MACLAURIN’S SERIES
Section A: Exponential Function
The Maclaurin’s expansion for e x is e x  1  x 
x2 x3

 ... The LHS is exponential but the RHS
2! 3!
(power series) is essentially made up of ‘polynomials’ with infinite number of terms.
The question is: “How can an exponential function be approximated by polynomials since they
are very different?”
The LiveMath template shows the graphs of the following approximations to y  e x :
y1  1  x
x2
y2  1  x 
2!
x2 x3
y3  1  x 

2! 3!

y10  1  x 
x2 x3
x 10

 ... 
2! 3!
10!
1.
State the (approximate) range of x where the graphs of y 2 and y  e x coincide.
2.
State the (approximate) range of x where the graphs of y 3 and y  e x coincide.
3.
State the (approximate) range of x where the graphs of y 4 and y  e x coincide.
4.
As the number of terms of y n increases, i.e. from y 2 to y 4 , what do you notice about the
range of x where the graphs of y n and y  e x coincide?
5.
When the numerical value of x becomes smaller, what can you say about the approximation
of y n to y  e x ? [Hint: See Q4]
6.
How would you expect the graph of y n to look like as n   ?
© Joseph Yeo
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Section B: Sine Function
Change the function on the LiveMath template to f ( x)  sin x and click anywhere in the template
(except on the graphs). Everything that is linked to this function will change automatically.
7.
Find, from the template, the Maclaurin’s expansion for y  sin x up to the term in x 7 and
write it in the space below.
8.
Since y 2  y1 , y 4  y3 , etc., then we consider only odd values of n. State the
9.
(approximate) range of x where the graphs of y n (for n = 1, 3, 5, 7) and y  sin x  coincide.
y1 :
_______________________________
y3 :
_______________________________
y5 :
_______________________________
y7 :
_______________________________
As the number of terms of y n (for odd n) increases, i.e. from y1 to y 7 , what do you notice
about the range of x where the graphs of y n and y  sin x coincide?
10.
Observe the number of turning points. Do you think y 3 or y 5 is a better approximation to
y  sin x ? Why?
11.
Is it always true that y n (for odd n) will always have more turning points when n increases?
Why or why not?
12.
Do you think y10 or y 20 is a better approximation to y  sin x ? Why?
Section C: Conclusion
13.
Write down two main lessons that you have learnt from this worksheet.
© Joseph Yeo
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ANSWERS
Section A: Exponential Function
1.
0.5  x  0.5 [Accept 0.5  x  0.75]
2.
1  x  1
3.
1.5  x  1.5 [Accept 1.5  x  1.75]
4.
The range increases.
5.
The approximation is better when the numerical value of x becomes smaller.
6.
y  ex
[Accept 1.25  x  1.5]
Section B: Sine Function
7.
sin x  x 
8.
y1 :
y3 :
y5 :
y7 :
1 3
1 5
1
x 
x 
x 7  ...
6
120
5040
0.75  x  0.75
1.5  x  1.5
2  x  2
[Accept 2.25  x  2.25]
3  x  3
[Accept 3.25  x  3.25]
9.
The range increases.
10.
y 5 is a better approximation to y  sin x than y 3 because y 5 has more turning points than
y 3 since y  sin x has an infinite number of turning points. [ y 3 has 2 turning points; y 5
has 4 turning points.]
11.
Not true. y 7 has fewer turning points than y 5 . [ y 7 has only 2 turning points.]
12.
y 20 is a better approximation to y  sin x than y10 because y 20 should have more turning
points than y10 .
[Or y 20 has more terms than y10 and so it is a better approximation: see Q8 and Q9.]
Section C: Conclusion
13.
Lesson 1: The approximation is better when the numerical value of x is smaller.
Lesson 2: The more terms y n has, the better is the approximation. [In the case of y  sin x ,
this is because there are generally more turning points as n becomes bigger.]
© Joseph Yeo
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