🌟 Maclaurin Series — Simplifying Functions into Infinite Polynomials

Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Series Expansion | Applications of Differentiation

---

🔹 What Is a Maclaurin Series?

The Maclaurin Series is a special form of the Taylor Series, where the expansion is centered at x=0x = 0x=0.

It expresses a smooth function as an infinite polynomial, allowing you to approximate complex functions like exe^xex, sin⁡x\sin xsinx, or ln⁡(1+x)\ln(1+x)ln(1+x) using simple powers of xxx.

In essence, the Maclaurin Series gives you a way to “rebuild” a function using only its derivatives at x=0x = 0x=0.

---

🔹 The Formula

If f(x)f(x)f(x) has derivatives of all orders at x=0x = 0x=0, its Maclaurin Series is: f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+f(4)(0)4!x4+⋯f(x) = f(0) + f'(0)x + \frac{f”(0)}{2!}x^2 + \frac{f”'(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdotsf(x)=f(0)+f′(0)x+2!f′′(0)​x2+3!f′′′(0)​x3+4!f(4)(0)​x4+⋯

or compactly: f(x)=∑n=0∞f(n)(0)n!xnf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^nf(x)=n=0∑∞​n!f(n)(0)​xn

This is simply a Taylor expansion about a=0a = 0a=0.

---

🔹 Why Is It Important?

• It provides powerful approximations for functions that are otherwise difficult to compute.
• It’s used in calculus, physics, computer science, and engineering.
• It forms the basis for scientific computing and numerical methods.
• And it’s a guaranteed topic in IB HL, AP Calculus BC, A Level P4, and STEP exams.

---

🔹 Step-by-Step Example 1 — exe^xex

f(x)=ex⇒f(n)(x)=ex⇒f(n)(0)=1f(x) = e^x \Rightarrow f^{(n)}(x) = e^x \Rightarrow f^{(n)}(0) = 1f(x)=ex⇒f(n)(x)=ex⇒f(n)(0)=1

Substitute into the formula: ex=1+x+x22!+x33!+x44!+⋯e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdotsex=1+x+2!x2​+3!x3​+4!x4​+⋯

✅ This series converges for all xxx — one of the most important functions in mathematics.

---

🔹 Step-by-Step Example 2 — sin⁡x\sin xsinx

f(x)=sin⁡x, f′(x)=cos⁡x, f′′(x)=−sin⁡x, f′′′(x)=−cos⁡xf(x) = \sin x, \ f'(x) = \cos x, \ f”(x) = -\sin x, \ f”'(x) = -\cos xf(x)=sinx, f′(x)=cosx, f′′(x)=−sinx, f′′′(x)=−cosx

At x=0x = 0x=0: f(0)=0, f′(0)=1, f′′(0)=0, f′′′(0)=−1f(0) = 0, \ f'(0) = 1, \ f”(0) = 0, \ f”'(0) = -1f(0)=0, f′(0)=1, f′′(0)=0, f′′′(0)=−1 sin⁡x=x−x33!+x55!−x77!+⋯\sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \cdotssinx=x−3!x3​+5!x5​−7!x7​+⋯

✅ Alternating signs and only odd powers appear — matching the symmetry of the sine curve.

---

🔹 Example 3 — cos⁡x\cos xcosx

f(x)=cos⁡x, f′(x)=−sin⁡x, f′′(x)=−cos⁡xf(x) = \cos x, \ f'(x) = -\sin x, \ f”(x) = -\cos xf(x)=cosx, f′(x)=−sinx, f′′(x)=−cosx

At x=0x = 0x=0: f(0)=1, f′(0)=0, f′′(0)=−1f(0) = 1, \ f'(0) = 0, \ f”(0) = -1f(0)=1, f′(0)=0, f′′(0)=−1 cos⁡x=1−x22!+x44!−x66!+⋯\cos x = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \cdotscosx=1−2!x2​+4!x4​−6!x6​+⋯

✅ Only even powers appear — another beautiful symmetry.

---

🔹 Example 4 — ln⁡(1+x)\ln(1+x)ln(1+x)

f(x)=ln⁡(1+x)f(x) = \ln(1+x)f(x)=ln(1+x) f′(x)=11+x, f′′(x)=−1(1+x)2, f′′′(x)=2(1+x)3f'(x) = \frac{1}{1+x}, \ f”(x) = -\frac{1}{(1+x)^2}, \ f”'(x) = \frac{2}{(1+x)^3}f′(x)=1+x1​, f′′(x)=−(1+x)21​, f′′′(x)=(1+x)32​

At x=0x = 0x=0: f(0)=0, f′(0)=1, f′′(0)=−1, f′′′(0)=2f(0) = 0, \ f'(0) = 1, \ f”(0) = -1, \ f”'(0) = 2f(0)=0, f′(0)=1, f′′(0)=−1, f′′′(0)=2 ln⁡(1+x)=x−x22+x33−x44+⋯\ln(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \cdotsln(1+x)=x−2×2​+3×3​−4×4​+⋯

✅ Valid for ∣x∣<1|x| < 1∣x∣<1 (convergence condition).

---

🔹 Maclaurin vs Taylor Series

Feature	Taylor Series	Maclaurin Series
Expansion Point	x=ax = ax=a	x=0x = 0x=0
Formula	f(a)+f′(a)(x−a)+⋯f(a) + f'(a)(x-a) + \cdotsf(a)+f′(a)(x−a)+⋯	f(0)+f′(0)x+⋯f(0) + f'(0)x + \cdotsf(0)+f′(0)x+⋯
Usage	Local expansion near x=ax = ax=a	Simplest case, centered at origin

---

🔹 Common Mistakes

1. ❌ Forgetting the factorial term n!n!n! in denominators.
2. ❌ Expanding about the wrong point — Maclaurin always means x=0x=0x=0.
3. ❌ Not checking convergence interval — some series only work near x=0x=0x=0.
4. ❌ Dropping signs in alternating series like sin⁡x\sin xsinx, ln⁡(1+x)\ln(1+x)ln(1+x).

---

🔹 Advanced Tip — Error Term (Remainder)

For better accuracy, the error after using nnn terms is given by: Rn(x)=f(n+1)(c)(n+1)!xn+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}Rn​(x)=(n+1)!f(n+1)(c)​xn+1

where ccc lies between 000 and xxx.

This helps in estimating how precise a truncated series is — crucial for IB Paper 3 and STEP reasoning questions.

---

🌟 Why It Matters

The Maclaurin Series is the language of approximation in calculus.
It connects infinite processes to finite expressions and builds the foundation for:

• Modern analysis and numerical computation
• Series solutions of differential equations
• Advanced modeling in physics and engineering
• Entrance tests like STEP, MAT, TMUA, and Olympiads

---

📘 Learn Beyond the Formula

At Math By Rishabh, the Maclaurin Series isn’t taught as memorization — it’s explored as a concept of creation and symmetry.

In the Mathematics Elevate Mentorship Program, you’ll:
✅ Learn to derive expansions from scratch,
✅ Apply them in real problems,
✅ Build conceptual mastery for higher exams.

🚀 Think like a mathematician — not a memorizer.
👉 Book your personalized session now at MathByRishabh.com