Answer to Question #50265 in Complex Analysis for nany

Question #50265

Find the sum of the power series

n from 0 to ∞ ∑ {(-1)^n / (2n)! } . ( z- { pi/2} ) ^2
afterward compute the sum of the series
n from 0 to ∞ ∑ 1 / (2n)!

Expert's answer

Download Answer

Learn more about our help with Assignments: Complex Analysis

Comments

Assignment Expert

14.01.15, 17:45

Dear vallle. We do not agree that ∑ 1 / (2n)! (the series with positive terms) can be directly obtained from the series ∑ [ (-1)^n. ( z-{pi/2})^n ] / [ (2n)!] (the series with alternating terms). We used the known expansion of hyperbolic cosine function.

Assignment Expert

14.01.15, 17:41

Dear vallle. Thank you for adding information.

vallle

06.01.15, 20:53

B: ∑ 1 / (2n)! means it substituted value (z) in above ∑ [ (-1)^n. ( z-{pi/2})^n ] / [ (2n)!] to get ∑ 1 / (2n)! So should find z value by by 1 / (2n)! = [ (-1)^n. ( z-{pi/2})^n ] / [ (2n)!] then only we should substitute this value in the above Sum to get the final sum of B

vallle

06.01.15, 20:53

For A i need begin from the starter Cos z= ∑ [ (-1)^n ( z)^2n ] / [ (2n)!] Goal :Get the Given power series ∑ [ (-1)^n. ( z-{pi/2})^n ] / [ (2n)!] Should do steps { on both side } but focus in right side to get the given power series, Such { replace z, or multiply ,or dividing or differentiate ,or integral ) each step { on both side } Solution : A Replace each z with : z-(pi /2) in both side, Cos [ z-(pi /2) ] = ∑ [ (-1)^n. ( z-{pi/2})^2n ] / [ (2n)!] then ??

Answer to Question #50265 in Complex Analysis for nany

Comments

Leave a comment

Related Questions