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I have Non-linear Fitting function like:
prate ~ (m1-((m1-m2)/(1+(IC50/(conc)))))
And a table:
[I] (µM) Max polymerization rate
25.00 3.08
12.50 3.30
6.13 4.44
and IC50 = 1.87
I want to create a function like the one above and use this data to make a plot. Is that possible?
This might help you get started.
You need to define your function using handles.
Say you have a variable PolymerRate you want to estimate and conc as input variable, I think in your case would be something like:
IC50 = 1.87;
prate = #(m,conc) (m(1)-((m(1)-m(2))/(1+(IC50/(conc)))));
m0 = [1 1];
[m,resnorm,~,exitflag,output] = lsqcurvefit(F,m0,conc,PolymerRate);
plot(conc,PolymerRate,'ro')
hold on
plot(conc,prate(m,conc))
Related
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syms k
S1 = symsum(1/((2*k)+1)^2,k,0,1000)
The result is this
59039372973827482202940226722672826425297321906316082356858983822169051832268260251807527611479190413293513875429587706186073872918905490907386679472102966658686481651660967093301512141946288248492833396616338323741632085379508599235923841007033467883638349122388806376761808189104503262045883240287482992169819848342303098664924237976221795758421152603069387903705445513260596627332283139648508194960733619500093010571517561429904500013876585156927070119332440687162376758374919870699278800835146651318465663183182583101377584105366558079836223068786457324044080570317649838092783113721959819118571747662368360095513856052974454509201490370810246175872510881504730747209788019551000695511879992198550955686739483474761130248789609061549535677663474218135370195381615899214931316241080337028498241295985409686314819267606796712968280842464845294917738460317179001491697993067157425958639996885239616893392960282441289069600101430806922004624472226999315951355963789249300352610312601262349650287009275097201871774652260892220551489305368617001974326978428202443548923140478853569492070442010110016068635424791389124439271253578545895132216218268847919848655110002938693346760862649668457282775860633067627110099340660770861888592018701206483696615682617062811616008107086256694453990688805738127607846586853460003073465075155412119309273843527076321601670400373937698518621100907936577387919537592519265365346619712200304996044229704602647674114176291753575322917531444831938509001759491229575945273985556769609288625450013634760596805884195325794441020339210402987018058377081579351119704065092777310976461961832919116412535470810011337916688085616171422473409544885864650134157327448050685723673514545806331081542320899927
It is a number in this form a/b
Why is this happening??
Do you know how to do this in octave too?
k = (0:1000);
k_sum = sum(1./((2*k)+1).^2);
disp(k_sum)
It's interesting that you jumped straight to using syms when the basic matlab functions work perfectly well for this problem. Why is that?
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The Matlab code I have plots the following curve:
X1= 1:600;
plot (X1,tmp)
basline = 0;% level
area(tmp,basline,'FaceColor','g');
how can I calculate the area in the red circle?
You need to find the 2nd and 3rd zero cross (z2 and z3). then do a sum over the tmp. Something like this:
X1= 1:600;
tmp = sin(0.03*X1);
plot (X1,tmp)
range = 209:314;
basline = 0;% level
area(tmp,basline,'FaceColor','g');
figure;area(tmp(range),basline,'FaceColor','g');
sum(tmp(range))
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I have been trying for a while to get this to work but I don't seem to be getting any closer to a solution.
I'm not sure of what the pattern is and what to write for the "d" value in a:d:b
Clearly you want to start from 0 (a) and go to 2*pi (b).
Now the question comes what is your step size (d)?
From your example you can see that you are changing from 0 to pi/n.
And from pi/n to 2*pi/n.
This means your step size is d=pi/n
Once you defined your n, e.g:
n=10;
you can do the rest like this:
x=0:(pi/n):(2*pi)
y=sin(x);
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I want to perform integration on a vector L but I don't know exactly what to use. I want to obtain a (the integral) as a vector that is the same size as NT.
clc;clear;
syms x
NT=input('NT=');
L=zeros(NT,1);
for i=1:NT
disp('Longeur de travée')
L(i)=input('L = ');
L(i)=L(i);
fa(i)=L(i).*x^2;
a(i)=int(fa)
end
An easy way would be to use trapz. If you have X and Y such that Y(i) = f(X(i)) (so Y contains the values of some function at the location X) then you simply do
I = trapz(X, Y)
In your case, you can do
I = trapz(L, fa)
I guess, looking at your code.
Note that you could use more advanced techniques, that will, in principle, give you a better result (because they are higher-order). This is just one method, but an easy one.
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Please help me out with this code:
for i = 1:n
u(t - a_i - td_i);
end
where:
u: step function
t: time vector with n elements
a_i, tau_i, and td_i: variables that change inside for loop
I guess I need to use zeros and ones, but how can I do this correctly?
Simple function to sum the value inside the loop:
func = #(t) sum(t > a + td)
func(t) will be the sum of the foor loop