De Moivre's theorem (1 Viewer)

[mathsbrain]

I'm confused...MUST you use induction to prove De Moivre's theorem(by the way how do you prove De moivre's theorem for negative integers?)?

I mean, isnt zw=r_1 r_2 cis(theta_1+theta_2)? then isnt it obvious that z^n=(r^n)cis(n theta)? so why induction?

 

[Carrotsticks]

I'm confused...MUST you use induction to prove De Moivre's theorem(by the way how do you prove De moivre's theorem for negative integers?)?

I mean, isnt zw=r_1 r_2 cis(theta_1+theta_2)? then isnt it obvious that z^n=(r^n)cis(n theta)? so why induction?

The most straightforward way to prove it for negative integers is to use induction.

It may seem 'obvious', but this is because in HSC Mathematics (not the subject, but Maths overall here), they generally lead you into a bad habit of thinking "Because it's true for the first couple of cases, it's probably true for the rest", which can be misleading.

For example, n^+n+17 appears to generate prime numbers for n=1, 2, 3, 4, 5, ... (try it!) So 'obviously' it must generate prime numbers!

But when you test n=16, you see that it doesn't work any more. However, it works for every positive integer up to and including 15.

Similarly, DMT seems to work for n=1, 2, 3, ... But what guarantee is there that it works for all positive integers?

The whole idea of this is that you shouldn't trust what you see at first sight until you have conclusive evidence of it... as 'obvious' as it may be.

Also, it is not at all obvious that DMT works for all n. How about n=1/2 ? Or even worse, n=pi !

Your intuition is often a good way to 'guess' a formula, but it isn't enough to prove it of course.

 

[anomalousdecay]

From memory in the inductive step you need to expand the cos and sin part where you should get something like cos (x(n+1)) + i sin (x(n+1)).

Check any textbook like Terry Lee, Cambridge, Fitzpatrick, etc and they will have the whole proof on De Moivre's theorem for HSC ONLY.

 

[mathsbrain]

The most straightforward way to prove it for negative integers is to use induction.

It may seem 'obvious', but this is because in HSC Mathematics (not the subject, but Maths overall here), they generally lead you into a bad habit of thinking "Because it's true for the first couple of cases, it's probably true for the rest", which can be misleading.

For example, n^+n+17 appears to generate prime numbers for n=1, 2, 3, 4, 5, ... (try it!) So 'obviously' it must generate prime numbers!

But when you test n=16, you see that it doesn't work any more. However, it works for every positive integer up to and including 15.

Similarly, DMT seems to work for n=1, 2, 3, ... But what guarantee is there that it works for all positive integers?

The whole idea of this is that you shouldn't trust what you see at first sight until you have conclusive evidence of it... as 'obvious' as it may be.

Also, it is not at all obvious that DMT works for all n. How about n=1/2 ? Or even worse, n=pi !

Your intuition is often a good way to 'guess' a formula, but it isn't enough to prove it of course.

Thanks and yes i see what you are saying.

But what i am saying is:
If we have zw=r_1 r_2 cis(\theta_1+\theta_2), then we can say z^2=r^2cis(\theta+\theta), and it is true that z^n=r(multiply itself n times) X cis(\theta(adding itself n times))=r^n cis(n \theta). Is that not proving DMT for POSITIVE integers of n? if so then why the ponderous induction or am i missing something?

 

[anomalousdecay]

Thanks and yes i see what you are saying.

But what i am saying is:
If we have zw=r_1 r_2 cis(\theta_1+\theta_2), then we can say z^2=r^2cis(\theta+\theta), and it is true that z^n=r(multiply itself n times) X cis(\theta(adding itself n times))=r^n cis(n \theta). Is that not proving DMT for POSITIVE integers of n? if so then why the ponderous induction or am i missing something?

You are missing the fact that how on earth can we just assume that (I have to use the cis notation but its cringe-worthy)



How can you assume this is true?

How is it possible to say that if we know for sure that it isn't true for cos or sin?

This is why you actually need to use induction. Its because we don't know if its true or not.

So you have the base case for n = 1 right, when you assume it is true for n equal some integer, say k, this is an assumption!!!

So you take the case for n = k+1 where you have to prove both sides are equal after expanding. This is just normal induction from there on.

Generally the point is that if you make an assumption you really need to crunch down and show that the assumption works for a specific case. It might work well for a few bits and pieces here and there but not for everything.

So you prove it by induction for the integers n. This way you can show that it really does work.

What Carrotsticks is saying that when you do this you can't really assume it is completely true without using induction (this is not HSC level but the definition from first principles of sin and cos is quite complex and when you get irrational numbers for n a lot of things could possibly change).

 

[Carrotsticks]

Thanks and yes i see what you are saying.

But what i am saying is:
If we have zw=r_1 r_2 cis(\theta_1+\theta_2), then we can say z^2=r^2cis(\theta+\theta), and it is true that z^n=r(multiply itself n times) X cis(\theta(adding itself n times))=r^n cis(n \theta). Is that not proving DMT for POSITIVE integers of n? if so then why the ponderous induction or am i missing something?

Induction is a way to 100% verify that something is true. What you've got there is a hypothesis, which needs proper verification.

 

[braintic]

I have to use the cis notation but its cringe-worthy.

Why cringeworthy?

 

[anomalousdecay]

Why cringeworthy?

Because the fact that HSC uses cis notation is why OP actually did get confused.

If it was taught in HSC to use:



notation then not as many students would get the misconceptions that cis notation can come along with (note OP's misconception).

So its much more understandable as to why cis notation is not really the clearest notation.

So basically I spent this whole semester getting rid of cis notation and replaced it with exp notation.

I can understand why in HSC methods cis notation is used but nevertheless, to me its cringe worthy.

 

[braintic]

Because the fact that HSC uses cis notation is why OP actually did get confused.

If it was taught in HSC to use:



notation then not as many students would get the misconceptions that cis notation can come along with (note OP's misconception).

So its much more understandable as to why cis notation is not really the clearest notation.

So basically I spent this whole semester getting rid of cis notation and replaced it with exp notation.

I can understand why in HSC methods cis notation is used but nevertheless, to me its cringe worthy.

I can't see any misconception by OP that has to do with the notation.
In fact I kind of agree with him. He is saying that before learning de Moivre's theorem, one has already learned that arg(zw) = argz + arg w, so de Moivre's theorem is an obvious consequence.