Answer:
[tex]\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n[/tex]
Step-by-step explanation:
Rn(x) →0
f(x) = 10/x
a = -2
Taylor series for the function f at the number a is:
[tex]f(x) = \sum^\infty_{n=0} \frac{f^{(n)}(a)}{n!} (x - a)^n[/tex]
[tex]f(x) = f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f"(a)}{2!} (x-a)^2 + ...[/tex] ............ equation (1)
Now we will find the function f and all derivatives of the function f at a = -2
f(x) = 10/x f(-2) = 10/-2
f'(x) = -10/x² f'(-2) = -10/(-2)²
f"(x) = -10.2/x³ f"(-2) = -10.2/(-2)³
f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴
f""(x) = -10.2.3.4/x⁵ f""(-2) = -10.2.3.4/(-2)⁵
∴ The Taylor series for the function f at a = -4 means that we substitute the value of each function into equation (1)
So, we get [tex]\sum^\infty_{n=0} - \frac{10(x+2)^n}{2^{n+1}}[/tex] Or [tex]\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n[/tex]
Taylor series is a power series that gives the expansion of a function f (x) in the neighborhood of a point.
Taylor series is, [tex]f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+........[/tex]
[tex]f(x)=\sum_{n=0}^{\infty }-\frac{10(x+2)^{n}}{2^{n+1}}[/tex]
Here, f(x) = 1/x and a = -2
Now find derivative,
f(x) = 10/x f(-2) = 10/-2
f'(x) = -10/x² f'(-2) = -10/(-2)²
f"(x) = 10.2/x³ f"(-2) = 10.2/(-2)³
f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴
Substituting above values in Taylor series expansion.
We get, [tex]f(x)=\sum_{n=0}^{\infty }-\frac{10(x+2)^{n}}{2^{n+1}}[/tex]
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