heureka

Please give at least the next 5 terms of this sequence...........

4, 7, 8, 11, 16, 19, 24, 27, 38, 41.........etc.

I asume:

\(\begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline &4&{\color{green}7}&8&11&16&19&24&2{\color{green}7}&38&41&{\color{grey}48}&{\color{grey}53}&{\color{grey}56}&{\color{grey}61}&{\color{grey}68} \\ \hline \text{Prime numbers}&2&3&{\color{blue}5}&{\color{blue}7}&{\color{blue}11}&{\color{blue}13}&{\color{blue}17}&{\color{blue}19}&23&29&31&37&41&43&47 \\ \text{Prime numbers}&{\color{red}3}&{\color{red}5}&7&11&13&17&19&23&{\color{red}29}&{\color{red}31}&{\color{red}37}&{\color{red}41}&{\color{red}43}&{\color{red}47}&{\color{red}53}\\ \hline \text{consecutive numbering}&{\color{red}1}&{\color{red}2}&{\color{blue}3}&{\color{blue}4}&{\color{blue}5}&{\color{blue}6}&{\color{blue}7}&{\color{blue}8}&{\color{red}9}&{\color{red}10}&{\color{red}11}&{\color{red}12}&{\color{red}13}&{\color{red}14}&{\color{red}15}\\ \hline \text{add red or add blue }& \\ \text{change with digit seven} &\\ \hline \end{array}\)

Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1.

\(\small{ \begin{array}{lrcll} &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } \\ \hline \text{Substitution : } & x &=& \sin{(u)} \qquad \rightarrow \qquad u = \arcsin{(x)}\\ & \ dx &=& \cos{(u)}\ du \\ \hline &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \qquad | \qquad 1-\sin^2{(u)} = \cos^2{(u)}\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{\cos^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \qquad | \qquad dx = \cos{(u)}\ du\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} }\cdot \cos{(u)}\ du } \\ & &=&\int \limits_{-1}^{1} { \ du } \\ & &=& [ u ]_{-1}^{1} \qquad | \qquad u = \arcsin{(x)} \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } & \mathbf{=} & \mathbf{ [~ \arcsin{(x)} ~]_{-1}^{1} }\\ \hline & \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } & = & [~ \arcsin{(x)} ~]_{-1}^{1} \\ & & = & [~ \arcsin{(1)}-\arcsin{(-1)} ~]\\ & & = & [~ \frac{\pi}{2} -\frac{-\pi}{2} ~]\\ & & = & [~ \frac{\pi}{2} +\frac{\pi}{2} ~]\\ & & = & \pi \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ \pi } \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ 3.14159265359 } \end{array} }\)

Siehe Bundeswettbewerb für Mathematik 2016

I am less than 500. All three digits are odd. All three digits are different. The sum of the digits is thirteen. The product of my digits is greater than 30. I am not divisible by 5. Who am I?

\(\begin{array}{lrcll} (1) & x + y + z &=& 13 \\ & z &=& 13-(x+y) \\ (2) & x\cdot 100 + y\cdot 10 + z &<& 500 \\ & x\cdot 100 + y\cdot 10 + 13-(x+y) &<& 500 \\ & 99x + 9y + 13 &<& 500 \\ & 9y &<& - 99x + 500 - 13 \\ & 9y &<& - 99x + 487 \\ && & \boxed{~ \mathbf{ y } ~ \mathbf{<}~ \mathbf{- 11x + \frac{487}{9} } \\ ~}\\ (3) & x\cdot y \cdot z &>& 30\\ & x\cdot y &>& \frac{30}{z}\\ &&& \boxed{~ \mathbf{x\cdot y } ~\mathbf{>}~ \mathbf{\frac{30}{13-(x+y)} }\\ ~} \end{array} \)

0,145kg Frischkäse kosten 1,44 Euro. Wie viel kostet 1kg?

\(\begin{array}{rcll} \frac{ 1,44 € } {0,145\ kg} \cdot 1\ kg &=& \frac{ 1,44 € } {0,145 } \cdot 1 \\ &=& \frac{ 1,44 € } {0,145 } \\ &=& \frac{ 1,44 € } {0,145 } \\ &=& 9.93103448276\ € \\ &=& 9.93\ € \qquad (\text{gerundet}) \end{array}\)

1. Given that cschx=-9/40, find the exact value of coshx and tanh2x

\(\begin{array}{rcll} csch(x) & =& -\frac{9}{40} \\\\ cosh(x) & =& \frac{ \sqrt{1+csch^2(x)} } { csch(x) }\\ cosh(x) & =& \frac{ \sqrt{1+ (-\frac{9}{40})^2 } } { -\frac{9}{40} }\\ cosh(x) & =& -\frac{40}{9} \cdot \sqrt{1+ (-\frac{9}{40})^2 } \\ cosh(x) & =& -\frac{40}{9} \cdot \sqrt{1+ \frac{81}{1600}) } \\ cosh(x) & =& -\frac{40}{9} \cdot \sqrt{ \frac{1681}{1600}) } \\ cosh(x) & =& -\frac{40}{9} \cdot \frac{41}{40} \\\\ cosh(x) & =& -\frac{41}{9} \\ \hline \\ sinh(x) & =& \frac{ 1 } { csch(x) }\\ sinh(x) & =& \frac{ 1 } { -\frac{9}{40} }\\ sinh(x) & =& -\frac{40}{9}\\ \hline \\ \end{array} \\\)

\(\begin{array}{rcll} sinh (2x) &=& 2\cdot sinh (x)\cdot cosh (x)\\ cosh (2x) &=& cosh^2(x) + sinh^2(x) = 2 \cdot cosh^2(x) — 1 = 1 + 2\cdot sinh^2(x)\\\\ tanh (2x) &=& \frac{ sinh(2x) } { cosh(2x) } \\ tanh (2x) &=& \frac{ 2\cdot sinh (x)\cdot cosh (x) } { 1 + 2\cdot sinh^2(x) } \\ tanh (2x) &=& \frac{ 2\cdot (-\frac{40}{9})\cdot (-\frac{41}{9}) } { 1 + 2\cdot (-\frac{40}{9})^2 } \\ tanh (2x) &=& \frac{ 2\cdot (\frac{40\cdot 41}{9\cdot 9}) } { 1 + 2\cdot (\frac{40^2}{9^2}) } \\ tanh (2x) &=& \frac{ 2\cdot (\frac{40\cdot 41}{9^2}) } { \frac{9^2 +2\cdot 40^2}{9^2} } \\ tanh (2x) &=& \frac{ 2\cdot 40 \cdot 41 } { 9^2 +2\cdot 40^2 } \\\\ tanh (2x) &=& \frac{ 3280 } { 3281 }\\ \hline \\ \end{array}\)

2. Solve the equation

\(x=tanh(~ \ln{(~ \sqrt{6x}~) } ~) \quad \text{ for } 0 < x < 1 \)

\(\begin{array}{rcll} x &=& tanh(~ \ln{(~ \sqrt{6x}~) } ~) \qquad & | \qquad tanh^{-1}() \\ tanh^{-1}(x) &=& \ln{(~ \sqrt{6x}~) } \qquad & | \qquad tanh^{-1}(x) = \frac12\cdot \ln{(~\frac{1+x}{1-x}~)} \quad \text{ for } \quad -1<x<1\\ \frac12\cdot \ln{(~\frac{1+x}{1-x}~)} &=& \ln{(~ \sqrt{6x}~) } \\ \ln{(~[\frac{1+x}{1-x}]^\frac12~)} &=& \ln{(~ \sqrt{6x}~) } \\ \left[\frac{1+x}{1-x} \right]^\frac12 &=& \sqrt{6x}\\ \sqrt{\frac{1+x}{1-x} } &=& \sqrt{6x} \qquad & | \qquad (\text{square both sides})\\ \frac{1+x}{1-x} &=& 6x \\ 1+x &=& 6x\cdot ( 1-x ) \\ 1+x &=& 6x - 6x^2 \\ 6x^2 -6x + x + 1 &=& 0 \\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

\(\small{ \begin{array}{rcll} 6x^2 -5x + 1 &=& 0 \qquad & a = 6 \qquad b = -5 \qquad c = 1\\ x_{1,2} &=& \dfrac{-(-5) \pm \sqrt{(-5)^2-4\cdot 6 \cdot 1} }{2\cdot 6} \\ x_{1,2} &=& \dfrac{ 5 \pm \sqrt{25 - 24} }{ 12 } \\ x_{1,2} &=& \dfrac{ 5 \pm \sqrt{1} }{ 12 } \\ x_{1,2} &=& \dfrac{ 5 \pm 1 }{ 12 } \\\\ x_1 &=& \dfrac{ 5 + 1 }{ 12 } \\ x_1 &=& \dfrac{ 6 }{ 12 } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{ 1 }{ 2 } }\\ \hline \\ x_2 &=& \dfrac{ 5 - 1 }{ 12 } \\ x_2 &=& \dfrac{ 4 }{ 12 } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{ 1 }{ 3 } } \\ \hline \end{array} } \)

{nl}

To: heureka and others..........

If you calculate the value of yk+1 in (1) and feed it into (2) and repeat the iterations at least 2-3 times, what does the result converge to and what is the rate of convegence, i.e., is it: quadratic, cubic, quartic.....etc. Also, what is the source of this algorithm? Thanks and have fun!.

Set a0 = 6−4√2 and y0 = √2−1 and k=0, 1, 2, 3................... Iterate:

yk+1 = 1−(1−y^4 k)^1/4 / 1 + (1−y^4 k)^1/4 ..............................(1)

and ak+1 = ak(1 + yk+1)^4 − 2^(2k+3)yk+1(1 + yk+1 + y^2 k+1)..(2)

\(\begin{array}{rcll} \text{Borwein's algorithm:}\\\\ \end{array} \\ \begin{array}{rcll} \text{Quartic algorithm (1985)}\\\\ \end{array} \\ \text{Start out by setting[1]}\\ \begin{array}{rcll} a_0 & =& 2\big(\sqrt{2}-1\big)^2 = 6-4\sqrt{2}\\ y_0 & =& \sqrt{2}-1 \end{array} \\ \begin{array}{rcll} \\ \text{Then iterate}\\ \end{array} \\ \\ \begin{array}{rcll} y_{k+1} & =& \frac{1-(1-y_k^4)^{1/4}}{1+(1-y_k^4)^{1/4}} \\ a_{k+1} & =& a_k(1+y_{k+1})^4 - 2^{2k+3} y_{k+1} (1 + y_{k+1} + y_{k+1}^2) \end{array} \\ \\ \text{Then }~a_k \text{ converges quartically against }~ 1/\pi; \\ \text{ that is, each iteration approximately quadruples the number of correct digits.} \)

Find the surface area of the figure. Round your answer to the nearest hundredth.

H; 18mm

R; 6mm

r; 3mm

\(\begin{array}{lrcll} \text{Area inside is a cylinder with radius r } & A_i &=& ( 2\pi r ) \cdot h \\ \text{Area outside is a cylinder with radius R } & A_o &=& ( 2\pi R ) \cdot h \\ \text{Area top is a ring } & A_{r_1} &=& \pi R^2 - \pi r^2 \\ \text{Area bottom is a ring } & A_{r_2} &=& \pi R^2 - \pi r^2 \\ \hline \\ \text{Area of the figure is the sum } \\ \end{array}\\ \begin{array}{lrcll} & A &=& A_i + A_o + A_{r_1} + A_{r_2} \\ & A &=& ( 2\pi r ) \cdot h + ( 2\pi R ) \cdot h + \pi R^2 - \pi r^2 + \pi R^2 - \pi r^2 \\ & A &=& ( 2\pi r ) \cdot h + ( 2\pi R ) \cdot h + 2\pi R^2 - 2\pi r^2 \\ & A &=& 2\pi ( r \cdot h + R \cdot h + R^2 - r^2 ) \\ & A &=& 2\pi [ h(r + R) + R^2 - r^2 ] \qquad & | \qquad R^2 - r^2 = ( R + r )(R - r)\\ & A &=& 2\pi [ h(r + R) + ( R + r )(R - r) ] \\ & A &=& 2\pi [ h(r + R) + ( r + R )(R - r) ] \\ & A &=& 2\pi (r + R)( h + R - r ) \\\\ & \mathbf{ A }& \mathbf{=} & \mathbf{ 2\cdot (r + R)\cdot ( h + R - r )\cdot \pi } \\ \end{array}\)

\(\begin{array}{lrcll} & A & = & 2\cdot (r + R)\cdot ( h + R - r )\cdot \pi \qquad r = 3\ mm \qquad R = 6\ mm \qquad h = 18\ mm \\ & A & = & 2\cdot (3 + 6)\cdot ( 18 + 6 - 3 )\cdot \pi \\ & A & = & 2\cdot 9\cdot 21 \cdot \pi \ mm^2 \\ & A & = & 378\cdot \pi \ mm^2 \\ & A & = & 1187.52202306 \ mm^2 \\ & \mathbf{A} & \mathbf{ = } & \mathbf{1187.52 \ mm^2 \qquad (\text{ rounded to the nearest hundredth}) }\\ \end{array}\)

or

\(\begin{array}{lrcll} & A & = & 1187.52202306 \ mm^2 \cdot \frac{1\ cm}{10\ mm} \cdot \frac{1\ cm}{10\ mm} \\ & A & = & \frac{1187.52202306}{100} \ cm^2 \\ & A & = & 11.8752202306 \ cm^2 \\ & A & = & 11.8752 \ cm^2 \\ \end{array}\)

3(x-4)=5 - (2x +3) has the solution x= 2 + 4/5 

change on term or sign to make the solution x=3

1.

\(\begin{array}{rcll} x &=& 2+ \frac45 \\ x +\frac15 &=& 2+ \frac45 +\frac15 \\ x +\frac15 &=& 3 \\\\ \bar{x} &=& x +\frac15 \qquad \text{ or } \qquad \\ \mathbf{x} &\mathbf{=}& \mathbf{\bar{x} - \frac15} \end{array}\)

2.

\(\begin{array}{rcll} 3(x-4) &=& 5 - (2x +3) \qquad & | \qquad \mathbf{x} = \mathbf{\bar{x} - \frac15} \\ 3(\mathbf{\bar{x} - \frac15}-4) &=& 5 - [2(\mathbf{\bar{x} - \frac15}) +3] \\ 3(\bar{x} - 4) -\frac35 &=& 5 - (2\bar{x}- \frac25 +3 ) \\ 3(\bar{x} - 4) -\frac35 &=& 5 - (2\bar{x} +3) + \frac25 \\ 3(\bar{x} - 4) &=& 5 - (2\bar{x} +3) + \frac25 +\frac35 \\ 3(\bar{x} - 4) &=& 5 - (2\bar{x} +3) + 1 \\ 3(\bar{x} - 4) &=& 6 - (2\bar{x} +3) \\\\ \mathbf{3( x - 4) } &\mathbf{=}& \mathbf{6 - (2x +3) \qquad \Rightarrow \qquad x=3 } \end{array}\)

What is the quadratic equation whose roots are -3, -1 and has a leading coefficient of 2 with X to represent the variable?

\(\begin{array}{rcll} 2\cdot (x+3)\cdot (x+1) &=& 0 \\ 2\cdot (x^2+x+3x+3) &=& 0\\ 2\cdot (x^2+4x+3) &=& 0\\ 2x^2+8x+6 &=& 0 \end{array}\)