Example 1
Maclaurin polynomial of eˣ
Every derivative of eˣ is eˣ, and e⁰ = 1, so each coefficient is 1/k!.
Answer
Build the Taylor polynomial of a function about any center, or a Maclaurin polynomial about 0. The calculator lists each derivative, its exact value at the center and the resulting coefficient, so you can follow the construction term by term.
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A Taylor polynomial approximates a smooth function near a point a using the function's derivatives at that single point. The degree-n polynomial matches f and its first n derivatives at a, so it hugs the graph closely near the center and usually drifts away further out.
Letting n grow without bound gives the Taylor series. When the center is a = 0, the polynomial and series are called Maclaurin polynomials and series. The calculator computes the polynomial up to the order you choose, which is what most homework and approximation problems need.
These Maclaurin series appear constantly. The calculator reproduces them, and you can check each against its output.
Converges for every real x.
Only odd powers, because sine is an odd function.
Only even powers, because cosine is even.
Converges only for |x| < 1.
Converges for −1 < x ≤ 1.
For some c between a and x; use it to bound the approximation error.
The calculator differentiates your function repeatedly using the same symbolic engine as the derivative calculator. It simplifies each derivative before taking the next, so expressions stay manageable.
Each derivative is evaluated at the center exactly: sin(π) = 0, e⁰ = 1, √4 = 2, with no rounding. Each value is then divided by k! to get the coefficient. Terms with a zero coefficient are skipped, and the rest are combined in increasing powers of (x − a).
If the function or one of its derivatives is undefined at the center (for example ln(x) at 0), there is no Taylor polynomial there, and the calculator says so instead of producing nonsense.
For the full details, including how answers are verified, see how the solver works.
Work through these by hand first, then enter them above to compare each step.
Example 1
Every derivative of eˣ is eˣ, and e⁰ = 1, so each coefficient is 1/k!.
Answer
Example 2
The derivatives cycle cos, −sin, −cos, sin. At 0 they give 1, 0, −1, 0, 1, 0, −1.
Answer
Example 3
The derivatives are 1/x, −1/x², 2/x³, −6/x⁴. At x = 1 they equal 1, −1, 2, −6, and ln 1 = 0.
Answer
Example 4
f(4) = 2, f′(x) = 1/(2√x) gives 1/4, and f″(x) = −1/(4x^{3/2}) gives −1/32. Divide the last by 2! to get −1/64.
At x = 4.1: 2 + 0.025 − 0.00015625 = 2.02484375. The true value is 2.0248457…, so the error is about 2 × 10⁻⁶.
Answer
The coefficient is f⁽ᵏ⁾(a)/k!, not f⁽ᵏ⁾(a). Missing the k! is the single most common error.
Away from a = 0, every term involves powers of (x − a). Expanding in powers of x mixes up the coefficients.
“Order 6” means the highest power is 6. For cos x that is only four nonzero terms, because the odd coefficients vanish.
The polynomial for 1/(1 − x) is useless at x = 2 because the series diverges for |x| ≥ 1. Stay inside the interval of convergence.
A Maclaurin series is simply a Taylor series centered at a = 0. Set the center to 0 in the calculator to get a Maclaurin polynomial.
It is the highest power of (x − a) kept. An order-5 polynomial includes terms up to (x − a)⁵, although some of those coefficients may be zero.
Very accurate close to the center and less accurate further away. The Lagrange remainder formula bounds the error, and the calculator reports a sample comparison between f and the polynomial near the center.
ln(x) is undefined at 0, so it has no derivatives there and no Taylor polynomial. Use a = 1, or expand ln(1 + x) about 0 instead.
Pick one of the worked examples, solve it on paper, then enter it in the calculator and compare your steps with each rule shown. Spotting where your work differs is the fastest way to improve.
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