Find Taylor series of the function?

Question

Find the Taylor series of the f(x)= sinx centered at a = (pie/2).
Q#2
a)Find the Taylor series of f(x) = (1/1-x) at a = 0. 
b)use the result to find taylor series of f(x) = 1/1-x^7 at a = 0 up to x^21.
c)estimate the value of  ∫ from -1 to 0 (1/1-x^7 dx )by approx. the integrand with a taylor polynomial obtained from part b.
d) Deduce from part a power series expansion for ln(1-x).
e) Using the result from part d to find the limit. lim as x approaces 0 (ln(1-x))/x.

kb · Accepted Answer

Q1)
f(x) = sin x ==> f(π/2) = 1
f'(x) = cos x ==> f'(π/2) = 0
f''(x) = -sin x ==> f''(π/2) = -1
f'''(x) = -cos x ==> f'''(π/2) = 0
f''''(x) = sin x ==> f''''(π/2) = 1
The derivatives start repeating themselves in cycles of 4.

Thus, 
sin x = 1 - (x - π)^2/2! + (x - π)^4/4! - ...
        = sum(n = 0 to ∞) (-1)^n (x - π/2)^(2n) / (2n)!.     

Q2) This is a long one...
a) 
f(x) = 1/(1 - x) ==> f(0) = 1
f'(x) = 1/(1 - x)^2 ==> f'(0) = 1
f''(x) = 2/(1 - x)^3 ==> f''(0) = 2
f'''(x) = 3!/(1 - x)^4 ==> f'''(0) = 3! = 6
...
f^(n)(x) = n!/(1 - x)^(n+1) ==> f^(n)(0) = n!.

Thus,
1/(1 - x) = Σ(n=0 to ∞) n! x^n/n! 
              = Σ(n=0 to ∞) x^n.

b) Replacing x with x^7:
1/(1 - x^7) = Σ(n = 0 to ∞) x^(7n).
               = 1 + x^7 + x^14 + x^21 + ...

c) ∫(-1 to 0) dx/(1 - x^7)
= ∫(-1 to 0) [Σ(n = 0 to ∞) x^(7n)] dx
= Σ(n = 0 to ∞) x^(7n+1)/(7n+1) {for x = -1 to 0}
= Σ(n = 0 to ∞) - (-1)^(7n+1)/(7n+1) 
= Σ(n = 0 to ∞) (-1)^n / (7n+1).
= 1 - 1/8 + 1/15 - 1/22 + ...

Using these first four terms, we have 0.89621...

d) Using a), integrating both sides from 0 to x yields
-ln(1 - x) = Σ(n = 0 to ∞) x^(n+1)/(n+1)
==> ln(1 - x) = (-1) * Σ(n = 0 to ∞) x^(n+1)/(n+1)

e) lim(x-->0) ln(1 - x) / x
= lim(x-->0) -(x + x^3/3 + ...) / x
= lim(x-->0) -(1 + x^2/3 + ...)  
= -1.

I hope this helps!

barnhart · Answer

sounds like a typo right here: 2-(x-a million) = 3-x not a million-x. yet to get the Taylor sequence at -a million, we could desire to have words of the form (x+a million), and a million-x does = 2-(x+a million). We use the geometric sequence right here: a million/(a million-u) = a million+u+u^2+u^3+.... the assumption is to rewrite a million/(a million-x) to look equivalent to that yet having a term of (x+a million): So a million-x = 2 - (x+a million) = 2[ a million - (x+a million)/2 ]. Now if we permit u = (x+a million)/2 we've 2[ a million - u ] So rewrite the preliminary difficulty as: 8*( a million / (2[a million-u])^3 ) = [a million/ (a million-u) ]^3 = [ a million+ u +u^2 +u^3 + ....]^3 = { a million + (x+a million)/2 + [(x+a million)/2]^2 + [(x+a million)/2]^3 + ....}^3

Anonymous · Answer

Please clarify where you have a problem with Taylor series.  Also, there are no pies in trig, just pi.

All questions are straightforward, no tricks involved.  Give it a try, post your answers and I'll tell you if they are correct or not.

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Find Taylor series of the function?

3 Answers