Taylor and Maclaurin series sit at the heart of the differential calculus module in VTU Engineering Mathematics 1 (BMATS101 under the 2022 scheme, 18MAT11 under 2018). They answer a question that comes up again and again in engineering: how do you approximate a complicated function by a polynomial, and how accurate is that approximation? Students who understand the idea can handle expansions, limits, and numerical estimates with the same method. Students who only memorise the formula usually struggle the first time a question changes the centre point.
The Idea Behind the Series
A polynomial is easy to differentiate and evaluate. If a function f(x) can be matched by a polynomial near a point a, you can study the function’s behaviour using the polynomial. The Taylor series does this by matching the value of f and all its derivatives at a:
f(x) = f(a) + f′(a)(x−a) + f″(a)(x−a)²/2! + f‴(a)(x−a)³/3! + …
Each term is built so that its derivative at x = a agrees with the corresponding derivative of f. The first term matches the value, the second matches the slope, the third matches the curvature, and so on. The series is therefore the polynomial that copies f as closely as possible at the point a.
When a = 0, the series is called the Maclaurin series:
f(x) = f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x³/3! + …
Most VTU questions use the Maclaurin form, because the expansion point is zero and the arithmetic is simpler. Questions that ask for a Taylor expansion about another point use the general formula with (x−a) in place of x.
Standard Maclaurin Series You Must Know
You should be able to write these from memory, because they appear inside larger questions as building blocks:
- eˣ = 1 + x + x²/2! + x³/3! + …
- sin x = x − x³/3! + x⁵/5! − …
- cos x = 1 − x²/2! + x⁴/4! − …
- ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … (valid for −1 < x ≤ 1)
- (1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + … (for any n, valid for |x| < 1)
Notice that sin x and cos x contain only odd or only even powers. This is a useful check: if your expansion of sin x has an x² term, something has gone wrong.
Worked Example 1: Expanding eˣ Up to x⁴
Start from the general formula. For f(x) = eˣ, every derivative is eˣ, and every derivative at zero is e⁰ = 1. Therefore:
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + …
Simplifying the factorials gives:
eˣ = 1 + x + x²/2 + x³/6 + x⁴/24 + …
This is the expansion you would use, for example, to evaluate e^0.1 to four decimal places without a calculator: substituting x = 0.1 gives 1 + 0.1 + 0.005 + 0.000167 + …, which is 1.1052 to four places.
Worked Example 2: Expanding ln(1 + x) Up to x⁴
Let f(x) = ln(1 + x). Compute the derivatives:
- f′(x) = 1/(1+x), so f′(0) = 1
- f″(x) = −1/(1+x)², so f″(0) = −1
- f‴(x) = 2/(1+x)³, so f‴(0) = 2
- f⁽⁴⁾(x) = −6/(1+x)⁴, so f⁽⁴⁾(0) = −6
Substitute into the Maclaurin formula:
ln(1+x) = 0 + x + (−1)x²/2! + 2x³/3! + (−6)x⁴/4!
Simplify: ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + …
Students often expect the coefficients to be 1/2, 1/3 and 1/4 without checking. The alternating signs come from the derivatives, and the pattern is worth recognising because it recurs in other questions.
Worked Example 3: Taylor Expansion About a Point Other Than Zero
Expand f(x) = eˣ about x = 1 up to the term in (x−1)³.
Since every derivative of eˣ is eˣ, each derivative at x = 1 equals e. So:
eˣ = e + e(x−1) + e(x−1)²/2! + e(x−1)³/3! + …
eˣ = e[1 + (x−1) + (x−1)²/2 + (x−1)³/6 + …]
Notice how the expansion is the same as the Maclaurin series with x replaced by (x−1) and an overall factor of e. That is a useful shortcut in exams: if the centre is a, write the Maclaurin series of eˣ, replace x by (x−a), and multiply by eᵃ.
Worked Example 4: Approximating √4.2
Let f(x) = √x and take a = 4, because 4 is a nearby number whose square root is easy to find. Compute:
- f(4) = 2
- f′(x) = 1/(2√x), so f′(4) = 1/4
- f″(x) = −1/(4x^(3/2)), so f″(4) = −1/32
The expansion with h = x − 4 = 0.2 gives:
√4.2 ≈ 2 + (1/4)(0.2) + (−1/32)(0.2)²/2
= 2 + 0.05 − 0.000625 = 2.049375
The true value is 2.04939 to five decimal places, so two terms of the series give an accurate answer. This is the practical reason the series matters: it turns a calculation you cannot do by hand into one you can.
Using Series to Evaluate Limits
Series expansions are the cleanest way to handle limits that give 0/0. Consider:
lim (x→0) (sin x − x)/x³
Direct substitution gives 0/0. Use the Maclaurin series of sin x: sin x − x = −x³/6 + x⁵/120 − … Dividing by x³ gives −1/6 + x²/120 − …, and as x → 0 the limit is −1/6.
This method avoids repeated applications of L’Hôpital’s rule, which can become long. In exams, using the series is often faster and less error-prone, provided you include enough terms to see the leading behaviour.
How to Decide How Many Terms to Write
VTU questions usually state the number of terms or the power of (x−a) they want. If not, write terms until the pattern is clear and the next term would be smaller than what matters for the question. For limits, include terms up to the power that determines the answer. For example, in the limit above, the x³ term in the numerator decides the result, so the series must be accurate through x³.
Common Mistakes That Cost Marks
Mistake 1: Forgetting the factorials. The coefficients are f⁽ⁿ⁾(a)/n!, not f⁽ⁿ⁾(a). Writing x³ with coefficient 1 instead of 1/6 for eˣ is a frequent slip.
Mistake 2: Using x where (x−a) is needed. If the question says “about x = 2”, every term contains (x−2), not x.
Mistake 3: Sign errors with ln(1+x) and sin x. The signs alternate in these series. Check the first two terms against your own derivative calculation.
Mistake 4: Stopping too early in a limit. If the leading terms cancel, you need more terms of the series to find the limit. Keep going until a non-zero term appears.
Practice Questions
- Expand cos x up to x⁴ using the Maclaurin series.
- Expand f(x) = 1/(1+x) up to x³.
- Use a Taylor expansion about x = 1 to find ln(1.1) to four decimal places.
- Evaluate lim (x→0) (eˣ − 1 − x)/x².
Checks: (1) 1 − x²/2 + x⁴/24. (2) 1 − x + x² − x³. (3) Expand ln(x) about x = 1 with h = 0.1: 0.1 − 0.005 + 0.000333 = 0.0953, which is correct to four decimal places (ln 1.1 = 0.0953). (4) Using eˣ = 1 + x + x²/2 + …, the numerator is x²/2 + …, so the limit is 1/2.
Another Limit: (1 − cos x)/x²
This limit is a standard result that students should be able to derive from the series rather than memorise. Direct substitution gives 0/0, so the series is the right tool. Using cos x = 1 − x²/2! + x⁴/4! − …, the numerator becomes 1 − cos x = x²/2 − x⁴/24 + … Dividing by x² gives 1/2 − x²/24 + …, and as x approaches 0 the limit is 1/2.
The same approach handles any limit where the leading terms cancel. Look at the power of x that first survives in the numerator and denominator, and include enough terms of each series to reach that power. Students often stop one term too early, which leaves a zero in the numerator and makes the limit look undefined.
Checking an Approximation
A series approximation is only useful if you know how large the error is. For an alternating series, such as the one for ln(1 + x) at a point where the terms decrease in size, the error after truncating is smaller than the first omitted term. For example, using ln(1.1) ≈ 0.1 − 0.005 + 0.000333 = 0.095333, the next term would be about −0.000025, so the true value lies close to 0.09531. The actual value, to five places, is 0.09531, which confirms the estimate.
When the terms do not alternate, or when you are approximating far from the centre, the error is harder to bound. In such cases, add terms until successive approximations agree to the accuracy the question asks for.
Connecting This Topic to the Full Module
Taylor and Maclaurin series belong to the differential calculus module of VTU Engineering Mathematics 1. They are used again in Leibnitz’s theorem, in curvature calculations, and in the partial differentiation part of the course. For a complete map of the module and the marks it usually carries, see our VTU Engineering Maths 1 Notes — Complete Module-Wise Guide (BMATS101).
Frequently Asked Questions
Q: Is the Maclaurin series the same as a Taylor series?
Yes, with a = 0. Every Maclaurin series is a Taylor series about zero.
Q: Do I need to memorise every standard series?
Memorise the common five: eˣ, sin x, cos x, ln(1+x), and (1+x)ⁿ. Derive others when needed, since the method is the same each time.
Q: How does this connect to the rest of the module?
The same derivative skills are used in Leibnitz’s theorem, partial differentiation, and the curvature formulas. The full module outline is in our VTU Engineering Maths 1 notes guide.
For more worked problems on this topic, see the Engineering Math Survival Kit, which includes solved problem sets for the current scheme. If you would like to discuss how this module is taught in batches, use the contact page.