P. Abbott’s Teach Yourself Calculus

January 14, 2021October 24, 2022

202101141227 Exercise 1 (Q10)

If $f(x)=2x^2$ , find expressions for:

1. $f(x+h)$ .

2. $f(x+h)-f(x)$ .

3. $\displaystyle{\frac{f(x+h)-f(x)}{h}}$ .

Solution.

1.

$\begin{aligned} f(x+h) & = 2(x+h)^2 \\ & = 2(x^2+2hx+h^2) \\ & = 2x^2+4hx+2h^2 \end{aligned}$

2.

$\begin{aligned} f(x+h)-f(x) & = 2(x+h)^2 - 2(x)^2 \\ & = 2(x^2+2hx+h^2) - 2x^2 \\ & = 4hx+2h^2 \\ \end{aligned}$

3.

$\begin{aligned} \frac{f(x+h)-f(x)}{h} & = \frac{2(x+h)^2-2(x)^2}{h} \\ & = \frac{2(x^2+2hx+h^2)-2x^2}{h} \\ & = \frac{4hx+2h^2}{h} \\ & = 4x+2h \\ \end{aligned}$

Remark.

The first step to derivation from first principles, is calculate the slope of tangent

$\displaystyle{\frac{f(x+h)-f(x)}{h}}$ ,

and put the limit $h\to 0$ , i.e.,

$\begin{aligned} f'(x) & =\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}x}} \\ & = \displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}} \end{aligned}$

a right-hand limit.

Afterword.

$\begin{aligned} f(x) & =2x^2 \\ f'(x) & = \frac{\mathrm{d}f}{\mathrm{d}x} \\ & = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & \stackrel{\textrm{(3)}}{=} \lim_{h\to 0} 4x+2h \\ & = 4x+2(0) \\ & = 4x \end{aligned}$

November 9, 2020October 24, 2022

202011091423 Exercise 1 (Q2)

If $f(x)=(x-1)(x+5)$ , find the values of $f(2)$ , $f(1)$ , $f(0)$ , $f(a+1)$ , $f\bigg(\displaystyle{\frac{1}{a}}\bigg)$ , $f(-5)$ .

Solution.

$\begin{aligned} f(2) & = (2-1)(2+5)=7 \\ f(1) & = (1-1)(1+5)=0 \\ f(0) & = (0-1)(0+5)=-5 \\ f(a+1) & = ((a+1)-1)((a+1)+5) = a(a+6) \\ f\bigg( \frac{1}{a}\bigg) & = \Bigg( \bigg( \frac{1}{a}\bigg) -1\Bigg) \Bigg( \bigg( \frac{1}{a}\bigg) +5 \Bigg) \\ f(-5) & = (-5-1)(-5+5) = 0 \end{aligned}$

On reflection.

Suppose you are given an unknown quadratic function $f(x;a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ such that

$\begin{aligned} 0 & \mapsto -5 \\ 1 & \mapsto 0 \\ 2 & \mapsto 7 \\ \end{aligned}$

and you are asked to make polynomial interpolation.

$\begin{cases} a_0 + a_1 (0) + a_2(0)^2 = f(0) = -5 \\ a_0 + a_1 (1) + a_2(1)^2 = f(1) = 0 \\ a_0 + a_1 (2) + a_2(2)^2 = f(2) = 7 \\ \end{cases}$
$\begin{cases} a_0=-5 \\ a_0 + a_1 + a_2 = 0\\ a_0 +2a_1 + 4a_2 = 7\\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 + a_2 = 5 \\ a_1 + 2a_2 = 6 \\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 = 4 \\ a_2 = 1 \\ \end{cases}$

And then is the interpolated polynomial $f(x)=x^2+4x-5\qquad \textrm{\scriptsize OR}\qquad (x+5)(x-1)$ .

November 5, 2020October 24, 2022

202011051527 Exercise 1 (Q1)

If $f(x)=2x^2-4x+1$ , find the values of $f(1)$ , $f(0)$ , $f(2)$ , $f(-2)$ , $f(a)$ , $f(x+\delta x)$ .

Solution.

Given $f(x)=2x^2-4x+1$ .

$\begin{aligned} f(1) & =2(1)^2-4(1)+1=-1 \\ f(0) & = 2(0)^2 - 4(0) +1 = 1 \\ f(2) & = 2(2)^2-4(2)+1 =1 \\ f(-2) & = 2(-2)^2-4(-2)+1=17\\ f(a) & = 2a^2 - 4a +1 \\ f(x+\delta x) & = 2(x+\delta x)^2 - 4 (x+\delta x) +1 \end{aligned}$

This exercise is done.

On reflection.

Suppose you are given the following conditions:

$\begin{aligned} x_0 = 0 & \qquad f(x_0) = 1 \\ x_1 = 1 &\qquad f(x_1) = -1 \\ x_2 = 2 &\qquad f(x_2) =1 \end{aligned}$

and you are asked to interpolate by Lagrange polynomials over the range $[0,2]$ .

$\begin{aligned} \mathcal{L}(x) & = (1)\bigg( \displaystyle{\frac{x-1}{0-1}} \bigg)\bigg( \displaystyle{\frac{x-2}{0-2}} \bigg) + (-1)\bigg( \displaystyle{\frac{x-0}{1-0}} \bigg) \bigg( \displaystyle{\frac{x-2}{1-2}} \bigg) + (1)\bigg( \displaystyle{\frac{x-0}{2-0}} \bigg) \bigg( \displaystyle{\frac{x-1}{2-1}} \bigg) \\ & = \displaystyle{\frac{(x-1)(x-2)}{2}} + x(x-2) + \displaystyle{\frac{x(x-1)}{2}} \\ & = \displaystyle{\frac{(x-1)(x-2)+2x(x-2)+x(x-1)}{2}} \\ & = \displaystyle{\frac{x^2-3x+2+2x^2-4x+x^2-x}{2}} \\ & = \displaystyle{\frac{4x^2-8x+2}{2}} \\ & = 2x^2-4x+1\\ \end{aligned}$

The interpolating polynomial $\mathcal{L}(x)$ checks with the original function $f(x)$ .