MATH SOLVE

2 months ago

Q:
# Use the definition of taylor series to find the taylor series (centered atc.for the function. f(x) = cos x, c = π 4

Accepted Solution

A:

Cos(x) is infinitely differentiable, can be expanded using Taylor's series.

The series about c can be expressed as

[tex]f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(c)}{n!}(x-c)^n[/tex]

Substituting cos(x), c= π /4,we have

[tex]cos(x)=\sum_{n=0}^{\infty}\frac{f^{n}(\frac{\pi}{4})}{n!}(x-\frac{\pi}{4})^n[/tex]

and since

[tex]f'(\frac{\pi}{4})=-sin(\frac{\pi}{4})=-\sqrt{2}/2[/tex]

[tex]f"(\frac{\pi}{4})=-cos(\frac{\pi}{4})=-\sqrt{2}/2[/tex]

[tex]f^{(iii)}(\frac{\pi}{4})=sin(\frac{\pi}{4})=\sqrt{2}/2[/tex]

[tex]f^{(iv)}(\frac{\pi}{4})=cos(\frac{\pi}{4})=\sqrt{2}/2[/tex]

we can simplify the expansion to

[tex]cos(x)=\frac{\sqrt{2}}{2}(1-(x-\frac{\pi}{4})/1!-(x-\frac{\pi}{4})^2/2!+(x-\frac{\pi}{4})^3/3!+(x-\frac{\pi}{4})^4/4!-(x-\frac{\pi}{4})^5/5!-...)[/tex]

Note that the sign pattern is - - + + - - + + - - ..... following the sign pattern of the derivatives of cos(x).

The series about c can be expressed as

[tex]f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(c)}{n!}(x-c)^n[/tex]

Substituting cos(x), c= π /4,we have

[tex]cos(x)=\sum_{n=0}^{\infty}\frac{f^{n}(\frac{\pi}{4})}{n!}(x-\frac{\pi}{4})^n[/tex]

and since

[tex]f'(\frac{\pi}{4})=-sin(\frac{\pi}{4})=-\sqrt{2}/2[/tex]

[tex]f"(\frac{\pi}{4})=-cos(\frac{\pi}{4})=-\sqrt{2}/2[/tex]

[tex]f^{(iii)}(\frac{\pi}{4})=sin(\frac{\pi}{4})=\sqrt{2}/2[/tex]

[tex]f^{(iv)}(\frac{\pi}{4})=cos(\frac{\pi}{4})=\sqrt{2}/2[/tex]

we can simplify the expansion to

[tex]cos(x)=\frac{\sqrt{2}}{2}(1-(x-\frac{\pi}{4})/1!-(x-\frac{\pi}{4})^2/2!+(x-\frac{\pi}{4})^3/3!+(x-\frac{\pi}{4})^4/4!-(x-\frac{\pi}{4})^5/5!-...)[/tex]

Note that the sign pattern is - - + + - - + + - - ..... following the sign pattern of the derivatives of cos(x).