A Manipulative Approach To Basel Problem

Basel Problem is one of the most intriguing problems in math which is quite famous for its mind-boggling result. The Basel Problem seeks to find a solution of the following series which is needless to say, converging to a very confusing and weird result.

$$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}$$

Which in terms of summation can be written as,

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

Quite a weird result! But as most of us must be familiar, this problem is “usually” solved using the Maclaurin Series to derive the eccentric result. Here, we are going to solve the series using only 10+2 mathematics. Often times we undervalue the tools that math provides us to the extent that we miss out on much of the unique applications of the methods that are taught to us in mathematics. This is going to be one such example because not only the solution feel like cheating after understanding it fully but it forces us to think that “How can a person come up with such a solution out of nowhere ?”. In the essence, we will delve deep into the motivations and properties of the ingenious solution that we present here.

The Solution is followed as,

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}$$

$$\ln(1+e^{ix}) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{e^{inx}}{n}$$

$$\ln(1+e^{ix}) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\cos{(n x)} + i \sin{(n x)}}{n}$$

Now for the left side of the equation,

$$1 + e^{ix} = 1 + \cos{x} + i\sin{x}$$

$$1 + e^{ix} = 2\cos^2{\left(\frac{x}{2}\right)} + 2i\sin{\left(\frac{x}{2}\right)}\cos{\left(\frac{x}{2}\right)}$$

$$1 + e^{ix} = 2\cos{\left(\frac{x}{2}\right)} \left( \cos{\left(\frac{x}{2}\right)} + i\sin{\left(\frac{x}{2}\right)} \right)$$

$$1 + e^{ix} = 2\cos{\left(\frac{x}{2}\right)} \cdot e^{i \frac{x}{2}}$$

Replacing the expression into the equation,

$$\ln(1+e^{ix}) = \ln \left( 2\cos{\left(\frac{x}{2}\right)} \cdot e^{i \frac{x}{2}} \right)$$

$$\ln(1+e^{ix}) = \ln{2} + \ln{\left(\cos{\left(\frac{x}{2}\right)}\right)} + i \frac{x}{2}$$

$$\ln{2} + \ln{\left(\cos{\left(\frac{x}{2}\right)}\right)} + i \frac{x}{2} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\cos{(n x)} + i \sin{(n x)}}{n}$$

The next step of the solution is quite self-explanatory. It is a common exercise that when 2 complex numbers are equal , we can and must equate their Real and Imaginary Parts of the equation to proceed with the solution. In particular, we shall equate the imaginary parts as the real parts are not quite fruitful to solve due to the generation of cosine term compounded by the natural logarithm.

$$\frac{x}{2} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin{(n x)}}{n}$$

$$\int_{0}^{\pi} \frac{x}{2} \, dx = \int_{0}^{\pi} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin{(n x)}}{n} \, dx$$

$$\int_{0}^{\pi} \frac{x}{2} \, dx = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} \int_{0}^{\pi} \sin{(n x)} \, dx$$

$$\int_{0}^{\pi} \sin{(n x)} \, dx = \left. -\frac{\cos{(n x)}}{n} \right|_{0}^{\pi} = \frac{1 - (-1)^n}{n}$$

$$\frac{\pi^2}{4} = \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} (-1)^{n+1} \frac{2}{n^2}$$

$$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$$

Huh, This looks extremely similar to the Basel Problem but rather we have ended up its little cousin with only the infinite sum of reciprocals of odd numbers. But, as many of you might have guessed already, we could use it to derive a solution for the actual solution to the Original Basel Problem.

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{1}{n^2} + \sum_{\substack{n=1 \\ n \text{ even}}}^{\infty} \frac{1}{n^2}$$

$$\sum_{\substack{n=1 \\ n \text{ even}}}^{\infty} \frac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2}$$

$$S = \sum_{n=1}^{\infty} \frac{1}{n^2}$$

$$S = \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{1}{n^2} + \frac{S}{4}$$

$$S - \frac{S}{4} = \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{1}{n^2}$$

$$S = \frac{4}{3} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{6}$$

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

And people, That is the solution of the famous Basel Problem using mathematics only involving the stuff we learn at 10+2 level.Now , already this is one of the most mind-boggling solutions which contain many forms of weird manipulations which seem to be motivated out of nowhere and just handed down as the word of God himself. But now, we shall go through each step in detail regarding what exactly motivated not just the solution but rather each step and the journey from Taylor Series Expansion to Complex Numbers to Integration to Series Manipulation which results in the magnificent result.

---

Part 1: Taylor Series of Natural Logarithm

The first step of picking the Taylor Series Expansion is rather weird but also understandable if one gives a long thought. In particular, let us look at the general taylor series expansion for any function.

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots$$

Notice the factorials in the denominator in the Taylor series expansion and now look at the expansion of ln(1+x).

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}$$

The matter of fact is that the expansion for ln(1+x) looks much more familiar and similar than it would for expansion for any other function due to the presence of factorials in the original general function expansion. Therefore, considering that this is the only series most similar to the Basel Problem, it is a natural foothold to start from. But the work is not completed yet.

$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \cdots + (-1)^{n+1} \frac{x^{n-1}}{n} + \cdots$$

Integrating on both sides like this would have resulted into a great way to calculate the series in its entire form due to the property of Integration of Algebraic expressions of bringing the power terms to the denominator but the left hand side of the integration is very hard to do. You can try it for yourself but it will definitely take much more than just the rules of Integration to integrate that part.

Part 2: Complex Numbers

So we are left empty handed, Until we realize that maybe we could perhaps find another way to convert the powers into another form which when integrated yields another sequenced constant in the denominator. That is exactly where complex numbers and their useful properties come into play and we can use them to pull the exponents of variable inside the trigonometric functions using the De Moivre's Theorem.

$$\left( e^{i\theta} \right)^n = e^{i n \theta} =\left( \cos{\theta} + i \sin{\theta} \right)^n = \cos{(n\theta)} + i \sin{(n\theta)}$$

Notice that how we can take the exponent and pull it inside the trigonometric functions which we can then integrate to yield us the same constant that we wanted back into its place , The Denominator. Thus, what we can do now is replace the Taylor Series Expansion of Natural Logarithm with a complex number with magnitude 1.

Here is an interesting caveat which could result in a discourse, But since this is out of our scope of the problem. I will be covering this very briefly. Notice how I selected a generic complex number of magnitude 1. Here lies an interesting thought that we could select a generic complex number of magnitude r such that r < 1 (because when r > 1 , the right hand side arguably diverges to infinity, I think but I am not sure). Considering that a person does take r < 1. It could lead to a lot of new and interesting results (such as a Geometric Progression Variant of Basel Problem) that are rather out of the scope of this discussion but still very interesting and worth mentioning it here.

And therefore, we continue to replace x with the generic complex number of magnitude 1.

$$\ln(1+e^{ix}) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\cos{(n x)} + i \sin{(n x)}}{n}$$

Here , the step of right hand side of the equation is naturally motivated due to the above reasons but due to the pursuit of this venture on the right side, we can clearly see that the left hand side has been messed up and therefore it requires simplification. Luckily, we have just the right kind of weird but very widely used manipulation for these sort of terms. There is a point to note that not only the positive version but also the negative version has a similar kind of manipulation that results in simplification like the following.

$$1 + e^{ix} = 2\cos{\left(\frac{x}{2}\right)} \cdot e^{i\frac{x}{2}} $$

$$1 - e^{ix} = -2i \sin{\left(\frac{x}{2}\right)} \cdot e^{i\frac{x}{2}}$$

These identities help out a lot not only in our case but in the usual case of many other manipulations taking complex numbers into consideration. Thus, we must step forward through our solution.

$$\ln{2} + \ln{\left(\cos{\left(\frac{x}{2}\right)}\right)} + i \frac{x}{2} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\cos{(n x)} + i \sin{(n x)}}{n}$$

Now that there is a clear separation between real part and imaginary part of the equation on both sides. A common and yet important maneuver is to equate real and imaginary parts of both sides. But WHY ? Let us take some time and ask ourselves why ? And the answer is that we have used nothing but identities to come all the way up to this point due to which we can equate the real and imaginary parts. And thus we can continue to move forward into our solution.

I particularly selected to equate the imaginary parts without even having a speck of thought about the real part because of the sheer simplicity of the terms and how close it looks to completion but I must say that equating the real part is also quite worth the effort as that could also result in many results that might be interesting but that is a venture I leave upon the reader.

Part 3: Manipulation of Series

After a long journey through seeing patterns and looking deep into properties of various things , we finally arrive at the stage where we are dangerously close to the final answer and thus I must say that there is barely anything left to explain regarding our intentions but rather our actions will speak more than anything else.

$$\frac{x}{2} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin{(n x)}}{n}$$

Now that everything is in place just as we would like it to be. We can finally execute our plan of integrating both sides while keeping the limits simple as from 0 to pi.

There is another caveat here that now we make another choice about the limits, I must say one thing and that is the choice of limits can also lead to many interesting results like choosing [0,pi/2] could lead to all sorts of weird infinite sums but due to the scope of this discussion, I only discuss [0,pi].

$$\int_{0}^{\pi} \frac{x}{2} \, dx = \int_{0}^{\pi} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin{(n x)}}{n} \, dx$$

Rest of the process is quite self-explanatory and mechanical.

$$\int_{0}^{\pi} \frac{x}{2} \, dx = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} \int_{0}^{\pi} \sin{(n x)} \, dx$$

$$ \frac{\pi^2}{4} = \sum_{\substack{n=1 \ n \text{ odd}}}^{\infty} (-1)^{n+1} \frac{2}{n^2}$$

$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$$

And as for the final trick of the show, we use the polar opposite of the result (and a very basic property of even numbers) we just got to convert all series into a simple linear equation.

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{1}{n^2} + \sum_{\substack{n=1 \\ n \text{ even}}}^{\infty} \frac{1}{n^2}$$

From here, we could proceed very easily by identifying that among the sum of squares of even numbers is hidden our own original series.

$$\sum_{\substack{n=1 \\ n \text{ even}}}^{\infty} \frac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2}$$

$$ S = \sum_{n=1}^{\infty} \frac{1}{n^2}$$

$$ S - \frac{S}{4} = \sum_{\substack{n=1 \ n \text{ odd}}}^{\infty} \frac{1}{n^2}$$

$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

And finally we arrive at the result.

Thus we have come to the end of the solution of famous Basel Problem using 10+2 level mathematics, bypassing the traditional use of the Maclaurin Series. By ingeniously employing Taylor Series, complex numbers, and integration, the solution derives the surprising result that the sum of reciprocals of positive integers squared equals (pi)²/6. Through meticulous series manipulation and integration steps, I have tried to reveal the underlying motivations and properties of this elegant approach, providing a comprehensive breakdown of how seemingly disparate mathematical concepts come together to produce the intriguing outcome. I sincerely hope that you have enjoyed, learnt and also found a few things to explore yourselves regarding the solution or the Basel Problem. I encourage the reader to explore several points mentioned through out the blog that are not elaborated and take it up upon yourself to embark upon a new journey to understanding math.