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Claudio Maggioni 2020-04-03 00:25:08 +02:00
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commit ded521e256
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midterm/midterm.tex
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@ -52,13 +52,13 @@ as $1|1000 0000 0000|000_F$
\subsection*{Point a)}
$$ \sqrt[3]{1 + x} - 1 = (\sqrt[3]{1 + x} - 1) \cdot
\frac{ \sqrt[3]{1 + x} + 1}{ \sqrt[3]{1 + x} + 1} = \frac{\sqrt[3]{(1 + x)^2} - 1}{\sqrt[3]{1 + x} + 1} =
\frac{\sqrt[3]{(1 + x)^2} - 1}{\sqrt[3]{1 + x} + 1} \cdot \frac{\sqrt[3]{(1 + x)^2} + 1}{\sqrt[3]{(1 + x)^2} + 1} = $$
$$\frac{(1 + x)\sqrt[3]{1 + x} - 1}{(\sqrt[3]{1 + x} + 1) \cdot (\sqrt[3]{(1 + x)^2} + 1)} $$
\frac{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1}{ \sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} = \frac{(1 + x) - 1}{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} =$$
$$ \frac{x}{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} $$
\subsection*{Point b)}
$$ \frac{1 - cos(x)}{sin(x)} = \frac{sin^2(x)cos^2(x) - cos(x)}{sin(x)} \cdot \frac{sin(x)}{cos(x)} \cdot \frac{cos(x)}{sin(x)} = (sin^2(x)cos(x) - 1)\cdot\frac{cos(x)}{sin(x)}$$
\subsection*{Point c)}
$$ \frac{1}{1-\sqrt{x^2-1}} = \frac{x}{x^2-\sqrt{x^4-x^2}}$$
$$ \frac{1}{1-\sqrt{x^2-1}} = \frac{1+\sqrt{x^2-1}}{(1-\sqrt{x^2-1})(1+\sqrt{x^2-1})} =
\frac{1+\sqrt{x^2-1}}{1 - (x^2-1)} = -\frac{1+\sqrt{x^2-1}}{x^2} $$
\subsection*{Point d)}
$$ x^3\cdot\left(\frac{x}{x^2-1}-\frac{1}{x}\right) = x^3\cdot\left(\frac{x^2-x^2+1}{x^3-x}\right) =
\frac{x^2}{x^2-1}$$