The Modulus Function
- No line with a modulus ever goes under the x-axis.
- Any line that does go below the x-axis, when modulated is reflected above it
Polynomials
- To find unknowns in a given identity:
- Substitute suitable values of x
- OR
- Equalize the given coefficients of like powers of x.
- Factor Theorem: If (x – t) is a factor of the function p(x), then p(t) = 0
- Remainder Theorem: If the function f(x) is divided by (x – t), then the remainder, R = f(t)
Binomial Series
- Expanding (1 + x)n, where |x| < 1
- Factor Case: If the constant is not 1, pull out a factor from the brackets to make it 1, and use the general equation. Do NOT forget the indices.
- Substitution Case: If the bracket contains more than one x term (e.g. 2 – x + x2), then make the last part u, expand and then substitute back in.
- Finding the Limit of x in the Expansion: E.g. (1 + ax)n, the limit can be found by substituting ax between the modulus sign in |x| < 1 and altering it to have only x in the modulus.
Partial Fractions
- Multiply through by (px + q), substitute x = -q/p and find A
- Multiply through by (rx + s), substitute x = -s/r and find B
- Multiply through by (px + q), substitute x = -q/p and find A
- Multiply through by (rx + s)2, substitute x = -s/r and find C
- Substitute any constant, e.g. x = 0, and find B.
- Multiply through by (px + q), substitute x = -q/p and find A
- Take A/px + q to the other side, subtract and simplify.
- The linear equation left at the top is equal to Bx + C
- Improper Fraction case
- If the numerator has x to the degree of power equivalent or greater than the denominator, then another constant is present.
- This can be found by dividing the denominator by the numerator and using the remainder.
Example
-
- Express the following in partial fractions:
- Expand the brackets:
- The greatest power of x is the same for the numerator and denominator, thus this is an improper case.
- Make it into a proper fraction:
- This is then written as:
- Now proceed with the normal case for the fraction:
- A(2x – 3) + B(x + 1) = 5 – 5x
- when x = -1
- -5A = 5 + 5
- ∴ A = -2
- when x = 3/2
- 5/2B = 5 – 15/2
- ∴ B = -1
- Thus the partial fraction is: