0
$\begingroup$

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,

  • 0
    try Lagrange's form of the error term with a linear approximation to $f(x)$ around $x = 2$.2011-11-22

3 Answers 3

3

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.

  • 0
    @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 $f(5/2)\geq 7/2$.2011-11-22
1

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.

1

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}} $