Recent content by Dick

I didn't worry about why they wrote it the way they did. Pay attention to the start of the sequence. They started with ##k=1##, I started with ##k=0##. That's the only difference. But they are still the same series.

Well, I think it would be more natural to write the series as ##\frac{1}{4}\, \sum \limits_{k=0}^\infty (\frac{3}{4})^k##. Does that look correct to you? But that's really the same series that the OP wrote. Just write out some terms to verify.

The Riemann sum is how you derive your power sum formula. If you draw your power sum as rectangles they are an upper sum for the curve ##y=x^{1.5}## and a lower sum for the curve ##y=(x+1)^{1.5}##. You integrate up to ##n## and average the two to get your formula. It's plenty accurate enough to...

Yes, that's what you get approximating by integrals. And it's not an equality, it's only a pretty good approximation to the sum.

It's not hard to get an approximate formula for ##\sum_{k=1}^n k^{1.5}##. Just replace the sum with an integral. You should also be able to get a subleading term by approximating the error you make by doing that.

How about ##q_n=\frac{1}{n}##?

Start by stating what the test says. It might get you started.

Think about how you might apply that test.

Use that ##(f\circ f) \circ f == f\circ (f \circ f)## (where ##\circ## is function composition) to conclude that ##f(x^2+2)=f(x)^2+2##. That let's you get more relations. SammyS is suggesting you guess values for ##f(1)## and see if they lead to contradictions. Try it. For example, can ##f(1)=1##?

Yes. A better definition of transitivity would say if ##(a,b)## and ##(b,c)## are elements of ##R##, then ##(a,c)## is an element of ##R##. It has to hold for ALL ##a## and ##c##, the case ##a=c## included.

If you look closely you'll notice that your series consists of two interlaced ordinary geometric series.

That is certainly untrue just by inspection. It's saying that ##\binom n k## only depends on ##n-k##.

You computed ##S^{-1}MS## not ##SMS^{-1}##, which is the one you want.

You can also count your ##3^{10}## directly. The sets ##G\subseteq F \subseteq H## can be identified by assigning a status to each element of ##H##. It can be either i) only in ##H##, ii) in ##F## and ##H## but not in ##G## or ii) in all three sets. That's three possibilities for each element of...

I don't know any good links. But approximations like this generally come from taking the first terms of the Taylor series. ##\ln(1+x)=x- \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots## Keeping just the first term gives the approximation. Similarly, ##\sin(x) \approx x## etc.