Respuesta :
Answer:
Required expression is,[tex]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!}[/tex].
Step-by-step explanation:
Given function is,
[tex]f(x)=\sqrt{x}[/tex]
we have to find Taylor series about x=c which is in formula,
[tex]T_n=\sum_{n=0}^{\infty}\frac{f^{(n)}}{n!}(x-c)^n[/tex]
[tex]=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)[/tex]
Now, in this problem,
[tex]f'(x)=\frac{1}{2}x^{-\frac{1}{2}}\implies f'(1)=-\frac{1}{2}[/tex]
[tex]f''(x)=-\frac{1}{4}x^{-\frac{3}{2}}\implies f''(1)=-\frac{1}{4}[/tex]
[tex]f'''(x)=\frac{3}{8}x^{-\frac{5}{2}}\implies f'''(1)=\frac{3}{8}[/tex]
[tex]f^{(iv)}=-\frac{15}{16}x^{-\frac{7}{2}}\implies f^{(iv)}(1)=-\frac{15}{16}[/tex]
and so on. After substitute these values in (1) we will get,
[tex]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!}[/tex]
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.
