# $13^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament <br> Saturday 20 February 2010 <br> Calculus Subject Test 

1. [3] Suppose that $p(x)$ is a polynomial and that $p(x)-p^{\prime}(x)=x^{2}+2 x+1$. Compute $p(5)$.

Answer: 50 Observe that $p(x)$ must be quadratic. Let $p(x)=a x^{2}+b x+c$. Comparing coefficients gives $a=1, b-2 a=2$, and $c-b=1$. So $b=4, c=5, p(x)=x^{2}+4 x+5$ and $p(5)=25+20+5=50$.
2. [3] Let $f$ be a function such that $f(0)=1, f^{\prime}(0)=2$, and

$$
f^{\prime \prime}(t)=4 f^{\prime}(t)-3 f(t)+1
$$

for all $t$. Compute the 4th derivative of $f$, evaluated at 0 .
Answer: 54 Putting $t=0$ gives $f^{\prime \prime}(0)=6$. By differentiating both sides, we get $f^{(3)}(t)=4 f^{\prime \prime}(t)-$ $3 f^{\prime}(t)$ and $f^{(3)}(0)=4 \cdot 6-3 \cdot 2=18$. Similarly, $f^{(4)}(t)=4 f^{(3)}(t)-3 f^{\prime \prime}(t)$ and $f^{(4)}(0)=4 \cdot 18-3 \cdot 6=54$.
3. [4] Let $p$ be a monic cubic polynomial such that $p(0)=1$ and such that all the zeros of $p^{\prime}(x)$ are also zeros of $p(x)$. Find $p$. Note: monic means that the leading coefficient is 1 .
Answer: $(x+1)^{3}$ A root of a polynomial $p$ will be a double root if and only if it is also a root of $p^{\prime}$. Let $a$ and $b$ be the roots of $p^{\prime}$. Since $a$ and $b$ are also roots of $p$, they are double roots of $p$. But $p$ can have only three roots, so $a=b$ and $a$ becomes a double root of $p^{\prime}$. This makes $p^{\prime}(x)=3 c(x-a)^{2}$ for some constant $3 c$, and thus $p(x)=c(x-a)^{3}+d$. Because $a$ is a root of $p$ and $p$ is monic, $d=0$ and $c=1$. From $p(0)=1$ we get $p(x)=(x+1)^{3}$.
4. [4] Compute $\lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n}|\cos (k)|}{n}$.

Answer: $\frac{2}{\pi}$ The main idea lies on the fact that positive integers are uniformly distributed modulo $\pi$. (In the other words, if each integer $n$ is written as $q \pi+r$ where $q$ is an integer and $0 \leq r<\pi$, the value of $r$ will distribute uniformly in the interval $[0, \pi]$.) Using this fact, the summation is equivalent to the average value (using the Riemann summation) of the function $|\cos (k)|$ over the interval $[0, \pi]$. Therefore, the answer is $\frac{1}{\pi} \int_{0}^{\pi}|\cos (k)|=\frac{2}{\pi}$.
5. [4] Let the functions $f(\alpha, x)$ and $g(\alpha)$ be defined as

$$
f(\alpha, x)=\frac{\left(\frac{x}{2}\right)^{\alpha}}{x-1} \quad g(\alpha)=\left.\frac{d^{4} f}{d x^{4}}\right|_{x=2}
$$

Then $g(\alpha)$ is a polynomial in $\alpha$. Find the leading coefficient of $g(\alpha)$.
Answer: $\frac{1}{16}$ Write the first equation as $(x-1) f=\left(\frac{x}{2}\right)^{\alpha}$. For now, treat $\alpha$ as a constant. From this equation, repeatedly applying derivative with respect to $x$ gives

$$
\begin{aligned}
(x-1) f^{\prime}+f & =\left(\frac{\alpha}{2}\right)\left(\frac{x}{2}\right)^{\alpha-1} \\
(x-1) f^{\prime \prime}+2 f^{\prime} & =\left(\frac{\alpha}{2}\right)\left(\frac{\alpha-1}{2}\right)\left(\frac{x}{2}\right)^{\alpha-2} \\
(x-1) f^{(3)}+3 f^{\prime \prime} & =\left(\frac{\alpha}{2}\right)\left(\frac{\alpha-1}{2}\right)\left(\frac{\alpha-2}{2}\right)\left(\frac{x}{2}\right)^{\alpha-3} \\
(x-1) f^{(4)}+4 f^{(3)} & =\left(\frac{\alpha}{2}\right)\left(\frac{\alpha-1}{2}\right)\left(\frac{\alpha-2}{2}\right)\left(\frac{\alpha-3}{2}\right)\left(\frac{x}{2}\right)^{\alpha-4}
\end{aligned}
$$

Substituting $x=2$ to all equations gives $g(\alpha)=f^{(4)}(\alpha, 2)=\left(\frac{\alpha}{2}\right)\left(\frac{\alpha-1}{2}\right)\left(\frac{\alpha-2}{2}\right)\left(\frac{\alpha-3}{2}\right)-4 f^{(3)}(\alpha, 2)$. Because $f^{(3)}(\alpha, 2)$ is a cubic polynomial in $\alpha$, the leading coefficient of $g(\alpha)$ is $\frac{1}{16}$.
6. [5] Let $f(x)=x^{3}-x^{2}$. For a given value of $c$, the graph of $f(x)$, together with the graph of the line $c+x$, split the plane up into regions. Suppose that $c$ is such that exactly two of these regions have finite area. Find the value of $c$ that minimizes the sum of the areas of these two regions.
Answer: $-\frac{11}{27}$ Observe that $f(x)$ can be written as $\left(x-\frac{1}{3}\right)^{3}-\frac{1}{3}\left(x-\frac{1}{3}\right)-\frac{2}{27}$, which has $180^{\circ}$ symmetry around the point $\left(\frac{1}{3},-\frac{2}{27}\right)$. Suppose the graph of $f$ cuts the line $y=c+x$ into two segments of lengths $a$ and $b$. When we move the line toward point $P$ with a small distance $\Delta x$ (measured along the line perpendicular to $y=x+c)$, the sum of the enclosed areas will increase by $|a-b|(\Delta x)$. As long as the line $x+c$ does not passes through $P$, we can find a new line $x+c^{*}$ that increases the sum of the enclosed areas. Therefore, the sum of the areas reaches its maximum when the line passes through $P$. For that line, we can find that $c=y-x=-\frac{2}{27}-\frac{1}{3}=-\frac{11}{27}$.
7. [6] Let $a_{1}, a_{2}$, and $a_{3}$ be nonzero complex numbers with non-negative real and imaginary parts. Find the minimum possible value of

$$
\frac{\left|a_{1}+a_{2}+a_{3}\right|}{\sqrt[3]{\left|a_{1} a_{2} a_{3}\right|}}
$$

Answer: $\sqrt{3} \sqrt[3]{2}$ Write $a_{1}$ in its polar form $r e^{i \theta}$ where $0 \leq \theta \leq \frac{\pi}{2}$. Suppose $a_{2}, a_{3}$ and $r$ are fixed so that the denominator is constant. Write $a_{2}+a_{3}$ as $s e^{i \phi}$. Since $a_{2}$ and $a_{3}$ have non-negative real and imaginary parts, the angle $\phi$ lies between 0 and $\frac{\pi}{2}$. Consider the function

$$
f(\theta)=\left|a_{1}+a_{2}+a_{3}\right|^{2}=\left|r e^{i \theta}+s e^{i \phi}\right|^{2}=r^{2}+2 r s \cos (\theta-\phi)+s^{2}
$$

Its second derivative is $\left.f^{\prime \prime}(\theta)=-2 \operatorname{rs}(\cos (\theta-\phi))\right)$. Since $-\frac{\pi}{2} \leq(\theta-\phi) \leq \frac{\pi}{2}$, we know that $f^{\prime \prime}(\theta)<0$ and $f$ is concave. Therefore, to minimize $f$, the angle $\theta$ must be either 0 or $\frac{\pi}{2}$. Similarly, each of $a_{1}, a_{2}$ and $a_{3}$ must be either purely real or purely imaginary to minimize $f$ and the original fraction.
By the AM-GM inequality, if $a_{1}, a_{2}$ and $a_{3}$ are all real or all imaginary, then the minimum value of the fraction is 3 . Now suppose only two of the $a_{i}$ 's, say, $a_{1}$ and $a_{2}$ are real. Since the fraction is homogenous, we may fix $a_{1}+a_{2}$ - let the sum be 2 . The term $a_{1} a_{2}$ in the denominator acheives its maximum only when $a_{1}$ and $a_{2}$ are equal, i.e. when $a_{1}=a_{2}=1$. Then, if $a_{3}=k i$ for some real number $k$, then the expression equals

$$
\frac{\sqrt{k^{2}+4}}{\sqrt[3]{k}}
$$

Squaring and taking the derivative, we find that the minimum value of the fraction is $\sqrt{3} \sqrt[3]{2}$, attained when $k=\sqrt{2}$. With similar reasoning, the case where only one of the $a_{i}$ 's is real yields the same minimum value.
8. [6] Let $f(n)=\sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k!}$. Calculate $\sum_{n=2}^{\infty} f(n)$.

Answer: $3-e$

$$
\begin{aligned}
\sum_{n=2}^{\infty} f(n) & =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^{n} \cdot k!} \\
& =\sum_{k=2}^{\infty} \frac{1}{k!} \sum_{n=2}^{\infty} \frac{1}{k^{n}} \\
& =\sum_{k=2}^{\infty} \frac{1}{k!} \cdot \frac{1}{k(k-1)} \\
& =\sum_{k=2}^{\infty} \frac{1}{(k-1)!} \cdot \frac{1}{k^{2}(k-1)}
\end{aligned}
$$

$$
\begin{aligned}
& =\sum_{k=2}^{\infty} \frac{1}{(k-1)!}\left(\frac{1}{k-1}-\frac{1}{k^{2}}-\frac{1}{k}\right) \\
& =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k \cdot k!}-\frac{1}{k!}\right) \\
& =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k \cdot k!}\right)-\sum_{k=2}^{\infty} \frac{1}{k!} \\
& =\frac{1}{1 \cdot 1!}-\left(e-\frac{1}{0!}-\frac{1}{1!}\right) \\
& =3-e
\end{aligned}
$$

9. [7] Let $x(t)$ be a solution to the differential equation

$$
\left(x+x^{\prime}\right)^{2}+x \cdot x^{\prime \prime}=\cos t
$$

with $x(0)=x^{\prime}(0)=\sqrt{\frac{2}{5}}$. Compute $x\left(\frac{\pi}{4}\right)$.
Answer: $\frac{\sqrt[4]{450}}{5}$ Rewrite the equation as $x^{2}+2 x x^{\prime}+\left(x x^{\prime}\right)^{\prime}=\cos t$. Let $y=x^{2}$, so $y^{\prime}=2 x x^{\prime}$ and the equation becomes $y+y^{\prime}+\frac{1}{2} y^{\prime \prime}=\cos t$. The term $\cos t$ suggests that the particular solution should be in the form $A \sin t+B \cos t$. By substitution and coefficient comparison, we get $A=\frac{4}{5}$ and $B=\frac{2}{5}$. Since the function $y(t)=\frac{4}{5} \sin t+\frac{2}{5} \cos t$ already satisfies the initial conditions $y(0)=x(0)^{2}=\frac{2}{5}$ and $y^{\prime}(0)=2 x(0) x^{\prime}(0)=\frac{4}{5}$, the function $y$ also solves the initial value problem. Note that since $x$ is positive at $t=0$ and $y=x^{2}$ never reaches zero before $t$ reaches $\frac{\pi}{4}$, the value of $x\left(\frac{\pi}{4}\right)$ must be positive. Therefore, $x\left(\frac{\pi}{4}\right)=+\sqrt{y\left(\frac{\pi}{4}\right)}=\sqrt{\frac{6}{5} \cdot \frac{\sqrt{2}}{2}}=\frac{\sqrt[4]{450}}{5}$.
10. [8] Let $f(n)=\sum_{k=1}^{n} \frac{1}{k}$. Then there exists constants $\gamma, c$, and $d$ such that

$$
f(n)=\ln (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+O\left(\frac{1}{n^{3}}\right)
$$

where the $O\left(\frac{1}{n^{3}}\right)$ means terms of order $\frac{1}{n^{3}}$ or lower. Compute the ordered pair $(c, d)$.
Answer: $\left(\frac{1}{2},-\frac{1}{12}\right)$ From the given formula, we pull out the term $\frac{k}{n^{3}}$ from $O\left(\frac{1}{n^{4}}\right)$, making $f(n)=$ $\log (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+\frac{k}{n^{3}}+O\left(\frac{1}{n^{4}}\right)$. Therefore,
$f(n+1)-f(n)=\log \left(\frac{n+1}{n}\right)-c\left(\frac{1}{n}-\frac{1}{n+1}\right)-d\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)-k\left(\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}}\right)+O\left(\frac{1}{n^{4}}\right)$.
For the left hand side, $f(n+1)-f(n)=\frac{1}{n+1}$. By substituting $x=\frac{1}{n}$, the formula above becomes

$$
\frac{x}{x+1}=\log (1+x)-c x^{2} \cdot \frac{1}{x+1}-d x^{3} \cdot \frac{x+2}{(x+1)^{2}}-k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}+O\left(x^{4}\right)
$$

Because $x$ is on the order of $\frac{1}{n}, \frac{1}{(x+1)^{3}}$ is on the order of a constant. Therefore, all the terms in the expansion of $k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}$ are of order $x^{4}$ or higher, so we can collapse it into $O\left(x^{4}\right)$. Using the Taylor expansions, we get

$$
x\left(1-x+x^{2}\right)+O\left(x^{4}\right)=\left(x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}\right)-c x^{2}(1-x)-d x^{3}(2)+O\left(x^{4}\right) .
$$

Coefficient comparison gives $c=\frac{1}{2}$ and $d=-\frac{1}{12}$.