Taylor polynomials

Janiffer · 2008-04-26 03:30:08

how do u find the taylor polynomial of degree 9 of f(x)= e^x about x=0?

luca-deltodesco · 2008-04-26 03:41:45

Taylor series, and expansions for a finite termed polynomial are generalisations of McLaurin series/expansions.

As you are taking it about x=0, what you are actually looking for is the mclauring polynomial of degree 9.

The mclaurin series for a function is found by:

where

denotes the r'th derivitive

The taylor series is a generalisation that takes it to be:

you can see the mclaurin series is just the taylor series with a=0

you see the wording ' about x=0 ' because the way these work is that at some x=a, x=0 for mclauring the series defines a polynomial who's n'th derivitive at x=a is equal to the n'th derivative of the function being evaluated, which for most function with a large enough polynomial can give exact values for the function for any x, some functions like logarithms are exceptions

-------

so for your question:

you should know that the derivitive of e^x is e^x, hence the n'th derivitive of e^x is e^x and e^0 = 1, so its mclaurin series is

which to the 9th degree is:

Last edited by luca-deltodesco (2008-04-26 03:42:32)

mathsyperson · 2008-04-26 06:25:45

Another way of thinking about the MacLaurin series is:

(Can't figure out how to stop the interval going on its own line )
That's also true if you alter it to be a Taylor series.
So the first part is the first n+1 terms of the MacLaurin sum, and the second term is the error in that approximation (which usually goes to 0 as n increases).

In the case of the 9th degree polynomial for e^x, that would be:

Janiffer · 2008-04-26 12:29:49

thnks a lot for the reply!!

Janiffer · 2008-04-27 01:26:25

Also if you had to find an approximate value for e... will 7^1/2 be useful??

luca-deltodesco · 2008-04-27 01:56:57

ehm, not really.

if you wanted to find an approximate value for e from an mclaurin or taylor expansion, substitute the value of x needed into the expansion. so from:

since e^1 = e

Math Is Fun Forum

#1 2008-04-26 03:30:08

Taylor polynomials

#2 2008-04-26 03:41:45

Re: Taylor polynomials

#3 2008-04-26 06:25:45

Re: Taylor polynomials

#4 2008-04-26 12:29:49

Re: Taylor polynomials

#5 2008-04-27 01:26:25

Re: Taylor polynomials

#6 2008-04-27 01:56:57

Re: Taylor polynomials

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