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From: Paul on 5 Aug 2010 15:34
"us " <us(a)neurol.unizh.ch> wrote in message <i3f2tp$sbs$1(a)fred.mathworks.com>...
> "Paul " <unicybsatr(a)gmail.com> wrote in message <i3f1ag$gfu$1(a)fred.mathworks.com>...
> > Walter Roberson <roberson(a)hushmail.com> wrote in message <i3evuf$khb$1(a)canopus.cc.umanitoba.ca>...
> > > Paul wrote:
> > >
> > > > Suppose, I have integral I(t)=int from 0 to inf x^-2
> > > > (1-0.8*i*t*x)^-5*erfc(1-logx/5)dx
> > > > Function
> > > > dblquad is used to calculate int from 0 to inf I(t)dt
> > > > But how can I use it to calculate int from 0 to inf I(t)*I(t)dt ?
> > > > Do I need here nested handles? If yes, then how should I apply them?
> > >
> > > I2 = @(t) I(t).^2;
> > >
> > > and dblquad I2
> > >
> > >
> > > Questions:
> > >
> > > - Is that log(x/5) or log(x)/5 ?
> > >
> > > - Is your log base 10 or natural log?
> > >
> > > - Is the 0.8 intended to be equivalent to 4/5 ?
> > >
> > > I am attempting some symbolic transforms to see if I can derive a nicer
> > > function, but when you start doing symbolic transforms, 4/5 versus 0.8 makes a
> > > difference, especially if the coefficient turns up as an exponent.
> >
> > Hello thanks for answer.
> > I2 = @(t) I(t).^2;
> > dblquad I2
> >
> > But how are you define I ?
>
> a hint:
> - the var I must live in the workspace when you create the function handle...
>
> us
can you give me the code, not hint.
I=@(x,t) x^-2*...... ?
From: Walter Roberson on 5 Aug 2010 15:44
Paul wrote:
> "us " <us(a)neurol.unizh.ch> wrote in message
> <i3f2tp$sbs$1(a)fred.mathworks.com>...
>> "Paul " <unicybsatr(a)gmail.com> wrote in message
>> <i3f1ag$gfu$1(a)fred.mathworks.com>...
>> > Walter Roberson <roberson(a)hushmail.com> wrote in message
>> <i3evuf$khb$1(a)canopus.cc.umanitoba.ca>...
>> > > Paul wrote:
>> > > > > > Suppose, I have integral I(t)=int from 0 to inf x^-2 > > >
>> (1-0.8*i*t*x)^-5*erfc(1-logx/5)dx
>> > > > Function
>> > > > dblquad is used to calculate int from 0 to inf I(t)dt
>> > > > But how can I use it to calculate int from 0 to inf I(t)*I(t)dt ?
>> > > > Do I need here nested handles? If yes, then how should I apply
>> them?
>> > > > > I2 = @(t) I(t).^2;
>> > > > > and dblquad I2
>> > > > > > > Questions:
>> > > > > - Is that log(x/5) or log(x)/5 ?
>> > > > > - Is your log base 10 or natural log?
>> > > > > - Is the 0.8 intended to be equivalent to 4/5 ?
>> > > > > I am attempting some symbolic transforms to see if I can
>> derive a nicer > > function, but when you start doing symbolic
>> transforms, 4/5 versus 0.8 makes a > > difference, especially if the
>> coefficient turns up as an exponent.
>> > > Hello thanks for answer.
>> > I2 = @(t) I(t).^2;
>> > dblquad I2
>> > > But how are you define I ?
>> a hint:
>> - the var I must live in the workspace when you create the function
>> handle...
>>
>> us
> can you give me the code, not hint.
> I=@(x,t) x^-2*...... ?

Take the function handle you have now for dblquad'ing x^-2 <etc>
and wrap another handle around it that takes the same parameters and calls the
first handle and squares the result.
From: Paul on 5 Aug 2010 16:11
Walter Roberson <roberson(a)hushmail.com> wrote in message <i3f4bk$r8e$1(a)canopus.cc.umanitoba.ca>...
> Paul wrote:
> > "us " <us(a)neurol.unizh.ch> wrote in message
> > <i3f2tp$sbs$1(a)fred.mathworks.com>...
> >> "Paul " <unicybsatr(a)gmail.com> wrote in message
> >> <i3f1ag$gfu$1(a)fred.mathworks.com>...
> >> > Walter Roberson <roberson(a)hushmail.com> wrote in message
> >> <i3evuf$khb$1(a)canopus.cc.umanitoba.ca>...
> >> > > Paul wrote:
> >> > > > > > Suppose, I have integral I(t)=int from 0 to inf x^-2 > > >
> >> (1-0.8*i*t*x)^-5*erfc(1-logx/5)dx
> >> > > > Function
> >> > > > dblquad is used to calculate int from 0 to inf I(t)dt
> >> > > > But how can I use it to calculate int from 0 to inf I(t)*I(t)dt ?
> >> > > > Do I need here nested handles? If yes, then how should I apply
> >> them?
> >> > > > > I2 = @(t) I(t).^2;
> >> > > > > and dblquad I2
> >> > > > > > > Questions:
> >> > > > > - Is that log(x/5) or log(x)/5 ?
> >> > > > > - Is your log base 10 or natural log?
> >> > > > > - Is the 0.8 intended to be equivalent to 4/5 ?
> >> > > > > I am attempting some symbolic transforms to see if I can
> >> derive a nicer > > function, but when you start doing symbolic
> >> transforms, 4/5 versus 0.8 makes a > > difference, especially if the
> >> coefficient turns up as an exponent.
> >> > > Hello thanks for answer.
> >> > I2 = @(t) I(t).^2;
> >> > dblquad I2
> >> > > But how are you define I ?
> >> a hint:
> >> - the var I must live in the workspace when you create the function
> >> handle...
> >>
> >> us
> > can you give me the code, not hint.
> > I=@(x,t) x^-2*...... ?
>
> Take the function handle you have now for dblquad'ing x^-2 <etc>
> and wrap another handle around it that takes the same parameters and calls the
> first handle and squares the result.

I tried this code: (function x*y^2)
And integration from 0 to 1:
I=@(x,y) x.^2*y;
I2=@(y) I(y).^2;
dblquad(I2,0,1,0,1);

But it's not working.

??? Error using ==> @(y) I(y).^2
Too many input arguments.

Please help!!!
From: Walter Roberson on 5 Aug 2010 16:39
Paul wrote:

> I tried this code: (function x*y^2)
> And integration from 0 to 1:
> I=@(x,y) x.^2*y;
> I2=@(y) I(y).^2;
> dblquad(I2,0,1,0,1);

Does that match what I said,

>> Take the function handle you have now for dblquad'ing x^-2 <etc>
>> and wrap another handle around it that takes the same parameters and calls
>> the first handle and squares the result.

Does your I2 have the *same* parameters as your I ?



I = @(x,y) x.^2*y;
I2 = @(x,y) I(x,y).^2;
dblquad(I2,0,1,0,1)

ans =
0.0666666666666667
From: Paul on 5 Aug 2010 16:50
Walter Roberson <roberson(a)hushmail.com> wrote in message <i3f7j2$35a$1(a)canopus.cc.umanitoba.ca>...
> Paul wrote:
>
> > I tried this code: (function x*y^2)
> > And integration from 0 to 1:
> > I=@(x,y) x.^2*y;
> > I2=@(y) I(y).^2;
> > dblquad(I2,0,1,0,1);
>
> Does that match what I said,
>
> >> Take the function handle you have now for dblquad'ing x^-2 <etc>
> >> and wrap another handle around it that takes the same parameters and calls
> >> the first handle and squares the result.
>
> Does your I2 have the *same* parameters as your I ?
>
>
>
> I = @(x,y) x.^2*y;
> I2 = @(x,y) I(x,y).^2;
> dblquad(I2,0,1,0,1)
>
> ans =
> 0.0666666666666667
Thanks!!!!!
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