How can I treat a symbolic function in this script related to lagrangian interpolation

2 次查看（过去 30 天）
Ricardo Boza Villar
评论： Bobby Fischer 2022-4-10
I have described what the script is intended to do with commentary (time spent to solve the problem 15~20 minutes max, I promise ;)).
The problem that I encounter is the following:
It says 'line 39'; I don't know how I can create the symbolic function in order to get the program running.
I wouldn't say it's really difficult and complicated. I think the difficulty lies in trying to understand what I have described. I've done it as thoroughly as I could, but you can ask me if you want some more information.
https://en.wikipedia.org/wiki/Lagrange_polynomial
% This function is intended to calculate the Lagrange polynomial that
% interpolates a one-variable function given by the user as input.
function []=a75lagrange(f,nodes)
% f : f is the function that the user wants to interpolate.
% nodes: nodes is a vector containing the nodes of interpolation, that is,
% the 'xvalues' of which the user has information because supposedly
% the user hasn't had access to any more of them. This may be caused by
% the fact that evaluating the function is very difficult or simply
% because the user doesn't have its expression.
% For example, I would write on the workspace:
% 'a75lagrange(@cos,[0,1/2,2,3])'
format compact, format shortg
% This format is practical, though it doesn't serve a particular purpose
% because the function doesn't have output.
syms x
% 'syms x' creates the symbolic variable 'x'
polynomial=0;
% I start the polynomial at zero because I want to define the variable
% before it enters de loop.
lon=length(nodes);
% How many nodes there are as input.
% 'lon' because in Spanish 'length' is 'longitud'.
for m=1:lon
% For example, if we have 4 nodes, we would have 4 terms L_1,L_2,L_3
% and L_4, and all of them will depend on x, the symbolic variable.
L(m)=1;
% I define each of them before they enter their loops.
yvalue(m)=f(nodes(m));
% This is the y value that the function takes when we evaluate it on
% its respective nodes.
for k=1:lon
% This loop is to develop the shape that each L_1,L_2,L_3 and L_4
% is going to take.
if m~=k
% m~=k is a condition given by the algorithm.
% If it wasn't this way one denominator below would cancel
% itself.
L(m)=L(m)*(x-nodes(k))/(nodes(m)-nodes(k));
end
% When exiting the loop we have 1 of the 4 expressions that
% we wanted at the beginning, corresponding to 1 Lagrangian term of
% the final interpolating polynomial.
polynomial=polynomial+yvalue(m)*L(m);
% After each loop, the polynomial takes with it another expression
% that contributes to its final shape.
end
end
s=(nodes(1)-5:0.01:nodes(end)+5);
a=axis;
% s is a vector of x values. My intention is to evaluate each of the
% Lagrangian terms separately, and plot them using subplot.
% Then, opening another figure, I would get the second graph, in which the
% real function and the whole interpolating polynomial would appear.
figure(1),close(1),figure(1)
hold on
for n=1:lon
q=sym2poly(L(m));
% As L(m) was a symbolic mathematical expression depending on x, using
% this I would get a vector of coefficients that MATLAB would give me.
y=polyval(q,s);
% I would use the command 'polyval' to evaluate that polynomial given
% by the vector of coefficients.
subplot(2,2,n)
plot(a(1:2),[0,0],':'), plot([0,0],a(3:4),':')
% With these I get the axes for the plots.
plot(nodes,zeros(size(nodes)),'o')
% With this I get the nodes.
plot(nodes(n),yvalue(n),'*')
% With this I get the y value of the n_th node.
plot(s,y)
% With this I get the graph of each of the Lagrangian terms.
end
hold off
figure(2),close(2),figure(2)
hold on
r=sym2poly(polynomial);
% As 'polynomial' was a symbolic expression, using sym2poly I get the
% coefficients of the polynomial.
yr=polyval(r,s);
% I evaluate that polynomial using the vector 's' defined earlier.
z=f(s);
% I get the y values for the real function.
plot(a(1:2),[0,0],':'), plot([0,0],a(3:4),':')
% With these I get the axes.
plot(nodes,zeros(size(nodes)),'o')
% With this I get the nodes.
plot(nodes,yvalue,'*')
% With this I get the y values of the nodes.
plot(s,z)
% With this I get the graph of the function given as input.
plot(s,yr,'r--')
% With this I get the graph of the interpolating polymial.
hold off
xlabel('Eje x')
ylabel('Eje y')

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采纳的回答

Walter Roberson
Walter Roberson 2016-5-17
L(m) = sym(1);
Remember the datatype is determined by the assignment, so by assigning plain 1 you were making the data type double rather than symbolic.

更多回答（2 个）

Ricardo Boza Villar
Thank you
Ricardo Boza Villar
编辑：Ricardo Boza Villar 2016-5-20
I was proud of the result, so I had to post it here. After some minor changes, this is what I got:
Thanks again.
% This function is intended to calculate the Lagrange polynomial that
% interpolates a one-variable function given by the user as input.
function [L,polynomial]=a75lagrange(f,nodes)
% f : f is the function that the user wants to interpolate.
% nodes: nodes is a vector containing the nodes of interpolation, that is,
% the 'xvalues' of which the user has information because supposedly
% the user hasn't had access to any more of them. This may be caused by
% the fact that evaluating the function is very difficult or simply
% because the user doesn't have its expression.
% For example, I would write on the workspace:
% 'a75lagrange(@cos,[0,1/2,2,3])'
format compact, format shortg
% This format is practical, though it doesn't serve a particular purpose
% because the function doesn't have output.
syms x
% 'syms x' creates the symbolic variable 'x'
polynomial=sym(0);
% I start the polynomial at zero because I want to define the variable
% before it enters de loop.
lon=length(nodes);
% How many nodes there are as input.
% 'lon' because in Spanish 'length' is 'longitud'.
yvalue=f(nodes);
for m=1:lon
% For example, if we have 4 nodes, we would have 4 terms L_1,L_2,L_3
% and L_4, and all of them will depend on x, the symbolic variable.
L(m)=sym(1);
% I define each of them before they enter their loops.
% This is the y value that the function takes when we evaluate it on
% its respective nodes.
for k=1:lon
% This loop is to develop the shape that each L_1,L_2,L_3 and L_4
% is going to take.
if m~=k
% m~=k is a condition given by the algorithm.
% If it wasn't this way one denominator below would cancel
% itself.
L(m)=L(m)*(x-nodes(k))/(nodes(m)-nodes(k));
end
end
% When exiting the loop we have 1 of the 4 expressions that
% we wanted at the beginning, corresponding to 1 Lagrangian term of
% the final interpolating polynomial.
polynomial=polynomial+yvalue(m)*L(m);
% After each loop, the polynomial takes with it another expression
% that contributes to its final shape.
end
s=nodes(1)-1:0.01:nodes(end)+1;
% s is a vector of x values. My intention is to evaluate each of the
% Lagrangian terms separately, and plot them using subplot.
% Then, opening another figure, I would get the second graph, in which the
% real function and the whole interpolating polynomial would appear.
figure(1),close(1),figure(1)
whitebg('k')
for n=1:lon
q=sym2poly(L(n));
% As L(m) was a symbolic mathematical expression depending on x, using
% this I would get a vector of coefficients that MATLAB would give me.
y=polyval(q,s);
% I would use the command 'polyval' to evaluate that polynomial given
% by the vector of coefficients.
subplot(2,2,n)
grid
hold on
plot([-1,4],[0,0],':','color',ones(3,1)), plot([0,0],[-2,2],...
':','color',ones(3,1))
% With these I get the axes for the plots.
plot(nodes,zeros(size(nodes)),'bo')
% With this I get the nodes.
plot(nodes(n),1,'b*')
% With this I get the y value of the n_th node.
% If the plot of this node is within the line of the graph belonging to
% the function, then the process has been properly done.
plot(s,y,'r')
% With this I get the graph of each of the Lagrangian terms.
axis([-1,4,-2,2])
hold off
end
figure(2),close(2),figure(2)
whitebg('k')
r=sym2poly(polynomial);
% As 'polynomial' was a symbolic expression, using sym2poly I get the
% coefficients of the polynomial.
yr=polyval(r,s);
% I evaluate that polynomial using the vector 's' defined earlier.
z=f(s);
% I get the y values for the real function.
grid
hold on
plot([-1,4],[0,0],':','color',ones(3,1)), plot([0,0],[-2,2],...
':','color',ones(3,1))
% With these I get the axes.
plot(nodes,zeros(size(nodes)),'bo')
% With this I get the nodes.
plot(nodes,yvalue,'b*')
% With this I get the y values of the nodes.
plot(s,z,'w')
% With this I get the graph of the function given as input.
plot(s,yr,'r--')
% With this I get the graph of the interpolating polymial.
xlabel('Eje x')
ylabel('Eje y')
hold off
%
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显示 1更早的评论
Bobby Fischer
Bobby Fischer 2022-4-10
Thanks guys. (I'm Ricardo (Boza, not Milos)). This reminds me of a sad moment in my life. But anyway... The important thing is to move on, stronger than before. Still alone, still alive, still unbroken.

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