Inspired by Otto Bretscher

Calculus Level 3

Evaluate the nested logarithm above.

Use the natural logarithm, IE base $e$ .

Inspiration

4 1 2 Does not exist

1 solution

Calvin Lin Staff
Apr 11, 2015

The issue is that we have to consider how this limit is defined.

We start off with $a_ 1 = \log_e 4$ , and then define the recurrence

$a_{n+1} = \log_{a_n} 4$

Calculating this, we get that

$a_1 = 1.3863\ldots$ and $a_2 = 4.2442 \ldots$ , $a_3 = 0.9590\ldots$ , $a _ 4 = -33.1217 \ldots$ , which makes $a_ 5$ undefined.

Thus, the limit does not exist.

For those who tried to say that:

Let the expression by equal to $Y$ . We have $\log_Y 4 = Y$ , or that $Y ^ Y = 4$ and hence $Y = 2$ .

The issue is that what you have shown is merely "If a limit exists (and is equal to $Y$ ), then the limit is equal to 2". Thus, the crux of the matter is that the limit does not exist in this particular case.

@Calvin Lin Beautiful example of why we always have to check if the limit exists! I made my problem expecting to introduce people to something new, and I am happy to see thoughts and ideas stemming from this.

Seth Lovelace - 6 years, 2 months ago

I believe that we should start with $a_1 = \log_4 4$ instead of $a_1 = \log 4$ (which doesn't give any justification on why $10$ is chosen, instead of any other number like $e$ or $2$ ).

Ivan Koswara - 6 years, 2 months ago

Agreed that there was no justification. I added that in subsequently.

Calvin Lin Staff - 6 years, 2 months ago

But when you said a1=log4, why did you take base 10, if you would have taken it less than 4, it would have existed

Archit Boobna - 6 years, 2 months ago

Agreed that 10 is somewhat arbitrary. If we used base $e$ (which arguably makes more sense), then the value would still not exist.

Calvin Lin Staff - 6 years, 2 months ago

Listen you never established in the original question that you were looking for the limit of a function. Your question said evaluate the imbedded log. See my point? Not worth your or my time at this point. Thanks for your response.

Julian Branker - 6 years, 1 month ago

@Julian Branker – Mr. Branker, is the question clear now that the dots have been placed on the sides, with the "e" at the bottom? Together we are trying to find the best way to pose this question...

Otto Bretscher - 6 years, 1 month ago

@Calvin Lin I started with $a_1 = 2$ and ended up selecting 2 as the answer.

$\log_2 4 = 2$ .

Where did I miss the bus!!

Janardhanan Sivaramakrishnan - 6 years, 2 months ago

I have the same answer. If we let $log_{log_{log_{log_{log_{...}4}4}4}4}4=x$ then

$log_x4=x$

$x^x=4$ and x can only be equal 2.

Roman Frago - 6 years, 1 month ago

You only proved that if the limit exists then it is equal to $2$ . You need to prove that the limit indeed exists. In fact, it doesn't exist, if the starting base is anything other than $2$ . (There is a starting base; the expression is defined as the limit of the sequence defined by Calvin.)

Compare this with the problem $\sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}$ , where the limit exists whatever (nonnegative) value acts as the first term of the sequence.

Ivan Koswara - 6 years, 1 month ago

@Ivan Koswara – I know. That was my outright answer. If we let $x=log_44$ then

$...log_{log_{log_{log_{log_x4}4}4}4}4...$ is undefined...

Assuming that $log_{log_{log_{log_{log_{...}4}4}4}4}4=...log_{log_{log_{log_{log_44}4}4}4}4...$ is a valid mathematical representation.

Roman Frago - 6 years, 1 month ago

@Roman Frago – Whoops, I think I selected the wrong person to reply. (Brilliant should make more levels for nested comments.) But yes, that's the point.

Ivan Koswara - 6 years, 1 month ago

@Ivan Koswara – How do you know that the limit exists in the case of $\sqrt{x+\sqrt{x+...}}$ , where $x$ is positive?

Otto Bretscher - 6 years, 1 month ago

@Otto Bretscher – The sequence $a_{n+1} = \sqrt{x + a_n}$ , for whatever starting value that doesn't make the sequence to immediately terminate (read: for any $a_1 \ge -x$ ), has two possibilities: either

$a_n$ is increasing, but bounded above by $\frac{1 + \sqrt{4x+1}}{2}$ , or
$a_n$ is decreasing, but bounded below by $\frac{1 + \sqrt{4x+1}}{2}$

The above can be proven rather easily (for example, if $a_n < \frac{1 + \sqrt{4x+1}}{2}$ , show that $a_{n+1} > a_n$ and $a_{n+1} < \frac{1 + \sqrt{4x+1}}{2}$ , and so you can finish by induction; likewise for the second). Since it's monotonic and bounded, this suffices to show that the limit exists.

Ivan Koswara - 6 years, 1 month ago

@Ivan Koswara – Perfect! I just wanted to make sure that everybody understood...

I wonder why you leave $a_1$ arbitrary , though. It seems to me that the natural definition of the expression $\sqrt{x+\sqrt{x+...}}$ is as the limit of the sequence $\sqrt{x},\sqrt{x+\sqrt{x}}, \sqrt{x+\sqrt{x+\sqrt{x}}},...$ , which would lead us to $a_1=\sqrt{x}$ . In this very simple case, it does not matter, but as soon as you have several equilibria, it does.

Otto Bretscher - 6 years, 1 month ago

@Otto Bretscher – The first term of the sequence is "hidden" in the ellipsis, so actually there's no particular reason of choosing one over the other, although I do agree $\sqrt{x}$ is a reasonable candidate and "more likely". Similar with this problem (as it's originally posed); the "initial" base is hidden, and so there's no particular reason of choosing $e$ , $10$ , $2$ , $4$ , or any number over the other. However, in this case, only one number gives a limit, while in the original, a lot of numbers give the same limit. (See also $\sqrt{0 + \sqrt{0 + \sqrt{0 + \ldots}}}$ ; with the starting term $0$ gives limit $0$ , while any positive number gives limit $1$ instead.)

Ivan Koswara - 6 years, 1 month ago

@Ivan Koswara – I do not agree that the first term is hidden in $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ , although this seems to be a common (mis)understanding. Placing the ellipsis to the right indicates that we are to read the expression from left to right, starting with the initial term $\sqrt{x}$ . In other words, the expression is defined as the limit of $\sqrt{x}$ , $\sqrt{x+\sqrt{x}}$ , ... , if that limit exists.

I do agree with your analysis of the nested log problem... that's why I suggested to write the expression with the initial term $\log_e{4}$ at the bottom.

Otto Bretscher - 6 years, 1 month ago

hey, no one told me i could use a calculator for this one!

Neil Yabut - 6 years, 1 month ago

You don't need a calculator to do this! If you use that $\ln{2}<0.7$ , you can easily show that $a_2>4, a_3<1$ and $a_4<0$ , so that $a_5$ is undefined.

Otto Bretscher - 6 years, 1 month ago

Sir, can you please give an elaborate algebraic proof that the limit does not exist as I just found a1,a2,a3 and got the answer.

Sukrut Waghmare - 6 years ago

What is a_5?

Calvin Lin Staff - 6 years ago

Sir but that we are finding out arithmetically, but if instead of 4, there was other number the hoe can we prove existence of limit algebraically?

Sukrut Waghmare - 6 years ago

How did calculated and derived a5 is undefined?? Please elaborate...

Puneet Pinku - 5 years, 1 month ago

Do the calculations directly. List them out as I did above. What is $a_1, a_2, a_3, a_4, a_5$ ?

Calvin Lin Staff - 5 years, 1 month ago

I thinked so: Let it equal to a, then log base a to 4=a then a^a=4. It exist only 1 a so that a^a=4 a=2.I dont know where i die a mistake.

Anton Amirkhanov - 1 year, 9 months ago

Did you read the last sentence of the solution?

Calvin Lin Staff - 1 year, 9 months ago

Inspired by Otto Bretscher

1 solution

0 pending reports