# Mathematical induction

• We'll show an alternative way of proving our formulas, that of Mathematical induction.
• It might be a little more incomfortable that little Gauß's proof, but it will give a streamlined approach to proving other formulas.
• Mathematical induction.
• Let $$m$$ be an integer, and let $$S_n$$ be a statement involving an integer $$n\ge m$$.
• Suppose that $$S_m$$ is true,
• and suppose that if $$S_n$$ is true, then so is $$S_{n+1}$$.
• Then $$S_n$$ is true for all $$n\ge m$$.
• Proof of $$\sum_{i=1}^ni=\frac{n(n+1)}{2}$$ using Mathematical Induction.
• We'll have $$m=1$$.
• $$S_1:\sum_{i=1}^1i=\frac{1\cdot 2}{2}=1$$.
• Suppose that $$S_n:\sum_{i=1}^ni=\frac{n(n+1)}{2}$$. Then to prove $$S_{n+1}:$$ $\sum_{i=1}^{n+1}i=\sum_{i=1}^ni+(n+1)=\frac{n(n+1)}{2}+(n+1)=\frac{n^2+n+2(n+1)}{2}=\frac{(n+1)(n+2)}{2}$ therefore $$S_{n+1}$$ is true.
• Therefore, Mathematical induction shows that $$S_n:\sum_{i=1}^ni=\frac{n(n+1)}{2}$$ is true for all $$n\ge1$$.
• Exercises. Appendix E: 38, 42, 50

# The area problem

• Let's try to solve the area problem.
• Suppose that $$f(x)\ge0$$ and the domain is $$[a,b]$$.
• Find the area of the region $$S$$ that lies under the curve $$y=f(x)$$ from $$a$$ to $$b$$.
• The solution is clear, if $$S$$ is a rectangle, a triangle, or a shape that can be decomposed to those ones, since they have area formulas.
• But what about, $$f(x)=x^2$$, over $$[0,1]$$?
• What you can do is to approximate the area the following way.
• Cut up the interval $$[a,b]$$ to $$n$$ subintervals $I_1=[a,a+\Delta x],\,I_2=[a+\Delta x,a+2\Delta x],\dotsc,I_n=[a+(n-1)\Delta x,b]$ of equal length.
• We denote the length of one piece by $$\Delta x=\frac{b-a}{n}$$.
• Then let's sum up the areas of the rectangles with base the subintervals $$I_i$$ and height the value of $$f(x)$$ at the rightmost point $$f(a+i\Delta x)$$.
• As $$n$$ increases, the sum seems to approach $$\frac13$$.
• Let us now prove that the area under $$y=f(x)=x^2$$ for $$0\le x\le1$$ is $$\frac13$$.
• We are approximating this area with the sum of the rectangles with base $$\Delta x=\frac{b-a}{n}=\frac{1}{n}$$, and height $$f(a+i\Delta x)=f(\frac{i}{n})=\frac{i^2}{n^2}$$.
• That is, the approximated area is $$R_n=\sum_{i=1}^n\frac{1}{n}\frac{i^2}{n^2}$$.
• We can calculate this using a formula we've seen: $R_n=\sum_{i=1}^n\frac{i^2}{n^3}=\frac{1}{n^3}\sum_{i=1}^ni^2=\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}.$
• Then we can take the limit of this: $\lim_{n\to\infty}R_n=\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}\frac{1(1+n^{-1})(2+n^{-1})}{6}=\frac13.$
• Exercise. Show that we get the same limit if we make the height of the rectangles the left endpoints $$f(a+(i+1)\Delta x)=f(\frac{i-1}{n})=\frac{(i-1)^2}{n^2}$$.