Question

Help with Maclaurin Series.

I am expanding the function:

f(x)=(x-2)^4
and I get (remembering that after the 4th derivative the value function becomes 0):

f(x)=16-32x-24x^2-8x^3+x^4

Which is right by wolfram alpha: http://www.wolframalpha.com/input/?i=series+expansion+%28x-2%29%5E4

Now I wish to check:
f(x)=(x-2)^4
f(5)=(5-2)^4
f(5)=81

f(x)=16-32x-24x^2-8x^3+x^4
f(5)=16-(32×5)-(24×5^2 )-(8×5^3 )+5^4
f(5)=-1119

Can someone please help me explain this discrepancy? I cannot logically explain it, as the function does not have any more derivatives.

Thanks in advance.

Answer 1

so you have a maclaurin series centered at 2

Answer 2

:s

Answer 3

can you give me the actual question because i can't tell what discrepancy you are asking for

Answer 4

Using the Maclauren series, find a four term polynomial approximation for the following:

f(x)=(x-2)^4

Use numerical values to check for the accuracy of your findings.

I used the numerical values and there was a huge discrepancy, by using x as 5. I cannot logically explain it.

Answer 5

you're forgetting what the sum of the function is

Answer 6

\[\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}|x-a|^n\]

Answer 7

you're missing the n!

Answer 8

:s I'm working with Maclaurin series, not Taylor Series.

I'm not sure. I'll just do it again without looking at any of my previous work and see how I go.

Answer 9

Maclaurin series are a specific Taylor series

Answer 10

Taylor series are centered at a whereas maclaurin's are centered at 0

Answer 11

so for a maclaurin series it is

\[\sum_{n=0}^\infty \frac{f^n(0)}{n!}|x-0|^n\]

Answer 12

I've got no idea :L

I can evaluate other functions using the same method I am trying to apply here, it's just not working.

Answer 13

for this function in particular.

Answer 14

let me see if i do it my way i get an approximate error

Answer 15

Okay :) thank you

Answer 16

Aaah, I found my error,

Thank you anyway. One of my expansions was negative rather than positive. (remember squaring a negative number becomes positive.

f(x)=16-32x-24x^2-8x^3+x^4
Should be:
f(x)=16-32x+24x^2-8x^3+x^4