Find derivative of function given the values

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To approximate the derivative of the error function erf at x=1, the discussion emphasizes using values close to 1, specifically erf(0.1) and erf(1). One participant calculated an average derivative using values from erf(0) and erf(0.01), resulting in an estimate of approximately 1.126. However, others pointed out that this approach is flawed since the values used are not near x=1, suggesting that the estimate is likely too high. The correct method should focus on values closer to x=1 for a more accurate approximation of erf'(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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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