### Video Transcript

Write the first four nonzero terms of the Maclaurin expansion for π of π₯ is equal to 11π₯ times π to the power of two π₯ in ascending powers of π₯.

In this question, we need to find the first four nonzero terms of the Maclaurin expansion of the function π of π₯ is equal to 11π₯ times π to the power of two π₯ written in ascending powers of π₯. This means that our exponents of π₯ will be increasing. First, letβs recall what the Maclaurin expansion or Maclaurin series for a function π of π₯ is. We recall π of π₯ is equal to the sum from π equals zero to β of the πth derivative of π of π₯ with respect to π₯ evaluated at zero divided by π factorial times π₯ to the πth power. This series is called the Maclaurin series for our function π of π₯.

To find successive terms in this series, we need to find successive derivatives of our function π of π₯. So, one way of finding the Maclaurin series for the function π of π₯ given to us in the question is to start differentiating, and we can do this. For example, we can see that π of π₯ is the product of two functions. Itβs 11π₯ times π to the power of two π₯, so we could differentiate this by using the product rule. However, although this would work and would give us the correct answer, this would get complicated very quickly. It would be very easy to make a mistake. So instead, weβre going to use a trick using what we know about Maclaurin series.

Instead of finding a Maclaurin series for the entire function π of π₯, letβs just find one for π to the power of two π₯. And we can actually find this easily from the Maclaurin series for the exponential function π to the power of π₯. We recall the Maclaurin series for π to the power of π₯ tells us π to the power of π₯ is equal to the sum from π equals zero to β of π₯ to the πth power divided by π factorial. In fact, this is valid for all real values of π₯. This is a very useful series which we should commit to memory. However, we couldβve also just formulated this directly from the definition of a Maclaurin series.

We want to use this to find a power series representation of π to the power of two π₯. Well, we have one for π to the power of π₯. So, what wouldβve happened if instead of having π₯, weβd written two π₯? So by replacing every instance of π₯ with two π₯, we get that π to the power of two π₯ is equal to the sum from π equals zero to β of two π₯ all raised to the πth power divided by π factorial. And because we know that our Maclaurin series for π to the power of π₯ is valid for all real values of π₯, this new power series for π to the power of two π₯ must also be valid for all real values of π₯.

And in fact, thereβs something extra we can determine about this power series for π to the power of two π₯. We, in fact, know the power series representation of a function centered at a point is unique. In fact, it must be equal to the Taylor series of that function. So in actual fact, this power, series representation for π to the power of two π₯ must be its Maclaurin series. So, weβve now found a Maclaurin series for π to the power of two π₯. But remember, π of π₯ is equal to 11π₯ times π to the power of two π₯. So letβs try and find the Maclaurin series for π of π₯.

To do this, weβll multiply our Maclaurin series of π to the power of two π₯ by 11π₯. Doing this, we get π of π₯ which is equal to 11π₯ times π to the power of two π₯ must be equal to 11π₯ times the sum from π equals zero to β of two π₯ raised to the πth power divided by π factorial. But now we need to remember our rules for series. We can treat 11π₯ a lot like a constant. We can just bring this inside of our series. This gives us π of π₯ is equal to the sum from π equals zero to β of 11π₯ times two π₯ raised to the πth power divided by π factorial.

And remember, the power series representation of a function centered at a point must be equal to its Taylor series centered at this point. So, in this case, this power series must be the Maclaurin series of our function π of π₯. So, we can use this to find the first four nonzero terms of our Maclaurin series for π of π₯ in ascending powers of π₯. We just need to find the first four terms of this series. Alternatively, we need to find the sum from π equals zero to three of 11π₯ times two π₯ to the πth power divided by π factorial. Letβs write this out term by term.

Letβs start when π is equal to zero. When π is equal to zero, we get 11π₯ times two π₯ raised to the zeroth power divided by zero factorial. And we can just simplify this expression. First by definition, zero factorial is equal to one. Next, any number raised to the zeroth power is equal to one, so two π₯ all raised to the zeroth power is just equal to one. And, of course, 11π₯ times one divided by one is just equal to 11π₯. So, our first term is just 11π₯.

Letβs now go on to the second term in our series when π is equal to one. We get 11π₯ times two π₯ raised to the first power divided by one factorial, and we can simplify this expression. First, we know one factorial is just equal to one. Next, we know that raising a number to the first power doesnβt change the number. In other words, two π₯ all raised to the first power is just equal to two π₯. So, our second term is 11π₯ times two π₯ divided by one. This is just equal to 22π₯ squared. So weβve simplified the second term to give us 22π₯ squared.

Letβs now move on to the third term in our series when π is equal to two. We get 11π₯ times two π₯ all squared divided by two factorial, and weβll simplify this just as we did before. First, two factorial is just equal to two. Next, by distributing the square over our parentheses, we get that two π₯ all squared is equal to four π₯ squared. So, our second term simplifies to give us 11π₯ multiplied by four π₯ squared all divided by two. We could simplify this to give us 22π₯ cubed.

Finally, we want our fourth term of the series when π is equal to three. We get 11π₯ times two π₯ all cubed divided by three factorial, and we simplify this in the same way we did before. First, three factorial is equal to three times two times one, which is, of course, equal to six. We then distribute the cube over our parentheses. We get two π₯ all cubed is equal to eight π₯ cubed. Finally, we just need to simplify 11π₯ times eight π₯ cubed all divided by six. If we do this, we get 44π₯ to the fourth power divided by three.

Therefore, we were able to show the first four nonzero terms in the Maclaurin expansion of the function π of π₯ is equal to 11π₯ times π to the power of two π₯ in ascending powers of π₯ is given by 11π₯ plus 22π₯ squared plus 22π₯ cubed plus 44π₯ to the fourth power divided by three.