display function does not show the exact value
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Mustafa Süve
le 27 Fév 2016
Commenté : Álvaro Sacristán Gonzalez
le 26 Sep 2021
I am new to MATLAB. I wrote code like this;
syms n
f(n) = 40*n.^1.5-875*n+35000;
fdiff(n) = diff(f(n));
x0 = 100;
xprev = 0;
while (abs(x0 - xprev)/x0)*100 >= 1
xprev = x0;
x0 = x0 - f(x0)/fdiff(x0);
end
display(x0);
When I tried to display the value of x0, it shows;
600/11 - (40*(600/11 - ((24000*11^(1/2)*600^(1/2))/121 - 140000/11)/((60*11^(1/2)*600^(1/2))/11 - 875))^(3/2) + (875*((24000*11^(1/2)*600^(1/2))/121 - 140000/11))/((60*11^(1/2)*600^(1/2))/11 - 875) - 140000/11)/(60*(600/11 - ((24000*11^(1/2)*600^(1/2))/121 - 140000/11)/((60*11^(1/2)*600^(1/2))/11 - 875))^(1/2) - 875) - ((24000*11^(1/2)*600^(1/2))/121 - 140000/11)/((60*11^(1/2)*600^(1/2))/11 - 875)
Instead of just;
62.6913
How can I make it to display only the final value?
2 commentaires
Andreas Sorgatz
le 21 Mar 2016
The answer is a representation of the "exact" value. However, you can get an approximation of this exact result using vpa(...) if you need high precision or using double(...) if hardware precision is fine.
Álvaro Sacristán Gonzalez
le 26 Sep 2021
I fixed it with disp(vpa(x)), which gives you 32 digits of precision. Newton-Raphson function analysis here 2 :D
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John BG
le 27 Fév 2016
Mustafa
If you are really after a numeric result, why don't start with numeric values?
x_range=10
x_step=.1
x=[0:x_step:x_range]
f=@(x) 40*x.^1.5-875*x+35000
y=f(x)
ydiff = diff(y)
x0 = 100
xprev = 0
while (abs(x0 - xprev)/x0)*100 >= 1
xprev = x0
x0 = x0 - f(x0)/ydiff(x0)
end
x0 =
-82.29
Play with x_range and x_step to center the curve wherever you need.
If you find this answer of any help solving this question, please click on the thumbs-up vote link,
thanks in advance
John
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