Sequences and Series

Arithmetic sequence

u_{n} = u_{1} + (n - 1) \cdot k

S_{n} - S_{n - 1} = u_{n}

S_{n} = \sum_{n = i}^{j} u_{1} + (n - 1) \cdot k = \frac{j - i + 1}{2} \cdot (u_{i} + u_{j})

Geometric sequence

u_{n} = u_{1} \cdot r^{n - 1}

S_{n} = \frac{u_{1} (1 - r^{n})}{1 - r}

S_{\infty} = \frac{u_{1}}{1 - r}

Binomial Expansion

(\begin{matrix} n \\ r \end{matrix}) = \frac{n!}{r! (n - r)!}

T_{r + 1} = n C r \cdot a^{r} \cdot b^{n - r}

(a x + b)^{n} = b^{n} \sum_{r = 0}^{\infty} (\begin{matrix} n \\ r \end{matrix}) {(\frac{a x}{b})}^{r}

| x | < | \frac{b}{a} |

Maclaurin Series

The Weierstrass Approximation Theorem

The Weierstrass Approximation Theorem shows that any continuous function on a compact interval can be uniformly approximated by polynomials.
$Let f : [a, b] \to R be a continuous function. Then, \forall ε > 0, there exists a polynomial P (x)$
$such that$

sup_{x \in [a, b]} | f (x) - P (x) | < ε .

Refer to Definitions of Supremum if unclear.
As such, we can define $ϵ$ to be infinitely close to 0 and have a polynomial closely fit $f (x)$ .

Thus suppose any function can be expressed in terms of a polynomial

f (x) \approx a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \dots

f (x) \approx \sum_{n = 0}^{\infty} a_{n} x^{n}

To find $a_{0}$ , we can sub $x = 0$ into $f (x)$ .
To find $a_{1}$ , we can differentiate $f (x)$ to obtain $f^{'} (x)$ . Substitute $x$ as 0, to obtain $a_{1}$ . Thus

f^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} \dots

\begin{aligned} f (0) & = a_{0} \\ f^{'} (0) & = a_{1} \\ f^{^{″}} (0) & = 2 \cdot a_{2} \\ f^{^{‴}} (0) & = 3 \cdot 2 \cdot a_{3} \\ f^{^{⁗}} (0) & = 4 \cdot 3 \cdot 2 \cdot 1 \cdot a_{4} \\ ⋮ \\ f^{(n)} (0) & = n! \cdot a_{n} \end{aligned}

Thus

a_{n} = \frac{f^{(n)} (0)}{n!}

Substituting it into the approximation equation above

f (x) \approx \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

Question

What is the Maclaurin Series expansion for $e^{x}$ ?

Solution: Since differentiating $e^{x}$ is always $e^{x}$ and $e^{0}$ is $1$ . Then

e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}

Maclaurin Series expansion for $\sin$ and $\cos$

\sin x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n + 1)!} x^{2 n + 1}

\cos x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n)!} x^{2 n}

Refer to Complex Numbers to see the link between trigonometry and exponential functions.

Fourier Series

Why does $\sin$ and $\cos$ form the orthonormal basis of functions?

The inner product of $\sin$ and $\cos$ is 0. This means the two functions are orthogonal to one another in a vector space. Thus, they can be the basis vectors of any coordinate system.

\int_{0}^{2 π} \sin x \cos x d x = 0

Thus any functions can be expressed as a sum of $\sin$ and $\cos$ waves.

⟨ \sin x, \cos x ⟩ = 0

Full explanation here