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How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$ prove that
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<\dfrac{5}{3}$$
my idea: use
$$3^k-2^k>2^k(k\ge 2)$$
then
$$\Longleftrightarrow 1+\sum_{k=2}^{n}\dfrac{k}{3^k-2^k}<1+\sum_{k=2}^{n}\dfrac{k}{2^k}<1+\sum_{k=2}^{\infty}\dfrac{k}{2^k}=1+\dfrac{3}{2}>\dfrac{5}{3}$$
use this methods,I can't prove it
other idea:
$$3^k-2^k>2\cdot 2^k(k\ge 3)$$
But This same can't prove it
$$\Longleftrightarrow \sum_{k=3}^{\infty}\dfrac{k}{3^k-2^k}<\sum_{k=3}^{\infty}\dfrac{k}{2^k}<\dfrac{2}{3}-\dfrac{2}{5}$$
so This problem have other nice methods? Thank you
|
By arithmetic/geometric means,
$$
3^k-2^k=3^{k-1}+3^{k-2}\cdot 2+\ldots +2^{k-1}< k\cdot 6^{(k-1)/2}.
$$
Hence
$$
\sum_{k\ge m}\frac{k}{3^k-2^k}<\sum_{k\ge m}\frac{1}{6^{(k-1)/2}}<\frac{1}{6^{(m-1)/2}}\frac{1}{1-\frac{1}{\sqrt{6}}}
$$
and it remains to choose $m$ as Raymond Manzoni advices.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?
|
$(a^3+b^3+c^3) = (a+b+c)(a^2 +b^2+c^2-ab-bc-ac)$
$1+\omega + \omega^2 = 0 \implies \omega+\omega^2 = -1$
$= (a+b+c)(a^2 + b^2 + c^2 +ab(\omega+\omega^2)+bc(\omega+\omega^2)+ac(\omega+\omega^2)) = (a+b+c)(a(a+b\omega^2+c\omega)+b\omega(a+b\omega^2+ c\omega)+ c\omega^2 (a+ \dfrac b\omega+c\omega)$
Now, note that $\dfrac{1}{\omega}= \omega^2$
And after substituting this we get the desired identity.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $(a^2-1)(b^2-1)=\frac{1}4 ,a,b\in \mathbb Q$ Does the equation $(a^2-1)(b^2-1)=\dfrac{1}4$ have solutions $a,b\in \mathbb Q$?
I search $0<p<1000,0<q<1000$, where $a=\dfrac{p}q$, but no solutions exist. I wonder is this equation solvable?
|
If there are rational points on your curve than there are rational points on the curve:
$$ (x^2-4)(y^2-1)=1, $$
so $(y^2-1)=\frac{p}{q}$ and $(x^2-4)=\frac{q}{p}$, with $(p,q)=1$, must hold.
This implies that both $\frac{p+q}{q}$ and $\frac{4p+q}{p}$ are squares of rational numbers, so $p=u^2,q=v^2$ and both $u^2+v^2$ and $4u^2+v^2$ are squares, say $u^2+v^2=q_1^2$ and $(2u)^2+v^2=q_2^2$. By the parametrization of primitive pythagorean triples, the second identity implies that $u=ab$ and $v=a^2-b^2$, with $(a,b)=1$ and $a,b$ not both odd. By substituting into the first identity we get that
$$ (ab)^2 + (a^2-b^2)^2 = (a^2+b^2)^2-3(ab)^2 = \frac{1}{4}(a^2+b^2)^2+\frac{3}{4}(a^2-b^2)^2$$
must be a square. So we started with $q_2^2=4p+q$ that belonged to both the quadratic forms $x^2+y^2$ (since $4p+q=(2u)^2+v^2$) and $x^2+3y^2$ (since $4p+q=q_1^2+3u^2$) and get that $q_2$ belong to the intersection of the integers represented by the same quadratic forms: this leads to an infinite descent and the impossibility to find rational points on the curve $(x^2-4)(y^2-1)=1$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x)
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:
*
*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?
*Are there trigonometric identities that I could have used to simplify either derivative?
|
Hint:
$$\begin{align}
0&=\cos x+\sin x\\
\sin x &= -\cos x \\
\text{Divide}&\text{ both sides by }\cos x\\
\tan x &= -1
\end{align}$$
Now you can solve for $x$ using inverse trigonometric functions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
proof that $f(x)=\sum_{n=1}^\infty \frac{\ln(x+n)}{x^2 + n^2}$ converges uniformily Prove that $f(x)=\sum_{n=1}^\infty \frac{\ln(x+n)}{x^2 + n^2}$ converges uniformly for $x\geq 0$.
This exercise is in a text about uniform convergence, I've tried to use Weierstrass test with no success. trying to quote with
$a_n=\sqrt{x+n}/(x^2 + n^2)$
so that
$\ln(x+n)/(x^2 + n^2)\leq\sqrt{x+n}/(x^2 + n^2)\,,$
but then I can't find a criteria to prove that $\sqrt{x+n}/(x^2 + n^2)$ converges (and if I prove this then $f(x)$ converges uniformly as Weierstrass test claims)
|
Hint: $(x+n)^2 \le 2(x^2 + n^2)$.
$$\dfrac{\sqrt{x+n}}{x^2+n^2} \le \dfrac{\sqrt[4]{2}\sqrt[4]{x^2+n^2}}{x^2+n^2} = \dfrac{\sqrt[4]{2}}{(x^2+n^2)^{3/4}} \le \dfrac{\sqrt[4]{2}}{n^{3/2}}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a_n}$ is bounded above or not, and prove your answer is correct.
I started to solve it in the following manner:
Let $a_n \geq a_{n+1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2(n+1)-1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2n+1}.$ Multiplying by the reciprocal of $a_n$, we have $1 \geq \frac{1}{2n+1}$ which shows that $a_n$ is increasing.
This is where I am stuck since I do not know how to proceed in showing it is bounded from what I have so far. Any assistance or criticism is welcome.
I am using the following textbook: Introduction to Analysis by Arthur Mattuck.
|
HINT: Let $b_n=\frac1{1\cdot3\cdot\ldots\cdot(2n-1)}$; for all $n\ge 1$ we have $b_{n+1}\le\frac13b_n$, so $$a_n=\sum_{k=1}^nb_k\le\sum_{k=1}^n\left(\frac13\right)^k\;.$$
Now use what you know about geometric series.
|
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|
Projections on the Riemann Sphere are antipodal Prove that given two points $z,w\in\mathbb{C}$, we have that their projections on the Riemann sphere are antipodal if and only if: $z\bar{w}=-1$.
|
We could of course just compute it using the explicit formula for the stereographic projection. But let's argue geometrically.
Let $z \in \mathbb{C}\setminus\{0\}$ and $\zeta$ the image of $z$ on the sphere under stereographic projection. Let $C$ be the great circle on the sphere passing through $\zeta$ and the north and south pole.
Let $\varphi$ be the angle between the plane and the line connecting $z$ with the north pole $N$. Since the radius of the sphere is $1$, the distance of $z$ from the intersection of the plane and the axis of the sphere (that is, $0 \in \mathbb{C}$) is $\cot \varphi$, i.e. $\lvert z\rvert = \cot \varphi$.
The line connecting $\zeta$ and its antipodal point $\alpha := -\zeta$ is a diameter of $C$, hence, by Thales' theorem, the triangle $\alpha N\zeta$ has a right angle at $N$. $\angle z 0N$ is a right angle, hence $\angle 0N\zeta = \angle 0Nz = \frac{\pi}{2}-\varphi$, and therefore $\angle 0N\alpha = \varphi$. But that means the line $N\alpha$ intersects the plane at a distance of $\tan\varphi$ from $0$, on the opposite side of $z$, and that means the point projected to $\alpha$ is $$w = -\frac{z}{\lvert z\rvert^2},$$
whence $z\overline{w} = -1$.
Conversely, if $z\overline{w} = -1$, and $w$ is projected to $\beta$ on the sphere, then $\lvert w\rvert = \tan \varphi$, whence $\angle \beta N0 = \varphi$, and since $w$ (and $\beta$) lie on the opposite side of $0$ from $z$, we have
$$\angle \beta N\zeta = \angle \beta N0 + \angle 0N\zeta = \varphi + \left(\frac{\pi}{2} - \varphi\right) = \frac{\pi}{2},$$
and by Thales, the line segment connecting $\beta$ and $\zeta$ is a diameter of the great circle $C$, hence $\beta = -\zeta$ is the antipodal point of $\zeta$.
Analytically, with the explicit formulae of the stereographic projection
$$\begin{align}
\varphi \colon z = x+iy &\mapsto \left(\frac{2x}{x^2+y^2+1},\,\frac{2y}{x^2+y^2+1},\,\frac{x^2+y^2-1}{x^2+y^2+1} \right)\\
\varphi^{-1} \colon (\alpha,\,\beta,\,\gamma) &\mapsto \frac{1}{1-\gamma}(\alpha + i \beta)
\end{align}$$
we obtain that the point being projected to the antipodal point of $z$'s projection is
$$\begin{align}
w &= \varphi^{-1} \left(\frac{-2x}{x^2+y^2+1},\,\frac{-2y}{x^2+y^2+1},\, \frac{1-x^2-y^2}{x^2+y^2+1}\right)\\
&= \frac{x^2+y^2+1}{2(x^2+y^2)}\frac{-2(x+iy)}{x^2+y^2+1}\\
&= \frac{-(x+iy)}{x^2+y^2}\\
&= -1/\overline{z}.
\end{align}$$
|
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|
Solving exponential complex equation. I need assistance in solving the following equation:
$$ e^{z^2}=1, $$
where $z$ is a complex number.. I can't seem to get the answer of $z=\sqrt{k\pi}(1\pm i)$. Thank you in advanced for your help.
|
We have $\displaystyle e^{z^2}=1=e^{2k\pi i}$ where $k$ is any integer
$\displaystyle\implies z^2=2k\pi i$
$\displaystyle\implies z=\sqrt{2k\pi i}=\sqrt{2k\pi }\sqrt i$
Method $1:$
By observation, $i=\frac{i^2+1+2i}2=\frac{(i+1)^2}2$
$\displaystyle\implies i^{\frac12}=\pm\frac{(1+i)}{\sqrt2}$
Method $2:$
Let $\sqrt{i}=x+iy$ where $x,y$ are real
Squaring we get, $i=(x+iy)^2=x^2-y^2+2xyi$
Equating the real & the imaginary parts we get $x^2-y^2=0$ and $2xy=1$
From the first relation $x=\pm y$
If $x=-y, 1=2xy=2(-y)y\implies y^2=-\frac12$ which is impossible as $y$ is real, $\implies y^2\ge0$
$\implies x=y$ and $1=2xy=2(y)y\implies y^2=\frac12\implies y=\pm\frac1{\sqrt2}$
$\implies \sqrt i=\pm\left(\frac1{\sqrt2}+i\frac1{\sqrt2}\right)=\pm\frac{1+i}{\sqrt2}$
Method $3:$
Let $i=r(\cos\theta+i\sin\theta)=r\cos\theta+ir\sin\theta$ where real $r\ge0$
Equating the real & the imaginary parts we get $r\cos\theta=0\ \ \ \ $ and $r\sin\theta=1\ \ \ \ (2) $
From $(1),$ either $r=0$ or $\cos\theta=0$
If $r=0,$ from $(2),1=r\sin\theta=0$ which is impossible
$\implies r>0$ and $\cos\theta=0\implies \sin\theta=\pm1$
If $\sin\theta=-1,1=r\sin\theta=-r\iff r=-1$ which is impossible as $r>0$
$\implies \sin\theta=1\implies \theta=2n\pi+\frac\pi2$ where $n$ is any integer
$\implies i=\cos(2n\pi+\frac\pi2)+i(2n\pi+\frac\pi2)=e^{(2n\pi+\frac\pi2)i}$ (using Euler's Formula)
Using de Moivre's formula,
$\implies i^{\frac12}=e^{\frac{(2n\pi+\frac\pi2)i}2}$ where $n=0,1$
|
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|
Prove that $1^3 + 2^3 + \cdots + n^3 < n^4$. I am trying to prove the following: $1^3 + 2^3 + \cdots + n^3 < n^4$ if $n \in \mathbb{N}, n>1$ by induction. From there, I am to prove that the sum is $< \frac{n^4}{2}$ if $n>2$. My attempt to do so is as follows:
First, we use $n = 2$ as a basis step since $n \in \mathbb{N}, n>1$ which shows that $1^3+2^3 < 2^4 \Rightarrow 9<16$ which is true. Now, for $n+1$, we have $n^3 + (n+1)^3 < n^4 + (n+1)^4$ by the induction hypothesis. By expanding the two binomials, we have
$2n^3+3n^2+3n+1< 2n^4+4n^3+6n^2+4n+1$. By simplifying our expression, we are left with
$0<n(2n^3+2n^2+3n+1)$ which would always be true with $n \in \mathbb{N}, n>1$.
I am not sure if this proof would be valid, or if it is logically correct. As for the second part of the question, I would not know where to begin. Could somebody please provide a hint or point out where this goes awry? Any suggestions and/or criticism is welcome.
I am using the textbook Introduction to Analysis by Arthur Mattuck.
|
As the others have said, you went wrong in the induction step. Here's one way to prove what we want without having to expand the binomials:
\begin{align*}
1^3 + 2^3 + \ldots + (n+1)^3 &= \left[1^3 + 2^3 + \ldots + n^3 \right] + (n+1)^3 \\
&< n^4 + (n+1)^3 \qquad\text{by the induction hypothesis} \\
&= n(n)^3 + 1(n+1)^3 \\
&< \color{red}{n}\color{blue}{(n+1)^3} \color{red}{+ 1}\color{blue}{(n+1)^3} \\
&= (\color{red}{n+1})\color{blue}{(n+1)^3} \qquad\text{factor out the common }(n+1)^3 \text{ term} \\
&= (n+1)^4 \\
\end{align*}
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate the sum $\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2n}}+\cdots+\frac{1}{\sqrt{n^2}}$ when $n$ grows? I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$
Thanks.
|
Notice for large $n$, we expect $\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}}
\text{ behave like }\int_1^n \frac{dx}{\sqrt{x}} \sim 2\sqrt{n}$. This suggests
$$\frac{1}{\sqrt{k}} \sim \int_{k-1/2}^{k+1/2} \frac{dx}{\sqrt{x}} \sim 2\left( \sqrt{k+\frac12} - \sqrt{k-\frac12}\right)$$ and the terms $\displaystyle \frac{1}{\sqrt{kn}}$
in the summands is close to something "telescopable". To make this idea concrete, we observe:
$$\begin{align}
\sum_{k=1}^n\frac{1}{\sqrt{kn}} \ge &
\sum_{k=1}^n \frac{2}{\sqrt{n}(\sqrt{k+1}+\sqrt{k})}
= \frac{2}{\sqrt{n}} \sum_{k=1}^n(\sqrt{k+1}-\sqrt{k})
= 2 \Big(\sqrt{1+\frac{1}{n}} - \frac{1}{\sqrt{n}}\Big)\\
\sum_{k=1}^n\frac{1}{\sqrt{kn}} \le &
\sum_{k=1}^n \frac{2}{\sqrt{n}(\sqrt{k}+\sqrt{k-1})}
= \frac{2}{\sqrt{n}}\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1}) = 2
\end{align}$$
As a result,
$$\left|\;\sum_{k=1}^n \frac{1}{\sqrt{kn}} - 2\;\right| \le 2 \left(1+\frac{1}{\sqrt{n}} -\sqrt{1+\frac{1}{n}}\right) < \frac{2}{\sqrt{n}}
\quad\implies\quad \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}} = 2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$S_n$ be the sum of areas of $y=\sin x, y=\sin {nx}$. Then $\lim_{n \to \infty}S_n=8/{\pi}$? Let $n$ be a natural number. Also let $S_n$ be the sum of the areas of the regions enclosed with the two curves $y=\sin x$ and $y=\sin {nx}$ in $0\le x\le \pi$.
It's easy to find $S_2, S_3$. For smaller $n$, we can also use wolfram to find $S_n$. Then, I got interested in the problems for larger $n$.
After my observation, I reached the following expectation:
My expectation: $$\lim_{n \to \infty}S_n=\frac{8}{\pi}.$$
I've tried to prove this, but I don't have any good idea with a tedious calculation. Then, here is my question.
Question: Could you show me how to find $\lim_{n \to \infty}S_n$ ?
|
Since I wasted some time solving this last night, I might as well post it:
The integrand changes sign at the zeros, unless the zero is a double zero, so we need to find the zeros. For that, we bring it in a form where the zeros are more easily determined:
$$\sin (nx) - \sin x = 2\sin \frac{(n-1)x}{2}\cos \frac{(n+1)x}{2}.$$
The zeros of the sine factor in $[0,\,\pi]$ are
$$a_k = \frac{2k\pi}{n-1},\quad 0 \leqslant k \leqslant \frac{n-1}{2},$$
and the zeros of the cosine factor are
$$b_k = \frac{(2k+1)\pi}{n+1}, \quad 0 \leqslant k \leqslant \frac{n}{2}.$$
We find that
$$a_k \leqslant b_k \iff k \leqslant \frac{n-1}{4},$$
where equality holds on one side iff it holds on both. We always have $b_k < a_{k+1}$. For the larger $k$, we have $b_k < a_k < b_{k+1}$.
So writing $n = 4j + r, \; 1 \leqslant r \leqslant 4$, we have that $\sin (nx) - \sin x$ is
*
*positive on $(a_k, b_k)$, for $0 \leqslant k \leqslant j$,
*negative on $(b_k, a_{k+1})$, for $0 \leqslant k < j$,
*negative on $(b_j, b_{j+1})$,
*positive on $(b_k, a_k)$, for $j+1 \leqslant k \leqslant \frac{n-1}{2}$,
*negative on $(a_k, b_{k+1})$, for $j+1 \leqslant k \leqslant \frac{n-1}{2}$.
Using $n a_k = 2k\pi + a_k$, $n b_k = (2k+1)\pi - b_k$, one finds
$$\begin{align}
\int_{a_k}^{b_k} \sin (nx) - \sin x\, dx &= \cos b_k - \cos a_k + \frac{\cos (na_k) - \cos (nb_k)}{n}\\
&= \cos b_k - \cos a_k + \frac{\cos a_k + \cos b_k}{n}\\
&= \left(1 + \frac1n\right)\cos b_k - \left(1 - \frac1n\right)\cos a_k
\end{align}$$
and similar expressions for the other intervals. Summing the integrals, we obtain
$$\begin{align}
\int_0^\pi \lvert \sin (nx) - \sin x\rvert\, dx &= \sum_{k=0}^j \left(\left(1+\frac1n\right)\cos b_k - \left(1-\frac1n\right)\cos a_k\right)\\
&\quad + \sum_{k=0}^{j-1} \left(\left(1+\frac1n\right)\cos b_k - \left(1-\frac1n\right)\cos a_{k+1}\right)\\
&\quad + \left(1+\frac1n\right)(\cos b_j - \cos b_{j+1})\\
&\quad + \sum_{k=j+1}^{2j+\lfloor r/2\rfloor-1} \left(\left(1-\frac1n\right)\cos a_k - \left(1+\frac1n\right)\cos b_k\right)\\
&\quad + \sum_{k=j+1}^{2j+\lfloor (r-1)/2\rfloor} \left(\left(1-\frac1n\right)\cos a_k - \left(1+\frac1n\right)\cos b_{k+1}\right)\\
&= \left(1+\frac1n\right)\left(\sum_{k=0}^j 2\cos b_k - \sum_{k=j+1}^{2j+\lfloor r/2\rfloor-1} 2\cos b_k -\delta(r)\cos b_{2j+(r+1)/2}\right)\\
&- \left(1-\frac1n\right)\left(\sum_{k=0}^j 2\cos a_k - \sum_{k=j+1}^{2j+ \lfloor r/2\rfloor-1} 2\cos a_k - 1 - \delta(r)\cos a_{2j+(r-1)/2}\right)
\end{align}$$
where $\delta(r) = 0$ if $r$ is even, and $\delta(r) = 1$ for $r$ odd.
Using the summation formulae for sines and cosines of arithmetic progressions, we obtain
$$\begin{align}
\sum_{k=0}^j 2\cos a_k &= \frac{\sin \frac{(2j+1)\pi}{n-1} + \sin \frac{\pi}{n-1}}{\sin \frac{\pi}{n-1}}\\
\sum_{k=j+1}^{2j} 2\cos a_k &= \frac{\sin \frac{(4j+1)\pi}{n-1} - \sin \frac{(2j+1)\pi}{n-1}}{\sin \frac{\pi}{n-1}}\\
\sum_{k=0}^j 2\cos b_k &= \cot \frac{\pi}{n+1}\left(\sin \frac{(2j+1)\pi}{n+1}+\sin\frac{\pi}{n+1}\right) - \left(\cos \frac{(2j+1)\pi}{n+1} - \cos \frac{\pi}{n+1}\right)\\
\sum_{k=j+1}^{2j} 2\cos b_k &= \cot \frac{\pi}{n+1}\left(\sin\frac{(4j+1)\pi}{n+1} - \sin \frac{(2j+1)\pi}{n+1}\right) - \left(\cos \frac{(4j+1)\pi}{n+1} - \cos \frac{(2j+1)\pi}{n+1}\right)
\end{align}$$
Inserting that and adding the correction terms due to the summation bound not always being $2j$, the formula for $n = 4j+1$ reduces nicely to
$$2\left(1+\frac1n\right)\cot \frac{\pi}{n+1} - 2\left(1-\frac1n\right)\cot \frac{\pi}{n-1} = \frac{8}{\pi} + O\left(\frac1n\right).$$
The cases $r \in \{2,3,4\}$ don't reduce quite as nicely, you get an expression with more terms left, but the integral is still (because $\cot z$ and $\frac{1}{\sin z}$ are both $\frac1z + O(z)$, and all other terms are $O(1/n)$)
$$\frac{8}{\pi} + O\left(\frac1n\right).$$
|
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|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
$$F(a)=\int-\frac{1}{1+ax^2}dx=-\frac{\arctan(\sqrt{a}x)}{\sqrt a}+c$$
$$\int_{-\infty}^{+\infty}-\frac{1}{1+ax^2}dx=-\frac{\pi}{\sqrt a}$$
so we take the derivaitve of $-\frac{\pi}{\sqrt a}$ with respect to $"a"$ then put a=1 so
$$\int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{\pi}{2}$$
|
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|
How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how to write the denominator because of the alternating signs in each term. As for the starting point, $n=?$, I think its going to be either $0$ or $1$, depending on the denominator - I don't want to be kicked from this site by attempting to divide by $0$ .
|
HINT
The denominators in every term, except the first, are the alternating sums of even powers.
\begin{array}
.x^2-1 &\equiv& x^2 - x^0 \\
x^4-x^2+1 &\equiv& x^4 - x^2 + x^0 \\
x^6-x^4+x^2-1 &\equiv& x^6-x^4+x^2-x^0
\end{array}
Putting these in to summation notation:
\begin{array}
.x^2-x^0 &\equiv& \sum_{k=0}^1 (-1)^{k}x^{2(1-k)} \\ \\
x^4-x^2+x^0 &\equiv& \sum_{k=0}^2 (-1)^{k}x^{2(2-k)} \\ \\
x^6-x^4+x^2-x^0 &\equiv& \sum_{k=0}^3 (-1)^{k}x^{2(3-k)} \\ \\
\cdots\cdots\cdots &\equiv& \sum_{r=1}^n\sum_{k=0}^r(-1)^kx^{2(r-k)}
\end{array}
|
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|
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is some polynomial whose solution is $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, but I have not shown it to be 1.
|
It is useful, when approaching a problem like this, to eliminate the more complex square roots. So given:
$$\sqrt{4 + 2 \sqrt{3}} - \sqrt{3} = 1$$
Rearrange to (so as to isolate the more complex square root on the LHS):
$$\sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$$
Square both sides:
$$4 + 2 \sqrt{3} = (\sqrt{3} + 1)(\sqrt{3} + 1)$$
Multiply out the RHS:
$$4 + 2 \sqrt{3} = 3 + 2\sqrt{3} + 1$$
So
$$4 + 2 \sqrt{3} = 4 + 2\sqrt{3}$$
This does not rely on you seeing the factorization, you just methodically work on simplifying the expression you were given.
|
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|
IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that the equation becomes $b^6+b^2=b^2(1+b^4)$ . Are there any other solutions?
|
Not quite a proof to the original IMO problem, but there definately is a very easy way to compute all possible answers.
Also it demonstrates that here, Vieta jumping is basically just using
symmetry to jump between the two solutions of the good old quadratic formula.
Nothing complicated below, but I haven't seen this explanation anywhere else.
Suppose:
$$
{{a^2+b^2} \over {1+ab}} = c
$$
Then using the quadratic formula to solve b (and a little rewriting) gives:
$$
b = {ac \pm \sqrt {a^2 (c^2-4) + 4c} \over 2}
$$
Using this we can compute all answers for any given c.
Eg: assume c = 4. Then we get:
$$
b = {a \cdot 4 \pm \sqrt {a^2 \cdot 12 + 16} \over 2}
$$
We know that a = 0 will always provide a solution. That gives us:
$$
a = 0 \Rightarrow b = {{0 \cdot 4\pm \sqrt {0 \cdot 12 + 16}} \over 2} = 2 \Rightarrow (a,b) = (0,2)
$$
But because of the symmetry, if (0,2) is a solution, then (2,0) must
also be a solution. And since the quadratic formula has 2 solutions,
it will produce another solution.
$$
a = 2 \Rightarrow \space b = {{2 \cdot 4 \pm \sqrt {4 \cdot 12 + 16}} \over 2} = 0 \space or \space 8 \Rightarrow (a,b) = (2,8)
$$
$$
a = 8 \Rightarrow \space b = {{8 \cdot 4 \pm \sqrt {64 \cdot 12 + 16}} \over 2} = {32 \pm 28\over 2} = 2 \space or \space 30 \Rightarrow (a,b) = (8,30)
$$
$$
a = 30 \Rightarrow \space b = {{30 \cdot 4 \pm \sqrt {900 \cdot 12 + 16}} \over 2} = {120 \pm 104\over 2} = 8 \space or \space 112 \Rightarrow (a,b) = (30,112)
$$
Same can be done for c = 9, 16, 25, etc.
|
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|
Summation of $\cos A+\cos(A+B)+ \ldots \cos(A+(n-1)B)$ How would I prove the following result?
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (n-1)B) = \frac{\sin\left(\frac{nB}{2}\right) \cos\left[A+\frac{(n-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
|
Induction on $n$.
Checking for $n=1$:
$$\cos A =\frac{\sin(\frac{B}{2})}{\sin(\frac{B}{2})}\cos A =\cos A $$
Induction hypothesis:
Suppose that the equality holds for $n=k$, thus
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
Induction step:
Proving for $n=k+1$
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B)+\cos(A+kB) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$
Now, we try to simplify the left right side:
$$\frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$
We have
$$\frac{\cos(A+kB)\sin (\frac{B}{2})+\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$\cos x \sin x = \frac{1}{2}[\sin(x+y)-\sin(x-y)]$$
$$\cos(A+kB)\sin (\frac{B}{2}) = \frac{1}{2}[\sin(A+kB+\frac{B}{2})-\sin(A+kB-\frac{B}{2})]$$
Or:
$$\cos(A+kB)\sin (\frac{B}{2}) = \frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]$$
Plugging it:
$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]+\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$ \sin x\cos x = \frac{1}{2}[\sin(x+y)+\sin(x-y)]$$
We will get:
$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]+\frac{1}{2}[\sin(A+\frac{B(2k-1)}{2})+\sin(-A+\frac{B}{2})]}{\sin \left(\frac{B}{2}\right)}$$
After simplifying more:
$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})]+\frac{1}{2}[+\sin(-A+\frac{B}{2})]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$\sin(x\pm y)=\sin x\cos y \pm \cos x \sin y $$
We will have:
$$\frac{\frac{1}{2}[\sin A \cos(\frac{B(2k+1)}{2})+\cos A\sin(\frac{B(2k+1)}{2})+\sin(\frac{B}{2})\cos(A)-\cos(\frac{B}{2})\sin(A)]}{\sin \left(\frac{B}{2}\right)}$$
Which equals to:
$$\frac{\frac{1}{2}[\sin( A)(\cos(\frac{B(2k+1)}{2})-\cos(\frac{B}{2}))+\cos(A)(\sin(\frac{B(2k+1)}{2})+\sin(\frac{B}{2}))}{\sin \left(\frac{B}{2}\right)}$$
Using the identities:
$$\cos x-\cos y=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$$
$$\sin x+\sin y=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$$
We will get:
$$\frac{\frac{1}{2}[-2\sin( A)(\sin(\frac{B(k+1)}{2})\sin(\frac{Bk}{2}))+2\cos(A)(\sin(\frac{B(k+1)}{2})\cos(\frac{Bk}{2}))}{\sin \left(\frac{B}{2}\right)}$$
After simplifying:
$$\frac{-\sin( A)(\sin(\frac{B(k+1)}{2})\sin(\frac{Bk}{2}))+\cos(A)(\sin(\frac{B(k+1)}{2})\cos(\frac{Bk}{2}))}{\sin \left(\frac{B}{2}\right)}$$
and
$$\frac{\sin(\frac{B(k+1)}{2})[\cos(A)\cos(\frac{Bk}{2})-\sin(A)\sin(\frac{B(k+1)}{2})]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$\cos x \cos y - \sin x \sin y=\cos(x+y)$$
We will get:
$$\frac{\sin(\frac{B(k+1)}{2})(\cos(A+Bk))}{\sin \left(\frac{B}{2}\right)}$$
Therefore
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B)+\cos(A+kB) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$
And we prove for $n=k+1$, thus the induction step completed, and the identity holds for any $n$.
|
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|
Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a])
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.
I wrote out some $m$ to try to conceive the proof. I observed:
$\bbox[5px,border:1px solid grey]{\text{Case 1: $m$ odd}}$ Odd numbers $\neq 2p$, thus the only choice is to put $p = 0$ and $k = m$.
$\bbox[5px,border:1px solid grey]{\text{Case 2: $m$ even and power of 2}}$ Then $p$ can be determined by division or inspection to "square with" the power of $2$. This requires $k = 1$. Is an explicit formula for $p$ necessary?
$\bbox[5px,border:1px solid grey]{\text{Case 3: $m$ even and NOT A power of 2}}$
$\begin{array}{cc}
\\
\boxed{m = 6}: 6 = 2^1 \cdot 3 \qquad \qquad & \boxed{m = 10}: 10 = 2^1 \cdot 5 \qquad \qquad & \boxed{m = 12}: 12 = 2^2 \cdot 3 \\
\boxed{m = 14}: 14 = 2 \cdot 7 \qquad \qquad & \boxed{m = 18}: 18 = 2^1 \cdot 9 \qquad \qquad & \boxed{m = 20}: 20 = 2^2 \cdot 5\\
\boxed{m = 22}: 22 = 2 \cdot 11 \qquad \qquad & \boxed{m = 24}: 24 = 2^3 \cdot 3 \qquad \qquad & \boxed{m = 26}: 26 = 2^1 \cdot 13
\end{array}$
Solution's Proof by Contradiction: $\color{#0073CF}{\Large{\mathbb{[}}}$Let $p$ be the greatest nonnegative integer
such that $2^p | m. \color{#0073CF}{\Large{\text{]}}}$ $\color{red}{\Large{\text{[}}}$ Then $m= 2^pk$ for some positive integer $k$. Necessarily $k$ is odd,
for otherwise this would contradicts the definition of $p$. $\color{red}{\Large{\text{]}}}$
$\Large{\text{1.}}$ Could someone please expound on the answer? It differs from my work above?
$\Large{\text{2.}}$ Is there a general formula/pattern for Case $3$?
I referenced 1. Source: Exercise 5.42(a), P125 of Mathematical Proofs, 2nd ed. by Chartrand et al
|
For any positive integer, $m$, there will be some non-negative integer, $r$ such that $m < 2^r$. This implies that there must be some non negative integer $k \le r$ such that $2^k \mid m$ and $2^{k+1} \nmid m$.$
For example $m = 96 \lt 128 = 2^7$ and $m = 2^5 \times 3$ where $5 \le 7$.
|
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|
Summation of series. If $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$
Then find the value of $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+..$$
The answer given in the book is $\frac{\pi^2}{8}$.
How can I find this and also how to find summation of fractions? Thank you!
|
Riemann's zeta function is given by, $$ \zeta (s)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+.... $$
Group even and odd number containing terms of this function as shown below,
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}....) $$
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+\frac{1}{2^s}(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+.... ) $$
The second bracket is nothing but the function itself, hence we can write,
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+\frac{1}{2^s}\zeta (s) $$
So, finally we get,
$$ \frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....=\zeta (s)(1-\frac{1}{2^s})$$
Hope this cleared your doubt.
Now, for you case s=2 . So, we get
$$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....=\zeta (2)(1-\frac{1}{2^2})$$
$$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....=\frac{\pi^2}{6}(1-\frac{1}{4})$$
$$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....=\frac{\pi^2}{8}$$
|
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|
How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$
My try:
and I find when $a=4,b=2$,then
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$
it maybe use Cauchy-Schwarz inequality
Thank you
|
Setting $x = a$ and $b = \frac{8}{x}$ we find that we want to minimise $$\sqrt{x^2 + 64} + \sqrt{\frac{64}{x^2} + 1} = \left (\frac{1}{x} + 1 \right ) \sqrt{x^2 + 64}$$
Differentiating, we obtain $$ \frac{x^3 - 64}{x^2 (x^2 + 64)} $$ Setting this equal to zero, we obtain a real extremum at $x=4$, corresponding to $a=4, b=2$ and a value of $5\sqrt{5}$. Analysis of the second derivative shows that this is a minimum.
|
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|
Simplifying large exponents in modular arithmetic like $1007$ in $4^{1007} \pmod{5}$ How would I rigorously prove that $4^{1007} \pmod{5} = 4$ and $4^{1008} \pmod{5} = 1$?
I was simplifying a larger modular arithmetic problem ($2013^{2014} \pmod{5}$) and got it down to $4^{1007} \pmod{5}$ and am wondering if there's a general approach to dealing with large exponents like $1007$.
In general, what approaches are there to simplify large exponents like $1007$ when doing modular arithmetic?
|
Another method for this is repeated squaring.
Take $n^d \mod b$ then
find all $a^{2^k} \mod b$ for all $2^k <= d$ where $d$ is $n^d$
$$\begin{align}
4^2 &= 16 = 1 \\
4^4 &= (4^2)^2 = 1^2 \\
4^8 &= (4^4)^2 = 1^2 \\
&... \\
4^{512} &= 1 \\
\end{align}$$
for $4^{1007}$ we have:
$$\begin{align}
4^{1007}
&= 4^{512} * 4^{256} * 4^{128} * 4^{64} * 4^{32} * 4^8 * 4^4 * 4^2 * 4 \\
&= 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 * 4 \mod 5 \\
&= 4 \mod 5
\end{align}$$
and for $4^{1008}$ we have:
$$\begin{align}
4^{1008}
&= 4^{512} * 4^{256} * 4^{128} * 4^{64} * 4^{32} * 4^{16} \\
&= 1 * 1 * 1 * 1 * 1 * 1 \\
&= 1 \mod 5 \\
\end{align}$$
Further details:
http://www.tricki.org/article/To_work_out_powers_mod_n_use_repeated_squaring
|
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|
Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
|
Multiplying by 81 on the top and the bottom gives
$\displaystyle\frac{16x^4-81y^4}{27(2x+3y)}=\frac{(4x^2-9y^2)(4x^2+9y^2)}{27(2x+3y)}=\frac{(2x-3y)(2x+3y)(4x^2+9y^2)}{27(2x+3y)}=$
$\frac{1}{27}(2x-3y)(4x^2+9y^2)$.
|
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|
Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers
|
I will look at
$x(x+1) = y(y+1)(y^2+k)$
for integral $k \ge 1$.
This becomes the original question
when $k = 2$.
I will show that
there are no solutions in
positive integral $x$ and $y$
for $y \ge k+2$.
Note that
$(x, y)
=(k^2-k, k-1)$
and
$(k^2+3k+1, k+1)
$
are solutions to this,
and there is no solution
with $y = k$.
These correspond to
the solutions
$(x, y) = (2, 1)$ and
$(11, 3)$
to the original equation.
This is essentially
my previous solution for
$k=2$
with slightly more complicated algebra.
$x(x+1) = y(y+1)(y^2+k)
=y(y^3+y^2+ky+k)
=y^4+y^3+ky^2+ky
$
Multiplying by 4,
$(2x+1)^2-1
=4y^4+4y^3+4ky^2+4ky
$
or
$(2x+1)^2
=4y^4+4y^3+4ky^2+4ky+1
$
My goal is to show algebraically
that this polynomial in $y$
is between two consecutive squares
for large enough $y$.
$(2y^2+y)^2
=4y^4+4y^3+y^2
$.
$\begin{align}
(2y^2+y+k)^2
&=4y^4+4y^3+y^2 +2k(2y^2+y)+k^2 \\
&=4y^4+4y^3+(4k+1)y^2+2ky+k^2 \\
\end{align}
$.
$\begin{align}
(2y^2+y+k-1)^2
&=4y^4+4y^3+y^2
+2(k-1)(2y^2+y)+(k-1)^2 \\
&=4y^4+4y^3+(4k-3)y^2+(2k-2)y+(k-1)^2 \\
\end{align}
$.
For
$(2x+1)^2$
to be between these consecutive squares,
we need
$(4k-3)y^2+(2k-2)y+(k-1)^2
<4ky^2+4ky+1
<(4k+1)y^2+2ky+k^2
$.
The first inequality is
$0
<3y^2+(2k+2)y-(k-1)^2+1
$
or
$y(3y+2k+2)
>k(k-2)
$
and this is certainly true for
$y \ge k$.
For the second inequality to be true,
we need
$4ky^2+4ky+1
<(4k+1)y^2+2ky+k^2
$
or
$y^2-2ky+k^2-1
> 0
$
or
$(y-k)^2-1
> 0$.
This is true for
$y \ge k+2$,
so the equation has no solution for $y \ge k+2$.
If
$y=k-1$,
the right side is
$k(k-1)((k-1)^2+k)
=(k^2-k)(k^2-k+1)
$,
so $x=k^2-k$, $y=k-1$
is a solution.
Similarly, if
$y=k+1$,
the right side is
$(k+1)(k+2)((k+1)^2+k)
=(k^2+3k+2)(k^2+3k+1)
$,
so $x=k^2+3k+1$, $y=k+1$
is a solution.
If $y=k$,
the equation is
$x(x+1)
= k(k+1)(k^2+k)
= (k^2+k)^2
$,
or
$(2x+1)^2-1
= (2k^2+2k)^2
$,
which has no solutions for
$k \ge 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the tangent line to a curve Find an equation for the tangent line to the curve
$$x\sin(xy-y^2)=x^2-1$$
through the point $(1,1)$.
|
When the curve is given by $y = f(x)$ then the slope of the tangent is $\frac{dy}{dx}$,then the equation of the tangent line can be found by the equation.$$y=\frac{dy}{dx}x+c$$the value of $c$ can be found with the help of the given point $(1,1)$. given:$$x\sin(xy-y^2)=x^2-1$$$$y^2-xy+arcsin\left(\frac{x^2-1}{x}\right)=0$$solve this quadratic equation in $y$ to get$$y=\frac{x+\sqrt{x^2-4arcsin\left(\frac{x^2-1}{x}\right)}}{2}$$the other root cancels as it does not satisfy $x=1,y=1$.find $\frac{dy}{dx}_{(1,1)}$$$\frac{dy}{dx}=\frac{1}{2}+\frac{2x-4\left(1+\frac{1}{x^2}\right)\left(\frac{1}{\sqrt{1-\frac{x^2-1}{x}}}\right)}{4\sqrt{x^2-4arcsin\left(\frac{x^2-1}{x}\right)}}$$$$\frac{dy}{dx}_{(1,1)}=-1$$therefore the equation of the tangent is$$y=2-x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Increasing and bounded sequence proof Prove that the sequence $a_n= 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln(n)$ is increasing and bounded above. Conclude that it’s convergent.
This what I got so far
Proof:
Part 1: Proving $a_n$ is increasing by induction.
Base:
$a_1=1$
$a_2=1+\frac 12= \frac 32$
$a_1≤a_2$
So the base case is established.
Induction step: We assume that $a_{n-1}≤a_n$. We will show that $a_n≤a_{n+1}$.
Since
$a_{n-1}≤a_n$
$$1+ \frac 12+ \frac 13+\cdots+ \frac{1}{(n-1)}-\ln(n-1) \leq 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln n$$
How should I continue?
|
This sequence is NOT increasing and is in fact DECREASING.
Proof:
$$a_n = 1+\frac{1}{2} + \cdots + \frac{1}{n} - \ln(n)$$
$$a_{n+1}=1+\frac{1}{2} + \cdots + \frac{1}{n}+\frac{1}{n+1} - \ln(n+1)$$
so
$$a_{n+1}=a_n + (\ln(n) +\frac{1}{n+1} - \ln(n+1)) = a_n + \frac{1}{n+1} - (\ln(n+1) - \ln(n))$$
$$\ln(n+1)-\ln(n) = \int_n^{n+1}\frac{1}{t}dt > \frac{1}{n+1}$$
To see the last line just draw a graph of $\frac{1}{t}$ between $n$ and $n+1$ and draw a box of width $1$ and height $\frac{1}{n+1}$ (i.e. the box touches $\frac{1}{t}$ at $t=\frac{1}{n+1}$).
NOTE: Your base case was not done properly since you forgot to subtract $\ln(1)$ and $\ln(2)$ (granted $\ln(1) = 0$) if you subtracted $\ln(2) \approx 0.693147 > 0.5$ you would have seen the base case not to hold.
|
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|
integration by parts !!!! PD: I did a little change in the denominator !!!!
I need to solve this integral using integration by parts.
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$
Thanks!
PS: I know that I can to do:
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}=\int\frac{(x-c)\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}+c\int\frac{dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$
but according to the book it is easier using integration by part.
|
You don't actually need integration by parts. First, observe the following:
$$\left[\sqrt{a^2+b^2+(x-c)^2}\right]'=\frac {x-c}{\sqrt{a^2+b^2+(x-c)^2}}$$
Integrating both sides and isolating the integral under question gives
$$\int\frac x{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx=\sqrt{a^2+b^2+(x-c)^2}+c\int\frac {\mathrm dx}{\sqrt{a^2+b^2+(x-c)^2}}$$
The latter integral can be evaluated using an Euler Substitution. A more detailed explanation of how to perform an Euler Substitution can be found on the Wikipedia page, but the substitution to make here is
$$t=x-c+\sqrt{a^2+b^2+(x-c)^2}$$
Differentiating gives
\begin{align*}
\mathrm dt & =1+\frac {x-c}{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx\\ & =\frac {x-c+\sqrt{a^2+b^2+(x-c)^2}}{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx\\ & =\frac t{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx
\end{align*}
As in
$$\frac {\mathrm dt}t=\frac {\mathrm dx}{\sqrt{a^2+b^2+(x-c)^2}}$$
The integral is then
\begin{align*}
& \int\frac x{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx\\ & =\sqrt{a^2+b^2+(x-c)^2}+c\int\frac {\mathrm dt}t\\ & =\sqrt{a^2+b^2+(x-c)^2}+c\log\left(x-c+\sqrt{a^2+b^2+(x-c)^2}\right)+K
\end{align*}
|
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|
When is $2^n \pm 1$ a perfect power Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?
I know that Catalan's conjecture (or Mihăilescu's theorem) gives the result directly, but I'm hopefully for a more elementary method.
I can show that it is never a square, except for $2^3 + 1$.
Proof: Cases $n=1, 2, 3$ are easily dealt with. Henceforth, let $n\geq4$.
$2^n -1 \equiv 3 \pmod{4}$ hence is never a square.
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2. Thus we must have $(x-1) = 2, (x+1) = 4$. This gives the solution of $2^3 + 1 = 3^2$.
Let's now do an odd prime.
Say $2^n +1 = x^p$. Then $2^n = x^p - 1 = (x-1)(x^{p-1}+x^{p-2} + \ldots +1)$, so both terms are powers of 2. We have $ x = 2^m+1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^p$. Then $2^n = x^p + 1 = (x+1)(x^{p-1} - x^{p-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x = 2^m -1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction.
|
I collected everything you may need about elementary methods solving special cases of Mihailescu theorem here (see cases (iii) and (vii) for this problem)
|
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|
Show $\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{k+1} $ I have been attempting to show $$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1} $$ and my work is
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{r!((k+1)-r)!} + \frac{(k+1)!}{(r+1)!((k+1)-(r+1))!}$$
$$=\frac{(k+1)!(r+1)}{(r+1)!((k+1)-r)!} + \frac{(k+1)!
\color{red}{(k-r)}}{(r+1)!((k+1)-(r+1))!\color{red}{(k-r)}}$$
$$=(k+1)!\frac{k+1}{(r+1)!((k+1)-r)!}$$
which appears correct. But I am having trouble seeing the final step; where does the $(k+2)$ come from?
Somewhere I've made a mistake but I just don't see it.
|
Without assuming much, we have (just by definition):
$\binom{n}{r}+\binom{n}{r-1}=\frac{n!}{(n-r)! r!}+\frac{n!}{(n-(r-1))! (r-1)!}$
$\frac{n!}{(n-r)! r!}+\frac{n!}{(n-(r-1))! (r-1)!}=\frac{n!}{(n-r)! r!}+\frac{n!}
{(n-r+1))! (r-1)!}$
$\frac{n!}{(n-r)! r!}+\frac{n!}
{(n-r+1))! (r-1)!}=\frac{n!}{(n-r)!(r-1)!}(\frac{1}{r}+\frac{1}{n-r+1})$
$\frac{n!}{(n-r)!(r-1)!}(\frac{1}{r}+\frac{1}{n-r+1})=\frac{n!}{(n-r)!(r-1)!}(\frac{n-r+1+r}{r.(n-r+1)})$
$\frac{n!}{(n-r)!(r-1)!}(\frac{n-r+1+r}{r.(n-r+1)})=\frac{n!}{(n-r)!(r-1)!}(\frac{n+1}{r.(n-r+1)})$
$\frac{n!}{(n-r)!(r-1)!}(\frac{n+1}{r.(n-r+1)})=\frac{(n+1)n!}{(n-r+1)(n-r)!.r(r-1)!}$
$\frac{(n+1)n!}{(n-r+1)(n-r)!.r(r-1)!}=\frac{(n+1)!}{(n-r+1)!r!}=\frac{(n+1)!}{((n+1)-r)!.r!}=\binom{n+1}{r}$
Thus, we conclude that $\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$
replacing $n$ by $n+1$ we get $\binom{n+1}{r}+\binom{n+1}{r-1}=\binom{n+2}{r}$
replacing $r$ by $r+1$ we get $\binom{n+1}{r+1}+\binom{n+1}{r}=\binom{n+2}{r+1}$
The reason i have proved for $n$ is just my comfort and nothing much :P
|
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|
How do I get the integral of $\frac{1}{(x^2 - x -2)}$ I'm working with this problem $$ \int \frac{1}{x^2 - x - 2}$$
I'm thinking I break up the bottom so that it looks like this $$\int \frac{1}{(x-2)(x+1)} $$
Then I do $$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiple both sides by the common denominator and come out with $$ A(x+1) + B(x-2) = x^2 - x - 2 $$
Which equals $$Ax + A + Bx - 2B = x^2 - x - 2$$
Or $$ (A+B)x + (A-2B) = x^2 - x - 2$$
After that I tried to get values for my A and B but it doesn't seem right since I don't have anything for the $x^2$
Did I mess up somewhere?
|
In your partial fraction,
$\dfrac{1}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}$
So,
$\dfrac{1}{(x-2)(x+1)} = \dfrac{A(x+1) + B(x-2)}{(x-2)(x+1)}$
and hence,
$1 = (A+B)x + (A-2B)$
I think you can take it from here.
|
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|
Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$ $\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: Let $\tan x = t$. Then $\sec^2 xdx = dt\Rightarrow \displaystyle dx = \frac{1}{1+\tan^2 t}dt\Rightarrow dx = \frac{1}{1+t^2}dt$
So $\displaystyle \int\frac{t+t^3}{1+t^3}\cdot \frac{1}{1+t^2}dt = \int\frac{t}{1+t^3}dt$
Now My Question is can we solve the Given Integral without Using Partial fraction Method
If Yes How can I solve
plz Help me , Thanks
|
Mathematica evaluates it as
$$\frac{\tan ^{-1}\left(\frac{2 \tan (x)-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{1}{6} \log \left(\tan ^2(x)-\tan (x)+1\right)-\frac{1}{3} \log (\tan (x)+1)$$
This can be done via partial fractions.
\begin{align*}
\int \frac{t}{1 + t^3} \; dt &= \frac 1 3 \int \frac{t+1}{t^2 - t + 1} dt - \frac 1 3 \int \frac{1}{1+t} \; dt\\
&= \frac 1 3 \int \frac{t+1}{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4} \; dt - \frac 1 3 \log(1+t)\\
&= \frac 1 3 \int \frac{t - \frac 1 2 + \frac 3 2 }{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4} \; dt - \frac 1 3 \log(1+t)\\
&= \frac 1 6 \log \left( \left ( t - \frac 1 2 \right )^2 + \frac 3 4 \right ) - \frac 1 3 \log(1+t) + \frac 1 2 \int \frac{1}{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4}\; dt \\
&= \frac 1 6 \log \left( \left ( t - \frac 1 2 \right )^2 + \frac 3 4 \right ) - \frac 1 3 \log(1+t) + \frac 1 {\sqrt{3}} \arctan \left( \frac{t - \frac 1 2 }{ \frac{\sqrt 3}{2 } } \right )
\end{align*}
|
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|
Calculus limit homework problem $$ \lim_{n→\infty} \frac1 n \left(\left(a + \frac 1 n\right)^2 + \left(a + \frac 2 n\right)^2 + ... + \left(a + \frac{n-1}{n}\right)^2\right)$$
$$ \text{hint: }\ 1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} $$
I can't figure out how to find this problem's limit. Does anybody have ideas?
|
\begin{align*}
& \; \; \; \lim_{n\to \infty}\frac 1 n \left( \left(a + \frac 1 n \right )^2 + \dots +\left(a + \frac {n-1} n \right )^2 \right ) \\
&= \lim_{n\to \infty} \frac 1 n \left( (n-1)a^2 + 2 a \left( \frac 1 n + \dots + \frac{n-1}{n} \right ) + \left( \frac {1^n}{n^2} + \dots + \frac{(n-1)^2}{n^2}\right )\right )\\
&= \lim_{n\to \infty}\frac{(n-1)a^2}{n} + 2a \frac{n(n-1)}{ 2 n^2} + \frac{n(n-1)(2n-1)}{6n^3}\\
&= a^2 + a + \frac 1 3 \\
\end{align*}
Also it can be evaluated via converting it into Integral noticing that as $n\to \infty $ as
$$\lim_{n\to \infty}\frac 1 n \left( \left(a + \frac 1 n \right )^2 + \dots +\left(a + \frac {n-1} n \right )^2 \right )= \lim_{n\to \infty } \frac 1 n \sum_{i=0}^{n-1} \left( a + \frac i n \right )^2= \int_0^1 \left( a + x \right )^2 dx = a^2 + a + \frac 1 3 $$
|
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|
Finding Volume using triple integrals Find the volume common to $y^2+z^2=4ax$ and $x^2+y^2=2ax$
I have problem establishing the common region. I know the region but unable to visualize it to get to the limits of the integral.
|
$x^2 + y^2 = 2 ax \implies (x-a)^2 + y^2 = a^2$ is a cylinder centered at $(a,0)$ and parallel to $z-$ axis.
$y^2+z^2=4ax \implies \frac 1 {4a} (y^2 + z^2) = x $ is a parabloid opening towards $+$ ve $x$-axis. Putting $z=0$, gives two curves are $(x-a)^2 + y^2 = a^2$ and $y^2 = 4ax$, and the circle completely lines inside the concave part of parabola.
That should give the integral (in Cartesian coordinate) as
\begin{align*}
& \; \; \; \int_{0}^{2a}\int_{-\sqrt{2ax -x^2}}^{\sqrt{2ax -x^2}} \int_{-\sqrt{4ax - y^2}}^{\sqrt{4ax - y^2}} dz dy dx \\
&= 4\int_0^{2a} \int_0^{\sqrt{2ax - x^2}} \int_0^{\sqrt{4 ax -y^2}} dz dy dx \\
&= 2 \left(\frac{8 a^3}{3}+\pi a^3\right)
\end{align*}
evaluated by Mathematica.
Integrate[1, {x, 0, 2 a}, {y, 0, Sqrt[2 a x - x^2]}, {z, 0,
Sqrt[4 a x - y^2]}]
On polar coordinates,
\begin{align*}
& \; \; \; \int_{-\frac \pi 2 }^{\frac \pi 2 } \int_{0}^{2a \cos(\theta) }\int_{-\sqrt{4a r \cos (\theta) - r^2 \sin^2(\theta)}}^{\sqrt{4a r \cos (\theta) - r^2 \sin^2(\theta)}} r \; dz dr d\theta \\
&= 4 \int_0^{\frac \pi 2} \int_0^{2a \cos \theta} \int_0^{\sqrt{4ar \cos\theta - r^2 \sin^2 \theta}} r \; dz dr d\theta
\end{align*}
It seems that both Matlab and Mathematica does not return the value of integral for variable $a$.
For $a=1,2, ... ,5$
Table[Integrate[
r, {\[Theta], -Pi/2, Pi/2}, {r, 0, 2 a Cos[\[Theta]]}, {z, 0,
Sqrt[4 a r Cos[\[Theta]] - r^2 Sin[\[Theta]]^2]} ], {a, 1, 5}]
yields,
$$\left\{\frac{8}{3}+\pi ,\frac{64}{3}+8 \pi ,9 (8+3 \pi ),\frac{64}{3} (8+3 \pi ),\frac{125}{3} (8+3 \pi )\right\}$$
which is half the value of our integral and seems consistent with the one evaluated from Cartesian coordinate.
|
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|
Asymptotic expansion of $(1+\frac{t}{n})^{-n-1}$ at $n \to \infty$ I'm reading through a proof in Analytic Combinatorics by Flajolet/Sedgewick and I have come across this:
We have the asymptotic expansion:
$(1+\frac{t}{n})^{-n-1}=e^{-(n+1)\log(1+\frac{t}{n})}=e^{-t}[1+\frac{t^2-2t}{2n}+\frac{3t^4-20t^3+24t^2}{24n^2}+...]$
How do I go from second expression to the third?
|
Using the Taylor expansion of the logarithm, we obtain
$$\begin{align}
(n+1)\log \left(1 + \frac{t}{n}\right) &= (n+1)\left(\frac{t}{n} - \frac{t^2}{2n^2} + \frac{t^3}{3n^3} + O\left(\frac{1}{n^4}\right)\right)\\
&= t + \frac{t}{n} - \frac{t^2}{2n} - \frac{t^2}{2n^2} + \frac{t^3}{3n^2} + \frac{t^3}{3n^3} - \frac{t^4}{4n^3} + O\left(\frac{1}{n^4}\right)\\
&= t - \frac{t^2-2t}{2n} - \frac{3t^2-2t^3}{6n^2} + O\left(\frac{1}{n^3}\right),
\end{align}$$
and hence, using $e^{x+y} = e^x\cdot e^y$ and the Taylor expansion of the exponential function:
$$\begin{align}
e^{-(n+1)\log \left(1+\frac{t}{n}\right)} &= \exp \left(-t + \frac{t^2-2t}{2n} + \frac{3t^2-2t^3}{6n^2} + O(n^{-3})\right)\\
&= e^{-t}\exp\left(\frac{t^2-2t}{2n} + \frac{3t^2-2t^3}{6n^2} + O(n^{-3})\right)\\
&= e^{-t}\left(1 + \frac{t^2-2t}{2n} + \frac{3t^2-2t^3}{6n^2} + \frac12\left(\frac{t^2-2t}{2n}\right)^2 + O(n^{-3})\right)\\
&= e^{-t}\left(1 + \frac{t^2-2t}{2n} + \frac{12t^2 - 8t^3 + 3(t^4 - 4t^3 + 4t^2)}{24n^2} + O(n^{-3})\right)\\
&= e^{-t}\left(1 + \frac{t^2-2t}{2n} + \frac{3t^4 - 20t^3 + 24t^2)}{24n^2} + O(n^{-3})\right).
\end{align}$$
|
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|
Breaking up an integral using substitution? I am faced with the following integral, and am required to use substitution to solve it, using $u = \frac{1}{x}$.
$$\int \dfrac{1}{x^2\sqrt{x^2+1}} \, dx$$
I know I can break up the equation in:
$$\int \dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{\sqrt{x^2+1}} \, dx.$$
How would I go about replacing the variable $x$ in the sqrt though?
|
$x=\tan u$ $\Rightarrow$ $dx=(1+\tan^{2}u)du$
So
\begin{align*}
\int\frac{1}{x^{2}\sqrt{x^{2}+1}} & =\int\frac{1}{\tan^{2}u\sqrt{1+\tan^{2}u}}(1+\tan^{2}u)du=\int\frac{\sqrt{1+\tan^{2}u}}{\tan^{2}u}du\\
& =\int\frac{\sec u}{\tan^{2}u}du
\end{align*}
$$
$$
after some simplification
\begin{align*}
\int\frac{\sec u}{\tan^{2}u}du & =\int\frac{\cos u}{\sin^{2}u}du
\end{align*}
make substution $s=\sin u\Rightarrow ds=\cos udu$
so
$$
\begin{align*}
\int\frac{\cos u}{\sin^{2}u}du & =\int\frac{ds}{s^{2}}=\frac{-1}{s}+c
\end{align*}
$$
by back substution $s=\sin u$
$$
\
=\frac{-1}{\sin u}+c
\
$$
and finally by $u=\arctan(x)$
we get
$$
\
=\frac{-1}{\sin (\arctan x)}+c
\
$$
|
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|
Writing a proposition as the conjunction of two conditional statements For each integer a, a is congruent to 3(mod 7) if and only if (a^2 + 5) is congruent to 3(mod 7).
A)Writing a proposition as the conjunction of two conditional statements
B) determine if the conditionals are true or false
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For completeness and because any deadline has probably passed for this homework-type quesiton, I will construct a full solution.
A) Let $A_a$ be defined as "$a$ is congruent to $3\bmod 7$" and $B_a$ defined as "$a^2+5$ is congruent to $3\bmod 7$." We are given $(\forall a \in \mathbb{Z})A_a \equiv B_a$, which by definition is composed of two conditionals $(\forall a \in \mathbb{Z})(A_a \rightarrow B_a) \wedge (B_a \rightarrow A_a)$ .
So, the full answer should be: For all integers $a$, (If $a$ is congruent to $3 \bmod 7$ then $a^2 + 5$ is congruent to $3 \bmod 7$) and (if $a^2 + 5$ is congruent to $3 \bmod 7$ then $a$ is congruent to $3 \bmod 7$).
B)
(1)Start with the first conditonal, $(\forall a)(A_a \rightarrow B_a)$: If $a$ is congruent to $3 \bmod 7$ then $a^2 + 5$ is congruent to $3 \bmod 7$. Suppose that ($\neg A$) $a$ is NOT congruent to $3 \bmod 7$ then the conditional is vacuously true, so we don't need to further check this case. Suppose ($A$ is true) $a$ is congruent to $3 \bmod 7$ this means that $7 \lvert a-3 $. We need to check that ($B$) $7\lvert ((a^2+5)-3)$.
$$\frac{a^2 + 5 - 3}{7} = \frac{a^2 +2}{7} = \frac{(a-3+3)(a-3+3)+2}{7} = \frac{(a-3)^2+6(a-3)+11}{7}$$The first two terms are clearly divisible by $7$ given that $7\vert a-3$ but the third is not. So, the consequent of the conditional is false and thus $(\forall a)(A_a \rightarrow B_a)$ is false.
(2) The second conditional says $(\forall a)(B_a \rightarrow A_a)$. As before if $B$ is false, then the conditional is vacuously true, but we must check what happens when $B$ is true. Suppose $7\vert (a^2 +2)$, then $a^2 \equiv -2 \ (\bmod 7) \Rightarrow a^2 \equiv 5 \ (\bmod 7)$ but the squares in the integers $ \mod 7$ are just $\{1,2,4\}$ so this equation has no solutions. The antecedent is then always false, so the conditional, $(\forall a) (B_a \rightarrow A_a)$ is true.
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How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$ \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4} $$
and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20
and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you
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We need to prove that
$$\sum_{cyc}\frac{1}{2a^2-6a+9}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or
$$\sum_{cyc}\left(\frac{18(a^2-3a+2)}{2a^2-6a+9}+1\right)\geq3$$ or
$$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}\geq\frac{3}{5}.$$
Now, by C-S we obtain:
$$\sum_{cyc}\frac{(2a-3)^2}{2a^2-6a+9}=\sum_{cyc}\frac{(2a-3)^2(a-8)^2}{(2a^2-6a+9)(a-8)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(2a-3)(a-8)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}=\frac{\left(\sum\limits_{cyc}(2a^2+5)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}.$$
Hence, it remains to prove that
$$\frac{\left(\sum\limits_{cyc}(2a^2+5)\right)^2}{\sum\limits_{cyc}(2a-3)^2(a-8)^2}\geq\frac{3}{5}.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, we need to prove a fourth degree polynomial inequality, id est, a linear inequality of $w^3$,
which says that it's enough to prove the last inequality for an extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$.
We need to prove that $(a-1)^2(34a^2-98a+73)\geq0$, which is obvious.
Done!
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Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
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$a+b, c+a$ is positive real,
consider $\frac{1+a}{2}= \frac{(a+b)+(c+a)}{2} \geq \sqrt{(a+b)(c+a)}$
i.e$(1+a)\geq2\cdot\sqrt{(1-c)(1-b)}$
and so..
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Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far.
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true
Induction: Assume $n^3 + (n + 1)^3 + (n + 2)^3 = k * 9$ // Why set it equal to $k * 9$? I know it works but why not just make the assumption => $n^3 + (n + 1)^3 + (n + 2)^3$ for some $n = k \ge 0$
Then, //and here's where I get lost
$(n + 1)^3 + (n + 2) + (n + 3)^3 = k * 9 + [(n + 3)^3 - n^3] = 9 (k + n^2 + 3n + 3)$
.
I've done similar examples but none like this. What am I not seeing?
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Set $f(n) = n^3 + (n+1)^3 + (n+2)^3$.
Hint: $f(n+1)-f(n) = (n+3)^3 - n^3 = 3 * 3 * n^2 + 3 * 3^2 * n + 3^3$.
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On Shanks' quartic approximation $\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)$ In Mathworld's "Pi Approximations", (line 58), Weisstein mentions one by the mathematician Daniel Shanks that differs by a mere $10^{-82}$,
$$\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)\color{blue}{+10^{-82}}\tag{1}$$
and says that $u$ is a product of four simple quartic units, but does not explicitly give the expressions. I managed to locate Shanks' Dihedral Quartic Approximations and Series for Pi (1982) Quartic Approximations for Pi (1980) online before so,
$$u = (a+\sqrt{a^2-1})^2(b+\sqrt{b^2-1})^2(c+\sqrt{c^2-1})(d+\sqrt{d^2-1}) \approx 1.43\times10^{13}$$
where,
$$\begin{aligned}
a &= \tfrac{1}{2}(23+4\sqrt{34})\\
b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\
c &= (429+304\sqrt{2})\\
d &= \tfrac{1}{2}(627+442\sqrt{2})
\end{aligned}$$
(Remark: In Shank's paper, the expressions for $a,b$ are different since he didn't express them as squares.)
A small tweak to $(1)$ can vastly increase its accuracy to $10^{-161}$,
$$\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2u)^6+24\big)\color{blue}{+10^{-161}}\tag{2}$$
I noticed the constant $u$ can also be expressed in terms of the Dedekind eta function $\eta(\tau)$ as,
$$u = \frac{1}{2}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-3502})}{\eta(\sqrt{-3502})}\right)^4\approx 1.43\times 10^{13}$$
which explains why $24$ improves the accuracy. Note that the class number of $d = 4\cdot3502$ is $h(-d)=16$, and $u$ is an algebraic number of deg $16$. Mathworld has a list of class numbers of d. However, we can also use those with $h(-d)=8$ such as,
$$x = \frac{1}{\sqrt{2}}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-742})}{\eta(\sqrt{-742})}\right)^2 \approx 884.2653\dots$$
which is a root of the 8th deg,
$$1 + 886 x + 1535 x^2 + 962 x^3 + 1628 x^4 - 962 x^5 + 1535 x^6 - 886 x^7 + x^8 = 0$$
Question: Analogous to $u$, how do we express $x$ as a product of two quartic units?
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Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then
$$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$
As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressions above are genuinely units. I wrote them in this form to conform with the previous example of the OP) The first unit (involving $a$) generates a degree four extension whose Galois closure has Galois group $D_8$, but the latter generates a Galois extension with Galois group $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$.
For those playing at home, this allows for the extra simplification
$$b + \sqrt{b^2 + 1} = (1 + \sqrt{2})^2 \cdot \frac{7 + \sqrt{53}}{2} .$$
|
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How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$ Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I don't have any idea on how to prove this.
Any help appreciated.
Thanks.
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If @Cocopops is correct, in that the +,- signs go like +,+,-,+,+,+,-,+,+,+, ... and the aperiodicity is just at the beginning, this is far less impressive.
Then if
$$x= \sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}} $$
then $$
y = \sqrt{5+x} = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}}},
$$
so the pattern for $y$ is +,+,+,-,+,+,+,-,+,+,+, ... and we can say
$$
(((y^2-5)^2-5)^2-5)^2-5 = -y.
$$
Numerically we should be able to find a root. However finding the analytic expression still seems hard.
I'd like to suggest that we pose this as a dual question, what if the signs DO follow +,+,-,+,+,+,-,+,+,+,+,-, ...
Does the expression have a closed form? In general, what about radicals of the form
$$
\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a- \ldots}}}}}}}}}?
$$
|
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A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$ Let
$$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$
its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$
$S$ can be represented in terms of the generalized hypergeometric function:
$$S={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)\cdot\frac12.\tag2$$
Let $\sigma$ be the closed-form expression constructed from integers and elementary functions as follows:
$$\sigma=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag3$$
where
$$\alpha=\frac{\sqrt[3]{3\,}}{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right),\tag4$$
$$\beta=\frac1{4\,\sqrt[3]{2\,}}\left(\sqrt[3]{25+3\,\sqrt{69}}+\sqrt[3]{25-3\,\sqrt{69}}\right)-\frac12,\tag5$$
$$\gamma=\frac1{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{57\,\sqrt{69}-459}-\sqrt[3]{57\,\sqrt{69}+459}\right)+\frac12\tag6$$
are the unique real roots of the following cubic equations:
$$8\,\alpha^3-2\,\alpha-1=0,\tag7$$
$$64\,\beta^3+96\,\beta^2+36\,\beta-23=0,\tag8$$
$$8\,\gamma^3-12\,\gamma^2+16\,\gamma+11=0.\tag9$$
Equivalently,
$$\sigma = \frac{3\,p}{2}\,\ln\big(p+1\big)-\frac{1}{2}\sqrt{\frac{3-p}{p}}\arccos\Big( \frac{p-6}{6p+2}\Big)\tag{10}$$
where $p$ is the plastic constant or the real root of
$$p^3-p-1=0\tag{11}$$
It can be numerically checked that the following inequality holds:
$$\Big|S-\sigma\Big|<10^{-10^5},\tag{12}$$
I conjecture that the actual difference is the exact zero, and thus $S$ has an elementary closed form:
$$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}\stackrel?=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag{13}$$
I am asking for you help in proving this conjecture.
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We show that the sum equals
$$
\int_0^1 \frac{2-3x}{1-x^2+x^3} dx.
$$
This integral is "elementary", but requires expanding the integrand
in partial fractions, which in turn requires all the solutions of
the cubic polynomial in the denominator; so if one insists on
writing everything in radicals then the answer is bound to be complicated.
The "conjecture" is surely correct ($10^5$ digits is more than enough
for moral certainty, especially since $\alpha,\beta,\gamma$ are all
in the field generated by the real root of $1-x^2+x^3$), though
it may be an unpleasant and unrewarding exercise to check that
the partial-fraction integration yields an equivalent answer.
(One also wonders how one could possibly "conjecture" such an answer
without some sense of where to look...)
The key is to write each term $n! (2n)! / (3n+2)!$ in terms of the
beta integral
$a!b!/(a+b+1)! = B(a+1,b+1) = \int_0^1 x^a (1-x)^b dx$. Here we write
$n! (2n)! / (3n+2)! = B(2n+1, n+2) / (n+1)$, and sum over $n$ to get
$$
\sum_{n=0}^\infty \frac{n! (2n)!} {(3n+2)!}
= \int_0^1 \sum_{n=0}^\infty \frac{(x^2-x^3)^{n+1}}{n+1} \frac{dx}{x^2}
= -\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}.
$$
(We easily justify the interchange of infinite sum and definite integral
because all integrands are positive on $0<x<1$.)
We can now integrate by parts to remove the logarithm:
$$
-\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}
= \int_0^1 \log(1-x^2+x^3) \phantom. d\left(\frac{1}{x}\right)
= \int_0^1 \frac1x d(\log(1-x^2+x^3)),
$$
in which the integrand simplifies to $(2-3x)/(1-x^2+x^3)$,
QED.
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If both the sum and sum of squares of two rationals are integers, the two rationals are integers too There are two rational numbers $\alpha, \beta$ such that $\alpha + \beta,\ \alpha^2 + \beta^2$ are both integers. Prove that $\alpha, \beta$ are integers.
I started off by assuming that $\alpha = \frac{a}{b}, \beta = \frac{c}{d}$ such that $\gcd(a, b) = \gcd(c, d) = 1$ and $b, d \neq 1$.
$\frac{a}{b} + \frac{c}{d}$ is an integer. Let $\frac{a}{b} - \lfloor\frac{a}{b}\rfloor = x , \frac{c}{d} - \lfloor\frac{c}{d}\rfloor = y$
$\implies x + y = 1$ and $\left(\lfloor\frac{a}{b}\rfloor+x\right)\left(\lfloor\frac{c}{d}\rfloor+y\right)$ is an integer
$\implies \lfloor\frac{a}{b}\rfloor \lfloor\frac{c}{d}\rfloor + \lfloor\frac{a}{b}\rfloor y + \lfloor\frac{c}{d}\rfloor x + xy$ is also an integer
Since the leading term is an integer, the rest must be an integer too. Substituting $y = 1-x$ we get
$\lfloor\frac{a}{b}\rfloor - \lfloor\frac{a}{b}\rfloor x + \lfloor\frac{c}{d}\rfloor x + x - x^2$
Again, the leading term is an integer, so we can remove it. Simplifying the rest, we get...
$x\left(\lfloor\frac{c}{d}\rfloor - \lfloor\frac{a}{b}\rfloor + 1 - x\right)$ is an integer.
I couldn't think of any other suitable transformation and couldn't hit a contradiction.
I would prefer a solution that elaborates on what I have done above.
|
Since $\alpha + \beta$ is an integer, $\alpha$ and $\beta$ have the same denominator; i.e. $\alpha = \frac{a}{b}$, $\beta = \frac{c}{b}$, in reduced form.
Then $2\alpha\beta = (\alpha + \beta)^2 - (\alpha^2 + \beta^2)$ is an integer, so $\frac{2ac}{b^2}$ is an integer. This means $b^2 \;\big| \; 2$, which means $b = 1$, so $\alpha$ and $\beta$ are integers.
|
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The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
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It can be seen that $\sqrt{x^2+\frac{1}{x^2}}$ is convex. Therefore we can use Jensen inequality:
$$
\frac{1}{3}\sqrt{a^2+1/a^2}+\frac{1}{3}\sqrt{b^2+1/b^2}+\frac{1}{3}\sqrt{c^2+1/c^2} \ge \sqrt{(\frac{a+b+c}{3})^2+(\frac{3}{a+b+c})^2}.
$$
On the other hand because $a+b+c<1$, we can see that $f(x)=\sqrt{\frac{x^2}{9}+\frac{9}{x^2}}$ for $x<1$ has its minimum at $x=1$ and hence:
$$
\frac{1}{3}\sqrt{a^2+1/a^2}+\frac{1}{3}\sqrt{b^2+1/b^2}+\frac{1}{3}\sqrt{c^2+1/c^2} \ge \sqrt{\frac{82}{9}}.
$$
which proves the inequality.
|
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Functional equation $f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$ Find all of the functions defined on the set of integers and receiving the integers value, satisfying the condition:
$$f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$$
for each pair of integers $(a,b)$.
|
An easier answer.
Using $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ and putting $x=f(a+b),y=-f(a),z=-f(b)$ we get that $x^3+y^3+z^3-3xyz=0$ so we have that either $x+y+z=0$ or $x^2+y^2+z^2-xy-yz-zx=0$. If $x+y+z=0$ then we have that $f(a+b)=f(a)+f(b)$ which gives us $f(a)=af(1)$. If the second one is true we get from it that $x=y=z$, which means that $-f(a+b)=f(a)=f(b)$. From which we can get that $f(x)=0$ fot evey real number x. So those are the two solutions.
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How to prove algorithm for solving a square congruence when p ≡ 5 (mod 8) I'm having trouble understanding why this algorithm works and where it comes from:
"Suppose p ≡ 5 (mod 8) is a prime and y is a square (mod p); that is, for some $ x, x^2
≡ y\ (mod\ p)$. This can be solved by the following algorithm
*
*Compute $d ≡ y^{(p−1)/4} \pmod {p}$. Show that $d^4 ≡ 1 \pmod {p}$
*Hence show $d ≡ ±1 \pmod {p}$ or $d^2 ≡ −1 \pmod {p}$.
*If $d ≡ 1 \pmod {p}$ compute $r ≡ y^{(p+3)/8} \pmod {p}$; if $d ≡ −1 \pmod {p}$ compute $r ≡
2y(4y)^{(p−5)/8}\pmod {p}$;
*Then (r, −r) are the square roots of y.
[We will return to the case $d^2 ≡ −1 \pmod {p}$ later.]"
Here are my attempts:
Since $ p \equiv 5 \pmod {8}$ then $p = 8k + 5$ for some integer $k$. With some algebra I found that $\frac{p-1}{2} = 4k +2 = 2(2k+1)$.
We can let $d \equiv y^{2k+1} \pmod {p}$. So $d^2 \equiv y^{2(2k+1)} \pmod {p}$. Then $d \equiv y^\frac{p-1}{4} \pmod {p}$. And $d^4 \equiv 1 \pmod {p}$ by Fermat's theorem.
For the second step, since $d^2 \equiv y^\frac{p-1}{2} \equiv (\frac yp) \equiv1 \pmod {p}$ by Legendre's theorem. So $d \equiv \pm 1 \pmod {p}$. I'm confused why $d^2$ would equal $-1$ though.
I'm even more confused about the third step. After playing around with it, I found that if we set $r \equiv y^{k+1} \pmod {p}$ and substitute $k = \frac{p-5}{8}$ (since $p = 8k + 5$) then we get the first r, but I have no idea where the $2y$ and the $4y$ came from in the second one.
Any advice, hints, or clarifications? Thanks
Edit: For the second step, since $d^4 \equiv 1 \pmod {p}$, then $d^2 \equiv \pm 1 \pmod {p}$. So either $d^2 \equiv 1 \pmod {p}$ which implies that $d \equiv \pm 1 \pmod {p}$ or $d^2 \equiv -1 \pmod {p}$.
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The $d^2 \equiv -1 \pmod{p}$ part is wrong. Presumably, the author mixed some parts of the $p \equiv 1 \pmod{8}$ case in, where one computes $d = y^{(p-1)/8}$.
For $p \equiv 5 \pmod{8}$, we always have $1 = \left(\frac{y}{p}\right) \equiv d^2 \pmod{p}$.
So if $d \equiv 1 \pmod{p}$, we choose $r \equiv y^{(p+3)/8}\pmod{p}$, and that yields
$$r^2 \equiv y^{(p+3)/4} \equiv y\cdot y^{(p-1)/4} \equiv y\cdot d \equiv y \pmod{p}.$$
For $d \equiv -1\pmod{p}$, choosing $r \equiv y^{(p+3)/8} \pmod{p}$ would lead to $r^2 \equiv y\cdot d \equiv -y \pmod{p}$ by the same computation as above, so we mix a square root of $-1$ in to obtain a square root of $y$. For $p \equiv 5 \pmod{8}$, we have $\left(\frac2p\right) = -1$, so $2^{(p-1)/4}$ is a square root of $-1$ modulo $p$. The author just wrote it in an obscure way, we have $2\cdot 4^{(p-5)/8} = 2\cdot 2^{(p-5)/4} = 2^{(p-1)/4}$.
|
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|
How do you factor $x^3 - 8 = 0$? I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$?
|
First of all, just a comment about the terminology. You can factor a polynomial, you can't factor an equation. What you are trying to do is solve the equation $x^3 - 8 = 0$.
As $2$ is a zero of the polynomial, $(x-2)$ is a factor by the factor theorem. So $x^3 - 8 = (x-2)p(x)$ for some polynomial. As the degree of a product of non-zero polynomials is the sum of their degrees, we see that $p(x)$ is a quadratic, so $p(x) = ax^2 + bx + c$. Comparing coefficients, we can then deduce the values of $a$, $b$, and $c$ which gives $x^3 - 2 = (x-2)(x^2+2x+4)$.
Now $x^3 - 8 = 0$ if $x - 2 = 0$ or $x^2 + 2x + 4 = 0$ by the null factor law. Solving the second equation (by using the quadratic formula for example), you find $x = -1\pm i\sqrt{3}$.
Alternatively, if you know some complex analysis (in particular, De Moivre's Theorem), you can obtain all three solutions at once.
|
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|
How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
|
Let $b^2=1-\delta$ where $\delta\ge 0$. Then, $a^2=2-c^2+\delta$. Now, $(a+b)^2=a^2+b^2+2ab=3-c^2+2\sqrt{1-\delta}\sqrt{2-c^2+\delta}$. The product of square roots is decreasing in $\delta$ (we can explicitly differentiate or just note that $1-\delta<1$ while $2-c^2+\delta>1$ so a small increase in $\delta$ decreases first square root by a greater percentage than it increases the second square root). Substituting $\delta=0$, $(a+b)^2\le 3-c^2+2\sqrt{2-c^2}<3+2\sqrt{2}=(1+\sqrt{2})^2$ so $a+b<1+\sqrt{2}$. Note strict inequality if $c>0$.
|
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|
Inclusion-Exclusion How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 32$ with $0 \leq x_i \leq 10$ for $i = 1, 2, 3, 4$?
How many integer solutions are there to the inequality
$$
y_1 + y_2 + y_3 + y_4 < 184
$$
with $y_1 > 0$, $0 < y_2 \leq 10$, $0 \leq y_3 \leq 17$, and $0 \leq y_4 < 19$?
Here are two problems. I can see this combination $(32+4-1,4-1)$ but I am having trouble finding out what to subtract from that.
Also, the second problem is looking for a more complicated solution. I understand that $y_1+y_2+y_3+y_4+y_5 = 182$ and $(182,4)$ for the combination but what do it have to subtract from that?
Thank you.
|
Generating functions seem to give a cleaner looking solution for the second question.
The generating function for number of ways to sum to $k$ is
$$
\overbrace{\ \ \frac{x\vphantom{x^1}}{1-x}\ \ }^{y_1\gt0}\overbrace{\frac{x-x^{11}}{1-x}}^{0\lt y_2\le10}\overbrace{\frac{1-x^{18}}{1-x}}^{0\le y_3\le17}\overbrace{\frac{1-x^{19}}{1-x}}^{0\le y_4\lt19}
$$
which is
$$
\begin{align}
&\frac{x^2}{(1-x)^4}\left(1-x^{10}-x^{18}-x^{19}+x^{28}+x^{29}+x^{37}-x^{47}\right)\\
&=\left(x^2-x^{12}-x^{20}-x^{21}+x^{30}+x^{31}+x^{39}-x^{49}\right)\sum_{k=0}^\infty\binom{k+3}{k}x^k\\
&=\sum_{k=0}^\infty\left[{\scriptsize\binom{k+1}{k-2}-\binom{k-9}{k-12}-\binom{k-17}{k-20}-\binom{k-18}{k-21}+\binom{k-27}{k-30}+\binom{k-28}{k-31}+\binom{k-36}{k-39}-\binom{k-46}{k-49}}\right]x^k
\end{align}
$$
and the sum of the coefficients for $k\lt184$ is
$$
\begin{align}
&\sum_{k=0}^{183}\left[{\scriptsize\binom{k+1}{k-2}-\binom{k-9}{k-12}-\binom{k-17}{k-20}-\binom{k-18}{k-21}+\binom{k-27}{k-30}+\binom{k-28}{k-31}+\binom{k-36}{k-39}-\binom{k-46}{k-49}}\right]\\
&=\scriptsize\binom{185}{4}-\binom{175}{4}-\binom{167}{4}-\binom{166}{4}+\binom{157}{4}+\binom{156}{4}+\binom{148}{4}-\binom{138}{4}\\[6pt]
&=547200
\end{align}
$$
|
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|
If $x=\prod^{27}_{n=1}(1+\frac{2}{n})$ then find $13x$ - Ramanujan Mathematics olmpiad 2013 I tried this: $$x=\prod^{27}_{n=1}(1+\frac{2}{n})=(1+\frac{2}{1})(1+\frac{2}{2})(1+\frac{2}{3})\ldots(1+\frac{2}{27})=\frac{3}{1}\cdot\frac{4}{2}\cdot\frac{5}{3}\cdots\frac{29}{27}$$ Then the terms cancel out. But I am not getting the correct answer.
|
As you noted, $$x=\frac31\cdot\frac42\cdot\frac53\cdots\frac{29}{27}=\frac11\cdot\frac12\cdot\frac{28}1\cdot\frac{29}1=406$$hence $$13x=13\cdot406=5278$$
|
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|
How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.
The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?
Thank you for your attention
|
Hint: $y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.
|
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|
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
|
$${ \left( \frac { a }{ b } \right) }^{ 2 }+1={ \left( \frac { c }{ b } \right) }^{ 2 }\\ \frac { c }{ b } +\frac { a }{ b } =k\\ \frac { c }{ b } -\frac { a }{ b } =\frac { 1 }{ k } \\ \frac { c }{ b } =\frac { { k }^{ 2 }+1 }{ 2k } \\ \frac { a }{ b } =\frac { { k }^{ 2 }-1 }{ 2k } \\ $$Say:$$a=({ k }^{ 2 }-1)n\\ b=2kn\\ { a }^{ 2 }-{ b }^{ 2 }={ n }^{ 2 }({ k }^{ 4 }-6{ k }^{ 2 }+1)=n^2(k^2-3)^2-8=d^2$$
Is the result a perfect square?
|
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|
Prove that $\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ Prove that
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$
What should I do for this equation? Should I focus on proving $\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}{k}2^k$?
|
I view that the purpose on the lefthand side of the eqn as:
Let $A = {x_{1},x_{2},...,x_{n}}$ and $B={y_{1},y_{2},...,y_{m}}$. where $n \ge m$
$ \left( \begin{array}{cc}
m \\
0 \\
\end{array} \right) $
$ \left( \begin{array}{cc}
n \\
m \\
\end{array} \right) $ accounts the number of ways to choose $m$ elements from $A$.
$ \left( \begin{array}{cc}
m \\
1 \\
\end{array} \right) $
$ \left( \begin{array}{cc}
n + 1 \\
m \\
\end{array} \right) $ accounts the number of ways to choose $m$ elements from $A$ plus one element from $B$. ( from the set {${x_{1},x_{2},...,x_{n},y_{i}}$} )
Keep on going until the last term,
$ \left( \begin{array}{cc}
m \\
m \\
\end{array} \right) $
$ \left( \begin{array}{cc}
n + m \\
m \\
\end{array} \right) $ accounts the number of ways to choose $m$ elements from $A$ and $B$. ( from the set {${x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{m}}$} )
$$ $$
Now for the right hand side of the eqn, count the same thing as before as :
Set the number of $x_{i}$'s that has been chosen is $j$, then the remaining $m-j$ are elements from $B$. That is
$ \left( \begin{array}{cc}
n \\
j \\
\end{array} \right) $
$ \left( \begin{array}{cc}
m \\
m-j \\
\end{array} \right) $ = $ \left( \begin{array}{cc}
n \\
j \\
\end{array} \right) $
$ \left( \begin{array}{cc}
m \\
j \\
\end{array} \right) $
About the $2^k$ i am still confused, but i hope this will help a little..
|
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|
Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
|
Here's another approach you might use, a sort of brute-force approach. You said you got as far as $$(n^2 + 3n)(n^2 + 3n + 2).$$ Continuing to multiply, you get: $$ n^4 + 6n^3 + 11n^2 + 6n.$$
The claim is that that this is one less than a perfect square, or equivalently, that $$n^4 + 6n^3 + 11n^2 + 6n + 1$$ is a perfect square. If this last expression is a perfect square, it must be the square of something of the form $n^2 + an+b$ for some $a$ and $b$. But what are $a$ and $b$?
Squaring $n^2 + an+b$, we get $$n^4 + 2an^3 +(a^2+2b)n^2 + (2ab)n + b^2$$
and equating the coefficients of the two polynomials we get $$\begin{align}
2a & = 6 \\
a^2+2b & = 11\\
2ab & = 6 \\
b^2 & = 1
\end{align}$$
which we can easily solve to obtain $a=3, b=1$. So putting together the various steps we have $$n(n+1)(n+2)(n+3) = (n^2 + 3n + 1)^2 - 1$$ which is what we were looking for.
|
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|
For what values of a and b does the following limit equal 0?
I understand I need to make the sum of the individual limits equal 0 - but I'm a little lost. I computed the limit of the first term to be -4/3 via L'Hospitals Rule but Wolframalpha contradicts me (http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin%282x%29%2F%28x%5E3%29%29).
'a' obviously should be left for last - and $b/($x^2) is some constant - so I get 0? Assuming 'b' is positive, or anything other than 0, I get that term is 0.
Thus, -4/3 + 0 + a = 0
Simple algebraic manipulating would lead me to believe a = 4/3 and then I'd plug that back in to find 'b'.
Questions
Why does Wolfram state what it does?
Is this solution correct? (I don't have the answer for the problem.)
|
First let us put this into a better form, with one variable term and one constant term:
$$ \lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a = 0$$
well if we evaluate the limit using L'Hopitals we get:
$$\lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} = \lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2} $$
$$\lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2} = \frac{2+b}{0} = (2+b) \infty $$
Since this technically equals infinity in all cases except when $(2+b) = 0$, we make the $b = -2$ so that our limit comes out to be zero. Now let's figure out the $a$
First I want to plug in the $b$ back into our original equation:
$$ \lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a = \lim\limits_{x \to 0} \frac{\sin(2x) - 2x}{x^3} + a = 0$$
then let's take the limit
$$ \lim\limits_{x \to 0} \frac{\sin(2x) - 2x}{x^3} + a = \lim\limits_{x \to 0} \frac{2\cos(2x) - 2}{3x^2} + a = \frac{2}{3}\lim\limits_{x \to 0} \frac{\cos(2x) - 1}{x^2} + a$$
and we can apply the L'Hospitals rule again:
$$ \frac{2}{3}\lim\limits_{x \to 0} \frac{\cos(2x) - 1}{x^2} + a = \frac{2}{3} \lim\limits_{x \to 0} \frac{-2\sin(2x)}{2x} + a = \frac{2}{3} \lim\limits_{x \to 0} \frac{-\sin(2x)}{x} + a$$
and once again with L'Hospitals:
$$ \frac{2}{3} \lim\limits_{x \to 0} \frac{-\sin(2x)}{x} +a = \frac{2}{3} \lim\limits_{x \to 0} {-2\cos(2x)} + a = \frac{-4}{3} + a$$
If we then set this equal to zero we get $a = \frac{4}{3} $
And so our final answer is:
$$ \lim\limits_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}= 0 , a = \frac{4}{3}, b = -2$$
|
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|
Prove : $\dfrac{a}{ac+1}+ \dfrac{b}{ab+1}+ \dfrac{c}{bc+1} \le \frac 12 (a^2+b^2+c^2)$ $a;b;c\in \mathbb{R}^{+}$ such that $abc=1$
Prove : $\frac{a}{ac+1}+ \frac{b}{ab+1}+ \frac{c}{bc+1} \leq \frac{1}{2}(a^2+b^2+c^2)$
|
$\frac{a}{ac+1} \leq \frac{a}{2\sqrt{ac}} = \frac{1}{2} a \sqrt{b}$. So it suffices to show that
$$a^2+b^2+c^2 \ge (abc)^{1/6} (a\sqrt{b} + b\sqrt{c} + c\sqrt{a}) = \sum_{cyc} a^{7/6}b^{2/3}c^{1/6}$$
This follows from adding up the AM-GM inequalities:
$$7a^2+4b^2+c^2 \ge 12 a^{7/6}b^{2/3}c^{1/6}$$
and its cyclic variants.
|
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|
If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
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$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
$$
\pars{a + b}^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = 1 - 3ab + 3a^{2}b + 3ab^{2}
=
1 + 3ab\pars{-1 + a + b}
$$
$$
x^{3} -3abx + 3ab - 1 = 0\quad\mbox{where}\quad x \equiv a + b
$$
$$
\pars{x - 1}\pars{x^{2} + x + 1} - 3ab\pars{x - 1} = 0
$$
One solution is $\color{#ff0000}{\large x = a + b = 1}$.
When $x \not=1$ $\pars{~a + b \not= 1~}$:
$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
0 = x^{2} + x + \pars{1 - 3ab} = x^{2} + x + 1 - 3a\pars{x - a}
=
x^{2} + \pars{1 - 3a}x + \pars{1 + 3a^{2}} = 0
\tag{1}
$$
This equation discriminant $\Delta$ is given by:
$$
\Delta = \pars{1 - 3a}^{2} - 4\pars{1 + 3a^{2}}
=
-3a^{2} - 6a - 3 = -3\pars{a + 1}^{2} \leq 0
$$
Since $x = \pars{a + b} \in {\mathbb R}$, this analysis doesn't yield any other solution when $a \not= -1$. When $a = -1$, Eq. $\pars{1}$ has the double root
$x_{\pm} = \pars{3a - 1}/2 = -2$. Then,
$$
\mbox{the solutions are}\quad
{\large\left\lbrace%
\begin{array}{rcl}
a + b & = & \phantom{-}1
\\[1mm]
\mbox{and}\quad a + b & = & -2 \quad\mbox{when}\quad a = -1
\end{array}\right.}
$$
|
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|
Derivative Arc-tangens We have to show that $(\arctan(x))' = \frac{1}{1+x^2}$, derived with the chain rule.
The hint given is that we should start with deriving $\tan(\arctan(x)) = x$; I am not sure though how this is helpful, since the derivative of $\arctan(x)$ is what we are looking for yet..
|
You are given $\tan(\tan^{-1} x) = x$. Take a derivative of that to get $\displaystyle \sec^2(\tan^{-1} x) \cdot \frac{d}{dx} \left(\tan^{-1} x\right) = 1$.
Now, consider a right triangle with adjacent length $1$ and opposite length $x$, so that the triangle has an angle $\theta$ such that $\tan \theta = x$. Now we take $\displaystyle \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{1}{\sqrt{x^2+1}}} = \sqrt{x^2+1}$. So, $\sec^2 (\tan^{-1} x) = \sec^2 \theta = x^2+1$. Plugging this back into the original equation, we get $$\frac{d}{dx} \left(\tan^{-1} x\right) = \frac{1}{x^2+1}.$$
|
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|
Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
|
You may also use the following formula valid for $x\neq 1$:
$\sum_{k=1}^{n-1}\frac{1}{(x-1)^2+4x\sin^2(\frac{k\pi}{n})}=\frac{n(x^n+1)}{(x^2-1)(x^n-1)}-\frac{1}{(x-1)^2}$.
The left hand side of the equation can be written as $\frac{x^{n+1}(n-1)-(n+1)x^n+x(n+1)-n+1}{(x-1)^3}\cdot \frac{1}{(x^{n-1}+x^{n-2}+\ldots +1)(x+1)}$.
You may use L' Hospital rule (for the limit $x\rightarrow 1$) for the first fraction (the second one gives a result of $2n$) and deduce that $\sum_{k=1}^{n-1}\frac{1}{4\sin^2(\frac{k\pi}{n})}=\frac{n^3-n}{6}\cdot \frac{1}{2n}=\frac{n^2-1}{12}$ and you are done.
|
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|
Finding a,b,c,d in a quartic expression Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far.
\begin{align}
a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51
\end{align}
Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)+p(-8)}{10}=\frac{24832+1216a+208b+4c+2d}{10}$$ $$=\frac{24832+202(6a+b)+(4a+4b+4c)+2b+2d}{10}$$ $$=\frac{19782+(36-4d)+2b+2d}{10}$$ $$=\frac{19818+2b-2d}{10}$$ How do I get rid of the $2b-2d$?
|
You have $p(x) = x^4 + ax^3 + bx^2 + cx + d$, and you're given that $a + b + c + d = 9$, $8a + 4b + 2c + d = 4$, and $27a + 9b + 3c + d = -51$
Now, $p(12) + p(-8) = 12^4 + 8^4 + (12^3 - 8^3) a + (12^2 + 8^2) b + (12 - 8) c + 2d = 24832 + 1216 a + 208 b + 4c + 2d = 24832 + 1216 a + 208 b + 2 (2c+d)$. You note that $2c + d = 4 - 8a - 4b$, and substitute that into the equation to get $p(12) + p(-8) = 24832 + 1216 a + 208 b + 8 - 16a - 8b = 24840 + 1200 a + 200b = 24840 + 200(6a + b).$
Plug in the $6a + b = -25$ and you get $p(12) + p(-8) = 24840 - 5000 = 19840$. Divide it by $10$ and you get $\displaystyle \frac{p(12) + p(-8)}{10} = 1984$.
Remember, $d + d \ne d$.
|
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|
Let $a,b,c \in \mathbb{R^+}$, does this inequality holds $\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$? Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours and I couldn't find a solution to this inequality. I tried to use Cauchy-Schwarz Inequality for fraction:
$$\frac{x^2}{a} + \frac{y^2}{b} + \frac{z^2}{c} \ge \frac{(z+x+y)^2}{a+b+c}$$
But I only get:
$$LHS \ge \frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{(n+k)(a+b+c)}$$
And then I can't prove that:
$$(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \ge 3(a+b+c)$$
$$a + b + c + 2\sqrt{ab} + 2\sqrt{ac} + 2\sqrt{bc} \ge 3(a+b+c)$$
$$\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \ge a+b+c$$
Which simplifyies to:
$$\sqrt{a}(\sqrt{b} - \sqrt{a}) + \sqrt{b}(\sqrt{c} - \sqrt{b}) + \sqrt{c}(\sqrt{a} - \sqrt{c}) \ge 0$$
But if $a\neq b \neq c$, one of the terms must be negative, so nothing here.
And at last I plug it into Wolfram Alpha, and for some random numbers $n$ and $k$ it's true. I wonder maybe the fact that the LHS is cyclic we make the LHS to have minimal values when $a=b=c$?
|
Assume $n,k\in\mathbb R^+$.
Necessary condition.- Notice that LHS-RHS is equal to $$\frac{(c-1)^2k(k-n)}{(k+n)(kc+n)(k+nc)}$$ for $a=1,b=1$. Therefore it is mandatory that $k\geq n$ for the inequality to hold.
Sufficient condition.- ...
|
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|
Vector norm Inequality proof Does anyone know how to start proving this inequality
$$
\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{4 \|x-y\|}{\|x\|+ \|y\|}
$$
The norm is a random norm on a vector space $V$
|
Here is a somewhat complicated proof, I do not know if there is a simpler
and more natural one.
Let $a=||x||,b=||y||,c=||x-y||$. The inequality is easy if $a=b$, so
assume that it is not the case. Since the problem is symmetric in
$x$ and $y$, we can assume $a < b$.
Then $2a < 2b$, so $a+b < 3b-a$, $ \frac{a+b}{3b-a}<1$, and hence
$\frac{b^2-a^2}{3b-a} < b-a$.
The triangle inequality shows that $||y|| \leq ||y-x||+||x||$ ; we deduce
$c \geq b-a > \frac{b^2-a^2}{3b-a}$. It follows that
$(3b-a)c > b^2-a^2$, $ 4bc > b^2-a^2+(a+b)c$ and hence
$\frac{4c}{a+b} > \frac{b-a+c}{b} $.
So it will suffice to show the following :
$$
\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{b-a+c}{b}
\tag{1}
$$
To show (1), we use the identity
$$
\bigg(\frac{x}{a}-\frac{y}{b}\bigg)=
\frac{c}{b}\bigg(\frac{x-y}{c}\bigg)+
\frac{b-a}{b}\bigg(\frac{x}{a}\bigg)
\tag{2}
$$
It follows from (2) that
$$
\bigg(\frac{x}{||x||}-\frac{y}{||y||}\bigg)=
\frac{c}{b}\bigg(\frac{x-y}{||x-y||}\bigg)+
\frac{b-a}{b}\bigg(\frac{x}{||x||}\bigg)
\tag{3}
$$
Applying the triangle inequality to (3), we see that
$$
\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\|
\leq \frac{c}{b}+\frac{b-a}{b} \tag{4}
$$
which concludes the proof.
|
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|
Compute the exact length of the curve $y = \frac{x^2}{4}-\frac{1}{2}\ln(x)$ from $x=1, x=2$ So here is my problem: $y = \frac{x^2}{4}-\frac{1}{2}\ln(x)$
Taking the derivative:
$$\frac{\textrm{d}y}{\textrm{d}x}=\frac{x}{2}-\frac{1}{2x}$$
And that simplifies further to:
$$\frac{\textrm{d}y}{\textrm{d}x}=\frac{x^2-1}{2x}$$
Since the formula for the curve is $$\int \sqrt{1+\left(\frac{\textrm{d}y}{\textrm{d}x}\right)^2}\textrm{d}x$$
I know I have to square my derivative:
$$1+\left(\frac{\textrm{d}y}{\textrm{d}x}\right)^2=1+\left(\frac{x^2-1}{2x}\right)^2$$
After expanding and adding $1$, I got the following:
$$\frac{5x^2-2x+1}{4x^2}$$
And I don't know what to do with it from here since the numerator doesn't seem to factor into a perfect square. Any suggestions?
|
\begin{align}
&\int_{1}^{2}\sqrt{1 + \left(\frac{{\rm d}y}{{\rm d}x}\right)^2}\,{\rm d}x
=
\int_{1}^{2}\sqrt{1 + \left(\frac{x^{2} - 1}{2x}\right)^2}\,{\rm d}x
=
\int_{1}^{2}\sqrt{1 + \frac{x^{4} - 2x^{2} + 1}{4x^{2}}}\,{\rm d}x
\\[3mm]&=
\int_{1}^{2}\sqrt{\frac{x^{4} + 2x^{2} + 1}{4x^{2}}}\,{\rm d}x
=
\int_{1}^{2}\sqrt{\left(\frac{x^{2} + 1}{2x}\right)^{2}}\,{\rm d}x
=
\int_{1}^{2}{\frac{x^{2} + 1}{2x}}\,{\rm d}x
=
\int_{1}^{2}\left(\frac{x}{2} + \frac{1}{2x}\right)\,{\rm d}x
\\[3mm]&=
\left[\frac{x^2}{4}+\frac{\ln\left(x\right)}{2}\right]_{1}^{2}
\end{align}
By applying your limits we get:
$$
l
=
\left[\frac{4}{4}+\frac{\ln\left(2\right)}{2}\right]
-
\left(\frac{1}{4} + 0\right)
=
\frac{3}{4}+\frac{\ln\left(2\right)}{2}
$$
Sorry but it is my first post. took me so much time to learn to use the symbols.
|
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|
Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
|
This result was a problem in American mathematical monthly in 50s.
Source:
H. F. Sandham and Martin Kneser, The American mathematical monthly, Advanced problem 4305, Vol. 57, No. 4 (Apr., 1950), pp. 267-268
\begin{align}S&=\sum_{n=1}^\infty \left(\frac{\text{H}_n}{n}\right)^2\\
&=\sum_{n=1}^\infty \frac{1}{n^2}\left(\int_0^1 \frac{1-t^n}{1-t}dt\right)\left(\int_0^1 \frac{1-u^n}{1-u}du\right)\\
&=\int_0^1 \int_0^1 \frac{\text{Li}_2(1)+\text{Li}_2( tu)-\text{Li}_2(t)-\text{Li}_2(u)}{(1-t)(1-u)}dtdu\\
&\overset{\text{IBP}}=\int_0^1 \int_0^1 \frac{\ln(1-t)\big(\ln(1-t)-\ln(1-tu)\big)}{t(1-u)}dtdu\\
&\overset{x=1-tu,y=\frac{1-t}{1-tu}}=\int_0^1\int_0^1 \frac{\ln y\ln(xy)}{(1-y)(1-xy)}dxdy\\
&\overset{w(x)=yx}=\int_0^1 \frac{\ln y}{y(1-y)}\left(\int_0^y \frac{\ln w}{1-w}dw\right)dy\\
&=\int_0^1 \frac{\ln y}{1-y}\left(\int_0^y \frac{\ln w}{1-w}dw\right)dy+\underbrace{\int_0^1 \frac{\ln y}{y}\left(\int_0^y \frac{\ln w}{1-w}dw\right)dy}_{\text{IBP}}\\
&=\frac{1}{2}\left(\int_0^1 \frac{\ln w}{1-w}dw\right)^2-\frac{1}{2}\int_0^1 \frac{\ln^3 w}{1-w}dw\\
&=\frac{1}{2}\times \left(\frac{\pi^2}{6}\right)^2-\frac{1}{2}\times -\frac{\pi^4}{15}\\
&=\boxed{\dfrac{17\pi^4}{360}}.
\end{align} NB:
I assume that:
\begin{align}\zeta(2)=\frac{\pi^2}{6},&&\zeta(4)=\frac{\pi^4}{90},&&\int_0^1 \frac{\ln w}{1-w}dw=-\zeta(2)=-\frac{\pi^2}{6},&&\int_0^1 \frac{\ln^3 w}{1-w}dw=-6\zeta(4)=-\frac{\pi^4}{15}\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{3}+k(\ln{x}+\ln{y}+\ln{z})=\dfrac{1}{3}$$
so let
$$f(x)=\dfrac{1}{2(x+1)^2+1}-k\ln{x}-\dfrac{1}{9}$$
$$\Longrightarrow f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}-\dfrac{k}{x}$$
let $f'(1)=0\Longrightarrow k=-\dfrac{8}{81}$
so
$$f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}+\dfrac{8}{81x}=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}$$
so note when $1>x>\dfrac{1}{2}$ then
$$f'(x)=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}<0$$
$x>1,f'(x)>0$
so
$$f(x)\ge f(1)=0$$
so if $x,y,z>\dfrac{1}{2}$ we have prove done.
But for other case,How prove it? Thank you
|
let $$x=\dfrac{bc}{a^2},y=\dfrac{ca}{b^2},z=\dfrac{ab}{c^2}$$
then we only prove follow inequality
$$\dfrac{a^4}{3a^4+2b^2c^2+4a^2bc}+\dfrac{b^4}{3b^4+2c^2a^2+4b^2ca}+\dfrac{c^4}{4c^4+2a^2b^2+4c^2ab}\ge\dfrac{1}{3}$$
By Cauchy-Schwarz inequality
\begin{align*}
\sum_{cyc}\dfrac{a^4}{3a^4+2b^2c^2+4a^2bc}&\ge\dfrac{(a^2+b^2+c^2)^2}{3(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2)+4abc(a+b+c)}\\
&\ge\dfrac{(a^2+b^2+c^2)^2}{3(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2b^2+b^2c^2+c^2a^2)}\\
&=\dfrac{(a^2+b^2+c^2)^2}{3(a^2+b^2+c^2)^2}=\dfrac{1}{3}
\end{align*}
|
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|
Expected number of rolls required to get sum greater than n for n faced die? Suppose a guy has a die with $n$ faces. He can go on rolling it as many times as possible and add the sum of each outcome. What is the expected number of rolls after which the sum is at least $n$?
|
I'm assuming the OP means what is the expected number of rolls until the sum is at least $n$ (as Alecos assumes in his answer).
Let $E(m)$ be the expected number of rolls until the sum is at least $n$, starting with a sum of $m$.
Then we have, for a start:
\begin{align*}
E(n) &=0 \\
E(n-1) &=1 \\
E(n-2) &=1+\frac{1}{n}E(n-1) = 1+\frac{1}{n} \\
E(n-3) &=1+\frac{1}{n}E(n-2)+\frac{1}{n}E(n-1) \\
&=1+\frac{1}{n}(1+\frac{1}{n}E(n-1))+\frac{1}{n} \\
&=1+\frac{2}{n} + \frac{1}{n^2} \\
E(n-4) &=1+\frac{1}{n}E(n-3) + \frac{1}{n}E(n-2)+\frac{1}{n}E(n-1) \\
&=1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\\
\end{align*}
We observe a pattern here, and you can prove with induction that
$$
E(n-k) = \sum_{i=0}^{k-1} \frac{\binom{k-1}{i}}{n^i}$$
for $1 \le k \le n$.
The value we seek is $E(0)$:
$$
E(0)=\sum_{i=0}^{n-1} \frac{ \binom{n-1}{i}}{n^i} = \left( 1+\frac{1}{n} \right)^{n-1}.
$$
Note that as $n \rightarrow \infty$, $E(0) \rightarrow e$.
|
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|
France Olympiad Team Selection Test 2005 In an international meeting of $n ≥ 3$ participants, $14$ languages are spoken. We know that:
- Any $3$ participants speak a common language.
- No language is spoken by more than half of the participants.
What is the least value of $n$?
|
Set up the standard incidence matrix.
Let the row be the $n$ participants and the 14 columns be the languages. Fill in the entry of 1 if the participant can speak the language.
We wish to count the number of column triples $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. Since every 3 participants speak a common language, this number is at least $ n \choose 3$. Since every language is spoken by half or less (my interpretation), there is at most $15 \times { \frac{n}{2} \choose 3}$ such triples.
Solving ${ n \choose 3} \leq 14{ \frac{n}{2} \choose 3}$, which is equivalent to $(n-1)(n-8) \geq 0 $, we get $ n \leq 1, n \geq 8 $. Hence, the least possible value of $n$ is 8.
It remains to show that $n=8 $ is possible. This is quite easy to do so, if you remember that equality must hold throughout. Hence, no 3 participants speak 2 common languages, is a necessary and sufficient condition. There is a natural choice, which works.
For the first 7 languages, use
$ \begin{array} { | l | l | l | l | l | l | l |}
\hline
1&1&1&1&0&0&0 \\ \hline
1&1&0&0&1&1&0 \\ \hline
1&0&1&0&1&0&1 \\ \hline
1&0&0&1&0&1&1 \\ \hline
0&1&1&1&0&0&0 \\ \hline
0&1&0&0&1&1&0 \\ \hline
0&0&1&0&1&0&1 \\ \hline
0&0&0&1&0&1&1 \\ \hline
\end{array} $
For the next 7 languages, use 1 minus entries in the above table.
|
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|
problem on solving equations with three variables Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously.
$x^3+y^3+z^3=x+y+z$
$x^2+y^2+z^2=xyz$
|
The equations do not have positive solutions. It follows directly from AM-GM inequalities.
Recall that $x,y,z> 0$ then
$x+y+z=x^3+y^3+z^3\geq3xyz=3(x^2+y^2+z^2)$
$3(x^2+y^2+z^2)-(x+y+z)^2=(x-y)^2+(y-z)^2+(z-x)^2\geq 0$
Hence $x+y+z\geq (x+y+z)^2$
then $x+y+z\leq 1$.
But $x,y,z>0$, so $x,y,z$ are strictly smaller than 1.
So $x^3<x$, $y^3<y$, $z^3<z$, which means $x^3+y^3+z^3<x+y+z$, a contradiction!
|
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|
The largest value of $f(n) $ Let us consider a function $ f:\mathbb{N}_0 \to \mathbb{N}_0 $ following the relations:
*
*$ f(0)=0$
*$f(n) = n+f\left(\left\lfloor \frac{n}{p} \right\rfloor\right)$ when the $n$ is not divisible by $p$
*$ f(np) = f(n) $
Here $p>1$ is a positive integer .$ \mathbb{N}_0 $ is the set of all non-negative integers and let $ a_k $ be the maximum value of $ f(n) $ for $ 0<n \leq p^k $ .
How can I find the value of $ a_k ?$
|
Proof by induction on $n$:
$$\begin{eqnarray}
f(p^{n+1}-1)=\\
f((p-1)(1+p+p^2+ \cdots p^{n-1}+p^n))=\\
f((p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1) p^{n-1} + (p-1)p^n)=\\
((p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1) p^{n-1} + (p-1)p^n)
+ f( \lfloor \frac{p-1}{p} + (p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1)p^{n-1} \rfloor)=\\
p^{n+1}-1 + f( (p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1)p^{n-1})=\\
p^{n+1} + p^{n} + \cdots + 1 -(n+2)=\\
\frac{p^{n+2}-1}{p-1} -n-2
\end{eqnarray}$$
Proof by induction on $n$:
$$0 \le a_i \lt p , \forall i: \;0 \le i \le n$$
$$\begin{eqnarray}
f(a_0+a_1p+a_2p^2+ \cdots a_{n-1} p^{n-1} + a_n p^n) = \\
(0 \; \text{if} \; a_0=0, \; \text{otherwise} \; a_0+a_1 p+a_2p^2+ \cdots a_{n-1} p^{n-1} + a_n p^n) + \\f( \lfloor \frac{a_0}{p}+a_1+a_2p+ \cdots a_{n-1} p^{n-2} + a_n p^{n-1} \rfloor) \\
\le
((p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1) p^{n-1} + (p-1)p^n) \\
+f( \lfloor \frac{p-1}{p} + (p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1)p^{n-1} \rfloor) =\\
f((p-1)+(p-1)p+(p-1)p^2+ \cdots (p-1) p^{n-1} + (p-1)p^n)= \\
f((p-1)(1+p+p^2+ \cdots p^{n-1}+p^n))=\\
f(p^{n+1}-1)
\end{eqnarray}$$
So the maximum on $\{0,\cdots , p^n\}$ is $f(p^n-1)= \frac{p^{n+1}-1}{p-1} -n-1$
|
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|
Exponentiation by squaring I need to calculate $7^{2012} \mod {13}$ by hand using exponentiation by squaring, but I cant seem to figure it out.
I started with this but I don't know for sure if its correct or where it's going.
\begin{align}
(7^2)^{1006} &\equiv 49^{1006} \pmod {13} \\
(10^2)^{503} &\equiv 100^{503} \pmod {13}
\end{align}
|
Looks good so far. Continuing, we find that:
\begin{align*}
100^{503} \pmod {13}
&\equiv 9^{503} \pmod {13} \\
&\equiv 81^{251} \cdot 9 \pmod {13} \\
&\equiv 3^{251} \cdot 9 \pmod {13} \\
&\equiv 9^{125} \cdot 3 \cdot 9 \pmod {13} \\
&\equiv 81^{62} \cdot 9 \cdot 1 \pmod {13} & \text{since $3 \cdot 9 = 27 = 2\cdot 13 + 1$}\\
&\equiv 3^{62} \cdot 9 \pmod {13} \\
&\equiv 9^{31} \cdot 9 \equiv 9^{32} \pmod {13} \\
&\equiv 81^{16} \pmod {13} \\
&\equiv 3^{16} \equiv 9^{8} \equiv 3^{4} \pmod {13} \\
&\equiv 9^2 \equiv 81 \equiv 3\pmod {13} \\
\end{align*}
|
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|
Find a minium value of a function Given x,y,z are positive real numbers such that
$$
x^2+y^2+6z^2=4z(x+y).
$$
Find the minimum value of the following function
$$
P=\frac{x^3}{y(x+z)^2}+\frac{y^3}{x(y+z)^2}+\frac{\sqrt{x^2+y^2}}{z}
$$
|
let $$a=\dfrac{x}{z},b=\dfrac{y}{z}$$
then $$a^2+b^2+6=4(a+b)$$
and we only find this minimum
$$\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}+\sqrt{a^2+b^2}$$
use Cauchy-Schwarz inequality,we have
$$\left(\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}\right)(ab(a+1)^2+ba(b+1)^2)\ge (a^2+b^2)^2$$
then we only find this follow minimum
$$\dfrac{(a^2+b^2)}{ab[(a+1)^2+(b+1)^2]}+\sqrt{a^2+b^2}$$
since
$$\dfrac{(a^2+b^2)}{ab[(a+1)^2+(b+1)^2]}\ge\dfrac{2}{a^2+b^2+2(a+b)+2} $$because use
$a^2+b^2\ge 2ab$
so we ony find
$$\dfrac{2}{a^2+b^2+2(a+b)+2}+\sqrt{a^2+b^2}=\dfrac{4}{3a^2+3b^2+10}+\sqrt{a^2+b^2}$$
let $x=\sqrt{a^2+b^2}\in[\sqrt{2},3\sqrt{2}]$
so
$$f(x)=\dfrac{4}{3x^2+10}+x$$
then follow is easy to find minimum.
|
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|
Power series help please! $$ \frac{2}{3-x} $$
I need to find a power series representation for this. I figured i'd pull the two out, but I can't figure out what to do with the three.
|
We know that $1 + r + r^2 + r^3......= \dfrac{1}{1-r}$. So we can write $\dfrac{2}{3-x} = \frac{2}{3}\dfrac{1}{1-\frac{x}{3}} = \frac{2}{3}(1+\frac{x}{3}+\frac{x^2}{9} + \frac{x^3}{27} + \frac{x^4}{81}.......)= \frac{2}{3}\sum_{n=0}^\infty(\frac{x}{3})^n$
|
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|
Average of 3 consecutive odd numbers
The average of $3$ consecutive odd numbers is $14$ more than one third of the first of these numbers, what is the last of these numbers?
$17/19/15/$data inadequate/none of these
Let three consecutive odd numbers be $a-2,a,a+2$. Their average is $a.$ So, $a=\frac13(a-2)+14\implies a=20.$ But that's not possible as $a$ is odd.
What am I missing?
|
We take $3$ odd numbers $a,b,c$ such that $a= (2n+1), b=(2n+3), c=(2n+5)$ for some integer $n$. From information given, we know $$\frac{(2n+1)+(2n+3)+(2n+5)}{3} = 14 + \frac{1}{3}(2n + 1) \\ \frac{6n+9}{3}=\frac{43+2n}{3} \\ \frac{4n-34}{3} = 0 \\ n = \frac{34}{4} \\ n=8.5$$ Since n is not an integer, the solution cannot exist.
|
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|
Someone who is calculating $43434343^2$. The answer is $18865ab151841649$, where the two digits $a$ and $b$ were lost. So I used congruence $\bmod 9$ and $\bmod 11$. First, $43434343^2 \pmod{11}$ is $5$ and I did the answer $18865ab15184169 \pmod{11}$ and got $-b+a+3 \pmod{11}$. I also did $\bmod 9$ and got $a+b+67 \pmod 9$. I am confused as to what to do next.
|
The simplest is probably to look modulo $99$.
Since $100 \equiv 1 \pmod {99}$,
$43434343^2 \equiv (43+43+43+43)^2 \equiv 172^2 \equiv 73^2 \equiv 5329 \equiv 82 \pmod {99} $
$18865ab151841649 \equiv 18+86+(50+a)+(10 b+1)+51+84+16+49 \equiv 355+ba \equiv 58+(10b+a) \pmod {99}$.
Hence $(10b+a) \equiv 24 \pmod {99}$. Luckily, we didn't get zero, so there is only one possibility : $b=2$ and $a=4$.
|
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|
Evaluate $\int\limits_0^\pi \frac{\sin^2x}{2-\cos x}\ \mathrm dx$ by complex methods find integral $$\int\limits_0^\pi \frac{\sin^2x}{2-\cos x}dx$$
what I had in mind is to use Euler formula, to turn it into a complex integral and change the limits of integration from $ -\pi$ to $\pi$ so that any odd parts of the integrand would go to zero. But does not seem to make problem easier. Substitution of $u=\tan(x/2)$ would work but the computation is very tedious. Any suggestions for a nicer approach?
|
Hint: Let $\gamma\colon [-\pi, \pi]\to \Bbb C, \theta \mapsto e^{i\theta}$. It is true that $\displaystyle \int \limits_0^ \pi\frac{(\sin(x))^2}{2-\cos(x)}\mathrm dx=\dfrac 1{4i}\int _\gamma \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}\mathrm dz$.
Further details: Note that for all $t\in \Bbb R$, $$\sin(t)=\dfrac{e^{it}-e^{-it}}{2i}\land \cos(t)=\dfrac{e^{it}+e^{-it}}{2}.$$
The map $x\mathop\longmapsto\dfrac{(\sin(x))^2}{2-\cos(x)}$ defined on $[-\pi, \pi]$ is even, thus $$\displaystyle \int \limits_{-\pi}^\pi\dfrac{(\sin(x))^2}{2-\cos(x)}\mathrm dx=2\int \limits_{0}^\pi\dfrac{(\sin(x))^2}{2-\cos(x)}\mathrm dx.$$
So,
$$\begin{align} \int \limits_{0}^\pi\dfrac{(\sin(x))^2}{2-\cos(x)}\mathrm dx&=\dfrac 1 2\int \limits_{-\pi}^\pi\dfrac{(\sin(x))^2}{2-\cos(x)}\mathrm dx\\
&=\dfrac 12\int \limits_{-\pi}^\pi\dfrac{\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2}{2-\frac{e^{ix}+e^{-ix}}{2}}\mathrm dx\\
\\&=\dfrac 12\int \limits_{-\pi}^\pi 1\cdot \dfrac{\frac{1}{-4}\left(e^{2ix}-2+e^{-2ix}\right)}{\frac{1}{2}\left(4-e^{ix}-e^{-ix}\right)}\mathrm dx\\
&=-\dfrac 1 {4}\int \limits_{-\pi}^\pi \dfrac{ie^{ix}}{ie^{ix}}\cdot \dfrac{e^{2ix}-2+e^{-2ix}}{4-e^{ix}-e^{-ix}}\mathrm dx\\
&=-\dfrac 1 {4i}\int \limits_{-\pi}^\pi \dfrac{\left(e^{2ix}-2+e^{-2ix}\right)ie^{ix}}{4e^{ix}-e^{2ix}-1}\mathrm dx\\ &=\dfrac 1 {4i}\int \limits_{-\pi}^\pi \dfrac{\left((e^{ix})^2-2+(e^{ix})^{-2}\right)ie^{ix}}{(e^{ix})^2-4e^{ix}+1}\mathrm dx\\
&=\dfrac 1 {4i}\int \limits_\gamma \dfrac{z^2-2+z^{-2}}{z^2-4+1}\mathrm dz\\
&=\dfrac 1 {4i}\int \limits_\gamma \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}\mathrm dz.\end{align}$$
The integral $\displaystyle \dfrac 1 {4i}\int \limits_\gamma \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}\mathrm dz$ can be found using residues.
It holds that for all $z\in \Bbb C$, $z^4-4z^3+z^2=z^2\left(z-(2-\sqrt 3)\right)\left(z-(2+\sqrt 3)\right)$.
Wolfram Alpha yields
$$\begin{align} &\operatorname{Res}\left(z\mapsto \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}, 0\right)&=4\\
&\operatorname{Res}\left(z\mapsto \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}, 2-\sqrt 3\right)&=-2\sqrt 3\\
&\operatorname{Res}\left(z\mapsto \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}, 2+\sqrt 3\right)&=2\sqrt 3.\end{align}$$
Since the winding numbers of the poles $2+\sqrt 3, 2-\sqrt 3$ and $0$ with respect to $\gamma$ are respectively $0, 1$ and $1$ it comes
$$\dfrac 1 {4i}\int \limits_\gamma \dfrac{z^4-2z^2+1}{z^4-4z^3+z^2}\mathrm dz=\dfrac 1{4i}\cdot 2\pi i\left(4-2\sqrt 3\right)=\pi(2-\sqrt 3).$$
|
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|
Prove that there are no such positive integers $a,b,c,d$ such that $a^2 + b^2 = 3(c^2 + d^2)$ Prove that there are no positive integers a, b, c, d such that $a^2 + b^2 = 3(c^2 + d^2)$. Hint: What can you say about divisibility of a and b by 3? Look at solution with smallest possible a.
|
By way of contradiction, suppose there exist a,b,c,d $\in \mathbb N$ such that $a^2 + b^2 = 3(c^2 + d^2)$.
(Note: see that the square of a natural number is also natural and the sum of natural numbers is also natural.)
Now let $a_o , b_o , c_o , d_o $ be the smallest solution set to the equation.
Then 3|$a_o$$^2 + b_o$$^2 \Rightarrow 3 | a_o$$^2$ and $ 3|b_o$$^2$ $\Rightarrow 3|a_o$ and $3|b_o$.
$\therefore$ $a_o = 3n$ and $b_o = 3m$, where m,n $\in \mathbb N$ such that n < $a_o$ and m < $b_o$.
Then $9(n^2 + m^2) = 3(c_o$$^2$ + $d_o$$^2)\Rightarrow 3(n^2 + m^2) = c_o$$^2 + d_o$$^2$,
Then by the similar argument above $3 | c_o$ and $ 3|d_o$.
$\therefore$ $c_o = 3r$ and $d_o = 3p$, where r,p $\in \mathbb N$ such that r < $c_o$ and p < $d_o$.
Now see that $9(n^2 + m^2) = 27(r^2 + p^2) \Rightarrow n^2 + m^2 = 3(r^2 + p^2)$, but that means n, m, r, p are a solution set to the equation $a^2 + b^2 = 3(c^2 + d^2)$ smaller than $a_o , b_o , c_o , d_o $, but this is a contradiction since $a_o , b_o , c_o , d_o $ is the smallest solution set, hence our assumption that there exist a,b,c,d $\in \mathbb N$ such that $a^2 + b^2 = 3(c^2 + d^2)$ is false.
|
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|
How many 5 digits numbers are there, whose digits sum to 22?
How many 5 digits numbers are there, whose digits sum to 22? Of course the first digit has to be larger than 0.
|
Using generating functions:
$$[x^{22}](x+x^2+\cdots+x^9)(1+x+\cdots +x^9)^4=\\
[x^{21}]\frac{1-x^9}{1-x}\cdot \frac{(1-x^8)^4}{(1-x)^4}=\\
[x^{21}](1-x^9)(1-x^8)^4(1-x)^{-5}=\\
[x^{21}]\sum_{i=0}^1 {1\choose i}(-x^9)^i\sum_{j=0}^4 {4\choose j}(-x^8)^j\sum_{k=0}^{\infty}{4+k\choose k}x^k=\\
\{(i,j,k)=(0,0,21),(0,1,11),(0,2,1),(1,0,12),(1,1,2)\}\\
{25\choose 21}-{4\choose 1}{15\choose 11}+{4\choose 2}{5\choose 1}-{16\choose 12}+{4\choose 1}{6\choose 2}=
5460.$$
WA answer.
|
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|
Prove that $a=2,b=2,c=2$ for a system of three equations. Let us consider a system equations :
\begin{gather}
a^3+b=3a+4\tag{i}\\
2b^3+c=6b+6 \tag{ii}\\
3c^3+a=9c+8\tag{iii}
\end{gather}
I have tried with the following steps :
$6\times(i)+3\times(ii)+2\times$(iii) gives :
$6(a^3+b^3+c^3)+6b+3c+2a=18(a+b+c)+58$. But I don't know how to solve in this way. By inspection method I have seen $a=b=c=2$.
|
Write the equations as
\begin{align*}
(a-2)(a+1)^2 = 2-b\\
2(b-2)(b+1)^2 = 2-c\\
3(c-2)(c+1)^2 = 2-a
\end{align*}
Note that $(a,b,c) = (2,2,2)$ is a valid solution.
$a>2 \Rightarrow b<2\Rightarrow c>2 \Rightarrow a<2. $ A similar logic precludes $a<2$. This forces $a=b=c=2$ which is the unique solution
|
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|
how to solve trigonometric inequalities? how does one solve trigonometric inequalities? Is there a method to this or is every solution done ad hoc?
simple equations of the type: $cos3x \leq 0$ when: $0\leq x \leq 2π$
The attempt at a solution: equating $cos 3x = 0$ yields $$ π /6 + 2\frac13πk\leq x \leq 2π -π /6 - πk /3 $$ as a general solution...what happens next?
|
I use Nghi H Nguyen's method to visually solve trig inequalities by the number unit circle.
Solve: $ F(x) = \cos (3x) < 0$
There are totally 6 end points at: $ \frac{\pi}{6}, \frac{\pi}{2}, 5\frac{\pi}{6}, 7\frac{\pi}{6}, 3\frac{\pi}{2} $ and $11\frac{\pi}{6} $that divide the unit circle into $6$ equal arc lengths.
Use the check point method to find the sign status of arc length $(\frac{\pi}{6}, \frac{\pi}{2})$.
Select $x = \frac{\pi}{3}.$
We get : $F(x) = \cos 3x = \cos \pi = -1. $Then, $f(x) < 0 $ inside this arc length.
Color it blue. Use endpoint's property to easily find other sign status of $F(x)$ for other arc lengths.
We have, starting from end point $\frac{\pi}{6}$ : Blue, red, blue, red, blue, red.
We get the solution set of $F(x) = \cos 3x < 0 $ (blue) that are the 3 open intervals:
$(\frac{\pi}{6}, \frac{\pi}{2})$ and $(5\frac{\pi}{6}, 7\frac{\pi}{6})$ and $(3\frac{\pi}{2}, 11\frac{\pi}{2})$
|
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|
Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substitution $x=a\sin t$ and $x=a\sec t$ Both have looked promising but I just can't seem to finish it out.
Thanks in advance for any help.
|
Putting $\displaystyle x=a\sin t,t=\arcsin\frac xa \implies -\frac\pi2\le t\le \frac\pi2$ as the principal value of inverse sine ratio lies $\displaystyle\left[-\frac\pi2,\frac\pi2\right]$
$\displaystyle\implies \cos t\ge0$
$\displaystyle\implies\sqrt{a^2-x^2}=\sqrt{a^2\cos^2t}=|a\cos t|=|a|\cos t$
$\displaystyle \int x^2\sqrt{a^2-x^2}dx=a^3|a|\int\sin^2t\cos^2tdt$
$$\text{Now as }\sin t\cos t=\frac{\sin2t}2, \sin^2t\cos^2t=\frac{\sin^22t}4$$
$$\text{Again as }\cos2u=\cos^2u-\sin^2u=1-2\sin^2u,\sin^22t=\frac{1-\cos4t}2$$
$$\int\sin^2t\cos^2tdt=\frac18\int(1-\cos4t)dt=\frac t8-\frac{\sin4t}{32}+C$$
Now as $\displaystyle x=a\sin t$ and $\sin4t=2\sin2t\cos2t=4\sin t\cos t\cos2t$
$\displaystyle\cos2t=1-2\sin^2t=1-2\left(\frac xa\right)^2=-\frac{2x^2-a^2}{a^2} $
As $\displaystyle\cos t\ge 0,\cos t=+\sqrt{1-\left(\frac xa\right)^2}=\frac{\sqrt{a^2-x^2}}{|a|}$
|
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|
Laurent Series of $(z^2-1)^{-2}$ I have problems to determine the Laurent series of the function $(z^2-1)^{-2}$ in the regions:
$$0<|z-1|<2$$
and
$$|z+1|>2$$.
My idea was as follows:
$$\frac{1}{z^2-1}=\frac{1}{4z(z-1)^2}-\frac{1}{4z(z+1)^2}$$
However, this procedure only makes it harder to make the series, and tried to do it by Taylor.
|
Make a partial fraction decomposition:
$$\begin{align}
\frac{1}{z^2-1} &= \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right)\\
\frac{1}{(z^2-1)^2} &= \frac{1}{4}\left(\frac{1}{(z-1)^2} - \frac{2}{z^2-1} + \frac{1}{(z+1)^2}\right)\\
&= \frac{1}{4}\left(\frac{1}{(z-1)^2} - \frac{1}{z-1} + \frac{1}{z+1} + \frac{1}{(z+1)^2}\right).
\end{align}$$
Each of these terms can be expanded easily in the regions as geometric series (possibly once differentiated), when they can't remain as is.
|
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Verifying a solution from Strauss' "Partial Differential Equations - An Introduction", 2nd edition In section 9.2 on page 241, question #12 is given as follows:
"Solve the three-dimensional wave equation in $\{r\ne0,t>0\}$ with zero initial conditions and with the limiting condition
\begin{equation*}
\lim_{r\to 0}4\pi r^{2}u_{r}(r,t) = g(t).
\end{equation*}
Assume that $g(0) = g^{\prime}(0) = g^{\prime\prime}(0) = 0$.
The text also provides the following solution:
\begin{equation*}
u = \begin{cases} -\frac{1}{4\pi r}g\left(t-\frac{r}{c}\right) &\mbox{if } t \ge \frac{r}{c} \\
0 & \mbox{if } 0\le t\le \frac{r}{c}. \end{cases}
\end{equation*}
For simplicity, let $c=1$. Now,
\begin{alignat*}{2}
u_{r} &= \frac{4\pi r^{2}g^{\prime}(t-r) + 8\pi rg(t-r)}{16\pi^{2}r^{4}} \\
&= \frac{g^{\prime}(t-r)}{4\pi r^{2}} + \frac{g(t-r)}{2\pi r^{3}} \Rightarrow \\
u_{rr} &= \frac{-4\pi r^{2}g^{\prime\prime}(t-r)-8\pi rg^{\prime}(t-r)}{16\pi^{2}r^{4}} - \left[ \frac{2\pi r^{3}g^{\prime}(t-r)+6\pi r^{2}g(t-r)}{4\pi^{2}r^{6}}\right] \\
&= -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} - \frac{g^{\prime}(t-r)}{\pi r^{3}} - \frac{3g(t-r)}{2\pi r^{4}}.
\end{alignat*}
However,
\begin{alignat*}{2}
u_{rr} + \frac{2}{r}u_{r} &= -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} - \frac{g^{\prime}(t-r)}{2\pi r^{3}} - \frac{g(t-r)}{2\pi r^{4}} \\
&\ne -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} \\
&= u_{tt}.
\end{alignat*}
So,
\begin{equation*}
u_{tt} \ne u_{rr} + \frac{2}{r}u_{r} = \Delta u \Rightarrow
\end{equation*}
$u$ does not solve the wave equation in 3D. What am I missing in my reasoning or calculation that can make the given solution for $u$ correct?
|
I prefer to write $u_{rr}+\frac{2}{r}u_r$ as $\frac{1}{r^2}(r^2 u_r)_r$, which has more meaning ($r^2 u_r$ is the flux through a sphere of radius $r$) and is easier to calculate with. Given
$$u=-\frac{1}{4\pi} r^{-1} g(t-r/c)$$
we get
$$u_r= \frac{1}{4\pi} r^{-2} g(t-r/c)+\frac{1}{4\pi c} r^{-1} g'(t-r/c)$$
hence
$$r^2u_r= \frac{1}{4\pi} g(t-r/c)+\frac{1}{4\pi c} r g'(t-r/c)$$
Take one more derivative:
$$\begin{split}(r^2u_r)_r&= -\frac{1}{4\pi c} g'(t-r/c)+\frac{1}{4\pi c} g'(t-r/c)-\frac{1}{4\pi c^2} r g''(t-r/c) \\ & = -\frac{1}{4\pi c^2} r g''(t-r/c)\end{split}$$
Finally,
$$\frac{1}{r^2}(r^2 u_r)_r=-\frac{1}{4\pi rc^2} g''(t-r/c)$$
No quotient rule at all. It has its uses, but when you can avoid it by using the negative power of a variable, do that.
|
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How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks
|
Let us utilize the fact that $a,b,c$ are real
Using this,
either $\displaystyle(i) ab+bc+ca=0$
$\displaystyle\implies bc-a^2=-(ab+ca)-a^2=-a(a+b+c)$
$\displaystyle\implies \frac a{(bc-a^2)^2}=\frac a{a^2(a+b+c)^2}=\frac{bc}{abc(a+b+c)^2} $
$\displaystyle\implies \sum_{\text{cyc}}\frac a{(bc-a^2)^2}=\frac{bc+ca+ab}{abc(a+b+c)^2}=\cdots $
or $\displaystyle(ii) a^2+b^2+c^2-ab-bc-ca=0$
$\displaystyle\implies (a-b)^2+(b-c)^2+(c-a)^2=0$
Now the sum of square of three real numbers is zero, so each individually must be equal to zero
Then $\displaystyle a=b=c\implies bc-a^2=0$ which is impossible
|
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|
On a ring $R$ where $x^2 = x$, we must have that $2x = 0$ and $R$ is commutative I want to show that on a ring $R$ where $x^2 = x$ for all elements $x$, we must have that $2x = 0$ and $R$ is commutative. Does my answer look sound?
Well, if $x = 0$ we have that $x^2 = x$ is trivially true.
If $x \neq 0$ then $x^2 = x$
$\Longleftrightarrow x \cdot x = x$
$\Longleftrightarrow x \cdot x \cdot x^{-1} = x \cdot x^{-1}$
$\Longleftrightarrow x = 1$
I.e. the only non-zero element that we can possibly find in $R$ is $1$.
So $R = \{0, 1\}$ and as $1 + 0 = 1$ the only way we can satisfy the additive inverse ring axiom is if $1 + 1 = 0$.
I.e. $2(1) = 0 \Longleftrightarrow 2x = 0$
$R$ must be commutative because if we take two elements from $R$, e.g. take $a = 0$ and $b = 0$ such that $ab \neq ba$ we will have that $ab$ is either $0$ or $1$.
$ab = 1$ is not possible as $a = b = 0$.
If $ab = 0 \implies ba = 1$ and this is not possible as again $a = b = 0$.
|
The party line on this problem, since I am such a good party member!: since for all $x \in R$ we have $x^2 =x$, it follows that
$(x + y)^2 = x + y, \tag{1}$
or
$x^2 + xy + yx + y^2 = x + y, \tag{2}$
or
$x + xy + yx + y = x + y, \tag{3}$
or
$xy + yx = 0, \tag{4}$
whence taking $y = x$,
$2x = x + x = x^2 + x^2 = 0, \tag{5}$
and thus we also have
$x = -x \tag{6}$
holding for all $x \in R$. Now we see that
$xy = -yx = yx \tag{7}$
follows from (4) and (6). Equations (4)-(7) express the requisite conclusions.QED
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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|
Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -> made this using the hypothesis.
so, i dont know if its right, or if its wrong, i dont know how to keep going from this, or if its the end.
Thanks.
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P(n) = (2^3n)-1
p(1) = (2^3x1)-1
= 8-1 = 7 is divisible by 7
Let n =k
then p(k) =( 2^3k)-1 = 7x [assume it is divisible by 7]
therefore 2^3k = 7x +1
Let n =k+1
then p(k+1) = (2^3(k+1))-1
= (2^3k+3)-1
= (2^3k x 2^3)-1
= ((7x +1)2^3 ) -1
= ((7x +1)8 ) -1
= (56x +8) -1
= 56x +8 -1
=56x +7
= 7(8x +1)
therefore it is divisible by 7
|
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|
Calculation of a square root of a big number
How can I calculate the following number:
$$ \sqrt{444 \cdots (2n \text{ digits}) + 111 \cdots (n+1 \text{ digits}) - 666 \cdots (n \text{ digits})}.$$
My trying :
I have tried to calculate these data by observing the pattern of $ 7^2 , 67 ^2 ,667^2 , 6667^2 , \dots$. But there is no promising news to solve it by this process.
|
Hint:
Use the fact that $111\dots 111 = \frac{10^n-1}{9}$ (where there are $n$ ones)
Then your number is
$$4\frac{10^{2n}-1}{9} + \frac{10^{n+1}-1}{9} - 6\frac{10^n-1}{9} = \frac{4\times 10^{2n}-4 +10^{n+1}-1-6\times 10^n+6}{9}$$
$$=\frac{4\times 10^{2n}+4\times 10^n+1}{9}$$
$$= \frac{(2\times 10^n+1)^2}{9}$$
|
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|
Inequality with cube roots $$(\sqrt{n}+1)^{1/3}-(\sqrt{n}-2)^{1/3} \geq (\sqrt{n+1}+1)^{1/3}-(\sqrt{n+1}-2)^{1/3}$$
$$n\in \mathbb{N}$$
I come upon this inequality when trying to use prove series declension for the Leibnit'z criterion.
I haven't been able to find a good solution to this.
|
Hint: Consider the function :$f(x)=(\sqrt{x}+1)^{\frac{1}{3}}-(\sqrt{x}-2)^{\frac{1}{3}}$ ,$ \forall x >4$.
$f'(x)=\frac{1}{6\sqrt{x}}(\frac{1}{\sqrt[3]{(\sqrt{x}+1)^2}} -\frac{1}{\sqrt[3]{(\sqrt{x}-2)^2}})=\frac{1}{6\sqrt{x}}\frac{\sqrt[3]{(\sqrt{x}-2)^2}-\sqrt[3]{(\sqrt{x}+1)^2}}{\sqrt[3]{(\sqrt{x}+1)^2}\times\sqrt[3]{(\sqrt{x}-2)^2}}<0 $ ,$\forall x >4$.
which gives us that $f(x)$ is monotonically decreasing.
so for $n<n+1$,we have $f(n)>f(n+1)$
hence,$$(\sqrt{n}+1)^{1/3}-(\sqrt{n}-2)^{1/3} \geq (\sqrt{n+1}+1)^{1/3}-(\sqrt{n+1}-2)^{1/3}$$
$$n\in \mathbb{N}$$
|
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|
Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
|
Let $m$ be the slope. Then,
$$m = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Find $m'(x)$ and set this to $0$.
Well, to find $m'(x)$, we use the Quotient Rule and we have $$ m'(x) = \dfrac {(1+x^4)(2x) - (x^2)(4x^3)}{(1+x^4)^2} = \dfrac {2x^5+2x-4x^5=2x-2x^5}{(1+x^4)^2}. $$Well, so the numerator must be $0$, so we have $$ 2x - 2x^5 = 0 \implies x \cdot (1-x^4) = 0. $$Thus, the only real solutions for $x$ are $x=0$ and $x=\pm1$. Finish up from here.
|
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|
Showing $(a+b+c)(x+y+z)=ax+by+cz$ given other facts $$x^2-yz/a=y^2-zx/b=z^2-xy/c$$
None of these fractions are equal to 0.We need to show that,
$(a+b+c)(x+y+z)=ax+by+cz$
This question comes from a chapter that wholly deals with factoring homogeneous cyclic polynomials.I multiplied the three sides of the first equality by $abc$ but that yields an unfactorizable polynomial.I haven't had much luck in manipulating the first equality.So I tried to understand what I was trying to prove by expanding the second equality.We are trying to prove that,
$$a(y+z)+b(x+z)+c(x+y)=0$$
But the LHS is still unfactorizable.I tried to manipulate the first equality more,but they yielded nothing.However,I did find the following equality:
$$bc(x^2-yz)-ac(y^2-zx)+ab(z^2-xy)=a(y+z)+b(x+z)+c(x+y) $$
I would appreciate a very small hint.
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Using Addendo formula $\displaystyle \frac Aa=\frac Bb=\cdots=\frac{A+B+\cdots}{a+b+\cdots},$
$$\frac{x^2-yz}a=\frac{y^2-zx}b=\frac{z^2-xy}c=\frac{x^2-yz+y^2-zx+z^2-xy}{a+b+c}\ \ \ \ (1)$$
Multiplying the numerator & the denominator of the first term by $x,$ and those of second by $y$ and those of third by $z$
$$\implies\frac{x^3-xyz}{ax}=\frac{y^3-xyz}{by}=\frac{z^3-xyz}{cz}=\frac{x^3+y^3+z^3-3xyz}{ax+by+cz}\ \ \ \ (2)\text{ (again applying Addendo)}$$
Equate $(1),(2)$ using Factorize the polynomial $x^3+y^3+z^3-3xyz$
|
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Ratio of binomial coefficients If the ratio of the coefficients of $x^6$ and $x^7$ in the expansion of $(2+ax)^{11}$ is $14:25$, find the value of $a$.
What I have so far:
$$\binom{11}{6} \cdot a^6 \cdot 2^5$$
And
$$\binom{11}{7} \cdot a^7 \cdot 2^4$$
|
You calculated the coefficients correctly. If $A$ is the coefficient of $x^6$ and $B$ is the coefficient of $x^7$, then you want to find $A/B$, and set it equal to $14/25$.
You calculated the following:
$$A = \binom{11}{6} \cdot a^6 \cdot 2^5$$
$$B = \binom{11}{7} \cdot a^7 \cdot 2^4$$
We can write out the $\binom{11}{6}$ and $\binom{11}{7}$ terms to get:
$$A = \binom{11}{6} \cdot a^6 \cdot 2^5 = \binom{11}{5} \cdot a^6 \cdot 2^5$$
$$A = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2} \cdot a^6 \cdot 2^5$$
$$B = \binom{11}{7} \cdot a^7 \cdot 2^4 = \binom{11}{4} \cdot a^7 \cdot 2^4$$
$$ B = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2} \cdot a^7 \cdot 2^4$$
Now we can calculate $A/B$:
$$\frac{A}{B} = \frac{\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2} \cdot a^6 \cdot 2^5}{\frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2} \cdot a^7 \cdot 2^4}$$
Lots of things conveniently cancel and we end up with
$$\frac{A}{B} = \frac{\frac{7}{5} \cdot a^6 \cdot 2^5}{a^7 \cdot 2^4}$$
$$\frac{A}{B} = \frac{7 \cdot 2}{5 \cdot a}$$
We want this to equal $14/25$.
$$\frac{A}{B} = \frac{14}{25} = \frac{7 \cdot 2}{5 \cdot a}$$
$$\frac{14}{25} = \frac{14}{5a}$$
$$25 = 5a$$
$$\boxed{a=5}$$
|
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How to proceed to show this hold by induction?
Show that $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{n+2}{n(n+1)(n+3)}=\frac16\left[\frac{29}6-\frac4{n+1}-\frac1{n+2}-\frac1{n+3}\right],\text{ for $n\in\mathbb N$}.$$
I try induction. For $n=1$, it is trivial and then let it is true for some $k\in\mathbb N.$ Then for $k+1$,we have to show $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{k+3}{(k+1)(k+2)(k+4)}=\frac16\left[\frac{29}6-\frac4{k+2}-\frac1{k+3}-\frac1{k+4}\right],$$ which is quivalent to show $$\frac16\left[\frac{29}6-\frac4{k+1}-\frac1{k+2}-\frac1{k+3}\right]+\frac{k+3}{(k+1)(k+2)(k+4)}=\frac16\left[\frac{29}6-\frac4{k+2}-\frac1{k+3}-\frac1{k+4}\right]$$ by induction hypothesis. Now is there any easier trick other than long and tedious computation?
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$$\begin{eqnarray*}\frac{n+2}{n(n+1)(n+3)}&=&\frac{1}{n(n+1)}\left(1-\frac{1}{n+3}\right)\\&=&\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(1-\frac{1}{n+3}\right)\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n(n+3)}+\frac{1}{(n+1)(n+3)}\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\\&=&\frac{\frac{2}{3}}{n}-\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{6}}{n+3}\end{eqnarray*}$$
(that can be done through the residue theorem, too), hence, by setting $H_n=\sum_{k=1}^{n}\frac{1}{k}$:
$$\begin{eqnarray*} \sum_{n=1}^{N}\frac{n+2}{n(n+1)(n+3)}&=&\frac{2}{3}\sum_{n=1}^{N}\frac{1}{n}-\frac{1}{2}\sum_{n=1}^{N}\frac{1}{n+1}-\frac{1}{6}\sum_{n=1}^{N}\frac{1}{n+3}\\&=&\frac{2}{3}\sum_{n=1}^{N}\frac{1}{n}-\frac{1}{2}\sum_{n=2}^{N+1}\frac{1}{n}-\frac{1}{6}\sum_{n=4}^{N+3}\frac{1}{n}\\&=&\left(\frac{2}{3}-\frac{1}{2}-\frac{1}{6}\right)H_N-\frac{1}{2}\left(\frac{1}{N+1}-H_1\right)-\frac{1}{6}\left(\frac{1}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}-H_3\right)\\&=&\frac{H_1}{2}+\frac{H_3}{6}-\frac{1}{6}\left(\frac{4}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}\right)\\&=&\color{red}{\frac{29}{36}-\frac{1}{6}\left(\frac{4}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}\right)} \end{eqnarray*}$$
as wanted.
|
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|
Apply the Frobenius theorem to a problem Given
$x^2y''+x(x-3)y'+3y=0$
In standard form, we have
$y''+\frac{x-3}{x}y'+\frac{3}{x^2}y=0$
So $p(x)=\frac{x-3}{x}$ and $q(x)=\frac{3}{x^2}$
Find the first and second solutions. I'm not sure how to start this.. I know I have to find the indicial roots and indicial equation.. but I don't know how.
|
Note that your differential equation is of the form
$$x^2y^{\prime\prime} + xp(x)y^{\prime}+q(x)y=0$$
So it follows that $p(x) = x-3$ and $q(x) = 3$ (the important thing to remember in these types of problems is that $p(x)$ and $q(x)$ are polynomials). Hence $p_0 = p(0) = -3$ and $q_0=q(0)=3$; thus the indicial equation is
$$\begin{aligned} r(r-1)+p_0r+q_0=0 &\implies r(r-1)-3r+3=0\\ &\implies r^2-4r+3=0\\ &\implies r_1=3\text{ and }r_2=1\end{aligned}$$
Now, suppose that the equation $$x^2y^{\prime\prime} +x(x-3)y^{\prime} + 3y = 0\tag{1}$$ has a solution of the form $\displaystyle y=x^r\sum_{n=0}^{\infty}c_nx^n = \sum_{n=0}^{\infty} c_n x^{n+r}$ where $c_0\neq 0$. It then follows that
$$y^{\prime} = \sum_{n=0}^{\infty}c_n(r+n)x^{r+n-1}\quad\text{and}\quad y^{\prime\prime} = \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2}$$
and thus substituting this into $(1)$ gives that
$$\begin{aligned}\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r} +& \sum_{n=0}^{\infty}c_n(n+r)x^{n+r+1}\\ -& 3\sum_{n=0}^{\infty}c_n(n+r)x^{n+r} + 3\sum_{n=0}^{\infty}c_nx^{n+r}=0 \\ \implies \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r} +& \sum_{n=1}^{\infty}c_{n-1}(n+r-1)x^{n+r}\\ -& 3\sum_{n=0}^{\infty}c_n(n+r)x^{n+r} + 3\sum_{n=0}^{\infty}c_nx^{n+r}=0\end{aligned}$$
after shifting the index on the second summation.
If $n=0$, then we get that $$(r(r-1) -3r+3)c_0 = 0$$ and since $c_0\neq 0$ by assumption, this yields the indicial equation we found previously (so we don't get anything new here). Now for $n\geq 1$, we have that
$$[(n+r)(n+r-1)-3(n+r)+3]c_n+c_{n-1}(n+r-1) =0\tag{2}$$
Note that $-3(n+r) +3 = -3(n+r-1)$ and thus $(2)$ becomes
$$[(n+r)(n+r-1)-3(n+r-1)]c_n+c_{n-1}(n+r-1) =0$$
$$\implies (n+r-3)(n+r-1)c_n = -c_{n-1}(n+r-1)$$
Therefore,
$$c_n = -\frac{c_{n-1}}{n+r-3}\tag{3}$$
for $n+r-1\neq 0$.
We now consider two different cases for solving this recurrence relation.
Case I: $r=r_1=3$. The recurrence relation $(3)$ simplifies to
$$c_n = -\frac{c_{n-1}}{n}$$
which has the solution $c_n = \dfrac{(-1)^nc_0}{n!}$ (Left for you to verify; should be pretty straightforward).
Therefore, the first Frobenius solution is
$$y_1(x) = c_0x^{3}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{n!} = c_0x^3e^{-x}.$$
Case II: $r=r_2=1$. If we take $r=1$ and $n=2$ in $(2)$, we get
$$0\cdot c_2+2c_1=0 \implies 0\cdot c_2 - 2c_0 = 0$$
and since $c_0\neq 0$, there is no way to chose $c_2$ such that this equation holds. Hence, there is no Frobenius solution corresponding to the root $r_2=1$.
Since we only found one Frobenius solution, we'll need to use reduction of order to find the second solution; in particular, if $y_1$ is a solution to the ODE
$$y^{\prime\prime}+P(x)y^{\prime}+Q(x)y = 0,$$
then the second solution is given by
$$y_2 = y_1\int\frac{\exp\left(-\int P(x)\,dx\right)}{y_1^2}\,dx$$
I assume you can take things from here?
|
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|
Logarithm / exponential equation, not sure what to make of this, (simple) Solve for $a:(2 \log_a x)(3 \log_{x^2} 4) = 3$
No idea how to approach this problem other than moving the 2 and the 3 into an exponent..
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$(2 \log_a x)(3 \log_{x^2} 4) = 3\iff( \log_a x^2)( \log_{x^2} 4) = 1$. Then, because of the well known formula $\log_xy\log_y x=1$ one has $$\log_4 x^2\log_{x^2}4=1\Rightarrow\log_4 x^2=\frac{1}{\log_{x^2}4}=\log_a x^2\Rightarrow4^{x^2}=a^{x^2}\iff\left(\frac4a\right)^{x^2}=1$$
Thus (since $g(x)=b^x$ is injective) we have $a=4$.
|
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|
calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
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$\large\tt\mbox{Just a hint:}$
Write
$$
\int{\cos\pars{x}\,\dd x \over \cos\pars{x}\sin^{3}\pars{x} + \cos^{4}\pars{x}}
=
\int{\dd z \over \root{1 - z^{2}}z^{3} + \bracks{1 - z^{2}}^{2}}
\quad\mbox{with}\quad z \equiv \sin\pars{x}
$$
Use an Euler substitution: $\root{1 - z^{2}} \equiv t + \ic z$ which yields
$1 - z^{2} = t^{2} + 2t\ic z - z^{2}$ such that
$\ds{z = {1 - t^{2} \over 2t\ic}}$:
\begin{align}
\root{1 - z^{2}}&=t + {1 - t^{2} \over 2t} = {1 + t^{2} \over 2t}
\\[3mm]
\dd z&= {\pars{-2t}\pars{2t\ic} - \pars{2\ic}\pars{1 - t^{2}} \over -4t^{2}}\,\dd t
=
\ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t
\end{align}
\begin{align}
\int&=\int{1 \over
\bracks{\pars{1 + t^{2}}/2t}\bracks{\pars{1 - t^{2}}/2t}^{3}\pars{-1/\ic}
+
\bracks{\pars{1 + t^{2}}/2t}^{4}}
\,\ic\,{t^{2} + 1 \over 2t^{2}}\,\dd t
\\[3mm]&=-8\int{t^{2} \over -\pars{1 - t^{2}}^{3} + \ic\pars{1 + t^{2}}^{3}}\,\dd t
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
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|
Series expansion of a function at infinity I know it is possible to expand an expandable function for a real, and for infinite by setting $x=\dfrac1y$ and then expanding for $0$.
But my question is, how do we do if the evaluation of the new function and its derivatives is not possible ?
I mean I find things like $\sqrt{\left(\dfrac1y\right)^2 - \dfrac1y +1}$, but I can't evaluate it at $0$ ...
Wolfram|Alpha says it can be expanded and gives me the result which works perfectly for the rest of my problem.
|
If you want $\sqrt{x^2 - x +1}\;$ for $x\rightarrow \infty,\;$ you have $x > 0\;$ and write
$$\sqrt{x^2 - x +1} = x \sqrt{1 - \frac{1}{x} +\frac{1}{x^2}}.$$
Now substitute $y=\frac{1}{x}$ and compute the series for $y\rightarrow 0$
$$\sqrt{1 - y + y^2} = 1-\frac{1}{2}y+\frac{3}{8}y^2 + \frac{3}{16}y^3 + O(y^4)$$
Reverse the substitution, multiply by $x$ and get for $x\rightarrow \infty$
$$\sqrt{x^2 - x +1} = x-\frac{1}{2}+\frac{3}{8}\frac{1}{x} + \frac{3}{16}\frac{1}{x^2} + O(\frac{1}{x^3})$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/595426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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|
How to find $\arccos(\cos\frac{15\pi}{11})$? How to find $\arccos(\cos\frac{15\pi}{11})$?
I'm completely lost. $\frac{15\pi}{11}$ isn't on the unit circle so how do I find the cosine of it?
|
As $\cos x=\cos A\implies x=2n\pi\pm A$ where $n$ is any integer
The general value of $$\arccos\left(\cos\frac{15\pi}{11}\right)=2n\pi\pm \frac{15\pi}{11}$$
Based on the definition of principal value inverse cosine ratio, $\displaystyle0\le \arccos x\le\pi$
Taking '+' sign, $\displaystyle0\le 2n\pi+\frac{15\pi}{11}\le\pi\implies 0\le 22n+15\le 11\implies0\le n<0$ which is impossible
Taking '-' sign, $\displaystyle0\le 2n\pi-\frac{15\pi}{11}\le\pi\implies 0\le 22n-15\le 11\implies 1\le n\le1\implies n=1$
So, the principal value is $\displaystyle 2\pi-\frac{15\pi}{11}=??$
Alternatively, as we know $\cos\left(\pi\pm y\right)=-\cos y$
$$\cos\left(\frac{15\pi}{11}\right)=\cos\left(\pi+\frac{4\pi}{11}\right)=-\cos\left(\frac{4\pi}{11}\right)=\cos\left(\pi-\frac{4\pi}{11}\right)=\cos\left(\frac{7\pi}{11}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/595790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
definite integral to evaluation (what am I doing wrong here 3) I was given the below definite integral to evaluate. below is the equation and my answer after I plug in the values but it is wrong. correct answer should be 29/6 what am I doing wrong? thank you
$$\int_1^2 \left(x + \frac 1x\right)^2 \,dx = \int_1^2 x^2 + \left(\frac 1x\right)^2 = \frac {x^3}{3} + \ln |x| = 3.0$$
(Source http://i.stack.imgur.com/V3UDP.jpg)
|
You expanded incorrectly: Recall
$$(a + b)^2 = a^2 + 2ab + b^2$$
That gives us
$$\left(x + \frac 1x\right)^2 = x^2 + 2\left(x \cdot \frac 1x\right) + \frac 1{x^2} = x^2 + 2 + \frac 1{x^2}$$
Now go ahead and apply the power rule to evaluate the integral.
$$\begin{align} \int_1^2 \left(x + \frac 1x\right)^2 \,dx & = \int_1^2 \left( x^2 + 2 + \frac 1{x^2}\right)\,dx \\ \\ & = \int_1^2 \left( x^2 + 2 + x^{-2}\right)\,dx \\ \\ &= \dfrac {x^3}{3} + 2x - \frac 1x {\Huge|}_1^2\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/597103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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|
Solve the Following First Order PDE Solve
$$F\frac{\partial F}{\partial x} - \frac{\partial F}{\partial y} = y$$
subject to $F(s,0) = s^2$.
This is the first time I am using the method of characteristics, so I would like to know if I have made any errors in my working. I have
$$
\frac{dx}{dt} = z, \quad \frac{dy}{dt} = -1, \quad \frac{dF}{dt} = y
$$
Solving these gives me
$$
y = -t + y_0, \quad F =-\frac{t^2}{2} + y_0t + F_0, \quad x = -\frac{t^3}{6}+\frac{y_0 t^2}{2} + F_0t + x_0
$$
Then assume without loss of generality that $y_0 = 0$, so
$$
y = -t, \quad F =-\frac{t^2}{2}+ F_0, \quad x = -\frac{t^3}{6} + F_0t + x_0
$$
The characteristic curves are therefore curves of the form
$$
t \mapsto \left(-\frac{t^3}{6} + F_0t + x_0,\quad-t,\quad -\frac{t^2}{2}+ F_0 \right)
$$
I can now use the initial condition $F(s,0) = s^2$. This means I substitute $x = s, y = 0$ and $F = s^2$ and solve for $x_0, F_0$ and $t$ . So
$$
y = 0 \implies t = 0, \quad s = x_0, \quad s^2 = F_0
$$
So the parametrisation of the solution surface is
$$
(s,t) \mapsto \left(-\frac{t^3}{6} + s^2t + s,\quad-t,\quad -\frac{t^2}{2}+ s^2 \right)
$$
To get this in terms of $x$ and $y$ I know $-y = t$, hence
$$
x = \frac{y^3}{6} - s^2y + s \implies s = \frac{1\pm\sqrt{1-4y\left(x-\frac{y^3}{6} \right)}}{2y}
$$
Thus
$$
F(x,y) = -\frac{y^2}{2} + \left( \frac{1\pm\sqrt{1-4y\left(x-\frac{y^3}{6} \right)}}{2y}\right)^2
$$
This function is not well defined when $y = 0$ or when the term inside the square root is negative.
Have I made a mistake somewhere or is the working fine?
|
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=-1$ , letting $y(0)=0$ , we have $y=-t$
$\dfrac{dF}{dt}=y=-t$ , letting $F(0)=F_0$ , we have $F=F_0-\dfrac{t^2}{2}=F_0-\dfrac{y^2}{2}$
$\dfrac{dx}{dt}=F=F_0-\dfrac{t^2}{2}$ , letting $x(0)=f_1(F_0)$ , we have $x=f_1(F_0)+F_0t-\dfrac{t^3}{6}=f_1\biggl(F+\dfrac{y^2}{2}\biggr)-\biggl(F+\dfrac{y^2}{2}\biggr)y+\dfrac{y^3}{6}=f_1\biggl(F+\dfrac{y^2}{2}\biggr)-\dfrac{y^3}{3}-Fy~,~\text{i.e.}~F=f\biggl(x+\dfrac{y^3}{3}+Fy\biggr)-\dfrac{y^2}{2}$
$F(s,0)=s^2$ :
$f(s)=s^2$
$\therefore F=\biggl(x+\dfrac{y^3}{3}+Fy\biggr)^2-\dfrac{y^2}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/597526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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|
A pack contains $n$ card numbered from $1$ to $n$ A pack contains $n$ card numbered from $1$ to $n$. Two consecutive numbered cards are removed from
the pack and sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on
the removing cards is $k$, Then $k$ is.
$\bf{My\; Try}::$ Let two consecutive cards be $k$ and $k+1,$ Then given sum of the number on the
remaining cards is $1224$ .So $\left(1+2+3+.........+n\right)-\left(k+k+1\right) = 1224$
So $\displaystyle \frac{n(n+1)}{2}-(2k+1) = 1224\Rightarrow n(n+1)-(4k+2) = 2448$
Now I did not understand how can i calculate value of $(n,k)$
Help Required
Thanks
|
$(n, k) = (50, 25)$
I think you are almost there. If I understand your question right you just solve for $k$ in terms of $n$.
\begin{align*}
& n(n+1) - (4k + 2) = 2448 \\
\implies & n(n+1) - 2 - 2448 = 4k \\
\implies & n(n+1) - 2450 = 4k \\
\implies & \frac{n(n+1) - 2450} 4 = k
\end{align*}
So $(n, k)$ can be written $$\left(n, \frac{n(n+1) - 2450} 4\right).$$
So then you just need to find an $n$ such that both $n$ and $k$ are both positive and integers (You can't have a fraction of a card).
This next part probably goes beyond pre-calc, but looking at your profile, I think you can handle it.
We need $n(n+1) - 2450 \equiv 0 \pmod 4$. This just means that $n(n+1) - 2450$ divided by $4$ has a remainder of $0$.
Modular arithmetic is a bit different than what you may be used to in precalc, but it goes like this (if I remember correctly, its been a while).
\begin{align*}
& n(n+1) - 2450 \equiv 0 \pmod 4 \\
\implies & n(n+1) \equiv 2450 \pmod 4 \quad \text{You can replace $2450$ with the remainder of $2450$ divided by $4$.}\\
\implies& n(n+1) \equiv 2 \pmod 4
\end{align*}
There are 4 potential answers to check, 0,1,2,3 (after that they just replace such that $n + 4 \equiv n + 8 \equiv n + 12\dotsb$)
At $n = 1 \implies 1\cdot(1+1) / 4$ the remainder is $2$
At $n = 2 \implies 2\cdot(2+1) / 4 = 6/4$ the remainder is $2$
Last part. You cannot have fewer than zero cards so $\frac{n(n+1) - 2450} 4 > 0$. This is a simple inequality. $n > 49$.
So the answer is $\left(n, \frac{n(n+1) - 2450} 4\right)$, where $n = 1 + 4x$, $n = 2 + 4x$ where $x$ is any positive integer and $n > 49$.
After looking at Steven Stadnicki's answer, I saw there is one more restriction on $n$. $n$ cannot be such that $n < k$. So you can set up an inequality,
$ n > \frac{n(n+1) - 2450} 4 \implies n \le 51$.
So given all our criteria for valid $n$ values, the only allowed n is:
$n=2 + 12 \cdot 4$
The solution for $(n, k)$ is thus $(50, 25)$.
That should work even if it's a little longer then the other answers proffered
|
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"url": "https://math.stackexchange.com/questions/599394",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$ This is my attempt:
$$
\begin{align}
& \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt]
& = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt]
& = (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt]
& = (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2
\end{align}
$$
But now I can't get rid of the cosines. How can I get rid of them?
|
You got off to a good start:
$$
\sin(A+B)\sin(A-B) = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B)-\cos(A)\sin(B))
$$
This is of the form $(x+y)(x-y)$ so
$$
\sin(A+B)\sin(A-B) = \sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B)
$$
Eliminate the cosines (since $\sin^2(x)+\cos^2(x)=1$, so $\cos^2(x)=1-\sin^2(x)$) and expand:
$$
\begin{align}
\sin(A+B)\sin(A-B) &= \sin^2(A)(1-\sin^2(B))-(1-\sin^2(A))\sin^2(B)\\
&= \sin^2(A)-\sin^2(A)\sin^2(B)-\sin^2(B)+\sin^2(A)\sin^2(B)\\
&=\sin^2(A)-\sin^2(B)
\end{align}
$$
as required.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/600681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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|
Why does $r = \cos \theta$ produce a circle? I am trying to do a double integral over the following region in polar coordinates:
I know that the limits of integration are:
$$\theta = -\frac{\pi}{2} \quad \to \quad \theta = \frac{\pi}{2} \\ r = 0 \quad \to \quad r = \cos \theta$$
However, I don't understand how $r = 0 \quad \to \quad r = \cos \theta$ works. Cosine is a function (not just a relation) meaning that it has only one value of $r$ for every value of $\theta$. However, it seems like the graph $r = \cos \theta$ has two values of $r$ for every value of $\theta$. Why does $r = \cos \theta$ produce a circle?
|
$$r = \cos \theta$$
$$\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}}$$
$$x^2 + y^2 = x$$
$$x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}$$
$$\left(x - \frac{1}{2}\right)^2 + (y - 0)^2 = \left(\frac{1}{2}\right)^2$$
This is the equation for a circle in Cartesian coordinates $(x, y)$ with center $({1 \over 2}, 0)$ and radius $1 \over 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/601727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 10,
"answer_id": 2
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|
Prove $1 + \tan^2\theta = \sec^2\theta$ Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$
I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
|
Here is an alternative using exponential forms:
$$
\begin{align*}
1+\tan^2 \theta &=1+\left( \frac{e^{i \theta}-e^{-i \theta}}{i\left( e^{i \theta}+e^{-i\theta} \right)} \right)^2 \\
&=1-\frac{\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{\left( e^{i \theta}+e^{-i\theta} \right)^2-\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\frac{4}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\
&=\left( \frac{2}{ e^{i \theta}+e^{-i\theta}} \right)^2 \\
&= \sec^2 \theta.
\end{align*}
$$
Here is an entirely different approach that focuses on the geometry of a right triangle.
Form a right triangle with angle $\theta$. Let $y$ be the side opposite $\theta$, $x$ be the side adjacent $\theta$, and label the hypotenuse $r$, where $r^2=x^2+y^2$ (by theorem of Pythagoras).
We can read trigonometric definitions right from the triangle as corresponding ratios of sides. Specifically, for angle $\theta$, $\tan \theta = \dfrac{y}{x}$, and $\sec \theta = \dfrac{r}{x}$. We can now write,
$$
\begin{align*}
1+\tan^2 \theta &= 1+\left( \frac{y}{x} \right)^2 \\
&=1+\frac{y^2}{x^2} \\
&=\frac{x^2+y^2}{x^2} \\
&=\frac{r^2}{x^2} \\
&= \left(\frac{r}{x}\right)^2 \\
&=\sec^2 \theta.
\end{align*}
$$
|
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"url": "https://math.stackexchange.com/questions/607269",
"timestamp": "2023-03-29T00:00:00",
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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.