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Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1}$ and so on? Like where does the $n-3$ come from or the $n-1$.
|
\begin{align}
\sum_{n=5}^\infty \frac{8}{n^2 -1}
&=\sum_{n=5}^\infty \frac{8}{(n -1)(n+1)}\\
&=\sum_{n=5}^\infty \left(\frac{4}{n-1}-\frac{4}{n+1} \right)\\
&=\left(\frac 44 - \color{red}{\frac 46} \right)
+ \left(\frac 45 - \color{blue}{\frac 47} \right)
+ \left(\color{red}{\frac 46} - \color{green}{\frac 48} \right)
+ \left(\color{blue}{\frac 47} - \frac 49 \right)
+ \left(\color{green}{\frac 48} - \frac{4}{10} \right) \dots \\
& =\frac 44 +\frac{4}{5} \\
&=\frac{9}{5}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1502309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Finding a sequence of elementary matrices So I have this matrix
A = $\begin{bmatrix}2 & 4\\1 & 1\end{bmatrix}$
I am tasked with finding all the elementary matrices such that Ek...E2E1A = I. Use this sequence to write both A and A-1 as products of elementary matrices/
I ended up getting four elementary matrices
E1 = $\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
E2 = $\begin{bmatrix}1 & 0\\-2 & 1\end{bmatrix}$
E3 = $\begin{bmatrix}1 & 0\\0 & 1/2\end{bmatrix}$
E4 = $\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}$
When I invert all the elementary matrices and multiply them it equals A but when I multiple All the elementary matrices by A I do not get the identity matrix
|
A matrix is elementary if it differs from the identity matrix by a single elementary row or column operation.
See for example, Wolfram MathWorld Elementary Matrix
The Gauss-Jordan reduction process is expressed as a sequence of elementary operations:
$$
%
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}
\end{array}
\right]
%
\qquad \Rightarrow \qquad
%
\left[
\begin{array}{c|c}
\mathbf{E_{A}} & \mathbf{R}
\end{array}
\right]
$$
First form the augmented matrix
$$
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}
\end{array}
\right]
=
\left[
\begin{array}{cc|cc}
2 & 4 & 1 & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
$$
Normalize row 1:
$$
\left[
\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cc}
2 & 4 & 1 & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|cc}
1 & 2 & \frac{1}{2} & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
$$
Clear column 1
$$
\left[
\begin{array}{rc}
1 & 0 \\
-1 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cc}
1 & 2 & \frac{1}{2} & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cr|rc}
1 & 2 & \frac{1}{2} & 0 \\
0 & -1 & -\frac{1}{2} & 1 \\
\end{array}
\right]
$$
Normalize row 2
$$
\left[
\begin{array}{cr}
1 & 0 \\
0 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cr}
1 & 2 & \frac{1}{2} & 0 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|cr}
1 & 2 & \frac{1}{2} & 0 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
$$
Clear column 2
$$
\left[
\begin{array}{cr}
1 & -2 \\
0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cr}
1 & 2 & \frac{1}{2} & 0 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|rr}
1 & 0 & -\frac{1}{2} & 2 \\
0 & 1 & \frac{1}{2} & -1 \\
\end{array}
\right]
$$
## Products of the elementary matrices
$$
%
\begin{align}
%
\mathbf{E}_{4} \, \mathbf{E}_{3} \, \mathbf{E}_{2} \, \mathbf{E}_{1} \mathbf{A} &= \mathbf{I}_{2} \\[3pt]
%
% four
\left[
\begin{array}{cr}
1 & -2 \\
0 & 1 \\
\end{array}
\right]
% third
\left[
\begin{array}{cr}
1 & 0 \\
0 & -1 \\
\end{array}
\right]
% second
\left[
\begin{array}{rc}
1 & 0 \\
-1 & 1 \\
\end{array}
\right]
% first
\left[
\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{cc|cc}
2 & 4 \\
1 & 1 \\
\end{array}
\right]
&=
\left[
\begin{array}{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right]
%
\end{align}
%
$$
$$
%
\begin{align}
%
\mathbf{E}_{4} \, \mathbf{E}_{3} \, \mathbf{E}_{2} \, \mathbf{E}_{1} \mathbf{I}_{2} &= \mathbf{A}^{-1} \\[3pt]
%
% four
\left[
\begin{array}{cr}
1 & -2 \\
0 & 1 \\
\end{array}
\right]
% third
\left[
\begin{array}{cr}
1 & 0 \\
0 & -1 \\
\end{array}
\right]
% second
\left[
\begin{array}{rc}
1 & 0 \\
-1 & 1 \\
\end{array}
\right]
% first
\left[
\begin{array}{cc}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{array}
\right]
\left[
\begin{array}{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right]
&=
\left[
\begin{array}{rr}
-\frac{1}{2} & 2 \\
\frac{1}{2} & -1 \\
\end{array}
\right]
%
\end{align}
%
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1504043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Complex conjugate and their product I have two complex numbers that are non real, $k$ and $z$. $k$ and $z$ are going to be complex conjugates if and only if the product $(x-k)(x-z)$ is a polynomial with real coefficients.
Here is my answer :
$$k=a+bi\\
z=c+di\\
(x-k)(x-z) = x^2-(k+z)x+kz$$
After that, I'm not sure what to do. I guess that I need to add $k+z$ and multiply $k*z$.
Any hints?
|
Okay, look at it this way.
Let $ax^2 + bx + c$ be a 2nd degree polynomial with real coefficients and no real roots. But it does have complex roots. What are the roots?
Well by the quadratic equation the are
$z = \frac {-b}{2a} + \frac {\sqrt{4ac - b^2}}{2a} = \frac {-b}{2a} + \frac {\sqrt{b^2 - 4ac}}{2a}i$ (as there are no real roots, we know $b^2 > 4ac$) and $k = \frac {-b}{2a} - \frac {\sqrt{b^2 - 4ac}}{2a}i$
Notice $z$ and $k$ are conjugates; Their real parts are equal and their imaginary parts are negatives of each other.
!!!!SO!!!!
IF $(x - z)(x -k) = ax^2 + by + c$ is a polynomial with only real coeficients and no real solutions, then z and k are conjugate complex numbers.
WE ARE HALF-WAY DONE!
!!!!NOW!!!! let's suppose z and k are conjugates. In other words, if $z = a + bi$ and $k = a - bi$, !!!THEN!!! z and k are the roots of the polynomial $(x -z)(x -k$. That polynomial is $(x - z)(x - k) = x^2 - (z + k)x + zk = x^2 - [(a + bi) + (a - bi)]x + (a + bi)(a - bi) = x^2 - 2ax + (a^2 - b^2)$ has coefficients, $1, 2a,$ and $(a^2 - b^2)$. These coefficients are all real numbers.
!!!SO!!!! IF x and z are conjugates THEN they are roots of a polynomial with real coefficients.
Now we have completed everything.
$z,$ and $k$ are complex roots to a second degree polynomial that has only real coefficients IF AND ONLY IF $z$ and $k$ are conjugates.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1507050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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|
Mistake in my NLP using Lagrange Multipliers?
I have the following NLP
\begin{matrix}\text{Minimize:} &x^2+y^2+z^2 \\
\text{Subject To:} & x+2y+z-1=0 \\ &2x -y -3z-4=0\end{matrix}
I need to solve this using the Lagrange Multiplier Method.
My attempt:
$$L(x,y,z,\lambda_1, \lambda_2) = x^2+y^2 +z^2 +\lambda_1(x+y+z-1) + \lambda_2(2x -y-3z-4)$$
Now we need to solve the following system of equations \begin{align}\nabla L(x,y,z,\lambda_1,\lambda_2) &= \bar{0} \\\therefore\begin{bmatrix}2x+\lambda_1 + 2\lambda_2 \\ 2y+2\lambda_1-\lambda_2 \\2z+\lambda_1-3\lambda_2 \\ x +2y+z-1 \\ 2x-y-3z-4\end{bmatrix}&=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}\end{align}
Now, I know that you can use matrix methods in order to solve this, but I tried to do it manually as follows:
From the first three lines we get \begin{align}x &=\frac{-\lambda_1-2\lambda_2}{2} \\ y &= \frac{-2\lambda_1 + \lambda_2}{2}\\ z &=\frac{-\lambda_1 + 3\lambda_2}{2}\end{align}
Now substituting this into the fourth equation \begin{align}\bigg(\frac{-\lambda_1-2\lambda_2}{2}\bigg)+2\bigg(\frac{-2\lambda_1 + \lambda_2}{2}\bigg) + \bigg(\frac{-\lambda_1 + 3\lambda_2}{2}\bigg)-1 &=0 \\ \therefore -6\lambda_1 + 3\lambda_2 -2 &=0 \\ \therefore \lambda_2 &= \frac{2+6\lambda_1}{3}\end{align}
Substituting $x,y,z$ into our fifth equation gives
\begin{align}2\bigg(\frac{-\lambda_1-2\lambda_2}{2}\bigg) - \bigg(\frac{-2\lambda_1 + \lambda_2}{2}\bigg) -3\bigg(\frac{-\lambda_1 + 3\lambda_2}{2}\bigg) -4 &=0 \\ \therefore 6\lambda_1 -23\lambda_2 -8 &=0\end{align}
Substituting for $\lambda_2$ as calculated earlier this becomes \begin{align}-40\lambda_1 - \frac{70}{3} &=0 \\ \therefore \lambda_1 &= -\frac{7}{12}\end{align}
Hence we also get that $$\lambda_2 = -\frac{1}{2}$$
Which we can then use to get our final answers of
$$(x,y,z)= \bigg(\frac{19}{24},\frac{1}{3},-\frac{11}{24}\bigg) $$
However, the problem is, when I substitute these values back into my original constraints, the first one is satisfied perfectly, however, the second equality contraint fails to hold. I cannot seem to think where I went wrong?
|
The right values (solving the system by row reduction) are
$$
x=\frac{16}{15},\ y=\frac13\, z=-\frac{11}{15},\ \lambda_1=-\frac{52}{75},\ \lambda_2=-\frac{18}{25}.
$$
Plugging the values you can check that you made a mistake in your second equation for the lambdas. It should have been
$$
3\lambda_1-14\lambda_2-8=0.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1508506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Algebra with complex numbers Let $m$ be the minimum value of $|z|$, where $z$ is a complex number such that
$$ |z-3i | + |z-4 | = 5. $$
Then $m$ can be written in the form $\dfrac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find the value of $a+b$.
I cannot seem to get rid of the absolute values, making it impossible to solve this. Is it possible for someone to give me some help?
Thank you!
|
Given $|z-4|+|z-3i|=5\;,$ Now Put $z=x+iy\;,$ We get $$\displaystyle \sqrt{(x-4)^2+y^2}+\sqrt{x^2+(y-3)^2}=5$$
Now Using Minkowski inequality, We get
$$\sqrt{(4-x)^2+y^2}+\sqrt{x^2+(3-y)^2}\geq \sqrt{[(4-x)-x]^2+[y-(3-y)]^2}=5$$
And equality Hold when $$\frac{4-x}{x}=\frac{y}{3-y}\Rightarrow 12-3x-4y+xy=xy$$
So we get $3x+4y=12$
So here we have equality condition is hold.
Now Using Cauchy Schwartz Inequality
$$(3^2+4^2)\cdot (x^2+y^2)\geq (3x+4y)^2\Rightarrow x^2+y^2\geq \frac{12^2}{5^2}\Rightarrow \sqrt{x^2+y^2}\geq \frac{12}{5}$$
So we get $$\min (z) = \min (\sqrt{x^2+y^2}) = \frac{12}{5}=\frac{a}{b}$$
So We get $$a+b = 12+5 = 17$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1508904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solving the equation $z^7=-1$
Solve the equation $z^7=-1$
My attempt:
$$z=x+yi$$
$$(x+yi)^7+1=0$$
$$(x^2+2yi-y^2)^3(x+yi)+1=0$$
but now it's start to look ugly.
I'm sure that there is a simple way
|
We can say that $$z^7=-1$$
or,$$z^7=\cos\pi+i\sin\pi$$
or,$$z^7=\cos(4n+2)\pi+i\sin(4n+2)\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
or,$$z=\left[\cos(4n+2)\pi+i\sin(4n+2)\pi\right]^{\frac{1}{7}} \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
or,$$z=\cos(\frac{4n+2}{7})\pi+i\sin(\frac{4n+2}{7})\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1511178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.
|
We have
$$9^{2n-2} = (7+2)^{2n-2} = \sum_{k=0}^{2n-2} \dbinom{2n-2}k 7^k2^{2n-2-k} = 2^{2n-2} + 7M$$
Hence, we have
\begin{align}
4^n+10 \cdot 9^{2n-2} & = 4^n + 10 \cdot (2^{2n-2}+7M) = 4^n+10\cdot 4^{n-1} + 70M = 4^{n-1}(4+10)+70M\\
& = 14(5M+4^{n-1})
\end{align}
Hence, in fact we have that $14$ divides $4^n+10 \cdot 9^{2n-2}$ for all $n \in \mathbb{Z}^+$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1512443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluating definite integral - Let's say we want to evaluate $$ \int_0^{2\pi} \frac{1}{a^2\cos^2x+b^2\sin^2x}dx$$
With substitution, one obtains $$ \frac{1}{ab} \arctan\left(\frac ba \tan x\right) $$
as antiderivate. For more details on how to do this, see this question.
Now my question is, why do I receive $0$ if I insert $0$ and $2\pi$ as integration bounds ? This obviously can't be true since the integrand is always positive.
What do I oversee ?
|
Notice, the property of the definite integral $\int_{0}^{2a}f(x)\ dx=2\int_{0}^{a}f(x)\ dx\ \ \ \ \ \ \forall \ \ \ f(2a-x)=f(x)$
$$\int_{0}^{2\pi} \frac{1}{a^2\cos^2 x+b^2\sin^2 x}\ dx$$$$=2\int_{0}^{\pi} \frac{1}{a^2\cos^2 x+b^2\sin^2 x}\ dx$$ $$=4\int_{0}^{\pi/2} \frac{1}{a^2\cos^2 x+b^2\sin^2 x}\ dx$$
$$=\frac{4}{b^2}\int_{0}^{\pi/2} \frac{\sec^2 x}{\left(\frac{a}{b}\right)^2+\tan^2 x}\ dx$$$$=\frac{4}{b^2}\int_{0}^{\pi/2} \frac{d(\tan x)}{\left(\frac{a}{b}\right)^2+\tan^2 x}$$
$$=\frac{4}{b^2}\frac{b}{a}\left(\tan^{-1}\left(\frac{bx}{a}\right)\right)_{0}^{\pi/2}=\frac{4}{ab}\tan^{-1}\left(\frac{\pi b}{2a}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1514980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to integrate a 4th power of sine and cosine? I'm having some trouble figuring out the right substitutions to make to integrate
$$\int \sin^4(\theta)d\theta$$
and
$$\int \cos^4(\theta)d\theta$$
Any hints or suggestions are welcome.
Thanks,
|
Notice that you can write $\sin^4(x)$ as follows:
\begin{align}
\sin^4(x) = &\ (\sin^2(x))^2 = \Big(\frac{1 - \cos(2x)}{2}\Big)^2 \\
= &\ \frac{1 -2\cos(2x) + \cos^2(2x)}{4} \\
= &\ \frac{1}{4} - \frac{\cos(2x)}{2} + \frac{1 + \cos(4x)}{8}.
\end{align}
For $\cos^4(x)$ we procede as above:
\begin{align}
\cos^4(x) = &\ (\cos^2(x))^2 = \Big(\frac{1 + \cos(2x)}{2}\Big)^2 \\
= &\ \frac{1 + 2\cos(2x) + \cos^2(2x)}{4} \\
= &\ \frac{1}{4} + \frac{\cos(2x)}{2} + \frac{1 + \cos(4x)}{8}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1520224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(c-b)(c+b)$$
Which is at least a product of linear factors.
However I am now stuck as to how to proceed.
Also I would prefer the $(b^2-a^2)$ and the $(c^2-b^2)$ factors of my second line of working to be in the form $(a^2-b^2)$ and $(b^2-c^2)$, for 'neatness' , if possible
|
Seeing it as a polynomial of one variable might help.
$$\begin{align}&-a^3b+a^3c+ab^3-ac^3-b^3c+bc^3\\&=(c-b)a^3+(b^3-c^3)a+bc^3-b^3c\\&=(c-b)a^3+(b-c)(b^2+bc+c^2)a+bc(c-b)(c+b)\\&=(c-b)(a^3-(b^2+bc+c^2)a+bc(c+b))\\&=(c-b)((-a+c)b^2+(-ac+c^2)b+a^3-ac^2)\\&=(c-b)((c-a)b^2+c(c-a)b+a(a-c)(a+c))\\&=(c-b)(c-a)(b^2+bc-a(a+c))\\&=(c-b)(c-a)((b-a)(b+a)+c(b-a))\\&=(c-b)(c-a)(b-a)(a+b+c)\\&=(a-b)(b-c)(c-a)(a+b+c)\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How many ways can $3$ regular polygons meet at a vertex? This is equivalent to the positive integer solutions to $$\frac{a-2}{a} + \frac{b-2}{b} + \frac{c-2}{c} = 2$$ with $3 \le a \le b \le c$.
Small solutions like $(6, 6, 6)$ and $(4, 8, 8)$ can be guessed, but other solutions such as $(4, 5, 20)$ exist.
|
Well, basically ${1\over a}+{1\over b} +{1\over c}={1\over2}$ is what you want.
Let $a\geq b\geq c$ we know since ${({1\over2})\over3}={1\over6}$, ${1\over c}\geq{1\over6}$ and hence $c\leq 6$.
This leaves us four possibilities.
(1) $c=3$ then ${1\over a}+{1\over b}={1\over 6}\implies ab=6a+6b\implies (a-6)(b-6)=36$
(2) $c=4$ then ${1\over a}+{1\over b}={1\over 4}\implies ab=4a+4b\implies (a-4)(b-4)=16$
(3) $c=5$ then ${1\over a}+{1\over b}={3\over 10}\implies 3ab=10a+10b\implies (3a-10)(3b-10)=100$
(4) $c=6$ then ${1\over a}+{1\over b}={1\over 3}\implies ab=3a+3b\implies (a-3)(b-3)=9$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1523647",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the Integral: $\int\frac{5x+1}{(2x+1)(x-1)}dx$ $$\int\frac{5x+1}{(2x+1)(x-1)}dx$$
$$\frac{A}{(2x+1)}+\frac{B}{(x-1)}$$
$$\frac{A(x-1)+B(2x-1)}{(2x+1)(x-1)}$$
I put $x=1$ to attain B: $5\ (1)+1=A(1-1)+B\ (2(1)+1)$
$$6=0+3B$$
$$B=2$$
However, I am unable to utilize the same method to attain A.
What should I do?
|
You had a mistake in your computations. See the first line below. You have written $B(2x-1)$ instead of $B(2x+1)$.
Consider the following steps
$$\eqalign{
& {{5x + 1} \over {(2x + 1)(x - 1)}} = {A \over {2x + 1}} + {B \over {x - 1}} = {{A\left( {x - 1} \right) + B\left( {2x + 1} \right)} \over {(2x + 1)(x - 1)}} \cr
& 5x + 1 = A\left( {x - 1} \right) + B\left( {2x + 1} \right) \cr
& x = 1\,\,\, \to \,\,\,6 = 3B\,\,\, \to \,\,\,B = 2 \cr
& x = - {1 \over 2}\,\,\,\, \to \,\,\, - {3 \over 2} = - {3 \over 2}A\,\,\, \to \,\,\,A = 1 \cr
& {{5x + 1} \over {(2x + 1)(x - 1)}} = {1 \over {2x + 1}} + {2 \over {x - 1}} \cr} $$
|
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|
Proof: $f(x)=x^2+x-4$ is continuous at $x=2$ Proof: Let $\epsilon>0$ and $\delta =\min\{1,\epsilon/6\}$.Then if $|x-2|<\delta$, $|x-2|<1$, so $|x+3|<6$.
Thus, $$|f(x)-f(c)|= |x^2+x-4-(2^2+2-4)| = |x^2+x-6|=|x-2||x+3|<6\delta \leq \epsilon$$
Doing this problem as practice. Thoughts? comments?
|
If $x \in \Bbb{R}$, then $|f(x) - f(2)| = |x^{2}+x-6| = |x-2||x+3|$; we have $|x-2| < 1$ only if $|x| - 2 \leq |x-2| < 1$, only if $|x| + 3 < 6$, only if $|x+3| \leq |x|+3 < 6$, and only if $|x-2||x+3| < 6|x-2|$; given any $\varepsilon > 0$, we have $6|x-2| < \varepsilon$ if $|x-2| < \varepsilon/6$. Hence we have proved: for every $\varepsilon > 0$, we have $|x-2| < \min \{ 1, \varepsilon/6 \}$ only if $|f(x)-f(2)| < \varepsilon$.
|
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|
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
|
Hint :
$$3^{3n+4}+2^{n+2}=27\times 3^{3n+1}+2\times 2^{n+1}=2\times(3^{3n+1}+2^{n+1})+25\times 3^{3n+1}$$
|
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|
$x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}$. Then what is the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\}$?
If the equation $x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}\;,$ Where $x_{1}<x_{2}<x_{3}$.
Then the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\} = \;,$ Where $\{x\}$ Represent fractional part of $x.$
$\bf{My\; Try::}$ Let $f(x) = x^3-3x+1\;,$ Then Using Second Derivative Test, We get
$f'(x) = 3x^2-3$ and $f''(x) = 6x\;,$ Now for $\max$ and $\min\;,$ Put $f'(x) =0\;,$ So we get
$3x^2-3=0\Rightarrow x = \pm 1.$ Now $f''(1) = 6>0$
So $x=1$ is a Point of $\min$ and $f''(-1) = -6<0$
So $x=-1$ is a Point of $\max.$
And Rough graph of $f(x) = x^3-3x+1$ is Like
So we get $-2<x_{1}<-1$ and $0<x_{2}<1$ and $1<x_{3}<2$
So $\lfloor x_{1}\rfloor = -2$ and $\lfloor x_{2}\rfloor = 0$ and $\lfloor x_{3}\rfloor = 1$
Now I did not Understand How can I calculate value of fractional part
Help me, Thanks
|
Note that
$$\{x\}=x-\lfloor x\rfloor$$
and that
$$x_1+x_2+x_3=0.$$
|
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|
What is the probability of getting exactly 3 two's OR three's when a die is rolled 8 times? What is the probability of getting exactly $3$ two's OR three's when a die is rolled $8$ times?
I know that $P(E) = |E| / |S|$.
I believe that $|S| = 36$, since there are $36$ different combinations when rolling a die. I am not sure how to get the probability of rolling exactly $3$ two's OR three's when rolling $8$ times.
|
Probability of exactly $3$ twos is $\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5$.
Probability of exactly $3$ threes is $\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5$.
Probability of exactly $3$ twos and exactly $3$ threes is $\binom{8}{3}\binom{5}{3}\left(\frac16\right)^6\left(\frac46\right)^2$
Thus, Inclusion-Exclusion says that the probability of exactly $3$ twos or $3$ threes is
$$
\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5+\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5-\binom{8}{3}\binom{5}{3}\left(\frac16\right)^6\left(\frac46\right)^2=\frac{7105}{34992}
$$
|
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|
How to solve the equation $t = \sqrt{x^2 - 1} - x$ for $x$?
Let the equation $t = \sqrt {x^2 - 1} - x$. Find $x$.
So I've tried the following:
$$t^2 = x^2 - 1 -2\sqrt {x^2-1}x + x^2 = 2x^2 - 2\sqrt{x^2-1}x - 1 $$
What should I do next?
|
Note that $$\left(\sqrt{x^2-1}-x\right)\left(\sqrt{x^2-1}+x\right) = -1,$$ so $$\frac{1}{t}=-\sqrt{x^2-1}-x$$ Adding we get $$t+\frac{1}{t}=-2x$$ Or:
$$x=\frac{-1}{2}\left(t+\frac{1}{t}\right)$$
|
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|
How do I complete the proof of proving the $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $ $$\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $$
Proof: Let $\epsilon > 0$
Then, $$ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -\frac { (x+1)^{ 2 } }{ (x+1)^{ 2 } } \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -\frac { x^{ 2 }+2x+1 }{ (x+1)^{ 2 } } \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -2x }{ (x+1)^{ 2 } } \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -2 }{ x^{ 3 }+2x^{ 2 }+x } \right| <\epsilon $$
$$\Longleftrightarrow \frac { 2 }{ \left| x^{ 3 }+2x^{ 2 }+x \right| } <\epsilon $$
Now, we will calculate the lower bound on $x$ to force $ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $ to hold true.
Without loss of generality, assume that $x>0$, then we get: $$\frac { 2 }{ x^{ 3 }+2x^{ 2 }+x } <\epsilon $$
$$\Longrightarrow 2<\epsilon (x^{ 3 }+2x^{ 2 }+x)$$
$$\Longrightarrow 2<\epsilon x^{ 3 }+\epsilon 2x^{ 2 }+\epsilon x$$
At this point I get stuck. How do I go from here? I can't find a way in which I could manipulate this to get it in the form $x>...$ A hint in the right direction would be beneficial.
|
Rewrite it as
$$
\lim_{x\to\infty}\frac{1+\dfrac{1}{x^2}}{1+\dfrac{2}{x}+\dfrac{1}{x^2}}
=1
$$
For $x>0$, the denominator is $>1$ and it's not restrictive to assume it.
The inequality to be solved becomes then
$$
\left|-\frac{2}{x}\right|<
\varepsilon\left(1+\dfrac{2}{x}+\dfrac{1}{x^2}\right)
$$
which is certainly satisfied if
$$
\frac{2}{x}<\varepsilon
$$
so for
$$
x>\frac{2}{\varepsilon}
$$
|
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|
How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the following function:
$$\lim _{ x\rightarrow \infty }{ \frac { \sqrt { x+1 } }{ \sqrt { x } +1 } =1 } $$
Steps I took:
$$\left| \frac { \sqrt { x+1 } }{ \sqrt { x } +1 } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { \sqrt { x+1 } }{ \sqrt { x } +1 } -\frac { \sqrt { x } +1 }{ \sqrt { x } +1 } \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { \sqrt { x+1 } -\sqrt { x } -1 }{ \sqrt { x } +1 } \right| <\frac { 1 }{ 3 } $$
How can I manipulate the function inside of the absolute value in order to simplify this and find a lower bound $x$ (the M)?
|
Hint for an intuitive solution:
$$\left|\frac{\sqrt{x+1}}{\sqrt{x}+1}-1\right|<\left|\frac{\sqrt{x+1}}{\sqrt{x}}-1\right|=\left|\sqrt{\frac{x+1}{x}}-1\right|=\left|\sqrt{1+\frac{1}{x}}-1\right|.$$
Now by letting that $x\to\infty$ then the result follows.
|
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|
Find the expected area of a randomly chosen triangle. The set of numbers $(x,y)$ are positive natural numbers such that $x+y=n$. 2 points are chosen from this set. What is the expected area of the triangle formed by the origin and the two points?
|
Note that the $x$ coordinate is supported on $[0:n]$. I'll assume that the points are selected uniformly and independently, and you should edit the question in case this is not true.
Suppose the first point is $(X_1, Y_1)$, and the second $(X_2, Y_2)$. It's easy to see that the area is
\begin{align}
A &= \frac{1}{2}|X_1 Y_2 - X_2 Y_1| &\text{(half the absolute value of the cross product)}\\
&= \frac{1}{2}|nX_1 - X_1X_2 - nX_2 + X_1 X_2| &\text{(Using $X_i + Y_i = n$)} \\
&= \frac{n}{2}|X_1 - X_2|
\end{align}
\begin{align}
\mathbb{E}[A] &= \frac{n}{2} \sum_{a = 0}^n \sum_{b = 0}^n \mathbb{P}\{X_1 = a, X_2 = b\} |a - b| \\
&\overset{(a)}= \frac{n}{2(n+1)^2} \sum_{a= 0}^n \left(\sum_{b = a+1}^n b - a + \sum_{b = 0}^a a - b \right) \\
&= \frac{n}{2(n+1)^2} \sum_{a = 0}^n \frac{(n-a)(n-a+1)}{2} + \frac{a(a+1)}{2} \\
&= \frac{n}{2(n+1)^2} \sum_{a = 0}^n a^2 + a\\
&= \frac{n}{2(n+1)^2} \left( \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right) \\
&= \frac{n^2(n+2)}{6(n+1)}
\end{align}
Where equality $(a)$ uses both uniformity and independence of $X_1$ and $X_2$.
|
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|
What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5? What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5?
I think that the only way to solve this would be to applying to the proposition that “the sum/product of congruence classes is equal to the congruence class of the sum/product", however I am unsure how to apply this properly, any suggestions?
|
We have $p=5$ is prime, then by Fermat's little theorem: for every integer $a$ such that $p \nmid a$ we have $$ a^{p-1} \equiv 1 \mod{p}$$
Now for any $x$ integer in $[1,100]$, if $5$ divides $x$ then the remainder is zero, if $5\nmid x$ , then $x^{4} \equiv 1 \mod{5}$, and so as $$y=1^6+ 2^6+3^6+...+99^6+100^6 = 1^4\cdot1^2+2^4\cdot2^2+3^4\cdot3^2+...+99^4\cdot99^2+100^4\cdot100^2 $$
then $$ y \equiv 1^2+ 2^2+3^2+4^2+6^2+...+99^2 $$
Now for any integer $x$, such that $5\nmid x$, we have $x^2 \equiv \pm 1 \mod{5}$. More precisely, if $ x \equiv 2,3 \mod{5}$ then $x^2 \equiv -1 \mod{5}$, and if $ x \equiv 1,4 \mod{5}$ then $x^2 \equiv 1 \mod{5}$. Thus summing the squares of any four consecutive numbers will be congruent to zero $\mod{5}$.
Hence summing the squares from $1$ to $99$, the remainder would be zero.
|
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|
The derivative of $ \ln\left(\frac{x+2}{x^3-1}\right)$ I know it is a simple question, but what would be the next few steps in this equation to find the derivative?
$$f(x)= \ln\left(\frac{x+2}{x^3-1}\right)$$
|
I would NOT use the chain rule before doing this:
\begin{align}
f(x) = \ln\frac{x+2}{x^3-1} & = \ln (x+2) - \ln(x^3-1) \\[10pt]
& = \ln(x+2) - \ln((x-1)(x^2+x+1)) \\[10pt]
& = \ln(x+2) - \ln(x+1) - \ln(x^2+x+1).
\end{align}
|
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|
Find the highest and the lowest value of the function $y=1+\sin x \cos x$
Please dont use the formula $2\sin x \cos x=\frac{1}{2}\sin 2x$. We have't learnt it yet.
The key is to return it into an inequation for example $-1 \le \sin x \le 1$
Then multiply with something to give an inequality
For example find the highest and lowest value of $y=\frac{1}{2}\sin x$
$$-1 \le \sin x \le 1$$
$$-\frac{1}{2} \le\frac{1}{2}\sin x \le \frac{1}{2}$$
The highest value is $\frac{1}{2}$, the lowest is $-\frac{1}{2}$
Could you solve $y=1+\sin x\cos x$ in this way after you have done equivalent changes?
|
We want to maximize/minimize $(\sin x)(\cos x)$, or equivalently $2(\sin x)(\cos x)$. Note that
$$2\sin x\cos x=(\sin^2 x+\cos^2 x)-(\sin x-\cos x)^2=1-(\sin x-\cos x)^2.$$
This attains a maximum when $(\sin x-\cos x)^2$ is as small as possible, namely $0$.
For the minimum, use the same idea, and
$$2\sin x\cos x=(\sin x+\cos x)^2 -(\sin^2 x+\cos^2 x).$$
|
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|
Showing that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$
Show that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$.
My reasoning is as follows:
Since the only ways $X^2$ and $X^3$ can be factored in $K[X^2,X^3]$ are $X^2 = X^2 * 1$ and $X^3=X^3*1$ it follows that $X^2$ and $X^3$ are irreducible.
To show that they are not prime in $K[X^2,X^3]$ I argue that $X^2\mid X^3*X^3$ but $X^2\nmid X^3$ and $X^3\mid X^4*X^2$ but $X^3\nmid X^4$ or $X^3\nmid X^2$.
Is that correct?
|
You can do as Thomas said, by appealing to unique factorization in the larger ring, but here it is simple enough to do it directly. By definition of product, the degrees of the factors must sum to the degree of the product. Now you can prove by induction that $K[X^2,X^3]$ does not have any element with a nonzero coefficient for $X^1$, or alternatively you can prove that $K + X^2 K[X]$ is a ring and hence contains $K[X^2,X^3]$ since it already contains $X^2,X^3$. Then clearly $X^2,X^3$ cannot have any factor of degree $1$, and so you are quite done. Similarly, you can justify that $X^2 \nmid X^3$ and $X^3 \nmid X^4$ because the other factor must have degree $1$.
|
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Alternate approaches to solve this Integral Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$
I have used parts taking first function as Integrand and second function as $1$ we get
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$
now $$\frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)=\frac{1}{2\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)} \times \left(\frac{\frac{1+\sqrt{x}}{-2\sqrt{x}}-\frac{1-\sqrt{x}}{2\sqrt{x}}}{(1+\sqrt{x})^2}\right)=\frac{-1}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})}$$ so
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\int\frac{xdx}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})} $$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int \frac{\sqrt{x}dx}{\sqrt{1-x}(1+\sqrt{x})}$$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}(1-\sqrt{x})dx}{(1-x)^{\frac{3}{2}}}$$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}-\frac{1}{2}\int\frac{xdx}{(1-x)^{\frac{3}{2}}}$$
Now $$\int\frac{xdx}{(1-x)^{\frac{3}{2}}}=\int\frac{(x-1+1)dx}{(1-x)^{\frac{3}{2}}}$$ and if we split is straight forward to compute.
Also $$\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}$$ can be evaluated using substitution $x=sin^2y$
I need any other better approaches to evaluate this integral.
|
$\sqrt x=\cos2t~$ seems like the most natural substitution, since there are well-known trigonometric formulas for $1\pm\cos2t,~$ and it is fairly obvious that $x\in\Big[0,1\Big]$.
|
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Minimize $x^2+y^2$, subject to... (optimal points, KKT conditions, dual theories) I am new to this. I am self learning to get ahead of my next years course and came across this question. I thought it would be a good question to look at due to it touching an many different aspects of optimization.
Minimize $x^2+y^2$
subject to $(x−1)^2 +(y−1)^2\leq 1$,
subject to
$(x−1)^2 +(y+1)^2\leq 1$, where $(x, y)\in\Bbb R^2$.
(i) What are the set of
feasible points for this problem? Using this, find the optimal point.
(ii) Write down the KKT conditions; are these conditions satisfied at
the optimal point? Describe why.
(iii) Write the Lagrange dual
problem, and find the optimal solution to the dual problem. Is this
optimum solution attained? Does the strong duality theorem for convex
programming apply to this scenario?
I have researched the basics of this course, and am trying to challenge myself with this question. Any help would be greatly appreciated.
|
I'm too lazy to use KKT, so I'm providing a non-KKT solution. Maybe, someone else will help with the KKT requirement.
With only the first constraint,
from $(x-1)^2+(y-1)^2\leq 1$, we have by AM-GM that $x^2+y^2+1\leq 2x+2y\leq 2\sqrt{2}\sqrt{x^2+y^2}$. Thus, if $r:=\sqrt{x^2+y^2}$, then $r^2-2\sqrt{2}r+1\leq 0$, which means $\sqrt{2}-1 \leq r\leq \sqrt{2}+1$. Therefore, $3-2\sqrt{2} \leq x^2+y^2\leq 3+2\sqrt{2}$. The LHS equality holds iff $x=y=1-\frac{1}{\sqrt{2}}$. The RHS equality holds iff $x=y=1+\frac{1}{\sqrt{2}}$.
With only the second constraint,
from $(x-1)^2+(y+1)^2\leq 1$, we have by AM-GM that $x^2+y^2+1\leq 2x-2y\leq 2\sqrt{2}\sqrt{x^2+y^2}$. Thus, if $r:=\sqrt{x^2+y^2}$, then $r^2-2\sqrt{2}r+1\leq 0$, which means $\sqrt{2}-1 \leq r\leq \sqrt{2}+1$. Therefore, $3-2\sqrt{2} \leq x^2+y^2\leq 3+2\sqrt{2}$. The LHS equality holds iff $x=-y=1-\frac{1}{\sqrt{2}}$. The RHS equality holds iff $x=-y=1+\frac{1}{\sqrt{2}}$.
With both constraints, $(x-1)^2+(y-1)^2\leq 1$ and $(x-1)^2+(y+1)^2\leq 1$ have only one feasible point $(x,y)=(1,0)$. Thus, $x^2+y^2=1$ under these two constraints.
There are also geometric solutions, if we talk only about one of the constraints. The extreme values of $\sqrt{x^2+y^2}$ are the radii of the circles centered at the origin $(x,y)=(0,0)$ that touch the constraint circle $(x-1)^2+(y-1)^2=1$ or the constraint circle $(x-1)^2+(y+1)^2=1$. (By symmetry, these values are the same for both constraint circles, so it suffices to work on only one constraint circle.) Thus, the point of tangency between a constraint circle and an optimal circle centered at $(0,0)$ must lie on the line connecting the centers of the two circles. The rest is quite easy.
|
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"timestamp": "2023-03-29T00:00:00",
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|
On an exercise with a series: $\sum_{n=1}^{\infty}n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)$ The series in question is the following:
$$\sum_{n=1}^{\infty}n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)$$
I want to study for which $\alpha >0$ does the series converge.
I tried the root test to not much avail, after I tried expanding with taylor the sine function (by the way I can taylor expand a function even if the input are only ration numbers of the form $\frac{1}{n^{\alpha}}$ right?) but I still haven't got anywhere.
We have not done the integral test, is that the way to go?
|
You can indeed do Taylor expansions for $\frac{1}{n^\alpha}$, as long as $\frac{1}{n^\alpha} \xrightarrow[n\to\infty]{} 0$. (That is, for any $\alpha > 0$. Note that the case $\alpha \leq 0$ is straightforward). Doing so, you obtain:
$$
\sin \frac{1}{n^\alpha} = \frac{1}{n^\alpha} + o\left(\frac{1}{n^{2\alpha}}\right)
$$
and
$$
\frac{1}{n^\alpha+1} = \frac{1}{n^\alpha}\frac{1}{1+\frac{1}{n^\alpha}} = \frac{1}{n^\alpha}\left(1-\frac{1}{n^\alpha}+ o\left(\frac{1}{n^{\alpha}}\right)\right)
= \frac{1}{n^\alpha}-\frac{1}{n^{2\alpha}}+ o\left(\frac{1}{n^{2\alpha}}\right)
$$
so
$$
\sin \frac{1}{n^\alpha} - \frac{1}{n^\alpha+1} =
\frac{1}{n^{2\alpha}}+ o\left(\frac{1}{n^{2\alpha}}\right)
$$
and finally
$$
n^2\left(\sin \frac{1}{n^\alpha} - \frac{1}{n^\alpha+1}\right) =
\frac{1}{n^{2\alpha-2}} + o\left(\frac{1}{n^{2\alpha-2}}\right)
$$
Can you conclude based on this?
|
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|
Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\frac{n}{2}\rfloor\\\lfloor\frac{n}{2}\rfloor\end{pmatrix} = F_{(n+1)}$$
Example: ${6\choose0} + {5\choose1} + {4\choose2} + {3\choose3} = 13 = F_{7}$
|
Use induction and a little cheat!
Consider $F_5=\dbinom50+\dbinom41+\dbinom32=1+4+3=8$
Now $\dbinom{n}{k-1}+\dbinom{n}{k}=\dbinom{n+1}{k}$, so for example $\dbinom41+\dbinom42=\dbinom52$.
So adding these like terms gives us the next Fibonacci number, except for the misplaced leading term.
And we have $\dbinom{n}{0}=1$ for all $n\in\mathbb{N}$ so we shift the leading term outwards, which is a bit like cheating really!
There's loads of pictures at Google.
|
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|
probability of getting at-least 2 heads Let assume, I through a fair coin three times, and I want to get at least two heads. I want to find the the probability.
To find that,
At first I evaluated the probability of getting 1 head which is:
$${3\choose1} \left(\frac 12\right)^{1} \times \left(\frac 12\right)^{3-1}= \frac 38 $$
the probability of getting 2 heads:
$${3\choose 2} \left(\frac 12\right)^{2} \times \left(\frac 12\right)^{3-2}= \frac 38 $$
The total probability of getting at least two heads is equal: $\frac 38 +\frac 38 =\frac 68 = \frac 34$
But the problem is that, in a text book they did it in different way and got the result $\left(\frac 12\right)$
|
At least two heads means "two or three heads". Let $X$ denote the number of heads in three tosses. In particular, $X\sim Binom(3, 1/2)$, hence
\begin{align}
P(X\ge 2) = P(X=2) + P(X=3) = \binom{3}{2}\frac{1}{2^3} + \binom{3}{3}\frac{1}{2^3} = \frac{1}{2} .
\end{align}
|
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|
Find all positive integers $n$ such that $2^8+2^{12}+2^n$ is a perfect square Find all positive integers $n$ such that $2^8+2^{12}+2^n$ is a perfect square.
For $n=2$ and $n=11$, $2^8+2^{12}+2^n$ is a perfect square.
How to find a closed form?
|
If $n\in\{1,2,\ldots,7\}$, then $n=2$ is the only solution. Let $n\ge 8$. Then $2^8\left(1+2^4+2^{n-8}\right)$ is a square iff $17+2^{n-8}$ is a square. Let $n-8=m$. You're solving $17+2^m=k^2$ in non-negative integers.
All the solutions are $m=3,5,6,9$. Now, the solutions for $m$ are relatively large, and assuming $m\ge 10$ and using mod $2^{k}$, $k\ge 10$ won't help, because $17$ will always be a quadratic residue.
This is very similar to Ramanujan–Nagell equation, namely $2^n-7=x^2$, which has the solutions $n=3,4,5,7,15$.
This equation has already been asked about on M.SE (see here and here), and the only solution someone could come up with uses Mordell's equations, so I'll use them here. We have three cases:
*
*$m=3t,\, t\ge 0$. Then $17+\left(2^t\right)^3=k^2$. But $17+a^3=b^2$ has $16$ solutions (see http://oeis.org/A081119 and http://oeis.org/A081119/b081119.txt), which can be found with a program or this table:
$$(a,b)=(-2,\pm 3), (-1,\pm 4), (2,\pm 5), (4,\pm 9),$$
$$ (8,\pm 23), (43,\pm 282), (52,\pm 375),
(5234,\pm 378661)$$
This gives $m=3,6,9$.
*$m=3t+1,\, t\ge 0$. Then $2^m\equiv 2\pmod{7}$, so $k^2\equiv 5\pmod{7}$, impossible, because $5$ is not a quadratic residue mod $7$.
*$m=3t+2,\, t\ge 0$. Then $272+\left(2^{t+2}\right)^3=(4k)^2$. But $272+a^3=b^2$ has $2$ solutions (http://oeis.org/A081119/b081119.txt), which are $(a,b)=(8,\pm 28)$. This gives $m=5$.
|
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|
Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
|
Multiplying $x$ to both sides, the identity is equivalent to
$$ \sum_{k=1}^{\infty} \frac{kx^k}{1+x^k} = \sum_{k=1}^{\infty} \frac{(2k-1)x^{2k-1}}{1-x^{2k-1}}. $$
Expanding and rearranging, each series can be written as
\begin{align*}
\sum_{k=1}^{\infty} \frac{kx^k}{1+x^k}
&= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (-1)^{j-1} k x^{jk} = \sum_{n=1}^{\infty} \Bigg( \sum_{d|n} (-1)^{d-1}\frac{n}{d} \Bigg) x^{n} \\
\sum_{k=1}^{\infty} \frac{(2k-1)x^{2k-1}}{1-x^{2k-1}}
&= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (2k-1)x^{j(2k-1)} = \sum_{n=1}^{\infty} \Bigg( \sum_{\substack{d | n \\ d \text{ odd}}} d \Bigg) x^{n}
\end{align*}
So it is sufficient to prove that
$$ \color{blue}{\sum_{d|n} (-1)^{d-1}\frac{n}{d} = \sum_{\substack{d | n \\ d \text{ odd}}} d} \quad \text{for } n = 1, 2, \cdots. \tag{1}$$
To this end, write $n = 2^e m$ for $e \geq 0$ and $m$ is odd. Then
$$ \sum_{d|n} (-1)^{d-1}\frac{n}{d}
= \sum_{d'|m} \underbrace{\sum_{i=0}^{e} (-1)^{2^i d'-1} 2^{e-i}}_{=1} \frac{m}{d'}
= \sum_{d'|m} \frac{m}{d'}
= \sum_{d|m} d
= \sum_{\substack{d | n \\ d \text{ odd}}} d $$
and hence $\text{(1)}$ is proved.
|
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|
Simplify $\sqrt[3]{162x^6y^7}$ The answer is $3x^2 y^2 \sqrt[3]{6y}$
How does $\sqrt[3]{162x^6y^7}$ equal $3x^2 y^2\sqrt[3]{6y}$?
|
$162x^6y^7=(3x^2y^2)^3\cdot6y$, such that
$$\sqrt[3]{162x^6y^7}=\sqrt[3]{(3x^2y^2)^3\cdot6y}=\sqrt[3]{(3x^2y^2)^3}\sqrt[3]{6y}=3x^2y^2\sqrt[3]{6y}$$
|
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|
Integrate. $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How should i solve this integral?
i know that it is the same question like here
Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$
but I've tried solve it for more then 3 hours and i still have no idea ho to solve it. Thank for help.
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
I tried use $$2x+7 = \frac{\sqrt{61}}{\cos u}$$
$$\int\frac{\frac{\sqrt{61}}{\cos u}-\frac72+1}
{(\frac{(\frac{\sqrt{61}}{\cos u})^2-61}{4})^3}dx$$
Now i have
$$\int \frac{61\sin u - 5\sqrt{61}\sin u \cos u}{4\cos u
\frac{226981\tan^6u}{65}
}$$
|
HINT:
$$\int\frac{x+1}{(x^2+7x-3)^3}\space\text{d}x=$$
$$\int\left(\frac{2x+7}{2(x^2+7x-3)^3}-\frac{5}{2(x^2+7x-3)^3}\right)\space\text{d}x=$$
$$\frac{1}{2}\int\frac{2x+7}{(x^2+7x-3)^3}\space\text{d}x-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$
Substitute $u=x^2+7x-3$ and $\text{d}u=(2x+7)\space\text{d}x$:
$$\frac{1}{2}\int\frac{1}{u^3}\space\text{d}u-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$
$$-\frac{1}{4u^2}-\frac{5}{2}\int\frac{1}{(x^2+7x-3)^3}\space\text{d}x=$$
$$-\frac{1}{4u^2}-\frac{5}{2}\int\frac{1}{\left(\left(x+\frac{7}{2}\right)^2-\frac{61}{4}\right)^3}\space\text{d}x=$$
Substitute $s=x+\frac{7}{2}$ and $\text{d}s=\text{d}x$:
$$-\frac{1}{4u^2}-\frac{5}{2}\int\frac{1}{\left(s^2-\frac{61}{4}\right)^3}\space\text{d}s$$
EDIT:
$$\int\frac{1}{(x^2-a)^3}\space\text{d}x=$$
Substitute $x=\sqrt{a}\sec(u)$ and $\text{d}x=\sqrt{a}\tan(u)\sec(u)\space\text{d}u$ so $u=\sec^{-1}\left(\frac{x}{\sqrt{a}}\right)$:
$$\sqrt{a}\int\frac{\cot^4(u)\csc(u)}{a^2}\space\text{d}u=$$
$$\frac{1}{a^{\frac{5}{2}}}\int\cot^4(u)\csc(u)\space\text{d}u=$$
$$\frac{1}{a^{\frac{5}{2}}}\int\csc(u)(\csc^2(u)-1)^2\space\text{d}u=$$
$$\frac{1}{a^{\frac{5}{2}}}\int\left(\csc^5(u)-2\csc^3(u)+\csc(u)\right)\space\text{d}u$$
From now on you can fix it!
|
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|
Olympiad problem algebra inequality I'm having trouble solving the following inequality problem:
If $n$ is positive integer greater than $1$, and $x>y>1$, then show that:
$\frac{x^{n+1}-1}{x(x^{n-1}-1)} > \frac{y^{n+1}-1}{y(y^{n-1}-1)}$
Any hints? Thanks.
|
Since $x^{n-1} > x^{n-2} > \dots > 1$, and $y^{n-1} > y^{n-2} > \dots > 1$,
we have from the rearrangement inequality
$$
x^{n-1} y^{n-1} + x^{n-2} y^{n-2} + \dots + 1
>
x^{n-1} + x^{n-2} y + \dots + y^{n-1}.
$$
Multiplying both side by $(xy - 1)(x - y)$, we get
$$
(x^n y^n - 1)(x - y) > (x^n - y^n)(xy - 1).
$$
Or
$$
(x^{n+1} - 1) (y^n - y)
>
(y^{n+1} - 1) (x^n - x).
$$
which is the required result after dividing both sides by $(y^n -y)(x^n -x)$.
|
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|
least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
So $$(2x-3)^2-4(2x^2+2x+8-K)\geq 0$$
So $$-4x^2-20x-23+4K\geq0\Rightarrow 4x^2+20x+23-4K\leq0$$
Now How can I Solve after that, Help me
Thanks
|
You have
$$4x^2+20x+23-4K\le 0.$$
So,
$$\begin{align}K&\ge x^2+5x+\frac{23}{4}\\&=\left(x+\frac 52\right)^2-\frac 12\\&\ge -\frac 12\end{align}$$
The equality is attained when $x=-\frac 52,y=4$.
|
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|
Finding two non-congruent right-angle triangles The map $g: B \to A, \ (x,y) \mapsto \left(\dfrac {x^2 - 25} y, \dfrac {10x} y, \dfrac {x^2 + 25} y \right)$ is a bijection where $A = \{ (a,b,c) \in \Bbb Q ^3 : a^2 + b^2 = c^2, \ ab = 10 \}$ and $B = \{ (x,y) \in \Bbb Q ^2 : y^2 = x^3 - 25x, \ y \ne 0 \}$.
Given the Pythagorean triple,
$$
\begin{aligned}
a&=\frac {x^2 - 25} y\\
b&=\frac {10x} y\\
c&=\frac {x^2 + 25} y
\end{aligned}
$$
so $a^2+b^2 = c^2$ and $y\neq 0$. We wish to set its area $G = 5$. Thus,
$$G = \frac{1}{2}ab = \frac{1}2 \frac {10x} y \frac {(x^2 - 25)} y = 5$$
Or simply,
$$x^3-25x = y^2\tag1$$
Can you help find two non-congruent right-angled triangles that have rational side lengths and area equal to $5$?
|
Update: It turns out that for rational $a,b,c$, if,
$$a^2+b^2 = c^2$$
$$\tfrac{1}{2}ab = n$$
then $n$ is a congruent number. The first few are $n = 5, 6, 7, 13, 14, 15, 20, 21, 22,\dots$
After all the terminology, I guess all the OP wanted was to solve $(1)$. He asks, "... find two non-congruent right-angled triangles that have rational side lengths and area equal to 5".
We'll define congruence as "... two figures are congruent if they have the same shape and size." Since we want non-congruent, then two such right triangles are,
$$a,b,c = \frac{3}{2},\;\frac{20}{3},\;\frac{41}{6}$$
$$a,b,c = \frac{1519}{492},\;\frac{4920}{1519},\;\frac{3344161}{747348}$$
However, since the elliptic curve,
$$x^3-25x = y^2\tag1$$
has an infinite number of rational points, then there is no need to stop at just two. Two small solutions to $(1)$ are $x_1 = 5^2/2^2 =u^2/v^2$ and $x_2 = 45$. We can use the first one as an easy way to generate an infinite subset of solutions. Let, $x =u^2/v^2$, so,
$$\frac{u^2}{v^6}(u^4-25v^4) = y^2$$
or just,
$$u^4-25v^4 = w^2\tag2$$
a curve also discussed in the post cited by J. Lahtonen. Here is a theorem by Lagrange. Given an initial solution $u^4+bv^4 = w^2$, then further ones are,
$$X^4+bY^4 = Z^2$$
where,
$$X,Y,Z = u^4-bv^4,\; 2uvw,\; (u^4+bv^4)^2 + 4 b u^4 v^4$$
Thus, using Lagrange's theorem recursively, we have the infinite sequence,
$$u,v,w = 5,\;2,\;15$$
$$u,v,w = 41,\;12,\;1519$$
$$u,v,w = 3344161,\; 1494696,\; 535583225279$$
and so on.
|
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|
Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
My question is, is this a double angle formula? And if so, how would I go about to solve it?
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
$$=2\sin{\pi \over 4}+\left(1-2\sin{\pi \over 4}\right)$$
Have I done this right so far? I feel I have not.
|
Notice, $$\left(\sin\frac{\pi}{8}+\cos\frac{\pi}{8}\right)^2$$
$$=2\left(\frac{1}{\sqrt2}\sin\frac{\pi}{8}+\frac{1}{\sqrt2}\cos\frac{\pi}{8}\right)^2$$
$$=2\left(\sin\frac{\pi}{8}\cos\frac{\pi}{4}+\cos\frac{\pi}{8}\sin\frac{\pi}{4}\right)^2$$
Using trig identity $\sin A\cos B+\cos A\sin B=\sin(A+B)$ $$=2\sin^2\left(\frac{\pi}{8}+\frac{\pi}{4}\right)$$
$$=2\sin^2\left(\frac{3\pi}{8}\right)$$
|
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|
Determinants: divisibility by 6 without remainder and $n\times n$ matrix
*
*Compute the determinant of \begin{pmatrix}
1 & 2 & 3 & ...& n\\
-1 & 0 & 3 & ...& n\\
-1 & -2 & 0 & ...& n\\
...& ...& ...& ...& \\
-1 & -2 & -3 & ...& n
\end{pmatrix}
after some elementary raw operation, one can reach:
\begin{pmatrix} 1 & 2 & 3 & ...& n\\
-2 & -2 & 0 & ...& 0\\
0& -2& -3& ...&0& \\
...& ...& ...& ...& \\
0 & 0 & 0 & 1-n & n
\end{pmatrix}
but I don't sure how to proceed.
*Why the det. of the following matrix is divisible by 6 without remainder?
\begin{pmatrix} 2^0 & 2^1 & 2^2 \\ 4^0 & 4^1 & 4^2\\ 5^0 & 5^1 & 5^2 \end{pmatrix}
So I know that I have to show that its det. is divisible by $2$ and $3$, or equivalently that the sum of its digits divisible by $3$ and last digit is even. But I don't sure how to start the process.
Thank you.
|
*
*Add the first row to the other rows:
$$
\pmatrix{
1 & 2 & 3 & 4 & 5 \\
-1 & 0 & 3 & 4 & 5 \\
-1 & -2 & 0 & 4 & 5 \\
-1 & -2 & -3 & 0 & 5 \\
-1 & -2 & -3 & -4 & 5 \\
}
\to
\pmatrix{
1 & 2 & 3 & 4 & 5 \\
0 & 2 & * & * & * \\
0 & 0 & 3 & * & * \\
0 & 0 & 0 & 4 & * \\
0 & 0 & 0 & 0 & 2\times 5 \\
}
$$
So the determinant is $2(n!)$.
*The matrix is a Vandermonde matrix and its determinant is $(4-2)(5-2)(5-4)=6$.
|
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|
$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $1^{\infty}$.
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}}=\exp \left[{\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}-1\right)\times \frac{3x^2+8}{x^2}}\right]=e^{-8}$$
But i want to know,cant it be found using Taylor/Maclaurin series?What are the limitations of Taylor/Maclaurin series?
When can we not use them?
Is it true that we can use them whenever L Hospital rule is applicable(in $\frac{0}{0}$ and $\frac{\infty}{\infty}$ cases.)Can we use them in $0^0,\infty^0,1^\infty$ cases.
Please help me.Thanks.
|
Using logarithms and Taylor expansions can help a lot. $$A=\Big(\frac{3x^2+2}{5x^2+2}\Big)^{\frac{3x^2+8}{x^2}}$$ $$\log(A)=\frac{3x^2+8}{x^2}\,\log\Big(\frac{3x^2+2}{5x^2+2}\Big)=\frac{3x^2+8}{x^2}\,\log\Big(1-\frac{2x^2}{5x^2+2}\Big)\approx -\frac{3x^2+8}{x^2}\times\frac{2x^2}{5x^2+2}$$
I am sure that you can take from here.
|
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|
Show that $\lim\limits_{x\rightarrow 0}f(x)=1$
Suppose a function $f:(-a,a)-\{0\}\rightarrow(0,\infty)$ satisfies $\lim\limits_{x\rightarrow 0}\left(f(x)+\frac{1}{f(x)}\right)=2$. Show that $$\lim\limits_{x\rightarrow 0}f(x)=1$$
Let $\epsilon>0$ , then there exists a $\delta>0$ such that $$\left(f(x)+\frac{1}{f(x)}\right)-2<\epsilon\;\;\;\text{and}\;\;\;|x|<\delta$$
Then
\begin{align}
&\left(f(x)+\frac{1}{f(x)}\right)-2\\
=&\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)<\epsilon\tag{1}
\end{align}
Squaring $(1)$ both sides gives $$\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2<\epsilon^2\tag{2}$$
Since $(f(x)-1)^2\leq\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2$, by using $(2)$, $(f(x)-1)^2<\epsilon^2\Rightarrow f(x)-1<\epsilon$; therefore, as $\epsilon$ is arbitrary, $\lim\limits_{x\rightarrow 0}f(x)=1$
Can someone give me a hint to do this question without using epsilon-delta definition? Thanks
|
$$
a + \frac 1 a = \left( a - 2 + \frac 1 a \right) +2 = \left( \sqrt a - \frac 1 {\sqrt a} \right)^2 + 2 = \text{square} + 2.
$$
So this is $\ge 2$ unless the square is $0$.
$a+ \dfrac 1 a$ cannot get close to $2$ unless $a$ gets close to $1$.
Even if you allow complex numbers (so that the inequality above doesn't hold), we have
$$
a + \frac 1 a = 2 \Longrightarrow a^2 + 1 = 2a
$$
and that is a quadratic equation whose only solution is $a=1$ (a double root).
|
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|
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is .... Problem :
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is
(a) $2^{105}-2^{121}$
(b) $2^{121}-2^{105}$
(c) $2^{120}-2^{104}$
(d) $2^{110}-2^{108}$
My approach :
Tried to find pattern if any in the given expansion, so expanded first three terms :
$(1-x)(1-2x)(1-2^2x) = (1-3x+2x^2)(1-4x) = (1-3x+2x^2 -4x+12x^2 -8x^3)$
$= (1-7x+14x^2 -8x^3)$
But didn't find any pattern of this series please suggest how to proceed will be of great help thanks.
|
Given $$(1-x)(1-2x)(1-2^2x)...........(1-2^{15}x) = 2^{0+1+2+3+....+15}x^{16}\cdot \left(\frac{1}{x}-1\right)\left(\frac{1}{2x}-1\right).......\left(\frac{1}{2^{15}x}-1\right)$$
So $$=2^{120}x^{16}\left[\left(1-\frac{1}{x}\right)\cdot \left(1-\frac{1}{2x}\right)\cdot \left(1-\frac{1}{2^2x}\right)..........\left(1-\frac{1}{2^{15}x}\right)\right]$$
So $$=2^{120}x^{16}\left[1-\frac{1}{x}\left(1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{15}}\right)+.....\right]$$
So Coefficient of $x^{15} $ in above expression
$$=-2^{120}\left(1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{15}}\right) = -2^{120}\left[1-\frac{1}{2^{16}}\right]\cdot \frac{1}{1-\frac{1}{2}}$$
So we get $$ = -2^{121}\left[1-\frac{1}{2^{16}}\right]=-2^{121}+2^{105}=2^{105}-2^{121}$$
|
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|
Roots of unity filter, identity involving $\sum_{k \ge 0} \binom{n}{3k}$ How do I see that$$\sum_{k \ge 0} \binom{n}{3k} = (1 + 1)^n + (\omega + 1)^n + (\omega^2 + 1)^n,$$where $\omega = \text{exp}\left({2\over3}\pi i\right)$? What is the underlying intuition behind this equality?
|
The first thing we need is the binomial theorem which says that
$$(1+x)^n = \sum_{k\geq 0}^n{n\choose k} x^k = \sum_{k\geq 0}{n\choose k} x^k$$
where the last equality follows since ${n\choose k} \equiv 0$ for all $k > n$.
Next we have that for any sequence $a_n$ we can split the sum of it over all the integers into a sum over their respective residue classes $\mod 3$ so that
$$\sum_{k\geq 0}a_k = \sum_{k\geq 0}a_{3k} + \sum_{k\geq 0}a_{3k+1} + \sum_{k\geq 0}a_{3k+2}$$
The last thing we need is that $\omega^3 = 1$ and $1+\omega + \omega^2 = 0$ when $\omega = e^{\frac{2\pi i}{3}}$. The first equality is obvious, and the second can be found from the sum of a geometrical series $1+\omega+\omega^2 = \frac{\omega^3-1}{\omega-1} = 0$.
With these ingredients we can now calculate
$$\matrix{(1+1)^n + (1+\omega)^n + (1+\omega^2)^n &=& \sum_{k\geq 0}{n\choose k}[1 + \omega^k + \omega^{2k}] \\&=& \sum_{k\geq 0}{n\choose 3k}[1 + \omega^{3k} + \omega^{6k}] \\&+& \sum_{k\geq 0}{n\choose 3k+1}[1 + \omega^{3k+1} + \omega^{6k+2}]\\&+& \sum_{k\geq 0}{n\choose 3k}[1 + \omega^{3k+2} + \omega^{6k+4}]\\&=&\sum_{k\geq 0}{n\choose 3k}[1 + 1 + 1]\\&+&\sum_{k\geq 0}{n\choose 3k+1}[1 + \omega + \omega^{2}]\\&+&\sum_{k\geq 0}{n\choose 3k+2}[1 + \omega^{2} + \omega]\\&=&3\sum_{k\geq 0}{n\choose 3k}}$$
The identity can be generalized. If we take $\omega = e^{\frac{2\pi i}{N}}$ then the same type of derivation as above gives us
$$2^n + (1+\omega)^n + \ldots + (1+\omega^{N-1})^n = N\sum_{k\geq 0}{n\choose Nk}$$
|
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|
Evaluate the integral $\int \frac{dx}{2 \sin x - \cos x + 5}$ I'm trying to evaluate the following integral:
$$ \int \frac{dx}{2 \sin x - \cos x + 5}.$$
This is in a set of exercises following a chapter on partial fractions, so I imagine there is a substitution we can make to get this into a rational function where we can use partial fraction decomposition. I can't seem to figure out what substitution to make in such a situation though.
|
Notice, $$\int \frac{1}{2\sin x-\cos x+5}\ dx$$
$$=\int \frac{1}{2\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}-\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+5}\ dx$$
$$=\int \frac{1+\tan^2\frac{x}{2}}{6\left(\tan^2\frac{x}{2}+\frac{2}{3}\tan\frac{x}{2}+\frac{2}{3}\right)}\ dx$$
$$=\int \frac{\sec^2\frac{x}{2}}{6\left(\left(\tan\frac{x}{2}+\frac{1}{3}\right)^2+\frac{5}{9}\right)}\ dx$$
$$=\frac{2}{6}\int \frac{d\left(\tan\frac{x}{2}+\frac{3}{2}\right)}{\left(\tan\frac{x}{2}+\frac{1}{3}\right)^2+\left(\frac{\sqrt 5}{3}\right)^2}$$
$$=\frac{1}{3}\frac{3}{\sqrt 5}\tan^{-1}\left(\frac{\tan\frac{x}{2}+\frac{1}{3}}{\frac{\sqrt 5}{3}}\right)+C$$
$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{\int \frac{1}{2\sin x-\cos x+5}\ dx=\frac{1}{\sqrt 5}\tan^{-1}\left(\frac{3\tan\frac{x}{2}+1}{\sqrt 5}\right)+C}}$$
|
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|
Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must factorize the denominator:
$$x^3+2x^2 = (x+2)\cdot x^2$$
Great. So now we write three fractions:
$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$
Eventually we conclude that
$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$
So now we look at what happens when $x = -2$:
$$C = 12$$
When $x = 0$:
$$A = -1$$
And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:
$$B = -1$$
We replace the values here:
$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$
Which results in
$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$
But I fear the answer actually is
$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$
Can you tell me what did I do wrong, and what should I have done?
|
Once you arrive at $$A(x+2) + B(x+2)x + C(x^2) = 5x^2 + 3x - 2,$$ this equation must hold true for all $x$. If you substitute $x = -2$, then $$C(-2)^2 = 5(-2)^2 + 3(-2) - 2 = 5(4) - 6 - 2 = 12,$$ or $4C = 12$, or $$C = 3.$$ You seem to be under the impression that you can disregard any expressions involving $x$ on the left-hand side. That is incorrect.
Next, substituting $x = 0$ gives $$A(0+2) = 5(0^2) + 3(0) - 2 = -2,$$ or $2A = -2$, or $$A = -1.$$ This you got right. Now finally, if you were to choose $x = 1$, you would obtain $$A(1+2) + B(1+2)(1) + C(1^2) = 5(1^2) + 3(1) - 2 = 6,$$ and the left-hand side simplifies to $3A + 3B + C$. To get $B$, we now have to substitute the values you previously obtained for $A$ and $C$: $$3(-1) + 3B + 3 = 6,$$ or $$B = 2.$$ Therefore, the solution is $$\frac{5x^2 + 3x - 2}{x^2(x+2)} = \frac{-1}{x^2} + \frac{2}{x} + \frac{3}{x+2}.$$
|
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Could someone check my solution for finding constant of a difference quotient? So the question was,
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be three times differentiable and $f'''$ is bounded, find constants $a,b,c$ such that $$f''(x) = \lim_{h\rightarrow 0} \frac{af(x-h)+bf(x)+cf(x+2h)}{h^2}$$
My Solution:
Expand $f
(x-h)$ and $f(x+2h)$ by Taylors theorem and solve for $f''(x)$:
$f(x-h) = f(x) - hf'(x) + 0.5h^2f''(x)$ so $f''(x) = 2f(x-h)-2f(x)+2hf'(x)$ and putting in the limit definition of $f'(x)$ we get $$f''(x) = \frac{2f(x-h)-4f(x)+2f(x+h)}{h^2}$$
Doing the same for $f(x+2h)$ we get $$\frac{4f''(x)}{3} =\frac{f(x+2h)+f(x)-2f(x+h)}{h^2}$$
Can I subtract $\frac{1}{3}$ of first one from the second to get $$f''(x) = \frac{f(x+2h)+\frac{7}{3} f(x) - \frac{2}{3} f(x-h) - \frac{8}{3} f(x+h)}{h^2}$$
And also can I say that $-f(x+h) = f(x-h)$?
|
We find the first three terms, in powers of $h$, of the Taylor expansion of the numerator.
The first term is
$$af(x)+bf(x)+cf(x).\tag{1}$$
The second term is
$$\left(-af'(x)+2cf'(x)\right)h.\tag{2}$$
The third term is
$$\frac{af''(x)+4cf''(x)}{2}h^2.\tag{3}$$
We will arrange for the first two terms to vanish, and for the third term to be $f''(x)h^2+o(h^2)$. This can be done by letting $a+b+c=0$, $-a+2c=0$, and $\frac{a+4c}{2}=1$. When we solve this system of equations, we get $a=2/3$, $b=-1$, and $c=1/3$.
|
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Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$
My questions:
$1)$ Is there an easy way to see that $x=-8$ is a root too?
$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$
|
Use the Distributive Property.
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2$$
$$=\underbrace{(x+2)(x-1)^2}_{\text{common factor}}\left(2(x-1)\right)-\underbrace{(x+2)(x-1)^2}_{\text{common factor}}(3(x+2))$$
$$=(x+2)(x-1)^2\left(2(x-1)-3(x+2)\right)$$
$$=(x+2)(x-1)^2(-(x+8))$$
|
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Is $K=\mathbb{Q}(\sqrt{-5+\sqrt{5}}) $a Galois extension? I am wondering if the number field $K=\mathbb{Q}(\sqrt{-5+\sqrt{5}})=\mathbb{Q}(\alpha)$ is a Galois extension. I think it is and I have the following argument, but it feels like a circulair argument:
I know that we have the subfield $\mathbb{Q}(\sqrt{5})$ so we have one automorphism $\sigma$ that sends $\sqrt{5}$ to $\sqrt{-5}$. The other roots of the minimal polynomial over $\alpha$ are:
\begin{align*}
-\alpha \\
-i\sqrt{5+\sqrt{5}} = \sigma(\alpha) \\
i \sqrt{5+\sqrt{5}} = \sigma(-\alpha)
\end{align*}
Clearly $-\alpha$ is also in $K$ and since $\sigma$ is an automorphism, $\sigma(\alpha)$ and $\sigma(-\alpha)$ must also. So we have a normal extension, hence a Galois extension. Is this a right argument or am I assuming something I cannot?
|
Let's start with a choice of radicals $x = i \sqrt{5 - \sqrt{5}}$. Note that
$x^2 = - (5-\sqrt{5})$ so $x^2 + 5 = \sqrt{5}$, $(x^2 + 5)^2 = 5 $, and finally
$$x^4 + 10 x^2 + 20 =0$$
an irreducible equation with roots $\pm i \cdot \sqrt{5 \pm\sqrt{5}}$. We see that the different choices of radicals gives isommorphic extensions.
We want to know whether $K =\mathbb{Q}(x)= \mathbb{Q}(i \sqrt{5 - \sqrt{5}})$ is normal, equivalently, whether $\mathbb{Q}(i \sqrt{5 - \sqrt{5}})$ contains $i \sqrt{5 + \sqrt{5}}$. Recall that $x^2 + 5 = \sqrt{5}$ so $\sqrt{5} \in K$.
Now, $i \sqrt{5 - \sqrt{5}} \cdot i \sqrt{5 + \sqrt{5}}= - \sqrt{ 25 - 5} = - \sqrt{20} = - 2 \sqrt{5} \in K$, so $i \sqrt{5 + \sqrt{5}}$ also in $K$. We conclude that $K$ is normal, hence the extension $K/\mathbb{Q}$ is Galois.
${\bf Added:}$. The roots of the equation $x^4 + 10 x^2 + 20 =0$ are $\pm x$ and $\pm ( 2x + \frac{10}{x})$. The Galois group acts transitively on the roots. If $\sigma(x) = -x$ then $\sigma( \pm( 2x + \frac{10}{x})) = \mp ( 2x + \frac{10}{x})$, so $\sigma$ is of order $2$. More interesting, consider $\sigma$ that takes $x$ to $2x + \frac{10}{x}$. Then $\sigma(2x + \frac{10}{x}) = 2 (2x + \frac{10}{x}) + \frac{10}{2x + \frac{10}{x}} $ and
$$ 2 (2x + \frac{10}{x}) + \frac{10}{2x + \frac{10}{x}} - (-x) = \frac{5 (10 x^4+ 10 x^2 + 20)}{x ( x^2+5)}=0$$
so $\sigma^2(x) = -x$. Hence, this $\sigma$ is of order $4$, and since the Galois group is of order $4$, it will be cyclic with this $\sigma$ as generator.
|
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Calculate $\sum_{n=0}^\infty$ $(n+1)(n+2)(\frac{i}{2})^{n-1}$ I want to calculate $\sum_{n=0}^\infty$ $(n+1)(n+2)(\frac{i}{2})^{n-1}$.
I tried to separate it into a sum of real numbers ($n=0,2,4,\dots$) and complex numbers that are not real numbers ($n=1,3,5,\dots$) but it didn't work.
So I did it another way, using Cauchy's integral theorem:
Let $f(z)=(\frac{z}{2})^{n+2}$. Then $4f''(i)$= $(n+1)(n+2)(\frac{i}{2})^{n-1}$, which is a term of the sum I started with. I don't know how to proceed from here.
What can I do? How do I solve this?
|
Taking the derivative of the geometric series twice and dividing by $z$ gives
\begin{align}
\frac{1}{1-z} &= \sum_{n=0}^\infty z^n
\\
\frac{1}{(1-z)^2} &= \sum_{n=1}^\infty nz^{n-1}
\\
\frac{2}{(1-z)^3} &= \sum_{n=2}^\infty n(n-1)z^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) z^n
\\
\frac{2}{z(1-z)^3} &= \sum_{n=0}^\infty (n+2)(n+1) z^{n-1}
\end{align}
|
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|
Convergence of infinite series question The sum of the series
$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$ equals?
The answer is $\frac{3e}{2}$.
But I dont Know How?
I have tried following:
$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$=$\sum\limits_{n=1}^{\infty} \frac{n(n+1)}{2n!}
$
I know that sequence of partial sum $S_n$=$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....+\frac{n(n+1)}{2n!}$ and $\sum\limits_{n=1}^{\infty} \frac{n(n+1)} {2n!}$ converges to same point.
Now how to find $S_n$ converges to which point?
|
Hint:
$$\frac{n(n+1)}{n!}=\frac{n+1}{(n-1)!}=\frac{(n-1)+2}{(n-1)!}=\frac{(n-1)}{(n-1)!}+2\frac{1}{(n-1)!}=\frac{1}{(n-2)!}+2\frac{1}{(n-1)!}$$
the both last ones with $n\to\infty$ gives us $e$
|
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Let $x^2+kx=0;k$ is a real number .The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set. Let $f(x)=x^2+kx;k$ is a real number.The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set.
The equation $x^2+kx=0$ has solutions $x=0,-k$.
So the solutions of the equation $f(f(x))=0$ should be $x=0,-k$.
$f(f(x))=(x^2+k x)^2+(x^2+k x)k=0$ gives
$x^4+k^2x^2+2kx^3+x^2k+x k^2=0$
$x^4+2kx^3+(k^2+k)x^2+xk^2=0$
$x(x^3+2kx^2+(k^2+k)x+k^2)=0$ has the solutions $x=0,x=-k$
Therefore $x^3+2kx^2+(k^2+k)x+k^2=0$ has the solution $x=-k$.
But putting $x=-k$ in the equation $x^3+2kx^2+(k^2+k)x+k^2=0$ gives me nothing.It just gives me $0=0$.
I am stuck here.Please help me.Thanks.
|
We have $f(x) = x(x+k)$. This means
$$f(f(x)) = f(x)(f(x)+k) = x(x+k)(x^2+kx+k)$$
Hence, we see that $0$ and $-k$ are solutions to both. For $x^2+kx+k$ not to have a real solution, we need $k^2-4k < 0 \implies 0 < k < 4$.
|
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How do you evaluate this sum of multiplied binomial coefficients: $\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} $? We have to find the value of x+y in:
$$\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} = \binom{x}{y} $$
My approach:
I figured that the required summation is nothing but the coefficient of $x^3$ is the following expression:
$$\sum_{r=1}^7 \frac{r(r+1)}{2}(1+x)^{10-r} $$
which looks like:
$$(1+x)^{10} + 3(1+x)^9 + 6(1+x)^8 + ..... + 36(1+x)^3$$
Hence we have a sequence in which the difference of the coefficients forms an AP.
So let the expression = $S$
So we have:
$$\frac {S}{1+x} = (1+x)^{9} + 3(1+x)^8 + 6(1+x)^7 + ..... + 36(1+x)^2$$
Subtracting $\frac{S}{1+x}$ from $S$, we get:
$$S(1-\frac{1}{1+x}) = (1+x)^{10} + 2(1+x)^9 + 3(1+x)^8 + ..... + 8(1+x)^3 - 36(1+x)^2$$
Here, all the terms except the last one are in AGP and hence we can find $S$ by applying the same method and isolating $S$ on one side. Then we can process to find the coefficient of $x^3$ but it got really lengthy. In the test that it was given, we have an average time of 3 minutes per question and not even the hardest question takes more than 7-8 minutes if you know how to do it.
Hence, I'm wondering if there is a better, shorter way to solve this question.
|
You can also do it with almost no computation if you make the right combinatorial argument. Let’s count the ways to choose $6$ numbers from the set $S=\{0,1,\ldots,12\}$. Let $r$ be the third-smallest of the $6$; how many ways are there to choose the other $5$ numbers?
There are $r$ members of $S$ less than $r$, so there are $\binom{r}2$ ways to choose the $2$ numbers smaller than $r$. Similarly, there are $12-r$ members of $S$ greater than $r$, so there are $\binom{12-r}3$ ways to choose the $3$ numbers larger than $r$. These choices are independent of each other, so there are $\binom{r}2\binom{12-r}3$ ways to choose the $6$ numbers so that the third-smallest is $r$.
Now sum over the possible values of $r$ to find that there are $\sum_{r=2}^9\binom{r}2\binom{12-r}3$ $6$-element subsets of $S$. Of course $|S|=13$, so there are $\binom{13}6$ such subsets, and we have
$$\sum_{r=2}^9\binom{r}2\binom{12-r}3\binom{13}6\;.$$
|
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|
Why series expansion in evaluating $\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}=\sqrt2$ is not working here? Prove that $\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}=\sqrt2$
I can evaluate this limit using L Hospital rule but i wonder why Series expansion method is failing here and L Hospital rule is working here.
$\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}$
As $\arccos(x)=\frac{\pi}{2}-x-\frac{x^3}{6}-\frac{3x^5}{40}-....$
$\arccos(1-x^2)=\frac{\pi}{2}-(1-x^2)-\frac{(1-x^2)^3}{6}-\frac{3(1-x^2)^5}{40}-....$
$\frac{\frac{\pi}{2}-(1-x^2)-\frac{(1-x^2)^3}{6}-\frac{3(1-x^2)^5}{40}-....}{x}$
And it is not simplifying.I cannot take $x$ common from the numerator to cancel out from the denominator.
Have i made some mistake or series expansion will not work here.
Are there some cases where series expansion fails and only L Hospital works.Please help me.Thanks.
|
As $x\to0,1-x^2>0$
$\arccos(1-x^2)=\arcsin\sqrt{1-(1-x^2)^2}=\arcsin\sqrt{2x^2-x^4}$
$$\lim_{x\to0^+}\dfrac{\arcsin\sqrt{2x^2-x^4}}x=\lim_{x\to0^+}\dfrac{\arcsin\sqrt{2x^2-x^4}}{\sqrt{2x^2-x^4}}\cdot\lim_{x\to0^+}\dfrac{x\sqrt{2-x^2}}x$$
Set $\arcsin\sqrt{2x^2-x^4}=y$ in the first
Can you take it home from here?
|
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|
Find $A$ when $A^2$ is a $2\times 2$ zero-matrix and $A$ is symmetric.
$A^2 =
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}$ and $A$ is symmetric. Find the matrix $A$.
Well, I know that $A = A^T$ and $A^2 =
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
$. I'm not sure how to continue from here. Any suggestions?
Btw, I thought of something: $A^2 = AA = A A^T = I$ but it is wrong.
|
$\left(\matrix{a & b\\ b & d}\right)^2=\left(\matrix{a^2+b^2 & b(a+d)\\b(a+d) & b^2+d^2}\right)=\left(\matrix{0 & 0\\ 0 & 0}\right)\to b=0$ and $a^2=d^2=0$ or $a=-d$ and $b^2=-a^2$, ie, A is zero
|
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|
Reducibility of polynomials modulo p
How do we show that $X^4-10X^2+1$ is reducible modulo every prime $p$?
I've managed to show it for all primes less than 10, for primes greater than 10 we have $X^4+(p-10)X^2+1$. Where do I go from here?
|
Write
$$\begin{align}
x^4-10x^2+1&=(x^2-1)^2-2(2x)^2\\
&=(x^2+1)^2-3(2x)^2\\
&=(x^2-5)^2-6\cdot 2^2
\end{align}$$
Now if $2$ is a quadratic residue $\pmod{p}$ then the first identity makes our polynomial the difference of two squares.
If $3$ is a quadratic residue $\pmod{p}$ then the second identity makes our polynomial the difference of two squares.
Remember that
$$\left(\frac2p\right)\left(\frac3p\right)=\left(\frac6p\right).$$
So if $2$ and $3$ are not quadratic residues $\pmod{p}$ then $6$ is and the third identity makes our polynomial the difference of two squares.
|
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|
Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$
Let $x$, $y$, $a$, $b$ be real numbers such that $a^2+b^2 \leq 1$ and $x^2+y^2 \leq 1$. Show that $$(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$$
I am unable to find a solution to this problem. My initial thoughts were to have a trigonometric substitution of variables, but that didn't lead me further. Please help.
Thank you.
|
Use C-S we have
$$(ax+by)^2\le (a^2+b^2)(x^2+y^2)\le 1
\Longrightarrow ax+by\le 1$$
we can rewrite the inequality
$$(1-ax-by)^2\ge (1-a^2-b^2)(1-x^2-y^2)$$
$$\Longleftrightarrow 1-(ax+by)\ge \sqrt{(1-a^2-b^2)(1-x^2-y^2)}$$
Let $A=a^2+b^2,B=x^2+y^2$,since
$$1-(ax+by)\ge 1-\sqrt{AB}$$it is enought to show that
$$1-\sqrt{AB}\ge \sqrt{(1-A)(1-B)}$$
Use C-S
we have
$$1=[(1-A)+A][(1-B)+B]\ge(\sqrt{(1-A)(1-B)}+\sqrt{AB})^2$$
By done.
|
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|
Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x) = x \frac{d}{dx}[f_0(x)] = \frac{1}{(1-x)^2} = 0 + x + 2x^2 + 3x^3 +\cdots$$
$$f_2(x) = x \frac{d}{dx}[f_1(x)] = \frac{x^2+x}{(1-x)^3} = 0^2 + 1^2x + 2^2x^3 + 3^2x^3+\cdots,$$
and I assume I'm supposed to be able to do something similar in this case, but things get trickier when it's bounded by n and I keep getting stuck.
|
Hint: The following perspective with focus on operator methods might also be useful.
We can successively apply the $\left(x\frac{d}{dx}\right)$-operator to a generating function
\begin{align*}
A(x)=\sum_{n=0}^{\infty}a_nx^n
\end{align*}
to obtain
\begin{align*}
\left(x\frac{d}{dx}\right)A(x)&=\sum_{n=0}^{\infty}na_nx^n\\
\left(x\frac{d}{dx}\right)^2A(x)&=\sum_{n=0}^{\infty}n^2a_nx^n
\end{align*}
Multiplication of $A(x)$ with $\frac{1}{1-x}$ results in summing up the coefficients $a_n$
\begin{array}{crl}
(a_n)_{n\geq 0}\qquad &\qquad A(x)=&\sum_{n=0}^{\infty}a_nx^n\\
\left(\sum_{k=0}^{n}a_k\right)_{n\geq 0}\qquad&\qquad\frac{1}{1-x}A(x)=&\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_k\right)x^n
\end{array}
It's also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a generating series.
Putting all together and applying it to the geometric series $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ we finally obtain
\begin{align*}
\sum_{k=0}^nk^2&=[x^n]\frac{1}{1-x}\left(x\frac{d}{dx}\right)^2\frac{1}{1-x}\\
&=[x^n]\frac{x(1+x)}{(1-x)^4}\\
&=\left([x^{n-1}+x^{n-2}]\right)\sum_{k=0}^{\infty}\binom{-4}{n}(-x)^n\tag{1}\\
&=\left([x^{n-1}+x^{n-2}]\right)\sum_{k=0}^{\infty}\binom{n+3}{3}x^n\tag{2}\\
&=\binom{n+2}{3}+\binom{n+1}{3}\\
&=\frac{1}{6}n(n+1)(2n+1)
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion and $[x^n]x^kA(x)=[x^{n-k}]A(x)$
*In (2) we use $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k=\binom{n+k-1}{n-1}(-1)^k$
|
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|
How do I get $ \int_0^1 \frac{dz}{\sqrt{z(z - 1\,)(z+1\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})}$? While reading physics papers I found a very interesting integral so I decided to write it down. Let $p(z) = z^ 3 - 3\Lambda^ 2 z$ where $\Lambda$ could be any number. If you want $\Lambda = 1$ and $p(z) = z^ 3 - 3z$. Then
$$ \int_0^{\sqrt{3}\Lambda} \frac{dz}{\sqrt{p(z)}} = \int_0^{\sqrt{3}\Lambda} \frac{dz}{\sqrt{z(z - \sqrt{3}\Lambda\,)(z+\sqrt{3}\Lambda\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})} (\sqrt{3}\Lambda)^{5/2}$$
The physics paper at least tell us $\int \propto \Lambda^{5/2}$ so we know the growth rate. It could be that $\Lambda = \frac{1}{\sqrt{3}}$ is even simpler than $\Lambda = 1$.
$$ \int_0^1 \frac{dz}{\sqrt{z(z - 1\,)(z+1\,)}} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{9}{4})}$$
This should be connected to the Riemann surface $y^ 2 = z(z^2 -1)$. And we are computing a period of the Riemann surface.
Contour integration is really important here. Checking on Wolfram Alpha gives a different answer:
$$ \int_0^1 \frac{dz}{x(x^2-1)} = - 2i\sqrt{\pi}\,\frac{ \Gamma(\frac{5}{4})}{\Gamma(\frac{3}{4})}$$
I am guessing this is either the same number or Mathematica is choosing a different contour. Physicists love contour integrals (taken from physics.stackexchange):
|
Since:
$$\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}\,dx = B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\tag{1}$$
through a change of variable we get:
$$ \int_{0}^{1} z^{-1/2}(1-z^2)^{-1/2}\,dz = \frac{1}{2}\int_{0}^{1}z^{1/4}(1-z)^{-1/2}\,dz = \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{2}\right)}{2\cdot \Gamma\left(\frac{7}{4}\right)}\tag{2} $$
and by using the reflection formulas for the $\Gamma$ function that simplifies to:
$$ \int_{0}^{1} \frac{dz}{\sqrt{z(1-z)(1+z)}} = \color{red}{\frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{2\pi}}}.\tag{3}$$
|
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|
find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$ If the equation of the tangent to the graph of the function $y=f(x)$ at $x=2$ is $4x-y-3=0$ and at this point tangent cuts the graph also,then find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$$
As the denominator is tending to zero,so numerator must br tending to zero.
So I applied L Hospital rule two times to get
$$\lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)}$$ for the first time and $$\lim_{x\to 2}\frac{4x^2f''(x^2-2)+2f'(x^2-2)-f'(f(x)-3).f''(x)-f''(f(x)-3).(f'(x))^2}{2}$$
As the equation of tangent at the point $x=2$ is $y=4x-3$
So slope of tangent at $x=2$ is $4$ i.e. $f'(2)=4$ and as the tangent cuts the graph at $x=2$,so put $x=2$ in $y=4x-3$ to get $y=5$, so $f(2)=5$ but
I don't know how to find $f''(2)$.
I am stuck here, please help me. Thanks.
|
In general, the equation of a tangent is $y-f(x_0)=f'(x_0)(x-x_0)$.
Rearrange $4x-y-3=0$, we have $y-5=4(x-2)$,
i.e. $f(2)=5$ and $f'(2)=4$.
As $t\rightarrow 2$, $f(x^2-2)-f(f(x)-3) \rightarrow f(4-2)-f(5-3)=0$.
Now
$\lim_{x\rightarrow 2} \frac{f(x^2-2)-f(f(x)-3)}
{(x-2)^2}
=\lim_{x\rightarrow 2} \frac{f'(x^2-2)\cdot 2x-f'(f(x)-3)\cdot f'(x)}
{2(x-2)} $
As $t\rightarrow 2$, $f'(x^2-2)\cdot 2x-f'(f(x)-3)\cdot f'(x) \rightarrow f'(2)\times 4-f'(5-3)\times 4=0$.
Then applying L'Hospital rule once more,
$\begin{eqnarray*}
&&
\lim_{x\rightarrow 2}
\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} \\
&=&
\lim_{x\rightarrow 2}
\frac{f''(x^2-2) \cdot (2x)^2+f'(x^2-2)\cdot 2-
f''(f(x)-3)\cdot [f'(x)]^2-f'(f(x)-3)\cdot f''(x)}{2} \\
&=&
\frac{f''(2)(4)^2+f'(2)\times 2-f''(2)[f'(2)]^2-f'(2)\times f''(2)}{2} \\
&=& 4-2f''(2)
\end{eqnarray*}$
Simply leaves $f''(2)$ in your answer.
|
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|
Least positive integer $n$ such that the digit string of $2^n$ ends on the digit string of $n$
What is the least positive integer $n$ such that the digit string of $2^n$ ends on the digit string of $n$:
$$ (2^n)_{10} = d_m \, d_{m-1} \cdots
d_{q+1} \, (n)_{10} \\ (n)_{10} = d'_{q} \cdots d_1' \\ d_i, d'_j \in
\{0, \ldots, 9 \} $$
As in $2^3$ would somehow end in 3, or $2^5$ would end in 5.
Frankly I don't even know where to start. Thanks in advance.
|
Here is a less brute-force method:
In order for the digit string of $2^n$ to end with the digit string of $n$, it is necessary (but not sufficient) that $2^n$ and $n$ have the same last digit, i.e. $2^n \equiv n \pmod{10}$.
It is easy to check that $\begin{cases}2^n \equiv 2\pmod{10} & \text{if} \ n\equiv 1 \pmod{4} \\ 2^n \equiv 4\pmod{10} & \text{if} \ n\equiv 2 \pmod{4} \\ 2^n \equiv 8\pmod{10} & \text{if} \ n\equiv 3 \pmod{4} \\ 2^n \equiv 6\pmod{10} & \text{if} \ n\equiv 0 \pmod{4}\end{cases}$.
Since $2^n$ is even, $n$ must also be even. So we have the following 2 possibilities:
$n \equiv 2\pmod{4}$ AND $n \equiv 2^n \equiv 4\pmod{10}$, i.e. $n \equiv 14 \pmod{20}$
$n \equiv 0\pmod{4}$ AND $n \equiv 2^n \equiv 6\pmod{10}$, i.e. $n \equiv 16 \pmod{20}$.
So we only need to check $n \equiv 14 \ \text{or} \ 16 \pmod{20}$.
Since $2^{14} = 16384$, $2^{16} = 65536$, $2^{34} = 17179869184$, and $2^{36} = 68719476736$, the smallest positive integer $n$ such that the digit string of $2^n$ ends with the digit string of $n$ is $n = 36$.
|
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|
How do you find the value of $\sum_{r=1}^{\infty} \frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})} $? How do you find the value of:
$$\sum_{r=1}^{\infty} \frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})} $$
I tried to apply partial fractions/divide numerator and denominator by $6^r$ and use the fact that $a^x \rightarrow 0$ as $x \rightarrow \infty$ for $0<a<1$, but couldn't use it because I couldn't form a telescoping series.
I can't think of anything else. Any help will be appreciated.
|
Hint:
$$\frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}
= \frac{2^r}{3^r-2^r} - \frac{2^{r+1}}{3^{r+1}-2^{r+1}}.$$
Edit:
$$\frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}
= \frac{A}{3^r-2^r} + \frac{B}{3^{r+1}-2^{r+1}}
= \frac{(3A+B)3^r - (2A+B)2^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}.$$
One obvious way to make the numerators equal is for $3A+B = 2^r$ and $2A+B=0$, which gives $A = 2^r$, $B = -2\cdot 2^r = -2^{r+1}$. (The other obvious possibility, $3A+B=0$ and $2A+B=-3^r$, gives a different decomposition which still telescopes, but where the "leftover" term at the end doesn't tend to zero).
|
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|
How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but had no luck. Thank you.
|
HINT: i think your inequality must be reversed, try $a=1,b=2,c=3$
We can prove it by BW
|
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|
If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem.
If x, y are nonzero natural
numbers with $x>y$ such that
$$x-y = 5y^2 - 4x^2,$$
prove that $x - y$ is perfect square.
What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) + y^2$$
$$x - y + 4(x-y)(x+y) = y^2$$
$$(x-y)(4x+4y+1) = y^2$$
So $4x+4y+1$ is a divisor of $y^2$.
I also take into consideration that $y^2$ modulo $4$ is $0$ or $1$ (I don't know if this can help.)
So how do I prove that $4x+4y+1$ is a perfect square (this would involve $x-y$ - a perfect square)? While taking examples, I couldn't find any perfect square with a divisor that is $M_4 + 1$ and is not perfect square.
If there are any mistakes or another way, please tell me.
Some help would be apreciated. Thanks!
|
Use Bill's observation that
Theorem: If $a \mid b^2 $ and $ a = \pm c^2 \pm kb^2$, then $ a = \pm (c, b) ^2$
Proof: $ a = \gcd(a, b^2) = \gcd(\pm c^2 \pm kb^2, b^2) = \gcd(c^2 , b^2 ) = \gcd(c, b)^2$.
Now,
*
*$ x -y > 0$ by assumption.
*OP has showed that $x-y \mid y^2$.
*$x-y = 5y^2 - 4x^2 = -(2x)^2 + 5y^2$ .
Hence, $ x - y = (2x, y)^2$.
|
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|
Solve this equation $xy-\frac{(x+y)^2}{n}=n-4$ Let $n>4$ be a given positive integer. Find all pairs of positive integers $(x,y)$ such that
$$xy-\dfrac{(x+y)^2}{n}=n-4$$
What I tried is to use
$$nxy-(x+y)^2=n^2-4n\Longrightarrow (n-2)^2+(x+y)^2=nxy+4$$
|
Note: For symmetry, it is best to do the minor change of variables $n=m+2$.
When dealing with integer solutions to quadratic forms, there may be a Pell equation lurking nearby. For any integer $m>2$, the equation,
$$xy-\frac{(x+y)^2}{m+2}=m-2$$
has an infinite number of positive integer solutions. Define $D = m^2-4$, then
$$x = (m^2-2)(p^2+Dq^2)-2Dmpq\tag1$$
and the variable $y$ as either,
$$y_1 = m(p^2+Dq^2)-2Dpq\tag{2a}$$
or,
$$y_2 =m(m^2-3)(p^2+Dq^2)-2D(m^2-1)pq\tag{2b}$$
where $p,q$ solve the Pell equation,
$$p^2-Dq^2=k,\;\;\text{where}\; k = 1\;\text{or}\; 4\tag3$$
The case $k=1$ is routine to solve. If we choose $k=4$, then $(3)$ has an infinite family of parametric solutions. For example: $p,q = m^2-2,\; m$ yields (after letting $m=n-2$),
$$x,\, y_1,\, y_2 = n^2-4n+2,\;(n^2-4n+1)(n-2),\;n-2$$
which is the first parameterization found by MacLeod in his answer while the next $p,q = m^3-3m,\; m^2-1$ gives,
$$x = n^4-8n^3+20n^2-16n+2$$
which is his second. Since $p^2-(m^2-4)q^2 = 4$ has an infinite number of parameterizations, then so does $x,y$ as MacLeod has suggested.
|
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|
How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5$$
where at least $2$ of the variables are $\geq 2$.
I found the total number of candidate solutions using the $\binom{B+N-1}{B} = 21$
However, only $9$ of them have two variables $\geq 2$.
\begin{align*}
2+0+3& =5\\
2+1+2& =5\\
3+0+2& =5\\
1+2+2& =5\\
3+2+0& =5\\
0+2+3& =5\\
0+3+2& =5\\
2+3+0& =5\\
2+2+1& =5
\end{align*}
I feel there is a connection to the Associated Stirling numbers of the second kind. But I can't place it :(
EDIT:
Here is my code for enumerating them all to count the number of ways of select B elements from a set of N (uniformly with replacement), such that you have at least C copies of K elements - also shows the output for this question I'm asking here as it's the core piece. Obviously can't be run for very large values of the parameters - that's why I'm here :) Code is here
Here is another example for B = 6, N = 3, C = 2 and K = 2 there are 16 solutions:
\begin{align*}
0+2+4& = 6\\
0+3+3& = 6\\
0+4+2& = 6\\
1+2+3& = 6\\
1+3+2& = 6\\
2+0+4& = 6\\
2+1+3& = 6\\
2+2+2& = 6\\
2+3+1& = 6\\
2+4+0& = 6\\
3+0+3& = 6\\
3+1+2& = 6\\
3+2+1& = 6\\
3+3+0& = 6\\
4+0+2& = 6\\
4+2+0& = 6\\
\end{align*}
There are a number of different and correct solutions below. I don't know which to accept.
|
Without loss of generalization, let us consider $x_1,x_2,\ldots,x_k \geq C$.
Then let
$$y_i = \begin{cases}x_i - C &i\leq k\\ x_i & i > k \end{cases}$$
Solving
$$\left\{\begin{array}[lll] xx_1 + x_2 + \ldots + x_n &=& N\\ x_i \geq 0& & \\ x_1,x_2,\ldots,x_k& \geq & C\end{array}\right.$$
is equivalent to just solving
$$\left\{\begin{array}[lll] yy_1 + y_2 + \ldots + y_n &=& N-kC\\ y_i \geq 0& &\end{array}\right.$$ The number of solutions excluding permutations is the number of partitions of $N-kC$ into $m$ integers where $m \leq n$.
For each of this solutions, adding $C$ to $k$ of them would yield a solution of your original problem. There are $n\choose k$ ways to do thus but we are overcounting some solutions if there are $i,j$ such that $y_i = y_j$ or if $y_i + C = y_j$. For each of this solutions we have at most $n!$ distinct solutions (including permutations), again we are overcounting some if there are $i,j$ such that $y_i = y_j$ or if $y_i + C = y_j$. This gives us a somewhat close upper bound on the amount of solutions:
$${n\choose k} n!\sum_{m=1}^n P_m(N-kC)$$
|
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|
Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1
\\
= &16k+4k+4 +1+4r+2+1
\\
=
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
|
You can prove this through two separate steps:
*
*$5^n \mod 8$ is 5 if $n$ is odd and 1 if $n$ is even.
*$2 \cdot 3^{n-1} \mod 8$ is 2 if $n$ is odd and 6 if $n$ is even.
Once you have the above two statements you have then concluded that in either case of $n$ odd/even it's equivalent to 0 modulo 8.
|
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|
Solve $x^n+y^n=2015$ Determine the natural numbers $x,y,n$ matching equality
$$x^n+y^n=2015.$$
I noticed for $n = 1$ the equation has solutions $(x, 2015-x), x$ integer.
For $n = 2$, given that $x$ and $y$ are different parities taking $x = 2k$ and $y=2m + 1$ we come to contradiction.
What must be done to $n\geq3$?
|
Piggybacking off user's answer that we need only concern ourselves with a few odd exponents, with $x\lt y$ also pretty small, let's use the fact that $x+y$ divides $x^n+y^n$ for odd $n$ and $x^p\equiv x$ mod $p$ for prime $p$.
Note that $2014$ is not an $n$th power for any $n\gt1$, so we can assume $1\lt x\lt y$. Note also that $2015=5\cdot13\cdot31$. Finally, $\sqrt[3]{2015}\lt13$ means we don't have to worry about potential solutions with $x+y\ge25$, which means we need only consider solutions with $x+y=5$ or $x+y=13$.
For $n=3$, we have $x+y\equiv x^3+y^3=2015\equiv2$ mod $3$, so $x+y=5$ is the only possibility, which means $x=2,y=3$ is the only possibility, but $2^3+3^3$ is way smaller than $2015$.
For $n=5$, we have $x+y\equiv x^5+y^5=2015\equiv0$ mod $5$, so again $x+y=5$ is the only possibility, and again $2^5+3^5$ is too small.
For $n=7$, we have $x+y\equiv x^7+y^7=2015\equiv6$ mod $7$, so $x+y=13$ is the only possibility. But $7^7$ is already way too big. (Alternatively, note that $\sqrt[7]{2015}\approx2.965\lt3$ already rules out any solutions.)
For $n=9$, working mod $3$, we have $x+y\equiv x^9+y^9\equiv2$ mod $3$, so again $2^9+3^9$ is the only possibility, which is too large. (Or, again, $\sqrt[9]{2015}\lt3$ rules out any solutions.)
|
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|
How to compute $\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$$
|
I thought it might be instructive to present a development that goes back to "the basics" and relies on elementary tools only. To that end, we recall from geometry that the sine function satisfies the inequalities
$$z \cos z \le \sin z\le z \tag 1$$
for $0\le z\le \pi/2$.
Rearranging $(1)$ we immediately see that $z\le \tan z$. Furthermore, from $(1)$ we see that $\sin^2 z\le z^2$ from which it is straightforward to show that $$\cos z\ge \sqrt{1-z^2}$$ for $0\le z\le 1$. Therefore, we have
$$z\le \tan z \le \frac{z}{\sqrt{1-z^2}} \tag 2$$
Now, letting $z=\sqrt{x-1}$ in $(2)$ reveals for $1 \le x\le 2$
$$\sqrt{x-1}\le \tan\left(\sqrt{x-1}\right)\le \frac{\sqrt{x-1}}{\sqrt{2-x}} \tag 3$$
Dividing both sides of $(2)$ by $\sqrt{x^2-1}$, we obtain for $1<x\le 2$
$$\frac{1}{\sqrt{x+1}}\le \frac{\tan\left(\sqrt{x-1}\right)}{\sqrt{x^2-1}}\le \frac{1}{\sqrt{x+1}\sqrt{2-x}} \tag 4$$
Applying the Squeeze Theorem to $(4)$ gives the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1^+}\left(\frac{\tan\left(\sqrt{x-1}\right)}{\sqrt{x^2-1}}\right)=\frac{1}{\sqrt{2}}}$$
and we are done!
|
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|
Maclaurin Series Representation for $f(z)=\frac{z}{z^4+9}$ I need help finding the Maclaurin series representation for $$f(z)=\frac{z}{z^4+9}$$
I first tried to factorize $z^4+9$, but am I missing something? I could not figure out how to factorize this. Is there another approach to this? I am open to any approach. Just help me figure this out through factorizing if possible.
|
The geometric series $1-x+x^2-x^3+\cdots$ converges to $\frac{1}{x+1}$ for all $|x|<1$.
Substituting $x=\frac{y^4}{9}$ tells us that
$$1-\frac{1}{9}\,y^4 + \frac{1}{9^2} \, y^8 - \frac{1}{9^3}\, y^{12} + \cdots = \frac{1}{\frac{y^4}{9}+1} \equiv \frac{9}{y^4+9}$$
for all $\left|\frac{y^4}{9}\right|<1$, i.e. for all $|y| < \sqrt{3}$.
To get the power series we need, we just divide by $9$ and multiply by $y$:
$$\frac{1}{9}\,y-\frac{1}{9^2}\,y^5 + \frac{1}{9^3} \, y^9 - \frac{1}{9^4}\, y^{13} + \cdots = \frac{y}{y^4+9}$$
for all $|y| < \sqrt{3}$.
|
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|
Find all $(x,y)$ satisfying $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$ Find all pairs $(x,y)$ of real numbers that satisfy the equation $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$
I supposed $a=\sin^2x$ and $b=\cos^2x$
So the equation becomes $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=12+\frac{1}{2}\sin y$
As $a+\frac{1}{a}\geq 2$ and $b+\frac{1}{b}\geq 2$
$12+\frac{1}{2}\sin y\geq 8$
$\sin y\geq -8$
I am stuck here.I could not solve further.Please help me.Thanks.
|
Expanding $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 = 12 + \sin y$
will get us $a^2 + 2 + \frac{1}{a^2} + b^2 + 2 + \frac{1}{b^2} = 12 + \sin y$.
And we can subtract 4 from both sides.
$a^2 + b^2 + \frac{1}{a^2} + \frac{1}{b^2} = 8 + \frac{1}{2}\sin y$
Now make y as a function of x:
$y = asin(2a^2 + 2b^2 + \frac{2}{a^2} + \frac{2}{b^2} - 16)$
Substitute back a and b to get
$y = asin(2sin^4x + 2cos^4x + 2sec^4x + 2csc^4x - 16)$
Because $asin(x)$ is only defined for $0 \leq x \leq 1$, we can say that
$0 \leq 2sin^4x + 2cos^4x + 2sec^4x + 2csc^4x - 16 \leq 1$ or $16 \leq 2sin^4x + 2cos^4x + 2sec^4x + 2csc^4x \leq 17$.
Now we can complete the square with $2sin^4x + 2cos^4x$ to get $16 \leq 2(sin^2x + cos^2x)^2 - 4sin^2xcos^2x + 2sec^4x + 2csc^4x \leq 17$.
This reduces to $16 \leq 2 - 4sin^2xcos^2x + 2sec^4x + 2csc^4x \leq 17$ which can be rewritten as $14 \leq 2sec^4x + 2csc^4x - 4sin^2xcos^2x \leq 15$.
Using some trigonometric identities we get $14 \leq 2sec^4x + 2csc^4x - sin^2(2x) \leq 15$.
So we have y as a function of x, we have the domain, and we have the range.
|
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|
Solve $ \int {\frac {(x+1)dx}{(x^2+x+2)(x^2+4x+5)}} $ $$ \int {\frac {(x+1)dx}{(x^2+x+2)(x^2+4x+5)}} $$
I know that the answer is $\frac{2\sqrt{7}}{21} arctg \frac {2x+1}{\sqrt{7}}-\frac{1}{3}arctg(x+2)+C$, but I have no idea what to do.
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We have: $$ \int {\frac {(x+1)dx}{(x^2+x+2)(x^2+4x+5)}} $$
$$(x^2+x+2):$$
$$\bigtriangleup = 1-4*2=-7$$
$$(x^2+4x+5):$$
$$\bigtriangleup = 16-4*5=-4$$
So: $$\frac{x+1}{(x^2+x+2)(x^2+4x+5)} = \frac{ax+b}{ x^2+x+2} +\frac{cx+d}{x^2+4x+5}$$
$$ x+1= (ax+b)(x^2+4x+5)+(cx+d)(x^2+x+2)$$
$$X+1= ax^3+4Ax^2+5ax+bx^2+4bx+5b+cx^3+cx^2+2cx+dx^2+dx+2d$$
$$\begin{cases}
0=a+c \\
0=4a+b+c+d \\
1=5a+4b+2c+d\\
1=5b+2d
\end{cases}$$
$$\begin{cases}
a=0\\
b=\frac{1}{3}\\
c=0\\
d=\frac{-1}{3}
\end{cases}$$
we finally have:
$$\frac{1}{3}\int \frac{dx}{x^2+x+2}-\frac{1}{3}\int \frac{dx}{x^2+4x+5}=$$
I use : $$a(x-p)^2+q$$
$p=\frac{-b}{2a}$ and $ q=\frac{-\bigtriangleup}{4a} $
Answer:
$$\frac{2\sqrt{7}}{21} arctg \frac {2x+1}{\sqrt{7}}-\frac{1}{3}arctg(x+2)+C$$
|
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|
Prove that $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc} \leq \frac{1}{abc}.$
Let $a,b,$ and $c$ be positive real numbers. Prove that $$\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc} \leq \dfrac{1}{abc}.$$
This question seems hard since we aren't given any information on $a,b,c$ except that they are positive. I tried cross multiplying by $abc$ and proving that $$\dfrac{abc}{a^3+b^3+abc}+\dfrac{abc}{b^3+c^3+abc}+\dfrac{abc}{c^3+a^3+abc} \leq 1.$$ Then we use AM-GM on the denominator. This didn't seem to help me.
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By AM-GM or Muirhead, $\frac{abc}{a^3+b^3+abc}\leq \frac{abc}{a^2b+ab^2+abc}=\frac{c}{a+b+c}$.
|
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|
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$,
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
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$$\int \frac { dx }{ 2x^{ 2 }+x+3 } =\int { \frac { dx }{ 2\left( { x }^{ 2 }+\frac { x }{ 2 } +\frac { 3 }{ 2 } \right) } =\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+\frac { x }{ 2 } +\frac { 1 }{ 16 } -\frac { 1 }{ 16 } +\frac { 3 }{ 2 } } } } =$$ $$\frac { 1 }{ 2 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }+\frac { 23 }{ 16 } } } =\\ =\frac { 1 }{ 2 } \int { \frac { dx }{ \frac { 23 }{ 16 } \left[ \frac { 16 }{ 23 } { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }+1 \right] } } =\frac { 8 }{ 23 } \int { \frac { dx }{ \left[ \frac { 16 }{ 23 } { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }+1 \right] } } =\frac { 8 }{ 23 } \frac { \sqrt { 23 } }{ 4 } \int { \frac { d\left( \frac { 4 }{ \sqrt { 23 } } x+\frac { 1 }{ \sqrt { 23 } } \right) }{ { \left( \frac { 4 }{ \sqrt { 23 } } x+\frac { 1 }{ \sqrt { 23 } } \right) }^{ 2 }+1 } } =\frac { 2 }{ \sqrt { 23 } } \arctan { \left( \frac { 4x+1 }{ \sqrt { 23 } } \right) +C } $$
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|
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
|
Hint: Let $a=\sqrt[3]{35-x^3}$. Then you're solving $ax(a+x)=30$.
$(a+x)^3=\left(a^3+x^3\right)+3ax(a+x)=35+3\cdot 30=5^3$.
$\iff a+x=5$. Then $(5-x)x(5)=30$. Solve this quadratic equation.
|
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|
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement.
|
$$
\begin{align}
&\left(\frac{a^3}b+\frac{b^3}c+\frac{c^3}a\right)-(ab+bc+ca)\\
&=\frac ab\left(a^2-b^2\right)+\frac bc\left(b^2-c^2\right)+\frac ca\left(c^2-a^2\right)\\[3pt]
&=\left(\frac ab-1\right)\left(a^2-b^2\right)+\left(\frac bc-1\right)\left(b^2-c^2\right)+\left(\frac ca-1\right)\left(c^2-a^2\right)\\
&=\frac{(a-b)^2(a+b)}b+\frac{(b-c)^2(b+c)}c+\frac{(c-a)^2(c+a)}a\\[6pt]
&\ge0
\end{align}
$$
This also shows that equality requires $a=b=c$.
|
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|
Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem :
Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$
My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$,
$$
\begin{align}
\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx &= \frac{105}{19}\int^{\pi/2}_0 \frac{\sin (4\pi -8x)}{\cos x}\ \text dx\\
&= \frac{105}{19}\int^{\pi/2}_0 -\frac{\sin 8x}{\cos x}\ \text dx
\end{align}
$$
But it seems it won't work please help thanks
|
Method $2$
Here are hints to another approach.
$1.$ You can observe from the De Moivre's formula that
$$\sin (8x) = 128\sin \left( x \right)\cos^7 {\left( x \right)} - 192\sin \left( x \right)\cos^5 {\left( x \right)} + 80\sin \left( x \right)\cos^3 {\left( x \right)} - 8\sin \left( x \right)\cos \left( x \right)$$
and hence
$$\frac{\sin(8x)}{\sin(x)}=128 \cos^7 {\left( x \right)} - 192 \cos^5 {\left( x \right)} + 80\cos^3 {\left( x \right)} - 8\cos \left( x \right)$$
$2.$ Next use this trick
$$\begin{align}
\cos^7 x= \cos^6 x \cos x = (1-\sin^2 x)^3 \cos x \\
\cos^5 x= \cos^4 x \cos x = (1-\sin^2 x)^2 \cos x \\
\cos^3 x= \cos^2 x \cos x = (1-\sin^2 x)^1 \cos x
\end{align}$$
$3.$ Finally, use the substitution
$$\sin x = u$$
|
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|
Problems & Solutions on Fermat Theorem of Multiple of 3 I am working on an assignment in elementary number theory, in which I have to come up with original problems and then work out their solutions on Fermat theorem of multiple of 3, that is, the equation
$$x^3 + y^3 = z^3$$
does not have integer solution for $xyz \neq 0$.
The first problem that I came up easily is one requiring readers to prove that $x^6 + y^6 = z^6$ does not have integer solution, which is very straightforward. But I need more than just one problem. I have searched around in the internet for ideas but could not find any. Therefore here is my question:
Do you have any ideas, suggestion, links or hints that perhaps can help me with writing couple more question related to the subject? Thank you for your time and help.
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Solve in integers $\left(a-b\right)\left(a^2-b^2\right)\left(a^3-b^3\right)=3c^3$.
To solve it, notice $(3c)^3=(a^2-2b^2+ab)^3 + (2a^2-b^2-ab)^3$,
so either $c=0$ or $a^2-2b^2+ab=0$ or $2a^2-b^2-ab=0$.
Here's where I found the problem.
$$(a-b)\left(a^2-b^2\right)\left(a^3-b^3\right)=(a-b)^3(a+b)\left(a^2+ab+b^2\right)$$
$a^2+ab+b^2=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge 0$ with equality if and only if $a=b=0$.
If $c=0$, then either $a=b$ or $a=-b$, so $(a,b,c)=(k,k,0),(k,-k,0)$, $k\in\mathbb Z$.
If $a^2-2b^2+ab=0$, then $(2a+b)^2=9b^2$, so $2a+b=\pm 3b$.
If $2a+b=3b$, then $a=b$, so $(a,b,c)=(k,k,0)$.
If $2a+b=-3b$, then $a=-2b$, so $(a,b,c)=\left(-2k,k,3k^2\right)$.
If $2a^2-b^2-ab=0$, then $b^2-2a^2+ab=0$ (symmetrical to the other case), so analogously $(a,b,c)=(k,k,0),\left(k,-2k,-3k^2\right)$.
All solutions are $(a,b,c)=(k,k,0),(k,-k,0),(-2k,k,3k^2),(k,-2k,-3k^2)$.
|
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|
Prove that for positive real numbers $a,b,c$ we have $\frac{a}{b+c}+ \frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}.$
Prove that for positive real numbers $a,b,c$ we have $$\dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}.$$
Attempt
I tried using AM-GM and got $ \dfrac{a}{2\sqrt{bc}}+\dfrac{b}{2\sqrt{ac}}+\dfrac{c}{2\sqrt{ab}} \geq \dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b}$ but that doesn't seem to help since that gives an upper not lower bound.
|
We have by AM-GM inequality twice: $\displaystyle \sum_{\text{cyclic}} \dfrac{a}{b+c}=\dfrac{1}{2}\displaystyle \sum_{\text{cyclic}} (a+b)\displaystyle \sum_{\text{cyclic}} \dfrac{1}{a+b}-3\geq \dfrac{1}{2}\cdot 3\cdot 3-3=\dfrac{3}{2}$
|
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|
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
|
$$f\left( x \right) =f\left( 0 \right) +{ f\left( 0 \right) }^{ \prime }x+\frac { { f\left( 0 \right) }^{ \prime \prime } }{ 2! } { x }^{ 2 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime } }{ 3! } { x }^{ 3 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime \prime } }{ 4! } { x }^{ 4 }+...\\ f\left( x \right) =\cos ^{ 6 }{ x } \\ f\left( 0 \right) =1\\ f^{ \prime }\left( x \right) =-6\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime }\left( 0 \right) =-6\cos ^{ 5 }{ 0 } \sin { 0 } =0\\ f^{ \prime \prime }\left( x \right) ={ \left( -6\cos ^{ 5 }{ x } \sin { x } \right) }^{ \prime }=30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x } } \Rightarrow f^{ \prime \prime }\left( 0 \right) =-6\\ f^{ \prime \prime \prime }\left( x \right) ={ \left( 30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x } } \right) }^{ \prime }=-120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime \prime \prime }\left( 0 \right) =0\\ f^{ \prime \prime \prime \prime }\left( x \right) ={ \left( -120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \right) }^{ \prime }=360\cos ^{ 2 }{ x } \sin ^{ 4 }{ x } -360\sin ^{ 2 }{ x } \cos ^{ 4 }{ x } -300\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +60\cos ^{ 6 }{ x } -180\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +36\cos ^{ 5 }{ x } \Rightarrow \\ f^{ \prime \prime \prime \prime }\left( 0 \right) =96\\ \cos ^{ 6 }{ x } =1-\frac { 6 }{ 2! } { x }^{ 2 }+\frac { 96 }{ 4! } { x }^{ 4 }+...=1-3{ x }^{ 2 }+4{ x }^{ 4 }+..\\ $$
|
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Walk me through some basic modular arithmetic? The problem is as follows:
Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?
The solution is as follows:
First we factor $abc + ab + a$ as $a(bc + b + 1)$, so in order for the number to be divisible by 3, either $a$ is divisible by $3$, or $bc + b + 1$ is divisible by $3$.
We see that $a$ is divisible by $3$ with probability $\frac{1}{3}$. We only need to calculate the probability that $bc + b + 1$ is divisible by $3$.
We need $bc + b + 1 \equiv 0\pmod 3$ or $b(c + 1) \equiv 2\pmod 3$. (I get it up to this point because of my lacking modular arithmetic knowledge). Using some modular arithmetic, $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$. The both cases happen with probability $\frac{1}{3} * \frac{1}{3} = \frac{1}{9}$ so the total probability is $\frac{2}{9}$.
Then the answer is $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$
Is there a rule relating that two things being multiplied (b and (c+1)), if you mod them and add the result, would equal the other side? As in, I don't know why $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$. Trying to find a pattern (2+0 = 2 and 1+1 = 2) lead me to that question. Also, where did the 2/3 come from?
I would appreciate someone walking me through the solution as I am not experienced with mod (and my brain is having a hard time functioning without food).
|
In this case you can think mod as an abreviate way for write the conditions. Concretely, you need $3|bc+b+1$. Now, as $3|3$, then $3|bc+b+1-3=bc+b-2$ (it means $b(c+1)\equiv 2(mod3)$). Now, $b$ only can be of the form $b=3n$ or $b=3n+1$ or $b=3n+2$. For the first one, non $c$ exists in order to get $3|bc+b-2$. If $b=3n+1$, then $c$ must be of the form $c=3m+1$ in order to get $3|bc+b-2$ (this means that $b\equiv 1$ and $c\equiv 1$). Analogously the other case.
|
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|
Solve $\sin 2x=-\cos x$ I'm working on solving the problem $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\pi }{2}-x +2\pi k \Leftrightarrow 3x=\frac{\pi }{2} +2\pi k\Leftrightarrow x_1=\frac{\pi }{6}+\frac{2\pi}{3} k\ $
$2x =\pi -(-\frac{\pi }{2}-x) +2\pi k \Leftrightarrow x_2=\frac{3\pi}{2}+2\pi k$
None of these answers are correct.
The correct answer is $x=\frac{\pi}{2}+k\pi$ and $x=-\frac{\pi}{2}\pm\frac{\pi}{3}+2 k\pi$
I don't know what I'm doing wrong in my attempt.
|
Hint:
$$
\sin (2x)=2 \sin x \cos x
$$
so your equation becomes :
$$
2\sin x \cos x + \cos x=0 \iff \cos x(2\sin x+1)=0
$$
|
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|
Nested Radicals Involving Primes How do you evaluate $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots } } } } } $ ?
This question appears to be rather difficult as there is no way to perfectly know what $p_{ n }$ is , if $p_{ n }$ denotes the $n$th prime.
It is simple to show that the value above is convergent. Bertrand`s Postulate implies that $p_{ n } \le 2^n$, which implies that $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots } } } } } \le \sqrt { 2+\sqrt { 4+\sqrt { 8+\sqrt { 16+\sqrt { 32+\dots } } } } } $, which is convergent, as seen here.
So it is pretty clear that $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots } } } } } $ is convergent.
However, in which fashion can you evaluate the value above? If there is no exact way to evaluate it, is it irrational or rational?
The value seems to be about $2.10359749633989726261993..$ as seen here.
Any help would be appreciated.
|
For the primes $(p_k):=(2,3,5,\dots)$ we have
$$\tag{1}0\le\dfrac{p_{n+1}}{p_1p_2\cdots p_n}\le2.$$
The inequality is trivial if $p_n\lt p_{n+1}\lt p_1p_2\cdots p_n$. Otherwise the number $1+p_1p_2\cdots p_n$ is a prime and the inequality holds.
For $a,x\gt0$ we have
$$0\lt\sqrt{a+x}={\sqrt{a}+\int_a^{a+x}\dfrac{d\xi}{2\sqrt{\xi}}}\le
{\sqrt{a}+\int_a^{a+x}\dfrac{d\xi}{2\sqrt{a}}}={\sqrt{a}+\dfrac{x}{2\sqrt{a}}}.$$
We obtain
$$\tag{2}\sqrt{p_1}\lt{\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}}}\le
{\sqrt{p_1}+\dfrac{1}{2\sqrt{p_1}}\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}}\le
{\sqrt{p_1}+\dfrac{1}{2}\sqrt{\dfrac{p_2}{p_1}}+\dfrac{1}{2^2\sqrt{p_1p_2}}\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}\le
{\sqrt{p_1}+\dfrac{1}{2}\sqrt{\dfrac{p_2}{p_1}}+\dfrac{1}{2^2}\sqrt{\dfrac{p_3}{p_1p_2}}+\dfrac{1}{2^3\sqrt{p_1p_2p_3}}\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}\le
{\sqrt{2}+\dfrac{1}{2}\sqrt{2}+\dfrac{1}{2^2}\sqrt{2}+\dfrac{1}{2^3}\sqrt{2}+\dots}\le{2\sqrt{2}}$$
and the limit
$$\sqrt{2}\lt\lim_{n\to\infty}{\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}}}\le{2\sqrt{2}}$$
exists. Though a strictly monotonic bounded sequence converges we prefer to know the error of the sum
$$s_{n-1}:={\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_{n-1}}}}}}}}\;.$$
Similar to above we have
$${\sqrt{p_n}}\le\sqrt{p_n+\sqrt{p_{n+1}+\sqrt{\dots}}}\le{\sqrt{p_n}+\dfrac{1}{2}\sqrt{\dfrac{p_{n+1}}{p_{n}}}+\dfrac{1}{2^2}\sqrt{\dfrac{p_{n+2}}{p_{n}p_{n+1}}}+\dots}$$
and with the inequality $(1)$ we obtain
$${\dfrac{\sqrt{p_n+\sqrt{p_{n+1}+\sqrt{\dots}}}}{2^{n-1}\sqrt{p_1p_2\cdots p_{n-1}}}}\le
{\dfrac{1}{2^{n-1}}\sqrt{\dfrac{p_{n}}{p_1p_2\cdots p_{n-1}}}+\dfrac{1}{2^{n}}\sqrt{\dfrac{p_{n+1}}{p_1p_2\cdots p_{n}}}+\dots}\le
{\dfrac{\sqrt{2}}{2^{n-1}}+\dfrac{\sqrt{2}}{2^{n}}+\dots}\le\dfrac{\sqrt{2}}{2^{n-2}}$$
or
$$\sqrt{p_n+\sqrt{p_{n+1}+\sqrt{\dots}}}\le2\sqrt{2p_1p_2\cdots p_{n-1}}.$$
We define
$$\tag{3}{s_{n-1}(\sigma):=\sqrt{p_1+\sqrt{p_2+\sqrt{\dots+\sqrt{p_{n-1}+\sigma}}}}}.$$
and obtain
$$s_{n-1}(0)\le s_{\infty}\le s_{n-1}(2\sqrt{2p_1p_2\cdots p_{n-1}}).$$
We get
$$s_5(0)=2.1028\dots\le s_{\infty}\le2.1046\dots=s_5(135.94\dots)$$
and
$$s_{10}(0)=2.10359748\dots\le s_{\infty}\le2.10359778\dots=s_{10}(227502.84\dots).$$
|
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Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$
I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But this does not seem to work here.I am stuck.What should i do?
|
you write the integral as $$I=\int \frac{1-\frac{1}{x}}{x^2\sqrt{1+(1-\frac{1}{x})^2}}$$ now put $1-\frac{1}{x}=y$
|
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|
Now am I doing induction correctly? Recursion: $L_n = L_{n-1} + n$ where $L_0 = 1$.
We guess that solution is $L_n = \frac{n(n+1)}{2} + 1$.
Base case: $L_0 = \frac{0(0+1)}{2} + 1 = 1$ is true.
Inductive step: Assume $L_n = \frac{n(n+1)}{2} + 1$ is true for some $n$. We will show that $L_{n+1} = \frac{(n+1)(n+2)}{2} + 1$ given that $L_n = L_{n-1} + n$ is true.
$L_{n+1} = \frac{(n+1)(n+2)}{2} + 1 = L_n + (n+1)$
$L_n = \frac{(n+1)(n+2)}{2} + 1 - (n+1)$
$L_n = \frac{(n+1)(n+2)}{2} + \frac{2}{2} - \frac{2n+2}{2} = \frac{n^2+3n+2 + 2 - 2n - 2}{2}$
$L_n = \frac{n^2+n+2}{2} = \frac{n^2+n}{2} + 1 = \frac{n(n+1)}{2} + 1$
This completes the proof.
Is everything in place for a correct induction proof? Is anything wrong? Backwards? Unclear? Awkward?
|
The induction schema is (base) $P(0)$ is true (might start elsewhere too), (induction) if $P(n)$ is true, we prove $P(n + 1)$ is true, (conclusion) $P(n)$ is true for all $n \in \mathbb{N}_0$.
You have to work from $P(n)$ to $P(n + 1)$ somehow.
In your case, the claim is that $L_n = n (n + 1) / 2 + 1$.
Base: For $n = 0$ you have that $L_0 = 1$ by definition, and the formula gives $\frac{0 (0 + 1)}{2} + 1 = 1$. This is true.
Induction: By hypothesis, for $n$ we have:
$\begin{align}
L_n = \frac{n (n + 1)}{2} + 1
\end{align}$
We know that:
$\begin{align}
L_{n + 1}
&= L_n + n + 1 \\
&= \frac{n (n + 1)}{2} + 1 + n + 1\\
&= \frac{(n + 2) (n + 1)}{2} + 1
\end{align}$
(the first step by the definition of $L_{n + 1}$, the second one uses the induction hypothesis, then just algebra)
This is the claim for $n + 1$, as we wanted to prove.
Conclusion: By induction, the claim is true for all $n \ge 0$.
|
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Minimum number of terms resulting from the product of two polynomials with a given number of terms Given two integers ($n$, $m$), what is the smallest number of terms that could result from the product of two polynomials with $n$ and $m$ non-zero terms respectively?
That is, what is the smallest number of non-zero terms that could result from the following product:
$$(a_1 x^{b_1} + \dots + a_n x ^ {b_n})(c_1 x^{d_1} + \dots + c_m x ^ {d_m})$$
where $a_i$ and $c_i$ are non-zero.
The answers to this question show that the answer cannot be $1$ in all cases, but it is obviously less than $n+m$, which should theoretically be the maximum.
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The minimum is 2 for all cases. For even number of terms, there is an elegant factorization solution, but for odd number of terms, my solution made use of complex roots.
Let $A$ have $a$ terms, $B$ have $b$ terms. $C=A\cdot B$.
We prove that for any $a,b\geq2$, we can always find $A,B$ so that $C$ has $2$ terms, which is obviously the minimum.
*
*$a$ or $b$ (or both) is even: assume $b$ is even. \begin{align*}
x^{\frac{ab}{2}}-1&=(x^a-1)\left[(x^a)^{\frac{b}{2}-1}+(x^a)^{\frac{b}{2}-2}+\dots+1\right] \\
&=(x^{a-1}+x^{a-2}+\dots+1)(x-1)\left[(x^a)^{\frac{b}{2}-1}+(x^a)^{\frac{b}{2}-2}+\dots+1\right]
\end{align*} Let $A=x^{a-1}+x^{a-2}+\dots+1$, $B=(x-1)\left[(x^a)^{\frac{b}{2}-1}+(x^a)^{\frac{b}{2}-2}+\dots+1\right]$. It's easy to check that $B$ has $b$ terms, since no terms in the expansion will cancel out.
*Both $a$ and $b$ odd: (I did not find a similar proof like (1), so I used complex roots.) We demonstrate with the example $a=3,b=7$. The general case only substitutes numbers as $a$s and $b$s.
We wish to construct $A$ to have $2$ roots and $B$ to have $b$ roots. Plot the $8$ complex number solutions of $x^8+1=0$ on the unit circle. Name them $z_1,z_2,\dots,z_8$.
Because $a,b$ is odd, $z_5=\overline{z_4},z_6=\overline{z_3},\dots,z_8=\overline{z_1}$ Let \begin{gather*}
A=(x-\mathcal{E}_1)(x-\overline{\mathcal{E}_1}) \\
B=(x-\mathcal{E}_2)(x-\overline{\mathcal{E}_2})(x-\mathcal{E}_3)(x-\overline{\mathcal{E}_3})(x-\mathcal{E}_4)(x-\overline{\mathcal{E}_4})
\end{gather*} It's easy to check that both $A$ and $B$ have non-zero co-efficients.
Note: the complex number solution can yield elementary factorization solutions for $a=3,b=3$. \begin{gather*}
\underbrace{x^4+4}_\text{2 terms}=(\underbrace{x^2+2x+2}_\text{3 terms})(\underbrace{x^2-2x+2}_\text{3 terms})\tag{*} \\
A=\left((x-\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\right)\left(x-\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\right)=x^2-\sqrt{2}x+1
\end{gather*} Similarly, we calculuate $$B=x^2+\sqrt{2}x+1$$ So $$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$$ Let $x=\frac{\sqrt{2}}{2}t$ and we'd get: $$\frac{t^4}{4}+1=\left(\frac{t^2}{2}+t+1\right)\left(\frac{t^2}{2}-t+1\right)$$ which is exactly (*).
For a handwritten version of this answer with diagrams, see photo 1.
|
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Find the values of a and k from the curve
The diagram below shows a curve with equation of the form ${y = kx(x +
a)^2}$, which passes through the points (-2, 0), (0, 0) and (1, 3).
What are the values of a and k.
I know my roots are x = -2, x = 0 and x = 3.
But as the y intercept is 0, I don't see how I can get any meaningful values for k:
${k(x + 2)(x - 3) = 0}$
How can I find the values of k and a?
|
Be careful, $x=3$ is not a root. The curve does not cross the $x$-axis when $x=3$. There are only two roots: $x=-2$ and $x=0$.
You know that $(-2,0)$, $(0,0)$ and $(1,3)$ are all points on the curve.
You should substitute $x=-2$ and $y=0$ to get one equation. Then substitute $x=0$ and $y=0$ to get a second equation. Finally substitute $x=1$ and $y=3$ to get a third equation. Then solve these simultaneously.
As you rightly say: $(x,y)=(0,0)$ doesn't give you anything.
When we substitute $(x,y)=(-2,0)$ into $y=kx(x+a)^2$ we get $0=-2k(-2+a)^2$. So, either $k=0$ or $a=2$. Obviously $k \neq 0$ or the whole question would collapse from $y=kx(x+a)^2$ down to $y=0$ which can't be right because the original graph isn't the line $y=0$. That means $a=2$.
When we substitute $(x,y)=(1,3)$ into $y=kx(x+a)^2$ we get $3=k(1+a)^2$. We know that $a=2$, so this becomes $3=k(1+2)^2$, which in turn becomes $3=9k$. This tells us that $k=\frac{1}{3}$.
Given that $a=2$ and $k=\frac{1}{3}$, the equation $y=kx(x+a)^2$ becomes
$$\color{red}{y=\frac{1}{3}x(x-2)^2}$$
|
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|
To find area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$ How to find the area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$ ? I am completely stuck ; please help . Thanks in advance
|
First, lets see where both surfaces intersect. In polar coordinates, we have
$$
z=3-r = 1+ r^2\quad \Rightarrow r=1.
$$
So the intersection is the circle $r=1$ at height $z=2$.
The area of your surface is the area $A_1$ of $z=3-\sqrt{x^2+y^2}$ above this circle (let $S_1$ be this surface), plus the area $A_2$ of $z=1+x^2+y^2$ beneath this circle (let $S_2$ be this surface).
You can parametrize $S_1$ with the vectorial function $f$:
$$
x=x, \quad y=y, \quad z=3-\sqrt{x^2+y^2},
$$
and $S_2$ with $g$:
$$
x=x, \quad y=y, \quad z=3+x^2+y^2,
$$
with $ (x,y)\in D=\{(r,\theta)\;|\;0\le r\le 1, 0\le \theta \le 2\pi\}$. Therefore, the total area equals
$$
A_1+A_2=\iint_D ||f_x\times f_y|| dA + \iint_D ||g_x\times g_y|| dA\\
=\sqrt{2}\int_0^{2\pi}\int_{0}^1rdrd\theta+\int_0^{2\pi}\int_{0}^1r\sqrt{4r^2+1}drd\theta\\
=\sqrt{2}\pi+\frac{5\sqrt{5}-1}{6}\pi
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that if $0 < a,b,c <1$, then $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$.
Prove that if $0 < a,b,c <1$, then $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$.
I think that using AM-GM might work. Thus we have $\dfrac{a+b+c}{3}+\sqrt{(1-a)(1-b)(1-c)} \geq \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}$ but I am not sure how to proceed. Maybe the maximum occurs when $a=b=c$ but how to prove that?
|
Note that $c<1$ and $1-c<1$ so :
$$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt{ab}+\sqrt{(1-a)(1-b)}$$
Now use AM-GM :
$$\sqrt{ab}+\sqrt{(1-a)(1-b)} \leq \frac{a+b}{2}+\frac{(1-a)+(1-b)}{2} =1$$ and so the conclusion follows .
|
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|
What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$
I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$.
|
The easiest method seems to be to use Lagrange multipliers.
$$
L = \sqrt{6} xy + 4 yz - \frac{1}{2} \lambda (x^2 + y^2 + z^2 - 1) \, ,
$$
for some $\lambda$. Looking at derivatives
$$
\frac{\mathrm{d} L}{\mathrm{d} x} = \sqrt{6} y - \lambda x = 0
$$
$$
\frac{\mathrm{d} L}{\mathrm{d} y} = \sqrt{6} x + 4z - \lambda y = 0
$$
$$
\frac{\mathrm{d} L}{\mathrm{d} z} = 4y - \lambda z = 0
$$
Another equation is the original constraint $x^2 + y^2 + z^2 = 1$. Now one has to solve the set of equations. After a few steps one gets $\lambda^2 = 22$ and $x = \pm \sqrt{3/22}$. From that it follows that $y = \pm \sqrt{1/2}$ and $z = \pm \sqrt{4/11}$.
The signs of $x,y,z$ are not independent, but the maximum of $\sqrt{6}xy + 4 yz$ is at $(\sqrt{3/22},\sqrt{1/2},\sqrt{4/11})$ and it is $\sqrt{11/2}$.
|
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|
Is it true that for every $k\in\mathbb N$ , there exist infinitely many $n \in \mathbb N$ such that $kn+1 , (k+1)n+1$ both are perfect squares ? Is it true that for every $k\in\mathbb N$ , there exist infinitely many $n \in \mathbb N$ such that $kn+1 , (k+1)n+1$ both are perfect squares ? What I have tried is that I have to necessarily solve $a^2-b^2=n$ for given $n$ ; but I can not proceed further . Please help . Thanks in advance
|
The answer is Yes.
I. (Update)
The solution to,
$$\begin{aligned}
kn+1 &= x^2\\
(k+1)n+1 &= y^2
\end{aligned}$$
is given by,
$$n = \frac{ -(\alpha^2 + \beta^2) + \alpha^{2(2m+1)}+\beta^{2(2m+1)} }{4k(k+1)}$$
where,
$$\alpha = \sqrt{k}+\sqrt{k+1}\\
\beta = \sqrt{k}-\sqrt{k+1}$$
For example, for $m=1,2,3,\dots$ we get,
$$\begin{aligned}
n &= 8 + 16 k \\
n &= 24 + 176 k + 384 k^2 + 256 k^3 \\
n &= 48 + 736 k + 3968 k^2 + 9472 k^3 + 10240 k^4 + 4096 k^5
\end{aligned}$$
and so on.
II. (Old answer)
$$\begin{aligned}
kn+1 &= x^2\\
(k+1)n+1 &= y^2
\end{aligned}\tag1$$
Eliminate $n$ between them and we get the Pell-like,
$$(k+1)x^2-ky^2 = 1$$
We can get an infinite number of solutions using a transformation (discussed in this post). Let $p,\,q = 4k+1,\;4k+3$, then,
$$x = p u^2 + 2 k q u v + k (k+1) p v^2$$
$$y = q u^2 + 2 (k+1)p u v + k (k+1) q v^2$$
and $u,\color{brown}\pm v$ solve the Pell equation,
$$u^2-k(k+1)v^2 = 1$$
This has initial solution,
$$u = 2 k+1,\quad v = 2$$
and an infinite more. Thus,
$$\begin{aligned}
n &= 8 + 16 k \\
n &= 24 + 176 k + 384 k^2 + 256 k^3 \\
n &= 48 + 736 k + 3968 k^2 + 9472 k^3 + 10240 k^4 + 4096 k^5 \\
n &= 80 + 2080 k + 20096 k^2 + 93952 k^3 + 235520 k^4 + 323584k^5 + 229376 k^6 + 65536 k^7
\end{aligned}$$
and so on for an infinite number of $n$ for any $k$.
|
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|
Given complex $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$ : when and what is $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
If $z_{1},z_{2},z_{3}$ are three complex number Such that $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$
Then $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
$\bf{My\; Try::}$ Let $z_{1}=2\left(\cos \alpha+i\sin \alpha\right)$ and $z_{2}=3\left(\cos \beta+i\sin \beta\right)$ and $z_{3}=4\left(\cos \gamma+i\sin \gamma\right)$
So $$f(\alpha,\beta,\gamma) = 58-\left[12\cos(\alpha-\beta)+24\cos(\beta-\gamma)+16\cos(\gamma-\alpha)\right]$$
Now How can I calculate $\max$ of $f(\alpha,\beta,\gamma)$
Help me
Thanks
|
$$\dfrac S4=3\cos(A-B)+6\cos(B-C)+4\cos(C-A)$$
$$=\cos A(3\cos B+4\cos C)+\sin A(3\sin B+4\sin C)+6\cos(B-C)$$
$$=\sqrt{25+24\cos(B-C)}\cos\left(A-\arccos\dfrac{3\cos B+4\cos C}{3\sin B+4\sin C}\right)+6\cos(B-C)$$
$$\le\sqrt{25+24\cos(B-C)}+6\cos(B-C)$$
If $\sqrt{25+24\cos(B-C)}=y, 1\le y\le7$ and $\cos(B-C)=\dfrac{y^2-25}{24}$
$$S\le4y+24\cdot\dfrac{y^2-25}{24}=4y+y^2-25=(y+2)^2-29\le-29$$
$$\implies f(\alpha+\beta+\gamma)\ge58+29$$
The equality occurs if $y=2$ and $A-\arccos\dfrac{3\cos B+4\cos C}{3\sin B+4\sin C}=2n\pi$ where $n$ is any integer
|
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|
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}{x+4} & \quad -4 < x < 0 \\
\frac{x}{x+4} & \quad x < −4
\end{array}
\right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
|
You need to analyze each case separately, and it is always wrong to round off in mathematics without explicitly stating that it is an approximation. Do not forget that to find the solutions you need to take into account the cases! Also, you missed the case of $x = 0$.
$\def\eq{\Leftrightarrow}$
Take any real $x \ne -4$.
If $0 \le x$:
$\left| \frac{x}{x+4} \right|<4 \eq \frac{x}{x+4} < 4 \eq x < 4x+16 \eq x > -\frac{16}{3}$.
If $-4 < x < 0$:
$\left| \frac{x}{x+4} \right|<4 \eq -\frac{x}{x+4} < 4 \eq -x < 4x+16 \eq x > -\frac{16}{5}$.
If $x < -4$:
$\left| \frac{x}{x+4} \right|<4 \eq \frac{x}{x+4} < 4 \eq x > 4x+16 \eq x < -\frac{16}{3}$.
Thus $\left| \frac{x}{x+4} \right|<4 \eq \left( 0 \le x \land x > -\frac{16}{3} \right) \lor \left( -4<x<0 \land x > -\frac{16}{5} \right) \lor \left( x<-4 \land x < -\frac{16}{3} \right)$.
Thus $\left| \frac{x}{x+4} \right|<4 \eq x \ge 0 \lor -\frac{16}{5}<x<0 \lor x<-\frac{16}{3}$.
["$\land$" means "and" and "$\lor$" means "or".]
|
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|
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(x)=f\left( \frac{x+1}{2} \right)+f\left( \frac{x-1}{2} \right)$.
But how to find all functions?
|
We can show that $f(x)=-f(-x)$, hence $f(x)$ must be an odd function. If $f(x)$ is a polynomial function then we can write it as:$$f(x)=\sum_{i=0}^\infty a_ix^{2i+1}\tag{1}$$
We can also show that:$$f(x^2)=x\,f(x)\tag{2}$$
Using (1) we get:$$f(x^2)=\sum_{i=0}^\infty a_ix^{4i+2}=a_0x^2+a_1x^6+a_2x^{10}+\cdots\tag{3}$$
and:$$x\,f(x)=\sum_{i=0}^\infty a_ix^{2i+2}=a_0x^2+a_1x^4+a_2x^6+\cdots\tag{4}$$
Using (2) we know that (3) and (4) must be equivalent which means $a_i=0$ for all $i\gt0$. Hence:$$f(x)=a_0x$$
|
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|
To obtain the condition for vanishing of the given determinant If $a,b,c$ are distinct real numbers obtain the condition under which the following determinant vanishes.
$$\left|
\begin{array}{cc}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{array}
\right|$$
My answer: After a little calculation I was able to show that $D=0$ reduces to $abc=-1$.
Is there a simpler one line answer to this? especially since this matrix looks suspiciously similar to the Vandermonde matrix?
|
This is NOT a one line solution you are expecting but it is an idea which is very useful in solving such determinants which may remind one of Vandermonde matrix.
I'm replacing $a$ by $x$ (for the sake of clarity). Using the linearity of determinants we get
$$\det=
\begin{vmatrix}
x & x^2 & 1+x^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{vmatrix}
=\begin{vmatrix}
x & x^2 & 1\\
b & b^2 & 1\\
c & c^2 & 1\\
\end{vmatrix}+\begin{vmatrix}
x & x^2 & x^3\\
b & b^2 & b^3\\
c & c^2 & c^3\\
\end{vmatrix}
=A(x)+B(x).
$$
The determinant on the left can be thought of as a third degree polynomial in $x$. Let us call the first determinant (on right) as $A(x)$ (a second degree polynomial in $x$) and the second determinant (on right) as $B(x)$ (a third degree polynomial in $x$).
First consider $A(x)$. Observe that for $x=b$ or $x=c$, this determinant is $0$. Thus both $x-b$ and $x-c$ are factors of this polynomial. Thus
$$A(x)=K(x-b)(x-c).$$
Moreover
$$A(0)=bc(c-b) = K(bc).$$
Thus $K=c-b.$ This means
$$A(x)=(c-b)(x-b)(x-c).$$
Likewise
$$B(x)=bc(c-b)x(x-b)(x-c).$$
Thus the given determinant is
$$\det=(c-b)(x-b)(x-c)[1+bcx].$$
Putting back everything in terms of $a$, we get
$$\det=(c-b)(a-b)(a-c)[1+abc].$$
Now you get all the conditions when this can be $0$, namely
$$a=b \quad \text{ or } \quad b=c \quad \text{ or } \quad c=a \quad \text{ or } \quad abc=-1.$$
|
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|
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
|
One of your examples is that $2^{11} + 1$ is divisible by $3$. We investigate as follows:
Let us consider instead raising $2$ to an even power and subtracting $1$. And then let us factor.
Example: $2^{10} - 1 = (2^5 - 1)(2^5 + 1)$.
Among any three consecutive integers, exactly one of them must be divisible by $3$.
Clearly $2^5$ is not divisible by $3$, so either its predecessor or successor is divisible by $3$.
That is, either $2^5 - 1$ or $2^5 + 1$ is divisible by $3$, whence their product is, as well.
Okay: Their product is $2^{10} - 1$, which we have now established is divisible by $3$.
This number is still divisible by $3$ after being doubled, and still divisible by $3$ when we add $3$ to it.
So: We have that $2(2^{10} - 1) + 3 = 2^{11} - 2 + 3 = 2^{11} + 1$ is divisible by $3$ as desired.
A similar bit of reasoning around $2^{2k} - 1$ yields the assertion at hand. "QED"
|
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|
Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n=2$,
$$
x^2+y^3=z^4
$$
I'm already pretty lost.
After this it gets even more complicated, has anyone encountered this problem before?
I want all the solutions, or at least infinitely many for every equation.
|
Here's a proof of infinitely many solutions $(x_1, x_2, x_3, \dots, x_n, z)$:
For $n = 1$ we have $(1, 1)$. For $n = 2$ we have $(3^3, 2 \cdot 3^2, 3^2)$.
For $n \ge 3$ we have $(2, 3, 1, 2, 2, 2, 2, \dots, 2)$.
(To verify: $2^2 + 3^3 + 1^4 + 2^5 + 2^6 + \dots + 2^n = 32 + 2^5 (1 + 2 + 4 + \dots + 2^{n-5}) = 32 + 2^5 (2^{n-4} - 1) = 2^{n+1}$, where we have used a geometric series to simplify the sum.)
Thus all $n \ge 1$ yields at least one solution in positive integers. To get infinitely many solutions, multiply term $x_k$ by $m^{\frac{(n+1)!}{k}}$ and $z$ by $m^{n!}$; then both sides of the Diophantine equation are multiplies by $m^{(n+1)!}$. Choosing arbitrary $m \ge 2$ gives infinitely many solutions, as desired.
|
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|
First derivative meaning in this case If we have a function:
$$f(x)=\frac{x}{2}+\arcsin{\frac{2x}{1+x^2}}$$
And it's first derivative is calculated as:
$$f'(x)=\frac{1}{2}+\frac{1}{\sqrt{1-\big(\frac{2x}{1+x^2}\big)^2}}\frac{2+2x^2-4x^2}{(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}\cdot(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}=$$
$$\frac{1}{2}+\frac{2\cdot sgn(1-x^2)}{1+x^2}$$
Why did my teacher say the critical points are at $x=\pm1$? Then she wrote:
$$f_+'(1)=-\frac{1}{2}$$
$$f_-'(1)=\frac{3}{2}$$
and I'm not sure what she meant by that eather, nor by the following:
$$f'(x)=
\begin{cases}
\frac{x^2-3}{2(x^2+1)}, & |x|>1 \\
\frac{x^2+5}{2(x^2+1)}, & |x|<1
\end{cases}$$
Thank you for your time.
|
You have obtained that $f'(x)=\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}.$ Note that this expression doesn't exist if $x=\pm 1.$ I don't think that critical point is the most adequate word but your teacher was saying that $f'(-1)$ and $f'(1)$ don't exist.
Now, $$f_+'(1)=\lim_{x\to 1^+}f'(x)=\lim_{x\to 1^+} \left(\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}\right)= \lim_{x\to 1^+} \left(\frac{1}{2}-\frac{2}{(1+x^2)}\right)=-\frac{1}{2},$$ where it is used that if $x>1$ then $\frac{1-x^2}{|1-x^2|}=-1.$ The value of $f_-'(1)$ can be obtained in a similar way.
Finally, note that if $|x|>1$ then $\frac{1-x^2}{|1-x^2|}=-1$ and if $|x|<1$ then $\frac{1-x^2}{|1-x^2|}=1.$ So, you have two different expressions for $f'(x):$ one if $|x|>1$ and other if $|x|<1.$
|
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|
Convert $f(x)=(\cos(x))^3$ to powers of x and find if converges. I started out by writing the Taylor series for $x_0=0$ (Maclaurin series) of $f(x)=(\cos(x))^3$.
If my calculations are correct $$f(x)=1-\frac{3x^2}{2!}+\frac{21x^4}{4!}-\frac{183x^2}{6!}+...$$
and after some simplifications $$f(x)=\sum\limits_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}a_n$$ where $a_n$ is a sequence of $a_0=1, a_1=3, a_2=21,a_3=183, ..$
Is there any way to find the $a_n$ sequence or a better way to solve the whole problem?
|
By using the identity $\cos(3x)=4\cos^3 x-3\cos x$ it follows
\begin{align}
\cos^3x&=\frac{1}{4}\cos (3x)+\frac{3}{4}\cos x\\
&=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(3x)^{2k}}{(2k)!}+\frac{3}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(x)^{2k}}{(2k)!}\\
&=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{3(3^{2k-1}+1)}{(2k)!}x^{2k}
\end{align}
|
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|
If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
If $\sin x+\sin^{2} x=1$, then the value of
$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
is equal to
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $
$\boxed{\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2\\
=\cos^{6} x\left(\cos^{6} x+3\cos^{4} x+3\cos^{2} x+1\right)+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\cos^{2} x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\sin x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin x[\sin x+1]\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin^{2} x+\sin x\right)^{3}+2\sin^{2} x+\sin x-2\\
=1+\sin^{2} x+\sin^{2} x+\sin x-2\\
=\sin^{2} x} $
I found this solution but considering the time as $1-2$ min to solve this question,
I am looking for a short and simple way.
I have studied maths up to $12$th grade
|
Hint: $\sin x = 1 - \sin^2 x = \cos^2x \Rightarrow \sin^2 x = \cos^4 x \Rightarrow \cos^4 x = 1 - \cos^2 x$. Thus you can express $\cos^{10}x , \cos^8 x, \cos^6 x, \cos^4x$ in term of $\cos^2 x$.
|
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|
Probability of almost correct order Let's say I'm a teacher handing tests back to seven students.
If I do it with my eyes closed, what's the probability I hand exactly 5 of the tests back to the correct students?
There are many possible ways this could happen, and if my understanding of probability is correct, the sum of the individual probabilities of all possibilities where I hand exactly 5 back to the correct student yields the answer I'm looking for.
One such probability (hand out the first two tests incorrectly):
$$\frac{6}{7}\cdot\frac{5}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
Another probability (hand out the last two tests incorrectly):
$$\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
One more:
$$\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
We can simplify these probabilities to:
$$\frac{6\cdot5}{7!}, \frac{1\cdot1}{7!}, \frac{4\cdot2}{7!} $$
More generally, we can think about all possible times when the two incorrect letters might occur, and deduce that the sum of all possible probabilities is:
$$\frac{6(5+4+3+2+1)+5(4+3+2+1)+4(3+2+1)+3(2+1)+2(1)+1(1)}{7!}$$
Is this really the answer? I really feel like I missed something that would simplify this answer.
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Or as another way way of reaching @Zubin Mukerjee correct answer:
There are $\binom{7}{5}$ ways to pick the 5 students who get their proper papers, and only one way to give the remaining two the wrong papers...
Edit to respond to OP's comment...
Well, consider the first way you enumerate.. Your first two terms give the total probability of getting the first two wrong, including times when the first two got papers from #3 through #7. But that's not possible, since you then give #3 through #7 their correct papers...
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"timestamp": "2023-03-29T00:00:00",
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Value of $\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$ To find the value of:
$\displaystyle\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$
The answer says: $1/2$
My attempt:
$=(\displaystyle\sum_{k=1}^\infty 1/{k^2})-1-(\sum_{k=1}^\infty 1/{k^3})+1+(\sum_{k=1}^\infty 1/{k^4})-1-\cdots$
$=(1+1/2^2+1/3^2+1/4^2+\cdots)-(1+1/2^3+1/3^3+1/4^3+\cdots)+(1+1/2^4+1/3^4+1/4^4+\cdots)-\cdots$
$=(1/2^2-1/2^3+1/2^4-\cdots)+(1/3^2-1/3^3+1/3^4-\cdots)+(1/4^2-1/4^3+1/4^4-...)+\cdots$
$=S_1+S_2+S_3+\cdots$
With:
$S_2=(1/2^2-1/2^3+1/2^4-\cdots)$
$S_3=(1/3^2-1/3^3+1/3^4-\cdots)$
$S_4=(1/4^2-1/4^3+1/4^4-\cdots)$
...
It looks like geometry series with alternating signs.
If I apply the geometric series value: $S=\frac{1}{1-z}$
Gives:
$S_2=2-1-1/2=1/2$
$S_3=3/2-1-1/3=1/2-1/3$
$S_4=4/3-1-1/4=1/3-1/4$
...
Original equation:
$=1/2+1/2-1/3+1/3-1/4+1/4-1/5+\cdots=1$
This is different than the answer (1/2) given.
Could any one help? Thanks,
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So I made calculation error when using the equation on geometric series.
$S_2=2/3-1+1/2=1/2-1/3$ (with z=-1/2)
$S_3=3/4-1+1/3=1/3-1/4$ (with z=-1/3)
$S_4=4/5-1+1/4=1/4-1/5$ (with z=-1/4)
Sum them up gives the value of original equation:
$=1/2-1/3+1/3-1/4+1/4-1/5+...=1/2$
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Fractions in Questions and Answers
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