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

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Discussion Overview

The discussion revolves around the normalization of a function, specifically how to adjust the function p(x) = β cos(πx) so that its maximum value is 1 and its minimum value is 0. The focus is on mathematical reasoning and proposed methods for achieving this normalization.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks how to alter the function p(x) = β cos(πx) to achieve a maximum of 1 and a minimum of 0.
  • Another participant confirms that the maximum value of p(x) is β and the minimum value is -β before any changes are made.
  • A different participant suggests a method to normalize the function by first dividing by 2β to adjust the range to -0.5 to 0.5, and then shifting the range up by 0.5 to achieve the desired normalization.

Areas of Agreement / Disagreement

Participants do not appear to have reached a consensus on the normalization method, as various approaches and confirmations of values are discussed without a definitive agreement.

Contextual Notes

The discussion does not clarify the specific values of β or any assumptions regarding its positivity or negativity, which may affect the normalization process.

maxtor101
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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)
 

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