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#1 Re: Help Me ! » Interesting Optics Problem » 2014-11-24 23:42:00

hi Rob,

Welcome to the forum.

First I made my own diagram using Sketchpad and took some measurements:

5q5DcPE.gif

Then I looked for a trigonometrical solution.  Couldn't spot an easy one, but there may be one lurking in there somewhere.  Meanwhile, here's the one I found that managed to calculate EH.  Once you have that, all the other lengths and angles are relatively ( smile ) easy to find.

Let angle BEH be alpha.

substitute this in (2)

and adding (1)

This re-arranges to

One solution is negative.  The other is EH ≈ 44.125 which agrees nicely with the Sketchpad answer.

Bob

#2 Re: Help Me ! » Can anybody help me with this equation pleaseeee?? » 2014-11-23 21:13:05

hi kroks1603

can help you win a girlfriend?

In my experience you cannot legally 'win' a girlfriend. 

Maybe it is advice to help you improve your chat up line.

This probably only works well if the girl is keen on math. 

If you don't know enough in advance about the girl, a more generally applicable version:

roflol

Bob

#3 Re: Help Me ! » minimum value at specific index is important » 2014-11-23 20:57:42

hi pari_alf

You need to supply a more detailed algorithm.

eg.

Find minimum values
If only one, state its index.

If more than one, what?

(a) state the one in the middle
(b) state the one nearest to index 4
(c) state ..........?

Bob

#4 Re: Help Me ! » trigonometry the period! » 2014-11-22 03:48:24

hi Whizzies,

g(2x)=f(x) <=> g(x)=f(x/2)

The first statement is true for all x.  So you can replace x by x/2 and it will still be true.

Similarly, if the second statement is true for all x, then replace x by 2x and it will still be true.

Here's a simple example.

f(x) = 6x + 7 = 3 times (2x) +7

So we can define g by:

g(2x) = 3 times (2x) + 7

which means g(2x) = f(x)

The definition of g may also be written:

g(x) = 3x + 7 = 6 times (x/2) + 7 = f(x/2)

So g(x) = f(x/2)

Bob

#5 Re: Help Me ! » Trigonometry » 2014-11-20 23:50:25

hi Whizzies,

I've not seen this way of defining point symmetry before so I've had to do a bit of experimenting.  When I come across a tricky bit of math, I try to simplify the numbers first, and also pick an example for what I'm trying to investigate.

For symmetry, I wouldn't choose a trig. function because the graphs have too many lines and points of symmetry, so its impossible to see what's going on.

I chose y = x^3, because it only has point symmetry around (0,0)

You'll find it is worth using the function grapher at

http://www.mathsisfun.com/data/function … c1=(1+x)^3

I've set this up with the function y = (1+x)^3.  Try some values:

x = -3 ,    y = -8
x = -2 ,    y = -1
x = -1 ,    y = 0
x = 0  ,    y = 1
x = 1  ,    y = 8

You can see from this that the graph had point symmetry around (-1,0)

Try changing that '1' into other numbers and you'll see that, with a = -1

And f(a-x) = -f(a+x)

Now let's try adding a constant.

y = x^3 + 2

You'll see that this shifts the graph up by 2, so the symmetry point is moved to (0,2)

Then

Here I've put square brackets around the function and introduced an additional '2' (either + 2 or -2) to balance the equation

so in general

You don't need the square brackets here, but I thought it would help you to compare with the line above if I left them in.

This function as point symmetry around (0,b)

Finally put both bits together:

So y = (1+x)^3 + 2

x = -3 , y = -6
x = -2 , y = 1
x = -1 , y = 2
x = 0 , y = 3
x = 1 , y = 10

Try it on the grapher as well.

You'll see the point symmetry is now around (-1,2)

Again you can vary the value of the constant.

So if I put in square brackets for the function again:

y = [(1+x)^3 + 2]

a = -1 and b = 2

and

so in general

with the point of symmetry (-a,b)

Bob

#6 Re: Help Me ! » trigonometry the period! » 2014-11-20 22:28:25

hi Whizzies

You'll probably find the MIF function grapher helpful:

http://www.mathsisfun.com/data/function … nc1=sin(x)

Enter sin(x/3) as the second graph and you'll see that the period gets bigger by a factor of 3, not smaller.

Think about it and maybe you can see why this is. 

If x = pi/2, we know sin(x) = 1.

On the graph of sin(x/3), and to get sin(something) = 1, you need x/3 = pi/2, so x = 3pi/2.  That is to say, you need an x value three times the size, to get a given y value on this graph.

eg.  On y = sin(x) then x = pi/6 gives y = sin(pi/6) = 1/2

On y = sin(x/3) we need x = 3pi/6 to get y = sin(3pi/6 over 3) = sin(pi/6) = 1/2

I'm also looking at you symmetry post.  I've never seen point symmetry written like this, so I'm having to work it out from scratch.  Hopefully I'll get there soon.  :)

Bob

#7 Re: Coder's Corner » Reducing the Expression of Time Complexity!!! » 2014-11-18 21:45:18

hi azair

01:   T(n)=T(n-1)+T(n-2)+4  and T(0)=T(1)=1
02:   Lets assume that T(n-1)~T(n-2)
03:   then, T(n)=2T(n-2)+c   where c=4
There's a line missing here that should make it clearer.
replace n by n-2 in line 03
03.5: T(n-2) = 2T(n-4) + c
Now replace T(n-2) in line 03 using line 03.5

04:   T(n) = 2{2T(n-4)+c}+c
and simplfy 
05:   =4T(n-4)+3c
Another missing line here
replace n by n-2 again
05.5: T(n-2) = 4(n-6) + 3c
but, from line 03,
T(n) = 2T(n-2) + c
and replace T(n-2) using line 05.5
And then I think there's a typo; my correction in red

06:  T(n) = 2{4T(n-6)+3c}+c
and simplify
06.5: = 8T(n-6) + 7c
and with n replaced by n-6 in line 03
06.75: T(n-6)=2T(n-8)+c
and replace T(n-6) in line 06.5
06.8: T(n) = 8(2T(n-8) + c) + 7c
and finally simplify to get

07:   T(n) = 16T(n-8)+15c

By now you can, hopefully, spot a pattern in the two numbers:

2,4,8,16,.......powers of 2

and 1,3,7,15,......powers of 2 minus 1

which is where the general formula comes from.

Hope that helps, smile

Bob

#8 Re: Maths Is Fun - Suggestions and Comments » Set Symbols » 2014-11-17 20:31:34

hi MathsIsFun,

Good page.

: is also used to mean 'such that'.
http://en.wikipedia.org/wiki/List_of_ma … al_symbols

The complement of a set only has meaning if the 'universe' is defined.
http://en.wikipedia.org/wiki/Complement_(set_theory)

so you could have:   if U, the universe, = {1,2,3,4,5,6,7,8,9} and A = {2,3,4,5,6,7,8} then Asuperscriptc = {1,9}

There does not appear to be a regularly used symbol for the irrationals.  Wolfram and Wiki do not have one.  Maths Exchange has several suggestions (none of which I've ever seen used) but R−Q would seem to do the job.

Bob

#9 Re: Help Me ! » Find the value » 2014-11-16 21:13:16

That's the one.  Viewing the film recommended.  smile

B

#10 Re: Computer Math » How to do it. » 2014-11-16 21:04:02

EM is an essential part of mathematics but the following example shows why you should beware.

Hypthesis:  the formula

is a prime number generator.

Investigate by trying n = 1, 2, 3 etc

Bob

#12 Re: Help Me ! » Find the value » 2014-11-16 04:52:37

smile

Someone will have to invent writing in the air, sort of like plane contrails.  Now that would be really useful.

Bob

#13 Re: Help Me ! » Find the value » 2014-11-16 04:37:01

hi bobbym,

Thank you.  I had the ideas in my head hours ago but I kept slipping up with the algebra.  Making the K and L substitutions made a big difference as I was able to get rid of all those horrible powers.  One of my early errors actually allowed the cancelling, leading me into thinking I was on track.  Then I spotted a slip and the more I corrected the further I got from the nice answer.  My table top is now covered with sheets of paper with, mostly, wrong versions.  More for the paper re-cycling.

Bob

#15 Re: Help Me ! » Find the value » 2014-11-16 02:32:24

hi Niharika,

This is my final and, I hope correct, version.

Consider the j-th term:

As this is a bit complicated I'm going to simplify by replacing 2^973 by K, and 2^j by L, so the expression becomes

Now there are an even number of terms in the sum, so find the one that is j from the end:

Add these two terms:

Thus you can pair up a term from the first half of the sum with another from the second half, and get the same j independent result every time.

There are 974 pairs so the sum is


Bob

#16 Re: Help Me ! » Advanced compound interest problem » 2014-11-15 06:19:43

hi pletharoe,

Welcome to the forum.

If I use P for the initial value of the house then the new value after n years is given by the standard compound interest formula:

If I've understood you correctly, the rent is found by multiplying this value by y.  And it's y every year. So the rental is given by

This is a geometric series, with first term Py, and common ratio (1+x), so using the sum of such a series we get

Let's check it out with your figures:

That looks good to me.  smile

Bob

#17 Re: Dark Discussions at Cafe Infinity » Dihydrogen Monoxide » 2014-11-14 23:28:03

hi Agnishom,

Yes, they are.  Back in the 70s they got a progressive rock band to write a song, reminding people to breathe:

https://www.youtube.com/watch?v=mrojrDCI02k

Bob

But health and safety is a world-wide concern:

health-and-safety-gone-mad.jpg

images?q=tbn:ANd9GcSjKphFHM3fmRcfJ35fmC0VdRXGFKbMC5sQbzkNhL2sdD-U5FhmJA

#18 Re: Dark Discussions at Cafe Infinity » Dihydrogen Monoxide » 2014-11-14 20:36:29

I think UK safety experts must be more aware of this than elsewhere.  During the building phase, every house is equipped with a mechanical device called a 'tap'.  If you find too much dihydrogen monoxide coming into the property, you just turn the tap and it stops.  There are also specially designed 'drains', to take away the excess.

Bob

#19 Re: Help Me ! » Triangle » 2014-11-12 20:22:43

hi Breeann522

Welcome to the forum.

In any triangle, the longest side is always opposite the largest angle; the middle sized side opposite the middle sized angle; and the shortest side opposite the smallest angle.

Bob

#20 Re: Help Me ! » Linear Programing - word problem confusing me ! » 2014-11-11 07:00:02

hi Martha,

I'm not sure what is meant by this.  The objective function is the expression you are trying to maximise.  I've been calling it the 'profit' line.  So the coefficients are the numbers in front of x, y and z.  Why these would have 'ranges' is a mystery.  Please post your solution (also the exact wording of this part of the question),  and I'll give it some more thought.

Bob

#21 Re: Help Me ! » Euclidean Algorithm Problem » 2014-11-11 02:31:15

hi SolarDevil

I've checked this by algebra and computation and get the same result.

I had to look up what the Euclidean Algorithm is (I knew it but not the name).

(3n-4) - (2n+7) = (n-11)

(2n+7)- (n-11) = (n+18)

(n+18)-(n-11) = 29

So the greatest common divisor is 29, which is a prime, so the only other divisor is 1.

Put 3n-4 = 2n+7 and n = 11.  So that is the first fraction that will simplify (from 29/29 to 1/1)

Thereafter, fractions that have a common factor of 29 occur every 29 cases, ie. at 29m+11 for whole numbered values of m.

m= 34 gives n = 997 and the fraction is 2001/2987 = 69/103,  the largest in range.

So fractions that will cancel are all those from m=0 (29/29) up to m = 34 (2001/2987)

Thus 35 in all.

As we are asked for the number that don't simplify in the range 2 to 1000 inclusive  that will be 999 - 35 = 964.

Bob

#22 Re: Help Me ! » Euclidean Algorithm Problem » 2014-11-09 10:53:44

hi SolarDevil

Hhhmmm.  I wonder how one is expected to do this.  It is looking like 997 at the moment, but I'm still trying to think of a rigorous approach to confirm that.

If I can come up with an intelligent method, I'll post again.

What is the Euclidean algorithm anyway ?

Bob

#23 Re: Maths Teaching Resources » Worksheets and stuff » 2014-11-08 23:08:52

hi PatternMan

in WORD 2007 you can do things like this:

Type equations in linear format

You can type most equations quickly by using Math AutoCorrect codes. For example, to align an equation array, you can use @ and &, as in the following:

\eqarray(x+1&=2@1+2+3+y&=z@3/x&=6)<space>

I cannot copy here the result (it's a word graphic) so you'll have to try it yourself.  smile

I just typed 'mathematical format' into WORD help.#


Bob

#25 Re: Help Me ! » hard algebra » 2014-11-08 00:43:03

hi thedarktiger,

Do you mean

It looks like you can use sum of roots = -b/a and product of roots = c/a with the usual ax^2 + bx + c = 0 quadratic.

Bob

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