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In binary you get the number $\underbrace{11\dots 11}_n$ which is the number that goes before $\underbrace{100\dots 00}_{n+1}=2^{n+1}$


Oops. The answer is trivially no - the functions are not linearly independent! (I've been thinking about why one of them is not in the closed span of the others; in fact one of them is in the span of the others.) If you're allowing $m=0$ and $n=0$ then $(-1) + (x) + (1-x)$ is a counterexample (all but three of the coefficients vanish). If you're assuming ...


HINT: $$z\cdot S=z\sum_{k=0}^\infty \frac{2^{2k}z^{2k-1}}{(2k)!}=z\sum_{k=0}^\infty\dfrac{(2z)^{2k}}{(2k)!}$$ Now $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$ $2\cosh(y)=e^y+e^{-y}=?$


\begin{align} S&=1+2+4+\dotsb+2^n\\ 1+S&=1+1+2+4+\dotsb+2^n\\ &=2+2+4+\dotsb+2^n\\ &=4+4+\dotsb+2^n\\ &=8+\dotsb+2^n\\ &\dotsb\\ &=2^n+2^n\\ 1+S&=2^{n+1}\\ S&=2^{n+1}-1 \end{align} Write it out for a specific example to understand this better.


One can write $$S=1+2+\cdots+2^n,$$ and note that $$S=2S-S.$$ Hence, since $$2S=2(1+2+\cdots+2^n)=2+4+\cdots+2^{n+1}=S-1+2^{n+1},$$ we have $$S=2S-S=S-1+2^{n+1}-S=2^{n+1}-1.$$


The series you present is not a Taylor series for G(z) about the origin. A Taylor series would be of the form $G(x)= a_0+a_1 x+a_2 x^2+...$. It would be easier to compute the radius of convergence of $1-2 G(x)=(1-4abz^2)^{-1/2}$ as it's the same radius as for $G(x)$. A special Taylor series discovered by Newton, is the generalized binomial theorem : If ...


Consider the series for $\cosh (2z)$ and divide it by $z$....


Let me just add one thought: We can view this as looking at the real analytic $f(x,y) = \sum a_{mn}x^my^n$ along the line $(t,1-t), t \in \mathbb R.$ Can such an $f$ vanish on a line without being identically $0,$ i.e., without having all $a_{mn} = 0?$ Sure, happens all the time. For example the function $xy$ vanishes on the axes. It's thus clear that for ...

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