Math Is Fun Forum

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Inversions: 0.9-1.11, 0.91-1.0981, 0.92-1.0864, 0.93-1.0749,...

To find an inversion is to let 1 divide it.

However, I've found out an easier approach to approximate the inversion of some number close to 1 with great precision. And the title demonstrates some examples of this method.

Have you guessed what's the method?
And the reason behind it??

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X'(y-Xβ)=0

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Inversions: 0.9-1.11, 0.91-1.0981, 0.92-1.0864, 0.93-1.0749,...

It looks like you're taking the amount that the thing you're inverting is less than one by, then adding it and its square to 1.

I don't think I could have explained that more badly if I'd tried.
As an example, say we have 0.94. This is less than one by 0.06, so you add that and its square (0.0036) to 1, which gives 0.0636.

I think this works because you're approximating 1/(1-x) by 1 + x + x², and I would guess that those are the first three terms of the MacLaurin expansion. You're neglecting all the terms after that, but because x is close to 0 and all of those terms would involve at least an x³, they don't affect the answer too much.

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Why did the vector cross the road?
It wanted to be normal.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Inversions: 0.9-1.11, 0.91-1.0981, 0.92-1.0864, 0.93-1.0749,...

Correct!

I need this approximation badly to work on my slide bar. The arcsine scale between 80° to 90° is too rough, making getting the angle back from its sine 0.985 to 0.99 imprecise. However, the arccot scale between 80° to 90° is detailed enough and it also provides a scale of √(1+x²). When x=tgx, √(1+x²)=1/sinx. So having developed this method, I can find the angle through its sine with great precision. I get the inverse of its sine first and find it on  √(1+x²) scale, then x scale represents its cot, arccot scale revealing its angle.

By this method arcsin0.95=71.6° and the calculator gives the result 71.805...°.
                        arcsin0.993= 83.22°                actual     83.216....°

Very close.

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Inversions: 0.9-1.11, 0.91-1.0981, 0.92-1.0864, 0.93-1.0749,...

Lucky that MacLaurin expansion doesn't get too complicated for 1/(1-Δx)

Last edited by George,Y (2007-05-14 02:24:22)

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X'(y-Xβ)=0