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 Complex Analysis Complex Analysis Math Forum

 March 10th, 2016, 04:59 AM #1 Newbie   Joined: Dec 2013 Posts: 4 Thanks: 0 Identity with binomial coefficients I have to show that $\displaystyle \sum \limits_{n=m}^\infty \binom{n}{m} x^{n-m} = \frac{1}{(1-x)^{m+1}}$ for $\displaystyle m \geq 0$ and $\displaystyle \forall x \in \mathbb{C}$ with $\displaystyle \|x\| <1$. If I compute the left side I get $\displaystyle \sum \limits_{n=m}^\infty \binom{n}{m} x^{n-m} = 1+\binom{m+1}{m}x+\binom{m+2}{m}x^2+...=1+(m+1)x+ \frac{(m+2)(m+1)}{2} x^2+\frac{(m+3)(m+2)(m+1)}{3*2}x^3+...$ but it doesn't help? Do you know what formula is needed to show the identity? March 10th, 2016, 06:35 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I would not expand the binomial like this. Instead change the index: let i= n-m. Then n= i+ m and when n= m, i= 0. The sum becomes $\displaystyle \sum_{i=0}^\infty \begin{pmatrix}i+ m \\ m \end{pmatrix} x^i$. Now, do you know a "formula" for simplifying $\displaystyle \begin{pmatrix}i+ m \\ m \end{pmatrix}$. Last edited by greg1313; March 10th, 2016 at 01:38 PM. March 10th, 2016, 07:24 AM #3 Newbie   Joined: Dec 2013 Posts: 4 Thanks: 0 Thank you. Then I have $\displaystyle \sum \limits_{i=0}^\infty \binom{i+m}{m}x^i= \sum \limits_{i=0}^\infty \binom{i+m}{i}x^i= ..?$ I know that $\displaystyle \sum \limits_{i=0}^\infty \binom{\alpha}{i}x^i=(1+x)^\alpha$ but my $\displaystyle \alpha$ in the sum above still depends on i which does not allow applying this equation. $\displaystyle ?...= \sum \limits_{i=0}^\infty ( \binom{i+m-1}{i-1}+\binom{i+m-1}{i} ) x^i$ doesn't help either or what formula did you mean? Last edited by hubolmen; March 10th, 2016 at 07:27 AM. March 10th, 2016, 01:02 PM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Start with the binomial series: $\displaystyle (1+z)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}z^k$ Let $z=-x$: $\displaystyle (1-x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}(-x)^k$ Let $\alpha=-m-1$: $\displaystyle \dfrac{1}{(1-x)^{m+1}}=\sum_{k=0}^\infty\binom{-m-1}{k}(-x)^k$ $\displaystyle \binom{-m-1}{k}=\dfrac{-(m+1)\cdot(-(m+1)-1)\dots(-(m+1)-(k-1))}{k!}=(-1)^k\binom{m+k}{k}$ $\displaystyle \dfrac{1}{(1-x)^{m+1}}=\sum_{k=0}^\infty\binom{-m-1}{k}(-x)^k=\sum_{k=0}^\infty\binom{k+m}{k}x^k$ Last edited by greg1313; March 10th, 2016 at 01:37 PM. March 22nd, 2016, 01:05 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 I see greg1313 has already done this. If you want to see it slightly differently: 1/(1-x)^n = 1 + nx + [n(n+1)/2!]x^2 + ... 1/(1-x)^(m+1) = 1 + (m+1)x + [(m+1)(m+2)/2!]x^2 + ...... 1/(1-x)^(m+1) = 1 + (m+1)Cmx + (m+2)Cmx^2 + ......... nCm = n!/m!(n-m)! (m+n)Cm=(m+n)!/m!n!= (m+1)(m+2)..(m+n)/n! Tags binomial, coefficients, identiti, identity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shikharras Calculus 4 January 1st, 2014 02:41 AM shikharras Calculus 2 January 1st, 2014 01:18 AM eddybob123 Number Theory 0 July 16th, 2013 03:44 PM shikharras Probability and Statistics 1 December 31st, 1969 04:00 PM shikharras Probability and Statistics 1 December 31st, 1969 04:00 PM

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