Title: muut

Calculate an estimate \(S_n=S_n(f,a,b)\approx \int_{a}^b f(x)\, \mathrm{d}x\) for the integral \[ \int_[value]^[value] \left( [value] \right)\, \mathrm{d}x \] using Simpson's rule, when the interval is divided into four sections. Calculate an error estimate by applying the formula \[\bigg| S_n-\int_{a}^{b} f(x) \, \mathrm{d}x \bigg| \leq C_4 \frac{(b-a)^5}{180n^4}, \] where \( C_4=\max_{t\in [a,b]} \left| f^{(4)}(t) \right|\).Laske likiarvo \(S_n=S_n(f,a,b)\approx \int_{a}^b f(x)\, \mathrm{d}x\) integraalille \[ \int_[value]^[value] \left( [value] \right)\, \mathrm{d}x \] Simpsonin säännöllä, kun integroimisväli jaetaan neljään osaan. Laske virhearvio käyttämällä kaavaa \[\bigg| S_n-\int_{a}^{b} f(x) \, \mathrm{d}x \bigg| \leq C_4 \frac{(b-a)^5}{180n^4}, \] missä \( C_4=\max_{t\in [a,b]} \left| f^{(4)}(t) \right|\).

EstimateLikiarvo \( S_[value]=~\) |
Estimate for errorVirhearvio \(C_4 \frac{(b-a)^5}{180n^4}=~\) |
Feedback: Model solution: Mallivastaus: The goal of the exercise is to calculate an estimate for the integral of the function \(f(x)=[value]\) using Simpson's rule. The boundaries for the integral are \(a=[value]\) and \(b=[value]\) and the interval is divided into \(n=[value]\) parts of equal length, i.e. each subinterval has a length of \(h=\frac{b-a}{n}=[value]\). In the following table you can see the evaluated function values at the endpoints of each subinterval.Tehtävänä on laskea funktion \(f(x)=[value]\) integraalille likiarvo Simpsonin säännöllä. Integrointirajat ovat \(a=[value]\) ja \(b=[value]\) ja integrointiväli jaetaan neljään saman pituiseen osaan eli \(n=[value]\), joten yhden osavälin pituus on \(h=\frac{b-a}{n}=[value]\). Seuraavaan taulukkoon on laskettu funktion arvot osavälien reunoilla. \(\begin{array}{rlllll} & x_0 & x_1 & x_2 & x_3 & x_4 \\ x: & [value] & [value] & [value] & [value] & [value] \\ f(x): & [value] & [value] & [value] & [value] & [value] \end{array}\) According to Simpson's rule, the estimate isSimpsonin säännön mukaan likiarvo on \begin{align*} S_4 & = \frac{h}{3} \left( f(x_0)+4\cdot f(x_1)+2\cdot f(x_2)+4\cdot f(x_3)+f(x_4)\right) \\ & = \frac{[value]}{3} \left( [value]+4\cdot [value]+2\cdot [value]+4\cdot [value]+[value] \right)\\ & \approx [value]. \end{align*} Estimate for the error isVirhearvio on \[\bigg| S_n-\int_{a}^{b} f(x) \, \mathrm{d}x \bigg| \leq C_4 \frac{(b-a)^5}{180n^4},\] wheremissä \[C_4=\max_{x\in[0,b]} |f^{(4)}(x)|\] \begin{align*} f'(x) & = [value] \\ f''(x) & = [value] \\ f^{(3)}(x) & = [value] \\ f^{(4)}(x) & = [value]. \end{align*} ThusSiis \[C_4=\max_{x\in [[value],[value]]} |[value]| = 0\] and the estimate for the error isja virhearvio on \[C_4 \frac{(b-a)^5}{180n^4}=0\cdot \frac{([value]-[value])^5}{180\cdot 4^4} =0.\] N.B. The fourth derivative of a third degree polynomial is always zero, so Simpson's rule provides a precise result for all third degree polynomials.Huom! Kolmannen asteen polynomin neljäs derivaatta on aina 0, joten Simpsonin sääntö on tarkka kolmannen asteen polynomeille.

Find the general solution \(y(t)\) for the differential equation \[ y''(t) = [value].\] Also, find the particular solution \(y_{e}(t)\), for which it holds that \(y([value])=[value]\) and \(y'([value])=[value]\). Note that the general solution needs to contain constants, as well. Hint: How would you write the characteristic polynomial for the corresponding homogeneous equation? Etsi yleinen ratkaisu \(y(t)\) differentiaaliyhtälölle \[ y''(t) = [value].\] Etsi lisäksi se erikoisratkaisu \(y_{e}(t)\), joka toteuttaa alkuarvot \[y([value])=[value] \text{ ja } y'([value])=[value]\]. Huomaa, että yleisessä ratkaisussa täytyy olla vakiotermejä. Vihje: Kuinka kirjoitat yhtälön homogeenisen osan karakteristisen polynomin?

\(y(t) = \)

\(y_{e}(t) = \)
Feedback: Model solution: Mallivastaus: We want to solve the differential equation \[ y''(t) = [value]. \] The equation in question is a second order differential equation. Since the equation is of suitable form, we can find the solution by integrating it twice.Tehtävänämme on ratkaista differentiaaliyhtälö \[ y''(t) = [value]. \] Kyseessä on toisen kertaluvun differentiaaliyhtälö. Koska yhtälö on sopivaa muotoa, voidaan se ratkaista integroimalla kahteen kertaan. Let us integrate both sides of the equation with respect to \(t\): \[\begin{array}{rll} \int y''(t) dt & = & \int \left([value] \right) dt \\ y'(t) & = & \int [value] dt + \int [value] dt \\ y'(t) & = & [value] + [value] + C \\ \end{array}\] We integrate once more: \[\begin{array}{rll} \int y'(t) dt & = & \int \left( [value] + [value] + C \right) dt \\ y(t) & = & \int [value] dt + \int [value] dt + \int C dt \\ y(t) & = & [value] + [value] + C t + D \\ \end{array}\] Now we have found the general solution for the differential equation, \[ y(t) = [value] + [value] + C t + D .\] Integroidaan puolittain muuttujan \(t\) suhteen: \[\begin{array}{rll} \int y''(t) dt & = & \int \left([value] \right) dt \\ y'(t) & = & \int [value] dt + \int [value] dt \\ y'(t) & = & [value] + [value] + C \\ \end{array}\] Integroidaan vielä toiseen kertaan: \[\begin{array}{rll} \int y'(t) dt & = & \int \left( [value] + [value] + C \right) dt \\ y(t) & = & \int [value] dt + \int [value] dt + \int C dt \\ y(t) & = & [value] + [value] + C t + D \\ \end{array}\] Nyt olemme siis saaneet selville differentiaaliyhtälön yleisen ratkaisun, \[ y(t) = [value] + [value] + C t + D .\] To find the particular solution, we need to determine the values for the constants \(C\) and \(D\) using the initial conditions. According to the first initial condition \[ y([value])=[value] ,\] so substituting in the equation \(y(t) = [value] + [value] + C t + D\) we get \[\begin{array}{rll} [value] & = & [value]+[value]+[value] + D \\ [value] & = & [value] \\ [value] & = & [value] . \end{array}\] According to the second initial condition \[ y'([value])=[value] ,\] so substituting in the equation \( y'(t) = [value] + [value] + C \) we get \[\begin{array}{rll} [value] & = & [value] + [value] + C \\ [value] & = & [value] \\ [value] & = & [value] . \end{array}\] Thus, the values of the constants are \(C = [value]\) and \(D = [value]\). Substituting these in the expression for the general solution we get \[\begin{array}{rll} y(t) & = & [value] + [value] + C t + D \\ & = & [value] + [value]. \end{array}\] The particular solution satisfying the initial conditions is therefore \[y_e(t) = [value] + [value].\] Erikoisratkaisua varten täytyy vakioiden \(C\) ja \(D\) arvot ratkaista alkuehtojen avulla. Ensimmäisen alkuehdon mukaan \[ y([value])=[value] ,\] joten lausekkeeseen \(y(t) = [value] + [value] + C t + D\) sijoittamalla saadaan \[\begin{array}{rll} [value] & = & [value]+[value]+[value] + D \\ [value] & = & [value] \\ [value] & = & [value] . \end{array}\] Toisen alkuehdon mukaan \[ y'([value])=[value] ,\] joten lausekkeeseen \( y'(t) = [value] + [value] + C \) sijoittamalla saadaan \[\begin{array}{rll} [value] & = & [value] + [value] + C \\ [value] & = & [value] \\ [value] & = & [value] . \end{array}\] Vakioiden arvot ovat siis \(C = [value]\) ja \(D = [value]\). Sijoittamalla nämä yleisen ratkaisun lausekkeeseen saamme \[\begin{array}{rll} y(t) & = & [value] + [value] + C t + D \\ & = & [value] + [value]. \end{array}\] Alkuehdot täyttävä erikoisratkaisu on siis \[y_e(t) = [value] + [value].\]

Geometrinen sarja

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 \[ \begin{align*} &\hphantom{=}\,\, \lim\limits_{x\to 0}\dfrac{[value]\sin(x^3)-x^2\sinh([value]x)}{x^4\ln(1+[value]x)}\\ &=\lim\limits_{x\to 0} \dfrac{[value](x^3-\dfrac{(x^3)^3}{3!}+\dfrac{(x^3)^5}{5!}-\dfrac{(x^3)^7}{7!}+\cdots)-x^2([value]x+\dfrac{([value]x)^3}{3!}+\dfrac{([value]x)^5}{5!}+\dfrac{([value]x)^7}{7!}+\cdots)} {\displaystyle{[value]x^5 -[value]x^6 + [value]x^7- \cdots}}\\ &=\lim\limits_{x\to 0} \dfrac{ [value]x^3-\dfrac{[value]x^9}{3!}+\dfrac{[value]x^{15}}{5!}-\dfrac{[value]x^{21}}{7!}+\cdots-([value]x^3+\dfrac{[value]^3x^5}{3!}+\dfrac{[value]^5x^7}{5!}+\dfrac{[value]^7x^9}{7!}+\cdots)} {\displaystyle{[value]x^5 -[value]x^6 + [value]x^7- \cdots}}\\ &=\lim\limits_{x\to 0} \dfrac{ (-\dfrac{[value]^3x^5}{3!}-\dfrac{[value]^5x^7}{5!}-\dfrac{[value]^7x^9}{7!}-\dfrac{[value]x^9}{3!}+\dfrac{[value]x^{15}}{5!}-\dfrac{[value]x^{21}}{7!}+\cdots):x^5} {(\displaystyle{[value]x^5 -[value]x^6 + [value]x^7- \cdots}):x^5}\\ &=\lim\limits_{x\to 0} \dfrac{ -\dfrac{[value]^3}{3!}-\dfrac{[value]^5x^2}{5!}-\dfrac{[value]^7x^4}{7!}-\dfrac{[value]x^4}{3!}+\dfrac{[value]x^{10}}{5!}-\dfrac{[value]x^{16}}{7!}+\cdots} {\displaystyle{[value] -[value]x + [value]x^2- \cdots}}\\ &=-\dfrac{[value]}{3![value]}=-[value]. \end{align*} \]