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        riddles >> putnam exam (pure math) >> natural logarithms and translations
        
(Message started by: Sjoerd Job Postmus on Apr 3^rd, 2005, 5:02am)

Title: natural logarithms and translations
Post by Sjoerd Job Postmus on Apr 3^rd, 2005, 5:02am

I hope this is the right topic. It's a purely mathematical question, so putnam seemed the right topic.
---Question:---
Ok, a few days ago, I was taking a test in Maths.

I found the following question (in my native language though).

You have formula f(x) = x ln x
When you multiply by 1/2 in regards to the y axle, you get formula g(x). This way, I believe we would get g(x) = 2x ln 2x

Now, according to the test, I could also write g(x) as a translation (a,b) of f(x).

g(x) = ((x-a) ln (x-a)) + b

Now I need to find the values a and b.

I couldn't answer it myself, and now I believe it is impossible. By the way, don't worry about me getting the answer from you and in turn solving it in time to get a higher mark for the test. That's not possible.
In contrast, if it can not be done, my mark would rise, as well as the marks of the other members of my class: If the teacher asks something that can not be done (with our current set of expertise), it's a false question and doesn't get counted anymore in the numbering.

Could you please tell me if it's possible or not, and if so, how you arrived to the answer.

Thank you in advance,
Sjoerd Job.

Title: Re: natural logarithms and translations
Post by Barukh on Apr 3^rd, 2005, 6:41am

Sjoerd,

on 04/03/05 at 05:02:10, Sjoerd Job Postmus wrote:

When you multiply by 1/2 in regards to the y axle, you get formula g(x). This way, I believe we would get g(x) = 2x ln 2x

I don't understand this. Doesn't your first statement mean that g(x) = 1/2 f(x) ?

Title: Re: natural logarithms and translations
Post by Sjoerd Job Postmus on Apr 3^rd, 2005, 6:53am

on 04/03/05 at 06:41:53, Barukh wrote:

Sjoerd,

I don't understand this. Doesn't your first statement mean that g(x) = 1/2 f(x) ?

No... that would be multiplication of .5 in regards to the x-axle. I talked about a multiplication of .5 in regards to the y-axle. For the line

y=x
You would get
.5y = x
y = 2x

So when multiplying in regards to the y-axle with a constant 'c', every 'x' gets replaced with 'x/c'...

Hope that helps a bit...

Title: Re: natural logarithms and translations
Post by towr on Apr 3^rd, 2005, 7:40am

I don't think so..

given y = f(x)
If you scale the y-axis by a factor A, then everything on the right hand side get's multiplied by A.
Or y= A*f(X)
Take f.i. y=2 if you want to shrink the y-axis by half, you want to get y=1 (which is 0.5*1)

If you scale the x-axis by B, the every occurence of x on the right hand side becomes (x/B)
Or y = f(x/B)
Take f.i. y=0 if x =/= 2 and y=1 if x==2
then if you scale the x by 1/2
you want to get y=0 if x =/= 1 and y=1 if x==1

In any case, a vector translation won't ever scale either axis. (Except when y or x are constant)

Title: Re: natural logarithms and translations
Post by Sjoerd Job Postmus on Apr 3^rd, 2005, 7:58am

What I mean in correspondance to 'multiplying in regards to the y-axle with c' is that if you have a line, every dot is c times as far from the y-axle as before.

take line "blue", for instance. Now I multiply in regards to the y-axle, with result line "red". (At least, that's what I've learned at school...)
---
In this case, we have f: y = x ln x
Multiply by .5
y = x ln x
y = 2x * ln 2x (this is what school taught me...)
When plotting on a graphical calculator, it seems that if I draw a line x = c ( constant, horizontal line ), and I label intersection f(x) with x=c 'A', and intersection with g(x) 'B', it seems that d(y=0,A) = 2d(y=0,B), which is what I want...

So the formula 2x ln 2x is correct = from what I know... But, I need to write it as (x+a) ln (x+a) + b (normal order of operators)...
That would be equiv to moving f(x) a to the left, and b upwards ( negative a, means abs(a) to the right, and negative b means abs(b) downwards, ofcourse )...

Title: Re: natural logarithms and translations
Post by towr on Apr 3^rd, 2005, 8:46am

on 04/03/05 at 07:58:01, Sjoerd Job Postmus wrote:

What I mean in correspondance to 'multiplying in regards to the y-axle with c' is that if you have a line, every dot is c times as far from the y-axle as before.

Ah, ok.. then it makes sense..

I suppose you could hack it, and take translation (x,0)
so (x+a) ln (x+a) + b = (x+x) ln (x+x) + 0 = 2x ln (2x)
(But that really isn't an translation)

If you only had y=ln x, then it wouldn't be hard
ln 2x = ln x + ln 2
So (0, ln 2) would do it (contradicting the last line of my last post; but we all make mistakes sometimes). But with x ln x I'm pretty sure it's impossible.

Title: Re: natural logarithms and translations
Post by Sjoerd Job Postmus on Apr 3^rd, 2005, 8:53am

on 04/03/05 at 08:46:10, towr wrote:

Yeah, with ln 2x, it wouldn't be too hard... but, I'm pretty sure that with x ln x => 2x ln 2x, it isn't...

Hope my teacher (5Havo NT(dutch system) ~ secondary school, targetted at Nature & Technology specific studies... before Calculus, I think...) can't solve it either, so I get higher mark!

Title: Re: natural logarithms and translations
Post by Barukh on Apr 3^rd, 2005, 9:15am

Sjoerd, you mentioned pre-Calculus, so using derivatives is not permitted probably, but let's try.

If g(x) = 2x ln(2x), h(x) = (x+a) ln(x+a) + b, then g(x) = h(x) implies all derivatives are also equal; so let us take the second derivative:

g''(x) = 2/x, h''(x) = 1/(x+a),
and g''(x) = h''(x) is not satisfied by any constant a.

Title: Re: natural logarithms and translations
Post by Sjoerd Job Postmus on Apr 3^rd, 2005, 9:30am

on 04/03/05 at 09:15:01, Barukh wrote:

Wasn't sure if derivaties was calculus already... so perhaps I have had some basic calculus. We do know derivs.

g(x) = 2x ln 2x and h(x) = (x+a) ln (x+a) + b

g'(x) = 2 ln 2x + 2x / x ( I think ) = 2 ln 2x + 2
g''(x) = [2 ln 2x + 2]' = [2 ln 2x]' = 2/x

g''(x) = h''(x) = 2/x = 1/(x+a) impossible...

Good to hear it's impossible after all...