Nice problem and solution, with an unexpected detour into continued fractions. Here is my solution, which includes a detailed discussion of the continued fractions method of approximating square roots. 2.25^2.25=(9/4)^(9/4) =[(3/2)²]^(9/4) =(3/2)^(9/2) =(3/2)^(4+½) =(3/2)⁴(3/2)^½ =81/16 √(3/2) =81√3/(16√2) =81√6/32 Now the idea is to approximate √6 using continued fractions, i.e. to generate a sequence of positive integers aₙ (though a₀ may be zero) and "remainders" rₙ (where 0≤rₙ1 instead of √6) √6=a₀+1/(a₁+1/(a₂+r₂)) etc. So that we get a (hopefully) convergent sequence √6=a₀+1/(a₁+1/(a₂+1/(a₃+1/(a₄+...)))) which we abbreviate as √6=[a₀;a₁,a₂,a₃,a₄,...] As 4
I arrived at 3 over the square root of 6, which I'm still not sure is correct or not, but either way I wasn't able to get a decimal approximation in my head. I'd love to learn more about the history behind the formula you used!
Absolutely loved this. Used to dread maths because of conventional teaching methodologies.
Please keep inspiring.
Nice problem and solution, with an unexpected detour into continued fractions.
Here is my solution, which includes a detailed discussion of the continued fractions method of approximating square roots.
2.25^2.25=(9/4)^(9/4)
=[(3/2)²]^(9/4)
=(3/2)^(9/2)
=(3/2)^(4+½)
=(3/2)⁴(3/2)^½
=81/16 √(3/2)
=81√3/(16√2)
=81√6/32
Now the idea is to approximate √6 using continued fractions, i.e. to generate a sequence of positive integers aₙ (though a₀ may be zero) and "remainders" rₙ (where 0≤rₙ1 instead of √6)
√6=a₀+1/(a₁+1/(a₂+r₂)) etc.
So that we get a (hopefully) convergent sequence
√6=a₀+1/(a₁+1/(a₂+1/(a₃+1/(a₄+...))))
which we abbreviate as
√6=[a₀;a₁,a₂,a₃,a₄,...]
As 4
Why not use the fact that sqr(1+t) is approximately 1+ .5t, for small values of t
I arrived at 3 over the square root of 6, which I'm still not sure is correct or not, but either way I wasn't able to get a decimal approximation in my head. I'd love to learn more about the history behind the formula you used!
Please see my comment for an explanation of the continued fraction formula.
I didn’t understand the part a^2 + b/a^2+b/etc
I don’t follow that radical approximation method
Please see my comment for an explanation of the continued fraction formula.
👏👏👏👏👏👏👏👏👏👏
I guesstimated it as 2π, or 6.28. I was very close. 🥧🥧
Start again
243/32
2,25^2,25=x => x=cln( 2,25 * ln(2.25)) => x=cln( 1,82459...) = 6,2....
Do not use commas. 2.25^2.25
omg
Good one! Now reveal a bit about yourself Math Window…. Who are you?
I suppose, as soon as 100.000 subscribers are reached we will get a "behind the scenes"- video answering all questions...
Put a comma right before "Math Window," because you are addressing him/her.