Mathematical Methods for Physics and Engineering
1. investigate a polynomial equation by watching [Page 6]
f(x) = x^7 + 5x^6 + x^4 − x^3 + x^2 − 2 = 0
(i) This is a seventh-degree polynomial equation; therefore the number of real roots is 1, 3, 5 or 7.
(ii) f(0) is negative whilst f(∞) = +∞, so there must be at least one positive root.
Notes:
Generalization.
You should also note that for an eighth-degree polynomial equation the number of real roots is 0, 2, 4, 6 or 8. The reason why this happened is just because the roots must be pairs adding up to the number of real roots at most can be the degree number of the polynomial equation. This means for a nth-degree polynomial equation you can get n, n-2, n-4, n-6... real roots (identical roots included).
Prove by.
As z can be written as z = z_0e^{i\theta} , we know (z^n)^*=z_0^ne^{-in\theta}=(z^*)^n
which means if z is one of the roots then z^* must also be one of them. This tells us the roots must be pairs.
2. investigate a polynomial equation by guessing [Page 7]
f(x) = a_nx^n + a_{n−1}x^{n−1} + · · · + a_1x + a_0, a_n\in\mathbb{R}
Polynomial equation can be converted to factorised polynomials:
f(x) = a_nx^n + a_{n−1}x^{n−1} + · · · + a_1x + a_0 = a_n(x − α_1)^{m_1} (x − α_2)^{m_2} · · · (x − α_r)^{m_r}
if you have known one root \alpha, thenf(x) = (x − α)f_1(x), we assume:
f_1(x)=b_{n−1}x^{n−1} + b_{n−2}x^{n−2} + b_{n−3}x^{n−3} + · · · + b_1x + b_0
Then
After finding all b_{n-1},b_{n-2}...b_0, repeat the procedure above we will be able to convert any equation to:
a_n(x − α_1)^{m_1} (x − α_2)^{m_2} · · · (x − α_r)^{m_r}. meanwhile find all its real or imaginary roots.
3. Properties of roots [Page 9]
f(x) = a_nx^n + a_{n−1}x^{n−1} + · · · + a_1x + a_0, a_n\in\mathbb{R}
Polynomial equation can be converted to factorised polynomials:
f(x) = a_nx^n + a_{n−1}x^{n−1} + · · · + a_1x + a_0 = a_n(x − α_1) (x − α_2) · · · (x − α_n)
The roots have some following properties (\alpha_nis not limited to real root):
Notes:
Example.
4. Useful trigonometric identities [Page 11-14,96]
Easy formulas easy to forget:
Somehow necessary:
Very useful while dealing with a complicated integral:
Very useful while dealing with complex number (note that z has unit modulus):
Arbitrary trigonometric Conversion:
de Moivre's theorem:
5. Coordinate geometry [Page 15-17]
Standard lines and conic curves:
Transformation of standard conic curves:
Parameterization of standard conic curves:
Simple definition of eccentricity but many don't know:
it is the locus of all points P whose distance to a fixed point F (called the focus) is a constant multiple e (called the eccentricity) of the distance from P to a fixed line L (called the directrix). For 0<e<1 we obtain an ellipse, for e=1 a parabola, and for e>1 a hyperbola.
Notes:
Example.
Find the equation of a parabola that has the line x = −a as its directrix and the point
(a, 0) as its focus. Using the definition of eccentricity:
For a ellipse, square eccentricity e^2 = \frac{a^2 − b^2} {a^2}. And if \alpha=\beta=0 then it has focus (\pm ae,0) , directrix \pm \frac{a}{e}.
6. Fast expand a partial fraction-1 [Page 20]
How to fast expand the following partial fraction:
to this kind of fractional polynomial:
Notes:
Let x=-1,-2,-3,-4,-5 respectively, which are roots of denominator, and ignore the referred factor, you can get:
Finally, the original partial fraction should be:
Here you can check the result.
But be careful, it can't be applied to \frac{a_{n-1}x^{n-1}+...}{(x+\alpha_1)^{m_1}...(x+\alpha_n)^{m_n}} which has repeated roots of denominator, and the index of numerator must be less than denominator. Both of these situations can be solved which will be introduced in the following sections.
7. Fast expand a partial fraction-2 [Page 21]
Expand a partial fraction which has an identical or greater index of numerator m than index of denominator n.
Then you can get a group of equations. Take an example:
Apply the method introduced in section 6 we can find:
Notes:
Simplify.
But I will not really recommend you solving the problem by this way. Using polynomial long division may be faster, simpler and actually they are completely the same way.
8. Fast expand a partial fraction-3 [Page 23]
Some additional rules:
Repeated roots in denominator:
For single root (x+\alpha), method of section 6 still works.
For (x^2+\alpha) in denominator:
Notes (further explanation, modified 2024/09/03):
Actually, for any (\sum\limits_{n=0}^{N}a_nx^n)^m in denominator, you will find:
Despite the higher degree of denominators among the expanded fractions, their degree of numerators are always less than or equal to N-1. Let's take an example for m=n=2:
There are 6 variables with only 4 equations. The most simple way to simplify the result is obvious: set higher degree coefficients to 0: c^{(2)}_3=c^{(2)}_2=0. In this way, the final result becomes:
which is consistent with the result: \frac{\sum\limits_{n=0}^{mN-1}b_nx^n}{(\sum\limits_{n=0}^{N}a_nx^n)^m}\rightarrow \frac{\sum\limits_{n=0}^{N-1}c^{(1)}_nx^n}{\sum\limits_{n=0}^{N}a_nx^n} +\frac{\sum\limits_{n=0}^{N-1}c^{(2)}_nx^n}{(\sum\limits_{n=0}^{N}a_nx^n)^2} +\cdots+ \frac{\sum\limits_{n=0}^{N-1}c^{(m)}_nx^n}{(\sum\limits_{n=0}^{N}a_nx^n)^m}.
9. Binomial expansion-1 [Page 26-29]
Basic formula:
Properties of binomial coefficients:
Notes:
Prove the last one:
equate coefficients of x^{p+q−r}y^r on the two sides of the equation:
Many may be confused that how finally get \sum\limits_{t=0}^{r}{}^{p}C_{r-t}{}^{q}C_t=\sum\limits_{t=0}^{r}{}^{p}C_{t}{}^{q}C_{r-t} from binomial expansion? You could first try yourself.
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10. Binomial expansion-2 [Page 29-30]
For negative values of n:
Important recurrence relation(Exception for x=-y,n<0):
11. Proof by induction and contradiction [Page 31-34]
Find \sum\limits_{r=1}^{n}r^3, and then prove it by induction:
Solve the problem by contadiction:
Suppose \sqrt{m} is not an integer first:
Prove there's no largest prime number:
Suppose prime number p_1,p_2,p_3,...,p_n, and p_n is the largest prime number. Then p_1p_2p_3...p_n+1 must also be a prime number, this is contradicted with "p_n is the largest prime number".
12. Necessary and sufficient condition [Page 34-35]
B is a sufficient condition for A: B\Rightarrow A
B is a necessary condition for A: A\Rightarrow B
B is a necessary and sufficient condition for A: A\Leftrightarrow B
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