Power Series

Mathematics · FE Reference Handbook section

Mathematics

1 formulas

10 exam-style examples

~47 min

All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

within the interval and is said to represent the function in that interval.
2. A power series may be differentiated term by term within its interval of convergence. The resulting series has the same
interval of convergence as the original series (except possibly at the end points of the series).
3. A power series may be integrated term by term provided the limits of integration are within the interval of convergence of
4. Two power series may be added, subtracted, or multiplied, and the resulting series in each case is convergent, at least, in the
interval common to the two series.
5. Using the process of long division (as for polynomials), two power series may be divided one by the other within their

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 — Power Series

A schedule of 31 monthly inspections starts at 6 units and increases by 4 units each month. What is the total over the 31 months?

Given

$a_{1} = 6$
$d = 4$
$n = 31$

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
$aₙ = 6 + (31 - 1)(4) = 126$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 31(6 + 126)/2 = 2,046$

Answer:

S = 2,046 units

Why the other options are there

186.0 (increment ignored)
3,906 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 2

Change of logarithm base — Power Series

Evaluate log₍2₎(555) using natural logarithms.

Given

$Base = 2$
$Argument = 555$

Find

log₍2₎(555)

Start with the thinking

Calculators carry ln and log₁₀ only; convert with the change-of-base rule.

Step-by-step solution

Change of base
$log_b(x) = \ln x / \ln b$
Substituting
$\log ₍2₎(555) = \ln (555)/\ln (2) = 6.3190/0.6931$
Result — 9.1163
Check
$2^9.116 = 555.0 ✓$

Answer:

9.116

Why the other options are there

0.110 (ratio inverted)
2.744 (base-10 log reported)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 3

Arithmetic progression sum — Power Series (2)

A schedule of 34 monthly inspections starts at 3 units and increases by 4 units each month. What is the total over the 34 months?

Given

$a_{1} = 3$
$d = 4$
$n = 34$

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
$aₙ = 3 + (34 - 1)(4) = 135$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 34(3 + 135)/2 = 2,346$

Answer:

S = 2,346 units

Why the other options are there

102.0 (increment ignored)
4,590 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 4

Change of logarithm base — Power Series (2)

Evaluate log₍10₎(222) using natural logarithms.

Given

$Base = 10$
$Argument = 222$

Find

log₍10₎(222)

Start with the thinking

Calculators carry ln and log₁₀ only; convert with the change-of-base rule.

Step-by-step solution

Change of base
$log_b(x) = \ln x / \ln b$
Substituting
$\log ₍10₎(222) = \ln (222)/\ln (10) = 5.4027/2.3026$
Result — 2.3464
Check
$10^2.346 = 222.0 ✓$

Answer:

2.346

Why the other options are there

0.426 (ratio inverted)
2.346 (base-10 log reported)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 5

Arithmetic progression sum — Power Series (3)

A schedule of 30 monthly inspections starts at 10 units and increases by 8 units each month. What is the total over the 30 months?

Given

$a_{1} = 10$
$d = 8$
$n = 30$

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
$aₙ = 10 + (30 - 1)(8) = 242$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 30(10 + 242)/2 = 3,780$

Answer:

S = 3,780 units

Why the other options are there

300.0 (increment ignored)
7,260 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 6

Change of logarithm base — Power Series (3)

Evaluate log₍2₎(792) using natural logarithms.

Given

$Base = 2$
$Argument = 792$

Find

log₍2₎(792)

Start with the thinking

Calculators carry ln and log₁₀ only; convert with the change-of-base rule.

Step-by-step solution

Change of base
$log_b(x) = \ln x / \ln b$
Substituting
$\log ₍2₎(792) = \ln (792)/\ln (2) = 6.6746/0.6931$
Result — 9.6294
Check
$2^9.629 = 792.0 ✓$

Answer:

9.629

Why the other options are there

0.104 (ratio inverted)
2.899 (base-10 log reported)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 7

Arithmetic progression sum — Power Series (4)

A schedule of 15 monthly inspections starts at 3 units and increases by 3 units each month. What is the total over the 15 months?

Given

$a_{1} = 3$
$d = 3$
$n = 15$

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
$aₙ = 3 + (15 - 1)(3) = 45$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 15(3 + 45)/2 = 360.0$

Answer:

S = 360.0 units

Why the other options are there

45 (increment ignored)
675.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 8

Change of logarithm base — Power Series (4)

Evaluate log₍10₎(290) using natural logarithms.

Given

$Base = 10$
$Argument = 290$

Find

log₍10₎(290)

Start with the thinking

Calculators carry ln and log₁₀ only; convert with the change-of-base rule.

Step-by-step solution

Change of base
$log_b(x) = \ln x / \ln b$
Substituting
$\log ₍10₎(290) = \ln (290)/\ln (10) = 5.6699/2.3026$
Result — 2.4624
Check
$10^2.462 = 290.0 ✓$

Answer:

2.462

Why the other options are there

0.406 (ratio inverted)
2.462 (base-10 log reported)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 9

Arithmetic progression sum — Power Series (5)

A schedule of 27 monthly inspections starts at 12 units and increases by 4 units each month. What is the total over the 27 months?

Given

$a_{1} = 12$
$d = 4$
$n = 27$

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
$aₙ = 12 + (27 - 1)(4) = 116$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 27(12 + 116)/2 = 1,728$

Answer:

S = 1,728 units

Why the other options are there

324.0 (increment ignored)
3,132 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Power Series

Example 10

Change of logarithm base — Power Series (5)

Evaluate log₍10₎(870) using natural logarithms.

Given

$Base = 10$
$Argument = 870$

Find

log₍10₎(870)

Start with the thinking

Calculators carry ln and log₁₀ only; convert with the change-of-base rule.

Step-by-step solution

Change of base
$log_b(x) = \ln x / \ln b$
Substituting
$\log ₍10₎(870) = \ln (870)/\ln (10) = 6.7685/2.3026$
Result — 2.9395
Check
$10^2.940 = 870.0 ✓$

Answer:

2.940

Why the other options are there

0.340 (ratio inverted)
2.940 (base-10 log reported)

Reference: FE Reference Handbook — Mathematics → Power Series

Power Series

Handbook notes for this section

Core formulas for this FE topic

!i3=0aix−ah! i_{3}= 0 ai ^ x - ah!i3​=0aix−ah

Worked exam-style examples