Example (Strang, p.179)

  m={{2,1,0,0,0},{1,4,1,0,0},{0,1,4,1,0},{0,0,1,4,1},{0,0,0,1,2}};
  MatrixForm[m]
  X={0,1,2,3,4};
  u={0,1,4,1,0};
  du={1,4,0,-4,-1}

  LinearSolve[m,3du]

Since leading up to (0, 0) the second derivative is 0, we will have a line, and we are leading up to (0, 0) with s0=0 (from above), we have a horizontal line at 0 up to (0, 0).

  spline0[x_]:=0

On the interval from x=0 to x=1 we have hermite[x] with a=u0=0, b=s0=0, c=u1=1, d=s1=3.

  Clear[x]
  spline1[x_]:=hermite[x,0,0,1,3];
  spline1[x]
  Simplify[%]

On the interval from x=1 to x=2 we have hermite[x-1] with a=u1=1, b=s1=3, c=u2=4, d=s2=0.

  spline2[x_]:=hermite[x-1,1,3,4,0];
  spline2[x]
  Simplify[%]

On the interval from x=2 to x=3 we have hermite[x-2] with a=u2=4, b=s2=0, c=u3=1, d=s3=-3.

  spline3[x_]:=hermite[x-2,4,0,1,-3];
  spline3[x]
  Simplify[%]

On the interval from x=3 to x=4 we have hermite[x-3] with a=u3=1, b=s3=-3, c=u4=0, d=s4=0.

  spline4[x_]:=hermite[x-3,1,-3,0,0];
  spline4[x]
  Simplify[%]

On the interval after x=4 we have a line with slope s4=0 so a horizontal line at u4=0.

  spline5[x_]:=0

  spline[x_]:=If[x<=0,spline0[x],
  				If[x<=1,spline1[x],
  					If[x<=2,spline2[x],
  						If[x<=3,spline3[x],
  							If[x<=4,spline4[x],spline5[x]]
  						]
  					]
  				]
  			]

  plot1=ListPlot[Table[{X[[i]],u[[i]]},{i,1,5}],PlotStyle->{RGBColor[1,0,0],PointSize[.02]}];
  plot2=Plot[spline[x],{x,-1,5}];
  Show[plot1,plot2];

Up to Best Bet: Cubic Spline Interpolation