[28hotshot]

um i have to know how to find X^X=Pi by tomorrow and know how to do it algebraically.. my chemistry teacher is asking this so just wondering if anyone knows how to do it.. like on the alculator or on paper.. i know the answer its 1.8 something but i need to know how to actually do it... thanks

[UG]

Err..√Pi = X

Have a read of: http://www.homeschoolmath.net/teaching/square-root-algorithm.php

[Borek]

Not x²=π but x^x=π.

[UG]

Oh dear me! Sorry....   

[UG]

I've been thinking about this and it is quite complicated, you need something like the Newton-Raphson method to solve it.

http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/Numerical/node6.html
http://www.sosmath.com/calculus/diff/der07/der07.html

The answer: x = ~1.854105968 

[cliverlong]

Not x²=π but x^x=π.

there isn't a "direct" solution to such a problem.

There are several "trial and improvement" techniques

Newton Raphson is one way - but it is a bit "technical" - however it is fast for "well-behaved" functions (I won't define well behaved at the moment)

"slower" but guaranteed and to any level of accuracy is bisection.
Step 1. If we try x=1 then we find 1¹ = 1
Step 2. If we try x=2 then we find 2² = 4

Now, π lies between 1 and 4. Can you think where to go from there?

Another alternative is to use a recursive technique. Has the original poster come across logarithms? Recursion has virtue of being faster than bisection but less "technical" than NR (some limitations on its application depending on "well-behavededness" of function if it is to work)

Clive

Edit: Fiddling with the recursive formula (one converges the other doesn't) and trying numbers in a calculator -  I think bisection as well as being simpler is quicker than recursion - probably due to the logarithm. I will give the details later after28hotshot has tried. UG's solution looks correct (at least where my attempts are heading).

[Borek]

TI-89 gives 1.85410596792

Using Open Office Calc to implement Newton method I got 1.85410596792103

[aldoxime_amine]

Graph it out using graphcalc or any program or use mathematica..

There is no closed form solution to this equation.