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# prove i^i is a real number watch

1. ????????????????????
2. 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.
3. (Original post by chrisbphd)
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.
fascinating stuff, no really, i mean it
4. 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!
5. (Original post by chrisbphd)
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.
Lol, now I know how to do one part of one of the questions on our Maths problem sheet.
6. (Original post by chrisbphd)
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.
You forgot to mension that assume the normal rule of indices apply here.
7. 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?
8. (Original post by chrisbphd)
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.
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.
9. (Original post by 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.
10. (Original post by 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?
11. (Original post by Camford)
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?
It went ok thanks you can post on this thread here i made: http://www.uk-learning.net/showthrea...highlight=bmo1

Let me know how you got on - I was disappointed not to get 5 having read a rather simple solution....
12. (Original post by theone)
It went ok thanks you can post on this thread here i made: http://www.uk-learning.net/showthrea...highlight=bmo1

Let me know how you got on - I was disappointed not to get 5 having read a rather simple solution....
I^I = 1
13. I^I is only equal to 1 if I = 1. i^i (i = sqrt(-1)) is NEVER, EVER 1, even on Sundays!
14. 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.
15. (Original post by 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.
16. (Original post by 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,
17. (Original post by rahaydenuk)
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,
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.
18. (Original post by 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.

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