How Do You Apply the Newton-Raphson Method to Find Intersections in Functions?

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Homework Help Overview

The discussion revolves around applying the Newton-Raphson method to find intersections of functions, specifically focusing on the function f(x) = 2x + ln(4x) and its derivative. Participants are exploring the implications of constants in logarithmic differentiation and the process of approximation in the Newton-Raphson method.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the differentiation of logarithmic functions and the relevance of arbitrary constants when finding intersections. There is a focus on the Newton-Raphson method, particularly the calculation of second approximations and the validity of the results obtained.

Discussion Status

Some participants are sharing their calculations and results, noting discrepancies in their approximations. There is an ongoing exploration of the relationship between the first and second approximations, with some guidance provided on the nature of the Newton-Raphson method and its approximations.

Contextual Notes

Participants are questioning the role of constants in logarithmic differentiation and how they affect the intersection calculations. The initial approximation used in the Newton-Raphson method is also under scrutiny, with specific values being tested.

semc
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alright there is this question which uses NR method to find intersect but i kept getting the wrong ans can anybody help me check which part i went wrong?:biggrin:

f(x)=2x + ln4x , f'(x)=2+ 1/x

btw,differentiate ln4x and lnx get the same result but the arbitary constant is different right?so do we consider the constant when finding the intersect?
 
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If you integrate 1/x, you get lnx + C. ln4x = ln4 + lnx, so C = ln4 in this case.
 
hmm...i see now so the constant changes although there's 1/x in both cases.

regarding the question, if i were to find the 2nd approx. , i am suppose to use 0.1 as 1st approx. however, 0.1-f(0.1)/f'(0.1) is not the ans:confused:
 
Why not? It won't be the exact solution to the equation, of course, but it certainly is a "second approximation" with first approximation 0.1. I get 0.16 as the second approximation.

By the way, the reason ln(4x) and ln(x) have the same derivative (1/x) is that ln(4x)= ln(x)+ ln(4).
 
hmm...i also get 0.16 guess the ans is wrong :biggrin: Thanks
 

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