Here are solutions for some Putnam problems that are in 8th Edition of Calculus of a Single Variable by Larson, Hostetler, and Edwards.
Plot and find the area of the region in the 2D plane constrained by \(|x|-|y|\leq1\) and \(|y|\leq1\).
The second inequality constrains the area to within a strip in the plane described by \(-1\leq y\leq1\). For the first inequality, we split it up into 4 cases, based on the sign of \(x\) and \(y\). When \(x\geq0\), we have \(|x|=x\) and when \(x<0\) we have \(|x|=-x\). This gives the following inequalities:
| Quadrant | Inequality |
|---|---|
| I: \(x\geq0,y\geq0\) | \(y\geq x-1\) |
| II: \(x<0,y\geq0\) | \(y\geq-x-1\) |
| III: \(x<0,y<0\) | \(y\leq x+1\) |
| IV: \(x\geq0,y<0\) | \(y\leq-x+1\) |
Next, we plot these in the 2D plane and can find the area geometrically.
The area can be found as the area of 2 trapezoids, a square and 4 triangles, or similar. The total area is 6.
Find all polynomials \(f(x)\) with real coefficients that satisfy \(f(g(x))=g(f(x))\) for all polynomials \(g(x)\) with real coefficients.
Let \(g(x)=c\) be a constant polynomial. Then \(g(f(x))=c=f(g(x))=f(c)\). Since this is true for any constant \(c\), we have \(f(c)=c\) for all \(c\). This can be equivalently written as \(f(x)=x\), which is that \(f(x)\) must be the identity function.
Find the value of \(h\) that makes the area of the rectangle and triangle equal. The radius of the circle is 1.
By using the center and the Pythagorean Theorem,
\[1=\left({b\over2}\right)^2+\left({h\over2}\right)^2\]Then equating the areas of the shapes,
\[bh={1\over2}b\left(1-{h\over2}\right)\]From the shape areas, we find \(h={2\over5}\). Combining this with the Pythagorean formula, we can also find \(b={4\over5}\sqrt{6}\).
A right circular cone has base of radius 1 and height 3. A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the side length of the cube?
A square face of the cube is on the cone base, and its 4 corners not on that face intersect the pointed surface of the cone. The diagram below shows the relevant parts, forming 2 right triangles. Here, we use \(s\) for the side length of the cube and \(t\) for the distance between a corner of the base face of the cube and the edge of the cone base.
Using the right triangles ratios, we have \({s\over t}=3\). Then substituting into \(1={s\over\sqrt{2}}+t\),
\[\begin{align}&1={s\over\sqrt{2}}+{s\over3}\Rightarrow3\sqrt{2}=3s+\sqrt{2}s \Rightarrow s(3+\sqrt{2})=3\sqrt{2}\\& \Rightarrow s={3\sqrt{2}\over3+\sqrt{2}}\cdot{3-\sqrt{2}\over3-\sqrt{2}} ={9\sqrt{2}-6\over7} \end{align}\]Prove or disprove: if \(x,y\in\mathbb{R}\), \(y\geq0\), and \(y(y+1)\leq(x+1)^2\), then \(y(y-1)\leq x^2\).
To get some idea of whether this is true or false, we can rewrite the 2 inequalities as the following hyperbolas and plot them.
\[\left(y+{1\over2}\right)^2-(x+1)^2\leq{1\over4},\quad \left(y-{1\over2}\right)^2-x^2\leq{1\over4}\]Then to plot these as functions of \(x\), we can write them as:
\[\left|y+{1\over2}\right|\leq\sqrt{{1\over4}+(x+1)^2},\quad \left|y-{1\over2}\right|\leq\sqrt{{1\over4}+x^2}\]When we take the negative of the absolute values, we end up with equations that are redundant because \(y\geq0\) is a stronger statement. To see why, consider than both of the square root terms are \(\geq{1\over2}\) and then work out the algebra. So we have to consider points with \(y\geq0\) and ask if \(y\leq-{1\over2}+\sqrt{{1\over4}+(x+1)^2}\) implies \(y\leq{1\over2}+\sqrt{{1\over4}+x^2}\). By plotting these portions of the hyperbolas, it appears that the statement must be true, and it can be proven with a few cases. For some of those proofs, it is easier to work with the inequalities in expanded form from the hyperbola equations, so we will be showing \(y^2+y\leq x^2+2x+1\) implies \(y^2-y\leq x^2\).
Suppose \(x\leq-{1\over2}\). Then we also have \(2x+1\leq\). We can derive \(y^2+y\leq x^2+2x+1\leq x^2\). Now subtract \(2y\) from each side and since \(y\geq0\), we find \(y^2-y\leq x^2-2y\leq x^2\).
Next, we need to prove it for \(x>-{1\over2}\). First, with the added constraint that \(2x+1\leq2y\). Subtract \(2y\) from each side in the first inequality so we have \(y^2-y\leq x^2+2x+1-2y\leq x^2\). Since we have \(2x+1-2y\leq0\) from the added constraint.
Finally, we have the case where \(x>-{1\over2}\) and \(2x+1>2y\). For this one, it is easier to do things a little differently. We have \(y<{1\over2}+x={1\over2}+\sqrt{x^2}\leq{1\over2}+\sqrt{{1\over4}+x^2}\). So it satisfies the required condition, completing the proof.
Find all polynomials \(P(x)\) such that \(P(x^2+1)=(P(x))^2+1\) and \(P(0)=0\).
If the constraint \(P(0)=0\) was removed, we would be able to find a family of solutions by starting with \(P(x)=x^2+1\) and using function composition. But that is not what the problem asks for, so we start with \(P(0)=0\). Let \(x_0=0\) and \(x_n=x_{n-1}^2+1\) for \(n\geq1\). Then we can show \(P(x_i)=x_i\) for each \(x_i\). The base case for \(x_0=0\) is true since \(P(0)=0\). Then \(P(x_{n-1}^2+1)=(P(x_{n-1}))^2+1\) which implies \(P(x_n)=x_{n-1}^2+1=x_n\). Therefore, \(P(x_i)=x_i\) for arbitrarily many \(x_i\) so we must have \(P(x)=x\) as the only solution.
Let \(f(x)=\sum_{i=1}^na_i\sin(ix)\) with each \(a_i\in\mathbb{R}\). Given \(|f(x)|\leq|\sin(x)|\) for all \(x\in\mathbb{R}\), prove that \(\left|\sum_{i=1}^nia_i\right|\leq1\).
The derivative is \(f'(x)=\sum_{i=1}^nia_i\cos(ix)\) and \(f'(0)=\sum_{i=1}^nia_i\), so we must show \(|f'(0)|\leq1\). Suppose that \(f'(0)=m\) with \(|m|>1\). Then by the definition of a limit \(\lim_{h\to0}{f(h)\over h}=m\). So we would be able to find arbitrarily small \(h>0\) such that \(\left|{f(h)\over h}\right|>1\), or \(|f(h)|>h\geq\sin(h)=|\sin(h)|\), so by contradiction the proof is complete.
We used that for \(h\geq0\), \(\sin(h)\leq h\). It is clearly true when \(h=0\). When \(0<h<1\), we can find \(c\in(0,h)\) so that \(\cos(c)={\sin(h)\over h}\), using the derivative of sine. So we have \(\left|{\sin(h)\over h}\right|\leq1\) since \(|\cos(c)|\leq1\), which lets us conclude \(\sin(h)\leq|\sin(h)|\leq h<1\). Finally, it is clearly true that \(\sin(h)\leq1\leq h\) when \(h\geq1\).
Let \(k\in\mathbb{Z}^+\). The \(n\)th derivative of \({1\over x^k-1}\) has the form \({P_n(x)\over(x^k-1)^{n+1}}\) where \(P_n(x)\) is a polynomial. Find \(P_n(1)\).
We can begin by evaluating some derivatives to get an idea of what to look for.
\[\begin{align}&{d\over dx}(x^k-1)^{-1}=-(x^k-1)^{-2}\cdot kx^{k-1}\\& {d^2\over dx^2}(x^k-1)^{-1}=2(x^k-1)^{-3}\cdot(kx^{k-1})^2 -(x^k-1)^{-2}\cdot k(k-1)x^{k-2}\\& =(x^k-1)^{-3}\left(2(kx^{k-1})^2-(x^k-1)\cdot k(k-1)x^{k-2}\right) =(x^k-1)^{-3}P_2(x) \end{align}\]Then the 3rd derivative will look like this. We do not need to fully compute \(P_3(x)\).
\[\begin{align}&{d^3\over dx^3}(x^k-1)^{-1}={d\over dx} (x^k-1)^{-3}P_2(x)=-3(x^k-1)^{-4}kx^{k-1}P_2(x) +(x^k-1)^{-3}P_2'(x)\\& (x^k-1)^{-4}\left(-3kx^{k-1}P_2(x)+(x^k-1)P_2'(x)\right)=(x^k-1)^{-4}P_3(x) \end{align}\]Since \(x^k-1=0\) when we put in \(x=1\), the value of \(P_2'\) is not needed. We find that \(P_3(1)=-3kP_2(1)\). This suggests the general formula \(P_n(x)=(-1)^nn!k^n\), which we can prove by induction. When \(n=0\), then \(P_0(1)=1\) so the base case is true. Now assume \(P_n(x)=(-1)^nn!k^n\) and compute the derivative.
\[\begin{align}&{d^{n+1}\over dx^{n+1}}(x^k-1)^{-1} ={d\over dx}(x^k-1)^{-(n+1)}P_n(x)\\& =-(n+1)(x^k-1)^{-(n+2)}kx^{k-1}P_n(x)+(x^k-1)^{-(n+1)}P_n'(x)\\& =(x^k-1)^{-(n+2)}\left(-(n+1)kx^{k-1}P_n(x)+(x^k-1)P_n'(x)\right) \end{align}\]The new polynomial is \(P_{n+1}(x)=-(n+1)kx^{k-1}P_n(x)+(x^k-1)P_n'(x)\) and \(P_{n+1}(1)=-(n+1)kP_n(1)=-(n+1)k\cdot(-1)^nn!k^n=(-1)^{n+1}(n+1)!k^{n+1}\) which completes the proof.