Now let's look at one more example where I'll use the technique of the shifting of index. But the problem is, for example, identify the following power series f of x given by the power series is the sum of a_n x^n satisfying f of 0. F of 0 is equal to minus 1. The coefficient satisfies the sum of the 1 to infinity of 2n a_n, x^n minus 1. Process sum of 0 to infinity of a_n x^n is equal to 0 for all real x. We'll like to identify to such a series in order to figure out what this identity mean. The first series has a term x^n minus 1. Second series has a term x^n. They are different so definitely we need the shifting of index. Look at this one, the first one. Raise those n by 1 then it will be a 2 times of n plus 1 times a_n plus 1 times x^n. To compensate that, lower this summation range by 1 so it'll be this one. They are the same thing by the shifting of index, and add this one. In fact, we get the summation 0 to infinity of 2 times of n plus 1 times a_n plus 1 plus a_n and x^n. This is identically 0 for all real values of x. That means what? Through the identity, that implies 2 times of n plus 1 times a_n plus 1 plus a_n. That is equal to 0 for all non-negative integer n. Or you can rewrite it as simply a_n plus 1 is equal to minus 1 over 2 times of n plus 1 times a_n for all n greater than equal to 0. Or rather we can write it as a_n is equal to minus 1 over 2n times a_n minus 1 for all n greater than equal to 0. Apply the same formula to the a's of n minus 1 then this will be minus 1 over 2_n. A_n minus 1 will be minus 1 over 2 times of n minus 1 times a_n minus 2. Keep going. Do the same thing. Then you will get, this is minus 1 over 2n times minus 1 over 2 times of n minus 1, dot, dot, dot minus 1 over. Finally, we will end by a_0. Then what? We look at the pattern. If we have a_n minus 1, then you have minus 1 over 2 times of n. Here we have a_0 then minus 1 over 2 times 1. You can write it as, how many negative 1s do we have? We have 1, 2, n minus 1, and n. That means we have n over minus 1, so minus 1^n. How many 2s do we have? 2^n. Now the remaining part will be 1 times, 2 times, and so on, finally n. That is n factorial. Times a_0. Because I'm lowering this n by 1, so it will be true for n greater than equal to 1. You get this identity again. What happen if n is equal to 0? If n is equal to 0, then minus 1^0 over 2^0 times 0 factorial, that is 1. We get a_0 is equal to a_0. What does that mean? That means this identity holds also for n is equal to 0. In fact, this identity holds for all n, including 0. Then what? Plugging this information into the original power series. The power series that we have is our f of x is equal to sum of 0 to infinity of a_n, x^n. Now using this, each a_n is equal to minus 1^n times 2^n, n factorial, times a_0, then times x^n. That is f of x. Now let's use this to the information. F of 0 is equal to negative 1. Negative 1 is equal to f of 0. Computing from this power series means when n is equal to 0, that is equal to a_0. Finally, f of x is equal to, a_0 is equal to negative 1, so this is the sum of the 0 to infinity of minus 1^n over 2^n, n factorial x^n. Can you recognize it? Let me write it here. Sum of 0 to infinity of 1 over n factorial. You have negative x over 2 and to the n. That's f of x. I hope you can recognize this power series. Look at this power series and think it over. This is exactly the Taylor series expansion of this negative and times e^negative x over 2. That's it, all right. That's the function f of x defined by the power series.
