P Passions Diamond Member Feb 17, 2000 6,855 3 0 Oct 4, 2002 #1 I need to solve this with precision: x^x=0.606530659 When I put it into my HP 48G, it solves x=0.367879479 But, when you put that back in to check, x=0.692200647 ARGGGGGGGG.....???
I need to solve this with precision: x^x=0.606530659 When I put it into my HP 48G, it solves x=0.367879479 But, when you put that back in to check, x=0.692200647 ARGGGGGGGG.....???
A Alphathree33 Platinum Member Dec 1, 2000 2,419 0 0 Oct 4, 2002 #2 The precision on a floating point variable is...
S spp Golden Member Jul 9, 2001 1,513 0 76 Oct 4, 2002 #3 hmm can't you take natural log of that and solve it?
S spp Golden Member Jul 9, 2001 1,513 0 76 Oct 4, 2002 #4 Originally posted by: spp hmm can't you take natural log of that and solve it? Click to expand... oops... yea use natural log and then use euler's method to keep approximating it?? you should be able to get to pretty precise answer
Originally posted by: spp hmm can't you take natural log of that and solve it? Click to expand... oops... yea use natural log and then use euler's method to keep approximating it?? you should be able to get to pretty precise answer
B Brian23 Banned Dec 28, 1999 1,655 1 0 Oct 4, 2002 #5 if x^2=.606530659 and e^(-.5)=.606530659 then x^2=e^(-.5) x=sqrt(x^(-.5)) and there you go! EDIT -- oops! I read the equation wrong. according to my TI 89, the CORRECT answer is this: x*ln(x)=-.5 I dunno if that helps
if x^2=.606530659 and e^(-.5)=.606530659 then x^2=e^(-.5) x=sqrt(x^(-.5)) and there you go! EDIT -- oops! I read the equation wrong. according to my TI 89, the CORRECT answer is this: x*ln(x)=-.5 I dunno if that helps
S spp Golden Member Jul 9, 2001 1,513 0 76 Oct 4, 2002 #7 Originally posted by: Brian23 if x^2=.606530659 and e^(-.5)=.606530659 then x^2=e^(-.5) x=sqrt(x^(-.5)) and there you go! Click to expand... dude.... it's x^x not x^2......
Originally posted by: Brian23 if x^2=.606530659 and e^(-.5)=.606530659 then x^2=e^(-.5) x=sqrt(x^(-.5)) and there you go! Click to expand... dude.... it's x^x not x^2......
P Passions Diamond Member Feb 17, 2000 6,855 3 0 Oct 4, 2002 #8 anyone with maple or mathmatica hook it up with a precise ans?
K KMurphy Golden Member May 16, 2000 1,014 0 0 Oct 4, 2002 #9 ans = .32447650802884661101302231982777+.31470495693373390251468816640960*i