# Different Taylor's Theorem form

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 $$f \left( 2 \right)$$.

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

• Ok thank you! How do I write down the taylor expansions for f(0) and f(2) from this please? Commented Apr 22, 2020 at 19:25
• Nevermind I think I understand! Commented Apr 22, 2020 at 19:29

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 $$\theta\in(0,1)$$.

• Appreciate the help! Meant to be a course on real analysis (continuity and differentiability) so not sure this is applicable but thanks for taking the time! Commented Apr 22, 2020 at 19:28