heureka

The function \(f(x) \)satisfies \(f(x) + f(x + 2y) = 6x + 6y - 8\)
for all real numbers \(x\) and \(y\)
Find the value of \(x\) such that \(f(x) = 0\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x) + f(x + 2y)} &=& \mathbf{6x + 6y - 8} \quad &| \quad y= 0\\\\ f(x) + f(x + 2\cdot 0) &=& 6x + 6\cdot 0 - 8 \\ f(x) + f(x) &=& 6x - 8 \\ 2f(x)&=& 6x - 8 \quad &| \quad : 2 \\ f(x)&=& 3x - 4 \quad &| \quad f(x) = 0 \\ 0 &=& 3x - 4 \\ 3x &=& 4 \quad &| \quad : 3 \\ \mathbf{x} &=& \mathbf{\dfrac{4}{3}} \\ \hline \end{array}\)

Given that \(A_k = \dfrac {k(k - 1)}2\cos\left( \dfrac {k(k - 1)\pi}2 \right)\),
find \(A_{19} + A_{20} + \cdots + A_{98}\).

\(\begin{array}{|rcll|} \hline &&\mathbf{ A_{19} + A_{20} + \cdots + A_{98}} \quad | \quad \small{98-18 = 80 \text{ terms}} \\\\ &=& \sum \limits_{k=19}^{98} \dfrac {k(k - 1)}2\cos\left( \dfrac {k(k - 1)\pi}2 \right) \\\\ &=&\dfrac{1}{2} \sum \limits_{k=19}^{98} k(k - 1)\cos\left( \dfrac {k(k - 1)\pi}2 \right) \\\\ &=&\dfrac{1}{2} \sum \limits_{k=19}^{98} k(k - 1)\cos(n\pi) \quad | \quad n = \dfrac{k(k - 1)}2 \\ && \boxed{\cos(n\pi)=\cos^n(\pi)\quad | \quad \cos(\pi) = -1 \\\cos(n\pi) = (-1)^{n} } \\ &=&\dfrac{1}{2} \sum \limits_{k=19}^{98} k(k - 1)(-1)^{n} \\\\ &=&\mathbf{\dfrac{1}{2} \sum \limits_{k=19}^{98} k(k - 1)(-1)^{\left(\dfrac{k(k - 1)}2\right)} } \\ \hline &=& \small{\dfrac{1}{2}\Big( -18\cdot 19+19\cdot 20+20\cdot 21-21\cdot 22-22\cdot 23+23\cdot 24+24\cdot 25-25\cdot 26 +- \ldots } \\ && \small{-94\cdot 95+95\cdot 96 + 96\cdot 97-97\cdot 98 \Big)} \quad | \quad \small{80 \text{ terms}} \\ &=& \small{\dfrac{1}{2}\Big( 19(-18+20)+21(20-22)+23(-22+24)+25(24-26) + \ldots }\\ && \small{+95(-94+96) +97(96-98)\Big)} \quad | \quad \small{40 \text{ terms}} \\ &=& \small{\dfrac{1}{2}\Big( 2\cdot 19 -2\cdot 21+2\cdot 23-2\cdot 25 +- \ldots +2\cdot 95 -2\cdot 97 \Big)} \quad | \quad \small{40 \text{ terms}} \\ &=& \small{\dfrac{2}{2}( 19 - 21+ 23- 25 +- \ldots + 95 - 97 )} \quad | \quad \small{40 \text{ terms}} \\ &=& \underbrace{ \underbrace{19-21}_{=-2} + \underbrace{23-25}_{=-2} + \underbrace{27-29}_{=-2} +- \ldots + \underbrace{91-93}_{=-2} + \underbrace{95-97}_{=-2}}_{ 20 \text{ terms} } \\ &=& -2\cdot 20 \\ &=& \mathbf{-40} \\ \hline \end{array}\)

Let \(f(x) = 5x+3\) and \(g(x)=x^2-2\). What is \(g(f(-1))\)?

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{5x+3} \quad | \quad x=-1 \\ f(-1) &=& 5(-1)+3 \\ f(-1) &=& -5+3 \\ \mathbf{f(-1)} &=& \mathbf{-2} \\ \hline && \mathbf{g(f(-1))} \quad | \quad f(-1)=-2 \\ &=& g(-2) \quad | \quad \mathbf{g(x)=x^2-2},\ x=-2 \\ &=& (-2)^2-2 \\ &=& 4-2 \\ &=& \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{g(f(-1))} = \mathbf{2} \)

SOLVE SECOND ORDER INITIAL VALUE PROBLEM.
\(y'' +6y' +5y = 0 ,\ y(0)=1 \text{ and } y' (0) = 1\)

\(\text{Let $y=e^{rx}$} \\ \text{Let $y'=re^{rx}$} \\ \text{Let $y''=r^2e^{rx}$} \)

\(\begin{array}{|rcll|} \hline \mathbf{y'' +6y' +5y} &=& \mathbf{0} \\\\ r^2e^{rx} +6re^{rx} +5e^{rx} &=& 0 \\ \underbrace{e^{rx}}_{\neq 0}\cdot \underbrace{ \left(r^2 +6r +5\right) }_{=0} &=& 0 \\ && \begin{array}{|rcll|} \hline r^2 +6r +5 &=& 0 \\ r &=& \dfrac{-6\pm \sqrt{36-4\cdot 5}}{2} \\ r &=& \dfrac{-6\pm \sqrt{16}}{2} \\ r &=& \dfrac{-6\pm 4}{2} \\\\ r_1 &=& \frac{-6+4}{2} \\ \mathbf{r_1} &=& \mathbf{-1} \\\\ r_2 &=& \frac{-6-4}{2} \\ \mathbf{r_2} &=& \mathbf{-5} \\ \\ \hline \end{array} \\\\ y &=& c_1\cdot e^{r_1\cdot x} + c_2\cdot e^{r_2\cdot x} \\ \mathbf{y} &=& \mathbf{c_1\cdot e^{-1\cdot x} + c_2\cdot e^{-5\cdot x} } & (1) \\ \mathbf{y'} &=& \mathbf{-c_1\cdot e^{-1\cdot x} -5\cdot c_2\cdot e^{-5\cdot x} } & (2) \\ \hline y &=& c_1\cdot e^{-x} + c_2\cdot e^{-5\cdot x} \quad | \quad y(0)=1,\ x=0,\ y=1\\ 1 &=& c_1\cdot e^{-0} + c_2\cdot e^{-5\cdot 0} \\ \mathbf{1} &=& \mathbf{c_1 + c_2} & (3) \\\\ y' &=& -c_1\cdot e^{-x} -5\cdot c_2\cdot e^{-5\cdot x} \quad | \quad y'(0)=1, \ x=0,\ y'=1\\ 1 &=& -c_1\cdot e^{-0} -5\cdot c_2\cdot e^{-5\cdot 0} \\ \mathbf{1} &=& \mathbf{-c_1 - 5c_2} & (4) \\\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \mathbf{1} &=& \mathbf{c_1 + c_2} \\ (4) & \mathbf{1} &=& \mathbf{-c_1 - 5c_2} \\ \hline (3)+(4): & 1+1 &=& c_1 + c_2+-c_1 - 5c_2 \\ & 2 &=& c_2 - 5c_2 \\ &-4c_2 &=& 2 \quad | \quad :(-4) \\ & \mathbf{c_2} &=& \mathbf{-\dfrac{1}{2}} \\\\ & 1 &=& c_1 + c_2 \\ & 1 &=& c_1 -\dfrac{1}{2}\\ & c_1 &=& 1+\dfrac{1}{2} \\ & \mathbf{c_1} &=& \mathbf{ \dfrac{3}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{ c_1\cdot e^{-1\cdot x} + c_2\cdot e^{-5\cdot x} } \\ \mathbf{y} &=& \mathbf{ \dfrac{3}{2}\cdot e^{-x} -\dfrac{1}{2}\cdot e^{-5\cdot x} } \\ \hline \end{array}\)

can you solve the initial value problem for:

\(\dfrac{dy}{dx}= xy-2x+y-2,\ y(2)= -1\)

\(\begin{array}{|rclcl|} \hline \mathbf{\dfrac{dy}{dx}} &=&\mathbf{ xy-2x+y-2 } \\\\ \dfrac{dy}{dx} &=& x(y-2)+ (y-2) \\\\ \dfrac{dy}{dx} &=& (y-2)(x+1) \\\\ \dfrac{dy}{(y-2)} &=& (x+1)dx \\\\ \large{ \int } \dfrac{1}{(y-2)}\ dy &=&\large{ \int } (x+1)\ dx \\\\ \mathbf{\ln(y-2) + c_1} &=& \mathbf{\dfrac{x^2}{2} +x + c_2 } \\ \hline && \ln(y-2) + c_1 &=& \dfrac{x^2}{2} +x + c_2 \quad | \quad y(2) = -1 \\ && \ln(-1-2) + c_1 &=& \dfrac{2^2}{2} +2 + c_2 \\ && \ln(-3) + c_1 &=& 4 + c_2 \quad | \quad \boxed{\mathbf{c_1 = -\ln(-3)}} \\ && \ln(-3) -\ln(-3) &=& 4 + c_2 \\ && 0 &=& 4 + c_2 \\ && \mathbf{c_2} &=& \mathbf{-4} \\ \hline \mathbf{\ln(y-2) -\ln(-3)} &=& \mathbf{\dfrac{x^2}{2} +x -4 } \\ \ln\left(\dfrac{y-2}{-3}\right) &=& \dfrac{x^2}{2} +x -4 \\ \dfrac{y-2}{-3} &=& e^\left(\dfrac{x^2}{2} +x -4 \right) \\ y-2 &=& -3e^\left(\dfrac{x^2}{2} +x -4 \right) \\ \mathbf{ y(x)} &=& \mathbf{2-3e^\left(\dfrac{x^2}{2} +x -4 \right)} \\ \hline \end{array} \)

Thank you, Melody !

Let\( f(x) = (x+2)^2-5\).
If the domain of \(f\) is all real numbers,
then \(f\) does not have an inverse function,
but if we restrict the domain of \(f\) to an interval \([c,\infty)\),
then \(f\) may have an inverse function.
What is the smallest value of \(c\) we can use here,
so that \(f\) does have an inverse function?

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x+2)^2-5} \\ \\ x &=& \Big(f^{-1}(x)+2\Big)^2-5 \quad | \quad \text{inverse function} \\ x+5 &=& \Big(f^{-1}(x)+2\Big)^2 \\ \Big(f^{-1}(x)+2\Big)^2 &=& x+5 \\ f^{-1}(x)+2 &=& \pm\sqrt{x+5} \\ \mathbf{f^{-1}(x)} &=& \mathbf{-2 \pm\sqrt{x+5}} \\ \hline \\ \sqrt{x+5} &=& 0 \quad | \quad \text{the smallest value of }x \text{ is } c \\ x+5 &=& 0 \\ x &=& -5 \quad | \quad \text{the smallest value is } -5 \\ \\ \hline \mathbf{[c,\infty)} &=& \mathbf{[-5,\infty)} \\ \hline \end{array}\)

Simplify
\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8\)( \(i\) is imaginary)

\(\text{Let $z = \dfrac{3 + i \sqrt{3}}{2}$} \\ \text{Let $\overline{z} = \dfrac{3 - i \sqrt{3}}{2}$} \)

\(\begin{array}{|rcll|} \hline z\cdot \overline{z} &=& \left(\dfrac{3}{2}\right)^2 +\left( \dfrac{\sqrt{3}}{2} \right)^2 \\ &=& \dfrac{9}{4} + \dfrac{3}{4} \\ \mathbf{z\cdot \overline{z}} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right) &=& \dfrac{3}{2} + \dfrac{i \sqrt{3}}{2}+ \dfrac{3}{2} - \dfrac{i \sqrt{3}}{2} \\ &=& 2\cdot \dfrac{3}{2} \\ \mathbf{\left(z+\overline{z} \right)} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right)^2 = 3^2 &=& z^2+2\cdot z\cdot \overline{z} + \overline{z}^2 \\ 3^2 &=& z^2 + \overline{z}^2 +2\cdot 3 \\ z^2 + \overline{z}^2 &=& 9-6 \\ \mathbf{z^2 + \overline{z}^2} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z^2+\overline{z}^2 \right)^2 = 3^2 &=& z^4+2\cdot z^2\cdot \overline{z}^2 + \overline{z}^4 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot \left(z\cdot \overline{z}\right)^2 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot 3^2 \\ z^4 + \overline{z}^4 &=& 9-18 \\ \mathbf{z^4 + \overline{z}^4} &=& \mathbf{-9} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z^4+\overline{z}^4 \right)^2 = (-9)^2 &=& z^8+2\cdot z^4\cdot \overline{z}^4 + \overline{z}^8 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot \left(z\cdot \overline{z}\right)^4 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot 3^4 \\ z^8 + \overline{z}^8 &=& 81-2\cdot 81 \\ \mathbf{z^8 + \overline{z}^8} &=& \mathbf{-81} \\ \hline \end{array} \)

\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8 = \mathbf{z^8 + \overline{z}^8} = \mathbf{-81} \)

Let \(a_n = \dfrac{10^n-1}{9}\).
Define \(d_n\) to be the greatest common divisor of \(a_n\) and \(a_{n+1}\).
What is the maximum possible value that \(d_n\) can take on?

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{\dfrac{10^n-1}{9}} \\ \hline \mathbf{a_{n+1}} &=& \mathbf{\dfrac{10^{n+1}-1}{9}} \\\\ &=& \dfrac{10^{n}10-1}{9} \\\\ &=& \dfrac{10^{n}10-(10-9)}{9} \\\\ &=& \dfrac{10^{n}10-10+9}{9} \\\\ &=& \dfrac{10\left( 10^{n} -1 \right) +9}{9} \\\\ &=& \dfrac{10\left( 10^{n} -1 \right)}{9} + 1 \\\\ &=& 10\cdot \left(\dfrac{ 10^{n} -1 }{9}\right) + 1 \quad | \quad \dfrac{10^n-1}{9} = a_n \\\\ \mathbf{a_{n+1}} &=& \mathbf{10a_n + 1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline d_n &=& \gcd( a_n,\ a_{n+1} ) \\ &=& \gcd( a_{n+1},\ a_n ) \quad | \quad a_{n+1} =10a_n + 1 \\ &=& \gcd( 10a_n+1,\ a_n ) \\ && \boxed{\text{Formula: }\gcd(a,b)=\gcd(a-b,b),\ \text{if } a > b } \\ &=& \gcd( 10a_n+1 - 10a_n,\ a_n ) \\ &=& \gcd( 1 ,\ a_n ) \\ && \boxed{\text{Formula: }\gcd(1,a)=1} \\ &=& \mathbf{1} \\ \hline \end{array}\)

The maximum possible value that \(d_n\) can take on is 1

Calculate a normal vector n to the plane through the points
triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below,
such that if
\(\mathbf{n} = \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}\),
and \(n_1 + n_2 + n_3 = 7\).

\(\text{Let $\vec{v} = \vec{B} - \vec{C}$ } \\ \text{Let $\vec{w} = \vec{A} - \vec{C}$ }\)

\(\begin{array}{|rcll|} \hline \vec{v} &=& \begin{pmatrix}-3 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{v}} &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \\\\ \vec{w} &=& \begin{pmatrix}1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{w}} &=& \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\ \hline \end{array}\)

1.

Vector cross product

\(\small{ \begin{array}{|lrcll|} \hline & \vec{n} =\begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}&=& \vec{v} \times \vec{w} \\\\ & &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \times \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\\\ & &=& \begin{vmatrix}1&1&1 \\ -4&0&2 \\ 0&-3&-1 \end{vmatrix} \\\\ & &=& \begin{pmatrix} 6 \\ -4 \\ 12 \end{pmatrix} \\ \hline n_1+n_2+n_3: & 6-4+12 &=& 14 \quad | \quad \cdot \dfrac{7}{14} \\\\ & 6\cdot\dfrac{7}{14}-4\cdot\dfrac{7}{14}+12\cdot\dfrac{7}{14} &=& 14\cdot\dfrac{7}{14} \\\\ &\mathbf{ 3-2+6 }&=& \mathbf{7} \\\\ & \vec{n} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array} }\)

2.

Vector dot product

\(\begin{array}{|lrcll|} \hline & \vec{n}\cdot \vec{v} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} &=& 0 \\ (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ \hline & \vec{n}\cdot \vec{w} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} &=& 0 \\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ \hline (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ \hline \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ &2n_3&=&4n_1 \quad | \quad :2 \\ & \mathbf{n_3} &=& \mathbf{2n_1} \\\\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ & \mathbf{n_3} &=& \mathbf{-3n_2} \\\\ & n_3= 2n_1 &=& -3n_2 \\ & 2n_1 &=& -3n_2 \quad | \quad :2 \\ & \mathbf{n_1} &=& \mathbf{-\dfrac{3}{2}n_2} \\\\ (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ & -\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\ & -\dfrac{3}{2}n_2 -2n_2 &=& 7 \quad | \quad \cdot (-1 )\\ & \dfrac{3}{2}n_2 +2n_2 &=& -7 \\ & \dfrac{7}{2}n_2 &=& -7 \quad | \quad \cdot \dfrac{2}{7} \\ & n_2 &=& -7\cdot \dfrac{2}{7} \\ & \mathbf{n_2} &=& \mathbf{-2} \\ \hline & n_3 &=& -3n_2 \\ & n_3 &=& -3\cdot (-2) \\ & \mathbf{n_2} &=& \mathbf{6} \\ \hline & n_1 &=& -\dfrac{3}{2}n_2 \\ & n_1 &=& \left(-\dfrac{3}{2}\right)\cdot(-2) \\ & \mathbf{n_2} &=& \mathbf{3} \\\\ & \mathbf{\vec{n}} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array}\)