# For any natural number, prove logarithmic result?

• May 21st 2009, 09:51 AM
fardeen_gen
For any natural number, prove logarithmic result?
For any natural number $\displaystyle n$, prove that:
$\displaystyle \frac{\log_{10} \left(1 + n\sqrt{2^{n - 1}}\right)}{n}\geq \log_{10} 2$

The question is from my sequences and series text.
• May 21st 2009, 08:08 PM
I-Think
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
$\displaystyle \frac{\log_{10} \left(1 + n\sqrt{2^{n - 1}}\right)}{n}\leq \log_{10} 2$
Test at values of $\displaystyle n>1$ 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
$\displaystyle 2^n\geq \ 1+n\sqrt{2^{n-1}}$

When we test $\displaystyle n=1$, we see that equality holds.
Test$\displaystyle n=2$and $\displaystyle n=3$and we see that the inequality holds. Assuming that this pattern will hold true for all natural numbers greater than $\displaystyle 1$
We now wish to prove

$\displaystyle 2^{n+1}>1+(n+1)(\sqrt{2^n})$
Let $\displaystyle p^2=2^n$
So we have
$\displaystyle 2p^2>(n+1)p+1$

With a little algebraic manipulation we shall now prove
$\displaystyle 2p^2-(n+1)p-1>0$

Assume that this is not true, i.e.
$\displaystyle 2p^2-(n+1)p-1<0$

Because the coefficient of $\displaystyle p^2$ 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 $\displaystyle \sqrt{b^2-4ac}<0$
Inserting values:
$\displaystyle \sqrt{(n+1)^2-4(2)(-1)}<0$
$\displaystyle \sqrt{n^2+2n+9}<0$

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

$\displaystyle n=\frac{-2\pm\sqrt{2^2-4(1)(9)}}{2}$

This value of $\displaystyle n$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 $\displaystyle 2p^2-(n+1)p-1>0$ must hold true for all natural numbers greater than $\displaystyle 1$.
Equality holds at $\displaystyle n=1$.

Hence, our original statement
$\displaystyle \frac{\log_{10} \left(1 + n\sqrt{2^{n - 1}}\right)}{n}\leq \log_{10} 2$
is true for all $\displaystyle n$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.

Hope this was useful
• May 21st 2009, 08:32 PM
fardeen_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, 07:31 AM
Isomorphism
Quote:

Originally Posted by fardeen_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?

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?

@I-Think: Writing it as a quadratic expression is a no-no(Shake)
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, 12:01 PM
I-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 $\displaystyle n=1$
So we shall prove that $\displaystyle 2^n>1+n\sqrt{2^{n-1}}$

Test $\displaystyle n=2$, we have: $\displaystyle 2^2>1+2\sqrt{2}$

Assume true for $\displaystyle n=k$: $\displaystyle 2^k>1+k\sqrt{2^{k-1}}$
If our assumption then : $\displaystyle 2^{k+1}>1+(k+1)\sqrt{2^k}$
From our assumption, by multiplying by $\displaystyle 2$we see that
$\displaystyle 2^{k+1}>2+2k\sqrt{2^{k-1}}$

Via algebra, we shall show that $\displaystyle 2+2k\sqrt{2^{k-1}}>1+(k+1)\sqrt{2^k}$
Subtract $\displaystyle 1+(k+1)\sqrt{k}$ from $\displaystyle 2+2k\sqrt{2^{k-1}}$
$\displaystyle (2+2k\sqrt{2^{k-1}}-(1+(k+1)\sqrt{2^k})$
$\displaystyle 2+2k\sqrt{2^{k-1}}-1-(k+1)\sqrt{2^k}$
$\displaystyle 1+2k\sqrt{2^{k-1}}-k\sqrt{2^k}-\sqrt{2^k}$
$\displaystyle 1+k(2\sqrt{2^{k-1}}-\sqrt{2^k})-\sqrt{2^k}$
$\displaystyle 1+k(\sqrt{2}\sqrt{2^k}-\sqrt{2^k})-\sqrt{2^k}$
$\displaystyle 1+k(\sqrt{2^k}(\sqrt{2}-1))-\sqrt{2^k}$
This statement is positive for all natural numbers $\displaystyle k>2$

Hence $\displaystyle 2+2k\sqrt{2^{k-1}}>1+(k+1)\sqrt{2^k}$
And if $\displaystyle 2^{k+1}>2+2k\sqrt{2^{k-1}}$
Then $\displaystyle 2^{k+1}>1+(k+1)\sqrt{2^k}$

And $\displaystyle 2^n>1+n\sqrt{2^{n-1}}$

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, 12:50 PM
Isomorphism
Quote:

Originally Posted by I-Think
Then $\displaystyle 2^{k+1}>1+(k+1)\sqrt{k}$

And $\displaystyle 2^n>1+n\sqrt{2^{n-1}}$

End of solution

This should be better, isomorphism?
So hopefully I won't have to learn to swim with rocks tied around my neck...

Hmm I wonder.. how does $\displaystyle 2^{k+1}>1+(k+1)\sqrt{k}$ imply $\displaystyle 2^n>1+n\sqrt{2^{n-1}}$

Quote:

Originally Posted by I-Think
Assume true for $\displaystyle n=k$: $\displaystyle 2^k>1+k\sqrt{2^{k-1}}$
If our assumption then : $\displaystyle 2^{k+1}>1+(k+1)\sqrt{k}$

The induction hypothesis should be $\displaystyle 2^{k+1}>1+(k+1)\sqrt{2^k}$ and not $\displaystyle 2^{k+1}>1+(k+1)\sqrt{k}$
• May 26th 2009, 02:44 PM
I-Think
Aaaah! More apologies necessary. Wherever there's $\displaystyle \sqrt{k}$, it's supposed to be $\displaystyle \sqrt{2^k}$, I'll edit it .

Quote:

Originally Posted by Isomorphism
Hmm I wonder.. how does $\displaystyle 2^{k+1}>1+(k+1)\sqrt{k}$ imply $\displaystyle 2^n>1+n\sqrt{2^{n-1}}$

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

Hence $\displaystyle 2+2k\sqrt{2^{k-1}}>1+(k+1)\sqrt{2^k}$
And if$\displaystyle 2^{k+1}>2+2k\sqrt{2^{k-1}}$
Then $\displaystyle 2^{k+1}>2+2k\sqrt{2^{k-1}}>1+(k+1)\sqrt{2^k}$
Therefore $\displaystyle 2^{k+1}>1+(k+1)\sqrt{2^k}$
We have shown that $\displaystyle 2^{k+1}>1+(k+1)\sqrt{2^k}$is true when $\displaystyle 2^k>1+k\sqrt{2^{k-1}}$ is true. And if$\displaystyle k=2$ is true, then by the Principle of Mathematical Induction, $\displaystyle 2^n>1+n\sqrt{2^{n-1}}$ is true for all natural numbers $\displaystyle n>2$
And our original statement
$\displaystyle \frac{\log_{10} \left(1 + n\sqrt{2^{n - 1}}\right)}{n}\leq \log_{10} 2$
is true.

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