# It is possible to demonstrate the taylor's formula (Peano) with the taylor's formula (Lagrange)?

I wonder if it is possible demonstrate the taylor's formula in the peano form of the reminder with the taylor's formula in the lagrange formula of the reminder.

In symbols: $$f(x)-F(x)=\frac{f(\xi)}{n!}(x-x_0)^n\implies f(x)-F(x)=o(x-x_0)^n$$ with F(x) the taylor's series.

I've tried different times but I have not been able to demonstrate.

• It's probably worth writing out your Taylor series, so we know to what degree you are approximating for instance – preferred_anon Jun 4 '15 at 21:55
• You are currently misquoting the Lagrange version. After correcting, please make sure the two formulas both give information about the remainder at the same place. After you do that, it will look as if things are simple but they are not. I don't believe one can get the Peano form directly from the Lagrange form. – André Nicolas Jun 4 '15 at 22:11

Taylor series with Lagrange's form of remainder: If $f$ is differentiable $n$ times in some neighborhood $I$ of $x_{0}$ then $$f(x) = P(x) + \frac{f^{(n)}(\xi)}{n!}(x - x_{0})^{n}$$ where $$P(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + \cdots + \frac{f^{(n - 1)}(x_{0})}{(n - 1)!}(x - x_{0})^{n - 1}$$ for all $x \in I$ and $\xi$ is a point between $x$ and $x_{0}$.
Taylor series with Peano's form of remainder: If $f$ is differentiable $n$ times at a point $x_{0}$ then there is a neighborhood $I$ of $x_{0}$ such that $$f(x) = Q(x) + o((x - x_{0})^{n})$$ for all $x \in I$ and $$Q(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + \cdots + \frac{f^{(n)}(x_{0})}{n!}(x - x_{0})^{n}$$ It should now be obvious that both the polynomials $P(x)$ and $Q(x)$ are different and the hypotheses of the theorems are different in each case.
Note however that if $f$ satisfies the hypotheses of the theorem with Lagrange form of remainder then it also satisfies the hypotheses of the other theorem. I guess this is the reason OP has asked the question whether Lagrange's form of remainder implies the Peano's form of remainder.
In order that this is possible we must be able to show that $$\frac{f^{(n)}(\xi)}{n!}(x - x_{0})^{n} = \frac{f^{(n)}(x_{0})}{n!}(x - x_{0})^{n} + o((x - x_{0})^{n})$$ i.e. $$f^{(n)}(\xi) - f^{(n)}(x_{0}) \to 0\text{ as }x \to x_{0}$$ Note that as $x \to x_{0}$ the number $\xi$ has no choice but to tend to $x_{0}$. However this necessarily does not imply that $f^{(n)}(\xi)$ also tends to $f^{(n)}(x_{0})$. This is guaranteed only when $f^{(n)}$ is continuous at $x_{0}$. So it does not seem possible to derive Peano's form of remainder via the use of Lagrange's form of remainder without any additional constraints.