Title: Potenssisarja ja Taylorin sarja

Tarkastellaan raja-arvon \[ \lim\limits_{x\to 0}\dfrac{x\sin([value])}{\cos([value])-e^{x^2}} \] laskemista potenssisarjojen avulla.

Määrää funktion \( x\sin([value])\) astetta kuusi oleva Maclaurinin polynomi.

Consider the evaluation of the limit \[ \lim\limits_{x\to 0}\dfrac{x\sin([value])}{\cos([value])-e^{x^2}} \] with power series.

Find the degree six Maclaurin polynomial for the function \( x\sin([value])\).

Polynomi on The polynomial is .

Laske potenssisarjojen avulla raja-arvo.

Use power series and evaluate the limit.

Raja-arvo on The limit is .
Feedback: Malliratkaisu: Potenssisarjan \(\sin(x)= x -\dfrac{x^3}{3!} + \dfrac{x^5}{5!}-\cdots\), \(x\in\mathbb{R}\), avulla saadaan Model solution: By using the series \(\sin(x)= x -\dfrac{x^3}{3!} + \dfrac{x^5}{5!}-\cdots\), \(x\in\mathbb{R}\) we get \begin{align*} x\sin([value])&=x([value] -\dfrac{([value])^3}{3!} + \dfrac{([value])^5}{5!}-\cdots)\\ &=[value] -\dfrac{[value]^3x^4}{3!} + \dfrac{[value]^5x^6}{5!}-\cdots\\ &=[value] -[value]x^4 + [value]x^6-\cdots. \end{align*} Tunnetaan potenssisarjat By using the known Maclaurin series \[ \begin{align*} x\sin([value])&= [value] -[value]x^4 + [value]x^6-\cdots,\ x\in\mathbb{R},\\ \cos(x)&= 1 -\dfrac{x^2}{2!} + \dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots,\ x\in\mathbb{R},\\ e^{x}&= 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}+\cdots,\ x\in\mathbb{R}, \end{align*} \] joten saadaan it follows \[ \begin{align*} &\hphantom{=}\,\, \lim\limits_{x\to 0}\dfrac{x\sin([value])}{\cos([value])-e^{x^2}}\\ &=\lim\limits_{x\to 0} \dfrac{[value] -[value]x^4 + [value]x^6-\cdots} {1 -\dfrac{([value])^2}{2!} + \dfrac{([value])^4}{4!}-\dfrac{([value])^6}{6!}+\cdots -(1 + x^2 + \dfrac{(x^2)^2}{2!} + \dfrac{(x^2)^3}{3!}+\cdots)}\\ &=\lim\limits_{x\to 0} \dfrac{[value] -[value]x^4 + [value]x^6-\cdots} {1 -\dfrac{[value]^2x^2}{2!} + \dfrac{[value]^4x^4}{4!}-\dfrac{[value]^6x^6}{6!}+\cdots -1 -x^2 - \dfrac{x^4}{2!} - \dfrac{x^6}{3!}-\cdots}\\ &=\lim\limits_{x\to 0} \dfrac{([value] -[value]x^4 + [value]x^6-\cdots):x^2} {[-(\dfrac{[value]^2}{2!}+1)x^2 - (\dfrac{[value]^4}{4!}+\dfrac{1}{2!} )x^4 -(\dfrac{[value]^6}{6!}+ \dfrac{1}{3!})x^6-\cdots]:x^2}\\ &=\lim\limits_{x\to 0} \dfrac{[value] -[value]x^2 + [value]x^4-\cdots} {-(\dfrac{[value]^2}{2!}+1) - (\dfrac{[value]^4}{4!}+\dfrac{1}{2!} )x^2 -(\dfrac{[value]^6}{6!}+ \dfrac{1}{3!})x^4-\cdots} =-\dfrac{[value]}{\dfrac{[value]^2}{2!}+1}=-[value]. \end{align*} \]

Tarkastellaan raja-arvon \[ \lim\limits_{x\to 0}\dfrac{x\sin([value])}{\cos([value])-e^{x^2}} \] laskemista potenssisarjojen avulla.

a) Määrää funktion \( x\sin([value])\) astetta kuusi oleva Maclaurinin polynomi.

Consider the evaluation of the limit \[ \lim\limits_{x\to 0}\dfrac{x\sin([value])}{\cos([value])-e^{x^2}} \] with power series.

a) Find the degree six Maclaurin polynomial for the function \( x\sin([value])\).

b) Laske potenssisarjojen avulla raja-arvo.

b)Use power series and evaluate the limit.

a) Polynomi on a) The polynomial is .

b) Raja-arvo on b) The limit is .
Feedback: Malliratkaisu: Potenssisarjan \(\sin(x)= x -\dfrac{x^3}{3!} + \dfrac{x^5}{5!}-\cdots\), \(x\in\mathbb{R}\), avulla saadaan Model solution: By using the series \(\sin(x)= x -\dfrac{x^3}{3!} + \dfrac{x^5}{5!}-\cdots\), \(x\in\mathbb{R}\), we get \begin{align*} x\sin([value])&=x([value] -\dfrac{([value])^3}{3!} + \dfrac{([value])^5}{5!}-\cdots)\\ &=[value] -\dfrac{[value]^3x^4}{3!} + \dfrac{[value]^5x^6}{5!}-\cdots\\ &=[value] -[value]x^4 + [value]x^6-\cdots. \end{align*} Tunnetaan potenssisarjat By using the known Maclaurin series \[ \begin{align*} x\sin([value])&= [value] -[value]x^4 + [value]x^6-\cdots,\ x\in\mathbb{R},\\ \cos(x)&= 1 -\frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}+\cdots,\ x\in\mathbb{R},\\ e^{x}&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+\cdots,\ x\in\mathbb{R}, \end{align*} \] joten saadaan it follows \[ \begin{align*} &\hphantom{=}\,\, \lim\limits_{x\to 0}\dfrac{x\sin([value])}{\cos([value])-e^{x^2}}\\ &=\lim\limits_{x\to 0} \dfrac{[value] -[value]x^4 + [value]x^6-\cdots} {1 -\frac{([value])^2}{2!} + \frac{([value])^4}{4!}-\frac{([value])^6}{6!}+\cdots -(1 + x^2 + \frac{(x^2)^2}{2!} + \frac{(x^2)^3}{3!}+\cdots)}\\ &=\lim\limits_{x\to 0} \dfrac{[value] -[value]x^4 + [value]x^6-\cdots} {1 -\frac{[value]^2x^2}{2!} + \frac{[value]^4x^4}{4!}-\frac{[value]^6x^6}{6!}+\cdots -1 -x^2 - \frac{x^4}{2!} - \frac{x^6}{3!}-\cdots}\\ &=\lim\limits_{x\to 0} \dfrac{([value] -[value]x^4 + [value]x^6-\cdots):x^2} {[-(\frac{[value]^2}{2!}+1)x^2 - (\frac{[value]^4}{4!}+\frac{1}{2!} )x^4 -(\frac{[value]^6}{6!}+ \frac{1}{3!})x^6-\cdots]:x^2}\\ &=\lim\limits_{x\to 0} \dfrac{[value] -[value]x^2 + [value]x^4-\cdots} {-(\frac{[value]^2}{2!}+1) - (\frac{[value]^4}{4!}+\dfrac{1}{2!} )x^2 -(\frac{[value]^6}{6!}+ \frac{1}{3!})x^4-\cdots} =-\dfrac{[value]}{\frac{[value]^2}{2!}+1}=-[value]. \end{align*} \]

Find the radius of convergence of the power series \[\sum_{n=1}^{\infty} \frac{[value]^n(x-[value])^{n}}{n^2}\] Määritä potenssisarjan \[\sum_{n=1}^{\infty} \frac{[value]^n(x-[value])^{n}}{n^2}\] suppenemissäde

R: .
Feedback: Model solution: Mallivastaus: We examine the convergence of the series by applying ratio test: Tarkastellaan sarjan suppenemista suhdetestin avulla: \begin{align*} \left|\frac{\frac{[value]^{n+1}(x-[value])^{n+1}}{(n+1)^{2}}}{\frac{[value]^{n}(x-[value])^{n}}{n^{2}}}\right| &= \left|\frac{[value]n^{2}(x-[value])}{(n+1)^{2}}\right| = \frac{[value]n^{2}\left|x-[value]\right|}{n^{2}+2n+1} \\ &= \frac{[value]\left|x-[value]\right|}{1+2/n+1/n^{2}} \to [value]\left|x-[value]\right|, \text{whenkun } n\to\infty\\ \end{align*} Since \([value]\left|x-[value]\right|<1\) as \(\left|x-[value]\right|<\frac{1}{[value]}\), then according to the ratio test the radius of convergence is \(R = \frac{1}{[value]}\).Koska \([value]\left|x-[value]\right|<1\), kun \(\left|x-[value]\right|<\frac{1}{[value]}\), suhdetestin perusteella suppenemissäde \(R = \frac{1}{[value]}\).

Määrää potenssisarjan \[ \sum\limits^{\infty}_{k=0} [value] x^k \] suppenemissäde \(R\).

Determine the radius of convergence \(R\) for the power series \[ \sum\limits^{\infty}_{k=0} [value] x^k. \]

Suppenemissäde \(R\) on The radius of convergence \(R\) is .
Feedback: Malliratkaisu: Lasketaan raja-arvo: Model solution: Evaluate the limit: \begin{align*} L &= \lim\limits_{k\to\infty} \Big|[value]:[value]\Big|\\ &= \lim\limits_{k\to\infty} [value]\cdot[value]\\ &= \lim\limits_{k\to\infty} [value] = \lim\limits_{k\to\infty}[value]\\ &=1. \end{align*} Suppenemissäde \(R=\frac{1}{L} =1\). The radius of convergence \(R=\frac{1}{L} =1\).

Let \(f(x)=\mathrm{e}^{-x^[value]}\). Determine the [value]th order Taylor polynomial of the function \(f\) about the point \(a=0\).Olkoon \(f(x)=\mathrm{e}^{-x^[value]}\). Laske funktion \(f\) Taylor-polynomi astetta [value], kun kehityskeskuksena on \(a=0\).

Hint: Make use of the familiar Taylor series of the function \(x\mapsto e^x\)! Vihje: Käytä apuna funktion \(x\mapsto e^x\) tuttua Taylor-kehitelmää!

\( P_[value](x)= \)
Feedback: Model solution: Mallivastaus: \(f(x)=\mathrm{e}^{-x^[value]}\). Taylor polynomial of order [value] about the point \(a=0\): \[e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!} \Rightarrow e^{-x^[value]}=\sum_{k=0}^{\infty} \frac{(-x^[value])^k}{k!}=\sum_{k=0}^{\infty} \frac{(-1)^k}{k!}x^{[value]k}. \] Taylorin polynomi astetta [value] pisteessä \(a=0\): \[e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!} \Rightarrow e^{-x^[value]}=\sum_{k=0}^{\infty} \frac{(-x^[value])^k}{k!}=\sum_{k=0}^{\infty} \frac{(-1)^k}{k!}x^{[value]k}. \] [value]th order Taylor terms, whose degree \(\leq [value]\): \[P_[value]f(x)=\sum_{k=0}^{3}\frac{(-1)^k}{k!}x^{[value]k}=1-x^[value]+\frac{1}{2}x^[value]-\frac{1}{6}x^[value]. \] [value]. asteen Taylorin termit, joiden aste \(\leq [value]\): \[P_[value]f(x)=\sum_{k=0}^{3}\frac{(-1)^k}{k!}x^{[value]k}=1-x^[value]+\frac{1}{2}x^[value]-\frac{1}{6}x^[value]. \]