Lecture on data values - or whatever algorithm you take, there is probably data that says you are handling it incorrectly.


Let's open Matlab

Type:
x=[0:6]*pi/3;
 plot(x,sin(x))
y=sin(x);

hold on;

xmany=[0:200]*pi/3;
ymany=sin(xmany);

plot(xmany,ymany,'r');

See the difference:
Could we have done better than "connect the dots':

With 3 points one can fit a quadratic. So let's take first 3 points and see what happens at 1.5. We create the matrix- A(The Vander Monde matrix)
whose columns have x^2  x and 1 with the first 3 points. Let y3= the first 3 points in y.
We want to solve Ac =y. and then determine
y(1.5) =x(1)^2*c(1)+x(1)*c(2) +c(3)
 in Matlab:

A = Vander([ x(1) x(2) x(3)])

c=A\[y(1) y(2) y(3)]'

y15=c(1)*1.5^2+c(2)*1.5+c(3)

plot (1.5, y15,'ro')

seems to be getting closer.

look at sin(1.5)

How about all 7 points:

A7=Vander(x);

Then set
c7=A\y'

calculate 

y715=c7(1)*1.5^6+ c7(2)*1.5^5 +c7(3)*1.5^4+ c7(4)*1.5^3 + c7(5)*1.5^2 + c7(6)*1.5 + c7(7)

which gets
y715 =

    1.0024

>> sin(1.5)

ans =

    0.9975

>> sin(1.5)-y715

ans =

   -0.0049
Does taking more points always work?-
If it did I would not be giving this lecture.

Go to http://www.mste.uiuc.edu/exner/ncsa/runge/

and see it depends where you sampled function

Thus polynomial interpolation is not even bitter than a linear fit.

Another possibility- use splines

Splines are

(1)  locally cubic

(2) Given any 3 points x1, x2, x3 and data values y1, y2, y3.
 let p1 = cubic polynomial  on (x1,x2] and p2 = cubic polynomial on [x2, x3)
then p1(x2)=p2(x2)=y2

(3)p1'(x2)=p2'(x2)     continuous derivative

p1''(x2)=p2''(x2)   continuous second derivative

In matlab

xx= 0:.25:6.28

yy=spline(x,y,xx)


plot(xx,yy,'r')

plot(xx,sin(xx),'g*')
>> z=yy-sin(xx);
>> max(z)

ans =

    0.0284

Look at www.mathworks.com/moler/interp.pdf p.14

On to taking derivatives:

We know the derivative of sin(x) = cos(x)

cos(x(2))

ans =

    0.5000

(sin(x(3))-sin(x(2)))/(x(3)-x(2))

ans =

  1.0602e-016
(sin(x(1))-sin(x(2)))/(x(1)-x(2))

ans =

    0.8270

df=(y(3)-y(1))/(2*(x(3)-x(1)))

df =

    0.2067
All wrong- ok big steps. Is it better with spline data

xx(5)

ans =

     1

>> (yy(5)-y(2))/(xx(5)-x(2))

ans =

    0.4665
So a bit better.

 Does differences always work with sin(x)


Consider
h=.5;
cos1=cos(1);
while h >1.e-15
   approx=(sin(1.+h)-sin(1.))/h;
   error=approx-cos1;
   disp([log(h) log(abs(error))])
   h=h*.1;
end

which gives
derivsin
   -0.6931   -1.4773

   -2.9957   -3.8510

   -5.2983   -6.1630

   -7.6009   -8.4665

   -9.9035  -10.7692

  -12.2061  -13.0718

  -14.5087  -15.3744

  -16.8112  -17.6915

  -19.1138  -18.0791

  -21.4164  -15.6261

  -23.7190  -13.7651

  -26.0216  -13.6596

  -28.3242   -8.7769

  -30.6268   -7.2171

  -32.9293   -4.2125

differentiation is an art not a science

Integration is easier:
divide intervals into trapezoidal regions and add the trapezoids:
int1=x(2)*(y(1)+2*y(2)+2*y(3)+y(4))/2

int1 =

    1.8138
true solution is 2.
working with spline data:

let's go until xx(13) = 3.
Integral from 0 to 13 is 1+cos(xx(13))
= 1.99
Doing trapezoids gives us
sum=yy(1)+yy(13)
for i=2:12
        sum=sum+2*yy(i);
end
integral1=sum*xx(2)/2

yielding
sum=yy(1)+yy(13)
for i=2:12
        sum=sum+2*yy(i);
end
   1.9909