Student Support Forum > General > > "Plotting non-linear values and solving"

 Post Reply: Name: Email Address: Please send email when my message is replied to. Url (optional): Subject: Message: view original message? Attachment (optional): Please answer this: 3+1 =

 Original Message (ID '85842') By Bill Simpson: In Response To 'Re: Re: Re: Re: Re: Re: Re: Re: Plotting non-li...' --------- Use your function and Sigma Plot's coefficients. In[8]:= fy2[x_]=(b-(Sqrt[b^2-2k^2 c x/s]))/(2k); k = 2900000; s = .84; In[9]:= fy2[.79333] Out[9]= 0.47222 In[10]:= fy2[10.313] Out[34]= 1.32 Those seem to match your calculations. Now a plot to see how good the fit is. In[11]:= Show[ListPlot[data,Joined->True], Plot[fy2[x],{x,0,10.313}], PlotRange->All] Out[11]= That fits the leading edge well, but nothing else. Try FindFit with Sigma Plot's starting values. In[12]:= k=.;s=.; fit=FindFit[data,y2,{{k,2900000},{s,.84}},x] Out[12]= {k-> -0.21075,s-> -0.184977} Those are very odd results, how good is the fit? In[13]:= k = -0.21074963872741664; s = -0.18497695239903436; fy2[x_] = b-(Sqrt[b^2-2 k^2 c x/s]))/(2k); Show[ListPlot[data, Joined->True], Plot[fy2[x], {x, 0, 10.313}], PlotRange->All] Out[16]= From this I think you might see why the first thing I always do is plot my fitted function over the data. At this point I don't know what else to say. If all we have to go on is your student thought the value was supposed to be about a million then I am at a loss. If you could somehow scale your y2 down by 25% then it might not be a bad approximation, but I don't have any basis for doing that. If we can somehow track down and point out a likely calculation error anywhere then I'd be happy to try to pinpoint where that is happening.