heureka

e^(1/x)-x=0

Root-Finding Algorithm with Newton's method: $$\small{\text{ $ \boxed{ x_{i+1} = x_i - \dfrac{f(x_i)}{f'(x_i)} } \quad f{(x)} = e^{\frac{1}{x}}-x $ and $ f'(x) = -\frac{ e^{\frac{1}{x}}}{x^2}-1 $ we have $ x_{i+1} = x_i + \dfrac{ e^{\dfrac{1}{x_i}}-x_i } {\frac{ e^{ \left(\dfrac{1}{x_i} \right) } } {x_i^2}+1} $ }$$

We start with:  

$$\small{\text{ $x_0 = 1$ }} \\ \small{\text{ $ x_1 = 1+\frac{e-1}{e+1}= 1.46211715726 $ }} \\ \small{\text{ $ x_2= 1.46211715726 + \frac{ 0.51955244051 }{1.92697260563 }= 1.73173824009 $ }} \\ \small{\text{ $ x_3= 1.73173824009 + \frac{ 0.04975957061}{1.59404698875}= 1.76295411450 $ }} \\ \small{\text{ $ x_4= 1.76295411450+ \frac{ 0.00042115233 }{1.56736524331}= 1.76322281532 $ }} \\ \small{\text{ $ x_5= 1.76322281532+ \frac{ 0.00000002982}{1.56714330612}= 1.76322283435 $ }} \\ \small{\text{ $ x_6= 1.76322283435+ \frac{ 0.000000000000000149544303}{1.56714329041}= 1.76322283435 $ }}$$

$$x\approx 1.76322283435$$

b = a*tan(alpha)+c*cos(alpha) führt zu Gleichungen 4. Grades.

I. $$\small{\text{ Setze $ \tan{\alpha}=\frac{2t}{1+t^2}$ und $ \cos{\alpha } = \frac{1-t^2}{1+t^2} $ }}\\ \small{\text{ dann erhalten wir $(c+b)t^4+2at^3-2ct^2+2at+c-b=0 $ mit t=\tan{(\frac{\alpha}{2})} $ }}$$

II. $$\small{\text{ Setze $ \tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}$ und $ \sin{\alpha } = \sqrt{1-\cos^2{\alpha}} $ }}\\ \small{\text{ dann erhalten wir $c^2t^4-2bct^3+(a^2+b^2)t^2-a^2=0 $ mit t=\cos{\alpha} $ }}$$

John computes the sum of the elements of each of the 21 two-element subsets of {1,2,3,4,5,6,7}.

What is the sum of these 21 sums ?

$$\small{\text{ $ sum = \frac{3}{2} * \left(\ 1*2+2*3+3*4+4*5+5*6+6*7\ \right) = \frac{3}{2} * \left(\ 2+6+12+20+30+42\ \right) $ }}$\\$ \small{\text{ $ = \frac{3}{2}*112=168 $ }}$\\\\$ \small{\text{ for n: $ sum = \frac{3}{2} * \left(\ 1*2+2*3+3*4+4*5+5*6+6*7+\dots + (n-1)*n\ \right) = 3*\left( \begin{array}{c} n+1\\3 \end{array} \right) $ }}$\\\\$ \small{\text{ $ = \dfrac{(n-1)n(n+1)}{2} $ }}$\\\\$ \small{\text{ for n=7: $ sum = 3*\left( \begin{array}{c} 8\\3 \end{array} \right) =3*\frac{6*7*8}{2*3}=21*8=168 $ }}$$

what is 5!

5! = 5 * 4! 

4! = 4 * 3!

3! = 3 * 2!

2! = 2 * 1!

1! = 1 * 0!

0! = 1

we have: 5! = 5 * 4 * 3 * 2 * 1 * 1 = 5 * 4 * 3 * 2 * 1

greatest common divisors (GCD) of 23^31 , 23^17

$$\small{\begin{array}{l|rclclclcl}\text{Number 1:} & 23^{31} &=& 23^{31} &&&&&& \\ \text{Number 2:} & 23^{17} &=& 23^{\textcolor[rgb]{1,0,0}{17}} && && && \\ \hline \text{The lowest Exponent}\\ \text{ of each prime number} &\text{GCD}&=& 23^{\textcolor[rgb]{1,0,0}{17}}&&&&&& \end{array}}}$$

 3^13*5^17,2^12*7^21 Least Common Multiple (LCM) ?

$$\small{\begin{array}{l|rclclclcl}\text{Number 1:} & 3^{13}*5^{17} &=& 2^0 &*& 3^{\textcolor[rgb]{1,0,0}{13}} &*& 5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^0 \\\text{Number 2:} & 2^{12}*7^{21} &=& 2^{\textcolor[rgb]{1,0,0}{12}}&*& 3^0 &*& 5^0 &*& 7^{\textcolor[rgb]{1,0,0}{21}}\\ \hline \text{The greatest Exponent}\\ \text{ of each prime number} &\text{LCM}&=& 2^{\textcolor[rgb]{1,0,0}{12}}&*&3^{\textcolor[rgb]{1,0,0}{13}} &*&5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^{\textcolor[rgb]{1,0,0}{21}}\end{array}}}$$

3^13*5^17,2^12*7^21 Greatest Common Devisor(GCD) ?

$$\small{\begin{array}{l|rclclclcl}\text{Number 1:} & 3^{13}*5^{17} &=& 2^{\textcolor[rgb]{1,0,0}{0}} &*& 3^{13} &*& 5^{17} &*& 7^{\textcolor[rgb]{1,0,0}{0}} \\\text{Number 2:} & 2^{12}*7^{21} &=& 2^{12}&*& 3^{\textcolor[rgb]{1,0,0}{0}} &*& 5^{\textcolor[rgb]{1,0,0}{0}} &*& 7^{21}\\ \hline \text{The lowest Exponent}\\ \text{ of each prime number} &\text{GCD}&=& 2^{\textcolor[rgb]{1,0,0}{0}}&*&3^{\textcolor[rgb]{1,0,0}{0}} &*&5^{\textcolor[rgb]{1,0,0}{0}} &*& 7^{\textcolor[rgb]{1,0,0}{0}} = 1*1*1*1=1 \end{array}}}$$

Least Common Multiples(LCM)  ?

A. 23^31,23^17

$$\small{ \begin{array}{l|rclclclcl} \text{Number 1:} & 23^{31} &=& 23^{\textcolor[rgb]{1,0,0}{31}} &&&&&& \\ \text{Number 2:} & 23^{17} &=& 23^{17} &&&&&&\\ \hline \text{The greatest Exponent} \\ \text{ of each prime number} &\text{LCM}&=& 23^{\textcolor[rgb]{1,0,0}{31}}&&&&&& \end{array} }}$$

B. 3^7*5^3*7^3,2^11*3^5*5^9

$$\small{ \begin{array}{l|rclclclcl} \text{Number 1:} & 3^{7}*5^3*7^3 &=& 2^0 &*& 3^{\textcolor[rgb]{1,0,0}{7}} &*& 5^3 &*& 7^{\textcolor[rgb]{1,0,0}{3}} \\ \text{Number 2:} & 2^{11}*3^5*5^9 &=& 2^{\textcolor[rgb]{1,0,0}{11}}&*& 3^5 &*& 5^{\textcolor[rgb]{1,0,0}{9}} &*& 7^0\\ \hline \text{The greatest Exponent} \\ \text{ of each prime number} &\text{LCM}&=& 2^{\textcolor[rgb]{1,0,0}{11}}&*&3^{\textcolor[rgb]{1,0,0}{7}} &*&5^{\textcolor[rgb]{1,0,0}{9}} &*& 7^{\textcolor[rgb]{1,0,0}{3}} \end{array} }}$$

C. 3^13*5^17,2^12*7^21

   $$\small{ \begin{array}{l|rclclclcl} \text{Number 1:} & 3^{13}*5^{17} &=& 2^0 &*& 3^{\textcolor[rgb]{1,0,0}{13}} &*& 5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^0 \\ \text{Number 2:} & 2^{12}*7^{21} &=& 2^{\textcolor[rgb]{1,0,0}{12}}&*& 3^0 &*& 5^0 &*& 7^{\textcolor[rgb]{1,0,0}{21}}\\ \hline \text{The greatest Exponent} \\ \text{ of each prime number} &\text{LCM}&=& 2^{\textcolor[rgb]{1,0,0}{12}}&*&3^{\textcolor[rgb]{1,0,0}{13}} &*&5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^{\textcolor[rgb]{1,0,0}{21}} \end{array} }}$$  

Find the sum of all positive integers less than 1000 ending in 3 or 7.

I.  

Sum of the Sequence: s = 3 + 7 + 13 + 17 + 23 + 27 + 33 + 37 + 43 + 47 + ... + 983 + 987 + 993 + 997

We can group in one series:

s= (3+7) + (13+17) + (23+27) + (33+37) + (43+47) + ... + (983+987) + (993+997)

s =           10 + 30 + 50 + 70 + 90 + ... + 1970 + 1990

s = 10 * (   1 + 3   + 5   + 7   + 9  + ... +  197  + 199 )

 The sum of the odd numbers (   1 + 3   + 5   + 7   + 9  + ... +  197  + 199 ) = n^2

2n-1 = 199  ->  n = 100

The sum of the odd numbers (   1 + 3   + 5   + 7   + 9  + ... +  197  + 199 ) = n^2 = 100^2 = 10000

Sum of the Sequence: s = 10 * 10000 = 100000