heureka

Note that

1111111=10^6+10^5+10^4+10^3+10^2+10+1

and

909091=10^6-10^5+10^4-10^3+10^2-10+1.

Compute the product of these two integers.

geometric series:

$\begin{array}{|rcll|} \hline a_n &=& a_1 \cdot r^{n-1} \\ s_n &=& a_1 \left( \frac{-1+r^n}{1-r} \right) \qquad \text{sum}\\ \hline \end{array}$

geometric series 1:

$\begin{array}{|rcll|} \hline && 10^6+10^5+10^4+10^3+10^2+10+1 \qquad | \qquad a_1 = 1 \qquad r = 10 \\\\ s_n &=& a_1 \left( \frac{-1+r^n}{1-r} \right) \\ s_n &=& 1\cdot \left( \frac{-1+10^n}{1-10} \right) \\ s_7 &=& 1\cdot \left( \frac{-1+10^7}{1-10} \right) \\ s_7 &=&\left( \frac{-1+10^7}{1-10} \right) \\ \hline \end{array}$

geometric series 2:

$\begin{array}{|rcll|} \hline && 10^6-10^5+10^4-10^3+10^2-10+1 \qquad | \qquad a_1 = 1 \qquad r = -10 \\\\ S_n &=& a_1 \left( \frac{-1+r^n}{1-r} \right) \\ S_n &=& 1\cdot \left( \frac{-1+(-10)^n}{1-(-10)} \right) \\ S_7 &=& 1\cdot \left( \frac{-1+(-10)^7}{1+10} \right) \\ S_7 &=&\left( \frac{-1-10^7}{1+10} \right) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && 1111111 \cdot 909091 \\ &=& (10^6+10^5+10^4+10^3+10^2+10+1) \cdot (10^6-10^5+10^4-10^3+10^2-10+1) \\ &=& s_7 \cdot S_7 \\ &=& \left( \frac{-1+10^7}{1-10} \right) \cdot \left( \frac{-1-10^7}{1+10} \right) \\ &=& \left[ \frac{(-1)^2-(10^7)^2}{1^2-(10)^2} \right] \\ &=& \frac{1-10^{14}}{1-10^2} \\ &=& \frac{10^{14}-1}{10^2-1} \\ &=& \frac{99999999999999}{99} \\ &=& 10101010101010 \\ \hline \end{array}$

Suppose the real numbers a,b,x, and y satisfy the equations

     ax+by=3,

     ax^2+by^2=7,

     ax^3+by^3=16,

     ax^4+by^4=42.  

Evaluate ax^5+by^5.

$\begin{array}{|rcll|} \hline && ax^5 + b y^5 \\ &=& ( \frac{49}{76}-\frac{457}{76 \cdot \sqrt{87}}) \cdot (-7-\sqrt{87})^5 +( \frac{49}{76}+\frac{457}{76 \cdot\sqrt{87}}) \cdot (\sqrt{87}-7)^5 \\ &=& 20 \\ \hline \end{array}$

An airplane makes a 990 km flight with a tailwind and returns,
flying into the same wind. The total flying time is 3 hours 20 minutes,
and the airplane’s speed in still air is 600 km/h. What is the speed of the wind?

velocity airplane:  $v_a$

velocity wind: $v_w$

time for the journey there: $t_1$

time for way back: $t_2$

distance: d = 990 km

total time: $t = t_1 + t_2$           

( t = 3 hours 20 minutes )

$\begin{array}{|rcll|} \hline d &=& (v_a+v_w)\cdot t_1 \quad \text{ for the journey there} \\ d &=& (v_a-v_w)\cdot t_2 \quad \text{ for way back} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline t_1 &=& \dfrac{d}{v_a+v_w} \\ t_2 &=& \dfrac{d}{v_a-v_w} \\\\ t &=& t_1 + t_2 \\\\ t &=& \dfrac{d}{v_a+v_w} + \dfrac{d}{v_a-v_w} \\\\ t &=& d\cdot \left( \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \right) \\\\ \dfrac{t}{d} &=& \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \\\\ \dfrac{t}{d} &=& \dfrac{v_a-v_w+v_a+v_w}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{v_a^2-v_w^2} \\\\ \dfrac{d}{t} &=& \dfrac{v_a^2-v_w^2}{2v_a} \quad &| \quad \cdot 2v_a\\ 2v_a\cdot \dfrac{d}{t} &=& v_a^2-v_w^2 \quad &| \quad +v_w^2\\ v_w^2 + 2v_a\cdot \dfrac{d}{t} &=& v_a^2 \quad &| \quad -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a^2 -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a\cdot \left( v_a - \dfrac{2d}{t} \right) \quad &|d=990 \quad v_a = 600 \quad t=3\frac{1}{3}=\frac{10}{3}\ \text{hours}\\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990}{\frac{10}{3}} \right) \\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990\cdot 3}{10} \right) \\ v_w^2 &=& 600\cdot \left( 600 - 6\cdot 99 \right) \\ v_w^2 &=& 600\cdot \left( 600 - 594 \right) \\ v_w^2 &=& 600\cdot 6 \\ v_w^2 &=& 3600 \quad &| \quad \sqrt{} \\ v_w & = & \sqrt{3600} \\ \mathbf{v_w} & \mathbf{=} & \mathbf{60} \\ \hline \end{array}$

The speed of the wind is $\mathbf{60 \ \frac{km}{h}}$ .

(a) Compute the sums of the squares of Rows 1-4 of Pascal's Triangle. That is, compute:

$\begin{array}{|lcll|} \hline \binom10^2 + \binom11^2 &=& \binom21 =2 \\ \binom20^2 + \binom21^2 + \binom22^2 &=& \binom42 = 6 \\ \binom30^2 + \binom31^2 + \binom32^2 + \binom33^2 &=& \binom63 = 20 \\ \binom40^2 + \binom41^2 + \binom42^2 + \binom43^2 + \binom44^2 &=& \binom84 = 70\\ \hline \end{array}$

Do these sums appear anywhere else in Pascal's Triangle?
Yes: as Central binomial coefficients

(b) Guess at an identity based on your observations from part (a). Your identity should be of the form

$\begin{array}{|rcll|} \hline \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 &=& \binom{2n}{n} \\ \hline \end{array}$

EVALUATE THE EXPRESSION:
$(2+1)(2^2+1)(2^4+1)...(2^{1024}+1)+1$

$\begin{array}{|rcll|} \hline (2^{\color{red}1}+1)(2^{\color{red}2}+1) &=& 2^{\color{red}3}+2^2+2^1+1 &| 1+2 = 3\\ (2^{\color{red}1}+1)(2^{\color{red}2}+1)(2^{\color{red}4}+1)&=& 2^{\color{red}7}+2^6+2^5+2^4+2^3+2^2+2^1+1 &| 1+2+4=7\\ \dots \\ (2^1+1)(2^2+1)(2^4+1)...(2^{1024}+1) &=& 2^{2047}+2^{2046}+\dots +2^1 + 1 \\ &&|1+2+4+\dots + 1024 = 2047 \\\\ & & 2^{2047}+2^{2046}+\dots +2^1 + 1 = 2^{2048}-1 \\\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ (2^1+1)(2^2+1)(2^4+1)...(2^{1024}+1)+1 } & \mathbf{=} & \mathbf{2^{2048}}\\ \hline \end{array}$

Eine kleine Spinne möchte ein Netz in einer rechteckigen Masche eines Gitterzauns bauen. Um ihr Netz zu befestigen zieht sie zunächst zwei stramm gespannte Fäden durch die Masche. Der eine geht von der linken Seite zur rechten unteren Ecke und hat eine Länge von 105 mm. Der andere Faden geht von der rechten Seite zur linken unteren Ecke und hat eine Länge von 87 mm. Sie stellt mit Hilfe ihres kleinen Spinnenlineals fest, dass sich die beiden Fäden 35 mm über der Unterkante ihrer Masche kreuzen.

Wie breit sind die Löcher im Gitterzaun und wie hoch sind sie mindestens?

Wenn ich die Aufgabe richtig verstanden habe,

so vermute ich das die Löcherbreite im Gitterzaun 63 mm beträgt

und sie mindestens 60 mm oder 84 mm hoch sind.

Actually, although I know how to do differentiation,
it's still a bit hard to get the original one from guessing by the
differentiated one.(I mean, doing integration)

How, actually, do I do integration by parts, for example,
I want to calculate

$\int{(\cos x)(e^{\sin x})dx}?$

$\begin{array}{|lcll|} \hline \int{\cos x\cdot e^{\sin x}\ dx} \ ? \\ \text{Substitute: } u = e^{\sin x} \qquad du = e^{\sin x} \cdot \cos x \ dx \quad dx = \frac{du}{u\cdot \cos x} \\ \hline \end{array}$

$\begin{array}{|lcll|} \hline && \int{\cos x\cdot e^{\sin x}\ dx} \\ &=& \int{\cos x\cdot u \cdot \frac{du}{u\cdot \cos x} } \\ &=& \int{\ du} \\ &=& u + c \\ \int{\cos x\cdot e^{\sin x}\ dx} &=& e^{\sin x} + c \\ \hline \end{array}$

4. Given

$a \equiv 1 \pmod{7}$,

$b \equiv 2 \pmod{7}$,

and

$c \equiv 6 \pmod{7}$,

what is the remainder when

$a^{81} b^{91} c^{27}$

is divided by 7?

$\begin{array}{|rclcl|} \hline a-1 &=& 7l \quad &\text{or}& \quad a =7l+1 \\ b-2 &=& 7m \quad &\text{or}& \quad b =7m+2 \\ c-6 &=& 7n \quad &\text{or}& \quad c =7n+6 \\ \hline \end{array}$

$\begin{array}{|rclcl|} \hline && a^{81} b^{91} c^{27} \pmod 7 \\ &\equiv& (7l+1)^{81}\cdot (7m+2)^{91}\cdot (7n+6)^{27} \pmod 7 \\\\ &\equiv& \left[\binom{81}{0}(7l)^{81}+ \binom{81}{1}(7l)^{80} +\dots + \binom{81}{81} 1^{81} \right] \\ && \cdot \left[\binom{91}{0}(7m)^{91}+ \binom{91}{1}(7m)^{90} +\dots + \binom{91}{91} 2^{91} \right] \\ && \cdot \left[\binom{27}{0}(7n)^{27}+ \binom{27}{1}(7n)^{26} +\dots + \binom{27}{27} 6^{27} \right] \pmod 7 \\\\ &\equiv& \binom{81}{81} 1^{81}\cdot \binom{91}{91} 2^{91} \cdot \binom{27}{27} 6^{27}\pmod 7 \\\\ &\equiv& 1^{81}\cdot 2^{91} \cdot 6^{27}\pmod 7 \\ &\equiv& 1 \cdot 2^{91} \cdot 6^{27}\pmod 7 \\ &\equiv& 2^{91} \cdot 6^{27}\pmod 7 \\ \hline \end{array}$

$\begin{array}{|lllcl|} \hline \text{Because } gcd(2,7) = 1 \quad \Rightarrow & 2^{\varphi(7)} \equiv 1 \pmod 7 \qquad \varphi(7) = 6\\ & \mathbf{2^{6} \equiv 1 \pmod 7} \\\\ \text{Because } gcd(6,7) = 1 \quad \Rightarrow & 6^{\varphi(7)} \equiv 1 \pmod 7 \qquad \varphi(7) = 6\\ & \mathbf{6^{6} \equiv 1 \pmod 7} \\ \hline \end{array}$

$\begin{array}{|rclcl|} \hline && a^{81} b^{91} c^{27} \pmod 7 \\ &\equiv& 2^{91} \cdot 6^{27}\pmod 7 \\ &\equiv& 2^{6\cdot 15+1} \cdot 6^{6\cdot 4+3}\pmod 7 \\ &\equiv& 2^{6\cdot 15}\cdot 2 \cdot 6^{6\cdot 4}\cdot 6^3 \pmod 7 \\ &\equiv& (2^6)^{15}\cdot 2 \cdot (6^6)^{4}\cdot 6^3\pmod 7 \qquad \mathbf{2^{6} \equiv 1 \pmod 7}\\ &\equiv& (1)^{15}\cdot 2 \cdot (6^6)^{4}\cdot 6^3\pmod 7 \qquad \mathbf{6^{6} \equiv 1 \pmod 7}\\ &\equiv& (1)^{15}\cdot 2 \cdot (1)^{4}\cdot 6^3\pmod 7 \\ &\equiv& 1 \cdot 2 \cdot 1 \cdot 6^3\pmod 7 \\ &\equiv& 2 \cdot 6^3 \pmod 7 \\ &\equiv& 2 \cdot 216 \pmod 7 \\ &\equiv& 432 \pmod 7 \\ &\equiv& \mathbf{5} \pmod 7 \\ \hline \end{array}$

The remainder is 5

x tends to 0 ((3+2x)^5-243)/(3x)

$\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} { \left( \frac{ (3+2x)^5-243 } {3x} \right) } \\\\ &=& \lim \limits_{x\to 0} { \left( \frac{ \binom{5}{0}3^5+ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 -243 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( \frac{ 3^5+ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 -3^5 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( \frac{ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( ~ \binom{5}{1}3^3\cdot 2+ \binom{5}{2}3^2\cdot 4x+ \binom{5}{3}3^1\cdot 8x^2+ \binom{5}{4}16x^3+ \binom{5}{5}\frac{32}{3}x^4 ~ \right) }\\\\ &=& \binom{5}{1}3^3\cdot 2 \\\\ &=& 5\cdot 27 \cdot 2 \\\\ &=& 270 \\\\ \hline \end{array}$

12 1\2+8 1\6+9 1\3=?

1) 20   2) 30

3) 40   4)50

$\begin{array}{|rcll|} \hline && 12 \frac 1 2 +8 \frac 1 6 +9 \frac 1 3 \\ &=& 12 + \frac 1 2 +8 + \frac 1 6 +9 + \frac 1 3 \\ &=& 12 + 8 + 9 + \frac 1 2 + \frac 1 6 + \frac 1 3 \\ &=& 29 + \frac 1 2 + \frac 1 6 + \frac 1 3 \\ &=& 29 + \frac{ 1 } { 2 }\cdot \frac{ 3 } { 3 } + \frac 1 6 + \frac {1}{ 3} \cdot \frac{ 2 } { 2 }\\ &=& 29 + \frac{ 3 } { 6 } + \frac 1 6 + \frac {2}{ 6}\\ &=& 29 + \frac{ 3+1+2 } { 6 } \\ &=& 29 + \frac{ 6 } { 6 } \\ &=& 29 + 1 \\ &=& 30 \\ && \mathbf{12 \frac 1 2 +8 \frac 1 6 +9 \frac 1 3 = 30} \\ \hline \end{array}$