# Finding $\pi$ through sine and cosine series expansions

I am working on a problem in Partha Mitra's book Observed Brain Dynamics (the problem was originally from Rudin's textbook Real and Complex Analysis, and appears on page 54 of Mitra's book). Unfortunately, the book I have does not contain any solutions... Here is the question:

By considering the first few terms of the series expansion of sin(x) and cos(x), show that there is a real number $x_0$ between 0 and 2 for which $\cos(x_0)=0$ and $\sin(x_0)=1$. Then, define $\pi=2x_0$, and show that $e^{i\pi/2}=i$ (and therefore that $e^{i\pi}=-1$ and $e^{2\pi i}=1$.

Attempt at a solution: In a previous problem I derived the series expansions for sine and cosine as

$$\sin(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{x^{2n+1}}{(2n+1)!} \right) \right]$$

$$\cos(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{x^{2n}}{(2n)!} \right) \right]$$

My thought is that you can show that $\cos(0)=1$ (trivially), and that $\cos(2)<0$ (less trivially). This then implies that there is a point $x_0$ between 0 and 2 where $\cos(x_0)=0$, since the cosine function is continuous. However, I do not understand how you could then show that $\sin(x_0)=1$ at this same point. My approach may be completely off here.

I believe that the second part of this problem ("Then, define $\pi=2x_0$...") will be easier once I get past this first part.

Thanks so much for the help. Also - I swear this is not a homework assignment. I am reading through this book on my own to improve my math.

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 I think the hint is to consider $\sin$ and $\cos$ not in terms of their series expansions, but as solutions to the differential equation $y'' + y = 0$. There's a pretty detailed discussion of this in George Simmons' book Differential Equations with Historical Notes. I will try to find the page number if you are interested. – MJD Sep 12 '12 at 2:39 The Simmons discussion, which you may enjoy, is on pages 115–118 of the book I mentioned. Among other familiar properties, he derives the property $\sin^2 x + \cos^2 x = 1$ from the differential equation I mentioned above. Since the differential equation property is implied by the power series for $\sin$ and $\cos$, you might find this interesting. – MJD Sep 12 '12 at 2:46 Thanks for the input! I'll try to check that out! – Alex Sep 12 '12 at 2:56 The development of $\pi$ that you are discussing appears on pages 182–183 of Principles of Mathematical Analysis by W. Rudin. It is part of the main text, not an exercise. Although I wouldn't say it is explained in detail, the development is sufficiently clear that it can be followed. If you're still interested but puzzled, you should have a look at it. – MJD Mar 19 at 4:38

## 1 Answer

How to show that $\sin (x_0)=1$ if $\cos (x_0)=0$? Quite simply:

$$\sin^2 x+\cos^2 x=1$$

(you may also want to specify that $\sin x$ is positive in the given range)

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But the question said By considering the first few terms of the series expansion of sin(x) and cos(x), show that [..] – user2468 Sep 12 '12 at 2:44
Yeah thats a good point... – Alex Sep 12 '12 at 2:46
I think you must be right Julien - you can show that cos(x)=0 by using the first few terms, and then use that identity you mentioned (which I derived in a separate problem). Thanks for the help! – Alex Sep 12 '12 at 2:51
@JenniferDylan Hm, yes I read it rather quickly. I understood the and as an and hence ... but now I see why this could be labeled as misinterpretation. It just seems doing them separately is complicating things for no good reason. – user39572 Sep 12 '12 at 2:56