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Malhar Rajpal If you ask most mathematicians what the most beautiful identity in maths is, they would likely say Euler’s identity, or more precisely: e^(iπ) + 1 = 0, where e refers to Euler’s number (~2.71828), π is the ratio of a circle’s circumference to its diameter (~3.14159) and i is the imaginary number, sqrt(-1). Since both e and π are irrational, it is easy to see why the equation is so beautiful: two irrational numbers and an imaginary number, with seemingly nothing in common, are put together in such a way that they evaluate to a simple constant: -1. In this article, I will be using the Taylor and Maclaurin series to explain why this identity is true. The Taylor series for an infinitely differentiable function is the sum of an infinite number of terms, with each of these terms being expressed with respect to the function’s derivatives (from the 0th derivative to the infinite derivative) at a point. The Taylor series is used to approximate the values of different mathematical functions and it is represented mathematically as: where f^(k)(a) is the k-t derivative of the function, evaluated at point a. Functions and their Taylor series are equal for most common functions. The Maclaurin series is a special case of the Taylor series centered at 0 (a=0). The Maclaurin series is therefore: As the sum of the Maclaurin series is approximated for values of k, the function approximated by the Maclaurin series tends closer to the true function, and when k=∞, the series is equal to the real function. This video explains the concept in an intuitive way. To see why e^(iπ) + 1 = 0, let us first evaluate the Maclaurin series of cos(x) and sin(x). To figure this out, we will try to find a pattern in the derivatives of both of these functions. Let’s start with cos(x): the first derivative of cos(x) = -sin(x), the second is -cos(x), the third = sin(x), and the fourth = cos(x). Hurrah! We have found a pattern, where every multiple of four derivatives of cos(x) = cos(x) and the derivatives which follow it repeat cyclically (in the pattern cos(x) → -sin(x) → -cos(x) → sin(x) for every subsequent derivative). If we evaluate these four distinct derivatives of cos(x) at 0 we get: And repeat. Since we have found a pattern, we can plug our values into the general Maclaurin formula. With cos(x), the Maclaurin series equals: where the powers increase by two each term and are all even, and the +/- signs alternate. If this series is extended to infinite terms, it is exactly equivalent to cos(x). By following the same process with sin(x), there is a similar pattern where every k-th derivative of sin(x) is the same the (k-4)th derivative of sin(x) (when k >=4): By following the general Maclaurin formula stated above, we can see that: where the powers increase by two each term and are all odd, and the +/- signs alternate. We can now do the same for e^x. Since the derivative of e^x is e^x, and since e^0 = 1, we can say the Maclaurin series of e^x = e^0 + e^0 x + (e^0 x^2)/2! +(e^0 x^3)/3! + ... = 1 + x + (x^2 / 2!) + (x^3 /3!) + ... where the power increases by one each term. Now that we have shown these series, we can try and find a connection. Initially it might seem hard, however if we replace the x in ex with ix, where i is the imaginary number, we start to see a link: Since i = sqrt(-1), i^2 = -1, i^3 = -i, i^4 = 1 and i^5 = i again, we get: Factoring out i gives: Do you see the connection? Yes! The first part of the expansion of e^(ix) = the Taylor expansion of cos(x) and the second part, with i factored out, = sin(x). Hence we can say that e^(ix)=cos(x) + isin(x), Euler’s formula! Neat! Now to derive Euler’s identity, all we have to do is substitute π for x in Euler’s formula to get: eiπ=cos(π) + isin(π)= -1+i*0 =-1. Voila: we get e^(iπ) = -1 and by adding 1 to both sides, we get Euler’s identity:
e^(iπ) + 1 = 0.
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2/8/2022 02:45:30 am
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