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> That would be 10.0011, wouldn't it?
> Still, rather neat.
>
> Hmm, actually 10.01 gives the same result.
>
> bisqwit@chii:~$ bc -l
> x=(1+sqrt(5))/2;
> x^1+x^-3+x^-4
> 1.99999999999999999999
> x^1+x^-2
> 1.99999999999999999999
Tut tut, youth of today, taking short cuts with numerical methods ... just
do the algebra and you'll see they are both *exactly* 2, not this
1.9999999999 mess ;-)
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scott wrote:
> Tut tut, youth of today, taking short cuts with numerical methods ...
> just do the algebra and you'll see they are both *exactly* 2, not this
> 1.9999999999 mess ;-)
...says the guy who thinks that "1/0" is a definite value. :-P
Hmm, let's see now...
φ = (1 + sqrt(5)) / 2
φ + 1 = (1 + sqrt(5)) / 2 + 1 = (3 + sqrt(5)) / 2
φ^2 = (1 + sqrt(5))^2 / 4 = (1 + 2 sqrt(5) + 5) / 4 = 1/4 + sqrt(5)/2 + 5/4
Nope, sorry, I'm not seeing it.
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> Hmm, let's see now...
>
> φ = (1 + sqrt(5)) / 2
>
> φ + 1 = (1 + sqrt(5)) / 2 + 1 = (3 + sqrt(5)) / 2
>
> φ^2 = (1 + sqrt(5))^2 / 4 = (1 + 2 sqrt(5) + 5) / 4 = 1/4 + sqrt(5)/2 +
> 5/4
>
> Nope, sorry, I'm not seeing it.
You just need to do a couple more steps:
1/4 + sqrt(5)/2 + 5/4
=
6/4 + sqrt(5)/2
=
3/2 + sqrt(5)/2
=
(3+sqrt(5))/2
:-)
Or you can just solve x+1=x^2 your favourite way.
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>> Nope, sorry, I'm not seeing it.
>
> You just need to do a couple more steps:
>
> 1/4 + sqrt(5)/2 + 5/4
> =
> 6/4 + sqrt(5)/2
> =
> 3/2 + sqrt(5)/2
> =
> (3+sqrt(5))/2
>
> :-)
I failed to notice the two constant terms at opposite ends of the
equation. Clearly I need to invent that CAS real soon now... o_O
> Or you can just solve x+1=x^2 your favourite way.
Yeah, I guess.
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