How can I find these values using polyfit command?

Alright , so i have the values of t time respectively first colum and R radioactive decay which corresponds to second colum (look below ) . The formula of radioactive decay is R=R0 * e^(-λt). Now I wanted to know if there is any way to find the values of R0 and λ for each case by using command polyfit ?

Does anybody have an idea how can I do this?

Thank you & Have a great day.

Values : 10       312
        11       328
        12       311
        13       290
        14       283
        15       285
        16       255
        17       291
        18       271
        19       272
        20       253
        21       255
        22       263
        23       246
        24       264
        25       213
        26       261
        27       241
        28       228
        29       242
        30       215
        31       192
        32       206
        33       182
        34       188
        35       197
        36       195
        37       180
        38       210
        39       176
        40       183
        41       199
        42       159
        43       171
        44       166
        45       145
        46       181
        47       177
        48       137
        49       149
        50       185
        51       148
        52       149
        53       142
        54       150
        55       140
        56       149
        57       123
        58       112
        59       118
        60       145
        61       146
        62       118
        63       113
        64       120
        65       126
        66       120
        67       125
        68       114
        69       114
        70       117
        71       103
        72       102
        73        96
        74        97
        75       102
        76       117
        77       114
        78        88
        79        94
        80       101
        81        93
        82       113
        83        74
        84        78
        85        81
        86       102
        87        94
        88        93
        89        81
        90        64
        91        88
        92        81
        93        96
        94        72
        95        82
        96        81
        97        71
        98        71

Answer 1

[Walter Roberson]

When R=R0 * e^(-λt) then log( R) = log(R0 * e^(-λt)) so log( R) = log(R0) + log(e^(-λt)) so log( R) = log(R0) - λ*t . This gives us a simple linear fit,

P = polyfit(t, log( R), 1);

and then λ = P(1) and log(R0) is P(2)

[Altin Guberi]

now these values are same for every case ? Or they differ? If there are more values how can I find them?

[Walter Roberson]

The whole point of polyfit() is to treat the entire series as being a (possibly noisy) approximation to a polynomial of the given order, and to find the coefficients of the polynomial that lead to the least error. If you think that the data might be two independent series each with the given model, then you need to fit differently, and when you do so you need to decide whether the join needs to be continuous or not (and if so, continuous only at the one point? A two point overlap? Continuous on the first derivative? ) Likewise if you think it might be the sum of two such series, you need to change how you fit.

If you think there might be more than 2 segments, then remember: the best fit will always be to put every point into a segment by itself. 100% perfect accuracy. So if you want the number of segments to be dynamically determined, then you need to specify some kind of penalty against subdividing into more and more segments.

[Walter Roberson]

You asked to fit a single decaying exponential line through the series of points. The result will be the two coefficients needed to express that single decaying exponential. Just like if you have a set of 100 points and you ask to fit a straight line through them, the result would just be the slope and the intercept, two values, not two values for each of the 100 points.

Perhaps what you are looking for is a list of times and projected values. Perhaps you are looking for a plot.

Values =  [10       312
      11       328
      12       311
      13       290
      14       283
      15       285
      16       255
      17       291
      18       271
      19       272
      20       253
      21       255
      22       263
      23       246
      24       264
      25       213
      26       261
      27       241
      28       228
      29       242
      30       215
      31       192
      32       206
      33       182
      34       188
      35       197
      36       195
      37       180
      38       210
      39       176
      40       183
      41       199
      42       159
      43       171
      44       166
      45       145
      46       181
      47       177
      48       137
      49       149
      50       185
      51       148
      52       149
      53       142
      54       150
      55       140
      56       149
      57       123
      58       112
      59       118
      60       145
      61       146
      62       118
      63       113
      64       120
      65       126
      66       120
      67       125
      68       114
      69       114
      70       117
      71       103
      72       102
      73        96
      74        97
      75       102
      76       117
      77       114
      78        88
      79        94
      80       101
      81        93
      82       113
      83        74
      84        78
      85        81
      86       102
      87        94
      88        93
      89        81
      90        64
      91        88
      92        81
      93        96
      94        72
      95        82
      96        81
      97        71
      98        71];

   p = polyfit(Values(:,1), log(Values(:,2)), 1);
   predicted = exp(p(2)) * exp(p(1)*Values(:,1));
   plot(Values(:,1), Values(:,2), 'r*', Values(:,1), predicted, 'b-');

Or if you prefer, you can use

   predicted = exp(polyval(p,Values(:,1));

Perhaps you would like to see error estimates:

t = Values(:,1);
R = Values(:,2);
[p, S, mu] = polyfit(t, log(R), 1);
[y, delta] = polyval(p, t, S, mu);
predictedR = exp(y);

plot(t, R, 'k*', t, predictedR, 'b-', t, exp(y-delta),'r-', t, exp(y+delta), 'r-');

The portion inside the red lines should contain 50% of the actual samples.

To go beyond this to get least-squared-fit of the actual data rather than least-squared fit of the log data, then you would need the Curve Fitting Toolbox, or the Stats Toolbox and use nlinfit()

But as for getting a pair of fitting values for each datapoint: NO. Not unless you are using a very different fit such as Piecewise Cubic Spline.

[Altin Guberi]

oh man , thank you for the long answer . I think that only two exponential coefficents are enough.