Asked by Marvin Gay
HELP! I'm in calculus class and they keep talking about Taylor series. I don't understand what's going on. Please explain what is Taylor series and what is Mclaurin series and how they are similar and how they are different.
Answers
Answered by
Marvin Gay #2
Essentially, the Taylor series is the default equation whereas the Mclaurin series is one in which the p can change. Whereas both are used to write nonzero terms, only one has different p values.
Answered by
Marvin Gay #3
Essentially the two series serve the same purpose. The only difference between the two is that in Mclaurin series the center is always zero.
Answered by
Reiny
Taylor series and the associated McLauran series are + x^5
series of infinite terms allowing us to find some common mathematical functions
eg
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + .... , x must be in radians
e^x = 1 + x + x^2/2! + x^3/3! + ....
ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...., -1 < x < 1
They must converge, even though some of them might diverge initially.
google Taylor series, there are some really good youtubes from the KhanAcademy
series of infinite terms allowing us to find some common mathematical functions
eg
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + .... , x must be in radians
e^x = 1 + x + x^2/2! + x^3/3! + ....
ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...., -1 < x < 1
They must converge, even though some of them might diverge initially.
google Taylor series, there are some really good youtubes from the KhanAcademy
Answered by
typo - Reiny
First sentence should say:
Taylor series and the associated McLauran series are series of infinite terms allowing us to find some common mathematical functions.
Taylor series and the associated McLauran series are series of infinite terms allowing us to find some common mathematical functions.
Answered by
Damon ?????
Both are the same idea
Taylor expands starting at some point where x = a
terms are something times (x-a)^n
Maclaurin assumes a = 0, terms are in x^n
thus Maclaurin is a special case of Taylor where we start at x = 0
Taylor expands starting at some point where x = a
terms are something times (x-a)^n
Maclaurin assumes a = 0, terms are in x^n
thus Maclaurin is a special case of Taylor where we start at x = 0
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