Taylor's Series

Mathematics · FE Reference Handbook section

Mathematics

4 formulas

10 exam-style examples

~53 min

All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

is called Taylor's series, and the function f (x) is said to be expanded about the point a in a Taylor's series.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Arithmetic progression sum — Taylor's Series

A schedule of 14 monthly inspections starts at 10 units and increases by 4 units each month. What is the total over the 14 months?

Given

$a_{1} = 10$
$d = 4$
$n = 14$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 10 + (14 - 1)(4) = 62$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 14(10 + 62)/2 = 504.0$

Answer:

S = 504.0 units

Why the other options are there

140.0 (increment ignored)
868.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 2

Arithmetic progression sum — Taylor's Series (2)

A schedule of 35 monthly inspections starts at 7 units and increases by 2 units each month. What is the total over the 35 months?

Given

$a_{1} = 7$
$d = 2$
$n = 35$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 7 + (35 - 1)(2) = 75$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 35(7 + 75)/2 = 1,435$

Answer:

S = 1,435 units

Why the other options are there

245.0 (increment ignored)
2,625 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 3

Arithmetic progression sum — Taylor's Series (3)

A schedule of 22 monthly inspections starts at 11 units and increases by 7 units each month. What is the total over the 22 months?

Given

$a_{1} = 11$
$d = 7$
$n = 22$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 11 + (22 - 1)(7) = 158$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 22(11 + 158)/2 = 1,859$

Answer:

S = 1,859 units

Why the other options are there

242.0 (increment ignored)
3,476 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 4

Arithmetic progression sum — Taylor's Series (4)

A schedule of 29 monthly inspections starts at 7 units and increases by 3 units each month. What is the total over the 29 months?

Given

$a_{1} = 7$
$d = 3$
$n = 29$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 7 + (29 - 1)(3) = 91$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 29(7 + 91)/2 = 1,421$

Answer:

S = 1,421 units

Why the other options are there

203.0 (increment ignored)
2,639 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 5

Arithmetic progression sum — Taylor's Series (5)

A schedule of 32 monthly inspections starts at 3 units and increases by 2 units each month. What is the total over the 32 months?

Given

$a_{1} = 3$
$d = 2$
$n = 32$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 3 + (32 - 1)(2) = 65$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 32(3 + 65)/2 = 1,088$

Answer:

S = 1,088 units

Why the other options are there

96 (increment ignored)
2,080 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 6

Arithmetic progression sum — Taylor's Series (6)

A schedule of 37 monthly inspections starts at 8 units and increases by 9 units each month. What is the total over the 37 months?

Given

$a_{1} = 8$
$d = 9$
$n = 37$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 8 + (37 - 1)(9) = 332$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 37(8 + 332)/2 = 6,290$

Answer:

S = 6,290 units

Why the other options are there

296.0 (increment ignored)
12,284 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 7

Arithmetic progression sum — Taylor's Series (7)

A schedule of 34 monthly inspections starts at 3 units and increases by 6 units each month. What is the total over the 34 months?

Given

$a_{1} = 3$
$d = 6$
$n = 34$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 3 + (34 - 1)(6) = 201$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 34(3 + 201)/2 = 3,468$

Answer:

S = 3,468 units

Why the other options are there

102.0 (increment ignored)
6,834 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 8

Arithmetic progression sum — Taylor's Series (8)

A schedule of 32 monthly inspections starts at 12 units and increases by 2 units each month. What is the total over the 32 months?

Given

$a_{1} = 12$
$d = 2$
$n = 32$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 12 + (32 - 1)(2) = 74$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 32(12 + 74)/2 = 1,376$

Answer:

S = 1,376 units

Why the other options are there

384.0 (increment ignored)
2,368 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 9

Arithmetic progression sum — Taylor's Series (9)

A schedule of 38 monthly inspections starts at 5 units and increases by 7 units each month. What is the total over the 38 months?

Given

$a_{1} = 5$
$d = 7$
$n = 38$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 5 + (38 - 1)(7) = 264$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 38(5 + 264)/2 = 5,111$

Answer:

S = 5,111 units

Why the other options are there

190.0 (increment ignored)
10,032 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Example 10

Arithmetic progression sum — Taylor's Series (10)

A schedule of 15 monthly inspections starts at 4 units and increases by 7 units each month. What is the total over the 15 months?

Given

$a_{1} = 4$
$d = 7$
$n = 15$

Find

Total (sum of the series)

Start with the thinking

Constant increment means an arithmetic series.
Get the last term first, then use the pair-average form.

Step-by-step solution

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 4 + (15 - 1)(7) = 102$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 15(4 + 102)/2 = 795.0$

Answer:

S = 795.0 units

Why the other options are there

60 (increment ignored)
1,530 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Taylor's Series

Taylor's Series

Handbook notes for this section

Core formulas for this FE topic

flahfmahfnahf l^a h f m^ a h f n ^a hflahfmahfnah

h^ hh

fxh=fah+1!x−ai+2!x−ai+...+n!x−ai+...f ^ x h = f ^a h + 1! _ x - a i + 2! _ x - a i + ... + n! _ x - a i + ...fxh=fah+1!x​−ai+2!x​−ai+...+n!x​−ai+...

Ifa=0,theTaylor′sseriesequationbecomesaMaclaurin′sseries.If a = 0, the Taylor's series equation becomes a Maclaurin's series.Ifa=0,theTaylor′sseriesequationbecomesaMaclaurin′sseries.

Worked exam-style examples