^2+b^2}=2$ and $\cos \alpha=\frac br=-\frac{\sqrt3}2$ and $\sin\alpha=\frac ar=\frac12\implies \alpha$ lies in the 2nd Quadrant, $\implies \alpha=150^\circ$
$$\text{So,} \frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}=2\sqrt{1^2+(-\sqrt3)^2}\cdot\frac{\sin(10^\circ+150^\circ)}{\sin(2\cdot10^\circ)}=2\cdot2=4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/172471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Laplace transform of a product of Modified Bessel Functions Working with a scalar field in 2 dimensions I've come to the following integral, from which I can extract the proper ultraviolet behavior ($a \ll 1$) of the theory:
$\int_0^\infty e^{-(4+a^2)x}\left[I_0(2x)\right]^2 ds$.
It is obvious to me that this is the Laplace transform of $\left[I_0(2x)\right]^2$ evaluated at $s = (4+a^2)$. From Wikipedia I got the formula
$\int_0^\infty e^{-sx} f(x)g(x) dx = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{c-iT}^{c+it} F(\sigma)G(s-\sigma) d\sigma$,
where $F(\sigma)$ and $G(\sigma)$ are the Laplace transforms of $f(x)$ and $g(x)$, respectively.
I'm encountering some trouble trying to get an analytical result for that integral.
What I actually need is its $a \approx 0$ behavior, but a full analytical answer would be great.
Thanks in advance for any help!
|
Given that $I_0(2 x)^2 = 1 + 2 x^2 + \frac{3}{2} x^4 + \frac{5}{9}x^6 + \mathcal{o}(x^6)$ we see that it is a hypergeometric function:
$$
I_0(2x)^2 = {}_1F_2\left(\frac{1}{2}; 1,1; 4 x^2\right) = \sum_{n=0}^\infty \frac{\left(\frac{1}{2}\right)_n}{(1)_n (1)_n} \frac{(4 x^2)^n}{n!} = \sum_{n=0}^\infty \left(\frac{x^{n}}{n!} \right)^2 \binom{2n}{n}
$$
Now, integrate term-wise:
$$
\int_0^\infty \mathrm{e}^{-k x} [ I_0(2x) ]^2 \mathrm{d} x = \sum_{n=0}^\infty \frac{(2n)!}{n!^4} \int_0^\infty x^{2n} \mathrm{e}^{-k x} \mathrm{d} x =
\sum_{n=0}^\infty \frac{(2n)!}{n!^4} \frac{(2n)!}{k^{2n+1}} = \frac{1}{k} \sum_{n=0}^\infty \left( \binom{2n}{n} \frac{1}{k^n} \right)^2
$$
The sum is again hypergeometric, since ratio of subsequent summands is $\frac{4 (2n+1)^2}{k^2 (n+1)^2} = \frac{16}{k^2} \frac{\left( n+ \frac{1}{2}\right)^2}{(n+1)^2} $, thus the sum equals:
$$
\int_0^\infty \mathrm{e}^{-k x} [ I_0(2x) ]^2 \mathrm{d} x = \frac{1}{k} \cdot {}_2F_1\left( \frac{1}{2}, \frac{1}{2}; 1; \frac{16}{k^2}\right) = \frac{2}{
\pi k} K\left(\frac{16}{k^2}\right)
$$
where $K(m)$ is a complete elliptic integral: $$K(m) = \int_0^{\pi/2} \frac{\mathrm{d} \phi}{\sqrt{1-m \cdot \sin^2(\phi)}} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/175936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt{2}}{4} + \frac{1}{2} }$
*$\sin(72^\circ) = \sqrt{ \frac{\sqrt{5}}{8} + \frac{5}{8} }$
*$\sin(75^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
*?
Is there is a list of known exact values of $\boldsymbol \sin$ somewhere?
Found a related post here.
|
Algebraically exact value of Sine for all integer angles is possible. Please visit https://archive.org/details/ExactTrigonometryTableForAllAnglesFinal for the list of exact values for Sine of integer angles in degrees.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
How do I proceed with these quadratic equations? The question is
$$ax^2 + bx + c=0 $$ and $$cx^2+bx+a=0$$ have a common root, if $b≠ a+c$, then what is $$a^3+b^3+c^3$$
|
If a=c, the equations become identical, we can hardly determine any relationship among a,b,c(=a).
If a≠c,
if y is a root of $ax^2+bx+c=0$, then observe that $\frac{1}{y}$ is a root of $cx^2+bx+a=0$.
For the common root, $y=\frac{1}{y}=>y=±1$, but y=-1 makes a-b+c=0 ⇔ b=a+c which is not acceptable according to the given condition (as André Nicolas has observed).
So, y=1 if a≠c
=>a+b+c=0
=>a+b=-c
Cubing both sides, $(a+b)^3=(-c)^3$
$=>a^3+b^3+3ab(a+b)=-c^3$
or, $=>a^3+b^3+3ab(-c)=-c^3$
$=>a^3+b+c^3=3abc$ if a≠c.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/180479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me the proof of
$$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$$
Thanks!
|
$$
\begin{align}
\sin(A)+\sin(B)+\sin(C)
&=\sin(A)+\sin(B)+\sin(\pi-A-B)\\[9pt]
&=\color{#C00000}{\sin(A)+\sin(B)}+\color{#00A000}{\sin(A+B)}\\[6pt]
&=\color{#C00000}{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}+\color{#00A000}{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)}\\
&=2\sin\left(\frac{A+B}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)\\
&=2\sin\left(\frac{A+B}{2}\right)\;2\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\\
&=4\cos\left(\frac\pi2-\frac{A+B}{2}\right)\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\\
&=4\cos\left(\frac{C}{2}\right)\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)
\end{align}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/180860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Using a circular contour integral I was having some problems preparing for an exam, and a friend of mine told me about this site :)
I have to prove this:
$$
\int_0^{2\pi} \frac{d\theta}{a + \cos\theta} = \frac{2\pi}{\sqrt{a^2 - 1}}
$$
Using
$$
z = e^{i\theta}\\
a>1
$$
and integrating over the unit circle $|z| = 1$.
I know there are proofs of this relationship, but I can't manage to do it using the unit circle contour.
Afterwards I also have to proof a similar relation, with the integrand squared:
$$
\int_0^{2\pi} \frac{d\theta}{( a + cos\theta)^2} = \frac{2a\pi}{(a^2 - 1)^{3/2}}
$$
I've tried to put up the equations, but as far as I can tell there are no poles ($z = -a$ lies outside of the unit circle ). Then I can rewrite
$$
\frac{1}{a + z} = \frac{1}{a + \cos\theta + i\sin\theta}
$$
But then I'm stuck :(
|
$$I=
\int _0^{2\pi} \frac{d\theta}{a + \cos\theta}
$$
$$
=2\int _0^{\pi} \frac{d\theta}{a + \cos\theta}
$$
as $\int _0^{2\pi} \frac{d\theta}{a + \cos\theta}=\int _0^{\pi} \frac{d\theta}{a + \cos\theta}+\int _\pi^{2\pi} \frac{d\theta}{a + \cos\theta}$
Now , $\int _\pi^{2\pi} \frac{d\theta}{a + \cos\theta}=-\int _\pi ^{0} \frac{dy}{a + \cos y}$ where $y=2\pi-\theta$
So, $\int _\pi^{2\pi} \frac{d\theta}{a + \cos\theta}=\int _0^\pi \frac{dy}{a + \cos y}$
$I= 2\int _0^{\pi} \frac{d\theta}{a + \frac{(1-tan^2\frac{\theta}{2})}{(1+tan^2\frac{\theta}{2})}} $
$= 2\int _0^{\pi} \frac{sec^2\frac{\theta}{2}d\theta}{(a+1) + (a-1)tan^2\frac{\theta}{2}} $
Now putting $tan\frac{\theta}{2}=z$, $sec^2\frac{\theta}{2}d\theta=2dz$
$=\int _0^∞ \frac{4dz}{(a+1) + (a-1)z^2}$
$=\frac{4}{a-1}\int _0^∞\frac{dz}{\frac{a+1}{a-1} + z^2}$
$=\frac{4}{a-1}\sqrt{\frac{a-1}{a+1}}{}tan^{-1}(\frac{z(a-1)}{a+1})| _0^∞$
$=\frac{4}{\sqrt{a^2-1}}\frac{\pi}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/180995",
"timestamp": "2023-03-29T00:00:00",
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|
Integrating $\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}$ I am bugged by this problem: how do I evaluate this?
$$\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}.$$
A closed form will be convenient and fine. Thanks (it does not seem particularly inpiring).
|
I like Bitrex's answer best, and my other answer next, but here is one without any substitution. Multiply by $\frac{(\sqrt{x^2+1}-x)^{99}}{(\sqrt{x^2+1}-x)^{99}}$ and you have $$\int\frac{(\sqrt{x^2+1}-x)^{99}}{1}\,dx=\int(\sqrt{x^2+1}-x)^{99}\,dx$$ The next part isn't fun to write out, but you could use the binomial theorem to expand the 99th power. $$\int\sum_{k=0}^{99}\binom{99}{k}(-x)^{99-k}(\sqrt{x^2+1})^k\,dx$$ For each $k$, you have a term that is easy to antidifferentiate.
$$\begin{align}
&\int\sum_{k=0}^{49}\binom{99}{2k}(-x)^{99-2k}(\sqrt{x^2+1})^{2k}\,dx+\int\sum_{k=0}^{49}\binom{99}{2k+1}(-x)^{99-2k-1}(\sqrt{x^2+1})^{2k+1}\,dx\\
=&-\int\sum_{k=0}^{49}\binom{99}{2k}x^{99-2k}(x^2+1)^{k}\,dx+\int\sum_{k=0}^{49}\binom{99}{2k+1}x^{98-2k}(x^2+1)^{k}\sqrt{x^2+1}\,dx\\
=&-\int\sum_{k=0}^{49}\binom{99}{2k}x^{99-2k}\sum_{j=0}^kx^{2j}\,dx+\int\sum_{k=0}^{49}\binom{99}{2k+1}x^{98-2k}\sum_{j=0}^kx^{2j}\sqrt{x^2+1}\,dx\\
=&-\sum_{k=0}^{49}\binom{99}{2k}\sum_{j=0}^k\frac{x^{2j+100-2k}}{2j+100-2k}+\sum_{k=0}^{49}\binom{99}{2k+1}\sum_{j=0}^k\int x^{2j+98-2k}\sqrt{x^2+1}\,dx
\end{align}$$
The last integral can be computed using a reduction formula for $x^n\sqrt{x^2+1}$ (effectively, integration by parts with $u=x^{n-1}$ and $dv=x\sqrt{x^2-1}\,dx$) and the antiderivative for $\sqrt{x^2+1}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
|
The function $$f(x)=\frac{1}{\sqrt{x}}$$ is convex. Applying Jensen as follows :
$$
a*f[a + b] + b*f[b + c] + c*f[c + a] >= (a + b + c)*f[\frac{
(a (a + b) + b (b + c) + c (c + a))}{(a + b + c)}] = \frac{(a + b + c)^{3/2}}{\sqrt{a^2 + a b + b^2 + a c + b c + c^2}}
$$
We need to prove $$\frac{(a + b + c)^{3/2}}{\sqrt{a^2 + a b + b^2 + a c + b c + c^2}}>=\frac{3}{\sqrt{2}}$$
This is equivalent to proving:
$$2 (a + b + c)^3 - 9 (a^2 + a b + b^2 + a c + b c + c^2)>=0$$
Which is an easy exercise.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/185825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
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|
On the number of possible solutions for a quadratic equation. Solving a quadratic equation will yield two roots:
$$\frac{-\sqrt{b^2-4 a c}-b} {2 a}$$
and:
$$\frac{\sqrt{b^2-4a c}-b}{2 a}$$
And I've been taught to answer it like:
$$\frac{\pm\sqrt{b^2-4a c}- b}{2 a}$$
Why does it yields only two solutions? Aren't there infinite solutions for that? Is there a proof on the number of possible solutions for a quadratic equation?
|
A more important question: Why should there be infinitely many solutions?
The roots/solutions of a quadratic equation $ax^2 + bx + c = 0$ are given by looking at when the graphs $y = ax^2 + bx + c$ cuts the $x$-axis. Remember that $a$, $b$ and $c$ are all fixed numbers, e.g. $y = x^2 + 2x + 1.$ Ask yourself this: Why should there be infinity many values of $x$ for which $x^2 + 2x + 1 = 0$? In fact, there is only one single value of $x$ for which $x^2 + 2x + 1 = 0$. This is because $x^2 + 2x + 1 = (x+1)^2$ and so $x = -1$ is the only solution. Find me another value of $x$ for which $x^2 + 2x + 1 = 0$ and I will give you US$1,000,000.
Draw yourself some sketches. Draw the one-parameter family of parabolae given by $y = x^2 - k$ where $k$ is a number we're going to play with. When $k > 0$, the parabola cuts the $x$-axis at two distinct points: $x = \pm \sqrt{k}$. When $k = 0$, the parabola cuts the $x$-axis at one point: $x = 0$. When $k < 0$, the parabola misses the $x$-axis altogether.
|
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|
For some integers $m,n$,proving that $p\mid m$ Assume $p$ is a prime number such that $p\equiv 1 \pmod3$, and $q=\lfloor \frac{2p}{3}\rfloor$.
If:
$$\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} =\frac{m}{n}$$
For some integers $m,n$, what is the proof that $p\mid m$
|
Let $p=3k+1$ and $H_n$ denote the n-th harmonic number. Since $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ (1) the sum in the question is $$\displaystyle \sum_{i=1}^q \frac{ (-1)^{i+1} }{i} = H_{2k}-H_k.$$
Working in mod $p$, we add $p$ to the denominator of each term in $H_k$ : $$ H_{2k} - H_k \equiv H_{2k} + \sum_{i=1}^k \frac{1}{p-i} = H_{p-1} =\frac{1}{2}\sum_{i=1}^{p-1} \left( \frac{1}{i} + \frac{1}{p-i}\right)\equiv \frac{1}{2}\sum_{i=1}^{p-1} \left( \frac{1}{i} + \frac{1}{-i}\right)=0.
$$
proving the result.
Using (1) we see that $$\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} \cdots + \frac{1}{q-1} - \frac{1}{q}.$$
The series is almost like a Harmonic number but it has alternating signs, so we add and subtract 2*even terms:
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} \cdots + \frac{1}{q-1} - \frac{1}{q} = \left( 1+ \frac{1}{2} + \cdots + \frac{1}{q} \right) -2 \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{q} \right).$$
The factor of 2 cancels with the denominator of every term in the second bracket. Remembering that $q=2k$ gives $$ \left( 1+ \frac{1}{2} + \cdots + \frac{1}{2k} \right) - \left( 1 + \frac{1}{2} + \cdots + \frac{1}{k} \right)= H_{2k} - H_k.$$
|
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|
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.
But how do I get the last step? Or am I already dead wrong?
|
Hint: From here:
$$\frac{2x^2 - 43x - 45}{2x(x + \frac{3}{2})} \lt 0$$
you could say that the fraction is less than zero if and only if either (1) the numerator is positive and the denominator is negative or (2) the numerator is negative and the denominator is positive. So for case (2), you solve:
$$
2x^2 - 43x - 45 <0 \quad \text{and}\quad 2x(x +\frac{3}{2}) > 0.
$$
You might already know this, but the way you solve each of these two inequalities is by finding possible roots. So for example for the $2x(x+ \frac{3}{2}) > 0$, the roots are $x = 0$ and $x = -\frac{3}{2}$. Now you just need to find the (constant) signs of the expression $2x(x+\frac{3}{2})$ on the intervals $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, 0)$, and $(0, \infty)$.
|
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|
Solving the cubic polynomial equation $x^3+3x^2-5x-4=0$
How can I solve the cubic polynomial equation $$x^3+3x^2-5x-4=0$$
I simplified it to:
$$x(x^2+3x-5)=4$$
But I don't know where to go from here.
|
You are given $x^3+3x^2-5x-4=0$
First, find one of the roots of the polynomial. By way of the rational root theorem, we find that -4 is one of the roots, therefore x + 4 is a factor. Next, split the polynomial in accordance with x + 4 as follows:
$x^3 + 4x^2 - x^2 -4x -x - 4 = 0$
Next factor each pair of terms from left to right as follows:
$x^2(x + 4) -x(x + 4) - 1(x + 4) = 0$
Since x + 4 is common to each factorization as expected, factor it out:
$(x + 4)(x^2 - x - 1)= 0$
By zero product property:
$x + 4 = 0$
$x^2 - x - 1 = 0$
Obviously $x = -4$ is one of the solutions. The other solution can be found by completing the square:
$x^2 - x = 1$
$x^2 - x + \left(\dfrac{1}{2}\right)^2 = 1 + \left(\dfrac{1}{2}\right)^2$
$\left(x - \dfrac{1}{2}\right)^2 = 1 + \dfrac{1}{4}$
$\left(x - \dfrac{1}{2}\right)^2 = \dfrac{5}{4} $
$x = \dfrac{\pm\sqrt{5} + 1}{2}$
|
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|
Numerically evaluating the limit of $\frac{x^4-1}{x^3-1}$ as $x\rightarrow 1$ What is the limit as $x \to 1$ of the function
$$ f(x) = \frac{x^4-1}{x^3-1} . $$
|
Note that $x^4-1=(x-1)(x^3+x^2+x+1)$ and $x^3-1=(x-1)(x^2+x+1)$.
Thus
$$f(x)=\frac{(x-1)(x^3+x^2+x+1)}{(x-1)(x^2+x+1)}.$$
When $x\ne 1$, the $x-1$ terms cancel. Now we can safely let $x$ approach $1$.
|
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|
Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?
I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.
Are they just a coincidence or is there a theory behind?
Is it about modular arithmetic?
|
Since you originally observed your pattern while doing proofs by induction, here is a proof by induction on $n$ that $a-b$ divides $a^n - b^n$ for all $n \in \mathbb{N}$:
The statement is clearly true for $n = 1$. Assume the statement is true for $n = m$ for $m \geq 1$. Thus, $a^m - b^m = (a-b)k$, for some $k \in \mathbb{Z}$. Then for $n = m + 1$,
\begin{equation}
a^{m+1} - b^{m+1} = a^{m+1} - b^mb = a^{m+1} + [(a-b)k - a^m]b = a^{m+1} - a^mb + (a-b)k = a^m(a - b) + (a-b)k = (a-b)(a^m + k).
\end{equation}
And voila, $a-b| a^{m+1} + b^{m+1}$, so you are done by induction.
|
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|
Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$ For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
|
By Cauchy-Schwarz inequality,
$$\begin{eqnarray*}
& & \left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2 \\
&=& \sqrt{\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2}\sqrt{\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(a+\frac{1}{b}\right)^2}\\
&\geq& \left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)\\
&=&3+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+ab+bc+ca+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}.
\end{eqnarray*}$$
Now using the condition $ab+bc+ca = 1$,
$$\begin{eqnarray*}
& & 3+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+ab+bc+ca+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \\
& = & \left(\frac{1}{ab}+9ab\right)+\left(\frac{1}{bc}+9bc\right)+\left(\frac{1}{ca}+9ca\right)+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) - 5 \\
& \geq & 6+6+6+3-5 = 16.
\end{eqnarray*}$$
We can also show that this inequality is strict, we can just put $(a, b, c) = \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.
(This minimizing case justifies our choice of $9ab+9bc+9ca$, combined with the equality condition of AM-GM inequality, and also of CS inequality.)
|
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|
Find a plane perpendicular to a plane passing by point In $\mathbb R^4$ I have: $$\pi: \begin{cases} x+y-z+q+1=0 \\ 2x+3y+z-3q=0\end{cases}$$
I have to find $\pi' \bot$ $ \pi $ and passing by $P=(0,1,0,1)$. How can I do that? Thanks a lot!
|
The equation of any plane$(\pi_1)$ passing through $P(0,1,0,1)$ is $a(x-0)+b(y-1)+c(z-0)+d(q-1)=0$ where $a,b,c,d$ are indeterminate constants,
If $\pi_1 \bot \pi $ , the sum of the product of the directional cosines will be $0$.
So, $(1)(a)+(1)(b)+(-1)(c)+(1)(d)=0\implies a+b-c+d=0$
and $(2)(a)+(3)(b)+(1)(c)+(-3)(d)=0\implies 2a+3b+c-3d=0 $
So, $2c=5a+6b$ and $2d=3a+4b$,
So, $\pi_1$ becomes $2a(x-0)+2b(y-1)+(5a+6b)(z-0)+(3a+4b)(q-1)=0$
Or, $a(2x+5z+3q-3)+b(2y+6z-2+4q-4)=0$
Or, $a(2x+5z+3q-3)+b(2y+6z+4q-6)=0$
If $ab≠0$, the equation of the plane $\pi_1$ will be $2x+5z+3q-3=0$ and $2y+6z+4q-6=0.$
|
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|
How to check if a point is inside a rectangle? There is a point $(x,y)$, and a rectangle $a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)$, how can one check if the point inside the rectangle?
|
area of the rectangle
Ai: areas of the triangles shown in the pictures. (i = 1, 2, 3, 4)
ai: lengths of the edges shown in the pictures. (i = 1, 2, 3, 4)
bi: lengths of the line segments connecting the point and the corners. (i = 1, 2, 3, 4)
If the point is inside the rectangle, the following equation holds:
$ \mathbf{A = A_1 + A_2 + A_3 + A_4} $
If the point is outside the rectangle, the following inequality holds:
$ \mathbf{A > A_1 + A_2 + A_3 + A_4} $
How do we calculate A, A1, A2, A3, A4, a1, a2, a3, a4, b1, b2, b3 and b4?
First we calculate the edge lengths:
$
a_1 = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\
a_2 = \sqrt{(x_2 - x_3)^2 + (y_2 - y_3)^2} \\
a_3 = \sqrt{(x_3 - x_4)^2 + (y_3 - y_4)^2} \\
a_4 = \sqrt{(x_4 - x_1)^2 + (y_4 - y_1)^2}
$
Next we calculate the lengths of the line segments:
$
b_1 = \sqrt{(x_1 - x)^2 + (y_1 - y)^2} \\
b_2 = \sqrt{(x_2 - x)^2 + (y_2 - y)^2} \\
b_3 = \sqrt{(x_3 - x)^2 + (y_3 - y)^2} \\
b_4 = \sqrt{(x_4 - x)^2 + (y_4 - y)^2}
$
Then we calculate the areas using Heron's Formula:
$
A \,\,\, = a_1a_2 = a_2a_3 = a_3a_4 = a_4a_1 \\
u_1 = \frac{a_1 + b_1 + b_2}{2} \\
u_2 = \frac{a_2 + b_2 + b_3}{2} \\
u_3 = \frac{a_3 + b_3 + b_4}{2} \\
u_4 = \frac{a_4 + b_4 + b_1}{2} \\
A_1 = \sqrt{u_1(u_1 - a_1)(u_1 - b_1)(u_1 - b_2)} \\
A_2 = \sqrt{u_2(u_2 - a_2)(u_2 - b_2)(u_2 - b_3)} \\
A_3 = \sqrt{u_3(u_3 - a_3)(u_3 - b_3)(u_3 - b_4)} \\
A_4 = \sqrt{u_4(u_4 - a_4)(u_4 - b_4)(u_4 - b_1)}
$
Finally you can do the area test to check if the point is inside or outside the rectangle.
|
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|
When is a sum of consecutive squares equal to a square? We have the sum of squares of $n$ consecutive positive integers: $$S=(a+1)^2+(a+2)^2+ ... +(a+n)^2$$ Problem was to find the smallest $n$ such, that $S=b^2$ will be square of some positive integer. I found an example for $n=11$. Now, I'm trying to prove, that if $2<n<11$ there is't any solution. So I need some help with this: If $n,a,b \in \mathbb{N}$ and $n<11$ prove that equation
$$
n\cdot a^2 + n(n+1)\cdot a + \frac{1}{6}n(n+1)(2n+1)=b^2
$$
can't be solved. Or if I'm wrong, find counterexample.
The only idea I have is: to consider the remains $\operatorname{Mod}[b^2,n]$ and $\operatorname{Mod}[\frac{1}{6}n(n+1)(2n+1),n]$. For each $2<n<11$, but it is very long.
|
Below is a reasonable (but not very illuminating) proof. Put $S_n(x)=\sum_{k=1}^{n} (x+k)^2$. Note that it is also true that for $2<n<11$, $S_n(x)$ is never a square modulo $n^2$, and $S_n(x)$ is also never a square modulo $900$. Perhaps this will inspire others to produce more intelligent proofs.
Here it goes :
$$
\begin{array}{ccll}
S_3(x) & \equiv & 3x^2+12x+14 \equiv 2 & {\sf mod} \ 3 \\
S_4(x) & \equiv & 4x^2+20x+30 \equiv 2 & {\sf mod} \ 4 \\
S_5(x) & \equiv & 5x^2+30x+55 \equiv 2 \ \text{or} \ 3 & {\sf mod} \ 4 \\
S_6(x) & \equiv & 6x^2+42x+91 \equiv 3 & {\sf mod} \ 4 \\
S_7(x) & \equiv & 7x^2+56x+140 \equiv 3,8,11 \ \text{or} \ 12 & {\sf mod} \ 16 \\
S_8(x) & \equiv & 8x^2+72x+204 \equiv 2,5,6 \ \text{or} \ 8 & {\sf mod} \ 9 \\
S_9(x) & \equiv & 9x^2+90x+285 \equiv 6 & {\sf mod} \ 9 \\
S_{10}(x) & \equiv & 10x^2+110x+385 \equiv 5,10 \ \text{or} \ 20& {\sf mod} \ 25 \\
\end{array}
$$
Each time, we see that $S_n(x)$ is never a square for the given modulus.
|
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|
Prove that: $ \int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} = 0$ How would you prove that?
$$ \int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} dx= 0$$
I'm looking for a solution at high school level if possible. Thanks.
|
$\cos 2x=1-2\sin^2x$
so, $$\frac{2x\sin x+\cos 2x-1}{2x^2}=\frac{2x\sin x-2\sin^2 x}{2x^2}=\frac{\sin x}{x}-\frac{\sin^2 x}{x^2}$$
so, $$\int_0^{\infty}\frac{2x\sin x+\cos 2x-1}{2x^2}dx=\int_0^{\infty}\frac{\sin x}{x}dx-\int_0^{\infty}\frac{\sin^2 x}{x^2}dx$$
Now, $$\int_0^{\infty}\frac{\sin x}{x}dx=\pi/2$$
and $$\int_0^{\infty}\frac{\sin^2 x}{x^2}dx=\frac{-\sin^2 x}{x}|_0^{\infty}+\int_0^{\infty}\frac{2\sin x\cos x}{x}dx=0+\int_0^{\infty}\frac{\sin 2x}{2x}d(2x)=\pi/2$$
Thus, $$\int_0^{\infty}\frac{2x\sin x+\cos 2x-1}{2x^2}dx=\pi/2-\pi/2=0$$
For proof of $\int_0^{\infty}\frac{\sin x}{x}dx=\pi/2$
visit Evaluating the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?
|
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|
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}
Thanks.
|
I don't have a rearrangement inequality proof yet, but I really like the following proof I got.
First note that $a+b \ge 2 \sqrt{ab} = 2$ by AM-GM.
$a^2 + 3 = a^2 + 3ab = a(a+3b) \ge a(2 + 2b) = 2ab(a + 1) = 2(a+1)$, and $b^2 + 3 = b^2 + 3ab = b(b+3a) \ge b(2 + 2a) = 2b(1+a)$.
Thus, we have
$$\frac{a}{a^2+3}+\frac{b}{b^2+3} \le \frac{a}{2(a+1)} + \frac{1}{2(a+1)} = \frac{1}{2}$$
and we're through!
|
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|
Compute $ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $ Compute
$$ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $$
I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.
|
In this answer I will make use of a Maclaurin series expansion for the term $\ln^2 (1 - x)$, which I show here to be
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},$$
and the well-known Euler sum of
$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$
several proofs for which can be found here.
From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as
\begin{align*}
\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \int_0^1 x^{n - 1} \ln x \, dx.
\end{align*}
The integral that appears to the right can be readily found by parts. The result is
$$\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2}.$$
Thus
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = -2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}.$$
From properties of harmonic numbers, since
$$H_n = H_{n - 1} + \frac{1}{n},$$
the integral becomes
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{1}{n^4} - 2 \sum_{n = 2}^\infty \frac{H_n}{n^3} = 2 \sum_{n = 1}^\infty \frac{1}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^3}.$$
As
$$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$
we have
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \zeta (4) - \zeta^2 (2) = - \frac{\pi^4}{180} = -\frac{1}{2} \zeta (4),$$
as expected.
|
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|
An intriguing definite integral: $\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$ I need some hints, suggestions for the following integral
$$\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$
Since it's a high school problem, I thought of some variable change, integration by parts, but I can't see yet how to make them work. I don't know where I should start from. Thanks!
|
Let $R\sin A=x^2-16$ and $R\cos A=8x\implies R=x^2+16$ and $\cos A=\frac{8x}{x^2+16}$
So, $(x^2-16)\sin x+8x\cos x$
$=(x^2+16)(\cos A \cos x+\sin A\sin x)$
$=(x^2+16)\cos (x-A)$
$=(x^2+16)\cos(x-\cos^{-1}(\frac{8x}{x^2+16}))$
Putting $y=x-\cos^{-1}(\frac{8x}{x^2+16})$, we get $\frac{dy}{dx}=\frac{x^2+8}{x^+16}$
$$\int\frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$
$$=\int\frac{x^2+8}{(x^2-16)\cos(x-\cos^{-1}(\frac{8x}{x^2+16}))} \ dx$$
$$=\int\frac{dy}{\mathrm{cos}y}=\int \sec ydy=\ln|\sec y+ \tan y|+C=\ln\tan (\frac{\pi}{4}+\frac{y}{2})+C$$ where C is the indeterminate constant for indefinite integral.
Now if $2B=\cos^{-1}(\frac{8x}{x^2+16}),\frac{8x}{x^2+16}=\cos2B$
$=\frac{1-\tan^2B}{1+\tan^2B}\implies \tan^2B=(\frac{4+x}{4-x})^2$
$y=x-2\tan^{-1}(\frac{4+x}{4-x})=x-2(\frac{\pi}{4}+\tan^{-1}(\frac{x}{4})) $
When $x=\frac{\pi}{2}, y=-2\tan^{-1}\frac{\pi}{8}$
When $x=\frac{\pi}{4}, y=-\frac{\pi}{4}-2\tan^{-1}\frac{\pi}{16}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Half Range Sine Series Question:
It is known that $f(x)=(x−4)^2$ for all $x\in [0,4]$.
Compute the half range sine series expansion for $f(x)$.
My answer :
Half range series: $p=8$, $l=4$, $a_0=a_n=0$.
$$b_n=\frac{2}{L}\int_{0}^{L}f(x)\sin\left(\frac{n\pi x}L\right)d(x)=\frac{2}{4}\int_{0}^{4}(x-4)^2\sin\left(\frac{n\pi x}4\right)d(x)$$
Partial Differentiation
Let $u=(x-4)^2$; $du=2(x-4)dx$; $v=\frac{-4}{n\pi}\cos(\frac{n\pi x}4)$
\begin{align}
b_n&=\frac{1}{2}[\frac{-4}{n\pi}cos(\frac{n\pi x}4)(x-4)^2+\frac{8}{n\pi}\int(x-4)\cos(\frac{n\pi x}4)d(x)]|^4_0\\
&=\frac{1}{2}[\frac{-4}{n\pi}\cos(\frac{n\pi x}4)(x-4)^2+\frac{8}{n\pi}[\frac{4}{n\pi}\sin(\frac{n\pi x}4)(x-4)-\frac{4}{n\pi}\int\sin(\frac{n\pi x}4)d(x)]]|^4_0\\
&=\frac{1}{2}[\frac{-4}{n\pi}\cos(\frac{n\pi x}4)(x-4)^2$+$\frac{8}{n\pi}[\frac{4}{n\pi}\sin(\frac{n\pi x}4)(x-4)-\frac{4}{n\pi}(\frac{-4}{n\pi}\cos\frac{n\pi x}4)]|^4_0\\
&=\frac{1}{2}[\frac{-4}{n\pi}\cos(\frac{n\pi x}4)(x-4)^2$+$\frac{8}{n\pi}(\frac{16}{n^2\pi^2}\cos\frac{n\pi x}4)]|^4_0
\end{align}
we know $cosn\pi=(-1)^n$
$\frac{1}{2}[(0+\frac{128(-1)^n}{n^3\pi^3})-(-\frac{64}{n\pi}+\frac{128}{n^3\pi^3})$
$b_n=\frac{64(-1)^n}{n^3\pi^3}-\frac{64}{n^3\pi^3}+\frac{32}{n\pi}$
My teacher told me my $b_n$ is wrong and it seems that my working also looks like wrong. Can someone please help me to recalculate my $b_n$ please. Just help me once only guys.If I get this right I would able to get full marks.I would be very very grateful to you guys if you guys able to help me. If can suggest me any ideas or simpler way to solve it if you can.Thanks
|
The integral defining your $b_n$ is correct, but the final answer is wrong. It differs from mine by ${64\over n\pi}$.
Here's my work from Mathematica:
Here's graphical verification that we are indeed computing the series correctly (I took the 5th, 10th, and 50th partial sum, respectively):
|
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|
Proving $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}\iff a^{2}-b$ is a square This is an exercise for the book Abstract Algebra by Dummit and Foote
(pg. 530):
Let $F$ be a field of characteristic $\neq2$ . Let $a,b\in F$ with $b$
not a square in $F$. Prove $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$ for
some $m,n\in F$ iff $a^{2}-b$ is a square in $F$.
I am having problem proving this claim, I tried to assume $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$
and I naturally squared both sides, to try and get $a^{2}$ I squared
both sides again and then reduced $2b$ from both sides and rearranged
to get $$a^{2}-b=(m+n+2\sqrt{mn})^{2}-2\sqrt{b}(a+\sqrt{b})$$ but
I don't see how I can use it.
Can someone please help me prove this claim ?
|
Let $F$ be a field of charcteristic different from 2. Let $a$ and $b$ be elements of the field $F$ with $b$ not a square in F. Prove that a necessary and sufficient condition for $\sqrt{a+\sqrt{b}}={\sqrt{m}+\sqrt{n}}$ for $m,n\in F$ is that $a^2-b$ is a square in $F.$
Solution.
$\Rightarrow:$ Suppose that $a^2-b$ is a square in $F$. Then $\sqrt{a^2-b}\in F.$ Let
$$m= \frac{a+\sqrt{a^2-b}}{2}$$ and $$n= \frac{a-\sqrt{a^2-b}}{2}.$$ Then $n,m\in F$ because $\textrm{char}\, F\neq 0.$
$\Leftarrow:$ Now
$$m =\frac{a+\sqrt{a^2-b}}{2} = \frac{(a+\sqrt{b})+2\sqrt{a^2-b}+(a-\sqrt{b})}{4} = \left( \frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2} \right)^2,$$ this means that $$\sqrt{m}=\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}.$$
Also
$$n =\frac{a-\sqrt{a^2-b}}{2} = \frac{(a+\sqrt{b})-2\sqrt{a^2-b}+(a-\sqrt{b})}{4} = \left( \frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2} \right)^2,$$ this means that $$\sqrt{n}=\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2}.$$
Thus $$\sqrt{m}+\sqrt{n} =\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}+\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2} = \sqrt{a+\sqrt{b}}.$$
|
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|
Give algebraic and geometric descriptions of the $\operatorname{Span} \{ a_1, a_2, a_3, a_4 \}$ Give algebraic and geometric descriptions of $\operatorname{Span} \{ a_1, a_2, a_3, a_4 \}$ where
$a_1 = (1, -1, -2), a_2 = (3, -3, -1), a_3 = (2, -2, -4), a_4 = (2, -2, 1)$
So far, I have:
$$
\begin{matrix}
\;\;\,1 & \;\;\,3 & \;\;\,2 & \;\;\,2 \\
-1 &-3 &-2 &-2\\
-2 & -1 &-4 & \;\;\,1 \\
\end{matrix}
$$
Though, I feel like I'm missing a column. What should this system be equal to?
|
$\begin{pmatrix}
1 & 3 & 2 & 2\\
-1& -3& -2& -2\\
-2& -2& 4& 1
\end{pmatrix}$
only use elementary row operation,we can get
$
\begin{pmatrix}
1 &0 &2 &-1 \\
0&1 & 0 &1 \\
0&0 &0 &0
\end{pmatrix}$
then,$a_1=2a_3$,and $a_4=-a_1+a_2$
$\operatorname{Span} \{ a_1, a_2, a_3, a_4 \}=\operatorname{Span} \{ a_1, a_2\}$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simple analytic geometry question I need help with Give the equation of a circle with the center $ (a,0) $ which is tangent to the line $ y = x $
I now have $ (x-a)^2 + y^2 = r^2 $ but I don't know how to continue.. please help!
|
Let's calculate the intersection of $y=x$ and the circle.
So, $(x-a)^2+x^2=r^2$ as $y=x$,
$\implies 2x^2-2ax+a^2-r^2=0$
As $y=x$ is a tangent of the circle, the roots of the above equation must be same, so that the two points of intersection coincide.
So, $(-2a)^2=4\cdot 2\cdot (a^2-r^2)$ (as the discriminant$(B^2-4AC)$ of $Ax^2+Bx+C=0$ must be $0$ for the roots to be equal)
$\implies r^2=\frac{a^2}{2}$
So, the equation of the circle :$$(x-a)^2+y^2=\frac{a^2}{2}$$
Observe that the value of $x$ is $\frac{2a}{2\cdot 2}=\frac a 2$ (as the equal roots of $Ax^2+Bx+C=0$ are $-\frac{B}{2A}$), which we did not need here.
The relationship between $r,a$ can attained at least in following 3 ways:
Differentiating, $(x-a)^2+y^2=r^2$ wrt $x$, $\frac{dy}{dx}=\frac{a-x}{y}$
(1)Observe that at the point of intersection the gradient is $1$.
So, $\frac{a-x}{y}=1\implies x+y=a$ , but $y=x$, So, $x=y=\frac{a}{2}$
$\implies r^2=(\frac{a}{2}-a)^2+(\frac{a}{2})^2=\frac{a^2}{2}$
(2)
$$\left(\frac{x-a}{r}\right)^2+\left(\frac{y}{r}\right)^2=1$$
So, the parametric equation of the circle $(a+r\cos t,r\sin t) $.
$a+r\cos t=r\sin t\implies a=r(\sin t-\cos t)\implies a^2=r^2(1-\sin2t)$
$\left(\frac{dy}{dx}\right)_t=-\cot t$
So, $-\cot t=1$ as at the point of intersection the gradient is $1$,
$\implies \tan t=-1\implies$
$$ \sin2t=\frac{2\tan t}{1+\tan^2t}=-1$$
$\implies a^2=r^2(1-(-1))=2r^2$
(3)Let the point of intersection be $(b,b)$.
The gradient of $y=x$ is 1.
The gradient of the line joining $(a,o), (b,b)$ is $\frac{b-0}{b-a}=\frac{b}{b-a}$
These two lines are perpendicular, so , $1\cdot \frac{b}{b-a}=-1\implies b=a-b\implies b=\frac a2$
So, $r^2=(\frac{a}{2}-a)^2+(\frac{a}{2})^2=\frac{a^2}{2}$
|
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|
Limit of the difference quotient of $f(x) = \frac{2}{x^2}$, as $x\rightarrow x_0$ Could someone please show me how to derive the limit of the difference quotient of $f(x) = \frac{2}{x^2}$, as $x\rightarrow x_0$
The difference quotient is just the expression: $(f(x+h)-f(x))/h$
So, it's easy to get an expression for this, but a little tricky to eliminate the 'h' from the denominator..
Would appreciate it if someone can show me how.
|
The trick is simply to add the two fractions together.
$$\lim_{h \to 0} \frac{\frac{2}{(x + h)^2} - \frac{2}{x^2}}{h} = \lim_{h \to 0} \frac{\frac{2[x^2 - (x+h)^2]}{x^2(x + h)^2}}{h} = \lim_{h \to 0} \frac{\frac{2[-2hx - h^2]}{x^2(x + h)^2}}{h} = \lim_{h \to 0} \frac{2[-2x - h]}{x^2(x + h)^2} = \frac{2(-2x)}{x^4} = \frac{-4}{x^3}$$
|
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|
how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
|
Let $\theta=\frac\pi{10}$ and $\tan\theta=x$. Then $5\theta=\frac\pi2$ so
\begin{align}
\tan4\theta&=\frac1x\tag{1}
\end{align}
By twice using the double-angle tan formula,
\begin{align*}
\tan2\theta&=\frac{2x}{1-x^2}\\
\tan4\theta&=\frac{2(\frac{2x}{1-x^2})}{1-(\frac{2x}{1-x^2})^2}\\
\frac1x&=\frac{4x(1-x^2)}{(1-x^2)^2-(2x)^2}\tag{by (1)}\\
(1-x^2)^2-4x^2&=x\times4x(1-x^2)\\
1-2x^2+x^4-4x^2&=4x^2-4x^4\\
5x^4-10x^2+1&=0\\
\implies x^2&=\frac{10\pm\sqrt{10^2-4\times5}}{2\times5}\\
&=1-\frac{\sqrt{5^2-5}}5\\
&=1-\frac{\sqrt{5-1}\sqrt5}5\\
&=1-\frac2{\sqrt5}\\
\implies x&=\sqrt{1-\frac2{\sqrt5}}
\end{align*}
signs being chosen because $0<x<1$ because $0<\theta<45^\circ$.
|
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|
Maclaurin expansion of $\arcsin x$ I'm trying to find the first five terms of the Maclaurin expansion of $\arcsin x$, possibly using the fact that
$$\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}.$$
I can only see that I can interchange differentiation and integration but not sure how to go about this. Thanks!
|
If I was doing this I would start with one of the very common series like $\sin(x)$, $e^x$, etc. and then use substitution. The one that fits best here in my opinion is:
$$
\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + \dotsc
$$
Through substitution, we can obtain:
$$
\frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \frac{35x^8}{128} + \dotsc
$$
Then by integration:
$$
\arcsin(x) = \int^x_0 \frac{1}{\sqrt{1-t^2}}\,dt = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Perfect squares Wonder whether anybody here can provide me with a hint for this one.
Is $c=1$ the only case in which the expression
$(c^2+c-1)(c^2-3(c-1))$
returns a perfect square?
|
Yes, $c=0$ is the only such value. For the proof, it is useful to let $c=x+1$. Then our expression becomes
$$(x^2+3x+1)(x^2-x+1).$$
Note that $x^2-x+1$ is always odd. Any common divisor of $x^2+3x+1$ and $x^2-x+1$ must divide the difference $4x$. But such a common divisor must be odd, so any common divisor must divide $x$. But then it must divide $1$.
Thus $x^2+3x+1$ and $x^2-x+1$ are relatively prime. Since $x^2-x+1$ is always positive, it follows that if their product is a perfect square, each must be a perfect square.
But that can only happen when $x=0$. To prove this, use the fact that for any integer $u$, there is no perfect square strictly between $u^2$ and $(u+1)^2$. Since you asked for a hint, I will, unless you request otherwise, leave out the rest of the argument. It is short.
|
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|
Find the inverse a matrix with trigonometic entries What is the inverse of
\[
\begin{pmatrix}
1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{pmatrix}
\]
Please help me to solve the above problem.
|
Let $A_x :=\left[ \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right]$, and let $R_x :=\left[ \begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array} \right]$ be the matrix of the rotation by angle $x$ in the plane (that is, for all ${\bf v}$ in $\mathbb R^2$, $\ R_x\cdot {\bf v}$ is the rotated version of $\bf v$), we have that
$$A_x = R_x\cdot \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]\ \text{ and }\ (R_x)^{-1} = R_{-x} = \left[ \begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array} \right]\text{, so} $$
$$(A_x)^{-1} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]\cdot R_{-x}$$
So, by easy matrix multiplication, one can verify that the additional $1$ in the additional dimension is not hurting much, ie. the requested inverse is:
$$ \left[ \begin{array}{ccc} 1&0&0\\ 0 &\cos x & \sin x \\ 0 & \sin x & -\cos x \end{array} \right] $$
|
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|
Determine the number of solutions of the equation $n^m = m^n$
Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$
Determine the number of solutions of the equation
$n^m = m^n$
where both m and n are integers.
|
Hint:
Since $m^n=n^m$, take logs and separate the variables:
$$
\frac{\log(m)}{m}=\frac{\log(n)}{n}
$$
This suggests considering the function $f(x)=\frac{\log(x)}{x}$.
$\hspace{2cm}$
Another Approach:
Start by comparing $n^{n+1}$ vs $(n+1)^n$. Divide both by $n^n$, to get $n$ vs $\left(1+\frac1n\right)^n$. We can use the binomial theorem to get
$$
\begin{align}
\left(1+\frac1n\right)^n
&=\sum_{k=0}^\infty\binom{n}{k}\frac1{n^k}\\
&=\sum_{k=0}^\infty\frac1{k!}\frac{n}{n}\frac{n-1}{n}\cdots\frac{n-k+1}{n}\\
&<\sum_{k=0}^\infty\frac1{k!}\\
&<1+\sum_{k=1}^\infty\frac1{2^{k-1}}\\
&=3
\end{align}
$$
Thus, for $n\ge3$, we have
$$
n\ge3>\left(1+\frac1n\right)^n
$$
Multiplying both sides by $n^n$ yields that for $n\ge3$
$$
n^{n+1}>(n+1)^n
$$
Taking the $n(n+1)$ root of both sides gives
$$
n^{1/n}>(n+1)^{1/(n+1)}
$$
So we have determined that $n^{1/n}$ is monotonically decreasing for $n\ge3$. What does that say about $m^n$ and $n^m$ when $n>m\ge3$?
Simpler Proof by Induction
I just noted that $n^{n+1}>(n+1)^n$ for $n\ge3$ can be also proven pretty simply by induction.
Note that $3^4=81>64=4^3$.
Suppose that $n^{n+1}>(n+1)^n$. Divide through by $n^n$ to get
$$
n>\left(1+\frac1n\right)^n
$$
Multiply through by $1+\frac1n$ to get
$$
n+1>\left(1+\frac1n\right)^{n+1}
$$
Since $1+\frac1n>1+\frac1{n+1}$ we get
$$
n+1>\left(1+\frac1{n+1}\right)^{n+1}
$$
Multiply through by $(n+1)^{n+1}$ to get
$$
(n+1)^{n+2}>(n+2)^{n+1}
$$
This finishes the induction.
|
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|
finding $\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$ If:
$$\frac{\cos x}{\cos y}=\frac{1}{2}$$ and $$\frac{\sin x}{\sin y}=3$$
How to find
$$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$$
|
Let $$\frac {\cos x}{1}=\frac {\cos y}{2}=a(say),\implies \cos x=a,\cos y =2a$$
and $$\frac{\sin x }{3}=\frac{\sin y }{1}=b(say),\implies \sin x=3b, \sin y =b$$
So, $$a^2+(3b)^2=1,(2a)^2+b^2=1\implies a^2=\frac 8{35}, b^2=\frac 3{35}$$
$$\implies \cos^2x=a^2=\frac 8{35},\sin^2y=b^2=\frac 3{35}$$
So, $$\frac{\sin 2x}{\sin 2y}=\frac{2\sin x\cos x}{2\sin y \cos y}=\frac{3b\cdot a}{b\cdot 2b}=\frac 3 2$$ as $ab \neq 0$ and
$$\frac{\cos 2x}{\cos 2y}=\frac{2\cos^2x-1}{1-2\sin^2y}=\frac{2\frac 8{35}-1}{1-2\frac 3{35}}=-\frac{19}{29}$$
So, $$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}=\frac 3 2-\frac{19}{29}=\frac{49}{58}$$
|
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|
Regd. Riemann Rearrangement Assume that A is an arbitrary set and there exists a bijection $\phi : B \rightarrow A$ and $x_{\alpha} \in [0,\infty]$, the book says that
$$\Sigma_{\alpha \in A} x_{\alpha}= \Sigma_{\beta\in B} x_{\phi(\beta)}$$
can fail if the series is not absolutely convergent and we are dealing with signed sums.
I fail to see why this is true.
|
If a series is not absolutely convergent, then it only converges because some positive and negative terms cancel in the right way as you continue summing. However, upon rearrangement, you may find that the positive and negative terms do not cancel the same way. Consider the series
$$ 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} + \ldots$$
You can see that this sum does not converge because the partial sums oscillate between $0$ and $1$. However, let us rearrange the terms in the following manner:
$$ 1 - \frac{1}{2} + \left(- \frac{1}{2} + \frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}- \frac{1}{8}\right) + \left( \frac{1}{4} - \frac{1}{8} - \frac{1}{8}\right) + \left( - \frac{1}{8} + \frac{1}{16} + \frac{1}{16} \right) - \ldots$$
If you check this you will find that the terms of this series are the same as the terms of the previous series, but in a different order. However, this second series does converge.
What we can see here, intuitively, is that in the first series, too many positive and negative terms are grouped together, so they cancel each other out in "huge chunks", causing the oscillation between $0$ and $1$. However, in the second series, the positive and negative terms are interleaved in such a way that they cancel each other out in small "small chunks", causing oscillations that exponentially decrease in size.
|
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|
Solve the inequality $|z^2|-|z|\ \Re(z)>0$ Determine the set of complex numbers $z$ such that
$|z|^2-|z|\ \Re(z)>0$
This is my process:
Putting $z=x+iy$, we have $\Re(z)=x$ (real part), $|z^2|=x^2+y^2$, $|z|=\sqrt{x^2+y^2}$, and:
$(x^2+y^2)-x\sqrt{x^2+y^2}>0\Rightarrow x^2\left(1+\frac{y^2}{x^2}\right)-x|x|\sqrt{1+\frac{y^2}{x^2}}>0$
If $x\geq0$ we have $x^2\left(1+\frac{y^2}{x^2}\right)-x^2\sqrt{1+\frac{y^2}{x^2}}>0$, and for $x\neq 0$:
$t-\sqrt{t}>0\ \ \ \ \ \left[t=\left(1+\frac{y^2}{x^2}\right)\right]$ from this: $t>1$ ($t<0$ is not acceptable) i.e. $1+\frac{y^2}{x^2}>1\Rightarrow \frac{y^2}{x^2}>0\Rightarrow y\neq 0$
Then $S_1=\left\{(x,y): x>0, y\neq 0\right\}$
If $x<0$ we have $x^2\left(1+\frac{y^2}{x^2}\right)+x^2\sqrt{1+\frac{y^2}{x^2}}>0$ always verified.
Then $S_2=\left\{(x,y): x<0\right\}$.
What do you think of my proceedings? Is my procedure right? I made a mistake?
thank you very much
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I think from $$(x^2+y^2)-x\sqrt{x^2+y^2}>0$$
$$\to (x^2+y^2) > x\sqrt{x^2+y^2}$$
If $x$ and $y$ are not both $0$, divide by $\sqrt{x^2+y^2}$ on both side
$$\sqrt{x^2+y^2} > x$$
which is always true as long as $y\neq 0$ or $x< 0$.
So the solution would be $z=\{x+iy: y\neq 0\ \text{ or } x <0\}$.
|
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|
Derivative of $\frac{1}{\sqrt{x+5}}$
I'm trying to find the derivative of $\dfrac{1}{\sqrt{x+5}}$
using $\displaystyle \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$
So,
$$\begin{align*}
\lim_{h\to 0} \frac{\dfrac{1}{\sqrt{x+h+5}}-\dfrac{1}{\sqrt{x+5}}}{h} &= \frac{\dfrac{\sqrt{x+5}-\sqrt{x+h+5}}{(\sqrt{x+h+5})(\sqrt{x+5})}}{h}\\\\
&= \frac{\dfrac{x+5-x-h-5}{(\sqrt{x+h+5})(\sqrt{x+5})}}{\dfrac{h}{\sqrt{x+5}}+\dfrac{h}{\sqrt{x+h+5}}}
\end{align*}$$
I do not know if this is correct or not.. please help. I'm stuck.
|
Do you know the derivative of logarithms?
$$
f(x)=\frac{1}{\sqrt{x+5}}\\
Lf(x)=\log f(x)=-\frac{1}{2}\log(x+5)\\
\frac{f'(x)}{f(x)}=\bigg( -\frac{1}{2}\log(x+5) \bigg)'_x\\
f'(x)=f(x)\bigg( -\frac{1}{2}\log(x+5) \bigg)'_x\\
L_1=\lim_{h \to 0}\frac{-\frac{1}{2}\log(x+5+h)+\frac{1}{2}\log(x+5)}{h}=-\frac{1}{2}\lim_{h \to 0}\frac{\frac{1}{2}\log(x+5+h)-\frac{1}{2}\log(x+5)}{h}
$$
Derivative of log function is available, e.g., here. Hence,
$$
L_1=-\frac{1}{2(x+5)}
$$
And, therefore
$$
f'(x)=f(x)L_1=-\frac{1}{2(x+5)^{\frac{3}{2}}}
$$
|
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|
How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
|
Hint: $x=2$ is a solution of the polynomial.
So $x^3-12x+16$ is divisible by $(x-2)$.
And $x^3-12x+16= (x-2)(x^2+2x-8)$.
The roots of $x^3-12x+16=0$ are $x=2$ and the roots of $x^2+2x-8=0$.
|
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|
Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ How to show the following equality?
$$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$
|
This is what I have from an essay I wrote. I don't know if there's a more elementary way (or if it's completely correct).
Consider $f(z) = \dfrac{\cot{\pi z}}{z^2 + k}$. This will have residues at $z = \pm i \sqrt{k}$, and at $z = n$ for $n \in \mathbb{Z}$. At $z = n$, we can compute the residues as
\begin{align*} \textrm{Res}_{z=n} f(z) & = \lim_{z \rightarrow n} \dfrac{(z-n) \cot{\pi z}}{z^2 + k} = \lim_{z \rightarrow n} \dfrac{(z-n)}{(z^2 + k) \tan{\pi z}} \\
& = \lim_{z \rightarrow n} \dfrac{1}{\pi (z^2 + k) \sec^2{\pi z} + 2z \tan{\pi z}} \\
& = \dfrac{1}{\pi (n^2 + k)}. \end{align*}
We can calculate the residues at $z = \pm i \sqrt{k}$: $\displaystyle \textrm{Res}_{z=i\sqrt{k}} f(z) = \lim_{z\rightarrow i\sqrt{k}}\dfrac{(z-i\sqrt{k})\cot{\pi z}}{z^2 + k}$.
This equals:
$\lim_{z \rightarrow i\sqrt{k}} \dfrac{\cot{\pi z}}{z + i\sqrt{k}} = \dfrac{\cot{\pi i\sqrt{k}}}{2i\sqrt{k}}.$
It can be shown that the residue at $z = -i \sqrt{k}$ is the same, because $\cot{\pi z}$ is an odd function. And so the residue contribution from the two poles at $z = \pm i \sqrt{k}$ is
$-\dfrac{\cot{\pi i \sqrt{k}}}{i\sqrt{k}} = -\dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1}$.
Hence, we have
$\displaystyle \int_\gamma f(z) dz = 2\pi i \left(\sum_{n \in \mathbb{Z}} \dfrac{1}{\pi(n^2 +k)} -\dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1}\right)$.
It is tempting for the left-hand side to go to zero, which we can arrange. Take the large square contour centered at the origin with sidelength $2R$. Observe that since
$\cot{z} = i\dfrac{e^{2iz} + 1}{e^{2iz}-1}$,
in the limit as $|z| \geq R \rightarrow \infty$, we will have $|\cot{z}| \rightarrow 1$ since the numerator and denominator of $\cot{z}$ grow equally fast. Moreover, we have that:
$|z^2 + k| \geq |z^2| \geq R^2$,
and so the maximum modulus of $f(z)$ on $\gamma$ is $1/R^2$. By the ML-inequality, we have that
$\left|\displaystyle \int_\gamma f(z) dz\right| \leq 8R \cdot \dfrac{1}{R^2}$.
So as $R \rightarrow \infty$, the integral goes to zero. And thus,
\begin{align*} \sum_{n \in \mathbb{Z}} \dfrac{1}{\pi(n^2 +k)} -\dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1} & = 0\\
\sum_{n \in \mathbb{Z}} \dfrac{1}{\pi(n^2 +k)} & = \dfrac{1}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1} \\
\sum_{n=1}^\infty \dfrac{1}{(n^2 +k)} & = \dfrac{\pi}{2\sqrt{k}} \dfrac{e^{2\pi \sqrt{k}} + 1}{e^{2\pi \sqrt{k}} - 1} - \dfrac{1}{2k}.
\end{align*}
Taking $k = a^2$, this formula becomes
$\dfrac{a \pi \coth{\pi a} -1}{2a^2}$.
Hmm.. not sure about -1 or +1.
|
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|
Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.
|
Note that
\begin{equation*}
4(5a^2 - 4ab - b^2 + 9) - 9(-21a^2-10ab+40a-b^2+8b-12) = (19a+5b-12)(11a+b-12).
\end{equation*}
|
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|
Findind the value of $x^4 + 1/x^4$ when $x = 2+3\sqrt{3}$ This question recently came up in a competitive exam and I have been struggling to find out the easiest way to solve it.
Given that $x = 2+3\sqrt{3}$, what is the value of $x^4 + 1/x^4$.
Could someone provide me with an approach other than the direct approach of multiplying $x$ to itself $4$ times and adding it to its reciprocal.
Thanks in advance!
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You the thing you need to calculate is $x^4$.
You start by isolating the square root $x = 2+3\sqrt{3} \Leftrightarrow x - 2 = 3\sqrt{3}$
Then you square both sides $(x - 2)^2 = (3\sqrt{3})^2 \Leftrightarrow x^2 - 4x + 4 = 27 \Leftrightarrow x^2 - 4x -23 = 0$
Let $P$ be $X^4$
Let $D$ be $X^2 - 4X - 23$
Now, you compute $R$, the remainder of the euclidean division of $P$ by $D$. I get $R = 248X+897$
If you name $Q$ the quotient, you have $P = D\times Q + R$. You don't need it but I get $X^2+4X+39$
But for your given $x$, $D$ evaluates to $0$ so $P$ evaluates to the same value as $R$. $( X^2+4X+39 ) \times (X^2 - 4X - 23)+ (248X+897) = X^4$
So $X^4 = 1393+744 \sqrt 3$
Then $X^4 + \frac{1}{X^4} = \frac{(X^4)^2+1}{X^4}$ and you finally get $\frac{2 \times (1800529+1036392 \sqrt 3))}{(1393+744\sqrt3)}$
|
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|
Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by doing this:
$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$
but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.
Does anyone have a better way of doing this?
|
For $b\not=0$ we have:
$$\begin{align*}\frac{1+a}{b} &= \frac{1+a+b}{1-a+b}\qquad&\iff \\
(1+a)(1-a+b) &= b(1+a+b)\qquad&\iff\\
1-a^2+b(a+1) &= b^2 + b(a+1)\qquad&\iff\\
a^2+b^2&=1
\end{align*}$$
So your equation is an alternate way to characterize $\cos^2 x+\sin^2 x = 1$
|
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find a point on ellipse closest to origin Find the points on the ellipse $2x^2 + 4xy + 5y^2 = 30$ closest and farthest from origin. How to do this problem? I know how to find a closest point if $z = f(x,y)$ is given, however, this is 2 dimensional.
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Another way is use Rotation of axes, to eliminate the $xy$ term.
Here $\cot 2\theta=\frac{2-5}4=-\frac 3 4$
$\frac {\cos 2\theta}3=\frac{\sin 2\theta}{-4}=\frac 1{\pm 5}$ (Using squaring & adding)
If $\cos 2\theta=\frac 3 5,\sin 2\theta=-\frac 4 5$
Using $\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta$, we get $\sin^2\theta=\frac 1 5$ and $\cos^2\theta=\frac 4 5\implies \cos\theta=\pm\frac 2{\sqrt 5},\sin\theta=∓\frac 1{\sqrt 5}$ as $\sin 2\theta<0$, the sign of $\cos\theta,\sin\theta$ will be opposite.
The equation of the ellipse in the rotated axes becomes, $$\frac{x'^2}{30}+\frac{y'^2}{5}=1$$
Now in orthogonal transformation, the distance between two points is one of the invariants and in rotation, the origin remains unchanged and any point of this curve can be P$(\sqrt{30}\cos \phi,\sqrt5 \sin \phi)$.
The distance of P from the origin is $\sqrt{30\cos^2\phi+5 \sin^2 \phi}$
$=\sqrt{30-25\sin^2 \phi}$
The distance will be minimum $(=\sqrt 5)$ if $\sin^2 \phi=1$ i.e., from $(0,\pm\sqrt 5)$ in the rotated co-ordinates.
The distance will be maximum $(=\sqrt {30})$ if $\sin^2 \phi=0$ i.e., from $(\pm\sqrt {30},0)$ in the rotated co-ordinates.
Now use $x=x'\cos\theta-y'\sin\theta$ and $y=x'\sin\theta+y'\cos\theta$.
We have already derived, $\cos\theta=\pm\frac 2{\sqrt 5},\sin\theta=∓\frac 1{\sqrt 5}$
For the minimum distance, if $\cos\theta=\frac 2{\sqrt 5},\sin\theta=-\frac 1{\sqrt 5}$
If $y'=\sqrt 5,x=-\sqrt 5(-\frac 1{\sqrt 5})=1$ and $y=\sqrt5(\frac 2{\sqrt 5})=2$
Similarly, if $y'=-\sqrt 5, (x,y)$ will be $(-1,-2)$
The maximum case can be handled in the same way.
|
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|
Mechanism Behind Dot Product and Least Square Sorry for my ignorance, but I want to know how the mechanism of finding the least square solutions or the closest points in Euclidean space works.
For example:
Find the closest point or points to $b =(−1,2)^T$ that lie on the line $x + y = 0$.
I know the answer is
$$\frac{\left( \begin{matrix} -1 \\ 2 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)}{\left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)} \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -\frac{3}{2} \\ \frac{3}{2} \end{matrix} \right)$$
But what does the dot product between $\left( \begin{matrix} -1 \\ 2\end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ tell you? If I commute it the answer is $-3$, but what exactly is $-3$? Also, the denominator is $2$, again, what exactly is $2$ telling us here?
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The projection onto a vector $x$ is give by $$Pv=\frac{xx^T}{||x||^2}v,$$ where $P \equiv \frac{xx^T}{||x||^2}$ is the projection operator, and $||x||^2=(x,x)$ (or $x*x$ in your notation) is the normalization factor.
|
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If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:
If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is
also an integer, does it implies that $k=1$?
This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$.
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Well notice that both the numerator and denominator are squares, so we may factor
\begin{equation}
\frac{4x^2 - 1}{4x^2 - y^2} = \frac{(2x+1)(2x-1)}{(2x-y)(2x+y)}.
\end{equation}
It is easy to see that $\gcd(2x + 1, 2x - 1) = 1$, so if $k$ is an integer, it must be the case that the denominator each divides one of the linear factors up top. Then there are only two cases for which this is possible: we must have either $y = 1$, or $2x-y = 1$ and $2x+y = (2x + 1)(2x - 1)$, since otherwise, the positive factor below will be larger than either of the numerator factors and thus cannot divide either of them.
Let's consider the case where $2x - y = 1$ and $2x + y = (2x + 1)(2x -1)$. From the first equation, we have $y = 2x + 1$, so
$$2x + y = 4x + 1 = 4x^2 - 1,$$
or
$$4x^2 - 4x - 2 = 0.$$
However, solving that for $x$ shows that $x$ is not an integer, so this case is impossible. Therefore, we are led to the conclusion that $k$ is also an integer when $y = 1$, and in that case, $k = 1$.
Edit: EuYu pointed out some details that I overlooked. Refer to the comments for the discussion.
|
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Number Theory: Prime modulus and Wolstenholme's congruence I apologize if my title is inappropriate, but I couldn't think of a better title name pertaining to this particular problem I have other than listing the section's name of this particular textbook I am using. I've been working on this number theory for a long time and can't figure out this problem. If it helps, the problem is found in Niven's Chapter 2 section 7, which talks about Wolstenholme's Congruence. It is the section after Hensel's Lemma.
Problem: Assume the prime $p \geq 3$. Let $\frac{a}{(p-1)!} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \cdots - \frac{1}{p-1}$. Prove that $a \equiv (2 - 2^p)/p \mod p$.
Here's my attempt at this problem. I said that it is equivalent in proving that $ap \equiv (2 - 2^p) \mod p^2$. However, this is equivalent in proving that ($2^p - 2) + ap \equiv 0 \mod p^2$. By Fermat's Little Theorem, we know that $(2^p - 2) + ap \equiv 0 \mod p$ holds. From here, I said that let $f(x) = x^p - x + ap$. Note that $f(x) \equiv 0 \mod p$, for any $x \in (\mathbb{Z}/p\mathbb{Z})$. But $f'(x) = px^{p-1} - 1 \equiv -1 \mod p$. So since $f(2) \equiv 0 \mod p$ but $f'(2) \equiv -1 \mod p$, by Hensel Lemma, there exists a unique $t$ in $(\mathbb{Z}/p\mathbb{Z})$ such that $f(2+tp) \equiv 0 \mod p^2$.
$f(2+tp) = (2+tp)^p - (2+tp) + ap \equiv 0 \mod p^2$. Equivalently, we have $-f(2+tp) = (2+tp) - (2+tp)^p - ap \equiv 0 \mod p$. Using the binomial theorem, we have that $(2+tp)^p = \displaystyle\sum_{0 \leq i \leq p}{p \choose i}(tp)^i2^{p-i} \equiv 2^p \mod p^2$. So we have that $(2+tp) - 2^p - ap \equiv 0 \mod p^2$. After rearranging the terms in the congruence, we have $(2^p - 2) + ap - tp \equiv 0 \mod p^2$. Now, all I need to show is that $t \equiv 0 \mod p$.
So I started with the expression $\frac{a}{(p-1)!} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \cdots - \frac{1}{p-1} = (\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{p-1}) - 2(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{p-1}) = (\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{p-1}) - (\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{\frac{p-1}{2}}) = \frac{1}{\frac{p+1}{2}} + \frac{1}{\frac{p+3}{2}} + \cdots + \frac{1}{\frac{p+(p-2)}{2}} = \frac{2}{p+1} + \frac{2}{p+3} + \cdots + \frac{2}{p + (p-2)}$. I guess where I am stuck is I could not get $a$ in the form where it can be useful in $(\mathbb{Z}/p\mathbb{Z})$. I have tried rearranging it only to get an expression I believe that does look ugly.
Any hint or any alternative way in how I can view this problem will be useful and appreciated. Thanks!
|
$$2^p=(1+1)^p=\sum_{k=0}^{p}\binom{p}{k}=2+\sum_{k=1}^{p-1}\binom{p}{k}.$$
Now $\binom{p}{k}/p=\frac{(p-1)\dots(p-(k-1))}{1\dots k}\equiv(-1)^{k-1}1/k$ mod $p$ (for $1\leq k\leq p-1$), so we are done.
|
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Let$\frac{\text{dy}}{\text{d}x}=\frac{3y+1}{x^2}$ ,What is $y(x)$? Let$$\frac{\text{dy}}{\text{d}x}=\frac{3y+1}{x^2}$$
What is $y(x)$?
I tried anti-differentiation,but it seems does not work. Is there any tricks to solve the problem?
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$$\begin{align*}\frac{1}{3y+1}dy&=x^{-2}dx\\
\frac{1}{3}\int \frac{3}{3y+1} dy&=\int x^{-2}dx\\
\frac{1}{3}\ln|3y+1|&=-x^{-1}+C\\
3y+1&=Ae^{-\frac{3}{x}}\text{, where }\ln A=3C\\
y&=\frac{1}{3}(Ae^{-\frac{3}{x}}-1)\\
y&=Be^{-\frac{3}{x}}-\frac{1}{3}\text{, where }B=\frac{1}{3}A
\end{align*}$$
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Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
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$$
\begin{eqnarray*}
y &=& x^2(2x+5)^{1/2} - 6 \\
\frac{dy}{dx} &=& \frac{d}{dx} \Big[ x^2(2x+5)^{1/2} \Big] - \frac{d}{dx}\Big[ 6 \Big], \qquad \textrm{Sum/Difference Rule}\\
&=& \frac{d}{dx}\Big[x^2\Big](2x+5)^{1/2} + x^2\frac{d}{dx}\Big[(2x+5)^{1/2}\Big] - 0, \qquad \textrm{Product Rule}\\
&=& 2x(2x+5)^{1/2} + x^2\left( \frac{1}{2}(2x+5)^{-1/2}\cdot 2\right),
\qquad \textrm{Chain Rule}\\
&=& 2x\sqrt{2x+5} + \frac{ x^2}{\sqrt{2x+5}}, \qquad \textrm{simplification.}
\end{eqnarray*}
$$
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For what value of m that the equation $y^2 = x^3 + m$ has no integral solutions? For what value of m does equation $y^2 = x^3 + m$ has no integral solutions?
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None of the solutions posted look right (I don't think this problem admits a solution by just looking modulo some integer, but possibly I'm wrong). Here is a proof.
First, by looking modulo $8$ one deduces we need $x$ to be odd.
Note that $y^2 + 1^2 = (x+2)(x^2 - 2x + 4)$. As the LHS is a sum of two squares, no primes $3 \pmod{4}$ divide it. This forces $x \equiv 3 \pmod{4}$ as if $x \equiv 1 \pmod{4}$ then $x+2$ obviously has a prime factor $3 \pmod{4}$. But then $x^2 - 2x + 4 \equiv 3 \pmod{4}$, implying $x^2 - 2x + 4$ has a prime factor $3 \pmod{4}$. But this is a contradiction, thus no $x,y$ can exist to satisfy this equation.
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Numerical method for finding the square-root. I found a picture of Evan O'Dorney's winning project that gained him first place in the Intel Science talent search. He proposed a numerical method to find the square root, that gained him $100,000 USD.
Below are some links of pictures of the poster displaying the method.
*
*Link 1
*Link 2
*Link 3
How does this numerical method work and what is the proof?
His method makes use of Moebius Transformation.
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The iteration to find $\sqrt k$ is
$f(x) = \frac{d x+k}{x+d}$
where $d = \lfloor \sqrt k \rfloor$.
The iterations start with $x = d$.
If $x$ is a fixed point of this,
$x = \frac{d x+k}{x+d}$,
or $x(x+d) = dx + k$
or $x^2 = k$,
so any fixed point must be the square root.
Now wee see if the iteration increases or decreases.
If $y = \frac{d x+k}{x+d}$,
$$y - x = \frac{d x+k}{x+d} - x
= \frac{d x+k - x(x+d)}{x+d}
= \frac{k - x^2}{x+d}
$$
so if $x^2 < k$, $y > x$
and if $x^2 > k$, $y < x$.
Also,
proceeding like analyses of Newton's method,
$y^2-k = \frac{(d x+k)^2}{(x+d)^2} - k
= \frac{d^2 x^2 +2 d x k + k^2 - k(x+d)^2}{(x+d)^2}
= \frac{d^2 x^2 +2 d x k + k^2 - k(x^2 + 2dx +d^2)}{(x+d)^2}
= \frac{d^2 x^2 +2 d x k + k^2 - kx^2 - 2dkx -kd^2)}{(x+d)^2}
= \frac{d^2 x^2 + k^2 - kx^2 -kd^2)}{(x+d)^2}
= \frac{d^2 (x^2-k) + k^2 - kx^2)}{(x+d)^2}
= \frac{d^2 (x^2-k) - k(x^2-k))}{(x+d)^2}
= \frac{(d^2-k) (x^2-k)}{(x+d)^2}
= (x^2-k)\frac{d^2-k}{(x+d)^2}
$.
Since $d = \lfloor \sqrt k \rfloor$,
$d < \sqrt k < d+1$
or $d^2 < k < d^2 + 2d +1$
or $-2d - 1 < d^2 - k < 0$,
so $|d^2-k| < 2d+1$.
Using this,
$|y^2-k|
< |x^2-k|\frac{2d+1}{(x+d)^2}|
= |x^2-k|\frac{2d+1}{x^2+2dx+d^2}
$,
so $|y^2-k|< |x^2-k|$,
and each iteration gets closer to the square root.
Since the starting iterate is $d$,
all following iterates exceed $d$
so
$|y^2-k|
< |x^2-k|\frac{2d+1}{(d+d)^2}|
< |x^2-k|\frac{2d+1}{4d^2}|
< |x^2-k|\frac{1+1/(2d)}{2d}|
\le 3|x^2-k|/4$
since $d \ge 1$.
This show that the iteration converges.
However, this does not show that it converges quadratically like Newton's,
only that it converges linearly.
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Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
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By the Cauchy-Schwarz Inequality: $$\frac{a^4}{a^2+ab} + \frac{b^4}{b^2 + bc} + \frac{c^4}{c^2 + ac} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc}$$
Now, I claim that $$\frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc} \ge \frac{ab+ac+bc}{2}$$
Expanding, it suffices to show: $$2a^4 + 2b^4 + 2c^4 + 4a^2b^2 + 4a^2c^2 + 4b^2c^2 \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b + 3a^2bc + 3b^2ac + 3c^2ab + a^2b^2 + a^2c^2 + b^2c^2$$
Using the inequality $\frac{3}{4}a^4 + \frac{1}{4}b^4 \ge a^3b$ which is a consequence of AM-GM, we can show: $$2a^4 + 2b^4 + 2c^4 \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$
Thus it suffices to show: $$4a^2b^2 + 4a^2c^2 + 4b^2c^2 \ge 3a^2bc + 3b^2ac + 3c^2ab + a^2b^2 + a^2c^2 + b^2c^2$$
Which is simply just: $$3a^2b^2 + 3a^2c^2 + 3b^2c^2 \ge 3a^2bc + 3b^2ac + 3c^2ab$$
By AM-GM one has $\frac{1}{2}a^2b^2 + \frac{1}{2}a^2c^2 \ge a^2bc$. From applying this a bunch one easily gets the desired result.
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