Question

Taylor Series question,
http://i.imgur.com/5HJYk3X.jpg
Find \[T_{5}(x)\], the Taylor polynomial of degree 5 of the function f(x) = \cos (x)\] at a = 0.
\[T_{5}(x)\] =

Find all value of x for which this approximation is within 0.000796 of the right answer. Assume for simplicity that we limit ourselves to |x| < or = 1
|x| < or = ?

Answer 1

f(x) ... not \(\int\)(x)

Answer 2

oh yeah whoops haha

Answer 3

there is a remainder setup for these .... that i vaguely recall:\[S_\infty=S_n+R_{n}\]
\[S_\infty-S_n=R_{n}\]

Answer 4

i knew it would be in there somewhere :)

Answer 5

the remainder part is rly complicated

Answer 6

it tends to boil down to:\[S_n+\int_{n+1}^{\inf}f(x)~dx~<~S~<~S_n+\int_n^{\inf}f(x)~dx\]

Answer 7

so we need to determine "n" such that the integration from (n to inf) is <= .000796

or the integration from (n+1 to inf) is >= .000796

Answer 8

\[\int_{n}^{inf}f(x)~dx=F(inf)-F(n)\]
\[F(inf)-F(n)<=.000796\]
\[F(inf)-.000796<=F(n)\]
\[F^{-1}[F(inf)-.000796]<=n\]
looks good in theory :)

Answer 9

another, more brute method, is simply to test out values and compare them ....

Answer 10

thats too brutal haha

Answer 11

\[cos(x)-(1-\frac12x^2+\frac1{24}x^4)<=.000796\]
\[cos(x)-1+\frac12x^2-\frac1{24}x^4<=.000796\]
\[cos(x)+\frac12x^2-\frac1{24}x^4<=1.000796\]

Answer 12

|x|<= 1 seems right to me :/

Answer 13

tried, and it said its incorrect :(

Answer 14

i noticed the red :)

Answer 15

\[\sum_3\frac{(-1)^n}{(2n)!}x^{2n}<=.000~796\]
when x=0 thats 0, when x=1 we have
\[\left|\sum_3\frac{(-1)^n}{(2n)!}\right|\approx .001~364~36\]
which is bigger than the error

Answer 16

.91366 seems to be it, but i cant see a nice algebraic way to get there

Answer 17

its correct but how did you get that?