heureka

Find the sum = \(\dfrac{1}{2} + 2*\dfrac{1}{2^2} + 3*\dfrac{1}{2^3} + 4*\dfrac{1}{2^4} + 5*\dfrac{1}{2^5}+ \ldots\)

\(\begin{array}{rcrl} s &=& 1*\dfrac{1}{2} +& 2*\dfrac{1}{2^2} + 3*\dfrac{1}{2^3} + 4*\dfrac{1}{2^4} + 5*\dfrac{1}{2^5}+ \ldots \\ \dfrac{1}{2}s &=& & 1*\dfrac{1}{2^2} + 2*\dfrac{1}{2^3} + 3*\dfrac{1}{2^4} + 4*\dfrac{1}{2^5}+ \ldots \\ \hline s-\dfrac{1}{2}s &=& 1*\dfrac{1}{2} +& 1*\dfrac{1}{2^2}+1*\dfrac{1}{2^3} +1*\dfrac{1}{2^4} +1*\dfrac{1}{2^5}+ \ldots \\ \end{array} \)

\(\begin{array}{rcll} \dfrac{1}{2}s &=& \dfrac{1}{2}* \left( 1+ \dfrac{1}{2^1}+1*\dfrac{1}{2^2} +1*\dfrac{1}{2^3} +1*\dfrac{1}{2^4}+ \ldots \right) \\ s &=& 1+ \dfrac{1}{2^1}+1*\dfrac{1}{2^2} +1*\dfrac{1}{2^3} +1*\dfrac{1}{2^4}+ \ldots \\ s &=& 1+ \dfrac{1}{2^1}+\dfrac{1}{2^2} +\dfrac{1}{2^3} + \dfrac{1}{2^4}+ \ldots \\ \end{array}\)

Now use the formula for the sum of an infinite geometric series.

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{1-\dfrac{1}{2}} \\ s &=& \dfrac{1}{ \dfrac{1}{2}} \\ \mathbf{s} &=& \mathbf{2} \\ \hline \end{array}\)

Two small semicircles are inscribed in a larger semicircle, and a circle is drawn such that it is internally tangent to all three semicircles, as seen above.

If the diameter of the biggest semicircle is 12, find the radius of the small circle.

\(\begin{array}{|rcll|} \hline \mathbf{(6-r)^2 +3^2} &=& \mathbf{(r+3)^2} \\ 36-12r+r^2+9 &=& r^2+6r+9 \quad | \quad -(r^2+9)\\ 36-12r &=& 6r\\ 18r &=& 36 \quad | \quad : 18 \\ \mathbf{r} &=&\mathbf{ 2 } \\ \hline \end{array}\)

The golden rectangle with the area of 8cm²  is inscribed in a circle.

What's the area of that circle?

\(\begin{array}{|rcll|} \hline \dfrac{x}{y} &=& \varphi \quad | \quad \boxed{\varphi =\dfrac{1+\sqrt{5}}{2}= 1.61803398875 } \\ x = y\varphi &\text{or}& x^2 = y^2\varphi^2 \\ \hline xy &=& 8 \\ y\varphi y &=& 8 \\ y^2 &=& \dfrac{8}{\varphi} \quad | \quad \dfrac{1}{\varphi} = \varphi - 1 \\ \mathbf{y^2} &=& \mathbf{8(\varphi - 1)} \\ \hline x^2 &=& y^2\varphi^2 \quad | \quad y^2=\dfrac{8}{\varphi} \\ x^2 &=&\dfrac{8}{\varphi} \varphi^2 \\ \mathbf{x^2} &=& \mathbf{8\varphi} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (2r)^2 &=& x^2+y^2 \\ (2r)^2 &=& 8\varphi+8(\varphi - 1) \\ 4r^2 &=& 16\varphi-8 \quad | \quad : 4 \\ \mathbf{r^2} &=& \mathbf{4\varphi-2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{area}_{\text{circle}} &=& \pi r^2 \\ &=& (4\varphi-2)\pi \quad | \quad \varphi = 1.61803398875 \\ &=& 4.47213595500\pi \quad | \quad \pi = 3.14159265359 \\ \mathbf{\text{area}_{\text{circle}}} &=& \mathbf{14.0496294621~ \text{cm}^2} \\ \hline \end{array}\)

The area of that circle is \(\approx \mathbf{14~ \text{cm}^2}\)

If the function f is defined by f(x) = 2x+sinx and the function g is the inverse of f, then g'(2)=

\(\begin{array}{|rcll|} \hline \mathbf{ g\Big(f(x)\Big) } &=& \mathbf{ x } \quad | \quad \text{chain rule} \\ g'\Big(f(x)\Big)f'(x) &=& 1\\ \mathbf{ g'\Big(f(x)\Big) } &=& \mathbf{ \dfrac{1}{f'(x)} } \\\\ && f(x) = 2 = 2x+\sin(x) \Rightarrow x = 0.68403665667782943943\ldots \text{ found with Newton's Method } \\ g'(2) &=& \dfrac{1}{f'(0.68403665667782943943)} \quad | \quad f'(x) = 2+\cos(x) \\ g'(2) &=& \dfrac{1}{2+\cos(0.68403665667782943943)} \\ \mathbf{g'(2)} &=& \mathbf{0.360356702385802856117...} \\ \hline \end{array}\)

Find constants A and B such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\) for all x such that \(x\neq -1\) and \(x\neq 2\).
Give your answer as the ordered pair \((A,B)\).

\(\text{Let $x^2 - x - 2=(x-2)(x+1)$ }\)

\(\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{x + 7}{x^2 - x - 2} } &=& \mathbf{ \dfrac{A}{x - 2} + \dfrac{B}{x + 1} } \quad | \quad x^2 - x - 2=(x-2)(x+1) \\\\ &\dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x-2)(x+1) \\\\ &\dfrac{(x + 7)(x-2)(x+1)}{(x-2)(x+1)} &=& \dfrac{A(x-2)(x+1)}{(x-2)} + \dfrac{B(x-2)(x+1)}{(x+1)} \\\\ & x + 7 &=& A(x+1) + B(x-2) \\\\ \mathbf{x=-1:}& -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& A\times 0 -3B \\ & 3B &=& -6 \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& -2 \\\\ \mathbf{x=2:}& 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3A +B\times 0\\ & 3A &=& 9 \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& 3 \\ \hline \end{array}\)

\(\mathbf{\boxed{(A,\ B) = (3,\ -2)}}\)

check:

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad A=3,\ B=-2 \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3}{x - 2} - \dfrac{2}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3x+3-2x+4}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{x+7}{(x-2)(x+1)} \ \checkmark \\ \hline \end{array}\)

A vector makes an angle of 30 degrees with the x-axis, and an angle of 80 degrees with the y-axis. 

Find the angle between the vector and the z-axis.

Formula: \(\boxed{ \cos{\alpha_x}+\cos{\alpha_y}+\cos{\alpha_z} = 1 }\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos{\alpha_x}+\cos{\alpha_y}+\cos{\alpha_z} } &=& \mathbf{ 1 } \quad | \quad \alpha_x = 30^\circ,\ \alpha_y = 80^\circ \\\\ \cos{30^\circ}+\cos{80^\circ}+\cos{\alpha_z} &=& 1 \\ \cos{\alpha_z} &=& 1-\cos{30^\circ}-\cos{80^\circ}\\ \cos{\alpha_z} &=& -0.03967358145\\ \mathbf{\alpha_z} &=& \mathbf{92.2737255129^\circ} \\ \hline \end{array}\)

Simplify

\(\tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ)\).

Formula:

\(\boxed{ \tan(x+y) = \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} \quad \text{or} \quad \tan(x)+\tan(y)= \Big(1-\tan(x)\tan(y)\Big)\tan(x+y) }\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) } \\\\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\tan(20^\circ+40^\circ)+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\tan(60^\circ)+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \quad | \quad \tan(60^\circ) = \sqrt{3} \\ &=&\Big(1-\tan(20^\circ)\tan(40^\circ)\Big)\sqrt{3}+ \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\sqrt{3}-\sqrt{3}\tan(20^\circ)\tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ) \\ &=&\mathbf{ \sqrt{3} } \\ \hline \end{array}\)

\(\tan(20^\circ) + \tan(40^\circ) + \sqrt{3}*\tan(20^\circ)*\tan(40^\circ)=\sqrt{3}\)

Thank you, CPhill !

One root of \(x^2+12x+k=0\) is twice the other root.

Find \(k\).

Let the roots are \(x_1\), \(x_2\).

\(\begin{array}{|rcll|} \hline x^2+12x+k=0 &=& (x-x_1)(x-x_2)= 0 \quad | \quad x_2 = 2x_1 \\\\ &=& (x-x_1)\Big(x-(2x_1)\Big) \\ &=& (x-x_1)(x-2x_1) \\ &=& x^2-2x_1x-x_1x+2x^2_1 \\ x^2+12x+k=0 &=& x^2-3x_1x+2x^2_1 \\ && \text{compare} \\ && \boxed{12=-3x_1\ \text{ or }\ x_1=-4 \\ k= 2x^2_1\ \text{ or }\ k=2*(-4)^2=32 } \\ \hline \end{array} \)

\(\boxed{k=32}\)

The sum of two of the three roots of \(x^3 + ax^2 + bx + c = 0 \) is \(0\). 

Write an equation expressing c explicitly in terms of \(a\) and \(b\).

Let the roots are \(x_1,\) \(x_2\) and \(x_3\).

\(\begin{array}{|rcll|} \hline x^3 + ax^2 + bx + c &=& (x-x_1)(x-x_2)(x-x_3) = 0 \quad | \quad x_1+x_2 = 0\ \text{ or }\ x_2 = -x_1 \\\\ &=& (x-x_1)\Big(x-(-x_1)\Big)(x-x_3) \\ &=& (x-x_1)(x+x_1)(x-x_3) \\ &=& (x^2-x^2_1)(x-x_3) \\ x^3 + ax^2 + bx + c &=& x^3-x_3x^2-x^2_1x+x^2_1x_3 \\ && \text{compare} \\ && \boxed{a=-x_3 \text{ or } x_3=-a \\ b=-x^2_1 \text{ or } x^2_1=-b \\ c= x^2_1x_3 \text{ or } c=(-b)(-a)=ab } \\ \hline \end{array}\)

\(\boxed{c=ab} \)