FP2: De Moivre's theorem and it's applications

So I just finished this chapter (AQA) and my brain is absolutely fried. There is so much to take in and it's so dense with information.

Does anybody have any helpful resources to help with it? I have used Examsolutions which is nice but I'm looking for more.

Reply 1

RhymeAsylumForever

https://youtu.be/P2m_u2HVrKs

Reply 2

diwangislucky

De Moivre's formula? I thought De Moivre's formula was a direct result of Euler's exponential representation of complex numbers.

Reply 3

ManLike007

Original post by darkforest

my brain is absolutely fried.

Lmao.

Anyway, building from De Moivre's Theorem.

You learnt the expansions of

\cos(n\theta)

and

\sin(n\theta)

in powers of

\sin\theta

and

\cos\theta

.

Let's skip ahead and assume you have proved the following,

\cos^{5}\theta = \frac{1}{16}(\cos5\theta + 5\cos3\theta + 10\cos\theta)

Now with Edexcel, there's a chapter called Integration by Reduction (in FP3), this is usually integrating a function involving

n, n \in \mathbb{Z^{+}}

usually being the power and is high. (IMO Integration by Reduction is faster but it's worth mentioning.)

You can integrate the function

\cos^{5}x

since you've expanded it,

i.e.

\displaystyle \int cos^{5}xdx \Rightarrow \frac{1}{16} \int (\cos5\theta + 5\cos3\theta + 10\cos\theta)d\theta

Alternatively, you'd use Integration by Reduction by setting the power as n, i.e.

\displaystyle \int \cos^{5}xdx \Rightarrow I_{n} = \int \cos^{n}x dx = \int \cos^{1}x \cdot \cos^{n-1}dx

then use integration by parts etc. (I won't get into it as this isn't related to complex numbers but it's the idea of integrating a function to a high power as shown).

Pretty much the only new thing I've learnt after A Level regarding De Moivre's Theorem. The rest is Euler's exponential representation.

(edited 6 years ago)

Reply 4

DFranklin

Original post by ManLike007

(IMO Integration by Reduction is faster but it's worth mentioning.)I know this is taking things even more off topic, but FWIW, to integrate

\cos^n x

where n is odd, the fastest method is to rewrite as:

\cos x (1-\sin^2 x)^{(n-1)/2}

and then integrate by recognition.

e.g.

\int \cos^5 x \,dx = \int \cos x (1-\sin^2 x)^2

=\int \cos x (1 - 2 \sin^2 x + \sin^4 x) \,dx = \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C

Reply 5

DFranklin

Original post by darkforest

I'm not sure what you really need to remember for this other than

(\cos x + i \sin x)^k = \cos kx + i \sin kx)

and the two main applications are:

you can expand

(\cos + i \sin x)^k

binomially to get expressions for cos kx and sin kx in terms of cos x and sin x.

Conversely, setting z = cos x + i \sin x, then z+(1/z) = 2 cos x and z-1/z = 2i sin x; you can use this and expand (z+1/z)^k (or (z-1/z)^k binomially to get expressions for cos^k x and sin^k x as a linear combination of basic trig expressions (e.g. cos^3 x = (cos 3x + cos x)/4), which is handy for reducing to something that's easy to integrate.

I'm not sure what else you think you need to know?

Reply 6

ManLike007

Original post by DFranklin

I know this is taking things even more off topic, but FWIW, to integrate

\cos^n x

where n is odd, the fastest method is to rewrite as:

\cos x (1-\sin^2 x)^{(n-1)/2}

and then integrate by recognition.

e.g.

\int \cos^5 x \,dx = \int \cos x (1-\sin^2 x)^2

=\int \cos x (1 - 2 \sin^2 x + \sin^4 x) \,dx = \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C

I'm assuming this also works for

\sin{x}

for odd values of n. It's interesting but I believe there is a doable limit you'd use this 'trick' as a power of

\frac{n-1}{2}

means any number, I mean if you had

\displaystyle \int \cos^{9}x \,dx \Rightarrow \int \cos x (1 -sin^{2}x)^{4} \,dx

, you know, it's not really fun to expand now is it? (Unless you can do Binomial expansion in your head then fair play to you sir)

(Apologies to OP for swaying this off even more)

(edited 6 years ago)

Reply 7

username3684524

Thanks for all of the replies!

It seems like I was really just struggling with the concept of the ‘n nth roots of complex numbers’.

I did a bit more work on it and I seem to get it now. :smile:

Reply 8

DFranklin

Original post by ManLike007

I'm assuming this also works for

\sin{x}

for odd values of n. It's interesting but I believe there is a doable limit you'd use this 'trick' as a power of

\frac{n-1}{2}

means any number, I mean if you had

\displaystyle \int \cos^{9}x \,dx \Rightarrow \int \cos x (1 -sin^{2}x)^{4} \,dx

, you know, it's not really fun to expand now is it? (Unless you can do Binomial expansion in your head then fair play to you sir)Well, yes, I can (and you should be able to for a number this small).

So integral is: S - 4S^3/3 +6S^5/5 -4S^7/7 + S^9 / 9, where S = sin x.

Which I literally wrote down.

Note that using a reduction formula is also not really fun when n is large (except for certain special cases where the limits mean it works out unusually nicely).

There are still times where you want to use a reduction anyhow (particularly when you want to examine the behaviour as n->infinity), but for a standard question, it saves a bit of time.

Reply 9

username3555092

Original post by DFranklin

I know this is taking things even more off topic, but FWIW, to integrate

\cos^n x

where n is odd, the fastest method is to rewrite as:

\cos x (1-\sin^2 x)^{(n-1)/2}

and then integrate by recognition.

e.g.

\int \cos^5 x \,dx = \int \cos x (1-\sin^2 x)^2

=\int \cos x (1 - 2 \sin^2 x + \sin^4 x) \,dx = \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C

PRSOM for the nifty method, but did you really think that A Level maths exams would afford you opportunities to use whichever method you wished? :biggrin:

If that were the case I'd probably have used modular arithmetic instead of induction for all the divisibility questions in FP1.

Attached to my post is how a similar integral would appear on an exam paper.

Capture.PNG21.2KB

Reply 10

Notnek

Original post by I hate maths

PRSOM for the nifty method, but did you really think that A Level maths exams would afford you opportunities to use whichever method you wished? :biggrin:

Well once part a) has already been done, the quickest way is to use it instead of using DFranklin's method :smile:

Reply 11

username3555092

Original post by Notnek

Well once part a) has already been done, the quickest way is to use it instead of using DFranklin's method :smile:

Which is true, but my point being that they often force part a on you.

Reply 12

DFranklin

Original post by I hate maths

Attached to my post is how a similar integral would appear on an exam paper.

I think you need to change "would" with "might".

But yeah, if they tell you to use a stupid method, you have to use a stupid method. (And I don't think there's much doubt that there are at least 2 methods *significantly* quicker than the one they tell you to use).

Reply 13

Notnek

Original post by DFranklin

But yeah, if they tell you to use a stupid method, you have to use a stupid method. (And I don't think there's much doubt that there are at least 2 methods *significantly* quicker than the one they tell you to use).

While this is true for a lot of A Level questions, I think for the one posted above it's really testing that a student can apply De Moivre's theorem. The integration is just a "nice" (in their opinion) follow up and isn't really testing that much integration ability.

Reply 14

DFranklin

Original post by Notnek

Yes sure. Given how the question is broken up, it's the natural way to do it. But if you look at it as "solve this integral, here's a hint for the first step" (i.e. if you assume the goal was to find the integral), then it's a reasonably inefficient way of going about it.

[More generally, I think the (z+1/z)^k trick is used very rarely outside of this kind of A-level question, so I find it frustrating that they try to pretend it's actually "equivalently useful" to the (c+is)^k trick. I'd guess I've used (c+is)^k ten times more often that the (z+1/z) trick with real problems.]