WEBVTT
00:00:02.590 --> 00:00:12.920
For the given function π of π₯ is equal to the inverse tan of two π₯, find a power series representation of π by integrating the power series for π prime.
00:00:13.590 --> 00:00:24.890
The question is asking us to integrate the power series for the function π prime so that we can find the power series representation of our function π of π₯ is equal to the inverse tan of two π₯.
00:00:25.180 --> 00:00:28.940
So letβs start by calculating what our function π prime of π₯ is.
00:00:29.080 --> 00:00:37.210
We know that the derivative with respect to π₯ of the inverse tan of ππ₯ is just equal to π divided by one plus π squared π₯ squared.
00:00:37.600 --> 00:00:44.150
We can use this to calculate that the function π prime of π₯ is equal to two divided by one plus two squared π₯ squared.
00:00:44.470 --> 00:00:48.170
And then, we can just simplify two squared π₯ squared to just be four π₯ squared.
00:00:48.790 --> 00:00:53.390
Next, the question tells us that we should find the power series for our function π prime of π₯.
00:00:53.420 --> 00:00:56.780
And we can do this by using what we know about geometric series.
00:00:57.190 --> 00:01:08.210
We know that if the absolute value of our ratio π is less than one, then the sum from π equals zero to β of π multiplied by π to the πth power is just equal to π divided by one minus π.
00:01:08.710 --> 00:01:18.120
Next, we can notice that our function π prime of π₯ is already in the form π divided by one minus π if we set π to equal two and π to equal negative four π₯ squared.
00:01:18.710 --> 00:01:43.860
So if we substitute this into what we know about geometric series, we get that if the absolute value of negative four π₯ squared is less than one, then we know that the sum from π equals zero to β of two multiplied by negative four π₯ squared all raised the πth power is equal to two divided by one minus negative four π₯ squared, where we can notice that our denominator, one minus negative four π₯ squared, is just one plus four π₯ squared.
00:01:44.100 --> 00:01:49.180
So this is equal to two divided by one plus four π₯ squared, which is equal to π prime of π₯.
00:01:49.390 --> 00:01:52.570
So weβve now found the power series for our function π prime.
00:01:52.740 --> 00:01:59.230
That is, itβs the sum from π equals zero to β of two multiplied by negative four π₯ squared raised to the πth power.
00:01:59.720 --> 00:02:03.950
So we can write the power series for our function π prime of π₯ into our top equation.
00:02:04.360 --> 00:02:06.490
Now, weβre ready to simplify this power series.
00:02:06.520 --> 00:02:09.640
Letβs start by distributing the exponent π over our parentheses.
00:02:09.670 --> 00:02:13.990
This gives us negative four to the πth power multiplied by π₯ squared to the πth power.
00:02:14.510 --> 00:02:18.750
Next, we can rewrite negative four as negative one multiplied by two squared.
00:02:19.190 --> 00:02:26.630
We can then distribute the exponent π over our first set of parentheses to get negative one to the πth power multiplied by two squared to the πth power.
00:02:27.020 --> 00:02:39.110
Finally, we can evaluate the exponents in our second and third set of parentheses, giving us a final answer of negative one to the πth power multiplied by two to the power of two π multiplied by π₯ to the power of two π.
00:02:39.480 --> 00:02:49.040
This gives us a new sum from π equals zero to β of two multiplied by negative one to the πth power multiplied by two to the power of two π multiplied by π₯ to the power of two π.
00:02:49.400 --> 00:03:05.870
Now, we can simplify this further by noticing that two multiplied by two to the power of two π is just equal to two to the power of two π plus one, giving us the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by π₯ to the power of two π.
00:03:06.380 --> 00:03:15.120
We now recall that the integral of our derivative function π prime of π₯ with respect to π₯ is equal to the function π of π₯ up to a constant of integration πΆ.
00:03:15.460 --> 00:03:21.040
At this point, we substitute our power series representation for our derivative function π prime of π₯.
00:03:21.420 --> 00:03:24.570
At this point, weβre going to move the integral symbol inside our sum.
00:03:25.090 --> 00:03:28.560
And we know we can do this because weβre integrating a power series.
00:03:28.890 --> 00:03:35.480
At this point, we can notice that two to the power of two π plus one and negative one to the πth power are a constant with respect to π₯.
00:03:35.600 --> 00:03:50.950
So integrating the term inside our summand is equivalent to just integrating π multiplied by π₯ to the power of two π with respect to π₯, which we know is equal to π multiplied by π₯ to the power of two π plus one divided by two π plus one plus our constant of integration πΆ.
00:03:51.160 --> 00:03:56.930
Instead of adding a constant of integration for every single term in our sum, weβre just going to add one right at the end.
00:03:57.350 --> 00:04:12.020
Evaluating the integral inside our sum gives us πΆ plus the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by π₯ to the power of two π plus one divided by two π plus one.
00:04:12.390 --> 00:04:15.550
We now have a power series representation for our function π of π₯.
00:04:15.550 --> 00:04:18.600
However, it still has an unknown variable which weβve labelled πΆ.
00:04:18.850 --> 00:04:26.450
To eliminate this unknown variable, we recall that this is a power series for our function π of π₯, which is equal to the inverse tan of two π₯.
00:04:26.580 --> 00:04:38.400
Therefore, if we were to substitute, say, π₯ is equal to zero into our function, we would get the inverse tan of zero is equal to π evaluated at zero is equal to our power series where we substitute π₯ equals zero.
00:04:38.430 --> 00:04:49.580
So πΆ plus the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by zero to the power of two π plus one divided by two π plus one.
00:04:49.930 --> 00:04:59.030
And we know we can justify using the substitution since itβs true if the size of negative four π₯ squared is less than one, which is true for a substitution π₯ is equal to zero.
00:04:59.510 --> 00:05:03.200
We see that, in this infinite sum, every term is being multiplied by zero.
00:05:03.200 --> 00:05:06.790
So this is just a sum of zeros and we can conclude is equal to zero.
00:05:07.460 --> 00:05:11.510
Therefore, we can conclude that the inverse tan of zero is equal to πΆ.
00:05:11.840 --> 00:05:15.800
And of course, we know the inverse tan of zero evaluates to just equal to zero.
00:05:16.470 --> 00:05:33.080
Therefore, what we have shown is a power series representation of the inverse tan of two π₯ is equal to the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by π₯ to the power of two π plus one divided by two π plus one.