4 ms·
EDIT: Ignore this. I tried doing this for x = sqrt(2) and it quickly approaches infinity. What is the actual solution for x?
by Houshalter 13y ago
EDIT: Ignore this.
I tried doing this for x = sqrt(2) and it quickly approaches infinity. What is the actual solution for x?
- ColinWright 13y agoI don't see what you're doing to make it approach infinity. Let r0=sqrt(2) and r(i+1)=sqrt(2)^r(i). Then r0 = 1.4142135624 r1 = 1.6325269194 r2 = 1.7608395559 r3 = 1.8409108693 r4 = 1.8927126968 r5 = 1.9269997018 r6 = 1.9500347738 r7 = 1.9656648865 ... This is clearly approaching 2. Not sure what you were doing. Edit: Ah! I see - you've got the association the wrong way round. The convention is that a^b^c is a^(b^c). This is because (a^b)^c can just as easily be expressed as a^(bc), so you get more expressive power.
- Houshalter 13y agoYou're right. I just completely abused math and did previous^x rather than x^previous. Fixed the code and it does become 2. Thanks.