soitgoes2019 said:Homework Statement
Find the Taylor Series for f(x)=1/x about a center of 3.
Homework Equations
The Attempt at a Solution
f'(x)=-x^-2
f''(x)=2x^-3
f'''(x)=-6x^-4
f''''(x)=24x^-5
...
f^n(x)=-1^n * (x)^-(n+1) * (x-3)^n
I'm not sure where I went wrong...
To find the Taylor series for 1/x around x=3, you can use the formula: f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ... + (f^n(a)/n!)(x-a)^n. In this case, a=3 and f(x)=1/x. So, the Taylor series for 1/x around x=3 becomes: 1/3 - (1/9)(x-3) + (2/81)(x-3)^2 - (5/729)(x-3)^3 + ... + ((-1)^n*n!)/3^(n+1)(x-3)^n.
The purpose of finding the Taylor series for 1/x around x=3 is to approximate the function 1/x near the point x=3. This can be useful in situations where the exact value of the function may be difficult to calculate, but an approximation is sufficient.
No, the Taylor series for 1/x around x=3 is only valid for values of x that are close to 3. As the distance from x=3 increases, the accuracy of the approximation decreases.
The number of terms needed in the Taylor series for 1/x around x=3 depends on the desired level of accuracy. Generally, the more terms used, the more accurate the approximation will be.
No, the Taylor series for 1/x around x=3 is an approximation and will never yield the exact value of 1/x. However, as more terms are added, the approximation will become closer to the exact value.