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Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? $$\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$$
Do you have an idea about this serie? If it converges what is the sum?
|
$\frac{2n+1}{n(n+1)}=\frac{1}{n}+\frac{1}{n+1}$
So,
$\frac{3}{1.2}-\frac{5}{2.3} +\frac{7}{3.4}...= \frac{1}{1}+\frac{1}{2}-\frac{1}{2} -\frac{1}{3}+\frac{1}{3}+\frac{1}{4}....$
That should give you the idea.
|
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|
Determining Complex Differentiability using Cauchy-Riemann Equations I need to find where $f(x+iy)=-6(\cos x+i\sin x)+(2-2i)y^3+15(y^2+2y)$ is complex differentiable.
I first rearranged the function into its real and imaginary parts:
$f(x+iy)=(-6\cos x+2y^3+15y^2+30y)+i(-6\sin x-6y^2)$
That means $u(x,y)=-6\cos x+2y^3+15y^2+30y$ and $v(x,y)=-6\sin x-6y^3$.
Then, if we take the partial derivative of u and v in terms of x and y:
$u_x=6\sin x$
$u_y=6y^2+30y+30$
$v_x=-6\cos x$
$v_y=-18y^2$
Then, by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$.
This means that: $6\sin x=-18y^2$ and $6y^2+30y+30=6\cos x$.
This is where I am stuck. How do I solve for x and y? I was thinking that I could proceed in this way:
$\sin^2 x + \cos^2 x=1 \Rightarrow (-3y^2)^2+(y^2+5y+5)^2=1 \Rightarrow 10y^4+10y^3+35y^2+50y+24=0$
However, from here, how do I solve for y and then solve for x? I'd appreciate any tips. Thanks for your help in advance!
|
We can simplify these equations into
$$ \sin{x} = -3 y^2, \quad y^2 + 5y + 5 = \cos{x} $$
Now, $x$ and $y$ are real, so the only way that the first equation can be satisfied is if $y \in [-1/\sqrt{3}, 1/\sqrt{3}]$, since otherwise $-3y^2$ will not be in the range of sine.
Now we turn to the second equation. The graph of $y^2 + 5y + 5$ achieves its minimum at $y = -5/2$; therefore it is monotone on the interval $[-1/\sqrt{3}, 1/\sqrt{3}]$. So we can simply check the endpoints to verify that $y^2 + 5y + 5 > 1$ when $y \in [-1/\sqrt{3}, 1/\sqrt{3}]$, and hence it can never equal $\cos{x}$ if the first equation is satisfied.
|
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|
Integrating a Rational Function I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will simplify it enough to solve in reasonably quick . I plugged it into wolfram and as usual it doesn't give a quick way either. If anybody has a way this integral can be solved quickly, as there will be a time crunch on my test.
$$\int\frac{x^3+x+2}{x^4+2x^2+1}dx$$
|
From partial fractions and two variable substitutions
$$\begin{align}
\int \frac{x^3+x+2}{x^4+2x^2+1}dx &= \int\frac{x}{x^2+1}dx + 2\int\frac{1}{x^2+1}\frac{1}{x^2+1}dx \\
&= \frac{1}{2}\int\frac{du}{u}+\int\frac{1}{(\tan v)^2+1}dv
\end{align}$$
where $u=x^2+1$ and $v=\arctan x$ (recall that $\frac{d}{dx}\arctan x = \frac{1}{1+x^2} $). Now since $\tan^2v+1 = \sec^2 v = 1/\cos^2 v$,
$$\begin{align}
\int \frac{x^3+x+2}{x^4+2x^2+1}dx &= \frac{\ln u}{2} + 2\int \cos^2(v)\, dv \\
&= \frac{\ln u}{2}+2\left(\frac{\sin v \cos v}{2} +\frac{v}{2}\right) \\
&= \frac{\ln \left(x^2+1\right)}{2}+\sin \left(\arctan x\right) \cos \left(\arctan x\right) +\arctan x \\
&= \frac{1}{2}\ln \left(x^2+1\right) + \frac{x}{x^2+1} + \arctan x
\end{align}$$
Hope this helps.
|
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|
Eigenvalues and Jordan form I have a $5\times 5$ matrix and I need to find the Jordan form and its inverse. I know how to find the inverse. But for the Jordan form I am screwed.
The matrix is $$\begin{bmatrix}3 & 0 & 0 & 0 & 0\\2 & 3 & 0 & 0 & 0\\1 & 0 & 2 & 2 & 0\\0 & 0 & 0 & 3 & 2\\0 & 0 & 0 & 0 & 2\end{bmatrix}.$$
Please help me to find the eigenvalues and number of Jordan blocks. Is there any easy method to know the eigenvalues other than solving $\det(A-\lambda I)=0$?
|
Note that
$$
M=\left[\begin{array}{cc|ccc}
3 & 0 & 0 & 0 & 0\\
2 & 3 & 0 & 0 & 0\\
\hline
1 & 0 & 2 & 2 & 0\\
0 & 0 & 0 & 3 & 2\\
0 & 0 & 0 & 0 & 2
\end{array}\right]
=\begin{pmatrix}A&0\\C&D\end{pmatrix}
$$
is lower block triangular. Therefore the eigenvalues of $M$ are precisely the eigenvalues of $A$ and $D$. Yet $A$ and $D$ are triangular matrices. So you can read off their eigenvalues $3,3,2,3,2$ immediately from their diagonals.
Having the eigenvalues, you may find the eigenvectors and the dimensions of the eigenspaces the "old" way (i.e. by inspecting the system $(M-\lambda I)x=0$. Alternatively, note that
$$
F=\begin{pmatrix}1&0\\0&0\\0&0\end{pmatrix}
\ \Rightarrow\ \begin{pmatrix}I&0\\-F&I\end{pmatrix}
\begin{pmatrix}A&0\\C&D\end{pmatrix}
\begin{pmatrix}I&0\\F&I\end{pmatrix}
=\begin{pmatrix}A&0\\0&D\end{pmatrix}.
$$
Therefore $M$ is similar to $A\oplus D$ and its Jordan form is just the direct sum of the Jordan forms of $A$ and $D$. Now the Jordan form of $A$ is clearly $J_2(3)$, the $2\times2$ Jordan block for the eigenvalue $3$. And the Jordan form of $D$ is $3\oplus J_2(2)$, because the nullity of $D-2I$ is $1$. Putting the pieces together, the Jordan form of $M$ is $J_2(3)\oplus 3\oplus J_2(2)$. You may confirm this using some online computer algebra system.
|
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|
Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2} \\
&= ab+bc+ca
\end{align}
$$
but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?
|
Hint: Multiply the original inequality by $a+b+c$ on the LHS and $3$ on the RHS, expand and eliminate common terms and you will arrive at something you have proved.
|
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|
How to solve this irrational equation? How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$
|
Note that, with $x=1$, $y=\sqrt{8x+1}$ takes the value $y=3$ and with $x=3$, $y=\sqrt{8x+1}$ takes the value $y=5$. Put the point $A(1, 3)$ and $B(3,5)$. The equation of the line passing two points $A$ and $B$ is $y = x + 2.$ And then we write $$\sqrt{8x + 1}-(x+2).$$ Similar to with $$(x+1- \sqrt{6x - 2}).$$
We write the given equation has the form
$$ \sqrt{8x + 1}-(x+2) + (x+1- \sqrt{6x - 2}) =2(x^2 - 4x + 3). $$
equavalent to $$ \dfrac{-(x^2 -4x + 3)}{ \sqrt{8x + 1}+(x+2)}+ \dfrac{x^2 -4x + 3}{x+1 + \sqrt{6x - 2}}=2(x^2 - 4x + 3).$$
Or
$$ (x^2 -4x + 3)\left(\dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 + \sqrt{6x - 2}}\right)=0.$$
With $x\geqslant \dfrac{1}{3}$, it is easy to see that
$$ \dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 + \sqrt{6x - 2}} >0.$$
Therefore we get $$ x^2 -4x + 3 = 0.$$
|
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|
How do I show that a matrix is injective? I need to determine whether this matrix is injective
\begin{pmatrix}
2 & 0 & 4\\
0 & 3 & 0\\
1 & 7 & 2
\end{pmatrix}
Using gaussian elimination, this is what I have done:
\begin{pmatrix}
2 & 0 & 4 &|& 0\\
0 & 3 & 0 &|& 0\\
1 & 7 & 2 &|& 0
\end{pmatrix}
Divide row1 by 2, and then minus row3 by values of row1:
\begin{pmatrix}
1 & 0 & 2 &|& 0\\
0 & 3 & 0 &|& 0\\
0 & 7 & 0 &|& 0
\end{pmatrix}
Divide row 2 by 3, divide row 3 by 7 and minus row 3 by row2:
\begin{pmatrix}
1 & 0 & 2 &|& 0\\
0 & 1 & 0 &|& 0\\
0 & 0 & 0 &|& 0
\end{pmatrix}
Am I doing this correctly? How do I show that the matrix is (not) injective? I was thinking along the lines of "$x + z \ne 0$."
|
Look at vectors of the form $\left(\begin{array}{c}2x\\ 0\\ -x\end{array}\right)$. What does your matrix do to these vectors? What does injective mean?
|
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|
Integral $\int_0^\pi \cot(x/2)\sin(nx)\,dx$ It seems that $$\int_0^\pi \cot(x/2)\sin(nx)\,dx=\pi$$ for all positive integers $n$.
But I have trouble proving it. Anyone?
|
Note that your question is same as proving
$$\int_{0}^{\frac \pi 2} \sin (2nx)\cot (x)\;\text{d}x = \frac \pi 2$$
Let
$$f_n=\int_{0}^{\frac \pi 2} \sin (2nx)\cot (x)\;\text{d}x$$
We will first show that $f_{n+1}-f_n=0$ for any $n\in \mathbb N$.
\begin{align*}
f_{n+1}-f_n&=\int_{0}^{\frac \pi 2}(\sin(2(n+1)x)-\sin(2nx))\cot(x)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\sin\left(\frac{2(n+1)x-2nx}2\right)\cot(x)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\sin(x)\frac{\cos (x)}{\sin (x)}\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\cos (x)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\cos\left(\frac{2(n+1)x-2nx}2\right)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} \cos(2(n+1)x)+\cos(2nx))\;\text{d}x\\
&=\frac 1{2(n+1)}\sin(2(n+1)x)+\frac 1{2n}\sin(2nx)\bigg|_{0}^{\frac \pi 2}\\
&=\frac 1{2(n+1)}\sin((n+1)\pi)+\frac 1{2n}\sin(n\pi)-\frac 1{2(n+1)}\sin(0)-\frac 1{2n}\sin(0)\\
&=0
\end{align*}
Now, just calculate $f_1=\frac \pi 2$ and complete the proof.
|
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|
Combination of logarithms and exponents I am given this question and told to solve for $a,b,c$:
$$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} = \frac{y^{3/2}\ln(y)}{3\ln(x)}$$
I tried to convert all the logarithms to $\ln$ and remove the $\frac{\ln(x)}{\ln(y)}$ term from both sides of the equation, but eventually I am stuck as this expression has 2 variables $x,y$ which are unknowns.
I got a final form $8ay^{8a-c-3/2} = 2x^{2-b}$, thus concluding that $a=1/4$, but from there I do not know how to continue on.
Any help is greatly appreciated. Thanks.
|
$$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} =\frac{y^{8a}}{y^c}\times\frac{x^b}{x^2}\times\frac{1}2\times\log_x (y^{8a})=y^{8a-c}\times x^{b-2}\times\frac{1}2\times\frac{\ln(y^{8a})}{\ln x}\\=y^{8a-c}\times x^{b-2}\times\frac{1}2\times \frac{8a\ln(y)}{\ln x}=y^{8a-c}\times x^{b-2}\times\frac{8a}2\times \frac{\ln y}{\ln x} $$ Now if we use the assumption we have: $$y^{8a-c}\times x^{b-2}\times\frac{8a}2\times \frac{\ln y}{\ln x}=\frac{y^{3/2}\ln(y)}{3\ln(x)}$$ so $8a-c=3/2$, $b-2=0$ and $8a/2=1/3$. This is what @pondy achieved.
|
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Further clarification needed on proof invovling generating functions and partitions (or alternative proof)
Show with generating functions that every positive integer can be written as a unique sum of distinct powers of $2$.
There are 2 parts to the proof that I don't understand. I will point them out as I outline the proof:
The generating function given is
$$
g^*(x) = (1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^k}) \cdots.
$$
Then I don't understand where the following comes from:
To show that every integer can be written as a unique sum of disticnt powers of $2$, we must show that the coefficient of every power of $x$ in $g^*(x) = 1 + x + x^2 + x^3 + \cdots = (1-x)^{-1} $ or equivalently, $(1-x)g^*(x)=1$.
Then they use the identity $(1-x^k)(1+x^k) = 1 - x^{2k}$ to manipulate $g^*(x)$:
$$\begin{align*}
(1-x)g^*(x) &= (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots\\
&= (1-x^2)(1+x^2) (1+x^4)(1+x^8)\cdots \\
&= (1-x^4)(1+x^4)(1+x^8)\cdots \\
&= \quad \quad \vdots\\
&= 1
\end{align*}$$
That is the the second part that I don't understand. If it's an infinite product, why can they "eventually eliminate all factors of $(1-x)g^*(x)$"?
Update: Alternative proofs are more than welcome.
|
The answer is binary: 0,1,10,11,100,101,110,111,1000,...
to prove every number has a unique binary expansion is easy. More generally every number has a unique base $b$ expansion: just iterate the division algorithm which uniquely splits any number $n$ into $n = m b + c$ with $0 \le c < b$.
The reason for using generating functions to show this is to get to grips with generating functions, rather than actually proving this result.
The most basic thing about these products is that whenever we have $$\cdots(a+b)(c+d)\cdots$$ multiplying it out will gives terms with $ac$ in it - or whatever choice we want. You should know the combinatorial interpretation of the coefficient of $x^k$ in $(1+x)^n$ for example: In the product $(1+x)(1+x)(1+x)$ we can get $x^2$ by choosing 1,x,x or x,1,x or x,x,1, so 3 ways.. so $3 x^2$ appears when you multiply it out.
So multiplying out $$(\color{blue}1+x)(1+\color{blue}{x^{2}})(1+\color{blue}{x^{4}})(\color{blue}1+x^{8})(\color{blue}1+x^{16})(1+\color{blue}{x^{32}})\cdots$$ will give you a sum with lots of terms in it, and it will surely contain the term $$1\cdot x^{2}\cdot x^{4} \cdot 1 \cdot 1 \cdot x^{32} = x^{2 + 4 + 32}$$ for example.
In fact multiplying out this product will give terms $x$ to the power of every sum if powers of two. So we know from the previous section that it's $$1 + x + x^2 + x^3 + x^4 + x^5 + \ldots$$ because each number can be written only once as a sum of powers of two.
It can be proved directly with the generating functions too, you should know this $$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \ldots$$ if not it is very important to understand that before continuing.
So take our product and divide it by that, if we get $1$ we prove they are equal:
$$\begin{array}{lll}
&& (1-x)(1+x)(1+x^{2})(1+x^{4})(1+x^{8})(1+x^{16})(1+x^{32})\cdots \\
&=& (1-x^2)(1+x^{2})(1+x^{4})(1+x^{8})(1+x^{16})(1+x^{32})(1+x^{64})\cdots \\
&=& (1-x^4)(1+x^{4})(1+x^{8})(1+x^{16})(1+x^{32})(1+x^{64})(1+x^{128})\cdots \\
&=& (1-x^8)(1+x^{8})(1+x^{16})(1+x^{32})(1+x^{64})(1+x^{128})(1+x^{256})\cdots \\
&=& \cdots \\
\end{array}$$
this clearly proceeds by induction.
How does this prove the product is 1? Well what is the coefficient of $x^{7}$ in there for example? Take the series $\mod x^{8}$ to find out. In the same way any $n > 1$ you will find the coefficient of $x^{n}$ is $0$.
|
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|
Find the smallest $x$ for $9x \equiv 3 \pmod {23}$ $9x \equiv 3 \pmod {23}$
How to derive the smallest $x$. I understand I can use the extended euclidean algorithm for eg $19x = 1 \pmod {35}$.
However, I not too sure how to work on it when it is $3 \pmod {23}$.
I am able to reach the step of $1 = -5(9) + 2(23$) after going thru the euclidean algorithm.
|
We employ an algorithm that was discussed here.
$\alpha-$Solve:
$\;9x \equiv 3 \pmod{23}, \text{ and } \; 23 = 9 \cdot 2 + 5,\quad -3 + 1 \cdot 9 = 6$
$\alpha-$Solve:
$\;5x \equiv 6 \pmod{9}, \text{ and } \; 9 = 5 \cdot 1 + 4,\quad -6 + 2\cdot 5 = 4$
$\alpha-$Solve:
$\;4x \equiv 4 \pmod{5}, \text{ ANS: } \; x = 1 \text{ is the least residue solution}.$
Propagating backward,
Solve $5x = 6 + 1 \cdot 9, \text{ ANS: } \; x = 3$
Solve $9x = 3 + 3 \cdot 23, \text{ ANS: } \; x = 8$
$\text{ANSWER: } x = 8 \text{ is the least natural number satisfying }$
$\quad 9x \equiv 3 \pmod{23}$
|
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|
Limits $\sqrt{2n^2-1}/(n+1)$ and $1 = 0.9999...$ $1)$ Find the limit (if it exists) of the following sequence:
$$\frac{\sqrt{2n^2-1}}{n+1} = x_n$$
Attempt: Rewrite as $$\frac{\sqrt{n^2(2 - \frac{1}{n^2})}}{n+1} = \frac{n\sqrt{(2-\frac{1}{n^2})}}{n+1} = \frac{\sqrt{2 - \frac{1}{n^2}}}{\frac{1}{n} + 1}$$ So as $n \rightarrow \infty,\,\,x_n \rightarrow \sqrt{2},$ using the limit law $\operatorname{lim}_{n \rightarrow \infty} \sqrt[n]{f(x)} = \sqrt[n]{\operatorname{lim}_{n \rightarrow \infty} f(x)} $and clearly $f(x) \geq 0 \,$for $n \neq 0$
Is the above calculation rigorous enough to determine the limit?
2)Interpret a decimal expansion $0.a_1a_2a_3....$ as $$0.a_1a_2a_3.... = \operatorname{lim}_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a_k}{10^k}.$$
Show that $1 = 0.9999...$
Attempt: $$0.9999... =\operatorname{lim}_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a_k}{10^k} = 9 \cdot \sum_{k=1}^{\infty} \frac{1}{10^k} = 9 \cdot \frac{1/10}{9/10} = 9 \cdot \frac{1}{9} = 1$$ Done?
In my book, they also include the following:Let $ y_n = \sum_{k=1}^{\infty} \frac{9}{10^k}$ and we want $|1-y_n| = |1 - \sum_{k=1}^{\infty} \frac{9}{10^k}|$ As shown above the RHS tends to 0 so $|1-y_n| \rightarrow 0 $as $n \rightarrow \infty$
What does this add and is it required?
Many thanks.
|
Another way for second question.
Let $x=0.999999999999\dots $
$10x=9.999999999999\dots$
subtract both equations
$9x=9$
$x=1;$
|
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|
Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
|
$$ \frac1{x-y} + \frac1{x+y} = \frac{2x}{x^2-y^2}$$
Proceeding in this fashion we would be left with
$$ \text{The sum } = \frac{16x^{15}}{x^{16}-y^{16}} + \frac{16x^{15}}{x^{16}+y^{16}} = \frac{32x^{31}}{x^{32}-y^{32}} $$
|
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|
Evaluation of $\int^1_0\frac{x^b-x^a}{\ln x} d x=\ln{{\frac{1+b}{1+a}}}$ $$\int^1_0\frac{x^b-x^a}{\ln x} d x =\ln{{\frac{1+b}{1+a}}}$$ The part inside $\ln$ is absolute value. A solution including integration under the integral can be found here. Which used the identity $$\int^1_0x^{\alpha} dx =\frac{1}{1+\alpha}$$. Multiplies both side by $d{\alpha}$ and integrates with respect to $\alpha$. Are there other methods to prove this identity?
Thank You.
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Define $$f(z) = \int^1_0 \frac{ x^{bz} - x^{az} }{\log x} dx.$$ Differentiating gives $$f'(z) = \int^1_0 \left(bx^{bz} - ax^{az} \right) dx = \frac{b}{bz+1} - \frac{a}{az+1}$$
and integrating this gives $$f(z) = \log(bz+1) - \log(az+1)+C$$ and setting $z=0$ indicates the constant is $0.$ So
$$\int^1_0 \frac{x^b-x^a}{\log x}dx=f(1) = \log \frac{b+1}{a+1}.$$
The integrand is not defined over the interval of integration, so I assume you wanted $$\lim_{p\to 0} \int^1_p \frac{x^b-x^a}{\log x} dx.$$
Define $$ f(z) = \int^1_p \frac{x^{bz}-x^{az}}{\log x} dx.$$ Then $$f'(z) = \frac{b(1-p^{bz+1})}{bz+1} - \frac{a (1-p^{az+1})}{az+1} $$ so $$f(z) = \log \frac{bz+1}{az+1} + \frac{1}{a} \text{Ei}( (az+1)\log p ) - \frac{1}{b} \text{Ei}( (bz+1)\log p) + \frac{a-b}{ab} \text{Ei}(\log p)$$
where $\text{Ei}$ is the Exponential Integral and the constant was found by letting $z=0.$ Since $\lim_{x\to -\infty} \text{Ei}(x)=0$ the result follows.
|
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|
How Can I Find The Value Of This Limit 10 find this limit:
$$\displaystyle\lim_{n\to+\infty}\left[\sum_{k=1}^{n}\left(\dfrac{1}{\sqrt{k}}- \int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt\right)-2\sqrt{n}\right]$$
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We have
$$\int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt=\frac{1}{\sqrt k}-\arctan\frac{1}{\sqrt k}$$
Now if we denote by
$$u_n=\sum_{k=1}^n\arctan\frac{1}{\sqrt k}-2\sqrt{n}$$
we have
$$u_n-u_{n-1}=\arctan\frac{1}{\sqrt n}-2\sqrt{n}+2\sqrt{n-1}\sim\frac{-7}{12n\sqrt{n}},$$
and since the series $\sum\frac{1}{n\sqrt{n}}$ is convergent then the sequence $(u_n)$ is also convergent to say $\ell$ and we we have
$$\sum_{k=n+1}^\infty u_k-u_{k-1}=\ell-u_n\sim\frac{-7}{12}\sum_{k=n+1}^\infty\frac{1}{k\sqrt{k}}\sim\frac{-7}{12}\int_{n+1}^\infty\frac{dx}{x\sqrt{x}}=\frac{-7}{6\sqrt{n}}$$
so we find the asymptotic equality
$$u_n=\ell+\frac{7}{6\sqrt{n}}+o(\frac{1}{\sqrt{n}})$$
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$\lim_{n \to \infty}\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$ Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$
For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge to $\infty$ as $n\rightarrow \infty$.
1 - How to find a bound for $f(n)$?
2- $f(n)\rightarrow\infty$, as $n\rightarrow\infty$?
Thank you for your time.
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Note that, with the substitution $x \mapsto nx$, we have
$$ \frac{f(n)}{n} = \int_{0}^{\infty} \log^2 \left( \frac{x + 1}{x + n^{-2}} \right) \, dx. $$
Now, let us denote
$$ h_n (x) = \log^2 \left( \frac{x + 1}{x + n^{-2}} \right) \quad \text{and} \quad h(x) = \log^2 \left( \frac{x + 1}{x} \right). $$
Then we easily observe that
*
*$h_n (x) \to h(x)$ pointwise and $h_n (x) \leq h(x)$. Indeed,
$$ 1 \leq \frac{x+1}{x+n^{-2}} \leq \frac{x+1}{x} $$
and the claim follows by observing that $x \mapsto \log^2 x$ is non-negative increasing on $x \geq 1$.
*$h(x)$ is integrable. It is a direct consequence of the following estimation:
\begin{align*}
\int_{0}^{\infty} h(x) \, dx
&= \int_{0}^{1} h(x) \, dx + \int_{1}^{\infty} h(x) \, dx \\
&= \int_{1}^{\infty} \frac{h(1/x)}{x^2} \, dx + \int_{1}^{\infty} h(x) \, dx \\
&= \int_{1}^{\infty} \frac{\log^2 (1+x)}{x^2} \, dx + \int_{1}^{\infty} \log^2 \left(1 + \frac{1}{x} \right) \, dx \\
&\leq \int_{1}^{\infty} \frac{\log^2 (1+x)}{x^2} \, dx + \int_{1}^{\infty} \frac{dx}{x^2}
< \infty.
\end{align*}
Then by dominated convergence we obtain
$$ \frac{f(n)}{n} = \int_{0}^{\infty} h_n(x) \, dx \xrightarrow[]{n\to\infty} \int_{0}^{\infty} h(x) \, dx. $$
(If you want to avoid the use of dominated convergence, which requires some basic measure theory, you can first show that $h_n(x) \uparrow h(x)$ as $n \to \infty$ pointwise. Then by Dini's theorem, on any compact subset of $(0, \infty)$ we have $h_n (x) \to h(x)$ uniformly. Now a suitable $\epsilon-\delta$ argument will give the same conclusion.)
This already shows that
$$ f(x) \sim cn $$
for some constant $c > 0$. To determine the constant $c$, we have to show that
$$ \int_{0}^{\infty} h(x) \, dx = \frac{\pi^2}{3}. $$
But this is also straightforward. Indeed, put $\displaystyle u = \frac{x}{x+1}$. Then $\displaystyle dx = \frac{du}{(1-u)^2}$ and therefore
\begin{align*}
\int_{0}^{\infty} h(x) \, dx
&= \int_{0}^{\infty} \log^2 \left( \frac{x + 1}{x} \right) \, dx
= \int_{0}^{1} \frac{\log^2 u}{(1-u)^2} \, du \\
&= \left[ \frac{u}{1-u} \log^2 u \right]_{0}^{1} - 2 \int_{0}^{1} \frac{\log u}{1-u} \, du \\
&= 2 \int_{0}^{\infty} \frac{t}{e^{t}-1} \, dt \qquad (u = e^{-t}) \\
&= 2 \zeta(2) = \frac{\pi^2}{3}.
\end{align*}
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Cauchy Integral Formula for Matrices How do I evaluate the Cauchy Integral Formula $f(A)=\frac{1}{2\pi i}\int\limits_Cf(z)(zI-A)^{-1}dz$ for a matrix $A=\left(\begin{array}{ccc}2&2&-5\\3&7&-15\\1&2&-4\end{array}\right)$ and a function $f(x)=3x^2+1$?
I have evaluated the function directly, using interpolation and using Jordan-Normal Form and want to show the solutions are equivalent.
The solution should be $f(A)=\left(\begin{array} \ 16&24&-60\\36&76&-180\\12&24&-56\end{array}\right)$.
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Choose the contour $C$ so that $|z| > \|A\|$ for all $z$ in $C$. Then note that $(zI -A)^{-1} = \frac{1}{z}(I-\frac{A}{z})^{-1}$, and for $|z| > \|A\|$, we have $(zI -A)^{-1} = \frac{1}{z} \sum_{k=0}^\infty \frac{A^k}{z^k}$. Since $|z| > \|A\|$ for all $z$ in $C$, the convergence is uniform so we may interchange integration and summation.
Also note that $f$ is analytic on $C$ and the 'inside' of $C$.
This gives
\begin{eqnarray}
f(A)&=&\frac{1}{2\pi i}\int\limits_Cf(z)(zI-A)^{-1}dz \\
&=& \frac{1}{2\pi i}\int\limits_Cf(z) \frac{1}{z} \sum_{k=0}^\infty \frac{A^k}{z^k} dz \\
&=& \frac{1}{2\pi i} \sum_{k=0}^\infty \left(\int\limits_Cf(z) \frac{1}{z^{k+1}} dz \right) A^k \\
&=& \sum_{k=0}^\infty \left( \frac{1}{2\pi i}\int\limits_Cf(z) \frac{1}{z^{k+1}} dz \right) A^k \\
&=& \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} A^k
\end{eqnarray}
(The specific value doesn't matter, but it is easy to compute $\|A\|_1 = 24$, so as long as $|z|> 24$ on $C$, the above formula holds.)
It follows that if $f(x) = \sum_{k=0}^n a_k z^k$, then $f(A) = \sum_{k=0}^n a_k A^k$. Hence, in this case, $f(A) = 3 A^2 +I$. Evaluating shows that it equals the answer above.
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Finding polynomial given the remainders Question: Find a polynomial $f(x) \in \mathbb{Q} (x)$ of minimal degree that has both the following properties:
When $f(x)$ is divided by $(x-1)^2$, the remainder is $2x$; and when $f(x)$ is divided by $(x-2)^3$, the remainder is $3x$.
Answer provided: $f(x)=(x-2)^3 \cdot (4x-3)+3x$
Work so far: Okay, so I know this problem shouldn't be difficult, but I've been stumped. I know that this is probably a simple application of the division algorithm, where $f(x)=q(x) \cdot d(x)+r(x)$ but I can't seem to get an answer. The polynomial's minimum degree should be $4$, intuitively.
So I have that $f(x)=(x-1)^2 \cdot q_1(x) + 2x$ and $f(x)=(x-2)^3 \cdot q_2(x)+3x$. Then I know that for the second equation, $q_2(x)$ should be of form $(ax+b)$ for some $a,b \in \mathbb{Q}$, also I also applied the remainder theorem to find that $f(1)=4$ and that $f(2)=6$, and so
$f(1)=4=(-1)^3 \cdot (a+b) +3(1) \rightarrow a+b=1$
so it makes sense that the answer is $q_2(x)=4x-3$, but I don't know how to get there after finding $a+b=1$.
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An alternative:
Note that if $f(x)$ satisfies the 2 properties, so does $f(x)-q(x-1)^2(x-2)^3, q \in \mathbb{Q}$. Thus the polynomial of minimal degree has degree 4. We have $f(x)=(x-2)^3(ax+b)+3x$. Note that $1$ is a repeated root of $f(x)-2x$, so $f(1)=f'(1)=2$.
$2=f(1)=(-1)^3(a+b)+3$ so $a+b=1$. $f'(x)=3(x-2)^2(ax+b)+a(x-2)^3+3$ so $2=f'(1)=3(-1)^2(a+b)+a(-1)^3+3=6-a$ so $a=4, b=-3$.
Thus $f(x)=(x-2)^3(4x-3)+3x$
|
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|
Finding the coefficient of $ x^n $ in the expansion of $ { ({\ log_e (1+x) })^2 } $ I've been trying to find the coefficient of $x^n$ in the expansion of $ { ({\log_e (1+x) })^2 } $.I wrote out the expansion of $ { ({\log_e (1+x) })^2 } $ explicitly and tried to generalize the terms involving $x^n$, but...so far no luck.
Is there any other alternative ?
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If $$y=\{\log(1+x)\}^2$$
So, $$\frac{dy}{dx}=2\frac{\log(1+x)}{(1+x)}=2 (1+x)^{-1} \log(1+x)$$
$$=2\left(1+\frac{x(-1)}{1!}+\frac{x^2(-1)(-2)}{2!}+\frac{x^3(-1)(-2)(-3)}{3!}+\cdots\right)$$
$$\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots\right)$$
$$=2 (1-x+x^2-x^3+\cdots )\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots\right)$$
$$=2\{x+x^2\left(-\frac12-1\right)+x^3\left(\frac13+\frac12+1\right)+x^4\left(-\frac14-\frac13-\frac12-1\right)+\cdots\}$$
So, $$ dy =2\sum_{1\le r<\infty} H_rx^r(-1)^{r-1} dx$$ where $H_n=\sum_{1\le r\le n}\frac1r$
Integrating both sides, $y= 2\sum_{1\le r<\infty} \frac{H_rx^{r+1}(-1)^{r-1}}{r+1}+C$ where $C$ is an arbitrary constant for indefinite integral.
As at $x=0, y=\{\log(1+x)\}^2=0\implies C=0 $
$$\implies\{\log(1+x)\}^2=2\sum_{1\le r<\infty} \frac{H_rx^{r+1}(-1)^{r-1}}{r+1}$$
|
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|
$x^4+y^4=2z^2$ has only solution, $x=y=z=1$ . How do I verify that the only solution in relatively prime positive integers of the equation $x^4+y^4=2z^2$ is $x=y=z=1$?
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Lemma I: If $(a,b,c)$ satisfies $a^2+b^2=c^2$, then $ab/2$ is not a square, nor twice a square.
Proof: By Euclid's formula, we can write $a=2pq$, $b=p^2-q^2$ $c=p^2+q^2$, so that $ab/2=pq(p+q)(p-q)$. If this is a square, then, as the four factors are pairwise coprime, there are $x,y,u,v$ such that $p=x^2$, $q=y^2$, $p+q=u^2$, and $p-q=v^2$. Then $2y^2=u^2-v^2=(u+v)(u-v)$. Since the g.c.d. of $u$ and $v$ is $2$, we conclude that one of $u+v,u-v$ is of the form $2r^2$, the other of the form $4s^2$. Hence $x^2=\frac{u^2+v^2}{2}=r^4+4s^4$ is another pythagorean triple with hypotenuse less than the original one, as $x^2=p^2<p^2+q^2$. So this completes the proof by descent.
On the other hand, if $ab/2$ is twice a square, then either $p=2x^2, q=y^2$, or $p=x^2, q=2y^2$. Moreover, $p+q=u^2$, and $p-q=v^2$, as above, with $u,v$ odd. Now $2p=\frac{u^2+v^2}{2}$, so $p$ must be odd. So $p=x^2$ and $q=2y^2$. Then $4y^2=2q=(u+v)(u-v)$. Thus $u+v=2r^2$ and $u-v=2s^2$ for some $r$ and $s$. Consequently $u=r^2+s^2$ and $v=r^2-s^2$. Finally, $x^2=p=\frac{u^2+v^2}{2}=r^4+s^4$ is another triple with area $2(\frac{rs}{2})^2$ and strictly less hypotenuse, hence again completing the proof.
Lemma II: No right-angled triangle with sides $a, b$, and hypotenuse $c$ can have $a,b$ both squares, nor $a,c$ both squares.
Proof: Suppose both $a$ and $b$ are squares, then we know that one of $a$ is divisible by $4$, and another odd, so that the area $ab/2$ is twice a square, contradicting the above lemma.
Suppose that $a$ and $c$ both are squares, then $a, c$ satisfy the equation $x^4-y^4=z^2$. If this equation has a non-trivial solution($x\neq y$), then, taking $p=x^2$ and $q=y^2$, we form a pythagorean triple $(2pq,p^2-q^2,p^2+q^2)$ with area $(xyz)^2$, again a contradiction.
Now your result follows easily: consider the triangle $(x^2z^2,\frac{x^4-y^4}{2},\frac{x^4+y^4}{2})$. By lemma II, neither $x^4+y^4$, nor $x^4-y^4$ can be twice a square, unless one is $0$, that is, $x=y=z=1$, which does not form a triangle. Q.E.D.
Ambiguity is never intended; the length is indeed a pain; so feel free to suggest improvements. Thanks in advance.
|
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Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$?
$Approach$:
$N$=$11^2$.$13^4$.$17^6$
$N^2$=$11^4$.$13^8$.$17^{12}$
This means $N$ has $(2+1) * (4+1) * (6+1) = 105$ factors and $N^2$ has $ (4+1) * (8+1) * (12+1) = 585 $ factors.
Therefore, there are 480 numbers that are not a factor of N. They are any combination of :
$11^3$,$11^4$,$13^5$,$13^6$,$13^7$,$13^8$,$17^7$,$17^8$,$17^9$,$17^{10}$,$17^{11}$,$17^{12}$.
But how many of these combinations are less than N? Not really sure how to do that in a easy way.
Please guide me how to do so.
Any help will be appreciated. Thanks.
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If a divisor $d$ of $N^2$ is smaller than $N$ then $\frac{N^2}{d}$ is a divisor of $N^2$ larger than $N$.
Let's say that there are $n$ divisors of $N^2$ that are smaller than $N$. Then there are $n$ divisors of $N^2$ that are larger than $N$.
In total there are 585 divisors of $N^2$ and each one is either smaller, larger or equal to $N$.
From these you should be able to find $n$.
Now $480$ divisors of $N^2$ do not divide $N$ and $n$ of them are larger than $N$ (some explanation here) therefore $480-n$ of them are smaller than $N$.
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Solving a system of congruences: CRT If $x \equiv 3 \pmod 6$ and $x \equiv 5\pmod 8$ then how is this $x \equiv 21\pmod {24}$?
I understand $24$ is $\operatorname{lcm}(6,8)$ but how to get from $3$ and $5$ to $21$?
This is is a Chinese Remainder Theorem problem.
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$x\equiv3\pmod 6\implies x\equiv0\pmod 3$ and $x\equiv 1\pmod 2$
Again, $x\equiv5\pmod 8\implies x\equiv1\pmod 2$
So, it suffices to use CRT with $x\equiv0\pmod 3$ and $x\equiv5\pmod 8$ which is legitimate as $(3,8)=1$
Alternatively,
So, $x$ can be written as $6a+3$ where $a$ is any integer
Similarly, $x=8b+5$ for some integer $b$
$\implies 6a+3=x=8b+5\implies 6a=8b+2,3a=4b+1=4b+4-3$
$\implies 3a+3=4b+4\implies a+1=\frac{4(b+1)}3$ which must be an integer as $a+1$ is
So, $3\mid 4(b+1)\implies 3\mid(b+1)$ as $(3,4)=1$
$\implies b=3c-1$ where $c$ is any integer
So, $x=8b+5=8(3c-1)+5=24c-3\equiv-3\pmod{24}\equiv21$
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Evaluate $\sum_{n=1}^{\infty}\frac{1}{n^3+3 n^2+2 n}$ Summing this series from $0$ to $\infty$, the result is $\frac{1}{4}$. I tried a lot, but I could not get this result. I think it´s wrong.
Can anybody help me?
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$\frac{1}{n(n+1)(n+2)} = \frac{1}{2(n+1)}(\frac{1}{n}-\frac{1}{n+2}) = $
$ = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}$
So the sum equals: $\displaystyle{\frac{1}{2\times1\times(1+1)} - \lim_{n\rightarrow +\infty}{\frac{1}{2n(n+1)}}} = \frac{1}{4}$
|
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Help with evaluating a sum I am trying to evaluate the following sum:
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n}$$
So far I have written the sum as
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n} = \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \frac{1}{5^n} = \sum_{n=1}^{\infty} \frac{1}{n5^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1)5^n}$$
I am stuck and I have not been able to find any similar examples. Wolfram Alpha gives the result
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n} = 1-4 \log(5/4)$$
I feel as though I should be writing the sum as an integral and evaluating the integral, but I do not know how to proceed. Any help appreciated. Thanks.
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$$\begin{align}
\sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=1}^\infty \frac{1}{(n+1)5^n} &= \sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=2}^\infty \frac{1}{n5^{n-1}} \\
&=\sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=1}^\infty \frac{1}{n5^{n-1}}+1\\
&=1+\sum_{n=1}^\infty \left(1-5\right)\frac{1}{n5^n}\\
&=1-4\sum_{n=1}^\infty \frac{1}{n5^n}\\
&=1-4\sum_{n=1}^\infty \frac{(1/5)^n}{n}\\
&=1+4\ln \left(1-\frac{1}{5}\right)\\
&=1+4\ln (4/5) = 1-4\ln (5/4)
\end{align}$$
|
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How can prove this inequality(8) Let $a,b,c,x,y,z >0$ and $A=a^2+b^2+c^2,\ B=x^2+y^2+z^2,\ C=ax+by+cz$. By Cauchy-Schwarz inequality, we always have $C^2\le AB$. If $C^2<AB$, prove that
$$
\frac{A}{C+\sqrt{2(AB-C^2)}}<\frac{a+b+c}{x+y+z}<\frac{C+\sqrt{2(AB-C^2)}}{B}.
$$
I created this inequality. Are there any nice proofs?
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It seems the following.
The inequality is not strict. For instance, put $a=b=c=x=1$, $y=z=0$. Then
$A=3$, $B=C=1$, and $C^2<AB$. But
$$\frac{a+b+c}{x+y+z}=3=\frac{C+\sqrt{2(AB-C^2)}}{B}.$$
The non-strict inequality can be easily proved by means of spherical trigonometry.
Consider the following vectors of unit length in $\mathbb R^3$. Let $u=(a,b,c)/\sqrt{A}$,
$v=(x,y,z)/\sqrt{B}$, and $w=(1,1,1)/\sqrt{3}$. Let $\alpha=\angle (v,u)$, $\beta=\angle(u,w)$,
and $\gamma=\angle(w,v)$. Since $a,b,$ and $c$ are positive, we can easily check that $\cos\beta=(a+b+c)/\sqrt{3A}=(a+b+c)/\sqrt{3(a^2+b^2+c^2)}\ge
1/\sqrt{3}$. Then $\tan\beta\le\sqrt{2}$. By spherical law of cosines, we have that
$\cos\gamma=\cos\alpha\cos\beta+\sin\alpha\sin\beta\cos\angle u$. Then
$\cos\gamma/\cos\beta\le \cos\alpha+\sin\alpha\tan\beta\le\cos\alpha+\sqrt{2}\sin\alpha$.
Substituting the cosines by the inner products, we obtain
$$\frac {(x+y+z)/\sqrt{3B}}{(a+b+c)/\sqrt{3A}}\le\frac{C}{\sqrt{AB}}+\sqrt{2-2\left(\frac{C}{\sqrt{AB}}\right)^2}.$$
The equality should hold only if $\cos\angle u=1$ (that is, the vectors $u,v,w$ and $0$ are
coplanar) and $\cos\beta=1/\sqrt{3}$, that is one of $a,b,$ and $c$ is equal to $1$, and the others
are equal to $0$.
From the another law of cosines
$\cos\beta=\cos\alpha\cos\gamma+\sin\alpha\sin\gamma\cos\angle v$,
we can similarly obtain the inequality
$$\frac{a+b+c}{x+y+z}\le \frac{C+\sqrt{2(AB-C^2)}}{B}.$$
PS. Maybe the $n$-dimensional version of the inequality holds, if we change in it "$2$" to "$n-1$".
|
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|
Laurent series expansions, complex variables Given $f(z)=\frac{1}{z^2(1-z)}$ I am to find two Laurent series expansions. There are two singularities, $z=0$ and $z=1$. So for the first expansion, I used the region $0<|z|<1$ and I got $\sum_{n=0}^\infty z^n+\frac{1}{z}+\frac{1}{z^2}$. The second expansion is for the region $1<|z|<\infty$. I don't know how to approach this, the explanation in my book is confusing. Any help?
Thanks
|
The problem is that for $|z|>1$ the absolute value of $z$ is greater than $1$ (ok that sound ridicolus but it is really the problem) as the geometric sequence wont work in the given form.
We just change it a bit.
$$\frac{1}{z^2(1-z)} =\frac{1}{z^2} \cdot \frac{1}{1-z} = \frac{1}{z^2} \cdot \frac{1}{\frac{z}{z}(1-z)}=\frac{1}{z^2} \cdot \frac{1}{z (\frac{1}{z}-1)}$$
This is equivalent to
$$\frac{1}{z^2(1-z)} =-\frac{1}{z^3} \cdot \frac{1}{1-\frac{1}{z}}$$
note that $\left|\frac{1}{z}\right|<1$ hence
$$\frac{1}{z^2(1-z)}=-\frac{1}{z^3} \cdot \sum_{k=0}^\infty \left(\frac{1}{z}\right)^k $$
This is the same as
$$-\sum_{k=0}^\infty \left(\frac{1}{z}\right)^{k+3}=-\sum_{n=3}^\infty \left(\frac{1}{z}\right)^n =-\sum_{n=3}^\infty z^{-n}=-\sum_{m=-\infty}^{-3} z^m$$
|
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|
How to prove $1Let $a,b>0$, $a\neq b$, $n\in\mathbb N$, $n>1$, and $a^n-b^n=a^{n+1}-b^{n+1}$.
How can we prove that
$1<a+b<\dfrac{2n}{n+1}$
Thank you everyone. I have proven it. My method:
$a+b-1=a+b-\dfrac{a^{n+1}-b^{n+1}}{a^n-b^n}=\dfrac{ab(a^{n-1}-b^{n-1})}{a^n-b^n}>0$
on the other hand
$$a+b-\dfrac{2n}{n+1}=a+b-\dfrac{2n}{n+1}\cdot\dfrac{a^{n+1}-b^{n+1}}{a^n-b^n}=\dfrac{(n+1)ab(a^{n-1}-b^{n-1})-(n-1)(a^{n+1}-b^{n+1})}{(n+1)(a^n-b^n)}$$
so
$$\Longleftrightarrow ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1}),n>1$$
let $a>b>0,t=\dfrac{a}{b}$
$$\Longleftrightarrow (n+1)t(t^{n-1}-1)<(n-1)(t^{n+1}-1),t>1 $$
follwing is very easy.
so we let $f(x)=(n-1)x^{n+1}-(n+1)x^n+(n+1)x-(n-1),x>1$
then we $f'(x)=(n-1)(n+1)x^n-n(n+1)x^{n-1}+(n+1),x>1$
$f''(x)=n(n-1)(n+1)x^{n-1}-n(n+1)(n-1)x^{n-2}=(n-1)n(n+1)x^{n-2}(x-1)>0,x>1$
so
$f'(x)>f'(1)=0$
then
$f(x)>f(1)=0$
so
$$ (n+1)t(t^{n-1}-1)<(n-1)(t^{n+1}-1),t>1 $$
I think there are other methods?
|
Firstly, a point of caution. In your solution, you had a step which went:
\begin{align}
& \frac{(n+1)ab(a^{n-1}-b^{n-1})-(n-1)(a^{n+1}-b^{n+1})}{(n+1)(a^n-b^n)}<0 \\
& \Longleftrightarrow ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1})
\end{align}
Note that in order to do this, you must first WLOG assume $a>b$, so that the denominator is positive and you don't need to worry about it.
Now, here's an alternative that continues from the step above. Since $a>b$, we divide by $a-b$ to get:
\begin{align}
& ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1}) \\
& \Longleftrightarrow (n+1)(a^{n-1}b+a^{n-2}b^2+ \ldots +ab^{n-1})<(n-1)(a^{n}+a^{n-1}b+ \ldots +b^{n}) \\
& \Longleftrightarrow (n-1)(a^n+b^n)>2(a^{n-1}b+a^{n-2}b^2+ \ldots +ab^{n-1})
\end{align}
By AM-GM,
$$a^n+b^n=\frac{(n-i)a^n+ib^n}{n}+\frac{ia^n+(n-i)b^n}{n} \geq a^{n-i}b^i+a^ib^{n-i}$$
Summing over $i=1, 2, \ldots . n-1$ gives the desired inequality. The inequality is strict because $a \not =b$.
|
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|
Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$ How to simplify the expression:
$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$
If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.
Thank you.
|
Put
$$x_0:=0,\quad x_1:=\sqrt{5},\quad x_2:=\sqrt{5+2\sqrt{5}},\quad x_3:=\sqrt{5+2\sqrt{5+2\sqrt{5}}}\ ,$$
and so on, which amounts to
$$x_0:=0,\qquad x_{n+1}:=\sqrt{5+2x_n}\quad(n\geq0)\ .$$
Then
$$x_{n+1}-x_n=\sqrt{5+2x_n}-\sqrt{5+2x_{n-1}}={2(x_n-x_{n-1}) \over \sqrt{5+2x_n}+\sqrt{5+2x_{n-1}}}\ .$$
As $x_1-x_0>0$ this shows that the sequence $(x_n)_{n\geq0}$ is momotonically increasing.
Furthermore $0\leq x_0<4$, and for any $n\geq0$ the statement $0\leq x_n< 4$ implies $$0\leq x_{n+1}<\sqrt{5+2\cdot 4}<4\ .$$ This shows that our sequence is as well bounded, so it has a limit $\xi\in[0,4]\ $. This limit satisfies the equation $x=\sqrt{5+2x}$ and is therefore given by $\xi=1+\sqrt{6}\doteq 3.45$.
|
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|
Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.
I know that for proving the $\gcd(a,b) = c$ you need to prove
*
*$c|a$ and $c|b$
*$c$ is the greatest number that divides $a$ and $b$
Number 2 is what I'm struggling with. Does anybody have any ideas?
|
Are you allowed to use Fermat's Little theorem?
$5^{\phi(14)} = 5^6 \equiv 1 \mod 14$
$5^{98} \equiv 5^{96}5^2 \equiv 25 \equiv -3 \mod 14$ so $14|5^{98} + 3$
and $5^{99} \equiv -3*5 \equiv -15 \equiv -1 \mod 14$ so $14|5^{99} + 1$
So $14$ is a common divisor.
Hmm, but how to show it is the greatest common divisor?
Well, we know $14|\gcd(5^{98} + 3, 5^{99} + 1)$
And we can see $5 \not \mid 5^{98} + 3$ nor $5^{99}+1$
so $\gcd(5^{98} + 3, 5^{99} + 1) = \gcd(5^{98} + 3, [5^{99} + 1] - 5[5^{98} + 3])$
$= \gcd(5^{98} + 3, [5^{99} + 1] - [5^{99} + 15]) = \gcd(5^{98} + 3, - 14)$
$\gcd(5^{98} + 3, 14)$
So $\gcd(5^{98} + 3, 5^{99} + 1)|14$ and $14|\gcd(5^{98} + 3, 5^{99} + 1)$.
So $14 = \gcd(5^{98} + 3, 5^{99} + 1)$.
|
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What is the remainder when $25^{889}$ is divided by 99? What is the remainder when $25^{889}$ is divided by 99 ?
$25^3$ divided by $99$ gives $26$ as a remainder.
$25*(25^3)$ divided by $99$ gives (remainder when $25*26$ is divided by $99$) as a remainder.
i.e. $25*(25^3)$ divided by $99$ gives $56$ as a remainder.
$(25^3)*(25^3)$ divided by $99$ gives (remainder when $26*26$ is divided by $99$) as a remainder.
i.e. $(25^3)*(25^3)$ divided by $99$ gives $82$ as a remainder.
|
Euler’s Number of $99$
= $99.\frac{2}{3}.\frac{10}{11}$
= $60$
From Fermat’s Theorem we know
$25^{60 × K} \mod {99} = 1$ (where K is any natural number)
Note that $25$ and $99$ are co-primes (∵ they don’t have any common factors other than $1$)
Putting $K =15$, we have, $25^{900} \mod {99} = 1$
Let’s assume $25^{899} \mod {99} = R$
∴ $25^{899} = 99N + R$ (where $N$ is a natural number)
$25^{900} \mod {99} = (25 × 25^{899}) \mod {99} = (25 × R) \mod {99} = 1$
we can conclude that $R=4$
|
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|
How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$
While solving one question, I am stuck, which looks obvious but without any feasible way to approach.
Few observations, not sure if it would help
$$
\begin{align}
\dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\
\dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7}
\end{align}
$$
|
Yes, This problem in 1963 IMO.http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908
|
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|
Series $\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z} \right)$ If $z$ is an integer, the sum of the series
$$\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)$$ is easy since it is a telescoping series. But if $z$ is a fraction, say $z=3/2$, I don't see why the series sums to $$\frac{8}{3}-\ln 4$$
Is there a formula for $z=m/n$, where $m,n$ are positive integers and $n\neq 1$?
|
For $\vert x \vert < 1$, if we define $f(x)$ as$$f(x)=\sum_{n=0}^{\infty} \left(x^n - x^{n+z}\right)$$ then $$\int_0^1 f(x) dx = \sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right)$$
But we have $$f(x) = (1-x^z) \sum_{n=0}^{\infty} x^n = \dfrac{1-x^z}{1-x}$$
Hence,
$$\sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right) = \int_0^1 \dfrac{1-x^z}{1-x}dx = I(z)$$
For $z=3/2$, we have
\begin{align}
I(3/2) & = \overbrace{\int_0^1 \dfrac{1-x^{3/2}}{1-x}dx = \int_0^1 \dfrac{1-t^3}{1-t^2} \cdot 2tdt}^{\text{Substitute }x = t^2} = \int_0^1 \dfrac{2t(1+t+t^2)}{1+t}dt\\
& = \int_0^1 \dfrac{2t((1+t)^2-t)}{1+t}dt = \int_0^1 2t(1+t) dt - 2 \int_0^1 \dfrac{t^2}{1+t} dt\\
& = \left(t^2 + \dfrac{2t^3}3\right)_{t=0}^{t=1} - 2 \int_0^1 \dfrac{t^2-1}{t+1}dt - 2 \int_0^1 \dfrac{dt}{1+t}\\
& = \dfrac53 - 2 \left(\dfrac{t^2}2 - t\right)_{t=0}^{t=1}-2\ln(2)\\
& = \dfrac53 - 2 \left(\dfrac12-1\right) - \ln(4) = \dfrac83 - \ln(4)
\end{align}
|
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|
Interpret the equation $17+28y+4x^2+4y^2=8x$ geometrically
Interpret the equation geometrically:
$$17+28y+4x^2+4y^2=8x$$
I have drawn the bend and now I got the expression $(y-2)=(1-x)^2$ but that is the wrong expression. What should it be?
|
Bring all terms onto one side. Club all the 'x' terms together and all the 'y' terms together.
You get:
$(4x^2 - 8x) + (4y^2 + 28y) + 17 = 0$
Simplify the above equation as :
$(4x^2 - 8x + 4) + (4y^2 + 28y + 49) + 13-49 = 0$
$4{(x-1)}^2 + 4{(y+7/2)}^2 = 36$
${(x-1)}^2 + {(y+7/2)}^2 = 3^2$
Hence it is a circle with center $(1,-7/2)$ and radius $3$.
|
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|
Evaluate $\int_{0}^{1000} \frac{e^{-10x}\sin x}{x} \text{d}x$ to within $\pm 10^{-5}.$ Evaluate $$\displaystyle \int_{0}^{1000} \frac{e^{-10x}\sin x}{x} \text{d}x$$ to within $\pm 10^{-5}$.
|
Thinking more about this, one don't need to approximate the integral by sending the upper limit to $\infty$ nor know how to evaluate the integral over $[0,\infty)$.
For $x > 0$, $|\frac{\sin x}{x}| < 1$ and $e^{-x}$ drops off to $0$ very quickly.
If we cutoff the integral at $1$ instead of $1000$. The error:
$$\left|\int_{1}^{1000} e^{-10x} \frac{\sin x}{x} dx\right| < \int_{1}^{1000} e^{-10x} dx < \frac{e^{-10}}{10} \sim 4.54 \times 10^{-6}$$
itself is small enough. For $x \in [0,1]$, one can Taylor expand $\frac{\sin x}{x}$ as:
$$\frac{\sin x}{x} = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!}$$
If one chop off the term for $k \ge 4$, we have:
$$\left|\frac{\sin x}{x} - (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!})\right| < 1/9! + 1/11! + \cdots \sim 2.781\times 10^{-6}$$
This implies
$$\begin{align}&\left|\int_{0}^{1} e^{-10x}\frac{\sin x}{x} dx - \int_{0}^{1}e^{-10x}(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!}) dx \right|\\ <& 2.781\times 10^{-6} \int_{0}^{1} e^{-10x} dx\\ \sim & 2.781 \times 10^{-7}\end{align}$$
Within an accuracy of $10^{-5}$, we have:
$$\int_0^{1000} e^{-10x} \frac{\sin x}{x} dx \sim \int_0^1 e^{-10x} (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!}) dx \\= \frac{20930417}{210000000}-\frac{50976061}{630000000} e^{-10} \sim 0.099665$$
|
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|
Laplace inverse transform of the complex expression I have an expression below which I need to do the laplace transform
Any help is highly appreicated.
The expression is :
$$
\frac{\exp\left(\frac{x}{2}\sqrt{(U/D)^2+4s/D}\right)}{s\sqrt{(U/D)^2+4s/D}}
$$
The integral of Ron Gordon's expression:
The first expression on the solution was extracted from Wolfram from the following link:
http://bit.ly/YFbLfH
|
I will reduce this to an integral over the real line. Consider the integral in the complex plane:
$$\displaystyle \oint_C \frac{dz}{z} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}} e^{t z}$$
where $C$ is the following contour
Note the point is the branch point. I will assert without proof that the integral vanishes along the sections $C_2$, $C_4$, and $C_6$ of $C$. This leaves $C_1$ (the ILT), $C_3$, and $C_5$. There is a pole within the contour at $z=0$, so by the residue theorem, we have
$$\frac{1}{i 2 \pi}\left [\int_{C_1} + \int_{C_3} + \int_{C_5} \right ]\frac{dz}{z} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}} e^{t z} = \frac{D}{U} e^{U x/(2 D)}$$
Along $C_3$, note that there is a branch point at $z=-U^2/(4 D)$. Thus we parametrize $z=-U^2/(4 D) + e^{i \pi} y$; the integral over $C_3$ becomes
$$-i \int_{\infty}^0 \frac{dy}{y+\frac{U^2}{4 D}} \frac{e^{i (x/2) \sqrt{y}}}{\sqrt{y}} e^{-t y}$$
Similarly, along $C_5$, let $z=-U^2/(4 D) + e^{-i \pi} y$; the integral over $C_5$ becomes
$$i \int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{e^{-i (x/2) \sqrt{y}}}{\sqrt{y}} e^{-t y}$$
Putting this all together as above, we get an expression for the ILT:
$$\frac{1}{i 2 \pi}\int_{c-i \infty}^{c+i \infty} \frac{ds}{s} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{s}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{s}{D}}} e^{t s} = \\ \frac{D}{U} e^{U x/(2 D)} - \frac{1}{\pi} \int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{\cos{[(x/2) \sqrt{y}]}}{\sqrt{y}} e^{-t y}$$
So evaluation of the posted ILT depends on the ability to evaluate
$$\int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{\cos{[(x/2) \sqrt{y}]}}{\sqrt{y}} e^{-t y}$$
This integral may be evaluated first by substituting $y=u^2$ and then applying the convolution theorem (or Parseval's theorem, depending on your mood). The substitution produces
$$\int_{-\infty}^{\infty} \frac{du}{u^2+b^2} e^{-a u^2} e^{i k u}$$
where $a = t$, $b^2=U^2/(4 D)$, and $k=x/2$. You may then use the convolution theorem on the Fourier transforms of the functions
$$\int_{-\infty}^{\infty} e^{-a u^2} e^{i k u} = \sqrt{\frac{\pi}{a}} e^{-k^2/(4 a)}$$
$$\int_{-\infty}^{\infty} \frac{du}{u^2+b^2} e^{i k u} = \frac{\pi}{b} e^{-b |k|}$$
Then
$$\int_{-\infty}^{\infty} \frac{du}{u^2+b^2} e^{-a u^2} e^{i k u} = \frac{1}{2 \pi} \sqrt{\frac{\pi}{a}} \frac{\pi}{b} \int_{-\infty}^{\infty} dk' e^{-(k-k')^2/(4 a)} e^{-b |k'|}$$
Frankly, the evaluation of this integral is straightforward but a mess, the derivation of which is not very instructive and will only serve to obfuscate the result. I leave it to the reader with assurances that I have done this out myself, by hand. The result is that
$$\int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{\cos{[(x/2) \sqrt{y}]}}{\sqrt{y}} e^{-t y} = \pi \frac{\sqrt{4 D}}{U} e^{U^2 t/(4 D)} \left [ e^{-U x/(4\sqrt{D})} \text{erfc}\left(-U \sqrt{\frac{t}{D}}+\frac{x}{4 \sqrt{t}}\right) + e^{U x/(4 \sqrt{D})} \text{erfc}\left(-U \sqrt{\frac{t}{D}}-\frac{x}{4 \sqrt{t}}\right) \right]$$
where erfc is the complementary error function. Plug this expression in the equation for the ILT and you are done.
|
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Residue Formula application Using the Residue formula, I've been trying to prove $$\int_0^{2\pi}\frac{1}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta=\frac{2\pi}{ab},\quad\quad a,b\in\Bbb R.$$First, it seems like the formula should be wrong (unless perhaps we assume $a,b\in\Bbb R^+$) since the right-hand side can be negative, but the integrand on the left is always non-negative. Currently I'm assuming the additional requirement $a,b>0$.
With that said, to approach it, I use Euler's formulas on the trig. functions in the denominator and make a change of variables, $$z=e^{i\theta},\quad \frac{1}{iz}\,dz=d\theta.$$Now, if I have calculated correctly, the integral reduces to $$\int_{|z|=1}\frac{1}{iz}\cdot\frac{1}{\frac{a^2}{4}\left(z+z^{-1}\right)^2-\frac{b^2}{4}\left(z-z^{-1}\right)^2}\,dz.$$We can factor $z^{-2}$ from the right-side denominator to get $$\int_{|z|=1}\frac{z}{i}\cdot\frac{1}{\frac{a^2}{4}\left(z^2+1\right)^2-\frac{b^2}{4}\left(z^2-1\right)^2}\,dz.$$Since the denominator is a difference of squares, we can factor the denominator as $$\int_{|z|=1}\frac{4z}{i}\cdot\left(\frac{1}{a(z^2+1)-b(z^2-1)}\right)\cdot\left(\frac{1}{a(z^2+1)-b(z^2-1)}\right)\,\,dz. $$This is where I really started running into trouble. I tried solving when the denominator of the right term vanished and I found $$z=\pm\sqrt{\frac{b+a}{b-a}}.$$ This didn't seem right because it doesn't always have to be inside the unit circle (I don't think), so I think I might have made an error in calculation.
Is my method so far correct, or is there a far better way to calculate this integral using the residue formula? This isn't homework, just prepping for an exam. Thanks!
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You can use trigonometric substitution to solve this problem.
Suppose $a<b$. Let $k=\frac{b}{a},t=\tan\frac{\theta}{2}$. Then $k>1$ and
$$ \sin\theta=\frac{2t}{t^2+1}, \cos\theta=\frac{t^2-1}{t^2+1},d\theta=\frac{2t}{t^2+1}dt $$
and hence
\begin{eqnarray*}
\int_0^{2\pi}\frac{1}{a^2\cos^2\theta+b^2\sin^2\theta}d\theta&=&\frac{2}{a^2}\int_{-\infty}^\infty\frac{t^2+1}{(t^2-1)^2+4k^2t^2}dt\\
&=&\frac{2}{a^2}\int_{-\infty}^\infty\frac{t^2+1}{t^4+2(2k^2-1)t^2+1}dt\\
&=&\frac{2}{a^2}\int_{-\infty}^\infty\frac{t^2+1}{(t^2+(2k^2-1))^2+4k^2(1-k^2)}dt\\
&=&\frac{2}{a^2}\int_{-\infty}^\infty\frac{t^2+1}{(t^2+\alpha)(t^2+\beta)}dt\\
&=&\frac{2}{a^2}\int_{-\infty}^\infty\left(\frac{A}{t^2+\alpha}+\frac{B}{t^2+\beta}\right)dt\\
&=&\frac{2}{a^2}(\frac{A\pi}{\sqrt{\alpha}}+\frac{B\pi}{\sqrt{\beta}})\\
&=&\frac{2\pi}{ab}
\end{eqnarray*}
where
$$ \alpha=(2k^2-1)+2k\sqrt{k^2-1},\beta=(2k^2-1)-2k\sqrt{k^2-1},A=\frac{\alpha-1}{\alpha-\beta}, B=\frac{1-\beta}{\alpha-\beta},$$
Suppose $a>b$. Let $k=\frac{b}{a},t=\tan\frac{\theta}{2}$. Then $k>1$ and hence
\begin{eqnarray*}
\int_0^{2\pi}\frac{1}{a^2\cos^2\theta+b^2\sin^2\theta}d\theta&=&2\int_0^{\pi}\frac{1}{a^2\cos^2\theta+b^2\sin^2\theta}d\theta\\
&=&2\int_0^{\pi}\frac{1}{a^2\frac{1+\cos2\theta}{2}+b^2\frac{1-\cos2\theta}{2}}d\theta\\
&=&4\int_0^{\pi}\frac{1}{(a^2+b^2)+(a^2-b^2)\cos2\theta}d\theta\\
&=&2\int_0^{2\pi}\frac{1}{(a^2+b^2)+(a^2-b^2)\cos\theta}d\theta\\
&=&\frac{2}{a^2-b^2}\int_0^{2\pi}\frac{1}{\alpha+\cos\theta}d\theta\\
&=&\frac{2}{a^2-b^2}\int_{-\infty}^{\infty}\frac{1}{\alpha+\frac{t^2-1}{t^2+1}}\frac{2}{t^2+1}dt\\
&=&\frac{2}{a^2-b^2}\int_{-\infty}^{\infty}\frac{2}{\alpha(t^2-1)+t^2-1}dt\\
&=&\frac{4}{a^2-b^2}\int_{-\infty}^{\infty}\frac{2}{(\alpha+1)t^2+\alpha-1}dt\\
&=&\frac{4}{(a^2-b^2)(\alpha+1)}\int_{-\infty}^{\infty}\frac{1}{t^2+\frac{\alpha-1}{\alpha+1}}dt\\
&=&\frac{4}{(a^2-b^2)(\alpha+1)}\sqrt{\frac{\alpha+1}{\alpha-1}}\pi\\
&=&\frac{2\pi}{ab}.
\end{eqnarray*}
where
$$ \alpha=\frac{a^2+b^2}{a^2-b^2}>1. $$
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the definite limit The following limit
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
evaluates to 1/12.
This is my progress so far:
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
$$\lim_{x\to 1}\frac{1 + \sqrt{x}}{2(1 - x)} - \frac{1 + \sqrt[3]{x} + \sqrt[3]{x^2}}{3(1 - x)}$$
$$\lim_{x\to 1}\frac{3(1 + \sqrt{x})- 2(1 + \sqrt[3]{x} + \sqrt[3]{x^2})}{6(1 - x)}$$
And that's as far as I go.
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OK, without l'Hôpital... do $x \mapsto u^6$:
$$
\begin{align*}
\lim_{x \to 1} \frac{1}{2 (1 - \sqrt{x})} - \frac{1}{3 (1 - \sqrt[3]{x})}
&= \lim_{u \to 1} \frac{1}{2 (1 - u^3)} - \frac{1}{3 (1 - u^2)} \\
&= \lim_{u \to 1}
\frac{3 (1 - u^2) - 2 (1 - u^3)}{6 (1 - u^2)(1 - u^3)} \\
&= \lim_{u \to 1}
\frac{1 - 3 u^2 + 2 u^3}{6 (1 - u)^2 (1 + u) (1 + u + u^2)} \\
&= \lim_{u \to 1}
\frac{(1 - u)^2 (1 + 2 u)}{6 (1 - u)^2 (1 + u) (1 + u + u^2)} \\
&= \lim_{u \to 1} \frac{1 + 2 u}{6 (1 + u) (1 + u + u^2)} \\
&= \frac{1}{12}
\end{align*}
$$
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|
How do you simplify $\tan 10A$ in terms of $5A$? How do you simplify $\tan 10A$ in terms of $5A$?
I just need a few steps to get me going. All help is appreciated.
Thanks!
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Using de Moivre's formula,
$$\cos 10x+i\sin 10x=(\cos x+i\sin x)^{10}$$
$$=\sum_{0\le r\le 10}\binom {10}r(\cos x)^{10-r}(i\sin x)^r$$
$$=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(i\sin x)^{2s}+\binom {10}{2s+1}(\cos x)^{2n-2s-1}(i\sin x)^{2s+1}$$
$$=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(\sin x)^{2s}(-1)^s+i\binom {10}{2s+1}(\cos x)^{2n-2s-1}(\sin x)^{2s+1}(-1)^s$$
Equating the real & the imaginary parts,
$$\cos10x=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(\sin x)^{2s}(-1)^s$$
$$=(\cos x)^{10}-\binom {10}2(\cos x)^8(\sin x)^{2}+\binom {10}4(\cos x)^6(\sin x)^4-\binom {10}6(\cos x)^4(\sin x)^6+\binom {10}8(\cos x)^2(\sin x)^8-(\sin x)^{10}$$
$$=(\cos x)^{10}\{1-(\tan x)^{2}+\binom {10}4 (\tan x)^4-\binom {10}6 (\tan x)^6+\binom {10}8 (\tan x)^8-(\tan x)^{10}\}$$
Simailrly,
$$\sin10x=(\cos x)^{10}\{\binom {10}1 (\tan x)-\binom {10}3 (\tan x)^3+\binom {10}5 (\tan x)^5-\binom {10}7 (\tan x)^7+\binom {10}9 (\tan x)^9\}$$
Divide to get $\tan10x$ in terms of $\tan x$
Validate here.
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|
Find all values x, y and z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes. Find all positive integers x, y, z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes.
It seems trivial that the only set of integers x, y and z that work are $(1^2 + 1)(2^2 + 1) = 3^2 + 1$, which is equivalent to $2 * 5 = 10$, but how would I go about proving this, or are there any other sets for which the following equation works out?
Thanks in advance.
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Suppose without loss of generality $x \ge y$. Then $z^2+1 \le (x^2+1)^2$, so $z < x^2+1$.
Note that $x^2+1 \mid (z^2+1)-(x^2+1)=(z-x)(z+x)$.
Since $(x^2+1)$ is a prime number and
$$0 < z-x < z+x < x^2+x+1<2(x^2+1),$$
we must have $x+z = x^2+1$ and $z-x=1$, from which we obtain $x=2$, and then $z=3, y=1$.
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|
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Hint: Follow the example in lectures to show that $2^n -1$ is not divisible by 3.
In lectures, the example showed the $3 \mid 2^n -1 \iff n$ is even. So what I have done is said all primes $>3$ are odd $\implies n = 2k + 1$ for some $k \in \mathbb{N}$. Checking $2^1$ and $2^3$, we see that $2^{2k+1} \equiv 2 \mod 3$. From the lecture notes, I know that $2^{2k} \equiv 1 \mod 3$. I can then write
$$2^{2k + 1} = \underbrace{2^{2}\times 2^{2} \times \cdots \times 2^{2}}_\text{k times} \times 2 \equiv (1 \times 1 \times \cdots \times 1) \times 2 \equiv 2 \mod 3.$$
So we get that
$$2^{n} - 1 \equiv (2 \mod 3) - 1 \equiv 1 \mod 3$$
and so $3$ does not divide $2^n - 1$ when $n$ is odd. How does this show for it not dividing by $7$?
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In fact we can prove more general result.
This holds true any number of the form $3k\pm1$ where $k$ is an integer
because $2^3=8\equiv1\pmod 7\implies 2^{3k}\equiv1$
$\implies 2^{3k+1}\equiv2\pmod 7\not\equiv1$
and $2^{3k-1}=2^{3(k-1)}\cdot2^2\equiv4\pmod 7\not\equiv1$
We know any prime $>3$ can be written as $6r\pm1$ (where $r$ is an integer) which is a proper subset of the numbers of the form $3k\pm1$
|
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|
Factorize $a^2-ab-bc\pm c^2$ I got this question in a test but it did not specify the variable with respect to which I was supposed to factorize
$$a^2-ab-bc\pm c^2$$
where it could be just $a(a-b)-c(b\pm c)$ but no common factor over all terms. I feel I may be missing something. The $\pm$ is there because I cannot remember whether the last sign was minus or plus.
Is there some trick to factorize this or is this question vacuous? What does it mean to factorize this?
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We observe that
\begin{eqnarray}
a^2-ab-bc+c^2&=&a^2+c^2-b(a+c)\\
&=&(a+c)^2-b(a+c)-2ac\\
&=&(a+c)^2-b(a+c)+\frac{b^2}{4}-\frac{b^2+8ac}{4}\\
&=&\left(a+c-\frac{b}{2}\right)^2-\frac{b^2+8ac}{4}.
\end{eqnarray}
Hence, if $b^2+8ac\geq 0$ then
$$
a^2-ab-bc+c^2=\left(a+c-\frac{b}{2}+\sqrt{\frac{b^2+8ac}{4}}\right)\left(a+c-\frac{b}{2}-\sqrt{\frac{b^2+8ac}{4}}\right)
$$
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|
Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$
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$\dfrac{S^{2}}{s}=(s-a)(s-b)(s-c)\leq \left(\dfrac{(s-a)+(s-b)+(s-c)}{3}\right)^{3}=\dfrac{s^{3}}{27}$
$\therefore$ $S\leq\dfrac{s^{2}}{3\sqrt{3}}=\dfrac{(a+b+c)^{2}}{12\sqrt{3}}\leq\dfrac{1}{12\sqrt{3}}\cdot 3(a^{2}+b^{2}+c^{2})$
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Summation of sequence $a_n - a_{n-1} = 2n$ $(a_1,a_2,a_3,..)$ be a sequence such that $a_1$ =2 and $a_n - a_{n-1} = 2n$ for $n \geq 2$. Then $a_1 + a_2 + .. + a_{20}$ is equal to?
$a_1$ = 2
$a_2$ = 2 + 2x2
$a_3$ = 6 + 2x3
$a_4$ = 12 + 2x4
$a_5$ = 20 + 2x5
$a_n$ = $b_n$ + $g_n$ ,here $g_n$=2xn
$b_3$-$b_2$=4
$b_4$-$b_3$=6
$b_5$-$b_4$=8
$b_6$-$b_5$=10
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As $a_n-a_{n-1}=2n, a_n$ can be at most quadratic.
Let $a_n=An^2+Bn+C$ where $A,B,C$ are arbitrary constants
So, $2n=a_n-a_{n-1}=A(2n-1)+B=2An+B-A$
Comparing the coefficients of $n,2A=2, A=1$
Comparing the constants $B-A=0\implies B=A=1$
So, $a_n= n^2+n +C$
$2=a_1=2+C\implies C=0\implies a_n= n^2+n $
So, $$\sum_{1\le r\le n}a_r= \sum_{1\le r\le n}r^2+ \sum_{1\le r\le n}r= \frac{n(n+1)(2n+1)}6+ \frac{n(n+1)}2=\frac{n(n+1)(n+2)}3$$
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Elementary Diophantine equation Solve $(x+y)(xy+1)=2^z$ in positive integres. My attempts is to use $x+y=2^a$, $xy=2^b-1$ and therefore $x,y$ are the roots of the quadratic equation $w^2-2^aw+2^b-1=0$. I try to analyze its dicriminant but it seems to be a dead end...
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ETA: Jyrki's answer and comments already says this it turns out, and so he really should get the bounty.
Let $x$ and $y$ be positive integers satisfying $x+y\ge 16$. Then $x+y$ must be a power of $2$. Write then, $x+y=2^a$ for some $a \ge 4$. Then $xy+1$ must also be a power of $2$, and as the strict inequality $xy+1 \ge x+y$ for all such $x,y$, it follows that
$$xy+1=(2^a-y)y+1 =2^b$$ $$= 2^ay -y^2+1,$$ where $b$ is an integer satisfying $b>a$. And so $2^a| (y^2-1)$. So from this it follows that $2^{a-1}$ divides one of $y+1, y-1$ [because $4$ can only divide one of $y+1,y-1$]. Likewise, $2^{a-1}$ divides one of $x-1,x+1$. So from this one can deduce WLOG that $2^{a-1}$ divides $y+1$ and $2^{a-1}$ divides $x-1$, as this is the only way $x$ and $y$ can sum to something that is $0$ mod $2^a$. As $x+y=2^a$ and both $x$ and $y$ are nonnegative integers, it follows that $x$ and $y$ must respectively satisfy either the equations $x=2^{a-1}+1$ and $y=2^{a-1}-1$, or $y=2^a-1$ and $x=1$.
However, for each positive integer $a \ge 4$, note that if on the one hand either $x$ and $y$ satisfy the respective equations $x=2^{a-1}+1$ and $y=2^{a-1}-1$, then $(x,y,z)$, with $z=3a-2$, is a solution: $$(x+y)(xy+1)=2^a((2^{2a-2}-1)+1) = 2^{3a-2}.$$ If on the other hand $x=1$ and $y=2^a-1$ then $(x,y,z)$, with $z=2a$ is also a solution: $$(x+y)(xy+1)=(1+y)(1+y)$$ $$=2^a×2^a =2^{2a}.$$
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In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
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You are looking for number of partitions of 12 in parts greater than 1. Because their number is not so big we can present all of them in a list bellow
$$\begin{array}{c|c}
2+2+2+2+2+2 & 1 \\
2+2+2+2+4 & 5 \\
2+2+2+3+3 & 10 \\
2+2+2+6 & 4 \\
2+2+3+5 & 12 \\
2+2+4+4 & 6 \\
2+3+3+4 & 12 \\
3+3+3+3 & 1 \\
2+2+8 & 3 \\
2+3+7 & 6 \\
2+4+6 & 6 \\
2+5+5 & 3 \\
3+3+6 & 3 \\
3+4+5 & 6 \\
4+4+4 & 1 \\
2+10 & 2 \\
3+9 & 2 \\
4+8 & 2 \\
5+7 & 2 \\
6+6 & 1 \\
12 & 1 \\
\end{array}$$
The partition turn in composition if we rearrange them. Number of arrangements is given on the right, so total number of compositions is $89$
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Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$ The question is basically in the title: Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$
I get how to do it from $7\mid $ and $7\mid y$ to $7\mid x^2+y^2$, but not the other way around.
Help is appreciated! Thanks.
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$x^2,y^2$ can be $0^2\equiv0, (\pm1)^2\equiv1,(\pm2)^2\equiv4, (\pm3)^2\equiv2\pmod 7$
Observe that for no combination except $0,0$ of $x^2+y^2 \equiv0\pmod 7$
Alternatively,
If $(7,xy)=1, x^2+y^2\equiv0\pmod 7\implies \left(\frac xy\right)^2\equiv-1\pmod 7$
But we know $-1$ is a Quadratic residue $\pmod p$ iff prime $p\equiv 1\pmod 4$
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solution set for congruence $x^2 \equiv 1 \mod m$ if $m$ is an integer greater than 2, and a primitive root modulo $m$ exists, prove that the only incongruent solutions of $x^2 \equiv 1 \mod m$ are $x \equiv \pm 1 \mod m$.
I know that if a primitive root mod $m$ exists, then $m = 1, 2, 4, p^m,$ or $2p^m$, where p is an odd prime and $m$ is a positive integer. Obviously I can check the cases where $m = 1, 2, 4$ by hand but I am having trouble proving the other two cases.
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$1,2,4$ can be dealt easily.
For $p^m$ divides $(x^2-1)=(x-1)(x+1)$
Now, $x+1-(x-1)=2\implies (x-1,x+1)$ divides $2$
So, either $p^m$ divides $(x-1)$ or $(x+1)$ resulting in exactly $2$ in-congruent solutions
If $2\cdot p^m$ divides $(x^2-1)=(x-1)(x+1)$
Clearly, $x$ is odd
So, $p^m$ divides $\frac{x-1}2\cdot\frac{x+1}2$
Now, $-\frac{x-1}2+\frac{x+1}2=1\implies (\frac{x+1}2,\frac{x-1}2)=1$
So, either $p^m$ divides $\frac{x-1}2$ or $\frac{x+1}2$ resulting in exactly $2$ in-congruent solutions
Alternatively, f $x^d\equiv1\pmod m$ and $m$ has a primitive root $a$
Taking Discrete logarithm, $d\cdot ind_ax\equiv ind_a1\pmod {\phi(m)}\equiv0$
Using Linear congruence theorem, it has exactly $(d,\phi(m))$ solutions as $(d,\phi(m))$ divides $0$
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Sums that are pythagorean and normal I noticed that
$3^2+4^2+15^2=9^2+13^2$
and also
$3+4+15=9+13$
Is there an easy way to find all pairs of disjoint sets of positive integers whose sum are the same and whose sum of squares are the same? How common are they?
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It is easy to generate two such sets of arbitrary length.
Let us choose $s_1,s_2,\ldots,s_{n-1}$ and $t_1,t_2,\ldots,t_{m-1}$ fairly freely and write
$$
\begin{array}{cc}
\sigma_1 = \sum_{i=1}^{n-1} s_i &~& \sigma_2 = \sum_{i=1}^{n-1} s_i^2 \\
\tau_1 = \sum_{i=1}^{m-1} t_i &~& \tau_2 = \sum_{i=1}^{m-1} t_i^2 \\
\end{array}
$$
and then try to see if we can choose $s_n,t_m$ to augment our lists and satisfy the desired criterion, that is
$$
\begin{align}
\sigma_1-\tau_1 &= t_m-s_n \\
\sigma_2-\tau_2 &= t_m^2-s_n^2 \\
&=(t_m-s_n)(t_m+s_n) \\
&=(\sigma_1-\tau_1)(t_m+s_n)
\end{align}
$$
The easiest case is if $|\sigma_1-\tau_1|=1$ and $\sigma_2-\tau_2$ is odd, then we can choose $s_n,t_m=\frac{1}{2}(|\sigma_2-\tau_2|\pm 1)$.
For example for $n=11,m=8$, let $\{s_i\}_{i=1}^{10} = \{1,2,3,5,8,13,21,34,55,89\}$, $\{t_i\}_{i=1}^7=\{4,9,16,25,36,49,93\}$ where we have picked $t_7=93$ to achieve $\tau_1=232=\sigma_1+1$ because we will have $\tau_2>\sigma_2$.
Then $\tau_2-\sigma_2 = 509$, so let $s_{11}=255,t_8=254$ to get
$$\small
1+2+3+5+8+13+21+34+55+89+255=4+9+16+25+36+49+93+254=486
$$
$$
1^2+2^2+3^2+5^2+8^2+13^2+21^2+34^2+55^2+89^2+255^2
\\ =4^2+9^2+16^2+25^2+36^2+49^2+93^2+254^2 = 77840
$$
In order to get every solution you'll have to consider any possible value of $d=|\sigma_1-\tau_1|$, and sets for which at least
*
*$d$ divides $|\sigma_2-\tau_2|$
*$(\sigma_2-\tau_2)/d \equiv d\pmod{2}$, since they sum to $2t_m$
*after adding $s_n,t_m$ the sets are still disjoint
and if you want to count without repetition then only include new solutions when $s_n$ and $t_m$ are the maximums of their respective sets.
|
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|
Finding the Matrix Power of a matrix and limit Find the matrix power, $A^k$, of
$$A=\begin{pmatrix}a & 1-a \\ b & 1-b\end{pmatrix}$$
$$D=P^{-1}AP$$
$$A^k=PD^kP^{-1}$$
I think that
$$P=\begin{pmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{pmatrix} \ \ \ \text{and}\ \ \ \ P^{-1}=\frac{b}{1+b-a}\begin{pmatrix}1&\frac{1-a}{b}\\-1&1\end{pmatrix}$$
I found $A^k$ to be $\begin{pmatrix}1& (a-1)/b(1-b)^k\\1&(a-b)^k\end{pmatrix}$.
What would the limit of $A^k$ be when $k \rightarrow \infty$?
|
A = $\begin{bmatrix}1-a \\ b & 1-b\end{bmatrix}$
Finding the eigenvalues and eigenvectors and writing the matrix in Jordan Normal Form yields:
$\displaystyle A = \begin{bmatrix}a & 1-a \\ b & 1-b\end{bmatrix} = P.D.P^{-1} = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & a-b\end{bmatrix} \cdot \begin{bmatrix}\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & -\frac{a-1}{{(-\frac{a}{b} +\frac{1}{b} +1})b} \\ -\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & \frac{1}{-\frac{a}{b} +\frac{1}{b} +1}\end{bmatrix}$
Now, raising $A^k = (P.D.P^{-1})^k$ is just a matter of raising the diagonal matrix to a power (which is straightforward), yielding:
$\displaystyle A^k = \begin{bmatrix}a & 1-a \\ b & 1-b\end{bmatrix}^k = (P.D.P^{-1})^k$
$\displaystyle = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & a-b\end{bmatrix}^k \cdot \begin{bmatrix}\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & -\frac{a-1}{{(-\frac{a}{b} +\frac{1}{b} +1})b} \\ -\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & \frac{1}{-\frac{a}{b} +\frac{1}{b} +1}\end{bmatrix}$
$\displaystyle = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & (a-b)^k\end{bmatrix} \cdot \begin{bmatrix}\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & -\frac{a-1}{{(-\frac{a}{b} +\frac{1}{b} +1})b} \\ -\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & \frac{1}{-\frac{a}{b} +\frac{1}{b} +1}\end{bmatrix}$
You can handle the rest by multiplying out the three matrices.
|
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|
Integer solutions of $n^3 = p^2 - p - 1$ Find all integer solutions of the equation, $n^3 = p^2 - p - 1$, where p is prime.
|
Rewrite the expression as $p(p-1)=(n+1)(n^2-n+1)$. Case 1: $p|n+1$. It follows that $p\leq n+1$. But then also $p-1 \geq n^2-n+1$, and from these we get $n \geq n^2-n+1$ which can only be true for $n=1$ which gives $p=2$.
Case 2: $p|n^2-n+1$. We write $\frac{n^2-n+1}{p}=k$. Plugging this into the equation we get $p=kn+k+1$. Return to the definition of $k$ and plug this in. We get $n^2-n+1=k(kn+k+1)$ which translates to the quadratic equation
$$n^2+n(-1-k^2)+(1-k-k^2)=0$$
Since these are all integers, the discriminant must be a perfect square, that is
$$(k^2+1)^2-4(1-k-k^2)=(k^2+3)^2+4k-12$$
It is trivial to conclude that this cannot be a square greater than $(k^2+3)^2$ so we can only have $k=3$. Which gives us $n=11,p=37$.
Therefore, the only solutions are $(n,p)\in \{(1,2),(11,37)\}$.
|
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|
Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$
Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$
No idea if I'm on the right track. Maybe distribute the $i$? Wondering if I can get some help.
|
This may be solved by complex variables. Put $z = e^{i\theta}$ to obtain
$$\int_0^{2\pi} \frac{d\theta}{1-a\cos\theta+a^2} =
\int_{|z|=1} \frac{1}{iz} \frac{dz}{1-a/2(z+1/z)+a^2} =
\int_{|z|=1} \frac{1}{iz} \frac{z dz}{z-a/2(z^2+1)+za^2} \\=
-i \int_{|z|=1} \frac{dz}{z-a/2(z^2+1)+za^2}$$
The two poles (use e.g. the quadratic formula) are at
$$z_{0,1} = \frac{1+a^2}{a} \pm \frac{\sqrt{1+a^2+a^4}}{a}.$$
Now assume that $a>1,$ so that the pole corresponding to the plus sign is clearly outside of the unit circle.
On the other hand $$z_1 =\frac{1+a^2}{a} - \frac{\sqrt{1+a^2+a^4}}{a} < 1$$
because it is equivalent to
$$ 1+a^2 -a < \sqrt{1+a^2+a^4}$$
which is $$ a^4 + 2 a^2(1-a) +(1-a)^2 < 1 + a^2 +a^4
\Leftrightarrow
2a^2 - 2a^3 + (1 - a)^2 < 1 + a^2\\
\Leftrightarrow
2a^2 < 2 a^3 + 2a
\Leftrightarrow a <a^2 + 1,$$
which holds trivially.
Now the residue of the integrand at $z_1$ is given by
$$\lim_{z\to z_1} \frac{1}{1-az+a^2} =
\frac{1}{1+a^2-1-a^2+\sqrt{1+a^2+a^4}} =
\frac{1}{ \sqrt{1+a^2+a^4} }.$$
It follows that the integral is given by
$$2\pi i \times -i \times \frac{1}{ \sqrt{1+a^2+a^4} } =
\frac{2\pi}{ \sqrt{1+a^2+a^4} }.$$
|
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|
Find the maximum value of $T=\frac{2}{3}(\cos 2A-\cos 2B)-\tan\frac{C}{2}$ Let $ABC$ be a triangle. Find the maximum value of
$$T=\frac{2}{3}(\cos 2A-\cos 2B)-\tan\frac{C}{2}$$
Please give me some hints. I don't know where to start
Thanks
|
Hint.
$$T=-\frac43\sin(A+B)\sin(A-B)-\frac{1}{\tan\frac{A+B}{2}}.$$
If $A+B\ge \pi/2$,
$$T\le \frac43\sin(A+B)-\frac{1}{\tan\frac{A+B}{2}},$$
where equality holds for $A-B=-\pi/2$.
If $A+B<\pi/2$,
$$T\le \frac43\sin^2(A+B)-\frac{1}{\tan\frac{A+B}{2}},$$
where equality holds for $A=0$.
Note that if you set $t=\tan\frac{A+B}{2}$, $$\sin (A+B)=\frac{2t}{1+t^2}.$$
Use derivatives with respect to $t$.
|
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|
Show abelian groups of order 3240? Show how to get all abelian groups of order $2^3 \cdot 3^4 \cdot 5$.
I just started learning this and was wondering how you would do this?
Is this correct?
$2^3 \cdot 3^4 \cdot 5 = 3240$. Therefore the number of abelian groups of order $3240$ is $3 \cdot 4 = 12$.
Is this the entire proof or do we need to do the table showing divisors like this?
Divisors:
$2^3 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^3 \cdot 3 \cdot 5$
|
You are implicitly using the Fundamental Theorem of Finite Abelian Groups. You should cite this by name.
Of order $8=2^3$, you can have $\mathbb{Z}_8$ or $\mathbb{Z}_2\times \mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$. Similar options exist for the $3^4$.
To fully explain, I would write out each one of the above (3 versions for $2^3$, 5 versions for $3^4$), to justify the $3\times 5$ calculation.
Edit: Corrected number of subgroups of order $3^4$.
|
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|
Differential Equation: Complex Eigenvalue For the following system
$$ x'=\left( \begin{array}{ccc}
\frac{-1}{2} & 1 \\
-1 & \frac{-1}{2} \end{array} \right)x $$
To find a fundamental set of solutions, we assume that $$ x = Ee^{rt}$$
According to the solution, this set of linear algebraic equations is obtained [1]:
$$ x'=\left( \begin{array}{ccc}
\frac{-1}{2} & 1 \\
-1 & \frac{-1}{2} \end{array} \right)
\left( \begin{array}{ccc}
E_1 \\
E_2 \end{array} \right) =
\left( \begin{array}{ccc}
0 \\
0 \end{array} \right)$$
For the eigenvalues and eigenvectors of A the characteristic equation is [2]
$$ \left| \begin{array}{ccc}
\frac{-1}{2}-r & 1 \\
-1 & \frac{-1}{2}-r \end{array} \right| = r^2 + r+\frac{5}{4} $$
Therefore the eigenvalues are $\frac{-1}{2} + i$ and $\frac{-1}{2} - i$
I don't understand how eqn [1] and [2] are obtained.
|
We are given:
$$ x'=\left( \begin{array}{ccc}
\frac{-1}{2} & 1 \\
-1 & \frac{-1}{2} \end{array} \right)x $$
To find the characteristic polynomial, we solve:
$$|A - \lambda I| = 0$$
So, for your matrix, we would write:
$$ \left| \begin{array}{ccc}
\frac{-1}{2} - \lambda & 1 \\
-1 & \frac{-1}{2} - \lambda \end{array} \right|= 0 $$
Taking the determinant results in the equation you show (you are using $r$ instead of the more traditional $\lambda$).
Here are the steps of finding the determinant and solving for the roots (eigenvalues):
$$\displaystyle \lambda^2 + \lambda + \frac{5}{4} = 0 \rightarrow \lambda_{1,2} = -\frac{1}{2} \pm i$$
Now that we know the eigenvalues, there are various ways to find the exponential solution using eigenvectors or other methods.
Is that all clear?
In the end, we would end up with (you used $E$ as the initial condition, where I call that vector $x_0$):
$$\displaystyle x(t) = x_0 e^{A t} = x_0 e^{-\frac{t}{2}}\begin{bmatrix} \cos t & \sin t\\- \sin t & \cos t\end{bmatrix}$$
|
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|
Is this determinant bounded? Let $D_n$ be the determinant of the $n-1$ by $n-1$ matrix such that the main diagonal entries are $3,4,5,\cdots,n+1$ and other entries being $1$. i.e.
$$D_n= \det \begin{pmatrix}
3&1&1&\cdots&1\\
1&4&1&\cdots&1\\
1&1&5&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots &n+1
\end{pmatrix}$$
Is the set $\{D_n/n! \ | \ n=2,3,4,\cdots\}$ bounded?
My attempt is to observe that $D_n= \det ((1)+ diag\{2,3,\cdots,n\})$, where $(1)$ is the matrix with all entries being $1$. We have $\det (1)=0$ and $\det (diag\{2,3,\cdots,n\})=n!$, which seems to be promising, but then I can't move along, could someone please helps.
|
Since adding a multiple of one row to another doesn't change a matrix's determinant, we can subtract the first row from each of the other rows to see that
$$D_n= \det \begin{pmatrix}
3&1&1&\cdots&1\\
-2&3&0&\cdots&0\\
-2&0&4&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-2&0&0&\cdots &n
\end{pmatrix}.$$
Then we subtract $1/(k+1)$ times the $k$th row from the first row for each $2\le k\le n-1$, yielding
$$D_n= \det \begin{pmatrix}
A&0&0&\cdots&0\\
-2&3&0&\cdots&0\\
-2&0&4&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-2&0&0&\cdots &n
\end{pmatrix}$$
with $A = 3 + \frac23 + \frac24 + \cdots + \frac2n = 2h_n$, where $h_n = \sum_{k=1}^n \frac1k$ is the $k$th harmonic number. Since the determinant of an upper triangular matrix is just the product of its diagonal entries, this computation shows that
$$
D_n = h_nn!.
$$
In particular, $D_n/n!$ is unbounded.
|
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|
Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({{x + a\ln a} \over a}\right) \cr
{1 \over y}{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \times x{a^x} \cr
{{dy} \over {dx}} &= {{x{a^{x + 1}} + {x^3}{a^x} + {a^{x + 1}}{x^2}\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {a^x} + {x^2}{a^{x - 1}} + {a^x}x\ln a \cr
{{dy} \over {dx}} &= {a^x}\left(1 + {x^2}{a^{ - 1}} + x\ln a\right) \cr
{{dy} \over {dx}} &= {a^x}\left({{a + {x^2} + x\ln a} \over a}\right) \cr} $$
However the answer in the back of the book is:
$${dy \over dx} = a^x (1 + x\ln a)$$
What have I done wrong?
|
$y=xa^x$
$$y'=x'a^x+x(a^x)'=a^x+xa^x\ln a=a^x(1+x\ln a)$$
|
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|
Is $\frac{\lfloor{x}\rfloor+1}{2} \le \lfloor\frac{x}{2}\rfloor + 1$ The answer seems to be yes.
Here's my reasoning:
For $x < 1$, $\frac{1}{2} < 1$
For $1 < x < 2$, $1 = 1$
For $2 \mid x$, $\frac{x}{2} + \frac{1}{2} < \frac{x}{2} + 1$
For $2 \mid x-1$, $\frac{x+1}{2} = \frac{x-1}{2} + 1$
For all other values, the value is equal to one of the above.
My reasoning is not very straight forward. What would be a standard way to demonstrate this?
Thanks,
-Larry
|
Set $n = \lfloor \frac{x}{2} \rfloor$. Then $x< 2n + 2$ and thus $\lfloor x \rfloor < 2n+2$ which implies $\lfloor x \rfloor \leq 2n+1$ as $\lfloor x \rfloor$ is an integer. This gives
$$\frac{\lfloor x \rfloor + 1}{2} \leq \frac{2n+2}{2} = n+1 = \lfloor \frac{x}{2} \rfloor + 1$$
|
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|
Finding normalised eigenvectors... I'm trying to find the eigenvector/eigenvalues of the $2\times2$ matrix:
\begin{pmatrix}4 & 2 \\ 2 & 3 \end{pmatrix}
This is my work:
$$\det(A-\lambda I) = \lambda^2-7 \lambda+8=0 \iff \lambda=\frac{7+\sqrt{17}}{2} \ \lor \ \lambda= \frac{7-\sqrt{17}}{2}$$
$x_1$ (eigenvector)=\begin{pmatrix} (1+\sqrt17)/4 \\ k \end{pmatrix} , where k is any number. How do I "NORMALISE" this eigenvector?
Can someone check my working because I'm getting weird answers.
|
If $\mathbf{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\mathbf{x}=\lambda\mathbf{x}$ and $(A-\lambda I)\mathbf{x}=\mathbf{0}$.
First, find the eigenvector corresponding to the eigenvalue $λ=\frac{7+\sqrt{17}}{2}$:
$$\begin{align*} &\quad\quad\quad\quad\left(\begin{array}{c|c}
A-\lambda I & 0
\end{array}\right)\quad\quad\text{insert your }A\text{ and }\lambda\\ &=\left(\begin{array}{cc|c}
4-\tfrac{7+\sqrt{17}}{2} & 2 & 0 \\
2 & 3-\tfrac{7+\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{compute the differences}\\ &\implies \left(\begin{array}{cc|c}
\tfrac{1-\sqrt{17}}{2} & 2 & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{multiply the first row by }\tfrac{4}{1-\sqrt{17}}\\ &\implies \left(\begin{array}{cc|c}
2 & \tfrac{8}{1-\sqrt{17}} & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{multiply the first fraction by }1+\sqrt{17}\\ &\implies \left(\begin{array}{cc|c}
2 & \tfrac{8(1+\sqrt{17})}{-16} & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{simplify the first fraction}\\ &\implies \left(\begin{array}{cc|c}
2 & \tfrac{-1-\sqrt{17}}{2} & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{subtract the first row from the second}\\ &\implies \left(\begin{array}{cc|c}
4 & -1-\sqrt{17} & 0 \\
0 & 0 & 0
\end{array}\right)\quad\quad\text{deduce the solution}\\ &\implies \mathbf{x}=k\pmatrix{1+\sqrt{17}\\4}\end{align*}$$
Now, normalize it by
$$\hat{\mathbf{x}}=\frac{\mathbf{x}}{||\mathbf{x}||}$$
and do the same thing for the second eigenvalue.
|
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|
Solve a cubic polynomial? I've been having trouble with this question:
Solve the equation, $$5x^3 - 24x^2 + 9x + 54 = 0$$ given that two of its roots are equal.
I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: $a$, $a$, $b$, but I am still unsuccessful.
Any help would be greatly appreciated.
|
By Vieta's formula, $$\begin{cases}2a+b=\dfrac{24}5\\a^2+2ab=a(a+2b)=\dfrac95\\a^2b=-\dfrac{54}5 \end{cases}$$
Notice that $0$ is not a root of the original equation, so both $a,b$ are nonzero. Divide the second equation by the third equation to get $\dfrac{a+2b}{ab}=\dfrac2a
+\dfrac1b=-\dfrac1{6}$. Multiplying by the first equation, we have $$(2a+b)(\dfrac2a
+\dfrac1b)=5+2(\dfrac ba+\dfrac ab)=-\dfrac45 $$
Let $\dfrac ba=t$, then the above equation becomes $$5+2(t+\dfrac1t)=-\dfrac45\implies 10t^2+29t+10=(5t+2)(2t+5)=0$$
Hence $t=-\dfrac25$ or $-\dfrac52$. If $t=-\dfrac25, 2a+b=2a+at=\dfrac85a=\dfrac{24}5,$ so $a=3$. On the other hand, if $t=-\dfrac52, 2a+b=-\dfrac12a=\dfrac{24}5,$ so $a=-\dfrac{48}5$. But from the third equation we must have $b\lt 0$, and then $a,b$ are both $\lt 0$, so $2a+b\lt 0$ and then the first equation cannot hold. Therefore $a=3, b=-\dfrac{6}5$ are the roots of the original cubic.
|
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|
$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$, $a_n\in \mathbb Q$ and not $a_n$ not monotonic How can I construct a sequence $\{a_n\}$ of positive rational numbers, which is not monotonic such that $$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$$
I thought a lot on this question but I always come up with a monotonic sequence. Any help will be appreciated.
|
From Madhava formula, we have
$$\pi = 4 - \dfrac43 + \dfrac45 - \dfrac47 \pm = \sum_{n=0}^{\infty}\dfrac{(-1)^n 4}{(2n+1)}$$
We can write $\dfrac1{2n+1} = \dfrac{k\sqrt{n}+1}{2n+1} - \dfrac{k\sqrt{n}}{2n+1}$. Hence, we have
$$\pi = \sum_{n=0}^{\infty} \left( (-1)^n\dfrac{\sqrt{n}+4}{2n+1} + (-1)^{n+1}\dfrac{\sqrt{n}}{2n+1}\right) = \sum_{n=0}^{\infty}(-1)^n a_n$$
Writing it out, we see
$$\pi = \dfrac{\sqrt{0} + 4}{2 \cdot 0 + 1} - \dfrac{\sqrt{0}}{2 \cdot 0 + 1} + \dfrac{\sqrt{1}}{2 \cdot 1 + 1} - \dfrac{\sqrt{1} + 4}{2 \cdot 1 + 1} + \dfrac{\sqrt{2} + 4}{2 \cdot 2 + 1} - \dfrac{\sqrt{2}}{2 \cdot 2 + 1} + \dfrac{\sqrt{3}}{2 \cdot 3 + 1} - \dfrac{\sqrt{3} + 4}{2 \cdot 3 + 1} + \cdots$$
where $a_n$'s are as desired, i.e.,
$$a_n = \begin{cases} \dfrac{\sqrt{2k}+4}{2 \cdot (2k)+1} & \text{ if $n = 4k$}\\ \dfrac{\sqrt{2k}}{2 \cdot (2k)+1} &\text{ if $n=4k+1$}\\ \dfrac{\sqrt{2k+1}}{2 \cdot(2k+1)+1} & \text{ if $n=4k+2$}\\ \dfrac{\sqrt{2k+1}+4}{2 \cdot(2k+1)+1} & \text{ if $n=4k+3$}\end{cases}$$
Below is the pictorial representation of the sequence $a_n$. The first $44$ values of this sequence are plotted. As you can see from the figure below, the sequence oscillates up and down and is not monotonic.
Below is the convergence of the partial sum $\displaystyle \sum_{n=0}^N a_n$ to $\pi$. The $\color{red}{\text{red line}}$ is the value of $\pi$ for reference. You can see how the partial sums converge to $\pi$.
I didn't see $a_n \in \mathbb{Q}$. Here is the answer if you want $a_n \in \mathbb{Q}$. $$\dfrac1{2n+1} = \dfrac2{2n+1} - \dfrac1{2n+1}$$
Hence,
$$\pi = \sum_{n=0}^{\infty} \left({(-1)}^n\dfrac{8}{2n+1} + (-1)^{n+1} \dfrac4{2n+1}\right)$$
$$\pi = \dfrac81 - \dfrac41 + \dfrac43 - \dfrac83 + \dfrac85 - \dfrac45 \pm \cdots$$
Now choose $a_n$'s as follows.
$$a_n = \begin{cases}\dfrac{8}{4k+1} & \text{ if $n=4k$}\\ \dfrac4{4k+1} & \text{ if $n=4k+1$}\\ \dfrac4{4k+3} & \text{ if $n=4k+2$}\\ \dfrac8{4k+3} & \text{ if $n=4k+3$}\end{cases}$$
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|
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$ Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$
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This uses the hints by Raymond Manzoni and Cameron Williams. I am not sure if the answer is correct, but this should give an idea of how to proceed. I believe the answer by Raymond Manzoni is correct.
$$ \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)} = \dfrac{2^n(2^{n+1})!\sqrt{\pi}}{4^{2^n}(2^n)!^2}$$
Hence,
\begin{align}
\log \left( \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)}\right) & = n \log(2) + \dfrac{\log(\pi)}2 - 2^{n+1} \log(2) + \log \left(\dbinom{2^{n+1}}{2^n}\right)
\end{align}
\begin{align}
S_n = \dfrac{\log \left( \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)}\right)}{2^n} & = \dfrac{n \log(2)}{2^n} + \dfrac{\log(\pi)}{2^{n+1}} - 2 \log(2) + \dfrac1{2^{n}}\log \left(\dbinom{2^{n+1}}{2^n}\right)
\end{align}
Hence,
$$\sum_{n=1}^{\infty}\dfrac{n \log(2)}{2^n} = 2 \log(2)$$
$$\sum_{n=1}^{\infty}\dfrac{\log(\pi)}{2^{n+1}} = \dfrac{\log(\pi)}2$$
Recall that
$$\log \left(\dbinom{2k}k\right) \sim 2k \log(2) - \dfrac{\log(\pi)}2 - \dfrac{\log(k)}2$$
This gives us
$$\log \left(\dbinom{2 \cdot 2^n}{2^n}\right) \sim 2^{n+1} \log(2) - \dfrac{\log(\pi)}2 - n \dfrac{\log(2)}2$$
$$\dfrac{\log \left(\dbinom{2 \cdot 2^n}{2^n}\right)}{2^n} \sim 2 \log(2) - \dfrac{\log(\pi)}{2^{n+1}} - n \dfrac{\log(2)}{2^{n+1}}$$
This gives us $S_n \sim \dfrac{n}{2^{n+1}} \log(2)$ and hence I would hope (this step needs more justification and is probably wrong in fact) $$\sum_{n=1}^{\infty} S_n \approx \log(2)$$
Hence, the answer you are looking for is approximately $$e^{\log(2)} = 2$$
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|
Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$
This should be fairly straightforward but the proof seems to be alluding me.
I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perhaps finals have fried my brain.
|
You can use the Taylor series of $\sin(x)$ about $x=0$:
$$\sin(x) = \sum_{n=0}^\infty {\frac {(-1)^n x^{2n+1}} {(2n+1)!}}$$
The first few terms are:
$$\sin(x) = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots$$
Thus, set:
$$x - \frac {x^3}{3!} < \sin x\\
x - \frac {x^3}{3!} < x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots\\
0 < \frac{x^5}{5!} - \frac {x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!}+ \dots\\
$$
For $x>0$, this is true.
Similarly, for $\sin(x) < x$:
$$x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots < x\\
\frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots > 0$$
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|
The unique root on (1,2) For any given $n\geq 2$, let $x^n=\sum\limits_{k=0}^{n-1}x^{k}$ be the equation, prove: there is only one real root which in (1,2).
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We have $x^n=\frac{1-x^n}{1-x}$, which rearranges to $0=\frac{1-x^n}{1-x}-x^n=\frac{1-x^n-x^n+x^{n+1}}{1-x}$. For this to be zero (on $x>1$), $f(x)=x^{n+1}-2x^n+1$ must be zero. We now have the following facts:
*
*$f(2)=2^{n+1}-2^{n+1}+1=1>0$.
*$f(1)=1-2+1=0$.
*$f'(x)=(n+1)x^{n}-2nx^{n-1}=0$ has two solutions, $x=0$ and $x=\frac{2n}{n+1}\in(1,2)$
*$f'(1)=(n+1)-2n=1-n<0$.
Combining (2) and (4) we know $f(x)<0$ immediately to the right of $1$. By (1) we know $f(x)>0$ at $2$. By IVT, $f(x)=0$ somewhere in $(1,2)$. Hence $f(x)$ has two zeroes, at least, in $[1,2)$. If there were a third one, then $f'(x)=0$ would have at least two solutions in $(1,2)$ by MVT. But by (3) there is only one such solution, so there is at most one zero of $f(x)$ in $(1,2)$.
|
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|
Find max of $x^7+y^7+z^7$
Find max of $x^7+y^7+z^7$ where $x+y+z=0$ and $x^2+y^2+z^2=1$
I tried to use the inequality:$$\sqrt[8]{\frac {x^8+y^8+z^8} 3}\ge\sqrt[7]{\frac {x^7+y^7+z^7} 3}$$ but stuck
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The condition gives $x^2+xy+y^2=\frac{1}{2}$.
Also $$x^7+y^7+z^7=x^7+y^7-(x+y)^7=-7xy(x+y)(x^2+xy+y^2)^2=$$
$$=-\frac{7}{4}xy(x+y)\leq\frac{7}{4}\sqrt{\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}}=\frac{7}{8\sqrt2}\sqrt{\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}}\leq$$
$$\leq\frac{7}{8\sqrt2}\sqrt{\frac{4}{27}}=\frac{7}{12\sqrt6}$$
because $\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}\leq\frac{4}{27}$ it's just $$(x-y)^2(x-z)^2(y-z)^2\geq0.$$
Done!
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|
Calculation of ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$ (1) Total no. of integer ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$
(2) Total no. of integer ordered pair $(x,y,z)$ in $x+yz = 1\;\;,y+zx = 1\;\;,z+xy = 1$
My Try:: (1) Clearly $ x = 0,y = 0,z = 0$ are the solution of given equation
and from three equation we observe that $x,y,z$ has same sign.
Now If $x\neq 0,y\neq 0$ and $z\neq 0,$ Then $x^2-y^2 =-z(x-y)\Leftrightarrow (x-y).(x+y+z) =0$
Means either $x=y$ or $x+y+z = 0$
$\bullet$ If $x = y$, The put in $z^2 = xy=x^2=y^2\Leftrightarrow z = \pm x = \pm y$
means $x = y =z$
So $(x,y,z) = (k,k,k)$ where $k\in \mathbb{Z}$
$\bullet$ If $x+y+z = 0$, Then put in third $z^2 = xy\Leftrightarrow x^2+y^2+xy = 0$
So $x^2+y^2+xy = x^2+y^2+(x+y)^2 = 0\Leftrightarrow x = 0,y = 0,x+y = 0$
So $(x,y,z) = (0,0,0)$
So Given equation has Infinite solution
My Question is I have Calculate Right or not
If not plz explain me.
Thanks
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Hint to (2): subtract 2nd equation out 1st:
$$(x-y)(1-z)=0.\ \ \ (1)$$
Addition:
Similarly
$$(x-z)(1-y)=0,\ \ \ (2)$$
$$(z-y)(1-x)=0.\ \ \ (3)$$
If $z=1$, then $x+y=1, xy=0$ (from your $1$st and $3$rd equations) whence we obtain two solutions
$$z=1, x=1, y=0,$$
$$z=1, x=0, y=1.$$
Similarly for $y=1$ we get one more solution from ($2$):
$$y=1, x=1, z=0.$$
Further, let $x,y,z\ne 1$. Then $x=y=z$ and $x^2+x-1=0$ from your $1$st equation. Then we have two solutions:
$$x=y=z=\frac{-1+\sqrt{5}}{2}$$
and
$$x=y=z=\frac{-1-\sqrt{5}}{2}.$$
|
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|
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$
I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.
Any help is appreciated! Thanks.
Edit: solution I require is $\frac{1}{2} (3^{n-1}+1-2^n)$
Edit2: Solutions to the homogeneous equation would be of the form $h_n = A(\alpha)^n + B(\beta)^n$, and $p_n$ will exist such that $s_n = h_n + p_n$
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Let $ t_n := \frac{s_n}{3^n} $. Hence, $ t_n = t_{n-1} + 3^{-n}\left(2^{n-2} - 1\right) $. Also, $ t_1 = 0 $. Hence, $$ t_n = \sum_{k = 2}^n \frac{1}{4}\left(\frac{2}{3}\right)^n - 3^{-n} $$ This is a geometric series and easily evaluated to arrive at $$ t_n = \frac{1}{2}\cdot 3^{-n} \cdot \left(1 - 2^{n}\right) + \frac{1}{6} \implies s_n = \frac{1 + 3^{n-1}- 2^{n}}{2} $$
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|
Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$ I need the closed-form for the $n$-th derivative ($n\geq0 $):
$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)$$
Thanks!
By following the suggestion of Hermite polynomials:
$$H_n(x)=(-1)^ne^{x^2}\frac{\partial^n}{\partial x^n}e^{-x^2}$$
and doing the variable change $x=\pi a y^{-\frac{1}{2}}$, I obtain:
$$\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n\frac{\partial^n}{\partial y^n}$$
and therefore
$$H_n(\pi a y^{-\frac{1}{2}})=(-1)^{n+1}e^{\frac{\pi^2a^2}{y}}2\left(\frac{\pi a}{y^{\frac{3}{2}}}\right)^n\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}$$
Finally
$$\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}=\frac{1}{2}e^{-\frac{\pi^2a^2}{y}}(-1)^{n+1}H_n(\pi a y^{-\frac{1}{2}})\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n$$
Is this correct?
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Here we show that OPs approach is regrettably not correct and provide an alternative to answer the problem.
This answer is based upon the $n$-th derivative of the composite of two functions. It is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:
Hoppe Form of Generalized Chain Rule
Let $D_x$ represent differentiation with respect to $x$ and $z=z(x)$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$. The following is valid for $n\geq 0$:
\begin{align*}
D_x^n f(z)=\sum_{k=0}^nD_z^kf(z)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_x^nz^j\tag{1}
\end{align*}
Let's denote for convenience only $c:=\pi^2a^2$.
The following is valid for $n\geq 1$
\begin{align*}
D_x^ne^{-\frac{c}{x}}=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{n-1}{k}k!c^{n-k}x^{-2n+k}\tag{2}
\end{align*}
Using Hoppes formula (1) we obtain by setting
\begin{align*}
f(z)=e^{z}\qquad\quad\text{and}\quad\qquad z=z(x)=-\frac{c}{x}
\end{align*}
the following for $n\geq 1$
\begin{align*}
D_x^ne^{-\frac{c}{x}}&=\sum_{k=0}^nD_z^ke^z\frac{(-1)^k}{k!}
\sum_{j=0}^k(-1)^j\binom{k}{j}\left(-\frac{c}{x}\right)^{k-j}D_x\left(-\frac{c}{x}\right)^j\tag{3}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}
\left(-\frac{c}{x}\right)^{k-j}n!c^j\binom{n+j-1}{j-1}\frac{(-1)^{n+j}}{x^{n+j}}\tag{4}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{n!}{k!}(-1)^nc^k\sum_{j=0}^k(-1)^j\binom{k}{j}\binom{n+j-1}{j-1}x^{-n-k}\tag{5}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{n!}{k!}(-1)^nc^k\binom{n-1}{k-1}(-1)^kx^{-n-k}\tag{6}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n-1}{k-1}c^kx^{-n-k}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^{k}\binom{n}{k}\binom{n-1}{k}c^{n-k}x^{-2n+k}\tag{7}\\
\end{align*}
and the claim follows.
Comment:
*
*In (3) we apply Hoppes formula (1)
*In (4) we use
\begin{align*}
D_z^ke^z=e^z=e^{-\frac{c}{x}}\quad\text{and}
\quad D_x\left(-\frac{c}{x}\right)^j=n!c^j\binom{n+j-1}{j-1}(-1)^{n+j}
\end{align*}
*In (5) we do some rearrangement
*In (6) we use the identity
\begin{align*}
\sum_{j=0}^k(-1)^j\binom{k}{j}\binom{n+j-1}{j-1}=\binom{n-1}{k-1}(-1)^k
\end{align*}
*
*In (7) we replace $k$ with $n-k$ and set the upper index limit to $n-1$
Small examples: $n=1,2,3$
Using formula (2) we obtain for $n=1,2,3$
\begin{align*}
D_x^1e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^2}\\
D_x^2e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^4}\left(c-2x\right)\tag{8}\\
D_x^3e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^6}\left(c^2-6cx+6x^2\right)\\
\end{align*}
Hermite polynomials:
If we consider Hermite polynomials in the form
\begin{align*}
H_n(x)=-e^{x^2}D_xe^{-x^2}
\end{align*}
and apply a substitution
\begin{align*}
x=x(y)=\sqrt{\frac{c}{y}}=\sqrt{c}y^{-\frac{1}{2}}
\end{align*}
in order to obtain
\begin{align*}
H_n(x(y))&=-e^{(x(y))^2}D_ye^{-(x(y))^2}\\
&=-e^{\frac{c}{y}}D_ye^{-\frac{c}{y}}
\end{align*}
we are in the same situation as above, since we have to calculate the $n$-th derivative of a composition of functions, which is not feasible with OPs approach. It could be done with Hoppes formula instead.
Let's look at a small example by taking $n=2$. In the following I omit most of the time the argument $y$ and write $xx^\prime$ instead of $x(y)x^\prime(y)$ to ease reading.
\begin{align*}
H_2(x(y))&=e^{x^2}D_y^2\left(e^{-x^2}\right)\\
&=-e^{x^2}D_y\left(-2xx^\prime e^{-x^2}\right)\\
&=-e^{x^2}\left(D_y\left(-2xx^\prime\right)e^{-x^2}+(-2xx^\prime)D_y\left(e^{-x^2}\right)\right)\\
&=-e^{x^2}\left(\left(-2{x^\prime}^2-2xx^{\prime\prime}\right)e^{-x^2}+4x^2{x^\prime}^2e^{-x^2}\right)\\
&=-2{x^\prime}^2-2xx^{\prime\prime}+4x^2{x^\prime}^2
\end{align*}
Since
\begin{align*}
x&=x(y)=\sqrt{c}\,y^{-\frac{1}{2}}\\
x^\prime&=x^\prime(y)=-\frac{1}{2}\sqrt{c}\,y^{-\frac{3}{2}}\\
x^{\prime\prime}&=x^{\prime\prime}(y)=\frac{3}{4}\sqrt{c}\,y^{-\frac{5}{2}}\\
\end{align*}
we finally obtain
\begin{align*}
H_2(x(y))&-2{x^\prime}^2-2xx^{\prime\prime}+4x^2{x^\prime}^2\\
&=-2\frac{1}{4}c\,y^{-3}-2\sqrt{c}\,y^{-\frac{1}{2}}\frac{3}{4}\sqrt{c}\,y^{-\frac{5}{2}}
+4cy^{-1}\frac{1}{4}c\,y^{-3}\\
&=-\frac{1}{2}c\,y^{-3}-\frac{3}{2}c\,y^{-3}+c^2\,y^{-4}\\
&=-2c\,y^{-3}+c^2\,y^{-4}\\
&=\frac{c}{y^4}\left(c^2-2y\right)
\end{align*}
in accordance with (8). So, we can apply Hoppes formula or perform some equivalent calculation.
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|
Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$
However, the answer is 3
|
Given: $\boxed{3\sqrt{x+13} = x+9}$
Isolate the radical, x's cancel:
$\dfrac{3\sqrt{x+13}}{3} = \dfrac{x+9}{3}$
$\sqrt{x+13} = \dfrac{x+9}{3}$
Square both sides:
$\left(\sqrt{x+13}\right)^2 =\left(\dfrac{x+9}{3}\right)^2$
$x+13 = \left(\dfrac{x+9}{3}\right)\left(\dfrac{x+9}{3}\right)$
Move the 9 to the other side:
$9(x+13) = \dfrac{x^2+18x+81}{9}$
Set the equation equal to zero and group like terms:
$9x+117 = x^2+18x+81$
$0 = x^2+18x-9x+81-117$
Factor:
$0 = x^2+9x-36$
$0=(x+12)(x-3)$
$0=x+12\implies \boxed{x=-12}$
$0=x-3\implies\boxed{x=3}$
We can check to see if these are solutions:
$3\sqrt{-12+13} = -12+9$
$3(1)=-3$
$\boxed{3 \neq -3}$
Therefore, -12 is not a solution to the original equation.
$3\sqrt{3+13} = 3+9$
$3(4)=12$
$\boxed{12= 12}$
Therefore, 3 is the solution to the original equation.
|
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If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n I'm reading elementary number theory and trying to understand the following problem: If $x^2\equiv 1 \pmod{n}$, $n=pq$, $p$ and $q$ are odd primes and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of $n$.
EDIT: Andreas Caranti wrote an updated, corrected version of my problem, so I wrote the definition again.
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If $n=p\cdot q$ divides $x^2-1=(x+1)(x-1)$
If odd prime $r$ divides $x-1$ and $x+1,$ it will divide $(x+1)-(x-1)=2$ which is impossible $\implies r$ can divide exactly one of $x-1,x+1$
Now let us check the possibilities
Case $1:$ $p$ divides $x-1$ and $q$ divides $x-1\implies (x-1)$ is divisible by lcm$(p,q)=pq=n\implies x\equiv1\pmod n$ which is not acceptable according to the given condition.
Similarly, Case $2:$ $p$ divides $x+1$ and $q$ divides $x+1$ is discarded
Case $3:$ $p$ divides $x-1$ and $q$ divides $x+1$
Then $x-1=p\cdot a,x+1=q\cdot b$ where $a,b$ are some integers
i.e., $q\cdot b-p\cdot a=2$ , we can always find such $a,b$ using Bézout's Lemma as $(p,q)=1$
If $q$ divides $a,$ the LHS will be divisible by $q$
$\implies 2$ must be divisible by $q$ which is impossible $\implies (q,a)=1$
Now, gcd$(x-1,n)=$gcd$(p\cdot a,p\cdot q)=p\cdot$gcd $(a,q)=q$ as gcd$(a,q)=1$
Case $4$ $p$ divides $x+1$ and $q$ divides $x-1$ can be handled similarly.
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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)
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In this case, the following trick also works: Dividing both the numerator and the denominator by $\cos^4 x$, we can use the substitution $ t = \tan x$ to obtain
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2}
&= \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x)^2} \sec^2 x \, dx \\
&= \int_{0}^{\infty} \frac{1 + t^2}{(a^2 + b^2 t^2)^2} \, dt \\
&= \frac{1}{a^2}\int_{0}^{\infty} \left( \frac{1}{a^2 + b^2 t^2} + \frac{(a^2 - b^2) t^2}{(a^2 + b^2 t^2)^2} \right) \, dt.
\end{align*}
The first one can be evaluated as follows: Let $bt = a \tan\varphi$. Then
$$ \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{1}{ab} \int_{0}^{\frac{\pi}{2}} d\varphi = \frac{\pi}{2ab}. $$
For the second one, we perform the integration by parts:
\begin{align*}
\int_{0}^{\infty} \frac{t^2}{(a^2 + b^2 t^2)^2} \, dt
&= \left[ - \frac{1}{b^2}\frac{1}{a^2 + b^2 t^2} \cdot \frac{t}{2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{2b^2}\frac{dt}{a^2 + b^2 t^2} \\
&= \frac{1}{2b^2} \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} \\
&= \frac{\pi}{4ab^3}.
\end{align*}
Putting together, the answer is
$$ \frac{(a^2 + b^2)\pi}{4(ab)^3}. $$
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Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$
Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c}
+ \frac{b}{c+a} + \frac{c} {a+b}$$
Using Nesbitt's inequality: $I \ge \frac 32$
I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c
}^{3}+{c}^{2}a-2\, \left( b+c \right) \left( c+a \right) \left( a+b
\right) \le 0 \\ \iff {a}^{3}+{b}^{3}+{c}^{3}\leq c{a}^{2}+{a}^{2}b+a{b}^{2}+c{b}^{2}+{c}^{2
}a+{c}^{2}b+abc
$$
and stuck.
EDIT: Look like prove $I \le 2$ not a good thinking :P
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This not a complete solution, just a 'case' of a problem.
When $\gcd(a+b,a+c,b+c)=1 \implies \text { } a,b,c \text { are not all odd or not all even}$
Your expression is
$\dfrac{a(c+a)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)}{(a+b)(b+c)(c+a)}$
Denoting $a+b=k, b+c=l$ and $c+a=m$
$$klm |akm+blk+cml \implies a=p_1l, b=p_2m, c=p_3k$$
$a=p_1l, b=p_2m,$ and $ c=p_3k$ is not possible. So, we have no solution if we consider $\gcd(a+b,a+c,b+c)=1$.
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Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$ Find the range of:
$$y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$$
What I tried:
Let:$$\log_e\frac{x^2+e}{x^2+1}=X,$$
then $$y=\sqrt {\sin X}+\sqrt{\cos X}$$
$$y_{max}at X=\pi/4$$
The rest is too complicated. I am stuck up.
Can someone please give a analytical solution.
Even Wolfram alpha doesn't help.
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A related problem. First, we study the expression
$$ \frac{x^2+e}{x^2+1}=1+\frac{e-1}{x^2+1} \longrightarrow_{|x|\to\infty} 1, $$
which implies $ y(x)\longrightarrow_{|x|\to \infty} 0 $. To find the maximum of the function $y(x)$, Let's study the function
$$ h(t)=\sqrt{\sin(t)}+\sqrt{\cos(t)}. $$
The maximum of the above function is attained at $t=\frac{\pi}{4}$ which can proved using derivative test. So, this implies our function attains its max when
$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} \implies x=0.6632987771, -0.6632987771. $$
Plugging back in the function y(x) gives the max which is $y=1.681792830$. So the range is
$$ 1 < y \leq 1.681792830.$$
Note: You can solve $ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} $ easily as,
$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4}\implies \frac{x^2+e}{x^2+1} =e^{\frac{\pi}{4}}\implies x^2+e= e^{\frac{\pi}{4}}x^2+ e^{\frac{\pi}{4}}=\dots.$$
I think you can finish it.
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Equation with high exponents I would appreciate any help with this problem:
$ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x^1+1x^0=0 $
I know that when $x$ isn't zero $x^0=1$ so the equation could be re-written as $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x+1=0 $. I am not sure what to do from here. I have tried using wolframalpha to get the solutions so I know real ones exist (2 of them, actually), but I have no idea how to get them.
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As the equation is Reciprocal Equation of the First type,
Divide either sides by $x^4,$ to get $$x^4+\frac1{x^4}+2\left(x^3+\frac1{x^3}\right)+2\left(x^2+\frac1{x^2}\right)+5\left(x+\frac1x\right)+3=0$$
Put $x+\frac1x=y$ to reduce the equation the degree $\frac82=4$
Can you take it form here?
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Extracting the coefficient from a generating function Consider (for fixed $r$) the following function:
$$f(z) = \frac{1}{1-z-z^2-\cdots-z^r} = \frac{1}{1-z\frac{1-z^r}{1-z}}=\sum_{j=0}^\infty\left(z\frac{1-z^r}{1-z}\right)^j$$
(Assume everything is ok with regards to convergence.)
The text I am reading claims that if we were to write $f(z)$ in the form $\sum_{n=0}^\infty A_n z^n$, that
$$A_n = \sum_{j,k} (-1)^k \binom{j}{k}\binom{n-rk-1}{j-1}$$
I have no idea where this comes from. Could anyone point me in the right direction? Thanks!
Edit: Also, the text states without explanation that in the case $r=2$, the coefficients are the Fibonnaci numbers:
$$\frac{1}{1-z-z^2} = 1 + z + 2z^2 + 3z^3 + 5z^4 + 8 z^5 + 13 z^6 + \cdots$$
Is this a non-trivial result, or am I just not seeing something?
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$$f(z) = \sum_{j=0}^\infty \left( z \frac{1 - z^r}{1-z} \right)^j = \sum_{j=0}^{\infty}z^j \left( \sum_{k=0}^j {j\choose k} (-z^r)^k \right) \left( \sum_{m=0}^{\infty} {m+j-1 \choose j-1}z^m \right)$$
Change the order of summation. Note that in order to get a term of $z^n$, we need $j + rk + m = n$, or that $m= n - rk - j$.
The coefficient of $z^n$ is thus
$$ \sum_{j,k} (-1)^k {j\choose k} {n-rk-1 \choose j-1}$$
Let $g(x)$ be the generating function of the Fibonacci numbers. Then,
$\begin{array}{l l l l l l}
g(x) &= 1 &+ x & + 2 x^2 & + 3x^3 & + \ldots \\
xg(x) & = & + x & + x^2 & + 2x^2 & + \ldots \\
x^2 g(x) & = & & + x^2 & + x^3 & + \ldots \\
\hline
(1-x-x^2)g(x) & = 1 \\
\end{array}$
Hence, $g(x) = \frac{1}{1-x-x^2}$.
Note: This is with $F_1=1, F_2=2$. Depending on how you want to start your Fibonacci sequence, some people use $g(x) = \frac{x}{1-x-x^2}$ as the generating function instead.
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Solving summation $2n+2^2(n-1)+2^3(n-2)+....+2^n$ Can anyone help me with this summation? I tried to use the geometric series on this, but I can't use that.
$$2n+2^2(n-1)+2^3(n-2)+....+2^n$$
I am trying to do this for studying algorithms. Can we get a closed form for this ?
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Notice that
$$S_n=2n+2^2(n-1)+2^3(n-2)+\cdots+2^{n-1}(2)+2^n(1)$$
$$S_{n-1}=2(n-1)+2^2(n-2)+2^3(n-3)+\cdots+2^{n-1}(1)+2^n(0)$$
differ by
$$S_n-S_{n-1}=2+2^2+2^3+\cdots+2^n=2(2^n-1)$$
Therefore, $$S_n=2(2^n-1) + S_{n-1} =2(2^n-1) + 2(2^{n-1}-1) + S_{n-2}$$and so on until
$$S_n=2(2^n-1 +2^{n-1}-1+2^{n-2}-1+\cdots+2^{1}-1)=2([2^1+2^2+\cdots+2^n]-n)$$
You should be able to figure out the answer.
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Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$ I was recently asked this question which stumped me.
How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?
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$$\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1}- n =\left(\dfrac{\sqrt{x_1}}{\sqrt{x_n}}-\dfrac{\sqrt{x_n}}{\sqrt{x_1}} \right)^2+\left(\dfrac{\sqrt{x_2}}{\sqrt{x_{n-1}}}-\dfrac{\sqrt{x_{n-1}}}{\sqrt{x_2}} \right)^2+...$$
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$n^5-n$ is divisible by $10$? I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$. However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!
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$$n^5-n=n(n^2+1)(n+1)(n-1)= n(n^2-4)(n+1)(n-1)+5n(n-1)(n+1)=(n-2)(n-1)n(n+1)(n+2)+5n(n-1)(n+1)$$
$(n-2)(n-1)n(n+1)(n+2)$ is even and divisible by 5, since it is the product of 5 consecutive integers.
$5(n-1)n(n+1)$ is also even and divisible by $5$.
Note: Both expressions are also divisible by $3$, so $n^5-n$ is actually divisible by $30$!
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Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$
$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$
$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$
$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$
$$=\dfrac{6x^3-2x^2}{9x^3-x}$$
This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$
Edit, missed a difference of squares:
$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$
Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$
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HINT: Pull out everything that you can: $6x^3-2x^2=2x^2(3x-1)$, and $9x^3-x=x(9x^2-1)$. Then notice that $9x^2=(3x)^2$, so that $9x^2-1=(3x)^2-1^2=(3x-1)(3x+1)$. Finally, do the cancellations that are now apparent.
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Why does $(k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor = 1,\;k\ge2\;\implies\;\text{isPrime}(k)$ Let $k$ be a integer such that $k\ge2$
Why does
$$(k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor = 1$$
only when $k$ is prime?
Example:
$$\pi(n) = \sum _{k=4}^n \left((k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor \right),\;n\ge4$$
where $k=4,$ since:
$$\pi(4)\quad=\quad(4-2)!-4 \left\lfloor \frac{4!}{(4-1) 4^2}\right\rfloor = 2$$
I've tried to evaluate it in different forms, and I am probably just overlooking something obvious; So if anyone has any information in regard to this, please share.
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If $k$ is not a prime, then the equation is $0$ modulo $k$. (We have $(k-2)! \equiv 0 \pmod{k}$.)
Conversely if $k$ is a prime, then by Wilson's theorem $(k-2)! \equiv 1 \pmod{k}$.
Edit: (To complete the argument.) If $k$ is a prime, then $(k-2)! = 1 + l k$ for some $l \ge 0$. Equivalently $l = \frac{(k-2)! - 1}{k}$ and we must show that $l = \left\lfloor \frac{k!}{(k-1)k^2} \right\rfloor$. But this is true since
$$\left\lfloor \frac{k!}{(k-1)k^2} \right\rfloor = \left\lfloor \frac{(k-2)!}{k} \right\rfloor$$
and the greatest number below $(k-2)!$ that is divisible by $k$ is $(k-2)! - 1$.
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How to find the Laurent expansion for $1/\cos(z)$ How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$
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Note $\frac{1}{\cos z}=-\frac{1}{\sin (z-\frac{\pi}{2})}$.
Let $t:=z-\frac{\pi}{2}$. Then $0<|t|<1$, $\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots$, so
$$
\begin{align}
\frac{1}{\sin t}&=\frac{1}{ t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots } \\
&=\frac{1}{t}\frac{1}{1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)}\\
&=\frac{1}{t}\left[1+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)^2+\dots\right]\\
&=t^{-1}+\frac{1}{3!}t+\left[\left(\frac{1}{3!}\right)^2-\frac{1}{5!}\right]t^3+\dots
\end{align}
$$
It seems no closed forms for higher terms.
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$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up getting
$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$
Dont know what to do...
EDIT: Thanks to Zev, what I did was:
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$
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You're on the right track so far; you've used the definition of $R_n$ to express the right side of the equation. Now just do this for the left side as well. You want to show that for any $n\geq 0$,
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
This can be done directly:
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
12(2^{n-2})-100(5^{n-2})&=42(2^{n-2})-140(5^{n-2})-30(2^{n-2})+40(5^{n-2})
\end{align*}$$
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|
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
|
\begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1\times0) + 1\\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\\
= &12
\end{align*}
|
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|
Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$ Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test?
If I remove the first term I can see $\dfrac{u_n}{u_{n+1}}=\dfrac{(2n+2)(2n+3)}{(2n+1)^2}
\\=\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^2}
\\={\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^{-2}}\\=\left(1+\dfrac{5}{2n}+\dfrac{3}{2n^2}\right)\left(1-\dfrac{1}{n}\ldots\right)\\=1+\dfrac{3}{2n}+O\left(\dfrac{1}{n^2}\right)$
So the series is convergent.
Is it a correcct attempt?
|
I'm not sure about your result; the ratio test is inconclusive anyway. The way I look at this, I express each term as
$$b_k = \frac{a_k}{2 k+1}$$
where
$$a_k = \frac{1}{2^{2 k}} \binom{2 k}{k}$$
Using, e.g., Stirling's formula, you may show that
$$a_k \sim \frac{1}{\sqrt{\pi k}} \quad (k \to \infty)$$
so that the ratio test on the coefficients provides
$$\frac{b_{k+1}}{b_k} \sim \left(1+\frac{1}{k}\right)^{-1/2} \left (1+\frac{2}{2 k+1}\right) \sim 1+\frac{1}{2 k}\quad (k \to \infty)$$
The limit test is inconclusive, as the limit if the ratio is $1$. However, we may see that the series converges by the comparison test because the terms in the sum behave as
$$b_k \sim \frac{1}{2 \sqrt{\pi}} k^{-3/2} \quad (k \to \infty)$$
That said, we know that the series converges to $\pi/2$ as follows: consider the generating function
$$f(x) = \sum_{k=0}^{\infty} a_k \frac{x^{2 k+1}}{2 k+1}$$
Then for all $x$ within the radius of convergence of the series on the right, we may differentiate to get
$$f'(x) = \sum_{k=0}^{\infty} a_k \, x^{2 k} = \frac{1}{\sqrt{1-x^2}}$$
Therefore
$$f(x) = \arcsin{x}$$
and the series has value $f(1) = \pi/2$.
|
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|
Did I derive this correctly? I derived this $$(2x+1)^2 \sqrt{4x+1}$$
and got $(8x+4)(\sqrt{4x+1})$+$\frac{2}{\sqrt{4x+1}}(2x+1)^2$
Is this correct?
I ask because Wofram Alpha gave me a different answer.
Thanks in advance.
|
Your answer:
$$\begin{aligned}
&(8x+4)(\sqrt{4x+1})+\frac{2}{\sqrt{4x+1}}(2x+1)^2
\\
=& \frac{(8x+4)(4x+1)}{\sqrt{4x+1}}+\frac{2(2x+1)^2}{\sqrt{4x+1}}
\\
=& \frac{32x^2 + 24x + 4}{\sqrt{4x+1}} + \frac{8x^2 + 8x + 2}{\sqrt{4x+1}}
\\
=& \frac{40x^2 + 32x + 6}{\sqrt{4x+1}}
\\
=& \frac{2(10x + 3)(2x+1)}{\sqrt{4x+1}}.
\end{aligned}$$
WolframAlpha's answer.
|
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|
Is This A Derivative? I am in a little over my head. This all began with my reading how each level of pascals triangle adds to $2^n$, where n=row# starting with n=0. I then though, "wouldn't it be clever if the rows added to something else--like say $3^n$ instead?" Or even better generalize it for any constant, $a^n$.
All that was needed was to Multiply any given number in pascal's triangle by $a^n/2^n$
$$
\begin{array}{rcccccccccc}
& & & & & & 1\\\
& & & & & \frac{a}{2} & & \frac{a}{2}\\\
& & & & \frac{a^2}{4} & & \frac{a^2}{2} & & \frac{a^2}{4}\\\
& & & \frac{a^3}{8} & & \frac{3*a^3}{8} & & \frac{3*a^3}{8} & & \frac{a^3}{8}\\\
& & \frac{a^4}{16} & & \frac{a^4}{4} & & \frac{3*a^4}{8} & & \frac{a^4}{4} & & \frac{a^4}{16}\\\
& \frac{a^5}{32} & & \frac{5*a^5}{32} & & \frac{5*a^5}{16} & & \frac{5*a^5}{16}& & \frac{5*a^5}{32} & & \frac{a^5}{32}\\\
& \frac{a^6}{64}& &\frac{3*a^6}{32}& &\frac{15*a^6}{64}& &\frac{5*a^6}{16}& &\frac{15*a^6}{64}& &\frac{3*a^6}{32}& &\frac{a^6}{64}\\\
& & ... & & & &... & & & & ... &
\end{array}
$$
Adding any row should give $a^n$. Placing said coefficients in front of a binomial expansion and solving for the binomial expression yields $$(a^n/2^n)*(x+y)^n$$
Letting a=2 makes pascals triangle, but every other value of "a" distorts every value and relation (except n=0 row for obvious reasons).
This triangle creates some interesting relations that are shared with Pascal's triangle and are immediately obvious: every term in the middle column may be divided by "a" to yield the term above and to the right or left of it--just like Pascals triangle (a=2).
Next is a really fascinating fluke:
$$ \frac{\partial \frac {a^4}{4}}{\partial a}=a^3
$$, which is the sum of the line above it.
$$
\frac{\partial \frac {a^2}{4}}{\partial a}=\frac {a}{2}
$$, which is found in the line above it.
I realize that "a" must have a definite value as a coefficient in order to have meaning, and it is not itself a function, but it seems curious that derivative relationship would show up in the relations between the coefficients of this modified triangle. This serendipitous relation fascinates me to the point of asking "whats up?" here. Is this a derivative? Have derivative relationships popped up organically elsewhere in function theory?
|
You may rewrite your $$\frac{a^n}{2^n}(1+1)^n$$ as $$\left(\frac{a}{2} + \frac{a}{2}\right)^n$$
Unfortunately neither of the two examples you mention appear to be specific examples of general patterns.
|
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|
How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
|
We use the identity: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
let $x=a-b ,y=b-c, z=c-a$ we see $x+y+z=0$ so $(a-b)^3+(b-c)^3+(c-a)^3-3(a-b)(b-c)(c-a)=0 $
Done!
|
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|
Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac{(x_1+2)^2-1}{4}=\frac{x_1^2+4x_1+3}{4}$
If we substitute $x_1^2=4k_1 + 1$, we end up with:
$k_2=k_1 + \sqrt{4k_1+1} + 1$
Therefore, finding solutions of $k=2(2^{n-3} - 3)$ is comparable to
finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$
Let $p=2(2^{n-3}-3)$
Therefore, $(\sqrt{4k+1})^2=(p-k-1)^2$, so
$(p-k)^2 = 2(p+k)$
Since $p$ has an infinite number of solutions $(p-k)^2=2(p+k)$ also has an infinite number of solutions, which implies the the original does also.
|
Here is another heuristic argument along probabilistic lines that there are finitely many solutions.
The distance between perfect squares near $n$ is approximately $2\sqrt{n}$. Thus, the probability of a given integer $n$ to be a perfect square is approximately $\frac1{2\sqrt{n}}$. Summing the probability that $2^n-23$ is a perfect square gives
$$
\begin{align}
\sum_{n=5}^\infty\frac1{2\sqrt{2^n-23}}
&\le\frac16+\sum_{n=6}^\infty\frac1{2\sqrt{2^{n-1}}}\\
&=\frac16+\frac{\sqrt2+1}8
\end{align}
$$
According to the Borel-Cantelli Lemma, this indicates that the probability that there are finitely mny solutions is $1$.
Of course, this really proves nothing, because the same argument suggests that there are finitely many perfect squares of the form $2^n$, when in fact $2^n$ is a perfect square whenever $n$ is even. However, in the absence of proof to the contrary, this provides a decent guess.
|
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|
How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$
Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$.
How many nice methods do you know for solving this problem? Thank you everyone.
My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so $$x^{2000}+\dfrac{1}{x^{2000}}=2\cos{\dfrac{2000\pi}{5}}=2.$$
Can you think of other nice methods? Or this problem has not used Euler's theorem: $(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}$
|
$$x+x^{-1}=\frac{\sqrt5+1}2$$
$$\implies 2x^2-x+2=\sqrt5x$$
$$\text{On squaring,} (2x^2-x+2)^2=5x^2$$
$$\implies 1-x+x^2-x^3+x^4=0$$
$$\text{ or, }x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2\cdot x\cdot\frac1x=\left(\frac{\sqrt5+1}2\right)^2-2=\frac{\sqrt5-1}2$$
$$\implies x+\frac1x-\left(x^2+\frac1{x^2}\right)=1\implies 1-x+x^2-x^3+x^4=0$$
which is a Geometric Series with common ratio $=-x$ and the first term being $=1$
$$\implies 1-x+x^2-x^3+x^4=\frac{1+x^5}{1+x}\implies 1+x^5=0$$
$$\text{ or } 1+x^5=(1+x)(1-x+x^2-x^3+x^4)=0$$
$\implies x^5=-1\implies x^{10n}=(x^5)^{2n}=(-1)^{2n}=1$
Put $n=200,-200$
|
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|
Factor $x^5-1$ into irreducibles in $\mathbb{F}_p[x]$ I have to factor the polynomial $f(x)=x^5-1$ in $\mathbb{F}_p[x]$, where $p \neq 5$ is a generic prime number.
I showhed that, if $5 \mid p-1$, then $f(x)$ splits into linear irreducible.
Now I believe that, if $5 \nmid p-1$ but $5 \mid p+1$, then $f(x)$ splits into three irreducible polynomial, one of degree $1$ and two of degree $2$.
Oterwise, if $5 \nmid p-1$ and $5 \nmid p+1$ (so $p \mid p^2+1$), then $f(x)$ splits into two irreducible factors, one of degree $1$ and one of degree $4$.
How can I prove this statements (if I'm right, well...)? Thank you!
|
Hints (for you to understand, complete/prove)
Over any field
$$x^5-1=(x-1)(x^4+x^3+x^2+ x +1)$$
Now, the roots of the second factor above are the roots of unity of order $\,5\,$ different from 1 itself. Since these roots (including $\;1\;$) form a cyclic group of order$\; 5\;$, we want to now when $\,5\mid p-1\,$ . In these cases we see the above pol. will split in different linear factors, for example:
$$p=11\;\implies x^5-1=(x-1)(x-4)(x-5)(x-9)(x-3)\in\Bbb F_{11}[x]$$
Also note that $\,a\neq 1\,$, otherwise the above equations have no solution (why?), so the other option for $\,b\,$ still's open. Try now to take it from here.
Note that $\,4=4^1\;,\;5=4^2\;,\;9=4^3\;,\;3=4^4\,$
It though can be the above quartic splits as the product of two irreducible quadratics:
$$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=$$
$$=x^4+(a+c)x^3+(d+ac+b)x^2+(ad+bc)x+bd$$
Comparing corresponding coefficients we get the equations
$$\begin{align*}a+c&=1\\ac+b+d&=1\\ad+bc&=1\\bd&=1\end{align*}$$
Thus
$$(1)+(3)\;\;a=1-c\;,\;d=\frac1b\implies (3)\;\;\frac ab+b(1-a)=1\implies$$
$$(1-a)b^2-b+a=0\;:\;\;\Delta=1-4a(1-a)=4a^2-4a+1=(2a-1)^2\implies$$
$$b_{1,2}=\frac{1\pm (2a-1)}{2(1-a)}=\begin{cases}\frac a{1-a}\\{}\\\;\;\;\;1\end{cases}$$
Now
$$b=1\implies d=1\;,\;\;ac=-1\;,\;c=1-a\implies a(1-a)=-1\implies$$
$$ a^2-a-1=0\implies a=\frac{1\pm\sqrt 5}2$$
Note that this choice requires $\,\sqrt 5\in\Bbb F_p\iff \binom5p=1\,$ ,but by quadratic reciprocity
$$1=\binom5p=\binom p5\iff p=1,4\pmod 5$$
|
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|
How to determine the rank and determinant of $A$? let $A$ be $$A_{a} = \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$$
How can I calculate the rank of $A$ by the Gauss' methode and $\det A$?
|
Notice that $A_a= J+aI_4$ with $$J= \left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \right)$$ Let $P_J(X)$ be the characteristic polynomial of $J$, then $\det(A_a)=P_J(1-a)$. By noticing:
$$\left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \right) \left( \begin{matrix} 1 \\ -1 \\ 1 \\ -1 \end{matrix} \right)= \left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \\ -1 \\ -1 \end{matrix} \right)=\left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \right) \left( \begin{matrix} 1 \\-1 \\ -1 \\ 1 \end{matrix} \right)=0$$
and
$$\left( \begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right)= 4 \left( \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right)$$
we can deduce that $P_J(X)=X^3(X-4)$, hence $\det(A_a)=-(1-a)^3(a+3)$. Because $A_a$ is trigonalizable over $\mathbb{C}$, we deduce that:
*
*If $a \neq 1$ and $a \neq -3$, $\mathrm{rank}(A_a)=4$,
*If $a=1$, $\mathrm{rank}(A_a)=1$,
*If $a=-3$, $\mathrm{rank}(A_a)=3$.
|
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|
proving $\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+$ Without Induction i proved that:
$$
\begin{align}
& {} \quad \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)\cdot 2n} \\[10pt]
& =\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\text{ for }n\in \mathbb{N}
\end{align}
$$
by induction.
i wonder if it can be done without using induction. if so, i'll appreciate if someone could show how.
thanks.
|
Let it be true for n
Then for $n+1$ we have,
$\displaystyle \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =(\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n-1)\cdot 2n})+\frac{1}{(2n+1)\cdot 2(n+1)}$(Using induction hypothesis for $n$)
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{1}{2n+1}-\frac{1}{2(n+1)}+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{4n+3}{(2n+1)2(n+1)}+\frac{1}{(2n+1)2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{4n+2}{(2n+1)2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{1}{(n+1)}$
$\displaystyle =\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}$
|
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|
Determine whether a multi-variable limit exists $\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$ I need to determine whether the next limit exists:
$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$$
Looking at the numerator $(-1-\frac{x^2}2)$ it immediately reminds me of maclaurin series of $\cos$, Where:
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...=1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)$$
So we can simplify the expression:
$\begin{align} \lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}&=\lim_{(x,y)\to(0,0)}\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)-1-\frac{x^2}2}{x^4+y^4}\\
&=\lim_{(x,y)\to(0,0)}\frac{-x^2+\frac{x^4}{4!}+O(x^6)}{x^4+y^4} \end{align}$
This limit does not exist.
So, after solving it the 'hard' way, I was thinking:
If I place $y=mx$ and then try to solve the limit I get:
$$\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4(1+m^4)}=\infty$$
Is that enough to prove that the limit does not exists? I mean, If I find at least one limit that does not 'converge' is it enough to say that a multi-variable function limit doesn't exist?
|
HINT
Find the limit as $x \to 0$ when $y=x$ and show that it is not equal the limit as $x\to 0$ when $y=0$.
|
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|
$4\times4$ matrices minimal polynomial Anyone can tell me how to find the minimal polynomial of $4\times4$ matrices belong to $M_4(\mathbb{R})$, because I know how to find $2\times2$ and $3\times3$ matrices, but never try to find $4\times4$. I think this should be linear algebra problem. Can anyone show me an example? Or can use the example from my notebook.
$$A= \begin{pmatrix}
\phantom{-}3&-4&\phantom{-}0&\phantom{-}2\\
-4&-5&-2&\phantom{-}4\\
\phantom{-}0&\phantom{-}0&\phantom{-}3&-2\\
\phantom{-}0&\phantom{-}0&\phantom{-}2&-1
\end{pmatrix}$$
Thanks so much!
|
$A=\begin{pmatrix}
3&-4&0&2\\
-4&-5&-2&4\\
0&0&3&-2\\
0&0&2&-1\\
\end{pmatrix}$
Minimal polynomial of A will be the minimal polynomial of
$X=\begin{pmatrix}
3&-4\\
-4&-5\\
\end{pmatrix}$ multiplied by the minimal polynomial of
$Y=\begin{pmatrix}
3&-2\\
2&-1\\
\end{pmatrix}$
Reason:We have $A^k=\begin{pmatrix}
X^k&C_k\\
0&Y^k\\
\end{pmatrix}$
|
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Solve $ax - a^2 = bx - b^2$ for $x$ Method 1
Solve for x
$$ax - a^2 = bx - b^2$$
Collect all terms with x on one side of the equation
$$ax - bx = a^2 -b^2$$
Factor both sides of the equation
$$(a -b)x = (a+b)(a - b)$$
Divide both sides of the equation by the coefficient of $x$ (which is $a-b$)
$$x = a + b$$ (where $a \neq b$ since this would mean dividing by $0$)
Method 2
Solve for $x$
$$ax - a^2= bx - b^2$$
Bring all the terms to one side of the equation
$$ax - a^2 -bx + b^2 = 0$$
Rearrange
$$ax - bx -(a^2-b^2)=0$$
Factor
$$(a - b)x - (a + b)(a - b) = 0$$
$$(a - b)( x - (a + b)) = 0$$
which is a true statement if $$a-b=0$$ $$a = b$$ or $$x-(a+b)=0$$ $$x = a + b$$
My question is I don't understand how this second method is consistent with the first in terms of the restriction on $a$ and $b$.
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If $a=b$, the left side of the equation $(a-b)(x - (a+b)) = 0$ is always equal to the right side, so $x$ can be anything.
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Compute this limit $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ using L'Hôpital's rule I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$ and I don't understand in this step how/why the $A-B$ and $A+B$ was replaced by $\frac{x^2}{2}$ and $\frac{x^2}{2}+\frac{1}{x}$ :
$$\lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x}= \lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}.$$
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Hint:its easy to prove $$\sin(x+y)+\sin(x-y)=2\sin(x)\cos(y)$$
then put $y=\frac{A+B}{2}$,$x=\frac{A-B}{2}$
$$ \sin(x)\sim x$$ $$\ cosx\sim1-\frac{x^2}{2}$$ because $$\lim_{x\to0}\frac{sinx}{x}=\lim_{x\to0}\frac{cosx}{1-\frac{x^2}{2}}=1$$
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
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A Lagrange multiplier proof:
Suppose $a^2b^2c^2$ is maximized. Then since $a + b + c = 0$, two of $a,b$, and $c$ are of one sign, and the third is of the other sign. (Clearly none are zero). Replacing $a,b,$ and $c$ by their negatives if needed, one can assume two are positive, and permuting the variables if necessary one can assume $a,b > 0$ and $c < 0$.
Then since $c = -(a + b)$, the equation $a^2 + b^2 + c^2 = 1$ becomes
$a^2 + b^2 + (a + b)^2 = 1$ or
$$a^2 + ab + b^2 = {1 \over 2}$$
And one wants to maximize $a^2b^2c^2$ subject to this contraint for $a, b > 0$. Since $c = -(a + b) < 0$ this is equivalent to maximizing $ab(a + b)$ subject to the constraint, for $a, b > 0$. The Lagrange conditions are
$$2ab + b^2 = \lambda(2a + b)$$
$$a^2 + 2ab = \lambda(a + 2b)$$
Since $2ab + b^2 = b(2a + b)$, the first equation says $\lambda = b$. Similarly, the second equation implies $\lambda = a$. So $a = b$, and then $a^2 + ab + b^2 = {1 \over 2}$ means $3a^2 = {1 \over 2}$ or $a = {1 \over \sqrt{6}}$. So $a = b = {1 \over \sqrt{6}}$. Then $c = -(a + b) = -{2 \over \sqrt{6}}$, and $a^2b^2c^2 = {4 \over 6^3} = {1 \over 54}$.
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How to find solutions to this equality $\; \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$ We have the following equality:
$$ \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$
Some of the solutions I found:
*
*$\mathrm{x} = 0$
*Also for $\mathrm{a}=0$, every $\mathrm{x}$ is a solution I believe
I tried getting everything out of the brackets but that just gave a nasty equality which I couldn't solve.
Let's take $-1 \leq \mathrm{a} \leq 1$. I was also wondering if there is a way of knowing how many solutions this equality has beforehand? Or do we just have to look at the cases $\mathrm{a} = 1$, $\mathrm{a} = -1$, $\mathrm{a} \neq 0$ and $\mathrm{a} = 0$ individually to find every solution (i.e. both of the bounds, and for a (not) equal to 0)?
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I've solve upto some extent
$$ x = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$
$$ {a^2x(1-x)(1-ax(1-x))-x=0}\implies x=0$$
$$ {a^2(1-x)(1-ax(1-x))-1=0}$$
$$ {a^2(x-1)(1+ax(x-1))+1=0}$$
put x-1 =t
$$ {a^2t(1+at(t+1))+1=0}$$
$$ {a^2t(1+at^2+at)+1=0}$$
$$a^3t^3+a^3t^2+a^2t+1=0$$
for this equation if we take a=1
$$t^3+t^2+t+1=0\implies t=-1,i,-i\implies x=0,1+i,1-i$$
so based on value of a there are different solution of t and x.
solution in continuity based on @maming's comment
$$a^3t^3+a^3t^2+a^2t+1=0$$
$$a^3t^3+1+a^3t^2+a^2t=0$$
$${(at)}^3+1^3+a^3t^2+a^2t=0$$
$$(at+1)(a^2t^2-at+1)+a^2t(at+1)=0$$
$$(at+1)(a^2t^2-at+1+a^2t)\implies (at+1)=0\;,(a^2t^2-at+1+a^2t)=0$$
$$t=\frac{-1}{a}\implies \mathbf{x=1-\frac 1a}\;\;,a^2t^2+t(a^2-a)+1=0$$
$$t=\dfrac {-(a^2-a)\pm\sqrt{{(a^2-a)}^2-4\cdot a^2\cdot 1}}{2a^2}$$
$$t=\dfrac {a\left((1-a)\pm\sqrt{{(a-1)}^2-4}\right)}{2a^2}$$
$$t=\dfrac {(1-a)\pm\sqrt{{(a-1)}^2-4}}{2a}\implies x=1+\dfrac {(1-a)\pm\sqrt{{(a-1)}^2-4}}{2a}$$
$$x=\dfrac {(a+1)\pm\sqrt{{(a-1)}^2-4}}{2a}$$
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Prove if $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} $ is a real number If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $
i found one link that had a similar problem.
Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$
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HINT:
Let $z=\cos A+i\sin A, w=\cos B+i\sin B$
$\implies z\cdot w=\cos(A+B)+i\sin(A+B)$
Using $\cos C-\cos D=-2\sin\frac{C-D}2\sin\frac{C+D}2$
and $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2$
$$\frac{z-w}{1-zw}=\frac{\cos A-\cos B+i(\sin A-\sin B)}{1-\{\cos(A+B)+i\sin(A+B)\}}$$
$$=\frac{-2\sin\frac{A-B}2\sin\frac{A+B}2+i2\sin\frac{A-B}2\cos\frac{A+B}2}{2\sin^2\frac{A+B}2-i2\sin\frac{A+B}2\cos\frac{A+B}2}$$
$$=\frac{2i\sin\frac{A-B}2\left(i\sin\frac{A+B}2+\cos\frac{A+B}2\right)}{-2i\sin\frac{A+B}2\left(i\sin\frac{A+B}2+\cos\frac{A+B}2\right)}$$
$$=\frac{\sin\frac{A-B}2}{-\sin\frac{A+B}2}$$
EDIT: The changed question can be addressed in the same way using
$\cos C+\cos D=2\cos\frac{C-D}2\cos\frac{C+D}2$
and $\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$
$$\frac{z+w}{1+zw}=\frac{\cos A+\cos B+i(\sin A+\sin B)}{1+\{\cos(A+B)+i\sin(A+B)\}}$$
$$=\frac{2\cos\frac{A+B}2\cos\frac{A-B}2+i2\sin\frac{A+B}2\cos\frac{A-B}2}{2\cos^2\frac{A+B}2+i2\sin\frac{A+B}2\cos\frac{A+B}2}$$
$$=\frac{2\cos\frac{A-B}2\left(\cos\frac{A+B}2+i\sin\frac{A+B}2\right)}{2\cos\frac{A+B}2\left(\cos\frac{A+B}2+i\sin\frac{A+B}2\right)}$$
$$=\frac{\cos\frac{A-B}2}{\cos\frac{A+B}2}$$
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Question about modular arithmetic and divisibility If $$a^3+b^3+c^3=0\pmod 7$$
Calculate the residue after dividing $abc$ with $7$
My first ideas here was trying the 7 possible remainders and then elevate to the third power
$$a+b+c=x \pmod 7$$
$$a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc=x^3\pmod 7$$
$$3(a+b+c)(ab+bc+ac)-3abc=x^3 \pmod 7$$
If I replace $x=0$ the result is immediate, $abc=0 \pmod7$. But with $x=1$
$$3(7n+1)(ab+bc+ac)-3abc=x^3 \pmod 7$$
$$3(ab+bc+ac)-3abc=x^3 \pmod 7$$
And there is nothing more to simplify. I know the LHS is a multiple of $3$, but what can i do with that? Is it necessary that $x^3$ or $7-x^3$ is a multiple of $3$? Any help is greatly appreciated
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If one of $a,b,c$ is divisible by $7,abc\equiv0\pmod 7$
Else
$n^3\equiv \begin{cases} 1 &\mbox{if } n \equiv 1,2,4\pmod 7 \\
-1 & \mbox{if } n \equiv 3,5,6\pmod 7 \end{cases} \pmod 7$
Observe that for no combination of $a,b,c$ $$a^3+b^3+c^3\equiv0\pmod 7$$
$$\implies a^3+b^3+c^3\equiv0\pmod 7\implies 7\text{ must divide at least one(or all three) of } a,b,c $$
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Given a matrix $A$ and what it maps two vectors to, is $0$ an eigenvalue of it? Studying for my Algebra exam, and this question popped out with no solution in a previous exam:
Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix}$.
(I) Is $0$ an eigenvalue of the matrix?
(II) Find a matrix like that, where the sum of its' eigenvalues is $0$.
So I (think) solved (I) but have no clue for (II).
Here's my solution for (I):
$A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = A \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix} = 0$, and then, the vector $v = \begin{pmatrix} 2 \\ -1 \\0 \end{pmatrix}$ supplies that $Av = 0v = 0$ meaning that $0$ is an eigenvalue of $A$ with an eigenvector $v$.
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The answer for (I) looks good.
Good for (II) might be to note that
$$
A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}2\\4\\-6\end{bmatrix}
$$
and
$$
A\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}
$$
which gives the first two columns of $A$:
$$
A=\begin{bmatrix}1&2&x\\2&4&y\\-3&-6&z\end{bmatrix}
$$
The trace is the sum of the eigenvalues. Thus, if the sum of the eigenvalues is $0$, we need $z=-5$. $x$ and $y$ are arbitrary.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.