#1 22. 12. 2016 12:11
Sequence
Determine the general term of the sequence,
.
Srdecne Vanok
The respect, the politeness are essential qualities...and also the willingness.
Do not judge the other one.
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#2 22. 12. 2016 23:39 — Editoval Pavel (22. 12. 2016 23:42)
Re: Sequence
we have 
for every 
.
. Let us define a sequence 
such that 
. Then![$
\sqrt 2\tanh v_{n+1}
&=\frac{4\sqrt 2\tanh v_n}{2(\tanh^2 v_n+1)}
=\frac{2\sqrt 2\cdot\frac{\sinh v_n}{\cosh v_n}}{\frac{\sinh^2 v_n}{\cosh^2 v_n}+1}
=\sqrt 2\cdot\frac{2\sinh v_n\cosh v_n}{\sinh^2 v_n+\cosh^2 v_n}
=\sqrt 2\cdot\frac{\sinh(2v_n)}{\cosh(2v_n)}
=\sqrt 2\tanh(2v_n)\\[.5\baselineskip]
\tanh v_{n+1}&=\tanh(2v_n)\\[.5\baselineskip]
\tanh v_n&=\tanh\left(2^nv_0\right)
$](/mathtex/63/630ea6b5ab79a2a0a45547c850c9dfeb.gif "kopírovat do textarea")
:![$
\frac{u_n}{\sqrt 2}&=\tanh\left(2^n\mathop{\mathrm{artanh}}\left(\frac{u_0}{\sqrt 2}\right)\right)\\[.5\baselineskip]
u_n&=\sqrt 2\tanh\left(2^n\cdot\frac 12\log\left(\frac{1+\frac{u_0}{\sqrt 2}}{1-\frac{u_0}{\sqrt 2}}\right)\right)
=\sqrt 2\tanh\left(\log\left(\frac{\sqrt 2+u_0}{\sqrt 2-u_0}\right)^{2^{n-1}}\right)
=\sqrt 2\cdot\frac{\exp\left(2\log\left(\frac{\sqrt 2+u_0}{\sqrt 2-u_0}\right)^{2^{n-1}}\right)-1}{\exp\left(2\log\left(\frac{\sqrt 2+u_0}{\sqrt 2-u_0}\right)^{2^{n-1}}\right)+1}
$](/mathtex/98/98db0261ea4e972719693c42936592b9.gif "kopírovat do textarea")
Backslash je v TeXu tak důležitý jako nekonečno při dělení nulou v tělesech charakteristiky 0.
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#3 23. 12. 2016 03:49
Re: Sequence
Hi ↑ Pavel:,
Yes, also look
, then 
.Srdecne Vanok
The respect, the politeness are essential qualities...and also the willingness.
Do not judge the other one.
Ak odpovedam na nejaku otazku. MOJ PRINCIP NIE JE DAT ODPOVED ALE UKAZAT AKO SA K ODPOVEDI DOSTAT
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