Taylor polynomial approximation- Help!

bcjochim07

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

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

bcjochim07

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

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

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