find under root 13 ans
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The underoot is 3.6......
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Since 13 is a prime number, there is no simpler form for its square root. √13 is an irrational number somewhere between 3=√9 and 4=√16.
Linearly interpolating, a reasonable first approximation would be:
√13≈3.6=185
We can get better approximations from our initial one (call it a0) using a Newton Raphson method.
A typical formula used to derive a more accurate approximation for √n would be:
ai+1=a2i+n2ai
I prefer to separate ai into numerator pi and denominator qi. So ai=piqi and we can iterate using the formulae:
{pi+1=p2i+nq2iqi+1=2piqi
In our example, n=13, p0=18, q0=5 and we find:
{p1=p20+13q20=324+13⋅25=649q1=2p0q0=180
If we stopped here our approximation would be:
√13≈649180=3.60¯5
Let's try one more iteration:
p2=p21+13q21=421201+13⋅32400=842401q2=2p1q1=233640
Stopping here, we have:
√13≈842401233640≈3.60555127547
Using a calculator:
√13≈3.60555127546398929311
Linearly interpolating, a reasonable first approximation would be:
√13≈3.6=185
We can get better approximations from our initial one (call it a0) using a Newton Raphson method.
A typical formula used to derive a more accurate approximation for √n would be:
ai+1=a2i+n2ai
I prefer to separate ai into numerator pi and denominator qi. So ai=piqi and we can iterate using the formulae:
{pi+1=p2i+nq2iqi+1=2piqi
In our example, n=13, p0=18, q0=5 and we find:
{p1=p20+13q20=324+13⋅25=649q1=2p0q0=180
If we stopped here our approximation would be:
√13≈649180=3.60¯5
Let's try one more iteration:
p2=p21+13q21=421201+13⋅32400=842401q2=2p1q1=233640
Stopping here, we have:
√13≈842401233640≈3.60555127547
Using a calculator:
√13≈3.60555127546398929311
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