Question

Let Tn(x) be the nth Maclaurin polynomial for f(x) = ex as given in the text. Using the error bound formula from section 9.4, determine a value of n so that |Tn(2) − e2| < 10−4.

Answer 1

ex = e^x?

Answer 2

e^x yes

Answer 3

k one sec I know how to do it im trying to multi task though so just wait up plz.

Answer 4

so it would be e^2 later on in the question

Answer 5

Ok. I'm back. Do you know the mclaurin expansion for e^x?

Answer 6

no

Answer 7

oh wait yes!! haha my book has it written down

Answer 8

Okay... well, I'll try to get you to get it.
You know each term is going to be coeff*x^n, and that the derivative of the expansion must equal the expansion again.

Answer 9

What I mean is...\[e^x = a_0 x^{n_0}+ a_1 x^{n_1}+ a_2 x^{n_2}+ a_3 x^{n_3}...\]and\[D_x(e^x) =D_x( a_0 x^{n_0}+ a_1 x^{n_1}+ a_2 x^{n_2}+ a_3 x^{n_3}...)\]\[e^x = a_0n_0x^{n_0-1}= a_0 x^{n_0}+ a_1 x^{n_1}+ a_2 x^{n_2}+ a_3 x^{n_3}...\]

Answer 10

Sorry, the last line should be:
\[e^x = a_0n_0x^{n_0-1}...= a_0 x^{n_0}+ a_1 x^{n_1}+ a_2 x^{n_2}+ a_3 x^{n_3}...\]

Answer 11

yes

Answer 12

can u figure it out yet?

Answer 13

have you worked it out yet? i think the answer is n=12 but i may have done it wrong

Answer 14

i can't give u the answer straight up but im trying to lead you to it.. get the basics down first plz... what's the expansion?

Answer 15

well the equation for Error is \[\left| T_{n}x - f(x) \right|\le K ((\left| x-a \right|^{n+1})\div (n+1)!)\]

Answer 16

so K= 7.389 because e^2=7.389

Answer 17

Well I was going to get there, eventually...

Answer 18

But watch out K has to be the MAXIMUM possible value over the interval (0, 2), so since e^2 is technically greater than 7.389, I'd go with 7.4

Answer 19

ok so then i get \[\left| 2 \right|^{n+1}\div (n+1)! < 1.351\times10^{-5}\]

Answer 20

well then solve.

Answer 21

so i was right with n=12

Answer 22

well im not gonna bother checking that but u can with alpha if u like

Answer 23

could you post a link?

Answer 24

wolframalpha.com