In the python file

Pmath$04$b$.$py$,2D$ 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.

$--------------------------------$

Pmath$04$b$.$py :

# $-*-$ coding: utf$-8-*-$

$"""$

$2-$dimensional interpolation. Based on the book:Numerical methods for engineers, $7$th$,$ by Chapra

$($page $522-523)$

$"""$

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 f$($X$)$:

x$=$X$[0]$; y$=$X$[1]$;

#return np$.$sin$($x$)*$np$.$sin$($x$)+$ np$.$sin$($x$)*$np$.$cos$($y$)+1$#x$*$x $+$ x$*$y $+2*$y$*$y $+2$

return x$**2+$y$+1$ #np$.$sin$($np$.$sqrt$($x$**2+$ y$**2))$

# xpoint$=[1,2]$

# print$(\text{'}$f$=\text{'},$f$($xpoint$))$

def myinterp$\_$bilin$\_4$ptval$($V$,$p$,$X$)$:#p is the new point; X contains the coord of $[$x$1,$y$1]$ and $[$x$2,$y$2]$

#V contains f values at $[$x$1,$y$1],[$x$1,$y$2],[$x$2,$y$1],[$x$2,$y$2]($the $4$ corners$)$

x$=$p$[0]$; y$=$p$[1]$; # x and y coordinate of the new point

p$1=$X$[0]$; p$2=$X$[1]$;

x$1=$p$1[0]$; y$1=$p$1[1]$;

x$2=$p$2[0]$; y$2=$p$2[1]$;

f$11=$V$[0]$; f$12=$V$[1]$;

f$21=$V$[2]$; f$22=$V$[3]$;

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

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

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

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

f$=$f$+($x$-$x$1)*($y$-$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$(\text{'}$Example from the book by Chapra. Function values at $4$ points are known:$\text{'})$

print$(\text{'}$f$(2,1)=69$; f$(2,6)=55$; f$(9,1)=57\mathrm{.}5$; f$(9,6)=70$;$\text{'})$

newp$=[5\mathrm{.}25,4\mathrm{.}8]$

V$=[60,55,57\mathrm{.}5,70]$

Xcoord$=[[2,1],[9,6]]$

print$(\text{'}$At the point',newp,$\text{'},$ f$=\text{'},$myinterp$\_$bilin$\_4$ptval$($V$,$newp,Xcoord$))$

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$,[1,2]))$

# xmat$=[[1,3],[5,4]]$#the $1$st row is $[$x$1,$y$1].$ The $2$nd row is $[$x$2,$y$2]$

# xnew$=[3,3\mathrm{.}5]$

# print$(\text{'}$ff$=\text{'},$myinterp$\_$bilin$($f$,$xnew, xmat$))$

def mydot$($a$,$b$)$:

na$=$len$($a$)$

nb$=$len$($b$)$

f$=0$

if na$!=$nb:

print$(\text{'}$Error$!$ The two input vectors should have the same length.$\text{'})$

else:

for i in range$(0,$na$)$:

f$=$f$+$a$[$i$]*$b$[$i$]$

return f

def p$($x$,$y$)$: # p$($x$,$y$)=$ sigma$\_$i sigma$\_$j x$**$i y$**$j

n$=4$

c$=$np$.$zeros$([$n$,$n$])$

for i in range$($n$)$:

for j in range$($n$)$:

c$[$i$][$j$]=$ c$[$i$][$j$]+$ y$**$j

c$[$i$]=$c$[$i$]*$x$**$i

return c

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