Maclaurin and Taylor Series Calculator: Partial Sums, Error and Radius of Convergence
See how well a power series approximates a function.
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How this calculator works
Choose the function, x and the number of terms.
The calculator lists the first terms, gives the partial sum and the true value, and shows the error.
It also gives how many terms give six decimal places and the radius of convergence.
Formulas used
- e^x = \sum_{k=0}^\infty\frac{x^k}{k!}
- \sin x = \sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}
- \ln(1 + x) = \sum_{k=1}^\infty\frac{(-1)^{k+1}x^k}{k}
- \frac{1}{1 - x} = \sum_{k=0}^\infty x^k
Worked example
With the default numbers in the calculator above:
- Series for e^x (general term, first terms): terms: 1, 1, 0.5, 0.16667, 0.041667, ...
- Partial sum with 6 terms: 2.71666667
- True value: 2.71828183
- Error (partial sum minus true value): -1.615e-3 (0.05942 percent of the true value)
- Terms needed for the error to be below 0.0000005 (six decimal places): 10 terms
- Radius of convergence: No limit: the series works for every x
Change the numbers in the calculator to see the full working, every step, for your own case.
A second worked example
A different case, with these numbers: Function sin x; The value of x (angles in radians) 1; Number of non-zero terms to add 3.
- Series for sin x (general term, first terms): terms: 1, -0.16667, 0.0083333
- Partial sum with 3 terms: 0.841666667
- True value: 0.841470985
- Error (partial sum minus true value): 1.957e-4 (0.02325 percent of the true value)
- Terms needed for the error to be below 0.0000005 (six decimal places): 5 terms
- Radius of convergence: No limit: the series works for every x
Common questions
What is a Maclaurin series?
A Taylor series about 0: a function written as a sum of powers of x with coefficients from its derivatives at 0.
What is the radius of convergence?
The distance from 0 within which the series adds up to the function. For e^x it is infinite, for ln(1 + x) it is 1.
Why does adding terms not help for large x?
Outside the radius of convergence the series diverges, so the partial sums do not settle on the function.
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Got a question about this? Ask an engineering question and get a plain-language answer with references.
Further reading: Any first-year calculus text (power series).
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