Mitglied: heureka

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:

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

$$x\approx 1.76322283435$$

heureka20.01.2015

+26387

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

heureka20.01.2015

+26387

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 ?

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

heureka20.01.2015

+26387

Part (a): Find the sum s = $a + (a + 1) + (a + 2) + \dots + (a + n - 1)$ in terms of $a$ and $n.$

$$s = a + (a+1) + ( a+2) + (a+3) + ... +(a+ (n-2)) + (a+(n-1))\\\\
s = \left[a + (a+(n-1))\right] *(\frac{n}{2} ) \\\\
s = \left[2a+(n-1))\right]*(\frac{n}{2} ) \\\\
\boxed{s = n*a+\frac{n(n-1)}{2}}$$

Part (b): Find all pairs of positive integers $(a,n)$ such that $n \ge 2$ and $a + (a + 1) + (a + 2) + \dots + (a + n - 1) = 100.$

$$\small{\text{$ 2\le n\le14 \text{ and } a > 0 $}}\\
\small{\text{
$ n= 2\quad a=49.500000 $ }} \\ \small{\text{
$ n= 3\quad a=32.333333 $ }} $\\$ \small{\text{
$ n= 4\quad a=23.500000 $ }} $\\$ \small{\text{
$\textcolor[rgb]{1,0,0}{n= 5\quad a=18.000000} $ }} $\\$ \small{\text{
$ n= 6\quad a=14.166667 $ }} $\\$ \small{\text{
$ n= 7\quad a=11.285714 $ }} $\\$ \small{\text{
$ \textcolor[rgb]{1,0,0}{n= 8\quad a=9.000000 }$ }} $\\$ \small{\text{
$n= 9\quad a=7.111111 $ }} $\\$ \small{\text{
$n=10\quad a=5.500000 $ }} $\\$ \small{\text{
$n=11\quad a=4.090909 $ }} $\\$ \small{\text{
$n=12\quad a=2.833333 $ }} $\\$ \small{\text{
$n=13\quad a=1.692308 $ }} $\\$ \small{\text{
$n=14\quad a=0.642857 $ }} $\\$ \small{\text{
The only 2 solutions for $(a,n)$ are $ (18,5),\ (9,8)$
}} $\\$
\small{\text{
$
\textcolor[rgb]{1,0,0}{18}+19+20+21+22 = 100 \quad $ and $\quad \textcolor[rgb]{1,0,0}{9}+10+11+12+13+14+15+16 = 100
$
}}$$

heureka20.01.2015

+26387

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

heureka19.01.2015

+26387

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

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

heureka19.01.2015

+26387

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) ?

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

heureka19.01.2015

+26387

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

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

heureka19.01.2015

+26387

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

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

heureka19.01.2015

+26387

Let $a_1$ , $a_2$ , . . . , $a_{10}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 + a_7 + a_9 = 17$

and $a_2 + a_4 + a_6 + a_8 + a_{10} = 15$ , then find $a_1$ .

1. arithmetic sequence: $$\small{\text{
$
a_1, a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10}
$
}}$\\$
\small{\text{
$
a_n=a_1+(n-1)*d $ and
$
s_n= \frac{n}{2}*[2a_1*(n-1)d]
$
}}
\small{\text{
$
\Rightarrow s_{10}= 17+15 = \frac{10}{2}*[2a_1+(10-1)d]
$
}} $\\$
\small{\text{
$
32 = 5*(2a_1+9d)
$
}} $\\$
\small{\text{
$
\textcolor[rgb]{0,0,1}{(1) \quad 32 = 10a_1+45d}
$
}}$$

2. arithmetic sequence:

$$\small{\text{
$
a_1,a_3,a_5,a_7,a_9$
}}$\\$
\small{\text{
$
a_i=a_1+(i-1)*(2d) $ and
$
s_i= \frac{i}{2}*[2a_1*(i-1)(2d)]
$
}}
\small{\text{
$
\Rightarrow s_5= 17 = \frac{5}{2}*[2a_1+(5-1)(2d)]
$
}} $\\$
\small{\text{
$
2*17 = 5*(2a_1+8d)
$
}} $\\$
\small{\text{
$
\textcolor[rgb]{0,0,1}{(2) \quad 34 = 10a_1+40d}
$
}}$$

$$\small{\text{
(1) - (2):
$\quad
32-34 = 10a_1-10a_1 + 45d-40d = 5d \qquad -2=5d \qquad \boxed{d = -\frac{2}{5}}
$
}}$$

$$\small{\text{
d into (2):
$
\quad
34 = 10a_1+40*(-\frac{2}{5}) = 10a_1 - 16 \qquad 10a_1 = 50 \qquad \boxed{a_1 =5 }
$
}}$$

heureka19.01.2015

Benutzername	heureka
Punkte	26387
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Antworten	5678