heureka

Gegeben sei die Funktion

\(f(x) = \dfrac{2}{1-x^2}\)
Zeigen Sie (mit vollständiger Induktion), dass die n-te  Ableitung von f von folgender Form ist:

\(f^{(n)} (x) = n! \left(\dfrac{1}{(1-x)^{n+1} }+ (-1)^n \dfrac{1}{(1+x)^{n+1}} \right)\)

Induktionsanfang für \(n_0\) gilt:

\(\begin{array}{|rcll|} \hline f^{(0)} (x) &=& 0! \left(\dfrac{1}{(1-x)^{0+1} }+ (-1)^0 \dfrac{1}{(1+x)^{0+1}} \right) \\\\ &=& 1\cdot \left(\dfrac{1}{1-x }+ 1\cdot \dfrac{1}{1+x} \right) \\\\ &=& \dfrac{1+x+1-x}{(1-x)(1+x) } \\\\ &=& \dfrac{2}{1-x^2}~ \checkmark \qquad f^{(0)} (x) = f (x)\\ \hline \end{array} \)

Induktionsvoraussetzung (I.V.):
Für ein beliebiges \(n \in \mathbb{Z}\) mit \(n \ge n_0\) gilt:

\(\begin{array}{|rcll|} \hline f^{(n)} (x) &=& n! \left(\dfrac{1}{(1-x)^{n+1} }+ (-1)^n \dfrac{1}{(1+x)^{n+1}} \right) \\ \hline \end{array}\)

Induktionsbehauptung: dann gilt auch für \((n+1)\)

\(\begin{array}{|rcll|} \hline f^{(n+1)} (x) &=& (n+1)! \left(\dfrac{1}{(1-x)^{(n+1)+1} }+ (-1)^{(n+1)} \dfrac{1}{(1+x)^{(n+1)+1}} \right) \\ \hline \end{array} \)

Induktionsschluss: \(f^{(n+1)} (x) = \left(f^{(n)}(x) \right)'\)

\(\begin{array}{|rcll|} \hline \left(f^{(n)}(x) \right)' &=& \left[ n! \left(\dfrac{1}{(1-x)^{n+1} }+ (-1)^n \dfrac{1}{(1+x)^{n+1}} \right) \right]' \\\\ &=& \Big[~ n! \left(~ (1-x)^{-(n+1)} + (-1)^n (1+x)^{-(n+1)} ~\right) ~\Big]' \\\\ &=& n! \Big[~\left(~ (1-x)^{-(n+1)} + (-1)^n (1+x)^{-(n+1)} ~\right) ~\Big]' \\\\ &=& n! \Big(~ -(n+1)(1-x)^{-(n+1)-1}\cdot (-1) + (-1)^n ( -(n+1))(1+x)^{-(n+1)-1}\cdot 1 ~\Big) \\\\ &=& n! \Big(~ (n+1)(1-x)^{-(n+1)-1} + (-1)^n (-(n+1))(1+x)^{-(n+1)-1} ~\Big) \\\\ &=& n!(n+1) \Big(~ (1-x)^{-(n+1)-1} + (-1)^n \cdot (-1)(1+x)^{-(n+1)-1} ~\Big) \\\\ &=& n!(n+1) \Big(~ (1-x)^{-(n+1)-1} + (-1)^n \cdot (-1)^1(1+x)^{-(n+1)-1} ~\Big) \\\\ &=& n!(n+1) \Big(~ (1-x)^{-(n+1)-1} + (-1)^{(n+1)}(1+x)^{-(n+1)-1} ~\Big) \\\\ &=& (n+1)! \Big(~ (1-x)^{-(n+1)-1} + (-1)^{(n+1)}(1+x)^{-(n+1)-1} ~\Big) \\\\ \left(f^{(n)}(x) \right)' &=& (n+1)! \left(~ \dfrac{1}{(1-x)^{(n+1)+1}} + (-1)^{(n+1)} \dfrac{1}{(1+x)^{(n+1)+1}} ~\right)~\checkmark \\ \hline \end{array}\)

Hallo,

ich bräuchte dringends Hilfe.

Die Aufgabe lautet: bestimmen Sie mit Hilfe der Partialbruchzerlegung den grenzwert der folge.

nun habe ich versucht die Partialbruchzerlegung zu rechnen es kommt mir aber falsch rüber und ich komm bei den Rest der Aufgabe auch nicht mehr klar.

Danke im Voraus.

Wie berechne ich den Grenzwert der Folge ?

\(\begin{array}{|rcll|} \hline && \mathbf{\displaystyle \sum \limits_{k=0}^{n}\dfrac{4}{(2k+1)(2k+3)} } \\\\ &=& \displaystyle 4\sum \limits_{k=0}^{n}\dfrac{1}{(2k+1)(2k+3)} \\\\ &=& \displaystyle 4\sum \limits_{k=0}^{n} \dfrac12\left( \dfrac{1}{2k+1}-\dfrac{1}{2k+3}\right ) \\\\ &=& \displaystyle 2\sum \limits_{k=0}^{n} \left( \dfrac{1}{2k+1}-\dfrac{1}{2k+3}\right ) \\\\ &=& 2 \Big[ \left(\dfrac11 - \dfrac13\right)+\left(\dfrac13 - \dfrac15\right)+ \left(\dfrac15 - \dfrac19\right)+ \ldots \\ && +\left( \dfrac{1}{2n+1} - \dfrac{1}{2k+3}\right) \Big] \\\\ &=& 2 \Big[\mathbf{1} + \left(-\dfrac13+\dfrac13\right)+ \left(-\dfrac15+\dfrac15\right) + \left(-\dfrac19+\dfrac19\right) + \ldots \\ && + \left(-\dfrac{1}{2n+1}+\dfrac{1}{2n+1} \right) \mathbf{- \dfrac{1}{2n+3}} \Big] \qquad \text{Teleskopreihe} \\\\ \mathbf{\displaystyle \sum \limits_{k=0}^{n}\dfrac{4}{(2k+1)(2k+3)} }&\mathbf{=}& \mathbf{2 \left(1 - \dfrac{1}{2n+3} \right) } \\ \hline \end{array}\)

Grenzwert der Folge:

\(\begin{array}{|rcll|} \hline && \lim \limits_{n\to \infty} 2 \left(1 - \dfrac{1}{2n+3} \right) \\\\ &=& \lim \limits_{n\to \infty} \left(2 - \dfrac{2}{2n+3} \right) \\\\ &=& \lim \limits_{n\to \infty} \left(2 - \dfrac{2}{2n+3} \cdot \left(\frac nn \right) \right) \\\\ &=& \lim \limits_{n\to \infty} \left(2 - \dfrac{\frac{2}{n}}{2+\frac{3}{n}} \right) \quad | \quad n > 0! \\\\ &=& 2 - \dfrac{0}{2+0} \\\\ &=& 2 - 0 \\\\ &\mathbf{=}& \mathbf{ 2 } \\ \hline \end{array}\)

Der Grenzwert der Folge ist 2

You are choosing between two different cell phone plans.

The first plan charges a rate of 21 cents per minute.

The second plan charges a monthly fee of $29.95 plus 8 cents per minute.

How many minutes would you have to use in a month in order for the second plan to be preferable?

Round up to the nearest whole minute.

\(\begin{array}{|rcll|} \hline 21\dfrac{\text{cent}}{\text{minute}}\times x &=& 2995~\text{cent} + 8\dfrac{\text{cent}}{\text{minute}}\times x \\\\ 21\dfrac{\text{cent}}{\text{minute}}\times x -8\dfrac{\text{cent}}{\text{minute}}\times x &=& 2995~\text{cent} \\\\ (21-8)\dfrac{\text{cent}}{\text{minute}}\times x &=& 2995~\text{cent} \\\\ 13\dfrac{\text{cent}}{\text{minute}}\times x &=& 2995~\text{cent} \\\\ x &=& \dfrac{2995~\text{cent}}{13} \cdot \dfrac{\text{minute}}{\text{cent}} \\\\ x &=& \dfrac{2995}{13} ~\text{minutes} \\\\ x &=& 230.38\ldots ~\text{minutes}\\\\ \mathbf{x} & \mathbf{=} & \mathbf{231 ~\text{minutes}} \\ \hline \end{array}\)

Find the sum of the first 30 terms of the following arithmetic sequence:

  -10, -18, -26, -34, ...

\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)d \\ a_n &=& -10 +(n-1)(-8) \\ a_n &=& -10-8n+8 \\ \mathbf{a_n} & \mathbf{=} & \mathbf{-2-8n } \\ a_{30} &=& -2-8\cdot 30 \\ \mathbf{a_{30}} & \mathbf{=} & \mathbf{-242} \\\\ s_n &=& \dfrac{a_1+a_n}{2}\cdot n \\ s_{30} &=& \dfrac{a_1+a_{30}}{2}\cdot 30 \\ s_{30} &=& \dfrac{-10-242}{2}\cdot 30 \\ s_{30} &=& -252 \cdot 15 \\ \mathbf{s_{30}} & \mathbf{=} & \mathbf{-3780} \\ \hline \end{array}\)

The sum of the first 30 terms is \(\mathbf{-3780}\)

A palindrome is an integer that reads the same forwards and backward.

How many positive 3-digit palindromes are multiples of 3?

\(\begin{array}{|r|r|r|r|} \hline & n & \text{palindrome } a(n) & \text{divisible by } 3 \\ \hline & 20& 1 01 \\ 1. & 21& 1 11 & \checkmark \\ & 22& 1 21 \\ & 23& 1 31 \\ 2. & 24& 1 41 &\checkmark \\ & 25& 1 51 \\ & 26& 1 61 \\ 3.& 27& 1 71 &\checkmark \\ & 28& 1 81 \\ & 29& 1 91 \\ \hline & 30& 2 02 \\ & 31& 2 12 \\ 4. & 32& 2 22 &\checkmark \\ & 33& 2 32 \\ & 34& 2 42 \\ 5. & 35& 2 52 &\checkmark \\ & 36& 2 62 \\ & 37& 2 72 \\ 6.& 38& 2 82 &\checkmark \\ & 39& 2 92 \\ \hline 7. & 40& 3 03 &\checkmark \\ & 41& 3 13 \\ & 42& 3 23 \\ 8. & 43& 3 33 &\checkmark \\ & 44& 3 43 \\ & 45& 3 53 \\ 9. & 46& 3 63 &\checkmark \\ & 47 & 3 73 \\ & 48& 3 83 \\ 10. & 49& 3 93 &\checkmark \\ \hline & 50& 4 04 \\ 11. & 51& 4 14 &\checkmark \\ & 52& 4 24 \\ & 53& 4 34 \\ 12. & 54& 4 44 &\checkmark \\ & 55& 4 54 \\ & 56& 4 64 \\ 13. & 57& 4 74 &\checkmark \\ & 58& 4 84 \\ & 59& 4 94 \\ \hline & 60& 5 05 \\ & 61& 5 15 \\ 14. & 62& 5 25 &\checkmark \\ & 63& 5 35 \\ & 64& 5 45 \\ 15. & 65& 5 55 &\checkmark \\ & 66& 5 65 \\ & 67& 5 75 \\ 16. & 68& 5 85 &\checkmark\\ & 69& 5 95 \\ \hline 17. & 70& 6 06 &\checkmark \\ & 71& 6 16 \\ & 72& 6 26 \\ 18. & 73& 6 36 &\checkmark \\ & 74& 6 46 \\ & 75& 6 56 \\ 19. & 76& 6 66 &\checkmark \\ & 77& 6 76 \\ & 78& 6 86 \\ 20. & 79& 6 96 &\checkmark \\ \hline & 80& 7 07 \\ 21. & 81& 7 17 &\checkmark \\ & 82& 7 27 \\ & 83& 7 37 \\ 22. & 84& 7 47 &\checkmark \\ & 85& 7 57 \\ & 86& 7 67 \\ 23. & 87& 7 77 &\checkmark \\ & 88& 7 87 \\ & 89& 7 97 \\ \hline & 90& 8 08 \\ & 91& 8 18 \\ 24. & 92& 8 28 &\checkmark \\ & 93& 8 38 \\ & 94& 8 48 \\ 25. & 95& 8 58 &\checkmark \\ & 96& 8 68 \\ & 97& 8 78 \\ 26. & 98& 8 88 &\checkmark \\ & 99& 8 98 \\ \hline 27. & 100& 9 09 &\checkmark \\ & 101& 9 19 \\ & 102& 9 29 \\ 28. & 103& 9 39 &\checkmark \\ & 104& 9 49 \\ & 105& 9 59 \\ 29. & 106& 9 69 &\checkmark \\ & 107& 9 79 \\ & 108& 9 89 \\ 30. & 109& 9 99 &\checkmark \\ \hline \end{array} \)

30  3-digit palindromes are divisible by 3.

​ Method of difference

Formula:

\(\begin{array}{|rcll|} \hline \displaystyle \sum \limits_{r=1}^{n}r^2 &=& \dfrac16n(n+1)(2n+1) \\\\ \displaystyle \sum \limits_{r=1}^{2n}r^3 &=& \dfrac1{2^2} (2n)^2(1+2n)^2 =n^2(1+2n)^2 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \displaystyle \sum \limits_{r=1}^{n}(6r-3)^2 } \\\\ &=& \displaystyle \sum \limits_{r=1}^{n}(36r^2-36r+9) \\\\ &=& \displaystyle 36\sum \limits_{r=1}^{n}r^2- 36 \sum \limits_{r=1}^{n}r+ 9\sum \limits_{r=1}^{n}1 \\\\ &=& 36\dfrac16n(n+1)(2n+1) -36\dfrac{(n+1)}{2}n +9n \\\\ &=& 6n(n+1)(2n+1) -18n(n+1) +9n \\\\ &=& 6n(n+1)(2n+1) -18n^2-18n+9n \\\\ &=& 6n(n+1)(2n+1) -18n^2-9n \\\\ &=& 3n\Big( 2(n+1)(2n+1) -6n-3 \Big) \\\\ &=& 3n\Big( 2(n+1)(2n+1) -3(2n+1) \Big) \\\\ &=& 3n(2n+1)\Big( 2(n+1) -3 \Big) \\\\ &=& 3n(2n+1)(2n-1) \\\\ &\mathbf{=}& \mathbf{3n(4n^2-1)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\displaystyle \sum \limits_{r=1}^{2n}r^3 - \sum \limits_{r=1}^{n}(6r-3)^2 } \\\\ &=& n^2(1+2n)^2 - 3n(4n^2-1) \\ &=& n^2(1+2n)^2 - 3n(2n-1)(2n+1) \\ &=& n(1+2n)\Big(n(1+2n) - 3(2n-1)\Big) \\ &=& n(1+2n)(n+2n^2 - 6n +3 ) \\ &=& n(1+2n)(2n^2 -5n +3 ) \\ &\mathbf{=}& \mathbf{ n(1+2n)(2n-3)(n-1) } \\ \hline \end{array}\)

Hab ein Problem eine Formel auf die relative Luftfeuchte umzustellen, da ein Logarithmus enthalten ist.
Es wäre super, wenn mir jemand helfen könnte.

237,3 * log10 (Relative Luftfeuchte) + 237,3 * log10(Sättigungsdruck/610,78)
============================================================== = Taupunkttemperatur
7,5 -  log10 (Relative Luftfeuchte) - log10 (Sättigungsdruck/610,78)

General Solution - Trigonometry

Sin Ø  - Sin 2Ø = Sin 4Ø - Sin 3Ø

\(\begin{array}{ rcl } \sin(\phi) - \sin(2\phi) &=& \sin(4\phi) - \sin(3\phi) \\ &\text{or} & \\ \sin(3\phi)+\sin(\phi) &=& \sin(4\phi) + \sin(2\phi) \\ \end{array} \)

Formula:

\(\begin{array}{|rcll|} \hline \underbrace{\sin(3\phi)+\sin(\phi)}_{=2\sin(2\phi)\cos(\phi)} &=& \underbrace{\sin(4\phi) + \sin(2\phi)}_{=2\sin(3\phi)\cos(\phi)} \\\\ 2\sin(2\phi)\cos(\phi) &=& 2\sin(3\phi)\cos(\phi) \\\\ \sin(2\phi)\cos(\phi) &=& \sin(3\phi)\cos(\phi) \\\\ \sin(3\phi)\cos(\phi) -\sin(2\phi)\cos(\phi) &=& 0 \\\\ \cos(\phi)\Big(\sin(3\phi)-\sin(2\phi) \Big) &=& 0 \\ \hline \end{array} \)

Formula:

\(\begin{array}{|rcll|} \hline \cos(\phi)\Big(\underbrace{\sin(3\phi)-\sin(2\phi)}_{=2\cos(\frac52\phi)\sin(\frac12\phi)} \Big) &=& 0 \\\\ \cos(\phi)2\cos(\frac52\phi)\sin(\frac12\phi) &=& 0 \quad | \quad :2 \\\\ \large{\mathbf{\cos(\phi)\cos(\frac52\phi)\sin(\frac12\phi)} }& \large{\mathbf{=}} & \large{\mathbf{0}} \\ \hline \end{array}\)

General Solution:

\(\begin{array}{|lrcll|} \hline 1. & \cos(\phi) &=& 0 \\ & \phi &=& 2n\pi\pm\arccos(0) \\ & \phi &=& 2n\pi\pm \frac{\pi}{2} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 2. & \cos(\frac52\phi) &=& 0 \\ & \frac52\phi &=& 2n\pi\pm\arccos(0) \\ & \frac52\phi &=& 2n\pi\pm \frac{\pi}{2}\quad | \quad \cdot \frac25 \\ & \phi &=& \frac45 n\pi\pm \frac{\pi}{5} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{5}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3. & \sin(\frac12\phi) &=& 0 \\ & \frac12\phi &=& n\pi+(-1)^n \arcsin(0) \\ & \frac12\phi &=& n\pi+(-1)^n \pi \quad | \quad \cdot 2 \\ & \phi &=& 2n\pi+(-1)^n 2\pi \quad \Rightarrow \quad \mathbf{\phi = 2n\pi} \\ \hline \end{array} \)

Adam has four short straws, five medium straws, and six long straws.
If he randomly draws two straws, one at a time without replacement,
what is the probability that both are medium straws?

\(\dfrac{5}{15}\times \dfrac{4}{14} = 0.09523809524\quad (9.5\ \%)\)

There are 32 students in the Classroom. There 12 boys and 20 girls.

She needs to select two students to represent her class .

She will put all students names on a piece of paper and randomly select two names, one after the other.

What is the probability that she will select a boy first and then a girl?

\(\dfrac{12}{32}\times \dfrac{20}{31} = 0.24193548387\quad (24.2\ \%) \)