Need help with the general formula for the Taylor series of $e^x+\sin(x)$ A while ago, I asked about ways to derive the sigma notation for the infinite series of $e^x+sin(x)$.
$$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$$
In this thread,
Clive Newstead brilliantly gave me two formulas.
There, he said that the pattern of the series, regarding the coefficient of the numerators, is as followed: 1,2,1,0,1,2,1,0
He said that the pattern is "is periodic with period 4"
This comment still puzzles me. I have tried to validate his formula and it is true, they are all correct, but I don't know why he can come up with succinct representation of the pattern:
$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$
How does one realize that the power is related to $4k, 4k+1$ and $4k+2$ and $4k+3$?
This seems to relate to finding the general formula for a sequence or series of number, for example, trivially
$1, 3, 5, 7, ...$ can be represented as $2k+1$
$0, 2, 4, 6, ...$ can be represented as $2k$
with $k$ as the position of the term in the sequence.
So my thought is you to count the terms that have the numerator's coefficient as $1$, $2$ and $0$
The ones that have $1$ is $1^{st}$, $3^{sd}$, $5^{th}$, $7^{th}$, etc.
The ones that have $2$ is $2^{sd}$, $6^{th}$, $10^{th}$, $14^{th}$, etc.
The ones that have $0$ is $4^{th}$, $8^{th}$, $12^{th}$, $16^{th}$, etc.
So the difference between the positions of the terms that contain the coefficient $1$ is $2$
So the difference between the positions of the terms that contain the coefficient $2$ is $4$
So the difference between the positions of the terms that contain the coefficient $0$ is $4$
I find that the general term for terms that contain $1$ is $2k+1$
I find that the general term for terms that contain $2$ is $4k-2$
I find that the general term for terms that contain $0$ is $4k$
But I don't know how to derive these into $4k$, $4k+1$, $4k+2$, $4k+3$
I realize that this is similar to finding patterns of a sequence on an IQ test, but I don't know how to derive it. One my friend said I should look into Lagrange's interpolation, but I don't why this technique has anything to do with writing down the general terms for sequences and series.
Could you help me on this?
Methinks, this is a simple question. I am not good at writing down general term of a sequence yet.
 A: $e^x = 1 + x + \frac {x^2}{2} + \frac {x^3}{3!} + \cdots$
It seem pretty natural to write this as
$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}$
But we could look at pairs of terms.
$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{2n}}{2n!}+\frac {x^{2n+1}}{(2n+1)!}\right)$
Or, even triples of terms.
$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{3n}}{3n!}+\frac {x^{3n+1}}{(3n+1)!}+\frac {x^{3n+2}}{(3n+3)!}\right)$
And do the same thing with $\sin x$
$\sin x = \sum_\limits{n=0}^\infty (-1)^n \frac {x^{2n+1}}{(2n+1)!} = \sum_\limits{n=0}^\infty \left(\frac {x^{4n+1}}{(4n+1)!} - \frac {x^{4n+3}}{(4n+3)!}\right)$
Now find a representation of $e^x$ that plays nicely with our representation of $\sin x$ add them together and you have what you show above.
A: Remember that for the series centered around the point b, the coefficient $a_{n} = \frac{f^{(n)}(b)}{n!}$. So if $f^{(4)}(x) = f(x)$ then  $f^{(n)}(b)$ is periodic with not necessarily prime period 4.
This is why you can be sure that the $f^{(n)}(b)$ is periodic with period 4 in your case.
Let $f(x) = e^{x}$, then $f'(x) = e^{x}$, and so continuing to take derivatives $f^{(4)}(x)=e^{x}$.
Let $g(x) = sin(x)$ then $g''(x)=-sin(x)$ and so $g^{(4)}(x) = sin(x)$
So for $h(x)=f(x) + g(x)$ we have $$h^{(4)}(x) = (f(x)+g(x))'''' = f^{(4)}(x) + g^{(4)}(x) = f(x) + g(x) = h(x)$$.
In this way you can see that the $f^{(n)}(b)$ has the period 4 structure.
A: The mod $4$ periodicity can perhaps be most easily understood using the formula $e^{ix}=\cos x+i\sin x$, so that $\sin x={e^{ix}-e^{-ix}\over2i}$, which gives
$$e^x+\sin x={1\over2i}\sum_{n=0}^\infty{(2i+i^n-(-i)^n)x^n\over n!}$$
Since $i^4=1$, the coefficients $2i+i^n-(-i)^n$ cycle through the values
$$\begin{align}
2i+i^0-(-i)^0&=2i+1-1=2i\\
2i+i^1-(-i)^1&=2i+i-(-i)=4i\\
2i+i^2-(-i)^2&=2i-1-(-1)=2i\\
2i+i^3-(-i)^3&=2i-i-i=0
\end{align}$$
