Extra (optional) exercises – Short Day

These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.

Extra exercise 1: Area under a parabola

Let \(N\) be a positive real number and \(n\) a natural number. We define \(d=N/n\). As you probably know from secondary school, the area between the parabola \(y=x^2\) and the first axis between \(x=0\) and \(x=N\) can be computed as a definite integral as follows:

\[\int_0^N x^2 dx =\frac13 N^3-\frac13 0^3=\frac13 N^3.\]

In this exercise we wish to investigate how well this area is approximated using the finite sum \(\sum_{k=1}^n d\cdot(kd)^2\). The situation is illustrated in the following figure:

Question a

Use the figure above to show that

\[\frac13 N^3 \le \sum_{k=1}^n d\cdot (kd)^2.\]

Question b

The following distributive rule holds true for all natural numbers \(n\) and all complex numbers \(a,b_1,\dots,b_n\):

\[a \cdot (b_1+\cdots+b_n)=a\cdot b_1+\cdots + a \cdot b_n.\]

Use this fact to show that

\[d^3\sum_{k=1}^n k^2=\sum_{k=1}^n d^3\cdot k^2=\sum_{k=1}^n d\cdot (kd)^2.\]

Question c

Use induction on \(n\) to show that \(\sum_{k=1}^n k^2=\frac13 n(n+1/2)(n+1)\) for all natural numbers \(n\).

Hint

In the induction step you will most likely at some point get to the formula

\[\frac13 (n-1)(n-1/2)n+n^2.\]

You could expand this expression by multiplying through and then check that it gives the correct result, but a lot of work can be saved by first factorising \(n\) outside of the brackets:

\[\frac13 (n-1)(n-1/2)n+n^2=n\left( \frac13 (n-1)(n-1/2)+n\right).\]

Now try to show that the expressions in the parentheses equal \(\frac13 (n+1/2)(n+1).\)

Answer

For the base case, when \(n=1\), the formula holds since \(\sum_{k=1}^1 k^2=1^2=1\) agrees with \(\frac13\cdot 1\cdot(1+1/2)(1+1)=\frac33=1\).

For the induction step, when \(n \ge 2\), if we assume the induction hypothesis which says that \(\sum_{k=1}^{n-1} k^2=\frac13 (n-1)(n-1/2)n\), we get

\[\sum_{k=1}^{n} k^2 = \sum_{k=1}^{n-1} k^2 +n^2=\frac13 (n-1)(n-1/2)n+n^2.\]

For example, by following the above hint it should be possible to see that \(\frac13 (n-1)(n-1/2)n+n^2=\frac13 n(n+1/2)(n+1)\), which means that the induction step holds. Hence, the induction principle implies that the formula \(\sum_{k=1}^n k^2=\frac13 n(n+1/2)(n+1)\) holds for all natural numbers \(n\).

Question d

Now show that \(\sum_{k=1}^n d\cdot (kd)^2=\frac{1}{3}N(N+d/2)(N+d)\).

Answer

From the previous question we have

\[\sum_{k=1}^n d\cdot (kd)^2=d^3\sum_{k=1}^n k^2=d^3\left(\frac13 n(n+1/2)(n+1)\right)=\frac{1}{3}N(N+d/2)(N+d).\]

For the last equality we used the fact that \(N=dn\).

Note that this equation implies that if \(N\) is held fixed while \(n\) tends towards infinity, then \(d\) tends towards zero, and the sum hence tends towards \(\frac13 N^3,\) which is precisely the area under the parabola.