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0

For $f(x)$ to be a frequency distribution, it requires that $\int_{-\infty}^{\infty}f(x)dx = 1$. So your question is moot.


0

ar first we have $y(1)=4+a+b=2$, then we obtain $y'(1)=8+a$ and this must be $5$


1

$$\prod_{p > 2}\left(1-\frac{1}{(p-1)^2}\right) > \prod_{n \ge 2}\left(1-\frac{1}{n^2}\right) = \frac12$$


1

$$\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)\geq\exp\left(-\frac{6}{5}\sum_{p>2}\frac{1}{(p-1)^2}\right)\geq\exp\left(-\frac{6}{5}(\zeta(2)-5/4)\right).$$


0

Not true. Consider $f(n) = 2^n,\ \ f(cn) = 2^{cn} \neq c_1 \cdot 2^n$ for arbitrary $c$. Therefore $f(cn) \notin \Theta(f(n))$


3

Planck's constant is a dimensionful quantity; its numerical value depends on which units you measure it in. Since the units are more or less arbitrary conventions, the precise numerical value is of no particular mathematical significance, nor does it have deep physical importance. In particular, one of the arbitrary unit choices that go into determining the ...


1

By definition, $h$ has a value only in a specified set of units. Since the units used are always arbitrary (the second and meter have no fundamental significance), there can be no special meaning of this constant in whatever units are chosen.



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