The square root of 2 can be written as an infinite continued fraction.
√2 = 1 + |
1
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2 + |
1
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2 + |
1
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2 + |
1
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2 + ... |
The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].
It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.
1 + |
1
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= 3/2 |
2
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1 + |
1
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= 7/5 | |
2 + |
1
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2
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1 + |
1
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= 17/12 | ||
2 + |
1
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2 + |
1
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2
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1 + |
1
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= 41/29 | |||
2 + |
1
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2 + |
1
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2 + |
1
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2
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Hence the sequence of the first ten convergents for √2 are:
What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
The first ten terms in the sequence of convergents for e are:
The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.
Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.
def func(result,i): t=i*result[1]+result[0] result[0]=result[1] result[1]=t return result def func1(x): if(x%3)==0: k=2*(x/3) else: k=1 return k n=100 result=[1,func1(n)] for i in range(n-1,1,-1): result=func(result,func1(i)) result[0]+=2*result[1] temp=str(int(result[0])) result=0 k=len(temp) for i in range(0,k): result+=int(temp[i]) print(result)