Taylor Series Calculator

The Taylor series calculator expands a function into an infinite polynomial series centred at a chosen point, allowing you to approximate complex functions with simple polynomial terms. It is used by maths and engineering students, physicists, and anyone needing high-accuracy function approximations. Enter your function, centre point, and the number of terms to generate the expansion.

How to Use the Taylor Series Calculator

1. Enter the function f(x) you want to expand.
2. Input the centre point a around which the series is built.
3. Choose the number of terms n to include in the expansion.
4. Click Calculate to see the Taylor series polynomial and each term's coefficient.
5. Compare the approximation to the true function value at a test point to check accuracy.

The Formula

The Taylor series of f(x) centred at x = a is:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

In summation form: f(x) = Sum from n=0 to infinity of [f^(n)(a) / n!] * (x - a)^n

Here, f^(n)(a) is the nth derivative of f evaluated at a, and n! is n factorial. Each term adds a higher-order correction, improving the approximation. When a = 0, this is called a Maclaurin series.

Real-World Example

Find the first four terms of the Taylor series for f(x) = e^x centred at a = 0.

1. f(0) = e^0 = 1, so the zeroth term is 1
2. f'(x) = e^x, f'(0) = 1, so the first term is x
3. f''(x) = e^x, f''(0) = 1, so the second term is x^2/2!= x^2/2
4. f'''(x) = e^x, f'''(0) = 1, so the third term is x^3/3! = x^3/6

Result: e^x approximately equals 1 + x + x^2/2 + x^3/6 near x = 0. At x = 0.5, this gives 1 + 0.5 + 0.125 + 0.0208 = 1.6458. The true value of e^0.5 is 1.6487, so just four terms gives an error of less than 0.2%.

Applications of Taylor Series

Taylor series appear throughout science and engineering. In physics, they simplify the equations of motion for pendulums and waves. In signal processing, they underpin the derivation of digital filters. Calculators and computers use Taylor series (or related Chebyshev approximations) to evaluate trigonometric and exponential functions internally. In statistics, they appear in the derivation of the normal distribution and in approximating likelihood functions. The more terms you include, the broader the interval over which the approximation remains accurate.

Frequently Asked Questions

What is the difference between a Taylor series and a Maclaurin series? A Maclaurin series is a Taylor series centred specifically at a = 0. All Maclaurin series are Taylor series, but Taylor series can be centred at any point a.

How many terms do I need for a good approximation? It depends on the function and how far x is from the centre a. Functions like e^x converge quickly, needing only a few terms near a = 0. Functions near their radius of convergence boundary require many more terms. Check the error by comparing successive partial sums.

What is the radius of convergence? The radius of convergence is the distance from the centre a within which the Taylor series converges to the true function. Outside this radius, adding more terms does not improve the approximation. For e^x, the radius is infinite; for 1/(1-x) centred at 0, it is 1.

Can Taylor series approximate discontinuous functions? No. Taylor series only work for functions that are infinitely differentiable at the centre point. Discontinuous functions or functions with corners cannot be represented by a Taylor series at those problem points.

Working the expansion by hand at x = 1

The example above uses x = 0.5, where four terms land within 0.2 percent of the true value. Move the test point to x = 1 and the same four terms fall a long way short, which shows how quickly a Taylor series loses ground as the evaluation point moves away from the centre.

For e^x centred at 0 the coefficients are all 1/n!, so the series runs 1 + x + x^2/2 + x^3/6 + x^4/24 and so on. The partial sums at x = 1 are:

| Terms used | Term added | Partial sum | Error against e = 2.7182818285 | | 1 | 1.0000000000 | 1.0000000000 | 63.212056% | | 2 | 1.0000000000 | 2.0000000000 | 26.424112% | | 3 | 0.5000000000 | 2.5000000000 | 8.030140% | | 4 | 0.1666666667 | 2.6666666667 | 1.898816% | | 5 | 0.0416666667 | 2.7083333333 | 0.365985% | | 6 | 0.0083333333 | 2.7166666667 | 0.059418% | | 7 | 0.0013888889 | 2.7180555556 | 0.008324% |

Six terms give 2.716667, which is 0.059 percent low. Seven give 2.718056, which is 0.008 percent low. The error falls by a factor of roughly five to seven with each extra term at this point, so another term adds little once the partial sum is close.

The same test on sin(x) at x = 1 behaves differently, because only the odd terms are non-zero:

| Degree | Partial sum | Error against sin(1) = 0.8414709848 | | 3 | 0.8333333333 | 0.967075% | | 5 | 0.8416666667 | 0.023255% | | 7 | 0.8414682540 | 0.000325% | | 9 | 0.8414710097 | 0.000003% |

Each extra pair of terms cuts the error by 42 between degree 3 and degree 5, by 72 between degree 5 and degree 7, and by 110 between degree 7 and degree 9. The reason is that each new term is the previous one divided by the product (2n)(2n+1), so the terms shrink faster as the order rises. Reading the sin(1) series this way gives a check on the tool: the degree 5 partial sum should sit close to 0.84167 and the degree 7 sum should sit close to 0.84147.

Choosing the centre point

The centre is the single largest lever on how many terms are needed. The series for ln(1 + x) about a = 0 converges slowly at x = 0.5:

| Terms | Partial sum | Error against ln(1.5) = 0.40546511 | | 1 | 0.50000000 | 23.31517% | | 2 | 0.37500000 | 7.51362% | | 3 | 0.41666667 | 2.76264% | | 4 | 0.40104167 | 1.09095% | | 5 | 0.40729167 | 0.45048% | | 6 | 0.40468750 | 0.19178% |

Eight terms are needed to push the error under 0.05 percent, against seven terms for e^x at a point twice as far from its centre. Centring the ln expansion at a = 0.5 instead would converge faster, because the distance from the centre to the evaluation point would be zero.

The error after n terms is bounded by the Lagrange form of the remainder: the absolute error is at most M times |x - a|^(n+1) divided by (n+1) factorial, where M is the largest value the (n+1)th derivative takes between a and x. For ln(1 + x) on the interval from 0 to 0.5 the derivatives do not stay small, which is what slows the convergence compared with e^x.

How the calculator obtains the derivatives

The tool does not differentiate symbolically. It estimates each derivative by repeated central differences with a step of 0.001, then divides by the factorial and adds the result to the running sum shown in its table.

That approach has a limit. Central differences carry a truncation error that grows with the order, and subtracting two nearly equal values adds a rounding error that the division by the step then doubles. For e^x at the centre 0 every true derivative equals 1, so the error is visible directly:

| Order | Numerical value | True value | Error | | 1 | 1.00000 | 1 | 0.0000167% | | 3 | 1.00000 | 1 | 0.0000499% | | 6 | 8.67362 | 1 | 767% | | 8 | -6938890 | 1 | 694000000% |

The estimate holds to about the fourth order and then breaks down, which is why the tool prints its own warning that accuracy decreases for high-order derivatives. Two habits follow from it. For functions whose derivatives are known by inspection, such as e^x, sin x and cos x, compare the coefficient column against the pattern the series should produce. For anything past the fourth or fifth order, work the derivatives out by hand and treat the tool's coefficient column as a check rather than a source.

The degree field accepts 1 to 10, so ten terms is the widest expansion available. The evaluation point x is free, and moving it away from the centre increases the number of terms needed in exactly the way the tables above set out.

Where the method comes from

Brook Taylor published the general expansion in Methodus Incrementorum Directa et Inversa in 1715. He had first stated the theorem in 1712, and its significance was not widely recognised until Lagrange treated it as the basic principle of differential calculus in 1772. The Maclaurin series is the same construction with the centre fixed at zero, and carries Colin Maclaurin's name.

What the table columns mean

The result table lists one row per term and six columns. Taking the e^x expansion at x = 0 with the centre at 0, the row for n = 3 reads as follows:

| Column | Value at n = 3 | What it holds | | n | 3 | The term index, counting from zero | | f^(n)(a) | 1 | The nth derivative evaluated at the centre | | n! | 6 | The factorial of the index, 3 x 2 x 1 | | Coefficient | 0.166667 | The derivative divided by the factorial | | (x - a)^n term | 0.166667 | The coefficient times (x - a)^n, which is 1 at x = 1 | | Running sum | 2.6666667 | Every term from n = 0 up to this row |

The coefficient column is the one to read first. Every derivative of e^x equals 1, so the column should run 1, 1, 0.5, 0.166667, 0.041667, 0.008333, 0.001389 as n goes from 0 to 6. That is the sequence 1/n!, and the running sum lands on 2.7180556 at the last row. A coefficient that departs from 1/n! on a later row means the numerical differentiation has run out of precision.

Also try these free tools:

• Maclaurin Series Calculator
• Linear Approximation Calculator
• Derivative Calculator