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In the previous lecture, we have created this data.

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These individual points here, and the reason why I want to start with these data points are not the

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function itself is that I want to avoid the possibility of decreasing the integration into law.

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So of course, what we could do when we take this method here is we could make these intervals smaller

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and smaller.

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So these these red rectangles, we could make them narrower.

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We take more points into account.

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And of course, then at some point, all the integral, all of the integration methods will give the

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same result and will have very good accuracy.

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However, quite often when you have very realistic data, then you don't know the function behind it.

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So you don't have the option to just decrease the width of these columns and to increase the data points.

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So we have to work with what we have just can work with these 13 data points, and now we will explore

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the methods that I have presented here.

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So we take the some of these data points and then we also take these trapezoidal methods into account.

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So let's get started with the sum and you I've written down that basically, this factor describes the

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size of each of these columns.

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So if each of these bars.

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And what I tried to do here is we have 13 data points and you see, if we really take 13 times the width

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of these bars, then we have a two large integration interval.

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So what I tried instead is I I decrease the, you know, the number of points and therefore decrease

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the weight of all of these points.

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And therefore, we have the integration interval correct from minus three to three.

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But still, these points at the edges, they will be over pronounced, which will give us a wrong answer.

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So maybe just to compare the analytical results was six point six.

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And now, since the edges are over pronounced and since they are positive, I would expect that now

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we will get two large values using this method.

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So let's get started.

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I define a function called integral some, and the only argument is to data and what we return is.

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And P dot sum.

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So it's a sum over the right and the array.

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Or yeah, well, basically just be the values of this function.

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So we have an odd sum of data and then the weichel components.

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So this will be the f of X I.

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And then we just have to consider this factor here so we can write data zero comma minus one minus data

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zero zero.

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So this means take the X value of the last point minus the value of the first point.

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So this will be three minus minus three.

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So this gives us six, which is the total interval for the integration.

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And now we can test two things.

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First, we can just divide by end, which would probably be the first idea.

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So we just right divided by the length of data one.

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So this will be the number of points that we have and then we can test what is the numerical result?

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Integral sum of data is 7.2, which is way too large.

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Yeah.

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So then we could also try to correct this with the minus one here to basically reduce the number of

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points.

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But then the results will get even worse.

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So you see, there is really a big problem that we have here.

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Of course, as I said, this method would give you a better result if you just increase the number of

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data points because then the effect of over pronouncing these two points becomes decreased.

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So then it's not that big of a deal anymore.

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However, what is much better is to use the trapezoidal methods and in fact, the trapezoidal method

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is equivalent to this methods.

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So we will first program this method and then we will.

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I'm sorry.

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First, we will program this method of trapezoidal methods and then we will program this methods and

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we will see that both of them give the same results.

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So the trapezoidal method basically means that we take our data and we do not take these rectangular

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bars anymore over which we add up, but instead we take.

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Is lights, so we have two points to neighboring points and we have this line, and then we can calculate

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just the area of this trapezoid and the area will be calculated by taking basically the average value

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of the Y coordinates.

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So somewhere between here this will be done the height and then we multiply by the width of this integral.

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So we have this interval.

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So what we do is we just right, define trap P. Zoidberg and then we can call it maybe, maybe call

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it better integral trapezoidal data.

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And now we first have this some term here.

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So of course, we could just take it and piss some of an array that we construct.

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But I think it's easier to just use loop.

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So we loop.

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For I in range length of data zero, which will be the number of our points, and then we subtract one

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because we have 13 data points, which means we have 13 individual intervals.

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This is why I write here minus one.

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And then here we add up.

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So we write is a plus some term, and in the end we return to value eight.

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And the thing over which we had this is this term here.

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So as I said, we take the average value of two neighboring points.

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So this would be data one comma pi plus one and then plus data one comma I.

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And then we divide by two and then we multiply by the size or by the width of these individual intervals.

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So in our case, we have equidistant points.

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But the trapezoidal methods works also for non equidistant data, which is really nice.

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So to make it more general, I will just write here data on zero.

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And then I plus one minus data zero comma.

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I thought we could have also taken just basically this one here would have worked in the same way.

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So I had to find this, and then I will calculate the results, so I will call integral trapezoidal

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data.

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And the result is six point sixty five.

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So the analytical result was six point six zero.

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And you see there is a small deviation.

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For example, here you can see the blue curve does look a bit different compared to the red curve.

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But overall, the result is pretty good and we only have an error of 0.05, which is below one percent.

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So the trapezoidal method works way better than this method, and it's quite a good approximation.

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So maybe one more comment.

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This trapezoidal method corresponds to integrating a linear spline function through two data points.

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So you see, we have linear functions that are defined piece wise, and then you can also show that

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when you rearrange the terms of the sum that we have just programmed that this method is equivalent

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to this method and this we will explore now by example.

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So we will program this method here, which relies on the fact that the starting and end point only

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contribute with half of their weight.

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Besides that, the method is the same as this one.

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So let's go ahead and program this one here for equidistant data.

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So what we can do to make our lives easier is to just copy the code from the integral sample.

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And I call this method.

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I don't know, maybe I call it therapy, he saw all the queue because then these two are equivalent

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and what we program here is this equation basically.

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So we have the prefecture that we also had before, which is basically the size of the interval, then

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we have here to some also the same as before.

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But the sum goes from the second point to the second last point and then the first and the last points.

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They are also added up, but with a weight factor of one half.

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So basically, you're right here that we have in sum and the sum of one, which will be the Y component.

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And now we don't start at zero, but we start at one, which will be a second value in the list and

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we go to minus one, which means we.

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So this is maybe a bit tricky in Python, I think, because this means excludes the last point.

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So this will be the second to the second last point.

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All right.

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And now we just have to add the two other terms.

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So we will have to add brackets here.

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And then we just write one half times data one from a zero, which is the first data point.

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Plus then our end piece sum and then plus one half times data, one comma and then the last point minus

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one.

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And now we can run this and we can call it.

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And you see the result is exactly the same.

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So this is because you can just take these individual components here and grouped them together.

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And you see here we have, for example, we take, for example, ie equal to three and equal to four.

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Then we have two times this term, which would be f of X for half and then we have the twice.

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So this gives us just f of X, and this is true for every I, except for the first and for the last

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data points, which we will only have once.

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So there we will have to factor one half.

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So you see, if we have equidistant data, then these two methods are really the same, even though

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from the graphics they look quite different.

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But adding up these areas in red here give you exactly the same result.