Do not worry about your difficulties in mathematics. I can assure you mine are still greater, Albert Einstein

All those who seem stupid, they are, and also so are half of those who do not, Quevedo

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

A Fourier series expands any periodic function f(t) (with period 2π) as an infinite sum of sines and cosines. The general form of a Fourier series is f is periodic with period 2π, is $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(nt)+b_nsin(nt)$ where $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$

Understanding the convergence of Fourier series is essential when representing functions using trigonometric series. The Fourier series allows us to express periodic functions as infinite sums of sines and cosines.

Definition. A function f(t) is called **even** if it satisfies the condition: f(-t) = f(t) for all t. Examples: x^{2}, cos(x), ∣x∣, x^{2} + 4. The graph of an even function is symmetric with respect to the y-axis.

A function f(t) is called **odd** if it satisfies the condition f(-t) = -f(t). Examples: x^{3}, sin(x), 2x^{3} −3x. The graph of an odd function has rotational symmetry around the origin.

The Fourier series for a general periodic function f(t), with period 2π, is given by: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt)+b_nsin(nt))$

Since cosine is an even function and sine is an odd function, the behavior of the Fourier coefficients a_{n} and b_{n} depends on whether the function f(t) is even, odd, or neither.

If a function f(t) is even, we know that f(-t) = f(t).

Evaluate f at −t:

$f(-t) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(n(-t))+b_nsin(n(-t)))$ =[cos(n(−t)) = cos(−nt) = cos(nt) (since cosine is even), sin(n(−t)) = sin(−nt) = −sin(nt) (since sine is odd).] $\frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt)-b_nsin(nt))$

Comparing this with the original Fourier series:

$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt)+b_nsin(nt))$

Since f(−t) = f(t) (because f is even) and because of the Uniqueness of Fourier Series, we conclude that all the sine terms must be zero, i.e., b_{n} = 0. Therefore, **the Fourier series for an even function contains only cosine terms**:

b_{n} = 0 ∀n, $f(t) = \frac{a_0}{2} + \sum_{n=0}^\infty a_ncos(nt)$ where the coefficients a_{n} are given by: $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt$

Because f(t)cos(nt) is also an even function (product of two even functions), the integral over [−π, π] simplifies to twice the integral over [0, π]:

$a_n=\frac{2}{π}\int_{0}^{π} f(t)cos(nt)dt$

This simplification reduces computation and is valid because the integrand is symmetric about the y-axis.

Similarly, if f(t) is an odd function, we know that f(-t) = -f(t).

Evaluate f at −t: f(-t) = $\frac{a_0}{2} + \sum_{n=0}^\infty (a_ncos(n(-t))+b_nsin(n(-t)))$ =[cos(n(−t)) = cos(−nt) = cos(nt) (since cosine is even), sin(n(−t)) = sin(−nt) = −sin(nt) (since sine is odd).] $\frac{a_0}{2} + \sum_{n=0}^\infty (a_ncos(nt))-b_nsin(nt)$

Comparing with Original Fourier Series: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt)+b_nsin(nt))$

Multiply both sides of the original equation by −1:-f(t) = $-\frac{a_0}{2} + \sum_{n=0}^\infty (-a_ncos(nt)-b_nsin(nt))$

If a function f is odd f(-t) = -f(t), then by the **Uniqueness of the Fourier Series** we conclude that all the cosine terms must be zero, i.e., a_{n} = 0. Therefore, the Fourier series for an odd function contains only sine terms:

$a_n = 0, b_n=\frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$

If f is odd, then f(n)sin(nt) is an even function (product of two odd functions), the integral over [−π, π] simplifies to twice the integral over [0, π] and the coefficients b_{n} are given by: $b_n = \frac{2}{π}\int_{0}^{π} f(t)sin(nt)dt.$

This simplification reduces computation and is valid because the integrand is symmetric about the origin.

**Examples**:

- f(t) = cos(t), cos(t) is even. Fourier series contains only cosine terms. $a_n=\frac{2}{π}\int_{0}^{π} cos(t)cos(nt)dt$
- f(t) = sin(t), sin(t) is odd. Fourier series contains only sine terms. $b_n=\frac{2}{π}\int_{0}^{π} sin(t)sin(nt)dt$

Theorem. If a function f(t) is periodic 2π and is continuous at a point t_{0}, then its Fourier series converges to the value of the function at that point: $f(t_0) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt_0)+b_nsin(nt_0))$.

If f(t) has a jump discontinuity at t_{0}, the Fourier series converges to the midpoint of the jump (the average of the left-hand and right-hand limits) at that location. This is known as the Gibbs phenomenon. It refers to the oscillatory behavior of the Fourier series and overshoots that occur near a jump discontinuity but still converges to the average value at the point of discontinuity.

Increasing the number of terms reduces the width of the overshoot but not its height.

If the jump is from f(t_{0}^{-}) to f(t_{0}^{+}), then the Fourier series converges to: $\frac{f(t_0^+)+f(t_0^-)}{2}$: $\lim_{N \to ∞}[\frac{a_0}{2} + \sum_{n=1}^N (a_ncos(nt)+b_nsin(nt))] = \frac{f(t_0^+)+f(t_0^-)}{2}$

This behavior is one of the key reasons Fourier series are widely used to represent piecewise continuous functions. Applications: modeling electrical signals (square waves, sawtooth waves); analyzing mechanical vibrations; solving partial differential equations (heat equation, wave equation).

When dealing with functions defined on a finite interval (0 , L), we often need to extend them periodically to apply Fourier series techniques. This section explains how to extend such functions and adjust the Fourier series accordingly.

Let’s consider a function f(t) that is periodic with period 2L. The goal is to express the Fourier series in terms of a normalized variable. Specifically, we want to map the interval [0, L] to [0, π] for easier computation and interpretation.

To do this, we perform a change of variables: $t = \frac{L}{π}u ↭ u = \frac{π}{L}t$. This maps t ∈ [0, L] to u ∈ [0, π].

Thus, the Fourier series expansion for a function with period 2L becomes: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(\frac{nπ}{L}t)+b_nsin(\frac{nπ}{L}t)$ where the Fourier coefficients are defined as: $a_n=\frac{1}{L}\int_{-L}^{L} f(t)cos(\frac{nπ}{L}t)dt, b_n = \frac{1}{L}\int_{-L}^{L} f(t)sin(\frac{nπ}{L}t)dt$

If f is even (i.e., f(-t) = f(t)) and periodic with period 2L, then the Fourier series will contain only cosine terms. In this case: $a_n=\frac{2}{L}\int_{0}^{L} f(t)cos(\frac{nπ}{L}t)dt, b_n = 0$, and the Fourier series simplifies to $f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(\frac{nπ}{L}t)$.

If f is odd (i.e., f(-t) = f(t)) and periodic with period 2L, then the Fourier series will contain only sine terms. In this case: $b_n=\frac{2}{L}\int_{0}^{L} f(t)sin(\frac{nπ}{L}t)dt, a_n = 0$, and the Fourier series simplifies to $f(t) = \sum_{n=1}^\infty b_nsin(\frac{nπ}{L}t)$.

We know that every periodic function has a Fourier series expansion. It is fairly common for functions arising from certain applications to be defined only on a finite interval 0 < x < L. In such cases, it’s common to extend the function periodically over the entire real line to make use of Fourier series. This can be done in two distinct ways:

We define F(t) to be a periodic extension of f is:

- F(t) = f(t) for 0 < t < L
- F(t) is periodic of period 2L.

There are two periodic extensions for f(t) that are very useful (Refer to Figure iii for a visual representation and aid in understanding it).

**Even Periodic extension**. Given any f(t), which is defined only for 0 < t < L, we define its even periodic extension, F^{e}(t), by the conditions that

- F
^{e}(t) = f(t) for 0 < t < L - F
^{e}(-t) = f(t) for -L < t < 0 (F^{e}is even). - F
^{e}(t) is a periodic function with period 2L.

**Odd Periodic extension**. We define its odd periodic extension, F^{o}(t), by the conditions that

- F
^{o}(t) = f(t) for 0 < t < L - F
^{o}(t) = -f(t) for -L < t < 0 (F^{o}is odd). - F
^{o}(t) is a periodic function with period 2L.

Since F^{e} is even, it has a Fourier cosine series. Similarity, since F^{0} is odd, it has a Fourier sine series.

Restrict to 0 < t < L, $f(t)=F^e(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(\frac{nπt}{L}) = F^0(t) = +\sum_{n=1}^\infty b_nsin(\frac{nπt}{L})$ where $a_n = \frac{2}{L}\int_{0}^{L} F^e(t)cos(\frac{nπt}{L}) = \frac{2}{L}\int_{0}^{L} f(t)cos(\frac{nπt}{L})$ and $b_n = \frac{2}{L}\int_{0}^{L} F^0(t)sin(\frac{nπt}{L})dt = \frac{2}{L}\int_{0}^{L} f(t)sin(\frac{nπt}{L})dt$

This shows that f(t) can be represented either by its cosine series (even extension) or sine series (odd extension) within (0, L).

For example consider the function f(x) = 1 defined on 0 < x < π. The even and odd periodic extensions F^{e}(x) and F^{0}(x) take the value 1 for all 0 < x < π and have period 2π (Refer to Figure ii for a visual representation and aid in understanding it).

$F^0(t) = \sum_{n=0}^\infty b_nsin(\frac{nπt}{L}) =[\text{2L=2π}] \sum_{n=0}^\infty b_nsin(nt)$ with $b_n = \frac{2}{L}\int_{0}^{L} F^0(t)sin(\frac{nπt}{L}) = \frac{2}{π}\int_{0}^{π} 1·sin(nt)dt = \frac{2}{π}\frac{-cos(nt)}{n}\bigg|_{0}^{π} = \frac{2}{πn}(-cos(nπ)+cos(0))$

Since cos(nπ)=(-1)^{n} and cos(0) = 1, we have:

$= \begin{cases} 0, &n,even \\ \frac{2}{nπ}(-(-1)+1) = \frac{4}{nπ}, &n,odd \end{cases}$

Using the calculated coefficients b_{n}, the odd periodic extension F^{0}(t) becomes:

$F^0(t) = \sum_{n=1, n~odd}^\infty \frac{4}{nπ}sin(nt)$. This gives us the Fourier sine series representation of the odd periodic extension.

$F^e(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(\frac{nπt}{L}) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(nt)$ where a_{n} are the Fourier cosine coefficients.

$a_n = \frac{2}{L}\int_{0}^{L} F^e(t)cos(\frac{nπt}{L}) = \frac{2}{π}\int_{0}^{π} 1·cos(nt)dt = \frac{2}{π}\frac{sin(nt)}{n}\bigg|_{0}^{π}=0$ because sin(nπ) = 0 for all n. Therefore, a_{n} = 0 for all n ≥ 1.

However, the constant term (the average value of the function over the interval) is: $a_0 = \frac{2}{π}\int_{0}^{π} 1dt = \frac{2}{π}·π = 2$

Thus, the Fourier cosine series for the even periodic extension F^{e}(t) is:

$F^e(t) = \frac{a_0}{2}+\sum_{n=0}^\infty a_ncos(nt) = \frac{a_0}{2} + 0 + 0 + ··· + 0 = 1$. This shows that the even periodic extension of f(x) = 1 is just the constant function F^{e}(t) = 1. There are no higher cosine terms since all a_{n} = 0 ∀ n ≥ 1.

- $f(t) = \begin{cases} t, &-π ≤ t < π \\ t-2π, &π ≤ t < 3π \\ …. \end{cases}$ (Refer to Figure iv for a visual representation and aid in understanding it)

For a function f(t) with period 2π, the Fourier series is generally expressed as: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(nt)+b_nsin(nt)$ where $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$

However, since f(t) is an odd function, we have:

- a
_{0}= 0: The average value over a period is zero. - a
_{n}= 0. All cosine coefficients vanish because cosine is an even function.

f(t) is periodic with a period 2π and odd. Therefore, we only need to compute the sine terms of the Fourier series. Specifically, $a_n = 0, b_n=\frac{2}{π}\int_{0}^{π} f(t)sin(nt)dt = \frac{2}{π}\int_{0}^{π} tsin(nt)dt$

We integrate from 0 to π because f(t) is odd, and integrating over [−π, π] would be redundant.

To do this, we apply integration by parts. Recall that the integration by parts formula is: ∫udv = uv −∫vdu. u = t ⇒ du = dt. dv = sin(nt)dv, so v = ^{-1}⁄_{n}cos(nt)

$b_n = \frac{2}{π}[-\frac{tcos(nt)}{n}]\bigg|_{0}^{π}$

$-\int_{0}^{π} \frac{-cos(nt)}{n}dt = \frac{sin(nt)}{n^2}\bigg|_{0}^{π} = 0$. This is because sin(nπ) = 0 and sin(0) = 0, so the entire integral vanishes.

- At t = π, cos(nπ) = (-1)
^{n}, $b_n = \frac{2}{π}·\frac{-π}{n}(-1)^n = \frac{2}{n}(-1)^{n+1}$ - At t = 0, $-\frac{0·cos(0)}{n} = 0$

Now that we have the Fourier coefficients, we can write the Fourier series for f(t): f(t) = $2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}sin(nt) = 2(sin(t)-\frac{1}{2}sin(2t)+\frac{1}{3}sin(3t)-\frac{1}{4}sin(4t)+···)$

The Fourier series tries to approximate the function f(t) not just at individual points, but over the entire interval [-π, π]. At points of discontinuity, such as t = −π, t = π, and t = (2n +1)π, the Fourier series converges to the midpoint of the jump (average of the left-hand and right-hand limits of f(t)).

For example, at t = π, there is a jump discontinuity, and the Fourier series will converge to the average value at the jump, which is: $\frac{f(π^+)+f(π^-)}{2} = \frac{(-π)+π}{2}=0$

In the discontinuity points, sin((2n +1)π) = 0, the Fourier series converges to zero (the mid point of the jump). This behavior is known as the Gibbs phenomenon, where the Fourier series oscillates near discontinuities but still converges to the average value of the jump.

Gibbs Phenomenon: The oscillatory behavior and overshoot of the Fourier series near points of discontinuity.

- $f(t) = \begin{cases} 0, &-π ≤ t < π/2 \\ 4, &-π/2 ≤ t < π/2 \\ 0, &π/2 ≤ t < π \end{cases}$

The function f(t) resembles a rectangular pulse centered at t = 0, with a height of 4 and width π. It repeats every 2π. Since f(−t) mirrors f(t) across the vertical axis, f(t) is even.

For a function f(t) with period 2π, the Fourier series is generally expressed as: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(nt)+b_nsin(nt)$ where $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$

f(t) is periodic with a period 2π and even. Therefore, we only need to compute the cosine terms of the Fourier series. Specifically, $b_n = 0, a_n=\frac{2}{π}\int_{0}^{π} f(t)cos(nt)dt$

$a_0 = \frac{2}{π}\int_{0}^{π} f(t)cos(0·t)dt = \frac{2}{π}[\int_{0}^{\frac{π}{2}} 4cos(0·t)dt+\int_{\frac{π}{2}}^{π} 0·cos(0·t)dt] = \frac{2}{π}4t\bigg|_{0}^{\frac{π}{2}} = \frac{2}{π}·4·\frac{π}{2} = 4.$

$a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt = \frac{2}{π}\int_{0}^{π} f(t)cos(nt)dt = \frac{2}{π}\int_{0}^{\frac{π}{2}} 4cos(nt)dt = \frac{8}{π}\frac{sin(nt)}{n}\bigg|_{0}^{\frac{π}{2}} = \frac{8}{π}\frac{sin(\frac{nπ}{2})}{n}$

- n = 2k (even n): $a_n = \frac{8}{π}\frac{sin(\frac{nπ}{2})}{n} = \frac{8}{π}\frac{sin(\frac{2kπ}{2})}{n} = \frac{8}{π}\frac{sin(kπ)}{n} = 0.$
- n = 2k -1 (odd n): $a_n = \frac{8}{π}\frac{sin(\frac{nπ}{2})}{n} = \frac{8}{π}\frac{sin(\frac{(2k -1)π}{2})}{n} = \frac{8}{π}\frac{(-1)^{k-1}}{n} =$[Since n = 2k-1 ⇒ k = $\frac{n+1}{2}, k-1 = \frac{n+1-2}{2} = \frac{n-1}{2}$] $\frac{8}{π}\frac{(-1)^{\frac{n-1}{2}}}{n}$
$k = 1, sin(\frac{(2k-1)π}{2}) = sin(\frac{π}{2}) = 1, k = 2, sin(\frac{(2k-1)π}{2}) = sin(\frac{3π}{2}) = -1, k = 3, sin(\frac{(2k-1)π}{2}) = sin(\frac{5π}{2}) = 1$ ···

f(t) = $\frac{4}{2} + \sum_{n=1, n~ odd}^\infty (\frac{8}{π}\frac{(-1)^{\frac{n-1}{2}}}{n})cos(nt) = 2 + \sum_{n=1, n~ odd}^\infty (\frac{8}{π}\frac{(-1)^{\frac{n-1}{2}}}{n})cos(nt)$

Expressing a_{n} in terms of k: $a_n = \frac{8}{π}\frac{(-1)^{k-1}}{2k-1}$. The Fourier Series becomes: $2 + \sum_{k=1}^\infty (\frac{8}{π}\frac{(-1)^{k-1}}{2k-1})cos((2k-1)t) = 2 + \frac{8}{π}\sum_{k=1}^\infty (\frac{(-1)^{k-1}}{2k-1})cos((2k-1)t)$

Discontinuities occur at t = ±^{π}⁄_{2} and every t = ±^{nπ}⁄_{2}, where n is an integer. At these points, the Fourier series converges to the average of the left-hand and right-hand limits of f(t).

Average value = $\frac{f(\frac{π}{2}^-) + f(\frac{π}{2}^+)}{2} = \frac{4+0}{2} = 2$. Therefore, the Fourier series converges to 2 at t = ^{π}⁄_{2}. Near the discontinuities at t = ±^{π}⁄_{2}, the Fourier series overshoots the function value but converges to the average value.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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