Video: Transcendental Functions as Power Series

The function cos π‘₯ can be represented by the power series βˆ‘_(𝑛 = 0)^(∞) ((βˆ’1)^(𝑛)/(2𝑛)!) π‘₯^(2𝑛). Use the first two terms of this series to find an approximate value of the cos 0.5 to two decimal places.

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Video Transcript

The function cos of π‘₯ can be represented by the power series the sum from 𝑛 equals zero to ∞ of negative one raised to the 𝑛th power divided by two 𝑛 factorial multiplied by π‘₯ to the power of two 𝑛. Use the first two terms of this series to find an approximate value of the cos of 0.5 to two decimal places.

The question tells us that the cos of π‘₯ can be represented by the power series given to us in the question. What this tells us is that, for the values of π‘₯ where the series converges, we have that the cos of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power divided by two 𝑛 factorial multiplied by π‘₯ to the power of two 𝑛. What this means is we can attempt to approximate the cos of π‘₯ by taking a partial sum. And we see that the question wants us to do this for the first two terms of our series to find an approximate value of the cos of 0.5 to two decimal places. So, we want our partial sum to only be two terms.

Substituting in this information gives us that the cos of 0.5 is approximately equal to the sum from 𝑛 equals zero to one of negative one to the 𝑛th power divided by two 𝑛 factorial multiplied by 0.5 to the power of two 𝑛. Our sum starts at 𝑛 equals zero, so the first term in our sum will be negative one to the zeroth power divided by two multiplied by zero factorial all multiplied by 0.5 to the power of two multiplied by zero. And our sum goes up to 𝑛 equals one. So, the second term in our sum is equal to negative one raised to the first power divided by two multiplied by one factorial all multiplied by 0.5 to the power of two multiplied by one.

We’re now ready to start evaluating. Any nonzero number raised to the zeroth power is just equal to one. So, we have negative one raised to the zeroth power is just one. We have that two multiplied by zero is equal to zero, and zero factorial is just equal to one, so the denominator of our first term is equal to one. And just as we had before, 0.5 raised to the zeroth power is just equal to one. Any number raised to the first power is just equal to itself. So negative one raised to the first power is just equal to negative one. Next, looking at the denominator in our second term, we have that two multiplied by one is equal to two. And then two factorial is just equal to two. So, we have a denominator of just two.

Finally, we can see that 0.5 raised to the second power is just equal to a quarter. This gives us one divided by one multiplied by one plus negative one over two multiplied by a quarter. Which we can evaluate to be equal to seven divided by eight, which, to two decimal places, is approximately 0.88. Therefore, we have shown by using the first two terms of our power series the sum from 𝑛 equals zero to ∞ of negative one raised to the 𝑛th power divided by two 𝑛 factorial multiplied by π‘₯ to the power of two 𝑛 that we can approximate the value of the cos of 0.5 to two decimal places as 0.88.

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