prove i^i is a real number

I don't know, but here it is again:

let y = i^i
Therefore ln(y) = i.ln(i)
i can be written as e^((1/2).i.pi))
meaning that ln(y) = i.ln(e^((1/2).i.pi)))
so ln (y) = i.i.pi.(1/2)
ln(y) = -pi/2
y = e^(-pi/2)
i^i = e^(-pi/2) ... which is a real number. QED.

Reply 2

byb3

chrisbphd

fascinating stuff, no really, i mean it

Reply 3

chrisbphd

I can't quite work out whether that is meant to be sarcasm. I hope not, because it was in response to somebody's question. And it is interesting! :smile:

Reply 4

Nylex

chrisbphd

Lol, now I know how to do one part of one of the questions on our Maths problem sheet.

Reply 5

Camford

chrisbphd

You forgot to mension that assume the normal rule of indices apply here.

Reply 6

chrisbphd

If they didn't apply, they would not be the LAWS of indices, would they? You do not state the laws of addition every time you add do you?

Reply 7

theone

chrisbphd

This isn't strictly true:

i^i is in fact multi-valued:

since i = e^i(pi/2+2kpi) i^i = e^-(pi/2 + 2kpi) and as such is multi-valued and real.

Reply 8

Rich

elpaw

????????????????????

The solution to this is similar to how one solves such questions as differentiate x^x or differentiate (sin(x))^x etc. Very popular maths interview questions.

Reply 9

Camford

chrisbphd

If they didn't apply, they would not be the LAWS of indices, would they? You do not state the laws of addition every time you add do you?

I remember that I had a situation where normally index rule does not apply, but I can't remember what it was now. I'll send it to you when I remember it, if you want it.

To Theone:
how did your BMO go yesterday?

Reply 10

theone

Camford

It went ok thanks you can post on this thread here i made: http://www.uk-learning.net/showthread.php?t=17853&highlight=bmo1

Let me know how you got on - I was disappointed not to get 5 having read a rather simple solution....

Reply 11

MC To Tha Zee

theone

I^I = 1

Reply 12

chrisbphd

I^I is only equal to 1 if I = 1. i^i (i = sqrt(-1)) is NEVER, EVER 1, even on Sundays!

Reply 13

Camford

Situation where normal rules of indices do not apply

e^i(pi)=-1
(e^i(pi))^2=1
e^2i(pi)=1
therefore
2i(pi)=0
i(pi)=0
therefore either i = 0 or pi = 0 which is neither.

Reply 14

theone

Camford

Situation where normal rules of indices do not apply

e^i(pi)=-1
(e^i(pi))^2=1
e^2i(pi)=1
therefore
2i(pi)=0
i(pi)=0
therefore either i = 0 or pi = 0 which is neither.

It is not that you're use of indices is invalid it's that e^2i(pi)=1 => 2i(pi)=0 is wrong, since the logarithmic function is multi-valued with complex numbers.

Reply 15

Rich

Camford

Situation where normal rules of indices do not apply

e^i(pi)=-1
(e^i(pi))^2=1
e^2i(pi)=1
therefore
2i(pi)=0
i(pi)=0
therefore either i = 0 or pi = 0 which is neither.

I'm sorry, but this has nothing to do with the 'normal rules' of indices not applying. There is a mistake in your logic. It's fine up until you take logs. The natural log of 1 is not single-valued. See the following:

1 = e^(2*pi*n*i) where n is an integer, hence ln(1) = 2*pi*n*i and is multi-valued as n ranges over the integers. Your mistake comes simply from you choosing the wrong value for n. You choose n = 0, when you need to choose n = 1 for this particular case (it is a particular case of the equation e^((1 + 2n)(i*pi)) = -1 for integral n). It's nothing to do with 'rules of indices'.

Regards,

Reply 16

Camford

rahaydenuk

I know it's not and I did not take logs. Rule of indices says that n^0=1. But in this case this particular rule does not applies since 2i(pi) <> 0.

Reply 17

Rich

Camford

I know it's not and I did not take logs. Rule of indices says that n^0=1. But in this case this particular rule does not applies since 2i(pi) <> 0.

Just because n^0 = 1, does not mean that if you have for example a^b = 1 that b has to be zero. Only that it could. So this is not even a valid application of the rule n^0 = 1.