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I was just wandering around equation and stuff,
After some calculation I get this
√(b^2 - 4c) = |(b^2-2c)/b|
It's so wierd , by it , I can calcultion a square root approximately.
As the number becomes greater , the result become more accurate
when I assume b^2-4c = 99999, I calculate the sqr99999 =316.22857
which by calculator it's 316.2261
when I assume b^2-4c=999999 , I get 999.9995 which is the same as the result come out of the calculator
Wierd man,.~
Last edited by Stanley_Marsh (2006-12-23 04:30:15)
Numbers are the essence of the Universe
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I guess the real question is *how* did you get that? Can you post your steps?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Just "powering" the equation:
Last edited by krassi_holmz (2006-12-23 10:31:13)
IPBLE: Increasing Performance By Lowering Expectations.
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Wow, awesome find! Can... uh... you provide a fully worked through example please?
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I think I get how this works.
Let's try to use it to find the square root of 1000. The square root of 1000 is an irrational number, and hard to calculate without some kind of computer or calculator. However, 1024 is much easier, because it is just 32².
We can say that √1000 = √(1024 - 24) = √(32² - 4x6)
Using Stanley's equation:
√(b^2 - 4c) = |(b^2-2c)/b|
We can see that in this case b = 32, and c = 6.
Substituting those values in gives an approximate square root of |(32² - 2x6)/32| = (1024 - 12)/32 = 31.625.
The actual square root of 1000 is 31.622..., so the approximation is accurate to 2 decimal places and much easier to work out on paper.
As Krassi says, the answer gets more accurate as b gets larger in relation to c, and so if you wanted to make your approximation better then you could instead work out the square root of 100000 and divide that answer by 10 afterwards.
Or, in general, find the square root of 1000k² and divide the answer by k afterwards. Increasing k should make the approximation more accurate but also make the calculation harder, because you'd be dealing with bigger numbers.
Incidentally, is there any way of tweaking the formula so that the error term, 4c²/b², gets smaller without making the calculation much harder? I have a feeling that this formula has potential to be developed.
Why did the vector cross the road?
It wanted to be normal.
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OMg , Thank you for solving my mystery! I get this from a x^2+bx+c=0 equation , the determinant of root b^2-4c . I think I see now .hehe
Numbers are the essence of the Universe
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And to how I get that , I get to find my notebook , it's been a long time .
Numbers are the essence of the Universe
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Isn't the determinant b²-4ac?
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Yep , but I assume a to be 1
Numbers are the essence of the Universe
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About Mathsy post - good point.
I suppose the best will be to use the first greater or equal to the number exact square.
And the approximation is really good.
As you can see form the image :
Last edited by krassi_holmz (2006-12-24 06:05:21)
IPBLE: Increasing Performance By Lowering Expectations.
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The Beginning Of All Things To End.
The End Of All Things To Come.
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Yes, the same.
Now, i'll uncover my results.
Let
Last edited by krassi_holmz (2006-12-24 06:01:49)
IPBLE: Increasing Performance By Lowering Expectations.
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b[sup]2[/sup]-4c≈b[sup]2[/sup]-4c+4(c/b)², when c/b is very small, or b is much larger than c.
That's how he got it.
X'(y-Xβ)=0
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b[sup]2[/sup]-4c+4(c/b)²= (b-2c/b)[sup]2[/sup]
X'(y-Xβ)=0
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But if b is much larger than c, simply b is a quicker result. Nevertheless, b minus 2c/b is more accurate.
X'(y-Xβ)=0
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This remids me about something...
Since:
Last edited by krassi_holmz (2006-12-25 21:47:21)
IPBLE: Increasing Performance By Lowering Expectations.
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f(x+Δx)≈f(x)+f'(x)Δx
X'(y-Xβ)=0
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yeap.
IPBLE: Increasing Performance By Lowering Expectations.
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