interpolation
in [Matlab]
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From: Matt J on 17 Jun 2010 13:06 Jason Bonior <jdbonior21(a)gmail.com> wrote in message <1d66dd84-f442-4a64-90f5-59e6ea5c6239(a)r27g2000yqb.googlegroups.com>... > I apologize but it seems I was not clear. Forget my simple example, my > actual problem is: > > P = gammainc(N/2, x) / gamma(N/2) > > N is an integer and x is my variable. I am looking for a way to find > the corresponding value of x given a desired value of P. For example > what value of x will give me P=0.9? ================= You weren't unclear. The same technique applies: y = @(x) gammainc(N/2, x) / gamma(N/2) Ptarget=0.9; x0=0; fzero(y-Ptarget,x0); The only issue is that, with this more complicated function, it's not entirely clear with what initial guess x0 to seed fzero(). I used x0=0 here only as an example. You could plot y(x) to get a reasonable initial guess.
From: John D'Errico on 17 Jun 2010 13:53 Jason Bonior <jdbonior21(a)gmail.com> wrote in message <1d66dd84-f442-4a64-90f5-59e6ea5c6239(a)r27g2000yqb.googlegroups.com>... > On Jun 17, 12:39 pm, "Matt Fig" <spama...(a)yahoo.com> wrote: > > y = @(t) 4*t.^2 > > fzero(@(t) y(t)-36,5) > > I apologize but it seems I was not clear. Forget my simple example, my > actual problem is: > > P = gammainc(N/2, x) / gamma(N/2) > > N is an integer and x is my variable. I am looking for a way to find > the corresponding value of x given a desired value of P. For example > what value of x will give me P=0.9? One thing I would point out, that this is NOT an interpolation problem. This is a root finding problem. This is why fzero is the appropriate solution. John
From: Jason Bonior on 17 Jun 2010 14:12 On Jun 17, 1:06 pm, "Matt J " <mattjacREM...(a)THISieee.spam> wrote: > Jason Bonior <jdbonio...(a)gmail.com> wrote in message <1d66dd84-f442-4a64-90f5-59e6ea5c6...(a)r27g2000yqb.googlegroups.com>... > > I apologize but it seems I was not clear. Forget my simple example, my > > actual problem is: > > > P = gammainc(N/2, x) / gamma(N/2) > > > N is an integer and x is my variable. I am looking for a way to find > > the corresponding value of x given a desired value of P. For example > > what value of x will give me P=0.9? > > ================= > > You weren't unclear. The same technique applies: > > y = @(x) gammainc(N/2, x) / gamma(N/2) > Ptarget=0.9; > x0=0; > > fzero(y-Ptarget,x0); > > The only issue is that, with this more complicated function, it's not entirely clear with what initial guess x0 to seed fzero(). I used x0=0 here only as an example. You could plot y(x) to get a reasonable initial guess. I appreciate the help, both in making this work in Matlab and the proper characterization of the problem. Would the consequences of choosing an a poor intial value simply be finding another less desirable root if one exists?
From: the cyclist on 17 Jun 2010 14:27 Jason Bonior <jdbonior21(a)gmail.com> wrote in message <aacdbd26-3616-4464-be03-3c2f69a431a4(a)k39g2000yqd.googlegroups.com>... > On Jun 17, 1:06 pm, "Matt J " <mattjacREM...(a)THISieee.spam> wrote: > > Jason Bonior <jdbonio...(a)gmail.com> wrote in message <1d66dd84-f442-4a64-90f5-59e6ea5c6...(a)r27g2000yqb.googlegroups.com>... > > > I apologize but it seems I was not clear. Forget my simple example, my > > > actual problem is: > > > > > P = gammainc(N/2, x) / gamma(N/2) > > > > > N is an integer and x is my variable. I am looking for a way to find > > > the corresponding value of x given a desired value of P. For example > > > what value of x will give me P=0.9? > > > > ================= > > > > You weren't unclear. The same technique applies: > > > > y = @(x) gammainc(N/2, x) / gamma(N/2) > > Ptarget=0.9; > > x0=0; > > > > fzero(y-Ptarget,x0); > > > > The only issue is that, with this more complicated function, it's not entirely clear with what initial guess x0 to seed fzero(). I used x0=0 here only as an example. You could plot y(x) to get a reasonable initial guess. > > I appreciate the help, both in making this work in Matlab and the > proper characterization of the problem. > > Would the consequences of choosing an a poor intial value simply be > finding another less desirable root if one exists? Usually, yes. You might consider getting a little more theoretical background on root finding. "Numerical Recipes" is a good reference to get you started.
From: John D'Errico on 17 Jun 2010 16:03 Jason Bonior <jdbonior21(a)gmail.com> wrote in message <aacdbd26-3616-4464-be03-3c2f69a431a4(a)k39g2000yqd.googlegroups.com>... > On Jun 17, 1:06 pm, "Matt J " <mattjacREM...(a)THISieee.spam> wrote: > > Jason Bonior <jdbonio...(a)gmail.com> wrote in message <1d66dd84-f442-4a64-90f5-59e6ea5c6...(a)r27g2000yqb.googlegroups.com>... > > > I apologize but it seems I was not clear. Forget my simple example, my > > > actual problem is: > > > > > P = gammainc(N/2, x) / gamma(N/2) > > > > > N is an integer and x is my variable. I am looking for a way to find > > > the corresponding value of x given a desired value of P. For example > > > what value of x will give me P=0.9? > > > > ================= > > > > You weren't unclear. The same technique applies: > > > > y = @(x) gammainc(N/2, x) / gamma(N/2) > > Ptarget=0.9; > > x0=0; > > > > fzero(y-Ptarget,x0); > > > > The only issue is that, with this more complicated function, it's not entirely clear with what initial guess x0 to seed fzero(). I used x0=0 here only as an example. You could plot y(x) to get a reasonable initial guess. > > I appreciate the help, both in making this work in Matlab and the > proper characterization of the problem. > > Would the consequences of choosing an a poor intial value simply be > finding another less desirable root if one exists? In one dimension, root finding is not as difficult as it is in higher dimensions. However, poor starting values can cause the solver to diverge into meaningless parts of the space, perhaps where your objective function is even undefined, or where the function exhibits numerical problems in evaluation. Or the solver may diverge to +/- inf. It may converge to a root that you are not interested in, when there are multiple roots. And at the very least, poor starting values will usually result in more effort (more time, more function evals) before the solver does converge. Good starting values are a good idea. John
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