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Taylor series of sin x near 0

A polynomial that hugs sin(x) closer and closer as you add more terms — the bridge between calculus and approximation.

−π−π/2π/2π-11sin xT₁(x) = xT₃(x) = x − x³/6T₅(x) = x − x³/6 + x⁵/120

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What this shows

The Taylor series of a smooth function f around 0 is the infinite polynomial whose nth coefficient is f⁽ⁿ⁾(0)/n!:

    f(x) ≈ f(0) + f'(0)·x + f''(0)·x²/2! + f'''(0)·x³/3! + ⋯

For sin x the derivatives at 0 cycle through 0, 1, 0, −1, 0, 1, …, so all even-power terms vanish and you get the famous series:

    sin x  =  x − x³/3!  + x⁵/5!  − x⁷/7!  + ⋯
          =  x − x³/6   + x⁵/120 − x⁷/5040 + ⋯

The figure overlays sin x (solid curve) with three of its partial sums:

    T₁(x) = x                      —  fits near x = 0 only
    T₃(x) = x − x³/6              —  fits roughly to |x| ≈ 1
    T₅(x) = x − x³/6 + x⁵/120    —  fits roughly to |x| ≈ 2

Each higher-degree polynomial extends the region of agreement further, but never quite catches the true sin function everywhere — for that you need the full infinite sum.

Where it shows up

Taylor series are how transcendental functions (sin, cos, eˣ, log) are actually computed inside calculators and CPUs. A finite truncation gives an answer to machine precision over a chosen interval, with the truncation error bounded by the next omitted term.

They are also the foundation of perturbation methods in physics (expanding the equations of motion in a small parameter and keeping the leading-order term), of error analysis in numerical methods (a forward-difference scheme is a Taylor-truncation error analysis), and of the proofs of L'Hôpital's rule and limit-of-derivatives theorems.

Frequently asked questions

Why does the series only contain odd powers of x?

Because sin is an odd function: sin(−x) = −sin(x). Any Taylor series of an odd function around 0 must itself be odd, which forces every even-power coefficient to be zero.

Does the series converge everywhere?

Yes — for every real (and in fact every complex) x, the series converges to sin(x). The radius of convergence is infinite. This is not true of every Taylor series; for instance, log(1 + x) only converges for |x| ≤ 1.

How many terms do I need for a given accuracy?

Use the alternating-series error bound: the error after stopping at a term is at most the absolute value of the next term. To get sin(x) to within 10⁻⁶ for |x| ≤ 1, you need the x⁹/9! term to be below that, which holds since 1/362880 ≈ 2.8·10⁻⁶ — so degree-9 suffices.

Is the Taylor series unique?

Yes — for any smooth function, the Taylor coefficients are uniquely determined by the derivatives at the centre. Two different power series can only equal the same function on an interval if their coefficients agree term by term.

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