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For what function?

∑ ((-1)^n*x^n)/n! is the Taylor series about zero for what function?
 
Answer: e^-x
 
 

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
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∑ (x^n)/n! = e^x, n from 0 to ∞
 
∑ ((-1)^n*x^n)/n!, n from 0 to ∞
= ∑ ((-x)^n)/n!, n from 0 to ∞
= e^-x

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When you learn Taylor expansion, ∑ (x^n)/n! = e^x, n from 0 to ∞ is generally one of the few examples. You can use the conclusion. But you can also prove it first.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
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Taylor Series at x = 0 is the Maclaurin Series.
 
It says that a function f(x) can be approximated by an infinite series:
f(x) = f(0) + f'(0) * x + f''(0) * x^2/2! + f'''(0) * x^3/3! + ...
 
If f(x) = e^x then all the derivatives are also e^x and at x = 0 are all 1.
 
So f(x) = 1 + x + x^2/2! + x^3/3! + ... = sum[n=0 to inf][x^n/n!] = e^x.
 
f(-x) = 1 - x + x^2/2! - x^3/3! + ... = sum[n=0 to inf][(-1)^n * x^n/n!] = e^(-x).