Mathematics for Electrical Science and Physical Science 1. Module 4, taylor and maclaurin series, fourier series and half-range expansions.

Taylor, Maclaurin, and Fourier Series

This notebook teaches how to represent smooth functions locally using Taylor and Maclaurin power series, and how to represent periodic and piecewise-smooth functions globally using Fourier trigonometric series. Learners will compute full and half-range Fourier expansions over arbitrary intervals and apply Dirichlet's convergence theorem to evaluate classical infinite numerical series.

Before you start: Higher-order derivatives of elementary functions; Definite integration and integration by parts; Trigonometric identities and values at integer multiples of pi.

Your path through this module

1. Local Approximation with Taylor and Maclaurin Series: Any infinitely differentiable function can be represented near a chosen center point as an infinite sum of powers whose coefficients are determined by the function's derivatives at that center.
2. Periodic Functions and Euler's Formulas for Fourier Series: A periodic, piecewise-smooth function on any interval of length 2L can be decomposed into an average value plus an infinite sum of sines and cosines whose coefficients are found by integrating the function against each harmonic.
3. Integration Shortcuts and Dirichlet's Convergence Theorem: By evaluating Euler integrals with tabular integration by parts and then plugging a specific x-value into Dirichlet's convergence formula, we can find exact closed-form sums for famous infinite numerical series.
4. Even and Odd Extensions and Half-Range Series: A function defined only on [0, L] can be mirrored evenly across the vertical axis to produce a pure cosine series, or mirrored oddly through the origin to produce a pure sine series.
5. Practice: Computing Maclaurin and Taylor Series: Find the general k-th derivative formula (or rewrite the expression in terms of a standard geometric series) to generate Taylor and Maclaurin expansions and identify their intervals of convergence.
6. Practice: Full Fourier Series on 2pi-Length Intervals: Break integrals across piecewise intervals on [-pi, pi] or [0, 2pi] and apply tabular integration by parts to find full Fourier expansions and sum classical series.
7. Practice: Full Fourier Series on Arbitrary Intervals [-L, L] and [0, 2L]: Replace nx with npix/L in Euler's formulas to handle arbitrary interval lengths, and combine evaluations at continuous and jump points to solve for multiple infinite series at once.
8. Practice: Half-Range Cosine Series on [0, L]: Extend a function on [0, L] as an even function to represent it purely with cosines, taking care to compute the n = 1 coefficient separately whenever a product-to-sum formula divides by n - 1.
9. Practice: Half-Range Sine Series on [0, L]: Extend a function on [0, L] as an odd function so that a0 = 0 and an = 0, leaving only a single sine integral bn to compute.

Local Approximation with Taylor and Maclaurin Series

You already know how to compute derivatives of all orders for elementary functions; now we use those derivatives at a single point to build an infinite polynomial that matches the function. This lets us replace complicated curves with simple addition and multiplication.

Imagine driving along a winding mountain road in heavy fog. Even if you only know your exact position, speed, and how fast you are turning the wheel right now, you can predict where the road goes for the next few meters. In mathematics, a smooth function is one you can differentiate infinitely many times without any sharp corners or breaks. An infinite power series is an endless polynomial built from powers of how far you step away from a starting point $x_0$.

To make our polynomial match the function $f(x)$ near the center point $x_0$, we force both to have the exact same value, slope, curvature, and higher derivatives at $x_0$. This rule produces the Taylor series:

In plain words, each term adjusts the polynomial's $k$-th derivative to match $f^{(k)}(x_0)$, while the factorial $k!$ cancels out the constants that drop down during repeated differentiation. To see this, draw a wavy curve $f(x)$ and mark a point $x_0$. Draw a straight tangent line at $x_0$ (using up to $k=1$); it hugs the curve right at $x_0$. Now draw a parabola that also matches the second derivative (up to $k=2$); it bends with the curve and stays close over a wider stretch.

When we choose the origin $x_0 = 0$ as our center, the formula simplifies and is called a Maclaurin series:

Rather than computing derivatives every time, engineers memorize four foundational Maclaurin expansions and their region of convergence, which is the set of $x$ values where the infinite sum actually equals the function. You can also substitute algebraic expressions directly for $x$ inside these series.

Worked example

Find the first three nonzero terms of the Taylor series for $f(x) = \ln(x)$ centered at $x_0 = 1$, and find the Maclaurin series for $g(x) = \frac{1}{1+x^2}$.

1. Evaluate derivatives at the center $x_0 = 1$ to find the Taylor coefficients: We need the function value and derivatives at $x_0 = 1$. Here $f(1) = \ln(1) = 0$, $f'(x) = x^{-1}$ so $f'(1) = 1$, $f''(x) = -x^{-2}$ so $f''(1) = -1$, and $f'''(x) = 2x^{-3}$ so $f'''(1) = 2$.
2. Assemble the Taylor series for $\ln(x)$: Plugging these numbers into the Taylor formula gives $\ln(x) = 0 + \frac{1}{1!}(x-1) + \frac{-1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 + \cdots = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \cdots$.
3. Use algebraic substitution for $g(x)$ to avoid taking derivatives: Rewrite $g(x)$ as $\frac{1}{1 - (-x^2)}$ to match the standard geometric form $\frac{1}{1-u}$ with $u = -x^2$.
4. Substitute into the known Maclaurin series: Replace $x$ with $-x^2$ in $\sum_{k=0}^{\infty} x^k$ to get $g(x) = 1 - x^2 + x^4 - x^6 + \cdots = \sum_{k=0}^{\infty} (-1)^k x^{2k}$, which converges when $|-x^2| < 1$, or $|x| < 1$.

Remember

• A Taylor series builds a polynomial around $x = x_0$ using coefficients $\frac{f^{(k)}(x_0)}{k!}$; a Maclaurin series is simply a Taylor series centered at $x_0 = 0$.
• Memorize the Maclaurin series for $e^x$, $\sin(x)$, $\cos(x)$ (valid for all $x$) and $\frac{1}{1-x}$ (valid for $|x| < 1$).
• Obtain new series quickly by substituting algebraic terms into the four standard series instead of differentiating from scratch.

Periodic Functions and Euler's Formulas for Fourier Series

In the previous lesson, we built Taylor series for smooth functions around a single point. Now we will represent repeating signals—even those with sharp corners or sudden jumps—using an infinite sum of sines and cosines across a whole interval.

Think of an electrocardiogram (ECG) trace or a blinking turn signal: the same pattern repeats over and over. A periodic function satisfies $f(x + T) = f(x)$ for all $x$, where the smallest positive shift $T$ is the fundamental period. For convenience in formulas, we write this period as $T = 2L$, where $L$ is the half-period. Any interval of length $2L$, written $[c, c+2L]$, captures one full cycle and is called a fundamental interval. On a graph, if you trace the curve from $x = -L$ to $x = L$ (or from $0$ to $2L$) and slide that cutout sideways by $2L$, it lands right back on the graph.

Real engineering signals, like a switch flipping on and off, cannot be expanded in Taylor series because derivatives fail at jumps. Instead, we only require the signal to obey Dirichlet's conditions on one period: it must have a finite number of maxima and minima, and be piecewise continuous. Piecewise continuous means the graph is made of unbroken smooth pieces separated by a finite number of jump discontinuities—places where the graph steps instantaneously from a finite left-hand value to a finite right-hand value without shooting to infinity.

Just as a musical chord can be separated into pure tuning-fork notes, any periodic function meeting Dirichlet's conditions can be written as a Fourier trigonometric series $S(x)$:

In words, the function is built from a constant baseline $\frac{a_0}{2}$ plus waves of frequency $\frac{n\pi}{L}$, called harmonics. To find how much of each wave is inside $f(x)$, we use Euler's formulas over any one period $[c, c+2L]$:

Here, $\frac{a_0}{2}$ is simply the average height of the graph over one cycle. On a standard interval like $[-\pi, \pi]$ or $[0, 2\pi]$, the half-period is $L = \pi$, so $\frac{n\pi x}{L}$ simplifies nicely to $nx$.

Worked example

Find the Fourier series for the square wave $f(x) = 0$ on $(-2, 0)$ and $f(x) = 4$ on $(0, 2)$, repeated with period $2L = 4$.

1. Identify the half-period and check Dirichlet's conditions: The fundamental interval is $[-2, 2]$ of length $2L = 4$, so $L = 2$. The graph has two flat pieces and finite jumps at $x = 0$ and $x = 2$, so Dirichlet's conditions hold.

2. Compute $a_0$ to find the average height: Set $c = -2$ and $L = 2$ in Euler's formula. Since $f(x) = 0$ on $(-2, 0)$, integrate only from $0$ to $2$:

   $$a_0 = \frac{1}{2}\int_{0}^{2} 4\,dx = \frac{1}{2}[4x]_0^2 = 4$$

3. Compute $a_n$ for the cosine weights:

   $$a_n = \frac{1}{2}\int_{0}^{2} 4\cos\frac{n\pi x}{2}\,dx = 2\left[\frac{2}{n\pi}\sin\frac{n\pi x}{2}\right]_0^2 = \frac{4}{n\pi}\sin(n\pi) = 0$$

4. Compute $b_n$ for the sine weights:

   $$b_n = \frac{1}{2}\int_{0}^{2} 4\sin\frac{n\pi x}{2}\,dx = 2\left[-\frac{2}{n\pi}\cos\frac{n\pi x}{2}\right]_0^2 = \frac{4}{n\pi}(1 - \cos(n\pi))$$

Assemble the series with $\frac{a_0}{2} = 2$:

Remember

• Always divide the full period $T$ by $2$ to get the half-period $L$ before writing Euler's formulas.
• The constant term in the series is $\frac{a_0}{2}$ (the average value over one period), so take care not to forget the factor of $\frac{1}{2}$.
• When $f(x)$ is defined in pieces, split each Euler integral into matching sub-intervals.

Integration Shortcuts and Dirichlet's Convergence Theorem

Now that you can set up Euler's integrals for a Fourier series, you need fast shortcuts to evaluate them and a rule to know what the infinite sum actually equals at any point, enabling you to sum famous numerical series.

When a periodic signal is a ramp or parabola, Euler's formulas require integrating a polynomial $u(x)$ times a wave $v(x)$. Doing integration by parts repeatedly is slow and error-prone. Instead, we use Bernoulli's generalized integration by parts (often called tabular integration), where you differentiate $u$ until it vanishes and repeatedly integrate $v$:

Here $u', u''$ are successive derivatives of $u$, and $v_1, v_2, v_3$ are successive integrals of $v$. In words: multiply each derivative of the polynomial by the next integral of the wave, alternating plus and minus signs.

Other signals decay like a plucked guitar string or beat like two tuning forks. To integrate a decaying wave $e^{ax}$ times $\cos(bx)$ or $\sin(bx)$, or the product of two waves, use these direct shortcuts:

When you plug integration limits like $\pi$ or $-\pi$ into your results for an integer $n$, always simplify using $\sin(n\pi) = 0$ and $\cos(n\pi) = (-1)^n$. On a unit circle, stepping by $n$ half-turns always lands on the horizontal axis, alternating between $-1$ and $+1$.

Once you build the series, does it equal $f(x)$? Imagine tracing a square wave: at a vertical cliff, the smooth waves cannot be in two places at once, so they split the difference. A jump discontinuity is a point $x_0$ where the left-hand limit $f(x_0^-)$ (value approached from the left) and right-hand limit $f(x_0^+)$ (value approached from the right) are finite but unequal. Dirichlet's convergence theorem states that for a periodic, piecewise-smooth function, the Fourier series at any $x_0$ converges to the midpoint of the left and right limits:

If you draw a vertical jump from height $1$ to height $3$ at $x_0$, place a solid dot at height $2$; the series lands on that dot. Where $f$ is continuous, both limits equal $f(x_0)$, so the series equals $f(x_0)$. Substituting a clever $x_0$ turns the function series into an exact sum for a numerical series.

Worked example

Given that $f(x) = x^2$ on $(-\pi, \pi)$ with period $2\pi$ has $a_0 = \frac{2\pi^2}{3}$ and $b_n = 0$, find $a_n$ and evaluate $\sum_{n=1}^{\infty} \frac{1}{n^2}$.

1. Integrate using Bernoulli's formula: To find $a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x^2 \cos(nx)\,dx$, set $u = x^2$ and $v = \cos(nx)$ so we avoid integrating by parts twice.

   $$a_n = \frac{1}{\pi}\left[ (x^2)\left(\frac{\sin nx}{n}\right) - (2x)\left(-\frac{\cos nx}{n^2}\right) + (2)\left(-\frac{\sin nx}{n^3}\right) \right]_{-\pi}^{\pi}$$

2. Simplify at the limits: Because $\sin(n\pi) = 0$ and $\cos(n\pi) = (-1)^n$, the sine terms vanish and the cosine term doubles:

   $$a_n = \frac{1}{\pi}\left[ \frac{2\pi(-1)^n}{n^2} - \frac{2(-\pi)(-1)^n}{n^2} \right] = \frac{4(-1)^n}{n^2}$$

3. Apply Dirichlet's theorem at $x = \pi$: The periodic extension of $x^2$ is continuous at $x = \pi$, since $f(\pi^-) = \pi^2$ and by periodicity $f(\pi^+) = f(-\pi^+) = (-\pi)^2 = \pi^2$. Plugging $x = \pi$ into the Fourier series gives:

   $$\pi^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} \cos(n\pi)$$

4. Solve for the numerical sum: Replace $\cos(n\pi)$ with $(-1)^n$, note that $(-1)^n(-1)^n = 1$, and isolate the sum:

   $$\frac{2\pi^2}{3} = 4 \sum_{n=1}^{\infty} \frac{1}{n^2} \implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

Remember

• Use Bernoulli's formula $\int uv\,dx = uv_1 - u'v_2 + u''v_3 - \cdots$ when $u$ is a polynomial.
• For any integer $n$, always simplify $\sin(n\pi) = 0$ and $\cos(n\pi) = (-1)^n$.
• By Dirichlet's theorem, a Fourier series converges to $f(x_0)$ at continuous points and to $\frac{f(x_0^-) + f(x_0^+)}{2}$ at jumps.

Even and Odd Extensions and Half-Range Series

In the previous lessons, you learned how to find full Fourier series by integrating both cosines and sines over a full period $[-L, L]$. Now, you will use symmetry to cut that work in half and represent a function defined only on $[0, L]$ using purely cosines or purely sines.

Imagine folding a sheet of paper along the vertical $y$-axis. An even function matches up with itself when folded across the $y$-axis, so $f(-x) = f(x)$, like $\cos(x)$ or $x^2$. On a symmetric interval $[-L, L]$, the signed area of an even function on the left equals the area on the right: $\int_{-L}^{L} f(x)\,dx = 2\int_{0}^{L} f(x)\,dx$. An odd function matches up only if you fold across the $y$-axis and then flip across the $x$-axis, so $f(-x) = -f(x)$, like $\sin(x)$ or $x$. Because its left and right areas have opposite signs, the symmetric integral of an odd function is zero: $\int_{-L}^{L} f(x)\,dx = 0$.

In engineering, a signal or a vibrating string of length $L$ is often defined only on $[0, L]$. Since we do not care what happens for negative $x$, we can invent the left half on $[-L, 0]$ to create symmetry. An even periodic extension reflects the graph across the $y$-axis onto $[-L, 0]$ and repeats it every $2L$, while an odd periodic extension reflects the graph across the $y$-axis and then flips it upside down across the $x$-axis before repeating.

Because the product of an even and an odd function is always odd, choosing an extension zeroes out half the Euler coefficients and yields a half-range series valid on $[0, L]$:

Worked example

Find the half-range sine series for $f(x) = 1$ on $[0, \pi]$.

1. Choose the extension: A half-range sine series requires an odd extension, so $a_0 = 0$ and $a_n = 0$ automatically.
2. Set up the sine coefficient $b_n$: With $L = \pi$, double the integral over the half-period $[0, \pi]$: $b_n = \frac{2}{\pi}\int_0^\pi 1 \cdot \sin(nx)\,dx$.
3. Integrate and simplify: Evaluating gives $b_n = \frac{2}{\pi}\left[-\frac{\cos(nx)}{n}\right]_0^\pi = \frac{2}{n\pi}(1 - (-1)^n)$. This equals $0$ for even $n$ and $\frac{4}{n\pi}$ for odd $n$.
4. Write the series: Summing over odd integers $n = 1, 3, 5, \dots$ gives $f(x) \sim \frac{4}{\pi}\left(\sin(x) + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5} + \dots\right)$.

Remember

• Even $\times$ Even = Even, Odd $\times$ Odd = Even, and Even $\times$ Odd = Odd.
• On $[-L, L]$, the integral of an odd function is $0$; the integral of an even function is $2\int_0^L$.
• On $[0, L]$, mirror evenly across the $y$-axis for a pure cosine series ($b_n = 0$), or flip oddly through the origin for a pure sine series ($a_0 = a_n = 0$).

Practice: Computing Maclaurin and Taylor Series

With the theory of Taylor and Maclaurin series in place, we now practice deriving power series expansions from higher-order derivative patterns and algebraic substitutions. You will find the general $k$-th derivative formula or rewrite an expression in terms of a standard geometric series to generate expansions and identify their intervals of convergence.

The idea

Writing out the first three or four terms of a Taylor or Maclaurin series is straightforward, but writing the infinite sum requires a general formula for the $k$-th derivative, $f^{(k)}(x)$. By computing the first few derivatives before plugging in the center $x_0$, you can spot how signs alternate, how factorials grow from repeated power-rule steps, or how trigonometric functions cycle every four derivatives. Once you write $f^{(k)}(x_0)$ in terms of $k$, dividing by $k!$ often cancels factorial factors cleanly.

For rational functions, repeated differentiation produces a geometric series whose terms are powers of $(x - x_0)$. Once you know the basic Maclaurin expansion $\frac{1}{1-u} = 1 + u + u^2 + u^3 + \dots$, which converges on the interval of convergence $|u| < 1$, you do not need to differentiate from scratch every time. You can factor out constants to create a leading $1$ in the denominator and substitute a rescaled variable $u$ directly into the known series. The series is valid wherever the substituted expression satisfies $|u| < 1$.

Worked example

Given: The exponential function $f(x) = e^x$ centered at $x_0 = 0$. Find: The Maclaurin series expansion of $f(x)$.

1. Evaluate the function at the center $x_0 = 0$: $f(0) = e^0 = 1$, which gives the constant term.
2. Compute the first few derivatives at $x = 0$: since $f'(x) = e^x$ and $f''(x) = e^x$, we get $f'(0) = 1$ and $f''(0) = 1$.
3. State the general $k$-th derivative pattern: because $e^x$ is its own derivative, $f^{(k)}(x) = e^x$, so $f^{(k)}(0) = 1$ for all $k \ge 0$.
4. Substitute $f^{(k)}(0) = 1$ into the Maclaurin formula $\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k$ to write out the expansion.

Answer: $1 + x + \frac{x^2}{2!} + \dots + \frac{x^k}{k!} + \dots$

More problems

Problem Q(i): Maclaurin series for the decaying exponential

Given: The function $f(x) = e^{-x}$ for all $x \in \mathbb{R}$. Find: The Maclaurin series of $f(x)$.

1. By the chain rule, each derivative of $e^{-x}$ multiplies by $-1$, giving $f^{(k)}(0) = (-1)^k$ (or substitute $-x$ into the series for $e^x$).
2. Substitute these coefficients into the Maclaurin series formula to get the alternating expansion.

Answer: $1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$

Problem Q(ii): Maclaurin series for the basic geometric fraction

Given: The function $f(x) = \frac{1}{1-x}$ in the region $|x| < 1$. Find: The Maclaurin series of $f(x)$.

1. Differentiate $f(x) = (1-x)^{-1}$ repeatedly using the chain rule to find $f^{(k)}(x) = \frac{k!}{(1-x)^{k+1}}$, so $f^{(k)}(0) = k!$.
2. Divide $f^{(k)}(0) = k!$ by $k!$ in the Maclaurin formula, leaving a coefficient of $1$ on every power $x^k$.

Answer: $1 + x + x^2 + x^3 + \dots$

Problem 1: Maclaurin series for the sine function

Given: The trigonometric function $f(x) = \sin x$. Find: The Maclaurin series of $f(x)$.

1. Evaluate the function and its first four derivatives at $0$: $f(0) = 0$, $f'(0) = \cos 0 = 1$, $f''(0) = -\sin 0 = 0$, $f'''(0) = -\cos 0 = -1$, and $f^{(4)}(0) = \sin 0 = 0$.
2. Group the derivatives into even and odd orders: for $k \ge 0$, all even derivatives vanish ($f^{(2k)}(0) = 0$) because $\sin 0 = 0$.
3. Write the odd derivative formula: $f^{(2k+1)}(x) = (-1)^k \cos x$, which gives $f^{(2k+1)}(0) = (-1)^k$ for $k \ge 0$.
4. Substitute the non-zero odd terms into the Maclaurin formula to obtain the odd-power series.

Answer: $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

Problem 2: Maclaurin series for the cosine function

Given: The trigonometric function $f(x) = \cos x$. Find: The Maclaurin series of $f(x)$.

1. Evaluate the derivatives of $\cos x$ at $x = 0$: odd derivatives contain $\sin 0$ and vanish, while even derivatives alternate as $f^{(2k)}(0) = (-1)^k$.
2. Substitute the even-order derivatives into the Maclaurin series formula.

Answer: $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}$

Problem 3: Maclaurin series for an alternating rational function

Given: The function $f(x) = \frac{1}{1+x}$ for $|x| < 1$. Find: The Maclaurin series of $f(x)$.

1. Compute the value and first two derivatives at $x = 0$: $f(0) = 1$, $f'(x) = \frac{-1}{(1+x)^2} \implies f'(0) = -1$, and $f''(x) = \frac{2}{(1+x)^3} \implies f''(0) = 2$.
2. Compute the third and fourth derivatives to see the factorial pattern: $f'''(x) = \frac{-2 \cdot 3}{(1+x)^4} \implies f'''(0) = -3!$ and $f^{(4)}(x) = \frac{2 \cdot 3 \cdot 4}{(1+x)^5} \implies f^{(4)}(0) = 4!$.
3. State the general $k$-th derivative at $x = 0$: $f^{(k)}(0) = (-1)^k k!$.
4. Substitute $f^{(k)}(0)$ into the Maclaurin formula and cancel $k!$ in the numerator and denominator.

Watch out. The $k$-th derivative of $(1+x)^{-1}$ has denominator $(1+x)^{k+1}$, not $(1+x)^k$; at $x = 0$ both equal $1$, so $f^{(k)}(0) = (-1)^k k!$ is unaffected.

Answer: $1 - x + x^2 - x^3 + \dots = \sum_{k=0}^{\infty} (-1)^k x^k$

Problem Q(1): Taylor series of the reciprocal function at x = 1

Given: The function $f(x) = \frac{1}{x}$ about $x_0 = 1$ in the region $|x-1| < 1$. Find: The Taylor series of $f(x)$ at $x = 1$.

1. Evaluate $f(x)$ and its first four derivatives at $x_0 = 1$: $f(1) = 1$, $f'(1) = -1$, $f''(1) = 2$, $f'''(1) = -3!$, and $f^{(4)}(1) = 4!$.
2. Write the general derivative value at $x_0 = 1$: $f^{(k)}(1) = (-1)^k k!$ for $k \ge 1$ (and $k = 0$).
3. Substitute $f^{(k)}(1)$ into the Taylor formula $\sum_{k=0}^{\infty} \frac{f^{(k)}(1)}{k!} (x-1)^k$ and cancel $k!$ to get powers of $(x-1)$.

Watch out. The $k$-th derivative of $x^{-1}$ is $\frac{(-1)^k k!}{x^{k+1}}$ (exponent $k+1$, not $k$), which still evaluates to $(-1)^k k!$ at $x = 1$.

Answer: $1 - (x-1) + (x-1)^2 - (x-1)^3 + \dots = \sum_{k=0}^{\infty} (-1)^k (x-1)^k$

Problem Q(2): Taylor series of a shifted rational function at x = 1

Given: The function $f(x) = \frac{1}{1+x}$ at $x_0 = 1$ in the region $|x-1| < 1$. Find: The Taylor series of $f(x)$ at $x = 1$.

1. Evaluate $f(x)$ and its first two derivatives at $x_0 = 1$: $f(1) = \frac{1}{2}$, $f'(x) = \frac{-1}{(1+x)^2} \implies f'(1) = -\frac{1}{4}$, and $f''(x) = \frac{2}{(1+x)^3} \implies f''(1) = \frac{2}{2^3}$.
2. Evaluate the third and fourth derivatives at $x_0 = 1$: $f'''(1) = \frac{-3!}{2^4}$ and $f^{(4)}(1) = \frac{4!}{2^5}$.
3. Write the general $k$-th derivative formula at $x_0 = 1$: from $f^{(k)}(x) = \frac{(-1)^k k!}{(1+x)^{k+1}}$, we have $f^{(k)}(1) = \frac{(-1)^k k!}{2^{k+1}}$ for $k \ge 0$.
4. Substitute $f^{(k)}(1)$ into the Taylor formula $\sum_{k=0}^{\infty} \frac{f^{(k)}(1)}{k!}(x-1)^k$ and cancel $k!$ in each term.

Answer: $\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{k+1}} (x-1)^k = \frac{1}{2} - \frac{1}{2^2}(x-1) + \frac{1}{2^3}(x-1)^2 - \dots$

Problem Q(3): Maclaurin series by algebraic factoring and substitution

Given: The function $f(x) = \frac{3}{2-x}$ in the region $|x| < 2$. Find: The Maclaurin series of $f(x)$.

1. Factor $2$ from the denominator so the constant term becomes $1$: $f(x) = \frac{3}{2}\left(\frac{1}{1 - x/2}\right)$.
2. Recall the standard geometric Maclaurin series $\frac{1}{1-u} = 1 + u + u^2 + u^3 + \dots$ for $|u| < 1$.
3. Check the convergence condition: since $|x| < 2$, the rescaled variable satisfies $|\frac{x}{2}| < 1$.
4. Replace $u$ with $\frac{x}{2}$ in the geometric series and multiply by $\frac{3}{2}$.

Answer: $\frac{3}{2} \left(1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^3 + \dots \right)$

Practice: Full Fourier Series on 2pi-Length Intervals

Having mastered power series at a point, we now practice computing full trigonometric Fourier series on standard $2\pi$ intervals for square waves and polynomials. The new skill here is breaking Euler's integrals across piecewise subintervals on $[-\pi, \pi]$ or $[0, 2\pi]$ and applying tabular integration by parts to find full Fourier expansions and sum classical infinite series.

The idea

When a function $f(x)$ has period $2L = 2\pi$, the harmonic factor $\frac{n\pi x}{L}$ simplifies directly to $nx$. Whether the base interval is $[-\pi, \pi]$ (where $c = -\pi$) or $[0, 2\pi]$ (where $c = 0$), Euler's formulas take the streamlined form:

Two integration techniques handle almost every problem on these intervals. First, if $f(x)$ is a piecewise-constant square wave with jump discontinuities, split each Euler integral into separate integrals over each subinterval where the formula for $f(x)$ is constant. Second, if $f(x)$ is a polynomial, use Bernoulli's integration by parts (tabular integration):

where $u$ is the polynomial differentiated until it vanishes, and $v_1, v_2, v_3, \dots$ are successive antiderivatives of $\cos(nx)$ or $\sin(nx)$.

Once you assemble the Fourier series $S(x) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)]$, you can sum numerical series by applying Dirichlet's convergence theorem: at any point $x$ where $f(x)$ is continuous, the series converges directly to the function value, $S(x) = f(x)$.

Worked example

Problem 2: Given: $f(x) = \begin{cases} -K, & -\pi < x < 0 \\ K, & 0 < x < \pi \end{cases}$ on $(-\pi, \pi)$ with $K > 0$. Find: The Fourier series of $f(x)$ and the sum of $1 - \frac{1}{3} + \frac{1}{5} - \dots$

1. Split the $a_0$ integral at the jump $x = 0$ because $f(x)$ changes formula there:

2. Split the $a_n$ integral at $x = 0$ and integrate $\cos(nx)$, which vanishes because $\sin(n\pi) = \sin(0) = 0$:

   $$a_n = \frac{1}{\pi}\left[\left[\frac{-K\sin(nx)}{n}\right]_{-\pi}^{0} + \left[\frac{K\sin(nx)}{n}\right]_{0}^{\pi}\right] = 0$$

3. Split the $b_n$ integral at $x = 0$ and integrate $\sin(nx)$ to get cosine terms:

   $$b_n = \frac{1}{\pi}\left[\left[\frac{K\cos(nx)}{n}\right]_{-\pi}^{0} + \left[\frac{-K\cos(nx)}{n}\right]_{0}^{\pi}\right]$$

4. Evaluate at the limits using $\cos(-n\pi) = \cos(n\pi) = (-1)^n$ to simplify $b_n$:

   $$b_n = \frac{K}{n\pi}[2 - 2\cos(n\pi)] = \frac{2K}{n\pi}(1 - (-1)^n)$$

5. Substitute $a_0$, $a_n$, and $b_n$ into $S(x)$, noting that $1 - (-1)^n$ equals $2$ for odd $n$ and $0$ for even $n$:

   $$f(x) = \frac{4K}{\pi}\left[\sin x + \frac{1}{3}\sin(3x) + \frac{1}{5}\sin(5x) + \dots\right]$$

6. Substitute $x = \frac{\pi}{2}$, where $f(x)$ is continuous with $f\left(\frac{\pi}{2}\right) = K$, to evaluate the alternating series:

   $$K = \frac{4K}{\pi}\left(1 - \frac{1}{3} + \frac{1}{5} - \dots\right) \implies 1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{\pi}{4}$$

Answer: $f(x) = \frac{4K}{\pi}\left[\sin x + \frac{1}{3}\sin(3x) + \frac{1}{5}\sin(5x) + \dots\right]$, and $1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{\pi}{4}$.