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Calculus/Taylor series

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Expert: - 3/10/2009


Question
I need help finding the Taylor series for f centered at 1 given f^(n)(1)= (-1/3)^n * (n+1)!/n^(3/2). Thanks for any help!

Answer

Taylor's Series
The Taylor's series around the point x=1 is this:
sum(0 to infinity)(x-1)^n*f^(n)(1)/n!.
Now (x-1)^n is (x-1) to the nth power.
f(n)(x) is the nth derivative of f evaluated at 1.
The number n! is n factorial (0!=1, by the way).  All other factorials can be found by n*(n-1)*(n-2) ....3*2*1.

Term 0: f(1);
Term 1: (x-1)*f'(1);
Term 2: (x-1)²*f"(1)/2;
Term 3: (x-1)³*f"'(1)/3! { yeah, that's f triple prime; 3!=6 };
Term 4: (x-1)^4*f""(1)/4! { 4th derivative at x=1; 4!=25 };
Term 5: (x-1)^5*f(5)(1)/5! { 5th derivative at x=1; 5!=120 };
Term 6: (x-1)^6*f(6)(1)/6! { 6th derivative at x=1; 6!=720 };
Term 7: (x-1)^7*f(7)(1)/7! { 7th derivative at x=1; 7!=5,040 };
etc.
I'm getting so I like sending pictures.  
They seem to get the point across.

That should be good.  You see how they go yet?

Scotto

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