Induction2 (1 Viewer)

[LaCe]

This question is pissing me off, i should be able to do this but i'm having a very poor day today with question 7s and 8s

x=cos@ + isin@

x+x^-1=2cos@

Prove that

Easy to prove for 1st term, then we have
2. Assume true for kth term
x^k+x^-k=2cosk@

3.Show statement is true for (k+1)th term

LHS=x^(k+1)+x^-(k+1)
RHS=2cos(k+1)@

I have gone through with it a bit but doesnt seem to work

im using (x1+x^-1)^(k+1) but u cant use that, and simplifying is getting me nowhere, what is the trick?

 

[FinalFantasy]

x=cos@ + isin@
u mean prove: x^n+x^-n=2cos n@
"2. Assume true for kth term
x^k+x^-k=2cosk@"
".Show statement is true for (k+1)th term"
i.e x^(k+1)+x^-(k+1)=2cos(k+1)@
LHS=cos(k+1)@+isin(k+1)@+cos(-(k+1)@)+isin(-(k+1)@)
but cos (a)=cos(-a) and sin(-b)=-sin(b)
.: LHS=cos(k+1)@+cos(k+1)@+isin(k+1)@-isin(k+1)@
=2cos(k+1)@

 

[FinalFantasy]

i just realised i didn't use the assume true for kth term..
so i dunno lol..
i haven't done harder induction:\

 

[Slidey]

LHS=(cis@)^(k+1)+(cis(-@)^(k+1)=cis([k+1]@)+cis(-[k+1]@)=2cos([k+1]@)=RHS...

I don't get what you mean?

 

[FinalFantasy]

i did x=cos@ + isin@
so x^(k+1)=(cos@ + isin@)^(k+1)=cos(k+1)@+isin(k+1)@ by de moivres theorem

 

[Slidey]

You don't need to use the assume true. Just assume it is true for k+1, and then if you can prove it's also true for some integer k+1, you're done. I think. Can you think of a reason why you MUST use the assume true?

 

[Slidey]

Yeah, I know. I did the same thing, FF.

 

[FinalFantasy]

oh goodie! i thought mine was wrong.. phew!

 

[Slidey]

Oh, no. Sorry, I am just slow to post. I didn't know you'd posted before me.

 

[LaCe]

So u dont have to use assumption??

I know its easy to use it without the assumption but r u allowed to do that?

answer: of course u are!

 

[yrtherenonames]

Are you sure that's 'harder induction'?

 

[FinalFantasy]

Are you sure that's 'harder induction'?

i dunno, i just call induction in 4unit "harder induction" lol