Question

Calculus formula question! I'm working on finding the Lagrange form of the remainder for a summation of a Taylor Series... Can someone explain to me how to work this out? (details inside)

Answer 1

Taylor series centered at c=5, using the definition of a tailor series for \[f(x)=\frac{ 3 }{ x^4 }\]
My answer:
\[\sum_{n=0}^{\infty} (-1)^n \frac{ (n+3)(n+2)(n+1) }{ 2*4^{n+4} }(x-4)^n\]
On the Interval of Convergence (0,8)

So now I need to find the lagrange form for the remainder, using the formula:\[R _{n}(x) = \frac{ f^{n+1}(z) }{ (n+1)! }(x-c)^{n+1}\]

Can anyone explain the process of using this to me please? the book doesnt make much sense...

Answer 2

c=4* not 5.

Answer 3

@jazzeh What you wrote there is the Taylor series without the reminder. Taylor's theorem for f(x) centered at c=4 looks like this : \[\sum_{k=0}^{n}(-1)^{k}\frac{ (k+3)(k+2)(k+1) }{ 2*4^{k+4} }(x-4)^{k}+R _{n}(x)\] With the Lagrange form of the remainder being : \[R _{n}(x)=\frac{ f ^{n+1}(z) }{ (n+1)! }(x-c)^{n+1}=(-1)^{n+1}\frac{ (n+4)(n+3)(n+2) }{ 2*z ^{n+5} }(x-4)^{n+1}\] for some z between x and 4.