FAQ for Maclaurin's Series

Below are the answers for Frequently Asked Questions (FAQ) on Maclaurin's Series.

Q: How do I know when to apply repeated differentiation and when to use the given expansions of the 4 common functions in the formula list?

A: The phrasing and/or structure of the question will hint to which method to use. For example, tutorial Q2 (By referring to the standard series in the formula list) and Q6 (By using the series expansion of e^x) hints of applying the given expansions in the formula list. In contrast, when phrases like by repeated / further differentiation of this result , by differentiating this result 3 times , etc or when the question first asks you to show some results involving the derivatives. All these hints of the 3-step procedure to find the Maclaurin series.


Q: How do I know when to link the expression for derivatives to the original function y?

A: There are no one-size-fit-all method. As fast as possible, try to link the answers for the higher derivatives to y or dy/dx, so that subsequent differentiation can be done neater by implicit differentiation.


Q: Is there a need to make each derivative the subject after every differentiation?

A: Unless required by the question, it is not needed for each derivative to be made the subject after every differentiation. Because, you would be substituting in numeric values anyway at a later stage.


Q: When applying the small angle approximation, how small is small?

A: For A-level purposes, there will not be trick or funny questions like cos(10000x). You will be sure that the question would be phrased in such a way that you know whether small angle can be applied. For example, cos(x) in Q11 can be approximated directly since x is small, but cos(x + pi/3) in Q7 can't be approximated directly since pi/3 is a big angle and thus (x + pi/3) is big.


Q: What does convergence mean?

A: Convergence basically means that for the valid domain of x values, the Maclaurin series will converge to the given function. For example, y = ln(1 + x) , as long as the value of x chosen is inside the valid domain of -1 < x < 1, say, pick x = 0.2, then convergence means that the Maclaurin series with x = 0.2, ie, (0.2) -( (0.2)^2)/2 + ((0.2)^3)/3 - ... will tend to the value of ln(1 + 0.2).


Q: How do I improve the approximation of the function by the Maclaurin series?

A: By increasing the number of terms in the Maclaurin expansion, you will obtain a better approximation. Consider the function y = e^x. When x = 1, e^x = 2.718282. Look at the improved accuracy of the expansion as more terms are included.
1 = 1
1 + x = 2
1 + x + (x^2)/2! = 2.5
1 + x + (x^2)/2! + (x^3)/3! = 2.666667
1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! = 2.708333
1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + (x^5)/5! = 2.716667
1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + (x^5)/5! + (x^6)/6! = 2.718056

Also, for any given Maclaurin series, it will give a better fit when the x chosen is closer to zero.
For example, consider e^x = 1 + x + (x^2)/2! + (x^3)/3! + ...
When x = 0, LHS = 1, RHS = 1
When x = 0.001, LHS = 1.001001, RHS = 1.0010005
When x = 0.01, LHS = 1.01005, RHS = 1.010050167
When x = 0.1, LHS = 1.105171, RHS = 1.105166667
When x = 1, LHS = 2.718282, RHS = 2.666666667
When x = 3, LHS = 20.08554, RHS = 13
Thus, the closer the value of x to zero, the better the accuracy of the Maclaurin series.


Q: I'm not convinced by the argument for Q6 last part.

A: Well, just do the algebra then! Substitute the expansion of e^x on the RHS to find the answer for 1 + 8e^x, up to the x^3 term. Then, expand y^2 on the LHS to obtain the answer. Both answers should be identical, hence verifying that y^2 = 1 + 8e^x (and thus, the expansion for y must be correct)

Here's the cross-multiplying (colour-coded for better reference):