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Revisiting 9/11?


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15 minutes ago, thplinth said:

Time is shorty andymac, we only have so much left to work this shit out.

i an pi are ripping the pish out of us it seems.

No one can pin them down!

I will dig out some stuff about i later today.

Edited by andymac
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14 hours ago, thplinth said:

Ok let's make it...

A^(BC) +1 = 0

Solve for B.

What is the answer? (please show your workings)

Aye, that would be the hard bit. I would need to go back to math school for at least 6 months before even trying to attempt that. It's been 30 years since I did any of this sort of stuff.

I just know that i squared = -1. 

It's like having an equation like x squared + 3x - 9 = 27 and asking what does 2 equal? 2 is defined as 2. i is defined as i.

 

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23 minutes ago, phart said:

Can you even solve for i, since i doesn't exist but i squared does?

Plus the simple euler equation is for a specific value in some Taylor series.

I just don't know what i'm doing with this.

I certainly can't.;)

I think you are probably correct. It's straight forward enough to "solve" these equations if we are talking about integral powers. But as soon as we start raising things to the power of irrational numbers and imaginary numbers we probably can't "solve" them in that sense. 

I'm happy to leave it at that, unless somebody can come up with a better explanation.

Where's Chicmac when we need him? I seem to remember that he knew about stuff like this?

 

Edited by Orraloon
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I pumped into this site and solved it for b and came out with the exact same answer as the Wind above... he knows his math it seems.

a^(bc) + 1 = 0 

https://quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#c=solve_stepssolveequation&v1=a%5E%28bc%29%20%2B%201%20%3D%200%20&v2=b

I am sure there will be some massive problem with doing it for Eulers one though. 

 

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On 9/16/2017 at 11:05 PM, Eisegerwind said:

e power( i pi )+1=0

e power (i pi)= -1

ln e power (i pi) =ln(-1)

i pi= ln(-1)

i= ln(-1)/pi

or

pi= ln(-1)/i

The previous get out clauses apply.

 

One problem with that is that natural logs are only defined for positive numbers, I think?

So, Your ln(-1) is meaningless or undefined in that equation.

So you would probably need to use complex logs, which another kettle of fish altogether.

 

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54 minutes ago, thplinth said:

I pumped into this site and solved it for b and came out with the exact same answer as the Wind above... he knows his math it seems.

a^(bc) + 1 = 0 

https://quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#c=solve_stepssolveequation&v1=a%5E%28bc%29%20%2B%201%20%3D%200%20&v2=b

I am sure there will be some massive problem with doing it for Eulers one though. 

 

 

Ditto.

I also get this wee message "Unable to show steps for equations of this type". Somebody else reluctant to show their workings?;)

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On 18/09/2017 at 11:45 AM, Orraloon said:

I certainly can't.;)

I think you are probably correct. It's straight forward enough to "solve" these equations if we are talking about integral powers. But as soon as we start raising things to the power of irrational numbers and imaginary numbers we probably can't "solve" them in that sense. 

I'm happy to leave it at that, unless somebody can come up with a better explanation.

Where's Chicmac when we need him? I seem to remember that he knew about stuff like this?

 

Commenting on Wee Ginger Dug's blog today.

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