我的实验得到一组数据,想拟合到一个公式:A=Ao*exp[(Bo/B)^(1/(1+d))] (d=1,2 or 3), Ao 和 Bo 是常数。 A 和 B 是实验数据, 想看到底d=1, 2还是3的时候拟合最好, 于是我两边区对数,画出 lnA---B^(-1/(1+d))的曲线,居然傻眼了,发现d=1,2和3的时候都可以得到很好的直线 ,问题到底出在哪里?请牛人指教,谢谢
Your plot is not right, Bo^(1/(1+d)) is in your formula but left out when you plot lnA- curves. Since you are varying d, you cannot ignore that term.
or 画出
【在 h****g 的大作中提到】 : 我的实验得到一组数据,想拟合到一个公式:A=Ao*exp[(Bo/B)^(1/(1+d))] (d=1,2 or : 3), Ao 和 Bo 是常数。 : A 和 B 是实验数据, 想看到底d=1, 2还是3的时候拟合最好, 于是我两边区对数,画出 : lnA---B^(-1/(1+d))的曲线,居然傻眼了,发现d=1,2和3的时候都可以得到很好的直线 : ,问题到底出在哪里?请牛人指教,谢谢
Of Course y=lnA, x=B^(-1/(1+d)); y=lnA0+B0^(1/(1+d))*x lnA0 and B0 are constant, you should get three straight lines with a same intercept y=lnA0 and three different slopes of B0^(1/2),^(1/3),^(1/4) you should plot lnA~B
or 画出
【在 h****g 的大作中提到】 : 我的实验得到一组数据,想拟合到一个公式:A=Ao*exp[(Bo/B)^(1/(1+d))] (d=1,2 or : 3), Ao 和 Bo 是常数。 : A 和 B 是实验数据, 想看到底d=1, 2还是3的时候拟合最好, 于是我两边区对数,画出 : lnA---B^(-1/(1+d))的曲线,居然傻眼了,发现d=1,2和3的时候都可以得到很好的直线 : ,问题到底出在哪里?请牛人指教,谢谢
【在 f*****y 的大作中提到】 : Of Course : y=lnA, x=B^(-1/(1+d)); : y=lnA0+B0^(1/(1+d))*x : lnA0 and B0 are constant, : you should get three straight lines with a same intercept y=lnA0 : and three different slopes of B0^(1/2),^(1/3),^(1/4) : you should plot lnA~B : : or : 画出
f*y
15 楼
If the d is just one of 1,2 or 3,(or any integer) you can plot lines with different d!then find out which line fits your data. If d is not an interger, you should find a way to regress your data by an ex ponential rule. you should get values of a, b and c in a regression equation like lnA=a+bB^c . A0=a,(-1/(1+d))=c and B0=-b/c.... Good luck
【在 h****g 的大作中提到】 : Then how can I determine d? : thanks for help
j*u
16 楼
It's very possible you get all straight lines if you don't have a broad rang e of data. Evaluating goodness of fit is a possible way to determine which o ne is better.
data. ex ^c
【在 f*****y 的大作中提到】 : If the d is just one of 1,2 or 3,(or any integer) : you can plot lines with different d!then find out which line fits your data. : If d is not an interger, you should find a way to regress your data by an ex : ponential rule. : you should get values of a, b and c in a regression equation like lnA=a+bB^c : . : A0=a,(-1/(1+d))=c and B0=-b/c.... : Good luck