Fungsi hiperbolik: Perbedaan antara revisi
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[[Berkas:Hyperbolic functions-2.svg|jmpl|ka|200px|Fungsi
'''Fungsi
== Definisi ==
[[Berkas:sinh cosh tanh.svg|thumb|<span style="color:#b30000;">sinh</span>, <span style="color:#00b300;">cosh</span>
[[Berkas:csch sech coth.svg|thumb|<span style="color:#b30000;">csch</span>, <span style="color:#00b300;">sech</span>
=== Definisi Eksponen ===
[[Berkas:Hyperbolic and exponential; sinh.svg|thumb|right|{{math|sinh ''x''}}
[[File:Hyperbolic and exponential; cosh.svg|thumb|right|{{math|cosh ''x''}}
Dalam istilah dari [[fungsi eksponensial]]:
* Hiperbolik sinus:
*:<math>\sinh x = \frac {e^x - e^{-x}} {2} = \frac {e^{2x} - 1} {2e^x} = \frac {1 - e^{-2x}} {2e^{-x}}.</math>
* Hiperbolik
*:<math>\cosh x = \frac {e^x + e^{-x}} {2} = \frac {e^{2x} + 1} {2e^x} = \frac {1 + e^{-2x}} {2e^{-x}}.</math>
* Hiperbolik
*:<math>\tanh x = \frac{\sinh x}{\cosh x} = \frac {e^x - e^{-x}} {e^x + e^{-x}}
= \frac{e^{2x} - 1} {e^{2x} + 1}</math>
* Hiperbolik
*:<math>\coth x = \frac{\cosh x}{\sinh x} = \frac {e^x + e^{-x}} {e^x - e^{-x}}
= \frac{e^{2x} + 1} {e^{2x} - 1}</math>
*
*:<math>\operatorname{sech} x = \frac{1}{\cosh x} = \frac {2} {e^x + e^{-x}}
= \frac{2e^x} {e^{2x} + 1}</math>
*
*:<math>\operatorname{csch} x = \frac{1}{\sinh x} = \frac {2} {e^x - e^{-x}}
= \frac{2e^x} {e^{2x} - 1}</math>
=== Definisi
- Dalam pengembangan -
<!--The hyperbolic functions may be defined as solutions of [[differential equation]]s: The hyperbolic sine and cosine are the unique solution {{math|(''s'', ''c'')}} of the system
Baris 69 ⟶ 42:
such that {{math|1=''f'' (0) = 1}}, {{math|1=''f'' ′(0) = 0}} for the hyperbolic cosine, and {{math|1=''f'' (0) = 0}}, {{math|1=''f'' ′(0) = 1}} for the hyperbolic sine.-->
=== Definisi
-Dalam pengembangan -
<!--Hyperbolic functions may also be deduced from [[trigonometric function]]s with [[complex number|complex]] arguments:
Baris 89 ⟶ 62:
The above definitions are related to the exponential definitions via [[Euler's formula]]. (See "Hyperbolic functions for complex numbers" below.)-->
==
- Dalam pengembangan -
<!--=== Hyperbolic cosine ===
Baris 156 ⟶ 129:
for the other functions.-->
===
:<math>\begin{align}
\sinh(x + y) &= \sinh x \cosh y + \cosh x \sinh y \\
Baris 163 ⟶ 136:
\end{align}</math>
terutama
:<math>\begin{align}
\cosh (2x) &= \sinh^2{x} + \cosh^2{x} = 2\sinh^2 x + 1 = 2\cosh^2 x - 1\\
Baris 174 ⟶ 147:
\sinh x + \sinh y &= 2 \sinh \left(\frac{x+y}{2}\right) \cosh \left(\frac{x-y}{2}\right)\\
\cosh x + \cosh y &= 2 \cosh \left(\frac{x+y}{2}\right) \cosh \left(\frac{x-y}{2}\right)\\
\end{align}</math>
===
:<math>\begin{align}
\sinh(x - y) &= \sinh x \cosh y - \cosh x \sinh y \\
Baris 184 ⟶ 156:
\end{align}</math>
:<math>\begin{align}
\sinh x - \sinh y &= 2 \cosh \left(\frac{x+y}{2}\right) \sinh \left(\frac{x-y}{2}\right)\\
Baris 190 ⟶ 162:
\end{align}</math>
=== Rumus
:<math>\begin{align}
\sinh\left(\frac{x}{2}\right) &= \frac{\sinh x}{\sqrt{2 (\cosh x + 1)} } &&= \sgn x \, \sqrt \frac{\cosh x - 1}{2} \\[6px]
Baris 198 ⟶ 169:
\end{align}</math>
:<math> \tanh\left(\frac{x}{2}\right) = \frac{\cosh x - 1}{\sinh x} = \coth x - \operatorname{csch} x </math>
=== Rumus kuadrat ===
:<math>\begin{align}
\sinh^2 x &= \frac{1}{2}(\cosh 2x -1) \\
Baris 211 ⟶ 183:
=== Pertidaksamaan ===
Pertidaksamaan berikut sangat berguna dalam statistik, yaitu <math>\operatorname{cosh}(t) \leq e^{t^2 /2}</math><ref>{{cite article|last1=Audibert|first1=Jean-Yves|title=Fast learning rates in statistical inference through aggregation|date=2009|publisher=The Annals of Statistics|page=1627}} [https://projecteuclid.org/download/pdfview_1/euclid.aos/1245332827] {{Webarchive|url=https://web.archive.org/web/20230726134112/https://projecteuclid.org/journalArticle/Download?urlId=10.1214%2F08-AOS623&isResultClick=False |date=2023-07-26 }}</ref>
== Fungsi invers sebagai logaritma ==
{{main|
:<math>\begin{align}
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\operatorname {arcoth} (x) &= \frac{1}{2}\ln \left( \frac{x + 1}{x - 1} \right) && |x| > 1 \\
\operatorname {arsech} (x) &= \ln \left( \frac{1}{x} + \sqrt{\frac{1}{x^2} - 1}\right) = \ln \left( \frac{1+ \sqrt{1 - x^2}}{x} \right) && 0 < x \leqslant 1 \\
\operatorname {arcsch} (x) &= \ln \left( \frac{1}{x} + \sqrt{\frac{1}{x^2} +1}\right) = \ln \left( \frac{1+ \sqrt{1 + x^2}}{x} \right) && x \ne 0
\end{align}</math>
<!--
Pembuktian <math>arcsinh x = ln (x + \sqrt{x^2 + 1})</math>!
: <math>y = arcsinh x</math>
: <math>x = arcsinh y</math>
: <math>x = \frac{e^y - e^{-y}}{2}</math>
: <math>2x = e^y - e^{-y}</math>
: <math>2e^yx = e^2y - 1</math>
: <math>(e^y)^2 - 2x(e^y) - 1 = 0</math>
: <math>e^y = \frac{2x + \sqrt{4x^2 + 4}}{2}</math>
: <math>e^y = x + \sqrt{x^2 + 1}</math>
: <math>ln e^y = ln (x + \sqrt{x^2 + 1})</math>
: <math>y ln e = ln (x + \sqrt{x^2 + 1})</math>
: <math>y = ln (x + \sqrt{x^2 + 1})</math>
: <math>arcsinh x = ln (x + \sqrt{x^2 + 1})</math>
-->
== Turunan ==
:<math>\begin{align}
\frac{d}{dx}\sinh x &= \cosh x \\
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<!-- from: http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/calculus/tableof.html and http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=2664 -->
==
- Dalam pengembangan -
<!--Sinh and cosh are both equal to their [[second derivative]], that is:
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\end{align}</math>
----
:<math>\begin{align}
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|}-->
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