Power Series

Mathematics · FE Reference Handbook section

Mathematics

1 formulas

10 exam-style examples

~47 min

All Mathematics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Power Series within Mathematics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what power series describes physically and when it applies.
State every one of the 1 relation the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: radians vs degrees — set the calculator before the first trig entry.

Lecture

Why this section exists. Power Series is the part of Mathematics that lets you connect an algebraic or calculus expression that must be evaluated exactly to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as short symbolic manipulations with one numeric evaluation at the end. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. radians vs degrees — set the calculator before the first trig entry. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: power series.
Capstone Studio instructional photograph

Mathematics — Power Series: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an algebraic or calculus expression that must be evaluated exactly. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 1 relation on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

! i3	Quantity produced by "! i3= 0 ai ^ x - ah" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

1. A power series, which is convergent in the interval –R < x < R, defines a function of x that is continuous for all values of x
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
the series.
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
common interval of convergence.

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

a₁ = 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
$a_{n} = a_{1} + (n - 1) d$
Substituting
$a_{n} = 6 + (31 - 1) (4) = 126$
Sum
$S = n (a_{1} + a_{n}) /2$
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
$l o g_{b} (x) = ln x / ln b$
Substituting
$lo g_{(} 2_{)} (555) = ln (555) / ln (2) = 6.3190/0.6931$
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

a₁ = 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
$a_{n} = a_{1} + (n - 1) d$
Substituting
$a_{n} = 3 + (34 - 1) (4) = 135$
Sum
$S = n (a_{1} + a_{n}) /2$
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
$l o g_{b} (x) = ln x / ln b$
Substituting
$lo g_{(} 1 0_{)} (222) = ln (222) / ln (10) = 5.4027/2.3026$
Result — 2.3464
Check
$1 0^{2} .346 = 222.0 ✓$

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

a₁ = 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
$a_{n} = a_{1} + (n - 1) d$
Substituting
$a_{n} = 10 + (30 - 1) (8) = 242$
Sum
$S = n (a_{1} + a_{n}) /2$
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
$l o g_{b} (x) = ln x / ln b$
Substituting
$lo g_{(} 2_{)} (792) = ln (792) / ln (2) = 6.6746/0.6931$
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

a₁ = 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
$a_{n} = a_{1} + (n - 1) d$
Substituting
$a_{n} = 3 + (15 - 1) (3) = 45$
Sum
$S = n (a_{1} + a_{n}) /2$
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
$l o g_{b} (x) = ln x / ln b$
Substituting
$lo g_{(} 1 0_{)} (290) = ln (290) / ln (10) = 5.6699/2.3026$
Result — 2.4624
Check
$1 0^{2} .462 = 290.0 ✓$

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

a₁ = 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
$a_{n} = a_{1} + (n - 1) d$
Substituting
$a_{n} = 12 + (27 - 1) (4) = 116$
Sum
$S = n (a_{1} + a_{n}) /2$
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
$l o g_{b} (x) = ln x / ln b$
Substituting
$lo g_{(} 1 0_{)} (870) = ln (870) / ln (10) = 6.7685/2.3026$
Result — 2.9395
Check
$1 0^{2} .940 = 870.0 ✓$

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

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an algebraic or calculus expression that must be evaluated exactly, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Power Series contains 1 relation; you must be able to find this page in under 15 seconds.
Exam style: short symbolic manipulations with one numeric evaluation at the end.
Unit rule: radians vs degrees — set the calculator before the first trig entry.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

radians vs degrees — set the calculator before the first trig entry
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Power Series

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

!i3​=0aix−ah

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section