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Given the taylor series, find the function it derives from.?
Taylor series (centered at -1) is given by:
SIGMA{(n+1)/n}*(x+1)^n
SIGMA (the 'sum of' sign) from n=1 to n=infinity
what function centered at -1 does this series represent?
for the keen: prove that its interval of convergence is (-2,0)
-also any hints on how to use mathematical notation on yahoo answers?
2 Answers
- kbLv 79 years agoFavorite Answer
Note that Σ(n = 1 to ∞) (n+1) (x+1)^n/n
= Σ(n = 1 to ∞) n(x+1)^n/n + Σ(n = 1 to ∞) 1(x+1)^n/n
= Σ(n = 1 to ∞) (x+1)^n + Σ(n = 1 to ∞) (x+1)^n/n.
The first term is geometric with initial term x+1 and common ratio r = x+1.
So, it converges to (x+1)/(1 - (x+1)) = -(x+1)/x when |x+1| < 1.
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The second term comes from integrating a geometric series:
Start with
Σ(n = 0 to ∞) (1+x)^n = 1/(1 - (1+x)) = -1/x for |x+1| < 1.
Subtract 1 from both sides:
Σ(n = 1 to ∞) (1+x)^n = -1/x - 1 = -(1+x)/x
Divide both sides by (x+1):
Σ(n = 1 to ∞) (1+x)^(n-1) = -1/x for |x+1| < 1.
Integrate both sides:
Σ(n = 1 to ∞) (1+x)^n/n = -ln |x| for |x+1| < 1.
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So, the series equals -(x+1)/x - ln |x| for |x+1| < 1 <==> x is in (0, 2).
I hope this helps!
- wysongLv 45 years ago
attempt Googling "information of Taylor's Theorem". it ought to probably be more desirable powerful. I have seen the information before. Its no longer enormously complicated, even if it really is nearly inauspicious to shop song of each and every of the expressions. various algebraic manipulations and calculus manipulations. There are distinct variations of the information, even if.
