How do you calculate f(7) and f(9)(7) for the series f(x)?

[harrietstowe]

Homework Statement

f(x) = the series (3^n * (x-7)^(2n+1))/(5+n) from n=2 to $$\infty$$
f(7)=?
f⁽⁹⁾(7) =?

Homework Equations

The Attempt at a Solution

I have no idea how to do this. I get how to do it if that 7 is a 0 but I am lost for this problem.

 

[Dick]

x-7 is a 0. You are evaluating at x=7. What's the ninth derivative of (x-7)^7 evaluated at x=7? Same question for (x-7)^8. Same question for (x-7)^9. Same question for (x-7)^10. Same question for (x-7)^11. If you got these right then you should see a pattern. There's only one term in your series that contributes to the ninth derivative.

 

[harrietstowe]

ok so is the answer to the second part 362880 because that is what the ninth derivative of (x-7)^9 equals? If not can you show me precisely how you would get the answer. At this point I just need to memorise how to do these

 

[Dick]

ok so is the answer to the second part 362880 because that is what the ninth derivative of (x-7)^9 equals? If not can you show me precisely how you would get the answer. At this point I just need to memorise how to do these

The answer to the ninth derivative of (x-7)^9 is 9!=362880. And if you answered 0 to the rest of my questions, then you should see the point. The answer to your question isn't that, because you still have the 3^n and the (5+n) to deal with. Don't memorize. Try to understand.

 

[harrietstowe]

ok so do I make that part (3^9)/(5+9) and multiply it by 9! for the answer?

 

[Dick]

ok so do I make that part (3^9)/(5+9) and multiply it by 9! for the answer?

Almost. Except n isn't 9. (2n+1)=9. What's n?