• Support PF! Buy your school textbooks, materials and every day products Here!

Find derivative of function given the values

  • Thread starter Painguy
  • Start date
  • #1
120
0

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?
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,508
730

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?
 
  • #3
33,106
4,801

Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
What is the HxL business?
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.
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.
 
  • #4
120
0
Sorry i meant to say at erf'(0).that was a typo on my part
 
Top