How can Newton's method be used to solve exponential equations iteratively?

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SUMMARY

The discussion centers on using Newton's method, specifically the Newton-Raphson root-finding algorithm, to solve the equation y=e^{ax}+e^{bx} for x iteratively. The user initially struggled with the application of this method due to a misunderstanding of its suitability for exponential equations. After correcting numerical errors, the user successfully implemented the algorithm, confirming its effectiveness for solving such problems.

PREREQUISITES
  • Understanding of Newton-Raphson method for root-finding
  • Familiarity with exponential functions and their properties
  • Basic calculus, including differentiation
  • Ability to manipulate and solve equations
NEXT STEPS
  • Study the derivation and application of the Newton-Raphson method in detail
  • Explore numerical methods for solving transcendental equations
  • Learn about convergence criteria for iterative methods
  • Investigate alternative root-finding algorithms, such as the bisection method or secant method
USEFUL FOR

This discussion is beneficial for mathematicians, engineers, and computer scientists who are interested in numerical methods for solving equations, particularly those involving exponential functions.

Gin
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Hi. I've been mulling over the problem of how to solve [tex]y=e^{ax}+e^{bx}[/tex] for x. I know it can't be solved numerically and have been trying to think of a way to solve it iteratively. The problem is, I can only find algorithms for polynomials and they don't seem to work for exponential problems. I'd rather not have to make an approximation for the exponential and was hoping someone might know an appropriate algorithm or have another means of approach.
 
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Try Newton Raphson root-finding method

x(n+1)=x(n)-f[x(n)]/f'[x(n)]

where f(x)=e^ax+e^bx-y
 
Hey thanks for that. I had tried Newton's method, but silly me messed up some simple numbers. :redface: I tried again and paid more attention and it worked. For some reason I thought that Newton's method wouldn't work for an exponential.
 

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