Using Taylor series to show all derivatives of sinx/x exist

[tomboi03]

Let f(x)= sinx/x if x $$\neq$$ 0 and f(0)=1
Find a polynomial pN of degree N so that
|f(x)-pN(x)| $$\leq$$ |x|^(N+1)
for all x.
Argue that f is differentiable, f' is differentiable, f" is differentiale .. (all derivatives exist at all points).

I'm not sure about this one at all. Can you guys help me out?

Thank You

 

[lurflurf]

$$f(x)=\int_0^1 \cos(x t) dt$$
so approximate cos first and the integral for f with cos approximated will approximate f.
The derivatives clearly exist and
|(D^n)f|<1/n+1

 

[tomboi03]

i still don't understand this, can you elaborate?

Thank You

 

[lurflurf]

Since cos(x t) is smooth the integral will be as well.
Since
Cos(x)~1-x^2/2+x^4/24-x^6/720+...
is a family of approximations of cosine (each member being a sum the first n=1,2,3,... terms) we may repace cosine by an approximation in the integral representation of f to see that
f~1-x^2/6+x^4/120-x^6/5040+...
are approximations of f.

You function f at zero has what is called a removable singularity, a ficticious singularity that is caused by the representation, not by actual properties of the function. By representing the function differently (such as using the integral representation I gave) the singularity and any problems it may cause vanish.

 

[HallsofIvy]

Did you consider taking the Taylor's series for sin x, around x= 0, and dividing each term by x? That seems to me to be far simpler than using the integral form.