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#126 Re: Help Me ! » I am useless at proving inequalities » 2008-08-09 21:26:41

uhm didnt work out hmm

(why do sometimes this forum post when i click preview??swear)

#128 Re: Puzzles and Games » Squares from sticks.. » 2008-07-08 02:03:35

How do you make four equilateral triangles with sides l, using 6 sticks with length l?

#129 Re: Puzzles and Games » Square » 2008-07-05 20:51:19

iirc this problem is considered one of the most difficult problems ever submitted in the IMO...:O

#130 Re: Puzzles and Games » integer and decimal product integral » 2008-06-27 01:37:46

Good work smile My solution was slightly different (with x'=u for easier reading):


substituting x with x=u+k, we have dx/du=1 -> dx=du, and if x=k=u+k, then u=0, and if x=k+1=u+k then u=1, so we get

#131 Puzzles and Games » integer and decimal product integral » 2008-06-25 06:39:47

Kurre
Replies: 2

Let [x] be the greatest integer less than x, and let x' be the fraction part of x, ie x'=x-[x]. Let n be any integer ≥2. Evaluate the following integral in terms of n:

#132 Re: Help Me ! » find x » 2008-06-15 06:45:49

Is 0° considered an acute angle? If so, that is another solution.

#134 Puzzles and Games » Sliding blocks » 2008-06-10 21:00:45

Kurre
Replies: 4

Triangle PQR with area A is given in a plane parallell to the gravitational acceleration g, with the length QR perpendicular to g, and the point P is above QR. Now we let two blocks (considered as points) slide down PQ and PR. This gives the different times t1 respective t2 for each block. If we assume there is no friction, find the minimum of T=t1+t2 and determine the triangle when this occurs.

#135 Re: Help Me ! » degree one » 2008-06-07 06:47:50

nlahr wrote:

Okay let's try this one more time. I need 3 quadraric equation examples with a, b, and c (coeffients of x^2, x and the constants as.
1 integers
2. rational numbers
3. irrational numbers
Thank You

The question is just to find any quadratic where a,b and c are irrational. So you just need to chosse any three irrational numbers. Do you know what an irrational number is?
http://mathworld.wolfram.com/IrrationalNumber.html

For example you can choose a=√2, b=pi. I leave for you to find a third irrational number for c.

#136 Re: Help Me ! » Modulus Inequality » 2008-06-07 05:06:34





Now letting y=x+1 we get the desired inequality.

edit: about your siolution, instead of setting equality between the two functions to see where they meet, keep the inequality sign and consider the same two cases.

#137 Puzzles and Games » Google US Puzzle Championship » 2008-05-29 05:46:26

Kurre
Replies: 0

I just want to tip you about this upcoming event smile
http://wpc.puzzles.com/
It's an internet based competition in logic puzzles.
How many here are going to participate?? smile

#138 Re: Exercises » Jane’s exercises II » 2008-05-25 05:25:29

smile
Its also possible to use the rearrangement inequality directly on the first inequality. However, I spotted an error on my solution to the second one:

#140 Re: Puzzles and Games » Jane’s puzzles » 2008-05-25 03:43:09

Still dont get it, doesnt the puzzle clearly state that the bar must be cut into 11 pieces??
Or is it my english that is failing? neutral

#142 Puzzles and Games » Two inequalities » 2008-05-19 06:11:04

Kurre
Replies: 0

Just two inequalities: smile

1. let p>1 be a real number, and given a sequence {ak} of n positive numbers whose sum is n, prove that:

2. let α,β,γ,δ be the angles of a quadrilateral. Prove that:

#143 Re: Help Me ! » Inequality a^p>b^p » 2008-05-17 20:47:20

Ricky wrote:

Is this question learning on your own, or is it part of a class?  If it's part of a class, then there should be some other way to do it.  How much have you covered in terms of real analysis (sometimes called "Advanced Calculus", which is a horrible name)?

Yes, im doing it on my own, and I dont think I have been doing a lot real analysis. Just the basic high school(?) calculus, like basic integration (substitution/parts), basic differential equations etc. In class I have not done anything close to investigating the definition of a limit or definition of a convergent serie hmm

#144 Re: Help Me ! » Logarithms » 2008-05-17 20:44:05

In both questions, use that logx+logy=logxy, logx-logy=log(x/y)

#145 Re: Help Me ! » Inequality a^p>b^p » 2008-05-17 09:58:05

Thanks for the reply Ricky!

Ricky wrote:

You need to use the definition of a convergent sequence here.

Unfortunately I have never worked with definitions of convergent series before, but I can give it a try. Could you please give a short startup/link to get me going?? smile

Ricky wrote:

Note that since a > b, a/b > 1, so (a/b)^p > 1 which is really a^p / b^p > 1.  Multiply both sides by b^p and you get a^p > b^p.

Now here is your challenge: why can't you use this?

I think that is the wrong bit, since wouldnt stating that "a/b>1 implies that (a/b)^p>1^p=1", use the theorem that we are trying to prove?

#146 Help Me ! » Inequality a^p>b^p » 2008-05-17 00:51:07

Kurre
Replies: 5

It struck me when reading about the axioms of inequalities, that this inequality wasnt that easy to straight ahead deduce from the axioms.
the inequality to be proven is given p>0, a>b>0, prove that a^p>b^p.
the axioms are:
1. for two real numbers x and y exactly one of the following cases i true:
a) x=y b) x<y c)x>y
2. if x<y, then for all z x+z<y+z
3. x<y, z>0, then xz<yz
4. if x<y and y<z, then x<z
proof:
lemma: for x,y,z,w>0, if x<y, w<z, then xw<yz.
By axiom 2 we have xw<yw, and again by axiom 2  wy<yz, and by axiom 4 xw<yz. Now assume p is natural, then for p=1 the inequaliy holds. Assume it holds for p=n, ie a^n>b^n. then by the lemma we can multiply left side with a, and right side with b and establishing a^(n+1)>b^(n+1), thus by the induction principle it holds for all p

N. It also follow by axiom 1, that if a^p>b^p for p[tex]\in[/tex]N, then a>b. Now write:

ad by earlier result, we get that a^p>b^p for rational p.
Now, let the sequence {p_k} of rational numbers be the numbers of the decimal expansion of the real number p, ie
to n decimal places.
thus we have:

thus a>b implies that a^p>b^p for all and p,a,b>0.

Now my questions are:
Is this proof correct? Im mostly concerned about the last step when introducing a limit. Do I need to state anything about continuity of the function a^p? Or is it sufficient to use the proved lemma, thus I can continue to multiply the inequality with a^(pk)>b^(pk) and continue to infinity to establish the inequality.
Also, thus the introduction of a limit need extra proofs of the limit, if I want to deduce the inequality straight down from the axioms?
I know that I can just state the the function a^x is strictly increasing for a,x>0 and thus a>b -> a^x>b^x, but that does need extra proofs conecrning derivatives etc, right? (and its not so much fun smile)
Or have I missed a really obvious way to prove this kind of intuitive inequality?rolleyes:
and last, a challenge for you, find a simplier proof. smile

#147 Re: Help Me ! » Circle Theorems » 2008-05-16 23:30:35

Hint: start by proving this theorem:
A circle and a point A is given. Let a line through A intersect the circle in points B and C, and draw a line PD, such that PD is a tangent to the circle with tangent point D. Then |AB|*|AC|=|AD|².
Your problem will follow quite easily from this theorem. smile

#148 Re: Exercises » Jane’s exercises II » 2008-05-16 01:42:30

#7


integration gives

same as

with solutions:

nice problem btw smile

#149 Re: Exercises » Jane’s exercises II » 2008-05-16 01:20:35

Using the inequality between arithmetic and geometric metdium yields:



equality holds if and only if ab=2bc=4ca

letting ab=1/c, bc=1/a, ca=1/b, yields
1/c=2/a=4/b -> b=2a and c=a/2
abc=1 ->a*2a*a/2=1 -> a=1, which gives that b=2, c=1/2

#150 Re: Help Me ! » Isosceles Triangle » 2008-05-14 08:29:52

It is possible to solve it exactly too.
as mathsy said, the area is a²sin150/2=9 -> a=√(18/sin150)=6
the ramaining side is 2*sin75*a (drawing the height gives two right triangles), so the perimeter is p=2*sin75*a +a +a = 2a(sin75+1)
but


so

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