Search found 77 matches
- Tue Jul 12, 2016 7:22 am
- Forum: Complex Analysis
- Topic: Contour Integration
- Replies: 1
- Views: 2535
Re: Contour Integration
Let \[f(z) = \frac{\sin(\pi z^2) + \cos(\pi z^2)}{z^2 - 3z+2}.\] Then \(f\) has exactly two simple poles inside the countour of integration, at \(z=1\) and at \(z=2\). So the integral \(I\) is equal to \(2\pi i (\mathrm{Res}(f;z=1) + \mathrm{Res}(f;z=2)).\) We have \[ \mathrm{Res}(f;z=1) = \lim_{z\t...
- Tue Jul 12, 2016 7:05 am
- Forum: Real Analysis
- Topic: Largest prime factor
- Replies: 2
- Views: 2325
Re: Largest prime factor
Note that \[ \frac{2012}{k(k+1)\cdots (k+2012)} = \frac{1}{k(k+1)\cdots(k+2011)} - \frac{1}{(k+1)(k+2)\cdots (k+2012)}.\] So the sum \[ \sum_{n=1}^{\infty} \frac{2012}{n(n+1)\cdots (n+2012)}\] is telescopic and equals \(1/(2012)!\). In particular the expression of the question is equal to \(2012!\) ...
- Sat Jul 09, 2016 9:01 am
- Forum: Real Analysis
- Topic: Maximum value of Ratio
- Replies: 1
- Views: 1760
Re: Maximum value of Ratio
By Hölder's inequality, and since \(f\) is positive, we have \[ \int f(x) \, dx \leqslant \left( \int f(x)^3 \, dx\right)^{1/3} \left( \int 1^{3/2} \, dx\right)^{2/3}.\] It immediately follows that \(R \leqslant 1\) and we can have equality when \(f\) is (for example) constant. [In fact, since we ar...
- Sat Jul 09, 2016 5:37 am
- Forum: Real Analysis
- Topic: A known exercise
- Replies: 1
- Views: 1599
Re: A known exercise
Making the substitution \(y = x^3\) we obtain \[ I = \frac{1}{3} \int_0^1 y^{-2/3}(1-y)^{-1/2} \, dy.\] It is well known that \[ B(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}.\] So \(I = \frac{\Gamma(1/3)\Gamma(1/2)}{3\Gamma(5/6)}.\) A theorem of Chebyshev says ...
- Thu Jul 07, 2016 1:51 pm
- Forum: Calculus
- Topic: \(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
- Replies: 3
- Views: 3258
Re: \(\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\)
We begin by observing that \[ \frac{1}{(2n-1)(3n-1)(4n-1)} = \frac{2}{2n-1} - \frac{9}{3n-1} + \frac{8}{4n-1}.\] We now define \[ f(x) = \sum_{n=1}^{\infty} \left( \frac{2x^{12n-6}}{2n-1} - \frac{9x^{12n-4}}{3n-1} + \frac{8x^{12n-3}}{4n-1} \right).\] This is a power series which converges absolutely...
- Thu Jul 07, 2016 1:42 pm
- Forum: Analysis
- Topic: A class of alternate infinite series
- Replies: 2
- Views: 3115
Re: A class of alternate infinite series
Let us begin by recalling that the Beta function satisfies \[ B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt = \frac{(x-1)!(y-1)!}{(x+y-1)!}\] whenever \(x,y\) are positive integers. In our case we have \[ \sum_{n \geqslant 1} \frac{(-1)^{n-1}}{n(n+1) \cdots (n+k)} = \frac{1}{k!} \sum_{n\geqslant 1} (-1...
- Thu Jul 07, 2016 1:18 pm
- Forum: Real Analysis
- Topic: Exponential Product in the Unit Cube
- Replies: 3
- Views: 2714
Re: Exponential Product in the Unit Cube
George, I have also thought about this. Ypur inequalities are correct. I only include an outline of the proof: Let \[U_m = \left( \frac{1}{m^{n-1}},m\right)^n \subseteq \mathbb{R}^n.\] Suppose we want to maximize \((1+x_1)^{1/x_1} \cdots (1+x_n)^{1/x_n}\) on \(\overline{U_m}\) subject to the conditi...
- Thu Jul 07, 2016 1:16 pm
- Forum: Real Analysis
- Topic: Exponential Product in the Unit Cube
- Replies: 3
- Views: 2714
Re: Exponential Product in the Unit Cube
The claim is false for every \(n \geqslant 4.\) Take \(x_1 = \cdots = x_{n-1} = \tfrac{1}{m}\) and \(x_n = m^{n-1}\) for some \(m\). Then \[ (1+x_1)^{\frac{1}{x_1}} \cdots (1+x_n)^{\frac{1}{x_n}} = \left(1 + \frac{1}{m} \right)^{(n-1)m}(1+m^{n-1})^{\frac{1}{m^{n-1}}}\] which tends to \(e^{n-1}\) as ...
- Thu Jul 07, 2016 12:29 pm
- Forum: Calculus
- Topic: Derivative of a Power Series
- Replies: 2
- Views: 2746
Re: Derivative of a Power Series
Just to add that the evaluation of the derivative at \(\pi/2\) is allowed since within the circle of convergence we can determine the derivative by differentiating term by term.
- Thu Jul 07, 2016 12:19 pm
- Forum: Analysis
- Topic: Non periodic function!
- Replies: 3
- Views: 3779
Re: Non periodic function!
I claim that \( \sin(p(x))\) is periodic if and only if \(p\) is linear (or constant). The "if" part is obvious. For the "only if" part we work similarly as my answer above. If it was periodic then its derivative \(p'(x)\cos(p(x))\) would also be periodic and since it is continuo...