Mathematics Challenge Marathon

[RuiAce]

Not hard; it just looks pretty

[winstondarmawan]

Not hard; it just looks pretty

Isn't this just equiangular?

[RuiAce]

Isn't this just equiangular?

Yes

[RuiAce]

Another one that's not hard but just looks pretty.

[RuiAce]

[RuiAce]

[zhen]

I’m not sure if I’ve structured this proof correctly. I’m honestly just having a bit of fun.

[RuiAce]

Let \(ABCD\) be a parallelogram. Prove that its diagonals bisect each other.

[RuiAce]

In this question, we fill in the gaps left in high school and prove properly that \( \frac{d}{dx} e^x = e^x\) and \( \frac{d}{dx} x^r = rx^{r-1} \) for any real \(r\). (It is adapted from sources easily obtainable online.)





Remember, the only derivative you know after doing Q1-Q3 is \( \frac{d}{dx} e^x = e^x\). You cannot assume anything about the derivative of log or any other function.

Hint for this part

Remember: \( (-1)^r\) is just a constant. Somehow borrow what you've used in Q4, or just regurgitate what you did in Q4 in a different manner.

[fun_jirachi]

Let \(ABCD\) be a parallelogram. Prove that its diagonals bisect each other.

Just gonna bump this thread (because i just found it) with the solution to this question

[RuiAce]

\[ \int_{-2}^3 \frac{1}{x^2}\,dx \]

[fun_jirachi]

I swear I'm going to get baited
Jumping straight in without thinking gets you this which seems a little off

But it seemed a bit too easy doing it like that, so I graphed the function on the cartesian plane and the area under the graph is infinite/indefinite/undefined because x is undefined as it approaches zero from both ends. so uh can someone that isnt braindead explain?

[RuiAce]

I swear I'm going to get baited
Jumping straight in without thinking gets you this which seems a little off

But it seemed a bit too easy doing it like that, so I graphed the function on the cartesian plane and the area under the graph is infinite/indefinite/undefined because x is undefined as it approaches zero from both ends. so uh can someone that isnt braindead explain?

Maybe wolframalpha can show you the way ;)

[fun_jirachi]

So do you just have two options:
- Give up and cry because the function isnt defined at x=0
- Use an approximation
Because that seems really sad

[RuiAce]

Pretty much the fact that the function being undefined at 0, in fact asymptotically approaching 0 is heavily related to this problem. The integral is said to be divergent because the area under the graph is actually approaching \(\infty\) as \(x\to 0\), both in the positive and in the negative direction. The interval \(-2\leq x \leq 3\) clearly contains the value \(x=0\), so when we try to compute the area under the curve we end up crossing this border, which leads to big problems.

That's why the question was a bait. A student who only thinks in terms of antiderivatives and not in terms of what the integral actually represents would obtain the wrong answer.

However, it may be worth noting that not all functions that have asymptotes will have divergent integrals. For example, \( \int_{-2}^3 \frac{1}{x^{2/3}}\,dx \) converges (it equals to some finite real number). Wolfram won't give the exact correct answer for this since it combines complex analysis into the mixture and gives the principal value of \( (-2)^{1/3} \), which is complex, but plugging into this integral calculator will give a value that's clearly not undefined.