Answer:
a) [tex]\sum_{n=0}^{3} \frac{f^{(n)}(x)}{n!} (x-a)^n =ln(9)+\frac{2}{9}(x-4)-\frac{2}{81}(x-4)^2 +\frac{8}{2187}(x-4)^3[/tex]
b) [tex]R_3 (x)=|f_n (x)-T_3 (x)|\leq \frac{0.019282}{4!}(3.7)^4 =0.150573[/tex]
Step-by-step explanation:
1) Part a
[tex]f(x)=ln(1+2x)[/tex] [tex]a=4,n=3[/tex]
The Taylor's approximation would be given byt this formula
[tex]\sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} (x-a)^n[/tex] (1)
Since we need the approximation T3 we just need to use this formula:
[tex]\sum_{n=0}^{3} \frac{f^{(n)}(x)}{n!} (x-a)^n[/tex] (2)
So now we proceed to find the terms for the T3 approximation:
[tex]f(4)=ln(1+8)= ln(9)[/tex]
[tex]f'(x)=\frac{2}{1+2x}[/tex] [tex]f'(4)=\frac{2}{1+8}=-\frac{2}{9}[/tex]
[tex]f''(x)=-\frac{4}{(1+2x)^2}[/tex] [tex]f''(4)=\frac{-4}{(1+8)^2}=-\frac{4}{81}[/tex]
[tex]f'''(x)=\frac{16}{(1+2x)^3}[/tex] [tex]f'''(4)=\frac{16}{(1+8)^3}=-\frac{16}{729}[/tex]
So then using equation (2) we have
[tex]\sum_{n=0}^{3} \frac{f^{(n)}(x)}{n!} (x-a)^n =ln(9)+\frac{2}{9}(x-4)-\frac{2}{81}(x-4)^2 +\frac{8}{2187}(x-4)^3[/tex]
2) Part b
We can estimate the accuracy of the approximation [tex]|f(x)\approxT_3 (x)|[/tex] with the Taylor's inequality given by:
[tex]R_n (x)=|f(x)\approxT_3 (x)|\leq \frac{M}{4!}(x-4)^4[/tex] (3)
On this case M is such that [tex]|f^{(4)} (x)|\leq M[/tex] for [tex]3.7\leq x\leq4.3[/tex]
So we have:
[tex]|f^{(4)} (x)|=\frac{96}{(1+2x)^4}\leq \frac{96}{[1+2(3.7)]^4}=\frac{96}{4978.7136}=0.019282[/tex]
Based on this we can take [tex]M=0.019282[/tex] and
[tex]|x-4|\leq3.7[/tex] so then [tex]|x-4|^4 \leq(3.7)^4[/tex]
So using equation (3) we got that
[tex]R_3 (x)=|f_n (x)-T_3 (x)|\leq \frac{0.019282}{4!}(3.7)^4 =0.150573[/tex]
3) Part c
In order to plot [tex]R_3 (x)[/tex] we just need to plot:
[tex]R_n (x)\leq \frac{0.019282}{4!}|x-4|^4[/tex]
And the graph obtained is attached below.
