Different Taylor's Theorem form

Question

Assume that $f: \mathbb{R} \to \mathbb{R}$ is such that $f^\prime$ and $f^{\prime \prime}$ exist for all $x \in \mathbb{R}$. Taylor's Theorem tells us that, for each $a,h \in \mathbb{R}$ there is a $\theta \in \left( 0 , 1 \right)$ such that $$f \left( a + h \right) = f\left(a\right) + hf^\prime\left( a\right) + \frac{h^2}{2} f^{\prime \prime} \left( a + \theta h \right)$$ Write down the taylor expansions of $f \left( 0\right)$ and $f \left( 2 \right)$ about the point $x \in \left[0,2 \right]$, using the above form of Taylor's Theorem, with a remainder involving $f^{\prime \prime}$.

Please can someone explain where this form of Taylor's Theorem comes from and how to write down the taylor expansions of $f \left( 0\right)$ and $f \left( 2 \right)$.

Answer 1

If you denote $x = a+h$ (and consequently $h = x-a$), the formula reads $$ f(x)=f(a)+ f'(a) (x-a) + \frac 12 f''(\xi) (x-a)^2, \quad \xi \in [a,x] $$

Answer 2

Recall Laplace transform for derivatives:

$$ \mathcal{L}\{f'(a+t)\}=s\mathcal{L}\{f(a+t)\}-f(a) $$

We can inductively prove that this formula gives Laplace transform for higher order derivatives:

$$ \mathcal{L}\{f^{(n)}(a+t)\}=s^n\mathcal{L}\{f(a+t)\}-\sum_{k=0}^{n-1}s^{n-k-1}f^{(k)}(a)$$

Now, we can solve for $\mathcal{L}\{f(a+t)\}$:

$$ \mathcal{L}\{f(a+t)\}={1\over s^n}\mathcal{L}\{f^{(n)}(a+t)\}+\sum_{k=0}^{n-1}{1\over s^{k+1}}f^{(k)}(a) $$

We know that $\mathcal{L}\{t^m\}=m!/s^{m+1}$ for all $m\in\mathbb{Z}$, and convolution theorem:

$$ \mathcal{L}\{(f*g)(t)\}=\mathcal{L}\{f(t)\}\mathcal{L}\{g(t)\} $$

Hence, we can perform inverse Laplace transform on both side and get

$$ f(a+t)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}t^k+\left({t^{n-1}\over(n-1)!}*f^{(n)}(a+t)\right)(t) $$

Since convolution is defined by $(f*g)(t)=\int_0^tf(u)g(t-u)\mathrm{d}u$, we can rewrite our function by

$$ f(a+t)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}t^k+\int_0^t{f^{(n)}(a+u)\over(n-1)!}(t-u)^{n-1}\mathrm{d}u $$

Let us now substitute $a+t$ with $x$, so we can rewrite $f(x)$ by

$$ f(x)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}(x-a)^k+\int_0^{x-a}{f^{(n)}(u+a)\over(n-1)!}(x-a-u)^{n-1}\mathrm{d}u $$

At last, we perform substitution: $\tau=u+a$

$$ f(x)=\sum_{k=0}^{n-1}{f^{(k)}(a)\over k!}(x-a)^k+\int_a^x{f^{(n)}(\tau)\over(n-1)!}(x-\tau)^{n-1}\mathrm{d}\tau $$

Now, we let $N=n-1$, so we can make the expression neater:

$$ f(x)=\sum_{k=0}^N{f^{(k)}(a)\over k!}(x-a)^k+\int_a^x{f^{(N+1)}(\tau)\over N!}(x-\tau)^N\mathrm{d}\tau $$

The integral we obtain on the RHS is the exact remainder for $f(x)$'s $N$-degree Taylor polynomial. Since $(x-\tau)^N$ does not change sign within the interval of $(a,x)$, we apply the second mean value theorem for integrals to get an alternative form for the remainder:

There exists a value $\xi\in(a,x)$ such that

$$ R(x)\triangleq\int_a^x{f^{(N+1)}(\tau)\over N!}(x-\tau)^N\mathrm{d}\tau={f^{(N+1)}(\xi)\over N!}\int_a^x(x-\tau)^N\mathrm{d}\tau $$

Expand the integral on the RHS will give us the alternative Taylor remainder expression:

$$ R(x)={f^{(N+1)}(\xi)\over(N+1)!}(x-a)^{N+1} $$

Now, let $h=x-a$ and $\xi=a+h\theta$, we obtain

$$ \mathrm{Remainder}={f^{(N+1)}(a+h\theta)\over(N+1)!}h^{N+1} $$

Because $\xi\in(a,x)$, we conclude $\theta\in(0,1)$.