AP Calculus BC Exam, 2004 BC 6, part (a)
Here, they ask you to find series for $\sin(5x+\frac{\pi}{4})$. The solution makes perfect sense to me; just a traditional method to find the series. But, can we also just use known series expansion of $\sin(x)$ and replace $x$ with $5x+\frac{\pi}{4}$ as well? I know this wouldn't help much in part (b), but I'm just wondering if this is also legit way to find the series. Here's an example:
AP Calculus BC Exam, 2011 BC 6, part (a)
Here, instead of finding the series manually, they just use known series of sin(x) and use it.
I'm asking bc if I replaced $x$ with $5x+\frac{\pi}{4}$, then i get totally different series. But graphing calculator says it's still legit way (although not identical to the polynomial in the solution) to estimate the given function.
The series using replacing:$$5x+\frac{\pi}{4}-\frac{(5x+\frac{\pi}{4})^3}{3!}$$The series using traditional way:$$\frac{\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}x-\frac{25\sqrt{2}}{2(2!)}x^2-\frac{125\sqrt{2}}{2(3!)}x^3$$
If the "replacing method" is wrong, how do I know I can replace or not? Is there any well-explained document? I'm using Stewart's textbook and I can't find any.
If the "replacing method" is right, does it mean it's possible to have multiple Taylor polynomials to estimate a same function?
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$\begingroup$Yes, the "replacing method" works in all cases, but you cannot use it the same way you use the "traditional way" with regards to approximation.
If you are using an approximation as you are here, the "traditional way" approximates $\sin(5x+\frac{\pi}{4})$ using $x^0,x^1 ,x^2, x^3$ and their true coefficients, but to get these true coefficients for the "replacing method" is difficult.
For example, using the replacing method each term will contribute to the coefficient of $x^1$, because each term will have $x^1$ as part of it and therefore $5x+\frac{\pi}{4}-\frac{(5x+\frac{\pi}{4})^3}{3!}$ does not have the true coefficients of $x^0,x^1 ,x^2, x^3$
Using the replacing method, the true coefficient of $x^1$ term would be $$5\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(2n+1)(\frac\pi4)^{2n} = 5\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(\frac\pi4)^{2n}$$which is equal to $$\frac{5\sqrt{2}}{2}$$that you obtained from the "traditional way." You can see that if you evaluate the above sum from just n = 0 to n = 2, you are already within 5 decimal places of the actual sum.
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