Is Taylor's Series the Key to Proving Differentiability for sinx/x?

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The discussion centers on proving the differentiability of the function f(x) = sin(x)/x for x ≠ 0 and f(0) = 1 using Taylor's series. Participants suggest approximating cos(xt) using its Taylor series expansion and integrating to derive f. The conversation emphasizes that f has a removable singularity at x = 0, and that all derivatives of f exist at all points. The integral representation and Taylor series are established as effective methods for demonstrating differentiability.

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  • Taylor series expansion of trigonometric functions
  • Understanding of removable singularities in calculus
  • Basic knowledge of differentiability and derivatives
  • Integral calculus, specifically integration of smooth functions
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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
 
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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
 
i still don't understand this, can you elaborate?

Thank You
 
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.
 
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.
 

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