Video: Differentiating Power Series

Find 2𝑦″ βˆ’ 7𝑦′ + 5𝑦 given 𝑦 = 1 + (π‘₯/1!) + (π‘₯Β²/2!) + (π‘₯Β³/3!) + ....

02:59

Video Transcript

Find two 𝑦 double prime minus seven 𝑦 prime plus five 𝑦 given that 𝑦 is equal to one plus π‘₯ divided by one factorial plus π‘₯ squared over two factorial plus π‘₯ cubed over three factorial and the sum continues indefinitely.

The question gives us 𝑦 as a power series of π‘₯. And it wants us to find an expression involving the derivative of 𝑦 and the second derivative of 𝑦. We might be tempted at this point to use our rules for differentiating a power series. However, there’s a simpler method in this case. We start by writing 𝑦 out term by term. We want to write this out in sigma notation. We can start by noticing that the powers of π‘₯ seem to be increasing by one each term. In fact, we can see this is true for all of our terms. π‘₯ is equal to π‘₯ to the first power and π‘₯ to the zeroth power is equal to one.

Next, we notice our denominators seem to be factorials which increase by one each time. In fact, these values seem to be the same as our exponents. We can continue this pattern for our first time as well by noticing that zero factorial is equal to one. Using this, we can change our first term to π‘₯ to the zeroth power divided by zero factorial. And now, we can see that 𝑦 is equal to the sum from 𝑛 is equal to zero to infinity of π‘₯ to the 𝑛th power divided by 𝑛 factorial. And this is the point we might be tempted to differentiate our power series term by term and this would work.

We could do this once to get a power series for our first derivative, and then we could differentiate this again to get a power series for our second derivative. However, our power series for 𝑦 is actually a special power series. We recall that the power series for 𝑒 to the power of π‘₯ is equal to the sum from 𝑛 is equal to zero to infinity of π‘₯ to the 𝑛th power divided by 𝑛 factorial for all real values of π‘₯. And this is an important power series which we should commit to memory.

In fact, we can see that the power series for 𝑦 is equal to the power series for 𝑒 to the power of π‘₯. So because our power series is true for all of the real values of π‘₯, we must have that 𝑦 is equal to 𝑒 to the power of π‘₯. And we know if 𝑦 is equal to 𝑒 to the power of π‘₯, then our first derivative of 𝑦 is equal to 𝑒 to the power of π‘₯ and our second derivative of 𝑦 is equal to 𝑒 to the power of π‘₯. So we can evaluate the expression two 𝑦 double prime minus seven 𝑦 prime plus five 𝑦 by using the fact that 𝑦 double prime, 𝑦 prime, and 𝑦 are all equal to 𝑒 to the power of π‘₯.

So this gives us two times 𝑒 to the power of π‘₯ minus seven 𝑒 to the π‘₯ plus five 𝑒 to the π‘₯, which we can evaluate to just give us zero. Therefore, we’ve shown if 𝑦 is equal to one plus π‘₯ over one factorial plus π‘₯ squared over two factorial plus π‘₯ cubed over three factorial and this sum goes on indefinitely, then two 𝑦 double prime minus seven 𝑦 prime plus five 𝑦 is just equal to zero.

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