Power Series Representation bad question!

[sour ledge]

How ?

[hard ruin]

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[sour ledge]

Please

[thin night]

you know the power series of e^x, so just plug that in and simplify

[sour ledge]

Simplify what

look

im like ababy

[thin night]

(e^x-1)/x

after subbing in e^x's series.

[sour ledge]

x^n / n!

i know that

[thin night]

no

$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

[hollow basin]

Omegabet_

[sour ledge]

yes!

[thin night]

e^x isnt x^n/n!

[sour ledge]

yes !!!!! Sigma!! notation!

[thin night]

but anyway, plug that in, and simplify to get (e^x-1)/x's series

[sour ledge]

I'm confused where to plug it in

[thin night]

there's only 1 e^x in the question...

[sour ledge]

yes

but

[thin night]

take a wild guess where you're going to replace e^x......

[sour ledge]

is it like:

series for e^x * -1/x ?

[thin night]

is a-b a*(-b)?

going back to elementary school

$\frac{e^x-1}{x}=\frac{-1+\sum_{n=0}^\infty \frac{x^n}{n!}}{x}$

[sour ledge]

.

[hollow basin]

Omegabet_

[sour ledge]

ok

[thin night]

now simplify

[sour ledge]

how

[thin night]

algebra

numerator is $\sum_{n=1}^\infty\frac{x^n}{n!}$

[hollow basin]

Omegabet_

[thin night]

since -1+(1+x+x^2/2+...)=x+x^2/2+...

then you divide by x

[sour ledge]

im confused how far to go with terms

hwo to deal with it

[sour ledge]

ok

i get that

[thin night]

you dont stop, there are infinitely many terms...

[sour ledge]

exactly

my confusion

[thin night]

review what an infinite series is then

[sour ledge]

Omega look. It's 3am where I am. I have been working on this HW for hours and i'm almost done. can't think anymore!

Can you do me a favor with this and have some patience with me

[sour ledge]

i.e last question

[thin night]

go to sleep then

[sour ledge]

It is due

[thin night]

ok, and we really can't have a constructive session if you're unable to use your brain

[sour ledge]

OK look:

Confusion: subtracting a finite number from an infinite series

[thin night]

same way you subtract functions.

$f:x\mapsto\sum_{n=0}^\infty\frac{x^n}{n!}$ is a function defined everywhere on the real line

[hollow basin]

Omegabet_

[thin night]

so is f(x)-1

alternatively.. it's the 1st term, it's tangible

mucking with the 1st umpteen terms has no impact on convergence

other than what it converges to.

[sour ledge]

You're saying basically I'm effectively subtracting one from every term for the series for e^x

makes sense

How do i express this?

[thin night]

no

im subtracting 1 from the series

not from every term

[sour ledge]

ok

so you can just deal with the first term

[thin night]

yes...

[sour ledge]

yumm

but dividing by x

whole series ?

[thin night]

duh

$\frac{e^x-1}{x}=\frac{1}{x}\sum_{n=1}^\infty\frac{x^n}{n!}$

[hollow basin]

Omegabet_

[sour ledge]

So do i just write out a few terms, differentiate them?

for the second step

[thin night]

dont jump ahead...

do the dividing by x.

[sour ledge]

x^(n-1) / n!

?

[thin night]

yes

[sour ledge]

Then differentiate that

[thin night]

that series, yes.

[sour ledge]

(n-1)x^(n-2) / (n-1)!

[thin night]

no

n! is a constant

it doesnt change

[sour ledge]

(n-1)x^(n-2) / n!

[thin night]

yes

[sour ledge]

How do I manipulate that to make it equal to:

n / (n + 1)!

[thin night]

well use common sense

is there any weird powers in n/(n+1)! ?

[sour ledge]

what do you mean weird

[thin night]

are there any powers...

[sour ledge]

no

[thin night]

so then you'll have to evaluate at x=1

since 1^(n-2) = 1.

[sour ledge]

but i need to manipulate bottom

to

[thin night]

that's just reindexing

[sour ledge]

(n + 1)!

[thin night]

$\dv{x}\frac{e^x-1}{x}=\dv{x}\sum_{n=1}^\infty\frac{x^{n-1}}{n!}=\sum_{n=2}^\infty\frac{(n-1)x^{n-2}}{n!}$

[hollow basin]

Omegabet_

[thin night]

since n=1 gives a 0 term for the rightmost series.

but anyway, n=2 makes n-1=1, so $=\sum_{m=1}^\infty\frac{mx^{m-1}}{(m+1)!}$

[hollow basin]

Omegabet_

[thin night]

(n=m+1 does the reindexing).

[sour ledge]

Ok, We still have a x there but it goes doesn

doesn't it need to be only n

.

[
\sum_{n=1}^{\infty} \frac{1}{(n-1)!}
]

[hollow basin]

Znova

[thin night]

did you fail to be literate?

[thin night]

.

[sour ledge]

[
\sum_{n=1}^{\infty} \frac{n}{(n+1)!} = 1
]

[hollow basin]

Znova

[thin night]

x=1 makes x^(m-1) disappear.

[sour ledge]

yes i remember

[thin night]

clearly you dont remember

[sour ledge]

I did but i was still confused

on

the part

where youre forced to make it like that

when it doesn't explicitly state it

[thin night]

so you're confused... cause you had to think for yourself?

[sour ledge]

small deal -> big deal. calm down! calm down!

[thin night]

Being confused cause you had to think is a big deal, not a small deal

[sour ledge]

New type of question! This is new! Calm down !

Thank you for the help! !

[thin night]

Sure, but anyway it's now a regular calculus problem

I assume you're able to do calc1?

[sour ledge]

yes

if i close this do i still have access to the channel?

[thin night]

dkdc

[sour ledge]

Do you have time for last question? I am not sure if i did this properly

[thin night]

you got help on that already

so no

[sour ledge]

Was still confused
here now
ask now

[thin night]

go back to that channel then.

[sour ledge]

ok.

i will go back to that channel.

thanks for using periods.

it makes it seem very serious.

[thin night]

Ok

[sour ledge]

+close