how do i use a taylor series centered at some x value to approximate the value of the function centered at a different x value?
for example, if im given some taylor series centered at 5 of f(x) but i want to find f(3), how do i do this? (btw that example is not my homework problem, i'm just trying to understand the method)
thanks alot!
It depends on the radius of convergence. If the series has a radius of convergence of ∞ such as the series for sin(x), cos(x), and exp(x), you can evaluate f(x) where x equals any defined value, even though the series are expanded about an arbitrary value x0. In the three examples cited above, they are expanded around 0.
If the radius of convergence is small, and the desired evaluation point is not within the radius of convergence, then you'll need to expand the series about an appropriate point. But then you'll also need to calculate f'(x0), f"(x0), .... but that's part of the game of series expansion.
To use a Taylor series centered at one value to approximate the value of a function at a different value, you can use a technique called Taylor series translation or Taylor series shift. Here's how you can do it:
1. Start with the Taylor series expansion of the function f(x) centered at the given x-value (in your example, centered at 5). Let's say the Taylor series is:
f(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...
2. Identify the known term in the Taylor series expansion. In this case, it is f(a), where a is the given x-value (in your example, a = 5).
3. Determine the difference between the desired x-value and the given x-value. In your example, the desired x-value is 3, and the given x-value is 5. So, the difference is (3 - 5) = -2.
4. Replace every occurrence of (x - a) in the Taylor series expansion with the difference calculated in the previous step. In this case, replace (x - 5) with -2:
f(x) = f(5) + f'(5)(x - 5) + f''(5)(x - 5)^2/2! + f'''(5)(x - 5)^3/3! + ...
becomes:
f(x) = f(5) + f'(5)(-2) + f''(5)(-2)^2/2! + f'''(5)(-2)^3/3! + ...
5. Simplify the expression. Replace each derivative term (like f'(5), f''(5), etc.) with the corresponding value obtained by evaluating the derivative at the given x-value. Use the known values from the Taylor series expansion to calculate the approximate value of f(x) at the desired x-value.
6. Evaluate the expression using the known values from the Taylor series expansion and the calculated values for the derivatives at the given x-value. This will give you an approximation for f(x) at the desired x-value.
Note that the accuracy of the approximation depends on how well the Taylor series represents the behavior of the function around the given x-value.
Remember, the Taylor series expansion is an infinite series, but for practical purposes, you can truncate the series after a certain number of terms if you only need a specific level of accuracy.