anonymous
  • anonymous
Oh I think I get the idea of stripping out a term now. Simply by taking out an \[a_1\] changes where the sum starts? Yes? \[\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^{\infty} a_n\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
so when I have \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=2}^{\infty}a_nnx^n\] but it's not n=0? Why?
anonymous
  • anonymous
I'm trying to start at n=0
anonymous
  • anonymous
\[\sum_{n=1}^{\infty}a_nnx^n \rightarrow \sum_{n=0}^{\infty}a_nnx^n\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I looked at example 1 a. http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx but I'm not trying to do an index shift :C
anonymous
  • anonymous
Well do you need to start at 0? because if you want n = 0 you can just make it at the bottom n = 0
anonymous
  • anonymous
why though?
anonymous
  • anonymous
why is that all I need to do?
anonymous
  • anonymous
\[\sum_{n =0}^{\infty}\]
anonymous
  • anonymous
because in your equation it would work for not getting an answer anyway
anonymous
  • anonymous
because multiplying by 0 just leaves the term out basically, so you can have it as n=0 but you won't get an actual term out.
anonymous
  • anonymous
if you are trying to get a term out there have to be different conditions met in your equation otherwise it won't work that way. But it looks fine for getting a term out if it's not n = 0.
anonymous
  • anonymous
by that do you mean that \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\] It leaves the \[a_1\]out?
anonymous
  • anonymous
no
anonymous
  • anonymous
:(
anonymous
  • anonymous
it's right the left part
anonymous
  • anonymous
but if you made that n = 0
anonymous
  • anonymous
on the left and n= 1 on the right
anonymous
  • anonymous
it would be \[\alpha _{0}*0*x^0\]
anonymous
  • anonymous
OMG! I think something just clicked!
anonymous
  • anonymous
Yeah, you see where im going with it
anonymous
  • anonymous
no wait, just a second
anonymous
  • anonymous
You can have it start at 0.. but it doesnt change anything if you start at 1 in this case
anonymous
  • anonymous
\[\sum_{n=0}^{\infty}a_nnx^n=0\]
anonymous
  • anonymous
it's like putting 0 + \[\sum_{n=1}^{\infty} (insert rest here)\]
anonymous
  • anonymous
for n = 0
anonymous
  • anonymous
I guess if you want to start at 0, it has to be fixed at 0 and you can't change that value for it so it's usually left out of summations and n starts at 1. There are situations where you can have n start at 0 but in you equation those conditions for having a different term at n = 0 is not met.
anonymous
  • anonymous
I'm a very visual learner. Where you said: "on the left and n= 1 on the right" \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\] ^^^Is wrong I know, but would you mind writing what you mean?
anonymous
  • anonymous
I'm honestly trying hard to get this...I think we're almost there...sorry =/
anonymous
  • anonymous
\[\sum_{n=0}^{\infty}a _{n}nx ^{n} = 0 + \sum_{n=1}^{\infty}a _{n}nx ^{n}\]
anonymous
  • anonymous
using the rule that 0 multiplied by anything is 0
anonymous
  • anonymous
Thank you, finally!
anonymous
  • anonymous
Lol it's ok, glad i could help

Looking for something else?

Not the answer you are looking for? Search for more explanations.