Error in find command....

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shoaib Ch
shoaib Ch 2019-3-14
编辑: shoaib Ch 2019-4-2
t=linspace(xq(1),xq(np),1001)
y2=polyval(pmon,t)
plot(t,y2,'k--',x)
for j=1:6
for i=1:1001;
t1(j)=find(y2(i)=pmon(j))
end
end
i want to find value of t(x axis) for which value of pmon(y axis) is given ??
  2 个评论
shoaib Ch
shoaib Ch 2019-3-14
it is not giving me value for t , i want to use it for finding roots
Adam Danz
Adam Danz 2019-3-14
Have you implemented the correction suggested in the answers section below?
If you've implemented that correction and 't' is empty, that means there were no matches between y2(i) and pmon(i).

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回答(1 个)

Adam Danz
Adam Danz 2019-3-14
One equal sign assigns a value to a variable.
Two equal signs perfoms a comparison between A and B and returns a logical value.
t1(j) = find(y2(i) = pmon(j)) %incorrect
t1(j) = find(y2(i) == pmon(j)) %correct
  12 个评论
shoaib Ch
shoaib Ch 2019-3-15
xq=[ -3, -2, -1, 2, 6, 7];
yq=[ 1, -2, 1, -2, 2, -2];
pmon=[-61/840, 49/60, -1663/840, -1693/420, 169/15, 73/7]; %interpolation values
t=linspace(xq(1),xq(6),1001) %values of xq
y2=polyval(pmon,t)
plot(t,y2,'k--')
Now i want to chect weather there exist a root between pmon values ,next program give me root values
for i=1:5;
if pmon(i)<0 & pmon(i+1)>0
pos_a(i)=pmon(i)
elseif pmon(i)>0 & pmon(i+1)<0
pos_a(i)=pmon(i)
else
end
now , pos_a is values of y axis and i want to find what be the values of t (x values) which give me these y values according to data ... for this i use that code but its not working
for i=1:5;
if pos_a(i)>= pos_a(i+1)
s0=t(pos_a(i))% ..statements to get the root in t(pos a(i)) to t(pos a(i+1))
s1=t(pos_a(i+1))
else
s1=t(pos_a(i)) % ..statements to get the root in t(pos a(i)) to t(pos a(i+1))
s0=t(pos_a(i+1))
end
end
And and Thanx again for your Kindness
shoaib Ch
shoaib Ch 2019-3-15
Figure 1.jpg
if these be pos_a values than what be t values ??

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