/sci/ - Science & Math » Thread #10870698

5KiB, 685x186, Question1.png

View Same Google iqdb SauceNAO

Downtime Math #1

Anonymous Wed 07 Aug 08:19:20 2019 No.10870698 View Reply Original Report

Quoted By: >>10870837 >>10870860 >>10870920 >>10870987 >>10870995 >>10871097 >>10871304

Hi /sci/
Thought it'd be fun to post simple downtime math problems every now and then.
If this thread goes well I'll try posting another later.

Anonymous

Anonymous Wed 07 Aug 2019 10:09:34 No.10870837 Report

Quoted By: >>10870987 >>10870995

>>10870698
This is as far as I'm going $\sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)^2}$

Anonymous

Anonymous Wed 07 Aug 2019 10:34:27 No.10870860 Report

Quoted By:

>>10870698
During my downtime I construct math problems in my head and solve them.

Anonymous

Anonymous Wed 07 Aug 2019 11:36:05 No.10870920 Report

Quoted By: >>10871031

>>10870698
You can use the expansion of 1/n into a series of 1/(n*(n+1)) terms, and then turn each of the 1/(n*(n+1)) terms into 1/(n+1)^2 + 1/(n*(n+1)^2) terms. Because this is absolutely convergent, we can rearrange these into most of the basel problem, and a series of 1/(n*(n+1)^2). So, we have 7/2 - (pi^2)/3 = series 2/(n*(n+1)^2), starting at n = 2. Again, we use the 1/n formula, but now we can subtract our new sum, so, our series is 1/2 - (7/2 - (pi^2)/3) = (pi^2)/3 - 3.

Anonymous

View Same Google iqdb SauceNAO CodeCogsEqn.gif, 4KiB, 592x144

Anonymous Wed 07 Aug 2019 12:08:19 No.10870987 Report

Quoted By:

>>10870837
>>10870698
first time messing around with online latex coder, sorry about that Anon

Anonymous

View Same Google iqdb SauceNAO CodeCogsEqn.png, 4KiB, 592x144

Anonymous Wed 07 Aug 2019 12:12:01 No.10870995 Report

Quoted By: >>10871015 >>10871031 >>10871034 >>10871042 >>10871054

>>10870837
>>10870698
First time messing around with latex coder, sorry about that anon

Anonymous

Anonymous Wed 07 Aug 2019 12:17:12 No.10871015 Report

Quoted By:

>>10870995
Anon, (pi^2)/4 - 19/8 is smaller than the second partial sum.

Anonymous

Anonymous Wed 07 Aug 2019 12:29:16 No.10871031 Report

Quoted By: >>10871042 >>10871054

>>10870920
This is correct! Great job.
>>10870995
Not quite, anon. Your approach to partial fractions was good though.

Anonymous

Anonymous Wed 07 Aug 2019 12:34:04 No.10871034 Report

Quoted By: >>10871075

>>10870995
Isn't the first term of the telescoping one 1/2 and not 1? That should fix it.

Anonymous

View Same Google iqdb SauceNAO Question1Hint.png, 6KiB, 1058x218

Anonymous Wed 07 Aug 2019 12:41:28 No.10871042 Report

Quoted By:

>>10870995
>>10871031
This should help.

Anonymous

View Same Google iqdb SauceNAO Question1Hint.png, 5KiB, 1056x194

Anonymous Wed 07 Aug 2019 12:49:19 No.10871054 Report

Quoted By:

>>10870995
>>10871031
This should help.

Anonymous

Anonymous Wed 07 Aug 2019 13:10:40 No.10871075 Report

Quoted By:

>>10871034
I messed up the Partial fraction part already. This kinna thing often take half my point in final-semester exam

Anonymous

View Same Google iqdb SauceNAO Question1Solution.png, 27KiB, 1634x944

Anonymous Wed 07 Aug 2019 13:22:58 No.10871097 Report

Quoted By: >>10871165 >>10871411

>>10870698
OP here.
This is the solution.
Thank you for everyone who attempted to solve this problem.
I think I'll be posting another one soon.

Anonymous

Anonymous Wed 07 Aug 2019 13:49:07 No.10871133 Report

Quoted By: >>10871259

Any more problems for us OP?

Anonymous

Anonymous Wed 07 Aug 2019 14:09:57 No.10871165 Report

Quoted By: >>10871204 >>10871234 >>10871259

>>10871097
How did you get that first line? It would never have occurred to me to split up like that, instead of one of the thousand ways you can do it.

Anonymous

Anonymous Wed 07 Aug 2019 14:29:13 No.10871204 Report

Quoted By:

>>10871165
One of the basic tricks when doing series problems is to guess a relevant partial sum formula, multiply the numerator and denominator by n+1 (or n+2, or whatever), and apply the distributive property, and then try to cancel out the n with something in the denominator. If you guessed well, then you should have one of your resulting series be relevant to your problem, and the other series should have a known formula (or better yet, its just your original guess formula, but with a different value plugged in).

Anonymous

Anonymous Wed 07 Aug 2019 14:45:54 No.10871234 Report

Quoted By: >>10871259

>>10871165
Apart from simple (= known) stuff like in OP's pic, there is a lot of literature on this topic, which is called "Partial fraction decomposition", if you want to dig deeper.

https://en.wikipedia.org/wiki/Partial_fraction_decomposition

Anonymous

Anonymous Wed 07 Aug 2019 14:55:15 No.10871259 Report

Quoted By:

>>10871133
I think I'll post another in a few hours or so.
>>10871165
From the denominator (n + 1)(n + 2)^2, (n + 2)^2 can be split as (n + 2) and (n + 2)^2, as factors with exponents require that the factors in the partial fractions are of each exponent from 1 and up.
Then with A/(n + 1), B/(n + 2) and C/(n + 2)^2, simultaneous equations can be used to solve for all numerators.
The wiki link from >>10871234 details this.

Anonymous

View Same Google iqdb SauceNAO nig-bike-keks12.jpg, 354KiB, 667x523

Anonymous Wed 07 Aug 2019 15:16:46 No.10871304 Report

Quoted By: >>10871359 >>10871360

>>10870698

Another problem pls

For 0<a<b, find the radius of convergence of the series

sum_{n=0}^{\infty} (a^n+b^n)x^n

Anonymous

Anonymous Wed 07 Aug 2019 15:46:35 No.10871359 Report

Quoted By: >>10871395

>>10871304
1/b

not related to these things tho

not related to these things tho Wed 07 Aug 2019 15:46:41 No.10871360 Report

Quoted By: >>10871398

>>10871304
is it a nagel point?

Anonymous

View Same Google iqdb SauceNAO you (2).jpg, 27KiB, 708x450

Anonymous Wed 07 Aug 2019 16:01:14 No.10871395 Report

Quoted By:

>>10871359

Anonymous

View Same Google iqdb SauceNAO DiCaprio-and-a-young-Greta-Thunb (...).jpg, 18KiB, 474x363

Anonymous Wed 07 Aug 2019 16:02:16 No.10871398 Report

Quoted By: >>10871416

>>10871360
>nagel point

Not sure how that fits into it. Expand pls

Anonymous

Anonymous Wed 07 Aug 2019 16:08:23 No.10871411 Report

Quoted By:

>>10871097
I feel the telescoping sum is fishy. The last term is left out, how can we be so sure that it does not affect the result of the sum? Yes, it goes to zero, but still.

Anonymous

Anonymous Wed 07 Aug 2019 16:11:01 No.10871416 Report

Quoted By: >>10871422

>>10871398

i thought maybe you were talking about this,

https://en.wikipedia.org/wiki/Nagel_point

or this

https://en.wikipedia.org/wiki/Gheorghe_?i?eica

like how do you calculate these things converging...

but totally am prolly wrong, i dint fullly understand what your Q was, and cant zoom in on .jpg cause am on public computer..

so not sure if this was your Q? like some kind of samurai-geometry puzzle

https://en.wikipedia.org/wiki/Sangaku

not sure.. so.. yeah

Anonymous

Anonymous Wed 07 Aug 2019 16:13:09 No.10871422 Report

Quoted By:

>>10871416

wait i found out what your Q? was, basel problems? yeah my answer was completely wrong

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches