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So here's my solution.

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If you had problems, it's really no problem.

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Maybe you can follow along with me and after each sell that I program, maybe you could stop and think

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about how you would continue from there.

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So the first thing that I will do, that's not really necessary, but I will do it here is that I will

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generate the three data sets.

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So for the three oscillators, the first one would be an array.

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And then for the X values, we have, the physical T values, which would be solution underscores three

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for C T and for the Y values we have solution on to score three or C Dot Y and then zero.

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And of course, for the other two datasets we have and the other two indices.

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So here one and two.

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So this is not a data, and I will show it to you first for data one, which is the oscillator with

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the index zero.

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So the very left one.

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So first of all, we must define, of course, the model function.

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So as before we write def function model like in the previous section, but not the previous one, but

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one of the earliest sections.

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And then we have here the argument, which is T and our case, and we have two parameters which will

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be a vector which we call a f and we return a zero times and p dot cosine.

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By the way, you could also use sine here.

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It doesn't matter because this is just a matter of the faced and we will fit anyway.

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So you could write here really cosine.

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It doesn't matter.

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Of course, I know sine doesn't matter.

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So then we multiply by T and we add the factor phase, which will be a one, and then we add the other

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two functions so it will copy and then fix it two to three.

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Four, three and five.

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OK.

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Looks good.

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No error, so this seems to work, and now we can define the error function, which the courts previously

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fit.

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And of course, the arguments are to function, which will be our model function.

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Then we have the coefficients, which will be eight and we have the data, which will be data one.

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So here we had some comments, which I will not write, which explained the parameters, but I think

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it's pretty clear.

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So we would just go ahead.

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And since it's a sum, we will start with the finding this to be zero and then looping for I in range

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length data zero.

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If you copy this from the previous notebook, then you already have this year and then the error will

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be the old error, plus the individual contributions.

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And so now we have to, you know, basically program this one here.

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So will be the brackets squared and then later I comma one.

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So this will be the other way around one comma.

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I will be the Y component minus F and then data zero comma I, which will be the X component and then

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has the argument for the function DX coefficients.

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OK.

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And then we print the summer so as not to print the picture.

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So as I told you earlier, you don't have to do it the loop.

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You can also define an array, of course, and then use to some command.

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So this will be our error.

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And now I want to just test if everything works.

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So I will just use some values for the starting parameters.

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I don't know.

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We could take random values, but maybe it's good here that we can compare with each other, that I

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just take some values.

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I don't know.

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And then I write error a fit.

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Function model, comma a comma data one.

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And this will be the era with these, with this value here as a starting value.

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So the thing is the arrow should probably be in a similar range for you, but it's probably not exactly

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the same value because we have taken random starting conditions for the simulation here.

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So your data will look a bit different compared to mine, which is why your number for the error may

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be a bit different.

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OK, now the next thing will be to define the gradient def arrow gradient or fits gradient we called

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it previously.

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Then we have here the same arguments.

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And once again, it's a song, so I wrote Total equal to zero.

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So as I told you previously, you probably copy this just from the other notebook and then you don't

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have to write all of what I'm writing right now.

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But I think it's better that I just write it one more time so that you can really see what's going on

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here in case you do not have the old notebook or you don't want to look it up.

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OK?

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So we basically have to add up now total the old total plus data one comma I minus half data, zero

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comma.

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I come out coefficients.

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So this is similar thing compared to here.

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So this is basically this one.

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And then we have to multiply only by this one here.

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So this I will call the FDA, which we have to define, of course, the FDA.

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And then in the end, we return minus two times total so that we have this factor here.

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OK, now we only have to define this one, the FDA, that's an array because it's a vector and you see

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what the vector is.

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It's just this one.

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So now it's a bit annoying.

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I just have to tag along and be sine omega one times data as well.

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B t so our T list theta zero comma I and then plus a one.

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And this we have several times.

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But first, let me program this one here.

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This will be minus a zero times and p dot sign of this whole thing.

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So and p told Sign and now we just have to copy this two more times and then fix, of course, the indices.

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All right.

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So I copy this.

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Here we have our first entries that we had before.

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Now, to the other ones, this will be omega two and a three.

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Here we have eight to omega to a three.

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And finally, Omega three, a five, a four, omega three and a five.

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OK, good.

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So now this is our gradient so we can call it arrow.

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Or maybe just let me copy this that I don't make any typos.

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Our fate gradient and then our function model a comma data one.

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And so once again, you probably will have different values here because you have a different dataset.

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And now I will test what happens if we do one iteration, one step.

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So we must defines a learning rate.

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So, h, we called it previously, it takes some small value and as I said, I will do one test step.

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So a the parameters will be updated according to a change of minus page times.

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And then what you just wrote?

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So we do this.

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And then afterwards, I will print the value of a profit.

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So basically this one.

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So right now, we have a row of 360, 2.5 four.

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Now we have an error of 295 Dot six or five, which is smaller when I rerun it.

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It will get smaller and smaller and smaller and smaller until eventually it will turn to zero.

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And then the parameters will hopefully look good so that I don't have to type enter over and over.

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I write iterations is equal to, I don't know.

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Here, of course, it depends sometimes a bit on the starting conditions, how many iterations you need.

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Maybe we try out one thousand.

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Let's see if it works.

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For I in range iterations.

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So it could also happen that in your case, you need more or less iterations because you have a different

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data.

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And also, if you rerun the whole notebook, it could also mean that you need a different number of

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iterations.

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So it's never bad to have too many iterations, but of course, and it takes much more time.

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So we just loop over what we've written here.

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OK.

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And then finally, we can look at the horrific gradient, and we can also look at the arrow fits, which

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can both be zero, approximately.

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And then we can have a look at the fit itself.

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It's not.

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I've run these two cells and for 1000 iterations took for me, maybe like three or four seconds.

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And you see, the arrow is now really, really small, close to zero.

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Maybe we could iterate a bit more, but hopefully it's OK.

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And you can also see the gradient is pretty small.

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All of them are below one, except for this one, which is still a bit high.

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So maybe 10000 would be better, but let's see what it looks like.

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So Peltier here, let me just scroll up to our previous plotting that we copy this, so we need basically

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this one.

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I would say

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it's curled up, and now I will change this year because we have used our data.

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We could, of course, leave it as it is, but I think it's nicer to just use data that we actually

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used for fitting some data one with the index, zero Theta, one with the index one.

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And now we want to plot our Yeah or Fitbit data.

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So I real right here t if all this was what we have used for the for the years over.

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So I want to show you this so that you believe me here.

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When we solved it numerically, we wrote in the integrates dot solve on the score IVP TVL is equal to

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00:12:32,350 --> 00:12:39,800
this Lind's based function here, so I will copy this go back down to if all is equal to this one.

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And now we can plot this Piatti dot plot for T underscore evolved from a function Model T underscore.

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You've all.

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Common a common threat.

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And this one, I will change to a scatterplot.

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And you see, this is the solution of the Fit's, which is the rent curve and the blue ones are our

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00:13:20,120 --> 00:13:22,510
data points that we have used for fighting.

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00:13:23,450 --> 00:13:30,400
You see, we have used a model function with, which is a superposition of three cosine functions.

154
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And indeed it worked.

155
00:13:32,580 --> 00:13:34,310
It's it's a pretty good fit.

156
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As I said, we could maybe improve it just a little bit more by using more iterations to decrease the

157
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error.

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But I think it looks pretty good now and it's a I think it's a remarkable, remarkable fit and really

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00:13:47,930 --> 00:13:53,720
a nice thing that we have seen when we just take our eigen frequencies, the three Agria frequencies

160
00:13:53,720 --> 00:14:00,710
that we have and we superimpose them, then we can recover the numerical results so we can have a look

161
00:14:00,710 --> 00:14:03,620
at the values of AA, which are the fitted parameters.

162
00:14:04,310 --> 00:14:07,940
And you see you, they are just some, some values.

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So this amplitude amplitude and another amplitude, and then we have here the face factors.

164
00:14:16,250 --> 00:14:22,310
So of course, now we can do the exact same thing with the oscillator number two.

165
00:14:22,640 --> 00:14:24,470
And this I will show to you in a second.

166
00:14:24,920 --> 00:14:28,500
So here you can see I have basically copied the code from above.

167
00:14:28,520 --> 00:14:36,950
So basically just starting once again with the first set of parameters and then looping over the gradient

168
00:14:36,950 --> 00:14:44,330
descent where our fit gradient would be this one with now the data set two and then I have plotted everything

169
00:14:44,330 --> 00:14:47,960
once again, and you see, it works very, very well.

170
00:14:48,860 --> 00:14:55,250
And when we look at the parameters, we see that all of them are reasonable values and this one here

171
00:14:55,700 --> 00:14:57,260
is actually very, very small.

172
00:14:58,100 --> 00:15:00,350
So it's much smaller than the other.

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So we could assume that it's probably zero.

174
00:15:03,830 --> 00:15:06,920
And when you look at it, what this means is to get the third entry.

175
00:15:07,250 --> 00:15:09,890
And if we go back, what was the third entry?

176
00:15:10,310 --> 00:15:12,530
This was this one.

177
00:15:13,190 --> 00:15:18,620
This was the amplitude for the second function, so the amplitude for the cosine function with omega

178
00:15:18,620 --> 00:15:19,100
two.

179
00:15:19,850 --> 00:15:25,580
And so indeed, this was the omega two frequency here.

180
00:15:25,910 --> 00:15:32,900
And you see, for the second oscillator, this one for the eigenvectors doesn't have an entry for this

181
00:15:32,900 --> 00:15:34,820
particular frequency.

182
00:15:35,530 --> 00:15:40,190
You can also see here in the full transform that the second oscillator does not have a peak here.

183
00:15:40,940 --> 00:15:47,900
So it makes sense that also in the fit, the second cosine function with the Omega two does not is not

184
00:15:47,900 --> 00:15:48,920
taken into account.

185
00:15:49,370 --> 00:15:51,110
And we didn't tell this to the program.

186
00:15:51,110 --> 00:15:56,270
We just said, please find the optimal fit using these three cosine functions.

187
00:15:56,270 --> 00:16:02,420
And it told us that the second cosine function or did we get the best fit if we don't take the second

188
00:16:02,420 --> 00:16:03,860
cosine function into account?

189
00:16:04,220 --> 00:16:09,350
And it's the same result as we got from the you transform and from the eigenvectors.

190
00:16:10,220 --> 00:16:18,320
So basically, we have now another way, or we have found another way that confirms our previous results.

191
00:16:18,680 --> 00:16:21,110
We have first solved the problem numerically.

192
00:16:21,590 --> 00:16:28,670
Then we have solved the eigenvalue problem by calculating the eigenvalues of the matrix that characterizes

193
00:16:28,670 --> 00:16:29,930
the equations of motion.

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00:16:30,350 --> 00:16:37,010
And from there, we have calculated the EIGEN frequencies, and now we have taken the cosine functions

195
00:16:37,010 --> 00:16:40,340
with the ion frequencies and have superimposed them.

196
00:16:40,880 --> 00:16:45,230
And we just had to fit the parameters a and fine.

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00:16:45,650 --> 00:16:48,110
And we have seen, OK, it really works.

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We can use three harmonic functions to recreate our solution, and everything agrees very, very well.