The 2013 STEP Prep Thread!

Are hyperbolic functions not covered in AHs then?

Hmmm... They're more useful in methods (and hence applied courses), so still fairly surprised they wouldn't be covered in a double maths qualification :-/

If somebody spent four hours a day working on their exam technique for English or French for six months, I'd be willing to bet they'd do very well. Conversely, many people who spend that much time on STEP don't do well.

Well at this stage I know no more about hyperbolic trig other than that it exists.

I can see it works - my real question is why make that substitution, rather than some arbitrary variable, a, substituting for (x-1)?

(a+1)a^1/2 is simple to integrate, or even not bother with the substitution if you can cope with it.


Thanks,

Well at this stage I know no more about hyperbolic trig other than that it exists.

I can see it works - my real question is why make that substitution, rather than some arbitrary variable, a, substituting for (x-1)?

(a+1)a^1/2 is simple to integrate, or even not bother with the substitution if you can cope with it.


Thanks,

The question asks for $\displaystyle\int x^2\sqrt{x^2-1}dx$, not $\displaystyle\int x\sqrt{x-1}dx$. If you mean substitute for a = x^2 - 1 then you are going to run into trouble with the change from dx to da.

Can't believe the irrationality of $e+\pi$ is an open question...

Can't believe the irrationality of $e+\pi$ is an open question...

You know what might help solve that? Infinite descen....oh wait, that doesn't help with anything ever.

It seems intuitively obvious I agree.

Consider two irrational numbers that "end" (as a CS applicant, I'm tempted to say the slice [n:] as a neater semantic) in a sequence of digits that sum to 1x10^-whatever.

Suddenly, you know that you are summing two rational numbers: you've just lopped the ends off them.

The solution, now, is quite clearly rational.


So such a proof could be based on knowing that the "ends" of e and pi do not sum to a whole decimal place, wherever it lies.

You know what might help solve that? Infinite descen....oh wait, that doesn't help with anything ever.

It seems intuitively obvious I agree.

Why not?

I guess I can take pride in having scarred you for life.



I don't know - just seems weird we have things like algebraic topology, string theory, but haven't managed to show that the sum of two of the most important numbers in mathematics can't be written as a fraction of integers...

Consider two irrational numbers that "end" (as a CS applicant, I'm tempted to say the slice [n:] as a neater semantic) in a sequence of digits that sum to 1x10^-whatever.

Suddenly, you know that you are summing two rational numbers: you've just lopped the ends off them.

The solution, now, is quite clearly rational.


So such a proof could be based on knowing that the "ends" of e and pi do not sum to a whole decimal place, wherever it lies.

Consider two irrational numbers that "end" (as a CS applicant, I'm tempted to say the slice [n:] as a neater semantic) in a sequence of digits that sum to 1x10^-whatever.

Suddenly, you know that you are summing two rational numbers: you've just lopped the ends off them.

The solution, now, is quite clearly rational.


So such a proof could be based on knowing that the "ends" of e and pi do not sum to a whole decimal place, wherever it lies.

Goldbach, abc (although watch this space!), Collatz...

irrational means they don't "end"

I don't know - just seems weird we have things like algebraic topology, string theory, but haven't managed to show that the sum of two of the most important numbers in mathematics can't be written as a fraction of integers...

I'm well aware -_-

That's why I used the quotes. By "end", I meant all (infinite) digits after an arbitrary point.

Yes but the irrationality property is more than just a sequence of digits that sum.

I don't see how this proves that pi+e can, or cannot, be written as a/b where a, b are integers. And that is what you have to prove or disprove, because that is the accepted definition of irrationality.

He means to prove that there exists an n such that the fraction part of $10^ne$ and $10^n\pi$ sum to an integer (which in fact actually has to be 1), hence $10^n\left(e + \pi\right)$ is an integer, iff $e + \pi$ is rational.

That was much more elegant, thank you.

Of course, it wouldn't be the only way - but one case. So I meant it as a conceptual reason that it could be rational, as I originally read LotF's comment as being "surely it would be irrational", rather than the "wow - no one has figured this out yet" that he clarified he meant.