# Proving an inequality with Taylor polynomials

This is a homework question I was asked to do

Of a twice differentiable function $f : \mathbb{R} \to \mathbb{R}$ it is given that $f(2) = 3, f'(2) = 1$ and $f''(x) = \frac{e^{-x}}{x^2+1}$ . Now I have to prove that $$\frac{7}{2} \leq f\left(\frac{5}{2}\right) \leq \frac{7}{2} + \frac{e^{-2}}{40} .$$ I tried this by computing the third Taylor polynomial of $f$ near $a=2$, setting $x = \frac{5}{2}$, which gave me $$f(5/2) \approx 7/2 + \frac{e^{-2}}{40} - \frac{ - e^{-5/2}}{48}$$, but now I don't know what to do next. I guess one has to do something with finding the error of the first and second order Taylor polynomials, but I'm not sure how to do so. Can you help me?

Thanks in advance,

• Please don't include a "signature" in your posts; see the FAQ. Thank you. – Arturo Magidin Nov 22 '11 at 20:59
• try Lagrange's form of the error term with a linear approximation to $f(x)$ around $x = 2$. – Zarrax Nov 22 '11 at 21:07

## 3 Answers

Using a local linear approximation (that is, a degree 1 Taylor polynomial approximation), we have that $$f(x) \approx f(2) + f'(2)(x-2) = x+1.$$

Using the Lagrange Error Bound (with $n=1$) we have that $$\left| f(x) - (x+1)\right| \leq \frac{M}{2!}|x-2|^2$$ where $\max|f''(x)|\leq M$ on the interval between $2$ and $x$.

On the interval $[2,2.5]$, the function $f''(x) = \frac{e^{-x}}{x^2+1}$ is decreasing (and positive), since the derivative is $$-\frac{(x^2+1)e^{-x} + 2xe^{-x}}{(x^2+1)^2},$$ so we can take $M=f''(2) = \frac{e^{-2}}{5}$. Thus, the bound at $x=\frac{5}{2}$ is $$\frac{M}{2}\left(\frac{1}{2}\right)^2 = \frac{e^{-2}}{10}\left(\frac{1}{2}\right)^2.$$

Plugging into the Lagrange Error Bound and resolving the absolute value gives: $$-\left(\frac{e^{-2}}{10}\right)\left(\frac{1}{2}\right)^2 \leq f\left(\frac{5}{2}\right) - \frac{7}{2}\leq \frac{e^{-2}}{10}\left(\frac{1}{2}\right)^2$$ from which you should be able to deduce what you want.

• Thanks a lot! Ok, I feel I should be able to deduce what I want but I can't figure it out completely. I can now find that f(5/2) <= e^-2/40 + 7/2 but I can't deduce the inequality on the left from the information you have provided me with. – Max Muller Nov 22 '11 at 22:17
• @Max: Since the second derivative is positive, the function is concave up; that means that the tangent lies under the graph of $f$. That means that the tangent line approximation is an underestimate of $f(x)$. Since the tangent line approximation for $f(5/2)$ is $(5/2)+1 = 7/2$, that means that $f(5/2)\geq 7/2$. – Arturo Magidin Nov 22 '11 at 22:38

The calculation can be done in the following way: $f'''(x)=(f''(x))'= -\frac{e^{-x}}{x^2+1} -\frac{2xe^{-x}}{(x^2+1)^2}$ which at $x=2$ yields $f'''(2)=-\frac{9e^{-2}}{25}$ and so \begin{eqnarray*} f(5/2) & = & f(2)+f'(2)(5/2-2) + f''(2)(5/2-2)^2/2! +f'''(2)(5/2-2)^3/3!+\dots \\ & = &3+1/2+e^{-2}/40+ \frac{-3e^{-2}}{50}+\dots \end{eqnarray*} and the remaining terms are smaller than $\frac{3e^{-2}}{50}$ so you obtain your inequalities.

Start with $f^\prime(x) = f^\prime(2) + \int_2^x f^{\prime\prime}(y) \mathrm{d} y = 1 + \int_2^x \frac{\exp(-u)}{1+u^2} \mathrm{d} u$. Then $$\begin{eqnarray} f(x) &=& f(2) + \int_2^x f^\prime(z) \mathrm{d} z = 3 + \int_2^x \left( 1 + \int_2^z \frac{\exp(-u)}{1+u^2} \mathrm{d} u \right) \mathrm{d} z \\ &=& 3 + (x-2) + \int_2^x \int_2^z \frac{\exp(-u)}{1+u^2} \mathrm{d} u \mathrm{d} z \end{eqnarray}$$

Since the double integral is a non-negative quantity (as an integral of non-negative function), it follows $f\left( \frac{5}{2} \right) \ge 3 + \left( \frac{5}{2} - 2\right) = \frac{7}{2}$.

On the other hand, since $\frac{\exp(-u)}{1+u^2}$ is decreasing for $u>0$: $$\begin{eqnarray} \int_2^\frac{5}{2} \int_2^z \frac{\exp(-u)}{1+u^2} \mathrm{d} u \mathrm{d} z &\le& \int_2^\frac{5}{2} \int_2^z \frac{\exp(-2)}{1+2^2} \mathrm{d} u \mathrm{d} z = \int_2^{\frac{5}{2}} \frac{\exp(-2)}{5} (z-2) \mathrm{d} z \\ &=& \frac{1}{5 \mathrm{e}^{2}} \cdot \left. \frac{1}{2} (z-2)^2 \right|_2^\frac{5}{2} = \frac{1}{5 \mathrm{e}^{2}} \cdot \frac{1}{8} = \frac{1}{40 \mathrm{e}^{2}} \end{eqnarray}$$ It, thus, follows that $$\frac{7}{2} \le f\left( \frac{5}{2} \right) \le \frac{7}{2} + \frac{1}{40 \mathrm{e}^{2}}$$