# Find the expression of the annual fraction of the Python ignition equation, sqrt (2)

Asked 2 weeks ago, Updated 2 weeks ago, 2 views

I'd like to write a code to get the circled a0, a1, a2... and a50.

``````i=0
for i in range(51):
j=i+1
xj=1/(xi-ai)
aj=int(xi)
print (aj)
``````

I thought we could do it roughly like this, but it's not possible. Masters, please help me.

python

2022-09-20 10:18

``````>>> b = [ 2**.5 ]
>>> a = []
>>> a
[]
>>> b
[1.4142135623730951]
>>> for n in range(51):
aa = int(b[n])
a.append(aa)
bb = 1/(b[n] - aa)
b.append(bb)

>>> a
[1, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 1, 1, 1, 3, 3, 1, 3, 1, 1,
2, 1809, 1, 2, 5, 2, 2, 1, 2, 1,
3, 3, 4, 1, 1, 3, 12, 2, 2, 10,
32]
``````

Put the beta n in the question in a list named b so that b, b, b, etc. are beta0, beta1, beta2, etc., and similarly put the alpha n in a list named a, a, a etc. represent alpha0, alpha1, and alpha2.

The value of the a list obtained in this way is the same as the result above.

However, this result differs from the annual fraction ([1:2,2,2,2,2...]) of sqrt(2). 1.414 not exactly sqrt(2)... Because we used an approximate value, we go back and get a value other than 2.

In the given formula, b0! = b1 but b1 == b2. Since the equation for obtaining b3 from b2 is the same as the equation for obtaining b2 from b1, b1 == b2 == b3 == b4 == ... You can see that it is. So the annual fraction of sqrt (2) can be obtained without a complex equation.

If you want to find the exact value of the annual fraction through formulas and operations... You need to think about it a little more.

2022-09-20 10:18

## If you have any answers or tips

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