Question

(using taylor polynomials) Calculate ln(2.5) to 7 correct digits.
(hint:use ln(e)=1, e is about 2.718)

Answer 1

expand about \(e\)?

Answer 2

\(f(3)=1\) is the constant, then \(f'(x)=\frac{1}{x}\) so \(f'(e)=\frac{1}{e}\) we start with
\[f(x)=1+\frac{1}{e}(x-e)\]

Answer 3

thats what i am thinkin x=e, but i dont know if my answer is correct. My prof, dosent believe in giving out the answers. even to check work

Answer 4

next term would be
\[f''(x)=-\frac{1}{x}\] so \[f''(e)=\frac{1}{e^2}\] we get
\[1+\frac{1}{e}(x-e)-\frac{1}{2e^2}(x-e)^2+...\]

Answer 5

if you ask me it is kind of silly since you cannot do this without a calculator anyway. might as well just find \(\ln(2.5)\)

Answer 6

can i ask a question? why do i need to approximate ln 2.5 if i get a number when i plug it into a calc?

Answer 7

like i said, this is kind of silly
you cannot compute \(\frac{1}{e}\) without a calculator
lets try an experiment

Answer 8

do you know something about graphing the remainder of a taylor poylnomial?

Answer 9

no

Answer 10

but you would say that is the right approach right? evalauting the taylor poylnomial at e?

Answer 11

you wouldnt have by any chance taken numerical anaylsis?

Answer 12

http://www.wolframalpha.com/input/?i=1%2B%281%2Fe%29%28-.218%29-%281%2F%282e^2%29%29%28-.218%29^2

http://www.wolframalpha.com/input/?i=ln%282.5%29

Answer 13

i think we got close
i am pretty sure that is what was in mind for this problem

Answer 14

we would have to carry the expansion out further to get closer

Answer 15

how many terms do you think?

Answer 16

not sure, but the derivatives are easy enough, so i will be ok to do a couple more

Answer 17

cool, thanks man.

Answer 18

\[\ln(x)\]
\[\frac{1}{x}\]
\[-\frac{1}{x^2}\]
\[\frac{2}{x^3}\]
\[-\frac{6}{x^4}\] for the first ones so
\[1,\frac{1}{e},-\frac{1}{e^2},\frac{2}{e^3},-\frac{6}{e^4}\] for the successive derivatives

Answer 19

i think i got it. it looks like it weill take 6 terms

Answer 20

\[1+\frac{1}{e}(x-e)-\frac{1}{2e^2}(x-e)^2+\frac{1}{3e^3}(x-e)^3-\frac{1}{4e^4}(x-e)^4\] i think is right

Answer 21

doing that gives us 0.9162907

Answer 22

comparing that to the ln 2.5 i think that 7 digits do match up

Answer 23

but whats giving me fits is that i have to use the taylor remainder to prove that i have accuracy up to that 7th digit

Answer 24

pattern becomes clear you get
yeah looks like it .9162907318 according to wolfram

Answer 25

what is the bound on the remainder?

Answer 26

isn't it just the next term? or am i mistaken?

Answer 27

yeah its, k+1

Answer 28

so if we take it out this far, next one would be
\[\frac{1}{5e^5}(2.5-e)^5\]

Answer 29

as long as this is close enough to zero, you are fine

Answer 30

http://www.wolframalpha.com/input/?i=1%2F%285e^5%29%282.5-e%29^5
looks ok

Answer 31

i guess so. Thanks for your help. You wouldnt be familair with maple would you?