Video: Find the Maclaurin Sereis for the Hyperbolic Cosine Function

Find the Maclaurin series of cosh 2π‘₯ = (𝑒^(2π‘₯) + 𝑒^(βˆ’2π‘₯))/2.

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Video Transcript

Find the Maclaurin series for the function the hyperbolic cos of two π‘₯ which is equal to 𝑒 to the power of two π‘₯ plus 𝑒 to the power of negative two π‘₯ all divided by two.

We recall that for an infinitely differentiable function 𝑓 of π‘₯, the Maclaurin series of 𝑓 of π‘₯ is equal to 𝑓 evaluated at zero plus the first derivative of 𝑓 evaluated at zero multiplied by π‘₯ plus the second derivative of 𝑓 evaluated at zero divided by two factorial multiplied by π‘₯ squared. And this sum goes on infinitely, giving us the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 evaluated at zero divided by 𝑛 factorial multiplied by π‘₯ to the 𝑛th power.

We might be tempted at this point to define 𝑓 of π‘₯ to be the cosh of two π‘₯ and then differentiate this function and try to find a pattern for the 𝑛th derivative evaluated at zero. And this would work. However, a slightly simpler method would be to use the cosh of π‘₯ and then substitute in to π‘₯ at the end. So let’s set our function 𝑓 of π‘₯ to be equal to the cosh of π‘₯, which is equal to 𝑒 to the power of π‘₯ plus 𝑒 to the power of negative π‘₯ all divided by two.

To find the first derivative of our function 𝑓 of π‘₯, we differentiate each term separately. First, we notice that the derivative of 𝑒 to the power of π‘₯ all divided by two is just equal to itself, 𝑒 to the power of π‘₯ all divided by two. Similarly, the derivative of 𝑒 to the power of negative π‘₯ all divided by two is negative 𝑒 to the power of negative π‘₯ all divided by two. We sometimes call this function the hyperbolic sin of π‘₯ or sinh π‘₯.

Now, to find the second derivative of our function 𝑓 of π‘₯, we take the derivative of our first derivative. So that’s the derivative of our function, the hyperbolic sign of π‘₯. The derivative of 𝑒 to the power of π‘₯ all divided by two is just equal to itself. And when we differentiate our second term, which has an 𝑒 to the power of negative π‘₯, we’ll get two negative ones, giving us plus 𝑒 to the power of negative π‘₯ all divided by two. So we have our second derivative of our function 𝑓 of π‘₯ with respect to π‘₯ is equal to the hyperbolic cos of π‘₯.

Therefore, what we have shown is if we differentiate our function 𝑓 of π‘₯ an odd number of times, we will get the hyperbolic sin of π‘₯. And if we differentiate our function 𝑓 of π‘₯ an even number of times, we will get the hyperbolic cos of π‘₯. We’re now ready to start evaluating these functions at zero. We have that our function 𝑓 evaluated at zero is equal to the hyperbolic cos evaluated at zero is equal to 𝑒 to the zeroth power plus 𝑒 to the negative zeroth power all divided by two. And since 𝑒 to the zeroth power and 𝑒 to the negative zeroth power are both just equal to one, we just have one plus one all divided by two which is equal to one.

We can then do the same to find the value of our first derivative function evaluated at zero, which is equal to hyperbolic sin evaluated at zero, which is just equal to zero. Now, using the fact that if we differentiate our function 𝑓 of π‘₯ an odd number of times with respect to π‘₯, we get the hyperbolic sin of π‘₯. Then if we evaluate this π‘₯ is equal to zero, we will just get zero. Similarly, since we’ve shown if we differentiate our function 𝑓 an even number of times, we’ll get the hyperbolic cos of π‘₯. If we evaluate this when π‘₯ is equal to zero, we’ll just get one.

In particular, what we’ve shown here is that for odd 𝑛 in our Maclaurin series we would just get zero in the series. So we can leave these out. So using our definition of Maclaurin series. We have the hyperbolic cos of π‘₯ is equal to the sum from 𝑛 equal zero to ∞ of the 𝑛th derivative evaluated at zero divided by 𝑛 factorial multiplied by π‘₯ to the 𝑛th power. We can split this series into two series, the first giving us all of the odd values of 𝑛 and the second one giving us all the even values of 𝑛.

As we discussed previously, every coefficient in our first series is equal to zero. So we can just remove the series. And in our second series, for our even values of 𝑛, we have that the evaluations of our function at zero is always equal to one. Giving us the Maclaurin series for the hyperbolic cos of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of two 𝑛 divided by two 𝑛 factorial.

Now, we want to use our Maclaurin series for the hyperbolic cos of π‘₯ to find a Maclaurin series for the hyperbolic cos of two π‘₯. And we can do this by just substituting two π‘₯ into our series. This gives us that the hyperbolic cos of two π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of two π‘₯ raised to the power of two 𝑛 all divided by two 𝑛 factorial. Therefore, we have shown that the Maclaurin series for the hyperbolic cos of two π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of two π‘₯ raised to the power of two 𝑛 all divided by two 𝑛 factorial.

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