Asymptotic tight bound question

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Homework Statement



Hi,

I just have a basic question regarding an asymptotic tight bound question.

The question is :
TRUE / FALSE

http://latex.codecogs.com/gif.latex?3^{n+1} \text{ belongs to } \Theta(3^{n})

By definition of big theta:

c_{1}g(n) \leq f(n) \leq c_{2}g(n) \text { } \forall n > n0

So in my case, g(n) = 3^{n} \text{ and } f(n)=3^{n+1}

Therefore to prove this true, I should show a set of values for c1, c2, and n for the definition to hold true.

Is that correct?
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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