Recent content by Sherlock1

$\displaystyle \frac{\cancel{\prod a_{k}}y}{\cancel{\prod a_{k}}}=y$ where $\displaystyle \prod a_{k}\ne0.$ Works beautifully for the big stuff now indeed. Good job Jameson as always. (Rock)

\rlap{/}x seems to work for stuff like $\displaystyle \frac{\rlap{/}x}{x\cdot \rlap{/}x} = \frac{1}{x}. $ Not sure about crossing out big stuff.

I'm looking forward to the Fourier transform method (unless you meant Laplace (Giggle)).

[FONT=Verdana] Let $\displaystyle I(\lambda) = \int_{0}^{\infty}\frac{\sin(\lambda x)}{x(1+x^2)}\;{dx} ~~ (\lambda > 0). $ Differentiating this w.r.t. $\lambda$ we have $\displaystyle I'(\lambda) = \int^{\infty}_{0} \frac{\cos(\lambda x) }{1+x^2}\;{dx}.$ Also $\displaystyle I''(\lambda) =...

Yes. (Rofl) It can be motivated this way. Basically we want the indexes to match as the first ranges over $[1,~ k+1]$ but the second ranges over $[0, ~k]$. So we range the first over $[0, ~1]$ and subtract the term at $i = 0$ and add the term at $i = k+1$. The fact that in this case both of...

$\displaystyle \binom{r}{k} = \left \{\begin{array}{cc} \frac{r(r-1)\cdots(r-k+1)}{k(k-1)\cdots 1} = \frac{r^{\underline{k}}}{k!}, &\mbox{ integer } & k \ge 0; \\0, & \mbox{integer} & k < 0\\ \end{array} \right.$ Where $r \in\mathbb{C}$. So $\displaystyle \binom{k-1}{-1} = 0$ and likewise...

I need to learn combinatorics. Combinatorial proofs come off as elegant and more informative than algebraic ones.

I've rechecked my steps and what I've written seems perfectly fine. Can you please point to where you exactly think there's a problem? :)

Call our sum $S$. Pascal's rule states that $\displaystyle {k\choose i} = {k-1\choose i} + {k-1\choose i-1}$, therefore $\displaystyle S = \sum_{i=0}^{k}{k-1\choose i}(-1)^i + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i$, putting $i \mapsto i-1$ for the first one: $$\begin{aligned} S & =...

This blows my mind: Buffon's needle. Specially the fact that you can get an estimate for $\pi$ by dropping needles!

Welcome. :] I fixed your latex. I hope the following is what you were after. Using the product formula...$\begin{aligned} \bigg(\sum_{n=0}^{\infty} (-1)^n n^{-3/4}\bigg) \bigg(\sum_{n=0}^{\infty} (-1)^n n^{-1/4}\bigg) = \sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^k k^{-3/4}(-1)^{n-k}(n-k)^{-1/4} =...

Thanks. Could you explain bit more as to how inverting the limits voids the negative, please? I can't see it. I know it cancels the immediate one before the integral, but in the end we still end-up with $-\pi$? Thanks!

Re: Something really weird What does it give when you enter $\displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}$ (both terms under the radical), though? Thanks.

Re: Something really weird You still don't get me. Let me illustrate what I mean by way of a solution: Let $x = \frac{7}{2}+t$ then we have:$\displaystyle \begin{aligned} I & = \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx} = \int_{-\frac{1}{2}}^{\frac{1}{2}}...

Re: Something really weird Sorry, the first post should have read: Wolfram agrees that the first integral is $\pi$ (see my link above). But as you have seen the second integral is $-\pi$. Why is this?