heureka

Let F(x) be the real-valued function defined for all real x except for x = 0 and x = 1 and satisfying

the functional equation

\(F(x) + F\left(\frac{x-1}x\right) = 1+x.\)

F(x) + F\left(\frac{x-1}x\right) = 1+x.

Find the F(x) satisfying these conditions.

Write F(x) as a rational function with expanded polynomials in the numerator and denominator.

\(\begin{array}{|lrclcl|} \hline & F(x) + F\left(\frac{x-1}x\right) &=& 1+x \qquad (1) \\\\ \text{Set in (1) }x=\frac{x-1}{x}: & F\left(\frac{x-1}x\right) + F\left(\frac1{1-x}\right) &=& 1+\frac{x-1}{x} \qquad (2) \\\\ \text{Set in (1) }x=\frac1{1-x}: & F\left(\frac1{1-x}\right) + F(x) &=& 1+\frac1{1-x} \qquad (3) \\\\ \hline \\ (1) - (2) + (3): & F(x) + F\left(\frac{x-1}x\right) \\ & - \Big(F\left(\frac{x-1}x\right) + F\left(\frac1{1-x}\right) \Big) \\ & + F\left(\frac1{1-x}\right) + F(x) &=& 1+x -(1+\frac{x-1}{x}) + 1+\frac1{1-x} \\\\ & F(x) + F\left(\frac{x-1}x\right) \\ & -F\left(\frac{x-1}x\right) - F\left(\frac1{1-x}\right) \\ & + F\left(\frac1{1-x}\right) + F(x) &=& 1+x -(1+\frac{x-1}{x}) + 1+\frac1{1-x} \\\\ & 2F(x) &=& 1+x -(1+\frac{x-1}{x}) + 1+\frac1{1-x} \\ & 2F(x) &=& 1+x -1-\frac{x-1}{x} + 1+\frac1{1-x} \\ & 2F(x) &=& 1+x -\frac{x-1}{x} +\frac1{1-x} \\ & 2F(x) &=& 1+x +\frac{1-x}{x} +\frac1{1-x} \\\\ & 2F(x) &=& \dfrac{(1+x)x(1-x)+(1-x)^2+x}{x(1-x)} \\\\ & 2F(x) &=& \dfrac{(1-x^2)x+1-2x+x^2+x}{x(1-x)} \\\\ & 2F(x) &=& \dfrac{(1-x^2)x+1-x+x^2}{x(1-x)} \\\\ & 2F(x) &=& \dfrac{ x-x^3 +1-x+x^2}{x(1-x)} \\\\ & 2F(x) &=& \dfrac{ 1+x^2-x^3}{x(1-x)} \\\\ & F(x) &=& \dfrac{ 1+x^2-x^3}{2x(1-x)} \\\\ & \mathbf{F(x)} & \mathbf{=} & \mathbf{\dfrac{ 1+x^2-x^3}{2x - 2x^2}} \\ \hline \end{array}\)

graph:

Use logarithmic differentation to find the derivative of the function.

\(y=\sqrt{x}{e}^{({x}^{2})}({{x}^{2}+1})^{10}\)

y=\sqrt{x}{e}^{(x^2)}(x^2+1)^{10}

Anyone who knows how to solve for the answer and can write down the steps?

Formula

\(\begin{array}{|rclcl|} \hline \left(~\ln f(x)~ \right)' = \dfrac{f'(x)}{f(x)} \\ \hline \end{array} \)

1. logarithm of both sides

\(\begin{array}{|rcll|} \hline y &=& \sqrt{x}{e}^{(x^2)}(x^2+1)^{10} \quad & | \quad \text{$\ln()$ both sides} \\ \ln(y) &=& \ln(\sqrt{x})+\ln({e}^{(x^2)}) + \ln( (x^2+1)^{10} ) \\ \ln(y) &=& \ln(x^{\frac12})+\ln({e}^{(x^2)}) + \ln( (x^2+1)^{10} ) \quad & | \quad \text{Formula: $\ln(a^b) = b\ln(a) $} \\ \ln(y) &=& \frac12\ln(x)+x^2\ln(e) + 10\ln(x^2+1) \quad & | \quad \text{Formula: $\ln(e) = 1 $} \\ \ln(y) &=& \frac12\ln(x)+x^2 + 10\ln(x^2+1) \\ \hline \end{array}\)

2. derivation of both sides

\(\begin{array}{|rcll|} \hline \ln(y) &=& \frac12\ln(x)+x^2 + 10\ln(x^2+1) \quad & | \quad \text{derivate both sides} \\ \Big(\ln(y)\Big)' &=& \left(\frac12\ln(x) \right)' + \left(x^2 \right)' + \left(10\ln(x^2+1) \right)' \\\\ \dfrac{y'}{y} &=& \frac12\cdot \frac1x + 2x + 10\cdot \frac{2x}{x^2+1} \\\\ \dfrac{y'}{y} &=& \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \quad & | \quad \cdot y \\\\ y' &=& y\cdot \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \quad & | \quad y = \sqrt{x}e^{(x^2)}(x^2+1)^{10} \\\\ y' &=& \sqrt{x}e^{(x^2)}(x^2+1)^{10} \cdot \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \\\\ y' &=& x^{\frac12}e^{(x^2)}(x^2+1)^{10} \cdot \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \\\\ y' &=& \frac{ x^{\frac12}{e}^{(x^2)}(x^2+1)^{10} }{2x} + x^{\frac12}e^{(x^2)}(x^2+1)^{10} \cdot 2x \\ &+& x^{\frac12}e^{(x^2)}(x^2+1)^{10} \cdot \frac{20x}{x^2+1} \\\\ y' &=& \frac{ e^{(x^2)}(x^2+1)^{10} }{2x^{1-\frac12}} + 2{e}^{(x^2)}x^{\frac12+1}(x^2+1)^{10}\\ &+& 20e^{(x^2)} x^{\frac12+1}(x^2+1)^{10} \cdot \frac{1}{(x^2+1)^1} \\\\ y' &=& \frac{ e^{(x^2)}(x^2+1)^{10} }{2x^{\frac12}} + 2{e}^{(x^2)}x^{\frac32}(x^2+1)^{10}\\ &+& 20e^{(x^2)} x^{\frac32}(x^2+1)^{10-1} \\\\ y' &=& \frac{ e^{(x^2)}(x^2+1)^{10} }{2\sqrt{x}} + 2{e}^{(x^2)}x^{\frac32}(x^2+1)^{10}\\ &+& 20e^{(x^2)} x^{\frac32}(x^2+1)^{9} \\ \hline \end{array} \)

\(\begin{array}{rcll} \mathbf{ y' } & \mathbf{=} & \mathbf{ \dfrac{ e^{(x^2)}(x^2+1)^{10} }{2\sqrt{x}} + 2{e}^{(x^2)}x^{\frac32}(x^2+1)^{10} + 20e^{(x^2)} x^{\frac32}(x^2+1)^{9} } \\ \end{array} \)

Suppose

\(f(x) \)

is a rational function such that  \(3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2\)

for all \(x \neq 0\). Find \(f(-2)\).

\(\begin{array}{|lrclcl|} \hline & 3f(\frac1x) + \frac2x f(x) &=& x^2 \\ & 3f(\frac1x) &=& x^2 - \frac2x f(x) \\ & f(\frac1x) &=& \frac{x^2}{3} - \frac{2}{3x} f(x) \qquad (1) \\ \\ \text{Set }x=\frac1x: & 3f(x) + 2x f(\frac1x) &=& \frac{1}{x^2} \\ & 2x f(\frac1x) &=& \frac{1}{x^2}-3f(x) \\ & f(\frac1x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \qquad (2) \\\\ \hline (1) = (2): & \frac{x^2}{3} - \frac{2}{3x} f(x) &=& \frac{1}{2x^3}-\frac{3}{2x}f(x) \\ & \frac{3}{2x}f(x) - \frac{2}{3x} f(x) &=& \frac{1}{2x^3} -\frac{x^2}{3} \\ & f(x)\left( \frac{3}{2x} - \frac{2}{3x} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{9x-4x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{6x^2} \right) &=& \frac{3-2x^5}{6x^3} \\ & f(x)\left( \frac{5x}{1} \right) &=& \frac{3-2x^5}{x} \\\\ & \mathbf{f(x)} & \mathbf{=} & \mathbf{\dfrac{3-2x^5}{5x^2}} \\ \hline \end{array} \)

The rational function is  \(f(x) = \dfrac{3-2x^5}{5x^2}\)

\(f(-2)=\ ?\)

\(\begin{array}{|rclcl|} \hline f(-2) &=& \frac{3-2(-2)^5}{5(-2)^2} \\ &=& \frac{3+2^6}{5\cdot 4} \\ &=& \frac{3+64}{20} \\ &=& \frac{67}{20} \\ \mathbf{f(-2)} & \mathbf{=} & \mathbf{3.35} \\ \hline \end{array}\)

\(\text{ $f(-2)$ is $3.35$ }\)

graph:

What is the remainder when

\(2001 \cdot 2002 \cdot 2003 \cdot 2004 \cdot 2005\)

is divided by 19?

\(\begin{array}{|rclcl|} \hline && 2001 \cdot 2002 \cdot 2003 \cdot 2004 \cdot 2005 \pmod{19} \\ &&&& 2001 \pmod{19} \\ &&&\equiv& {\color{red}6} \pmod{19} \\\\ &&&& 2002 \pmod{19} \\ &&&\equiv& {\color{red}7} \pmod{19} \\\\ &&&& 2003 \pmod{19} \\ &&&\equiv& {\color{red}8} \pmod{19} \\\\ &&&& 2004 \pmod{19} \\ &&&\equiv& {\color{red}9} \pmod{19} \\\\ &&&& 2005 \pmod{19} \\ &&&\equiv& {\color{red}10} \pmod{19} \\\\ &\equiv& 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \pmod{19} \\ &\equiv& 30240 \pmod{19} \\ &\equiv& {\color{red}11} \pmod{19} \\ \hline \end{array}\)

The remainder is 11

I'm the challenger, I post questions every week!

Week 1:

Find the units digit of 3^17*7^23?

\(3^{17} * 7 ^{23} \pmod{10} = \ ? \)

\(\begin{array}{|rclcl|} \hline && 3^{17} * 7 ^{23} \pmod{10} \\ &\equiv& (3^2)^{8}*3^1*(7^2)*7^1 \pmod{10} \\\\ &&&& 3^2 \pmod{10} \\ &&&\equiv& 9 \pmod{10} \\ &&&\equiv& 9-10 \pmod{10} \\ &&&\equiv& {\color{red}-1} \pmod{10} \\\\ &&&& 7^2 \pmod{10} \\ &&&\equiv& 49 \pmod{10} \\ &&&\equiv& 9 \pmod{10} \\ &&&\equiv& 9-10 \pmod{10} \\ &&&\equiv& {\color{red}-1} \pmod{10} \\\\ &\equiv& ({\color{red}-1})^{8}*3^1*({\color{red}-1})*7^1 \pmod{10} \\ &\equiv& 3 *(-7) \pmod{10} \\ &\equiv& -21 \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& {\color{red}9} \pmod{10} \\ \hline \end{array}\)

Hallo Freunde!

Damit es euch in den Ferien nicht zu langweilig wird, kommt hier was zum ausknobeln.

Dies sind Gleichungen, drei wagerecht, drei senkrecht.

         I          II            III

IV  a b b  - a c d  =   c b

          :           +          +  

 V    e g    e c  =  a a c

       ________________

VI    e b  +  a h c =  a i a

Einzige Lösung:  \( a = 2 \\ b = 8 \\ c= 4 \\ d = 0 \\ e = 1 \\ g = 6 \\ h = 5 \\ i = 7 \\\)

\(\begin{array}{rrcrcr} & I && II && III \\ IV & 288 &-& 240 &=& 48 \\ & : && + && + \\ V & 16 &\times& 14 &=& 224 \\ \\ \hline \\ VI & 18 &+& 254 &=& 272 \\ \end{array}\)

Let $n$ be the inverse of $2\pmod{17}$.

That is, let $n$ be the integer $0\leq n < 17$ for which $2n \equiv 1 \pmod{17}$.
What is $\left(2^n\right)^2 - 2 \pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive.

\(\begin{array}{|rcll|} \hline 2n &\equiv& 1 \pmod {17} \quad & | \quad : 2 \\\\ n &\equiv& \frac12 \pmod {17} \\ \mathbf{n} &\mathbf{\equiv}& \mathbf{ 2^{-1} \pmod {17}} \\ \hline \end{array} \)

\(\text{ Solve $2^{-1} \pmod {17}$ }\)

\(\text{Because gcd(2,17)=1 we can continue.}\)

\(\begin{array}{|rcll|} \hline 2^{-1} \pmod {17} &\equiv& 2^{\varphi(17)-1} \pmod {17} \quad & | \quad \varphi(p) = p - 1 \quad \text{$p$ is a prime number} \\\\ && \varphi(17) = 16 \\\\ &\equiv& 2^{16-1} \pmod {17} \\ &\equiv& 2^{15} \pmod {17} \\ &\equiv& 32768 \pmod {17} \\ 2^{-1} \pmod {17}&\equiv& 9 \pmod {17} \\ \mathbf{n} &\mathbf{=}& \mathbf{9} \\ \hline \end{array} \)

\(Proof:\\ 2\cdot 9 \equiv 1 \pmod{17} \ \checkmark \)

\(\begin{array}{|rcll|} \hline && (2^n)^2 - 2 \pmod{17} \quad & | \quad n = 9 \\ &\equiv& (2^9)^2 - 2 \pmod{17} \\ &\equiv& 2^{18} - 2 \pmod{17} \\ &\equiv& 262144- 2 \pmod{17} \\ &\equiv& 262142 \pmod{17} \\ &\equiv& 2 \pmod{17} \\ \hline \end{array} \)

\(\begin{array}{rcll} \mathbf{(2^n)^2 - 2 \pmod{17}} &\mathbf{=}& \mathbf{ 2} \\ \end{array}\)

Hallo Melody,

this is right, but i think \(m =5 \) is a value for \(m\) too.

Here my table:

\(\begin{array}{|r|r|r|r|r|} \hline m & & & =1 & \ne 1 \\ \hline 1 & 9*9=81\equiv 0 \pmod 1 & 3*3=9\equiv 0 \pmod 1 & & \checkmark \\ 2 & 9*9=81\equiv 1 \pmod 2 & 3*3=9\equiv 1 \pmod 2 & \checkmark & \\ 4 & 9*9=81\equiv 1 \pmod 4 & 3*3=9\equiv 1 \pmod 4 & \checkmark & \\ \color{red}5 & 9*9=81\equiv 1 \pmod 5 & 3*3=9\equiv 4 \pmod 5 & \checkmark & \checkmark \\ 8 & 9*9=81\equiv 1 \pmod 8 & 3*3=9\equiv 1 \pmod 8 & \checkmark & \\ \color{red}10 & 9*9=81\equiv 1 \pmod{ 10} & 3*3=9\equiv 9 \pmod{ 10} & \checkmark & \checkmark \\ \color{red}16 & 9*9=81\equiv 1 \pmod{ 16} & 3*3=9\equiv 9 \pmod{ 16} & \checkmark & \checkmark \\ \color{red}20 & 9*9=81\equiv 1 \pmod{ 20} & 3*3=9\equiv 9 \pmod{ 20} & \checkmark & \checkmark \\ \color{red}40 & 9*9=81\equiv 1 \pmod{ 40} & 3*3=9\equiv 9 \pmod{ 40} & \checkmark & \checkmark \\ \color{red}80 & 9*9=81\equiv 1 \pmod{ 80} & 3*3=9\equiv 9 \pmod{ 80} & \checkmark & \checkmark \\ \hline \end{array}\)

So the possible values of m are  5,10,16,20,40, and 80       There are 6 of them

Hallo Melody.

In mathematics, in particular the area of number theory, a modular multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m. In the standard notation of modular arithmetic this congruence is written as

\( {\displaystyle ax\equiv 1{\pmod {m}}}\)

\(\text{We have $a = 9$ and $x =$ the inverse of integer $a$ } \\ {\displaystyle 9x\equiv 1{\pmod {m}}} \)

So "9 is its own inverse" means that \(\text{ $x=a=9 $}\)

Finally

\(\begin{array}{|rcll|} \hline 9\cdot 9 &\equiv& 1 \pmod m \\ \hline \end{array}\)

Jenny's grandmother has 24 cats.

Seventeen of the cats do not catch mice.

Ten of the cats have black fur.

What is the smallest possible number of cats that do not catch mice that have black fur?

\(\large{1.} \\ \begin{array}{r|r|r|r} & \text{mice} & \overline{\text{mice}} \\ \hline \text{black fur} & & x & \color{red} 10 \\ \hline \overline{\text{black fur}} & & & \small{24-10} \\ \hline & \small{24-17}& \color{red}17 & \color{red}24 \\ \end{array}\)

\(\large{2.} \\ \begin{array}{r|r|r|r} & \text{mice} & \overline{\text{mice}} \\ \hline \text{black fur} & \small{10-x } & x & 10 \\ \hline \overline{\text{black fur}} & \small{7-(10-x) =} & \small{17-x} & 14 \\ & \small{x-3 } & & \\ \hline & 7 & 17 & 24 \\ \end{array} \)

\(\large{3.} \\ \begin{array}{r|r|r|r} & \text{mice} & \overline{\text{mice}} \\ \hline \text{black fur} & \small{10-x } & x & 10 \\ & \small{\text{min: } 10-x=0 } & & \\ & \Rightarrow x_{min} = \color{red}10 & & \\ \hline \overline{\text{black fur}} & \small{x-3 } & \small{17-x} & 14 \\ & \small{\text{min: } x-3=0} & \small{\text{min: } 17-x=0 } & \\ & \Rightarrow x_{min} = \color{red}3 & \Rightarrow x_{min} = \color{red}17 \\ \hline & 7 & 17 & 24 \\ \end{array}\)

\(\large{4.} \\ \begin{array}{r|r|r|r} & \text{mice} & \overline{\text{mice}} \\ \hline \text{black fur} & \small{10-x } & x & 10 \\ & \Rightarrow x_{min} = \color{red}10 & = min({\color{red}10},{\color{red}3},{\color{red}17}) & \\ & & = 3 & \\ \hline \overline{\text{black fur}} & \small{x-3 } & \small{17-x} & 14 \\ & \Rightarrow x_{min} = \color{red}3 & \Rightarrow x_{min} = \color{red}17 \\ \hline & 7 & 17 & 24 \\ \end{array} \)

\(\large{5.} \\ \begin{array}{r|r|r|r} & \text{mice} & \overline{\text{mice}} \\ \hline \text{black fur} & \small{7 } & \color{red}3 & 10 \\ \hline \overline{\text{black fur}} & \small{0 } & \small{14} & 14 \\ \hline & 7 & 17 & 24 \\ \end{array}\)