In order to make a better approximation the most natural thing to do is to find

# In order to make a better approximation the most

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In order to make a better approximation, the most natural thing to do is to find the simplest available function that has the same value, derivative, and second derivative all at the same time: P ( a ) = f ( a ) P 0 ( a ) = f 0 ( a ) P 00 ( a ) = f 00 ( a ) .
3.2. TAYLOR SERIES APPROXIMATIONS 59 Now we begin solving the IVP with the second derivative held at a constant value: P 00 ( a ) = f 00 ( a ) P 0 ( x ) = Z f 00 ( a ) dx = f 00 ( a ) x + C P ( x ) = Z f 00 ( a ) x + C dx = f 00 ( a ) x 2 2 + Cx + D Now we plug in our two initial values to find C and D . Collecting terms, we get f ( x ) P ( x ) = f 00 ( a ) 2 ( x - a ) 2 + f 0 ( a )( x - a ) + f ( a ) . This can be called either a quadratic approximation or a second-order ap- proximation. Let’s take a moment to notice a pattern. It’s very easy, once we have P ( x ) written down, to check that P ( a ) = f ( a ) , P 0 ( a ) = f 0 ( a ) , P 00 ( a ) = f 00 ( a ) because of the power rule. Notice how the 1 2 up front is ready to cancel with the 2 in the exponent upon differentiation. Let’s move to a third-order approximation. The lower-order terms should be the same, so we only need to think about the new x 3 term. When we differentiate three times, we want to be left with f 000 ( a ). Since ( x 3 ) 000 = 6, we get that P 3 ( x ) = f 000 ( a ) 6 ( x - a ) 3 + f 00 ( a ) 2 ( x - a ) 2 + f 0 ( a )( x - a ) + f ( a ) where we labeled the approximation by the order (i.e. the degree). For derivatives of order higher than 3, we usually write the number in paren- theses, like f (4) ( x ) for the 4th derivative. We can also write d n f dx n . Notice that d n dx n ( x n ) = n ! . So, for an n th-order approximation, we compute P n ( x ) = n X k =0 f ( k ) ( a ) k ! ( x - a ) k . (3.4) where f (0) ( x ) is just f ( x ). These are known as Taylor approximations .
60 CHAPTER 3. APPROXIMATIONS WITH CALCULUS 3.2.1 Examples Example 1: Find the N th order approximation to f ( x ) = e x at x = 0. This one is the easiest! Since d dx e x = e x and e 0 = 1, we have e x N X k =0 1 k ! x k = 1 + x + x 2 2 + x 3 6 + x 4 24 + · · · + x N N ! . (3.5) It’s standard to write approximations like 1 + x + x 2 2 instead of x 2 2 + x + 1 since we are usually considering x to be small, which makes powers of x even smaller. This is just like writing decimals. You will be expected to remember the series for approximations at x = 0 for many basic functions. There will be a list later. Example 2: Find the 4th order approximation to f ( x ) = cos x at x = 0. Before beginning computations, take a quick look at the function. What do we know about cos x ? It’s bounded, it’s even, its derivatives are easy... that’s already plenty to process. Being bounded will be important later, but we’ll set that aside for the moment. The fact that it’s even saves us some work. The coefficients in the approximation on all of the odd terms must be 0 since otherwise we’d have even + odd which is neither even nor odd. We want an even approximation to our even function. This pans out if you go through mechanically since sin 0 = 0 (check). The “easy derivatives” comment was our best observation: d dx cos x = - sin x d 2 dx 2 cos x = - cos x d 3 dx 3 cos x = sin x d 4 dx 4 cos x = cos x d n dx n cos x = d n - 4 dx n - 4 cos x for n > 4 .