About Singularity Problem

Oguzhan Ayrancioglu

unread,

Jun 3, 2016, 3:00:09 AM6/3/16

to

Hi,

First of all I am new on Matlab. I searched some topics and "Matlab Help" section but didn't find that how can i solve problems which have singularity issue. There are some cases in the "help" menu and it says if there is a singularity u should split function boundaries. But i just have function that is "F(x)=1/x" and interval "0-1" so as you see, when x gets value of "0" function goes to infinity..

Could someone help me what is the approach i need to follow for this situation?

Thank you

Torsten

unread,

Jun 3, 2016, 4:30:08 AM6/3/16

to

"Oguzhan Ayrancioglu" wrote in message <nir9tj$17i$1...@newscl01ah.mathworks.com>...

What are you trying to do with the function that causes the error message ?

Best wishes
Torsten.

Oguzhan Ayrancioglu

unread,

Jun 3, 2016, 6:19:09 AM6/3/16

to

> What are you trying to do with the function that causes the error message ?
>
> Best wishes
> Torsten.

I just want to integrate "1/x" between 0,1. When "x=0" i got error which is "Warning: Infinite or Not-a-Number value encountered." i gave Tolerance (like "RelTol" and "AbsTol") nothing changed.

Than use this;

[int_2, err]=quadgk(fun,inf,1,'RelTol',1e-8,'AbsTol',1e-12,'MaxIntervalCount',1024)

at intervals, "inf" is "0" here. cause like i said before when x=0 function is infinite. so i need to have good approximation to "0" and it has to stop before x hits to "0" to avoid being infinite. But the output "err" gave me 3.7. And said i need to more iteration for interval counting.. The warning is like below;

"Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on
error is 3.7e+00. The integral may not exist, or it may be difficult to approximate
numerically. Increase MaxIntervalCount to 1504 to enable QUADGK to continue for another
iteration."

How do i solve this?

Thanks for your help
Oguzhan

Bruno Luong

unread,

Jun 3, 2016, 6:28:08 AM6/3/16

to

"Oguzhan Ayrancioglu" wrote in message <nirlio$pnk$1...@newscl01ah.mathworks.com>...

> How do i solve this?
>

Easy:

int2 = Inf

Nasser M. Abbasi

unread,

Jun 3, 2016, 6:41:30 AM6/3/16

to

On 6/3/2016 5:19 AM, Oguzhan Ayrancioglu wrote:
>> What are you trying to do with the function that causes the error message ?
>>
>> Best wishes
>> Torsten.
>
> I just want to integrate "1/x" between 0,1.
>When "x=0" i got error which is "Warning: Infinite or Not-a-Number value encountered."
>

> How do i solve this?
>
> Thanks for your help
> Oguzhan
>

int(1/x,x) is ln(x). When x=0, ln(x)=infinity.

You could integrate starting from some epsilon after zero.

int(1/x,x,eps('double'),1)
52*log(2)
double(ans)
36.0437

some integrals with singularities can be integrated
using 'PrincipalValue' option. For example

syms x
int(1/x,x,-1,1)

NaN

but

int(1/x,x,-1,1,'PrincipalValue',true)
0

--Nasser

Oguzhan Ayrancioglu

unread,

Jun 3, 2016, 6:46:08 AM6/3/16

to

> Easy:
>
> int2 = Inf

Thank you for ur post but, int_2=inf???

That is my problem indeed Bruno. Before it goes infinite i should hold it at some tolerance level giving me result. But i can't hold it. x goes every time to zero and Function goes infinite either.. So integral as u said equals to "inf" I need approximation before reaching value of zero.

Can i explain myself? Cause maybe i couldn't cause of my pretty bad english..

Best Wishes,
Oguzhan

Bruno Luong

unread,

Jun 3, 2016, 6:53:09 AM6/3/16

to

"Nasser M. Abbasi" wrote in message <nirmsj$1pff$1...@gioia.aioe.org>...

> On 6/3/2016 5:19 AM, Oguzhan Ayrancioglu wrote:

>
> int(1/x,x) is ln(x). When x=0, ln(x)=infinity.
>

Sign is reversed:

ln(x=0+) = -Inf.
But integral on [0,1] 1/x dx = log(1) - log(0+) = 0 - (-Inf) = +Inf