Question

Taylor Series Expansion.

I need to find the taylor expansion of e^(sinx) to an error of x^4.

I though I knew the answer, where sinx is plugged into the taylor series of e^x, but wolfram doesn't give me the answer I expected to see.

Can some one please show me the steps to get to it?

Answer 1

take 5 derivatives of e^(sinx) is one idea

Answer 2

im assuming "to an error of x^4" means that they want the first 5 terms?

Answer 3

its up to the term with x^3 then + O(x^4) which is the remainder (error).

but taking the derivatives is just too messy. i think im supposed to just plug it in the taylor of e^x which is \[\sum_{0}^{\infty}\frac{ x^{n} }{ n! }\]giving me:
\[\sum_{0}^{\infty} \frac{ \sin^{n}(x) }{ n! }\]

but idk why wolfram omitted the n = 3 term and why the fourth term looks the way it does

http://www.wolframalpha.com/input/?i=taylor+series+e^%28+sin+x%29

Answer 4

\[e^u\]
\[u'e^u\]
\[u''e^u+(u')^2e^u\]
\[u'''e^u+3u''u'e^u+(u')^3e^u\]

Answer 5

\[e^{sin0}=1\]
\[cos(0)e^{sin0}=1\]
\[-sin(0)e^{sin0}+(cos0)^2e^{sin0}=1\]
\[-cos0e^{sin0}-3sin0cos0e^{sin0}+(cos0)^3e^{sin0}=0\]

Answer 6

makes a lot of sense :) but what is wolfram showing me on the second line?

Answer 7

sin^n (x) seems fine, but trying to power a power series might seem a bit difficult to me

Answer 8

line 2 is the series expansion starting at f+f'+f''+f'''

Answer 9

\[\sum_0\frac{f^{(n)}(a)}{n!}(x-a)^n\]let a=0
\[\frac{f^{0}(0)}{0!}(x)^0+\sum_1\frac{f^{(n)}(a)}{n!}(x-a)^n\]
\[\frac{f^{0}(0)}{0!}(x)^0+\frac{f^{1}(0)}{1!}(x)^1+\sum_2\frac{f^{(n)}(a)}{n!}(x-a)^n\]
\[\frac{f^{0}(0)}{0!}(x)^0+\frac{f^{1}(0)}{1!}(x)^1+\frac{f^{2}(0)}{2!}(x)^2+\sum_3\frac{f^{(n)}(a)}{n!}(x-a)^n\]
\[\frac{f^{0}(0)}{0!}(x)^0+\frac{f^{1}(0)}{1!}(x)^1+\frac{f^{2}(0)}{2!}(x)^2+\frac{f^{3}(0)}{3!}(x)^3+\sum_3\frac{f^{(n)}(a)}{n!}(x-a)^n\]
\[1+x+\frac{1}{2!}x^2+\frac{0}{3!}x^3+...\]

Answer 10

why is f^3(0) = 0?

Answer 11

notice that there seems to be a nice convergence of e^x and e^sinx centered at x=0
http://www.wolframalpha.com/input/?i=y%3De%5Ex%2C+y%3De%5E%7Bsinx%7D

f'''(0) = 0, thats just the way the successive derivatives played out

Answer 12

at x=0, f''' = -1-0+1

Answer 13

i see now. thank you very much for the help! :)

Answer 14

youre welcome