Let "phi" be the golden mean. Can you show that 0 < (phi^2) /5 - (pi/6) < 10^(-5)?

φ²/5 - π/6

= (1 + √5)²/20 - π/6

= (18 + 6√5 - 10π)/60

I used Generalized Continued Fraction for √5, with a couple of nested fractions, I get

√5 ≈ 2 + 1/(4 + 1/(4 + 1/4)) = 682/305

It's trivial that 682/305 < √5

==> 18 + 6√5 > 18 + 6(682/305) = 9582/305 = 31.41639

==> 18 + 6√5 > 10(3.141639) > 10π, because π < 355/113 = 3.1415929

==> 18 + 6√5 > 10π

==> 18 + 6√5 - 10π > 0

==> (18 + 6√5 - 10π)/60 > 0

==> 0 < φ²/5 - π/6 .................... [1]

I used Generalized Continued Fraction once more but with more nested fractions to get

√5 ≈ 51841/23184

It's trivial that √5 < 51841/23184, and that π > 3.14159

==> 18 + 6√5 < 18 + 6(51841/23184) = 121393/3864 = 31.4164

==> 18 + 6√5 - 10π < 31.4164 - 10(3.14159) = 5 * 10^(-4)

==> (18 + 6√5 - 10π)/60 < (5/6) * 10^(-5) < 10^(-5)

==> φ²/5 - π/6 < 10^(-5) .................... [2]

From [1] and [2]:

0 < φ²/5 - π/6 < 10^(-5) ∎

EDIT1:

We can also prove that the both lower bound and upper bound can be improved to

7.68 * 10^(-6) < φ²/5 - π/6 < 8.33 * 10^(-6)

By showing that (31.41639 - 10(3.14159))/60 > 7.68 * 10^(-6) and the line written above:

... < (5/6) * 10^(-5)

EDIT2:

You can also rephrase this question to

Prove that

0 < 6 + 6φ - 5π < 10^(-3)

Because φ² = φ + 1

Confirmation:

http://www.wolframalpha.com/input/?i=approximate+%...

Press the "Approximate form" button

@gianlino: none of the calculations I used requires a calculator.

You can prove that (682/305 < √5) and (51841/23184 > √5) by contradiction. Suppose 682/305 >= √5. Then you square both sides of the equation, take out the denominator, then you get some absurd inequality.

18 + 6(682/305) = 121393/3864

18 + 6(51841/23184) = 121393/3864

are true and easily proven if you have the patience to multiply and add large numbers.

So does

355/113 = 3.1415929

EDIT2: Oh I can't help you there... =(

Phi is an infinite series in the form of sqrt(1+sqrt(1+sqrt(1.......))...))

Phi^2 is therefore 1+phi. Also, phi is mathematically equal to (1+sqrt(5))2

So, this becomes 1/5+phi/5-pi/6 which becomes

(3+sqrt(5))/10-pi/6.

This is kind-of an awkward problem now, because it's asking for a specific arbitrary sum, and the easiest way to do it is just to evaluate it by comparing the expansions of both. It's asking to show that

sqrt(5)/10-pi/6 is only slightly more than -.3

Now, noting that all values have denominators with a factor of 2, let's multiply.

sqrt(5)/5-pi/3 is slightly greater than -.6

And, sqrt(5)/5 is another way of saying, 1/sqrt(5)

So, we must show that 1/sqrt(5)-pi/3 is slightly>-.6

This is where my interest is piqued. You may have heard of Rogers–Ramanujan continued fractions, and in them, these numbers hold some significance.

According to his work,

e^(-2pi/sqrt(5))/

((sqrt(5)/(1+5^(3/4)(phi-1)^(5/2)-1)^(1/5))-phi)=1+e^(-2pi(sqrt(5))/(1+e^(-2pi(sqrt(5)).....))

This is somewhat interesting because sqrt(5) is actually just 2+1/(4+1/(4+1/...)). The best thing to do is to make both have the same denominator.

1/sqrt(5)-pi/3=(3sqrt(5)-5pi)/15 is slightly>-.6

Flip it with the reciprocals.

15/(3sqrt(5)-5pi) is slightly<-5/3

Divide both sides by 5 and multiply by 3.

9/(3sqrt(5)-5pi) is slightly<-1

Flip it again.

(3sqrt(5)-5pi)/9 is slightly>-1

Let's adjust this to be <1 instead.

(5pi-3sqrt(5))/9 is slightly<1

We've come a long way from the original. Also, now sqrt(5) is in the denominator, which is nice because it can be expanded. From our earlier expansion, this becomes (5pi-6-3/(4+1/(4/1...))/9

Let's add 2/3 to both sides and scale it so that only one side has irrational numbers. This is somewhat unnecessary, but it makes the continuation more visible. 5/3*pi-1/... is slightly less than 5.

Let x=1/(4+1/(4/1...))

It can be rewritten as

1/(4+x)=x

Additionally, 5/3pi=5/12*(1-1/3+1/5...)

5/12*(1-1/3+1/5...)-x is slightly less than 5.

x, however, can also be written as a long series of fractions, but it's more useful to represent it as a continued fraction. If you rewrite the multiplication as a continued fraction, you get that,

pi*5/3=

5+1/(4+1/(4+1/(4+1/(1+3/(19....))))).

After subtracting 5 from both sides, you find that as a result of the fact that it is 1/ a number less than 4 after the streak of 4's, it causes the fraction of pi to be slightly larger. The fraction 5/3 was conveniently chosen because it allows for a value which is easy to multiply out.

So, it is demonstrable by comparing the continued fractions of reduced forms of the numbers. The difference ends up being about 1/12465, which can be found if going through the work of subtracting the two on their own, as that is the first denominator of the continued fraction of the answer.

So, tl;dr:

Pi and phi are both irrational. This means that they can be represented as an infinite series of receding fractions. By multiplying and dividing these fractions, people can find fractions that make two irrational numbers similar. In this case, they happened to be convenient numbers.

Random C&A Peoples Survey: (>^.^)-b Do you regularly provide Thumbs UP right here in C&A!? d-(^.^<) definite? <(?-?)> No? (>o.o)>... Why?....<(0-o)> Monkeys?... - definite :) === Random Anime and different Questions === (>^.^)>[a million] What replaced into the final Anime you have Watched? What approximately Manga? - Anime: Getbackers (for the third time i'm gazing it) - Manga: Hitori De Yare Yo (oneshot yaoi manga) (>^.^)>[2] Do you watch Anime on line? If definite, what's your determination: computer or computer!? - definite, i might prefer to visual demonstrate unit on a computer, yet i do no longer very own one so it must be computer. :( (>^.^)>[3] Which Room of your place are you Answering this question from genuine now? - My mom and dad room (it relatively is the place the computer is regrettably). (>^.^)>[4] Do the "(>^.^)>[#]" 'type sign Holders" Annoy you? - No (>^.^)>[5] what's the colour of your famous Anime character's Hair and Eyes? - Hair is black and eyes are brown i think of. XD

Please extend the time on this question, to give the rest of us a chance to try to figure out how to do this one. Very interesting.

Edit: Christopher Ricci has worked out an interesting "squaring of the circle" based on this "classic approximation" between phi and pi. See link. I don't think it answers your question, however, it's a place to start.