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#102 Re: This is Cool » Scrabble Probability: » 2009-02-11 04:26:34

Scrabble has no replacement, and an easy modification to make to solve the problem you've pointed out, is to say:

'i' iterates over all 7 letter words with unique letter combinations:

Which would make POT TOP and OPT in your example treat as the same word since the use the same combination of letters.

or alternatively:

'i' iterates over all combinations of 7 letters that form valid words. Allowing repeated letters obviously.

#103 This is Cool » Scrabble Probability: » 2009-02-10 22:34:19

luca-deltodesco
Replies: 5

Challenge a friend has given me; he has been given it as a homework assignment, i'm just doing it for fun:

Calculate the probablity of drawing your first 7 tiles from a full bag, and getting a 7 letter word (He has been given a full list of valid scrabble words, i'm just doing the maths bit for fun)

So far I've found the probablity given that you totally ignore the blank tiles: What i've gotten is this:

Where T is the total number of tiles in the bag
O(z[sub]j[/sub]) denotes how many of each tile is in a full bag
and 'n[sub]j[/sub]' is the number of each tile in the word
'i' iterates over all valid 7 letter words

I'm struggling on trying to include blank tiles in the calculation though! (I'm not asking for help, since this is someone else's homework assignment)

#104 Re: Help Me ! » Log » 2009-02-10 08:26:20

simply because logarithms on the reals are not defined for negative numbers:

since ln(-n) is not defined for reals, neither is log_(-n)x although as you have shown there are special cases where you can define it

---

if you venture into the complex realm, then you can define logarithms for negative bases:

natural logarithm of a negative real is given by the multivalued function

with principal branch:

so you could then have for some negative n

which gives for n = 5 and x = 25

and then, although this certainly is NOT '2' the expected answer, it is still of use since

Infact, you 'can' get 2 as the result if you do not use the principal branch, and instead define the logarithm as:

which gives

you can see this since:

although note that only the principal branch version is guaranteed to retain the property:


for all n,x

#107 Re: Help Me ! » Power Set Algorithm » 2009-02-04 05:26:25

You seem to be saying that we havn't proved pi is irrational george?

#109 Re: Help Me ! » commutitive group » 2009-01-26 07:41:36

That looks a little messy, i think it's easier to write:

(xy)(xy) = e  (since xy is also in G)
x(yx)y = e
x(yx)y² = ey
x(yx) = y
x²(yx) = xy
yx = xy

#110 Re: Help Me ! » Matrix Question - FP1 » 2009-01-21 06:50:42

basicly, you have the line y = mx, and you have to find the values of 'm' such that every single point on the line y = mx, would be mapped to a point that is also on the line itself.

#111 Re: Help Me ! » Matrix Question - FP1 » 2009-01-21 05:23:46

x' = 5x + 10y
y' = -3x - 8y

y = mx, so substitute

x' = 5x + 10mx
y' = -3x - 8mx

you want the transformed point (x', y') to lie on the line y = mx, so

-3x - 8mx = m(5x + 10mx)
-3 - 8m = 5m + 10m²
10m² + 13m + 3 = 0
m = (-13 ± 7)/20

check my working out.

#112 Re: Help Me ! » i forgot something » 2009-01-20 10:13:55

mathsyperson wrote:

That's something that not everyone agrees on.

The solutions of x² - k² = 0 are definitely x = ± k, but √(k²) is generally taken to be just k.

Does the symbol √ not refer specificaly to the 'principal' square root, aka the positive root, wherein simply 'square root' gives no such specifics, aka (x[sup]2[/sup])[sup]0.5[/sup] = ±x = ±√(x[sup]2[/sup])

#113 Re: Help Me ! » integration » 2009-01-20 04:30:32

This is just a simple polynomial integration:

except for the special case when n = -1

in your case, n = -2

#114 Re: Help Me ! » circular motion under gravity » 2009-01-15 11:43:18

I never even questioned that that bit could be wrong as it gave me the correct answer for the first part of the question tongue I just need to be more careful in doing these things to make sure i don't leave myself prone to causing them.

#115 Re: Help Me ! » circular motion under gravity » 2009-01-15 10:48:21

aha! thank god i wasn't going crazy then! thanks jane smile

#116 Re: Help Me ! » circular motion under gravity » 2009-01-15 10:28:56

although seemingly, using that equation involving k', i don't get 6.4 for v if plug in 150 for theta... what is wrong there?
seemingly there must be something wrong with the maths in my previous post.

#117 Re: Help Me ! » circular motion under gravity » 2009-01-15 10:24:31

I still don't understand where my thinking has gone wrong... (maybe i'm just very tired)

but, i get the equation

whch i integrate with respect to theta to get


i calculate k' with a v,theta pair as you usually would after an integration, and that should allow me to calculate the centripetal force for any theta? equating that to the equation for centripetal force with tension, i find the tension such that i never have to calculate 'v' itself?

#118 Re: Help Me ! » circular motion under gravity » 2009-01-15 10:02:18

I understand how you found T, but I still don't understand what is wrong with my method, i have integrated to get an equation for the centripetal force in terms of theta, which because of integrating has the constant which i find by plugging in a known pair for the centripetal force and theta.

#119 Help Me ! » circular motion under gravity » 2009-01-15 09:19:26

luca-deltodesco
Replies: 10

One end of a light inextensible string is attached to a fixed point O. The other end of the string is attache4d to a particle P of mass 0.2kg which moves in a vertical circle of radius 0.3 m. Air resistance is ignored. When the particle has speed 4m[sup]-1[/sup] the string makes an acute angle θ with the downward vertical. At this instanct the magnitude of the transvese component of acceleration of P is 6.3 ms[sup]-2[/sup]. Show that θ = 40° approximately.



At a later instant the string makes an angle of 30° with the upward vertical. Calculate the tension in the string at this instant.




v = 4, θ = 40°


The answer however says it should be 2.57

#120 Re: Help Me ! » changing orbits » 2009-01-15 03:45:22

from circular motion equations:



  (just me checking the equation you had was correct)

then potential energy:


the change in energy from increasing it's orbit then is the combination of the change in potential, and the change in kinetic



i'm really stumped how to get the answer they want

#121 Re: This is Cool » Imaginary numbers! » 2009-01-15 03:30:32

square root of -1, is named the imaginary unit 'i' there are an infinite amount of imaginary numbers, since you can have any real number, and multiply it by 'i'  so you can have 3i 4.5i πi -7.8i etc.

you then have the complex numbers which are a superset of the real and imaginary so that you can have for example 3+4i  7-8i etc

#122 Re: This is Cool » 0/0 » 2009-01-14 21:56:23

0/0 is equally undefined, who told you that it wasn't?

#123 Re: Dark Discussions at Cafe Infinity » Microsoft begins Windows 7 push » 2009-01-13 10:09:59

The system resources quoted to be needed to run Vista have always put me off of it straight away, i don't have the most powerful computer as it is, and i really don't want to have my programs run any slower than they already do.

#124 Re: This is Cool » 0.9999....(recurring) = 1? » 2009-01-11 07:18:04

you're assuming ∞+1 =/= ∞, to show that ∞+1 =/= ∞, that is not allowed, i can equally show that pi = 1, by assuming pi = 1 to show that pi = 1, it shows nothing.

#125 Re: Jokes » Jokes galore...... » 2009-01-06 19:52:04

Tigeree wrote:

but i don't get the medieval joke

"proving that the squire of the high pot and noose is equal to the sum of the squires of the other two sides."

Playing with the words a bit based on the pronunciation of them and the context.

Proving that the square of the hypotenuse is equal to the sum of the squares of the other two sides

Which is the pythagorean theorem.

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