# Taylor polynomial approximation- Help!

## Main Question or Discussion Point

Use Taylor's theorem to determine the degree of the Maclaurin polynomial required for the error in the approximation of the function to be less than .001.

e^.3

I really, really don't know what to do for this one, and I have a quiz tomorrow. I have read through the section in the book, but I am frustrated and can't figure out what kind of method I should use to solve these types of problems.

lurflurf
Homework Helper
Do you know the remainder?
exp(x)=1+x+x^2/2+x^3/6+x^4/24+...+x^n/n!+R
R is the error and can be written different ways for different situations
ie
R=f^(n+1)(u)/(n+1)!
ie the formula can be made exact by evaluating the final term at x* with 0<=u<=x for some x* in the case of exp we have terms like x^k/k! so the remainder looks like
u^(n+1)/(n+1)!
it is usually hard to find u so we instead chose n so that u des not mater ie
we want n so that
u^(n+1)/(n+1)!<.001
we do not know u but we know (since exp is increasing on [0,.3])
u^(n+1)/(n+1)!<.3^(n+1)/(n+1)!
so that if
.3^(n+1)/(n+1)!<.001
then
u^(n+1)/(n+1)!<.001
so we do not need u, but we may end up with n larger than needed
solve
.3^(n+1)/(n+1)!<.001
for n
then the maclaurin polynomial of degree n has error<.001

So is the procedure to take the derivatives and plug in 0 (since c=0) and find an expression for the n+1 derivative?

f'(c) = 1 f''(c)=1 f'''(c) =1 ......

so the n+1 derivative is 1

So Rn= 1/(n+1)! * (.3) ^(n+1)

Then I set up an equality to find n so that Rn < .001

and n = 3 ???

lurflurf
Homework Helper
yes
we also want n an integer, so just round up
that is take the first n so that
.3^(n+1)/(n+1)!<.001

do you see how you might do
say
approximate
1/(1-x)
at x=1.1 to an error less than .005
or aproximate
sin(x) at x=.001 to an error less than .00001

lurflurf
Homework Helper
sorry we should have

Do you know the remainder?
exp(x)=1+x+x^2/2+x^3/6+x^4/24+...+x^n/n!+R
R is the error and can be written different ways for different situations
ie
R=f^(n+1)(u)/(n+1)!
ie the formula can be made exact by evaluating the final term at x* with 0<=u<=x for some x* in the case of exp we have terms like x^k/k! so the remainder looks like
e^u/(n+1)!
it is usually hard to find u so we instead chose n so that u des not mater ie
we want n so that
e^u/(n+1)!<.001
we do not know u but we know (since exp is increasing on [0,.3])
u^(n+1)/(n+1)!<e^.3/(n+1)!
so that if
e^.3/(n+1)!<.001
then
e^u/(n+1)!<.001
so we do not need u, but we may end up with n larger than needed
solve
e^u/(n+1)!<.001
for n
then the maclaurin polynomial of degree n has error<.001