A better approximate Pythagoras

Calculus Level 5

In the problem Approximate Pythagoras you can see that adding one-half of the shorter leg to the longer leg of a right triangle gives a pretty good estimate for the hypotenuse. b + a / 2 ≈ c b+a/2 \approx c

Now the task is to improve the parameter one-half

Find x x on the interval [ 0 , 1 ] [0,1] such that the maximum percentage error (larger or smaller) of using x a + b xa+b to approximate a 2 + b 2 \sqrt{a^{2}+b^{2}} is minimized.


The answer is 0.3363572758.

Jeremy Galvagni by Jeremy Galvagni , 48, USA

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1 solution

Jeremy Galvagni
Oct 9, 2018

Define the function f ( x ) = x + a 1 + x 2 f(x)= \frac{x+a}{\sqrt{1+x^{2}}} as the proportional error. 1 − f ( x ) 1-f(x) is the percentage difference.

Using the derivative you can show the maximum value of f f occurs at x = 1 a x=\frac{1}{a} and f ( 1 a ) = 1 + a 2 f\left(\frac{1}{a}\right)=\sqrt{1+a^{2}} which tends to 1 1 as x x goes to infinity.

The minimum of f f is 1 1 if a ≥ 2 − 1 a \ge \sqrt{2}-1 otherwise it's f ( 1 ) = 1 + a 2 f(1)=\frac{1+a}{\sqrt{2}}

We seek to minimize the maximum error so that the error from being over is the same as the error from being under: f ( 1 a ) − 1 = 1 − f ( 1 ) f\left(\frac{1}{a}\right) - 1 = 1 - f(1) or 1 + a 2 + 1 + a 2 = 2 \sqrt{1+a^{2}}+\frac{1+a}{\sqrt{2}}=2

Wolfram alpha gives the solution x = 1 − 2 2 + 2 4 − 2 2 ≈ 0.3363572758 x=1-2\sqrt{2}+2\sqrt{4-2\sqrt{2}} \approx \boxed{0.3363572758}

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