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So we have just seen that the central difference, this method is much more accurate than the forward

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and backward diferencias methods for calculating the first order derivative of the function F.

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We have also seen that we can prove this analytically.

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And the reason was basically that certain error terms cancel out when we calculate the difference of

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these two terms.

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Then these two terms, he will vanish.

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And this would lead to a first order error term and h.

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And since they cancel out, the order of the error will be square, which arises from this term, which

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will be the next higher term.

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But from this idea, you can already see that it's possible to calculate the derivative even more accurately.

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And for this, we of course, have two methods.

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The first one is pretty similar to the mathematical way we just decreased the value of H.

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So if we have a continuous function like in our case and we know already what the function is, then

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this is not such a big problem.

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We can just decrease the value of H.

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But for example, if we would just have some data points and we would not know the precise function,

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then this is of course, not an option because then H is basically fixed.

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So what it must tend do is we must find ways to calculate the derivative with a higher accuracy.

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And for this, we must take into account even more terms, of course.

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So even more points of our list.

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So you see, for example, here we take forward the four differences method the current point and the

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next point.

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And for the back, what differences the current point and the previous point.

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And here for the central differences, we take the next and the previous point.

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So an idea would be to take an account even the second next and second previous points.

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And this is exactly what you can do to get a higher accuracy.

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And this is based on this Richardson equation for the first derivative term.

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You see, we take into account the yeah, the term with a minus two h minus one h positive one h and

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positive to h.

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So this is like a bit of a central difference this method, but even more refined because we take into

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account even the further terms.

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But you also see that in this weighted sum here, the the terms with it x minus h index plus h have

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a much higher weight than the terms with the two h.

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But anyway, let's defined as richesse and function and see how it performs compared to our central

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differences methods.

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So we write and define and then I just give it a name and I would call the D1 Richardson.

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00:02:57,270 --> 00:02:59,670
You could, of course, give it a different name as well.

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00:03:00,660 --> 00:03:03,090
And now we must find some arguments.

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Of course, we need the function itself.

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We need the argument and we need the step size.

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So let me write this down just in case we forget that later on after will be the function X will be

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00:03:17,070 --> 00:03:18,690
the argument.

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00:03:18,830 --> 00:03:21,150
So these here are just comments, of course.

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So you could leave them out.

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But it's always helpful to have something like this in your notebook so that later on you remember what

46
00:03:29,190 --> 00:03:30,300
these arguments mean.

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00:03:31,350 --> 00:03:36,330
And then our function should just return a number and the number is given by this equation.

48
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So it's pretty easy to implement which is the right one over 12 h, and then we must write down these

49
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four terms.

50
00:03:44,940 --> 00:03:48,150
So the first term will be f of X minus two h.

51
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Then we have another term, which is minus eight times half of x minus h.

52
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Then we have plus eight times of x plus h, and then the last term will be minus one times f of x minus

53
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two times h.

54
00:04:10,650 --> 00:04:13,230
And so this is our Richardson formula.

55
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And now we can compare how this method performs compared to the other methods.

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00:04:19,740 --> 00:04:24,600
So I will just copy this one here and.

57
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And another plot, so I will just add the plot.

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First of all, we must create a list, of course, so I write Richardson.

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The list is given by the one Richardson.

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Our new function and the function will be after the one that we have to find previously.

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X will be our list of X coordinates and H will be h, which was zero point one that we have to find

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previously.

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And now we can add another plot here where we subtract this new list from the analytical solution and

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for the color, it's just black.

65
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And so here we see that there seems to be a problem and an arrow because the newly added function the

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Richardson method gives the largest arrow, which is of course not what we expect.

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So let's check where it is the error.

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So this one is correct.

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And see here I wrote accidentally a minus two h, but it's a plus to age, of course.

70
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So let's redefine and let's rerun it.

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And we see this time the black function is almost equal to zero.

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So we know that these first two functions to forward and backward method, they are not so accurate.

73
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So let's at them out so that we only have green and black.

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And you see, black is a much, much better.

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So for green, we have an aura of the power of 10 to the minus two, and for green we have an error

76
00:06:00,640 --> 00:06:02,200
of 10 to the minus five.

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So it's approximately 1000 times better.

78
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And this is, of course, because here we did not only consider the neighboring points, but also the

79
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second neighbors.

80
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And so this allowed us to construct a superposition that has more of these error terms cancel out and

81
00:06:23,170 --> 00:06:26,110
therefore gives us a much better accuracy.

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And an even better accuracy can, of course, be achieved if we take more and more neighbors into account.

83
00:06:33,700 --> 00:06:40,870
So that in adding several of these Taylor expansions, more and more of these error terms will cancel

84
00:06:40,870 --> 00:06:46,450
out and then the error will have a higher and higher order and the result will be better and better.

85
00:06:47,080 --> 00:06:55,300
And this is what is called the richest in interpolation or rich iteration formula.

86
00:06:56,140 --> 00:07:02,350
And so the iteration formula is as every iteration formula based on the previous results.

87
00:07:02,350 --> 00:07:05,890
So you repeat this process over and over again.

88
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So we start from D1, which is the result of our D1 Richardson given by this equation.

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00:07:14,470 --> 00:07:19,870
And then we can improve the results by calculating D2, according to this equation.

90
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So in this case, it would be and would be one.

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00:07:24,850 --> 00:07:29,960
So we would have here two to the power of two, which is four than we have here.

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Our previous results for the step with H.

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Then we have here our previous results for the step with two h.

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And here we have four minus one, which is three.

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00:07:41,170 --> 00:07:47,650
And then we can iterated another time that we can go to the third order term and we can repeat this

96
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as many times as we want.

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And the result will get better and better because more of these terms in its halo expansion will cancel

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00:07:55,420 --> 00:07:55,750
out.

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00:07:56,830 --> 00:08:05,650
So I want to show you how we can, how we can implement such an iteration formula in our Python notebook.

100
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And of course, here it's a bit tricky because you see we have here the previous term for Step with

101
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H and here, the previous term for step was to H.

102
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So if we go to the second order, so which would be here an equal one so that we have here two, then

103
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we would need D1 at h and we want it to h if we go one step further.

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If we go to an equal to which we have here the three, then we would need D two of H and the two of

105
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two to, uh, two h.

106
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And for the two of two h, we would need D one of two h and the one for H.

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So you see, we need more and more First Order terms, the higher order we have, the higher order we

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want to consider.

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So let me just maybe show you the solution or my idea how I would implement it.

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So of course, we'll be a bit similar to our D one Richardson.

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So I ride to D1 and Richardson, and we need to define.

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Here's some order.

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So this would be the order not of the derivative, but order of iteration for first degree.

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Of it.If.

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So this time, we will not return this command here, but we will store this in some intermediate value.

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So maybe something like this.

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However, I also told you that we do not only need the first auditorium at age, but also at 2H.

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And if we go to a higher order, we even need it for age and then at eight h.

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So this depends on the order.

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So for our zero, I will define an array and this array will of course, be given by this term.

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But here we multiply.

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Or basically we replace all of the ages by age times two to the power of J.

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So let me do this right here at times two to the power of J.

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And then I do this for every age.

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And the value of G goes for J in range from zero to end.

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So this, of course, depends on the order of iteration that we want to consider.

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So this is now an array.

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So of course, what?

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Maybe then it becomes more clear what I want to say here.

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Of course, we could have also written down here is that the zero is just such an empty dot array for

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j and range.

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But the entry itself is just given by the Yvonne Richardson.

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F comma, x comma and then h will be modified by H Times two to the power of G.

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But the thing is, I do not really like to call functions inside other functions because maybe at some

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point we want to copy this to a different notebook and then we must remember that we also copy this

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function.

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So I think it's better to just use this more general approach here, even though it looks a bit more

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difficult.

139
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But this is just the same thing that we have written down in the Richardson function.

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All right.

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So this will be our starting point.

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So to say our D one and then to calculate D two, so which means an equal one, we need to do the following

143
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we need to update this we need to see I right.

144
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It's already in such a loop here for I in range one to NW, we do the following just for testing purposes,

145
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I would say we print, first of all, d zero so that we see the value at the current stage of iteration.

146
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And then we update the value of D um, which we will, by the way, further out outputs later.

147
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So.

148
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So return D will be this thing that we do at the very end.

149
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But now we update to B and P dot array.

150
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And here we were at the command, which would basically be this equation here for Jay can range.

151
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And now basically we must reduce the length with every iteration.

152
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So I just try to length D0 minus one.

153
00:13:00,170 --> 00:13:01,970
And here we write them this equation.

154
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So there will be two to the power of two times.

155
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I yes, because um yeah, maybe I should have called it a bit differently.

156
00:13:21,530 --> 00:13:25,400
So maybe you let me do it like this.

157
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I hope this is sort of it's less confusing to you.

158
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I called the order M or I could also call it, and max and Max is good.

159
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And here and we have to replace this one with and Max.

160
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And here it goes to and Max and then here we can use and so that we have the same one as here.

161
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So then we can ride like this.

162
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Yeah, I like this much better so that we can vote two to the power of two end times.

163
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The solution of the for argument h.

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So this would be a D0 and Jayco component.

165
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So of course, for the first element of the list of the NRA, this will be zero zero, which will be

166
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the first term here, which will be the end of h.

167
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And then we rides minus D zero j plus one and then we divide by the denominator.

168
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So this will be two to the power of two times J.

169
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Two times and sorry.

170
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And then the line is one.

171
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Exactly.

172
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So now we have Jay here.

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Iterator for the index of the zero and An, which is the looping index for this and from this equation,

174
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I think this looks good.

175
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And of course, then we have to update our D0 with the current D.

176
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So when I run the function, I think you will see what happens here.

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So I hope I didn't make a typo.

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This would be bad, but let's see what happens.

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So I just call the function one.

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And Richardson and Max let's, I don't know, use five f will be our function from previously and I

181
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want to do it at the position, let's say three

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three and I run this, OK?

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And so this these will be the intermediate results from the print command.

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So you see in the first loop, we get this result in the second loop, we get this one and the third

185
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loop, this one, fourth, this one and then the fifth loop.

186
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Finally, we get our result, which will be this value.

187
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And now we can compare this to the D1 Richardson.

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And for this, I will write D1 Richardson, for example.

189
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This value and this value corresponds to this value.

190
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So let me show you something.

191
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I think that becomes more clear what I did above.

192
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I will copy this five times and then I will write you two times h four times h, eight times h, 16

193
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times h.

194
00:16:37,910 --> 00:16:41,840
And these five values are exactly the values that we get here.

195
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So these are the values that are in this and P Todd's array.

196
00:16:48,830 --> 00:16:54,710
So, yeah, basically we have here this iterated J going from zero to and max.

197
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And so we have five terms in this array.

198
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And then comes the first iteration where we calculate where we update our D, according to this equation.

199
00:17:06,079 --> 00:17:11,300
And so here we calculate basically D to.

200
00:17:12,450 --> 00:17:19,710
Based on this one, and we need the data for four different values we needed for age four to age four

201
00:17:19,710 --> 00:17:21,210
four eight four eight each.

202
00:17:21,960 --> 00:17:28,710
Then comes these three and then we need the three for age to age and for age, the four we need for

203
00:17:28,710 --> 00:17:29,760
age and to age.

204
00:17:30,150 --> 00:17:32,520
And then from this, we can calculate DeFife.

205
00:17:33,870 --> 00:17:40,980
And now the only thing left to do is to see if this really is an approved is it is an improvement.

206
00:17:41,190 --> 00:17:49,380
So I have to write three times dot and P Dot Cosine, because that's the analytical solution.

207
00:17:49,650 --> 00:17:55,110
So you see, I tested this before with an argument of two, but this time I choose three, so we have

208
00:17:55,110 --> 00:17:55,890
to update this.

209
00:17:57,620 --> 00:18:02,580
Um, I'm not really sure why I did this before, but that's updated to three.

210
00:18:03,150 --> 00:18:14,820
And then let's right and p dot sign of three minus three over 100 times three squared.

211
00:18:16,720 --> 00:18:24,700
So this will be the value that we get analytically and now we can compare this value and we can just

212
00:18:24,700 --> 00:18:29,350
subtract the value from the Richardson formalism.

213
00:18:29,830 --> 00:18:40,000
So when I called his one and then the output from the return disappears and will instead be stored here.

214
00:18:40,600 --> 00:18:49,540
And so then we can subtract a two year deal on end, and we see the error with respect to the analytical

215
00:18:49,540 --> 00:18:51,610
solution would be a very, very small.

216
00:18:51,610 --> 00:18:55,810
It will be to tend to the power of minus nine.

217
00:18:55,810 --> 00:19:01,780
And previously from the first or two terms, we had an error of 10 to the power of minus five.

218
00:19:03,070 --> 00:19:10,900
So this was just to show you that there are methods to calculate the derivative much more accurately.

219
00:19:11,530 --> 00:19:17,500
And the trick is to combine more and more terms and further and further nearest neighbors for these

220
00:19:17,500 --> 00:19:18,670
individual data points.

221
00:19:19,030 --> 00:19:22,360
So it had several hours and a Taylor function cancel out.

222
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Of course, in our course, we will not always do this difficult stuff here.

223
00:19:28,600 --> 00:19:34,390
We will not always want to have such a precise derivative.

224
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Most often, we will just use the central difference as methods.

225
00:19:39,190 --> 00:19:44,140
Well, we'll just use the left and the right neighbors and calculate the derivative from here.

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And maybe sometimes we need a bit of a better accuracy.

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So we take the Richardson formula.

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But if you want to have a really precise derivative, then this is possible and you can take this iteration

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formula.

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And of course, here we have taken and max equal to five.

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But of course, you could take year 100 and then the derivative will be more and more precise.

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But most of the time, it's really a question is it worth it?

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Because you see to calculate this one derivative?

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We had to calculate here five plus four plus three plus two plus one values to finally have it.

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So you really have to think, is it worth it to spend the time to calculate the derivative more precisely?

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Or is it OK to just stick with the central differences?

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And then most of the cases, it is OK to just use central differences.