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dengwei_19

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ÒýÓûØÌû:
7Â¥: Originally posted by dbb627 at 2012-04-13 14:43:37:
Êǵģ¬Õâ¸ö¾ÍÊÇmatlabÇó³öµÄÊýÖµ½â£¬×öͼҲ¿ÉÒÔ¿´³öÀ´ÔÚ80¸½½ü¡£

>>  fun=@(x)-5.4*(pi^(1/2)*erf(10*2^0.5./x))/2-2.8*(pi^(1/2)*erf(40*2^0.5./x))/2+2.6;
x=0.1:0.1:100;
y=feval(fun,x);
plot(x,y)
[x,feval,flag]=fsolve(fun,80)
Õâ¸öÄÜÏê½âϹý³ÌÂð£¿
9Â¥2012-04-13 15:07:09
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²é¿´È«²¿ 15 ¸ö»Ø´ð

onesupeng

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2Â¥2012-04-13 13:47:46
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dengwei_19

ľ³æ (ÖøÃûдÊÖ)

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ÒýÓûØÌû:
2Â¥: Originally posted by onesupeng at 2012-04-13 13:47:46:
Õâ¸ömatlabÓеãÐü°É£¿ÎÒ¾õµÃ×Ô¼º±à³ÌÐòËã¿ÖŸüºÃÄØ

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3Â¥2012-04-13 13:53:05
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dbb627

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¡ï ¡ï ¡ï ¡ï ¡ï ¡ï
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dengwei_19: ½ð±Ò+6, ¡ï¡ï¡ïºÜÓаïÖú 2012-04-13 14:36:22
dengwei_19: »ØÌûÖö¥ 2012-04-13 15:40:01
>> syms x
>> syms w;int(exp(-x^2),0,w)

ans =

(pi^(1/2)*erf(w))/2

>> syms w;int(exp(-x^2),0,-w)

ans =

-(pi^(1/2)*erf(w))/2
===========
>> fun=@(x)-5.4*(pi^(1/2)*erf(10*2^0.5./x))/2-2.8*(pi^(1/2)*erf(40*2^0.5./x))/2+2.6;
>> x=0.1:0.1:100;
>> y=feval(fun,x);
>> plot(x,y)
>> [x,feval,flag]=fsolve(fun,80)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.




x =

   81.4787


feval =

-9.0775e-010


flag =

     1

» ±¾ÌûÒÑ»ñµÃµÄºì»¨£¨×îÐÂ10¶ä£©

The more you learn, the more you know, the more you know, and the more you forget. The more you forget, the less you know. So why bother to learn.
4Â¥2012-04-13 14:19:25
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