Difficulty: Intermediate to Advanced | Prerequisites: Taylor/Maclaurin polynomials notes, power series convergence notes
Once you know the standard Maclaurin series (e^x, cos x, sin x, 1/(1 - x), tan^(-1) x), you can substitute, multiply, differentiate, and integrate them to handle a huge range of problems: finding new series, evaluating tricky limits, approximating definite integrals to a given error, and converting infinite sums back into closed-form functions.
Substitution into a known series
Replacing x in a standard Maclaurin series with another expression (e.g. replacing x with -3x in e^x to get the series for e^(-3x)).
Think of it as plugging one function into another's blueprint.
Term-by-term differentiation
Differentiating a power series by differentiating each term individually. Valid inside the interval of convergence.
Term-by-term integration
Integrating a power series by integrating each term individually. Valid inside the interval of convergence. This is how you approximate integrals that have no elementary antiderivative.
Alternating series estimation theorem (ASET)
For an alternating series that satisfies the AST conditions, the error from using the first n terms is at most the absolute value of the (n+1)-th term. This gives a concrete error bound for approximations.
Closed-form sum
A known function that equals a given infinite series (e.g. recognising a series as the derivative of a geometric series to write it in terms of 1/(1-x)^2).
Find the first four non-zero terms, the full series, and the interval of convergence for f(x) = 6e^(-3x).
Build a derivative table:
n | f^(n)(x) | f^(n)(0) |
|---|---|---|
0 | 6e^(-3x) | 6 |
1 | -18e^(-3x) | -18 |
2 | 54e^(-3x) | 54 |
3 | -162e^(-3x) | -162 |
First four non-zero terms: 6 - 18x + 27x^2 - 27x^3.
The pattern gives the full series: 6e^(-3x) = sum of 6(-3x)^k / k! for k = 0 to infinity.
Since e^u converges for all u, the interval of convergence is (-infinity, infinity).
Find the first four non-zero terms of 9x tan^(-1)(x^2).
Start from the known series for tan^(-1)(u) = u - u^3/3 + u^5/5 - u^7/7 + ...
Substitute u = x^2:
tan^(-1)(x^2) = x^2 - x^6/3 + x^10/5 - x^14/7 + ...
Multiply by 9x:
9x tan^(-1)(x^2) = 9x^3 - 3x^7 + (9/5)x^11 - (9/7)x^15 + ...
First four non-zero terms: 9x^3 - 3x^7 + (9/5)x^11 - (9/7)x^15.
Find lim as x tends to 0 of [cos(6x) - 1 - 18x^2] / x^2.
Expand cos(6x) using the Maclaurin series for cos(u) with u = 6x:
cos(6x) = 1 - (6x)^2/2! + (6x)^4/4! - ... = 1 - 18x^2 + 54x^4 - 64.8x^6 + ...
Substitute into the numerator:
cos(6x) - 1 - 18x^2 = (1 - 18x^2 + 54x^4 - ...) - 1 - 18x^2 = -36x^2 + 54x^4 - ...
Divide by x^2:
[-36x^2 + 54x^4 - ...] / x^2 = -36 + 54x^2 - ...
As x tends to 0, all terms with x vanish, leaving -36.
Find the closed form of sum from k = 1 to infinity of (-1)^k times k times x^(k+1) / 3^k.
Factor out x^2 and recognise the remaining series as the derivative of a geometric series:
The series equals x^2 times sum of (-1)^k times k times (x/3)^(k-1), starting at k = 1.
Recognise sum of k r^(k-1) = d/dx [sum of r^k] = d/dx [1/(1 - r)] for the geometric series with r = -x/3.
sum of (-x/3)^k from k = 0 to infinity = 1/(1 + x/3) = 3/(3 + x).
d/dx [3/(3 + x)] is not quite right; instead, d/dx [-x/(3 + x)] = -3/(x + 3)^2.
Multiply by x^2 to get -3x^2 / (x + 3)^2.
Approximate the integral from 0 to 0.3 of e^(-x^2) dx with error less than 0.001.
Start from e^u = 1 + u + u^2/2! + u^3/3! + ... and substitute u = -x^2:
e^(-x^2) = 1 - x^2 + x^4/2 - x^6/6 + x^8/24 - ...
Integrate term by term from 0 to 0.3:
Integral = [x - x^3/3 + x^5/10 - x^7/42 + x^9/216 - ...] evaluated from 0 to 0.3.
Compute partial sums and check the error using the alternating series estimation theorem:
a_1 = 0.3
a_2 = (0.3)^3 / 3 = 0.009
a_3 = (0.3)^5 / 10 = 0.000243
Since a_3 = 0.000243 < 0.001, stopping at S_2 gives sufficient accuracy.
S_2 = 0.3 - 0.009 = 0.291.
The calculator check confirms: the integral is approximately 0.2912.
Standard Maclaurin series to memorise:
Function | Series | Interval |
|---|---|---|
e^x | sum of x^n / n! | (-infinity, infinity) |
sin x | sum of (-1)^n x^(2n+1) / (2n+1)! | (-infinity, infinity) |
cos x | sum of (-1)^n x^(2n) / (2n)! | (-infinity, infinity) |
1/(1 - x) | sum of x^n | (-1, 1) |
tan^(-1) x | sum of (-1)^n x^(2n+1) / (2n+1) | [-1, 1] |
ln(1 + x) | sum of (-1)^(n+1) x^n / n | (-1, 1] |
Alternating series estimation theorem:
For an alternating series satisfying the AST conditions, the error |S - S_n| is at most |a_{n+1}|.
Students often forget to adjust the interval of convergence after substitution. If you replace x with 3x in a series that converges for |x| < 1, the new series converges for |3x| < 1, i.e. |x| < 1/3.
When evaluating limits, students sometimes expand too few terms. You need enough terms to cancel whatever is in the denominator; expanding one or two terms may leave you with 0/0.
For integral approximations, students sometimes apply the ASET to a series that is not alternating. Verify that the integrated series actually alternates before using the error bound.
When finding closed-form sums, students confuse d/dx with d/dr. Differentiate with respect to the variable inside the series (the one playing the role of r).
"Find the first four non-zero terms" is a nearly guaranteed question. Know how to do it both by direct computation and by substitution.
Limit problems where L'Hopital is messy or recursive are solved cleanly with series. Expand, cancel, and read off the answer.
Approximating integrals with error bounds is a classic show-your-work problem. Show the term-by-term integration, identify the first omitted term, and verify it is below the error threshold.
Closed-form sum problems test whether you can work backwards from a series to a known function.
Fill in the blank: The Maclaurin series for cos(u) starts 1 - u^2/2! + ___. (u^4/4!.)
True or false: To get the series for e^(-x^2), you substitute u = -x^2 into the series for e^u. (True.)
Fill in the blank: By the alternating series estimation theorem, the error from using n terms is at most ___. (|a_{n+1}|.)
True or false: If you substitute x with 5x in the series for 1/(1 - x), the new radius of convergence is 5. (False, it is 1/5.)
True or false: You can differentiate a power series term by term inside its interval of convergence. (True.)
Q: Find the first four non-zero terms of the Maclaurin series for 6e^(-3x).
A: 6 - 18x + 27x^2 - 27x^3.
Q: Use series to evaluate lim as x tends to 0 of [cos(6x) - 1 - 18x^2] / x^2.
A: Expand cos(6x) = 1 - 18x^2 + 54x^4 - ... Numerator becomes -36x^2 + 54x^4 - ... Dividing by x^2 gives -36 + 54x^2 - ..., which tends to -36.
Q: Approximate the integral from 0 to 0.3 of e^(-x^2) dx with error < 0.001. How many terms of the integrated series do you need?
A: Two terms suffice (S_2 = 0.291), because the third term of the integrated series is (0.3)^5/10 = 0.000243 < 0.001.
Q: Find the first four non-zero terms of 9x tan^(-1)(x^2) using the known series for tan^(-1)(u).
A: Substitute u = x^2 into tan^(-1)(u), multiply by 9x: 9x^3 - 3x^7 + (9/5)x^11 - (9/7)x^15.
This material ties directly to the Taylor polynomials notes (finite truncations of these series) and the power series convergence notes (the interval of convergence determines where the series is valid). The technique of integrating a series term by term also previews topics in differential equations, where series solutions to ODEs use the same machinery.
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Maclaurin series, Taylor series applications, series substitution, term-by-term differentiation, term-by-term integration, alternating series estimation theorem, ASET, error bound, closed-form sum, limit by series expansion, integral approximation, M 2414, integral calculus, UT Austin