Math0Phile

  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

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.