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non game related math question


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#1 CarlFGauss   Members   -  Reputation: 122

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Posted 27 June 2001 - 06:45 AM

assume there is non-negative numbers a(1),a(2),... satisfying a(m+n)<=a(m)+a(n) for every m,n=1,2,3,... is the following correct? a(kn)<=ka(n) for every k,n=1,2,3,... anyone who is good in math please help me. an answer ''Yes'' or ''No'' will be sufficient if you don''t want to type out the proof.

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#2 jmg3   Members   -  Reputation: 122

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Posted 27 June 2001 - 02:41 PM

i don''t understand your notation. what are a(1) and a(2) supposed to mean?
do you mean assume m & n are non-negative integers? reals? what is the < for?
i get the feeling you''re asking about if a function is linear or not...

#3 HyprLogik   Members   -  Reputation: 122

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Posted 27 June 2001 - 02:59 PM

Correct, this does deal with linearity. So assuming these are functions that you mention of, the conjecture you present is deemed true iff the functions are linear.

#4 CarlFGauss   Members   -  Reputation: 122

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Posted 27 June 2001 - 04:23 PM

the notation "<=" means "less than or equal to"

#5 Anonymous Poster_Anonymous Poster_*   Guests   -  Reputation:

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Posted 28 June 2001 - 01:09 AM

a(kn) = ka(n) always, not smaller than or equal to.
and a(m+n) = a(m) + a(n) always as well, also not just smaller than, it is equal to.

Cheers,

#6 mossmoss   Members   -  Reputation: 326

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Posted 28 June 2001 - 02:09 AM

There is a way in mathematics to solve certain kinds of problems called "proof by induction". It works in three steps:

1) Show that a base case is true. That is, your solution works when b is input.

2) Assume your theory works when x is input.

3) Given your assumption in #2 and all other attributes of your system, show that your theory works when x+1 is input.

This is only suitable for discrete (ie, integer) problems, not continuous. Let's try it here.

We know: a(m+n) <= a(m) + a(n) for m,n = 1,2,3...
Want to prove: a(kn) <= ka(n) for k,n = 1,2,3...

Step 1, Base case: k = 1
Show that: a(kn) <= ka(n)
Easy. With k=1, we get a(n) <= a(n). True.


Step 2, Assumption: k = x
Assume that: a(xn) <= xa(n)


Step 3, Proof: k = x+1
Show that a((x+1)n) <= (x+1)a(n)
Here we go... keep in mind our precondition about a(m+n)... Also, I use <?= below here for what we want to test/show, while <= is used for known truths.

We want to prove this:

a((x+1)n) <?= (x+1)a(n)

Apply distribution rule on both sides.

a(xn + n) <?= xa(n) + a(n)

Use our precondition on the left. Note how that we can move the <?= to the right, since the <= relation on left is known. This is an ordering, whereby if we prove a more restrictive case, the less restrictive case will automatically be true.

a(xn + n) <= a(xn) + a(n) <?= xa(n) + a(n)

So can we show this:

a(xn) + a(n) <?= xa(n) + a(n)

Subtract a(n) from both sides:

a(xn) <?= xa(n)

But this is our assumption in step #2 above! Proven, Q.E.D.


You have to be careful sometimes with this method; it can easily be abused to show things that aren't true. You also have to be careful that your preconditions do match your situation, otherwise you're proving the wrong theorem. I'm pretty sure I've been careful here and not missed anything.

Some may say that I'm using the theory to prove the theory, and so it's not a proof at all. But look again at the three steps. I made an assumption at x and had to prove x+1. That alone is insufficient proof, UNLESS you have a starting point. I proved it worked for k=1. Because of that, the rest of the proof shows it will work for k=2. Hey, because it works for k=2, the proof also shows it works for k=3. And so on, and so forth, so the theory is proven for k=1,2,3... Doesn't matter what n is.



---- --- -- -
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Edited by - mossmoss on June 28, 2001 9:13:29 AM

#7 Dean Harding   Members   -  Reputation: 546

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Posted 28 June 2001 - 02:14 AM

quote:
Original post by Anonymous Poster
a(kn) = ka(n) always, not smaller than or equal to.
and a(m+n) = a(m) + a(n) always as well, also not just smaller than, it is equal to.

Cheers,


not true. take a(x) = 2 - x. This is a line with slope -1 and y-intercept 2. Clearly a linear .

Now,

a(1) = 1
a(2) = 0
a(3) = -1

and

a(1 + 2) = -1 < a(1) + a(2) = 1

I think you''re confusing this problem with one of linear transformations i.e. if a() was a linear transformation from R->R then you can say that a(tx0 + tx1) = t((a(x0) + a(x1)), but the original problem never said that a() was a linear transformation.

As for the original question, I don''t think you can say anything about whether that is true or not. If k <= 2 then obviously you could, but if you take a(3n) or a(n + n + n) you can say that that is <= a(n + n) + a(n) but you can''t say anything about a(n) + a(n) + a(n), since a(n) + a(n) is >= a(n + n) (be definition). If you can see what I''m saying...


War Worlds - A 3D Real-Time Strategy game in development.

#8 Anonymous Poster_Anonymous Poster_*   Guests   -  Reputation:

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Posted 28 June 2001 - 04:43 AM

a(x)=2-x is not linear. Any linear function has a(0)=0.
Also note that a(x+y)=2-x-y != a(x)+a(y) = 4-x-y.
It is, however, an affine function.

#9 Anonymous Poster_Anonymous Poster_*   Guests   -  Reputation:

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Posted 28 June 2001 - 05:03 AM

Also, a(m+n)<=a(m)+a(n) is true for many other functions.
For example, take a(1)=2,a(2)=1,a(m)=0 m>2.
Another example: take a(x)=x^.5
It is always true sqrt(m+n)=1.

The proof by induction was correct, but here is a simpler proof, actually this is the exact same proof but I''ll rewrite it to make it a bit clearer:

Let k,n be positive integers.
Then a(kn)=a(n+(n+...+n))<=a(n)+a(n+...+n)
(k-1 times) (k-1) times

=a(n)+a(n+(n+...n))<=a(n)+a(n)+a(n+...n)
(k-2 times) (k-2 times)

etc...
<=a(n)+...+a(n)=na(k), i.e. a(kn)<=ka(n).
(k times)


#10 Anonymous Poster_Anonymous Poster_*   Guests   -  Reputation:

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Posted 28 June 2001 - 05:06 AM

Doh!
Line 4 above should read:
sqrt(m+n)=1.

#11 CarlFGauss   Members   -  Reputation: 122

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Posted 28 June 2001 - 05:09 AM

thnx for the reply

#12 grhodes_at_work   Moderators   -  Reputation: 1361

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Posted 28 June 2001 - 07:36 AM

quote:
Original post by Anonymous Poster
a(x)=2-x is not linear. Any linear function has a(0)=0.



It is indeed a linear function of x, since x is not raised to any power (except to the power of 1), and the coefficient of x is a constant (coefficient = 1).

If linear functions always satisfied a(0) = 0, then all lines would pass through the origin, which clearly isn''t true.


Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

#13 Graylien   Members   -  Reputation: 160

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Posted 28 June 2001 - 03:07 PM

Beware anonymous cowards who post silly mathematical comments. I speak from experience here. I was once dragged into a pointless debate in the GDNet Lounge that ended up turning into a flame war because anonymous posters tend to be good trollers. (The debate was whether or not the infinite decimal 0.9999... is equal to 1).

#14 grhodes_at_work   Moderators   -  Reputation: 1361

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Posted 29 June 2001 - 02:29 AM

Good point and thanks for the advice Graylien!

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

#15 Anonymous Poster_Anonymous Poster_*   Guests   -  Reputation:

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Posted 29 June 2001 - 01:28 PM

Hey grhodes_at_work, I am the anonymous poster again.
Since you are a senior scientist I thought you would be smarter than to make your erroneous correction to my previous post.

A linear function is not defined as a line. A linear function, in the most common definition, is a function f:V->V, where V is a vector space, such that
f(x+y)=f(x)+f(y) for all x,y in V.
This means that f(0)=f(0+0)=f(0)+f(0), i.e. that f(0)=0.

Just because the function a(x)=2-x is not a linear function does not mean that the graph of a(x) is not a line. These are two completely different things. a(x) here is an affine function, as I said before, and the graph of all affine functions are lines.






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