How could I make an exponential function which has a limit of around 1.53?

mpatryluk
Messages
46
Reaction score
0
I'm modelling a variable output Y which has a value of 1 at x=0.

I've noticed that in the system I'm modelling, as x increases, y increases at an exponentially decreasing rate, up until a limit of around 1.53. I view this as changes in x causing the Y value to increase by a max of 53%.

The only problem is I've been working at it but I don't know where to begin in modelling a function with such a limit.

Can anyone think of a solution?

Thanks!
 
Mathematics news on Phys.org
You could use something like y = 1.53 - 0.53/(a*x+1), where a >= 0. I'm assuming x >= 0 as well here.

Adjust a to determine how quickly it approaches 1.53, for example determine the x at which y is say 1.35 (2/3rds), in which case a = 2/x.
 
  • Like
Likes mpatryluk
Lord Crc said:
You could use something like y = 1.53 - 0.53/(a*x+1), where a >= 0. I'm assuming x >= 0 as well here.

Adjust a to determine how quickly it approaches 1.53, for example determine the x at which y is say 1.35 (2/3rds), in which case a = 2/x.

Good solution, Thanks!
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top