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## #26 2006-03-07 01:32:57

ashwil
Member
Registered: 2006-02-27
Posts: 121

### Re: *** Problems

***5

Ok, here goes, maybe a little clumsy and probably not 100% proof, but here goes:

for values of a,b,c greater than or equal to 1:-

If a, b & c all = 1, then (a + 1)^7 * (b+1)^7 * (c+1)^7 = 2,097,152 and 7^7* a^4 * b^4 * c^4 = 823,543
any increase in a,b or c can be viewed simply as a comparison between the functions (a+1)^7 and a^4. For all positive numbers greater than 1, this only increases the divergence. Therefore, it is only necessary to consider values of a,b,c between 0 & 1.

As a,b,c tend to zero, [(a+1)^7....] tends to an minimum value of 1, while [7^7.....] tends to zero. Again, as a,b,c increase above zero, the effect on the comparison is the same as for values above one. Namely, any increase in (a+1)^7 will be greater than the corresponding increase in a^4.

Obviously, negative values would reverse this, but for all positive values, the expression must be true, though I cannot actually see a point at which the expressions could ever be equal!

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## #27 2006-03-07 04:33:47

ganesh
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Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

ashwil, I shall post the solution to ***3 tomorrow. I was occupied with the other topics today. (Particularly, my post on quotes of mathematicians in 'Members only').

The solution to ***5, as you admitted, is not a 100% proof. You have considered certain values, and I think your reasoning is good, although it does not constitute a mathematical proof. I shall post the proof by the weekend.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #28 2006-03-07 06:53:10

ashwil
Member
Registered: 2006-02-27
Posts: 121

### Re: *** Problems

I'll settle for my reasoning being good. When you don't spend your time actually doing mathematical proofs, you do forget the notation, the methodology and the formulae, but reasoning powers can still get you a long way!

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## #29 2006-03-07 07:24:15

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: *** Problems

***5

To simplify it, I'm going to represent a, b, and c by x, because you have to repeat this three times (once for each letter).

Thus, it becomes apparent we must show:

Where

If x < 1, then

Since all of the negative powers of x must be greater than or equal to 1.

The same reasoning goes for x > 2,

But I can't seem to get the numbers in between 1 and 2.

Edit:

How's this for an argument?

The graph

is continuous, above, and never intersects with 94 for all positive values x less than 2.  Thus, it must always be greater there.

Last edited by Ricky (2006-03-07 07:36:19)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #30 2006-03-07 07:36:52

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

Logaritms?

IPBLE:  Increasing Performance By Lowering Expectations.

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## #31 2006-03-07 07:47:38

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

Can someone do this:
7(ln(a + 1) + ln(b + 1) + ln(c + 1) - ln 7) ≥ 4ln abc

IPBLE:  Increasing Performance By Lowering Expectations.

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## #32 2006-03-07 08:00:57

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

Well done, Ricky!
This was great idea!
Your function has local minimum between 1 and 2 at x=4/3.

IPBLE:  Increasing Performance By Lowering Expectations.

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## #33 2006-03-07 08:21:51

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

***5
Something stinks here.
And where's the ricky's post?

IPBLE:  Increasing Performance By Lowering Expectations.

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## #34 2006-03-07 08:23:24

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

Let

(ricky's function)

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## #35 2006-03-07 08:25:00

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

Here's a plot:

Last edited by krassi_holmz (2006-03-07 08:25:44)

IPBLE:  Increasing Performance By Lowering Expectations.

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## #36 2006-03-07 08:28:51

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

We need to find the minimum for x>0 (actually we don't need the other side because a,b,c are positive).
How?
Calculus, of course!:

Last edited by krassi_holmz (2006-03-07 08:29:21)

IPBLE:  Increasing Performance By Lowering Expectations.

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## #37 2006-03-07 08:40:39

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

The main thing is that f'(x) is factorizible:

So clearly the roots of f'(x)=0 are x=-1 or x=4/3.
leaving -1. So f(x) has minimum at x=4/3!!!
And here's where the stinky comes:
f(4/3)=823543/6912=119.147...
So for ALL x>0 f(x)>=119.147...!!!
But then
f(a)f(b)f(c)>=119.147...^3=558545864083284007 / 330225942528=1691405.16...
But the main question was to prove that
f(a)f(b)f(c)>=823543 {/*this is 7^7*/}
And ricky an I got that:
f(a)f(b)f(c)>=1691405.16
and the minimum is at {a,b,c}={4/3,4/3,4/3}.
Am I wrong?

IPBLE:  Increasing Performance By Lowering Expectations.

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## #38 2006-03-07 19:17:19

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

Ganesh, please reply.

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## #39 2006-03-07 19:25:37

ganesh
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Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

Sure, krassi_holmz. The problem is being approached in a much different way than I expected. The solution I had to the problem didn't involve differentiation and maximum/minimum. I shall wait for other responses, particularly from mathsyperson/irspow/John. Please wait for the solution for a day more.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #40 2006-03-07 19:43:40

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

But I'm confused about this: f(a)f(b)f(c)>=7^7
Are there a,b,c for which f(a)f(b)f(c)==7^7?
If there are, so I have a mistake.

Last edited by krassi_holmz (2006-03-08 06:02:48)

IPBLE:  Increasing Performance By Lowering Expectations.

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## #41 2006-03-08 03:34:22

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

Solution to ***3

Consider the series 3² /1! + 5² /3! + 7² /5! +..........
the nth term = (2n+1)² /(2n-1)! = (4n² +4n+1)/(2n-1)!
= [(2n-1)(2n-2)+10n-1]/(2n-1)! = 1/(2n-3)! + (10n-1)/(2n-1)!
= 1/(2n-3)! + 5. 1/(2n-2)! + 4. 1/(2n-1)!
Sum to n terms would be

or
5e (on simplification).
Since we started the series as 3² /1! + 5² /3! + 7² /5! +..........,
1 should be added to the sum.

Hence the sum to infinity is 1+5e.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #42 2006-03-08 03:55:35

ashwil
Member
Registered: 2006-02-27
Posts: 121

### Re: *** Problems

Many thanks. I had forgotten the technique of considering the nth term. All now clear.

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## #43 2006-03-08 16:10:32

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

***6

Prove that

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #44 2006-03-08 19:14:17

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: *** Problems

***6 beautiful!

so:

Let reduce the left side:

Last edited by krassi_holmz (2006-03-08 19:23:57)

IPBLE:  Increasing Performance By Lowering Expectations.

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## #45 2006-03-09 03:20:12

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #46 2006-03-09 19:22:51

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

***7

Let

and

if z be a complex number such that

then prove that |z - 7 -9i| = 3√2.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #47 2006-03-10 17:43:21

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

***8

Given that p, q are the roots of the equation Ax²-4x+1=0 and r,s are the roots of the equation Bx²-6x-1=0, find the values of A and B if p, r, q, and s are in Harmonic Progression.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #48 2006-03-11 18:11:47

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

***9

Show that if a,b>0, then

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #49 2006-03-12 02:23:36

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

***10

Show that the curves

and
cut orthogonally if
.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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## #50 2006-03-15 04:24:56

ganesh
Moderator
Registered: 2005-06-28
Posts: 18,127

### Re: *** Problems

***11

Find the value of a such that the vectors

are coplanar.

The divergent series are the invention of the devil, and its a shame to base on them any demonstration whatsoever - Neils Henrik Abel.
Knowledge is of two kinds. We know a subject ourselves, or we know where we can find information upon it - Samuel Johnson.

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