/sci/ - Science & Math » Thread #12846407

11KiB, 422x115, Screenshot from 2021-03-17 14-04-06.png

View Same Google iqdb SauceNAO

Anonymous Wed 17 Mar 18:05:22 2021 No.12846407 View Reply Original Report

Quoted By: >>12846571 >>12846585 >>12846722

I cannot figure out the infinite sum of this series. I have been trying for 2 hours. Should I drop out of CS and rope?

Anonymous

Anonymous Wed 17 Mar 2021 18:48:56 No.12846571 Report

Quoted By:

>>12846407
? = 1, 2/3, 3/9, 4/27...
?/3 = 1/3, 2/9, 3/27...
?-?/3 = 1, 1/3, 1/9, 1/27...

you can finish from here?

Anonymous

Anonymous Wed 17 Mar 2021 18:52:50 No.12846585 Report

Quoted By: >>12846589 >>12846596 >>12846695

>>12846407
using
$\sum_{k=1}^\infty z^k = \frac{z}{1-z}$
and
$\sum_{k=1}^\infty kz^k = \frac{z}{\left(1-z\right)^2}$
rewrite your series as
$1+\sum_{k=1}^\infty \left(k+1\right)\left(\frac{1}{3}\right)^k$
rewrite your series as
$=1+\sum_{k=1}^\infty \left(\frac{1}{3}\right)^k+\sum_{k=1}^\infty k\left(\frac{1}{3}\right)^k$
and applying the formulas above
$=1+\frac{\frac{1}{3}}{1-\frac{1}{3}}+\frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2}=\frac{9}{4}$

Anonymous

Anonymous Wed 17 Mar 2021 18:54:00 No.12846589 Report

Quoted By:

>>12846585
last line should be
$=1+\frac{\frac{1}{3}}{1-\frac{1}{3}}+\frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2}=\frac{9}{4}$

Anonymous

Anonymous Wed 17 Mar 2021 18:55:29 No.12846594 Report

Quoted By: >>12846695

the sum can be written as (n+1)/(3^n). separate this into two geometric series. take the derivative of the n/(3^n) sum and multiply both sides by n. add to get 9/4.
wala

Anonymous

Anonymous Wed 17 Mar 2021 18:56:03 No.12846596 Report

Quoted By:

>>12846585
last line should be
$=1+\frac{\frac{1}{3}}{1-\frac{1}{3}}+\frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2}=\frac{9}{4}$

Anonymous

Anonymous Wed 17 Mar 2021 19:28:07 No.12846695 Report

Quoted By: >>12846712 >>12846715

I finally solved it thanks to your help.. But I still don't fully get it.

>>12846585
I've never seen the formula for kx^k . Is that supposed to be the derivative of the z^k formula?

>>12846594
Isn't the derivative of n/3^n difficult with an answer including the natural log? What would I be differentiating in respect to? n?

This has taken 3 hours of my day. How badly have I failed the IQ test?

Anonymous

Anonymous Wed 17 Mar 2021 19:35:58 No.12846712 Report

Quoted By: >>12846715

>>12846695
n/3^n is a geometric function of the form sum(x^n), which sums to 1/(1-x). if you take the derivative of both sides, you will now have sum(nx^(n-1)). Multiplying both sides by x yields (nx^n), which is the form of n/3^n, with x=1/3.
apologies for formatting, i do not know latex, let me know if this makes sense

Anonymous

Anonymous Wed 17 Mar 2021 19:37:18 No.12846715 Report

Quoted By: >>12846730

>>12846695
i see now that you say you havent seen the form kx^k. that can be derived with the steps i gave in >>12846712
i hope that clears things up

Anonymous

Anonymous Wed 17 Mar 2021 19:39:07 No.12846722 Report

Quoted By:

>>12846407
Sum (n >= 0) : (n + 1) / 3^n
I believe.
I dunno I was just glancing at it.
also fuck jannie niggers, just generally
1 + 1/2 + 3/4 or something.
I dunno.
just did a final today, pretty tired.

Anonymous

Anonymous Wed 17 Mar 2021 19:42:15 No.12846730 Report

Quoted By:

>>12846715
Yes, it makes sense. Thank you.

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