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Hi People... 2 Questions Here ...

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Masoud Ghanbari
Masoud Ghanbari 2013 年 3 月 28 日
Hi Every One
1- Suppose We Have 1.7321 . Which Command returns the non-numerical value? (I mean sqrt(3) )
2-Run This
% Interactive script to fit a cubic to data points
clf;clc;clear all;
grid on
hold on
axis([0 100 0 100]);
diff = 10;
xold = 68;
i = 0;
xp = zeros(1); % data points
yp = zeros(1);
while diff > 2
[a b] = ginput(1);
diff = abs(a - xold);
if diff > 2
i = i + 1;
xp(i) = a;
yp(i) = b;
xold = a;
plot(a, b, 'k*')
end
end
p = polyfit(xp, yp, 3 );
plotfunction=poly2sym((p))
x = 0:0.1:xp(length(xp));
y=p(1)*x.^3 + p(2)*x.^2 + p(3)*x + p(4);
plot(x,y), title( 'cubic polynomial fit'), ...
ylabel('y(x)'), xlabel('x')
hold off
end
The Output has a large amount of number... how can i simplify it???
  2 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2013 年 3 月 28 日
Can you explain question 1?
Masoud Ghanbari
Masoud Ghanbari 2013 年 3 月 29 日
I mean you have the 1.7321 and you want to radical(?)=1.7321

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採用された回答

Walter Roberson
Walter Roberson 2013 年 3 月 28 日
編集済み: Walter Roberson 2013 年 3 月 28 日
For #1: if you have access to the full version of Maple, such as through the old Extended Symbolic Mathematics Toolbox, then you can use the Maple command "identify". If you do have that access (it is not the current Symbolic Toolbox, which is MuPAD rather than Maple), then try
feval(symengine, 'identify', 1.7321)
or
maple(sprintf('identify(%g)', 1.7321))
MuPAD (the current Symbolic Toolbox) does not appear to have a similar function.
You could try using
simplify(plotfunction)
to get a more compact expression. The result might still have a lot of rational numbers in it. If you want the rational numbers converted to fixed point numbers, then use
vpa(plotfunction)
  5 件のコメント
Jan
Jan 2013 年 4 月 8 日
if strcmpi(userans, 'y')
Masoud Ghanbari
Masoud Ghanbari 2013 年 4 月 11 日
Thanks Dear Friends...

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