All Answers 358514 Answers

If it emits the power equally in all directions then by the time it reaches the Earth it will be spread out over the surface of a sphere of radius 19800 km.  The surface area of this is 4*pi*(1.98*10^7)^2 m^2.  Divide the 20W by this area to get the power in Watts per square metre (it will be a very small number!).

Thanks Rom,

I just hunted high and low too find this answer.

I knew that you had answered a question because I saw your icon in the answer list but i could not jump straight into it because it does not recognise answers that are pure LaTex.

To get over this problem could I please request that you copy  the original question into your post before you start answering.

If you do that it will no longer be a problem.

Thank you   

Arrhenius equation.  Does this help?

Hi Brodude,

Welcome back to the forum, it is great to see you again   

I have had a lot of problems with editing LaTex on the new forum but I just went back into an old post and edited the LaTex.

So you can edit it some of the time at least.

It was very unresponsive.  I had to click it many times to get the yellow box to turn blue.  It has to go blue before you have any chance of editing.

Anyway, if you are persistent enough maybe you will be able to edit your LaTex too.

GOOD LUCK   !!

Nicely done, you nailed it

Strategy that I came up with:

Starting with the exponent at 5 onwards, if you subtract it by 1 then divide by 4 and get a whole number then it equals i. Otherwise if not a whole number then do the process again, subtract the original exponent by 2 then divide by 4 and see if you get a whole number, if you do then it equals -1 ...... if you have to subtract it by 3 it equals -i ...... if you have to subtract it by 4 then it equals 1.

Using this strategy: the exponent of i^10 is 10 soooo:

(10-1)/4 =/= whole so reject i

(10-2)/4 = 2 which is a whole number therefore i^10 = -1.

Now tell me what i^2015 equals =)

Well, i^2015 is -i 

Thanks Maximillian,

It is good to be able to preview the LaTex   

If the sqare root of -1 is "i", then what is i^2, 3, and 4. Do you see the pattern? Now, following this knowledge, what is i^10? (you can do it!)

See the imaginary coordinate system. The real axis in x (-1,+1) and the imaginary axis in y ( -i, + 1).

Set a Point on ( 1,0) for \(i^0\) so \(i^0 = 1\)

Move the Point counter clockwise 90 degrees you have the Point (0,i) for \(i^1\) so \(i^1 = i\)

Move the Point again counter clockwise 90 degrees you have the Point (-1,0) for \(i^2\) so \(i^2 = -1\)

Move the Point again counter clockwise 90 degrees you have the Point (0,-i) for \(i^3\) so \(i^3 = -i\)

Move the Point again counter clockwise 90 degrees you have the Point (1,0) for \(i^4\) so \(i^4 = 1\)

etc.

\(i^{10}=-1\)

Dont know if it used to be like this, but having a preview for the latex code that you type out has definitely got to be the best feature here

\(i=i \quad i^2=-1\quad i^3=-i\quad i^4=1\\i^5=i....\)

Strategy that I came up with:

Starting with the exponent at 5 onwards, if you subtract it by 1 then divide by 4 and get a whole number then it equals i. Otherwise if not a whole number then do the process again, subtract the original exponent by 2 then divide by 4 and see if you get a whole number, if you do then it equals -1 ...... if you have to subtract it by 3 it equals -i ...... if you have to subtract it by 4 then it equals 1.

Using this strategy: the exponent of i^10 is 10 soooo:

(10-1)/4 =/= whole so reject i

(10-2)/4 = 2 which is a whole number therefore i^10 = -1.

Now tell me what i^2015 equals =)

I do agree with CPhill's last point.  My computer is very slow to load and the preveiw step is usually unnecessary.  

Plus it makes it that I sometimes leave a page thinking that I have posted when I have not.

I do like having the option of a preview but there should be an "immediate post" button to the right of it as well.

I take it you are posting this as a challenge question to other students?  I mean, you already know how to do it.?  

I intend to draw even our young member's and guest's attention to this question so I think I better explain a little.

You cannot find the square root of a negative number because nothing times by itself can be negative.  

Right ?

Well sort of right.   There is NO answer in the real number system and many of you have not yet heard of imaginary numbers.  BUT

There is also a complex number system that has imaginary numbers in it.

\(The\;\;\sqrt{-1}\;\;\mbox{is the basic unit of all imaginary numbers, it is called }i\\ so\\ i^2=\sqrt{-1}\times \sqrt{-1} = \sqrt{-1*-1}=\sqrt{1}=1\\ i^3=i^2*i=1*i=i \)

and so on.   See if you can work out  i^4 

Heureka has given you the answer below but imaginary numbers are fun to play with  

ok That is good    :D

6+(x^3+4) (x^3-4)

\(6+(x^3+4) (x^3-4)\\ = (x^3+4) (x^3-4) \;\;\;\;+6\)

See how yow you have the same thing in both brackets but one is plus and one is minus?

This is a difference ot 2 squares.

and

\((a-b)(a+b) =  a^2-b^2\)

so

\(= (x^3+4) (x^3-4) \;\;\;\;+6\\ = (x^3)^2\;-\;(4)^2 \;\;\;\;+6\\\)

Andre you can try finishing it now :)

Thankyou i understand :D

Hi  andreEee,

It is nice to meet you   :)

(x+2) (x-3) (x+4) =0

ok, think about this       abc=0

What can a,b or c be?

Well the only way that 3 things can multply together to give 0 is if one (or more than one) of them is 0 to start with.

SO

if  abc=0 then either a=0  or  b=0   or c=0      (or more than one of them can be 0)

It is the same for your problem.

If    (x+2) (x-3) (x+4) =0

then

either         x+2=0     or     x-1=0     or     x+4=0

so either      x= -2    or       x=1       or      x = -4

Thanks to you too, heureka........that's two methods that I have to learn from......!!!!!

I know this question is old but I wanted to look at it and this is the first chance I have had.

Alan has given a great answer but I wanted to do it without using a computer graphing program to help me.

(I did cheat and use some of Alan's results but that was just to save the necessity of number crunching)

\(f(x)=2xsinx   \qquad   0\le x\le \pi\\ f'(x)=2(sinx+xcosx)\\ f''(x)=2(cosx+cosx-xsinx) = 4cosx-2xsinx\\ \mbox{Now turning points will occur when f'(x)=0}\\ 2(sinx+xcosx)=0\\ sinx+xcosx=0\\ xcosx=-sinx\\ x=-tanx\\ \mbox{Now if I plot y=x and y=tanx then the point of intersection will be where this is true.}\\ \mbox{I can do a rough graph of this without the help of any graphing calculators..}\\ \mbox{I can see that the point of intersection is between } x=\pi/2 \;\;and x=\pi\\ \mbox{So like Alan said x=2 is a good first estimate.} \)

Now you use Newton's method to get closer and closer approximations.

I can never remember the formula so I have to derive it each time but with a little practice this is easy to do.

anyway, it is

\(\boxed{ x_1=x_0-\frac{f(x_0)}{f'(x_0)} }\\ Where \;\;x_0 \;\;\mbox{is the first estimate and }x_1 \mbox{ is the next estimate}\)

This is tedious but it can certainly be done by hand.   (you only need to do it 3 times.)

Anyway, Alan got the answer x=2.0288158.

Now you do need to ascertain what kind of a stationary point this is 

so you need to plug this valu of x into the second derivative.

It will give you   f''(2.0288158) < 0 

I have not done it because I cannot be bothered but I know this is true because Alan has kinly plotted f(x)for me and I can see that the concavity of the curve at the stat point is negative.

There you go.  That is how you do it without the need of graph aids.

I have graphed it just to help you understand what I was tallking about.  

I did graph it too but only roughly with a pen and paper.

\(2.56 = \dfrac {256}{100} \mbox{ is the ratio of two integers so it is a rational number.} \\ -\dfrac{22}{9} \mbox{ is also the ratio of two integers and so also a rational number.}\)

Thanks , again.....I haven't seen this before and the "fixed point" concept is a new one on me......your explanation was excellent....!!!!!

[BTW......I think Melody mentioned to me that the "new and improved" forum will not let you edit LaTex......!!!!........bummer....!!!! ]

Totally should be a minus, my bad. Glad I could help =)

Edit: Seems I cant edit my original post =/

OK,  Brodudedoodebrodude............got it!!!!........Thanks!!!!

[Shouldn't the " + "  sign in Step 2 of your original answer be a " -"   instead??? ]

Not a problem! The "fixed point" refers to the value(s) of \(f_{n-1}\) that when you work out \(f_n\), you get the same value that \(f_{n-1}\) was, this putting you in a never ending loop and hence is a fixed point (or equilibrium).

Here's the method for finding the fixed point (explained pretty loosely):

-> By definition the fixed point means that the next number should be the same as the current one which is equivalent to saying \(...=f_{n-2}=f_{n-1}=f_n=f_{n+1}=...\)  This means we can forget about the subscripts and treat everything as the same unknown \(f\).

-> This means at the fixed point the equation \(f_n=3f_{n-1}-60\)  transforms into \(f=3f-60\).

-> Now you just solve for \(f\) and that will be your fixed point

In this example, \(f = 30\) and substituting it into the original difference equation we get:

\(f_{n-1}=30\\f_n=3f_{n-1}-60\quad=>\quad f_n=3(30)-60=30\quad=>\quad f_n=f_{n-1}\)

so this is a fixed point.

f(1)=20 , f(n)=3*f(n-1)-60 find the 10th term

\(\small{ \begin{array}{lrcl} & f_n &=& 3\cdot f_{n-1} - 60 \qquad a=3 \qquad c=-60\\ \text{or general}: & f_n &=& a\cdot f_{n-1} + c \\ \hline \\ & f_1 &=& a\cdot f_0 + c \\\\ & f_2 &=& a\cdot f_1 + c \\ & f_2 &=& a\cdot [a\cdot f_0 + c] + c \\ & f_2 &=& a^2\cdot f_0 + a\cdot c + c \\ & f_2 &=& a^2\cdot f_0 + c\cdot(1+a) \\\\ & f_3 &=& a\cdot f_2 + c \\ & f_3 &=& a\cdot [a^2\cdot f_0 + c\cdot(1+a)] + c \\ & f_3 &=& a^3\cdot f_0 + a\cdot c\cdot(1+a) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(a+a^2) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(1+a+a^2) \\\\ & f_4 &=& a\cdot f_3 + c \\ & f_4 &=& a\cdot [a^3\cdot f_0 + c\cdot(1+a+a^2)] + c \\ & f_4 &=& a^4\cdot f_0 + a\cdot c\cdot(1+a+a^2) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(a+a^2+a^3) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(1+a+a^2+a^3) \\\\ \hline \text{general solution}: & f_n &=& a^n\cdot f_0 + c\cdot (1+a+a^2+a^3+...+ a^{n-1}) \\ & && 1+a+a^2+a^3+...+ a^{n-1} = \frac{1-a^n}{1-a}\\ & f_n &=& a^n\cdot f_0 + c\cdot \frac{1-a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 + c\cdot \left( \frac{1}{1-a} - \frac{a^n}{1-a} \right) \\ & f_n &=& a^n\cdot f_0 + c\cdot\frac{1}{1-a} - c\cdot\frac{a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 - c\cdot\frac{a^n}{1-a} + c\cdot\frac{1}{1-a} \\ & && \boxed{~ \begin{array}{lrcl} & f_n &=& a^n\cdot \left( f_0 - c\cdot\frac{1}{1-a} \right) + c\cdot\frac{1}{1-a} \\ & f_n &=& x + \beta \cdot a^n \\ &&& x = c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - x \end{array} ~}\\ \hline \\ \text{general}: & f_n &=& a\cdot f_{n-1} + c \\ \text{general solution}: & f_n &=& x + \beta \cdot a^n \qquad x = \frac{c}{1-a} \qquad \beta = f_0 - x \\\\ & f_n &=& 3\cdot f_{n-1} -60 \qquad a=3 \qquad c=-60\\\\ & f_1 &=& 3 \cdot f_0 - 60 \\ & 3 \cdot f_0 &=& f_1 + 60 \\ & f_0 &=& \frac{ f_1 + 60 } {3} \qquad f_1 = 20\\ & f_0 &=& \frac{ 20 + 60 } {3} \\ & f_0 &=& \frac{ 80 } {3} \\\\ & x &=& \frac{c}{1-a} \\ & x &=& \frac{-60}{1-3} \\ & x &=& \frac{-60}{-2} \\ & x &=& 30 \\ \\ & \beta &=& f_0 - x \\ & \beta &=& \frac{ 80 } {3} - 30 \\ & \beta &=& -\frac{ 10 } {3} \\\\ & f_n &=& 30 -\frac{ 10 } {3} \cdot a^n \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 3^{10} \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 59049 \\ & f_{10} &=& 30 -196830\\ & \mathbf{f_{10} }& \mathbf{=} &\mathbf { -196800 } \end{array} }\)

well

1*14=14

2*7=14

that is it.

so the factors or 14  are 1,2,7,and 14    ta-da