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https://github.com/danbulant/notes
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vault backup: 2024-09-24 09:39:22
This commit is contained in:
Daniel Bulant
parent
bc699aceef
commit
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5 changed files with 143 additions and 38 deletions
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@ -12,7 +12,7 @@ imagePrefix: 'data/'
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- [[cjl/jazyky/jazyky|jazyky]]
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- [[cjl/jména|jména]]
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- [[cjl/literatura/literatura|literatura]]
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- [[Maturita|Maturita]]
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- [[cjl/Maturita|Maturita]]
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- [[cjl/Pedagog a didaktik|Pedagog a didaktik]]
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- [[cjl/Povinné knihy|Povinné knihy]]
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- [[cjl/Slohové práce/Slohové práce|Slohové práce]]
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@ -62,13 +62,94 @@ $-15/2=2a_3$
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$a_3=-15/4$
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$a_{n-1}=3n+2a_{n+1}-{a_n}$
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$a_5=1;a_6=7$
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$a_4=3*5+2*7-1$
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$a_4=15+14-1=28$
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$a_3=3*4+2*1-28$
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$a_3=12+2-28=-14$
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$a_2=3*3+2*28+14$
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$a_2=9+56+14=79$
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$a_5=3*6+2*a_7-7$
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$1=18+2a_7-7=11+2a_7$
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$-2a_7=10$
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$a_7=-5$
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$a_6=3*7+2*a_8+5$
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$a_6=21+2a_8+5=26+2a_8$
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$7=26+2a_8$
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$2a_8=7-26=-19$
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$a_8=-9.5$
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---
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Dokaž že posloupnost je nebo není monotónní; jak?
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$a_n=\frac{n+1}n \rightarrow a_{n+1}=\frac{n+1+1}{n+1}$
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$a_{n+1}-a_n=\frac{n+2}{n+1}-\frac{n+1}{n}=\frac{(n+2)n}{n(n+1)}-\frac{(n+1)n}{n(n+1)}$
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$\frac{n^2+2n}{n(n+1)}-\frac{(n^2+n)}{n(n+1)}\lt0\forall n \in \mathbb{N}$
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klesajici
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$a_n=\frac{2n+1}{n+2}$
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$a_{n+1}=\frac{2(n+1)+1}{n+1+2}$
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$\frac{2n+1}{n+2}-\frac{2(n+1)+1}{n+1+2}$
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$\frac{(2n+1)(n+1+2)}{(n+2)(n+1+2)}-\frac{(n+2)(2(n+1)+1)}{(n+2)(n+1+2)}$
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$\frac{(2n+1)(n+3)-(n+2)(2(n+1)+1)}{(n+2)(n+3)}$
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$\frac{2n^2+6n+n+3-(2n(n+1)+n+4(n+1)+2)}{(n+2)(n+3)}$
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$\frac{2n^2+6n+n+3-(2n^2+2n+n+4n+4+2)}{(n+2)(n+3)}$
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$\frac{2n^2+7n+3-(2n^2+7n+6)}{(n+2)(n+3)}$
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$\frac{2n^2+7n+3-2n^2-7n-6}{(n+2)(n+3)}$
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Odečteno ve špatném pořadí! Proto obrácené znaménko
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$a_{n+1}-a_n$
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$\frac{3}{(n+2)(n+3)}>0\forall n\in\mathbb{N}$
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---
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$\{n(-1)^n\}^\infty_1$
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Přímý důkaz:
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$a_n=n(-1)^n$
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$a_{n+1}=(n+1)(-1)^{n+1}$
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$(n+1)(-1)^{n+1}-n(-1)^n$
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$(n+1)(-1)(-1)^n-n(-1)^n$
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$(-1)^n[-n-1-n]$
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$\pm$ vlevo
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$-$ vpravo
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Sporem:
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$a_1=-1$
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$a_2=2$
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$a_3=-3$
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Není monotóní - sporem jsme dokázali neplatnost podmínek pro monotónost
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## Funkce
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### Vlastnosti
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#### Monotónost
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- Rostoucí
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- Nerostoucí
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- Klesající
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- Neklesající
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- Rostoucí $\forall n:a_n<a_{n-1}; a_{n+1}-a_n>0$
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- Nerostoucí $a_{n+1}-a_n\le0$
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- Klesající $\forall n:a_n>a_{n-1};a_{n+1}-a_n<0$
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- Neklesající $a_{n+1}-a_n\ge0$
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#### Prostota
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- Prostá - neopakují se hodnoty, dá se sestavit inverzní funkce
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#### Parita
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