We now have all the arguments for splice4(). To find x given s(x), use
polyeval(spl4y,ky*(s(x)-yy2))
where ky = 1/hy.
This inverse polynomial is not the true inverse of the original splice. It is simply a polynomial which
passes through the same three points as the splice, and has the same derivatives at the endpoints. If
the splice is used for a narrow interval, the accuracy can be very good. Before using the inverse splice,
you should check the accuracy. This is easily accomplished by graphing the error in the estimated
x-values, found from
error = si(s(x)) - x
where s(x) is the splice function and si(x) is the inverse splice function. For the example shown above,
this is the error in x:
The error in x is about ±2.4E-10, so this inverse is accurate to about 9 significant digits. Note that the
error is zero at the endpoints and in the center, as it should be.
The function spli4inv(), shown below, finds the inverse splice by applying the steps described above.
spli4inv(f1,f2,x2,h,c)
Func
© ("f1(x)","f2(x)",x2,h,coeff_list)
© output: {a,b,c,d,e,y2,hy}
© 4th-order splice inverse polynomial
© 10may02/dburkett@infinet.com
© calls math\spli4x(), math\splice4()
© Input arguments:
©
© f1 f1(x), passed as a string, for example "y1(x)"
© f2 f2(x), passed as a string, for example "y2(x)"
© x2 splice midpoint for s(x)
© h splice half-width for s(x)
© c list of splice coefficients for s(x)
© Output: {a,b,c,d,e,y2,hy}
© where si(y) = a*ys^4 + b*ys^3 + c*ys^2 + d*ys + c
© and ys =(y-y2)/hy
© so find x = polyeval({a,b,c,d,e},(y-y2)/hy)
local yy1,yy2,yy3,yd,x1,x2,x3,xp1,xp3,hy
©
© x1 the original left interval bound; treated as si(y1)
© x2 s(x) such that x2=si(yy2)
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