Taylor/Maclaurin Series & Power Series of ln(1+x)
What if you could find the 2025th derivative of a function without differentiating even once? In this lesson, you’ll learn how to turn \(\ln(1+x)\) into an infinite polynomial using a clever integration trick, and then use that polynomial to instantly read off any derivative you want. Along the way, you’ll also see the beautiful cycle that happens when you keep taking derivatives of sine and cosine. This is where algebra and calculus start to feel like superpowers.
Power series let us turn complicated functions into infinite polynomials we can work with easily. The function \(\ln(1+x)\) shows up everywhere:
- Computer science: algorithms measure efficiency using logarithms — understanding \(\ln(1+x)\) for small \(x\) helps approximate running times
- Finance: when interest rates are small, \(\ln(1+r) \approx r\) — bankers use this shortcut daily
- Physics: sensors measure tiny changes, and \(\ln(1+x) \approx x\) for small \(x\) simplifies calculations enormously
- Engineering: signal processing uses logarithmic series to analyze audio and radio waves
- Biology: population models use \(\ln(1+\text{growth rate})\) to predict how species change over time
Today we’ll discover how to build the power series for \(\ln(1+x)\) from a geometric series, and we’ll use Maclaurin series to find ridiculously high-order derivatives without actually differentiating!
Topics Covered
- Derivatives of \(\sin x\) and \(\cos x\) from Euler’s formula
- Periodicity of higher-order derivatives of \(\sin x\) and \(\cos x\)
- Geometric series \(\frac{1}{1+x}\) as a power series
- Integrating a power series term by term to get \(\ln(1+x)\)
- Determining the integration constant by plugging in \(x = 0\)
- Maclaurin coefficients: if \(f(x) = \sum a_k x^k\), then \(a_n = \frac{f^{(n)}(0)}{n!}\)
- Using Maclaurin expansions to find higher derivatives (e.g., the 2025th derivative of \(\ln(1+x)\) at \(x = 0\))
- Recognizing derivatives of known functions: \(f'(x) = -\frac{1}{1+x}\) matches the derivative of \(\ln(1+x)\)
Lecture Video
Key Frames from the Lecture
What You Need to Know First
Euler’s formula connects exponentials with trig functions using the imaginary number \(i\) (where \(i^2 = -1\)):
\[e^{i\theta} = \cos\theta + i\sin\theta\]
From this, we can extract \(\sin\) and \(\cos\) by looking at the power series of \(e^{i\theta}\) and separating the real and imaginary parts:
\[\cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots\]
\[\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots\]
These power series are what let us differentiate \(\sin\) and \(\cos\) term by term!
A geometric series is a sum where each term is a fixed multiple of the previous one:
\[\sum_{k=0}^{\infty} r^k = 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r} \quad \text{(when } |r| < 1\text{)}\]
For example, with \(r = \frac{1}{2}\): \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 2\).
In this lesson we’ll use the substitution \(r = -x\) to get a series for \(\frac{1}{1+x}\).
A Maclaurin series expresses a function as an infinite polynomial centered at \(x = 0\):
\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\]
The key insight is that the coefficient of \(x^n\) equals \(\frac{f^{(n)}(0)}{n!}\). If you already know the power series, you can read off derivatives at zero without differentiating!
The natural logarithm \(\ln(x)\) is the inverse of \(e^x\). It answers: “what power of \(e\) gives \(x\)?”
Key facts:
- \(\ln(1) = 0\) because \(e^0 = 1\)
- \(\ln(e) = 1\) because \(e^1 = e\)
- \(\frac{d}{dx}\ln(x) = \frac{1}{x}\)
So \(\frac{d}{dx}\ln(1+x) = \frac{1}{1+x}\) by the chain rule.
Key Concepts
Derivatives of sin and cos from Euler’s Formula
Last time we saw the Maclaurin series for \(\sin x\) and \(\cos x\). By differentiating those series term by term, we can prove:
Differentiating the power series for sine and cosine term by term proves these two fundamental results. Notice the minus sign for cosine – it shows up because cosine decreases wherever sine is positive.
\[\boxed{\frac{d}{dx}(\sin x) = \cos x} \qquad \boxed{\frac{d}{dx}(\cos x) = -\sin x}\]
Why the minus sign for cosine? Differentiate the series for \(\cos x\):
\[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\]
\[\frac{d}{dx}(\cos x) = 0 - \frac{2x}{2!} + \frac{4x^3}{4!} - \frac{6x^5}{6!} + \cdots = -x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots\]
That’s exactly \(-\sin x\)!
Explore – see \(\sin x\), \(\cos x\), and their derivatives:
Drag the slider for \(a\). The dashed line shows the tangent to \(\sin x\) – its slope is always \(\cos(a)\), the height of the red curve!
The Derivative Cycle: Higher Derivatives of sin
Something beautiful happens when you keep differentiating \(\sin x\):
| Derivative | Result |
|---|---|
| \(f(x) = \sin x\) | \(\sin x\) |
| \(f'(x)\) | \(\cos x\) |
| \(f''(x)\) | \(-\sin x\) |
| \(f'''(x)\) | \(-\cos x\) |
| \(f^{(4)}(x)\) | \(\sin x\) |
After 4 derivatives, we’re back to \(\sin x\)! The cycle repeats forever:
\[\sin x \;\to\; \cos x \;\to\; -\sin x \;\to\; -\cos x \;\to\; \sin x \;\to\; \cdots\]
So to find the \(n\)th derivative, just divide \(n\) by 4 and look at the remainder:
| Remainder (\(n \bmod 4\)) | \(f^{(n)}(x)\) |
|---|---|
| 0 | \(\sin x\) |
| 1 | \(\cos x\) |
| 2 | \(-\sin x\) |
| 3 | \(-\cos x\) |
For example, the 100th derivative of \(\sin x\): \(100 \div 4 = 25\) remainder \(0\), so \(f^{(100)}(x) = \sin x\).
Geometric Series as a Power Series
Recall the geometric series formula:
\[\frac{1}{1-r} = 1 + r + r^2 + r^3 + \cdots \quad \text{for } |r| < 1\]
Now substitute \(r = -x\):
\[\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - \cdots = \sum_{k=0}^{\infty} (-1)^k x^k\]
This is valid for \(|x| < 1\). We’ve turned \(\frac{1}{1+x}\) into a power series!
Explore – see how partial sums of the series approximate \(\frac{1}{1+x}\):
Increase \(n\) with the slider and watch the red dashed curve match the blue curve better and better – but only for \(|x| < 1\)!
From Geometric Series to ln(1+x)
Here’s the key idea: we know that \(\frac{d}{dx}\ln(1+x) = \frac{1}{1+x}\). So if we integrate the geometric series term by term, we should get \(\ln(1+x)\)!
\[\int \frac{1}{1+x}\,dx = \int \left(1 - x + x^2 - x^3 + \cdots\right) dx\]
\[= C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]
Determining the constant \(C\): plug in \(x = 0\):
\[\ln(1+0) = C + 0 - 0 + 0 - \cdots\] \[\ln(1) = C\] \[0 = C\]
So \(C = 0\), and we get the beautiful result:
By integrating the geometric series for \(\frac{1}{1+x}\) term by term, we turn the natural logarithm into an infinite polynomial. This series lets you compute logarithms using only addition, subtraction, and division.
\[\boxed{\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}\, x^k}{k}}\]
This is valid for \(-1 < x \le 1\).
Explore – see the power series for \(\ln(1+x)\) converge:
Increase \(n\) and watch the series (red dashed) hug the real \(\ln(1+x)\) curve (blue) more closely, especially between \(-1\) and \(1\).
Recognizing Derivatives of Known Functions
Suppose someone gives you a function and tells you \(f'(x) = -\frac{1}{1+x}\). How do you find \(f(x)\)?
You recognize that:
\[\frac{d}{dx}\ln(1+x) = \frac{1}{1+x}\]
So \(-\frac{1}{1+x}\) is the derivative of \(-\ln(1+x)\). Therefore:
\[f(x) = -\ln(1+x) + C\]
This is the art of “working backwards” from a derivative – a preview of integration!
Matching Maclaurin Coefficients
This is one of the most powerful tricks in the course. If a function has a Maclaurin series:
\[f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots\]
then each coefficient is related to a derivative at zero:
If you already know a function’s power series, you can instantly find any derivative at \(x = 0\) without doing any differentiation at all. Just read off the coefficient and multiply by \(n!\).
\[\boxed{a_n = \frac{f^{(n)}(0)}{n!}}\]
This means: if you know the power series, you automatically know every derivative at \(x = 0\). You can rearrange:
\[f^{(n)}(0) = n! \cdot a_n\]
Finding Crazy-High Derivatives Without Differentiating
This is where things get amazing. Suppose we want the 2025th derivative of \(\ln(1+x)\) at \(x = 0\).
Differentiating \(\ln(1+x)\) two thousand and twenty-five times by hand? No way! But we know the power series:
\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]
The coefficient of \(x^n\) is \(a_n = \frac{(-1)^{n+1}}{n}\).
Using \(f^{(n)}(0) = n! \cdot a_n\):
\[f^{(2025)}(0) = 2025! \cdot \frac{(-1)^{2026}}{2025} = 2025! \cdot \frac{1}{2025} = \frac{2025!}{2025} = 2024!\]
So the 2025th derivative of \(\ln(1+x)\) at \(x = 0\) is \(2024!\). We never differentiated even once!
Why does \((-1)^{2026} = 1\)? Because 2026 is even, and \((-1)^{\text{even}} = 1\).
Let’s try another: the 4th derivative of \(\ln(1+x)\) at \(x = 0\):
\[f^{(4)}(0) = 4! \cdot \frac{(-1)^{5}}{4} = 24 \cdot \frac{-1}{4} = -6\]
You can verify this by differentiating \(\ln(1+x)\) four times the hard way – you’ll get the same answer!
Cheat Sheet
| What you want | Formula |
|---|---|
| Derivative of \(\sin x\) | \(\frac{d}{dx}(\sin x) = \cos x\) |
| Derivative of \(\cos x\) | \(\frac{d}{dx}(\cos x) = -\sin x\) |
| Derivative cycle of \(\sin\) | \(\sin \to \cos \to -\sin \to -\cos \to \sin\) (period 4) |
| Geometric series | \(\frac{1}{1+x} = \displaystyle\sum_{k=0}^{\infty} (-1)^k x^k = 1 - x + x^2 - x^3 + \cdots\) |
| Power series for \(\ln(1+x)\) | \(\ln(1+x) = \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^k}{k} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\) |
| Maclaurin coefficient formula | \(a_n = \frac{f^{(n)}(0)}{n!}\), so \(f^{(n)}(0) = n!\cdot a_n\) |
| Integration constant trick | Plug in \(x = 0\) after integrating to find \(C\) |
The Big Chain of Ideas
\[\frac{1}{1+x} \;\xrightarrow{\text{integrate}}\; \ln(1+x) \;\xrightarrow{\text{read off } a_n}\; \frac{(-1)^{n+1}}{n} \;\xrightarrow{\;n!\cdot a_n\;}\; f^{(n)}(0)\]