Taylor Series of f(x) = 1/(1-6x) at c=6

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The discussion focuses on finding the Taylor Series for the function f(x) = 1/(1-6x) centered at c=6. The initial attempt includes a proposed formula for the nth derivative, but there is confusion regarding the correct expression for the Taylor series. Participants suggest starting from the definition of the Taylor series and recommend writing out the first few terms to identify a pattern. Clarification is needed on the function itself, as the original statement incorrectly refers to f(x) as 1/(1-6). The conversation emphasizes the importance of correctly applying the Taylor series formula and understanding the derivatives involved.
Soccerdude
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Homework Statement



Find the Taylor Series for f(x) = 1/(1-6) centered at c=6

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

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?
 
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Soccerdude said:

Homework Statement



Find the Taylor Series for f(x) = 1/(1-6) centered at c=6
Did you mean: $$f(x) = \frac{1}{1-6x}$$ ... from below, it appears so.

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
What leads you to believe that?
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?
Start from the definition of the Taylor series.
Try writing out the 1st 3-4 terms and see if you spot a pattern.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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