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Solve the composition $f \circ g=?$ and $g\circ f=?$ a) $f(x) = \sqrt[3]{x}\ $ and $g(x) = x^3$
find $f\circ g=?$ and $g\circ f=?$
I have $f\circ g = f(g(x)) = f(x^3) = \sqrt[3]{x^3} = x$
So basically, first we replace $g(x)$ with its value, then we replace $x$ with the value of $f(x)$ inside $g(x)$?
$g\circ f = g(f(x)) = g( \sqrt[3]{x} ) = \sqrt[3]{x^3} $ OR $ \sqrt[3]{x}^3 $ ?
b) $f(x) = \sin x$ and $g(x) = x^2$
$$(f\circ g) = f(g(x)) = f(x^2) = \sin (x^2)$$
$$(g\circ f) = g(f(x)) = g(\sin x) = (\sin x)^2 = \sin^2x$$
Am I doing it right? I'm kinda confused whether I have to replace x with the original value or not.
|
so there is two different case and different result,for example let us suppose that
$f(x)=x^3$
and $g(x)=x^2$
then multiplication is directly $x^3*x^2=x^5$
for composition like $f\circ g=f(g(x))=(x^2)^3$
$g\circ f=g(f(x))=(x^3)^2$
|
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|
Finding the Sum of the square of two positive integers.
Write the following equation as the sum of the square of two integers, $a^2 +b^2$.
$$(8^2+5^2)(13^2+7^2)$$
I remember that you are supposed to do something with complex numbers or at least that is what my teacher did in class.
|
From identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$, that is easy to check, we get:
$$(8^2+5^2)(13^2+7^2)=(8\cdot13+5\cdot7)^2+(8\cdot7-5\cdot13)^2=139^2+9^2.$$
|
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|
Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
|
Using the trick mentioned in a comment, observe that $0\leq x \leq 2$ for this to make sense, so substitute $x = 2\cos(t)$. Squaring and rearranging, $$4\cos^2(t) - 2 = \sqrt{2-\sqrt{2+2\cos(t)}}.$$ Apply the double angle formula and square again to get $$4\cos^2(2t) - 2 = -\sqrt{2+2\cos(t)}$$ whence $$4\cos(4t)^2-2 = 2\cos(t)$$ and hence $$\cos(8t) = \cos(t).$$
The solutions to this are $8t = \pm t + 2n\pi$, giving $\displaystyle t = \frac{2n\pi}{9}$ or $\displaystyle t = \frac{2n\pi}{7}$. Restricting $t$ to the range $[0,\pi)$ one finds that only $\displaystyle 2\cos(\frac{2\pi}{9})$ satisfies the original equation. To check this one need only make sure that the sign of the left hand side matches the appropriate sign on the right at each step. For instance from the original, $\cos(t)$ must be positive, excluding $\displaystyle t =\frac{8\pi}{9}$ among others, and from the first simplification, $\cos(2t)$ needs to be positive, which excludes, say, $\displaystyle t = \frac{2\pi}{7}$. There is exactly one solution since the right hand side of the original is decreasing between $x=0$ and $x=2$ with value $\sqrt{2}$ at $x=2$; this is enough to verify that $\displaystyle 2\cos(\frac{2\pi}{9})$ works after excluding the others.
|
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|
Mixed number fractions vs regular fractions? $3\frac{1}{6}-1\frac{11}{12}$ I just passed Calculus 2 in college with an A and I'm rather embarrassed that I'm asking this question. My wife is taking an intermediate Algebra course in college and they gave her the below problem.
$$3\frac{1}{6}-1\frac{11}{12}$$
Well I guess I gave her bad advice because I told her that this problem is equivalent to multiplying the whole numbers $3$ and $1$ by the fraction simplifying the problem to:
$\frac{3}{6}-\frac{11}{12}$ or $\frac{1}{2}-\frac{11}{12}$
After solving I got an answer of $-\frac{5}{12}$ which is not the correct answer according to her book.
Is true that the mixed fraction gives a different result than a normal fraction multiplied by a number?
|
$3 \dfrac16$ denoted $3+\dfrac16 = \dfrac{19}6$ and not $3 \cdot \dfrac16$.
Similarly, $1\dfrac{11}{12}$ denotes $1+\dfrac{11}{12} = \dfrac{23}{12}$ and not $1 \cdot \dfrac{11}{12}$.
Hence,
$$3 \dfrac16 - 1\dfrac{11}{12} = \dfrac{19}6 - \dfrac{23}{12} = \dfrac{38-23}{12} = \dfrac{15}{12} = \dfrac54$$
|
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|
positive integer ordered pairs $(x,y,z)$ in $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$ Total no. of positive integer ordered pairs of $(x,y,z)$ that satisfy the equation $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$
$\bf{My\ Try:}$ Using Simple Guess $x=2\;,y=3\,z=6.$ satisfy $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$
Now I have tried is there is any ordered pairs of $(x,y,z)$ that satisfy the above equation or not.
So Given $\displaystyle \frac{1}{z} = 1-\frac{1}{x}-\frac{1}{y} = 1-\left(\frac{x+y}{xy}\right)=\frac{xy-x-y}{xy}\Rightarrow z=\frac{xy}{xy-x-y} = 1+\frac{x+y}{xy-x-y}$
Now I did not understand How can i solve for Integer ordered pairs.
Help me
Thanks
|
The only solutions are $(2,3,6); (2,4,4); (3,3,3)$ (and of course its permutations). This can be argued as follows (in fact you are almost there):
As you have it, we need $\dfrac{x+y}{xy-x-y}$ to be an integer. Without loss of generality we can take $x \leq y$. We have $x+y \leq 2y$ and $xy-x-y \geq (x-2)y$. Hence, for $x>2$, we have
$$\dfrac{x+y}{xy-x-y} \leq \dfrac2{x-2}$$ which is not an integer for $x>4$. Hence, this bounds $x$ between $2$ and $4$. Hence, only three cases to check:
*
*$x=2$. We need $\dfrac{y+2}{y-2}$ to be an integer and $y \geq x$. This gives us $y=3,4,6$.
*$x=3$. We need $\dfrac{y+3}{2y-3}$ to be an integer and $y \geq x$. This gives us $y=3,6$.
*$x=4$. We need $\dfrac{y+4}{3y-4}$ to be an integer and $y \geq x$. This gives us $y=4$.
Now put this together to get the set of solutions as $(2,3,6); (2,4,4); (3,3,3)$ and its permutations.
|
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|
max and min $x y - \ln(x^2 + y^2)$ Find max and min
$$x y - \ln(x^2 + y^2) , 1/4 \le x^2 + y^2 \le 4$$
Problem: With this hairy expression as my partial derivatives, I do not know what to do.
Attempt:
|
$$y-\frac{x}{x^2+y^2}=0,x-\frac{y}{x^2+y^2}=0\Rightarrow x^2=y^2\Rightarrow x=\pm y$$
You have to minimize $$xy-\log (x^2+y^2)$$
with given conditions
$$1/4 \le x^2 + y^2 \le 4$$
If $1/4 \le x^2 + y^2 \le 4$ can you find what are the bounds for $x^2$??
If you know what are bounds for $x^2$,
take its minimum value and substitute that in: $xy-\log(x^2+y^2)$
Can you do that now?
|
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|
Prove that $|\sin n|+|\sin (n+1)| > 2\sin(1/2)$ for all $n\in \mathbb N$
Show that
$$|\sin{(n)}|+|\sin{(n+1)}|>2\sin{\dfrac{1}{2}},n \ge 1,n\in \mathbb N$$
My try: let
$$F(n)=|\sin{(n)}|+|\sin{(n+1)}|$$
then
$$F(n+\pi)=|\sin{(n+\pi)}|+|\sin{(n+\pi+1)}|=|\sin{(n)}|+|\sin{(n+1)}|=F(n)$$
and
$$F(\pi-n)=|\sin{(\pi-n)}|+|\sin{(\pi-n+1)}|=|\sin{n}|+|\sin{(n-1)}|\neq F(n)$$
so we must prove when $n\in (0,\pi)$,
have
$$F(n)>2\sin{\dfrac{1}{2}}$$
when
$n\in (0,\pi-1)$,then
$$F(n)=\sin{n}+\sin{(n+1)}=\sin{n}(1+\cos{1})+\sin{1}\cos{n}$$
and
$n\in(\pi-1,\pi)$,then
$$F(n)=\sin{n}-\sin{(n+1)}$$
How prove it this two case have $F(n)>2\sin{\dfrac{1}{2}}$? Thank you
and I know this well know inequality
$$|\sin{x}|+|\sin{(x+1)}|+|\sin{(x-1)}|\ge 2\sin{1},x\in \mathbb R$$
|
Using induction and the formula $$\sin(nx)=2^{n−1}\Pi_{k=0}^{n−1}\sin(x+\frac{\pi k}{n})$$
which is not very hard to prove by induction (again) for instance, $\sin(2x)=2\sin(x)\cos(x)=2\sin(x)\sin(x+\pi/2)$.
1). $n=1$ case is shown above $|\sin{(1)}|+|\sin{(1+1)}|=\sin(1)(1+2\sin(1+\frac{\pi}{2})>2\sin{\frac{1}{2}}$;
2). $n=k$ assumption case, $|\sin{(k)}|+|\sin{(k+1)}|=2^{k-1}|\sin(1)||\sin(1+\frac{\pi}{k})|\dots|\sin(1+\frac{\pi(k-1)}{k})|(1+2|\sin(1+\frac{k\pi}{k+1}))>2\sin{\frac{1}{2}}$
3). $n=k+1$ then
$$|\sin{(k+1)}|+|\sin{(k+2)}|=2^k|\sin(1)||\sin(1+\frac{\pi}{k})|\dots|\sin(1+\frac{\pi(k-1)}{k})||(1+2\sin(1+\frac{(k+1)\pi}{k+2}))|$$=$\left\{|\sin{(k)}|+|\sin{(k+1)}|\right\}\left\{2\frac{|\sin(1+\frac{k\pi}{k+1})|+2|\sin(1+\frac{(k+1)\pi}{k+2})|}{1+2|\sin(1+\frac{k\pi}{k+1})|}\right\}>
....>2\sin{\dfrac{1}{2}}$
Here $\dots$ means we need to prove $\left\{2\frac{|\sin(1+\frac{k\pi}{k+1})|+2|\sin(1+\frac{(k+1)\pi}{k+2})|}{1+2|\sin(1+\frac{k\pi}{k+1})|}\right\}>1$.
Note $$\left\{2\frac{|\sin(1+\frac{k\pi}{k+1})|+2|\sin(1+\frac{(k+1)\pi}{k+2})|}{1+2|\sin(1+\frac{k\pi}{k+1})|}\right\}=2\left\{\frac{1+2\frac{|\sin(1+\frac{(k+1)\pi}{k+2})|}{|\sin(1+\frac{k\pi}{k+1})|}}{\frac{1}{|\sin(1+\frac{k\pi}{k+1})|}+2}\right\}>2\left\{\frac{1+2}{\frac{1}{|\sin(1+\frac{k\pi}{k+1})|}+2}\right\}>1$$.
Here $\frac{|\sin(1+\frac{(k+1)\pi}{k+2})|}{|\sin(1+\frac{k\pi}{k+1})|}>1$ and $\min(\sin(1+\frac{k\pi}{k+1}))>\frac{1}{4}$ are used due to sin function's increasing characteristics from $[0,\frac{\pi}{2}]$ where $1$ is within it.
Suppose now, it is proved $|\sin{(n)}|+|\sin{(n+1)}|>2\sin{\dfrac{1}{2}},n \ge 1,n\in \mathbb N$.
|
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|
Trouble solving $\int\sqrt{1-x^2} \, dx$ I am trying to learn how to solve integrals and I've got the hang out of a lot of examples, but I haven't got the slightest idea how to solve this example, this is how far I've got:
$$
\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} - 2\int\frac{x^2}{\sqrt{1-x^2}} \, dx
$$
Can you please help me solve it, and also some tips concerning the integration are welcome.
Thank you
|
subsititute $ x =sin\theta $, this gives,
$ dx = cos\theta d\theta$
hence, $ \int \sqrt{(1-x^2)} dx = \int (\sqrt{1 - sin^2\theta} ). cos\theta d\theta = \int cos^2 \theta d\theta $
Remember that, $cos(\alpha + \beta) = cos \alpha .cos \beta - sin \alpha . sin \beta $,
substituting, $\alpha = \beta = \theta$,
*
*$ (cos^2\theta - sin^2\theta) = cos(2\theta)$, as well
*$ (cos^2\theta + sin^2\theta) = 1 $
Adding equation 1. and 2. yields,
$2 cos^2 \theta = 1 + cos2\theta ~ \implies ~ cos^2\theta = \frac{1}{2}(1+cos2\theta)$
hence going back to integral,
$\int cos^2\theta d\theta = \frac{1}{2}\int (1+cos2\theta) d\theta = \frac{1}{2}(\theta + \frac{sin2\theta}{2}) + C$
By reverse substitution, $\theta = cos^{-1}x = arccos(x) $ and $sin2\theta = 2.sin\theta . cos\theta$
The solution of indefinite integral is ,
$ \int \sqrt{(1-x^2)} dx = \frac{1}{2} ( \theta + \frac{2.sin\theta.cost\theta}{2} )+ C \implies \frac{1}{2}(\theta + sin\theta. cos\theta) + C $
$ \int \sqrt{(1-x^2)} dx \implies \frac{1}{2}(arccos(x) + x\sqrt{1-x^2}) + C $
|
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|
calculating mod 7 Problem: Calculate $10^{10^{10}} \pmod 7$
According to Fermat's little theorem: $a^{p-1}\equiv1 \pmod p$, so $10^6\equiv1 \pmod 7$ and $10^n\equiv4 \pmod 6$, $n$ being any integer, why can we write $10^{10^{10}}\equiv10^4 \pmod 7$?
Similarly, if it's $10^n\equiv1 \pmod 3$, n being any integer, then wouldn't the equation become $10^{10^{10}}\equiv10^1\equiv3\pmod 7$?
I'm confused on the translation phase.
Thanks
|
I believe your argument is correct. To compute $10^{10^{10}} \pmod{7}$ you can first compute $10^{10} \pmod{6}$. In turn, you can do this modulo $2$ and $3$, so you get that $10^{10} \equiv 4 \pmod{6}$, and $10^{10^{10}} \equiv 3^{4} \equiv 2 \cdot 2 \equiv 4 \pmod{7}$.
|
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|
How prove this equation $A^2+B^2=C^2+D^2$ define:
plane $W:Ax+By+Cz+D=0$ and the
hyperboloid of one sheet $U:x^2+y^2-z^2=1$
if $$W\bigcap U=l_{1},W\bigcap U=l_{2}$$
where $l_{1},l_{2}$ be two straight lines
show that :$$A^2+B^2=C^2+D^2$$
My try: since
$$\begin{cases}
Ax+By+Cz+D=0\\
x^2+y^2-z^2=1
\end{cases}$$
then
$$(Ax+By+D)^2=C^2(x^2+y^2-1)$$
then
$$(A^2-C^2)x^2+(B^2-C^2)y^2+2ABxy+2BDy+2ADx+D^2+C^2=0$$
Follow is user44197 idea
$$(A^2-C^2)x^2+(2ABy+2AD)x+(B^2-C^2)y^2+D^2+C^2=0$$
$$\Delta (y)=(2ABy+2AD)^2-4(A^2-C^2)[(B^2-C^2)y^2+D^2+C^2]$$
$$\Delta=4A^2B^2y^2+8A^2BDy+4A^2D^2-4(A^2B^2-A^2C^2-B^2C^2+C^4)y^2-4(A^2D^2+A^2C^2-C^2D^2-C^4)$$
so
$$\Delta(y)=4[C^2(A^2+B^2-C^2)y^2+2A^2BDy+C^2(C^2+D^2-A^2)]$$
then I can't.Thank you very much!
|
You need to make use of the fact that the intersection is a pair of straight lines.
Look at the last equation as a quadratic in $x$, i.e of the form $a x^2+b x +c=0$
Its discriminant is $b^2-4ac$ is given by
$$
\Delta(y) = 4\,{C}^{2}\,\left( {D}^{2}+2\,y\,B\,D-{y}^{2}\,{C}^{2}+{C}^{2}+{y}^{2}\,{B}^{2}+{y}^{2}\,{A}^{2}-{A}^{2}\right) $$
which is a quadratic in $y$. Now taking its discriminant (as a quadratic in $y$) gives
$$
\Delta =
64\,{C}^{4}\,\left( C-A\right) \,\left( C+A\right) \,\left( {D}^{2}+{C}^{2}-{B}^{2}-{A}^{2}\right) $$
This has to be zero to get the intersection to be straight lines.
Ruling $C=A$ and $C=-A$ and $C=0$ you get the desired answer.
|
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|
show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^3}{\sqrt[7]{(1-t^3)^6}} \, dt-\int_0^1 \sqrt[7]{1-x^3} \, dx$$
By parts,we have
$$\int_0^1 \sqrt[7]{1-x^3} dx=\dfrac{3}{7}\int_0^1 \frac{x^3}{\sqrt[7]{(1-x^3)^6}} \, dt$$
so
$$I=0$$
this problem maybe have more other nice methods!Thank you
|
Both parts of the integral express the area under the curve given by $x^7+y^3=1$.
|
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|
Prove that $\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots.$ Question. Let
$$
f(x)=\!\left\{\,\,\,
\begin{array}{ccc}
\displaystyle{\left\lfloor{1\over x}\right\rfloor}^{-1}_{\hphantom{|_|}}&\text{if} & 0\lt x\le 1, \\ & \\
0^{\hphantom{|^|}} &\text{if} & x=0.
\end{array}\right.
$$
Is f(x) Riemann integrable on $[0,1]$? If it is Riemann integrable, then what is the value of the integral $\,\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}dx$?
An attempt: Since $f$ is increasing, non-negative and bounded the integral does exist. Choosing the partition $P=\big\{0,\frac{1}{n},\frac{1}{n-1},...,1\big\}$, we have the following upper and lower sums
$$
U(f,P)=\sum_{i=1}^{n-1}{1\over n-i}\left [ {1\over n-i}-{1\over n-i+1}\right] - {1\over n^2},\\L(f,P)=\sum_{i=1}^{n-1}{1\over n-i+1}\left[{1\over n-i} - {1\over n-i+1} \right].
$$
Simplifying we obtain
$$
U(f,P)= \sum_{i=1}^n{1\over i^2}+ {1\over n} -1, \quad
L(f,P)= 2-{1\over n}-\sum_{i=1}^n{1\over n^2}.
$$
As $n\to\infty$, $U(f,P)\to{\pi^2\over 6}-1$ and $L(f,P)\to2-{\pi^2\over 6}$. Therefore $2-{\pi^2\over 6}\le\int_0^1f(x)\,dx\le{\pi^2\over 6}-1$. Direct calculation using MATLAB shows $\int_0^1f(x)\,dx={\pi^2\over 6}-1$.
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We observe that: if $x\in\big(\frac{1}{k+1},\frac{1}{k}\big]$, then $\frac{1}{x}\in[k,k+1)$, thus $\left\lfloor{1\over x}\right\rfloor=k$ and hence
$$
\left\lfloor{1\over x}\right\rfloor^{-1}=\frac{1}{k}, \quad \text{whenever}\,\, x\in\Big(\frac{1}{k+1},\frac{1}{k}\Big].
$$
Therefore
$$
\int_{1/n}^1{\left\lfloor{1\over x}\right\rfloor}^{-1}dx=\sum_{k=1}^{n-1}
\int_{1/(k+1)}^{1/k}{\left\lfloor{1\over x}\right\rfloor}^{-1}dx=\sum_{k=1}^{n-1}
\int_{1/(k+1)}^{1/k}\frac{1}{k}dx=\sum_{k=1}^{n-1}\frac{1}{k}\cdot\frac{1}{k(k+1)},
$$
and thus
$$
\int_0^1
{\left\lfloor{1\over x}\right\rfloor}^{-1}dx=\lim_{n\to\infty}
\int_{1/n}^1{\left\lfloor{1\over x}\right\rfloor}^{-1}dx=
\sum_{n=1}^\infty \frac{1}{n^2(n+1)}.
$$
Meanwhile
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)}=\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)=1
\quad\text{and}\quad \frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)}.
$$
Hence, finally
\begin{align}
\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\cdots&=\sum_{n=1}^{\infty}\frac{1}{n^2}-1
=\sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{n(n+1)}\\&=\sum_{n=1}^\infty
\frac{1}{n^2(n+1)}=\int_0^1
{\left\lfloor{1\over x}\right\rfloor}^{-1}dx.
\end{align}
|
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|
$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
|
Every integer $n$
can be written in the form
$3m+k$,
where $m$ is a non-negative integer
and $k = 0, 1, $ or $2$.
(This is a particular case of
the result that
for any positive integer $j$,
every integer $n$
can be written in the form
$jm+k$, where $m$ is a non-negative integer
and $k$ is an integer such that
$0 \le k < j$.)
Therefore
$n^2+1
=(3m+k)^2+1
=9m^2+6mk+k^2+1
=3(3m^2+2mk)+k^2+1
$.
If $3 \mid n^2+1$,
then $3 \mid k^2+1$.
But
the possible values of
$k^2+1$ are
$1, 2, $ and $5$
(for $k = 0, 1, 2$, respectively),
and $3$ does not divide any of them.
Therefore $3$ does not divide $n^2+1$.
|
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|
How prove this $e=\frac{2}{1}\left(\frac{4}{3}\right)^{\frac{1}{2}}\left(\frac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\cdots$ I see this nice equation:
$$e=\dfrac{2}{1}\left(\dfrac{4}{3}\right)^{\frac{1}{2}}\left(\dfrac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\left(\dfrac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots$$
and this equation who first found it? and How prove it.
Thank you very much!
|
This (PDF) has the proof of your equation.
New Wallis- and Catalan-Type Infinite Products for $\pi,e$, and $\sqrt{2+\sqrt 2}$
by Jonathan Sondow and Huang Yi
|
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|
How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$ let $a,b,c,d,e$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$
My try:
I have solved follow Four-variable inequality:
let $a,b,c,d$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge 0$$
poof:since
\begin{align*}
&\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}=\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}-4\\
&=(a+c)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)+(b+d)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)-4
\end{align*}
By Cauchy-Schwarz inequality we have
$$\dfrac{1}{b+c}+\dfrac{1}{d+a}\ge\dfrac{4}{(b+c)+(d+a)},\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{(c+d)+(a+b)}$$
so we get
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge\dfrac{4(a+c)}{(b+c)+(d+a)}+\dfrac{4(b+d)}{(c+d)+(a+b)}-4=0$$
Equality holds for $a=c$ and $b=d$
By done!
But for Five-variable inequality,I can't prove it.Thank you
|
I would have written this in a comment, but I don't have enough 'reputation' to do so...
@Macavity I'd just like to point out that $$\sum \frac{a_i}{b_i} \ge \frac{\big(\sum a_i \big)^2}{\sum a_i b_i}$$ holds when $a_i b_i$ is positive for all $i$. (To use Cauchy-Schwarz, you need to square-root and then square those terms) A counterexample for this might be $a_1=-1, a_2=1, b_1=2$, and $b_2=4$.
Also, this is a conjecture proposed by Vasile Cirtoaje many years ago in Old and New Inequalities (page 60). Inequalities with more variables were discussed here
|
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|
How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=\dfrac{(-1)^nn!}{2ai}\left(\dfrac{1}{(x-ai)^{n+1}}-\dfrac{1}{(x+ai)^{n+1}}\right)$$
so
let$$x=a\cot{\theta},0<\theta<\pi,$$
then
$$x\pm ai=a(\cos{\theta}\pm i\sin{\theta})/\sin{\theta}$$
so
$$\dfrac{1}{(x\pm ai)^{n+1}}=\dfrac{\sin^{n+1}{\theta}}{a^{n+1}}[\cos{(n+1)\theta}\mp i\sin{(n+1)\theta}]$$
so$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
Question:
Have other methods?
Because this is important reslut,so I think this have other methods? Thank you
|
Mathematical induction and trigonometric function relations and derivatives should do it.
I'll assume $x/a\in (0,\pi).$
Step: $n = 0$
$$\frac{1}{x^2+a^2}=\frac{\sin{[\mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{1/2}}$$ as
$$\sin{[\mathrm{arccot}{(x/a)}]} = \frac{a}{(a^2+x^2)^{1/2}}. $$
Step: $n \Rightarrow n+1$
Assume
$$\left(\frac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\frac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
it is true, and prove
$$\left(\frac{1}{x^2+a^2}\right)^{(n+1)}=(-1)^{(n+1)}(n+1)!\frac{\sin{[(n+2)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+2)/2}}.$$
We now need to show:
$$\left((-1)^{(n)}n!\frac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}\right)^\prime = (-1)^{(n+1)}(n+1)!\frac{\sin{[(n+2)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+2)/2}}.$$
This is due to:
$$(f/g)^\prime = (f^\prime g - fg^\prime)/g^2$$
and
$$ \sin(x+y) = \sin x\cos y + \sin y \cos x$$
and
$$ \left(\mathrm{arccot}{(x)}\right)^\prime =-\frac{1}{1+x^2}. $$
|
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|
How find minimum of $\sqrt{(a+7)^2+(a+10)^2}+\sqrt{(a-b)^2+(a+5)^2}+\sqrt{(c-a)^2+(c-3)^2}+\sqrt{(c-6)^2+(c-8)^2}$ let $a,b,c\in R$,then Find this mimimum of the value
$$\sqrt{(a+7)^2+(a+10)^2}+\sqrt{(a-b)^2+(a+5)^2}+\sqrt{(c-a)^2+(c-3)^2}+\sqrt{(c-6)^2+(c-8)^2}$$
My try: since
$$\sum_{i=1}^{n}\sqrt{a^2_{i}+b^2_{i}}=\sqrt{(\sum_{i=1}^{n}a_{i})^2+(\sum_{i=1}^{n}b_{i})^2}$$
But can't use full,Thank you
|
Hint:
An alternative to Minkowski - Consider a path through the points $(b, a), (a, -5), (-7, a+5), (c-10, c+5), (-4, 13)$. The LHS is the length of this path, which obviously gets minimised when all these points are on a straight line, in order. Can you find $a, b, c$ so that these are on a straight line?
|
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|
Minimum value of: $x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$
$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
$$x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$$
I put it in the form $x^6y +x^6z+y^6x+y^6z+z^6x +z^6y$. I tried AM-GM but it's not helping.
|
Step:$1$ $$x^6y +x^6z+y^6x+y^6z+z^6x +z^6y \geq 6x^2y^2z^2(xyz)^{\frac{1}{3}}$$
Step:$2$$$\frac{\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}}{3} \geq \frac{1}{(xyz)^{\frac{2}{3}}}$$
$$(xyz)^{\frac{2}{3}} \geq 3$$
Next step : Substitute result of step:$2$ in step:$1$
|
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|
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
|
Set $a=\sqrt{x-4}$ and $b=\sqrt{x-5}$ for $-2\leq x\leq +2$. Equivalently we have $a^2=x-4$, $b^2={x-5}$ for $-2\leq x\leq +2$. It's implies for $-2\leq x\leq +2$,
\begin{align}
a-b=-1 \implies & (a-b)(a+b)=-1\cdot(a+b)\\
\implies & a^2-b^2=-(a+b)\\
\implies & (x-4)-(x-5)=-(a+b)\\
\implies & -1=-(a+b)\\
\implies & a+b=+1.\\
\end{align}
The same calculations and similar implications $a+b=+1\implies a-b=+1$. Then $ x $ is a root of $a+b=-1$, if only if,
\begin{cases}
a+b=&+1\\
a-b=&-1\\
\end{cases}
Here, $a=0$ and $b=1$ implies $a^2=x-4=0$ and $b^2={x-5}=1$. In both cases, $x\in\!\!\!\!\!/[-2,+2]$.
So the equation
$$
\sqrt{x-4}-\sqrt{x-5}+1=0
$$
has no solution.
|
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|
Compute $\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$ Compute the value of the following expression
$$\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$$
The answer is $\boxed{2n-\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )}$.
I've been trying to do it but I've been failed. Any ideas ?
Wolfram's test
|
You can prove the claim by induction. For $n=1$ you have $1=1$ which is clearly true.
For $n+1$ you can compute
$$
\left(1+\frac{1}{2}+\ldots+\frac1n+\frac{1}{n+1}\right)^2 + \left(\frac{1}{2}+\ldots+\frac1n+\frac{1}{n+1}\right)^2 + \ldots + \left(\frac{1}{n}+\frac{1}{n+1}\right)^2 + \left(\frac{1}{n+1}\right)^2
$$
using the binomial theorem as
$$
\left(1+\frac{1}{2}+\ldots+\frac1n\right)^2 + \frac{2}{n+1}\left(1+\frac{1}{2}+\ldots+\frac1n\right)+\left(\frac{1}{n+1}\right)^2+\\ + \left(\frac{1}{2}+\ldots+\frac1n\right)^2 + \frac{2}{n+1}\left(\frac{1}{2}+\ldots+\frac1n\right)+\left(\frac{1}{n+1}\right)^2+\\+\ldots+\\+\left(\frac1{n}\right)^2+\frac{2}{n+1}\frac1n+\left(\frac1{n+1}\right)^2+\\+\left(\frac1{n+1}\right)^2
$$
The first terms in each row (except the last) can be collectively computed used the induction hypothesis as
$$
2n-\left(1+\frac12+\ldots+\frac1n\right);
$$
the last terms from each row (including the last) give
$$
(n+1)\left(\frac{1}{n+1}\right)^2=\frac{1}{n+1}.
$$
The middle terms give
$$
\frac{2}{n+1}\left(1+\left[\frac12+\frac12\right] + \ldots +\left[\frac1n + \ldots + \frac1n\right] \right) = \frac{2}{n+1}\left(n\times 1\right) = 2\frac n{n+1}.
$$
In total we thus obtain
$$
2n-\left(1+\frac12+\ldots+\frac1n\right) + 2\frac n{n+1} + \frac{1}{n+1}\\ = 2n-\left(1+\frac12+\ldots+\frac1n\right) + \frac {2n+1}{n+1}\\=2n-\left(1+\frac12+\ldots+\frac1n\right) + \frac {2(n+1)-1}{n+1}\\=2n-\left(1+\frac12+\ldots+\frac1n\right) + \frac {2n+1}{n+1}\\=2(n+1)-\left(1+\frac12+\ldots+\frac1n+\frac{1}{n+1}\right),
$$
from which the claim follows.
|
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|
Evaluating the primitive $\int \frac{\mathrm dx}{e^{2x} + e^x + 1} $ Could someone help me evaluate this?
$$\int \frac{\mathrm dx}{e^{2x} + e^x + 1} $$
I tried to solve it for hours with no success.
I tried Wolframalpha but it's giving a step by step solution that is too long that in an exam I won't even have the time to write the solution.
Thanks in advance.
|
Let $u = e^x$. Then, $du = e^x\,dx$, or, equivalently, $dx = \frac{1}{u}du$. Thus, the integral becomes:
$$
\int\frac{dx}{e^{2x}+e^x+1} = \int\frac{du}{(u^2+u+1)u}$$
Now, hit it with partial fractions:
$$\begin{align}
\frac{1}{(u^2+u+1)u} &= \frac{Au+B}{u^2+u+1} + \frac{D}{u}\\
&=\frac{Au^2+Bu+Du^2+Du+D}{(u^2+u+1)u}\\
&=\frac{(A+D)u^2+(B+D)u+D}{(u^2+u+1)u}
\end{align}$$
Thus:
$$A+D = 0\\
B+D = 0\\
D = 1$$
So, the integral is:
$$\int\frac{du}{(u^2+u+1)u} = \underbrace{\int \frac{-u-1}{u^2+u+1} du}_{\text{integral 1}} + \underbrace{\int \frac{1}{u}du}_{\text{integral 2}}$$
Integral $2$ is trivial, so I won't write it out. For integral $1$, we apply the substitution $w = u^2+u+1$, so $dw = (2u + 1)du$.
$$\begin{align}
\int \frac{-u-1}{u^2+u+1} du &= \frac{-1}{2}\int \frac{2u+1}{u^2+u+1}du + \frac{-1}{2}\int\frac{1}{u^2+u+1}du\\
&= \frac{-1}{2}\int \frac{1}{w}dw - \frac{1}{2} \int \frac{1}{u^2 + u+1}du\\
&= \frac{-1}{2}\ln|w| - \frac{1}{2} \underbrace{\int \frac{1}{u^2 + u+1}}_{\text{integral 3}}
\end{align}$$
For Integral $3$, we use our knowledge of the derivative of arctangent. We have:
$$\begin{align}
\int \frac{1}{u^2 + u+1}du &= \int \frac{1}{\left(u+\frac{1}{2}\right)^2 + \frac{3}{4}}du\\
&= \int \frac{1}{\left(\frac{u+\frac{1}{2}}{\sqrt{3}/2}\right)^2 + 1}du\\
&= \int \frac{1}{\left(\frac{2u+1}{\sqrt{3}}\right)^2 + 1}du\\
&= \frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right) +C
\end{align}$$
Thus, our overall answer is:
$$\begin{align}
\int\frac{dx}{e^{2x}+e^x+1} &= I_1 + \ln|e^x| \\
&= \frac{-1}{2}\ln|e^{2x}+e^x+1| - \frac{1}{\sqrt{3}}\arctan\left(\frac{2e^x+1}{\sqrt{3}}\right) + x + C
\end{align}$$
|
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|
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
|
Let $S = 1+2^1+2^2+...+2^{10}$
Multiply by 2
$2S = 2^1+2^2+2^3+...+2^{11}$
Substract the former from the latter:
$S = 2^{11}-1$
$1+2^1+2^2+...+2^{10} = 2^{11}-1$
|
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|
Prove that for every $x\in[0,1]$, this is true $\sqrt{1+2x}\geq x+1-(1/(2x^2))$ Prove that for every $x\in[0,1]$, this is true $\sqrt{1+2x}\geq x+1-(\frac{1}{2x^2})$
i proved that $x+1-\sqrt{1+2x}>0$ by: $(x+1)^2 -1-2x=x^2$ so $x+1>\sqrt{1+2x}$ but then don't know how to proceed for this question
Thank you in advance
|
Each of the following are equivalent
$$
\begin{align}
\sqrt{1+2x}&\ge x+1-\frac{1}{2x^2}\tag{1}\\
\sqrt{1+2x}-(x+1)&\ge-\frac{1}{2x^2}\tag{2}\\
\frac{-x^2}{\sqrt{1+2x}+(x+1)}&\ge-\frac{1}{2x^2}\tag{3}\\[6pt]
\sqrt{1+2x}+(x+1)&\ge2x^4\tag{4}
\end{align}
$$
Reversible Operations:
$(2)$: subtract $x+1$ from both sides
$(3)$: rewrite the left side by multiplying by $\frac{\sqrt{1+2x}+(x+1)}{\sqrt{1+2x}+(x+1)}$
$(4)$: both sides are negative, so taking negative reciprocals preserves order
Now the left hand side of $(4)$ is $\ge2$ and the right hand side is $\le2$.
A far simpler approach
Note that $\sqrt{1+2x}$ and $x+1-\dfrac{1}{2x^2}$ are increasing functions on $[0,1]$ and
on $[0,\frac58]$, $\quad\sqrt{1+2x}\ge1\quad$ while $\quad x+1-\dfrac{1}{2x^2}\le\frac{69}{200}\quad$ and
on $[\frac58\!,1]$, $\quad\sqrt{1+2x}\ge\frac32\quad$ while $\quad x+1-\dfrac{1}{2x^2}\le\frac32$
|
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|
convergence of series, exercise I must prove convergence or divergence of the following series:
$$
\sum_{n=1}^{\infty}
\left[\arctan\left(1 \over n\right) - {n \over n^{2} + 1}\right]
$$
I've tried the integral test:
$$
\int_{1}^{\infty}x\,\arctan\left(1 \over x\right)\,{\rm d}x$$ and this is $<1$ so I would conclude convergence.
However this integral test requires a monotone decreasing function. Could someone give me some pointers to how to prove this?
|
Hint.
$$(\arctan t)'=t-\frac{t^3}{3}+\frac{t^5}{5}+\cdots$$
thus
$$
\left|\arctan\frac{1}{n}-\frac{1}{n}\right|\le \frac{1}{3n^3}+\frac{1}{5n^5}+\cdots\le
\frac{1}{3n^3}\left(1+\frac{1}{n^2}+\cdots\right)=\frac{1}{3n^3}\frac{n^2}{n^2-1}\le\frac{1}{n^3},
$$
for $n\ge 2$, and
$$
\left|\,\frac{n}{n^2+1}-\frac{1}{n}\,\right|=\left|\,\frac{n^2}{n(n^2+1)}-\frac{n^2+1}{n(n^2+1)}\,\right|\le \frac{1}{n^3}.
$$
Thus
$$
\left|\,\arctan\frac{1}{n}-\frac{n}{n^2+1}
\,\right|\le\frac{2}{n^3},
$$
for $n\ge 2$.
|
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|
Finding order of pole I have some problems with the following excersice: Find the order of the pole of:
$$\frac{1}{(2\cos z -2 + z^2)^2}$$ at $z=0$. I thought it is maybe better to work here with $1/f$ and find the order of the zero at $z=0$. But I don't know how to continue.
Thanks in advance!
|
$$2\cos z-2+z^2=2\left(1-\frac{z^2}2+\frac{z^4}{24}-\ldots\right)-2+z^2=$$
$$=\frac{z^4}{12}-\frac{2z^6}{6!}+\ldots=z^4\left(\frac1{12}-\frac{2z^2}{6!}+\ldots\right)\implies$$
$$\left(2\cos x-2+z^2\right)^2=z^8\left(\frac1{144}+\frac{z^2}{3\cdot 6!}+\ldots\right)\implies$$
$$\frac1{(2\cos z-2+z^2)^2}=\;\;\ldots\ldots$$
|
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|
Inequality $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$ Let $x,y,z$ be real numbers such that $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.
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Set
$$u:=\cos x , v:=\cos y , w:=\cos z$$
We have
$$u + v + w = 0$$
and
$$\cos 3x + \cos 3y + \cos 3z = 4u^3 - 3u + 4v^3 - 3v + 4w^3 - 3w = 4u^3 + 4v^3 + 4w^3 = 0$$
So
$$u^3 + v^3 + w^3 = 0$$
Now consider
$$\cos 2x * \cos 2y * \cos 2z = \left(2u^2-1\right)\left(2v^2-1\right)\left(2w^2-1\right)$$
Further, we have
$$\left(u+v+w\right)^3 - u^3 - v^3 - w^3 = 3\left(u^2v+u^2w+v^2u+v^2w+w^2u+w^2v+2uvw\right) = 0$$
$$u^2v+u^2w+u^3+v^2u+v^2w+v^3+w^2u+w^2v+w^3+2uvw = 0$$
$$2uvw = 0$$
WLOG u = 0 , v=-w
As $2v^2-1=2w^2-1$ and $2u^2-1 < 0$, the proof is completed.
|
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Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| < \epsilon $.
Suppose $ n>4 $. Then $ n+6 < 7n $ and $ n^2 -6 > \frac{1}{2} n^2 $. So
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} $.
Consider $ K = \max\{4,\displaystyle \frac{14}{\epsilon}\} $ and suppose $ n> K $. Then $ n > \displaystyle \frac{14}{\epsilon} $. This implies that $ \epsilon > \displaystyle \frac{14}{n} $. Therefore
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} < \epsilon $.
Thus $ \lim\limits_{n \rightarrow \infty } \displaystyle \frac{n+6}{n^2-6}=0 $.
Is this proof correct? What are some other ways of proving this? Thanks!
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Personally (assuming I wasn't supposed to use sharper tools), I would tackle it by first noting that the fraction should be about $1/n$ when $n$ is very large. So pull out a factor $1/n$, and simplify the rest, and you get
$$\frac{n+6}{n^2-6}=\frac1n\cdot\frac{1+6/n}{1-6/n^2}.$$
Now you just need to find a good bound for the fracton on the right, when $n$ is large. To make life really simple, you could require $n\ge6$, which implies $1+6/n\le2$ and $1-6/n^2\ge5/6$, so that
$$\frac{n+6}{n^2-6}\le\frac{12}5\frac1n\qquad\text{for $n\ge6$}.$$
The rest is now trivial.
|
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Solving a challenging differential equation How would one go about finding a closed form analytic solution to the following differential equation?
$$\frac{d^2y}{dx^2} +(x^4 +x^2+x+c)y(x) =0 $$
where $c\in\mathbb{R}$
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You can have a closed form solution in terms of HeunT (the Heun Triconfluent function)
$$ y( x ) ={C_1}\,{{\rm e}^{\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}}{\it HeunT} \left(-\frac{{3}^{2/3}\sqrt[3]{2} \left(4\,c-1\right)}{8}, \frac{3\,i}{2}, -\frac{{2}^{2/3}\sqrt[3]{3}}{2}, \frac{i\sqrt [3]{2}\,{3}^{2/3}}{3}x \right)
+{ C_2} {{\rm e}^{-\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}} {\it HeunT } \left( -\frac{{3}^{2/3}\sqrt[3]{2} \left(4\,c-1\right)}{8} ,-\frac{3\,i}{2}, -\frac{{2}^{2/3}\sqrt[3]{3}}{2}, -\frac{i\sqrt [3]{2}\,{3}^{2/3}}{3}x\right).$$
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Determine the number of divisors in $K[x]$ of $1 + x^{15}$ and of $1+x^{120}$ where $K[x]$ is the set of all polynomials where coefficients are elements of $K$ $(0,1)$
Is this related to the problem of finding how many cyclic linear codes there are if $n = 15$ and $120$?
I've seen the following corollary in a book about the subject but I'm not sure if this can help me:
Let $n = 2^rs$, where $s$ is odd and let $1 + x^s$ be the product of $z$ irreducible polynomials. Then there are $(2^r +1)^z$ linear cyclic codes of length $n$ and $(3^r+1)^z - 2$ proper linear cyclic codes of length $n$.
I've seen different versions of the problem with: $x^{30}$ instead of $x^{120}$, $1+x^{15}$ and $1 + x^{60}$ or $1+x^{21}$ and $1+x^{42}$. It's always that the degree of the $1^{st}$ polynomial divides that of the $2^{nd}$ so, I suppose, there must be some property that takes that into account.
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Notice that over the binary field $K$, $(x^{15}+1)^8 = x^{120} + 1$ by repeated squarings. So if we find the irreducible factorization of $x^{15} + 1$, we have that of $x^{120} + 1$ as an immediate consequence.
Since $15 = 3\cdot 5$, we already know a couple of easy factorizations of $x^{15} + 1$, as the sum of third and fifth powers, respectively. Let's pick the first of these and refine it:
$$ x^{15} + 1 = (x^5 + 1)(x^{10} + x^5 + 1) \tag{1}$$
where the minus signs that normally appear are replaced without error by plus signs over $K$.
Now we further factor:
$$ x^5 + 1 = (x + 1)(x^4 + x^3 + x^2 + x + 1) \tag{2}$$
This earlier Question elicits all the irreducible polynomials in $K[x]$ up to degree 5, and the Answers there inform us that the second factor in (2) is indeed irreducible.
It remains to get irreducible factors of the second factor in (1). We know there must be an irreducible factor $x^2 + x + 1$ hiding in there, because of the overall factor $x^3 + 1$ we did not pursue, and:
$$ x^{10} + x^5 + 1 = (x^2 + x + 1)(x^8+x^7+x^5+x^4+x^3+x+1) \tag{3}$$
Finally the last of these can be factored into irreducibles:
$$ x^8+x^7+x^5+x^4+x^3+x+1 = (x^4+x^3+1)(x^4+x+1) \tag{4}$$
Piecing all of this together gives us that $x^{15} + 1$ is the product of five distinct irreducible factors over $K$. Raising it to the eighth power gives $x^{120} + 1$.
The distinct divisors of $x^{15} + 1$ correspond to subsets of its five distinct irreducible factors, with the empty set corresponding to $1$ and the total set corresponding to $x^{15} + 1$ itself. So, in all there are $2^5 = 32$ divisors of $x^{15} + 1$.
The distinct divisors of $x^{120} + 1$ can be counted similarly, except that instead of having just the two choices for each distinct irreducible (to include or not), we choose between zero and eight of them. Thus in all there are $9^5 = 59049$ divisors of $x^{120}+1$.
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minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$ How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$
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Starting with $F=\cos(a-b)+\cos(b-c)+\cos(c-a)$ expand the cosines via the addition formula for cosine, and get
$$\cos a \cos b + \cos b \cos c + \cos c \cos a \\
+\sin a \sin b + \sin b \sin c + \sin c \sin a.$$
Then after applying $(u+v+w)^2-u^2-v^2-w^2=2uv+2vw+2wu,$ the double $2F$ of our objective function may be seen to be
$$2F=(\cos a+\cos b +\cos c)^2+(\sin a +\sin b + \sin c)^2-3.$$
Note we have combined the terms e.g. $-\cos^2 a -\sin^2 a=-1$ to obtain the final $-3.$
Thus $2F \ge -3$ i.e. $F \ge -3/2.$ Since there are values of $a,b,c$ which achieve $F=-3/2$ this finishes a proof.
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limit problem - can't get rid of $0$. I am trying to evaluate limit:
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}$$
I tried to use known limit in denominator to get:
$$\lim_{x\rightarrow 0}\frac{\frac{\arcsin}{x} - \frac{\arctan x}{x}}{x( \frac{e^x-1}{x}\cdot \frac{1}{x}+\frac{1-\cos x}{x^2} -1 -\frac{1}{x})}$$
But I still get $1-1=0$ in numerator. I also tried to use L'Hôpital rule, but it didn't help.
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Just use L‘hospital multiple times:
$$ \lim_{x \rightarrow 0} \frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}
= \lim_{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}} - \frac{1}{x^2+1}}{e^x+\sin x -2x-1}
= \lim_{x \rightarrow 0} \frac{\frac{x}{\sqrt{(1-x^2)^3}} + \frac{2x}{(x^2+1)^2}}{e^x+\cos x -2}
= \lim_{x \rightarrow 0} \frac{\frac{3x^2}{\sqrt{(1-x^2)^5}} + \frac{1}{\sqrt{1-x^2)^3}} - \frac{8x^2}{(x^2+1)^3} \frac{2}{(x^2+1)^2}}{e^x-\sin x}
= \frac{3}{1} =3.
$$
Edit: I edited my pot. My original answer was just wrong.
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Dice probability where sum of seven doesn't precede sum of eight. Given the question:
A pair of fair dice is rolled until the first sum of 8 appears. What is the probability that a sum of seven doesn't precede sum of 8?
My solution is:
P(sum of 7 doesn't precede 1st sum of 8) = 1 - P(sum of 7 precedes 1st sum of 8) $= 1 - \left(\frac{6}{36}\right)\left(\frac{5}{36}\right) = \frac{211}{216}$
This solution seems right to me but I thought why can't it be solved using conditional probability: 1 - P(first sum of 8 comes given sum of seven is rolled) = 1 - P(8|7) = 1 - P(8 and 7)/P(7) = 1 - (5/36)(6/36)/6(/36) = 1 - (5/36) = 31/36 or is that just for 8 after 7 on any roll following 7, that is not necessarily 8 right after 7?
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Sums other than $7$ or $8$ are irrelevant: if we do not record anything that is not a $7$ or $8$, the game ends in one "step." The ordinary probability of a $7$ is $\frac{6}{38}$, and the probability of an $8$ is $\frac{5}{36}$. Thus if we record only a $7$ or $8$, the probability of $8$ is $\frac{5}{11}$.
More formally, the probability that a toss is an $8$ given that it is a $7$ or an $8$ is, by the usual formula for conditional probabilities, equal to $\frac{\frac{5}{36}}{\frac{6}{36}+\frac{5}{36}}$, that is, $\frac{5}{11}$.
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Solve $\frac{dy}{dx}=\frac{y^2-1}{x^2-1}$ with initial condition $y(2)=2$
Solve the following differential equation: $\dfrac{dy}{dx}=\dfrac{y^2-1}{x^2-1}$, with the initial condition $y(2)=2$.
My attempt:
I notice that this is a separable differential equation, so I try to get it into the form $p(y) dy=f(x) dx$.
$\dfrac{dy}{dx}=\dfrac{y^2-1}{x^2-1}\\\implies \dfrac{1}{y^2-1} dy=\dfrac{1}{1-x^2}dx\\\implies\displaystyle\int\dfrac{1}{y^2-1} dy=\displaystyle\int\dfrac{1}{1-x^2}dx\\\implies \ln(y^2-1)=\ln(x^2-1)$
Now I tried to apply the initial condition of $y(2)=2$:
$\ln(2^2-1)=\ln(2^2-1)\implies\ln3=\ln3$
I'm fairly new to solving differential equations, so I don't know if I'm doing a step incorrectly or what the problem is. If anyone could shed any light on this, it would be appreciated. Thanks in advance.
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Since you are considering the boundary condition $y(2)=2,$ it is only fair to restrict the domain of the function to $x\gt1.$ Therefore, we have that $$\frac{y'(x)}{y(x)^2-1}=\frac1{1-x^2}\implies\frac{2y'(x)}{y(x)^2-1}=\frac2{1-x^2}$$ $$\implies\left[\frac1{y(x)-1}-\frac1{y(x)+1}\right]y'(x)=\frac1{1+x}+\frac1{1-x}=\frac1{x+1}-\frac1{x-1}.$$ This of course assumes that $y(x)\neq-1$ and $y(x)\neq1,$ with these corresponding to the trivial solutions. Since $y(2)=2,$ it is the case that $y(x)\gt1,$ so $$\ln\left[\frac{y(x)-1}{y(x)+1}\right]=\ln\left(\frac{x+1}{x-1}\right)+C.$$ Let $x=2,$ hence $$\ln\left(\frac13\right)=\ln(3)+C,$$ implying $$C=\ln\left(\frac19\right).$$ Therefore, $$\ln\left[\frac{y(x)-1}{y(x)+1}\right]=\ln\left(\frac{x+1}{x-1}\right)+\ln\left(\frac19\right).$$ This is equivalent to $$\frac{y(x)-1}{y(x)+1}=\frac{x+1}{9x-9},$$ which is equivalent to $$[y(x)-1](9x-9)=[y(x)+1](x+1)=(9x-9)y(x)-(9x-9)=(x+1)y(x)+(x+1),$$ which is equivalent to $$(8x-10)y(x)=10x-8,$$ which simplifies to $$y(x)=\frac{10x-8}{8x-10}=\frac{5x-4}{4x-5}.$$
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Stuck with a tricky existence proof Show that there exists a continuous function $f: [-1, 1] \rightarrow \mathbb{R}$ such
$f(0) = 1$ and
$f(x) = \frac{2-x^2}{2} \cdot f(\frac{x^2}{2-x^2})$
$\forall x \in [-1, 1]$
I tried putting in $x = 1$ and $x = -1$ in the second condition to find that $f(1) = f(-1) = 0$.
I also took the derivative of the second equation to find that:
$f'(x) = x (f'(\frac{x^2}{2-x^2})\frac{2}{2-x^2}-f(\frac{x^2}{2-x^2}))$
This gives me
$f'(0) = f'(1) = f'(-1) = 0$ but now I'm stuck. Anybody see a way?
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I recall solving this problem before... was it on a Putnam? Sadly, I don't remember what leap of intuition led me to the answer.
One may take $f(x) = \sqrt{1-x^2}$. It is clear that $f$ is continuous on $[-1,1]$ and that $f(0) = 1$, and one sees that
$$\frac{2-x^2}{2} \cdot \sqrt{1-\frac{x^4}{(2-x^2)^2}} = \frac{2-x^2}{2} \cdot \sqrt{\frac{x^4-4x^2+4-x^4}{x^4-4x^2+4}} \\
= \frac{2-x^2}{2} \cdot \sqrt{\frac{4-4x^2}{(2-x^2)^2}} = \sqrt{1-x^2},$$
so that $f(x) = \frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$.
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Is there a trick to evaluating a matrix multiplied by the sum of three vectors? I am looking at some old linear algebra test material along with some sample answers. I don't understand a bit of computation in the sample answer. Here's the exam question:
Examine the recurrence equation $x_n = Ax_{n-1} + u$, where
$$A = \begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & a \\\ 0 & a & a^2 \end{pmatrix}$$
Now suppose $x_n = α_np + β_nq + γ_nr$ and $u = cp - acr + bq$ for some set of vectors $p, q, r, u$ ($α, β, γ, a, b, c$ are constants). Express $α_n, β_n, γ_n$ in terms of $α_{n-1}, β_{n-1}, γ_{n-1}$.
The answer proceeds:
$$x_n = Ax_{n-1} + u = \begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & a \\\ 0 & a & a^2 \end{pmatrix}\begin{pmatrix} \\\ α_{n-1}p + β_{n-1}q + γ_{n-1}r \\\ \end{pmatrix} + u$$
$$x_n = \begin{pmatrix} 1 & 0 & 0 \\\ 0 & 1 & a \\\ 0 & a & a^2 \end{pmatrix}\begin{pmatrix} \\\ (a^2 + 1)α_{n+1}p + β_{n-1}q \\\ \end{pmatrix} + u$$
$$x_n = ((a^2 + 1)α_{n+1} + c)p + (β_{n-1} + b)q + acr$$
*
*$α_n = (a^2 + 1)a_{n+1}c$
*$β_n = β_{n-1} + b$
*$γ_n = ac$
Unfortunately, I can't understand the second step in the solution. Where did $r$ go? Where did $a^2 + 1$ and $α_{n+1}$ come from? Is the answer using some sort of matrix-vector product shortcut I don't know about?
In case it's relevant, in an earlier part of the problem the values of $p, q, r$ are calculated as:
$$p = \frac{1}{a^2 + 1}\begin{pmatrix} 0 \\\ 1 \\\ a \end{pmatrix}\space,\space\space\space q = \begin{pmatrix} 1 \\\ 0 \\\ 0 \end{pmatrix}\space,\space\space\space r = \frac{1}{a^2 + 1}\begin{pmatrix} 0 \\\ -a \\\ 1 \end{pmatrix}$$
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There appear to be some typos.
Yes the previous computations are relevant. As in the comment by Avitus,
$Ap = (a^2 + 1)p , Aq=q, Ar = 0$. Then
$$ \begin{align}x_n = \alpha_n p + \beta_n q + \gamma_n r
&= A(\alpha_{n-1}p + \beta_{n-1}q + \gamma_{n-1}r ) + u\\
&= (\alpha_{n-1}(a^2 + 1) + c)p + (\beta_{n-1} + b)q - acr
\end{align}$$
Then by comparing the coefficients of $p,q$ and $r$ you see that
$\alpha_n = \alpha_{n-1}(a^2 + 1) + c$, $\beta_n = \beta_{n-1} + b$ and $\gamma_n = -ac$.
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ordered pair of unequal positive integer solution of $x+y+z+w = 20$ [1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$\bf{My\; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,z\in \mathbb{Z^{+}}$
$\bullet $ If $x=1$, Then $y+z=9$, So $(y,z) = (2,7)\;,(3,6)\;,(4,5)\;(5,4)\;,(6,3)\;,(7,2)$
$\bullet $ If $x=2$ Then $y+z=8$, So $(y,z) = (1,7)\;,(3,5)\;,(5,3)\;(7,1)$
$\bullet $ If $x=3$ Then $y+z=7$, So $(y,z) = (1,6)\;,(2,5)\;,(5,2)\;(6,1)$
$\bullet $ If $x=4$ Then $y+z=6$, So $(y,z) = (1,5)\;,(5,1)$.
$\bullet $ If $x=5$ Then $y+z=5$, So $(y,z) = (1,4)\;,(2,3)\;,(3,2)\;(4,1)$
$\bullet $ If $x=6$ Then $y+z=4$, So $(y,z) = (1,3)\;,(3,1)$
$\bullet $ If $x=7$ Then $y+z=3$, So $(y,z) = (1,2)\;,(2,1)$
So Total unordered pair is $ = 24$
My Question is , is there is any other Method to calculate the ordered pair in less complex way
because above is very Lengthy method
Help Required
Thanks.
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For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) \le S
$$
so $3x+3 \le S$, i.e., $x\le (S/3)-1$.
As well, $y+(y+1)\le S-x$, i.e., $2y+1 \le S-x$, i.e. $y \le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=\sum_{x:1 \le x \le \frac{S}{3}-1} \sum_{y:x+1\le y \le \frac{S-x-1}{2}} 1
= \sum_{x:1 \le x \le \frac{S}{3}-1} \left( \left\lfloor \frac{S-x-1}{2}\right\rfloor -x\right)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=\mathrm{round}\left( \frac{(S-3)^2}{12} \right)$$
and I've verified that for $6\le S\le 10000$, so it's pretty surely correct. A proof is probably among the many references at that OEIS entry.
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differentiate arctan (maclaurin?) I have this assignment:
Differentiate this expression:
$$ f(x) =\arctan \frac{x-1}{x+1} $$
There is also known that $-1 < x$ (Why is that important?). I do not know how to solve this problem... By using Maclaurin I can come up with this:
$$ f(x) = \arctan (g(x)) $$
$$ g(x) = \frac{x-1}{x+1} $$
$$ \arctan g(x) = g(x) - \frac{g(x)^3}{3} + \frac{g(x)^5}{5} + O(g(x)^7) $$
I am confused by the inner function g(x) when using Maclaurin, should I just differentiate g(x) or should I not do that?
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The stating of $x \gt -1$ merely defines a valid domain for inner function so that differentiation is possible, since at $x=1$ we would have a singularity.
Differentiate $f(x)$ using the chain rule, where $$f'(z)=(\tan^{-1} (z))'=\frac{1}{1+z^2}.$$
Thus, if $g(x) = \frac {x-1}{x+1}$, we have by the chain rule: $$\begin{equation}\begin{split}\left[f(g(x))\right]'=f'(g(x))\cdot g'(x)&=\left(\tan^{-1} \left(\frac {x-1}{x+1}\right)\right)'\\ & =\frac{1}{1+\left(\frac {x-1}{x+1}\right)^2}\cdot\left(\frac {x-1}{x+1}\right)'\\& =\frac{1}{1+\frac {\left(x-1\right)^2}{\left(x+1\right)^2}}\cdot\frac {2}{\left(x+1\right)^2}\\&=\frac{2}{\left(x+1\right)^2+\left(x-1\right)^2}\\&=\frac{2}{2x^2+2}=\frac{1}{1+x^2}\end{split}\end{equation}$$
|
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|
Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer Consider the differential equation
$$x^2y''+3(x-x^2)y'-3y=0$$
$(a)$ Find the recurrence equation and first three nonzero terms of the series solution in powers of $$ corresponding to the larger root of the indicial equation.
$(b)$ What would be the form of a second linearly independent solution of this differential equation?
I found the indicial equation to be $r(r-1)+3r-3 = 0$, so the two roots are $r_1=-3$ and $r_2=1$.
And the recurrence relation is $a_n = \dfrac{3(n-1+r)a_{n-1}}{(n+r)(n+r-1)+3(n+r)-3}$
set $r=1$, then $a_n = \dfrac{3na_{n-1}}{(n+3)(n+1)-3}$. And I can figure out first linearly independent solution.
But for $r=-3$,then $a_n = \dfrac{3(n-4)a_{n-1}}{(n-3)(n-1)-3}$, if let $a_0$ be an arbitrary constant, then $a_1 =\dfrac{3(-3)a_0}{(-2)(0)}$, which doesn't work.
Then how do I figure out the second linearly independent solution?
|
Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}$
$\therefore x^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+3(x-x^2)\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}-3\sum\limits_{n=0}^\infty a_nx^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r}+\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r}-\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r+1}-\sum\limits_{n=0}^\infty 3a_nx^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r+3)(n+r-1)a_nx^{n+r}-\sum\limits_{n=1}^\infty3(n+r-1)a_{n-1}x^{n+r}=0$
$(r+3)(r-1)a_0x^r+\sum\limits_{n=1}^\infty((n+r+3)(n+r-1)a_n-3(n+r-1)a_{n-1})x^{n+r}=0$
The indicial equation is $(r+3)(r-1)=0$ , the two roots are $r_1=-3$ and $r_2=1$ .
The recurrence relation is $(n+r+3)(n+r-1)a_n-3(n+r-1)a_{n-1}=0$ , i.e. $(n+r+3)(n+r-1)a_n=3(n+r-1)a_{n-1}$
Each root from the indical equations may not always only find one group of the linearly independent solutions, sometimes we can find more than one group of the linearly independent solutions at the same time.
When we choose $r=-3$ , the recurrence relation becomes $n(n-4)a_n=3(n-4)a_{n-1}$ , which has a value of $n$ at $n=4$ so that it is independent to the recurrence relation. Besides $a_0$ can chooes arbitrary, $a_4$ can also chooes arbitrary. This makes the effect that we can find two groups of the linearly independent solutions that basing on $n=0$ to $n=3$ and $n=4$ to $n=+\infty$ respectively.
|
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|
The difference of two consecutive perfect squares is always odd I am working on another homework assignment about proofs. The question is:
Prove or find counterexample: the difference of two consecutive perfect squares is odd?
There is no counterexample correct? I am thinking this is always true. If I were to do 7^2-6^2 the answer is odd. I am unsure of how to start the proof though. I am new to proofs and not sure what to really do
|
Well of course it is.
Consider $n^2$ and the next greater perfect square would be $(n+1)^2$, which factors to $n^2+2n+1$. Now what would the difference between them be?
$n^2+2n+1-n^2=2n+1$.
All odd integers conform to 2n+1 where n is an integer.
There is also this to consider.
$1=1^2$
$1+3=2^2$
$1+3+5=3^2$
$1+3+5+7=4^2$
$1+3+5+7+9=5^2$
$1+3+5+7+9+11=6^2$
$1+3+5+7+9+11+13=7^2$
...and so on
|
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|
How to prove $\left(|a+b|^p+|a-b|^p\right)^{1/p}\ge 2^{1/p}\left(a^2+(p-1)b^2\right)^{1/2}$ For real numbers $a, b$ and all $1\le p\le 2$, how to prove $$\left(|a+b|^p+|a-b|^p\right)^{1/p}\ge 2^{1/p}\left(a^2+(p-1)b^2\right)^{1/2}?$$
|
let $m=|a+b|,n=|a-b|$
first we prove $\left (\dfrac{m^p+n^p}{2} \right)^2 \ge \left(p\left(\dfrac{m-n}{2}\right)^2+mn\right)^p$
WLOG, let $ m\ge n , x=\dfrac{m}{n}\ge 1 \iff \left (\dfrac{x^p+1}{2} \right)^2 \ge \left(p\left(\dfrac{x-1}{2}\right)^2+x\right)^p \iff f(x)=2\ln{\left (\dfrac{x^p+1}{2} \right)}-p\ln{\left ( \left( \dfrac{x-1}{2}\right)^2+x \right)} \ge 0$
$f'(x)=\dfrac{2px((2-p)x^p+px^{p-1}-px-2+p)}{x^p(px^3+(4-2p)x^2+px)+px^3+(4-2p)x^2+px} $
$g(x)=(2-p)x^p+px^{p-1}-px-2+p, \\g'(x)=p(2-p)x^{p-1}+p(p-1)x^{p-2}-p,\\g''(x)=p(2-p)(p-1)(x-1)x^{p-2} \ge 0 \implies g'(x) \ge g'_{min}(x)=g'(1)=0 \implies g(x)\ge g(1)=0 \implies f'(x)\ge 0 \implies f(x) \ge f(1)=0$
it remains
$p\left(\dfrac{m-n}{2}\right)^2+mn \ge a^2+(p-1)b^2 \iff \dfrac{p(a^2+b^2)+(2-p)|a^2-b^2|}{2} \ge a^2+(p-1)b^2 \iff (a^2\ge b^2 \cap (a^2+(p-1)b^2 \ge a^2+(p-1)b^2) , (a^2<b^2) \cap (b^2+(p-1)a^2 \ge a^2+(p-1)b^2 \iff (2-p)b^2 \ge (2-p)a^2)$
QED
|
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|
Find complex Fourier coefficients
*
*let $f(x) = \sum^{10}_{m=1}(-1)^m \sin(2^m x)$.
denote complex Fourier coefficients of $f(x)$ over $[-\pi, \pi]$ as $c_n = \frac{1}{2\pi} \int _{-\pi}^\pi f(x) e^{-inx}\,dx.$
Calculate $c_n$ for any $n$.
I don't have any idea how to even start.
Can you please explain me? thanks in advance.
|
The coefficient $c_{0}=0$, because $f(x)$ is odd
\begin{equation*}
f(-x)=\sum_{m=1}^{10}(-1)^{m}\sin (-2^{m}x)=-\sum_{m=1}^{10}(-1)^{m}\sin
(2^{m}x)=-f(x).
\end{equation*}
For $n<0$, $n=-\left\vert n\right\vert $. Then
\begin{eqnarray*}
c_{n} &=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)e^{-inx}\,dx=\frac{1}{2\pi }
\int_{-\pi }^{\pi }f(x)e^{i\left\vert n\right\vert x}\,dx \\
&=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)\overline{e^{-i\left\vert
n\right\vert x}}\,dx=\frac{1}{2\pi }\int_{-\pi }^{\pi }\overline{
f(x)e^{-i\left\vert n\right\vert x}}\,dx=\overline{c}_{\left\vert
n\right\vert }=\overline{c}_{-n},
\end{eqnarray*}
which means that the negative coefficients are easily computed from the
positive ones. For $n>0$, interchanging the order of the integration and the
summation in
\begin{eqnarray*}
c_{n} &=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)e^{-inx}\,dx \\
&=&\frac{1}{2\pi }\int_{-\pi }^{\pi }\sum_{m=1}^{10}(-1)^{m}\sin
(2^{m}x)e^{-inx}\,dx,
\end{eqnarray*}
we conclude that
\begin{equation*}
c_{n}=\frac{1}{2\pi }\sum_{m=1}^{10}(-1)^{m}I(m,n),
\end{equation*}
where
\begin{eqnarray*}
I(m,n) &=&\int_{-\pi }^{\pi }\sin (2^{m}x)e^{-inx}\,dx,\qquad e^{-inx}=\cos
(nx)-i\sin (nx) \\
&=&\int_{-\pi }^{\pi }\sin (2^{m}x)\cos (nx)\,dx-i\int_{-\pi }^{\pi }\sin
(2^{m}x)\sin (nx)\,dx.
\end{eqnarray*}
The first integral
\begin{equation*}
\int_{-\pi }^{\pi }\sin (2^{m}x)\cos (nx)\,dx=0,
\end{equation*}
because $g(x)=\sin (2^{m}x)\cos (nx)$ is odd ($g(-x)=-g(x)$). Hence $I(m,n)$
reduces to
\begin{equation*}
I(m,n)=-iJ(m,n)=-i\int_{-\pi }^{\pi }\sin (2^{m}x)\sin (nx)\,dx.
\end{equation*}
To evaluate this integral we can use the following trigonometric identities
\begin{eqnarray*}
\sin (2^{m}x)\sin (nx) &=&\frac{1}{2}\left[ \cos \left( (2^{m}-n)x\right)
-\cos \left( (2^{m}+n)x\right) \right] \\
\sin ^{2}(2^{m}x) &=&\frac{1-\cos \left( 2\cdot 2^{m}x\right) }{2},
\end{eqnarray*}
to obtain the well known orthogonal relation (or in Orthonormality/Fourier Series)
\begin{equation*}
J(m,n)=\int_{-\pi }^{\pi }\sin (2^{m}x)\sin (nx)\,dx=\left\{
\begin{array}{c}
\pi \\
0
\end{array}
\begin{array}{c}
\text{if} \\
\text{if}
\end{array}
\begin{array}{c}
2^{m}=n \\
2^{m}\neq n
\end{array}
\right.
\end{equation*}
Consequently,
\begin{equation*}
I(m,n)=\left\{
\begin{array}{c}
-i\pi \\
0
\end{array}
\begin{array}{c}
\text{if} \\
\text{if}
\end{array}
\begin{array}{c}
n=2^{m} \\
n\neq 2^{m}.
\end{array}
\right.
\end{equation*}
*
*For $1\leq n\neq 2^{m}$ as well as for $n=2^{m}$, $m>10$, $c_{n}=0=
\overline{c}_{\left\vert n\right\vert }=\overline{c}_{-n}$. For $n=2^{m}$, $
1\leq m\leq 10,$ I've computed the following values :
\begin{eqnarray*}
c_{2} &=&\frac{1}{2\pi }(-1)^{1}I(1,2)=-\frac{1}{2\pi }\left( -i\pi \right) =
\frac{1}{2}i=\overline{c}_{-2}, \\
c_{4} &=&\frac{1}{2\pi }(-1)^{2}I(2,4)=\frac{1}{2\pi }\left( -i\pi \right) =-
\frac{1}{2}i=\overline{c}_{-4} \\
c_{8} &=&\frac{1}{2\pi }(-1)^{3}I(3,8)=-\frac{1}{2\pi }\left( -i\pi \right) =
\frac{1}{2}i=\overline{c}_{-8} \\
&&\cdots \\
c_{2^{2k-1}} &=&\frac{1}{2}i=\overline{c}_{-2^{2k-1}},\quad c_{2^{2k}}=-
\frac{1}{2}i=\overline{c}_{-2^{2k}},\quad 1\leq k\leq 5.
\end{eqnarray*}
|
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|
Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\
&= \frac{1}{\cos(x) \sin(x)}\\
&= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\
&=\frac{1}{\frac{1}{\sec \csc}}\\
&=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\
&= \sec(x) \csc(x)
\end{align}
$$
|
For acute $\theta$, there's this trigonograph:
$$\sec\theta \cdot \csc\theta \;=\; 2\,|\triangle OPQ| \;=\; 1\cdot( \tan\theta +\cot\theta)$$
(Showing that this works in any quadrant is straightforward.)
|
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|
How prove $3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$ let $a,b,c>0$ and such $abc=1$,show that
$$3(a^4+b^4+c^4)+2(a+b+c)\ge 5\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$$
my idea: maybe can use AM-GM inequality,
$$3(a^4+b^4+c^4)a^2b^2c^2+2(a+b+c)(a^2b^2c^2)\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
$$\Longleftrightarrow 3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
then I can't,Thank you very much
|
Here is a straightforward (but not very elegant) solution : let us put
$$
D=3(a^4+b^4+c^4)+2(a+b+c)abc - 5(a^2b^2+b^2c^2+a^2c^2)
$$
We may assume without loss that $a\leq b\leq c$.
Then, if we put $u=b-a$ and $v=c-b$, we have
$$
D=u^4 + (8a + 2v)u^3 + (10a^2 + 12va + 13v^2)u^2 + (10va^2 + 28v^2a + 12v^3)u + (10v^2a^2 + 12v^3a + 3v^4)
$$
and we are done since all the coefficients are positive.
|
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|
Pseudo-pythagorean theorem Pythagoras' theorem is a special case of the Cosine theorem for a angle of $90°$. But also for an angle of 60° and 120°, "aesthetical" special cases derive:
$c^2=a^2+b^2\pm ab$
First question:
Are there further angle $x°$ with a rational number $x$, so that $\cos x$ is rational as well, thus creating "aesthetical" special cases?
Second question:
Does anybody know some internet sources to the equivalent of pythagorean triplets (Integers $a,b,c$, so that $c^2=a^2+b^2\pm ab$)?
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If $\cos\frac {2m\pi}n$ is rational with $\gcd(m,n)=1$, then the primitive $n$th root of unity $\zeta=\cos \frac{2m\pi}{n}+i\sin\frac{2m\pi}n$ is a root of the rational polynomial $X^2-2\cos \frac{2m\pi}{n}X+1$. On the other hand, we know that $\zeta$ is a root of $X^n-1$ and any factorization of this over the rationals can be resaled to a factorization over the integers. Since $X^2-2\cos \frac{2m\pi}{n}X+1$ is monic this implies that $2\cos\frac{2m\pi }{n}$ is already an integer of absolute value $\le 2$.
This leads to the cases
*
*$c^2=a^2+b^2+2ab$ for $\gamma = \pi$
*$c^2=a^2+b^2+ab$ for $\gamma = \frac 23\pi$ or $\gamma = \frac 43\pi$
*$c^2=a^2+b^2$ for $\gamma =\frac 12\pi$ or $\gamma = \frac 32\pi$
*$c^2=a^2+b^2-ab$ for $\gamma=\frac13\pi$ or $\gamma =\frac 53\pi$
*$c^2=a^2+b^2-2ab$ for $\gamma =0$
and that's all with $\gamma \in[0,2\pi)\cap \pi\mathbb Q$.
Regarding your second question: Pythagorean triples are best viewed as numbers $z=a+bi\in\mathbb Z[i]$ with norm $z\bar z$ a perfect square, which leads to a partitioning of the set of primes into those with $p\equiv -1\pmod 4$ (which can only occur as factor of $c$ if they are also factors of $a$ and $b$) and those with $p\equiv 1\pmod 4$ (which can be written as sums of squares and lead to primitive pythagorean triangles; for example $5=2^2+1^2=(2+i)(2-i)$ give us $\color{red}5^2=(2+i)^2(2-i)^2=(\color{red}3+\color{red}4i)(3-4i)$ and hence the most famous Pythagorean triangle) and the special prime $2$. Similarly, the numbers aou are after should be viewed as elements of $\mathbb Z[\omega]$ where $\omega=-\frac12+\frac i2\sqrt 3$. It turns out that one gets a similar partitioning of the primes, this time based on the remainders modulo $3$. Those are very interesting number theoretic questions indeed.
|
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|
Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariable calculus for this problem. (I've only taken single variable)
The only solution I've seen uses the Cauchy-Schwarz inequality.
|
Eliminating $b,$ we have $$a^2+(2-a-c)^2+c^2=12$$
$$\implies 2a^2-2a(2-c)+2c^2-4c-8=0\iff a^2-a(2-c)+c^2-2c-4=0$$ which is a Quadratic Equation in $a$
As $a$ is real, the discriminant must be $\ge0$
$$\implies (2-c)^2-4(c^2-2c-4)\ge0$$
$$\implies -3c^2+4c+20\ge0\iff3c^2-4c-20\le0$$
Now, we know if $(x-\alpha)(x-\beta)\le0$ with $\alpha\le\beta$
we can write $\alpha\le x\le\beta$
If $\alpha,\beta(>\alpha)$ are the roots of $3c^2-4c-20=0,$ we need $\displaystyle\beta-\alpha=+\sqrt{\left(\beta+\alpha\right)^2-4\beta\alpha}$
we have $\displaystyle\beta+\alpha=\frac43$ and $\displaystyle\beta\alpha=-\frac{20}3$
|
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|
if $(a+\sqrt{a^2+1})$ and $(b+\sqrt{b^2+1})$ are converse then prove that a and b are opposites $(a+\sqrt{a^2+1})\,(b+\sqrt{b^2+1})=1$ is supposed to equal: $b = -a$ but how do i get that? I've been trying to solve for like 2 days now.
|
If
$$
(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$$
we have
$$
a+\sqrt{a^2+1}=\sqrt{b^2+1}-b \ \ \ \ \ (1)
$$
$$
b+\sqrt{b^2+1}=\sqrt{a^2+1}-a \ \ \ \ \ (2)
$$
$(1)+(2)$ then $a=-b$.
|
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|
For $a,b,c>0$, prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}$
For $a,b,c>0$, prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}.$$
(Source : Balkan Mathematical Olympiad 2005)
My work:
From the given inequality, we can have,
$$\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\ge \frac{4(a-b)^2}{a+b+c}$$
This can be re-written as,
$$(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\right)\ge 4(a-b)^2$$
Now, what am I supposed to do? I am doubtful whether the question is correct or not. Please help.
EDIT: After checking quite a few cases I conclude that the inequality is actually working. So there might not be any mistakes at all.
|
from your last one:
LHS$\ge (|a-b|+|b-c|+|c-a|)^2 \ge (|a-b|+|b-c+c-a|)^2=4(a-b)^2$
|
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|
Proves of identities in inverse trigonometry Can someone please help me prove the following results from inverse trigonometry?
$$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x+y}{1-xy}( x>0, y>0, xy>1)$$
and
$$\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x+y}{1-xy} ( x<0, y< 0, xy>1)$$
I know the prove for $\tan^{-1}x + \tan^{-1}y =\tan^{-1}\frac{x+y}{1-xy} ( xy<1)$ but cant prove the other two. Please do help.
Thanks in advance :)
|
Apply $\;\tan\;$ to both sides and use trigonometric identities:
$$\begin{align*}\bullet&\tan (\arctan x+\arctan y)=\frac{\tan\arctan x+\tan\arctan y}{1-\tan\arctan x\tan\arctan y}=\frac{x+y}{1-xy}\\{}\\\bullet&\tan\left(\pi+\arctan\frac{x+y}{1-xy}\right)=\tan\arctan\frac{x+y}{1-xy}=\frac{x+y}{1-xy}\end{align*}$$
|
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Expressing a holomorphic function as an infinite sum I want to show that $\frac{1}{\sin z}=\frac{1}{z}+\sum_{n=1}^\infty\frac{(-1)^n2z}{z^2-n^2\pi^2}$. So it's easy to see that the L.H.S. minus the R.H.S. is an entire function, using the fact that the residue of the meromorphic function $\pi\csc\pi z$ at $n\in\mathbb{Z}$ is $(-1)^n$. The problem is to show that L.H.S. minus R.H.S. is bounded, so that we can conclude using Liouville's theorem that it is actually a constant. Then by letting $z\rightarrow 0$ we conclude that the constant is actually 0. But it seems not easy to show the boundedness.
|
If you can't show that the difference is bounded, how about showing its derivative is $\equiv 0$? Let
$$g(z) = \frac{1}{\sin z} - \frac{1}{z} - \sum_{n=1}^\infty (-1)^n\left(\frac{1}{z - n\pi} + \frac{1}{z+n\pi}\right).\tag{1}$$
Then
$$g'(z) = -\frac{\cos z}{\sin^2 z} + \sum_{n\in\mathbb{Z}} \frac{(-1)^n}{(z-n\pi)^2}.\tag{2}$$
Now $\lvert \cos z\rvert^2 = \lvert \cos x \cosh y - i\sin x \sinh y\rvert^2 = \cos^2 x\cosh^2 y + \sin^2 x\sinh^2 y = \cos^2 x + \sinh^2 y$, and $\lvert \sin z\rvert^2 = \lvert \sin x\cosh y + i\cos x \sinh y\rvert^2 = \sin^2 x \cosh^2 y + \cos^2 x\sinh^2 y = \sin^2 x + \sinh^2 y$, so
$$\left\lvert \frac{\cos z}{\sin^2 z}\right\rvert \leqslant \frac{\cosh y}{\sinh^2 y} \xrightarrow{\lvert y\rvert \to \infty} 0$$
uniformly in $x$. The sum in $(2)$ is obviously $2\pi$-periodic, and on $\lvert \operatorname{Re} z\rvert \leqslant \pi$, it converges to $0$ as $\lvert \operatorname{Im} z\rvert \to \infty$ uniformly in $\operatorname{Re} z$: Let $\varepsilon > 0$. Then there is an $N\geqslant 2$ with
$$\sum_{n = N}^\infty \frac{1}{(n-1)^2} < \frac{\varepsilon}{4}\pi^2.$$
For $\lvert\operatorname{Re} z\rvert \leqslant \pi$ and $\lvert n\rvert\geqslant N$, we have $\lvert z-n\pi\rvert \geqslant (\lvert n\rvert - 1)\pi$, thus
$$\left\lvert \sum_{\lvert n\rvert \geqslant N} \frac{(-1)^n}{(z-n\pi)^2}\right\rvert \leqslant \frac{2}{\pi^2}\sum_{n=N}^\infty \frac{1}{(n-1)^2} < \frac{\varepsilon}{2},$$
and for $\lvert \operatorname{Im} z\rvert \geqslant 2\sqrt{\frac{N}{\varepsilon}}$ we have
$$\left\lvert \sum_{\lvert n\rvert < N} \frac{(-1)^n}{(z-n\pi)^2}\right\rvert \leqslant \sum_{\lvert n\rvert < N} \frac{1}{\lvert \operatorname{Im} z\rvert^2} < \frac{2N}{\lvert\operatorname{Im} z\rvert^2} \leqslant \frac{\varepsilon}{2}.$$
Altogether, $g'$ is an entire $2\pi$-periodic function with $g'(z) \to 0$ for $\lvert\operatorname{Im} z\rvert\to 0$ uniformly in $\operatorname{Re} z$. That implies $g' \equiv 0$, hence $g$ is constant, and since one can read off $(1)$ that $g$ is odd, it follows that $g \equiv 0$ too.
|
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Eliminate the parameter from the parametric equations $$x=\frac{3t}{1+t^3} , y=\frac{3t^2}{1+t^3} , t \neq -1,$$
and hence find an ordinary equation in x and y for this curve, The parameter t can be interpreted as the slope of the line joining the general point $(x,y)$ to the origin. Sketch the curve and show that the line $x+y=-1$ is an asymptote.
|
$$ x^3 + y^3 =\frac{27t^3}{(1+t^3)^3} +\frac{27t^6}{(1+t^3)^3}
=\frac{27t^3(1+t^3)}{(1+t^3)^3}=\frac{27t^3}{(1+t^3)^2}$$
$$ x^3 + y^3 = 3xy$$
|
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Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$. Show that the sequence is upper bounded.
My idea: since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$
then
$$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$
so
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$
since
$$a_{n+1}>a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}<\dfrac{1}{a_{1}+i^2}$$
so
$$\dfrac{1}{a_{n+1}}>\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)$$
But the RHS might be $\lt0$ for a sufficiently large starting value; for instance, with $a_{1}=\dfrac{99}{100}$
then
$$\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)<0,n\to\infty$$
see:
so this method won't let me bound the series and I don't know what else to do.
|
Now,today I have solve this problem,I post my methods,I hope is not wrong?
since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$
then
$$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$
so
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$
so
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{1}}-\left(\dfrac{1}{a_{1}+1^2}
+\dfrac{1}{a_{2}+2^2}+\cdots+\dfrac{1}{a_{n}+n^2}\right)$$
by induction,we have
$$a_{n}>nt^{n+1},t=\sqrt{a_{1}}\in (0,1)$$
because
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}>nt^{n+1}+t^{2n+2}$$
we only prove
$$nt^{n+1}+t^{2n+2}>(n+1)t^{n+2},t=\sqrt{a_{1}}\in (0,1)$$
$$\Longleftrightarrow n+t^{n+1}-(n+1)t>0$$
use this Bernoulli inequality:
$$(1+x)^n\ge 1+nx,x>-1,n>1$$
then we have
$$t^{n+1}=(1+t-1)^{n+1}>1+(n+1)(t-1)=1+(n+1)t-(n+1)$$
Now $$\dfrac{1}{a_{n+1}}>\dfrac{1}{t^2}-\left(\dfrac{1}{t^2+1}+\dfrac{1}{2t^3+2^2}+\cdots+\dfrac{1}{nt^{n+1}+n^2}\right)$$
other hand,Use AM-GM inequality,we have
\begin{align*}
\dfrac{1}{t^2+1}+\dfrac{1}{2t^3+2^2}+\cdots+\dfrac{1}{nt^{n+1}+n^2}&=\dfrac{1}{1(1+t^2)}+\dfrac{1}{2(1+1+t^3)}+\cdots+\dfrac{1}{n(1+1+\cdots+1+t^{n+1})}\\
&<\dfrac{1}{2t}+\dfrac{1}{2\times 3t}+\cdots+\dfrac{1}{n(n+1)t}\\
&=\dfrac{1}{t}\left(1-\dfrac{1}{n+1}\right)\\
&<\dfrac{1}{t}
\end{align*}
so
$$\dfrac{1}{a_{n+1}}>\dfrac{1}{t^2}-\dfrac{1}{t}$$
$$\Longrightarrow a_{n+1}<\dfrac{t^2}{1-t}$$
|
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|
Prove this trigonometry inequality I'm having difficulty proving that tan(26°) < 0.5 < tan(27°) . Any idea ? Thanks.
|
OK, here is the solution. Feel free to ask if you don't understand any of the steps.
Let's first prove the left side of the inequality:
$\tan26 < \frac{1}{2} \Rightarrow 2sin26<cos26 \Rightarrow 4sin^2{26}<cos^2{26}$
$\Rightarrow 4sin^2{26} < 1-sin^2{26} \Rightarrow sin^2{26}<\frac{1}{5} \Rightarrow cos^2{52}>\frac{3}{5} \Rightarrow cos^2{52}>\frac{9}{25} $
$\Rightarrow cos104>\frac{-7}{25} \Rightarrow sin14< \frac{7}{25}$ --- But sin14 < sin15
also $sin\frac{\pi}{12}<\frac{\pi}{12}<\frac{7}{25} \Rightarrow\Rightarrow tan26<\frac{1}{2}$
Now let's prove the right side :
$tan27>\frac{1}{2} \Rightarrow 4sin^2{27}>cos^2{27} \Rightarrow sin^2{27}>\frac{1}{5}$
$\Rightarrow cos54<\frac{3}{5} \Rightarrow sin36<{3}{5} \Rightarrow sin^2{36}<\frac{9}{25} \Rightarrow cos72>\frac{7}{25} \Rightarrow sin18>\frac{7}{25}$
But $sin18=\frac{\sqrt{5}-1}{4} > \frac{1.2}{4} > \frac{7}{25}$
So $sin18>\frac{7}{25} \Rightarrow \Rightarrow tan27>\frac{1}{2}$
|
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Prove that the sequence $\{\int_0^{\pi/2} \sin(t^n)\, dt\}$ converges to 0. Prove that the sequence
$$
\left\{\int_0^{\pi/2} \sin(t^n)\, dt :n\in\mathbb N\right\}
$$
converges to 0.
|
First split the integral at $t=1$. Estimating the first half is simple; since $0 \leq \sin x \leq x$ for $0 \leq x \leq 1$ we have
$$
0 \leq \int_0^1 \sin(t^n)\,dt \leq \int_0^1 t^n \,dt.
$$
For the second half, make the change of variables $t^n = (\pi/2)^n u$ to get
$$
\int_1^{\pi/2} \sin(t^n)\,dt = \frac{\pi}{2n} \int_{(2/\pi)^{\Large n}}^1 \sin\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\,du.
$$
Integrating by parts yields
$$
\begin{align}
&\frac{\pi}{2n} \int_{(2/\pi)^{\Large n}}^1 \sin\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\,du \\
&\qquad = \left.- \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\right|_{(2/\pi)^{\Large n}}^1 \\
&\qquad\qquad + \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \\
&\qquad = - \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \cos\left[\left(\frac{\pi}{2}\right)^n\right] + \frac{\cos(1)}{n} \\
&\qquad\qquad + \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \tag{$*$}
\end{align}
$$
Finally we use the fact that $|\cos(x)| \leq 1$ to estimate the remaining integral;
$$
\begin{align}
\left| \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \right| &\leq -\left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 u^{1/n-2}\,du \\
&= \left(\frac{\pi}{2}\right)^{n-1} - 1 \\
&< \left(\frac{\pi}{2}\right)^{n-1}.
\end{align}
$$
It follows that
$$
\left| \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \right| < \frac{1}{n}.
$$
Each term in $(*)$ tends to zero as $n \to \infty$, so the result follows.
|
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|
What is the last digit of this expression: $4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$ The expression:
$4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$
I know that the last digit of each of the sumands is 6, but I have trouble proving that. I tried to prove it using induction, but then I realized that I don't have a clue how to write it properly.
Also, because of the fact that the last digit of every sumand is 6, the last digit of the given expression is the last digit of $2013 * 6$ which is 8.
|
$4^2\equiv 6 \mod 10$ and $6^2\equiv 6\mod 10$, which is equivalent to $4(^2)^2=4^4$.
With the $(6 \mod 10)$ relation, we can continue doubling the exponent of $4$ with ease.
$(4^4)^2=4^8\equiv 6^2=6 \mod 10, \dots$
|
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|
Probability Questions! Alex, Bret, and Chloe repeatedly take turns tossing a fair die. Alex begins; Bret always follows Alex; Chloe always follows Bret; Alex always follows Chloe, and so on. Find the probability that Chloe will be the first one to toss a six.
|
The third player in line wins if the first player fails to roll a six ($5/6$) and then the second player in line wins. So $p_3 = \frac{5}{6}p_2$. Similarly, $p_2 = \frac{5}{6}p_1$. Finally, the first player in line wins if he rolls a six immediately ($1/6$) or if he fails to roll a six ($5/6$) and then the third player in line wins:
$$
p_3 = \frac{5}{6}p_2 = \frac{5}{6}\left(\frac{5}{6}p_1\right)=\frac{5}{6}\left(\frac{5}{6}\left(\frac{1}{6}+\frac{5}{6}p_3\right)\right)=\frac{25}{216}+\frac{125}{216}p_3.
$$
Solving this gives
$$
p_3=\frac{25}{216-125}=\frac{25}{91};
$$
then
$$
p_1=\frac{1}{6}+\frac{5}{6}\cdot\frac{25}{91}=\frac{91+125}{6\cdot 91}=\frac{36}{91}
$$
and
$$
p_2=\frac{5}{6}\cdot\frac{36}{91}=\frac{30}{91}.
$$
These satisfy $p_1+p_2+p_3=1$, as they must.
|
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Find the possible values of $k$, if the equation has equal roots. The equation $x^2+5k=kx+x+19$ has equal roots. Find the possible values of $k$.
Um having problem in rearranging the equation;
$x^2+5k-kx-x-19=0$
$x^2+k(5-x)-x-19=0$
What is the next step?
|
Try this
$$ x^2 - (k +1)x + (5k - 19) = (x - a)^2 = x^2 -2a + a^2 = 0$$
So now by inspection , $$5k - 19 = a^2 $$ $$(k+1) = 2a$$
$$ k = 2a -1 $$
$$5(2a -1) - 19 = a^2$$
$$a^2 - 10a + 24 = 0$$
$$(a -6)(a-4) = 0$$
$$a = 6 \ , \ a = 4$$
$$k = 2(6)-1 \ , \ k = 2(4)-1 $$
$$k= 11 \ , \ k = 7 $$
|
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Have I done something wrong in solving the following pair of equations? Question given:
Solve,$3x^2-5y^2-7=0\\3xy-4y^2-2=0$
What I have done so far:
$$
3xy-4y^2-2=0\\\frac{3xy}{y}-\frac{4y2}{y}-\frac{2}{y}=0\\3x-4y-\frac{2}{y}=0\\3x=4y+\frac{2}{y}\\x=\frac{(4y+\frac{2}{y})}{3}
$$By substituting $x=\frac{(4y+\frac{2}{y})}{3}$ in the first equation,$$3x^2-5y^2-7=0\\3(\frac{4y+\frac{2}{y}}{3})^2-5y^2-7=0\\3(\frac{16y^2+\frac{4}{y^2}+16}{9})-5y^2-7=0\\16y^2+\frac{4}{y^2}+16-15y^2-21=0\\y^2+\frac{4}{y^2}-5=0\\y^4-5y^2+4=0$$Let $a=y^2$,$$(y^2)^2-5(y^2)+4=0\\a^2-5a+4=0\\a=\frac{5\pm\sqrt{25-4.1.4}}{2}\\a=\frac{5\pm\sqrt{25-16}}{2}\\a=\frac{5\pm3}{2}\\y^2=\frac{5\pm3}{2}\\y=\pm\sqrt{\frac{5\pm3}{2}}$$
Have I done something wrong when solving these pair of equations ? If so, please correct me.
Best Regards !
|
Hint: Heroic! Let us work more simply. Our two equations are $3x^2-5y^2=7$ and $3xy-4y^2=2$. Multiply the first through by $2$, the second by $7$, and subtract. We get $6x^2-21xy +18y^2=0$, or more simply $2x^2-7xy+6y^2=0$. Now we can solve for $y$ in terms of $x$. One could use the Quadratic Formula, but we get lucky, the thing factors simply. Carry on!
|
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|
Proof review: cauchy sequences Problem statement:
a) Let $(s_n)$ be a sequence such that $\mid s_{n+1} - s_{n} \mid < 2^{-n}$ for all $n \in \mathbb{N}$. Prove that $(s_n)$ is a Cauchy sequence and hence a convergent sequence.
b) Is the result in (a) treu if we only assume that $\mid s_{n+1} - s_{n} \mid < \frac{1}{n}$ for all $n \in \mathbb{N}$?
My proof for (a):
Take any $\epsilon > 0$. Let $N = \log_2{2/\epsilon}$. Take $m, n > N$. Without loss of generality, we may take $m > n$. Let $\delta = m-n$.
$\mid s_{n+1} - s_{n} \mid < \frac{1}{2^n}$
$\mid s_{n+2} - s_{n+1} \mid < \frac{1}{2^{n+1}}$
...
$\mid s_{n+\delta} - s_{n+\delta-1} \mid < \displaystyle{\frac{1}{2^{n+\delta-1}}}$
Adding up the inequalities and using the triangle inequality:
$\mid s_{n+\delta} - s_n \mid = \mid s_{m} - s_{n} \mid < \displaystyle{\frac{1}{2^{n}} + \frac{1}{2^{n+1}} ... \frac{1}{2^{n+\delta-1}}}$
Using the geometric series sum formula:
$\displaystyle{\frac{1}{2^{n}} + \frac{1}{2^{n+1}} ... \frac{1}{2^{n+\delta-1}}} = \frac{1}{2^{n-1}} - \frac{1}{2^{n+\delta-1}} < \frac{1}{2^{n-1}} - \frac{1}{2^n} < \frac{1}{2^n-1} < \frac{1}{2^{N-1}} = \epsilon$
Therefore,
$\mid s_m - s_n \mid < \epsilon$.
For (b), it doesn't work because the argument is that $\sigma \frac{1}{n}$ does not converge, but I'm not sure how to flesh that out.
|
For part (b), try this counter-example:
$$s_n = \sum_{k=1}^{n} \frac{1}{n}.$$
Then
$$|s_{n+1} - s_{n}| = \frac{1}{n+1} < \frac{1}{n},$$
but $s_n \to \infty$.
|
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|
Problem with a simple exercise So I have to solve the equation $$y^2=4\tag{1.9.88 unit 3*}$$
I did this: $$y^2=4 \text{ means } \sqrt{y^2}=\sqrt{4}=>y=2$$
But I have a problem, $y$ can be either negative or positive so I need to do: $$\sqrt{y^2}=|y|=2=>y=2- or- y=-2$$
Is it right?
|
For this type of problem, I always think, "How would it appear on a graph? What are the x-intercepts?" Of course, I do not have time to actually draw it out. But I can think, "Let $y^2$ be $x^2$. How can $x^2=4$ be turned into a function? I can just move $4$ to the left by subtracting $4$ on both sides, and replacing the $0$ with $f(x)$."
So, what are the x-intercepts of $f(x)=x^2-4$? Hopefully you can immediately identify the x-intercepts as $\pm \ 2$. So that means $x=\pm \ 2$ if $x^2=4$. Just replace $x^2$ with $y^2$. If $y^2=4$, then $y=\pm \ 2$.
Another much quicker way to solve the problem is factoring. Subtract $4$ from both sides to get $y^2-4 = 0$. Then factor the equation using difference of squares. $(y+2)(y-2)=0$. The equation now splits into two different cases. You know that if $ab=0$, then either $a=0$, $b=0$, or $a$ and $b$ both equal $0$.
Case 1: $y+2=0$
$y+2=0$
$y=-2$
Case 2: $y-2=0$
$y-2=0$
$y=2$
So, the answer is $y = \pm \ 2$.
Extra Information Below:
You may be wondering, "Why can you just replace the $0$ in $x^2-4=0$ with $f(x)$ in the first method of solving the equation?" You must know the difference between $f(x)=x^2-4$ and $0=x^2-4$. $f(x)=x^2-4$ is the graph of all the points that satisfy the condition that each point on the curve has coordinates $(x, f(x))$. $x^2-4=0$ is a special case of the function $f(x)=x^2-4$. Here, $0=f(x)$. So, what points on the curve are in the form $(x, 0)$? Obviously, the points are $(2, 0)$ and $(-2, 0)$.
You can also think of the equation $x^2-4=0$ as a system of equations like:
$$\begin{cases} f(x)=x^2-4 \\ f(x)=0 \\ \end{cases}$$
The solution to the system will be the intersections of the function $f(x)=x^2-4$ and the line $f(x)=0$. The line $f(x)=0$ is the x-axis, so the question becomes, "Where does $f(x)=x^2-4$ intersect with the x-axis?" Obviously, they intersect at the points $(2, 0)$ and $(-2, 0)$.
|
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Please explain where the $2x$ came from in this cubic identity: The question is to factorize the difference of cube identity $(x + 1)^3 - y^3$
I obviously want to put it in the form of $(a - b)(a^2 + ab + b^2)$
My working out:
$(a - b) = (x + 1) - (y) = (x + 1 - y)$
$(a^2 + ab + b^2) = (x^2 + 1) + (x + 1)(y) + (y^2) = (x^2 + 1 + xy + y + y^2) $
Therefore,
$=(x + 1 - y)(x^2 + 1 + xy + y + y^2)$
Much to my dismay, the correct answer is..
$=(x + 1 - y) (x^2 + 2x + 1 + xy + y + y2)$
Any help would be much appreciated, regards.
|
You made an error while substituting into $(a^2 + ab + b^2)$ .
|
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|
Increasing function on small interval given positive derivative Suppose that $f'(0)>0$. Does it imply that there exists a $\delta > 0 $ such that $f$ is increasing on $[0,\delta]$?
I think this is false and I've been trying to think of a counter example. I was thinking using the function
$f(x) = x^2sin(1/x)$ if $x \neq 0 $ and $f(x) = 0 $ if $x=0$, but I'm not sure if this example works?
I'd appreciate any help!
|
I came across this problem as well. The solution I found dealt with decreasing functions, but the problem is equivalent to the increasing case (just take $-f$ as defined below).
Consider the function $f$ on $[ 0 , 2/\pi ]$:
$$
f(x)=\begin{cases}
x^2\sin^2 \left( \frac 1x \right) -\frac 12 x&, \text{if } x\neq 0 \\
-\frac 12 x &, \text{if } x = 0
\end{cases}
$$
Its derivative is:
$$
f'(x) = \begin{cases}
2x\sin^2 \left(\frac{1}{x} \right) - \sin \left( \frac{2}{x} \right) - \frac 12 &, \text{if } x\neq 0 \\
-\frac 12 &, \text{if } x = 0
\end{cases}
$$
After doing some fiddling with the identities, it is possible to find a sequence $\langle a_n \rangle_{n=1}^\infty$ such that $f'(x)>0$. Letting $\langle a_n \rangle_{n=1}^\infty=\left\langle \left( \frac{5\pi}{8} + n\pi \right)^{-1} \right\rangle_{n=1}^\infty$, then $a_n \to 0 $ as $n \to \infty$. Therefore, in any open interval $[0, \delta]$ for positive $\delta$ one can find an element of this sequence contained within it.
Consider:
\begin{align*}
f'(a_n) &= 2a_n\sin^2 \left(\frac{1}{a_n} \right) - \sin \left( \frac{2}{a_n} \right) - \frac 12 \\
& = \frac{2}{\frac{5\pi}{8} + n\pi}\sin^2 \left(\frac{5\pi}{8} + n\pi \right) - \sin \left( \frac{5\pi}{4} + 2n\pi \right) - \frac 12\\
& = \frac{2}{\frac{5\pi}{8} + n\pi}\sin^2 \left(\frac{5\pi}{8} \right) - \left( \frac{-1}{\sqrt{2}} \right) - \frac 12 \\
& = \frac{2}{\frac{5\pi}{8} + n\pi}\sin^2 \left(\frac{5\pi}{8} \right) + \frac{\sqrt{2}-1}{2} \\
&>0
\end{align*}
But we know that a differentiable function $f$ is decreasing (strictly or not) on an interval $I$ only if $\forall x \in I: f'(x) \le 0$. But since for any interval $[0,\delta]$, we can find an $n$ such that $f'(a_n) > 0$, we see that for no $\delta > 0$ is $f$ decreasing on $[0,\delta]$, just as you had conjectured.
|
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|
Showing that $n$ is pseudoprime to the base $a$ Show that if $n=\frac{a^{2p}-1}{a^2-1},$ where $a$ is an integer, $a>1$, and $p$ is an odd prime not dividing $a(a^2-1)$, then $n$ is pseudoprime to the base $a$.
Let $n=\frac{a^{2p}-1}{a^2-1}=1+a^2+(a^2)^2+(a^2)^3+\cdots$. Then $n-1=\frac{a^{2p}-1}{a^2-1}-1=\frac{a^{2p}-a^2}{a^2-1}=a^2\frac{(a^2)^{p-1}-1}{a^2-1}.$
I have kind of hit a roadblock with this proof. I'm stuck trying to show that $n-1$ is even and that $p | (a^{2p}-1)\equiv 0 \mod p$. Is there some obvious steps that I'm missing
|
Write
$$\begin{align}
n-1 &= \frac{a^{2p}-a^2}{a^2-1}\\
&= \frac{(a^p-a)(a^p+a)}{a^2-1}\\
&= a\frac{a^{p-1}-1}{a^2-1}(a^p+a).
\end{align}$$
Since $p$ is odd, $a^2-1$ divides $a^{p-1}-1$, so we have written $n-1$ as a product of three integers, and the last one, $a^p+a$ is even, whether $a$ is even or odd. Hence $n-1$ is even.
Also, since $p\nmid a$, we have $a^{p-1}\equiv 1\pmod{p}$, and since by assumption $p\nmid (a^2-1)$, we further have
$$p \mid \frac{a^{p-1}-1}{a^2-1}.$$
Together, $2p\mid n-1$, and so $(a^{2p}-1) \mid (a^{n-1}-1)$, which implies $n\mid (a^{n-1}-1)$.
We do not have $p\mid (a^{2p}-1)$, since $a^{2p} = a^{2(p-1)+2} \equiv a^2 \not\equiv 1\pmod{p}$, but that is not anything we would need.
|
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|
Express in terms of $x$ and $y$ when the values of $x$ and $y$ are given. Given,
$x=1+3a+6a^2+10a^3+\ldots$
$y=1+4b+10b^2+20b^3+\ldots$
$s=1+3ab+5(ab)^2+7(ab)^3+\ldots$
Express $s$ in terms of $x$ and $y$.
My work:
I could see how the first sequence works, but could not find how the second sequence works, until I wrote down the Pascal's Triangle. Then I realised that,
$x=1+{3 \choose 1}a+{4 \choose 2}a^2+{5 \choose 3}a^3+\ldots$
$y=1+{4 \choose 1}b+{5 \choose 2}b^2+{6 \choose 3}b^3+\ldots$
And $s$ sequence was easy to see. It was just an arithmetico-geometric progression.
$s~~~~~~~=1+3ab+5(ab)^2+7(ab)^3+9(ab)^4+\ldots$
$s(ab)=~\cdot+~~ab+3(ab)^2+5(ab)^3+7(ab)^4+\ldots$
$s(1-ab)=1+2\{ab+(ab)^2+(ab)^3+\ldots\}$
But, I do not see how does that help me. Please help.
|
Assume, $|a|,|b|<1$
You can find s by using formula for infinite GP in the last step.
$s=\frac{1+\frac{2ab}{1-ab}}{1-ab}=\frac{1+ab}{(1-ab)^2}$
Also note that taylor series for $\frac 1 {(1-x)^n}$ is :
$ 1+{n\choose 1} x +{n+1\choose2}x^2 + {n+2\choose3}x^3... $
Hence $x$ stands for $n=3$ and $y$ for $n=4$
Now find $x$ and $y$ in terms of $a$ and $b$ and put in $s$.
|
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|
Prove that any palindrome with an even number of digits is divisible by 11 Confusing myself here, need some clarification..
First, we consider the palindrome $abccba$. We can see this can be written as
$$a(10^5 + 10^0) + b(10^4 + 10^1) + c(10^3 + 10^2) = a(10^5 + 1) + 10b(10^3 + 1) + 100c(10 + 1)$$ So essentially we see that all palindromes of even digits can be written in the form $x(10^{2k+1} + 1)$, i.e. we must show that any number of the form $(10^{2k+1} + 1)$ is divisible by $11$.
Base case: $10^{2(0)+1} + 1 = 10 + 1 = 11$, which is clearly divisible by $11$.
Induction hypothesis: Assume that $(10^{2k+1} + 1)$ is divisible by $11$, we work to show that $(10^{2(k+1)+1} + 1)$ is divisible by $11$. That is, $(10^{2k+3} + 1) = (10^2\cdot10^{2k+1} + 1) = \dots$
Where am I going wrong?
|
Hint: Notice that $10^k=(-1)^k \pmod{11}$. So, for $k$ odd and $k'$ even, $10^{k} + 10^{k'}=0\pmod{11}$.
|
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|
If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
|
since
$$\sum_{cyc}\dfrac{a}{b+c^2}=\sum_{cyc}\dfrac{a^2}{ab+ac^2}$$
Use Cauchy-Schwarz inequality,we have
$$\left(\sum_{cyc}\dfrac{a^2}{ab+ac^2}\right)\cdot\sum_{cyc}(ab+ac^2)\ge\left(\sum_{cyc}a\right)^2=1$$
so we only prove this following inequality
$$\dfrac{1}{\sum_{cyc}(ab+ac^2)}\ge\dfrac{9}{4}$$
$$\Longleftrightarrow 4\ge9\sum_{cyc}ab+9\sum_{cyc}ac^2$$
$$\Longleftrightarrow 4(\sum_{cyc}a)^2\ge 9\sum_{cyc}ab+9\sum_{cyc}ac^2$$
$$\Longleftrightarrow 4\sum_{cyc}a^2\ge \sum_{cyc}ab+9\sum_{cyc}ac^2$$
since
$$\sum_{cyc}a^2\ge \sum_{cyc}ab$$
so we only prove this
$$3\sum_{cyc}a^2\ge 9\sum_{cyc}ac^2$$
$$\Longleftrightarrow \sum_{cyc}a^2\ge3\sum_{cyc}ac^2$$
$$\Longleftrightarrow \sum_{cyc}a\cdot\sum_{cyc}a^2\ge3\sum_{cyc}ac^2$$
$$\Longleftrightarrow \sum_{cyc}a^3+\sum_{cyc}ab^2\ge 2\sum_{cyc}ac^2$$
since Use AM-GM inequality
$$\sum_{cyc}a^3+\sum_{cyc}ab^2=\sum_{cyc}(a^3+ab^2)\ge\sum_{cyc}(2a^2b)=2\sum_{cyc}ac^2$$
By done!
|
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|
How do I prove this trigonmetric identity? I need to prove that the following identity is true:
$$
\frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x
$$
This isn't homework; just a practice exercise. But I keep getting stuck! Thanks much.
|
Let's start with our original expression:
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}$$
We need to prove that:
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x$$
Step 1: Recall that $\tan x = \dfrac{\sin x}{\cos x}$. This means that $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\dfrac{\cos^2 x - \sin^2 x}{\left(1-\dfrac{\sin^2 x}{\cos^2 x}\right)}$$
Step 2: Simplify the denominator by letting $1=\dfrac{\cos^2 x}{\cos^2 x}$.
$$\dfrac{\cos^2 x - \sin^2 x}{\left(1-\dfrac{\sin^2 x}{\cos^2 x}\right)}=\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x}{\cos^2 x}-\dfrac{\sin^2 x}{\cos^2 x}\right)}$$
$$\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x}{\cos^2 x}-\dfrac{\sin^2 x}{\cos^2 x}\right)}=\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x}\right)}$$
Step 3: Recall that $\dfrac{a}{c}=a\left(\dfrac{1}{c}\right)$.
$$\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x}\right)}=\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)$$
$$\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)$$
$$\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x$$
$$\displaystyle \boxed{\therefore \dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x}$$
|
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|
Can this be converted into a polynomial equation for $x$? I came upon this monster equation while fiddling with the area of pentagons:
$$\sqrt{(a+b+c-x)(a+b-c+x)(a+c-b+x)(b+c-a+x)}+\sqrt{(c+d+x)(c+d-x)(c-d+x)(d-c+x)}=4T$$
Where $a,b,c,d,e,T$ are known quantities. So my question is:
*
*Can this be converted into a polynomial equation for $x$?
*If so how can we do it, and what would be its degree?
Any help is appreciated :)
|
If you have an equation of the form $\sqrt a + \sqrt b = c$ then by taking squares we get $2\sqrt {ab} = c^2-a-b$, and then $4ab = (c^2-a-b)^2$.
You can also develop the product in the equation $(\sqrt a + \sqrt b - c)(\sqrt a - \sqrt b - c)(- \sqrt a + \sqrt b - c)(- \sqrt a - \sqrt b - c) = 0$, which gives you (again) $4ab = (c^2-a-b)^2$.
|
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|
Integrate $(x^2+1)^\frac{1}{3}$ I tried many times to integrate this function using integration by parts, substitution as $x^2+1=t$ without any conclusion... is it integrable? If it is integrable, how can I integrate it?
Thank you in advance
|
Let $u=(x^2+1)^\frac{1}{3}$ ,
Then $x=(u^3-1)^\frac{1}{2}$
$dx=\dfrac{3u^2}{2(u^3-1)^\frac{1}{2}}du$
$\therefore\int(x^2+1)^\frac{1}{3}~dx=\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
Case $1$: $|u|\leq1$ , i.e. $\left|(x^2+1)^\frac{1}{3}\right|\leq1$
Then $\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
$=\int\dfrac{3u^3}{2i(1-u^3)^\frac{1}{2}}du$
$=-\int\dfrac{3u^3i}{2}\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{3n}}{4^n(n!)^2}du$
$=-\int\sum\limits_{n=0}^\infty\dfrac{3i(2n)!u^{3n+3}}{2^{2n+1}(n!)^2}du$
$=-\sum\limits_{n=0}^\infty\dfrac{3i(2n)!u^{3n+4}}{2^{2n+1}(n!)^2(3n+4)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{3i(2n)!(x^2+1)^{n+\frac{4}{3}}}{2^{2n+1}(n!)^2(3n+4)}+C$
Case $2$: $|u|\geq1$ , i.e. $\left|(x^2+1)^\frac{1}{3}\right|\geq1$
Then $\int\dfrac{3u^3}{2(u^3-1)^\frac{1}{2}}du$
$=\int\dfrac{3u^3}{2u^\frac{3}{2}(1-u^{-3})^\frac{1}{2}}du$
$=\int\dfrac{3u^\frac{3}{2}}{2}\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{-3n}}{4^n(n!)^2}du$
$=\int\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{\frac{3}{2}-3n}}{2^{2n+1}(n!)^2}du$
$=\sum\limits_{n=0}^\infty\dfrac{3(2n)!u^{\frac{5}{2}-3n}}{2^{2n+1}(n!)^2\left(\dfrac{5}{2}-3n\right)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{3(2n)!}{4^n(n!)^2(6n-5)u^{3n-\frac{5}{2}}}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{3(2n)!}{4^n(n!)^2(6n-5)(x^2+1)^{n-\frac{5}{6}}}+C$
|
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|
Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\right)}{1+x^2}}\,dx$$
|
First Approach - Using Euler sums
We begin by finding the Maclaurin series expansion for $\arctan^2 x$.
Since
$$\arctan x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1}, \qquad |x| < 1,$$
applying the Cauchy product to the product between two inverse tangent functions leads to:
\begin{align}
\arctan^2 x &= \arctan x \cdot \arctan x\\
&= \left (\sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1} \right ) \cdot \left (\sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1} \right )\\
&= \sum_{n = 1} \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) x^{2n}, \qquad |x| \leqslant 1.
\end{align}
Here $H_n$ is the $n$th harmonic number.
Making use of this result, the integral becomes
\begin{align}
I &= \int_0^1 \arctan^2 x \, dx\\
&= \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) \int_0^1 x^{2n} \, dx\\
&= \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n (2n + 1)} \left (H_{2n} - \frac{1}{2} H_n \right )\\
&= \sum_{n = 1}^\infty (-1)^{n + 1} \left (\frac{1}{n} - \frac{2}{2n + 1} \right ) \left (H_{2n} - \frac{1}{2} H_n \right )\\
&= \frac{1}{2} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} - \sum_{n = 1}^\infty \frac{(-1)^n H_n}{2n + 1} - \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n} + 2 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{2n+1}.\tag1
\end{align}
Values for each of the above four alternating Euler sums can be found here. Theirs values are:
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} = \frac{1}{2} \ln^2 2 - \frac{\pi^2}{12}, \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_n}{2n + 1} = \mathbf{G} - \frac{\pi}{2} \ln 2,$$
$$\sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n} = \frac{1}{4} \ln^2 2 - \frac{5\pi^2}{48}, \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{2n + 1} = -\frac{\pi}{8} \ln 2.$$
On substituting into (1) one obtains
$$\int_0^1 \arctan^2 x \, dx = \frac{\pi^2}{16} + \frac{\pi}{4} \ln 2 - \mathbf{G},$$
as desired.
Second Approach - Integral of a function and its inverse
Let $f(x) = \arctan^2 x$. As $f$ is a monotonically increasing function on the interval $[0,1]$ an inverse given by $f^{-1} (x) = \tan (\sqrt{x})$ exists on the interval $[0,\frac{\pi^2}{4})$. Making use of the result:
$$\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1} (x) \, dx = b \cdot f(b) - a \cdot f(a),$$
noting that for our $f$ we have $f(a) = f(0) = 0$ and $f(b) = f(1) = \frac{\pi^2}{16}$, one sees that
$$\int_0^1 \arctan^2 x \, dx + \int_0^{\frac{\pi^2}{16}} \tan (\sqrt{x}) \, dx = \frac{\pi^2}{16}.$$
Thus
\begin{align}
\int_0^1 \arctan^2 x \, dx &= \frac{\pi^2}{16} - \underbrace{\int_0^{\frac{\pi^2}{16}} \tan (\sqrt{x}) \, dx}_{x \, \mapsto \, x^2}\\
&= \frac{\pi^2}{16} - 2 \int_0^{\frac{\pi}{4}} x \tan x \, dx\\
&= \frac{\pi^2}{16} - 2 \left (-x \ln (\cos x) \Big{|}_0^{\frac{\pi}{4}} + \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx \right ) \qquad (\text{by parts})\\
&= \frac{\pi^2}{16} - \frac{\pi}{4} \ln 2 - 2 \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx\\
&= \frac{\pi^2}{16} - \frac{\pi}{4} \ln 2 - 2 \left (\frac{\mathbf{G}}{2} - \frac{\pi}{4} \ln 2 \right )\tag1\\
&= \frac{\pi^2}{16} + \frac{\pi}{4} \ln 2 - \mathbf{G},
\end{align}
as expected. Note that in ($1$) we have made use of the result
$$\int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx = \frac{\mathbf{G}}{2} - \frac{\pi}{4} \ln 2,$$
several proofs for which can be found here.
|
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|
Cramer's Rule Question Use Cramer's rule to solve this system for z:
$$2x+y+z=1$$
$$3x+z=4$$
$$x-y-z=2$$
so my work is:
$$\frac{\left|\begin{matrix}
2 & 1 & 1\\
3 & 0 & 4\\
1 & -1 & 2
\end{matrix}\right|}{\left|\begin{matrix}
2 & 1 & 1\\
3 & 0 & 1\\
1 & -1 & 1
\end{matrix}\right|}$$
which gives $\frac{3}{-3}$ or $-1$. The solution is $1$, can someone tell me what I am doing wrong?
|
The determinant in the denominator is incorrect. It should be
$$\left|\begin{matrix}
2 & 1 & 1\\
3 & 0 & 1\\
1 & -1 & {\color{red} {-1}}
\end{matrix}\right|$$
which evaluates to (according to Sarrus' Rule):
$$(2\cdot0\cdot-1) + (1\cdot1\cdot1) + (1 \cdot 3 \cdot -1) - (1\cdot0\cdot1) - (1\cdot3\cdot-1) - (1\cdot2\cdot-1)\\
= 0 +1 - 3 - 0 +3 + 2 \\
= 3$$
|
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|
Solving an equality in 2 variables I need to prove that
$$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$
if $a+b = 1$ and $a b \le 1/4$
I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get
$a b \le 4$ in the inequality ?
|
$$
\begin{align}
\hspace{-1cm}\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2
&\ge2\left(a+\frac1a\right)\left(b+\frac1b\right)\tag{1}\\
&=\frac2{ab}\left(a^2+1\right)\left(b^2+1\right)\\
&\ge8\left(a^2+1\right)\left(b^2+1\right)\tag{2}\\
&=8\small\left(\left(a-\tfrac12\right)^2+\left(a-\tfrac12\right)+\tfrac54\right)\left(\left(a-\tfrac12\right)^2-\left(a-\tfrac12\right)+\tfrac54\right)\\
&=8\left[\left(\left(a-\tfrac12\right)^2+\tfrac54\right)^2-\left(a-\tfrac12\right)^2\right]\\
&=8\left[\left(a-\tfrac12\right)^4+\tfrac32\left(a-\tfrac12\right)^2+\tfrac{25}{16}\right]\\[4pt]
&\ge\frac{25}{2}\tag{3}
\end{align}
$$
Inequalities:
$(1)$: $(x-y)^2\ge0\implies x^2+y^2\ge2xy$
$(2)$: $ab\le\tfrac14$
$(3)$: $x^2\ge0$
|
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|
Find the integral : $\int\frac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$ Find the integral : $\int\dfrac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$
Please guide which substitution fits in this I am not getting any clue on this .. thanks..
|
After using the substitution posted you'll have to do a bit of work but should get this at your final answer
$$2\sqrt{x}+\ln(\sqrt{x}-1)-\ln(\sqrt{x}+1)-\ln(x-1)-3x^\frac{1}{3}+\ln(x^\frac{2}{3}+x^\frac{1}{3}+1)-2\ln(x^\frac{1}{3}-1)-\ln(x^\frac{1}{3}+x^\frac{1}{6}+1)+2\ln(x^\frac{1}{6}-1)+\ln(x^\frac{1}{3}-x^\frac{1}{6}+1)-2\ln(x^\frac{1}{6}+1)+6x^\frac{1}{6}$$
|
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|
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?
|
Substitute $t= \sin2x$
\begin{align}
&\int \cos 2x\ln \frac{\cos x+\sin x}{\cos x-\sin x}dx\\
=&\>\frac12\int \tanh^{-1}t\>dt\overset{ibp}=\frac12
t \tanh^{-1}t+\frac14\ln(1-t^2)+C
\end{align}
|
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|
Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}\ge{12}$$
|
Applying AM-GM on $2ab$ and $a^2+b^2$ we find $2ab(a^2+b^2) \leq \left(\frac{(a+b)^2}{2} \right)^2$ or $a^2+b^2 \leq \frac{(a+b)^4}{8ab}$. Plugging this in, we find
$$
\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{(a+b)^3}{\sqrt[3]{2(a+b)\frac{(a+b)^4}{8ab}}} = \frac{(a+b)^3\sqrt[3]{4ab}}{\sqrt[3]{(a+b)^5}} = 2^{2/3} (a+b)^{4/3} (ab)^{1/3}.
$$
Applying AM-GM on $a$ and $b$ we find $a+b \geq 2\sqrt{ab}$. Hence
$$
\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq 2^{2/3} (a+b)^{4/3} (ab)^{1/3} \geq 2^{2/3} (2\sqrt{ab})^{4/3} (ab)^{1/3} = 4ab.
$$
It follows that the given expression is at least $4(ab+bc+ca) = 24 \geq 12$.
|
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|
Find Max value of this expression: $P=(x-2yz)(y-2zx)(z-2xy)$ Let $x$, $y$ and $z$ be positives and satisfying $x^2+y^2+z^2=2xyz+1$ .
Find a maximum of this expression:
$$P=(x-2yz)(y-2zx)(z-2xy).$$
|
Set $y=x$ and $z=1$. Then $x^2+y^2+z^2=2x^2+1=2xyz+1$ and
$$(x-2yz)(y-2zx)(z-2xy)=(-x)(-x)(1-2x^2)=x^2(1-2x^2)$$
which clearly has no minimum...
|
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|
standard Taylor series using substitution Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?
Determine a range of validity for this series.
|
Hint
Start with Taylor expansion $$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+O\left(y^6\right)$$ Rise to the third power to obtain $$\frac{1}{(1+y)^3}=1-3 y+6 y^2-10 y^3+15 y^4-21 y^5+O\left(y^6\right)$$ Now, replace $y$ by $\frac{4 x}{5}$ to get your result.
I am sure that you can take from here.
|
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|
Analysis problem from Romanian Contest - 2 sequences which forms another one Let $a,b$ be 2 real numbers, and the sequences $(a_n)_{n \geq 1}, (b_n)_{n \geq 1}$ defined by $a_{1}=a$, $b_{1}=b$, $a^2+b^2 <1$ and
\begin{cases}
a_{n+1}=\frac{1}{2}\left(a_{n}^{2}-\frac{b_{n}^{2}}{n^{2}}\right), \mbox{ }(\forall) n \geq 1\\
b_{n+1}=-\left(1+\frac{1}{n}\right)a_{n}b_{n}, \mbox{ }(\forall) n \geq 1.
\end{cases}
Prove that the sequence $\displaystyle x_{n}=\frac{a_{n}\cdot b_{n}}{n}, (\forall) n \geq 1$ is convergent and calculate $\lim\limits_{n \to \infty}{x_{n}}.$
Seems hard... Thanks for your help!
|
i try to find the regular of this sequence :
$a_{2}=\frac{1}{2}$$(a^2-b^2)$$,$$b_{2}=-2ab$
$a_{3}=\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4})$$,$$b_{3}=(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{3}{2})$
$a_{4}=\frac{1}{2}((\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4}))^{2}$$-(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{3}{2})^{2}\cdot\frac{1}{9})$$,$
$b_{4}=\frac{1}{2}$$((\frac{1}{2}$$(a^2-b^2))^2-\frac{(-2ab) ^2}{4})$$\cdot$$(\frac{1}{2}$$(a^2-b^2)(-2ab))(-\frac{4}{3})\cdot(-\frac{3}{2})$
$......$
it seems that $b_{n}$ is always less than $1$
then, $limx_{n}=lim\frac{a_{n}\cdot b_{n}}{n}$$\longrightarrow$$\frac{b_{n+1}}{n}$
if $b_{n}\le1$ , it is obviously that $x_{n}$ is convergent, but i am not sure about how to use mathematical deduction to prove $b_{n}$ is always less than $1$ ?
i guess that, the hint is :
$ab\ge-b^{2}-a_{2}-b_{2}$
$a_{2}b_{2}\ge-b_{2}^{2}/4-a_{3}-b_{3}$
$a_{3}b_{3}\ge-b_{3}^{2}/9-a_{4}-b_{4}$
$......$
here i use the deduce method to find :
$a_{n}b_{n}\le\frac{1}{2^{2n-1}}$ $,$ $b_{n}\le{\frac{n}{2^{n}}}$$\Longrightarrow$$a_{n+1}b_{n+1}\le\frac{1}{2^{2n+1}}$ $,$ $b_{n+1}\le{\frac{n+1}{2^{n+1}}}$
$\Longrightarrow$$a_{n}+b_{n}\le{a_{n-1}+b_{n-1}}\le{a_{n-2}+b_{n-2}}\le......\le1$$\Longrightarrow$$b_{n}^{2}\le{b_{n-1}^{2}}\le{b_{n-2}^{2}}\le......\le{b^{2}}$
since, $\frac{n+1}{2^{n}}<1$
is it helpful? thank you !
|
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|
The MacLaurin series of $e^{\frac{x^2}{2}}$ Could someone please explain to me how you derive the MacLaurin series for $e^{ \frac{x^2}{2}}$?
I understand how it is derived from the MacLaurin series for $e^x$ where it is
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum_{n=1}^\infty\frac{x^{2n}}{2^nn!}$
and you just substitute $\frac{x^2}{2}$ for $x$ but when I try to do it using differentiation I get
$e^{\frac{x^2}{2}} = 1 + \frac{x^2}{2!} + \frac{x^4}{3!} + ... = \sum_{n=1}^\infty\frac{x^2n}{n!}$ which is not the correct answer.
Any help/comments would be appreciated.
|
$$e^x=\sum^\infty_{k=0}\dfrac{x^k}{k!}\\
\implies e^{x^2/2}=\sum^\infty_{k=0}\dfrac{(x^2/2)^{2}}{k!}=\sum^\infty_{k=0}\dfrac{x^{2k}}{2^kk!}$$
Differentiate:
$$e^{x^2/2}|_{x=0}=1\\
e^{x^2/2}x|_{x=0}=0\\
(e^{x^2/2}+x^2e^{x^2/2})|_{x=0}=1$$
E.t.c. Every differential $\dfrac{d^{2n}}{dx^{2n}}e^{x^2/2}=1$, and every differential $\dfrac{d^{2n+1}}{dx^{2n+1}}e^{x^2/2}=0$. Thus,
$$e^{x^2/2}=\sum^\infty_{k=0}\dfrac{f^{(k)}(0)}{k!}(x)^k=\sum^\infty_{k=0}\dfrac{x^{2k}}{2^kk!}$$
Both are the same.
|
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|
Find constants of function I have this equality :
$$f(x)=\frac{9}{(x-1)(x+2)^2}$$
I am required to find the constants A, B and C so that,
$$f(x) = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^{2}} $$
How do we go about solving such a question?
I am not sure on how to solve such questions. Approach and Hints to solve these kinds of questions are welcomed. :)
Thank you!
Edit:
Wow, saw all the answers! Didn't know that there were a lot of different ways to solve this question. Math is such a fascinating thing!
|
From
\begin{equation}
\frac{9}{(x-1)(x+2)^2} = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^2}
\end{equation}
it follows, after multiplying each side with $(x-1)(x+2)^2$:
\begin{equation}
9=A(x+2)^2+B(x+2)(x-1)+C(x-1)
\end{equation}
which, after regrouping, gives
\begin{equation}
9=(A+B)x^2 + (4A+B+C)x + (4A-2B-C)
\end{equation}
There are two polynomials, one on each side of this equation, and two polynomials $P_a(x)=\sum a_nx^n$ and $P_b(x)=\sum b_nx^n$ are equal if and only if $a_n=b_n \forall n$.
Therefore, three equations follow from the previous one:
\begin{eqnarray}
A+B &=& 0 \\
4A+B+C &=& 0 \\
4A-2B-C &=& 9
\end{eqnarray}
Solving this system of equations gives the solution: $A=1$, $B=-1$ and $C=-3$.
|
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|
How to prove this trig identity? If $A+B+C=\pi$ then prove:$$\sin^2A+\sin^2B+\sin^2C=2-2\cos A\cos B\cos C$$
I am completely lost on this, please help.
|
$A+B+C = \pi \implies A + B = \pi - C \implies \sin(A+B)=\sin(C) \land \cos(A+B)=-\cos(C)$
$\sin^2(A) + \sin^2(B) + \sin^2(C) \\ = \sin^2(A) + 1-\cos^2(B) + 1 − \cos^2(A+B) \\ =2+(\sin^2(A)-\cos^2(B))-\cos^2(A+B) \\ = 2+\cos(B-A)\cos(A+B)-\cos(A+B)\cos(A+B) \\ = 2+(\cos(B-A)+\cos(A+B))\cos(A+B) \\ = 2 - 2\cos(A)\cos(B)\cos(C)$
|
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|
Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$. Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.
I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.
|
[I'm assuming by function you mean cubic polynomial, otherwise this is simple.]
Observe that the first three equations are satisfied by the function $2x$.
Hence, the polynomial $f(x) - 2x$ has roots $x=1, 2, 3$, or that $f(x) = 2x + A(x-1)(x-2)(x-3)$.
Finally, set $x=4$, we get $\pi = 8 + A\times 3\times 2 \times 1$, or that $ A = \frac{\pi-8}{6}$. Hence,
$$f(x) = 2x + \frac{\pi - 8} {6} ( x-1)(x-2)(x-3).$$
Lagrange Interpolation would be the general method to approach problems like this. This solution exploits a property from the selected values.
|
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|
Using EigenValues to form a diagonal matrix After going through my Linear algebra note, I know if for any matrix $A$, we find the eigenvalues and eigenvectors , we can construct a matrix P, such that $P^{-1}AP$ is a diagonal matrix.
Now for this matrix for example
$\left( \begin{array}{ccc}
1 & -2 \\
1 & 4 \end{array} \right)$
I found that that the EigenValues are $3,2$ and the eigenvectors are;
$\left( \begin{array}{ccc}
1 \\
-1 \end{array} \right)$
, $\left( \begin{array}{ccc}
1 \\
-2 \end{array} \right)$
so now using these eigenvectors as columns, I construct matrix $P$;
$\left( \begin{array}{ccc}
1 & 1 \\
-1 & -2 \end{array} \right)$ and find that $P^{-1}$ is $\left( \begin{array}{ccc}
-1 & -1 \\
2 & 1 \end{array} \right)$.
So going by the examples online and my lecture notes $P^{-1}AP$ should give me a diagonal matrix with the diagonal entries being the eigenvalues.
Like this $\left( \begin{array}{ccc}
2 & 0 \\
0 & 3 \end{array} \right)$.
But when I multiply the matrices Ive got :
$P^{-1}AP$=$\left( \begin{array}{ccc}
-1 & -1 \\
2 & 1 \end{array} \right)$
$\left( \begin{array}{ccc}
1 & -2 \\
1 & 4 \end{array} \right)$ $\left( \begin{array}{ccc}
1 & 1 \\
-1 & -2 \end{array} \right)$ = $\left( \begin{array}{ccc}
2 & 0 \\
3 & 3 \end{array} \right)$.
Can anyone guide me to where I went wrong ?
|
It was noted in the comments that the eigenvector for $2$ is wrong and the OP needs further help in finding an eigenvector for $2$.
Let $\begin{pmatrix} v_1\\ v_2\end{pmatrix}$ be an eigenvector for $2$. It holds that $\begin{pmatrix} 1 & -2\\ 1 & 4\end{pmatrix}\begin{pmatrix} v_1\\ v_2\end{pmatrix}=2\begin{pmatrix} v_1\\ v_2\end{pmatrix}$ and this is equivalent to $\begin{cases} v_1-2v_2&=2v_1\\ v_1+4v_2&=2v_2\end{cases}$, which in turn is equivalent to $v_1+2v_2=0$ (or $v_1=-2v_2$). Since $v$ is an eigenvector, it is not null, so take for instance $v_2=1$ to get $\begin{pmatrix} -2\\ 1\end{pmatrix}$.
|
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|
How find this $\frac{3x^3+125y^3}{x-y}$ minimum value
let $x>y>0$,and such $xy=1$, find follow minimum of the value
$$\dfrac{3x^3+125y^3}{x-y}$$
My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequality.But I can't.
Thank you very much
|
continuation:
let $x-y=t$, $xy=1$, we get:
$$x=\frac{1}{2}t+\frac{1}{2}\sqrt{t^2+4},\\y=-\frac{1}{2}t+\frac{1}{2}\sqrt{t^2+4}$$
then we have
$$f=\frac{3x^3+125y^3}{x-y}\\=\frac{-61t^3+64t^2\sqrt{t^2+4}-183t+64\sqrt{t^2+4}}{t}$$
we calculate the differential:
$$g=\frac{df}{dt}=-\frac{2(61t^3\sqrt{t^2+4}-64t^4-128t^2+128)}{t^2\sqrt{t^2+4}}$$
let $g=0$, we get the good solution:
$$t=\frac{4}{5}\sqrt{5}$$
let $t=\frac{4}{5}\sqrt{5}$, we have
$$f(\frac{4}{5}\sqrt{5})=25$$
|
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|
Proof that $y^2=x^3+x$ has a unique integer solution Prove that the equation $y^2=x^3+x$ has only one integer solution, namely $x=y=0$.
|
Write $y^2=x(x^2+1)$. If $x=0$ then $y=0$ and if $y=0$ then $x=0$. We can then assume that $xy\not=0$, and replace $x$ with $-x$ or $y$ with $-y$ so that $x>0$, $y>0$. Observe that $x$ and $x^2+1$ are coprime, so since $x(x^2+1)$ is a square, so is $x$ and so is $x^2+1$. This is impossible because $x^2+1=a^2$ is impossible as $x>0$ and $x,a\in \mathbb{Z}$.
|
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|
how to factor $x^4+2x^3+4x^2+3x+2$ I'm trying my hand on these types of expressions.
How to factorize $x^4+2x^3+4x^2+3x+2$ into two (or more) polynomials with rational coefficients. please write step by step solution.
|
I think the best way is by intuition.
$x^4+2x^3+4x^2+3x+2$
$=(x^4+x^3+x^2)+(x^3+x^2+x)+(2x^2+2x+2)$
$=x^2(x^2+x+1)+x(x^2+x+1)+2(x^2+x+1)$
$=(x^2+x+1)(x^2+x+2)$
|
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|
have trouble with this limit question
a) By considering the areas of the triangle OAD, the sector OAC and the triangle OBC,
show that
$(\cos \theta)(\sin \theta) < \theta < \frac{\sin\theta}{\cos\theta}$
I find out:
Area of OAD=$\frac{1}{2}OD\cdot AD \cdot \sin \theta$
Area of OAC=$\frac{1}{2}OC^2 \theta$
Area of OBC=$\frac{1}{2}OC\cdot BC \cdot\sin\theta$
Now I'm stuck at how to apply this to prove
How to prove?
(b) Use (a) and the Squeeze Theorem to show that
$\displaystyle\lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}= 1$
|
The geometry has been discussed in comments. So we have
$$\frac{1}{2}\cos\theta\lt \frac{1}{2}\theta.$$
Multiply through by $2$. We get $\cos\theta\sin\theta\lt \theta$.
Divide both sides by $\theta\cos\theta$. We get
$$\frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{1}$$
We also got from the geometry that
$$\frac{1}{2}\theta\lt \frac{1}{2}\frac{\sin \theta}{\cos\theta}.$$ Multiplying by $2$, and rearranging, we get $\theta\cos\theta\lt \sin\theta$. Divide through by $\theta$. We get
$$\cos\theta\lt\frac{\sin\theta}{\theta}.\tag{2}$$
Now putting (1) and (2) together, we get
$$\cos\theta\lt \frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{3}$$
Finally, let $\theta\to 0$. The left wall and the right wall in (3) both approach $1$, and poor $\frac{\sin\theta}{\theta}$ is therefore forced to approach $1$.
|
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"url": "https://math.stackexchange.com/questions/738858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Can we solve this using stars and bars? The number of ways of distributing 12 identical oranges among 4 children so that every child gets at least one and no child more than 4 is 31.
My try: First of all give each child 1 orange and we are left with 8 oranges.
Then
3 3 2 0
3 3 1 1
3 2 2 1
2 2 2 2
and permuting each, we get total 31. But can we solve this using stars and bars?
|
As SandeepThilakan said in the comments above, the coefficient of $x^{12}$ in the expansion of the polynomial $(x^1+x^2+x^3+x^4)^4$ counts the way that between 1 to 4 objects distributed to 4 groups gives a total of 12 objects.
Using the binomial expansion theorem:
$$(x+x^2+x^3+x^4)^4 \\ = ((x+x^2)+x^2(x+x^2))^4 \\ = x^4(1+x)^4(1+x^2)^4 \\ = x^4 \left(\sum\limits_{i=0}^4 {4\choose i}x^i \right)\left(\sum\limits_{j=0}^4 {4\choose j}x^{2j} \right) \\= x^4 \sum\limits_{i=0}^4 \sum\limits_{j=0}^4 {4 \choose i}{4 \choose j} x^{i+2j}$$
So the coefficients of $x^{12}$ must be $${4 \choose 0}{4 \choose 4}+{4 \choose 2}{4 \choose 3}+{4 \choose 4}{4 \choose 2} = 31$$
|
{
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"url": "https://math.stackexchange.com/questions/739198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How do I find the $x$ intercepts for $-x^2-3x+3$ How do I find the $x$ intercepts? $-x^2-3x+3$?
I converted the function into vertex form but I am stuck at $3/4= -(x^2+1.5)$.
Can someone give me and idea what I can do?
|
Alternatively, you can complete the square.
$x^2 + 3x - 3 = (x + \frac{3}{2})^2 - 3 - \frac{9}{4} = 0$. So, $(x + \frac{3}{2})^2 = \frac{21}{4} \Rightarrow x = -\frac{3}{2} \pm \frac {\sqrt{21}}2.$
Thus the x-intercepts are $(\frac{-3+\sqrt{21}}{2},0)$ and $(\frac{-3-\sqrt{21}}{2},0)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/740161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
|
Looking at $(x+1)^4 + (x-1)^4=16$, I recall that $16=2^4$. So I look at the equation trying to see if I could get a $2^4$ in there. This sugguests that we guess $x=1$ to make the left term a power of $2$.
Plugging in $x=1$,
$$ (1+1)^4 + (1-1)^4 = 2^4 + 0^4 = 16 $$
So $x=1$ is a solution.
This problem has a symmetry property. Notice that sending $x\rightarrow -x$ doesn't change the expression on the left.
$$ (x+1)^4 + (x-1)^4 \rightarrow (-x+1)^4 + (-x-1)^4 = (x-1)^4+(x+1)^4$$
So we know that $x=-1$ will also be a solution.
Some thought should convince you the only possible solutions are between $-1$ and $1$ since any large values of $x$ will immediately make the expression on the left bigger than $16$.
Looking at the derivative of the expression on the left,
$$ y' = 4(x+1)^3 + 4(x-1)^3, $$
we see that the derivative is $0$ at $x=0$ and monotonically increasing for $x > 0$ which means that there are no other solutions since our graph can only intersect the horizontal line $y=16$ once.
|
{
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"url": "https://math.stackexchange.com/questions/741051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that $\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ As the title states, trying to solve $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$
|
Hints: Expand the denominator as a geometric series in $-x^4$:
$$\frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}=1-x^4+x^8-x^{12}+...$$
Multiply by $1+x^2$ to obtain a series form for the integrand.
$$\frac{1+x^2}{1+x^4}=\frac{1}{1+x^4}+\frac{x^2}{1+x^4}=1+x^2-x^4-x^6+x^8+x^{10}-x^{12}-x^{14}+...$$
All that's left is to integrate term-by-term.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the lengths of the sides of a triangle given 3 angles only. If a right $\triangle ABC$ has $\angle A= 90^\circ$, $\angle B=45^\circ$, $\angle C=45^\circ$. Is there a way of finding the lengths of the sides $a$, $b$ & $c$ without knowing any of their lengths ? Normally we use $(\cos{x})^2+(\sin{x})^2=1$ as the hypotenuse.
I have an equation that says: $$(\sin {x}\times \sin {x}\times \cos {x})^2+(\cos {x}\times \cos {x}\times \sin {x})^2=(\sin {x}\times \cos {x})^2$$
$$\frac{(\sin {x}\times \sin {x}\times \cos {x})^2}{(\sin {x}\times \cos {x})^2}=(\sin{x})^2$$
$$\frac{(\cos {x}\times \cos {x}\times \sin {x})^2}{(\sin {x}\times \cos {x})^2}=(\cos{x})^2$$
|
Proof:if x,y and z belong to real numbers and$x<y<z$ then
$(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times((1-\frac{x}{z})\times\sqrt\frac{(x+z)}{(z-x)})=\sin A \tag{1}$
$(\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})-(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times(\frac{x}{z})=\cos A \tag{2}$
$\sqrt\frac{(z-y)}{z}=\cos B \tag{3}$
$\sqrt\frac{y}{z}=\sin B \tag{4}$
$\frac{x}{z}=\cos C \tag{5}$
$((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\sin C \tag{6}$
$(\sqrt{\frac{y}{z}}\times\frac{x}{z})+\sqrt\frac{z-y}{z}\times((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\sin A \tag{7}$
$(-\sqrt{\frac{z-y}{z}})\times\frac{x}{z})+\sqrt{\frac{y}{z}}\times((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\cos A \tag{8}$
$((((\frac{\sin B}{\sin C})\cos C)+\cos B)\sin C)=\sin A \tag{9}$
$(\frac{\sin B}{\sin C})-((((\frac{\sin B}{\sin C})\cos C)+\cos B)\cos C)=\cos A \tag{10}$
$((((\frac{\sin A}{\sin C})\cos C)+\cos A)\sin C)=\sin B \tag{11}$
$(\frac{\sin A}{\sin C})-((((\frac{\sin A}{\sin C})\cos C)+\cos A)\cos C)=\cos B \tag{12}$
find $\cos C$ and $\sin C$ in terms of $\sin A, \cos A, \sin B, \cos B$ just like equations $(9)$ to $(12)$?
Notice that the last four lines, equations $(9)$ to $(12)$, are symmetric with respect to $A,B,C.$ The missing equations are
$ ((((\frac{\sin B}{\sin A})\cos A)+\cos B)\sin A) = \sin C, \tag{13}$
$ (\frac{\sin B}{\sin A})-((((\frac{\sin B}{\sin A})\cos A)+\cos B)\cos A) = \cos C. \tag{14}$
The reason is that $\, A + B + C = \pi \,$ because they are angles of a triangle. As just one example, let $\, x=1, \, y=2, \, z=4. \,$ Then the angles are
$\, A \approx 59.4775^\circ, \, B = 45^\circ, \, C \approx 75.5224^\circ. \,$
The unusual parametrization of the triangle, given $\, x,y,z \,$ and $\, R, \,$ the radius of the circumscribed circle, is
$$\, b = 2R \sin B. \sin B=\sqrt{y/z}, \quad c = 2R \sin C. \sin C=\sqrt{1 - x^2/z^2},
\quad a = 2R \sin A = b \cos C + c \cos B. \,$$
Assuming A,B,C form a triangle.
*
*cos(A+B+C)=-1, cos(A+B)cos(C)-sin(A+B)sin(C)=-1.
*sin(A+B+C)=0, sin(A+B)cos(C)+cos(A+B)sin(C)=0.
*sin(A+B)=sin(A)cos(B)+cos(A)sin(B).
*cos(A+B)=cos(A)cos(B)-sin(A)sin(B).
Knowing cos(A) etc., use 3. and 4. to get cos(A+B) and sin(A+B).
1. and 2. become simultaneous linear equations in cos(C) and sin(C) - readily solved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Indeterminate limit that is supposed to be solved with De L'Hospital's rule Last week my Maths teacher gave the class this exercise taken from our text book. We are working on De L'Hospital's rule at the moment and this exercise is from that part of the book so everybody assumed that was the right procedure to solve it. One week later and nobody has been able to get to the right solution (even using other procedures).
This is the exercise:
Solve
$$
\lim_{x \to 0} \frac{1}{x^{2}} - \cot^{2}{x}
$$
According to the book and the mighty WolframAlpha the solution is $\frac{2}{3}$ but I can't get anywhere near it. The only solution I were able to get was a $-\infty$, which I got by transforming the $\cot^{2} x $ in $\frac{\cos^{2} x}{\sin^{2} x}$ and using De L'Hospital's.
EDIT: I'm asking just out of curiosity, I don't have to turn in this exercise and I'm not asking to avoid doing my homework myself.
|
\begin{eqnarray*}
\frac{\sin ^{2}x-x^{2}\cos ^{2}x}{x^{2}\sin ^{2}x} &=&\frac{\left( \sin
x-x\cos x\right) \left( \sin x+x\cos x\right) }{x^{2}\sin ^{2}x} \\
&=&\frac{\frac{\left( \sin x-x+x-x\cos x\right) }{x^{3}}\frac{\left( \sin
x+x\cos x\right) }{x}}{\frac{x^{2}\sin ^{2}x}{x^{4}}} \\
&=&\frac{\left( \frac{\sin x-x}{x^{3}}+\frac{1-\cos x}{x^{2}}\right) \left(
\frac{\sin x}{x}+\cos x\right) }{\left( \frac{\sin x}{x}\right) ^{2}}
\end{eqnarray*}
$\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\overset{HRule}{=}%
\lim_{x\rightarrow 0}\frac{\cos x-1}{3x^{2}}\overset{HRule}{=}%
\lim_{x\rightarrow 0}\frac{-\sin x}{6x}\overset{HRule}{=}\lim_{x\rightarrow
0}\frac{-\cos x}{6}=\frac{-1}{6}.$
$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}\overset{HRule}{=}%
\lim_{x\rightarrow 0}\frac{\sin x}{2x}\overset{HRule}{=}\lim_{x\rightarrow 0}%
\frac{\cos x}{2}=\frac{1}{2}$
$\lim_{x\rightarrow 0}\frac{\sin x}{x}\overset{HRule}{=}\lim_{x\rightarrow 0}%
\frac{\cos x}{1}=1.$ It follows that
$$
\lim_{x\rightarrow 0}\frac{\sin ^{2}x-x^{2}\cos ^{2}x}{x^{2}\sin ^{2}x}=%
\frac{\left( \frac{-1}{6}+\frac{1}{2}\right) \left( 1+1\right) }{\left(
1\right) ^{2}}=\frac{2}{3}.
$$
|
{
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"url": "https://math.stackexchange.com/questions/744172",
"timestamp": "2023-03-29T00:00:00",
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|
Integrating a function with substitution Totally forgot how to integrate.
$$ \int \frac{1}{x^2 \sqrt{x^2+4}}dx$$
Just need a tip, for this what would I use to substitute?
|
$$\int \frac{1}{x^2 \sqrt{x^2+4}}dx = \int \frac{1}{8(\frac{x}{2})^2 \sqrt{(\frac{x}{2})^2+1}}dx= \int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}dx$$
$$=\int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}dx\frac{\frac{d(tan^{-1}(\frac{x}{2}))}{dx}}{\frac{d(tan^{-1}(\frac{x}{2}))}{dx}}$$
$$=\int \frac{1}{8tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}\frac{d(tan^{-1}(\frac{x}{2}))}{\frac{1}{2}\frac{1}{(\frac{x}{2})^2+1}}$$
$$=\int \frac{(\frac{x}{2})^2+1}{4tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}d(tan^{-1}(\frac{x}{2})$$
$$=\int \frac{tan^2(tan^{-1}(\frac{x}{2}))+1}{4tan^2(tan^{-1}(\frac{x}{2})) \sqrt{tan^2(tan^{-1}(\frac{x}{2}))+1}}d(tan^{-1}(\frac{x}{2})$$
$$=\int \frac{sec^2(tan^{-1}(\frac{x}{2}))}{4tan^2(tan^{-1}(\frac{x}{2})) sec(tan^{-1}(\frac{x}{2}))}d(tan^{-1}(\frac{x}{2})$$
$$=\int {\frac{1}{4}cot(tan^{-1}(\frac{x}{2}))csc(tan^{-1}(\frac{x}{2}))}d(tan^{-1}(\frac{x}{2})$$
$$=-\frac{1}{4}csc(tan^{-1}(\frac{x}{2}))+C$$
skipping ahead
$$=-\frac{1}{4}\sqrt{1+\frac{1}{tan^2(tan^{-1}(\frac{x}{2}))}}+C$$
$$=-\frac{1}{4}\sqrt{1+\frac{1}{x^2/4}}+C$$
$$=-\frac{\sqrt{x^2+4}}{4x}+C$$
Does everyone see now why we use substitutions?
|
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|
What's wrong with this recursion of counting codes of length $n$ formed by $a$, $b$, and $c$ such that no three consecutive letters are distinct I found the following problem in a combinatorics book and gave it a try.
Let $B_n$ denote the set of codes of length $n$ formed by using the letters $a$, $b$, and $c$, none of which contains three consecutive letters that are distinct (so at least two of the three letters are the same). Express $|B_n|$ as a recursion.
Let $X_n$, $Y_n$, and $Z_n$ denote the sets of codes of length $n$ ending with $a$, $b$, and $c$ respectively. Then $|B_n| = |X_n| + |Y_n| + |Z_n|$.
A code of length $n$ ending with $a$ can be formed from a code of length $n - 2$ by the following three ways.
*
*Appending $aa$, $ba$, and $ca$ after a code ending with $a$.
*Appending $aa$ and $ba$ after a code ending with $b$.
*Appending $aa$ and $ca$ after a code ending with $c$.
Hence $|X_n| = 3 \cdot |X_{n - 2}| + 2 \cdot |Y_{n - 2}| + 2 \cdot |Z_{n - 2}|$. By symmetry, $|Y_n| = 2 \cdot |X_{n - 2}| + 3 \cdot |Y_{n - 2}| + 2 \cdot |Z_{n - 2}|$ and $|Z_n| = 2 \cdot |X_{n - 2}| + 2 \cdot |Y_{n - 2}| + 3 \cdot |Z_{n - 2}|$.
But, $|X_n| = |X_{n - 2}| + 2 \cdot (|X_{n - 2}| + |Y_{n - 2}| + |Z_{n - 2}|) = |X_{n - 2}| + 2 \cdot |B_{n - 2}|$. Again, by symmetry, $|Y_n| = |Y_{n - 2}| + 2 \cdot |B_{n - 2}|$ and $|Z_n| = |Z_{n - 2}| + 2 \cdot |B_{n - 2}|$.
Hence, $|B_n| = (|X_{n - 2}| + |Y_{n - 2}| + |Z_{n - 2}|) + 6 \cdot |B_{n - 2}| = 7 \cdot |B_{n - 2}|$ with $|B_1| = 3$ and $|B_2| = 9$. But the book's answer is $|B_{n}| = 2 \cdot |B_{n - 1}| + |B_{n - 2}|$. What did I get wrong?
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You have two cases to consider: strings that end in $x x$ and strings that end in $x y$ ($x \ne y$). Call the numbers of each of length $n$ respectively $u_n$ and $v_n$, set up recurrences for both. You are interested in $u_n + v_n$.
*
*$\ldots xx$: You can add a new $x$ (1 possibility) to get a new $\ldots xx$,
or anything else to get an $\ldots xy$ (2 options)
*$\ldots xy$: If you add an $x$, you get an $\ldots xy$ (1 option), if you add a $y$ you get a $\ldots xx$ (1 option)
This leads to the system of recurrences:
\begin{align}
u_{n + 1} &= u_n + v_n \\
v_{n + 1} &= 2 u_n + v_n
\end{align}
As starting points for $\ldots xx$ you have $u_2 = 3$, for $\ldots xy$ it is $v_2 = 6$, by "running the recurrences backwards" you get $u_0 = - 3$ and $v_0 = 6$. These values are pure fiction, the real starting points are $u_3 = 9$ and $v_3 = 12$.
Subtracting you get $v_{n + 1} - u_{n + 1} = u_n$, so $v_{n + 1} = u_{n + 1} + u_n$; substituting in $u_{n + 2}$ gets you $u_{n + 2} = 2 u_{n + 1} + u_n$. But from the first recurrence the value you are really interested in, $u_n + v_n$, is nothing more than $u_{n + 1}$, which satisfies the same recurrence. Thus you have that
$$
\lvert B_{n + 2} \rvert = 2 \cdot \lvert B_{n + 1} \rvert + \lvert B_n \rvert
$$
as was requested.
Now use generating functions. Define $U(z) = \sum_{u \ge 0} u_n z^n$ and $V(z) = \sum_{u \ge 0} v_n z^n$, multiply the recurrences by $z^n$, sum each over $n \ge 0$, and express in terms of the generating functions:
\begin{align}
\frac{U(z) - u_0}{z} &= U(z) + V(z) \\
\frac{V(z) - v_0}{z} &= 2 U(z) + V(z)
\end{align}
This gives:
\begin{align}
U(z) &= - \frac{3 - 9 z}{1 - 2 z - z^2} \\
V(z) &= \frac{6 - 12 z}{1 - 2 z - z^2}
\end{align}
But you are interested in
$$
B(z) = U(z) + V(z) = 3 \frac{1 - z}{1 - 2 z - z^2}
$$
Next step is to expand as partial fractions, and read off the coefficients from the resulting terms. Sadly, the zeros of the denominator are $-1 \pm \sqrt{2}$, so this gets ugly. In the immortal words of all texbook writers, "the details are left as an exercise for the reader."
|
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|
Find sequence function and general rule the function $$a_{n+2}=3a_{n+1}-2a_n+2$$
is given, and $$a_0=a_1=1, (a_n)_{n\ge0}$$
multiplying everything by $$/\sum_{n=0}^\infty x^{n+2}$$
also adding $$\sum_{n=0}^\infty (a_{n+2}x^{n+2}+a_1x+a_0)-a_1x-a_0=\sum_{n=0}^\infty (3a_{n+1}x^{n+2}+a_0)-a_0-\sum_{n=0}^\infty 2a_nx^{n+2}+\sum_{n=0}^\infty 2x^{n+2}$$
we get $$R(X)-2x-1=3x(R(X)-1)-2x^2R(X)+\frac{2x^2}{1-x}$$
$$R(X)(1-3x+2x^2)+x-1=\frac{2x^2}{1-x}$$
$$R(X)(1-3x+2x^2)=\frac{3x^2-2x+1}{1-x}$$
$$R(X)=\frac{3x^2-2x+1}{(1-x)^2(1-2x)}=\frac{A}{(1-x)^2}+\frac{B}{1-2x}$$
so $$3x^2-2x+1=A(1-2x)+B(1-x)^2$$
$$3x^2-2x+1=x^2(B)+x(-2A-2B)+A+B$$
B=3
-2A-2B=-2
A+B=1$$\quad\Longrightarrow\quad A=-2, B=3$$
$$R(X)=\frac{-2}{(1-x)^2}+\frac{3}{1-2x}=-2\sum_{n=0}^\infty nx^{n-1}+3\sum_{n=0}^\infty (2x)^2=$$
how to proceed?
|
The generating function I'm getting is:
$
\begin{align}
R(x) &= \dfrac{1-3\, x+4\, x^2}{1-4\, x+5\, x^2-2\, x^3}\\
&= \dfrac{2}{1-2 \, x} + \dfrac{1}{1-x} - \dfrac{2}{{\left(1-x\right)}^{2}}\\
&= \sum_{n\ge 0} \left(2^{n+1}-2\, n-1\right)\, x^n
\end{align}
$
|
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|
Find two triangles of longest side length 25? I'm using the quadratic Diophantine equations to solve for two integer triangles of longest side $25$. It's been shown that for $a^2+b^2=c^2$, which goes to $x^2+y^2=1$ where $x=\frac ac, y=\frac bc, a=t^2−1, b=2t, c=t^2+1$.
If I want to solve for $c=25$, how will I go about doing this?
Just letting $c=25$ gives me an irrational answer for $b$ after having solved for $t$. I'm a little confused.
Thanks for your help.
|
All Pythagorean triples, i.e. triplets of positive integers such that $a^2+b^2=c^2$, can be expressed as $a=k(m^2-n^2)$, $b=2kmn$, $c=k(m^2+n^2)$; where $m$ and $n$ are coprime, $m>n$ and $m$ and $n$ have opposite parity. (Up to interchanging of $a$ and $b$.)
So in your case you want $k(m^2+n^2)=25$. This gives you the possibilities:
$k=1$ and $m^2+n^2=25$
$k=5$ and $m^2+n^2=5$
$k=25$ and $m^2+n^2=1$
For such small numbers, it is easy to find all expressions as a sum of two squares by hand, you will get $25=0^2+5^2=3^2+4^2$, $5=1^2+2^2$ and $1=1^2+0^2$.
Since you are interested only in non-zero values, you get only two possibilities for $(k,m,n)$ namely $(1,4,3)$ and $(5,2,1)$. From the first one you get $a=7$, $b=24$, $c=25$. The second one gives you $a=15$, $b=20$, $c=25$.
|
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|
Prove the inequality $({1+\frac{a}b})^n$ + $(1+\frac{b}a)^n$ $\geq$ $2^{n+1}$ Let $a$ and $b$ be positive real numbers and let $n$ be a natural number
prove that
$$\left({1+\frac ab}\right)^n+\left(1+\frac ba\right)^n\ge2^{n+1}.$$
|
$a,b$ are positive, so
$$\dfrac{a^k}{b^k}+\dfrac{b^k}{a^k}\ge2\Longleftrightarrow a^{2k}+b^{2k}\ge2a^kb^k\Longleftrightarrow\big(a^k-b^k\big)^2\ge0,$$
and using the binomial theorem (three times total) we get
$$\left(1+\frac ab\right)^n+\left(1+\frac ba\right)^n=\sum_{k=0}^n\binom nk\left(\dfrac{a^k}{b^k}+\dfrac{b^k}{a^k}\right)\ge2\sum_{k=0}^n\binom nk=2\,(1+1)^n=2^{n+1}.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/753593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help is appreciated
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When numbers are so small you can proceed by attempt.
However $a^2+b^2=6$ doesn't have no integer solution: wlog you can take $a,b\geq0$.
So try: $a=0\Rightarrow b=\sqrt6$ (not valid in $\mathbb Z$).
$a=1\Rightarrow b=\sqrt5$ (not valid as above)
$a=2\Rightarrow b=\sqrt2$ (idem).
$a\geq3\Rightarrow b^2<0$.
But maybe you mean to find $a,b\in\mathbb R$.
In this case just notice that $14=a^3+b^3=x(6-y)$ and $6=a^2+b^2=x^2-2y$, where $x=a+b$ and $y=ab$. Solving the system you'll find $x=\frac{14}{6-y}$ and $y^3-9y^2+10=0$, which is solvable using the well known formulas for the 3rd degree polynomials.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/756124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
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You have
$$(2^x+2^y)^2=4^x+4^y+2^{x+y+1}=2^{x+1}+2^{y+1}+2^{x+y+1}$$
If $a=2^x$, $b=2^y$, then
$$(a+b)^2=2(a+b+ab)>2(a+b)$$
so $a+b>2$.
I'm looking for an upper bound. If I find it, I'll edit this answer.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/756213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$ I would like to show that
$$ \sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64} $$
I've been working on this for a few days. I've used product-to-sum formulas, writing the sines in their exponential form, etc. When I used the product-to-sum formulas, I'd get a factor of $1/64$, I obtained the same with writing the sines in their exponential form. I'd always get $1/64$ somehow, but never the $\sqrt{13}$.
I've come across this: http://mathworld.wolfram.com/TrigonometryAnglesPi13.html, (look at the 10th equation). It says that this comes from one of Newton's formulas and links to something named "Newton-Girard formulas", which I cannot understand. :(
Thanks in advance.
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For positive integer $n$
If $\sin(2n+1)x=0, (2n+1)x=m\pi\iff x=\frac{m\pi}{2n+1} $ where $m$ is any integer
From $(3)$ of this,
$\displaystyle \sin(2n+1)x=2^{2n}s^{2n+1}+\cdots+(2n+1)s=0$ where $s=\sin\frac{m\pi}{2n+1}$
So the roots of $\displaystyle 2^{2n}s^{2n+1}+\cdots+(2n+1)s=0 $ are $\sin\frac{m\pi}{2n+1}; 0\le m\le2n$
So the roots of $\displaystyle 2^{2n}s^{2n}+\cdots+(2n+1)=0 $ are $\sin\frac{m\pi}{2n+1}; 1\le m\le2n$
Using Vieta's formula
$\displaystyle\prod_{m=1}^{2n}\sin\frac{m\pi}{2n+1}=\frac{2n+1}{2^{2n}}$
Now using $\displaystyle\sin(\pi-y)=\sin y,\sin\left(\pi-\frac m{2n+1}\pi\right)=\sin\frac{(2n+1-m)\pi}{2n+1}$
$\displaystyle\prod_{m=1}^{2n}\sin\frac{m\pi}{2n+1}=\prod_{m=1}^n\sin^2\frac{m\pi}{2n+1}$
Now for $\displaystyle, 1\le m\le n;0<\frac{m\pi}{2n+1}<\frac\pi2\implies\sin\frac{m\pi}{2n+1}>0$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/756489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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