Find derivative of function given the values

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Homework Help Overview

The discussion revolves around approximating the derivative of the error function, erf(x), specifically at the point x = 1, using given values of the function at various points. Participants are tasked with explaining their reasoning and calculations based on these values.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • One participant attempts to calculate the derivative by averaging the slopes between points near erf(0) and erf(0.1), while another questions the appropriateness of using values far from x = 1 for this approximation. There is also a mention of a potential typo regarding the intended point of evaluation.

Discussion Status

The discussion is ongoing, with participants providing calculations and questioning the validity of using certain values. Some guidance has been offered regarding the selection of points for approximation, but no consensus has been reached on the correctness of the estimates provided.

Contextual Notes

Participants are working under the constraint of using specific function values provided in the original problem statement. There is a noted concern about the relevance of the chosen points for estimating the derivative at x = 1.

Painguy
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Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
erf (0)= 0
erf (1) = 0.84270079
erf(.1)= 0.11246292
erf(.01) = 0.01128342
Use this information to approximate the value of erf '(1). Explain how you arrived at your answer.

Homework Equations





The Attempt at a Solution


I basically took the average from both sides of erf(.01)
(((erf(0)-erf(.01))/.01)+((erf(.01)-erf(.1))/(-.09)))/2 = (1.128342 + 1.124216667)/2 =1.126279335

I ended up with 1.126279335 so I'm assuming its something close to that like 1.12? Am I right or wrong?
 
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Painguy said:

Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
erf (0)= 0
erf (1) = 0.84270079
erf(.1)= 0.11246292
erf(.01) = 0.01128342
Use this information to approximate the value of erf '(1). Explain how you arrived at your answer.

Homework Equations





The Attempt at a Solution


I basically took the average from both sides of erf(.01)
(((erf(0)-erf(.01))/.01)+((erf(.01)-erf(.1))/(-.09)))/2 = (1.128342 + 1.124216667)/2 =1.126279335

I ended up with 1.126279335 so I'm assuming its something close to that like 1.12? Am I right or wrong?

Why would you use values near 0 when you are interested in erf'(1)? Did you type the last two data points correctly?
 
Painguy said:

Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
What is the HxL business?
Painguy said:
erf (0)= 0
erf (1) = 0.84270079
erf(.1)= 0.11246292
erf(.01) = 0.01128342
Use this information to approximate the value of erf '(1). Explain how you arrived at your answer.

Homework Equations





The Attempt at a Solution


I basically took the average from both sides of erf(.01)
(((erf(0)-erf(.01))/.01)+((erf(.01)-erf(.1))/(-.09)))/2 = (1.128342 + 1.124216667)/2 =1.126279335
I can't see that this will do you much good. To get an estimate of the derivative at x = 1, you need values that are reasonably close to x = 1. The two that are closest are x = .1 and x = 1.
Painguy said:
I ended up with 1.126279335 so I'm assuming its something close to that like 1.12? Am I right or wrong?
Your estimate is too large by at least a factor of 2.
 
Sorry i meant to say at erf'(0).that was a typo on my part
 

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