Taylor & Maclaurin Series Expander
Expand sin, cos, eˣ, ln(1+x), 1/(1−x) and polynomials about center a into the first n terms, with coefficients and a sample comparison.
Input Data
Results
At a glance:The Taylor series writes a function as a power series about center a: f(x)=Σ_{k=0}^∞ f⁽ᵏ⁾(a)/k!·(x−a)ᵏ; at a=0 it is the Maclaurin series. This tool gives exact coefficients for common functions and compares the series value with the exact value at a sample point to observe convergence.
Formula
General: f(x) ≈ Σ_{k=0}^{n−1} [f⁽ᵏ⁾(a)/k!] (x−a)ᵏ.
eˣ = 1 + x + x²/2! + x³/3! + …
sin x = x − x³/3! + x⁵/5! − …
cos x = 1 − x²/2! + x⁴/4! − …
ln(1+x) = x − x²/2 + x³/3 − … (|x|<1)
1/(1−x) = 1 + x + x² + x³ + … (|x|<1)
$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k$$$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$How to Use
- Pick a function (sin/cos/eˣ/ln(1+x)/1/(1−x) or a custom polynomial).
- Enter center a (a=0 = Maclaurin) and term count n.
- Enter a sample x to compare series vs exact.
- In polynomial mode enter coefficients c₀–c₄.
Common Maclaurin series (a=0)
| Function | Series | Converges |
|---|---|---|
| eˣ | Σ xᵏ/k! | all x |
| sin x | Σ (−1)ᵏ x^{2k+1}/(2k+1)! | all x |
| cos x | Σ (−1)ᵏ x^{2k}/(2k)! | all x |
| ln(1+x) | Σ (−1)^{k−1} xᵏ/k | |x|<1 |
| 1/(1−x) | Σ xᵏ | |x|<1 |
More terms → better approximation, but ln(1+x) and 1/(1−x) converge only for |x|<1.
Case Studies
sin x at a=0, 5 terms
= x − x³/6 + x⁵/120 − x⁷/5040 + x⁹/362880.
x=0.5: series ≈ 0.479427, exact sin0.5 ≈ 0.479426 (match).
eˣ at a=0, 6 terms
= 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120.
x=1: series ≈ 2.716667, exact e ≈ 2.718282.
cos x at a=0, 5 terms
= 1 − x²/2 + x⁴/24 − x⁶/720 + x⁸/40320.
ln(1+x) at a=0, 5 terms
= x − x²/2 + x³/3 − x⁴/4 + x⁵/5 (converges |x|<1).
x=0.5: series ≈ 0.407292, exact ln1.5 ≈ 0.405465.
1/(1−x) at a=0, 5 terms
= 1 + x + x² + x³ + x⁴ (geometric, |x|<1).
eˣ at a=1 (Taylor, not Maclaurin)
= e·[1 + (x−1) + (x−1)²/2 + (x−1)³/6 + …].
Centered at a=1; radius of convergence still ∞.
FAQ
Taylor vs Maclaurin?
The Maclaurin series is the Taylor series at center a=0. Same form, different expansion center.
Why does ln(1+x) converge only for |x|<1?
ln(1+x) has a singularity at x=−1, so the radius of convergence is 1; for x≥1 or x≤−1 the series diverges.
More terms always better?
Within the convergence domain, yes. Near the boundary (x close to 1) convergence is slow and needs many terms.
Where do coefficients f⁽ᵏ⁾(a)/k! come from?
They make the series match the function's derivatives up to order k at a. This tool gives closed forms directly, no step-by-step differentiation needed.
Is polynomial expansion meaningful?
Yes. A polynomial is itself a finite Taylor series (higher coefficients 0); expanding about a shows the shifted form.
Series for 1/(1−x)?
The geometric series Σ xᵏ = 1 + x + x² + …, convergent for |x|<1; it underlies many expansions.
Can a ≠ 0?
Yes. The Taylor series uses (x−a); this tool supports any center a, e.g. eˣ expanded at a=1.
Series far from exact?
If the sample is outside the convergence domain (e.g. ln at x>1) or n is too small, the gap is large; increase n or pick x inside the domain.
Related Tools
References
Content review: Calculatorism Science Team. Taylor coefficients for sin/cos/eˣ/ln/1/(1−x) and polynomials verified. Results are for reference only.