Using Maclaurin Polynomials to Evaluate Trigonometric Functions at f(0.1)

[francisg3]

I am having some difficulty with a homework problem I was recently assigned. The problem says to "Replace each trigonometric function with its third Maclaurin polynomial and then evaluate the function at f(0.1)"

This is what I have done so far:

f(x)=(x cosx- sinx)/(x-sin⁡x)
1st trig function f(x)=x cos⁡x
f'(x) =cos⁡x-x sin⁡x
f''(x) =-sin⁡〖x-sin⁡x-x cos⁡x 〗
f'''(x) =-cos⁡〖x-cos⁡x-cos⁡x 〗+x sin⁡x f'''(0) =-3

3rd Maclaurin Polynomial-3/3! x^3

2nd trig functionf(x)=-sinx
f'(x)=-cosx
f''(x) =sinx
f'''(x) =cosx = f'''(0) =1

3rd Maclaurin Polynomial(1/3!) x3


Replacing those values in the original function I get:
(-3/3! x^3+1/3! x^3)/(1+1/3! x^3 )

I should get an answer around -2 I believe however this is not the case. Any help would be appreciated.

Thanks!

 

[SammyS]

I am having some difficulty with a homework problem I was recently assigned. The problem says to "Replace each trigonometric function with its third Maclaurin polynomial and then evaluate the function at f(0.1)"

This is what I have done so far:

f(x)=(x cosx- sinx)/(x-sin⁡x)
1st trig function f(x)=x cos⁡x
f'(x) =cos⁡x-x sin⁡x
f''(x) =-sin⁡〖x-sin⁡x-x cos⁡x 〗
f'''(x) =-cos⁡〖x-cos⁡x-cos⁡x 〗+x sin⁡x f'''(0) =-3

3rd Maclaurin Polynomial-3/3! x^3

2nd trig functionf(x)=-sinx
f'(x)=-cosx
f''(x) =sinx
f'''(x) =cosx = f'''(0) =1

3rd Maclaurin Polynomial(1/3!) x3


Replacing those values in the original function I get:
(-3/3! x^3+1/3! x^3)/(1+1/3! x^3 )

I should get an answer around -2 I believe however this is not the case. Any help would be appreciated.

Thanks!

Hi Francis.

I put in bold an important instruction in the above quote.

\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}-\dots

\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots

f(x)= \frac{x \cos(x)- \sin(x)}{x-\sin(x)}

You probably meant third degree MacLaurin polynomial, in which case, use only the first two terms in each of the above.

 

[francisg3]

it works, thank you very much.