# Answer to Question #1751 in Calculus for Imee Carla

Question #1751

How can I use the concept in differentials to approximate a certain number which is not a perfect square?

Expert's answer

You can expand the function of square root in a series :

We can represent the number as a sum of nearest perfect square root R and the difference of the number and this sq.root (N-R), it can be both positive and negative.

f(x) = √N = √(R + (N-R)) = √R √(1 + (N-R)/R).

Denote (N-R)/R as α , |α|<1. The function √(1 +α) can be represented as a sum:

√(1 +α)= 1 + 1/2 α - 1/8 α

Thus the final fomula would be as

√N = √R (1 + 1/2 α - 1/8 α

We can represent the number as a sum of nearest perfect square root R and the difference of the number and this sq.root (N-R), it can be both positive and negative.

f(x) = √N = √(R + (N-R)) = √R √(1 + (N-R)/R).

Denote (N-R)/R as α , |α|<1. The function √(1 +α) can be represented as a sum:

√(1 +α)= 1 + 1/2 α - 1/8 α

^{2 }+ 1/16 α^{3}- 5/128 α^{4}+ ... This expression was obtained by expanding the function √(1 +α) in a Teylor series in neiborhood of 1. √(1 +α) = sum from n=0 to infinity ( f^{(n)}(1)/n! α^{n}), where f^{(n) }(1) the n-th derivative of the function f(a)=√x at the point x = 1.Thus the final fomula would be as

√N = √R (1 + 1/2 α - 1/8 α

^{2 }+ 1/16 α^{3}- 5/128 α^{4}+...), where α = (N-R)/RNeed a fast expert's response?

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