3. Compute the Taylor series of f(x) = √ x about c = 1. Hint: The first two terms don’t fit the pattern of the remaining terms, so you will need to write those two terms out before writing the rest of the series compactly as an infinite series. This one is a bit messier than most of the examples we’ve seen up to this point.


Answer:

Required expression is,T_n=1+\frac{(x-1)}{2}-\frac{(x-1)^2}{2}+\sum_{n=3}^{\infty}\frac{(-1)^{n+1}(x-1)^n(2n-3)}{2\times n!}.

Step-by-step explanation:

Given function is,

f(x)=\sqrt{x}

we have to find Taylor series about x=c which is in formula,

T_n=\sum_{n=0}^{\infty}\frac{f^{(n)}}{n!}(x-c)^n

=f(c)+\frac{f'(c)}{1!}(x-c)^1+\frac{f''(c}{2!}(x-c)^2+\frac{f'''(c)}{3!}(x-c)^3+........\hfill (1)

Now, in this problem,

f'(x)=\frac{1}{2}x^{-\frac{1}{2}}\implies f'(1)=-\frac{1}{2}

f''(x)=-\frac{1}{4}x^{-\frac{3}{2}}\implies f''(1)=-\frac{1}{4}

f'''(x)=\frac{3}{8}x^{-\frac{5}{2}}\implies f'''(1)=\frac{3}{8}

f^{(iv)}=-\frac{15}{16}x^{-\frac{7}{2}}\implies f^{(iv)}(1)=-\frac{15}{16}

and so on. After substitute these values in (1) we will get,

T_n=1+\frac{(x-1)}{2}-\frac{(x-1)^2}{2}+\sum_{n=3}^{\infty}\frac{(-1)^{n+1}(x-1)^n(2n-3)}{2\times n!}

The first two terms don't fit the pattern of the remaining terms, since because of derivative the exponents are changes for every derivative of f(x). Hence the result follows.


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