# Real Analysis Of Taylor Polynomial

I have a random function $$f(x)$$. I know nothing about the shape of the function, but I know

• $$f'(x)$$ exists, $$x \in \mathbb{R}$$
• $$f(3)=1$$, $$f'(3)=5$$, $$f''(3)=7, f''''(3)=3$$
• If $$x\in (3,6)$$, then $$f'(x) \in[1,5]$$
• $$-3\leq f''(x)\leq 9, \forall x \in \mathbb{R}$$
• $$-4 \leq f'''(x) \leq -1, \forall x \in \mathbb{R}$$

So having this information, how do I determine the second degree taylor polynomial, $$T_{2, 3}$$ for this function? I know the value of $$f'(3)$$ up to $$f''''(3)$$, which is perfect for this question. But the taylor polynomial theorem requires the function is differentiable everywhere in order for this to work right? Do I need to prove that first, and how should I prove it?

Also, how do we find the upper bound for the error using $$T_{2,3}$$ for f(5)? Using taylor's remainder theorem is fine but which points should I choose for $$f'''(c)$$?

• comment: It is more common to use notation $f^{(4)}(x)$ instead of $f''''(x)$ to denote $4th$ derivative of $f(x)$ Dec 9 '20 at 23:40
• That is true. Also ignore "If x∈(3,6), then f′(x)∈[1,5]", this does not have much to do with the question, as I figured out. Dec 10 '20 at 0:11
• Can someone plz help me with this question? Dec 10 '20 at 4:38

Second degree taylor polynomial: $$1+5(x-3)+\frac{7}{2}(x-3)^2$$

Upper bound on error for $$f(5)$$: $$\frac{16}{3}$$

We already know the function is differentiable everywhere, because $$f'(x)$$ exists $$x \in R$$

Because the function is differentiable, we can write out its Taylor series expansion (centered at 3 because that's where we know the derivatives, and up to a degree of 2 because we're only trying to find the second degree Taylor polynomial).

$$f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2=1+5(x-3)+\frac{7}{2}(x-3)^2$$

To find the upper bound on the error, we use Taylor's Remainder Formula:

if $$\left|f^{(k+1)}(z)\right|\leq M$$ for all z between x and a, then $$\left|R_k(x)\right|\leq \frac{M}{(k+1)!}|x-a|^{k+1}$$ Where $$R_x$$ is the error in the approximation

So in our case $$k=2$$, $$\left|f^{(3)}(z)\right|\leq M = 4$$ (because $$-4 \leq f'''(x) \leq -1, \forall x \in \mathbb{R}$$), $$x=5$$, and $$a=3$$, so $$\left|R_2(x)\right|\leq \frac{4}{(2+1)!}|5-3|^{2+1}=\frac{16}{3}$$