Requesting guidance

[Agent Smith]

The graph is the result of a computation in google sheets. How should I describe the behavior of this graph? I'd say we see an "extremely rapid increase" and then plateauing of the curve. Is there more that can be said? Within the limits of google sheet's capabilities and my own the value tends towards (converges to?) 2.169797697... but I'm not sure.

 

[Dr.Peterson]

The graph appears to present strong evidence (but, of course, not proof) that your function approaches some number in the vicinity of 2.17 asymptotically; also, it is monotonically increasing (as far as we can see).

If I were to guess, it could possibly have the form A - Be^(Cx). For example,

​

But presumably you know either the equation or a process that produced it, so I don't need to guess. Proof of the limit might be either easy or difficult, depending on the details.

 

[mario99]

View attachment 38710
The graph is the result of a computation in google sheets. How should I describe the behavior of this graph? I'd say we see an "extremely rapid increase" and then plateauing of the curve. Is there more that can be said? Within the limits of google sheet's capabilities and my own the value tends towards (converges to?) 2.169797697... but I'm not sure.

I don't think so.
But if you are interested to represent this graph by an infinite series, here is one:

$\displaystyle \sum_{k = 1}^{1} 2^{k} + \sum_{k = 1}^{1} \frac{1}{10^k} + 69797\sum_{k = 1}^{\infty}\frac{1}{10^{5k + 1}}$

 

[fresh_42]

I don't think so.
But if you are interested to represent this graph by an infinite series, here is one:

$\displaystyle \sum_{k = 1}^{1} 2^{k} + \sum_{k = 1}^{1} \frac{1}{10^k} + 69797\sum_{k = 1}^{\infty}\frac{1}{10^{5k + 1}}$

What did you use to find that? Are the first two upper limits typos or did you mean only one term?

 

[mario99]

What did you use to find that?

Just a pencil and paper.

What did you use to find that? Are the first two upper limits typos or did you mean only one term?

Just one term. I was trying to write the first two terms in a fancy way. I think that they would be more fancy if I wrote them in this way:

$\displaystyle \sum_{k = 1}^{1}\frac{20^k + 1}{10^k}$

 

[fresh_42]

Just one term. I was trying to write the first two terms in a fancy way.

You mean such as $\displaystyle{\sum_{k=1}^\infty }\dfrac{1}{2^{k-1}}$ and $\displaystyle{\sum_{k=1}^\infty }\dfrac{1}{11^k}\;\;$?

(Sorry, I couldn't resist. )

 

[mario99]

You mean such as $\displaystyle{\sum_{k=1}^\infty }\dfrac{1}{2^{k-1}}$ and $\displaystyle{\sum_{k=1}^\infty }\dfrac{1}{11^k}\;\;$?

(Sorry, I couldn't resist. )

wOW!

This is even fancier than what I did. You are a Genius professor fresh_42.

 

[Agent Smith]

I don't think so.
But if you are interested to represent this graph by an infinite series, here is one:

$\displaystyle \sum_{k = 1}^{1} 2^{k} + \sum_{k = 1}^{1} \frac{1}{10^k} + 69797\sum_{k = 1}^{\infty}\frac{1}{10^{5k + 1}}$

I've been following the conversation and I wonder if you could simplify that to one simple summation/etc. Doe it sum to 2.169797697...? And, have you seen this number 2.169797697... anywhere?

 

[Agent Smith]

Desmos

@Dr.Peterson

 

[mario99]

I've been following the conversation and I wonder if you could simplify that to one simple summation/etc.

The best you can do is to include what professor fresh_42 has given us:

$\displaystyle \sum_{k=1}^{\infty}\left(\frac{1}{2^{k-1}} + \frac{1}{11^k} + \frac{69797}{10^{5k+1}}\right)$

Or you can have it with a common denominator. (But I think that it will look ugly!)

$\displaystyle \sum_{k=1}^{\infty}\frac{2^{4k+2} \times 5^{5k+1} + 10^{5k+1} \times 11^{-k} + 69797}{10^{5k+1}}$

Doe it sum to 2.169797697...?

Absolutely:

$\displaystyle \sum_{k=1}^{\infty}\left(\frac{1}{2^{k-1}} + \frac{1}{11^k} + \frac{69797}{10^{5k+1}}\right) =2.169797697\cdots = \frac{1084888}{499995} = 2.1\overline{69797}$

And, have you seen this number 2.169797697... anywhere?

Why do you think that this number is special? I have solved $1000$ infinite series and I might have seen it.


Note: I would not be surprised if there was a simple infinite series which converges to that number. I just cannot think of it right now.

 

[mario99]

Last attempt:

$\displaystyle \frac{1084888}{500000}\sum_{k=0}^{\infty}\frac{1}{10^{5k}} = 2.169797697\cdots$

 

[Agent Smith]

Most kind of you. I was under the impression that it's an irrational, but your computation is nice. Gracias @mario99 @fresh_42

 

[mario99]

Most kind of you@mario99 @fresh_42

And it will look prettier if it starts at $k = 1$.

$\displaystyle \frac{1084888}{5}\sum_{k=1}^{\infty}\frac{1}{10^{5k}} = 2.169797697\cdots$

I was under the impression that it's an irrational, but your computation is nice. Gracias @mario99 @fresh_42

Why do you think that it is an irrational number? Of course, we can express 2.169797697... as a fraction!

Why do you think that this number is special?

Also, you did not answer my question.

 

[Agent Smith]

Also, you did not answer my question.

It just popped up in my calculations in google sheets. Not sure if it appears elsewhere in math.

If I use trig ratios does it mean the result of the computation is likely an irrational. Not clear as to whether cancellations occur or not.

 

[fresh_42]

If I use trig ratios does it mean the result of the computation is likely an irrational. Not clear as to whether cancellations occur or not.

Real numbers are strange. Algebraic calculations usually lead to rational numbers although "most" real numbers are irrational. The values of trig functions and whether they are rational or not depend strongly on the angles. Only "a few" angles lead to rational numbers, "some more" to roots, and "mostly" to even weirder values.

The situation becomes even more complicated if you distinguish between algebraic numbers (zeros of polynomial expressions) and transcendental numbers like $\pi\, , \,\mathrm{e} \, , \,\sin(1)$ or $2^{\sqrt{2}}.$

 

[mario99]

It just popped up in my calculations in google sheets. Not sure if it appears elsewhere in math.

If I use trig ratios does it mean the result of the computation is likely an irrational. Not clear as to whether cancellations occur or not.

Whatever were your calculations, since $\displaystyle 2.169797697\cdots = \frac{1084888}{499995}$, it is not an irrational number.