Poglejmo, kaj imajo skupnega naslednje naloge:
- Določi vrednost izraza
![Rendered by QuickLaTeX.com \[ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}},\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-a4c7a4698efb998a3a79fb308ede73fb_l3.png)
- Določi vrednost izraza
![Rendered by QuickLaTeX.com \[ \sqrt{2\sqrt{2\sqrt{2\sqrt{2\dots}}}},\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-67c79f1c23e1dd6bf5dc34eaf4b0843a_l3.png)
- Določi vrednost verižnega ulomka
![Rendered by QuickLaTeX.com \[1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-45a428a1d41fe0c82d80eb1a22a1d4c1_l3.png)
- Določi vrednost nadomestnega upora v neskončni verigi uporov na skici, če je
in
.
Rešitev: Vse naloge so take, da lahko nadomestimo del izraza s celotnim
- Označimo
![Rendered by QuickLaTeX.com \[x=\sqrt{2+\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}}_{x}},\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-45b48cbb296efd722c7361fc87f3153c_l3.png)
Če celoten izraz izrazimo z x, dobimo enačbo
![Rendered by QuickLaTeX.com \[x=\sqrt{2+x}\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-3ed0b1a1701c655982cfc4c6e62a4375_l3.png)
. Rešitev te enačbe, ki ustreza, je

- Ravno tako označimo
![Rendered by QuickLaTeX.com \[x= \sqrt{2\underbrace{\sqrt{2\sqrt{2\sqrt{2\dots}}}}_{x}},\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-5af37c9e1353e932592d0446255c4f2b_l3.png)
dobimo enačbo
![Rendered by QuickLaTeX.com \[x^2=2x\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-d524fbc8f5c180e2d16d7fb8fd7d2895_l3.png)
, kateri ustreza rešitev

- Podobno označimo
![Rendered by QuickLaTeX.com \[x=1+\frac{1}{\underbrace{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}_{x}}\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-655fdbbe0a7482d0ba7b77452db7b87a_l3.png)
. Tokrat dobimo enačbo
![Rendered by QuickLaTeX.com \[x=1+\frac{1}{x}\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-88fd6b69feca562c8998b1e34fd779fe_l3.png)
oziroma kvadratno enačbo
![Rendered by QuickLaTeX.com \[x^2-x-1=0,\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-6ad47e2c7ed1e67505767bacf9510867_l3.png)
ki ima za rešitev zlato število
![Rendered by QuickLaTeX.com \[\varphi=\frac{1+\sqrt{5}}{2}.\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-ab4d37d437ff0bfc041cee62f199e0b1_l3.png)
- Opazimo, da to vezje lahko nadomestimo z vezjem
ki ima nadomestni upor
![Rendered by QuickLaTeX.com \[R=2R_1+\frac{R_2R}{R_2+R}.\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-2fd929889e3ab190c2bc23893f2d05da_l3.png)
Od tod dobimo kvadratno enačbo
![Rendered by QuickLaTeX.com \[R^2-2R_1R-2R_1R_2=0,\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-1682c0752f719c8eed7328f0b261e31c_l3.png)
ki ima rešitev
![Rendered by QuickLaTeX.com \[R=R_1+\sqrt{R_1^2+2R_1R_2}=30\Omega.\]](https://vincenc.petruna.com/wp-content/ql-cache/quicklatex.com-ab6e632df71696c362da5857834fdf25_l3.png)
“Opazimo, da to vezje lahko nadomestimo z vezjem”
he, he, ‘opazimo’ – zanimiv način reševanja …