Math Is Fun Forum

Hi Alive,

You're welcome.  Whoops I seem to have called you Alice earlier.  Apologies.

B

Hi Alice,

An asymptote is a straight line that the curve approaches (tends to) without ever meeting (and therefore crossing).

(a)  As you track towards the right along this curve it just carries on getting higher and higher without limit so no asymptote there.
But, as you track left the curve gets flater and flater; 2^x tends to zero as x tends to minus infinity; so the asymptote is Y = -4

(b)  Interesting the way your graph software has handled negative x values; normally I wouldn't expect to see any curve left of the y axis for no real log values exist there. [You may not have met 'complex' numbers yet, in which case don't worry.]

Again, there's no asymptote to the right as log curves just carry on getting bigger, but much more slowly than exponentials (a).
As x tends to zero, y tends to minus infinity, so the asymptote is the y axis itself.

(c) and (d) are similar but the asymptote is translated 4 units right.  Don't be put off by the apparent flattening out of these curves for large positive or negative x.  It might look as though the curves will eventually level off but they don't.  If you can alter the x scale to show much higher values of x, you'll see that the curve in (d) crosses the x axis eventually and goes on getting bigger and bigger.

Bob

You have been given that

P(x) = (x-a)(x-b)(x-c)(x-d)Q(x) for a, b, c, d all integers and Q is some polynomial of lower power than P.

Now assume 'to the contrary' that n exists,

ie  P(x) = (x-n)R(x) + 13   where R is also a polynomial of lower order than P

Put x = n in both expressions

(n-a)(n-b)(n-c)(n-d)Q(n) = 0 + 13

This is a contradiction as 13 cannot have these many (distinct) integer factors, so the assumption was false.

Hope this helps,

Bob

Hi

Because of the way this question is phrased, I'm assuming there's an analytical solution in there somewhere.

You can sub X = x^2 but then what?  The 'answer' seems to point towards arctan but still then what?

I tried to find factors (by inspection!) and got nowhere.  I haven't got Mathematica which I guess would help.

If you've got a factorisation please post it.

Thanks,

Bob

hi Heirot,

My diagram software (Sketchpad) is a vector geometry program so it's fairly accurate for playing about with this problem.

My three points post (again below), has b = 2.021 whereas the previous concentric post had b = 1.934 or less.

I think the best solution is to buy an open ended spanner (wrench). 

Bob

Hi Haftakhan,

A kite has two pairs of adjacent (next to) sides equal so SR must be 6.9 as well.

The line QS must be a line of symmetry and the diagonals cross at 90 degrees.

(i) Start by drawing a line and mark Q at one end.  (see diagram)

(ii) Measure with a protractor 62 degrees each side of this line and measure 3.6 to mark P and R.

(iii) Set a compass to 6.9 and with centre P draw an arc to cut the first line at S.

If your questioner wants a measured answer for the diagonals, that's now straight forward.

If you want a calculated answer, call the point where the diagonals cross T.

Use trigonometry PT/PQ = sine 62 to calculate PT and double it to get PR.

Then calculate QT ( = PQ x cosine62).

And TS by Pythagoras (you know PS and PT now).

Add QT + TS to get QS.

Bob

Hi

4 points?  Very much doubt it ... all to do with 60 and 90 degree symmetries.  I haven't got a rigorous proof though.

Bob

Sorry jk22,

I cannot see these pictures at all.  What I do is (i) screen shot of the pic (ii) paste into 'Paint' and edit (iii) save as a jpg file (iv) upload using the forum upload facility.

I can give more details if any of these steps is unclear to you.

Bob

Draw square BCED and circle, centre B, radius BD (=BC).
Construct line DA so that angle EDA = 15, and A lies on the circle.
ADB = 90 – 15 = 75
Triangle ADB is isosceles as AB = BD = radius.
So, BDA = BAD = 75, and therefore ABD = 30.
Therefore, ABC = 60.
But BA = BC (= radius) so triangle  BAC is isosceles,
So BAC = BCA = (180-60)/2 = 60
So triangle BAC is equilateral.
Draw a line of symmetry  through A, parallel to DB,
CAE is another isosceles triangle (75,75,30)
So ADE is the 15, 15 150 triangle.

hi

For the rotations question, combinations are made by matrix multiplication.  I find myself confused by your use of the term 'linear combination'.

I think, if a and b are vectors then pa + qb is a linear combination (p and q scalars).

What do you think a linear combination of two matrices would be like?

Meanwhile, for your second question:

I am assuming you are looking at 2x2 matrices that square to give
          -1  0
           0  -1

Let such a matrix be
           a   b
           c   d

Then squaring and setting equal to the above gives

a^2 + bc = -1 and d^2 + bc = -1  => a = + or - d

and replacing one a with -d gives

-ad +bc = -1     =>     ad - bc = 1 so the determinant is always 1.

A generator for all such matrices is

                                    a                        b

                         (-1 -a^2)/b    -a             -a

for all a and b (would you allow complex values here?)

Thus the following examples:

  (i)                                        3       5

                                            -2      -3

  (ii)                                        i       0

                                              0       i

  (iii)                                  1 +  i          2i

                                    (-1-2i)/2i       -1  -i

Since you can choose my real valued matrix and also yours as I1 and I2 you cannot generate any complex values from these.

Is that what you are after?

Bob

If the area scale factor is x N then the length scale factor is x (root N)

hi  jk22,

'the cardinality of the group of rotation of 90 around xyz which i suppose to be 48'

Unit vectors along the axes would be:
1     0     0
0     1     0
0     0     1

The origin is invariant so any transformation of these axes will have the form

x1     y1     z1
x2     y2     z2
x3     y3     z3
where

one of the xs is either 1 or -1 and the rest are zeros and similarly for the ys and the zs.

eg one transformation would be

1     0     0
0     0     -1
0     1     0

This is a rotation around the x axis of +90.

So how many such transformations are there?

Choose one from x1, x2, x3 (3 choices) and then whether it's +1 or -1 (2 choices)

Then choose from only two out of y1, y2, y3. (You cannot choose from the row that you chose your x from) (2 choices) and the whether it's +1 or -1. (2 choices).

Finally you have no choice about which of z1, z2, z3 since you must pick from the row that hasn't been chosen yet but you may chose from +1 and -1 (2 choices).

Altogether this gives 3 x 2 x 2 x 2 x 1 x 2 choices = 48.

The above rotation is one example.

But, another would be

-1     0     0
0      1     0
0      0     1

and this isn't a rotation; it's a reflection in the YZ plane.

I suspect that half the 48 are rotations and half are reflections.

So the cardinality for rotations alone would be 24.

You can prove it by making a 24 x 24 'multiplication' table of all the elements.

Here's a start:

                I         X         X^2      Y
______________________________

I              I          X         X^2      Y
X             X         X^2      X^3      P
X^2         X^2    X^3        I
Y              Y

where P is

0     0     1
1     0     0
0     1     0

You're going to need a large piece of paper and be prepared for lots of matrix multiplication.

Warning.  It isn't generally commutative ( XY not = YX )

If you find 24 elements in a closed set (ie. no more elements generated by those 24 in any combination) you've got your result.

Good luck.

ps. Not sure what your subsequent question means.  Can you provide one example of such a 2 x 2 matrix?

Bob