Title: Potenssisarja ja Taylorin sarja

Laske 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 \(R=\frac{1}{L} =1\).

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 \(R = \frac{1}{[value]}\).

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 \(\leq [value]\): \[P_[value]f(x)=\sum_{k=0}^{3}\frac{(-1)^k}{k!}x^{[value]k}=1-x^[value]+\frac{1}{2}x^[value]-\frac{1}{6}x^[value]. \]

a) Tarkastellaan sarjaa \(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{2^n}(x-5)^n\).

- Sarjan kehityskeskus on .
- Kerroin \(a_n = \) .
- Suhdetesti \(\displaystyle \lim_{n\to \infty}\left|\frac{u_{n+1}}{u_n}\right|\) antaa tulokseksi , kun \(\displaystyle u_n = \frac{n!}{2^n}(x-5)^n\).
- Suppenemissäde \(R = \) .
- Milloin sarja suppenee? \(x = \) .

b) Tarkastellaan sarjaa \(\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k\cdot 10^k}(x-2)^k\).

- Kehityskeskus on .
- Suppenemissäde \(R = \) .
- Sarja suppenee, kun \(x \in \Big(\)\(,\)\(\Big]\).
(päätepisteet on tarkasteltu erikseen)

c) Tarkastellaan sarjaa \(\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1}10^k}{k!}(x-10)^k\).

- Sarjan kehityskeskus on .
- Suhdetesti \(\displaystyle \lim_{n\to \infty}\left|\frac{u_{n+1}}{u_n}\right|\) antaa tulokseksi \(0\), kun \(\displaystyle u_n = \frac{(-1)^{k+1}10^k}{k!}(x-10)^k\).
- Tällöin \(R = \) .
- Sarja suppenee, kun \(x\in \Big(\)\(,\)\(\Big)\).

Huom! \(\infty = \) inf.
Feedback: Malliratkaisu: a) Sarjan kehityskeskus on \(5\) ja \(\displaystyle a_n = \frac{n!}{2^n}\). Suhdetesti \(\displaystyle \lim_{n\to \infty}\left|\frac{\frac{(n+1)!}{2^{n+1}}\cdot (x-5)^{n+1}}{\frac{n!}{2^n}(x-5)^n}\right| = \lim_{n\to \infty} \frac{(n+1)!2^n}{n!2^{n+1}}\frac{|x-5|^{n+1}}{|x-5|^n} = \lim_{n\to \infty} \frac{n+1}{2}|x-5| = \infty\). \(\displaystyle R = \frac{"1"}{\infty} = 0\). Siispä sarja suppenee vain kehityskeskuksessa \(x = 5\). b) Kehityskeskus on \(2\). Suppenemissäde \(\displaystyle R = \lim_{k\to \infty} \left| \frac{a_k}{a_{k+1}} \right| = \lim_{k\to \infty} \left| \frac{\frac{1}{k\cdot 10^k}}{\frac{1}{(k+1)\cdot 10^{k+1}}} \right| = \lim_{k\to \infty} \left| \frac{(k+1)\cdot 10^{k+1}}{k\cdot 10^k} \right| = \lim_{k\to \infty} \frac{10(k+1)}{k} = \lim_{k\to \infty} 10+\frac{1}{k} = 10\). Sarja suppenee, kun \(|x-2|<10\) eli \(-8<x<12\). Päätepisteet tulee tutkia erikseen. c) Kehityskeskus on \(10\). Suppenemissäde \(\displaystyle R = \lim_{k\to \infty} \left| \frac{a_k}{a_{k+1}} \right| = \lim_{k\to \infty} \left| \frac{\frac{10^k}{k!}}{\frac{10^{k+1}}{(k+1)!}} \right| = \lim_{k\to \infty} \left| \frac{10^k(k+1)!}{10^{k+1}k!} \right| = \lim_{k\to \infty} \frac{k+1}{10} = \infty\). Sarja suppenee kaikilla \(x\in \mathbb{R}\) eli \(x\in (-\infty, \infty)\).

Tutkitaan potenssisarjan \[ \sum\limits^{\infty}_{k=0} \frac{[value]}{[value]} x^k \] suppenemista.

a) Laske potenssisarjan suppenemissäde \(R\).

b) Suppeneeko potenssisarja pisteessä \(x=1\)?

c) Suppeneeko potenssisarja pisteessä \(x=[value]\)?

Consider the convergence of the power series \[ \sum\limits^{\infty}_{k=0} \frac{[value]}{[value]} x^k. \]

a) Determine the radius of convergence \(R\) for the power series. (2/3 p)

b) Does the power series converge at the point \(x=1\)? (1/6 p)

c) Does the power series converge at the point \(x=[value]\)? (1/6 p)

a) Suppenemissäde on a) The radius of convergence is .

b) Väite: Potenssisarja suppenee pisteessä \(x=1\). b) Claim: The power series converges at the point \(x=1\).

c) Väite: Potenssisarja suppenee pisteessä \(x=[value]\). c) Claim: The power series converges at the point \(x=[value]\).
Feedback: Malliratkaisu: \(\textrm{a) }\) Lasketaan raja-arvo: Model solution: \(\textrm{a) }\) Evaluate the limit: \[ \begin{align*} L &= \lim\limits_{k\to\infty} \Big|\frac{[value]}{[value]}:\frac{ [value]}{[value]}\Big|\\ &= \lim\limits_{k\to\infty}\frac{[value]}{[value]}\cdot\frac{[value]}{[value]}\\ &= \lim\limits_{k\to\infty}\frac{[value]}{[value]} = \lim\limits_{k\to\infty}\frac{1}{e^[value]}=\frac{1}{e^[value]}. \end{align*} \] \(\hphantom{a) }\) Suppenemissäde \(R=\frac{1}{L} =e^[value]\). \(\hphantom{a) }\) The radius of convergence \(R=\frac{1}{L} =e^[value]\). \(\textrm{b) }\) Koska potenssisarjan suppenemissäde \(R=e^[value]\), potenssisarja suppenee varmasti, kun \( | x | < e^[value] \) eli \( -e^[value] < x < e^[value] \). Piste \(x=1\) kuuluu tähän väliin, joten potenssisarja suppenee pisteessä \(x=1\). \(\textrm{b) }\) Because the radius of convergence \(R=e^[value]\) for the power series, the power series surely converges when \( | x | < e^[value] \) i.e. \( -e^[value] < x < e^[value] \). The point \(x=1\) belongs to this interval and hence the power series converges at the point \(x=1\). \(\textrm{c) }\) Potenssisarja suppenee varmasti, kun \( -e^[value] < x < e^[value] \). Piste \(x=[value]\) ei kuulu tähän väliin, joten potenssisarja hajaantuu pisteessä \(x=[value]\). \(\textrm{c) }\) The power series surely converges when \( -e^[value] < x < e^[value] \). The point \(x=[value]\) doesn't belong to this interval and hence the power series diverges at the point \(x=[value]\).

Given the function \(f(x) = \frac{[value]}{[value]-[value]x}\), find the constant multipliers \(a_k\) of its second degree Taylor polynomial \(P_2(x;0) = \sum_{k=0}^{2} a_k x^k\).Määritä funktion \(f(x) = \frac{[value]}{[value]-[value]x}\) toisen asteen Taylor-polynomin \(P_2(x;0) = \sum_{k=0}^{2} a_k x^k\) kertoimet \(a_k\).

\(a_0 =~\)

\(a_1 =~\)

\(a_2 =~\)
Feedback: Model solution: Mallivastaus: Find the constant multipliers \(a_k\) of the second degree Taylor polynomial of the function \(f(x) = \frac{[value]}{[value]-[value]x}\).Määritä funktion \(f(x) = \frac{[value]}{[value]-[value]x}\) toisen asteen Taylorin polynomin \(P_2(x;0) = \sum_{k=0}^{2} a_k x^k\) kertoimet \(a_k\). The constants in the Taylor polynomial are of the form \(a_k = \frac{f^{(k)}(x_0)}{k!}\).Taylorin polynomin kertoimet ovat muotoa \(a_k = \frac{f^{(k)}(x_0)}{k!}\). The first constant can be found by a simple substitution.Ensimmäinen kerroin saadaan suoraan sijoittamalla \(a_0 = \frac{f(0)}{0!} = \frac{\frac{[value]}{[value]-[value]\cdot 0}}{1} = [value]\). For the remaining constants, we need to determine the first and second derivatives of the function \(f\). These can found by applying the chain rule.Lasketaan jäljellä olevia kertoimia varten funktion ensimmäinen ja toinen derivaatta käyttämällä yhdistetyn funktion derivointikaavaa \(f'(x) = [value]\) (Outer function: \(\frac{[value]}{x} = [value]x^{-1}\), inner function: \([value]-[value]x\)Ulkofunktio: \(\frac{[value]}{x} = [value]x^{-1}\), sisäfunktio: \([value]-[value]x\)) \(f''(x) = [value]\) (Outer function: \(\frac{[value]}{x^2} = [value]x^{-2}\), inner function: \([value]-[value]x\)Ulkofunktio: \(\frac{[value]}{x^2} = [value]x^{-2}\), sisäfunktio: \([value]-[value]x\)). Now substituting these we getNyt sijoittamalla saadaan \(a_1 = \frac{f'(0)}{1!} = [value]\) \(a_2 = \frac{f''(0)}{2!} = [value]\).

Let \(f(x)=\ln(x)\). Determine the fourth order Taylor polynomial of the function \(f\) about the point \([value]\).Olkoon \(f(x)=\ln(x)\). Laske funktion \(f\) Taylor-polynomi astetta neljä, kun kehityskeskuksena on \([value]\).

\(T_4 f(x)=~\)
Feedback: Model solution: Mallivastaus: \(f(x)=\ln(x)\). Taylor polynomial of order 4 around the point \(x_0=[value]\): Taylorin polynomi astetta 4 pisteessä \(x_0=[value]\): \[T_{4}f(x)=\sum_{k=0}^{4}\frac{1}{k!}f^{(k)}(x_0)(x-x_0)^k \] \[f'(x)=\frac{1}{x}, \ \ \ f^{(2)}(x)=-x^{-2}, \ \ \ f^{(3)}(x)=2x^{-3}, \ \ \ f^{(4)}(x)=-6x^{-4} \] \[ T_4 f(x)= \frac{1}{0!}\ln([value])+\frac{1}{1!}[value]^{-1}(x-[value])-\frac{1}{2!}[value]^{-2}(x-[value])^{2}+\frac{1}{3!}\cdot 2\cdot[value]^{-3}(x-[value])^3-\frac{1}{4!}\cdot 6\cdot [value]^{-4}(x-[value])^4 \] \[= [value]. \]

Let \(f(x) = e^{[value]x}\).Olkoon \(f(x) = e^{[value]x}\)

DetermineLaske

\(f(0)\): |
\(f'(0)\): |
\(f''(0)\): |
\(f'''(0)\): |

Using the previous results and the formula for Taylor polynomial \[P_{n}(x;x_{0}) = \sum_{k=0}^{n} \frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}\] find the third degree Taylor polynomial of function \(f\): Edellisiä tuloksia ja Taylorin polynomin kaavaa \[P_{n}(x;x_{0}) = \sum_{k=0}^{n} \frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}\] avuksi käyttäen, laske \(f\):n kolmannen asteen Taylor-polynomi:

\(P_{3}(x;0)\)
Feedback: Model solution: Mallivastaus: To differentiate the function \(f(x) = e^{[value]x}\), we apply the chain rule \(D(f(g(x)) = f'(g(x))g'(x)\) for outer function \(e^{x}\) and inner function \([value]x\).Funktio \(f(x) = e^{[value]x}\) derivoidaan soveltamalla yhdistetyn funktion derivointisääntöä \(D(f(g(x)) = f'(g(x))g'(x)\) ulkofunktiolle \(e^{x}\) ja sisäfunktiolle \([value]x\). We get: Näin saadaan: \(f'(x) =\) | [value] \(f''(x) =\) | [value] \(f'''(x) =\) | [value] Substituting x=0: Sijoittamalla x=0: \(f(0) =\) | [value] \(f'(0) =\) | [value] \(f''(0) =\) | [value] \(f'''(0) =\) | [value] Using the previous results and the Taylor polynomial \[P_{n}(x;x_{0}) = \sum_{k=0}^{n} \frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k},\] we can find the third degree Taylor polynomial of function \(f\): Edellisiä tuloksia ja Taylorin polynomin kaavaa \[P_{n}(x;x_{0}) = \sum_{k=0}^{n} \frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}\] avuksi käyttäen, saadaan laskettua \(f(x)\):n kolmannen asteen Taylor-polynomi: \begin{align*} P_{3}(x;0) &= \sum_{k=0}^{3}\frac{f^{k}(0)}{k!}x^{k}\\ &= f(0)+f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} \\ &= [value] \end{align*}

Tarkastellaan raja-arvon \[ \lim\limits_{x\to 0}\dfrac{[value]\sin(x^3)-x^2\sinh([value]x)}{x^4\ln(1+[value]x)} \] laskemista potenssisarjojen avulla.

a) Määrää funktion \( x^4\ln(1+[value]x)\) astetta seitsemän oleva Maclaurinin polynomi.

b) Laske potenssisarjojen avulla raja-arvo.

Consider the evaluation of the limit \[ \lim\limits_{x\to 0}\dfrac{[value]\sin(x^3)-x^2\sinh([value]x)}{x^4\ln(1+[value]x)} \] with power series.

a) Find the degree seven Maclaurin polynomial for the function \( x^4\ln(1+[value]x)\).

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: \(\textrm{a) }\) Potenssisarjan \( \ln(1+x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3}-\cdots\), \(\left|x\right|<1\), avulla saadaan Model solution: \(\textrm{a) }\) By using the series \( \ln(1+x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3}-\dfrac{x^4}{4}-\cdots\), \(\left|x\right|<1\), we get \begin{align*} x^4\ln(1+[value]x)&=x^4([value]x -\dfrac{([value]x)^2}{2} + \dfrac{([value]x)^3}{3}-\cdots)\\ &=x^4([value]x -\dfrac{[value]^2x^2}{2} + \dfrac{[value]^3x^3}{3}- \cdots)\\ &=[value]x^5 -[value]x^6 + [value]x^7- \cdots. \end{align*} \(\textrm{b) }\) Tunnetaan potenssisarjat \(\textrm{b) }\) By using the known Maclaurin series \[ \begin{align*} x^4\ln(1+[value]x)&=[value]x^5 -[value]x^6 + [value]x^7- \cdots,\\ \sin(x)&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,\\ \sinh(x)&=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots, \end{align*} \] joten saadaan it follows \[ \begin{align*} &\hphantom{=}\,\, \lim\limits_{x\to 0}\dfrac{[value]\sin(x^3)-x^2\sinh([value]x)}{x^4\ln(1+[value]x)}\\ &=\lim\limits_{x\to 0} \dfrac{[value](x^3-\dfrac{(x^3)^3}{3!}+\dfrac{(x^3)^5}{5!}-\dfrac{(x^3)^7}{7!}+\cdots)-x^2([value]x+\dfrac{([value]x)^3}{3!}+\dfrac{([value]x)^5}{5!}+\dfrac{([value]x)^7}{7!}+\cdots)} {\displaystyle{[value]x^5 -[value]x^6 + [value]x^7- \cdots}}\\ &=\lim\limits_{x\to 0} \dfrac{ [value]x^3-\dfrac{[value]x^9}{3!}+\dfrac{[value]x^{15}}{5!}-\dfrac{[value]x^{21}}{7!}+\cdots-([value]x^3+\dfrac{[value]^3x^5}{3!}+\dfrac{[value]^5x^7}{5!}+\dfrac{[value]^7x^9}{7!}+\cdots)} {\displaystyle{[value]x^5 -[value]x^6 + [value]x^7- \cdots}}\\ &=\lim\limits_{x\to 0} \dfrac{ (-\dfrac{[value]^3x^5}{3!}-\dfrac{[value]^5x^7}{5!}-\dfrac{[value]^7x^9}{7!}-\dfrac{[value]x^9}{3!}+\dfrac{[value]x^{15}}{5!}-\dfrac{[value]x^{21}}{7!}+\cdots):x^5} {(\displaystyle{[value]x^5 -[value]x^6 + [value]x^7- \cdots}):x^5}\\ &=\lim\limits_{x\to 0} \dfrac{ -\dfrac{[value]^3}{3!}-\dfrac{[value]^5x^2}{5!}-\dfrac{[value]^7x^4}{7!}-\dfrac{[value]x^4}{3!}+\dfrac{[value]x^{10}}{5!}-\dfrac{[value]x^{16}}{7!}+\cdots} {\displaystyle{[value] -[value]x + [value]x^2- \cdots}}\\ &=-\dfrac{[value]}{3![value]}=-[value]. \end{align*} \]

Q: Etsi potenssisarjan \(\displaystyle [value] \) suppenemissäde. Valitse lisäksi oikea vaihtoehto sarjan suppenemisvälille ja hae oikeat arvot välin päätepisteille \(a\) ja \(b\).

Suppenemissäde \(R = \) .

Alla olevista vaihtoehdoista
1: \((a,b)\)
2: \([a,b)\)
3: \((a, b]\)
4: \([a,b]\)
sarjan suppenemisväliä vastaa väli numero , missä \(a = \) ja \(b = \)
Feedback: Mallivastaus: Selvitetään suppenemissäde \(R\) suhdetestillä. Lasketaan \( \displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} [value] = [value] \). Sarja siis suppenee itseisesti, kun \( \displaystyle [value] < 1 \), eli kun \([value] < [value] \), ja suppenemissäde on siten \(R = [value] \). Ehdosta \([value] < [value] \) saadaan lisäksi suppenemisvälin päätepisteiksi \(a = [value] \) ja \(b = [value] \). Tarkastellaan vielä sarjan suppenemista erikseen välin päätepisteitä. Koska päätepisteissä sarjan yleinen termi on muotoa \( [value] \) tai \( [value] \), sarja ei suppene, ja siten suppenemisväli on \( ([value],[value]) \).
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Q: Muodosta funktiolle \(f(x) = [value]\) \([value]\). asteen Taylorin polynomi \(P_[value](x)\) kehityskeskuksena \([value]\).
Feedback: Mallivastaus: \( [value] \). asteen Taylorin polynomi funktiolle \(f(x) = [value] \) kehityskeskuksena \( [value] \) saadaan Taylorin sarjan osasummana \[ \displaystyle \sum_{k=0}^[value] \frac{f^{(k)}([value])}{k!}(x - [value])^k, \] missä \(f^{(k)}([value]) \) on funktion \(f\) \(k\). derivaatta pisteessä \(x = [value]\). Kysytty Taylorin polynomi on siten \[ \displaystyle P_[value](x) = [value]. \]
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Q: Olkoon \(f(x) = [value] \). Etsi se \([value].\) asteen polynomi, joka käyttäytyy origon ympäristössä mahdollisimman samoin kuin \(f(x)\).

\(P_[value](x) \) =
Feedback: Mallivastaus: Kysytty polynomi saadaan Taylorin sarjan \([value].\) osasummana. Kysytty polynomi on siis \([value]\) asteen Taylorin polynomina \( \displaystyle P_[value](x) = \sum_{k=0}^[value] \frac{f^{(k)}(0)}{k!}x^k = [value] \). Oheisesta kuvasta nähdään, että polynomi (kuvassa vihreällä) approksimoi funktiota \(f(x)\) (kuvassa sinisellä) varsin hyvin origon lähellä, mutta huonosti kauempana origosta.
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