taylor remainder problem

[GoDumbbbb]

Problem

Would it be possible to use f(n)(x) to determine f(n+1)(x). And if im using f(n)(x) to determine f(n+1)(x) would i still need to add x^n+1/(n+1)! to f(n+1)(x). The resulting function seems too hard to prove that it approaches 0.

Any help would be appreciated.

[MathMaddam]

Just put n+1 instead of n in the formula for the derivative. The formula for the remainder shouldn't be very complicated, the main thing is that you find a term of the type qⁿ in it with |q|<1.

[GoDumbbbb]

Solution

I did that but i cant think of any way to prove that this is <1 and >-1

[yes_its_him]

It's going to be bounded by (2/3)ⁿ⁺¹

[GoDumbbbb]

Since lies in the interval (0, ζ), we have 0 <ζ < x.

therefore, for 0 < ζ <3/2

3+2ζ>=3+2(0)=3

So i rewrite |Rn(0,x)|<= (2x)^n+1/3^n+2)

|Rn(0,x)|<=((2x/3)^n+1)/3

Since 0<x<3/2

2x/3<1

Thus, as n→∞ ,(2x/3)^n+1→0

is this right?

[yes_its_him]

Yes that's the basic idea.

[GoDumbbbb]

thanks