Problems

---

Solution:

1. Consider \(f(x) = x - \ln (1+ x)\), then \(f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0\) if \(x >0\). Therefore \(f(x)\) is strictly increasing on the positive reals. Since \(f(0) = 0\) we must have \(f(x) > 0\) for all positive \(x\), ie \(x - \ln(1+x)\) is positive for all positive \(x\). \begin{align*} \sum_{k=1}^n \frac1k &\underbrace{>}_{x > \ln(1+x)} \sum_{k=1}^n \ln \left (1 + \frac1k \right ) \\ &= \sum_{k=1}^n \ln \left ( \frac{k+1}{k} \right ) \\ &= \sum_{k=1}^n \left ( \ln (k+1) - \ln (k) \right) \\ &= \ln (n+1) - \ln 1 \\ &= \ln (n+1) \end{align*}
2. Let \(g(x) = x + \ln (1-x)\) ,then \(g'(x) = 1 - \frac{1}{1-x} = \frac{-x}{1-x} < 0\) if \(0 < x < 1\) and \(g(0) = 0\). Therefore \(g(x)\) is decreasing and hence negative on \(0 < x < 1\), in particular \(x < -\ln(1-x) \) \begin{align*} \sum_{k=2}^n \frac1{k^2} &\underbrace{<}_{x < -\ln(1+x)} \sum_{k=2}^n - \ln \left (1-\frac1{k^2} \right) \\ &= -\sum_{k=2}^n \ln \left ( \frac{k^2-1}{k^2}\right) \\ &= \sum_{k=2}^n \l 2 \ln k - \ln(k-1) - \ln(k+1) \r \\ &= \ln n - \ln(n+1) - \ln 0+\ln 2 \\ &= \ln 2 + \ln \frac{n}{n+1} \end{align*} as \(n \to \infty\) we must have \(\displaystyle \sum_{k=2}^{\infty} \frac1{k^2} < \ln 2\) ie \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2\]

---

Solution:

1. \(\,\) \begin{align*} && f(x) &= (1+x^2)e^x \\ && f'(x) &= 2xe^x + (1+x^2)e^x \\ &&&= (1+x)^2e^x \geq 0 \end{align*}
   

   + - ↺
   As \(x \to -\infty\), \(f(x) \to 0\) and as \(x \to \infty\), \(f(x) \to +\infty\) and since \(f\) is strictly increasing we have exactly one solution to \(f(x) = k\) on \((0,\infty)\). Since \(f(x) > 0\) there are no solutions if \(k \leq 0\).
2. Considering the function \(g(x) = (e^x-1)-k\tan^{-1} x \) then \(g'(x) = e^x - \frac{k}{1+x^2}\) therefore \(g'(x)\) has exactly one turning point when \(k > 0\) and \(0\) otherwise at the root of \(f(x) = k\) Notice also that \(g(0) = 0\) so we already have one solution, and \(g'(0) = 1 - k > 0\). Notice from our sketch that if \(0 < k < 1\) the root for \(f(x) = k\) has \(x \leq 0\), so our turning point is to the left of the origin and we are interested in the behaviour of \(g(x)\) as \(x \to -\infty\). (ie do we cross the axis again). \(\lim_{x \to -\infty} \left [ e^x - 1 - k \tan^{-1} x \right] = 0 - 1 +k \frac{\pi}{2} = k \frac{\pi}{2} - 1\). if this is positive, ie if \(k > \frac{2}{\pi}\) there are two solutions, otherwise there is only one real root.

---

Solution:

1. Notice that \(f(x) = x - \tanh x\) has \(f'(x) = 1-\textrm{sech}^2 x = \tanh^2 x > 0\) so \(f(x)\) is strictly increasing on \((0, \infty)\) and \(f(0) = 0\) therefore \(f(x)\) is positive for all \(x\) positive
2. Let \(f(x) = x\sinh x-2\cosh x+2\) then \(f'(x) = \sinh x +x \cosh x - 2 \sinh x = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0\) by the first part. \(f(0) = 0\) so \(f(x)\) is positive for all \(x\) positive.
3. Let \(f(x) = 2x\cosh2x-3\sinh2x+4x\) then \begin{align*} f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\ &= 4( x\sinh 2x-\cosh 2x +1) \\ &= 4(x2\cosh x \sinh x -2\cosh^2x ) \\ &= 8 \cosh^2 x (x - \tanh x) \end{align*} Which is always positive when \(x\) > 0, \(f(0) = 0\) so \(f(x) > 0\) for all positive \(x\).

Let \(f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}\) then \begin{align*} f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\ &= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\ \end{align*} which from the earlier part is always negative.