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

Click For Summary
SUMMARY

This discussion focuses on modeling a variable output Y that approaches a limit of approximately 1.53 as x increases. The proposed function is y = 1.53 - 0.53/(a*x+1), where a is a non-negative parameter that controls the rate of approach to the limit. The discussion emphasizes adjusting the value of a to determine specific output values, such as y reaching 1.35 at a defined x. This approach effectively captures the behavior of Y as it increases at an exponentially decreasing rate.

PREREQUISITES
  • Understanding of exponential functions and their properties
  • Familiarity with mathematical modeling techniques
  • Knowledge of parameter adjustment in functions
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Research the behavior of asymptotic functions in mathematical modeling
  • Learn about parameter tuning in exponential decay functions
  • Explore the concept of limits in calculus
  • Study applications of exponential functions in real-world scenarios
USEFUL FOR

Mathematicians, data scientists, and anyone involved in modeling variable outputs with limits, particularly those interested in exponential functions and their applications.

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!
 
Physics 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   Reactions: 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!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
9K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K