How to evaluate A Taylor series at a specific value

05122018, 04:00 PM
Post: #12




RE: How to evaluate A Taylor series at a specific value
Bernard's example used an extra, undocumented 4th paramter: polynom. This appears to return the expansion without the O term. If we use his syntax, it works:
taylor(sin(x),x,5,polynom) = x1/6*x^3+1/120*x^5 taylor(sin(x),x,5,polynom)x=(1/2) = 1841/3840 = 0.479427083333 Ceci n'est pas une signature. 

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