Difference between revisions of "User:Grogg007"

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 * [https://artofproblemsolving.com/wiki/index.php/AMC_historical_results#USAMO_.28March_19.2C_2025.29| AMC Historical Results] (2025 USAJMO/USAMO Stats)
-==Proving the infinite sum of reciprocal squares:==
-The derivative of a function <math>f(x)</math> is defined as <math>f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.</math> So the derivative of <math>f(x) = e^x</math> is
-<cmath>\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}.</cmath>
-Let <math>e^h - 1 = y.</math> Then as <math>h \to 0</math>, <math>y \to 0,</math> and <math>h = \ln(1+y).</math>
-<cmath>e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \lim_{y \to 0} \frac{y}{\ln(1+y)} = e^x \lim_{y \to 0} \frac{1}{\ln(1+y)^{1/y}} = e^x \cdot \frac{1}{\ln \left( \lim_{y \to 0} (1+y)^{1/y} \right)} = e^x \cdot \frac{1}{\ln e} = e^x.</cmath>
-Since the derivative of <math>e^x</math> is <math>e^x</math> itself, the <math>n</math>th derivative of <math>e^x</math> will be <math>e^x.</math>
-Consider the infinite sum <math>P(x) = \sum_{n=0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + \dots</math>
-The derivatives at <math>x=0</math> are <math>P(0) = a_0,</math> <math>P'(0) = a_1,</math> <math>P''(0) = 2a_2,</math> <math>P'''(0) = 3 \cdot 2a_3,</math> and so on, with <math>P^{(n)}(0) = n!a_n.</math>
-Let this sum equal the function <math>f(x) = e^x.</math> Then we must have <math>f^{(n)}(0) = P^{(n)}(0),</math> which means <math>n!a_n = f^{(n)}(0) = e^0 = 1,</math> so <math>a_n = \frac{1}{n!}.</math>
-The series for <math>e^x</math> is thus <math>e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots</math>
-Now, from Euler's Formula, we have <math>e^{ix} = \cos(x) + i\sin(x).</math>
-Using the infinite series expansion and substituting <math>ix</math> for <math>x,</math> we get:
-<cmath>e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \dots</cmath>
-<cmath>= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} + \dots</cmath>
-<cmath>= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \right) + i \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \right).</cmath>
-By equating the real and imaginary parts of the series with Euler's formula, we find the series expansions for sine and cosine:
-<cmath>\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots</cmath>
-<cmath>\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots</cmath>
-Since <math>\sin(n\pi) = 0</math> for all integers <math>n,</math> the roots of the function <math>\sin(x)/x</math> are at <math>x = \pm \pi, \pm 2\pi, \pm 3\pi, \dots</math>
-We can factor the polynomial expansion in terms of its roots:
-<cmath>\sin(x) = x( 1 - \frac{x^2}{\pi^2})( 1 - \frac{x^2}{(2\pi)^2})( 1 - \frac{x^2}{(3\pi)^2}) \dots</cmath>
-Now, consider the coefficient of the <math>x^3</math> term in the expansion of <math>\sin(x).</math>
-From the infinite product, the coefficient of the <math>x^3</math> term is found by multiplying the <math>x</math> with each term with the <math>-(x^2/n^2\pi^2):</math>
-<cmath>\text{Coefficient of } x^3 = -\left( \frac{1}{\pi^2} + \frac{1}{(2\pi)^2} + \frac{1}{(3\pi)^2} + \dots \right) = -\frac{1}{3!}</cmath>
-Equating the two expressions for the <math>x^3</math> coefficient, we get:
-<cmath>-\frac{1}{6} = -\frac{1}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2}</cmath>
-<cmath>\implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.</cmath>
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