# How to show $f,g$ are equal up to n'th order at $a$.

I need help with this problem.

Let two real-valued functions $f,g$ be equal to the $n'$th order at $a$ if

$$\lim_{h\rightarrow 0}\frac{f(a+h)-g(a+h)}{h^n} = 0$$

If $f'(a),\ldots,f^n(a)$ exist, show that $f$ and the function $g$ be given by

$$g(x) = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i$$

are equal up to the $n$'th order at $a$.

I'm also given the following hint. The limit

$$\lim_{x\rightarrow a}\frac{f(x) -\sum_{i=0}^{n-1}\frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}$$

may be evaluated using L'Hospital's rule.

Can't I just plug in $g(a+h)$ and apply L'Hopital $n$ times? I'm not sure what to do with this hint.

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If you plug $x-a$ into $h$ in the definition of $n$th order, you end up evaluating that limit. You can then apply L'Hopital's Rule enough times to evaluate the limit.

UPDATE:

The sum only goes up to $n-1$. That's because you can separate out the $n$th term to get

$$-\frac{\frac{f^{(n)}(a)}{n!}(x-a)^n}{(x-a)^n}=-\frac{f^{(n)}(a)}{n!}$$

for $x\neq a$.

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I want to know how I'm supposed to use the hint I'm given. – Fred May 25 '12 at 1:19
@Fred My answer says that you can evaluate the limit that proves they are $n$th order by substituting in $x-a$ for $h$. Then you will have the limit in the hint. The hint says to use L'Hopitals to evaluate this new limit that is equal to the old limit. – Joe Johnson 126 May 25 '12 at 1:21
The sum goes up to $n-1$ though and not $n$. That's where I'm confused. – Fred May 25 '12 at 1:26
@Fred I added to my solution. – Joe Johnson 126 May 25 '12 at 1:46

The definition you are given is equivalent to

$$\lim_{x \to a} \frac{f(x)-g(x)}{(x-a)^n}=0$$

We are given $g$ defined as the sum

$$g(x) = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i$$

so we want to check wether or not

$$\lim_{x \to a} \frac{f(x)-\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}=0$$

Since the last term has $(x-a)^n$ in it, we can take it out and get

$$\lim_{x \to a} \frac{f(x)-\sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}-\frac{f^{(n)}(a)}{n!}=$$

The sum is a polynomial in $(x-a)$ of degree $n-1$, which, as well as $f$, has its $k^{th}$ derivative defined for $k=0,1,\dots,n$, so we can use L'Hôpital Bernoulli to find the indeterminate limit. Applying it $n-1$ times gives

$$\frac 1 {n!}\mathop {\lim }\limits_{x \to a} \frac{{{f^{\left( {n - 1} \right)}}(x) - {f^{\left( {n - 1} \right)}}\left( a \right)}}{{\left( {x - a} \right)}} - \frac{{{f^{(n)}}(a)}}{{n!}} =$$

And by definition of the derivative you get

$$\frac{{{f^{(n)}}(a)}}{{n!}} - \frac{{{f^{(n)}}(a)}}{{n!}} = 0$$

Note we can apply L'Hôpital in every $n-1$ step because it is the case in every step we face a $\frac 0 0$ indeterminate form.

What you are asked to prove is usually stated as

$$f(x) = T_n^a(x)+o((x-a)^n)\text{ ; when } x \to a$$

and means the error is of greater order (it goes to zero faster) than $(x-a)^n$ when $x \to a$. Though confusing, the meaning of "greater order" might be cleared up with this example: $(x-a)^{n+1}$ is of greater order than $(x-a)^n$, and thus goes to $0$ "faster" than it for $x \to a$, since

$$\mathop {\lim }\limits_{x \to a} \frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{{{\left( {x - a} \right)}^n}}} = \mathop {\lim }\limits_{x \to a} \left( {x - a} \right) = 0$$

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