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From: Diane pham on 28 Apr 2010 14:20
Hi i have the function
((((ERFC(x/(2*(SQRT((0.0101)*t))))))-((EXP((8*x)/6.7)+((64*(0.0101)*t)/(6.7^2)))*((ERFC((x/(2*(SQRT((0.0101)*t))))+((8*(SQRT((0.0101)*t)))/6.7))))))= T(x,t)-Ti/To-Ti

Can I use matlab to solve?
From: Alan Weiss on 28 Apr 2010 14:28
On 4/28/2010 2:20 PM, Diane pham wrote:
> Hi i have the function
> ((((ERFC(x/(2*(SQRT((0.0101)*t))))))-((EXP((8*x)/6.7)+((64*(0.0101)*t)/(6.7^2)))*((ERFC((x/(2*(SQRT((0.0101)*t))))+((8*(SQRT((0.0101)*t)))/6.7))))))=
> T(x,t)-Ti/To-Ti
>
> Can I use matlab to solve?

Are you solving a one-dimensional problem or a multidimensional problem?
I mean, is x fixed and you want to find t, or do you need to find both x
and t? Do you have known constants for Ti, To?

Assuming you want to solve a one-dimensional problem, use fzero:
http://www.mathworks.com/access/helpdesk/help/techdoc/math/bsgpq6q-44.html

If you want to solve a two-dimensional problem, use fsolve from the
Optimization Toolbox(TM):
http://www.mathworks.com/access/helpdesk/help/toolbox/optim/ug/fsolve.html

Alan Weiss
MATLAB mathematical toolbox documentation
From: Walter Roberson on 28 Apr 2010 15:17
Diane pham wrote:
> Hi i have the function
> ((((ERFC(x/(2*(SQRT((0.0101)*t))))))-((EXP((8*x)/6.7)+((64*(0.0101)*t)/(6.7^2)))*((ERFC((x/(2*(SQRT((0.0101)*t))))+((8*(SQRT((0.0101)*t)))/6.7))))))=
> T(x,t)-Ti/To-Ti
>
> Can I use matlab to solve?

You have an unmatched '(' at the very beginning, and the () pair right after
that appears to be redundant as it appears to be joining two summands into a
form (A+B) = C when A + B = C would be equivalent.

Please cross-check that the bracketing is as you would want it in the
following expression:

((erfc(x / (2 * (sqrt((101/10000) * t)))))) - ((exp((8 * x) / (67/10)) + ((64
* (101/10000) * t) / (((67/10)^2))) * ((erfc((x / (2 * (sqrt(101/10000) *
t)))) + ((8 * (sqrt((101/10000) * t))) / (67/10)))))) = T(x,t)-Ti/To-Ti;

The question someone posed about what you are solving _for_ is a good one. The
difficulty of solving is going to depend upon the complexity of T(x,t) . With
some trial values, I can see that you are not going to get nice neat formula,
but what you do get is not too bad to solve numerically. The number of roots
you will get will depend upon what you are solving for.
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