calculuseducationintuition
Recently I stumbled upon someone who said he wanted to understand why
$\arctan x = \int\dfrac{dx}{1+x^2}$
At first I was confused. This is an easy result in any integral calculus course. But then he explained that although he understood the proof, he wanted to understand it "intuitively". He wanted to see why it was in terms of arclength and addition and subtraction.
My question is: Is there an "intuitive" way to explain this identity?
Out of a lack of modesty, I am going to post my two cents worth as an answer rather than a comment, as others more qualified than me have done. Let the roots of an equation be A, B, C, etc. We are told that the unsolvability of the general quintic equation is related to the unsolvability of the associated Galois group, the symmetric group on five elements. I think I can tell you what this means on an intuitive level.
For three elements A, B, and C, you can create these two functions:
AAB + BBC + CCA
ABB + BCC + CAA
These functions have the interesting property that no matter how you reshuffle the letters A, B and C, you get back the same functions you started with. You might reverse them (as you would if you just swap A and B) or they might both stay put (as they would if you rotate A to B to C) but either way you get them back.
For four elements, something similar happens with these three functions:
AB + CD
AC + BD
AD + BC
No matter how you reshuffle A, B, C and D, you get these three functions back. They might be re-arranged, or they might all stay put, but either way you get them back.
For five elements, there exists no such group of functions. Well, not exactly...there is a pair of huge functions consisting of sixty terms each that works, similar to the ones I drew out for the cubic equation...but that's it. There are no groups of functions with three or especially four elements, which is what you would actually want.
(EDIT: There is also a set of six functions that map to each other under permutations, but these don't help you either. We had an intersting follow-up about Dummit's Resolvents in this discussion here Resolvent of the Quintic...Functions of the roots)
If you try to create functions on five letters with this symmetry property, you'll convince yourself that it's impossible. But how can you prove it's impossible? You probably need a little group theory for that, which I haven't yet written it up in a presentable form (but I think I can). I've written more about this on my blogsite in a series of articles starting here. You'll see I left it hanging in midstream about a year ago, but I think I'm going to finish it off soon.
I listened to the podcast which Christopher Ernst linked to in the comments, and I didn't think it was very good. Yes, it's all about the symmetries, but just because a guy is talking with an English accent doesn't mean he's profound. I'm not even sure that the stuff he said about re-shuffling five sets of 24 elements even makes sense. Anyhow, stuff on the level that Stewart is talking about was already understood long before Galois... Lagrange (most notably) had worked out all those symmetries fifty years earlier. There's an exceptionally good article about these things on a website by one Fiona Brunk which you can read here.
EDIT: I'm going to expand on the answer I posted the other day, because I think I really have identified the "intuitive" reason the quintic is unsolvable, as opposed to the "rigorous" reason which involves a lot more group theory. For the third degree equation, I identified these functions:
AAB + BBC + CCA = p
ABB + BCC + CAA = q
A, B and C are the roots of a cubic, but p and q are the roots of a quadratic. You can see that because if you look at pq and (p+q), the elementary symmetric polynomials in p and q, you will see they are symmetric in A, B and C. So they are easily expressible in terms of the coefficients of our original cubic equation. And that's why p and q are the stepping stone which gets us to the roots of the cubic.
Similarly, for the fourth degree, we identified these functions:
AB + CD = p
AC + BD = q
AD + BC = r
You can rewrite the previous paragraph word for word but just take everything up a degree, and it remains true. A, B, C, and D are the roots of a quartic, but p,q and r are the roots of a cubic. You can see they must be because if you look at the elementary symmetric polynomials in p, q and r, you will see they are symmetric in A, B, C and D. So they are easily expressible in terms of the coefficients of our original quartic equation. And that's why they are the stepping stone which gets us to the roots of the quartic.
And the intuititve reason why the fifth degree equation is unsolvable is that there is no analagous set of four functions in A, B, C, D, and E which is preserved under permutations of those five letters. As I mentioned earlier, I think Lagrange understood this intuitively fifty years before Galois. You probably needed a little more group theory to make it completely rigorous, but that's another question.
I think Lagrange would have understood the algebraic tricks whereby you went from, say, A B and C to p and q. It involves taking linear functions which mix A B and C with the cube roots of unity and examining the cube of those functions. Its a reversible process, so you can work backward the other way (by taking cube roots of functions in p and q) to solve the cubic. A very similar trick works for the fourth degree. I think Lagrange was able to show conclusively that the same trick does not work for the fifth degree...that's the "intuitive" proof. The "rigorous" proof would have had to show that in the absence of the obvious tricks (analogous to the 3rd and 4th degree), there was no other possible tricks that you could come up with.
This is not really an answer, but it was getting too long to be a comment.
Mathematics draws much of its power from deep, sometimes mysterious dualities between geometry and algebra, so I do not think there is any way to understand the relationship between geometric intuition and symbol manipulation in general.
Diophantine equations are one of the least geometrically intuitive areas of mathematics, though it is certainly worth mentioning that geometric insights (albeit very difficult ones) are behind some of the deepest results in the field, like Fermat's Last Theorem.
One can sometimes make arguments from intuitive principles that give partial answers. To take your example of $(xy-7)^2 = x^2 + y^2$, one can observe immediately that this is the intersection in the plane of two curves, one of degree $2$ and one of degree $4$, that have no components in common. By Bézout's theorem, there are at most $8$ intersection points (in fact, we can conclude that there are exactly $8$ solutions, if we know how to count them).
This tells us that the integer solutions, in particular, are a finite, possibly empty set. But it doesn't really give a good a priori method for finding them. That requires some symbol pushing.
To some extent, we can gain a lot of intuition for pushing symbols around. But this can be a lifelong process, and I think any mathematician will tell you that there are advantages to being able to manipulate equations without completely understanding what you are doing. There is some truth to the famous quotation that mathematics is about getting used to things, and for myself I can say that diligent practice has been far more valuable than theory when it comes to developing comfort with difficult problems.
The theory and intuition comes eventually, and with specific examples you may get little boosts to your insight here and there from helpful mentors, but there is no singular approach that will allow you to "plan".
Best Answer
The geometric picture is as follows. Let $O = (0, 0), A = (1, 0), X = (1, x)$. Then $\arctan x = \angle AOX$. We want to understand why $\angle AOY \approx \arctan x + \frac{h}{1 + x^2}$ where $Y = (1, x + h)$ and $h$ is small; equivalently, we want to understand why $\angle XOY \approx \frac{h}{1 + x^2}$. Since this angle is small, we equivalently want to understand why $\sin \angle XOY \approx \frac{h}{1 + x^2}$.
Now $\triangle XOY$ evidently has area $\frac{h}{2}$. On the other hand, it has area $\frac{1}{2} |OX| |OY| \sin \angle XOY$ where $|OX| = \sqrt{1 + x^2}$ and $|OY| \approx |OX|$. The result follows.
(The derivative follows, anyway. The integral follows by dividing up $AX$ into little pieces and drawing a bunch of lines to $O$, then summing up all of the contributions.)