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Is there two sequences of real numbers $a_i$ and $b_i\neq 8$, not depending on $x$, such that $x^8=\sum_{k=1}^{\infty}a_kx^{b_k}$ for all $x$?

If $\displaystyle\sum_{k=1}^{\infty}a_kx^{b_k}=\sum_{k=1}^{\infty}c_kx^{d_k}$ for all $x>1$, and all coefficients are real and $b_k>b_{k+1}$, and $d_k>d_{k+1}$, is there a way to prove that $a_k=c_k$ and $b_k=d_k$ for all $k$?


If $\displaystyle\sum_{k=1}^{\infty}a_kx^{b_k}=0$ for all $x>1$ and $b_i\neq b_j$, and all coefficients are real, must $a_k=0$?

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For the second: what happens if you perform the usual algorithm for generating Taylor series on $x^8$? –  J. M. Nov 20 '11 at 16:27
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Well, "Taylor series" usually refers to power series in which the exponents are integers, so you should probably edit your title to avoid confusions. Two convergent power series series (with non-negative integral exponents) which converge to the same function are equal term by term; this is easily proved by considering the iterated derivatives, for example. –  Mariano Suárez-Alvarez Nov 20 '11 at 16:28
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Also, your first sentence is quite confusing: what exactly is the condition? –  Mariano Suárez-Alvarez Nov 20 '11 at 16:31
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@MarianoSuárez-Alvarez Ok, if we allow non-integers, still only 1 series? –  user1708 Nov 20 '11 at 16:32
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@J.M. Then you get an x^8 term? –  user1708 Nov 20 '11 at 16:44
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2 Answers

Note that these questions are the same as asking whether non-zero $a_i$, $b_i$ can satisfy $\sum_{k=1}^\infty a_kx^{b_k}=0$ for all $x$ (or for $x>1$). This assumes the series converge absolutely, so that we can combine them. If we're not assuming absolute convergence, then can prove absolute convergence assuming that the $b_i$ are positive (this can be relaxed a bit, but not done away with).

Assuming the $b_i$ are strictly decreasing, we can answer it as follows. Assume the series converges absolutely for $x\ge A$ (that such an $A> 1$ exists will be shown at the end). Without loss of generality, assume $a_1\ne 0$. For $x\ge A$, we have $$0=a_1 x^{b_1}+\sum_{n=2}^\infty a_nx^{b_n}$$ So we solve for $a_1$ and take absolute values $$|a_1|\le x^{-b_1}\sum_{n=2}^\infty |a_n|x^{b_n}=x^{-b_1+b_2}\sum_{n=2}^\infty |a_n|x^{b_n-b_2}$$ Since the $b_i$ are strictly decreasing, both $-b_1+b_2$ and $b_n-b_2$ are negative (or zero, for $n=2$), thus for $x\ge A$, $x^{b_n-b_2}\le A^{b_n-b_2}$. Therefore, for $x\ge A$, $$|a_1|\le x^{-b_1+b_2}\sum_{n=2}^\infty |a_n|A^{b_n-b_2}$$ Since $x$ can be arbitrarily large, this implies that $a_1=0$. Contradiction.

We now prove the existence of such an $A$. Assume the series converges at $x_0$. Fix $\epsilon>0$ and set $$x_n=x_0e^{\log(n^{-(1+\epsilon)})/b_n}$$ Then $$|a_n|x_n^{b_n}=|a_n|x_0^{b_n}(x_n/x_0)^{b_n}\le (|a_n|x_0^{b_n})n^{-(1+\epsilon)}$$ Set $A=\lim\sup_n x_n$. Since the series converges at $x_0$, the terms $|a_n|x_0^{b_n}$ are bounded, hence the series will converge absolutely for $x\ge A$. A priori, $A$ could be infinite. But this won't happen as the $b_n$ are decreasing and positive.

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Why do the series in the intermediate steps converge? –  user7530 Nov 21 '11 at 10:05
    
I'll edit the answer to make that clear. Thanks! –  B R Nov 21 '11 at 10:27
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