Questions Math
How do we find power series when given a function?
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f(x)=(e^(-x)^2) -1+(x^2) = 1 - x^2 + x^4/2! - x^6/3! + ... -1 + x^2 = x^4/2! - x^6/3! + x^8/4! - ...
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No, take a look at the definition of the Maclaurin series. It starts with f(0). very power series
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e^1 --> substitute 1 wherever you see x: e^1=1 + 1/1! + 1^2/2! + 1^3/3!+⋯,-∞<x<∞ Add as many
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Maclaurin series of function f(x) is a Taylor series of function f(x) at: a = 0 f(x) = f(0) + [
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Yes, your approach for part (b) is correct. To find the power series for f(x) = 1/(1+x)^3 using the
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To find the Taylor series about 0 for the function f(x) = (1+x)^(-3/5), we can use the binomial
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recall that ln(x) = -∑(-1)^k (x-1)^k/k so ln(7-x) = -∑(-1)^k (6-x)^l/k
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taking various derivatives, it is clear that 1/(1+x^2) = 1 - x^2 + x^4 - x^8 + ... for |x|<1 now,
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These two articles should point the way for you
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