Proof by induction
Set out proofs as Step 1 (prove claim true for n=1) and Step 2 (prove claim infectious, i.e. if it’s true for n=k, then it’s true for n=k+1), plus a final line: “Step 1 + Step 2 ⇒ true for all n, by induction”.
Don’t set them out as four steps (Basis, Assumption, Induction, Conclusion) like the book says. There are only two steps in proofs by induction.
Do as in this example for finding roots of f(x)=x2–2=0
f(1.4)=1.962−2=−0.038<0 and f(1.5)=2.25−2=0.25>0
∴ change of sign ∴ root in [1.4,1.5]
∴ root in [1.4,1.45]
Don’t do a table as in the book. Much too long and complicated. If you do revision questions from the book, only do two interval bisections for each question. Not more.
Do a diagram like this:
and calculate Δ from similar triangles: ED/EA=CB/CA. Then your next guess is a+Δ
In the book they also draw similar triangles, but in a way that makes a more complicated calculation.
For complex numbers, swapping between
x,y [or Re(z), Im(z)] and
r,θ [or |z|, arg(z)] formats
On your calculator:
Use Pol( , ) to convert from x,y [Re(z), Im(z)] to r,θ [|z|, arg(z)]
Use Rec( , ) to convert from r,θ [|z|, arg(z)] to x,y [Re(z), Im(z)]
Always draw a diagram to check
Don’t use r=|z|=√(x2+y2) and θ=arg(z)=tan–1(y/x) and x=r cosθ and y=r sin θ as in book. Those equations are correct, and you should know them. But they are more complicated. Also, tan–1 has two values for any given (y/x). Your calculator shows only one of them, and you have to do extra work to find whether to use the number your calculator gives you or add π or subtract π.
Solving cubics and quartics given one root
If necessary, divide through the equation so that the term with the highest power (x3 or z3 for a cubic, x4 or z4 for a quartic) has coefficient 1. Then use the facts that:
- the conjugate of a root is also a root (so if a+bi is a root, a−bi is too)
- sum of roots = r1+r2+r3=minus coefficient of second term (i.e. coefficient of x2 in a cubic, coefficient of x3 in a quartic).
- product of roots = r1r2r3=minus constant term (in a cubic) or plus constant term (in a quartic) [it’s minus if if you have an odd number of roots, and plus if you have an even number].
Don’t use long division, as in the book. It is never simpler, and usually more complicated and longer.
Square roots of complex numbers
Example: find √(3+4i)
Two methods, both better than the book
One: via Pol(,)
Pol(3,4) ⇒ 5 cis 0.92729521800161
Square-root modulus, halve argument
⇒ √5 cis 0.46364760900081
Rec(√5,0.46364760900081) = 2 + i
and other square root must be 2 − i
Two: sum and difference of squares
Call the square root a+bi
So, squaring, a2+b2=5
So, equating real parts, a2−b2=3
Solving those two equations, a2=4 and b2=1
a=±2 and b=±1
If the number you’re finding the square root of has a positive imaginary part, then the signs of the real and imaginary parts of the square roots have to be the same; if it has a negative imaginary part, then the signs of the real and imaginary parts of the square roots have to be opposite.
Therefore the roots are 2+i and −(2+i).
[If we were looking for the square roots of 3−4i, they would be 2−i and −(2−i). The square roots of the conjugate are the conjugates of the square roots].
Converting between matrices (tables of numbers) and linear transformations
Nothing the book says is wrong, but this basic rule for converting between linear transformations (scalings, rotations, reflections, shearings) and matrices (tables of numbers) will help. Draw a picture of where the matrix moves (1,0) and (0,1). Remember also that (0,0) stays where it is. And those three facts tell you what it does to all shapes!
If the linear transformation moves (1,0) to (a,c) and (0,1) to (b,d), then the matrix is
If the matrix is
then it moves (1,0) to (a,c) and (0,1) to (b,d).