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shoaib Ch
shoaib Ch on 14 Mar 2019
Edited: shoaib Ch on 2 Apr 2019
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 Comments
shoaib Ch
shoaib Ch on 14 Mar 2019
it is not giving me value for t , i want to use it for finding roots
Adam Danz
Adam Danz on 14 Mar 2019
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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Answers (1)

Adam Danz
Adam Danz on 14 Mar 2019
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 Comments
shoaib Ch
shoaib Ch on 15 Mar 2019
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 on 15 Mar 2019
Figure 1.jpg
if these be pos_a values than what be t values ??

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