A kinda crazy Facebook-like math problem

You know that stupid math problems with food that 50-old people share on Facebook? I’m kinda use the answer is yes; those problems are obviously ridiculously easy even if they’re usually presented with captions like “only 1% of the population of the entire multiverse can solve this”. Of course the various math-themed communities on Facebook started to meme on this fact coming out with pearls like the “only 0,000000000001% of the population of the world can solve this” associated with an image featuring the Riemann Hypothesis. Here’s a common food problem:


Nothing too hard: it’s clear that apple=10,\quad banana=4,\quad coconut=2; easy. Parodies of such “problems” exist are usually feature an array of fruit linear equations and an integral right below with fruit in it that say “now solve me, you 1% of population genius”. However I’ve always laughed over those parodies-images and never tried to actually solve one till yesterday; and holy shit that was painful. I want to report here the whole thing since I found it really challenging and fun (well, if you’re masochist enough) and I also want to see how WordPress’ LaTeX can handle this:


First let’s do the system of linear equations: let’s set pear=p,\quad apple=a,\quad banana=b and rewrite the whole this as:

\displaystyle \begin{cases}2p-2a+b=2\\-3p+2b=1\\3p+a-b=3\end{cases}

This is not so hard so I’m not going to explain what I did step by step, but I can tell you that I used the substitution method starting by finding a in the third equation. The results are not so good-looking:

\displaystyle \begin{cases}p=\frac{17}{13}\\a=\frac{20}{13}\\b=\frac{32}{13}\end{cases}

I personally expected more plain results like 1, 3, 5 and even numbers like 6, 8, 12; those fractions made me realize that I was going to suffer a lot to break down that integral. Let’s rewrite it without fruit, the lemon will be our x:

\displaystyle \int_{a-b}^{b-p}\frac{2\sin^2(x)+cos(2x)}{x^4+1}\, dx

Let’s now solve the indefinite integral first:

\displaystyle \int\frac{2\sin^2(x)+\cos(x)}{x^4+1}\, dx

This thing initially scared me, trig always does, but then I realized that we can use the trigonometric identities \cos(2\theta)=2\cos^2(\theta)-1 and \sin^2(theta)=1-\cos^2(\theta) to simplify it:

\displaystyle \int\frac{2\sin^2(x)+\cos(x)}{x^4+1}\, dx = \int\frac{2-2\cos^2(x)+2\cos^2(x)-1}{x^4+1}\, dx=\int\frac{1}{x^4+1}\, dx

So the trig went away but this new integral is still problematic: we need to decompose this fraction (if you have any, this is the time to play some epic music like this). This implies that we have to factor x^4+1 and that’s not banal because it have no real roots, it have complex roots but I don’t know if dealing with them is useful. However I found that if we rewrite the thing as x^4 + 2x^2 - 2x^2 + 1 we can then substitute u = x^2 to get u^2 + 2u - 2u + 1; we can now see that u^2 + 2u + 1 easy factor to (u + 1)^2 and thus the whole thing is just (u + 1)^2 - 2u: a difference of squares! A difference of squares a^2 - b^2 factor to (a+b)(a-b) so (u + 1)^2 - u^2 equals (u + 1 + \sqrt{2u})(u + 1 - \sqrt{2u}), undoing the substitution: (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1). So the factoring is done:

\displaystyle \int\frac{1}{x^4+1}\, dx=\int\frac{1}{(x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)}\, dx

It’s now time to decompose this fraction:

\displaystyle\frac{1}{(x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)}=\frac{Ax+B}{x^2 + \sqrt2x + 1} + \frac{Cx+D}{x^2 - \sqrt2x + 1}

We need to find values for A,\, B,\, C,\, D  so:

\displaystyle\frac{1}{(x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)}=\frac{Ax^3-\sqrt2Ax^2+Ax+Bx^2-\sqrt2Bx+b+Cx^3+\sqrt2Cx^2+Cx+Dx^2+\sqrt2Dx+D}{(x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)}

We can now solve this system of linear equations to get our answers:

\displaystyle \begin{cases}A + C = 0\\B + D = 1\\-\sqrt2A+B+\sqrt2C+D=0\\ A-\sqrt2B+C+\sqrt2D=0\end{cases}

Playing with the first two equations and substituting things is easy to come up with these solutions:

\displaystyle \begin{cases}A=\frac{\sqrt2}{4}\\B=\frac{1}{2}\\C=-\frac{\sqrt2}{4}\\D=\frac{1}{2}\end{cases}


\displaystyle \frac{Ax+B}{x^2+\sqrt2x+1}=\frac{\sqrt2x+2}{4(x^2+\sqrt2x+1)}=\frac{x+\sqrt2}{\sqrt{2^3}(x^2+\sqrt2x+1)}

\displaystyle \frac{Cx+D}{x^2+\sqrt2x+1}=-\frac{\sqrt2x-2}{4(x^2+\sqrt2x+1)}=-\frac{x-\sqrt2}{\sqrt{2^3}(x^2+\sqrt2x+1)}

We can now rewrite the integral applying linearity:

\displaystyle\int\frac{1}{x^4+1}\, dx=\frac{1}{\sqrt{2^3}}\int\left(\frac{x+sqrt2}{x^2+\sqrt2x+1}-\frac{x-\sqrt2}{x^2-\sqrt2x+1}\right)\, dx

The first thing I noticed at this point is that \frac{d}{dx}[x^2+\sqrt2x+1]=2x+\sqrt2 and that \frac{d}{dx}[x^2-\sqrt2x+1]=2x-\sqrt2; that’s very similar to what we have as numerators of our fractions: we just have to multiply the whole integral by 2/2:

\displaystyle\frac{1}{2\sqrt{2^3}}\int\left(\frac{2x + 2\sqrt2}{x^2+\sqrt2x+1}-\frac{2x-2\sqrt2}{x^2-\sqrt2x+1}\right)\, dx

Considering that 2\sqrt2=\sqrt2+\sqrt2 we can write the integral as:

\displaystyle \frac{1}{\sqrt{2^5}}\left[\int\left(\frac{2x+\sqrt2}{x^2+\sqrt2x+1}\right)\, dx - \int\left(\frac{2x-\sqrt2}{x^2-\sqrt2x+1}\right)\, dx + \sqrt2\int\left(\frac{1}{x^2+\sqrt2x+1}+\frac{1}{x^2-\sqrt2x+1}\right)\, dx\right]

The first two integrals are easy: they are in the form \int\frac{f'(x)}{f(x)}\, dx that easily equal \log\left|f(x)\right|, where \log(x) is the natural logarithm of x, so:

\displaystyle \int\frac{2x+\sqrt2}{x^2+\sqrt2x+1}\, dx=\log\left|x^2+\sqrt2x+1\right|

\displaystyle \int\frac{2x-\sqrt2}{x^2-\sqrt2x+1}\, dx=\log\left|x^2-\sqrt2x+1\right|

The third one is a bit tricky; we have to complete the square in order to go on (in doing so we can even apply linearity and split it in two):

\displaystyle \sqrt2\int\left(\frac{1}{x^2+\sqrt2x+1}+\frac{1}{x^2-\sqrt2x+1}\right)\, dx = \sqrt2\left[\int\frac{1}{\left(x+\frac{\sqrt2}{2}\right)^2+\frac{1}{2}}\, dx +\int\frac{1}{\left(x-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}}\, dx\right]

Now we can use the rule \int\frac{1}{x^2+a^2}\, dx = \frac{1}{a}\arctan(\frac{x}{a}) to solve those two integrals and get that:

\displaystyle \sqrt2\left[\int\frac{1}{\left(x+\frac{\sqrt2}{2}\right)^2+\frac{1}{2}}\, dx +\int\frac{1}{\left(x-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}}\, dx\right] = \sqrt2\left[\sqrt2\arctan(\sqrt2x+1)+\sqrt2\arctan(\sqrt2x-1)\right]

Putting everything together we finally have a solution for our initial integral \int\frac{1}{x^4+1}\, dx, let’s call it I:

\displaystyle I=\dfrac{\log\left|x\left(x+\sqrt{2}\right)+1\right|-\log\left|x\left(x-\sqrt{2}\right)+1\right|+2\left(\arctan\left(\sqrt{2}x+1\right)+\arctan\left(\sqrt{2}x-1\right)\right)}{\sqrt{2^5}}

One last step: we need to calculate [I]_{a-b}^{b-p}. Due to previous results this ask us to calculate [I]_{-\frac{12}{13}}^{+\frac{15}{13}} that apparently equals:

\displaystyle \dfrac{\ln\left(39{\cdot}2^\frac{5}{2}+313\right)-\ln\left(39{\cdot}2^\frac{5}{2}-313\right)+2\left(\arctan\left(\frac{3{\cdot}2^\frac{5}{2}+13}{13}\right)+\arctan\left(\frac{3{\cdot}2^\frac{5}{2}-13}{13}\right)+\arctan\left(\frac{15\cdot\sqrt{2}+13}{13}\right)+\arctan\left(\frac{15\cdot\sqrt{2}-13}{13}\right)\right)+\ln\left(195\cdot\sqrt{2}+394\right)-\ln\left(195\cdot\sqrt{2}-394\right)-2\ln\left(-1\right)}{2^\frac{5}{2}}

That we can approximate to 1.76; probably there are better ways to solve it but I couldn’t figured them out since the last time I did integrals seriously was probably some months ago. However that was fun (and probably also one of the hardest I’ve ever done) and challenging; maybe I’m going to try some other parody-images math problems in the future. If you like math I’ll suggest you to try those things out, they can be pretty cool.



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