heureka

Given that k  is a positive integer less than 6,

how many values can k take on such that

\(3x \equiv k \pmod{6}\) has no solutions in x ?

Rewrite:

\(\begin{array}{|lrcll|} \hline 3x \equiv k \pmod{6} \\ \text{rewrite...} \\ \begin{array}{|rcll|} \hline 3x-k &=& n\cdot 6 \quad &| \quad n \in \mathbf{Z} \qquad (n \text{ is a integer}) \\ 3x &=& n\cdot 6 +k \\ x &=& \frac{n\cdot 6 +k}{3} \\ x &=& 2n+\frac{k}{3} \quad &| \quad \frac{k}{3} \text{ is a integer, if } k = 0 \text{ or } k = 3 \\ & & \quad &| \quad \frac{k}{3} \text{ is not a integer, if } k = 1,\ k = 2,\ k=4,\ k= 5 \qquad 0 < k < 6 \\ \hline \end{array} \\ \hline \end{array} \)

What line is perpendicular to y=-3/4x+5?

\(y=\frac43x\)

generally: \(y=\frac43x + b\)

Integrate the following:  ∫x^2 sin^3 (x) dx. Also, please show steps of solution.

I thank you for any help.

Formula \(\sin^3(x)\):

\(\begin{array}{|rcll|} \hline \sin(3x) &=& \sin(2x+x) \\ &=& \underbrace{ \sin(2x) }_{=2\sin(x)\cos(x)}\cdot \cos(x)+\underbrace{ \cos(2x) }_{=\cos^2(x)-\sin^2(x)}\cdot \sin(x) \\ &=& 2\sin(x)\cos^2(x) + \cos^2(x)\sin(x)-\sin^3(x) \\ &=& 3\sin(x)\cos^2(x) -\sin^3(x) \\ &=& 3\sin(x) \Big( 1-\sin^2(x) \Big) -\sin^3(x) \\ &=& 3\sin(x) -3\sin^3(x) -\sin^3(x) \\ \sin(3x)&=& 3\sin(x) -4\sin^3(x) \\\\ 4\sin^3(x) &=& 3\sin(x) - \sin(3x) \\ \mathbf{ \sin^3(x) } & \mathbf{=} & \mathbf{ \frac14 \Big( 3\sin(x) - \sin(3x) \Big) } \\ \hline \end{array}\)

Double Integration by parts:

\(\begin{array}{|rcll|} \hline \int u\cdot v' &=& u\cdot \int v' - \int (u'\cdot \int v') \quad & | \quad \text{Integrate by parts}\\ \int (u'\cdot \int v') &=& u'\cdot \iint v' - \int (u''\cdot \iint v') \quad & | \quad \text{Integrate by parts} \\\\ \int u\cdot v' &=& u\cdot \int v' - \Big( u'\cdot \iint v' - \int (u''\cdot \iint v') \Big) \\ \mathbf{ \int u\cdot v' } & \mathbf{=} & \mathbf{ u\cdot \int v' - u'\cdot \iint v' + \int (u''\cdot \iint v') } \\ \hline \end{array} \)

\(\begin{array}{|llll|} \hline \int x^2 \sin^3 (x)\ dx \qquad & u = x^2 \qquad u' = 2x \qquad u'' = 2 \\ \qquad & v' = sin^3(x)\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{ \int u\cdot v' } & \mathbf{=} & \mathbf{ u\cdot \int v' - u'\cdot \iint v' + \int (u''\cdot \iint v') } \\ \int x^2\cdot sin^3 (x) &=& x^2\cdot \int sin^3(x) - 2x\cdot \iint sin^3(x) + 2\cdot \iiint sin^3(x) \\ \hline \end{array} \)

\(\int \sin^3(x)\ dx\)

\(\begin{array}{|rcll|} \hline \int \sin^3(x) &=& \frac14 \Big( \int 3\sin(x)-\int \sin(3x) \Big) \\ \int \sin^3(x) &=& \frac14 \Big( -3\cos(x)+\frac13 \cos(3x) \Big) \\ \hline \end{array}\)

\(\iint \sin^3(x)\ dx\)

\(\begin{array}{|rcll|} \hline \iint \sin^3(x) &=& \frac14 \Big(\int -3\cos(x)+ \int \frac13 \cos(3x) \Big) \\ \iint \sin^3(x) &=& \frac14 \Big( -3\sin(x)+\frac19 \sin(3x) \Big) \\ \hline \end{array} \)

\(\iiint \sin^3(x)\ dx\)

\(\begin{array}{|rcll|} \hline \iiint \sin^3(x) &=& \frac14 \Big(\int -3\sin(x)+ \int \frac19 \sin(3x) \Big) \\ \iiint \sin^3(x) &=& \frac14 \Big( 3\cos(x)-\frac{1}{27} \cos(3x) \Big) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ \int x^2\cdot \sin^3 (x) } & \mathbf{=} & \mathbf{ x^2\cdot \int sin^3(x) - 2x\cdot \iint sin^3(x) + 2\cdot \iiint sin^3(x) } \\ \int x^2\cdot \sin^3 (x) &=& x^2\cdot \frac14 \Big[ -3\cos(x)+\frac13 \cos(3x) \Big] - 2x\cdot \frac14 \Big[ -3\sin(x)+\frac19 \sin(3x) \Big] + 2\cdot \frac14 \Big[ 3\cos(x)-\frac{1}{27} \cos(3x) \Big] \\ \int x^2\cdot \sin^3 (x) &=& \frac14 x^2 \Big[ -3\cos(x)+\frac13 \cos(3x) \Big] - \frac12 x\Big[ -3\sin(x)+\frac19 \sin(3x) \Big] + \frac12 \Big[ 3\cos(x)-\frac{1}{27} \cos(3x) \Big] \\ \hline \end{array} \)

My proffesor gave me this to problem to resolve it, but I really do not know how to do this 

8444...4...455= divisible to 19

Between the 8 and the two 5 are 2017 fours

8444...4...455 is divisible to 19, if 8444...4...455 modulo 19 = 0

• We calculate 8444...4...455 modulo 19:
  We divide 8444...4...455 in parts of 3-digits. ( The partition is arbitrary ).
  But the first part is 8444.

So the Number is:

\(\begin{array}{lcll} \underbrace{8444}_{\text{ Fist Part }}\ \underbrace{444}_{\text{ Second Part }}\ 444\ 444\ 444\ 444\ 444\ 444\ 444\ \ldots \underbrace{444}_{\text{ Part 672 }} \ \underbrace{455}_{\text{ Last Part }} \qquad \text{2017 fours} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline && \text{first part} \pmod {19} \\ &\equiv& 8444\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The last remainder } = \color{red}8 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \text{Second part} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{444}_{\text{ Second part }}\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The new last remainder } = 8 \\ \hline \end{array} \)

\(\cdots \)

\(\begin{array}{|rcll|} \hline && \text{Part 672} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{444}_{\text{ Part 672 }}\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The new last remainder } = 8 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \text{Last part} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{455}_{\text{ Last part }}\pmod {19}\\ &=& \color{red}0 \color{black}\qquad (8455=445\cdot 19 + \color{red}0 \color{black}) \\\\ && \text{The new last remainder } = 0 \\ \hline \end{array} \)

8444...4...455 is divisible to 19, because 8444...4...455 modulo 19 = 0

n=5*(10!)

no of zeros does n end with?

Legendre's Theorem - The Prime Factorization of Factorials

In mathematics,

Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial 

\({\displaystyle n!}\)

1. The exponent of 2^x

We calculate 10 in base 2:

\(\begin{array}{rcll} 10_{10} &=& 1010_2 \\ \end{array}\)

The sum of the digits in base 2 is \(\begin{array}{rcll} 1+0+1+0 = 2 \end{array}\)

The exponent is: \(\begin{array}{|rcll|} \hline x &=&\frac{10-(\text{sum of the standard base-}\mathbf{2}\text{ digits of 10})}{\mathbf{2}-1} \\ x &=&\frac{10-2}{2} \\ x &=&4\\ \hline \end{array}\)

2. The exponent of 5^x

We calculate 10 in base 5:

\(\begin{array}{rcll} 10_{10} &=& 20_5 \\ \end{array}\)

The sum of the digits in base 5 is \(\begin{array}{rcll} 2+0 = 2 \end{array}\)

The exponent is: \(\begin{array}{|rcll|} \hline x &=&\frac{10-(\text{sum of the standard base-}\mathbf{5}\text{ digits of 10})}{\mathbf{5}-1} \\ x &=&\frac{10-2}{4} \\ x &=&2\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 10! &=& 2^4\cdot 5^2\ldots \\ n=5\cdot 10! &=& 2^4\cdot 5^3\ldots \\ \hline \end{array}\)

Number of zeros does n end with ?:

\(\begin{array}{|rcll|} \hline && 2^4\cdot 5^3 \\ &=& 2^3\cdot5^3\cdot 2 \\ &=& (2\cdot 5)^3 \cdot 2 \\ &=& 2\cdot 10^\mathbf{\color{red}3} \\ \hline \end{array}\)

n ends with 3 zeros.

Find the length of a pendulum that oscillates with a frequency of 0.19 Hz.

The acceleration due to gravity is 9.81 m/s 2 .

Formula:

For small amplitudes, the period of such a pendulum can be approximated by:

\(\begin{array}{|lrcll|} \hline & T &=& 2\pi\sqrt{\frac{L}{g}} \\ \qquad L~ \text{expresses the pendulum length in meters } \\ \qquad g~ \text{expresses the acceleration of gravity}\approx 9.81 \frac{m}{s^2} \\ & T &=& \frac{1}{f} \\ \qquad f~ \text{expresses the frequency in Hz} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \frac{1}{f} &=& 2\pi\sqrt{\frac{L}{g}} \\ \frac{1}{2\pi f} &=& \sqrt{\frac{L}{g}} \\ \Big(\frac{1}{2\pi f}\Big)^2 &=& \frac{L}{g} \\ L &=& g\cdot \Big(\frac{1}{2\pi f}\Big)^2 \\ L &=& 9.81\cdot \left(\frac{1}{2\pi\cdot 0.19}\right)^2 \\ L &=& 9.81\cdot \left(0.83765759522\right)^2 \\ L &=& 9.81\cdot 0.70167024683 \\ \mathbf{L} &\mathbf{=}& \mathbf{6.88338512141\ m} \\ \hline \end{array}\)

 The length of the pendulum is 6.88 m

(cosx / sinx) + (2sinx / cosx) + (sin^3x / cos^3x)

How to simplify the expression with the LCM?

\(\begin{array}{|rcll|} \hline && \frac{\cos(x)}{\sin(x)} + \frac{2\sin(x) }{ \cos(x) } + \frac{ \sin^3(x) } { \cos^3(x) } \\ &=& \frac{\cos(x)}{\sin(x)}\cdot \frac{\cos^3(x)}{\cos^3(x)} + \frac{2\sin(x) }{ \cos(x) } \cdot \frac{\sin(x)\cos^2(x)}{\sin(x)\cos^2(x)} + \frac{ \sin^3(x) } { \cos^3(x) }\cdot \frac{ \sin(x) } { \sin(x) } \\ &=& \frac{\cos(x)\cos^3(x)+ 2\sin(x)\sin(x)\cos^2(x) + \sin^3(x)\sin(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\cos^4(x)+ 2\sin^2(x)\cos^2(x) + \sin^4(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\Big( \sin^2(x)+\cos^2(x) \Big)^2 } {\sin(x)\cos^3(x) } \quad & | \quad \sin^2(x)+\cos^2(x) = 1 \\ &=& \frac{ 1^2 } {\sin(x)\cos^3(x) } \\ &=& \frac{ 1 } {\sin(x)\cos^3(x) } \\ \hline \end{array} \)

However, the Wolfram Alpha expressions don't seem to include all possible solutions as far as I can see.

In particular they don't give the 5.268s result!

Yes the Wolfram Alpha expressions gives all solutions also 5.268s.

Calculation of 5.268s with Wolfram Alpha:

Formula (Wolfram Alpha):

Let \(c_ 1 = 0\)

\(\begin{array}{|rcll|} \hline t &=& 6.3662(6.28319\cdot c_1 + 0.82757)\ \text{for} \ c_1 \in \mathbb{Z} \\ t &=& 6.3662(6.28319\cdot c_1 + 0.82757) \quad & | \quad c_1=0 \\ t &=& 6.3662(6.28319\cdot 0 + 0.82757) \\ t &=& 6.3662(0.82757) \\ t &=& 5.268\ldots \\ \hline \end{array}\)

see:

OK

or

\(\begin{array}{|rcll|} \hline t &=& 12.732(3.1416\cdot n+0.41379)\ \quad \ n \in \mathbb{Z} \\ t &=& 12.732(3.1416\cdot n + 0.41379) \quad & | \quad n=0 \\ t &=& 12.732(3.14169\cdot 0 + 0.41379) \\ t &=& 12.732(0.41379) \\ t &=& 5.268\ldots \\ \hline \end{array}\)

see:

Hi, ich habe eine Frage: Bestimmen Sie ein ganzzahliges Intervall [x;x+1], in welchem das Gefälle von f den Wert -0,25 annimmt. f(x)= 6xe^(-0,25x)

Es gibt zwei Intervalle:

1. Intervall [4; 5] mit x = 4,515344

2. Intervall [17; 18] mit x = 17,610364

Man kann diese beiden Werte für x mit Hilfe der Lambert'schen W Funktion (W) finden.

\(x = \dfrac{1-W(-\frac{e}{24} )} {0,25}\)

Let x1(t) and x2 (t) be functions that represent the horizontal displacement of two simple pendulums from their central position at time t in seconds.

If both pendulums start at the same displacement at time t= 0,

at what time are they both next at the same displacement?

WolframAlpha (Real Solutions) :

\(\begin{array}{|l|rcll|} \hline x(t)=0 & t &=& 80\cdot c_1 \quad & | \quad c_1 \in \mathbb{Z} \\ & t &=& 40\cdot(2c_1-\frac12) \quad & | \quad c_1 \in \mathbb{Z} \\ & t &=& 40\cdot(2c_1+\frac12) \quad & | \quad c_1 \in \mathbb{Z} \\ & t &=& 40\cdot(2c_1+1) \quad & | \quad c_1 \in \mathbb{Z} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline x(t)\ne 0 & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 -2\cdot \arctan(\sqrt{0.1928427661680941676676430}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 -0.8275700335453974790842315\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 +2\cdot \arctan(\sqrt{0.1928427661680941676676430}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 +0.8275700335453974790842315\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 -2\cdot \arctan(\sqrt{1.102624388267996959021457}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 -1.6196234867202913859018615\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 +2\cdot \arctan(\sqrt{1.102624388267996959021457}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 +1.6196234867202913859018615\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 -2\cdot \arctan(\sqrt{6.348976379766929903229600}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 -2.3859824690979305397688904\ldots ) \\\\ & t &=& \frac{20}{\pi} \cdot \Big( 2\pi c_1 +2\cdot \arctan(\sqrt{6.348976379766929903229600}) \Big) \quad & | \quad c_1 \in \mathbb{Z} \\ & &=& \frac{20}{\pi} \cdot ( 2\pi c_1 +2.3859824690979305397688904\ldots ) \\\\ \hline \end{array} \)

WolframAlpha: