approximation with Taylor polynomial

[sambellamy]

I have the question:

f⁽ⁿ⁾(4) = (-1)ⁿ·n! / 3ⁿ(n+1)

"The Taylor series of f centered at 4 converges to f(x)for all x in the interval of convergence. Show that the fifth-degree Taylor polynomial approximates f(5) with an error of less that 0.0002."

I think the instructions here are telling me that this is f⁽ⁿ⁾(a) when a=4. I feel like "for all x in the interval of convergence" is redundant. I think they are asking me to find f⁽⁵⁾(5) and compare it to f⁽⁵⁾(4), and show that they are very close in value - am I reading this correctly?

I found f⁽⁵⁾(4) = (-1)⁵·5! / 3⁵·6

= -120/458
≈ -0.082305

How do I find f⁽⁵⁾(5)?

 

[HallsofIvy]

No, they are NOT asking you to find \(\displaystyle f^{(5)}(5)\). This specifically says "Show that the fifth-degree Taylor polynomial approximates f(5) with an error of less that 0.0002." Use the "error formula" for a fifth degree Taylor polynomial.

(And "The Taylor series of f centered at 4 converges to f(x)for all x in the interval of convergence." is NOT redundant. There exist functions whose Taylor's polynomial converge but NOT to the function value.)

 

[Ishuda]

I have the question:

f⁽ⁿ⁾(4) = (-1)ⁿ·n! / 3ⁿ(n+1)

"The Taylor series of f centered at 4 converges to f(x)for all x in the interval of convergence. Show that the fifth-degree Taylor polynomial approximates f(5) with an error of less that 0.0002."

I think the instructions here are telling me that this is f⁽ⁿ⁾(a) when a=4. I feel like "for all x in the interval of convergence" is redundant. I think they are asking me to find f⁽⁵⁾(5) and compare it to f⁽⁵⁾(4), and show that they are very close in value - am I reading this correctly?

I found f⁽⁵⁾(4) = (-1)⁵·5! / 3⁵·6

= -120/458
≈ -0.082305

How do I find f⁽⁵⁾(5)?

Do you know what a Taylor series expansion of a function is? It says, without the convergence criterion, etc.,
f(x) = \(\displaystyle \Sigma_{n=0}\frac{f^{(n)}(x_0)}{n!} (x-x_0)^n\)

There are several things you can use to estimate the error in stopping at a particular place in the (possibly) infinite series. Since this is an alternating series for x=5, you can use the alternating series test or you could use the integral bounds or ... The page
http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx
talks about several of these. It appears as though you could have been on your way to the comparison test but stopped way short.

 

[sambellamy]

OK, so I am familiar with the Taylor series expansion. For "Show that the fifth-degree Taylor polynomial approximates f(5)...", I am to add up the first five in the Taylor sequence, from a₁ to a₅, correct? And then show that adding a₆ to that sum is less than 0.0002 different in value?

I guess I also still do not understand how to get from f⁽ⁿ⁾(4) to f⁽ⁿ⁾(5).

 

[sambellamy]

OK, I now see that in the Taylor polynomial notation, f⁽ⁿ⁾(4) refers to f⁽ⁿ⁾(a), and they want me to use 5 for x in the (x-a) in the Taylor expansion. From then I can use the ASET to show that if the magnitude of the n=6 term is smaller than 0.0002, we're good.

Thanks for your help!