# Question Video: Transcendental Functions as Power Series Mathematics • Higher Education

The function sin 𝑥 can be represented by the power series ∑_(𝑛 = 0)^(∞) ((−1)^(𝑛)/(2𝑛 + 1)!) 𝑥^(2𝑛 + 1). Use the first two terms of this series to find an approximate value of the sin 0.5 to 2 decimal places.

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### Video Transcript

The function sin 𝑥 can be represented by the power series the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power divided by two 𝑛 plus one factorial times 𝑥 to the power of two 𝑛 plus one. Use the first two terms of this series to find an approximate value of the sin of 0.5 to two decimal places.

The question tells us that the function the sin of 𝑥 can be represented by the following power series. We need to use the first two terms of this series to find an approximation of the sin of 0.5 to two decimal places.

First, we’re told the following power series represents the function the sin of 𝑥. It’s worth pointing out that, in fact, this is the Maclaurin series for the sin of 𝑥. And we know this is valid for all real values of 𝑥. But we don’t need this piece of information to answer our question. Instead of using an infinite series , we can use a finite number of terms of this series to approximate the sin of 𝑥.

The question tells us to approximate the sin of 𝑥 by using the first two terms of this series. Since our series starts at 𝑛 is equal to zero and we want two terms in our series, the values of 𝑛 should range from zero to one. So now we have the sin of 𝑥 is approximately equal to the sum from 𝑛 equals zero to one of negative one to the 𝑛th power divided by two 𝑛 plus one factorial times 𝑥 to the power of two 𝑛 plus one. In other words, the sin of 𝑥 is approximately equal to the sum of the first two terms of our series.

Let’s now expand our series. Our series only has two terms. To find these two terms, we need to substitute 𝑛 is equal to zero into our summand and then add onto this the value we get substituting 𝑛 is equal to one. Let’s start with finding the first term of this series expansion. That’s when 𝑛 is equal to zero. We get negative one to the zeroth power divided by two times zero plus one factorial multiplied by 𝑥 to the power of two times zero plus one.

And we do exactly the same to find the second term in our series. This time, we substitute 𝑛 is equal to one into our summand. This gives us negative one to the first power divided by two times one plus one factorial multiplied by 𝑥 to the power of two times one plus one. And we can evaluate both of these expressions.

First, negative one raised to the zeroth power is equal to one. Next, we have two times zero plus one is equal to one. This appears twice in our expression: once in the denominator as a factorial and as the exponent of 𝑥. So our first term simplifies to give us one divided by one factorial multiplied by 𝑥 to the first power. Well, one factorial is equal to one, and 𝑥 to the first power is just equal to 𝑥. So our first term is just 𝑥.

We can do something similar to evaluate our second term. First, negative one raised to the first power is just equal to negative one. Next, we have two times one plus one is equal to three. And again, this appears twice in our expression: once in the denominator as a factorial and once as the exponent of 𝑥. So our second term simplifies to give us negative one divided by three factorial times 𝑥 cubed. And remember, three factorial means we take three, multiply it by two, and then multiply it by one. This gives us six.

So what we’ve shown is the sin of 𝑥 is approximately equal to 𝑥 minus one-sixth 𝑥 cubed. And we can then use this to approximate the value of the sin of 0.5. We just substitute 𝑥 is equal to 0.5 into our expression. Substituting 𝑥 is equal to 0.5, we get that the sin of 0.5 is approximately equal to 0.5 minus one-sixth times 0.5 cubed. And if we calculate this to two decimal places, we get 0.48.

Therefore, by using the first two terms of the Maclaurin series for the sin of 𝑥, we were able to show that the sin of 0.5 is approximately equal to 0.48.