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Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

Sagot :

MrRoyalidn

Answer:

[tex]\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n[/tex]

Step-by-step explanation:

Given

[tex]f(x) = \sin x\\[/tex]

[tex]c = \frac{3\pi}{4}[/tex]

Required

Find the Taylor series

The Taylor series of a function is defines as:

[tex]f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n[/tex]

We have:

[tex]c = \frac{3\pi}{4}[/tex]

[tex]f(x) = \sin x\\[/tex]

[tex]f(c) = \sin(c)[/tex]

[tex]f(c) = \sin(\frac{3\pi}{4})[/tex]

This gives:

[tex]f(c) = \frac{1}{\sqrt 2}[/tex]

We have:

[tex]f(c) = \sin(\frac{3\pi}{4})[/tex]

Differentiate

[tex]f'(c) = \cos(\frac{3\pi}{4})[/tex]

This gives:

[tex]f'(c) = -\frac{1}{\sqrt 2}[/tex]

We have:

[tex]f'(c) = \cos(\frac{3\pi}{4})[/tex]

Differentiate

[tex]f"(c) = -\sin(\frac{3\pi}{4})[/tex]

This gives:

[tex]f"(c) = -\frac{1}{\sqrt 2}[/tex]

We have:

[tex]f"(c) = -\sin(\frac{3\pi}{4})[/tex]

Differentiate

[tex]f"'(c) = -\cos(\frac{3\pi}{4})[/tex]

This gives:

[tex]f"'(c) = - * -\frac{1}{\sqrt 2}[/tex]

[tex]f"'(c) = \frac{1}{\sqrt 2}[/tex]

So, we have:

[tex]f(c) = \frac{1}{\sqrt 2}[/tex]

[tex]f'(c) = -\frac{1}{\sqrt 2}[/tex]

[tex]f"(c) = -\frac{1}{\sqrt 2}[/tex]

[tex]f"'(c) = \frac{1}{\sqrt 2}[/tex]

[tex]f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n[/tex]

becomes

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

Rewrite as:

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

Generally, the expression becomes

[tex]f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n[/tex]

Hence:

[tex]\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n[/tex]

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