Video: Find the Maclaurin Series for the Exponential Function and Find Its Radius of Convergence

Consider the function 𝑓(π‘₯) = 𝑒^(π‘₯). Find 𝑓′(π‘₯). Find 𝑓^(𝑛) (π‘₯), where 𝑓^(𝑛) (π‘₯) represents the 𝑛th derivative of 𝑓 with respect to π‘₯. Hence, derive the Maclaurin series for 𝑒^(π‘₯). What is the radius of convergence 𝑅 of the Maclaurin series for 𝑒^(π‘₯)?

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯. Find the function 𝑓 prime of π‘₯. Then, find the function 𝑓 𝑛 of π‘₯, where 𝑓 𝑛 of π‘₯ represents the 𝑛th derivative of 𝑓 with respect to π‘₯. Hence, derive the Maclaurin series for 𝑒 to the π‘₯. What is the radius of convergence 𝑅 of the Maclaurin series for 𝑒 to the π‘₯?

The first part of our question is asking us to find the function 𝑓 prime of π‘₯ where 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯. We know that the function 𝑓 prime of π‘₯ is just the derivative with respect to π‘₯ of our function 𝑓, which we’re told is 𝑒 to the π‘₯. And we know the derivative of 𝑒 to the π‘₯ with respect to π‘₯ is just equal to 𝑒 to the π‘₯ itself. So we have shown that the function 𝑓 prime of π‘₯ is just equal to 𝑒 to the power of π‘₯. Next, the question asks us to find the function 𝑓 𝑛 of π‘₯, where 𝑓 𝑛 of π‘₯ represents the 𝑛th derivative of 𝑓 with respect to π‘₯. If 𝑓 𝑛 of π‘₯ represents the 𝑛th derivative of 𝑓 with respect to π‘₯, then 𝑓 one of π‘₯ represents the first derivative of 𝑓 with respect to π‘₯. And the first derivative of of 𝑓 with respect to π‘₯ is just the derivative of 𝑒 to the π‘₯ with respect to π‘₯, which is just 𝑒 to the π‘₯.

Now, 𝑓 two of π‘₯ will be the second derivative of 𝑓 with respect to π‘₯, which is just equal to the derivative of 𝑓 one with respect to π‘₯. And we showed above that 𝑓 one of π‘₯ was just equal to 𝑒 to the π‘₯. So 𝑓 two of π‘₯ is the derivative with respect π‘₯ of 𝑒 to the π‘₯. And, of course, we know that the derivative with respect to π‘₯ of 𝑒 to the π‘₯ is just equal to 𝑒 to the π‘₯. We can see that this pattern will carry on. 𝑓 𝑛 of π‘₯ will be the derivative of 𝑓 𝑛 minus one of π‘₯. And if we choose the value of 𝑛 greater than or equal to two, then our function 𝑓 𝑛 minus one of π‘₯ came from the line before and it’s just equal to 𝑒 to the π‘₯. Therefore, we have that 𝑓 𝑛 of π‘₯ is equal to the derivative of 𝑒 to the π‘₯ with respect to π‘₯, which is just equal to 𝑒 the π‘₯. So we’ve just shown that 𝑓 𝑛 of π‘₯ is equal to 𝑒 to the power of π‘₯.

Now, the question is asking us to derive the Maclaurin series for 𝑒 to the π‘₯. We recall that for a function 𝑓 of π‘₯, the Maclaurin series gives us that 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 plus the third derivative of 𝑓 evaluated at zero divided by three factorial multiplied by π‘₯ cubed. And we keep going and going. We can actually simplify this slightly by using the notation given to us in the question. That 𝑓 𝑛 of π‘₯ represents the 𝑛th derivative of 𝑓 with respect to π‘₯. So we could rewrite 𝑓 prime of zero multiplied by π‘₯ as 𝑓 of one evaluated at zero multiplied by π‘₯ all divided by one factorial.

We could then do the same with the third term in our series to get 𝑓 two evaluated at zero multiplied by π‘₯ squared all divided by two factorial. And we could do the exact same thing for all the later terms in our series. Now, we just have to change the first term in our series. And we can do this by letting 𝑓 zero of π‘₯ be equal to 𝑓 of π‘₯. Then the first term in our Maclaurin series will be equal to 𝑓 of zero evaluated at zero divided by zero factorial. This is because zero factorial is just equal to one. We see that each term in our series seems to have an increasing power of π‘₯. If we change the π‘₯ in our second term to be an π‘₯ to the first power, then we can make this trend continue. In fact, since π‘₯ to the zeroth power is just equal to one, we can make this trend true for all of our terms.

So what we have shown is that we can rewrite our Maclaurin series for 𝑓 of π‘₯ as the sum from 𝑛 equals zero to infinity of 𝑓 𝑛 evaluated at zero multiplied by π‘₯ to the 𝑛th power all divided by 𝑛 factorial. So let’s try to evaluate this new series that we have for our Maclaurin series of 𝑒 to the π‘₯.

We showed previously that for 𝑛 greater than or equal to one, our 𝑛th derivative function is just 𝑒 to the power of π‘₯. So if 𝑛 greater than or equal to one, we have 𝑓 𝑛 of zero is equal to 𝑒 to the zeroth power. And, of course, we can evaluate 𝑒 to the zeroth power to just be equal to one. This just leaves us to evaluate 𝑓 zero at zero. Well, we defined 𝑓 zero to just be the function 𝑓 of π‘₯. So this is equal to 𝑓 evaluated at zero. And, of course, our function 𝑓 is just equal to 𝑒 to the π‘₯. So we get this is all equal to 𝑒 to the zeroth power. And we evaluate this to just get one.

So what we have just shown is that 𝑓 𝑛 evaluated at zero is equal to one for all 𝑛 greater than or equal to zero. So because we just showed that 𝑓 𝑛 evaluated at zero is equal to one whenever 𝑛 is greater than or equal to zero, we can rewrite the numerator in our series as just π‘₯ to the 𝑛th power. This gives us the sum from 𝑛 equals zero to infinity of π‘₯ to the 𝑛th power divided by 𝑛 factorial. Therefore, we have shown that our Maclaurin series for our function 𝑒 to the π‘₯ is equal to the sum from 𝑛 equals zero to infinity of π‘₯ to the 𝑛th power divided by 𝑛 factorial.

The last part of our question asks us to find the radius of convergence 𝑅 of the Maclaurin series for 𝑒 to the π‘₯. To find the radius of convergence of the Maclaurin series for 𝑒 to the π‘₯, we recall the ratio test. Which says that if the limit as 𝑛 approaches infinity of the absolute value of the ratio of successive terms is less than one. Then the sum from 𝑛 equals zero to infinity of π‘Ž 𝑛 converges absolutely. Since we want to calculate the radius of convergence for our Maclaurin series of 𝑒 to the π‘₯, we’ll set our sequence π‘Ž 𝑛 to be equal to π‘₯ to the 𝑛th power divided by 𝑛 factorial. So we now need to evaluate the limit as 𝑛 approaches infinity of the absolute value of the ratio of the successive terms.

First, we know that π‘Ž 𝑛 plus one is just equal to π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one factorial. And we divide this by π‘Ž 𝑛 which is equal to π‘₯ to the 𝑛th power divided by 𝑛 factorial. Instead of dividing one fraction by another fraction, we can flip the second fraction and multiply. Flipping the second fraction and multiplying gives us π‘₯ to the power of 𝑛 plus one multiplied by 𝑛 factorial divided by 𝑛 plus one factorial multiplied by π‘₯ to the 𝑛th power. We can now start removing shared factors from the numerator and the denominator. If we remove the shared factors of π‘₯ from both the numerator and the denominator, then we end up with one factor of π‘₯ left in the numerator.

At this point, we can notice that 𝑛 plus one factorial is actually equal to 𝑛 plus one multiplied by 𝑛 factorial. So if we remove the shared factor of 𝑛 factorial in the numerator and the denominator, then we end up with just a factor in the denominator of 𝑛 plus one. This gives us the limit as 𝑛 approaches infinity of the absolute value of π‘₯ divided by 𝑛 plus one. At this point, we can notice that changing 𝑛 does not change the value of π‘₯, i.e., π‘₯ is a constant with respect to 𝑛. So we can just take π‘₯ out of our limit. Giving us the absolute value of π‘₯ multiplied by the limit as 𝑛 approaches infinity of the absolute value of one divided by 𝑛 plus one.

Finally, as 𝑛 is getting larger and larger, one divided by 𝑛 plus one is getting smaller and smaller. In fact, this limit is equal to zero. So we get zero multiplied by the absolute value of π‘₯, which is just equal to zero. Since we have that the limit as 𝑛 approaches infinity of the absolute value of the ratio of successive terms in our Maclaurin series is equal to zero for any value of π‘₯. By the ratio test, we must have that the Maclaurin series for 𝑒 to the π‘₯ converges for all values of π‘₯. And so, the radius of convergence 𝑅 is equal to plus infinity.

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