Taylor Series — Approximating a Function with a Polynomial
Your calculator does not "know" sin 0.5 or e0.5, and it does not look them up in a table. It adds up a few terms of a polynomial — and that polynomial is the Taylor series. The same idea sits behind every small-angle approximation in physics. This article builds the series from scratch, derives the first terms for ex, sin x and cos x, and then shows in plain numbers how much better the answer gets with each extra term.
The idea in one sentence
Near a chosen point, a smooth curve looks almost like a straight line. Add a squared term and it looks like a parabola that also matches the bending. Add a cubed term and it matches how the bending changes. Keep going and the polynomial hugs the curve more and more closely. Taylor's formula says exactly which coefficients to use:
In compact form: f(x) = Σn=0∞ f(n)(a) (x − a)n / n!
| Symbol | What it means |
|---|---|
| a | The centre — the point the approximation is built around and is most accurate near |
| f(n)(a) | The n-th derivative of f, evaluated at the centre a — a plain number |
| n! | Factorial: 0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040 |
| (x − a)n | How far you have moved from the centre, raised to the n-th power |
When the centre is a = 0 the formula is usually called a Maclaurin series, and it simplifies to f(x) = f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + … All three series below are Maclaurin series.
Deriving the series for ex
f(x) = ex. Every derivative of ex is ex again, and e0 = 1, so every coefficient f(n)(0) equals 1:
f(0) = 1, f′(0) = 1, f″(0) = 1, f‴(0) = 1, …
Substituting into the Maclaurin formula:
ex = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + …
Sanity check: differentiating term by term turns each term into the one before it, so the series is its own derivative — as ex must be.
Deriving the series for sin x and cos x
For f(x) = sin x the derivatives cycle in fours:
f(x) = sin x → f(0) = 0
f′(x) = cos x → f′(0) = 1
f″(x) = −sin x → f″(0) = 0
f‴(x) = −cos x → f‴(0) = −1
f⁗(x) = sin x → f⁗(0) = 0 (and the cycle repeats)
Only the odd powers survive, with signs alternating:
sin x = x − x³/3! + x⁵/5! − x⁷/7! + …
For f(x) = cos x the same cycle starts one step earlier: f(0) = 1, f′(0) = 0, f″(0) = −1, f‴(0) = 0, f⁗(0) = 1. Only the even powers survive:
cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …
Both require x in radians, because the derivative of sin x is cos x only when x is measured in radians.
Notice that the odd-power terms of ex look like the sine series and the even-power terms like the cosine series. That is no coincidence: putting ix in place of x in the exponential series gives Euler's relation eix = cos x + i sin x.
How fast does it actually converge? e0.5 term by term
The true value is e0.5 = 1.6487212707. Here is each term added in turn, with the running total and the remaining error:
| Terms used | Term added | Running total | Error |
|---|---|---|---|
| 1 | 1 | 1.0000000 | 0.6487213 |
| 2 | x = 0.5 | 1.5000000 | 0.1487213 |
| 3 | x²/2 = 0.25/2 = 0.1250000 | 1.6250000 | 0.0237213 |
| 4 | x³/6 = 0.125/6 = 0.0208333 | 1.6458333 | 0.0028879 |
| 5 | x⁴/24 = 0.0625/24 = 0.0026042 | 1.6484375 | 0.0002838 |
| 6 | x⁵/120 = 0.03125/120 = 0.0002604 | 1.6486979 | 0.0000234 |
Read the error column downwards: 0.65 → 0.15 → 0.024 → 0.0029 → 0.00028 → 0.000023. Each extra term cuts the error by roughly a factor of ten, and six terms already give five correct decimal places.
The same test for sin 0.5 and cos 0.5
True values: sin 0.5 = 0.4794255386 and cos 0.5 = 0.8775825619 (0.5 radians ≈ 28.65°).
| Series | Terms | Running total | Error |
|---|---|---|---|
| sin 0.5 | x | 0.5000000 | 0.0205745 |
| − x³/6 = −0.0208333 | 0.4791667 | 0.0002589 | |
| + x⁵/120 = +0.0002604 | 0.4794271 | 0.0000015 | |
| − x⁷/5040 = −0.0000016 | 0.4794255 | 0.0000000054 | |
| cos 0.5 | 1 | 1.0000000 | 0.1224174 |
| − x²/2 = −0.1250000 | 0.8750000 | 0.0025826 | |
| + x⁴/24 = +0.0026042 | 0.8776042 | 0.0000216 | |
| − x⁶/720 = −0.0000217 | 0.8775825 | 0.000000097 |
Four terms of the sine series give eight correct decimal places at x = 0.5. This is why the method is used inside real calculators.
Worked example — the approximation is only good near the centre
Use the same ex series to estimate e³. True value: e³ = 20.0855369.
1 → 1
+ 3 → 4
+ 3²/2 = 4.5 → 8.5
+ 3³/6 = 4.5 → 13
+ 3⁴/24 = 81/24 = 3.375 → 16.375
+ 3⁵/120 = 243/120 = 2.025 → 18.4
+ 3⁶/720 = 729/720 = 1.0125 → 19.4125
+ 3⁷/5040 = 2187/5040 = 0.4339 → 19.8464
After eight terms the error is still 20.0855 − 19.8464 = 0.2391 — about 1.2%. Compare that with six terms at x = 0.5, which were already accurate to 0.0000234.
The lesson: the further x is from the centre a, the more terms you need. These three series do converge for every x, but "eventually" can mean a great many terms. If you need e³ accurately, re-centre the series at a = 3.
Worked example — a series that stops working entirely
Expand ln(1 + x) about a = 0 and use it to find ln 1.2.
f(x) = ln(1 + x) → f(0) = 0
f′(x) = 1/(1 + x) → f′(0) = 1
f″(x) = −1/(1 + x)² → f″(0) = −1, so the term is −x²/2! = −x²/2
f‴(x) = 2/(1 + x)³ → f‴(0) = 2, so the term is +2x³/3! = +x³/3
ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + …
Put x = 0.2: 0.2 − 0.02 + 0.0026667 − 0.0004 = 0.1822667. The true value is ln 1.2 = 0.1823216, so four terms are already accurate to 0.0000549 ✔
But try x = 2 (that is, ln 3): the terms are 2, −2, +2.667, −4, +6.4 … they grow instead of shrinking, and the sum never settles. This series is valid only for −1 < x ≤ 1. Every series has an interval where it works — always check it.
Why physics and chemistry care
Almost every "approximately equal" in a physics chapter is a truncated Taylor series:
- Small-angle approximation. Keeping only the first term gives sin θ ≈ θ. At θ = 0.1 rad, sin 0.1 = 0.0998334, so the error is about 0.17% — small enough that the simple-pendulum equation becomes solvable. Keeping cos θ ≈ 1 − θ²/2 alongside it is what turns the pendulum into simple harmonic motion.
- Non-relativistic kinetic energy. Expanding (1 − v²/c²)−1/2 ≈ 1 + v²/(2c²) turns the relativistic energy mc²/√(1 − v²/c²) into mc² + ½mv² — the familiar ½mv² is literally the second Taylor term.
Common mistakes that cost marks
- Using degrees. sin x = x − x³/6 + … is a radian formula. Putting x = 30 (meaning 30°) gives −4470, not 0.5.
- Dividing by n instead of n!. The fourth term of ex is x³/6, not x³/3. The factorial is what makes the terms shrink fast enough.
- Forgetting (x − a) when the centre is not 0. A series about a = 2 is built from powers of (x − 2), never powers of x.
- Assuming more terms always fixes accuracy. Outside the interval of convergence, as in the ln(1 + x) example at x = 2, extra terms make the answer worse.
- Dropping the sign pattern. sin and cos alternate + − + −; ex does not. Mixing them up is a common slip under time pressure.
Where this appears in exams
| Exam / subject | Typical use |
|---|---|
| CBSE/ICSE Class 11–12 | Binomial expansion for small quantities, approximate values, error estimates |
| JEE Main & Advanced | Expansions used to resolve limits and to approximate roots |
| Class 11–12 Physics | Small-angle approximation, simple harmonic motion, error and uncertainty work |
| BSc / IIT-JAM level | Maclaurin and Taylor expansions, radius of convergence, remainder term |
Build and test a series in seconds. Enter the function, the centre and the number of terms, and the Taylor Series tool returns the polynomial and its value — so you can watch the error shrink exactly as the tables above do.
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