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yan_fei

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dbb627

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¡ï ¡ï
xiegangmai: ½ð±Ò+2, ר¼Ò¿¼ºË, ÐÁ¿àÁË£¡ 2012-07-30 21:50:33
»òÕßÕâôдҲ¿ÉÒÔ
function yan
global s phi lam alpha
s=3.2;
phi=1;
lam=3;
alpha=0.6;
[X,FVAL,EXITFLAG,OUTPUT]=fsolve(@sumt,10)

function y=sumt(t)
global s phi lam alpha
y=0;
% for N=0:floor(s*t-phi)
%     y=y+exp(-lam*t)*(lam*t)^N/gamma(N+1);
% end
N=0:floor(s*t-phi);
yN=exp(-lam*t)*(lam*t).^N./gamma(N+1);
y=sum(yN);
y=y-alpha;
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¥2012-07-30 21:15:00
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
²é¿´È«²¿ 16 ¸ö»Ø´ð

yan_fei

Òø³æ (СÓÐÃûÆø)

ûÓÐÅóÓѻظ´°¡£¬Êǽð±ÒÉÙÂð£¿Ð¡µÜ½ð±ÒÓÐÏÞ£¬ÊÇÿÌìÐÁ¿àÇ©µ½»ýÔܵģ¬ÔÙ×·¼Ó100¸ö½ð±Ò£¬ÇëÖªµÀµÄÅóÓѰïæ£¬ÔõôÓÃmatlab±à³Ì°ÑÕâ¸ö·½³Ì½â³öÀ´£¬¹«Ê½ÇóºÍ·ûºÅÉÏÃæµÄ¡¾¡¿ÊÇÈ¡ÕûÊýµÄÒâ˼¡£
2Â¥2012-07-30 20:39:08
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

dbb627

ÈÙÓþ°æÖ÷ (ÖøÃûдÊÖ)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ...
¸Ðл²ÎÓ룬ӦÖúÖ¸Êý +1
xiegangmai: ½ð±Ò+3, ллӦÖú£¡ 2012-07-30 21:50:21
yan_fei: ½ð±Ò+130, ¸ÐлÕâλÅóÓѵĻظ´£¬½ð±ÒÈ«²¿¸øÄ㣡²»¹ý²»ÖªµÀΪʲô³õÖµ²»Í¬£¬½á¹û²î±ðºÜ´ó£¬ÎÒÏëÖ»¸Ä±äalphµÄÖµÇóÏàÓ¦µÄt£¬Ëæ±ãÒ»¸ö³õÖµ¶¼ÄÜÇó³ö׼ȷµÄ½â£¬ÈçÖªµÀÍû¸æÖª£¡¸ÐлÄúµÄ°ïÖú~~~ 2012-07-30 22:39:56
CODE:
function yan
global s phi lam alpha
s=3.2;
phi=1;
lam=3;
alpha=0.6;
[X,FVAL,EXITFLAG,OUTPUT]=fsolve(@sumt,10)

function y=sumt(t)
global s phi lam alpha
y=0;
for N=0:floor(s*t-phi)
    y=y+exp(-lam*t)*(lam*t)^N/gamma(N+1);
end
y=y-alpha;

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 =

   10.0881


FVAL =

-9.6716e-010


EXITFLAG =

     1


OUTPUT =

       iterations: 2
        funcCount: 6
        algorithm: 'trust-region dogleg'
    firstorderopt: 2.0551e-010
          message: [1x695 char]
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.
3Â¥2012-07-30 21:09:48
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

yan_fei

Òø³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
4Â¥: Originally posted by dbb627 at 2012-07-30 21:15:00
»òÕßÕâôдҲ¿ÉÒÔ
function yan
global s phi lam alpha
s=3.2;
phi=1;
lam=3;
alpha=0.6;
=fsolve(@sumt,10)

function y=sumt(t)
global s phi lam alpha
y=0;
% for N=0:floor(s*t-phi)
%     y=y+ ...

Ê×ÏȸÐлÄúµÄ»Ø¸´£¬ÎÒ¸ÕÊÔÁËÏÂÄúµÄ³ÌÐò£¬ºÃÏñ¶Ô³õʼֵµÄÒÀÀµÐÔºÜÇ¿£¬Ã»´Î¸ø²»Í¬µÄ³õʼֵÇóµÄ½á¹û¶¼²»Ò»Ñù£¬¶øÇÒ½á¹û¼¸ºõºÍ¸øµÄ³õʼֵ²î²»¶à¡£ÎÒÊÇÏë¸Ä±ä²»Í¬µÄalphaµÄÖµÇó³ö¶ÔÓ¦µÄt. ³ÌÐòÔËÐкóµÄÌáʾΪ£º

No solution found.

fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the default value of the function tolerance.



fsolve stopped because the relative norm of the current step, 4.694166e-013, is less than
max(options.TolX^2,eps) = 1.000000e-012. However, the sum of squared function values,
r = 9.759502e-003, exceeds sqrt(options.TolFun) = 1.000000e-003.

Optimization Metric                                               Options
relative norm(step) =  4.69e-013                max(TolX^2,eps) =  1e-012 (default)
r =  9.76e-003                                    sqrt(TolFun) = 1.0e-003 (default)


fsolve stopped because the relative norm of the current step, 4.694166e-013, is less than
max(options.TolX^2,eps) = 1.000000e-012. However, the sum of squared function values,
r = 9.759502e-003, exceeds sqrt(options.TolFun) = 1.000000e-003.

Optimization Metric                                               Options
relative norm(step) =  4.69e-013                max(TolX^2,eps) =  1e-012 (default)
r =  9.76e-003                                    sqrt(TolFun) = 1.0e-003 (default)
5Â¥2012-07-30 21:53:23
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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