How Do You Normalize a Function to Have a Maximum of 1 and Minimum of 0?

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To normalize the function p(x) = β cos(πx) so that its maximum is 1 and minimum is 0, first determine its original range, which is from -β to β. The maximum value is β and the minimum value is -β. To adjust this, divide the function by 2β to shrink the range to -0.5 to 0.5, resulting in p'(x) = 0.5 cos(πx). Finally, shift the function upward by adding 0.5, yielding the normalized function p''(x) = 0.5 (cos(πx) + 1). This process effectively transforms the function to meet the desired criteria.
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Hi all,

Say if I had a function for example p(x) = \beta \cos(\pi x)

And I wanted to alter it such that the max value of p(x) is 1 and its minimum value is 0.

How would I go about doing this?

Thanks for your help in advance!
Max
 
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maxtor101 said:
Hi all,

Say if I had a function for example p(x) = \beta \cos(\pi x)

And I wanted to alter it such that the max value of p(x) is 1 and its minimum value is 0.

How would I go about doing this?

Thanks for your help in advance!
Max
Do you know the minimum and maximum values of p(x) = \beta \cos(\pi x) (before changing p(x))?
 
Well yes, the maximum value would be \beta and the minimum value would be - \beta..
 
Well a very simple way to do it would be to first "shrink" your range from being -β to β, and making it 1. You can do this by dividing by 2β, and you get p'(x) = 0.5 cos(\pix)
Now your function covers -0.5 to 0.5 so what you have to do now is move its range "up" by 0.5... so you get p''(x) = 0.5 (cos(\pix) + 1)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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