Recent content by gumi_kr

So you have 2 sets of data: 1) calculations from your program 2) handmade calculations and you want to check it's accuracy, right? My thoughts: (i'm not advanced in statistics though): 1) define what's 'accurate' (5% error?) 2) If you want to then define 'Confidence interval' and do some...

Try typing 'finding eigenvalues' in google and you will find your answer there. It is very common thing and you shall find it in any advanced 'linear algebra' book. (ctrl+f + eigenvalue) The general idea of 'spectral decomposition' is to find eigenvalues and vectors associated with it...

i was too excited to write it properly ;) 1) choose a group of (2M+2-k) people (from 2M+2) 2) choose a leader between them (multiply B(2m+2,2M+2-k) by 2M+2-k) 3) keep a leader and 'give' rest of the people to him (number of people 2M+2 - (2M+2-k) = k) 3) now you've got k people and a leader...

Sorry if i misunderstood what you've written but I think you are trying to prove wrong equality. Your proof is correct for {n\choose k} but we want to proove it for k{n\choose k}.btw1.Notice that because in our example n is even, we've got odd numbers of elements in this row of Pascal's...

Oh! Thats simple! \sum_{p=M+2}^{2M+2}\binom{2M+2}{p}p = \sum_{p=0}^{M+1}\binom{2M+2}{p}p We could create a simple bijection between this two sets. Let's simplify it: We know that: \binom{2M+2}{2M+2 -k}(2M+2-k) = \binom{2M+2}{k+1}(k+1) Why?: 1) choose...

about chaoseverlasting idea: We want to proove, that: \sum_{p=0}^{M}\binom{2M+2}{p}(M+1-p)=(2M+1)\binom{2M}{M} 1) Of course: (2M+1)\binom{2M}{m} = \frac{1}{2}(M+1)\binom{2M+2}{M+1} So let's change it and multiple both sides by 2...

"you can do this because the terms from 0 to m+1 are equal to the terms from m+2 to 2m+2, hence the sum on the right is twice the sum from 0 to m+1." could you explain it to me. I know it's true before differentiating, but why after? (* well it's not true before differentiating ;), my mistake...

Hi! Big thanks! i will check it later but it looks correct ("you can do this because the terms from 0 to m+1 are equal to the terms from m+2 to 2m+2!" I'm not sure about it yet) It really took me a lot of time trying to solve it. (as i thought changing this expresion /*with R_M^P)/ to...

Hi! Does anybody know how to solve the following problem: \sum_{p=0}^{M}\binom{2M+2}{p}(M+1-p)=? Well, actually i know that the solution is: (2M+1)\binom{2M}{m}=\frac{2^{2M+1}\Gamma(\frac{3}{2}+M)}{\sqrt{\pi}\Gamma(M+1)} but i cannot prove it (mathematica calculates the first...

It was an usual exercise in probability theory course. It looks easy but I'm trying to solve this for a long time without success. 1) I know that trying simply to count it - isn't the right way (even using some advanced properties of binomial coefficient).2) It could be connected with Banach's...

Homework Statement Let R ^{M} _{P}= \sum_{s=0}^{P} {M+1 \choose s}, for 0 \leqslant P \leqslant M , P,M\in \mathbb{N}. Proove that: \sum_{q=0}^{M}R^{M}_{q}\cdot R^{M}_{M-q}=(2M+1) {2M \choose M} and give it's combinatorical idea. I'm trying to solve this for 3 days - please help..