(e^(-x²))...from there I went to e^(-x²)≡ [(-x²)^n/n!]= [(-1)^n*x^(2n)/n!] so (e^(-x²))= [!]= [(-1)^n*x^(2n)/n!] dx]...but I am stuck on where to go from here if there is route

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- March 31st 2008, 06:15 PMMathstud28Ok I am drawing a blank
(e^(-x²))...from there I went to e^(-x²)≡ [(-x²)^n/n!]= [(-1)^n*x^(2n)/n!] so (e^(-x²))= [!]= [(-1)^n*x^(2n)/n!] dx]...but I am stuck on where to go from here if there is route

- March 31st 2008, 06:16 PMMathstud28Sorry
It should read (e^(-x²))= [(-1)^n*x^(2n)/n!] dx]...

- March 31st 2008, 06:17 PMKrizalid
Well it's just a direct application of Maclaurin series for

- March 31st 2008, 06:18 PMMathstud28This is embarrassing
I've learned it but forgotten (Giggle) could you elaborate

- March 31st 2008, 06:20 PMTheEmptySet
- March 31st 2008, 06:21 PMMathstud28Just checking
but that holds true because n! and (-1)^n are constants and you just yanked them out evaluated your integral and then reintroduced them?