Intro to Proof by Induction (Edexcel A Level Further Maths: Core Pure)

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Intro to Proof by Induction

What is proof by induction?

Proof by induction is a way of proving a result is true for a set of integers by showing that if it is true for one integer then it is true for the next integer
It can be thought of as dominoes:

All dominoes will fall down if:

The first domino falls down
Each domino falling down causes the next domino to fall down

What are the steps for proof by induction?

STEP 1: The basic step

Show the result is true for the base case
This is normally n = 1 or 0 but it could be any integer

In the dominoes analogy this is showing that the first domino falls down

STEP 2: The assumption step

Assume the result is true for n = k for some integer k

In the dominoes analogy this is assuming that a random domino falls down

There is nothing to do for this step apart from writing down the assumption

STEP 3: The inductive step

Using the assumption show the result is true for n = k + 1
The assumption from STEP 2 will be needed at some point

In the dominoes analogy this is showing that the random domino that we assumed falls down will cause the next one to fall down

STEP 4: The conclusion step

State the result is true
Explain in words why the result is true
It must include:

If true for n = k then it is true for n = k + 1
Since true for n = 1 the statement is true for all n ∈ ℤ, n ≥ 1 by mathematical induction

The sentence will be the same for each proof just change the base case from n = 1 if necessary

What type of statements might I be asked to prove by induction?

There are 4 main applications that you could be asked
- Formulae for sums of series
- Formulae for recursive sequences
- Expression for the power of a matrix
- Showing an expression is always divisible by a specific value
Induction is always used to prove de Moivre's theorem
It is unlikely that you will be asked unfamiliar applications in your exam but induction is used in other areas of maths
- Proving formulae for nth derivative of functions
- Proving formulae involving factorials

Proving de Moivre's Theorem by Induction

How is de Moivre’s Theorem proved?

When written in Euler’s form the proof of de Moivre’s theorem is easy to see:
- Using the index law of brackets: ${{(r e^{i θ})}^{n} = r}^{n} e^{i n θ}$
However Euler’s form cannot be used to prove de Moivre’s Theorem when it is in modulus-argument (polar) form
Proof by induction can be used to prove de Moivre’s Theorem for positive integers:
- To prove de Moivre’s Theorem for all positive integers, n
- $[r (\cos θ + isin θ)]^{n} = r^{n} (\cos n θ + isin n θ)$
STEP 1: Prove it is true for n = 1
- $[r (\cos θ + isin θ)]^{1} = r^{1} (\cos 1 θ + isin 1 θ) = r (\cos θ + isin θ)$
- So de Moivre’s Theorem is true for n = 1
STEP 2: Assume it is true for n = k
- $[r (\cos θ + isin θ)]^{k} = r^{k} (\cos k θ + isin k θ)$
STEP 3: Show it is true for n = k + 1
- $[r (\cos θ + isin θ)]^{k + 1} = ([r (\cos θ + isin θ)]^{k}) ([r (\cos θ + isin θ)]^{1})$
- According to the assumption this is equal to
  - $(r^{k} (\cos k θ + isin k θ)) (r (\cos θ + isin θ))$
- Using laws of indices and multiplying out the brackets:
  - $= r^{k + 1} [\cos k θ \cos θ + icos k θ \sin θ + isin k θ \cos θ + i^{2} \sin k θ \sin θ]$
- Letting i² = -1 and collecting the real and imaginary parts gives:
  - $= r^{k + 1} [\cos k θ \cos θ - \sin k θ \sin θ + i (\cos k θ \sin θ + \sin k θ \cos θ)]$
- Recognising that the real part is equivalent to cos(kθ + θ ) and the imaginary part is equivalent to sin(kθ + θ ) gives
  - $r^{k + 1} [\cos (k + 1) θ + isin (k + 1) θ]$
- So de Moivre’s Theorem is true for n = k + 1
STEP 4: Write a conclusion to complete the proof
- The statement is true for n = 1, and if it is true for n = k it is also true for n = k + 1
- Therefore, by the principle of mathematical induction, the result is true for all positive integers, n
De Moivre’s Theorem works for all real values of n
- However you could only be asked to prove it is true for positive integers

Worked example

Show, using proof by mathematical induction, that for a complex number $z = r (\cos θ + isin θ)$ and for all positive integers, n,

$z^{n} = [r (\cos θ + isin θ)]^{n} = r^{n} (\cos n θ + isin n θ)$

1-9-3-ib-aa-hl-proof-of-de-moivres-theorem-we-solution

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