Rekker: Potens-rekker

En potens-rekke omkring $x = a$ er en rekke på formen:

\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1 (x-a) + c_2 (x-a)^2 + \cdots

der $c_n$ er til koeffisienten til det n-te leddet og $a$ er en konstant.

Konvergensradius, $r$, er et tall slik at:

  • Rekken konvergerer for alle $x$ når $|x-a| < r$
  • Rekken divergerer for alle $x$ når $|x-a| > r$.

+ Eksempel 1

For hvilke $x$-verdier konvergerer rekken?

\sum_{n=0}^{\infty} 3(x-3)^n

Ofte bruker vi forholdstesten for å finne konvergensradiusen:

L = \lim_{n\to\infty} \left| \frac{\textcolor{red}{a_{n+1}}}{\textcolor{blue}{a_n}} \right| 
= \lim_{n \to \infty} \left| \frac{\textcolor{red}{3(x-3)^{n+1}}}{\textcolor{blue}{3(x-3)^n}}\right| 
= \lim_{n \to \infty} \left| \frac{\cancel{3}(x-3)\cancel{(x-3)^n}}{\cancel{3}\cancel{(x-3)^n}}\right| 
= |x-3|

Forholdstesten gir:

  • Dersom $L = |x-3| < 1$, konvergerer rekken, dvs. $2 < x < 4$
  • Dersom $L = |x-3| > 1 $, divergerer rekken, dvs. $x < 2$ og $x > 4$.

Og, vipps, vet vi at konvergensradiusen er 1.

+ Eksempel 2

For hvilke $x$-verdier konvergerer rekken?

\sum_{n=0}^{\infty} \frac{3}{n}(x-2)^n

Forholdstesten brukes for å finne konvergensradiusen:

\begin{aligned}
L = \lim_{n\to\infty} \left| \frac{\textcolor{red}{a_{n+1}}}{\textcolor{blue}{a_n}} \right| 
& = \lim_{n \to \infty} \left| \frac{\textcolor{red}{\frac{3}{n+1}(x-2)^{n+1}}}{\textcolor{blue}{\frac{3}{n}(x-2)^n}}\right|  \\
\textnormal{Vil fjerne småbrøkene: } \quad & L = \lim_{n \to \infty} \left| \frac{\frac{3}{\textcolor{green}{n+1}}(x-2)^{n+1} \cdot \textcolor{green}{n(n+1)}}{\frac{3}{\textcolor{green}{n}}(x-2)^n \cdot \textcolor{green}{n(n+1)}}\right|  \\
\textnormal{Forkorter: } \quad & L = \lim_{n \to \infty} \left| \frac{\frac{3}{\cancel{n+1}}(x-2)^{n+1} \cdot n\cancel{(n+1)}}{\frac{3}{\cancel{n}}(x-2)^n\cdot \cancel{n}(n+1)}\right|  \\
\textnormal{Rydder: } \quad & L = \lim_{n \to \infty} \left| \frac{3(x-2)^{n+1} \cdot n}{3(x-2)^n\cdot (n+1)}\right|  \\
\textnormal{Forkorter: } \quad & L = \lim_{n \to \infty} \left| \frac{\cancel{3}(x-2)\cancel{(x-2)^n} \cdot n}{\cancel{3}\cancel{(x-2)^n}\cdot (n+1)}\right|  \\
\textnormal{Deler på høyeste potens: } \quad & L = \lim_{n \to \infty} \left| \frac{(x-2) \cdot n \cdot \textcolor{green}{\frac{1}{n}}}{(n+1) \cdot \textcolor{green}{\frac{1}{n}}}\right|  \\
\textnormal{$\frac{1}{n}$ inn i parentesen: } \quad & L = \lim_{n \to \infty} \left| \frac{(x-2) }{1 + \frac{1}{n}}\right|  \\
\textnormal{Lar $n \to \infty$: } \quad & L = |x-2|
\end{aligned}

Forholdstesten gir:

  • Dersom $L = |x-2| < 1$, konvergerer rekken, dvs. $1 < x < 3$
  • Dersom $L = |x-2| > 1 $, divergerer rekken, dvs. $x < 1$ og $x > 3$.

Og, vipps, vet vi at konvergensradiusen er 1.

+ Eksempel 3

For hvilke $x$-verdier konvergerer rekken?

\sum_{n=0}^{\infty} \frac{3^n}{n^3 + 3} (x-3)^n

Forholdstesten brukes for å finne konvergensradiusen:

\begin{aligned}
L = \lim_{n\to\infty} \left| \frac{\textcolor{red}{a_{n+1}}}{\textcolor{blue}{a_n}} \right| 
& = \lim_{n \to \infty} \left| \frac{\textcolor{red}{\frac{3^{n+1}}{(n+1)^3 + 3}(x-3)^{n+1}}}{\textcolor{blue}{\frac{3^n}{n^3 + 3}(x-3)^n}}\right| \\
\textnormal{Vil fjerne småbrøkene: } \quad & L = \lim_{n \to \infty} \left| \frac{\frac{3^{n+1}}{\textcolor{green}{(n+1)^3+3}}(x-3)^{n+1} \cdot \textcolor{green}{(n^3+3)((n+1)^3+3)}}{\frac{3^n}{\textcolor{green}{n^3+3}}(x-3)^n \cdot \textcolor{green}{(n^3+3)((n+1)^3+3)}}\right|  \\
\textnormal{Forkorter: } \quad & L = \lim_{n \to \infty} \left| \frac{\frac{3^{n+1}}{\cancel{(n+1)^3+3}}(x-3)^{n+1} \cdot (n^3+3)\cancel{((n+1)^3 +3)}}{\frac{3^n}{\cancel{n^3+3}}(x-3)^n \cdot \cancel{(n^3+3)}((n+1)^3+3)}\right|  \\
\textnormal{Rydder: } \quad & L = \lim_{n \to \infty} \left| \frac{3^{n+1} (x-3)^{n+1} \cdot (n^3+3)}{3^n (x-3)^n \cdot ((n+1)^3+3)}\right|  \\
\textnormal{Forkorter: } \quad & L = \lim_{n \to \infty} \left| \frac{3 \cdot \cancel{3^n} (x-3) \cancel{(x-3)^n} \cdot (n^3+3)}{\cancel{3^n} \cancel{(x-3)^n} \cdot ((n+1)^3+3)}\right|  \\
\textnormal{Deler på høyeste potens: } \quad & L = \lim_{n \to \infty} \left| \frac{3 (x-3)\cdot (n^3+3) \cdot \textcolor{green}{\frac{1}{n^3}}}{((n+1)^3+3) \cdot \textcolor{green}{\frac{1}{n^3}} }\right|  \\
\textnormal{$\frac{1}{n^3}$ inn i parentesene: } \quad & L= \lim_{n \to \infty} \left| \frac{3 (x-3) \cdot \Big(1+\frac{3}{n^3}\Big) }{ \Big(1 + \frac{1}{n}\Big)^3+ \frac{3}{n^3} }\right|  \\
\textnormal{Lar $n \to \infty$: } \quad & L = |3(x-3)|
\end{aligned}

Forholdstesten gir:

  • Dersom $L = |3(x-3)| < 1$, dvs. $|x-3| < \frac{1}{3}$, konvergerer rekken, dvs. $3 – \frac{1}{3}< x < 3 + \frac{1}{3}$.
  • Dersom $L = |3(x-3)| > 1 $, dvs. $|x-3| > \frac{1}{3}$, divergerer rekken, dvs. $x < 3 – \frac{1}{3}$ og $x > 3 + \frac{1}{3}$.

Og, vipps, vet vi at konvergensradiusen er $\frac{1}{3}$.

+ Eksempel 4

For hvilke $x$-verdier konvergerer rekken?

\sum_{n=0}^{\infty} \frac{n^3}{n^3 + 3} (x-3)^n

Forholdstesten brukes for å finne konvergensradiusen:

\begin{aligned}
L = \lim_{n\to\infty} \left| \frac{\textcolor{red}{a_{n+1}}}{\textcolor{blue}{a_n}} \right| 
& = \lim_{n \to \infty} \left| \frac{\textcolor{red}{\frac{(n+1)^3}{(n+1)^3 + 3}(x-3)^{n+1}}}{\textcolor{blue}{\frac{n^3}{n^3 + 3}(x-3)^n}}\right|  \\
\textnormal{Vil fjerne småbrøkene: } \quad & L = \lim_{n \to \infty} \left| \frac{\frac{(n+1)^3}{\textcolor{green}{(n+1)^3+3}}(x-3)^{n+1} \cdot \textcolor{green}{(n^3+3)((n+1)^3+3)}}{\frac{n^3}{\textcolor{green}{n^3+3}}(x-3)^n \cdot \textcolor{green}{(n^3+3)((n+1)^3+3)}}\right|  \\
\textnormal{Forkorter: } \quad & L = \lim_{n \to \infty} \left| \frac{\frac{(n+1)^3}{\cancel{(n+1)^3+3}}(x-3)^{n+1} \cdot (n^3+3)\cancel{((n+1)^3+3)}}{\frac{n^3}{\cancel{n^3+3}}(x-3)^n \cdot \cancel{(n^3+3)}((n+1)^3+3)}\right|  \\
\textnormal{Rydder: } \quad & L = \lim_{n \to \infty} \left| \frac{(n+1)^3 (x-3)^{n+1} \cdot (n^3+3)}{n^3 (x-3)^n \cdot ((n+1)^3+3)}\right|  \\
\textnormal{Forkorter: } \quad & L = \lim_{n \to \infty} \left| \frac{(n+1)^3 (x-3) \cancel{(x-3)^n} \cdot (n^3+3)}{n^3 \cancel{(x-3)^n} \cdot ((n+1)^3+3)}\right|  \\
\textnormal{Deler på høyeste potens: } \quad & L = \lim_{n \to \infty} \left| \frac{(n+1)^3 \cdot \textcolor{green}{\frac{1}{n^3}} (x-3)\cdot (n^3+3) \cdot \textcolor{green}{\frac{1}{n^3}}}{n^3 \cdot \textcolor{green}{\frac{1}{n^3}} \cdot ((n+1)^3+3) \cdot \textcolor{green}{\frac{1}{n^3}} }\right|  \\
\textnormal{$\frac{1}{n^3}$ inn i parentesene: } \quad & L = \lim_{n \to \infty} \left| \frac{\Big(1+\frac{1}{n}\Big)^3 \cdot (x-3) \cdot \Big(1+\frac{3}{n^3}\Big) }{ \Big(1 + \frac{1}{n}\Big)^3+ \frac{3}{n^3} }\right|  \\
\textnormal{Lar $n \to \infty$: } \quad & L = |x-3|
\end{aligned}

Forholdstesten gir:

  • Dersom $L = |x-3| < 1$, konvergerer rekken, dvs. $2 < x < 4$
  • Dersom $L = |x-3| > 1 $, divergerer rekken, dvs. $x < 2$ og $x > 4$.

Og, vipps, vet vi at konvergensradiusen er 1.

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