Honga

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We can rewrite the modular equations as:

\(n-1162=1657k,\\ n-453=1162m.\)

Now we are able to remove n from the equation:

\(1657k+1162=1162m+453,\)

we can split the first term to tidy things up:

\(1162k+495k+1162-1162m=453\Rightarrow\\ 1162(k+1-m)+495k=453\Rightarrow\\ 1162(k+1-m)+495k-495=453-495\Rightarrow\\ 1162(k+1-m)+495(k-1)=-42.\)

The first term is even and the outcome is even, hence the second term must be even as well. We conclude that 495(k-1) is a multiple of 10. In order to get a negative number (k+1-m) must be negative as well. Since the second term doesn't contribute to the last digit of the sum we know the last digit of the first term must also be 2. As a result (k+1-m) must have a last digit equal to 1 or 6.

Basically you do the same thing, your formula will look something like:

\(v(x)=\left\{\begin{array}{c l}v_1(x)&\text{for: }0\le x\le x_\text{max},\\v_\text{max}&\text{for: }x>x_\text{max}. \end{array}\right.\)

\(\text{The first solution that comes to mind is choosing }n=7\\ \text{and }k=2\text{. This allows the product }mn\text{ to consist of}\\ \text{even powers which results in a natural number when }\\ \text{taking the second power root. Now we need to check if}\\ \text{this solution is indeed the smallest possible value for }n.\\ \text{We know that a smaller solution can be acquired only}\\ \text{if we don't need to add an extra factor 7. This requires}\\ \text{343 being divisible by }k\text{. We know }343=7^3\text{ so }k\text{ must}\\ \text{equal some power of 7 for this to be the case. 1980 and}\\ \text{384 are not divisible by 7, 1694 however is. To correct for}\\ \text{this }n\text{ needs to equal 6 (because 1981 and 385 are divisible}\\ \text{7). As 6 is smaller than 7 are solution becomes: }n=6,\\ k=7;n+k=13.\\ \text{If this explanation is still a bit hazy, allow me to show it in}\\ \text{formula form:}\\ \)

\(n=2^\alpha\cdot3^\beta\cdot5^\gamma\cdot7^\delta\Rightarrow\\ mn=2^{1980+\alpha}\cdot3^{383+\beta}\cdot5^{1694+\gamma}\cdot7^{343+\delta}\Rightarrow\\ 2^{\frac{1980+\alpha}k}\cdot3^{\frac{383+\beta}k}\cdot5^{\frac{1694+\gamma}k}\cdot7^{\frac{343+\delta}k}\in\mathbb{N}\Rightarrow\\ \frac{1980+\alpha}k,\frac{383+\beta}k,\frac{1694+\gamma}k,\frac{343+\delta}k\in\mathbb{N}.\)

\(\text{In contrast to what has been said before the union of two linear subspaces of }V\\ \text{ is in general not a linear subspace itself. Take for example the linear vector space}\\ \mathbb{R}^2\text{ of two dimensional vectors. We can now define the two following subspaces:}\\ M=\{(a,0);a\in\mathbb{R}\},\\ N=\{(0,b);b\in\mathbb{R}\}.\\ \text{It should be clear that each individual subspace is a linear subspace of }\mathbb{R}^2. \text{The}\\ \text{union of the two is not a linear subspace because the vector (1,1) is not an element}\\ \text{of the union (which it should be since (1,0) and (0,1) are elements of the union).}\\ \text{Hence the given statement is disproven by means of counterexample.}\)

\(\text{We know that the velocity doubles every interval }\Delta t\text{, hence the velocity}\\ \text{must adhere to the general formula: }v(t)=v_0\cdot2^{t/\Delta t}\text{, with }v_0\text{ the initial}\\ \text{velocity. The distance travelled is of course the integral of the velocity}\\\text{over the time spent travelling:}\\ \Delta x=\int^{t_e}_0v(t)\text{d}t=v_0\int^{t_e}_02^{t/\Delta t}\text{d} t=\frac{v_o\Delta t}{\ln 2}(2^{t_e/\Delta t}-1).\\ \text{We require the time it takes to travel this distance }(t_e)\text{ so we rewrite the}\\ \text{formula to extract }t_e:\\ t_e=\frac{\Delta t}{\ln 2}\ln\left(\frac{\Delta x\ln 2}{v_0\Delta t}+1\right).\\ \text{In order to test our answer we try }\Delta x=0\text{ this should give us }t_e=0,\\ \text{and indeed it does.}\)

\(\text{An intuitive approach of solving discrete convolution is sliding an inverted version of }\\f\text{ over }h. \text{ For }n=0\text{ we have:} \)

\(\begin{array}{c| c c} m&-1&0&1&2&3&4&5\\\hline h&0&0&I_1&I_2&I_3&I_4&I_5\\ f&c&b&a&0&0&0&0\\\hline f*h[0]&0&0&aI_1&0&0&0&0 \end{array}\)

\(\text{the convolution product in this case is the product of each column.}\\ \text{Now we slide }f\text{ three places:}\)

\(\begin{array}{c| c c} m&-1&0&1&2&3&4&5\\\hline h&0&0&I_1&I_2&I_3&I_4&I_5\\ f&0&0&0&c&b&a&0\\\hline f*h[3]&0&0&0&cI_2&bI_3&aI_4&0 \end{array}\)

\(\text{and we have the same solution.}\)

\(\text{The definition of the discrete convolution is:}\)

\(f*h[n]=\sum\limits^\infty_{m=-\infty}f[m]g[n-m].\)

\(\text{We require the solution for the case }n=3:\)

\(f*h[3]=\sum\limits^1_{m=-1}f[m]g[3-m],\)

\(\text{since } f[m]=0, \text{for: } m\neq-1,0,1.\)

\(\text{Our solution is:} \)

\(\begin{align*} f*h[3]&=f[-1]g[4]+f[0]g[3]+f[1]g[2]\\ &=aI_4+bI_3+cI_2. \end{align*}\)

The first thing we do is move the a₇/7! to the other side:

\(\frac57-\frac{a_7}{7!}=\frac{5\cdot6!}{7!}-\frac{a_7}{7!}=\frac{5\cdot6!-a_7}{7}\frac1{6!},\)

we require the first fraction to be an integer. This is the case for a₇=2, now we can bring the next fraction to the left:

\(\frac{514}{6!}-\frac{a_6}{6!}=\frac{514-a_6}{6}\frac{1}{5!},\)

for the first fraction to be an integer a₆ needs to equal 4. We bring the next fraction to the left:

\(\frac{85}{5!}-\frac{a_5}{5!}=\frac{85-a_5}{5}\frac1{4!},\)

which requires a₅ to be zero. We bring the next fraction to the left:

\(\frac{17}{4!}-\frac{a_4}{4!}=\frac{17-a_4}{4}\frac1{3!},\)

which requires a₄ to be 1. We continue on our course:

\(\frac{4}{3!}-\frac{a_3}{3!}=\frac{4-a_3}3\frac12,\)

obviously a₃ should be one. We are left with:

\(\frac12=\frac{a_2}{2!},\)

and we get a₂=1, case solved. Checking our answer with the calculator:

\(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{4}{6!}+\frac{2}{7!}-\frac57=0.0000000000000000143,\)

which is purely a machine error, since 1/7! equals 0.0001984126984127 which is far larger than our error.

You should try the log button. ;)

$$If: $$tan(x)=a$$ then:$$tan^{-1}(a)=x$$

$$slope=\frac{y_a-y_b}{x_a-x_b}=\frac{y-10}{3--5}=\frac{y-10}{8}=-\frac{3}{8}$$

Now solve for y.

Good luck! :)

First you need to calculate the mean:

$${\frac{\left({\mathtt{1.25}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.25}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{6}}}{{\mathtt{5}}}} = {\mathtt{1.2}}$$

No you need to calculate the square of the deviation of each value to the mean:

$$\\(1.2-1.25)^2=(-0.05)^2\\(1.2-1)^2=(0.2)^2\\(1.2-1.5)^2=(-0.3)^2\\(1.2-1.25)^2=(-0.05)^2\\(1.2-1)^2=(0.2)^2$$

Now we need the mean of the squared deviations:

$${\frac{\left({\left(-{\mathtt{0.05}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{0.2}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{0.3}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{0.05}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{0.2}}\right)}^{{\mathtt{2}}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{7}}}{{\mathtt{200}}}} = {\mathtt{0.035}}$$

The square root of this mean is equal to the standard deviation:

$${\sqrt{{\mathtt{0.035}}}} = {\mathtt{0.187\: \!082\: \!869\: \!338\: \!697\: \!1}}$$

Hope this helps!

Hallo Hamidaik, 

Welkom op 't forum. Je weet dat zowat niemand hier Nederlands kan, he?

Gr,

Honga

Well done! :)