# 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?

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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

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.

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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.

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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 $$\frac{7}{2} \le f\left( \frac{5}{2} \right) \le \frac{7}{2} + \frac{1}{40 \mathrm{e}^{2}}$$

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