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Converting Trig equation in Maple $$\frac{\sin^4(3x)}{18\cos^6(3x)} + \frac{\sin^4(3x)}{36\cos^4(3x)} = \frac{\tan^6(3x)}{18} + \frac{\tan^4(3x)}{12}$$
The answer on the left is what I got when I punched in the problem in Maple: $\operatorname{int}(\tan^3(3x)\cdot\sec(3x)^4, x)$
The answer on the right is my own using u-sub and a trig identity. I need to convert the Maple answer into the same form as the answer I came up with. Whether someone can show me how to do that in Maple somehow or manually would be greatly appreciated :). I tried to convert it manually and got to:
$$\frac{\tan^4(3x)\cdot(2+\cos^2(3x)))}{36\cos^2(3x)}$$
I'm not sure how to proceed past this point or even if this equation is correct so far.
|
Let's do this manually, since it's actually not that bad! I'll write in a condensed notation with $s=\sin(3x)$ and $c=\cos(3x)$ so I can type this more easily.
\begin{align*}
\frac{s^4}{18c^6} + \frac{s^4}{36c^4} &=\frac{s^4}{18c^6}-\frac{s^4}{18c^4} + \frac{s^4}{18c^4} + \frac{s^4}{36c^4} \\
&=\frac{s^4(c^2+s^2)}{18c^6} -\frac{s^4c^2}{18c^6}+\frac{s^4}{12c^4} \\
&= \frac{s^6}{18c^6} +\frac{s^4}{12c^4}
\end{align*}
Which is your answer!
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many numbers are there which only contain digits $4$ and $7$ in them? I wanna know how many numbers $n$ are there which only contain digits $4$ and $7$ in them, where $1 ≤ n ≤ 10^9$.
Ex: $4, 7, 44, 47, 74, 77, ...$
I am trying to find a general equation to compute the numbers, given how many digits, which is $2$ in this case, and the range, which is $1 ≤ n ≤ 10^9$ in this case.
|
There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+\cdots+2^9=2^{10}-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+\cdots+2^9)=(2^2+2^3+\cdots+2^{10})-(2+2^2+2^3+\cdots+2^9)=2^{10}-2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function
Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function
I have done the following:
$f(a-x)=f(a+x)$....(1)
or, $f(-(x-a))=f(x+a). $
Putting $x=x+a$, we get
$f(-x)=f(x+2a) ....(i)$
Similarly,
$f(b-x)=f(b+x).....(2) $
Putting $x=x+b,$ we get
$f(-x)=f(x+2b)......(ii)$
Equating $(i)$ & $(ii)$, we get:
$f(x+2a)=f(x+2b).$
Given, $a>b$
Therefore, $2a=2b+h.$
Putting $x=x-2a$, we get
$f(x+h)=f(x).$
Is this mathematically correct? Also, what other ways are there to solve it?
|
I guess that $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$ hold for every $x$.
In particular, evaluating at $x+a$ instead of $x$,
$$
f(-x)=f(x+2a)
$$
Similarly $f(-x)=f(x+2b)$. Therefore
$$
f(x+2a)=f(x+2b)
$$
for every $x$. Evaluating at $x-2b$ instead of $x$, we obtain
$$
f(x+2a-2b)=f(x)
$$
and, setting $T=2a-2b$,
$$
f(x)=f(x+T)
$$
for every $x$.
Good job!
|
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|
Tricks for estimating $ \lim_{x\rightarrow 0} \frac{d}{dx} \bigl(-\frac{1}{x} \ln\bigl(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \bigr) \bigr)$ I'm trying to find a Taylor approximation of $ f(x) =-\frac{1}{x} \ln\left(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \right) $ at $x = 0$.
For the derivation part Wolfram returns a quite a tedious expression to the problem, taking the limit of which does result in the answer I'm looking for.
As I found this problem in a textbook which is not focused on honing derivation/approximation skills, I'm curious, whether there are some known tricks that would help easying up the calculation, or is this just a technical exercise?
Apologies if this is a dumb question.
|
$\begin{array}\\
f(x)
&=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+O(x^2)) (-bx+O(x^2))}{-x+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{abx^2+O(x^3)}{-x+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - abx+O(x^2) \right)\\
&=\frac{1}{x}(abx+O(x^2))\\
&=ab+O(x)\\
\end{array}
$
Take more terms
(e.g.,
$e^{-x} = 1-x+x^2/2+O(x^3)
$)
to get more precision.
Here is a try to get
one more term.
$\begin{array}\\
f(x)
&=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+a^2x^2+O(x^3)) (-bx+b^2x^2/2+O(x^3))}{-x+x^2/2+O(x^3)} \right)\\
&=-\frac{1}{x} \ln\left(1 + x^2\dfrac{(a-a^2x+O(x^2)) (b-b^2x/2+O(x^2))}{-x+x^2/2+O(x^3)} \right)\\
&=-\frac{1}{x} \ln\left(1 + x\dfrac{(a-a^2x+O(x^2)) (b-b^2x/2+O(x^2))}{-1+x/2+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - x\dfrac{ab-(a^2b+ab^2)x+O(x^2)}{1-x/2+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - x(ab-(a^2b+ab^2)x+O(x^2))(1+x/2+O(x^2)) \right)\\
&=-\frac{1}{x} \ln\left(1 - abx(1-(a+b)x+O(x^2))(1+x/2+O(x^2)) \right)\\
&=-\frac{1}{x} \ln\left(1 - abx(1-(a+b+1/2)x+O(x^2)) \right)\\
&=-\frac{1}{x} \left( - abx(1-(a+b+1/2)x+O(x^2)) +((a+b+1/2)x+O(x^2))^2/2\right)\\
&=-\frac{1}{x} \left( - abx+ab(a+b+1/2)x^2+O(x^3)) +x^2((a+b+1/2)+O(x))^2/2\right)\\
&=-\frac{1}{x} \left( - abx+ab(a+b+1/2)x^2+O(x^3)) +x^2((a+b+1/2)^2+O(x))/2\right)\\
&=-\frac{1}{x} \left( -x\left(ab+ab(a+b+1/2)x+O(x^2) +x((a+b+1/2)^2+O(x))/2\right)\right)\\
&=-\frac{1}{x} \left( -x\left( ab +x(ab(a+b+1/2)+(a+b+1/2)^2\right)/2+O(x^2)\right)\\
&= ab +x(ab(a+b+1/2)+(a+b+1/2)^2)/2+O(x^2)\\
&= ab +x(ab(a+b+1/2))(1+(a+b+1/2))/2+O(x^2)\\
&= ab +xab(a+b+1/2)(a+b+3/2)/2+O(x^2)\\
\end{array}
$
You better check my algebra,
because
Prob(error) > 1/e.
|
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|
Prove that gcd$(n^2+1, (n+1)^2+1)$ is either 1 or 5 My try to solve this question goes as follows:
$g=gcd(n^2+1, (n+1^2)+1) = gcd(n^2+1, 2n+1) = gcd(n^2-2n, 2n+1)$.
By long division:
$$n^2-2n = -2n(2n+1) + 5n^2$$
Since $g$ divides $n^2-2n$ and $g$ divides $(2n+1)$, $g$ divides $5n^2$. However, I need to show that $g$ is prime in order to show that it divides $5$ or $n^2$. I am stuck at this point.
Any Ideas,,
|
Clearly $$\gcd(n^ 2+1,(n+1)^2+1)\mid \bigl((n+1)^2+1\bigr)-\bigl(n^ 2+1\bigr)=2n+1,$$
hence also
$$\gcd(n^ 2+1,(n+1)^2+1)\mid (2n+1)n-2(n^2+1)=n-2 $$
as well as
$$\gcd(n^ 2+1,(n+1)^2+1)\mid (2n+1)n-2(n^2+1)=2n+1-2(n-2)=5. $$
This leaves only the possibilites $1$ and $5$ (which are both possible as can be seen by checking the cases $n=1$ and $n=2$)
|
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|
$ (5+2\sqrt6)^{\sin x} +(5-2\sqrt6)^{\sin x} = 2\sqrt3 $ , where $ 0 ≤ x≤ 360 $ There is something I haven't picked up on, a hint would be appreciated
Given that $(\sqrt3+\sqrt2)^2 = (5+2\sqrt6)$ and $ (\sqrt3-\sqrt2)^2 = (5-2\sqrt6)$
Find the values of x
for which$
(5+2\sqrt6)^{\sin x}
+(5-2\sqrt6)^{\sin x}
= 2\sqrt3 $
, where $
0
0
≤ x≤ 360
$
|
Hint: Use that $$(5+2\sqrt{6})(5-2\sqrt{6})=1$$
Then you will get $$(5+2\sqrt{6})^{\sin(x)}+\frac{1}{(5+2\sqrt{6})^{\sin(x)}}=2\sqrt{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve recurrence relation! I was working with this recurrence relation :
$$\begin{cases}A(n,k) = A(n-1, k-1)+A(n-2, k-1)+A(n-1, k)\\
A(n, 0) = 1\\
A(n, 1) = 2n
\end{cases}$$
Generating function : $(1+x)/(1-x-x*y-x^2*y)$
Now since this involves two parameters, I tried changing it to a single parameter by fixing $k$ and iterating over $n$ for $n=k$, $n=k+1$, $n=k+2$.. and so on.
What I found was that $A(k, k)=2$, $A(k+1, k)=4k$, $A(k+2, k)=4k^2+2$ and the later results were too haphazard to write i.e. they did not form a pattern to the previous formulas. I tried to use the above equation to find characteristic roots and use $A(n) = a(x^n)+b(y^n)$ where $x$ and $y$ are roots. However, I wasn't able to compute it correctly.
What am I missing here? Am I tackling the problem in a complete wrong way? Is there any other method to solve such recurrences? Any help is appreciated. I assume the formula for $A(n)$ could be pretty cumbersome but I still wish to find it.
|
We look at the generating function $\frac{1+x}{1-x-xy-x^2y}$ and derive the coefficients $A(n,k)$. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
\color{blue}{A(n,k)}&=[x^ny^k]\frac{1+x}{1-x-xy-x^2y}\\
&=[x^ny^k]\frac{1+x}{1-x-x(1+x)y}\\
&=[x^ny^k]\frac{1+x}{1-x}\cdot\frac{1}{1-\frac{x(1+x)}{1-x}y}\\
&=[x^n]\frac{1+x}{1-x}[y^k]\sum_{j=0}^\infty\left(\frac{x(1+x)}{1-x}\right)^jy^j\tag{1}\\
&=[x^n]\frac{1+x}{1-x}\left(\frac{x(1+x)}{1-x}\right)^k\tag{2}\\
&=[x^{n-k}]\frac{(1+x)^{k+1}}{(1-x)^{k+1}}\tag{3}\\
&=[x^{n-k}](1+x)^{k+1}\sum_{j=0}^\infty\binom{-k-1}{j}(-x)^j\tag{4}\\
&=[x^{n-k}](1+x)^{k+1}\sum_{j=0}^\infty\binom{k+j}{j}x^j\tag{5}\\
&=\sum_{j=0}^{n-k}\binom{k+j}{j}[x^{n-k-j}](1+x)^{k+1}\tag{6}\\
&=\sum_{j=0}^{n-k}\binom{k+j}{j}\binom{k+1}{n-k-j}\tag{7}\\
&\,\,\color{blue}{=\sum_{j=0}^{n-k}\binom{n-j}{k}\binom{k+1}{j}}\tag{8}
\end{align*}
Comment:
*
*In (1) we do a geometric series expansion with respect to $y$.
*In (2) we select the coefficient of $y^k$.
*In (3) we do some simplifications and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.
*In (4) we do a binomial series expansion.
*In (5) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (6) we select the coefficient of $x^{n-k}$ and restrict the upper bound of the sum by $n-k$ since other terms do not contribute to $[x^{n-k}]$.
*In (7) we select the coefficient of $x^{n-k-j}$.
*In (8) we do a final rearrangement by changing the order of summation $j\to n-k-j$.
We use the formula (8) to check the boundary conditions. We obtain
\begin{align*}
A(n,0)&=\sum_{j=0}^n\binom{n-j}{0}\binom{1}{j}=\sum_{j=0}^n\binom{1}{j}=\begin{cases}
1&\qquad\qquad\qquad\qquad n=0\\
2&\qquad\qquad\qquad\qquad n>0
\end{cases}\\
A(n,1)&=\sum_{j=0}^{n-1}\binom{n-j}{1}\binom{2}{j}=\sum_{j=0}^{n-1}(n-j)\binom{2}{j}=\begin{cases}
1&\qquad n=1\\
4(n-1)&\qquad n>1
\end{cases}
\end{align*}
We observe, the generating function $\frac{1+x}{1-x-xy-x^2y}$ does not follow OPs stated boundary conditions $A(n,0)=1\ (n\geq 0)$ and $A(n,1)=2n\ (n\geq 1)$ and this is a reason for the difficulties OP has to cope with.
Note: We find with some help of Wolfram Alpha a series expansion
\begin{align*}
\frac{1+x}{1-x-xy-x^2y}&=1+(2+y)x+(2+4y+y^2)x^2+(2+8y+6y^2+y^3)x^3\\
&\qquad+(2+12y+18y^2+8y^3+y^4)x^4+\cdots
\end{align*}
The corresponding sequence of the coefficients $A(n,k)$ starting with
\begin{align*}
&1;\\
&\color{blue}{2},\color{blue}{1};\\
&\color{blue}{2},\color{blue}{4},1;\\
&\color{blue}{2},\color{blue}{8},6,1;\\
&\color{blue}{2},\color{blue}{12},18,8,1;\ldots
\end{align*}
can be found as A113413 in OEIS. We can find there OPs stated generating function $\frac{1+x}{1-x-xy-x^2y}$ as well as the recurrence relation $A(n,k) = A(n-1, k-1)+A(n-2, k-1)+A(n-1, k)$, but we have different boundary conditions (marked above in $\color{blue}{\mathrm{blue}}$).
|
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|
Solve $9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}$ $$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$
The equation states solve for $x$.
What I first did was put like bases together.
$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$
Then I factored $3^{2x}$ and $2^x$
$$3^{2x}(1+\frac{1}{3})=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}}),$$
then I got
$$\frac{3^{2x}}{2^x}=9\sqrt{2}.$$
From here I took $\log$s, but the answer wasn't nice.
What to do?
|
$$9^x-2^{x+\frac12}=2^{x+\frac72}-3^{2x-1}$$
$$\to\frac43(9^x)=9(2^{x+\frac12})$$
Hence we form the two iterates:
$$x_{n+1}=\log_9{\bigg[\frac{27}{4}(2^{x_n+\frac12})\bigg]}$$
and
$$x_{n+1}=\log_2{\bigg[\frac{4}{27}(9^{x_n})\bigg]}$$
The first gives the solution $x=\frac32$, the second diverges to $-\infty$, and thus there is no second solution
|
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|
Find $c$ and $n$ such that $\frac{x^3 \arctan x}{x^4 + \cos x +3} \sim cx^n$ as $x \to 0$ Where is my mistake in the below:
$$\frac{1}{c} \lim_{x \to 0} \frac{x^3 \arctan x}{x^{4+n} + x^n \cos x + 3x^n} = \frac{1}{c} \lim_{x \to 0} \frac{x^4}{x^{4+n}+x^n \cos x + 3x^n} \\ =\frac{1}{c} \lim_{x \to 0} \frac{1}{x^n + x^{n-4} \cos x + 3x^{n-4}}$$
Looking at it now it feels like it is the step that I am about to do, because it appears as though the denominator becomes smaller and smaller going to infinity. However, my reasoning was that since it is given that there can be an equivalence between $cx^n$ and our function, i.e. the limit is finite and goes to $1$ as $x \to 0$, then it also must be true that if we invert the fraction, then the limit is still finite and is $1/1=1$ as $x \to 0$. So, next step:
$$c \lim_{x \to 0} \left(x^n + x^{n-4}\cos x + 3x^{n-4} \right)$$
$$c \lim_{x \to 0} (x^{n-4}(\cos x -1))=1$$
$$-c \lim_{x \to 0} \left(x^{n-4} \frac{x^2}{2}\right) = 1$$
We need $x^{n-4} = x^{-2}$ $\implies n=2$
$$-c \frac{1}{2} =1 \implies c=-2$$
|
We have that
$$\frac{x^3 \arctan x}{x^4 + \cos x +3}\sim\frac{x^4}{x^4+4}\sim \frac14 x^4$$
and the guess is correct indeed
$$\frac{\frac{x^3 \arctan x}{x^4 + \cos x +3}}{\frac14 x^4}\to 1$$
As an alternative
$$\frac{\frac{x^3 \arctan x}{x^4 + \cos x +3}}{c x^n}=\frac1c\frac{x^3 \arctan x}{x^{4+n} + x^n\cos x +3x^n}=\frac1c\frac{x^4 +o(x^4)}{x^{4+n} + x^n+o(x^{n}) +3x^n}=$$$$=\frac1c\frac{1 +o(1)}{4x^{n-4}+o(x^{n-4})} \to 1$$
from which we obtain $n=4$ and $c=\frac14$.
|
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|
Find all pairs of positive integers $(a,b)$ such that $2^a+5^b$ is a perfect square. How do you solve such questions when they appear?
I know that this problem involves quadratic residues. Moreover, I also know that a=2,b=1 is possible. It may also be the only solution
I tried to take $\pmod{5}$ of this equation but it didn't really help me. I would appreciate if someone could post a solution to this problem.
Thank You
|
If $2^a+5^b = m^2$, then $m^2\equiv 0, 1, 4\pmod{5}$. $0$ is impossible since $2^a$ is never divisible by 5, so $2^a+5^b\equiv 2^a \equiv 1, 4\pmod{5}$. This implies that $a$ is even, say $a=2k$. This gives $5^b = m^2-2^{2k} = (m-2^k)(m+2^k)$, which implies that both $m-2^k$ and $m+2^k$ are powers of $5$, say
$$m-2^k = 5^r,\ m+2^k=5^s, r<s\quad\Rightarrow\quad 2^{k+1} = 5^s-5^r.$$
But $5^s-5^r$ is divisible by 5 for $r>0$, so it cannot be a power of two unless $r=0$.
Thus any solution must have $m=2^k+1$, $m+2^k = 5^b$, so that $5^b = 2^{k+1}+1$, or $5^b-2^{k+1}=1$. By Mihăilescu's theorem, either $b\le 1$ or $k+1\le 1$. Since $b$ is a positive integer, we get $b=1$, whence $2^{k+1}=4$ and $k=1$, so $a=2$.
|
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|
Partial Fraction of $\int \frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)}$ If
$$\int\frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)} = \int \frac{A \ \mathrm{d}x}{(\sin x)} + B \int\frac{\sin x \ \mathrm{d}x}{ 1 + \sin^2 x} + C \int \frac{\mathrm{d}x}{1 + \sin^2 x}$$
then, which of the following is correct?
*
*$A+B+C=4$
*$A+B+C=2$
*$A+BC=1$
*$A+B+C=5$
My approach is as follows:
$$\int\frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(2 - \cos^2 x)(\sin x)}= \int \frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(1 + \sin^2 x)(\sin x)} =$$
$$= \int \frac{\cos x \ \mathrm{d}x}{( 1 + \sin^2 x)(\sin x)} + \int \frac{\sin 2x \ \mathrm{d}x}{(1 + \sin^2 x )(\sin x)} =$$
$$= \int \frac{ \cos x \ \mathrm{d}x}{(1 + \sin^2 x )(\sin x)} + \int \frac{ 2 \cos x \ \mathrm{d}x}{(1 + \sin^2 x )} =$$
$$= \int\frac{- \sin \cos x \ \mathrm{d}x}{ 1 + \sin^2 x } + \int\frac{\cos x \ \mathrm{d}x}{ \sin x } + \int \frac{2 \cos x \ \mathrm{d}x}{1 + \sin^2 x}$$
After this step I am not able to approach
|
Write your Integrand in the form
$$\frac {\csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$
and then multiply numerator and denominator by
$$\frac{csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$
and we get
$$\newcommand{\dx}{\; \mathrm{d}}x\int \frac{\cot(x)(\csc(x)+2)\csc(x)}{\csc^2(x)+1}\dx$$
now Substitute
$$u=\csc(x)$$
then
$$du=-\cot(x)\csc(x)dx$$
and we obtain
-$$\int\frac{u+2}{u^2+1}du$$
Can you finish?
|
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|
Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$ A while ago I encountered this integral $$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$$ To be fair I spent some time with it and solved it in a heuristic way, I want to avoid that way so I won't show that approach, but the result I got is $\frac{\pi}{6} \ln(2+\sqrt 3)$ so I bet it can be shown in a nice way.
Solving other integrals I also encountered this one: $$J=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx=\pi \ln(2+\sqrt 3)$$ Which is pretty easy to compute, so most of my time I tried to show that $J=6I$, however without explictly evaluating them I had no luck.
Also I tried to use partial fractions: $$I=\frac12 \left(\int_0^1 \frac{\ln(1+x)}{x^2+\sqrt 3x +1}dx - \int_0^1 \frac{\ln(1+x)}{x^2-\sqrt 3 x+1}dx\right) $$
Considering: $$K(t) =\int_0^1 \frac{\ln(1+x)}{x^2-2\cos(t)x+1}dx$$
We have that $I=\frac12 \left(K\left(\frac{5\pi}{6}\right)-K\left(\frac{\pi}{6}\right)\right) $ and since: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$\small \Rightarrow K(t)=\frac12 \left(\frac{1}{\sin \left(\frac{5\pi}{6}\right)}\sum_{n=0}^\infty \sin\left(\frac{5\pi}{6} (n+1)t \right) + \frac{1}{\sin \left(\frac{\pi}{6}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{6} (n+1)t \right)\right)\int_0^1 x^n \ln(1+x)dx$$
$$=2\sum_{n=0}^\infty \sin\left(\frac{\pi}{6} (n+1)t \right)\int_0^1 x^n \ln(1+x)dx$$
Now I don't know how to deal with the integral and the sum combined.
|
I thought about the comment you left the other day, and it has led me to this alternative approach, which is more direct than my other answer. Thanks a lot, I have learned so much from you!
Let $\omega=\sqrt[3]{-1}$, and let Ti2 denote the inverse tangent integral function:
\begin{align*}
I&=\int_0^1\frac{\left(1+x^2\right)\log(1+x)}{x^4-x^2+1}dx\\
&=\int_0^1\left(\frac{\omega}{1+\omega^2 x^2}+\frac{\bar{\omega}}{1+\bar{\omega}^2 x^2}\right)\log\left(1+x\right)dx\\
&=\left.\left(\arctan(\omega x)+\arctan(\bar{\omega}x)\right)\log(1+x)\right|_0^1-\int_0^1\frac{\arctan(\omega x)+\arctan(\bar{\omega} x)}{1+x}dx\\
&=\frac{\pi}{2}\log 2-\operatorname{Ti_2}(\omega,\omega)-\operatorname{Ti_2}(\bar{\omega},\bar{\omega})
\end{align*}
A result (3.27) from Polylogarithms and Associated Functions reads:
$$\operatorname{Ti_2}(a,a)+\operatorname{Ti_2}(a^{-1},a^{-1})=\frac{\pi}{2}\log\left(\frac{2}{\sqrt{1+a^2}}\right)+\arctan(a)\log(a)
$$
Thus letting $a=\omega$, we get:
$$I=\frac{\pi}{2}\log\left(\sqrt{1+\omega^2}\right)-\arctan(\omega)\log(\omega)=\frac{\pi}{6}\log\left(2+\sqrt{3}\right)
$$
This approach also produces a general result, for when $-\frac{\pi}{2}<\varphi<\frac{\pi}{2}$:
$$\int_0^1\frac{\left(1+x^2\right)\log(1+x)}{x^4+2\cos(2\varphi)x^2+1}dx=\frac{\pi}{8\cos\varphi}\log(2\cos\varphi)+\frac{\varphi}{4\cos\varphi}\log\tan\left(\frac{\varphi}{2}+\frac{\pi}{4}\right)
$$
|
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|
Finding the centroid of a tetrahedron I have four points to form a tetahedron
$$A=(0,-\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\B=(0,\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\C=(-\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})
\\D=(\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})$$
it looks like this :
I'm asked to find the centroid of the tetrahedron formed by these points.
I looked up in the internet and there is an easy way to find it:
$$G=(\frac{x_1+x_2+x_3+x_4}{4}+\frac{y_1+y_2+y_3+y_4}{4}+\frac{z_1+z_2+z_3+z_4}{4})$$
I tried to check it with the triple intregrals. In order to do it I find the plane at which A, D and B lie, and that would be my first integration limit and the plane formed by A,B and C.
The planes :
$$\overline{AB}=(0,1,0)\\
\overline{AD}=(\frac{1}{2},\frac{1}{2},\frac{1}{2}\sqrt{\frac{1}{2}})$$
And after takind the cross product of both, and then using $a(x-x_0)+b(y-y_0)+c(z-z_0)$ to obtain the plane I get $$\frac{1}{2}\sqrt{\frac{1}{2}}x-\frac{1}{2}z=\frac{1}{8}\sqrt{\frac{1}{2}}$$ and the other one containing A,B,C is $$ \frac{1}{2}\sqrt{\frac{1}{2}}x+\frac{1}{2}z=\frac{1}{8}\sqrt{\frac{1}{2}}$$
I divide it into this two planes, and then see how it looks like in the $x,y$ plane
And I get the integrals:
$$ \int_{0}^{\frac{1}{2}}\int_{-x+\frac{1}{2}}^{\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{-\frac{1}{4}\sqrt{\frac{1}{2}}+x\sqrt{\frac{1}{2}}} zdz dy dx dx+\int_{\frac{1}{2}}^{0}\int_{0}^{-x+\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{-\frac{1}{4}\sqrt{\frac{1}{2}}+x\sqrt{\frac{1}{2}}}z dz dy dx $$
and the other part
$$ \int_{0}^{\frac{1}{2}}\int_{x+\frac{1}{2}}^{\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{\frac{1}{4}\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}}} zdz dy dx dx+\int_{\frac{1}{2}}^{0}\int_{0}^{-x-\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{\frac{1}{4}\sqrt{\frac{1}{2}}-x\sqrt{\frac{1}{2}}}z dz dy dx $$
I don't get 0.
I can't see why. Is the formula I use not valid?
|
Your equation for the plane that passes through $\{A, B, C\}$ should be
$\displaystyle \frac12\sqrt{\frac12} x + \frac12 z = -\frac18 \sqrt{ \frac12 }$.
Namely, you were off by one negative sign starting from there.
As for the integrals, guessing your intention from the way you set them up, it looks like you also need the planes that pass through $\{A,C,D\}$ and $\{B,C,D\}$.
Currently you have only the equations for the two lower planes. You also need the two upper planes.
That is, even if you plug in the correct equation for the $A$-$B$-$C$ plane and get zero (something like $\frac{-1}{768}+\frac1{768}+\frac{-1}{768}+\frac1{768}$), that would be a coincidence.
The formula of taking the arithmetic mean of the four points is valid here. Even though your tetrahedron is not a regular tetrahedron (which all four faces are equilateral triangles), it still has the relevant symmetry to make the centroid be at the "average".
|
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|
how is that expression is generated
I have no idea that how is red arrowed mammoth term is generated from the yellow arrowed term
Please explain
|
This is just a substitution. You have from Taylor Series
$$
\sin(t) = t- \dfrac{t^3}{3!} + \dfrac{t^5}{5!}- \dfrac{t^7}{7!} + \cdots= t- \dfrac{t^3}{6} + \dfrac{t^5}{120}- \dfrac{t^7}{5040} + \cdots
$$
But you know also the Taylor Series for $e^x-1$:
$$
e^x-1 = x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots
$$
which the book says to call this $t$. But then we plop this Taylor Series where we see $t$ in the Taylor Series for $\sin t$:
$$
\begin{split}
\sin(e^x-1)&= \sin(t) \\
&= t- \dfrac{t^3}{3!} + \dfrac{t^5}{5!}- \dfrac{t^7}{7!} + \cdots \\
&= t- \dfrac{t^3}{6} + \dfrac{t^5}{120}- \dfrac{t^7}{5040} + \cdots \\
&=\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)- \dfrac{\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)^3}{6} + \dfrac{\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)^5}{120}- \dfrac{\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)^7}{5040} + \cdots
\end{split}
$$
|
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|
Solving $z^{1+i}=4$. Solution:
Let $z=re^{i\theta}=r(\cos \theta + i\sin \theta)$. Then $z^{1+i}=e^{i\theta(1+i)}=e^{i\theta -\theta}=4$. So $e^{i\theta-\theta}=e^{-\theta}e^{i\theta}=e^{-\theta}(\cos \theta + i\sin \theta)=4(\cos 0 +i\sin 0) \Longleftrightarrow e^{-\theta}=4 \quad \text{and}\quad \theta =0$. But, $e^0 =1$. What am I doing wrong?
EDIT 10/26/18
Would like to provide a more systematic and traditional approach:
\begin{align}
r^{1+i}e^{i\theta(1+i)}&=r^{1+i}e^{-\theta}(\cos\theta + i\sin\theta)=4(\cos 0 +i\sin 0)=4\\
r^{1+i}e^{-\theta}&=4 \quad \text{and}\quad \theta=0\\
\log_e(r^{1+i}e^{-\theta})&=\log_{e}(4)\\
\log_(r^{1+i})+\log(e^{-\theta})&=\log_{e}(4)\\
(1+i)\log_{e}(r)-\theta&=\log_{e}(4)\\
\log_{e}(r)&=\frac{\log_e(4)}{1+i}\\
\log_{e}(r)&=\log_{e}(4^{\frac{1}{1+i}})\\
r&=4^{\frac{1}{1+i}}\\
r&=4^{\frac{1-i}{2}}=[4^\frac{1}{2}]^{1-i}=2^{1-i}
\end{align}
|
$$4 = 2^2 = z^{1 + i}$$
$$2 \log 2 = (1+i) \log z$$
$${2 \over 1+i} \log 2 = \log z$$
$$2^{2 \over 1 + i} = z$$
$$2^{2(1-i) \over (1+i)(1-i)}$$
$$2^{2(1-i) \over 2} = z$$
$$2^{1 - i} = z$$
Check:
$$\left( 2^{1-i}\right)^{1+i} = 2^{1+1} = 2^2 = 4$$
|
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|
$S(r,s) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+r+1)(2k+s+1)}$ for $r>s\geq0$ and even integers $S$ generates $\pi$ or rational values. \begin{align}
S(r,s) = \sum_{k=0}^{\infty} = \frac{(-1)^k}{(2k+r+1)(2k+s+1)}
\end{align}
Assuming $r>s\geq 0$ and both $r,s$ even integers.
A few values, using wolfram:
\begin{align}
&S(2,0) = \pi/4-1/2\\
&S(4,0) = 1/6\\
&S(4,2) = 5/6 - \pi/4\\
&S(6,0) = \pi/12-13/90\\
&S(6,2) = 1/30\\
&S(6,4) = \pi/4 - 23/30\\
&S(8,0) = 19/210\\
&S(8,2) = 181/630 - \pi/12
\end{align}
For the rational values of $S(r,s)$ we have:
\begin{align}
S(r,s) = \frac{1}{r-s} \left( \frac{1}{s+1}-\frac{1}{s+2}+\dots+\frac{1}{r-1} \tag{1} \right)
\end{align}
I think for the irrational values must be something similar to $(1)$ except we will have to add $\pi$ after the sum.
What is the general formula for $S$ (or just the formula for irrational values) $?$
|
Write
$$H_z=\psi(z)+\gamma$$
where $\psi(z)$ is the digamma function. Then
$$H_{(z+1)/2}-H_{z/2}=2\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+z}$$
$$H_{z+1}=\frac{1}{z}+H_z$$
for $z$ nonzero and not a negative integer (DLMF 5.7.7, DLMF 5.5.2). Additionally, if $0<p/q<1$ is a rational number in lowest terms, then
$$H_{p/q}=-\ln q-\frac{\pi}{2}\cot\left(\frac{\pi p}
{q}\right)+\frac{1}{2}\sum_{k=1}^{q-1}\cos\left(\frac{2\pi kp}{q}\right)\ln%
\left(2-2\cos\left(\frac{2\pi k}{q}\right)\right)$$
(DLMF 5.4.19).
You should be able to get a closed form for your $S(r,s)$ from there.
|
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|
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
|
Your calculations are likely to be correct when $x>1$. When $x <-1$ write $\sqrt {\frac {x+1} {x-1}}$ as $\sqrt {\frac {-x-1} {-x+1}}=\frac {\sqrt {-x-1}}{\sqrt {-x+1}}$ and now compute the derivative as before. You will get the correct answer now.
|
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|
Using the Residue Theorem to Prove that $\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{3/{2}}}.$ How do you evaluate the following integral? Here we take $a>1$.
$$\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{\frac{3}{2}}}.$$
I know I have to use the Residue Theorem, however, I am stuck on which contour to use, and also how to find the pole of the function. Any hints are greatly appreciated.
|
Use the contour $|z| = 1$
First change the cosines into exponential forms.
$$\large\int \frac {1}{(a+\frac{e^{it}}{2} + \frac{e^{-it}}{2})^2} \ d\theta$$
$$z = e^{i\theta}\\ d\theta = \frac {1}{iz} dz$$
$$\large \oint_{|z| = 1} \frac {1}{iz(a+\frac{z}{2} + \frac{z^{-1}}{2})^2} \ dz$$
Which simplifies to:
$$\large \oint_{|z| = 1} \frac {4z}{i(z^2 + 2az+ 1)^2} \ dz\\
\oint_{|z| = 1} \frac {4z}{i(z+ a + \sqrt {a^2-1})^2(z+ a - \sqrt {a^2-1})^2} \ dz$$
Has one pole inside the contour.
The residual at $z = -a+\sqrt {a^2 - 1} = 2\pi i \frac {d}{dz} \frac {4z}{i(z+ a + \sqrt {a^2-1})^2}$ evaluated at $z = -a+\sqrt {a^2 - 1}$
|
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|
For any integer $p \ge 3$, the largest integer $r$ such that $(x-1)^r$ is a factor of the polynomial $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$
For any integer $p \ge 3$, the largest integer $r$ such that $(x-1)^r$ is a factor of the polynomial $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$
I have tried the following:
If $(x-1)^r$ is a factor of $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$ $\implies$
$(x-1)^r.f(x)=2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$. Now $x-1$ is definitely a factor of $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$ as putting $x=1$ in the equation we get $2-p(p+1)+2(p^2-1)-p(p-1)=0$. Now how do I draw a similar approach for $(x-1)^r$ . I was thinking of induction but I am stuck here. Can anyone help. The answer is 3 i.e $for \space r_{max}=3$ $(x-1)^r$ is a factor of the polynomial $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$
|
For $(x-1)^r$ to be a divisor of $f(x)$, we should have $f^{(r-1)}(1)=0$ (the $r-1$-st derivative).
So consider
$$f'(x)=2(p+1)x^{p}-2p(p+1)x+2(p^2-1).$$
Then $f'(1)=2(p+1)-2p(p+1)+2(p^2-1)=0.$ This means $(x-1)^2$ is a factor of $f$.
Now consider,
$$f''(x)=2p(p+1)x^{p-1}-2p(p+1).$$
Thus $f''(1)=2p(p+1)-2p(p+1)=0$. This means $(x-1)^3$ is a factor of $f$.
Now consider,
$$f'''(x)=2p(p-1)(p+1)x^{p-2}.$$
It is clear that $f'''(1) \neq 0$. Thus $(x-1)^3$ is the highest power of $(x-1)$ that divides $f(x)$.
|
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|
Convergence or Divergence of $\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$ How to determine convergence of this series. $$\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$$
I've tried using comparison test:
$\sin(\pi/n) \leq \pi/n $, so:
$$(3^{1/n}-1)\sin(\pi/n)<(3^{1/n}-1)\pi/n < (3^{\frac{1}{n}})\frac{\pi}{n}$$
By comparison test if $\sum(3^{\frac{1}{n}})\frac{\pi}{n}$ is convergent, so would be initial.
But $\sum(3^{\frac{1}{n}})\frac{\pi}{n}$ is divergent.
I also know that $\sin(\pi/n)$ is divergent. How would it help? Can you give a hint?
|
Your series is absolutely convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{\log 3}{n}\cdot\frac{\pi}{n}=\frac{\pi^3}{6}\log(3)$.
If you like explicit bounds, you may notice that
$$ \frac{3}{2} = \frac{2n+1}{2n}\cdot\frac{2n+2}{2n+1}\cdot\ldots\cdot\frac{2n+n}{3n-1} ,\qquad 2=\frac{n+1}{n}\cdot\frac{n+2}{n+1}\cdot\ldots\cdot\frac{n+n}{2n-1}$$
$$ 3 = \prod_{k=1}^{n}\frac{(2n+k)(n+k)}{(2n+k-1)(n+k-1)}=\prod_{k=1}^{n}\frac{(2n+k)(2n-k+1)}{(2n+k-1)(2n-k)} $$
and by the AM-GM inequality
$$ 3^{1/n} \leq \frac{1}{n}\sum_{k=1}^{n}\frac{(2n+k)(2n-k+1)}{(2n+k-1)(2n-k)}=\frac{1}{n}\sum_{k=1}^{n}\frac{(4n+1)^2-(2k-1)^2}{(4n-1)^2-(2k-1)^2} $$
or
$$ 3^{1/n}-1 \leq \sum_{k=1}^{n}\frac{16}{(4n-1)^2-(2k-1)^2}=\sum_{k=1}^{n}\frac{4}{(2n-k)(2n+k-1)}. $$
By letting $H_n = \sum_{k=1}^{n}\frac{1}{k}$ the previous line gives
$$ 3^{1/n}-1 \leq \frac{4}{4n-1}\sum_{k=1}^{n}\left(\frac{1}{2n-k}+\frac{1}{2n+k-1}\right)=\frac{4}{4n-1}\left[H_{3n-1}-H_{n-1}\right] $$
and by the Hermite-Hadamard inequality it follows that
$$ 3^{1/n}-1 \leq \frac{4}{4n-1}\left(\log 3+\frac{4}{3(4n-1)}\right)\leq \frac{1}{n}\left(\log(3)+\frac{1}{n}\right). $$
In particular the given series is bounded by $\frac{\pi^3}{6}\log(3)+\pi\zeta(3)$.
|
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|
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)} $$
I do not know how to get a telescoping series from here to cancel terms.
|
You are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $3$ or greater than $n$.
$$
\begin{aligned}
& \sum_{k=1}^n\frac{-1}{2k} + \sum_{k=1}^n\frac{3}{k+1} - \sum_{k=1}^n\frac{5}{2}\frac{1}{k+2} \\
& = \left[-\frac{1}{2} - \frac{1}{4} + \frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{3}{2} + \frac{1}{2}\sum_{k=2}^n\frac{6}{k+1}\right] - \left[\frac{1}{2}\sum_{k=1}^n\frac{5}{k+2}\right] \\
& = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n+1}\frac{6}{k}\right] - \left[\frac{1}{2}\sum_{k=3}^{n+2}\frac{5}{k}\right] \\
& = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n}\frac{6}{k} + \frac{6}{2}\frac{1}{n+1}\right] - \left[\frac{1}{2}\sum_{k=3}^{n}\frac{5}{k} + \frac{5}{2}\frac{1}{n+1} + \frac{5}{2}\frac{1}{n+2}\right] \\
& = \frac{3}{4} + \frac{1}{2}\sum_{k=3}^{n}\left[\frac{-1 + 6 - 5}{k}\right] + \frac{6}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+2} \\
& = \frac{3}{4} + \frac{1}{2(n+1)} - \frac{5}{2(n+2)}
\end{aligned}
$$
|
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|
Fractional Linear Transformation of the Image of the Line $y=4-x$
I am trying to find the image of the line $y=4-x$ under the fractional linear transformation $$w=\frac{8}{z-2-2i}.$$
My method is as follows:
Rearranging yields $$z=\frac{8}{w}+2+2i.$$
Now, \begin{align}
z=x+iy&=\frac{8}{u+iv}+2+2i \\
&=\frac{8+(u+iv)(2+2i)}{u+iv}\times\frac{u-iv}{u-iv} \\
&=\frac{8u+2u^2+2v^2}{u^2+v^2}+i\frac{2u^2-8v+2v^2}{u^2+v^2}
\end{align}
Equating real and imaginary components yields
$$x=\frac{8u+2u^2+2v^2}{u^2+v^2} \ \ \text{and} \ \ \ y=\frac{2u^2-8v+2v^2}{u^2+v^2}.$$
Hence, $$\frac{2u^2-8v+2v^2}{u^2+v^2}=4-\frac{8u+2u^2+2v^2}{u^2+v^2}\implies v=u.$$
Is my method valid?
|
Your calculation is correct.
Here is a shorter way using the properties of the given mapping:
*
*Note that your mapping is a so called Möbius transformation. So, it maps generalized circles (circles and lines) onto generalized circles.
*To find the image of the line $y= 4-x$ which is the same as $z=x + (4-x)i$ you need only the image of $3$ points:
$$x = 0 \Rightarrow z = 4i \mapsto w = -2(1+i)$$
$$x = 4 \Rightarrow z = 4 \mapsto w = 2(1+i)$$
$$\mbox{Finally } z = \infty \mapsto w = \lim_{z \to \infty}\frac{8}{z-2-2i} = 0$$
So, the image is the line passing through $-2(1+i), 2(1+i), 0$. Hence, you get
$$\boxed{y = x}$$
|
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|
Telescoping $\sum_{n\ge2} \ln(1-\frac1{n^2})$ leads to wrong result I know that this series converges to $-\ln2$. I'm not looking for someone to show me that here. I've used a way of telescoping the series that gives a result of 0. I want to know what's wrong about my process, I don't care (I already know) about other ways to solve this.
$$\sum_{n=2}^\infty \ln\left(1-\frac1{n^2}\right) = \sum_{n=2}^\infty \ln\left(\frac{n^2-1}{n^2}\right) = \sum_{n=2}^\infty \left( \ln\left(n^2-1\right) - \ln\left(n^2\right)\right)$$
From that we can start telescoping:
$$= \ln3 - \ln 4 + \ln 8 - \ln 9 + \ln 15 - \ln 16 + \ln 24 - \ln 25 + \ln 35 - \ln 36 + \ln 48 - \ln49 + \ln 63 - \ln 64 + \ln 80 - \ln 81 + \,\,...$$
I separate the odd and even ones and make pairs:
$$= (\ln3 - \ln 9) + (\ln 15 - \ln 25) + (\ln 35 - \ln 49) + (\ln 63 - \ln 81) + \,\,... \\+ (-\ln 4 + \ln 8) + (-\ln 16 + \ln 24) + (-\ln36 + \ln48) + (-\ln 64 + \ln80) + \,\,... $$
$$= \ln\frac13 + \ln\frac35 + \ln\frac57 + \ln\frac79 + \,\,...\\+\ln 2 + \ln\frac32 + \ln\frac43 + \ln\frac54 + \,\,...$$
And they all cancel out from that.
$$=\ln\left(\frac{1\cdot 3\cdot 5\cdot 7\cdot\, ...}{3\cdot 5\cdot 7\cdot 9\cdot\, ...}\right) + \ln\left(\frac{2\cdot 3\cdot 4\cdot 5\cdot\, ...}{2\cdot 3\cdot 4\cdot\, ...}\right)$$
We end up with $\ln 1 + \ln 1$ which is 0. Any explanation? What did I do wrong?
|
Your argument should work if you carefully analyze the partial sum:
\begin{align*}
\sum_{k=1}^{2n} \log\left(1-\frac{1}{k^2}\right)
&= \log\left(\frac{1\cdot3\cdots(2n-1)}{3\cdot5\cdots(2n+1)}\right) + \log\left(\frac{2\cdot3\cdots(n+1)}{1\cdot2\cdots n}\right) \\
&= -\log(2n+1) + \log(n+1) \\
&= \log\left(\frac{n+1}{2n+1}\right) \\
&\xrightarrow[n\to\infty]{} \log\frac{1}{2}.
\end{align*}
|
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|
calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$ So I multiplied by the conjugate and got $$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$
and this is where I got stuck.
|
Here's a completely elementary solution.
For $x > 1$
$\begin{array}\\
d(x)
&=x + \sqrt[3]{1-x^3}\\
&=x - \sqrt[3]{x^3-1}\\
&=x(1 - \sqrt[3]{1-1/x^3})\\
&\gt 0\\
\end{array}
$
Also,
since
$(1-z)^3 < 1-z$
for $0 < z < 1$,
$\sqrt[3]{1-z} > 1-z$
so that
$\sqrt[3]{1-1/x^3}
\gt 1-\frac1{x^3}
$
for $x > 1$.
Therefore
$\begin{array}\\
d(x)
&=x(1 - \sqrt[3]{1-1/x^3})\\
&<x(1 - (1-\frac1{x^3}))\\
&= x(\frac1{x^3})\\
&=\frac1{x^2}\\
\end{array}
$
Therefore
$0
\lt d(x)
\lt \frac1{x^2}
$
so
$\lim_{x \to \infty} d(x)
=0
$.
|
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|
$2 < (1 + \frac{1}{n})^n < 3$ for all $n > 2$ Show by induction that $2 < (1 + \frac{1}{n})^n < 3$ for all natural n > 2.
*
*Induction base. For $n = 3$ the inequality is obviously true.
*Assume that for $n = k$ inequality is true, than i can prove $(1 + \frac{1}{n+1})^{n +1} < 3$.
$$\frac{(n + 1)^{n}}{n^{n}} < 3 \\
1 < \frac{3n^{n}}{(n+1)^{n}} \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < \frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}}, but \\
\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3, then \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < 3
$$
How can i prove, that $(1 + \frac{1}{n+1})^{n +1} > 2$
|
I'd like to point out an error in the proof for $\left(1+\frac1n\right)^n < 3$.
The step that says $\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3$ is wrong (I plugged $n=3$ into the windows calculator and got a term slightly above 3). That term is equal to
$$\frac{3(n^2 + 2n)^n(n+2)}{(n^2+2n+1)^n(n+1)} = 3\frac{(n^2 + 2n)^n}{(n^2+2n+1)^n}\frac{n+2}{n+1}$$
so the product of $3$, a term sligthly below $1$ and a term slightly above $1$. I suspect it will be $>3$ for all cases, but that might not be easy to proof.
|
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|
Find a locus of points Given a triangle $ABC$, $A'$ and $B'$ halves $BC$ and $AC$. We have a variable point on a line $AB$. Parallel to $AA'$ and $BB'$ through $P$ cuts $AC$ in $E$ and $BC$ in $F$. Now line $EF$ cuts $AA'$ in $M$ and $BB'$ in $N$. Lines $A'N$ and $B'M$ meet at $Q$. What is a locus of $Q$?
When playing in Geogebra I found it could be an ellipse which goes through the midpoints of a segments $AG$, $BG$ and $CG$ where $G$ is a gravity center. Also, this ellipse seem to touches the sides of triangle at their midpoints. But when I was trying to write down equations of intersection points I get a headeacke due to huge formulas I get. Does anyone have nice idea how to avoid this equations?
|
Since the construction uses only affine-independent properties (incidence, collinearity, parallelism, midpoints) we can choose a convenient form of $\triangle ABC$. If we can show that the locus is the Steiner Inellipse for that form, then the locus is the Inellipse for every triangle.
We'll position vertex $C$ at the origin. Treating $A$ and $B$ as position vectors, we find the midpoints $A^\prime$ and $B^\prime$ as ...
$$A^\prime = \frac12(B+C) = \frac12 B \qquad B^\prime = \frac12(A+C) = \frac12 A \tag{1}$$
Define
$$P := \frac12(A+B)+\frac{p}{2}(B-A) \tag{2}$$
(where $p$ affects displacement from the midpoint of $\overline{AB}$), from which we derive
$$E = \frac{1}{4}(3-p)A \qquad F = \frac{1}{4}(3+p)B \tag{3}$$
and
$$\begin{align}
M &= \frac13( 2E+\phantom{2}F) = \,\,\frac16(3 - p) A + \frac1{12} (3 + p) B \\[6pt]
N &= \frac13(\phantom{2}E+2F) = \frac1{12}(3 - p) A + \frac16 (3 + p) B
\end{align} \tag{4}$$
(Note: The $EF$-representation shows that $M$ and $N$ trisect $\overline{EF}$.) And, finally, that
$$Q = \frac{1}{6(p^2+3)}\;\left(\;(p - 3)^2 A + (p + 3 )^2 B \;\right) \tag{5}$$
Now, let's locate $A$ and $B$ so that $\triangle ABC$ is an equilateral triangle of inradius $r$ (and circumradius $2r$), with the $x$-axis as an axis of symmetry:
$$A,B = 2r\sqrt{3}\left(\cos\frac{\pi}6,\pm\sin\frac{\pi}{6}\right) \tag{6}$$
Then
$$Q = \frac{r}{p^2+3} \left(p^2+9, -2 p \sqrt{3}\right) = 2r (1,0) + \frac{r}{p^2+3}\left(3-p^2, -2p\sqrt{3}\right) \tag{7}$$
and we see that
$$\left(Q_x-2r\right)^2+Q_y^2 = \frac{r^2}{\left(p^2+3\right)^2}\left((3-p^2)^2 + 12 p^2\right) = r^2 \tag{8}$$
That is, the locus of $Q$ for this particular $\triangle ABC$ is the incircle, which is, in fact, the Steiner Inellipse for that form. $\square$
|
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|
Permutation of the string Ronald Mcdonald I am checking to see if I am on the right track for my Discrete Math class.
Given the string: RONALDMCDONALD
Vowels: AAOO
How many permutations do NOT have consecutive vowels in them?
I first visualized the problem: _R_N_L_D_M_C_D_N_L_D_
I answered this question by taking the total number of choices: ${11 \choose 4}$
Then the total number of vowel arrangements: ${11 \choose 2}$
Finally, the total number of constant arrangements $\frac{10!}{2!2!3!}$
I multiplied these together for a final answer of: ${11 \choose 4}$ * ${11 \choose 2}$ * $\frac{10!}{2!2!3!}$
Am I right to assume the total number of vowel arrangements is equal to 11 choose 2?
How many permutations contain the vowels in order (A's before O's)?
Would the answer to this problem be ${11 \choose 4}$?
How many permutations contain the substring OLAND?
If we take OLAND as a super-letter then the letters we have left to permutate are
RNDMCDOAL
Within this string we have 1 R, 1 N, 2 D, 1 M, 1 C, 1 O, 1 A, 1 L
Therefore the answer is $\frac{10!}{2!}$
|
How many permutations of the string RONALDMCDONALD do not have consecutive vowels in them?
The ten consonants consist of $1$ R, $2$ Ns, $2$ Ls, $3$ Ds, $1$ M, $1$ C. The four vowels consist of $2$ As and $2$ Os.
Your idea of arranging the consonants, then separating the vowels by placing them between the consonants or at the ends of the row is sound. Let's correct your count.
We arrange the ten consonants first. We have ten positions to fill. Choose three of them for the Ds, 2 of the remaining seven positions for the Ls, and two of the remaining five positions for the Ls. This leaves us with three positions for three distinct consonants. The C, R, and M can be placed in those positions in $3!$ ways. Hence, the number of arrangements of the ten consonants is
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! = \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{2!3!} \cdot 3! = \frac{10!}{3!2!2!}$$
As you observed, arranging the ten consonants creates eleven spaces, nine between successive consonants and two at the ends of the row. To separate the vowels, we must select four of these eleven spaces in which to place a vowel, as you did. We must then select which two of these four spaces will be filled with As. The remaining two must be filled with Os. Hence, the number of ways of selecting spaces for the vowels and arranging them in the selected spaces is
$$\binom{11}{4}\binom{4}{2} = \frac{11!}{7!4!} \cdot \frac{4!}{2!2!} = \frac{11!}{7!2!2!}$$
The factor of $7!$ in the denominator results from the fact that the seven spaces that are left blank spaces are indistinguishable. The factors of $2!$ in the denominator represent the fact that the two As are indistinguishable and the two Os are indistinguishable.
Hence, the number of permutations of the string RONALDMCDONALD that do not have consecutive vowels in them is
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! \cdot \binom{11}{4}\binom{4}{2}$$
How many permutations of the string RONALDMCDONALD contain the vowels in alphabetical order?
If we ignore the restriction, we have $14$ letters to arrange. Of these, we have $1$ R, $2$ Ns, $2$ Ls, $3$ Ds, $1$ M, $1$ C, $2$ As, and $2$ Os. They can be arranged in
$$\binom{14}{3}\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3! = \frac{14!}{3!2!2!2!2!}$$
distinguishable ways.
The four vowels can be arranged in $\binom{4}{2} = 6$ distinguishable ways. Of these six ways, only one leaves the vowels in alphabetical order. By symmetry, the number of arrangements of the string RONALDMCDONALD in which the vowels appear in alphabetical order is
$$\frac{1}{6}\binom{14}{3}\binom{11}{2}\binom{7}{2}\binom{5}{2}3!$$
How many permutations of the string RONALDMCDONALD contain the substring OLAND?
If we treat OLAND as a block, we have ten objects to arrange. They include OLAND, $1$ R, $1$ N, $1$ L, $2$ Ds, $1$ M, $1$ C, $1$ A, $1$ O. Choose two of the ten positions for the two Ds. The remaining eight distinct objects can be arranged in the remaining eight places in $8!$ orders. This gives a preliminary count of
$$\binom{10}{2}8! = \frac{10!}{2!}$$
which is what you have obtained.
However, notice that it is possible to have two substrings of the form OLAND. We have counted these strings twice, once for each way we could have designated one of the substrings as the OLAND in the string. We only want to count them once, so we must subtract them from the total.
If we have two blocks of OLANDs, we also have $1$ R, $1$ D, $1$ M, and $1$ C. Thus, we have six objects to arrange, including two identical blocks and four distinct letters. We choose two of the six positions for the blocks, then arrange the four distinct letters in the remaining four positions, which can be done in
$$\binom{6}{2}4! = \frac{6!}{2!}$$
distinguishable ways.
Hence, the number of permutations of the string RONALDMCDONALD that contain the string OLAND is
$$\binom{10}{2}8! - \binom{6}{2}4!$$
|
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|
Determinate of symmetric $82\times 82$ matrix. I am trying to calculate the determinant of the matrix:
$
\left(\begin{matrix} -6 & -5 & & \dots & -5\\
-5 & -6 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -6 & -5\\
-5 & \dots & & -5 & -6
\end{matrix}\right)\in \mathbb{R}^{82\times 82}
$
How am I supposed to do this with a matrix so large? I tried looking for a pattern of Gauss Elimination but I got stuck. For clarification, the diagonal entries are $-6$ and the rest of the entries are $-5$.
|
Let $M = \left(\begin{matrix} -6 & -5 & & \dots & -5\\
-5 & -6 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -6 & -5\\
-5 & \dots & & -5 & -6
\end{matrix}\right)$.
Then $M=\left(\begin{matrix} -5 & -5 & & \dots & -5\\
-5 & -5 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -5 & -5\\
-5 & \dots & & -5 & -5
\end{matrix}\right)+\left(\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right)$
$=
\left(\begin{matrix} 5\\
5\\
\vdots \\
5\\
5
\end{matrix}\right)\left(\begin{matrix} -1 & -1 & & \dots & -1\\
\end{matrix}\right)+\left(\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right)$.
By the matrix determinant lemma, $\det M=\left(1+\left(\begin{matrix} -1 & -1 & & \dots & -1\\
\end{matrix}\right)\left(\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right)\left(\begin{matrix} 5\\
5\\
\vdots \\
5\\
5
\end{matrix}\right)\right)\cdot \left\vert\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right\vert=(1+82\cdot5)\cdot1=411.$
|
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|
Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$?
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$
I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few combinatorial proofs but not able to succeed this one.
Help would be appreciated.
|
Brute force:
$27n^{3} - 27n^{2}+ 6n = 27n^{3} - 27n^{2} + 6n$
$27n^{3} - 9n^{2} - 18n^{2} + 6n = 3n^{3} - 3n^{2} - 6n^{2} + 6n + 18n^{3} - 18n^{2} + 6n^{3}$
$(9n^{2} - 3n)(3n - 2) = (3n^{2} - 3n)(n - 2) + 18n^{2}(n-1) + 6n^{3}$
$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^{3}$
$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6 + (6n)n(n-1)/2 + n^{3}$
$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$
|
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|
Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is wrong
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3=((\frac{\alpha}{\alpha +1})+(\frac{\beta}{\beta +1})+(\frac{\gamma}{\gamma +1}))^3-3((\frac{\alpha}{\alpha +1})(\frac{\beta}{\beta +1})(\frac{\gamma}{\gamma +1}))$
And then I broke the terms to get
$(\frac{13}{5})^3-\frac{36}{5}$
but this is $\ne 44$, which should be the answer.
I know where I was wrong, the formula is $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$
|
I think your way is not really practical. Here it is a shortcut:
$$ \begin{array}{|c|c|}\hline \text{Roots}&\text{Polynomial}\\
\hline \alpha,\beta,\gamma & x^3+2x^2+3x+3\\
\hline \frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma} & x^3+x^2+\frac{2}{3}x+\frac{1}{3}\\
\hline 1+\frac{1}{\alpha},1+\frac{1}{\beta},1+\frac{1}{\gamma} & x^3-2x^2+\frac{5}{3}x-\frac{1}{3}\\
\hline \frac{\alpha}{\alpha+1},\frac{\beta}{\beta+1},\frac{\gamma}{\gamma+1}&x^3-5x^2+6x-3 \\ \hline
\end{array} $$
Now by setting $A=\frac{\alpha}{\alpha+1},B=\frac{\beta}{\beta+1},C=\frac{\gamma}{\gamma+1}$ we have $A+B+C=5$ and $AB+AC+BC=6$ by Vieta's formulas, hence $A^2+B^2+C^2=25-12=13$. Additionally for any $z\in\{A,B,C\}$ we have $z^3=5z^2-6z+3$, hence
$$ A^3+B^3+C^3 = 5\cdot 13-6\cdot 5+3\cdot 3 = 65-30+9 = \color{red}{44}.$$
An equivalent way is to compute the trace of $\left(M(M+I)^{-1}\right)^3=(I+M^{-1})^{-3}$, where $M$ is the companion matrix of $x^3+2x^2+3x+3$.
|
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"url": "https://math.stackexchange.com/questions/2983790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct? Is only one correct? Is there an exception that I am missing?
|
We can also proceed as follow
$$x(1 - ax) \le \frac{1}{4a} \iff ax^2- x+\frac{1}{4a}\ge 0$$
which holds when $a>0$ since
$$\Delta=1-4\cdot \frac{1}{4a}=1-1=0$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove $\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$ without the Beta function I know how to prove
$$\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$$
by tackling it with the beta function.
I was actually wondering if there is a proof of this fact without using the property of the Beta function $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
|
We have
\begin{align}
f_n&=\color{blue}{\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{1}{k}}\\
&=\sum_{k=1}^{n}(-1)^{k-1}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)\frac{1}{k}\\
&=f_{n-1}+\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}\frac{1}{k}\\
&=f_{n-1}-\frac{1}{n}\sum_{k=1}^{n}(-1)^k\binom{n}{k}\\
&=f_{n-1}-\frac{1}{n}\left((1-1)^n-1\right)\\
&=f_{n-1}+\frac{1}{n}\\
&\,\,\color{blue}{=H_n}
\end{align}
Note: This can be found for instance as Example 3, section 1.2 in Combinatorial Identities by John Riordan.
Another approach is based upon generating functions and the Euler transform of a generating series $A(z)$ which is given as
\begin{align*}
A(z)=\sum_{n= 0}^\infty a_nz^n&\quad\longrightarrow\quad
\frac{1}{1-z}A\left(\frac{z}{1-z}\right)=\sum_{n= 0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_k\right)z^n\tag{1}\\
a_n\quad&\quad\longrightarrow\quad\qquad
\sum_{k=0}^n\binom{n}{k}a_k&
\end{align*}
Applying the Euler transform (1) to the left-hand side of the identity
\begin{align*}
\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n
\end{align*}
we set $a_k=\frac{(-1)^{k-1}}{k}$ and obtain as generating function
\begin{align*}
A(z)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}z^k=\ln(1+z)\tag{2}
\end{align*}
The Euler transform of (2) is
\begin{align*}
\frac{1}{1-z}\ln\left(1+\frac{z}{1-z}\right)&=\sum_{n=1}^\infty\left(\color{blue}{\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k}}\right)z^n
\end{align*}
On the other hand we have
\begin{align*}
\frac{1}{1-z}\ln\left(1+\frac{z}{1-z}\right)&=\frac{1}{1-z}\ln\left(\frac{1}{1-z}\right)\\
&=-\frac{1}{1-z}\ln(1-z)\\
&=\left(\sum_{k=0}^\infty z^k\right)\left( \sum_{l=1}^\infty \frac{1}{l}z^l\right)\\
&=\sum_{n=1}^\infty\left(\sum_{{k+l=1}\atop{k\geq 0,l\geq 1}}^n \frac{1}{l}\right)z^n\\
&=\sum_{n=1}^\infty\sum_{k=1}^n\frac{1}{k}z^n\\
&=\sum_{n=1}^\infty\color{blue}{H_n}z^n\\
\end{align*}
and the claim follows.
Note: A proof of the Euler transformation formula can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a=b+c$ prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer
If $a=b+c$, and $a$,$b$,$c\in \Bbb N$, prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer.
Source: a list of problems used in the preparation to math contests.
My attempt:
By making the substitution $a=b+c$ in $S$ and developing $(b+c)^4$, it is easy to show that
$$S=2(b^4+c^4+bc(2b^2+3bc+2c^2))$$
an expression with the form $S=2K$. The problem now is how to prove that K is a square of a positive integer. I also tried to use Newton Identities but with no luck (Note: later, after a hint, I found a way to solve using this approach, see below).
Hints and answers are welcomed. Sorry if this is a dup.
|
We have:
$$S=(b+c)^4+b^4+c^4=2b^4+4b^3c+6b^2c^2+4bc^3+2c^4$$
$$=2(b^4+2b^3c+3b^2c^2+2bc^3+c^4)=2 (b^2+bc+c^2)^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simultaneous real solution of $x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$ I am trying to solve the following system of non-linear equations in real numbers:
$x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$, with $x,y$ real.
I can only see that $xy\ne 0$.
I have no clue whether a solution exists or not and how to find any solution. I cannot seem to be able to separate $x,y$ or write it in a good parametric form.
Please help.
|
Not a very satisfactory answer since it involves the cubic formula...
Set $u = xy$. Multiply the first equation by $x^3$ and the second by $x$. Then we get
$x^6+u^3+x^3+6ux^3 = 0$
$u^2+u+x^3 = 0$
Solving for $x^3$ in the second and substituting into the first, we get
$u^4+2u^3+u^2+u^3-u^2-u+6u(-u^2-u)=0$
$u^4+2u^3+u^2+u^3-u^2-u-6u^3-6u^2=0$
$u^4-3u^3-6u^2-u=0$
$u^3-3u^2-6u-1=0$
Which can be solved by the cubic formula. Once this is done, $u$ can be substituted back in to find $x$. Then we can find $y$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
|
One can see that $1/2$ is a solution of $\frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. Then it is easy to see that also $5/2$ is a solution.
Can you proceed ?
|
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"timestamp": "2023-03-29T00:00:00",
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|
What value does $\frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots(2n)}$ tend to? I need to find where this sequence tends to:
$$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$
My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.
|
By ratio root criterion we have
$$a_n=\frac{n(n+1)(n+2)...(2n)}{n^n} \quad b_n=\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}=\sqrt[n] a_n$$
then
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}\frac{n^n}{(n+1)^{n+1}}=\frac{(2n+2)(2n+1)}{n(n+1)}\frac{1}{\left(1+\frac1n\right)^n}\to \frac 4 e$$
which implies that $b_n \to \frac 4 e$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2992724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that $\det\left[\begin{smallmatrix}1&\cos a&\cos b\\ \cos a&1&\cos(a+b) \\ \cos b&\cos(a+b)&1 \end{smallmatrix}\right]=0$ I am unable to show - without expanding, by using determinant properties - that
$$\det\begin{bmatrix} 1 &\cos a &\cos b\\
\cos a &1 &\cos(a+b) \\
\cos b &\cos(a+b) &1
\end{bmatrix}=0$$
I am using trigonometric identities to solve this but I don't understand what would be the next step.
|
To flesh out my comment from earlier,$$\Delta:=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
\cos b & \cos\left(a+b\right) & 1
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & 0\\
\cos a & 1 & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|.$$We may as well also simplify the second row/column:
$$\Delta=\left|\begin{array}{ccc}
1 & 0 & 0\\
\cos a & \sin^{2}a & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left|\begin{array}{ccc}
1 & 0 & 0\\
0 & \sin^{2}a & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|.$$Well, now we just have an easy $2\times 2$ determinant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2994980",
"timestamp": "2023-03-29T00:00:00",
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|
What am I doing wrong finding the limit of $\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$? $$\lim_{x\to\infty}\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$$
Divide by $x^2$, get
$$\lim_{x\to\infty}(1,5)^\infty=\infty$$
The answer in the book is $0$.
I've also tried substituting $x$ for $\frac{1}{h}$ where $h$ tends to zero and using some form of $(1+\frac{1}{n})^n$, and using the exponent rule, but everything lands me at infinity.
Do I misunderstand something fundamentally?
|
Because the natural logarithm is injective, it turns out that: $\lim_{x \to \infty} \frac{3x²-x+1}{2x²+x+1}^{\frac{x^3}{1-x}}=(\lim_{x \to \infty} \frac{3x²-x+1}{2x²+x+1})^{(\lim_{x \to \infty} \frac{x^3}{1-x})}$. Thus the limit is $(\frac{3}{2})^{-\infty}=\frac{1}{1.5^{\infty}}=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2998322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), on the right hand side of the page. Am I incorrect with my use of the modulus operator after I integrated each side then?
$$\begin{align*}x\frac{dy}{dx}&=y-\frac{1}{y}\quad\quad(-1<y<1)\\
\int\frac{1}{y-\frac{1}{y}}\ dy&=\int\frac{1}{x}\ dx\\
\int\frac{y}{y^2-1}\ dy&=\int\frac{1}{x}\ dx\\
\frac{1}{2}\ln{|y^2-1|}&=\ln{|x|}+c\\
\sqrt{|y^2-1|}&=A|x|\ \text{, where $A=e^c$}\\
|y^2-1|&=A^2x^2\\
\implies y^2-1&=A^2x^2&1-y^2&=A^2x^2\\
y^2&=A^2x^2+1&y^2&=1-A^2x^2\\
\\
\therefore y&=\pm\sqrt{Kx^2+1}&y&=\pm\sqrt{1-Kx^2}\ \text{, where $K=A^2$}\\
\end{align*}$$
Thanks :)
|
If $-1<y<1$ then $|y^2-1|=1-y^2$.
Another approach: Write the equation as
$$\dfrac{x\ dy-y\ dx}{x^2}=\dfrac{-1}{x^2y}\ dx$$
or
$$\left(\dfrac{y}{x}\right)d\left(\dfrac{y}{x}\right)=\dfrac{-1}{x^3}\ dx$$
and
$$\left(\dfrac{y}{x}\right)^2=\dfrac{1}{x^2}+C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post is someone said that this integral cannot be solved without using special functions.
Another alternative solutions will be appreciated.
|
$$I=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{x\to 1-x}=\int_0^1 \frac{\ln(1+x-x^2)}{1-x}dx$$
Averaging the two integrals from above gives us:
$$I=\frac12 \int_0^1 \frac{\ln(1+x-x^2)}{x-x^2}dx=\frac12I(1)$$
Where we considered, in order to apply Feynman's trick, the following integral:
$$I(a)=\int_0^1 \frac{\ln(1+a(x-x^2))}{x-x^2}dx\Rightarrow I'(a)=\int_0^1 \frac{1}{1+a(x-x^2)}dx$$
$$=\frac1a\int_0^1 \frac{1}{\left(\frac1a+\frac14\right)-\left(x-\frac12\right)^2}dx=\frac{2}{\sqrt{a(4+a)}}\ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)$$
$$I(0)=0\Rightarrow I=\frac12 (I(1)-I(0))=\int_0^1 \frac{1}{\sqrt{a(4+a)}}\ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)da$$
$$\text{let } \ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)=x\Rightarrow \frac{1}{\sqrt{a(4+a)}}da=dx$$
$$\Rightarrow I=\int_0^{\ln(\varphi^2)} x dx=\frac12 \ln^2( \varphi^2)=2\ln^2 \varphi$$
|
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|
Seeking methods to solve $\int_{0}^{\infty} \frac{x - \sin(x)}{x^3\left(x^2 + 4\right)} \:dx$ I recently asked for definite integrals that can be solved using the Feynman Trick. One of the responses is the following integral:
$$I = \int_{0}^{\infty} \frac{x - \sin(x)}{x^3\left(x^2 + 4\right)} \:dx$$
I employed Laplace Transforms with Feynman's Trick (as per a response below) to solve it, but am curious about other approaches that can be taken (in particular if using Feynman's Tricks).
|
My approach:
Let
$$I(t) = \int_{0}^{\infty} \frac{xt - \sin(xt)}{x^3\left(x^2 + 4\right)} \:dx$$
Where $I = I(1)$
Taking the first derivative:
$$ \frac{dI}{dt} = \int_{0}^{\infty} \frac{x - x\cos(xt)}{x^3\left(x^2 + 4\right)} \:dx = \int_{0}^{\infty} \frac{1 - \cos(xt)}{x^2\left(x^2 + 4\right)} \:dx$$
Taking the second derivative:
$$ \frac{d^2I}{dt^2} = \int_{0}^{\infty} \frac{x\sin(xt)}{x^2\left(x^2 + 4\right)} \:dx = \int_{0}^{\infty} \frac{\sin(xt)}{x\left(x^2 + 4\right)} \:dx$$
Now, take the Laplace Transform w.r.t $t$:
\begin{align}
\mathscr{L}\left[ \frac{d^2I}{dt^2}\right] &= \int_{0}^{\infty} \frac{\mathscr{L}\left[\sin(xt)\right]}{x\left(x^2 + 4\right)} \:dx \\
&= \int_{0}^{\infty} \frac{x}{\left(s^2 + x^2\right)x\left(x^2 + 4\right)}\:dx \\
&= \int_{0}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 4\right)}\:dx
\end{align}
Applying the Partial Fraction Decomposition we may find the integral
\begin{align}
\mathscr{L}\left[ \frac{d^2I}{dt^2}\right] &= \int_{0}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 4\right)}\:dx \\
&= \frac{1}{s^2 - 4} \int_{0}^{\infty} \left[\frac{1}{x^2 + 4} - \frac{1}{x^2 + s^2} \right]\:dx \\
&= \frac{1}{s^2 - 4} \left[\frac{1}{2}\arctan\left(\frac{x}{2}\right) - \frac{1}{s}\arctan\left(\frac{x}{s}\right)\right]_{0}^{\infty} \\
&= \frac{1}{s^2 - 4} \left[\frac{1}{2}\frac{\pi}{2} - \frac{1}{s}\frac{\pi}{2} \right] \\
&= \frac{\pi}{4s\left(s + 2\right)}
\end{align}
We now take the inverse Laplace Transform:
$$ \frac{d^2I}{dt^2} = \mathscr{L}^{-1}\left[\frac{\pi}{4s\left(s + 2\right)} \right] = \frac{\pi}{8}\left(1 - e^{-2t} \right) $$
We now integrate with respect to $t$:
$$ \frac{dI}{dt} = \int \frac{\pi}{8}\left(1 - e^{-2t} \right)\:dt = \frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) + C_1$$
Now
$$ \frac{dI}{dt}(0) = \int_{0}^{\infty} \frac{1 - \cos(x\cdot 0)}{x^2\left(x^2 + 4\right)} \:dx = 0 = \frac{\pi}{8}\left(\frac{1}{2} \right) + C_1 \rightarrow C_1 = -\frac{\pi}{16}$$
Thus,
$$ \frac{dI}{dt} = \int \frac{\pi}{8}\left(1 - e^{-2t} \right)\:dt = \frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) - \frac{\pi}{16}$$
We now integrate again w.r.t $t$
$$ I(t) = \int \left[\frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) - \frac{\pi}{16} \right] \:dt = \frac{\pi}{8}\left(\frac{t^2}{2} - \frac{e^{-2t}}{4} \right) - \frac{\pi}{16}t + C_2 $$
Now
$$I(0) = \int_{0}^{\infty} \frac{x\cdot0 - \sin(x\cdot0)}{x^3\left(x^2 + 4\right)} \:dx = 0 = \frac{\pi}{8}\left( -\frac{1}{4} \right) + C_2 \rightarrow C_2 = \frac{\pi}{32}$$
And so we arrive at our expression for $I(t)$
$$I(t)= \frac{\pi}{8}\left(\frac{t^2}{2} - \frac{e^{-2t}}{4} \right) - \frac{\pi}{16}t + \frac{\pi}{32}$$
Thus,
$$I = I(1) = \frac{\pi}{8}\left(\frac{1}{2} - \frac{e^{-2}}{4} \right) - \frac{\pi}{16} + \frac{\pi}{32} = \frac{\pi}{32}\left(1 - e^{-2}\right)$$
|
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|
Find $\alpha$,$\beta$ if $\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x -\beta] = 0$ Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = \frac {-b±\sqrt {b^2-ac}}{a} = \frac {-b±\sqrt D}{a}$$
Hence;
$$\sqrt {ax^2+2bx+c} = \sqrt {(x+\frac {b-√D}{a})(x+\frac {b+√D}{a})}$$
$$=x\sqrt {(1+\frac {b-√D}{ax})(1+\frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$\sqrt {ax^2+2bx+c} ≈ x(1+\frac {b-√D}{2ax})(1+\frac {b+√D}{2ax})$$
$$=x + \frac {b}{a} + \frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x - \beta ] = 0,$$
gives;
$$x+\frac {b}{a} - \alpha x - \beta = 0,$$
or;
$$(1-\alpha)x + (\frac {b}{a} - \beta) = 0$$
As $x→∞$, $1-\alpha$ must be $0$ for the former term to vanish, hence,
$$\alpha = 1, \beta = \frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
|
You want $$ \lim _{x\to \infty }ax^2+2bx+c -(\alpha x +\beta )^2=0$$
That implies $$ a= \alpha ^2, c=\beta ^2, b=\alpha \beta$$
|
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|
Solving $\sin^2 x +1=2x$ How do I solve this equation?
$$\sin^2 x +1=2x$$
I have no idea how to attack the problem.
Thanks!
|
Just for the fun of the approximation.
Using the double angle formula Rewrite the equation as $$\cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
$$\cos(t)+2t=3$$
Now, using the approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad (-\frac \pi 2 \leq t\leq\frac \pi 2)$$ we get the cubic equation
$$2 t^3-7 t^2+2 \pi ^2 t-2 \pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $t\approx 1.428167$ that is to say $x\approx 0.714083$ while the "exact" solution is $x\approx 0.714836$.
We can even do better building the $[2,2]$ Padé approximant around $x=\frac \pi 4$
$$\cos(2x)+4x-3=\frac{(\pi -3)+2 \left(x-\frac{\pi }{4}\right)+\left(2-\frac{2 \pi }{3}\right)
\left(x-\frac{\pi }{4}\right)^2 } {1-\frac{2}{3} \left(x-\frac{\pi }{4}\right)^2 }$$ Solving the quadratic
$$x=\frac \pi 4+\frac{6-\sqrt{252-144 \pi +24 \pi ^2}}{4 \pi -12}\approx 0.714837$$
To even avoid solving the quadratic equation, building instead the $[1,3]$ Padé approximant around $x=\frac \pi 4$ would lead to
$$x=\frac \pi 4 -\frac{(\pi -3) \left(15-6 \pi +\pi ^2\right)}{4 \left(12-6 \pi +\pi ^2\right)}\approx 0.714837$$
|
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|
Integrate $\frac{1}{ \sqrt{T^2 - \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} - \frac{b^2}{4 (T-\tau )}\right)$ I want to compute the integral
$$
\int_t^T \int_{-\infty}^\infty \frac{1}{ \sqrt{\tau - t} (T-\tau)} \exp\left(-\frac{(z-x)^2}{2(\tau - t)} -\frac{(z-v)^2 + (z-w)^2}{2(T-\tau)} \right) d z d \tau.
$$
First integrating with respect to $z$ I get
$$
\int_t^T \frac{\sqrt{2\pi}}{ \sqrt{T - \tau}\sqrt{-2 t + \tau + T}}\exp\left(-\frac{(v + w - 2 x)^2}{4 (-2 t + \tau + T)} - \frac{(v-w)^2}{4 (T-\tau )}\right) d \tau.
$$
So to go further I think I should be able to integrate
$$
\int_0^T \frac{1}{ \sqrt{T^2 - \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} - \frac{b^2}{4 (T-\tau )}\right) d \tau
$$
|
$$I=\underbrace{\int_0^{T}\frac{d\tau}{\sqrt{T^2-\tau^2}}\exp\left(-\frac{a^2}{4(T+\tau)}-\frac{b^2}{4(T-\tau)}\right)}\\{\tau\rightarrow \frac{T(1-y^2)}{1+y^2}}$$
$$=\exp\left(-\frac{a^2+b^2}{8T}\right)\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\frac{a^2y^2+\frac{b^2}{y^2}}{8T}\right)$$
Let's parametrize the integral in order to build two ODE: $$ f(\alpha,\beta)=\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\alpha^2y^2-\frac{\beta^2}{y^2}\right)$$
$$g(\alpha,\beta)=f(\alpha ,\beta)-\frac{1}{2\alpha}\frac{\partial f}{\partial\alpha}=\int_{0}^{1}\exp\left(-\alpha^2y^2-\frac{\beta^2}{y^2}\right)dy$$
Now, let's use $g(\alpha,\beta)$ to solve the remaining integral: $$\alpha g(\alpha,\beta)+\frac{1}{2}\frac{\partial g}{\partial\beta}=\exp\left(2\alpha\beta\right)\int_{0}^{1}\left(\alpha-\frac{\beta}{y^2}\right)\exp\left[-\left(\alpha y+\frac{\beta}{y}\right)^2 \right]dy=-\frac{\sqrt{\pi}}{2}\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$
$$\frac{\partial g}{\partial\beta}+2\alpha g(\alpha,\beta)=-\sqrt{\pi}\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$
$$\mu_1(\beta)=\exp\left(\int 2\alpha d\beta\right)=\exp\left(2\alpha\beta\right)$$
$$\exp\left(2\alpha\beta\right)\frac{\partial g}{\partial\beta}+2\alpha\exp\left(2\alpha\beta\right) g(\alpha,\beta)=-\sqrt{\pi}\exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$
$$\left(\exp\left(2\alpha\beta\right)g(\alpha,\beta)\right)'=-\sqrt{\pi}\int \exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)d\beta=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\operatorname{erf}\left(\alpha-\beta\right)\right)$$
$$g\left(\alpha,\beta\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)\right)+c$$
$$g\left(\alpha,0\right)=\int_{0}^{1}\exp\left[-\left(\alpha y\right)^2 \right]dy=\frac{\sqrt{\pi}}{2\alpha}\ \operatorname{erf}\left(\alpha\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\ \operatorname{erfc}\left(\alpha\right)-\ \operatorname{erf}\left(\alpha\right)\right)+c\\ c=\frac{\sqrt{\pi}}{4\alpha}$$
$$\boxed{g\left(\alpha,\beta\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)}$$
Using this result in the first ODE: $$f(\alpha ,\beta)-\frac{1}{2\alpha}\frac{\partial f}{\partial\alpha}=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)$$
$$\frac{\partial f}{\partial\alpha}-2\alpha f(\alpha ,\beta)=\frac{\sqrt{\pi}}{2}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)$$
$$\mu_2(\alpha)=\exp\left(-\int 2\alpha d\alpha\right)=\exp\left(-\alpha^2\right)$$
$$\left(\exp\left(-\alpha^2\right)f(\alpha,\beta)\right)'=\frac{\sqrt{\pi}}{2}\int\left(\exp\left(-\alpha^2+2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-\alpha^2-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-\exp\left(-\alpha^2\right)\right)d\alpha$$
$$=-\frac{\operatorname{erf}\left(\alpha\right)\sqrt{\pi}}{2}+\frac{e\sqrt{\pi}}{2}\int\left(\underbrace{\exp\left(-\left(\alpha-\beta\right)^2\right)\ \operatorname{erfc}\left(\alpha+\beta\right)}_{IBP}-\exp\left(-\left(\alpha+\beta\right)^2\right)\ \operatorname{erf}\left(\alpha-\beta\right)\right)d\alpha$$
$$=-\frac{\operatorname{erf}\left(\alpha\right)\sqrt{\pi}}{2}+\frac{e\sqrt{\pi}}{2}\left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\int\left(\underbrace{\operatorname{erf}\left(\beta-\alpha\right)+\operatorname{erf}\left(\beta+\alpha\right)}_{0}\right)\exp\left(-(\alpha+\beta)^2\right)d\alpha\right)$$
$$f(\alpha ,\beta)=-\frac{\sqrt{\pi}\operatorname{erf}\left(\alpha\right)e^{\alpha^2}}{2}-\frac{\pi e^{a^2+1}}{4}\left(\operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)\right)+k$$
$$f(0,0)=\int_0^1\frac{dy}{1+y^2}=\frac{\pi}{4}=k$$
$$\boxed{f(\alpha,\beta)=-\frac{\sqrt{\pi}e^{\alpha^2}}{2}\operatorname{erf}\left(\alpha\right)-\frac{\pi e^{\alpha^2+1}}{4}\left(\operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)\right)+\frac{\pi}{4}}$$
Therefore, the original integral is equal to: $$I=\exp\left(-\frac{a^2+b^2}{8T}\right)f\left(\frac{a}{2\sqrt{2T}},\frac{b}{2\sqrt{2T}}\right)=$$
$$ \bbox[5px,border:2px solid red] { \frac{\pi}{4}\exp\left(-\frac{a^2+b^2}{8T}\right)-\frac{\sqrt{\pi}}{2} \exp\left(-\frac{b^2}{8T}\right)\ \operatorname{erf}\left(\frac{a}{2\sqrt{2T}}\right)-\frac{\pi}{2} \exp\left(-\frac{b^2}{8T}+1\right)\ \operatorname{erf}\left(\frac{b-a}{2\sqrt{2T}}\right)\ \operatorname{erfc}\left(\frac{b+a}{2\sqrt{2T}}\right) } $$
|
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|
Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1) $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\log_2 x > \sum\limits_{i \le x}\frac{1}{i}$
(3) $(x+1)^{x+1} > {{x+1}\choose{x+1}}x^{x+1} + {{x+1}\choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$
(4) $(x+1)\log_2\left(\frac{x+1}{x}\right) > 1$
(5) $\log_2(x+1) - \log_2(x) > \frac{1}{x+1}$
(6) $\log_2(x+1) - \log_2(x) + \log_2(x) = \log_2(x+1) > \sum\limits_{i \le x+1}\frac{1}{i}$
|
We can use induction as well. If we assume that it's true for $n$:
$$1+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}<\log_2(n)+\frac{1}{n+1}$$
So we need to prove that
$$\log_2(n)+\frac{1}{n+1} < \log_2(n+1)$$
$$\frac{1}{n+1}< \log_2\left(1+\frac{1}{n}\right)$$
$$2^{\frac{1}{n+1}}<1+\frac{1}{n}$$
$$2<\left(1+\frac{1}{n}\right)^{n+1}$$
$$2<\left(1+\frac{1}{n}\right) \left(1+\frac{1}{n}\right)^n$$
And it's true, because
$$1+\frac{1}{n}>1$$
And by Bernoulli's inequality:
$$\left(1+\frac{1}{n}\right)^n \geq 1+n \frac{1}{n}=2$$
So we just need to find a good starting $n$. Let's check it for $n=6$:
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<\log_2(6)$$
$$\frac{9}{20}<\log_2({6/4})$$
$$2^{\frac{9}{20}}<\frac{6}{4}$$
And by Bernoulli's inequality,
$$2^{\frac{9}{20}}<1+\frac{9}{20}<1+\frac{10}{20}=\frac{6}{4}$$
So it's true $\forall n\geq 6 \land n \in \mathbb{N}$ (it's true for $n=5$ as well, but Bernoulli is not helping in that case).
|
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|
Solution to infinite product $\prod_{p-primes}^{\infty} \frac{p}{p-1}$ I want to find the $\prod_{p-primes}^{\infty} \frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$$
of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $\sum_0^\infty \frac{1}{2^n}$$\sum_0^\infty \frac{1}{3^n}$$\sum_0^\infty \frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $\sum_1^\infty log(a_n)$ seems to converge, but hey I could be wrong.
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${p\over p-1}=1+{1\over p-1}>1+\frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.
|
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|
Find the remainder of the division of polynomials $x^{2007}$ divided by $x^2-x+1$.
I consider to solve this problem, should I break the $x^{2007}$ to find the formula $x^2-x+1$?
|
Doing long division:
$$\frac{x^{2007}}{x^2-x+1}=\frac{x^{2008}+x^{2007}}{x^3+1}=\\
\frac{(x^3+1)(x^{2005}+x^{2004}-x^{2002}-x^{2001}+x^{1999}+x^{1998}-\cdots-x^4-x^3+x+1)-x-1}{x^3+1}=\\
a(x)-\frac{x+1}{x^3+1}=\\
a(x)+\frac{\color{red}{-1}}{x^2-x+1}.$$
|
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|
Closed form of $\int{\lfloor{x}\rfloor}dx$ I calculated $\int{\lfloor{x}\rfloor}dx$ and i got this result:
$$\int{\lfloor{x}\rfloor}dx = \frac{x^2-x}{2}+\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2+c$$
Do you know if this series have a closed form?
We found a nice identity!
$$\int{\lfloor{x}\rfloor}dx = \frac{\pi^2\left(x^2-x\right)+\Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right]}{2\pi^2}+c$$
So
$$\int_0^x{\lfloor{t}\rfloor}dt = \frac{\pi^2\left(x^2-x\right)+\Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right]}{2\pi^2}+\frac{1}{12}$$
Also, using $$\Re\left[ \mathrm{Li_2}\left(e^{ix}\right)\right] =\sum_{k\ge1}\frac{\cos(kx)}{k}=\frac{x^2}{4}+\frac{\pi x}{2}+\frac{\pi^2}{6}, \forall x\in\left[0,2\pi\right]$$
We get
$$\forall x\in\left[0,1\right] \forall \alpha\in\mathbb{Z},\ \ \ \ \ \Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right] =\pi^2\left((x-\alpha)^2-(x-\alpha)+\frac{1}{6}\right)$$
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Interpret the integral as a Riemann-Stieltjes integral, you can integrate it by part.
For $y > 0$, you get something like
$$\begin{align}\int_0^y \lfloor x \rfloor dx
&=
\int_{0-}^{y+} \lfloor x \rfloor dx =
y \lfloor y \rfloor - \int_{0-}^{y+} x d\lfloor x \rfloor\\
&= y \lfloor y \rfloor - \sum_{k=0}^{\lfloor y\rfloor} k
= y \lfloor y \rfloor - \frac12 \lfloor y \rfloor(\lfloor y \rfloor + 1)\\
&= \frac12y(y-1) + \frac12 \{y\}(1-\{y\})
\end{align}
$$
where $\{y\} = y - \lfloor y \rfloor$ is fractional part of $y$.
This suggests
$$\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2 = \frac12 \{x\} (1-\{x\})$$
One can verify this by computing the Fourier series of the periodic function on RHS.
Notice
$$\begin{align}
a_0 &= \frac12\int_0^1 x(1-x) dx = \frac{1}{12} = \frac{1}{2\pi^2}\frac{\pi^2}{6}\\&= \frac{1}{2\pi^2}\sum_{k=1}^\infty \frac{1}{k^2}\\
a_k &= \frac12\int_0^1 x(1-x)\cos(2\pi k x)dx
= \frac{1}{4\pi k}\int_0^1 x(1-x) d\sin(2\pi k x)\\
&= \frac{1}{4\pi k}\left\{
\left[x(1-x) \sin(2\pi k x)\right]_0^1
+ \int_0^1 (2x-1)\sin(2\pi k x) dx
\right\}\\
&= -\frac{1}{8\pi^2 k^2}\int_0^1 (2x-1)d\cos(2\pi k x)\\
&= -\frac{1}{8\pi^2 k^2} \left\{
\left[(2x-1)\cos(2\pi k x)\right]_0^1 - 2\int_0^1 \cos(2\pi k x) dx
\right\}\\
&= -\frac{1}{4\pi^2k^2}
\end{align}$$
and by symmetry, $\displaystyle\;b_k = \frac12\int_0^1 x (1-x) \sin(2\pi k x)dx = 0$ for all $k$. This leads to
$$\begin{align}\frac12\{x\}(1 - \{x\}) &=
a_0 + 2\sum_{k=1}^\infty (a_k \cos(2\pi k x) + b_k\sin(2\pi kx))\\
&= \frac{1}{2\pi^2}\sum_{k=1}^\infty \frac{1 - \cos(2\pi k x)}{k^2}\\
&= \sum_{k=1}^\infty \left(\frac{\sin(\pi k x)}{k \pi}\right)^2\end{align}$$
|
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|
System of 3 equations problem
If
$$x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1$$
Then how much is
$$x^4+y^4+z^4$$
So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.
I did this:
$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$
And from here I can see:
$x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$
But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$
I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.
I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine
|
Hint:
$$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$
$$(x+y+z)^2=?$$
$$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$
|
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|
Is there another mathematical way to approach this problem?
$$f(x)-f(x-1) = x-5$$
$$f(16) = 74$$
Compute $f(1)$.
So, this is basically a linear function. I, however, calculated it without taking the easier way to approach this problem.
If
$$f(16) - f(15) = 11$$
Then
$$74-f(15) = 11 \implies f(15)=63$$
This will take so long as it seems. Since we're given a linear function, how would we compute it in an other mathematical way? Furthermore, is there a general rule for what you advise?
|
First, note that $f$ isn't linear at all (if it were linear, then $f(x)-f(x-1)$ would be constant. However, that difference does suggest something quadratic: its second difference will be constant. Thus, we'll try to find $a$, $b$, and $c$ such that $f(x) = ax^2 + bx + c$ satisfies this. For this purpose, we garner some equations relating $a$, $b$, and $c$:
First, $f(16) = 74$ tells us that $256a + 16b + c = 74$. Second, $f(x)-f(x-1) = x - 1$ gives us $a(x^2 - (x-1)^2) + b(x - (x-1)) = x - 1$. Simplifying that gives us $a(2x-1)+b=x-1$, for all $x$. Taking $x = 0$ gives us $b - a = -1$, $x = 1$ give $a + b = 0$, so $b = -a$, and the $x = 0$ case gives $2a = 1$, so $a = \frac{1}{2}$, and $b = \frac{-1}{2}$. Finally, $f(16)=74$ gives us $120 + c = 74$, so $c = -46$. Thus, $f(x) = \frac{1}{2}x^2 - \frac{1}{2}x - 46$.
You can quickly go back and check that this satisfies all relevant conditions, if you like (it does), and therefore $f(1) = \frac{1}{2}(1^2) - \frac{1}{2}(1)-46 = -46$.
|
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|
Proving a three variables inequality Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+c\ge3\sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)\ge63$$The last thing I tought about was that I have both $a+b+c\ge9$ and $a+b+c\ge3\sqrt[3]{abc}$ so if I somehow related them I would have $\sqrt[3]{abc} \ge 3 \rightarrow abc\ge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
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Since $ab+bc+ca= abc$ and $$ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}\implies abc\geq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1\geq 55+3\sqrt[3]{27} = 64$$
|
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|
How to factor equations with the form $x^2 + xy + y^2$? Title says it all. I'm having a hard time factoring anything looking like the equation $x + xy + y$. Especially trig ones, like $9\sin^2x + 12\sin x\cos x + 4\cos^2x$.
Thanks!
|
If you want to factor expressions of the form $\alpha x^2+\beta xy+\gamma y^2$, observe that $$\begin{align*}\alpha x^2+\beta xy+\gamma y^2&=\alpha y^2\left((xy^{-1})^2+\beta\alpha^{-1} (xy^{-1})+\gamma\alpha^{-1}\right)\\&=\alpha y^2\left(xy^{-1}-\frac{-\beta+ \sqrt{\beta^2-4\alpha\gamma}}{2\alpha}\right)\left(xy^{-1}-\frac{-\beta- \sqrt{\beta^2-4\alpha\gamma}}{2\alpha}\right)\\&=\boxed{\left(\sqrt{\alpha}x-\frac{-\beta+ \sqrt{\beta^2-4\alpha\gamma}}{2\sqrt{\alpha}}y\right)\left(\sqrt{\alpha}x-\frac{-\beta- \sqrt{\beta^2-4\alpha\gamma}}{2\sqrt{\alpha}}y\right)}\end{align*}$$
Of course, the usual warnings involving principal square roots apply.
In the case of $x^2+xy+y^2$ this gives us the factorization $$x^2+xy+y^2=\left(x-\tfrac{-1+i\sqrt{3}}{2}y\right)\left(x-\tfrac{-1-i\sqrt{3}}{2}y\right)$$
|
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|
Conjecture: if $a$, $b$ and $c$ have no common factors, dividing each of them by their sum yields at least one irreducible fraction
Let $a$, $b$ and $c$ be $3$ integers with no common factors.
I conjecture that at least one of the three fractions:
$$\frac{a}{a+b+c},\quad\frac{b}{a+b+c},\quad\frac{c}{a+b+c}$$
is irreducible.
I know that they are not necessarily all irreducible.
For instance, taking $(a,b,c)=(2,7,15)$, it is verified that $2$, $7$ and $15$ have no common factors.
We have $a+b+c=24$ so
$$\dfrac{2}{2+7+15}=\dfrac{2}{24}\quad(\text{reducible)}$$
$$\dfrac{7}{2+7+15}=\dfrac{7}{24}\quad(\text{irreducible)}$$
$$\dfrac{15}{2+7+15}=\dfrac{15}{24}\quad(\text{reducible)}$$
I have yet to find any counter-example to the initial statement so I'm pretty certain it must be true. I believe I managed to prove the analogous statement for $2$ integers but I would like to generalize it to $3$, and eventually $n$, integers.
Proof for two integers $a$ and $b$ with no common factors
Suppose $k$ is a factor of $a$ and that $a'$ is an integer such that $a=ka'$. Then:
$$\frac{a}{a+b}=\frac{ka'}{ka'+b}=\frac{ka'}{k\left(a'+\dfrac{b}{k}\right)}=\dfrac{a'}{a'+\dfrac{b}{k}}$$
But $k$ is a factor of $a$ and $a$ and $b$ have no common factors by hypothesis, so $k$ does not divide $b$. So $\dfrac{b}{k}$ is irreducible. Since $a'$ is an integer and $\dfrac{b}{k}$ isn't (assuming $k\neq 1$), then $\dfrac{a'}{a'+\dfrac{b}{k}}$ is not a 'fraction' in the sense that it is not a ratio of two integers, making the original fraction $\dfrac{a}{a+b}$ irreducible.
This proves (correct me if I'm wrong) that at least one of the the fractions $\dfrac{a}{a+b}$ or $\dfrac{b}{a+b}$ is irreducible.
For the case with $3$ integers, a similar argument doesn't seem to work. Taking $(4,5,11)$ as an example:
$$\frac{4}{4+5+11}=\frac{4}{4\left(1+\dfrac{5}{4}+\dfrac{11}{4}\right)}=\frac{1}{1+\dfrac{16}{4}}=\frac{1}{5}$$
the fact that there are now $3$ numbers allows for some fractions to combine into an integer. Because of this, I'm having difficulty proving the statement further.
I'm not particularly good with number theory proofs and I apologize if this is not as elegant as it should be (or worse, if there are mistakes). I'm welcoming any tips and possible improvements. Thanks in advance!
|
The smallest counterexample is $(2,3,25)$. An infinite family of counterexamples is $(2,3,25+30n)$:
$$
\begin{align*}
\frac{a}{a+b+c} &= \frac{2}{30+30n} = \frac{1}{15+15n}
\\ \frac{b}{a+b+c} &= \frac{3}{30+30n} = \frac{1}{10+10n}
\\ \frac{c}{a+b+c} &= \frac{25+30n}{30+30n} = \frac{5+6n}{6+6n}
\end{align*}
$$
There are several other infinite family of counterexamples.
|
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|
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$$\displaystyle =\displaystyle \lim\limits _{x\rightarrow \infty }\dfrac{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }} =$
$\displaystyle =\lim\limits _{x\rightarrow \infty }\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}$
What is the next step should be? Please help!
|
Now, use that
$$\sqrt{x^4+1}-x^2=\frac{1}{\sqrt{x^4+1}+x^2}.$$
|
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|
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.
I tried to use the Ratio test.
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}\right)\cdot \left(\frac{n^2 + 2n + 2}{n^2 + 3n + 3}\right)^{2n +1}$$
Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.
|
Another approach using the Root test.
Let $L = \lim_{n\to\infty} \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} $
$$ \begin{align}
L &= \lim_{n\to\infty} \exp \log \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} \\
&= \exp \lim_{n\to\infty} \frac{\log \left(\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}\right)}{n} \\
&= \exp \lim_{n\to\infty} \frac{n^2 \log \frac{n^2+1}{n^2+n+1}}{n} \\
&= \exp \lim_{n\to\infty} n \log \frac{n^2+1}{n^2+n+1} \tag{1} \\
&= \exp(-1) \\
\end{align}$$
and since $L = \exp(-1) < 1$ the series is absolutely convergent by the root test.
* (1) can be shown by the fact that the Laurent expansion for $n\log\frac{n^2+1}{n^2+n+1}$ at $n=\infty$ is $-1 + \textrm{O}(\frac{1}{n})$.
|
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|
Find the value of $Q(x)$ at $x= -1$, knowing some of its properties.
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
\frac{Q(2x)}{Q(x+1)}= 8 -\frac{56}{x+7},
\quad \forall x \ne -7 \text{ and } Q(x+1)\neq 0\,.
$$
Find $Q(-1)$.
|
Rewrite the equation as
$$(x+7)\,Q(2x)=8x\,Q(x+1)\,.\tag{*}$$
Thus, $2x\mid Q(2x)$ or $x\mid Q(x)$, and $x+7\mid Q(x+1)$ or $x+6\mid Q(x)$. That is, $Q(x)=x(x+6)\,R(x)$ for some polynomial $R(x)$. Show that
$$(x+3)\,R(2x)=2(x+1)\,R(x+1)\,.$$
Ergo, $2x+2\mid R(2x)$ or $x+2\mid R(x)$, and $x+3\mid R(x+1)$ or $x+2\mid R(x)$. Thence, $R(x)=(x+2)\,S(x)$ for some polynomial $S(x)$. Then, prove that
$$S(2x)=S(x+1)\,,$$
which implies that $S(x)$ is a constant polynomial. Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.
You can also notice that $\lim\limits_{x\to\infty}\,\dfrac{Q(2x)}{Q(x+1)}=8=\left(\dfrac{2}{1}\right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0\,,$$ and so on.
|
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|
Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$
Solve this equation:
$$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\log_{4}x}}{\sqrt{3}} = \sqrt{x}$$
$$\Leftrightarrow 3^{\log_{4}x}.3+ 3^{\log_{4}x} = \sqrt{3x}$$
$$\Leftrightarrow 4.3^{\log_{4}x}= \sqrt{3x}$$
At this step, I can't continue. Please help me!
|
First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x \geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + \log_4(3)\log_4(x) = \frac{\log_4(3)}{2} + \frac{\log_4(x)}{2}$$
$$\log_4(x)\left(\log_4(3) - \frac{1}{2}\right) = \frac{\log_4(3)}{2} - 1$$
$$\log_4(x) = \frac{\frac{\log_4(3)}{2} - 1}{\log_4(3) - \frac{1}{2}} = \frac{\log_4(3)-2}{2\log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$\large x = \large 4^{\frac{\log_4(3)-2}{2\log_4(3)-1}}$$
|
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|
Limit of sum as definite integral I don't understand why $$\displaystyle \sum_{k=1}^n \dfrac{n}{n^2+kn+k^2} < \lim_{n\to \infty}\sum_{k=1}^n \dfrac{n}{n^2+kn+k^2}$$
whereas
$$\displaystyle \sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2} > \lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2} $$
I know that $$\lim_{n\to \infty}\sum_{k=1}^{n} \dfrac{n}{n^2+kn+k^2}=\dfrac{\pi}{3\sqrt{3}}$$ $$\lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2}=\dfrac{\pi}{3\sqrt{3}}$$
I saw somewhere on the internet that $$\displaystyle \dfrac{1}{n}\sum_{k=0}^{n-1} f\left(\dfrac{k}{n}\right) > \int_0^1 f(x)dx > \dfrac{1}{n}\sum_{k=1}^{n} f\left(\dfrac{k}{n}\right)$$ Why is this true?
|
Consider the function
$$f(x)=\frac{1}{1+x+x^2}$$
and note that the your inequality holds because $f$ is strictly decreasing.
Indeed for $n\geq 1$, $k\geq 0$, and $x\in [\frac{k}{n},\frac{k+1}{n}],$
$$f(\frac{k}{n})> f(x)> f(\frac{k+1}{n}).$$
By integrating over the interval $[\frac{k}{n},\frac{k+1}{n}]$, we get
$$\frac{f(\frac{k}{n})}{n}=\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(\frac{k}{n})dx> \int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx> \int_{\frac{k}{n}}^{\frac{k+1}{n}}f(\frac{k+1}{n})dx=\frac{f(\frac{k+1}{n}) }{n}.$$
Finally we take the sum for $k=0,\dots,n-1$,
$$\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})>\int_0^1 f(x)\,dx
>\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k+1}{n}).$$
Note that the last sum on the right is equal to
$$\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})=
\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})+\frac{f(1)-f(0)}{n}.$$
P.S. Once we have the double inequality, we may conclude that
$$\lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2}=\lim_{n\to \infty}\sum_{k=1}^{n} \dfrac{n}{n^2+kn+k^2}=\int_0^1\frac{dx}{1+x+x^2}=\dfrac{\pi}{3\sqrt{3}}.$$
|
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|
Finding matrix $A^n,$ when $\lim_{n \to \infty}$
Finding $\lim_{n\rightarrow \infty}\begin{pmatrix}
1 & \frac{x}{n}\\ \\
-\frac{x}{n} & 1
\end{pmatrix}^n$ for all $x\in \mathbb{R}$
Try:
Let $$ A = \begin{pmatrix}1&\frac{x}{n}\\\\-\frac{x}{n}&1\end{pmatrix}.$$
Then $$ A^2 = \begin{pmatrix}1-\frac{x^2}{n^2}&\frac{2x}{n}\\\\-\frac{2x}{n}&1-\frac{x^2}{n^2}\end{pmatrix}$$
And then $$A^3 = \begin{pmatrix}1-3\frac{x^2}{n^2}&\frac{3x}{n}-\frac{x^3}{n^3}\\\\-3\frac{x}{n}+\frac{x^3}{n^3}&1-\frac{x^2}{n^2}\end{pmatrix}$$
So by using same way and taking $\lim_{n\rightarrow \infty}A^n = \begin{pmatrix}1&0\\\\0&1\end{pmatrix}$
But answer given as $$\begin{pmatrix}\cos x &\sin x\\\\-\sin x&\cos x\end{pmatrix}$$
Could some help me where I am missing and also explain how to solve it?
|
We have
$$A = \sqrt{1 + \frac {x^2}{n^2}}\begin{pmatrix}\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}&\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}\\\\-\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}&\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\end{pmatrix}$$
So if $\theta_n$ is such that $\cos \theta_n=\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}$ and $\sin \theta_n = -\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}$, and $\rho_n=\sqrt{1 + \frac {x^2}{n^2}}$, then
$$A=\rho_n\begin{pmatrix}\cos \theta_n&-\sin \theta_n\\\\\sin \theta_n&\cos \theta_n\end{pmatrix}$$
Therefore, $A$ is the product of the scaling factor $\rho_n$ with the rotation matrix of angle $\theta_n$. So powers of $A$ are
$$A^n=\rho_n^n\begin{pmatrix}\cos n\theta_n&-\sin n\theta_n\\\\\sin n\theta_n&\cos n\theta_n\end{pmatrix}$$
Now, using Taylor expansions, $$\lim_{n\rightarrow +\infty}\rho_n^n=1$$ and $$\cos(n\theta_n)=\cos\left(n\arccos\bigg(\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\bigg)\right)\rightarrow\cos(x)$$
$$\sin(n\theta_n)=\sin\left(n\arcsin\bigg(\frac {-\frac {x} n} {\sqrt{1 + \frac {x^2}{n^2}}}\bigg)\right)\rightarrow-\sin(x)$$
|
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|
How to solve the equation $2z^{2}+i=-2$?
How to solve the equation $2z^{2}+i=-2$?
Assuming that $k=-1-i/2$ we have $|k|=\sqrt{5}/2$ and
$$
\cos \phi=(-2\sqrt5)/5,\quad \sin \phi= -\sqrt5/5.
$$
How should I find $\phi$?
|
You won't find the exact value of $\phi$.
Here is a general method for solving $z^2=p+qi$, where $p$ and $q$ are real.
If $z=a+bi$ where $a$ and $b$ are real numbers, then $z^2=a^2-b^2+2abi=p+qi$, hence $p=a^2-b^2$ and $q=2ab$. You can also notice that $\vert z\vert^2=a^2+b^2=\sqrt{p^2+q^2}$, so you can find $a^2$ and $b^2$, and since you know the sign of $ab$, you can find two couples of solutions $(a,b)$.
In your case, $z^2=-1-\dfrac i2$, so if $z=a+bi$, then $$a^2-b^2=-1$$ $$2ab=-\dfrac 12$$ and $$a^2+b^2=\sqrt{\left(-1\right)^2+\left(-\dfrac 12\right)^2}=\dfrac{\sqrt 5}2\,.$$
Hence $$a^2=\dfrac{\dfrac{\sqrt 5}2-1}2=\dfrac{\sqrt{5}-2}4$$ and $$b^2=\dfrac{\dfrac{\sqrt 5}2+1}2=\dfrac{\sqrt{5}+2}4$$
Since $ab<0$, $a$ and $b$ have opposite signs, and we find that $$(a,b)=\left(\pm\dfrac{\sqrt{\sqrt{5}-2}}2,\mp\dfrac{\sqrt{\sqrt{5}+2}}2\right)$$ and so
$$z=\pm\dfrac{\sqrt{\sqrt{5}-2}-i\sqrt{\sqrt{5}+2}}2$$
|
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Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $\sum_{\text{cyc}}\frac{1}{a^3(b+c)}\geq \frac32$ I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}\geq \frac32$$
First, substitute $\frac1a = x, \frac1b = y, \frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$\frac{x^2}{z+y} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \geq \frac32$$ Now, let $f(x)=\dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=\dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,z\in \mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$\begin{alignat*}{2}\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} & =\frac{x^2}{S-x} + \frac{y^2}{S-y} + \frac{z^2}{S-z} \\
&\geq 3f\left(\frac{x+y+z}{3} \right) \\
&= 3\frac{\left(\frac{x+y+z}{3}\right)^2}{S-\frac{x+y+z}{3}} = \frac{1}{3}\frac{(x+y+z)^2}{\frac{2x+2y+2z}{3}} = \frac13 \frac{(x+y+z)^2}{\frac23 (x+y+z)} \\
&= \frac12 (x+y+z) \geq \frac32\end{alignat*}$$
The final inequality follows by AM-GM ($x+y+z\geq 3\sqrt[3]{xyz}=3$)
|
I think it's better to end your proof by C-S and AM-GM:
$$\sum_{cyc}\frac{x^2}{y+z}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(y+z)}=\frac{1}{2}(x+y+z)\geq\frac{3}{2}.$$
|
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|
Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve
$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$
through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration.
Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$
Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$
|
Decompose the integrand to express the integral as
\begin{align}
I=&\int_{-\infty}^\infty \frac1{1+x^8}dx \\
=& \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2+x^2}{x^4+\sqrt2 x^2+1}dx
+ \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2-x^2}{x^4+\sqrt2 x^2+1}dx\\
\overset{x\to \frac1x}=&
\frac1{2\sqrt2}\int_{-\infty}^\infty \frac{(\sqrt2+1)x^2}{x^4+\sqrt2 x^2+1}dx
+ \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{(\sqrt2-1)x^2}{x^4+\sqrt2 x^2+1}dx\\
=& \frac{\sqrt2+1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{(x-\frac1x)^2+2+\sqrt2}dx
+ \frac{\sqrt2-1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{(x-\frac1x)^2+2-\sqrt2}dx\\
\end{align}
Then, apply the Glasser master theorem $\int_{-\infty}^\infty f(x-\frac1x)dx= \int_{-\infty}^\infty f(x)dx$
\begin{align}
I =& \frac{\sqrt2+1}{2\sqrt2}\int_{-\infty}^\infty \frac1{x^2+(2+\sqrt2)}dx
+ \frac{\sqrt2-1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{x^2+(2-\sqrt2)}dx\\
=& \frac{\sqrt2+1}{2\sqrt2}\cdot \frac{\pi}{\sqrt{2+\sqrt2}}
+ \frac{\sqrt2-1}{2\sqrt2}\cdot \frac{\pi}{\sqrt{2-\sqrt2}}
=\frac\pi2 \sqrt{1+\frac1{\sqrt2}}=\frac\pi4 \csc\frac\pi8
\end{align}
|
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|
If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried:
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$ divides through both $xy$ and $x^2 + y^2$
This then allows me to introduce another equation $1=ax+by$ for some $a, b.$
But I then get stuck ...
|
Rational Algebraic Integer Approach
Suppose that
$$
\frac{x^2+y^2}{xy}=\frac xy+\frac yx\in\mathbb{Z}\tag1
$$
Note that if $q=\frac xy\in\mathbb{Q}$ and $q+\frac1q=n\in\mathbb{Z}$, then
$$
\left(q-\frac1q\right)^2=n^2-4\in\mathbb{Z}\tag2
$$
This means that $z=q-\frac1q$ is a rational solution to $z^2-(n^2-4)=0$; that is, $z$ is a rational algebraic integer. Thus, $z\in\mathbb{Z}$ (see this answer). Therefore, $(n+z)(n-z)=4$ is an integer factorization of $4$ where both factors have the same parity. That is, $n+z=n-z=\pm2$, which means $n=\pm2$ and $q-\frac1q=z=0$. Thus, $\frac{x^2}{y^2}=q^2=1$, and therefore, $x=\pm y$.
Bezout Approach
Let $d=(x,y)$ and $u=x/d$ and $v=y/d$. Then, there exist $a,b$ so that $au+bv=1$. Suppose that
$$
\begin{align}
n
&=\frac{x^2+y^2}{xy}\\
&=\frac{u^2+v^2}{uv}\\
&=\frac{b^2u^2+(1-au)^2}{bu(1-au)}\\
&=\frac{\left(a^2+b^2\right)u^2-2au+1}{bu-abu^2}\tag3
\end{align}
$$
Then
$$
\frac1u=n(b-abu)+2a-\left(a^2+b^2\right)u\in\mathbb{Z}\tag4
$$
Thus, $u\cdot\frac1u=1$ is an integral factorization of $1$. That is, $u=\pm1$. Similarly, $v=\pm1$.
Therefore, $x=\pm d$ and $y=\pm d$, which means that $x=\pm y$.
|
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|
Determining properties of a polynomial $f$ satisfying $f(x^2)-xf(x) = x^4(x^2-1)$ for $x \in\Bbb R^+$
Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x \in\Bbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $\displaystyle\lim_{x\to \infty} \frac{f(x)}{x^3}=1$
D) $\displaystyle\lim_{x\to \infty} \left(\frac{f(x)}{x^2}-x \right)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
|
You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=\frac{f(x^2)-x^4(x^2-1)}{x} \tag{*}
$$
and so
$$
f(-x)=\frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $\deg f(x)=3$. Now
$$
\frac{f(x^2)-xf(x)}{x^6}=\frac{f(x^2)}{(x^2)^3}-\frac{1}{x^2}\frac{f(x)}{x^3}=\frac{x^2-1}{x^2}=1-\frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
\frac{f(x)}{x^2}-x=\frac{f(x^2)}{x^3}-x(x^2-1)-x=\frac{f(x^2)}{x^3}-x^3=
x\left(\frac{f(x^2)}{(x^2)^2}-x^2\right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
|
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|
With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.
|
The reverse triangle inequality implies that $|z^2+1|\geq ||z^2|-1|$ so that if $|z|^2\geq 1$ we see that
$$|z|\geq |z|^2-1$$
By solving this inequality using the quadratic formula, we see that $|z|\leq \frac{1}{2}(1+\sqrt{5})$, so the maximum must be less than $\frac{1}{2}(1+\sqrt{5})$. We see that this maximum is attained by setting $z=\frac{1}{2}(1+\sqrt{5})$.
|
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|
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
|
Sketch of a solution rewriting the given system in terms of linear algebra:
*
*Set $\boxed{a = \binom{x}{y}}$. Then the given system looks as follows:
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases} \Longleftrightarrow \boxed{a +\frac{1}{||a||^2}\begin{pmatrix}3 & -1 \\ -1 & -3 \end{pmatrix}a = \binom{3}{0}}$$
*Set $\boxed{t = \frac{1}{ ||a||^2}}$ and $\boxed{A = \begin{pmatrix}3 & -1 \\ -1 & -3 \end{pmatrix}} \Rightarrow (I + tA)a = \binom{3}{0} \Rightarrow \boxed{a = (I + tA)^{-1}\binom{3}{0}}$.
*So, calculate $a$ by inverting $(I + tA)$:
$$\boxed{a =} (I + tA)^{-1}\binom{3}{0} = \begin{pmatrix}1+3t & -t \\ -t & 1-3t \end{pmatrix}^{-1}\binom{3}{0} = \ldots = \boxed{\frac{3}{1-10t^2}\binom{1-3t}{t} }$$
*Find $t >0 $ such that $t = \frac{1}{ ||a||^2} \Rightarrow \boxed{t= \frac{1}{5}, \frac{1}{2}}$
*Plug these values into $a = \frac{3}{1-10t^2}\binom{1-3t}{t}$:
$$\Rightarrow \begin{cases} t= \frac{1}{2} \Rightarrow \boxed{a= \binom{1}{-1}} \\ t= \frac{1}{5} \Rightarrow \boxed{a= \binom{2}{1}} \end{cases}$$
|
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|
Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\frac12(a+b)\cos\frac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.
We first expand
$$\sin a + \sin b = 2 \sin\frac12(a+b) \cos\frac12(a-b)$$
Then we add $\sin(a+b)$, which is $\sin a \cos b + \cos a \sin b$.
We now have:
$$2 \sin\frac12(a+b) \cos\frac12(a-b) + \frac12 \left(\sin(a-b) + \sin (a+b)\right) + \frac12 \left( \sin(b-a) + \sin (a+b)\right)$$
From there, we can't see how to obtain the right-hand side of the equation which is
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$
|
Use that $$\sin(A)+\sin(B)=2\cos\left(\frac{A-B}{2}\right)\sin\left(\frac{A+B}{2}\right)$$ and $$\sin(A+B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
|
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|
Proof of polynomial divisibility without using complex numbers? My question is the same as polynomial of degree n and its divisor except I want a solution that does not make use of complex numbers
Problem: Find all positive integers $n$ such that $x^2+x+1\mid (x+1)^n+x^n+1$
Using wolframalpha, I can see that a number of solutions work, such as $2,4,7$. I tried substituting various values of $x$ in and then working in specific mod cases, but this doesn't really seem to work because it only gives possible values, not 'actual' values. Polynomial long division here is also quite unwieldy...
|
First I will prove that: $$n \in \{2,4\} \pmod{6}$$ is a necessary condition:
Just substitute x for the value 2 so that the equation results:
$$7 | 3 ^ n + 2 ^ n + 1$$
$$3 ^ n + 2 ^ n + 1 \equiv 0 \pmod{7}$$
whose all solutions are:
$$n \in \{2,4\} \pmod{6}$$
Now I am going to prove that these two conditions are sufficient:
As mentioned above:
$$ P = x^2+x+1 $$
$$(x + 1) ^ 6 \equiv 1 \pmod{P}$$
$$x ^ 6 = 1 \pmod{P}$$
If $n \equiv 2 \pmod {6}$:
$$(x + 1) ^ {6a+2} + x ^ {6a+2} + 1 \pmod{P}$$
$$(x + 1) ^ {6a} (x ^ 2 + 2x + 1) + x ^ {6a} x ^ 2 + 1 \pmod{P}$$
$$(P + x) + x ^ 2 + 1 \pmod{P}$$
$$P + x + x ^ 2 +1 \pmod{P}$$
$$2P \pmod{P}$$
$$ 0 \pmod{P} $$
With which is a sufficient condition.
If $n \equiv 4 \pmod {6}$:
$$(x + 1) ^ {6a+4} + x ^ {6a+4} + 1 \pmod{P}$$
$$(x + 1) ^ {6a} (x ^ 4 + 4x ^ 3 + 6x ^ 2 + 4x + 1) + x ^ {6a} x ^ 4 + 1 \pmod{P}$$
$$(x ^ 4 + 4x ^ 3 + 6x ^ 2 + 4x + 1) + x ^ 4 + 1 \pmod{P}$$
$$2x^4 +4x^3+6x ^ 2 + 4x + 2 \pmod{P}$$
$$2(x^2+x+1)^2 \pmod{P}$$
$$ 0 \pmod{P} $$
With which is a sufficient condition.
It is proved that the condition is necessary and sufficient.
|
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|
Radius of largest circle in an ellipse Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.
What I tried:
Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)
Then equation of ellipse is $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ where $a=5,b=4$
Then $\displaystyle b^2=a^2(1-e^2)$ . getting $\displaystyle e=3/5.$
coordinate of focus is $(\pm ae,0)=(\pm 3,0)$
Let equation of circle is $(x\pm 3)^2+y^2=r^2$
How do I solve it from here?
|
You want to minimize $r^2=(x-3)^2+y^2$ subject to $\frac{x^2}{25}+\frac{y^2}{16}=1$:
$$L(x,y,\lambda)=(x-3)^2+y^2+\lambda\left(1-\frac{x^2}{25}-\frac{y^2}{16}\right)\\
\begin{cases}L_x=2x-6-\frac{2x\lambda}{25}=0\\ L_y=2y-\frac{y\lambda}{8}=0\\ L_{\lambda}=1-\frac{x^2}{25}-\frac{y^2}{16}=0\end{cases} \Rightarrow (x,y)=(\pm5,0);\\
r^2(5,0)=2^2 \ \text{(min)}$$
Notes:
1) WolframAlpha answer.
2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $\frac{x^2}{25}+\frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.
3) Foci $(\pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 \Rightarrow c=\pm 3$.
|
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|
If $-1\leq x, y \leq 1$ and $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1$, find $x^2+y^2$ Let $x,\, y\in\mathbb
R,\ -1\leq x,\, y\leq 1$ such that $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1.$
Find the sum $S = x^2+y^2.$
|
Let $x=\sin\alpha$ and $y=\sin\beta$, where $\{\alpha,\beta\}\subset[-\frac{\pi}{2},\frac{\pi}{2}].$
Thus, the condition gives $\sin(\alpha+\beta)=1.$
Can you end it now?
Also, we can make squaring twice.
Indeed, the condition gives
$$x^2(1-y^2)+y^2(1-x^2)+2xy\sqrt{(1-x^2)(1-y^2)}=1$$ or
$$2xy\sqrt{(1-x^2)(1-y^2)}=1-x^2-y^2+2x^2y^2,$$ which after squaring again gives
$$1+x^4+y^4-2x^2-2y^2+2x^2y^2=0$$ or
$$(x^2+y^2)^2-2(x^2+y^2)+1=0$$ or
$$x^2+y^2=1.$$
|
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|
Find the interval of convergence of the series $ \sum^{\infty}\limits_{k=0} ((-1)^k+3)^k(x-1)^k $ I wish to find the interval of convergence of the following series
\begin{align} \sum^{\infty}_{k=0} ((-1)^k+3)^k(x-1)^k \end{align}
PROOF
Wittingly,
\begin{align} \left[(-1)^k+3\right]^k= \begin{cases} 0,&\text{if}\;j=0;\\4^j,&\text{if}\;j=2k;\\2^j,&\text{if}\;j=2k+1.\end{cases} \end{align}
Thus,
\begin{align} \limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}&=|x-1|\limsup_{k\to\infty}\sqrt[k]{ 2^{2k+1}}\\&=|x-1|\limsup_{k\to\infty} 2^{(2k+1)\times \frac{1}{k}} \\&=4|x-1| \end{align}
The series converges absolutely when $\limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}<1,$ i.e.,
\begin{align} |x-1|< \dfrac{1}{4}\iff \dfrac{3}{4}<x<\dfrac{5}{4} \end{align}
QUESTION:
Why must \begin{align} \limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}= \limsup_{k\to\infty} \sqrt[k]{\left| 2^{2k+1}(x-1)^k\right|}\end{align}
as stated in the book before me and not
\begin{align} \limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}= \limsup_{k\to\infty} \sqrt[k]{\left| 4^{2k}(x-1)^k\right|}\;?\end{align}
|
Your justification of the equality$$\limsup_{k\to\infty}\sqrt[k]{((-1)^k+3)^k(x-1)^k}=\lvert x-1\rvert\limsup_{k\to\infty}\sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $\bigl((-1)^{2k+1}+3\bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$\sqrt[2k+1]{2^{2k+1}}=2.$$But, since $\bigl((-1)^{2k}+3\bigr)^{2k}=4^{2k}$ and$$\sqrt[2k]{4^{2k}}=4,$$you have$$\limsup_{k\to\infty}\sqrt[k]{((-1)^k+3)^k(x-1)^k}=4\lvert x-1\rvert.$$
|
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|
Find the $\frac mn$ if $T=\sin 5°+\sin10°+\sin 15°+\cdots+\sin175°=\tan \frac mn$ It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.
$$T=\sin(5^\circ) +\sin(10^\circ) + \sin(15^\circ) + \cdots +\sin(175^\circ) =\tan \frac mn$$
Find the $\frac mn=?$, where $m$ and $n$ are positive integer numbers.
Attepmts:
$$T=2\Big(\sin(5°)+\sin(10°) + \cdots + \sin(85°)\Big) + 1 = 2\Big((\sin(5°) + \cos(5°))+(\sin(10°)+ \cos(10°))+\cdots + (\sin(40°)+\cos(40°))\Big)+1 = 2\Big(\sqrt 2((\sin50°+\sin55°)+\cdots+\sin(80°))\Big)+1.$$
and then I can not see an any way...
|
By geometric series,
$$T=\sin 5°+\sin10°+\sin 15°+...+\sin175° = \\
\operatorname{Im} (\sum_{n=1}^{35}\exp(i n 5 \pi/180)) =\\
\operatorname{Im} \exp(i 5 \pi/180) \frac{\exp(i 35 \cdot 5 \pi/180)-1}{\exp(i 5 \pi/180)-1} =\\
\operatorname{Im} \exp(i 5 \pi/180) \exp(i 34 \cdot 5 \pi/360) \frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} =
\\
\operatorname{Im} \exp(i 36 \cdot 5 \pi/360) \frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} =
\\ \frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} \\
$$
and
$$
\arctan(\frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} ) = \\
\arctan(\frac{\sin( 35 \cdot 5 \pi/360)}{\cos( \frac{\pi}{2} - 5 \pi/360)} ) = \\
\arctan(\frac{\sin( 35 \cdot 5 \pi/360)}{\cos( 35 \cdot 5 \pi/360)} ) =
\frac{35 \pi}{72}
$$
or, in degrees, $\frac{35 \cdot 180}{72} =\frac{175}{2} = 87.5 $
|
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|
In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$
If in a triangle ABC, $b+c=3a$, then $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
\cot\dfrac{B}{2}.\cot\dfrac{C}{2}=\frac{\cos\frac{B}{2}.\cos\frac{C}{2}}{\sin\frac{B}{2}.\sin\frac{C}{2}}=\frac{\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})}{\cos(\frac{A-B}{2})-\cos(\frac{A+B}{2})}
$$
|
In the standard notation we obtain: $$\sin\beta+\sin\gamma=3\sin(\beta+\gamma)$$ or
$$2\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}=6\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}$$ or
$$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}+\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=3\left(\cos\frac{\beta}{2}\cos\frac{\gamma}{2}-\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\right)$$ or
$$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}=2\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$$ or
$$\cot\frac{\beta}{2}\cot\frac{\gamma}{2}=2.$$
Also, we can use the following way:
$$\cot\frac{\beta}{2}\cot\frac{\gamma}{2}=\frac{a+b+c}{b+c-a}=\frac{4a}{2a}=2.$$
|
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|
prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac{25}{29}=\frac{20}{29}\\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
|
It is $$\tan(\alpha/2)=\frac{r}{s-a}$$ and $$\tan(\gamma/2)=\frac{r}{s-c}$$ where $$s=\frac{a+b+c}{2}$$ so we get $$\frac{5}{6}(s-a)=\frac{2}{5}(s-c)$$
Can you finish now?
|
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|
Proving prime divisibility relation between $a^2-a+3$ and $b^2-b+25$. Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:
Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.
I had the idea of multiplying everything by $9$ because $3\times 9\approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.
According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).
|
$a^2-a+3\equiv0\pmod p$
$\iff(2a-1)^2\equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2\equiv-99\pmod p$
$\implies(2b-1)^2\equiv3^2(2a-1)^2\pmod p$
$\iff2b-1\equiv\pm3(2a-1)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3062320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
How to divide ${2k^3+3k^2+k-2j^3+3j^2-j}$ with $(k+1-j)$? The question I had was calculating $$\displaystyle\frac{1}{k+1-j}\sum_{i=j}^k i^2$$
Because I didn't know how to do a variable change, I did
$$\frac{1}{k+1-j}\sum_{i=j}^k i^2 = \frac{1}{k+1-j}\left(\sum_{i=1}^k i^2 - \sum_{i=1}^{j-1} i^2\right) = \frac{2k^3+3k^2+k-2j^3+3j^2-j}{6(k+1-j)}$$
The solution given used variable change and was able to cancel out $(k+1-j)$ very easily. Finally the answer was $$\frac{2k^2 + 2j^2 + 2kj - j + k}{6}$$ I have verified that these two solutions are the same by multiplying the second one with $(k+1-j).$
My question is:
How do you divide an expression like ${(2k^3+3k^2+k-2j^3+3j^2-j)}$ with $(k+1-j)$? Is it even worth it in this case to get the most simplified solution?
|
You could use synthetic division fairly easily here.
First, we consider $2k^3+3k^2+k-2j^3+3j^2-j$ as a polynomial in $k,$ and write the $4$ coefficients $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\ & & & &\\\hline & & & &\end{array}$$ Next, we invert the coefficients of the divisor $k+1-j$ to get $-1k+j-1,$ and write the constant term in as $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & & &\\\hline & & & &\end{array}$$ We bring down the first term to get $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & & &\\\hline & 2 & & &\end{array}$$ Now, we multiply what we've just brought down by $j-1$ and write in the result as $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & 2j-2 & &\\\hline & 2 & & &\end{array}$$ Now, adding the numbers in that column gets us $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & 2j-2 & &\\\hline & 2 & 2j+1 & &\end{array}$$ Multiply that result by $j-1$ and write it in to get $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & 2j-2 & 2j^2-j-1 &\\\hline & 2 & 2j+1 & &\end{array}$$ Adding again gives us $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & 2j-2 & 2j^2-j-1 &\\\hline & 2 & 2j+1 & 2j^2-j &\end{array}$$ Multiplying again by $j-1$ and then adding once more, we get $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\j-1 & & 2j-2 & 2j^2-j-1 & 2j^3-3j^2+j\\\hline & 2 & 2j+1 & 2j^2-j &0\end{array}$$ Translating back into terms of a polynomial in $k,$ this means that $$\frac{2k^3+3k^2+k-2j^3+3j^2-j}{k+1-j}=2k^2+(2j+1)k+2j^2-j+\frac{0}{k+1-j},$$ or more simply, $$\frac{2k^3+3k^2+k-2j^3+3j^2-j}{k+1-j}=2k^2+2j^2+2jk-j+k,$$ as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3062708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
}
|
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$
Use this identity to (4) too
and simplify (3),
we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$
And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$
But got stuck...
|
We are going to show that $(a,b)$ belongs to one of the two lines with equations $b=\sqrt{a}$ and $b=\frac{1}{\sqrt{a}}$ as displayed on the following figure. It will give the answer, due to the symmetry of the system of equations with respect to the group of variables $(a,b)$ vs. $(c,d)$. Moreover, we will establish (see (*) at the bottom) that the last equation is superfluous.
Here is the explanation :
Let :
$$S_1:=a+b, \ \ S_2:=c+d, \ \ P_1:=ab, \ \ P_2:=cd$$
The system constituted by the first three equations can be written, with these variables, using classical transformations :
$$\begin{cases}
(A) \ &S_1S_2&=&1& \ &\\
(B) \ &(S_1^2-2P_1)(S_2^2-2P_2)&=&9 & \ \implies \ & (C) \ 1-2(P_1S_2^2+P_2S_1^2)+4(P_1P_2)=9\\
(D) \ &(S_1^3-2P_1S_1)(S_2^3-2P_2S_2)&=&7 & \ \implies \ & (E) \ 1-3S_1S_2(P_1S_2^2+P_2S_1^2)+9(P_1P_2)=7.
\end{cases}$$
(equations (C) and (E) are obtained by expansion of (B) and (D) resp., using relationship (A)).
Setting
$$\alpha := P_1P_2 \ \text{and} \ \beta := P_1S_2^2+P_2S_1^2,$$
equations (C) and (E) become :
$$\begin{cases}
(C) & \ 2\alpha-\beta&=&4\\
(E) & \ 3\alpha-\beta&=&2
\end{cases} \ \ \implies \ \ \alpha=-2 \ \text{and} \ \beta=-8.$$
Using the fact that $S_1S_2=1$ and $\alpha=P_1P_2=-2$, equation $\beta=-8$ becomes :
$$P_1 \frac{1}{S_1^2} - \frac{2}{P_1}S_1^2 = -8$$
i.e.,
$$(F) \ \ \ \ P_1^2 + 8 P_1S_1^2 - 2 S_1^4 =0,$$
which can be considered as a quadratic equation in variable $P_1$ giving two solutions. Due to classical condition
$$(a+b)^2 \geq 2ab \ \iff \ S_1^2 \geq 2P_1,$$
only one of these solutions is eligible :
$$P_1=(-4+3\sqrt{2})S_1^2 \ \ \ \iff \ \ \ ab=(-4+3\sqrt{2})(a+b)^2 \ \ \ \iff \ \ \ (b-\sqrt{2}a)(b-\frac{\sqrt{2}}{2}a)=0$$
whence the result corresponding to the figure.
The parametric equations of the two lines are
$$(a,b)=(p,p \sqrt{2}) \ \ \text{and} \ \ (a,b)=(p,p \frac{\sqrt{2}}{2}), \ \ \text{for any} \ \ p \neq 0$$
Due to the symmetry of equations, we have as well, for any $q \neq 0$ :
$$(c,d)=(q,q \sqrt{2}) \ \ \text{and} \ \ (c,d)=(q,q \frac{\sqrt{2}}{2}).$$
A quick glance at any of the four equations show that necessarily $q=\frac{1}{p}$. We find back in this way all the solutions given by @Claude Leibovici and @A. Pongrácz .
(*) In fact, the fourth equation is a consequence of the first three. Here is why :
First of all, relationship (F) is equivalent to :
$$(G) \ \ \ \ S_1^4=\frac12P_1^2+4P_1S_1^2.$$
As the fourth equation can be written :
$$(H) \ \ \ \ (S_1^4+2P_1^2-4P_1S_1^2)(S_2^4+2P_2^2-4P_2S_2^2)=25,$$
using (G) in (H), we get :
$$\frac52P_1^2 \frac52P_2^2=25,$$
which is a tautology due to the fact that $\alpha=P_1P_2=-2.$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
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|
What is $\int_0^{\pi/2}\sin^7(\theta)\cos^5(\theta)d\theta$ I have to integrate the following:
$\int_0^\limits\frac{\pi}{2}\sin^7(\theta)\cos^5(\theta)d\theta$
I decided to use a $u$ substitution of $u=\sin^2(\theta)$, and $\frac{du}{2}=\sin(\theta)\cos(\theta)$
and arrived at this integral
$\int_\limits{0}^{1}u^3(1-u)^2du$
From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$
I get the following:
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\int_\limits{0}^{1}(u^2*(1-u)^3)du$$
Repeated again $g=u^2$, and $dv=(1-u)^3du$
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\int_\limits{0}^{1}u(1-u)^4$$
Repeating again $g=u$, and $dv=(1-u)^4$
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\biggl[\frac{u(1-u)^5}{5}\biggr]_0^1-\frac{1}5\int_\limits{0}^{1}(1-u)^5$$
and I get
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\biggl[\frac{u(1-u)^5}{5}\biggr]_0^1-\frac{1}{30}\biggl[(1-u)^6\biggr]_0^1$$
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I think you complicated the last part, after all you are integrating a polynomial.
$\displaystyle \int_0^1 u^3(1-u)^2\mathop{du}=\int_0^1 (u^3-2u^4+u^5)\mathop{du}=\left[\frac{u^4}4-2\frac{u^5}5+\frac{u^6}6\right]_0^1=\frac 14-\frac 25+\frac 16=\frac 1{60}$
Also you dropped the coeff $\dfrac 12$ from $\dfrac{du}2$, the result should be $\dfrac 1{120}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3064190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Integral over recurrence relationship I'm interested in evaluating the following definite integral
\begin{equation}
I_n = \int_0^{\gamma} F_n(x)\:dx
\end{equation}
Where $\gamma \gt 0$ and $F_n(x)$ is based on the recurrence relationship:
\begin{equation}
F_{n + 1}(x) = \frac{1}{1 + F_n(x)}
\end{equation}
Here $F_0(x) = f(x)$ where $f$ is a continuous function on $\left[0,\gamma\right]$. My first task was to find a general solution for $F_n(x)$ and this is where I've become unstuck.
I started by letting $F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)}$. Applying it to the recurrence relationship we have:
\begin{equation}
F_{n + 1}(x) = \frac{\alpha_{n+1}(x)}{\beta_{n+1}(x)} = \frac{\beta_n(x)}{\alpha_n(x) + \beta_n(x)}
\end{equation}
And so we have a recurrence relationship over both $\alpha_n(x)$ and $\beta_n(x)$ with $F_0(x) = f(x) = \frac{\alpha_0(x)}{\beta_0(x)}$.
To begin with I'm focused on $f(x) = \sec(x)$ with $\alpha_0(x) = 1$ and $\beta_0(x) = \cos(x)$ and $\gamma = \frac{\pi}{2}$.
With a few iterations via Wolframalpha the pattern that emerges is that $F_n(x)$ takes the form:
\begin{equation}
F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{a_n\cos(x) + b_n}{c_n \cos(x) + d_n}
\end{equation}
Where $a_n, b_n, c_n, d_n \in \mathbb{N}$ Here I would like to be able to solve for each but I'm not sure how to start.
Does anyone have any good starting points/references that I can use to begin?
Edit: Changed the definition of $I_n$ to have generalised upper limit of $\gamma$.
Update Thanks to hypernova's comment's below, it can be seen that $\beta_n(x)$ follows a Fibonacci Sequence:
\begin{equation}
\beta_{n + 1}(x) = \beta_n(x) + \alpha_n(x) = \beta_n(x) + \beta_{n - 1}(x)
\end{equation}
And so we can now represent $F_n(x)$ as:
\begin{equation}
F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{\beta_{n-1}(x)}{\beta_n(x)}
\end{equation}
for $n \geq 2$.
For the specific example above we have:
\begin{equation}
F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}
\end{equation}
Where $a_n$ and $b_n$ are Fibonacci Sequences with $b_0 = 0$, $b_1 = 1$ and $a_0 = 1$, $a_1 = 1$. We see that $a_n \gt b_n$ (this will be important later)
So, we may now evaluate the integral
\begin{equation}
I_n = \int_0^{\tfrac{\pi}{2}}\frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}\:dx
\end{equation}
I will here employ the Weierstrass substitution $t = \tan\left(\frac{x}{2} \right)$:
\begin{align}
I_n &= \int_0^{\tfrac{\pi}{2}}\frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}\:dx = \int_0^1 \frac{a_{n-1}\frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n \frac{1 - t^2}{1 + t^2} + b_n}\frac{2\:dt}{1 + t^2} \\
&= 2\int_0^1 \frac{a_{n - 1}\left(1 - t^2\right) + b_{n - 1}\left(1 + t^2\right)}{\left(1 + t^2\right)\left(a_n\left(1 - t^2\right) + b_n\left(1 + t^2\right)\right)}\:dt \\
&= 2\int_0^1 \frac{\left(b_{n - 1} - a_{n-1}\right)t^2 + \left(b_{n-1} + a_{n-1}\right)}{\left(1 + t^2\right)\left(\left(b_n - a_n\right)t^2 + \left(b_n + a_n\right)\right)}\:dt \\
&= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right) \int_0^1 \frac{t^2 - \theta_{n - 1}}{\left(1 + t^2\right)\left(t^2 + \theta_n\right)}\:dt
\end{align}
Where
\begin{equation}
\theta_n = \frac{b_n + a_n}{b_n - a_n}
\end{equation}
As $a_n \gt b_n \geq 0$ we see that $\theta_n \lt 0$. Hence:
\begin{align}
I_n &= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right) \int_0^1 \frac{t^2 - \left|\theta_{n - 1}\right|}{\left(1 + t^2\right)\left(t^2 - \left| \theta_n\right)\right|}\:dt \\
&= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left[ \frac{1}{\left|\theta_n\right| + 1} \left[\left(\left|\theta_{n-1}\right| + 1 \right) \arctan(x) + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{x}{\sqrt{\left|\theta_n\right|}} \right)\right]\right]_0^1 \\
&=2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left(\frac{1}{\left|\theta_n\right| + 1} \right)\left[\left(\left|\theta_{n-1}\right| + 1 \right) \frac{\pi}{4} + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{1}{\sqrt{\left|\theta_n\right|}} \right)\right]
\end{align}
Note $b_n = a_{n - 1}$ and thus:
\begin{align}
I_n &=2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left(\frac{1}{\left|\theta_n\right| + 1} \right)\left[\left(\left|\theta_{n-1}\right| + 1 \right) \frac{\pi}{4} + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{1}{\sqrt{\left|\theta_n\right|}} \right)\right] \\
&= \frac{a_{n - 1}}{a_n}\frac{\pi}{2} + \left[1 - \frac{a_{n + 1}\left(a_{n - 1} - a_{n - 2} \right)}{a_n\left(a_n - a_{n - 1} \right)} \right]\sqrt{\frac{a_n - a_{n - 1}}{a_{n + 1}}}\operatorname{arccoth}\left(\sqrt{\frac{a_{n + 1}}{a_n - a_{n - 1}}} \right)
\end{align}
|
This is copied from
another answer of mine:
Let
$f_1 = \frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=\frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
\to \dfrac{\sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$\begin{array}\\
f_n(x)
&=\frac{1}{1+\frac{1}{1+f_{n-2}(x)}}\\
&=\frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\\
&=\frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\\
\end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$\frac12 < f_n(x)
\lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
\lt 1$.
$\begin{array}\\
f_n(x)-f_{n-2}(x)
&=\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\\
&=\dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\\
&=\dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\\
\end{array}
$
$\begin{array}\\
f_n(x)+f_n^2(x)
&=\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\\
&=\dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+\dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
&=\dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
&=\dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
\text{so}\\
1-f_n(x)-f_n^2(x)
&=\dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\\
&=\dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\
\end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
\lt \frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
\to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = \frac{\sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x \ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
\to x_0$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Maximizing $f$ in $\mathbb{R}^3$
Find the domain and the maximum value that the function
$$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
may attain in its domain.
I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-2 x y-3 x z+y^2+z^2}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_y=\frac{2 x^2-x y+z (2 z-3 y)}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_z=\frac{3 \left(x^2+y^2\right)-z (x+2 y)}{\left(x^2+y^2+z^2\right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?
|
By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)\geq(x+2y+3z)^2,$$
which gives
$$-\sqrt{14}\leq\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}\leq\sqrt{14}.$$
The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3065678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find maximum value of $\frac xy$
If $x^2-30x+y^2-40y+576=0$, find the maximum value of $\dfrac xy$.
First I completed the squares and got $(x-15)^2+(y-20)^2=7^2$, which is the equation of a circle.
I think I need to use some properties but I don't know what to do next.
|
An alternative method, uses calculus. Similar to Sauhard Sharma's answer.
First note, this is the equation of a circle centred on $(15,20)$ of radius $7$, so it does not reach the $x$ axis (otherwise you'd be able to get arbitrarily large values for $x/y$). The value of $x/y$ is obviously going to be maximised on the lower half of the circle, since these have lower values of $y$, with the same values of $x$ as the corresponding points in the upper half of the circle. So we can write $$y=20-\sqrt{49-(x-15)^2}$$
We want to maximise $$f=\frac xy=\frac{x}{20-\sqrt{49-(x-15)^2}}$$
The range of $x$ is $[8,22]$. First we can check if this function has any maxima (or equivalently, if its reciprocal has a minimum).
$$\frac{d(1/f)}{dx}=\frac{d}{dx}\left(\frac{20}x-\sqrt{\frac{49}{x^2}-\left(1-\frac{15}x\right)^2}\right)=0\\-\frac{20}{x^2}-\frac12\left(-\frac{49\cdot2}{x^3}-2\left(1-\frac{15}x\right)\left(\frac{15}{x^2}\right)\right)\left(\frac{49}{x^2}-\left(1-\frac{15}x\right)^2\right)^{-1/2}=0\\\left(\frac{49}{x^3}+\frac{15}{x^2}-\frac{225}{x^3}\right)^2=\frac{400}{x^4}\left(\frac{49}{x^2}-1+\frac{30}{x}-\frac{225}{x^2}\right)\\\left(15x-176\right)^2=400\left(-x^2+{30}x-176\right)\\625x^2-17280x+101396=0$$This has solutions $$x_\pm=\frac{1728}{125}\pm\frac{2\sqrt{112771}}{125}$$ To maximise $x/y$, we choose the largest value of $x$, giving $x_+=19.197\dots$, which corresponds to $y=14.3977\dots$. This gives $$\frac xy=1.3333\dots\approx\frac43$$
|
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|
How to solve pell type equation Example $x^2-7y^2=2$
Find $x,y$ in general when $x,y$ are integer
I don't know how to solve it , I need the method to help to solve it
|
Apply the PQa algorithm for D=7, determining that $\sqrt{7}=[2,\overline{1,1,1,4}]$
$
\begin{array}{c|c|c|c|c|c}
j & P_j & Q_j & a_j & A_j & B_j \\
\hline
0 & 0 & 1 & 2 & 2 & 1\\
1 & 2 & 3 & 1 & \textbf{3} & \textbf{1}\\
2 & 1 & \textbf{2} & 1 & 5 & 2\\
3 & 1 & 3 & 1 & \textbf{8} & \textbf{3}\\
4 & 2 & \textbf{1} & 4 & 37 & 14\\
\end{array}
$
From this (bold highlighted cells), we see that $3^2-7 \cdot 1^2 = 2$ and $8^2-7 \cdot 3^2=1$.
All solutions $(x_n,y_n)$ can then be found via
$$x_n+y_n\sqrt{7} = (3+\sqrt{7}) \cdot (8+3\sqrt{7})^n$$
or via
$$\begin{pmatrix}x_n&y_n\\7y_n&x_n\end{pmatrix} = \begin{pmatrix}3&1\\ 7&3\end{pmatrix} \begin{pmatrix}8&3\\ 21&8\end{pmatrix}^n$$
Or, equivalently (taking products of this type of matrices is commutative, and removing the redundancy)
$$\begin{pmatrix}y_n\\x_n\end{pmatrix} = \begin{pmatrix}8&3\\ 21&8\end{pmatrix}^n\begin{pmatrix}1\\ 3\end{pmatrix} $$
Which is the same (but upside down) what Jagy mentions.
Of course, this only works for $x^2-Dy^2 = k$ when $k < 2\sqrt{D}$. For general $k$, other methods are required (see Jagy's answer).
|
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|
Solve the limit using Taylor seris with Big-O notation I have a limit
$$
\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2}
$$
I've tried to solve it like this:
\begin{align}
&\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2} =\\
&\lim_{x \to 0} \frac{x^2 - \sin x}{x^2\sin x} = \\
&\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}{6} - \frac{x^5}{120} - \mathcal{O}\left(x^7\right) }{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + x^2\mathcal{O}(x^7)}
\end{align}
My questions:
*
*How am I supposed to evaluate $x^2\mathcal{O}(x^7)$ in the denominator? I'd say it will be $\mathcal{O}(x^9)$, but I'm not sure about it.
*How am I supposed to evaluate the whole limit with respect to 0 considering Big-O notation? Could you, please, provide some intuitive explanation?
|
Just as you did
$$y=\frac{1}{\sin x} - \frac{1}{x^2}=\frac{x^2 - \sin x}{x^2\sin x}=\frac{x^2-\left( x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)\right) } {x^2\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) \right) }$$ that is to say
$$y=\frac{-x+x^2+\frac{x^3}{6}-\frac{x^5}{120}+O\left(x^7\right) } {x^3-\frac{x^5}{6}+\frac{x^7}{120}+O\left(x^9\right) }$$ Now, long division to get
$$y=-\frac{1}{x^2}+\frac{1}{x}+\frac{x}{6}+\frac{7 x^3}{360}+O\left(x^4\right)$$ which shows the limit and how it is approached.
It also gives you a way to estimate the value of $y$ without using the trigonometric functions. Suppose $x=\frac \pi 6$ (which is really far away from $0$). The exact value would be $\approx -1.64756$ while the above expansion would give $\approx-1.64765$. Doing the same with $x=\frac \pi {12}$ and get respectively $\approx -10.7265$ and $\approx -10.7265$ !
|
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|
Inequality in 3 variables with a constraint condition To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers
Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.
|
If we arrange
$$
\frac{ab+bc+ac}{abc} \ge \frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =\sum_{\text{cyc}}a^2b^2 +2abc\ge abc(2+3(ab+bc+ac))=2abc +3\sum_{\text{cyc}}a^2b^2c.
$$ We need to show
$$
\sum_{\text{cyc}}a^2b^2\ge 3\sum_{\text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
\sum_{\text{cyc}}a^2b^2(a+b+c)\ge 3\sum_{\text{cyc}}a^2b^2c.
$$ By rearrangement, we have
$$
\sum_{\text{cyc}}a^3b^2\ge \sum_{\text{cyc}}a^2b^2c,
$$and
$$
\sum_{\text{cyc}}a^2b^3\ge \sum_{\text{cyc}}a^2b^2c.
$$ This proves the inequality.
Another approach: We need to show
$$
\sum_{\text{cyc}}\frac{a}{b}+\sum_{\text{cyc}}\frac{a}{c}\ge \frac{2}{ab+bc+ca}.
$$ By C-S, we have
$$
\sum_{\text{cyc}}\frac{a}{b}\sum_{\text{cyc}}ab\ge \left(\sum_{\text{cyc}}a\right)^2 =1,
$$ and
$$
\sum_{\text{cyc}}\frac{a}{c}\sum_{\text{cyc}}ac\ge \left(\sum_{\text{cyc}}a\right)^2 =1.
$$
|
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|
Find all integer solutions to $x^2+xy+y^2=((x+y)/3 +1)^3$
Find all ordered pairs of integers $(x,y)$, that satisfy the following relation:
$$x^2+xy+y^2=((x+y)/3 +1)^3$$
I tried bashing at first. Then I assumed $x+y = 3k$ for some integer $k$ so that $y=3k-x$, substituted in the given equation and got a cubic polynomial involving $k$ and $x$ if I had solved it right. But I dont know how to proceed further. Please help. Any other solution is also welcome.
|
Put $a = x+y, b = x -y \implies x^2+y^2 +xy = \dfrac{(x+y)^2+(x-y)^2}{2}+ \dfrac{(x+y)^2-(x-y)^2}{4}= \dfrac{a^2+b^2}{2}+\dfrac{a^2-b^2}{4}= \dfrac{3a^2+b^2}{4}$. The equation becomes: $\dfrac{3a^2+b^2}{4} = \dfrac{(a+3)^3}{27}\implies 81a^2+ 27b^2=4(a+3)^3$. Observe that $3 \mid a$, thus put $a = 3n$, and we have a new equation:
$3(3n)^2+b^2= 4(n+1)^3\implies 4n^3-15n^2+12n+4 = b^2\implies (4n+1)(n-2)^2 = b^2\implies 4n+1 = m^2\implies n = k^2+k, m = 2k+1, k \in \mathbb{Z}\implies b = m(n-2)= (2k+1)(k^2+k-2)\implies a = 3n = 3k^2+3k\implies x = \dfrac{a+b}{2}= \dfrac{(2k+1)(k^2+k-2)+3k^2+3k}{2}, y = \dfrac{a-b}{2}= \dfrac{-(2k+1)(k^2+k-2)+3k^2+3k}{2}, k \in \mathbb{Z}$. Note that $k \neq 0$ for the solutions to "work".
|
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|
What is the maximum value of $(a+ b+c)$ if $(a^n + b^n + c^n)$ is divisible by $(a+ b+c)$ where the remainder is 0?
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$\ge b\geq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n \ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.
Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then $(a^n + b^n + c^n) = (a +b+ c) $.
|
The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n \text{ } \forall \text{ } n \ge 1 \tag{1}\label{eq1}$$
Since $a, b, c \ge 1$, then $a + b + c \ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thus,
$$a + b + c \equiv a^n + b^n + c^n \equiv 0 \pmod d \tag{2}\label{eq2}$$
As this must hold for all $n \ge 1$, consider $n = 3$ and substitute
$$c \equiv -a - b \pmod d \tag{3}\label{eq3}$$
into the middle part of the congruence in \eqref{eq2} to get
$$a^3 + b^3 + \left(-a - b\right)^3 \equiv 0 \pmod d \tag{4}\label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ab\left(a + b\right) \equiv 0 \pmod d \tag{5}\label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b \equiv -c \pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b \text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a \equiv 0 \pmod d$. Thus, $b + c \equiv 0 \pmod d$, i.e., $b \equiv -c \pmod d$. Now, use $n = 2$ in the congruence in the middle part of \eqref{eq2} to get
$$a^2 + b^2 + c^2 \equiv 0 + \left(-c\right) + c^2 \equiv 2c^2 \equiv 0 \pmod d \tag{6}\label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$\left(a, b, c\right) = \left(1, 1, 4\right) \tag{7}\label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n \equiv 0 \pmod 6$ for all $n \ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n \equiv 1^n \equiv 1 \pmod 3$, so $3$ divides it as well. I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track. I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
|
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|
Compute $S = \sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor$ I want to compute the following sum
$$S = \sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor.$$
Here is what I tried:
$$ S = \sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rfloor.$$
If $m= 2t$ then
$$S =\sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rfloor = \frac{t(t+1)}{2} + \frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = \sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rfloor = \frac{t(t+1)}{2} + \frac{t(t+1)}{2}= t(t+1).$$
But I am not sure if this is correct. Perhaps someone could give an indication.
|
Yes, you are correct. You may also write the result as a more compact formula:
$$\sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor=
\begin{cases}
t^2&\text {if $m=2t$}\\
t(t+1)&\text {if $m=2t+1$}\\
\end{cases}=\left\lfloor \frac{m^2}{4}\right\rfloor.$$
Indeed, if $m=2t$ then
$$\left\lfloor \frac{m^2}{4}\right\rfloor=\left\lfloor t^2\right\rfloor=t^2$$
and if $m=2t+1$ then
$$\left\lfloor \frac{m^2}{4}\right\rfloor=\left\lfloor t^2+t+\frac{1}{4}\right\rfloor=t(t+1).$$
|
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|
Finding maxima of a function $f(x) = \sqrt{x} - 2x^2$ without calculus My question is how to prove that $f(x) = \sqrt x - 2x^2$ has its maximum at point $x_0 = \frac{1}{4}$
It is easy to do that by finding its derivative and setting it to be zero (this is how I got $x_0 = \frac{1}{4}$). But the task is to do that without using any calculus tools and I`m stuck at it.
My idea was to introduce $t = \sqrt x$ to get $f(t) = t - 2t^4$ and then find its maximum because I know how to do that with parabola which can be transformed to $a(x-x_0)^2 + y_0$. So I was trying to turn $f(t) = t - 2t^4$ into $f(t) = a(x-x_0)^4 + y_0$ but it seems impossible.
My second thought was to use the definition of rising function on an interval so I supposed we have $a, b \in \left(0; \frac{1}{4}\right) \text{and } a < b$. Then I prove that $$\sqrt a - 2a^2 < \sqrt b - 2b^2$$ $$2(b - a)(a + b) < \sqrt b - \sqrt a$$ $$ 2(\sqrt a + \sqrt b)(a + b) < 1$$ which is true since $0 < a < b < \frac{1}{4}$
Exactly the same way I prove that for every $a, b \in \left(\frac{1}{4}; +\infty\right) \text{and } a < b$ $$\sqrt a - 2a^2 > \sqrt b - 2b^2$$
So we have that $f(x)$ is rising on $\left(0; \frac{1}{4}\right)$ and declining on $\left(\frac{1}{4}; +\infty\right)$, consequently at $\frac{1}{4}$ we have a maximum of $f$
But I guess it is not fair to use derivatives to find the maximum and then simply prove that this value is correct. Is there an even better solution? What are your thoughts about my proof?
UPD: there are a lot of solutions which are formally OK but first we have to guess $x_0 = \frac{1}{4}$ or $y_0 = \frac{3}{8}$. My goal is to find a solution which does not rely on guessing and I think the inequality above is the best way to do that
|
$f\left(\frac{1}{4}\right)=\frac{3}{8}.$
We'll prove that we got a maximal value.
Indeed, we need to prove that
$$\sqrt{x}-2x^2\leq\frac{3}{8},$$ which is true by AM-GM:
$$2x^2+\frac{3}{8}=2x^2+3\cdot\frac{1}{8}\geq4\sqrt[4]{2x^2\left(\frac{1}{8}\right)^3}=\sqrt{x}.$$
We can get a value $\frac{3}{8}$ by the following way.
Let $$\max_{x\geq0}f=k.$$
Thus, $$\sqrt{x}-2x^2\leq k$$ or
$$2x^2+k\geq\sqrt{x}.$$
Now, since for $x\geq0$ we have
$$\sqrt{x}=\sqrt[4]{x^2},$$ we need four addends if we want to use AM-GM.
And indeed, by AM-GM
$$2x^2+k=2x^2+3\cdot\frac{k}{3}\geq4\sqrt[4]{2x^2\left(\frac{k}{3}\right)^3}.$$
Id est, we need $$4\sqrt[4]{2x^2\left(\frac{k}{3}\right)^3}=\sqrt{x}$$ or
$$512k^3=27$$ or
$$k=\frac{3}{8}.$$
The equality occurs for $$2x^2=\frac{k}{3}$$ or $$x=\frac{1}{4},$$ which says that we got a maximal value.
|
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|
Solving with L'Hôpital's rule. What is wrong? L'Hôpital's rule can be used infinitely many times as the limit remains $0/0$. However, this problem does not work with L'Hôpital's rule (I may have counted wrong).
Here it is:
$$\lim _{x\to 0}\left(\frac{\left(e^x+\sin x\right)x+e^x-\cos x}{x^2}\right)$$
We use L'Hôpital's rule once:
$$\lim _{x\to 0}\left(\frac{2\sin x\left(x\right)+x\left(\cos\left(x\right)+e^x\right)+2e^x}{2x}\right)$$
One more time:
$$\left(\frac{−x\sin\left(x\right)+3\cos\left(x\right)+\left(x+3\right)e^x}{2}\right)$$
This will equal to $3$. But the real answer is $1$:
$$\lim _{x\to 0}\left(\frac{\left(e^x+\sin x\right)x+e^x-\cos x}{x^2}\right)=1$$
$3$ is not equal to $1$
Thoughts?
|
You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
\begin{align}
(e^x+\sin x)x+e^x-\cos x
&=(1+x+x+o(x))x+1+x+\frac{x^2}{2}-1+\frac{x^2}{2}+o(x^2) \\
&=x+2x^2+x+x^2+o(x^2)\\
&=2x+3x^2+o(x^2)
\end{align}
Thus we see that the given limit cannot be finite (and indeed it is $-\infty$ from the left and $\infty$ from the right).
It's also difficult to understand how the given solution could be $1$, unless we do a sign switch:
\begin{align}
(e^x+\sin x)x-e^x+\cos x
&=(1+x+x+o(x))x-1-x-\frac{x^2}{2}+1-\frac{x^2}{2}+o(x^2) \\
&=x+2x^2-x-x^2+o(x^2)\\
&=x^2+o(x^2)
\end{align}
This proves that
$$
\lim_{x\to0}\frac{(e^x+\sin x)x-e^x+\cos x}{x^2}=1
$$
|
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|
Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $\zeta$-function, but I was looking for an explicit calculation.
|
This will require some calculations but no special functions. Making the change of variables $x = -x$ on $\mathbb R^-$, we obtain
$$\int_{\mathbb R} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx =
\underbrace {\int_{\mathbb R^+} \frac {2 x} {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx}_
{= I} +
\int_{\mathbb R^+} \frac x {(x^2 + 1)^2} dx, \\
I = 2 \pi i \sum_{k \geq 1}
\operatorname*{Res}_{x = i k} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} +
\frac 1 {2 (x^2 + 1)} \bigg\rvert_{x = 0}^\infty = \\
2 \pi i \frac {4 \pi^2 + 3} {96 \pi i} -
2 \pi i \sum_{k \geq 2} \frac k {2 \pi i (k^2 - 1)^2} -
\frac 1 2 = \\
\frac {4 \pi^3 - 21} {48} - \frac 1 4 \sum_{k \geq 2}
\left( \frac 1 {(k - 1)^2} - \frac 1 {(k + 1)^2} \right) = \\
\frac {4 \pi^3 - 21} {48} - \frac 1 4 \left( 1 + \frac 1 4 \right) =
\frac {\pi^2 - 9} {12}.$$
|
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|
Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.
How many solutions does
$$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$
have between $-90^\circ$ and $90^\circ$?
I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt{3}}{2}\right).$$
Since $a$ is between $-90^\circ$ and $90^\circ$, then $2a$ is between $-180^\circ$ and $180^\circ$. The RHS can be $60^\circ$, $120^\circ$, $-240^\circ$, and $-300^\circ$. Only $60^\circ$ and $120^\circ$ fit the criteria, but the answer is 4 solutions.
Where did I go wrong?
|
$\sin(2a-45) = \frac{\sqrt{3}}{2}$
$2a-45 = 60 + 360n \Rightarrow a = 52.5 + 180n$
$2a-45 = 120+ 360n \Rightarrow a = 82.5 + 180n$
Either $a = 52.5 $ or $a = 82.5$. If you square
$\sin(4a) = \frac{-1}{2}$
$4a = 210 + 360n \Rightarrow a = 52.5 + 90n $
$4a = 330 + 360n \Rightarrow a = 82.5 + 90n $
First two solution
$a = 52.5 $
$a= 52.5 - 90 = -38.5$
The other two
$a = 82.5 , 82.5 - 90 = -7.5$
In both cases the second solution is rejected see
Checking $\sin(-15) - \cos(-15) = \frac{-\sqrt{6}}{2}$ which is wrong same for the other
|
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|
$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$ evaluation using expansion series $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
This is the limit which I got from the book of Joseph Edwards' Differential Calculus for Beginners the second exercise of the First Chapter Page -10 question number 19. Whose Answer was given as $\frac 1 2$.
So my try was as follows:
$$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
$$\Rightarrow \lim \limits _{x \rightarrow 0} [ \frac{x^3(1 + \frac{x^4}{4} + \frac{x^8}{32}...) - (x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}...)^{\frac 3 2} }{x^7} ]$$
Now, this is the quite a big problem, as we can see that sine expansion series has the power $\frac 3 2$ which is nearly impossible to evaluate.
The expansion series given in the book are: $(1+x)^n$ and $(1-x)^{-n}$ which are not useful here as there are infinite series where those are 2-terms.
Please anyone help me , I've just started Calculus :P
|
You are on the right track. By using little-o notation you may control the expansions in a better way. We have that as $x\to 0^+$,
$$\frac{x^3(1 + \frac{x^4}{4} + o(x^4)) - (x^2 - \frac{x^6}{3!} + o(x^6))^{\frac 3 2} }{x^7}=\frac{1 + \frac{x^4}{4} + o(x^4) - (1 - \frac{x^4}{3!} + o(x^4))^{\frac 3 2} }{x^4}$$
Now all you need is that $(1+t)^{3/2}=1+\frac{3t}{2}+o(t)$ this can be shown by verifying that,
$$\lim_{t\to 0}\frac{(1+t)^{3/2}-1}{t}=
\lim_{t\to 0}\frac{(1+t)^{3}-1}{t((1+t)^{3/2}+1)}=\lim_{t\to 0}\frac{3+o(1)}{((1+t)^{3/2}+1}=\frac{3}{2}.$$
|
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|
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)? I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.
Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?
|
Let $p(a,b)$ be the probability that team A is the first to win 4 games, given that team A has already won $a$ games and team B has already won $b$ games. By conditioning on the outcome of the next game, we see that $p$ satisfies the following recurrence relations:
\begin{align}
p(4,b) &= 1 \\
p(a,4) &= 0 \\
p(a,b) &= \frac{1}{2} p(a+1,b) + \frac{1}{2} p(a,b+1) &&\text{if $a<4$ and $b<4$}
\end{align}
You want to compute $p(1,2)$. Now
\begin{align}
p(1,2)&=\frac{1}{2} p(2,2) + \frac{1}{2} p(1,3)\\
&=\frac{1}{2} \left(\frac{1}{2} p(3,2) + \frac{1}{2} p(2,3)\right)
+ \frac{1}{2} \left(\frac{1}{2} p(2,3) + \frac{1}{2} p(1,4)\right)\\
&=\frac{1}{4} p(3,2) + \frac{1}{2} p(2,3) + \frac{1}{4} \cdot 0\\
&=\frac{1}{4} \left(\frac{1}{2} p(4,2) + \frac{1}{2} p(3,3)\right) + \frac{1}{2} \left(\frac{1}{2} p(3,3) + \frac{1}{2} p(2,4)\right) \\
&=\frac{1}{8} \cdot 1 + \frac{3}{8} p(3,3) + \frac{1}{4} \cdot 0 \\
&=\frac{1}{8} + \frac{3}{8} \left(\frac{1}{2} p(4,3) + \frac{1}{2} p(3,4)\right) \\
&=\frac{1}{8} + \frac{3}{8} \left(\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot 0\right) \\
&=\frac{1}{8} + \frac{3}{16} \\
&=\frac{5}{16}
\end{align}
|
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|
Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}
\lim_{x \to 1}
\frac{x-x^x}{1-x+\ln x}&=\lim_{x \to 1}\frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2)]}\\
&=\lim_{x \to 1}\frac{-(x-1)^2-o((x-1)^2)}{-\frac{1}{2}(x-1)^2+o((x-1)^2)}\\
&=2.
\end{align*}
Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?
|
Why no?
$$\lim_{x\rightarrow1}\frac{x-x^x}{1-x+\ln{x}}=\lim_{x\rightarrow1}\frac{1-x^x(1+\ln{x})}{-1+\frac{1}{x}}=\lim_{x\rightarrow1}\frac{-x^x(1+\ln{x})^2-x^x\cdot\frac{1}{x}}{-\frac{1}{x^2}}=2$$
|
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|
Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me your proof.
Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.
|
We have:
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=\frac{15k^7+21k^5+70k^3-k}{3\cdot 5\cdot 7}
$$
To prove this is an integer we need that:
$$15k^7+21k^5+70k^3-k\equiv 0 \pmod{3\cdot 5\cdot 7}$$
According to the Chinese Remainder Theorem, this is the case iff
$$\begin{cases}15k^7+21k^5+70k^3-k\equiv 0 \pmod{3} \\
15k^7+21k^5+70k^3-k\equiv 0 \pmod{5}\\
15k^7+21k^5+70k^3-k\equiv 0 \pmod{7}\end{cases} \iff
\begin{cases}k^3-k\equiv 0 \pmod{3} \\
k^5-k\equiv 0 \pmod{5}\\
k^7-k\equiv 0 \pmod{7}\end{cases}$$
Fermat's Little Theorem says that $k^p\equiv k \pmod{p}$ for any prime $p$ and integer $k$.
Therefore the original expression is an integer.
|
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|
How do I find the order of a bijection? Question: List all bijections (permutations) from $\{1, 2, 3\}$ onto $\{1, 2, 3\}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
\begin{Bmatrix}
1 & 2 & 3 \\
1 & 2 & 3 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
1 & 2 & 3 \\
2 & 1 & 3 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{Bmatrix}
$,$
\begin{Bmatrix}
1 & 2 & 3 \\
3 & 1 & 2 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
\end{Bmatrix}
$
I understand that order of a permutation $\sigma$ is the smallest possible integer $k$ such that $\sigma^k = \epsilon$, where $\epsilon$ is the identity permutation.
But I am confused by the definition of "identity permutation". If my identity is:
$$
\begin{Bmatrix}
1 & 2 & 3 \\
1 & 2 & 3 \\
\end{Bmatrix}
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
\begin{Bmatrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{Bmatrix}
$, order $= 1$, sign $= -1$. And for $
\begin{Bmatrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{Bmatrix}
$ order $= 2$, sign $= 1$.
But if my first bijection from $\{1, 2, 3\}$ onto $\{1, 2, 3\}$ is:
$$
\begin{Bmatrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{Bmatrix}
$$
then order $= 0$ and sign $= 1$. So depending on which projection I chose as an identity, the order differs. Can someone clarify?
|
The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $\epsilon^1=\epsilon$ (the order must be positive). The order of $
\sigma_1=\begin{Bmatrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{Bmatrix}
$ is $2$ because $\sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $\sigma_1^2=\epsilon$. The order of $\sigma_2=\begin{Bmatrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{Bmatrix}$ is $3$ because $\sigma_2\neq \epsilon$ and $\sigma_2^2\neq \epsilon$, but $\sigma_2^3=\epsilon$.
|
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|
Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6\over 3}-2x^2={2x^6\over3}$
As for the numerator $\arctan(2x^3)\sim2x^3$ and ${2x^7+x^8\over 3x^2+x^4}\sim{2\over3}x^5$.So if we expand $\sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(\sin x)^2\sim2x^3-{2\over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $\sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
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The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ \frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 \cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $\log(1+t),$ and for $\arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$ \frac{1}{x^2 + 3} = \frac{1}{3} - \frac{x^2}{9} + \frac{x^4}{27} - \frac{x^6}{81} + \cdots $$
$$ \frac{1}{x^4 + 3x^2} = \frac{1}{3x^2} - \frac{1}{9} + \frac{x^2}{27} - \frac{x^4}{81} + \cdots $$
$$ \frac{x^8 + 2x^7}{x^4 + 3x^2} = \frac{2x^5}{3} + \frac{x^6}{3} - \frac{2x^7}{9} - \frac{x^8}{9} + \cdots $$
$$ 2x \sin^2 x = 2 x^3 - \frac{2x^5}{3} + \frac{4x^7}{45} - \cdots $$
$$ \arctan (2x^3) = 2 x^3 - \frac{8x^9}{3} - \cdots $$
The numerator is $$ \frac{x^6}{3} - \frac{2x^7}{15} + \cdots $$
The denominator is $$ \frac{2x^6}{3} + \frac{2x^{10}}{5} + \cdots $$
giving limit $1/2$
|
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$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
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Prove: for all $n$: $a^{2n+1}+b^{2n+1}=c^{2n+1}+d^{2n+1}$
If $a+b\ne 0$ then we get $$a^2-ab+b^2 =c^2-cd+d^2\implies (a+b)^2-3ab = (c+d)^2-3cd$$
so $$ab = cd$$
Induction step: $n-1,n\to n+1$
By I. H. we have $$a^{2n-1}+b^{2n-1} = c^{2n-1}+d^{2n-1}\;\;\; $$
now we multiply this with $a^2+b^2 = c^2+d^2$ we get $$ a^{2n+1}+\color{red}{b^{2n-1}a^2+ a^{2n-1}b^2}+b^{2n+1} =c^{2n+1}+\color{red}{c^2d^{2n-1} +c^{2n-1}d^2}+d^{2n+1}\;\;\; $$
Since (again by I.H.)$$ a^2b^2(\color{red}{b^{2n-3}+ a^{2n-3}}) =c^2d^2(\color{red}{d^{2n-3} +c^{2n-3}})$$
and we are done.
|
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|
Differential Equations - Is the solution to $(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$, $y=\ln(-5(e^x+1)^5+c)$? This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$\frac{-e^x}{(e^x+1)^6}dx=\frac{e^y}{(e^y+1)^2}dy$$
$$\int{\frac{-e^x}{(e^x+1)^6}dx}=\int{\frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$\int{\frac{e^y}{(e^y+1)^2}dy}=\int{\frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$\int{u^{-2}dy}=-\int{v^{-6}}dx$$
$$-u^{-1} + c_1=\frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-\frac{1}{e^y+1}+c_1=\frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-\frac{1}{e^y+1}=\frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$\ln(-5(e^x+1)^5+c_3)=\ln(e^y)$$
$$\ln\left(-5(e^x+1)^5+c_3\right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
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The step from:
$$ -\frac{1}{e^y + 1} = \frac{1}{5\left(e^x+1\right)^5} +c_1 \tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 \tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -\frac{1}{e^y + 1} = \frac{1}{5\left(e^x+1\right)^5} +c_3 $$
$$ \frac{1}{5(e^x +1)^5+c_2} = \frac{1}{5\left(e^x+1\right)^5} +c_3 $$
$$ 1 = \frac{5(e^x +1)^5+c_2}{5\left(e^x+1\right)^5} +c_3\big[5(e^x +1)^5+c_2\big] $$
$$ 1 = 1+\frac{c_2}{5\left(e^x+1\right)^5} +c_3\big[5(e^x +1)^5+c_2\big] $$
$$ 0 = \frac{c_2}{5\left(e^x+1\right)^5} +c_3\big[5(e^x +1)^5+c_2\big] $$
it is a polynomial equation if $u=e^x+1$, and the equation has a finite set of solutions of $u$ (thus $x$), and the set solution isn't a interval in $\Bbb{R}$, thus the implication from (1) to (2) is false, becaus it doesn't have sense.
|
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|
Relation between Roots and Coefficients Question If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.
Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.
|
The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -\sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-\sqrt[3]{r}$ is a root.
Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p \ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.
For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.
This means that we should consider $-\sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - \sqrt{3})/2$ and $5(2 + \sqrt{3})/2$. Hence we get $a = 5(2 - \sqrt{3})/2$ and $b = 2 + \sqrt{3}$.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.