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3

Gauss sums Your sum is strongly related to the Gauss sum. The usual trick is to compute the modulus. This works particularly smoothly over $\mathbf{Z}/p\mathbf{Z}$ as with usual Gauss sums, but essentially it works here too: If $S = \sum_{k=0}^{n-1} e^{ik^2},$ then \begin{align} |S|^2 &= \sum_{k=0}^{n-1} \sum_{k'=0}^{n-1} e^{i(k'^2 - k^2)}\\ &= ...


1

Hint Use that $a^n\cdot a^m=a^{n+m}$ to simplify $a).$ You should obtain $x^{N(N+1)/2}.$ Use that $\log_a a^x=a^{\log_a x}=x$ to simplify $b),c),d).$ You should obtain $6,280$ and $N,$ respectively.


1

I think what are you supposed to do is to sinplify them. For example $$x\cdot x^2=x^3$$ So you could do the same thing for the first expression, then go on with the others.


0

How about $e^{-x}\leq c(\gamma)(\frac{1}{1+x})^\gamma$ for non-negative $x$?


0

$A^{2x} = (A^x)^2 = 4 \implies A^x = \pm 2$ ${A^{3x}-A^{-3x}\over A^x - A^{-x}}= {\pm 8 - {\pm 1 \over 8}\over \pm 2 - {\pm 1 \over 2}} = {64 - 1 \over 16 - 4} = {63 \over 12}$


1

Two ways: 1)$A^{2x}=4$ and $A>0$ implies $A^x=2$. Then ${A^{3x}-A^{-3x}\over A^x - A^{-x}}=\dfrac{8-\frac{1}{8}}{2-\frac{1}{2}}=\frac{21}{4}$ 2)${A^{3x}-A^{-3x}\over A^x - ...


3

Hint: $\frac{u^3-v^3}{u-v}=u^2+uv+v^2$. Let $u=A^x$ and $v=A^{-x}$.


2

Hint: Multiply numerator and denominator by $A^x$. $$\begin{align} \frac{A^{3x}-A^{-3x}}{A^x-A^{-x}}\frac{A^x}{A^x} & = \frac{A^{4x}-A^{-2x}}{A^{2x}-1}\\ & = \frac{(A^{2x})^2-1/A^{2x}}{A^{2x}-1}\\ & = \frac{4^2-1/4}{4-1}\\ & = \frac{16-1/4}{3}\\ & = \frac{64-1}{12}\\ & = \frac{63}{12} \end{align}$$


0

Here is a very simple C-program: http://ideone.com/GKvh6J. Its result (see the "stdout" section): total population: 35017 percentage of population over 13: 41.18% total fertile female population: 5578



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