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Mathematical Artifacts: Exploring Ancient Mathematical Discoveries

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  Introduction: Mathematics is often perceived as an abstract and intangible discipline, existing solely in the realms of equations and theorems. However, throughout history, mathematical ideas have found expression in tangible artifacts, providing us with glimpses into the ancient civilizations' profound understanding of numbers, geometry, and complex calculations. These mathematical artifacts serve as valuable clues, shedding light on the innovative mathematical discoveries of our predecessors. In this article, we embark on a fascinating journey to explore the mathematical artifacts left behind by ancient civilizations, uncovering their mathematical prowess and the enduring impact of their discoveries. 1. The Antikythera Mechanism: Our exploration begins with a truly astonishing artifact, the Antikythera Mechanism. Discovered in a shipwreck off the Greek island of Antikythera in 1901, this ancient analog computer dates back to the 2nd century BCE. While its purpose remained a my...
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PUTNAM 1985: PROBLEM B-1

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PROBLEM Let k be the smallest positive integer for which there exist distinct integers `m_1,\m_2,\m_3,\m_4,\m_5` such that the polynomial  p(x) = `(x-m_1)(x-\m_2)(x-\m_3)(x-\m_4)(x-\m_5)` has exactly k nonzero coefficients. Find, with proof, a set of integers `m_1,\m_2,\m_3,\m_4,\m_5` for which this minimum k is achieved. SOLUTION p(x) = `(x-m_1)(x-\m_2)(x-\m_3)(x-\m_4)(x-\m_5)`           = `x^5+Ax^4+Bx^2+Cx+D` where A, B, C, D are integers Now we want to minimise the number of nonzero coefficients that p(x) can have. So starting from the smallest positive integer, can k be 1 ? Well definitely no, because then p(x) = `x^5` `\Rightarrow` `m_1=m_2=m_3=m_4=m_5=0` Well then,  can k be 2 ? If k is 2, then p(x)  = `x^5+Cx` or p(x) = `x^5+D` as for any other case `x^2`| p(x)`\Rightarrow`any two of the `m_i` are zero and thus not distinct. If p(x)  = `x^5+Cx`, then p(x) = 0 will have one root as zero and the other four roots will come from ...
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SOME BEAUTIFUL 2D CURVES AND THEIR EQUATIONS

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  THE LOVE EQUATION `x^2+{(y-x^{2/3})}^2=1` THE POLAR ROSE General Equation:  `r=a\sin(k\theta)`  or  `r=a\cos(k\theta)` For positive integral values of k, the rose will have 2k petals if k is even, and k petals if k is odd. (For k = 1, it will simply be a circle).    `r=\sin(4\theta)` `r=\sin(5\theta)` For non-integral value of k, we get beautiful graphs like : `r=\sin(2.5\theta)` `r=\sin\left(\pi\theta\right)` THE ARCHIMEDIAN SPIRAL General Equation:  `r=a\theta`   ('a' is a constant) `r=\theta` THE LOGARITHMIC SPIRAL General Equation:  `r=ae^{b\theta}`  `r=e^{0.5\theta}` THE BRACELET x = cos t y = sin t + 0.1 cos 6.2t THE LEMNISCATE OF BERNOULLI General Equation:  `\left(x^2+y^2\right)^2=2c^2\left(x^2-y^2\right)` `\left(x^2+y^2\right)^2=2\left(x^2-y^2\right)` THE BATMAN LOGO
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MEAN VALUE THEOREM

PROBLEM:  Using Mean Value Theorem prove the inequality:   `0<\frac1x\log\left(\frac{e^x-1}x\right)<1` SOLUTION: Let f(x) = `e^x`. f(x) is continuous in [0, x] and differentiable in (0, x). So, by Langrange's Mean Value Theorem we can say that `\exists` at least one value of c in (0, x) such that  f '(c) = `\frac{f(x)-f(0)}{x-0}`  `\Rightarrow` `e^c` = `\frac{e^x-1}{x}` Now, 0 < c < x  `\Rightarrow` 1 < `e^c` < `e^x`  `\Rightarrow\1\<\frac{e^x-1}x\<\e^x\Rightarrow\0\<\log(\frac{e^x-1}x)\<\x` `\Rightarrow`  `0<\frac1x\log\left(\frac{e^x-1}x\right)<1` Q.E.D.
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