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So as we have seen, expanding the exponential function works quite well.

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And the reason is that the exponential function is probably the most easy function to expand besides

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polynomials, because the here the derivative of every position is the same.

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So the first derivative, second derivative and in general, the derivative, it's always the exponential

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function at this particular position.

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It's not will continue with the second example, which is to sign function, and we will expand this

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at the position.

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X0 is equal to zero.

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And here the derivatives change, but they are quite similar.

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So in fact, we only have zeros, ones and minus ones in the derivatives.

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So this is because the actual value of sign of the same function is zero for an argument of zero.

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The first derivative is cosine of zero, which is one second derivative is minus zero minus sign of

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zero, which is zero.

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And then the third derivative is minus cosine of zero, which is minus one.

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And then starting from the fourth derivative, all of the derivatives will repeat.

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So then the fourth derivative will again be the function itself sine of X, and so sine of zero will

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be zero again.

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So you see the first, fifth, ninth and so on.

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The river two and zero will be one.

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And the third, seventh and the 11th and so on derivative will be minus one.

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So we can expand the sine function in this Taylor expansion that is given here where the sign of terms

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change.

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And we have only the odd terms here.

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And when we change our index where the index is not really meant to be describing the power of the polynomial.

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But when we change it to this two and plus one where and equals zero corresponds to one and equal to

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corresponds.

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All right.

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And a quick one corresponds to three and equal to corresponds to five.

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Then we can express it in this series here.

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So now we can go ahead and define this, and I don't want to type everything again.

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So I will copy the code from the exponential function and I will write sine Taylor.

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And one thing that we have to take care of is that when we want to calculate it analytically, where

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the derivatives are zero, one and minus one.

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This only works at the position.

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Zero is equal to zero.

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We will later on define a very general Taylor expansion function for Python, where we will rule out

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this problem here.

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But for now, we cannot consider X0 because zero is always zero.

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So I will delete these two lines and now we can go ahead and we basically just have to change what is

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written down here.

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And I will start with the first thing I thought to be elite this.

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And then we can write minus one to the power of and then we can multiply and have this power, this

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x to the power of two times and plus one.

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And then we just have to divide by nine factorial.

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So we write and p dot dot factorial and the material will be calculated off to and plus one.

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And then we can plot this.

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So once again, I take the code for plotting the Taylor expansion of the exponential function.

48
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But of course, I will change it a bit.

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So this time I will plot from minus 10 to 10.

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I have tested this before and also I will plot from minus two to two for the UI range.

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And then, of course, we have to change it.

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This one to and pitot sine.

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This will be our reference plot.

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And then for the actual plots, we just have to replace this one.

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We have to write sign Taylor sine Taylor, sine Taylor.

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And for the and Max, we will write explicitly what we want.

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So here we don't need to position anymore.

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So we will just have to define and max and I want to use three, six and nine to show you the difference.

59
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So we will always expand around to Position X. Zero is equal to zero, but we take a different number

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of terms into account.

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When you look at the plots, you will see that the blue function or basically first of all, you could

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see that all the functions reproduced assign function very well from PI over to minus 2+.

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So from here to here, then the red and the green function they even work from.

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Yeah, you could say maybe minus PI to PI.

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So a whole period of the function and then the green function even works from minus two Pi two plus

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two PI.

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00:05:07,090 --> 00:05:10,450
So this is already when you take into account nine terms.

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So you can imagine if you go one step further or several steps further and you right here black for

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the color and you take into account, I don't know, 18 terms.

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Then you can see the sine function is a very well reproduced.

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So this works pretty, pretty well.

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And we can, of course, also check this numerically by calculating the accuracy of the term sine of

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ten point five.

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00:05:38,110 --> 00:05:45,700
This is just some arbitrary number, which gives some, yeah, some real number, and we can now compare

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00:05:45,700 --> 00:05:51,850
this to our sine Taylor expansion, its position ten point five.

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00:05:52,210 --> 00:05:59,770
And if we now go very high with the order, for example, 50, then you see the error is 10 to the power

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of minus 13.

78
00:06:01,010 --> 00:06:03,160
So it's, yeah, basically zero.

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So once again, you see that this works quite well and that we can approximate even these more complicated

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functions like assign function.