Calculating Error Function for f(x)=tan(x) with Mathematica/Fortran/C

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The discussion revolves around calculating the error function for f(x)=tan(x) over the interval [-1.25, 1.25] using linear, quadratic, and cubic shape functions. Participants clarify that the task involves interpolation with varying degrees of polynomial approximations based on the number of elements (10, 100, and 1000). The process includes evaluating tan(x) at specified points, fitting polynomial curves, and calculating the error between these approximations and the actual function. There is some confusion regarding the terms used, particularly "nodes" and the degree of polynomials, but the consensus is that more sample points lead to potentially better approximations. The conversation emphasizes the importance of understanding the relationship between the number of points and the accuracy of the interpolation methods.
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Hello!
don t understant the question.
Calculate the error function for f(x)=tan(x), [-1.25,1.25] for linear, quadratic and cubic shape functions and carry out the calulation using Mathematica or Fortran, C for 10,100 and 1000 elements.
WHat does mean for linear, quadratic and cubic shape functions
It refers to interpolation or means f(x)=tan(x) in 1, 2, u 3 degree
As i understand first i have to replace the function with polinomial in n degree (10, 100 or 1000 by problem specification) and then some how calculate the aproximation
please, help
 
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I don't quite get it either, but how about this:

For n=10:
Divide the range [-1.25, 1.25] by n and evaluate tan(x) for those 10 values.
Fit a straight line through the 10 tan(x) values.
Calculate the error between the line and the actual tan(x).
Fit quadratic line through the 10 tan(x) values.
Calculate the error between the quadratic line and the actual tan(x).
Fit cubic line through the 10 tan(x) values.
Calculate the error between the cubic line and the actual tan(x).

Repeat for n=100 and n=1000

my 2 cents
 
Is´nt degree of interpolation is amount of nodes (10, 100 and 1000)?
 
I am not sure what you mean by nodes. If you mean degrees of the polynomial...I seriously doubt it...I mean, who in the world would like a 1000 degree polynomial?

I think what they mean is that you get more and more (10, 100, 1000) samples, out of the actual function, which you can then curve-fit a line through...the more points you have, presumably, the more accurate your curve-fit...

...then again, for example, there is only so much accuracy a straight line is going to provide no matter how many more point you have...whereas a cubic could certainly benefit from more points...

what do you think? does it make sense to you?
 
i'm confused now, need to figure things out-) thanks for helping
 
GoYu said:
Hello!
don t understant the question.
Calculate the error function for f(x)=tan(x), [-1.25,1.25] for linear, quadratic and cubic shape functions and carry out the calulation using Mathematica or Fortran, C for 10,100 and 1000 elements.
WHat does mean for linear, quadratic and cubic shape functions
It refers to interpolation or means f(x)=tan(x) in 1, 2, u 3 degree
As i understand first i have to replace the function with polinomial in n degree (10, 100 or 1000 by problem specification) and then some how calculate the aproximation
please, help

Tell you what, I'd interpret it as is and then just run with it and hope for the best:

So error function for tan(x)?

That:

f(x)=\text{erf}(\tan(x))=\int_{0}^{\tan(x)} e^{-z^2}dz

Shape function? How about just linear, quadratic, and cubic approximations to that function so for the linear approximation, let:

e^{-x^2}=a+bx

Split the domain[-1.25, 1.25] into 10 parts so:

x1=-1.25
x2=-1

Then compute a and b for:

e^{-x1^2}=a+b(x1)
e^{-x2^2}=a+b(x2)

Now, compute the integral:

\int_{\tan(x1)}^{\tan(x2)} (a+bx)dx

do that over the entire interval and sum them then compare to the actual value. Then do the quadratic and cubic over the interval.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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