fitting triple exponential term function to data

Question

The function, I am trying to fit to data is: $$y(x) = −(A+B)e^{−x/a_1} + e^{−x/a_2} + Be^{−x/a_3}$$

this function is a little bit different to Is it possible to find initial parameters when fitting triple exponential term function to data?,

Here, we have $5$ parameters. In the post by JJacquelin he mentioned he prepared a "Triple exponential.docx". Does anyone know where I can find this document? And how can I get these paremeters?

Thanks.

Below is one example of data:

experiment data

1: https://i.stack.imgur.com/GqsxS.jpg 1

x y

0 1662

1 1661

2 1662

3 1660

4 1660

5 1662

6 1662

7 1662

8 1661

9 1660

10 1660

11 1661

12 1663

13 1663

14 1661

15 1660

16 1661

17 1663

18 1661

19 1660

20 1661

21 1661

22 1663

23 1662

24 1660

25 1661

26 1662

27 1662

28 1664

29 1659

30 1660

31 1659

32 1663

33 1662

34 1662

35 1661

36 1660

37 1662

38 1664

39 1661

40 1662

41 1660

42 1662

43 1663

44 1664

45 1662

46 1661

47 1661

48 1662

49 1665

50 1662

51 1660

52 1662

53 1662

54 1664

55 1661

56 1662

57 1663

58 1671

59 1681

60 1688

61 1695

62 1700

63 1706

64 1709

65 1714

66 1717

67 1720

68 1724

69 1726

70 1728

71 1727

72 1727

73 1730

74 1730

75 1731

76 1728

77 1728

78 1726

79 1728

80 1728

81 1724

82 1722

83 1720

84 1722

85 1722

86 1720

87 1717

88 1713

89 1714

90 1712

91 1713

92 1708

93 1709

94 1709

95 1706

96 1706

97 1703

98 1702

99 1699

100 1699

101 1699

102 1697

103 1696

104 1694

105 1693

106 1694

107 1693

108 1692

109 1690

110 1689

111 1689

112 1689

113 1686

114 1686

115 1684

116 1686

117 1686

118 1682

119 1679

120 1680

121 1682

122 1682

123 1680

124 1678

125 1679

126 1679

127 1680

128 1681

129 1677

130 1675

131 1676

132 1676

133 1677

134 1676

135 1674

136 1673

137 1675

138 1676

139 1673

140 1672

141 1671

142 1675

143 1673

144 1673

145 1670

146 1670

147 1673

148 1673

149 1671

150 1670

151 1669

152 1670

153 1671

154 1671

155 1669

156 1669

157 1670

158 1671

159 1671

160 1669

161 1666

162 1668

163 1669

164 1668

165 1668

166 1668

167 1668

168 1669

169 1670

170 1670

171 1667

172 1666

173 1668

174 1670

175 1667

176 1667

177 1666

178 1666

179 1667

180 1667

181 1667

182 1665

183 1665

184 1667

185 1667

186 1665

187 1665

188 1666

189 1666

190 1667

191 1667

192 1663

193 1665

194 1666

195 1667

196 1665

197 1665

198 1664

199 1664

200 1666

201 1665

202 1664

203 1663

204 1665

205 1665

206 1666

207 1665

208 1663

209 1663

210 1664

211 1665

212 1665

213 1664

214 1663

215 1664

216 1666

217 1666

218 1665

219 1663

220 1664

221 1667

222 1666

223 1664

224 1664

225 1663

226 1665

227 1665

228 1665

229 1663

230 1665

231 1665

232 1665

233 1663

234 1663

235 1661

236 1662

237 1664

238 1665

239 1663

240 1662

241 1664

242 1666

243 1663

244 1662

245 1664

246 1662

247 1665

Answer 1

By chance, I had this document.

As a very quick answer, I should try @JJacquelin's method just to get an idea about parameters $(a_1,a_2,a_3)$.

I should keep them fixed at these values and perform the linear regression $$y(x) = A\left(e^{−x/a_2}-e^{−x/a_1}\right) + B\left(e^{−x/a_3}-e^{−x/a_1}\right)$$ to get $(A,B)$.

Now, use all of that for a nonlinear regression.

Please, let me know if it works.

Answer 2

Sorry for my late answer.

The function you are trying to fit to data : $$y(x) = −(A+B)e^{−x/a_1} + e^{−x/a_2} + Be^{−x/a_3} \tag 1$$

is far to be convenient. Your data states $y(0)=1662$ while the above equation gives $y(0)=0$.

Obviously a constant is missing. One can suppose that the function might be : $$y(x) =C −(A+B)e^{−x/a_1} + e^{−x/a_2} + Be^{−x/a_3}\tag 2$$

The regression involves six parameters. This is not an easy problem. Nevertheless, this can be carried out thanks to the method explained in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . In fact the case of equation on the form $(2)$ is not treated in the paper. But it is possible to extend the case treated pp.71-72 in the paper to the case of equation $(2)$.

I updated my software for equation $(2)$ and tested it with your data. The analysis of the result convinced me that the equation $(2)$, even much better than $(1)$, is still far to be convenient.

Of course a sum of exponential terms must be used, but on a different manner to combine them. The function $(3)$ allows a much better fitting : $$y(x)=a+\frac{1}{b\:e^{\:p\:x}+c\:e^{\:q\:x}} \tag 3$$ The equation involves five parameters $p,q,a,b,c$. In case of data such as given by the OP one must have $p>0$ and $q<0$.

Unfortunately this kind of function is not theoretically studied and not applied in the paper referenced above. As a consequence one cannot take profit of the easy approach developed in the paper. The usual non-linear methods of regression are not easy with so many parameters especially for guessing the initial values to start an iterative process.

The result below was obtained with a non-linear regression method. The approximate values of the parameters are : $$\begin{cases} p\simeq 0.02670\\ q\simeq -0.27125\\ a\simeq 1661.6486\\ b\simeq 0.0018619\\ c\simeq 419482.7 \end{cases}$$ The root mean square error is $\quad\text{RMSE }\simeq 2.123\quad$ which is beautifully small compared to the order of magnitude (about $1700.$ ) of $y$ .