Re: Taylor Series

Clearbluesky a écrit :
exp(10^3) = (approx) 1 + 10^3 + (1/2)*10^6 + (1/6)*(10^9) +......

How many terms would i need to obtain exp(10^3) with error less than
1%?


You have Taylor-Lagrange inequality for positive x:

|exp(x)-(1+x+...+x^n/n!)| <= exp(x)x^(n+1)/(n+1)!

thus relative error is less than x^(n+1)/(n+1)!

You can try to find a correct n+1 using Stirling formula as a rough approximation of n!

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Mû
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