For any natural number , prove that:

The question is from my sequences and series text.

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- May 21st 2009, 10:51 AMfardeen_genFor any natural number, prove logarithmic result?
For any natural number , prove that:

The question is from my sequences and series text. - May 21st 2009, 09:08 PMI-ThinkPossible Answer
Good night, fardeen gen. (Or good afternoon, evening, morning, I'm not sure what time of day it is in India)

First off, I believe that you have copied the equation wrong. It should be

Test at values of you should see that you need to flip the sign (unless I'm wrong)

Onto the solution

Solution

Step 1

Cross-multiply and then remove logs. The equation becomes

When we test , we see that equality holds.

Test and and we see that the inequality holds. Assuming that this pattern will hold true for all natural numbers greater than

We now wish to prove

Let

So we have

With a little algebraic manipulation we shall now prove

Assume that this is not true, i.e.

Because the coefficient of is positive, the curve has a max. turning point and because of our assumption it means that the entire curve is below the x-axis, i.e. it has no real roots.

For this equation to have no real roots the discriminant

Inserting values:

To determine the values for n under which our assumption holds, we use the quadratic formula

This value of is complex, however the question asks us to prove this statement of natural numbers.

Because our assumption holds true only when n is complex, we must reject our assumption due to question constrains.

Therefore, the statement must hold true for all natural numbers greater than .

Equality holds at .

Hence, our original statement

is true for all natural numbers.

End of solution

I'm not sure how valid, so I (and fardeen_gen) would appreciate it if a more experienced forum member critique this solution

Also, I would appreciate it if someone could critique my method of proof writing and give me tips on how to better my proof-writing skills.

Merci beaucoup in advance.

Hope this was useful - May 21st 2009, 09:32 PMfardeen_gen
Its morning here :) And Good morning/evening/night to you too :D

I feel that the proof is in order(a bit unconventional though). I copied the question correctly. Printer's Devil for the book in all probability.

Any other ideas, anyone? - May 26th 2009, 08:31 AMIsomorphism
You might have copied the question correctly, but in problems like these, one can check the validity by substituting values. So assume you want to do it by induction. Lets start inducting from n=2.... What do you see?

Writing it as a quadratic expression is a no-no(Shake)**@I-Think:**

Because in a quadratic expression the coefficients are constants. Here your co-efficient of p is depending on n and p itself is again dependent on 'n'. Thus its not a quadratic expression... - May 26th 2009, 01:01 PMI-Think
My most humble apologies for leading innocents astray. May a millstone be tied around my neck and I be thrown into the sea...

But before we get too carried away, I present as a sign of my remorse another solution. Take and critique, hopefully I won't have to walk the plank now.

Solution:

It has already been shown that equality holds when

So we shall prove that

Test , we have:

Assume true for :

If our assumption then :

From our assumption, by multiplying by we see that

Via algebra, we shall show that

Subtract from

This statement is positive for all natural numbers

Hence

And if

Then

And

End of solution

This should be better, isomorphism?

So hopefully I won't have to learn to swim with rocks tied around my neck... - May 26th 2009, 01:50 PMIsomorphism
- May 26th 2009, 03:44 PMI-Think
Aaaah! More apologies necessary. Wherever there's , it's supposed to be , I'll edit it .

More editing to be done. I'll show them here. The final statements are supposed to be

Hence

And if

Then

Therefore

We have shown that is true when is true. And if is true, then by the Principle of Mathematical Induction, is true for all natural numbers

And our original statement

is true.

Am I safe now? I see sharks circling...