triclino

prove that the following sequence converges:

[math]x_{k+2} =\frac{2}{x_{k+1} + x_{k}} ,x_{1},x_{2}>0[/math]

So if somebody is sick and needs help he/she must do not go to the Doctor but try to fix him/her self

Thread moved to Homework Help, where it belongs. Triclino, you're very lucky our experts feed you the answers with a spoon; but I remind you that the proper action to do in cases where you get stuck on a question is to ask for help, rather than order people around.

 

Asking for help includes showing what you've already tried and failed, and why you think you need help. We aren't in the habit of feeding the answers back, or acting as online calculators. Please take that into account next time.

 

~moo

Mooeypoo if i was to ask your experts, for example, for a formal proof of a high school question they would loose their spoon .

There are a lot of things that your experts do not know.

If for example i was to ask your experts for a formal proof in Analysis probably i would wait centuries for an answer .

Boasting around is not a good thing.

I can even ask questions in Analysis (not formal ) that your experts wont know the answer.

So if you are so sure about your experts let us try .

I am not ordering anybody around.

Yes that substitution led us to infinite solutions

You don't need a proof for everything, triclino. Suppose a mathematician, after some arduous calculations in a math paper, arrived at

 

[math]x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2 = 0[/math]

 

Nobody would blink an eye if the next sentence in the paper was "and thus the only solution in the reals is the trivial solution."

And if he was asked to prove it ,then he would realize that it is an impossible task

Can you prove it??

inequalities like this are countless .The question is, is there a general way to tackle them??

Solve the following inequality:

[math] xy+(xy)^2 +(xy)^3>14[/math]

needless to say i have no idea how even to start this inequality

So i take it that you cannot formally analyze your own proof??

If i were to prove that your proof is not correct we will have to formally analyze it

But can you formally analyze your own proof??

For a start ,if i were to say that your method 2 is wrong what would you say??

For case 3, let [math]|a| > |b|[/math].

 

[math]-ab < a^2[/math], correct? Both are positive, and [math]|b| < |a|[/math], so it must be true.

 

You can do it likewise for [math]|b| > |a|[/math], and [math]|b|=|a|[/math] is trivial.

 

This should let you complete the proof.

.................................WRONG.....................................................

This proof is completely wrong .Go over your proof again .Ask a teacher a Doctor ,anyone . This proof is wrong

No. I want to see you try it.

 

First try it for the case where a and b are both greater than 0. That should be trivial.

O.k teacher here we go:

case 1 : ab>0 ,then -ab<0 and since [math]0\leq a^2+b^2[/math] we have:

[math] -ab\leq a^2+b^2[/math]

case 2 : ab=0 ,then -ab=0 and since [math]0\leq a^2+b^2[/math] we have:

[math] -ab\leq a^2+b^2[/math]

case 3 : ab<0 ,then -ab>0. And here is the problem. Now we cannot say :

[math] -ab\leq a^2+b^2[/math].

Because if you have two Nos x>0 and y>0 you cannot say x>y

No, No you curry on and finish the proof because i want stop posting in this forum

D.H, why you throw your problems on me .I think this is unfair.

Besides i accept advice and teaching from anybody as long as it is correct.

Perhaps your dedication to learning mathematics should be redirected to mastering your English?

And you should learn to read the whole thread before you get involved in a discussion and come into worthless and impolite conclusions

D H is not proving [math]a^2+ab+b^2\geq 0[/math]. He is stating that it is equivalent to [math](a+b)^2\geq ab[/math]. If you prove the latter, you have proven the former. Try proving the latter.

If you prove the latter i will stop posting in this forum.

The latter is unprovable (unless you do the same mistake with D.H to use [math]a^2+ab+b^2\geq 0[/math] in your proof)

It would be good if you paid attention to that valuable advice.

 

You do not have to assume [math]a^2+ab+b^2\ge 0[/math] to prove [math](a+b)^2\ge ab[/math]. These two statements say exactly the same thing. This means of course that if you do prove the former you have automatically proven the latter. However, if you can prove the latter you have automatically proven the former. I thought you might have an easier time proving the latter statement.

I suggest you go and read (perhaps in a book of logic) what is the double implication proof .Check the following example:

0x = 0 <=> 0x+x = 0+x <=> x(0+1) = 0 +x <=> x= 0+x <=> x=x

We want to prove 0x = 0 (for all real ,x) ,so by double implication we arrive at a well known fact x=x ,then we can except 0x=0 as true.

So to go from [math]a^2+ab+b^2\geq 0[/math] to [math] (a+b)^2\geq ab[/math] you do the following:

[math]a^2+ab+b^2\geq 0\Longleftrightarrow a^2+ab+b^2+ab\geq ab\Longleftrightarrow (a+b)^2\geq ab[/math]

Is [math](a+b)^2\geq ab[/math] a well known fact??

In post #7 i proved the inequality in concern .Read it if you like.

The hole proof is just a line .

I would like to comment on your proof ,but as usually the moderators

will intervene and give me valuable advice on how to behave e.t.c

Now,

on your comment that i have problems with proofs i can say the following:

To really say whether a proof is correct or not you must analyze the proof by writing a formal issue of it.

Then i will except you as capable of criticizing my proofs.

Can you write a formal proof of any high school theorem?

No, for two reasons.

 

1. This looks too much like homework. You need to figure this out on your own.

2. You already did it in post #9.

Now i can see you have problems with inequalities. In post # 9 i showed exactly the opposite.

I showed that if ab>0 then you cannot prove [math](a+b)^2\geq ab[/math]

Exactly.

 

Adding the same quantity to both sides of an inequality does not change the inequality.

Please, write down the inequality and the quantity that you add to both sides of the inequality

Oh, please.

 

[math]a^2+ab+b^2 = a^2 + 2ab + b^2 - ab = (a+b)^2 - ab[/math]

If ab>0 => -ab<0 .But[math] (a+b)^2 >0[/math] .How can you add these two inequalities to get :

[math](a+b)^2 -ab>0[/math] and consequently [math](a+b)^2>ab[/math] ??

There are no other solutions if x, y, and z are restricted to the reals.

How can you say that without a proof??

If you complete the square you should get

 

[math](a+b)^2 \ge ab[/math]

 

.

The method that i know from high school in completing the square is the following :

[math]a^2+ab+b^2=a^2+2a(\frac{b}{2})+\frac{b^2}{4}-\frac{b^2}{4}+b^2[/math] which is equal to:

[math] (a+\frac{b}{2})^2 +\frac{3b^2}{4}[/math]. Which is definitely greater than or equal to zero for all values of a and b.

On the other hand to get : [math] (a+b)^2\geq ab[/math] you must assume [math]a^2+ab+b^2\geq 0[/math] ,which is what you want to prove

Treating the left-hand expression as a quadratic in a, verify that the determinant is never positive.

If we treat the L.H.S as quadratic we have:

[math]a = \frac{-b+\sqrt{b^2-4b^2}}{2}= \frac{-b+\sqrt{3}bi}{2}[/math]

.............................or.......................................................

[math]a = \frac{-b-\sqrt{b^2-4b^2}}{2}= \frac{-b-\sqrt{3}bi}{2}[/math]

Which makes, a ,a complex No

Merged post follows:

Hint: You should be able to readily prove that this is the case when ab<0. The cases where ab=0 are even easier. That leaves ab>0.

 

Complete the square.

Yes, i completed the square and it works.

But what is all this about ab??

If x, y, and z are limited to the reals, what is the only solution to equation (2)?

D.H the solution x=y=z=0 is the solution to every system of equations that are equal to 0.

This i suppose is well known to every one of us .

Here we want solutions other than those equal to zero

Moved to homework help.

 

triclino, please tell us what you've tried and where you got stuck. We would love to help, but we're not in the habit of feeding answers with a spoon.

 

~moo

I do not know how even to start with this system of equations .

Do you .

If yes please show me.

On the other hand answers like the one given by shyvera is a clever way of saying i do not know