Finding Taylor series using known series expansion.

AP Calculus BC Exam, 2004 BC 6, part (a)

https://www.dropbox.com/s/8w60eyuc3smwasw/2004%20BC%206.pdf?dl=0

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) https://www.dropbox.com/s/ibtf8tah3dkajt0/2011%20BC%206.pdf?dl=0

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?

• If I understand what you're doing correctly, note that $\sin\left(x\right)$ does not normally have the same value as $\sin\left(5x + \frac{\pi}{4}\right)$ for any particular $x$, so I wouldn't expect the Taylor series, expressed in terms of $x$ for each, to be the same. However, for each case, if you plug a value of $x$ into the Taylor expansion, it should approximate the appropriate value of $\sin$, whether that be of $5x + \frac{\pi}{4}$ or just $x$. – John Omielan Feb 14 at 8:18

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.
• interesting. So although third-degree polynomials would be different depending on which "method" I use, the $\infty$-degree polynomials would be eventually the same when all terms are expanded and evaluated? – Harry Hong Feb 14 at 9:58
• Technically the third-degree polynomials are the same. $$5x+\frac{\pi}{4}-\frac{(5x+\frac{\pi}{4})^3}{3!}$$ is the not correct third-degree polynomial as you have truncated the series and lost subsequent terms in the series which contribute to $x^0,x^1,x^2,x^3$ terms. – Akash Patel Feb 14 at 16:15