heureka

1. Let $f(x) = Ax - 2B^2$ and $g(x) = Bx$, where $B \neq 0$. If $f(g(1)) = 0$, what is $A$ in terms of $B$?

2. Suppose that $f$ is a function and $f^{-1}$ is the inverse of $f$. If $f(1)=2$, $f(2) = 6$, and $f(3)=5$, then what is $f^{-1}(f^{-1}(6))$?

$\begin{array}{|lrcll|} \hline x = 1 : & g(x) &=& g(1) \\ & g(1) &=& B\cdot 1 \\ & g(1) &=& B \\\\ & f(g(1)) &=& f(B) \\ x=B: & f(B) &=& A\cdot B - 2B^2 \\ f(B) = 0: & A\cdot B - 2B^2 &=& 0 \\ & A\cdot B&=& 2B^2 \\ & A &=& \dfrac{2B^2}{B} \quad & | \quad B \ne 0~ ! \\ &\mathbf{ A }& \mathbf{=}& \mathbf{ 2B } \\ \hline \end{array}$

Given {a_n}, a_n=4/(2n-1)^2 and given {bn}, bn=(1-a_1)(1-a_2)...(1-a_n),

please use mathematical induction to prove that b_n=(2n+1)/(1-2n ). 

$\begin{array}{|rcll|} \hline a_n &=& \dfrac{4}{(2n-1)^2} \\\\ b_n &=& (1-a_1)(1-a_2)\ldots (1-a_n) \\\\ b_n &=& \dfrac{2n+1}{1-2n} \\ \hline \end{array}$

$\mathbf{n = 1}$

$\begin{array}{|rcll|} \hline a_1 &=& \dfrac{4}{(2\cdot 1-1)^2} \\\\ &=& \dfrac{4}{1} \\\\ &=& 4 \\\\ b_1 &=& \dfrac{2\cdot 1+1}{1-2\cdot 1} \\\\ &=& \dfrac{3}{-1} \\\\ &=& -3 \\\\ b_1 &=& (1-a_1) \\\\ &=& 1-4 \\ &=&\ -3\ \checkmark \\ \hline \end{array}$

$\mathbf{n = k}$

$\begin{array}{|rcll|} \hline a_k &=& \dfrac{4}{(2k-1)^2} \\\\ b_k &=& (1-a_1)(1-a_2)\ldots (1-a_k) \\\\ b_k &=& \dfrac{2k+1}{1-2k} \qquad \mathbf{b_{k+1} = \dfrac{2(k+1)+1}{1-2(k+1)}} \\ \hline \end{array}$

$\mathbf{k+1}$

$\begin{array}{|rcll|} \hline a_{k+1} &=& \dfrac{4}{(2(k+1)-1)^2} \\ &=& \dfrac{4}{(2k+2-1)^2} \\ &=& \dfrac{4}{(2k+1)^2} \\\\ b_{k+1} &=& (1-a_1)(1-a_2)\ldots (1-a_k)(1-a_{k+1}) \\ &=& b_k(1-a_{k+1}) \\ &=& b_k \left(1- \dfrac{4}{(2k+1)^2} \right) \quad & | \quad b_k = \dfrac{2k+1}{1-2k} \\\\ &=& \left(\dfrac{2k+1}{1-2k}\right) \left(1- \dfrac{4}{(2k+1)^2} \right) \\\\ &=& \left(\dfrac{2k+1}{1-2k}\right) \left( \dfrac{(2k+1)^2-4}{(2k+1)^2} \right) \\\\ &=& \dfrac{\left(2k+1\right)\Big( (2k+1)^2-4 \Big)}{\left(1-2k\right)(2k+1)^2} \\\\ &=& \dfrac{(2k+1)^2-4}{\left(1-2k\right)(2k+1)} \\\\ &=& \dfrac{4k^2+4k+1-4}{\left(1-2k\right)(2k+1)} \\\\ &=& \dfrac{4k^2+4k-3}{(1-2k)(2k+1)} \\\\ &=& -\dfrac{(2k+3)(1-2k)}{(1-2k)(2k+1)} \\\\ &=& \dfrac{(2k+3)}{-(2k+1)} \\\\ &=& \dfrac{2k+3}{-2k-1} \\\\ &\mathbf{=}& \mathbf{\dfrac{2(k+1)+1}{1-2(k+1)}}\ \checkmark \\ \hline \end{array}$

Suppose that $f$ is a function and $f^{-1}$ is the inverse of $f$. If $f(3)=4$, $f(5)=1$, and $f(2)=5$,
evaluate $f^{-1}\left(f^{-1}(5)+f^{-1}(4)\right)$.

$\begin{array}{|r|r|r|r|} \hline x & f(x) \\ \hline 2 & 5 & f(2) = 5 & f^{-1}(5) = 2 \\ 3 & 4 & f(3) = 4 & f^{-1}(4) = 3 \\ 5 & 1 & f(5) = 1 & \\ \hline f^{-1}(x) & x \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && f^{-1}\Big(f^{-1}(5)+f^{-1}(4)\Big) \\ &=& f^{-1}\left(2+3\right) \\ &=& f^{-1}(5) \\ &=& 2 \\ \hline \end{array}$

The 1st line is the original expression:

 
In line three, there is a mistake:

The binom

$(2a-b)^2 = (2a\cdot 2a -2ab-2ab+(-b)\cdot(-b) )\\=(4a^2-4ab+b^2)$

The line four:

$\begin{array}{|rcll|} \hline && 9(9a^2+6ab+b^2)-4(4a^2-4ab+b^2) \\ &=& 81a^2 +54ab+9b^2-16a^2+16ab-4b^2 \\ &=& 81a^2-16a^2 +54ab+16ab+9b^2-4b^2 \\ &=& 65a^2+70ab+5b^2 \\ &=& 5\cdot(13a^2+14ab+b^2) \\ &=& 5\cdot(a+b)(13a+b) \\ \hline \end{array}$

Hallo Max,

On the right side(RHS) 2n is an even number, so the left side(LHS) must also be even!

Ich und mein Vater haben eine veränderte version des Spiels 10.000 gespielt. statt mit 6 würfeln haben wir mit 10 gespielt. Mein Vater hatte einen unglaunblichen Wert von 50 millionen mit einem einzigen wurf gehabt (also 7 mal die augenzahl 5)

Berechnung mit dem Computer,

bei insgesamt 6^{10} = 60466176 verschiedenen Würfen, haben genau 15000 Würfe 7 mal die Augenzahl 5.

Die Wahrscheinlichkeit 7 mal die gleiche Augenzahl 5 zu würfeln, bei 10 Würfeln beträgt:

$\begin{array}{|rcll|} \hline P(X=7) &=& \\ &=& \dfrac{15000}{6^{10}} \\\\ &=& \dfrac{15000}{60466176} \\\\ &=& 0.00024807258 \quad (0.024807258\ \%)\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline P(X=7) &=& \text{binompdf}\left( 10,\dfrac16,7 \right) \\\\ &=& \dbinom{10}{7} \left( \dfrac16 \right)^7 \left( \dfrac56 \right)^3 \\\\ &=& \dfrac{15000}{6^{10}} \\ \hline \end{array}$

Suppose we have a house (with finitely many rooms) in which every room has an even number of doors.

Prove that the number of doors from the house to the outside world is also even.

Point P  is inside rectangle ABCD . Show that
PA^2 +PC^2 =PB^2 + PD^2.

Point P  is inside rectangle ABCD . Show that

PA^2 +PC^2 =PB^2 + PD^2.

$\begin{array}{|rcll|} \hline PA^2 &=& x_1^2+y_1^2 & PC^2 = y_2^2+x_2^2 \\\\ PB^2 &=& y_2^2+x_1^2 & PD^2 = x_2^2+y_1^2 \\\\ \hline PA^2 + PC^2 &=& x_1^2 + y_1^2 +y_2^2+x_2^2 \\ &=& x_1^2 + y_1^2 +x_2^2 + y_2^2 \\\\ PB^2 + PD^2 &=& y_2^2 + x_1^2 +x_2^2+y_1^2 \\ &=& x_1^2 + y_1^2 +x_2^2 + y_2^2 \\ \hline \end{array}$