Title: Kertaus

Calculate the indefinite integral of the function

\[\displaystyle f(x) = \frac{x+[value]}{x(x-[value])(x+[value])} \]

For natural logarithm, use the command ln(a), e.g. ln(2). For absolute value, use the command abs(x), respectively.

Laske funktion

\[\displaystyle f(x) = \frac{x+[value]}{x(x-[value])(x+[value])} \]

integraali. Luvun a luonnollinen logaritmi merkitään ln(a), esim. ln(2). Itseisarvo x merkitään abs(x).

\(\displaystyle\int \frac{x+[value]}{x(x-[value])(x+[value])} \, \mathrm{d}x=~\)
Feedback: Model solution: Mallivastaus: \[\int \frac{x+{[value]}}{x(x-{[value]})(x+{[value]})} \, \mathrm{d}x \] The partial fraction decomposition of the integrand isIntegroitavan lausekkeen osamurtokehitelmä on \[\begin{array}{rll} \frac{x+[value]}{x(x-[value])(x+[value])} & = & \frac{A}{x}+\frac{B}{x-[value]}+\frac{C}{x+[value]} \\ & = & \frac{A(x-[value])(x+[value])+Bx(x+[value])+Cx(x-[value])}{x(x-[value])(x+[value])} \\ & = & \frac{(A+B+C)x^2+([value]+[value]B-[value]C)x-[value]A}{x(x-[value])(x+[value])}, \end{array}\] wheremissä \[\begin{cases} A+B+C &= 0 \\ [value]+[value]B-[value]C &= 1 \\ -[value]A &= [value]. \end{cases} \Leftrightarrow \begin{cases} A &= [value] \\ B+C &= [value] \\ [value]B-[value]C &= [value] \end{cases} \Leftrightarrow \begin{cases} A &= [value] \\ B &= [value] \\ C &= [value]. \end{cases} \] ThereforeSiten \[\begin{array}{rll} \int \frac{x+[value]}{x(x-[value])(x+[value])} \, dx & = & \int \left( [value]\cdot \frac{1}{x}+[value]\cdot \frac{1}{x-[value]}+[value]\cdot \frac{1}{x+[value]} \right) \, \mathrm{d}x \\ & = & [value]. \end{array}\]

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

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 \(a_1 = \frac{f'(0)}{1!} = [value]\) \(a_2 = \frac{f''(0)}{2!} = [value]\).