The beauty of Approach

Lights out Alice!

Today, i came across an answer in Quora by Dr. Richard Muller who highlighted the beauty of approach when it comes to solving equations.

Note: The answer was compiled by Dr. Richard Muller and it uses the answers of Physics grad Prahar Mitra, engineer Jafar Alaa and mathematician Alon Amit.


How do you solve:  3 x2=94x+6
Dr. Muller writes in his answer that if you look at these three solutions, the solution that you find intrigues you will help you decide whether you want to be an engineer, a mathematician or a physicist.



Prahar's answer (personal favourite):

As many other answers have alluded, there is a solution for positive real x$x$ that is really close to 8 so that the real solution is something like x=8.000000⋯000#$x=8.000000\cdots 000\mathrm{\#}$. Can we estimate how close it actually is? For instance, can determine the number of zeros we have before we get to a non-zero digit?

Yes, we can!

The idea is to use Taylor expansions. Now, we know that the actual solution is really close to 8, so let us write x=8+ϵ$x=8+ϵ$. The equation is then

3(8+ϵ)2=38(8+ϵ)+6$3^{(8+ϵ{)}^2}=3^{8(8+ϵ)}+6$

Expanding both sides in ϵ$ϵ$ (which is legitimate since we know ϵ≪1$ϵ\ll 1$, we find to leading order

364(1+16ϵlog3)=364(1+8ϵlog3)+6$3^{64}(1+16ϵ\log 3)=3^{64}(1+8ϵ\log 3)+6$

which implies

ϵ=14×363log3$ϵ=\frac{1}{4\times 3^{63}\log 3}$

To estimate how small this really is, take a log10${\log }_{10}$ on both sides and we find

log10ϵ=63log103−log104−log10(log3)≈−30.7≈−31${\log }_{10}ϵ=63{\log }_{10}3-{\log }_{10}4-{\log }_{10}(\log 3)\approx -30.7\approx -31$

This implies that ϵ≈10−31$ϵ\approx {10}^{-31}$

Thus, we find that the actual solution in positive reals is something that is almost 8 followed by 30 zeroes.

PS - If you want to be precise and have a calculator with you, you can compute that ϵ=14×363log3=1.988×10−31$ϵ=\frac{1}{4\times 3^{63}\log 3}=1.988\times {10}^{-31}$

This implies

x≈8.0000000000000000000000000000001988

Jafar's answer:

From the graph we see one intersection for x<0. Since 3^(x^2) grows faster than 3^(8x)+6, there will eventually be another intersection for x>0. The 2nd solution will have too many decimal places due to the large value of the functions at the 2nd intersection.

Alon's answer (Salute):

As with any search for an answer or solution to anything, the first step is to know what the question is (or to accept that the question is vague, and that figuring it out is part of the journey).

But this question seems perfectly clear, no? Well, no, it isn't, in at least two ways.

First of all, what is x$x$ supposed to be? An integer? A real number? A complex number? This needs to be made clear since it makes a huge difference. The variable name x$x$ is often implicitly assumed to refer to a real number, so let's stick with that.

But then, what do we mean by “solve”? Most real numbers don't have a name. We can't pinpoint them using any finite description. “Solving” an equation involving an unknown x$x$may, therefore, mean one of several things:

1. Find an explicit expression for x$x$ using standard elementary expressions, such as x=log(23−−√+8)$x=\log (\sqrt{23}+8)$. Beware, there may not be such an expression at all, even if the equation has a perfectly good solution.
2. Find an explicit expression involving not-so-standard expressions, thereby connecting the solution of this problem with the solution of some other problem. For example, “x$x$is half the Feigenbaum constant”.
3. Find a numerical approximation for x$x$, as in x=17.2964123…$x=17.2964123\dots$
4. Prove that the equation has a solution, or that it has a unique solution, or that it has exactly five solutions. Often times this is all we can hope for.

Problems that are given as exercises in problem solving classes or calculus classes usually have a solution of the first kind, a nicely explicit one. I don't expect this is the case here. Exponentials and sums don't mix very well, and if there is a solution to an equation of this type, it's usually because someone specifically designed it to have such a solution. It's possible to develop some intuition around whether this is so. My intuition tells me this isn't the case here, but I could of course be wrong.

This reduces us to searching for solutions of type 2 (unexpected here) or 3 (very doable). There are many ways to numerically solve equations such as this. A good starting point is to guess approximate solutions, as follows.

The equation can be rewritten as 3x2=38x+6$3^{x^2}=3^{8x}+6$. If x$x$ is reasonably large, the exponents will be huge relative to the measly +6$+6$, so we guess that there must be a solution near x=8$x=8$, since this makes x2$x^2$ and 8x$8x$ the same.

Indeed, for x=8$x=8$, we have 364$3^{64}$ on the left and 364+6$3^{64}+6$ on the right. These numbers aren't equal, of course, but saying they are “close” would be an understatement worthy of the Black Knight (“all right, we’ll call it a draw”).

At x=8$x=8$, the right side is (barely) winning, while at x=9$x=9$ it is clearly losing as badly as the aforementioned knight. Therefore, somewhere in between, and very close to the 8$8$, there's an actual solution, since continuous functions can't switch sides without meeting. In the decimal expansion of this solution, I'd be surprised if there are fewer than ten 0’s after the decimal point before anything interesting happens.

We expect another solution when x$x$ is small, in fact negative: when x=−1$x=-1$ the RHS wins, while at x=−2$x=-2$ it loses. With a bit of care we can show that those are precisely the only two real solutions.

This, I suspect, is about as much as can be said about this problem.

If you do read closely through, the beauty shines through. I will let the answers talk for me.

Lights out Alice!