0
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I have the following data and im trying to get an exponential fit. Ive tried a variety of different tools for this, which all seem to give quite a large error margin at the top of the curve.

Plotting the data in here http://www.zizhujy.com/en-us/Plotter using exponential fit gives an error margin of around 20 at the top of the curve.

Im wondering is there any way of correcting this in order to fit the data better?

Do I need to add something to x in the following equation to do this?

enter image description here

0   0.00000001
1   0.11
2   0.23
3   0.35
4   0.46
5   0.58
6   0.7
7   0.82
8   0.94
9   1.06
10  1.19
11  1.31
12  1.43
13  1.56
14  1.69
15  1.82
16  1.95
17  2.08
18  2.21
19  2.34
20  2.48
21  2.61
22  2.75
23  2.89
24  3.03
25  3.17
26  3.31
27  3.45
28  3.6
29  3.74
30  3.89
31  4.04
32  4.19
33  4.35
34  4.5
35  4.66
36  4.81
37  4.97
38  5.13
39  5.3
40  5.46
41  5.63
42  5.8
43  5.97
44  6.14
45  6.31
46  6.49
47  6.67
48  6.85
49  7.03
50  7.22
51  7.41
52  7.6
53  7.79
54  7.99
55  8.18
56  8.39
57  8.59
58  8.8
59  9.01
60  9.22
61  9.43
62  9.65
63  9.88
64  10.1
65  10.3
66  10.6
67  10.8
68  11
69  11.3
70  11.5
71  11.8
72  12
73  12.3
74  12.6
75  12.9
76  13.1
77  13.4
78  13.7
79  14
80  14.3
81  14.6
82  14.9
83  15.2
84  15.6
85  15.9
86  16.2
87  16.6
88  16.9
89  17.3
90  17.7
91  18.1
92  18.5
93  18.9
94  19.3
95  19.8
96  20.2
97  20.7
98  21.1
99  21.6
100 22.1
101 22.7
102 23.2
103 23.8
104 24.4
105 25
106 25.7
107 26.3
108 27
109 27.8
110 28.6
111 29.4
112 30.3
113 31.2
114 32.2
115 33.3
116 34.5
117 35.7
118 37.1
119 38.6
120 40.3
121 42.3
122 44.5
123 47.1
124 50.4
125 54.5
126 60.3
127 70
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  • $\begingroup$ Hey, the 5th order polynomial fit seems to look a bit better on the same plotter, if that helps... $\endgroup$ Apr 15, 2014 at 9:53
  • $\begingroup$ Yep 5th order is ok, I really wanted to try and get a perfect fit though, well as perfect as possible :/ $\endgroup$
    – Ke.
    Apr 15, 2014 at 10:38
  • $\begingroup$ The link you give does not contain your data. Please update in order we see them. $\endgroup$ Apr 16, 2014 at 9:05
  • $\begingroup$ The data inside my post, gets pasted into the box on the left on the zizhujy.com site. Unfortunately, there is no way of taking you straight there with the data input already, you have to paste it in yourself. $\endgroup$
    – Ke.
    Apr 16, 2014 at 10:15

1 Answer 1

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The large discrepancy between the computed data and the experimental data is due to the supposed relationship $y=\alpha e^{\beta x}$ which is not convenient. You cannot correctly fit this function with your data. This is obvious when drawing $\ln(y)$ as a function of $x$. The curve is far from a straight line $\ln(y)=a+\beta x$ where $a=\ln(\alpha)$ as it is visible on the joint figure.

Correcting $x$ and/or $y$ by adding some adjusted constants will not solve the problem, but will reduce the error on one side and increase the error on the other side. If a better fitting is necessary, one have to find another relationship between $x$ and $y$. If the question comes from an physical problem, the modeling has to be sufficiently correct to give a more convenient kind of function.

enter image description here

If there is no reliable physical model available, on a pure mathematical viewpoint there are a infinity of functions likely to be convenient, but they are more sophisticated than $y=\alpha e^{\beta x}$. For example the sum of a polynomial and an exponential leads to an accurate fit (Figure below)

enter image description here

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  • $\begingroup$ Thanks so much JJacquelin. You have a great way of explaining things, to help understand what is happening. Really helpful :) $\endgroup$
    – Ke.
    Apr 19, 2014 at 20:34

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