Here we explore Taylor things for the function

Let us say right from the start that this function is not weird or ugly, it
is a nice function. After all, the expression in the definition is just a composition of an exponential
(the best function around) and a power, there is nothing fishy about
*e*^{-x-2}.*f* at 0 in such a way
that it becomes continuous. It looks like this.

When we ask about the usual properties, we find that this function pretty much has them all, it is about the nicest function possible, since it has continuous derivatives of all orders everywhere. Because this will be important below, we will prove this assumption and actually give a better specification of this result.

Claim.

For every positive integerkthere is a polynomialof degree at most P_{k}(x)2 such that for non-zerok- 2xwe have

Such a statement is usually proved using mathematical induction.

*k* = 1: We find the first derivative in the usual way and just
out of curiosity, we also show the second derivative.

In both cases it works, the form is as prescribed in the Claim.

Now we should prove the induction step. We assume that this Claim is true
for *k* and we will show that then it must be also true for
*k* + 1.

By our assumption,
*P*_{k}) 2*k* - 2,*P*_{k}') 2*k* - 3*P*_{k+1}) 2*k* = 2(*k* + 1) - 2,

The estimate on the degree of the numerator is important for the following
reason. We actually just need to know that its degree is less than the
degree in the denominator, so when we divide the polynomial, there will only
be terms of the form
*x*^{n}.*Q*_{k}
such that the
*k*-th*Q*_{k}(1/*x*)*e*^{-1/x2}.

Claim.

For every positive integerkwe have

Indeed, it is enough to apply substitution

Then repeated l'Hôpital's rule yields zero, see
scale of powers,
similarly we work out the limit at 0 from the left.
This substitution can also simplify differentiating *f*, see Remark at
the end.

This claim shows that the function *f* also has
derivatives of all orders at the origin and they are zero. This means that
the graph of this function is extremely flat at the origin. Recall that if
we want to find the Taylor polynomial/series with center 0, we use exactly
these derivatives to find the coefficients. Thus we get the following
conclusion.

Fact.

The functionfhas Taylor polynomials of all orders with centerand the Taylor series with center a= 0and a= 0

What does this mean? First, since the given function is not zero except
at the origin, it follows that (apart from the center 0) the Taylor
polynomials do not approximate the original function *f* well, in fact
they do not approximate it at all. This is despite the fact that *f*
is smooth, with derivatives of all orders everywhere, so from the usual
point of view it is "nice".

Concerning the series, the Taylor polynomials converge - as partial sums of
the series - uniformly on the whole real line to *T*. Thus
the series is again the best possible, but it is definitely not
equal to the original function *f*. Recall that we had a theorem that
guaranteed good convergence of series if *f* has uniformly bounded
derivatives on some neighborhood. This is obviously not true here, there
is no neighborhood of 0 on which all derivatives would be uniformly bounded.

This shows that the problem of approximation of functions by power series can be quite tricky even for nice functions.

**Remark:**
The substitution
*y* = 1/*x*

*y*' = -1/*x*^{2} = -*y*^{2}.

Now we just apply the chain rule to find that

*f* '(*x*) = [*e*^{-y2}]' = (-2*y**e*^{-y2})·*y*' = 2*y*^{3}*e*^{-y2}.

Similarly

*f* ''(*x*) = [2*y*^{3}*e*^{-y2}]' = (6*y*^{2} - 4*y*^{4}*e*^{-y2})·*y*' = (4*y*^{6}-6*y*^{4})*e*^{-y2}.

When we substitute
*y* = 1/*x**k*-th*Q*_{k}(*y*)*e*^{-y2}*k*.