In the python file

Pmath$04$b$.$py$,2$D bicubic interpolation is done with the assumption that both x and y are $=$

$0,1,2,3.$ The code does not work if the unknown point is outside the $0,3$ domain. Please write a bicubic

interpolation function $($without using derivatives$)$ similar to the function myinterp$\_$bilin$($ x$,$y$,$z$,$p in this file.

That is$,$ the user can specify the x and y vectors with the corresponding z$=$f$($x$,$y$).$ And the function value of the

new point p can be found if p is within the specified x$,$y domain.

$*$Importance$*$ I need the complete code!!! And please check the code is correct : when print$($myinterp$\_$bilin$($x$,$y$,$z$,[1,2])),$ it should output $"-0\mathrm{.}93532843"$

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import numpy as np

from scipy import interpolate as interp #$.$interpolate import interp$2$d

x $=$ np$.$arange$(-5\mathrm{.}01,5\mathrm{.}01,0\mathrm{.}25)$

y $=$ np$.$arange$(-5\mathrm{.}01,5\mathrm{.}01,0\mathrm{.}25)$

xx$,$ yy $=$ np$.$meshgrid$($x$,$ y$)$

z $=$ np$.$sin$($xx$**2+$yy$**2)$ #based on length of x$,$ y$,$ construct matrix z $($size: len$($x$)$ by len$($y$))$

# finterp $=$ interp.interp$2$d$($x$,$ y$,$ z$,$ kind$=$'cubic'$)$

finterpL $=$ interp.interp$2$d$($x$,$ y$,$ z$,$ kind$=$'linear'$)$

print$(\text{'}$Use scipy interpolate interp$2$d function, linearly interpolated value $=\text{'},$finterpL$(1,2))$

def myinterp$\_$bilin$($x$,$y$,$z$,$p$)$:#p is the new point; x and y are grid points; z$=$f$($x$,$y$)$

xn$=$p$[0]$; yn$=$p$[1]$

if xnx$[-1]$ or yny$[-1]$:

print$(\text{'}$Out of range!$\text{'})$

return $0$

else:

for i in range$(1,$len$($x$))$:#find the x position of the new point

if x$[$i$]>$xn:

indx$=$i

break

for i in range$(1,$len$($y$))$:#find the y position of the new point

if y$[$i$]>$yn:

indy$=$i

break

x$1=$x$[$indx$-1]$; x$2=$x$[$indx$]$;

y$1=$y$[$indy$-1]$; y$2=$y$[$indy$]$;

f$11=$z$[$indx$-1][$indy$-1]$;

f$12=$z$[$indx$-1][$indy$]$;

f$21=$z$[$indx$][$indy$-1]$;

f$22=$z$[$indx$][$indy$]$;

den$=($x$1-$x$2)*($y$1-$y$2)$;

f$=($xn$-$x$2)*($yn$-$y$2)*$f$11/$den #$(($x$1-$x$2)*($y$1-$y$2))$

f$=$f$+($xn$-$x$2)*($yn$-$y$1)*$f$12/(-$den$)$#$(($x$1-$x$2)*($y$2-$y$1))$

f$=$f$+($xn$-$x$1)*($yn$-$y$2)*$f$21/(-$den$)$#$(($x$2-$x$1)*($y$1-$y$2))$

f$=$f$+($xn$-$x$1)*($yn$-$y$1)*$f$22/($den$)$#$(($x$2-$x$1)*($y$2-$y$1))$

#print$(\text{'}$x$1$y$1$x$2$y$2\text{'},$x$1,$y$1,$x$2,$y$2)$

return f

print$(\text{'}$Use myinterp$\_$bilin:$\text{'},$myinterp$\_$bilin$($x$,$y$,$z$,[1,2]))$