Did you mean: $$f(x) = \frac{1}{1-6x}$$ ... from below, it appears so.Soccerdude said:Homework Statement
Find the Taylor Series for f(x) = 1/(1-6) centered at c=6
What leads you to believe that?Homework Equations
∞
Ʃ Fn(a)(x-a)/n!
n=0
The Attempt at a Solution
I believe that the nth derivative of 1/(1-6x) is
(-6)n-1n!/(1-6x)n+1
Start from the definition of the Taylor series.So i figured that the taylor series at c=6 would be
(-6)n-1(x-6)n/(1-6x)n+1
What am I doing wrong here?
The Taylor Series of f(x) = 1/(1-6x) at c=6 is an infinite series that represents the function at a specific point, c=6. It is used to approximate the function with a polynomial of infinite degree.
The formula for the Taylor Series of f(x) = 1/(1-6x) at c=6 is:
∑n=0 (fn(c)/n!) * (x-c)n
In this case, fn(c) represents the nth derivative of f(x) evaluated at c=6.
The Taylor Series of f(x) = 1/(1-6x) at c=6 can be used to approximate the function by evaluating the polynomial at a specific value of x. The more terms included in the series, the more accurate the approximation will be.
The interval of convergence for the Taylor Series of f(x) = 1/(1-6x) at c=6 is the range of x-values for which the series will converge and give an accurate approximation of the function. In this case, the interval of convergence is -1/6 < x < 1/6.
The Taylor Series of f(x) = 1/(1-6x) at c=6 is derived from the original function and represents it at a specific point, c=6. As more terms are included in the series, the approximation becomes closer to the original function. In the limit, as the number of terms approaches infinity, the series will be equal to the original function.