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Why it is mathematically impossible to always do your best!

FearoftheNight

Diamond Member
I would like to thank my calc instructor. <3
Imagine a 2-d plane of x/y axis. Imagine a curve designating your performance in life designated by the function f(x). When someone says you're best they are referring to the point of f(x) where your y value is the global maximum. Which is impossible to have a striaght line curve defining your output. Now say your actual performance is denoted by funtion g(x). Assuming you do absolutely to the best of your raw potential your achievement in life would be denoted by the integral between 0 - infinity (your death) as the area under curve f(x). Now the best you can do it maximize the distance that your g(x) actual output is to f(x). So how much of your best is denoted by the area between the two function denoted by the integral of 0 to infinity of the f(x) - g(x). So all you really have to do is try to minimize the distance between the two curve and reduce the area between. Them. See? Now you know what to say anyone says you're not doing your best.

Good luck on finals guys! 😀

Edit: Of course you can also calculate how far you are from your max potential at point x by taking the limit as that point reaches (delta x) reaches 0 (finding the slope of the tangent line) at that instantaneous moment ala the derivative.
 
Originally posted by: Howard
The function is continuous up until your death, so it's not the integral from 0 to infinity.

i know i said (death) but i put infinity to undicate the uncertainty of that point
 
Originally posted by: FearoftheNight
Originally posted by: Howard
The function is continuous up until your death, so it's not the integral from 0 to infinity.

i know i said (death) but i put infinity to undicate the uncertainty of that point
Why not just call it 'a' (or 'd' 😉), because using infinity to represent a finite number is ridiculous.
 
oh what the hell why not...ur the only person that read my post anyway and should call it b since b is the typical notation to denote upperbounds of integrals
 
Originally posted by: Howard
Originally posted by: FearoftheNight
Originally posted by: Howard
The function is continuous up until your death, so it's not the integral from 0 to infinity.

i know i said (death) but i put infinity to undicate the uncertainty of that point
Why not just call it 'a' (or 'd' 😉), because using infinity to represent a finite number is ridiculous.

ouchy
 
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