Elementary my dear Watson.

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
It is more often said than not that the number one raised to any power at all will always result in one. That is to say,..one squared is one. One cubed is one. One raised to the power of one billion is one. Lets not forget negative numbers,...one raised to the power of negative one is one,...to negative 1000 is one....how frustrating. I know...what about zero...one raised to the power of zero....you guessed it...is one. :(

Is this always the case? Well the answer in short is no... :D

What if we raise one to the power of the square root of negative one? Now the square root of negative one is imaginary,...not make believe,..it's as real as you and I and denoted by 'i' if your a mathematician or 'j' if your an electrical engineer.

So...1^i = 1^j...what does it equal?

Ron.

P.S : This is the hardest and most complex question I have asked so far, and I dare say rather unfair,..falling within the category of advanced engineering mathematics, but it is wonderful to the learn, I hope you will enjoy.
 
SydneyRoverP6B said:
It is more often said than not that the number one raised to any power at all will always result in one. That is to say,..one squared is one. One cubed is one. One raised to the power of one billion is one. Lets not forget negative numbers,...one raised to the power of negative one is one,...to negative 1000 is one....how frustrating. I know...what about zero...one raised to the power of zero....you guessed it...is one. :(

Is this always the case? Well the answer in short is no... :D

What if we raise one to the power of the square root of negative one? Now the square root of negative one is imaginary,...not make believe,..it's as real as you and I and denoted by 'i' if your a mathematician or 'j' if your an electrical engineer.

So...1^i = 1^j...what does it equal?

Ron.
Click to expand...

I would need to check this up...

Are you sure that 1^(0+j1) is not 1?

lets take a general view...

a^(b + jc), a, b and c are real numbers...

a = exp ^ (ln (a))

[exp ^ (ln (a))]^(b + jc) = [exp ^ (b x ln (a))] x [exp ^ j(c x ln (a))] = exp [b x ln (a)] x [cos (c x ln(a)) + jsin (c x ln(a)]

b x ln (a) = ln (a^b), {also, c x ln (a) = ln (a^c)} (Laws of logs!)

exp [b x ln (a)] = exp [ln (a^b)] = a^b

a^(b+jc) = a^b x [cos (ln a^c) + jsin (ln a^c)]

In our case, we want 1^(0+j1), so a = 1, b = 0 and c =1

1^0 x [cos (ln {1^1}) + j sin (ln {1^1})]

1^0 = 1, 1^1 =1

1 x [cos (ln {1}) + j sin (ln {1})]

ln (1) = 0

1 x [cos (0) + j sin (0)]

cos (0) = 1, sin (0) = 0

So.. 1 x [1+j0] = 1 x 1 = 1

If my understanding is correct, 1^j = 1 also!
 
I found this on the internet...

power11.gif


In our case, a = 1, b = 0, c = 0, d =1

(a^2 + b^2) = 1,

c/2 = 0, 1^0 = 1

d=1, but arg (a+ib) = 0, so -d x arg (a+ib) = -1 x 0 = 0

exp [-d x arg (a+ib)] = exp (0) = 1

The top most part is 1, i.e. [(a^2 + b^2) ^ (c/2)] x {exp [-d x arg (a+ib)]} = 1 x 1 = 1

c x arg (a+ib) = 0 x 0 = 0

(d/2) x [ln (a^2 + b^2)] = (1/2) x [ln (1^2 + 0^2)] = (1/2) x [ln (1)] = (1/2) x 0 = 0

{c x arg (a+bi)} + {(d/2) x [ln {a^2 + b^2}]} = 0 + 0 = 0

What we are left with basically is...

1 x (cos (0) + i x sin (0)). Since cos (0) = 1, and sin (0) = 0

1 x (1 + i0) = 1 x 1 = 1

Even this rather large expression ends up stating 1^j = 1

Unless I am missing something really fundamental/complicated here... :oops:
 
i need to have a good read of my engineering mathematics books to have a chance with this... Unfortunately(or fortunately;-) i'm off on holiday on tuesday morning so preparations are currently taking precedence.... For some reason in my brain it can be more than one value - could be repeating, infinity, or positive or negative... Like i say i need a little research first....;-)

rich
 
rockdemon said:
i need to have a good read of my engineering mathematics books to have a chance with this... Unfortunately(or fortunately;-) i'm off on holiday on tuesday morning so preparations are currently taking precedence.... For some reason in my brain it can be more than one value - could be repeating, infinity, or positive or negative... Like i say i need a little research first....;-)

rich
Click to expand...

You sparked my memory!

1 = exp (j x 2 x n x pi), where n = integer i.e. n = -Inf .....-5, -4, -3, - 2, -1 , 0, 1, 2, 3, 4, 5.... Inf

1^j = [exp (j x 2 x n x pi)] ^ j = exp (j x j x 2 x n x pi) = exp (-2 x n x pi). Since we can go from - inf to + inf,

1^j = exp (2 x n x pi),

1^j = 1, exp (2 x pi), exp (-2 x pi), exp (4 x pi), exp (-4 x pi), exp (6 x pi), exp (-6 x pi), etc...

Or, the other way...

1^j =, say, p

ln (1^j) = ln (exp (j x 2 x n x pi)^j) = j x ln (exp (j x 2 x n x pi)) = j x j x 2 x n x pi = -2 x n x pi = ln (p)

p = 1^j = exp (-2 x n x pi) = exp (2 x n x pi) (since n spans - inf to + inf)

1^j = exp (2 x n x pi)

I knew I was missing something fundamental! :oops: :LOL:
 
Whilst wandering aimlessly through Baker Street the other day I passed number 221B and espied the famous detective, Sherlock Holmes, painting his front door.

All doors in Baker Street are white, all window frames are white, and all soffits and other external woodwork is white.

I stopped and asked Sherlock the reason why he was painting HIS door a glaring Yellow and he turned and repiled,

"It's a lemon entry my dear Quattro."

8)

That's all I can offer to this thread :LOL:

I wonder how he knew who I was?
 
Rather an interesting puzzle isn't it, and what an amzing result. The internet will largely tell you that the answer is 1 which as Darth found is incorrect,..so lets take a look.

1^i can be expressed using the exponential (hence the elementary in the title..in mathematics exponentials and natural logs are known as elementary functions) as e^lni

now looking at lni we express 'i' which is the imaginary operator in exponential form as a complex number...1e^i0
so lni = ln(1e^io)
= ln(1) + lne^i(0 + 2PIk)
= i(2PIk)

So 1^i = e^i(ln1 + lne^i(0 + 2PIk))
= e^i(ln1 + i(0 + 2PIk))
= e^iln1.e^-2PIk
= e^-2PIk,..........where k is an integer including zero.

How cool is that as they say,...the principal value of the argument is equal to one, but all other values of which there is an infinite number are not. What is even more amazing is that the results are all real even though we raised one to an imaginary number.

Ron.
 
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