how on god's holy earth do you derive a function with sigma

[gloomy prism]

i'm scared

[storm oak]

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[gloomy prism]

n belongs to N btw

[vale abyss]

Well it's a sum with a finite number of terms

so you can differentiate each term of the sum

[gloomy prism]

what did you say

can i change the indexes like this bro

am i cooking

[vale abyss]

I mean, think about a polynomial for example

which is exactly what that sum in the sigma is

P(x) = x^5 + 4x^4 - 2x^3 + 5x^2 + x - 1

how do you differentiate that?

$P(x) = x^5 + 4x^4 - 2x^3 + 5x^2 + x - 1$

[undone leaf]

Rion

[gloomy prism]

uh

one sec

$P'(x) = 5x^4 + 16x^4 - 6x^2 + 10x +1$

[undone leaf]

#longliveyoavmal

[vale abyss]

Yes, and how did you do that?

[gloomy prism]

idk

[vale abyss]

As in, the method

[gloomy prism]

uh

[vale abyss]

I'm sure you didn't just apply the limit definition of the limit

[gloomy prism]

remove one from the power and multiply by the og power value idk

[vale abyss]

You differentiated each term

of the sum

[gloomy prism]

you can do that?

[vale abyss]

you could, if you really wanted to die

[gloomy prism]

yeah i guess those are the terms

[vale abyss]

[gloomy prism]

haha i'm already dying

[vale abyss]

Anyway look

[gloomy prism]

anywyas

[vale abyss]

Here

The sigma thing is just a sum

so it's perfectly legal for you to differentiate each term individually

[gloomy prism]

yeah but

[vale abyss]

just like you did with a polynomial

[gloomy prism]

n+1

it's not like a real number

[vale abyss]

the variable is x

[gloomy prism]

god knows what n+1 can be

[gloomy prism]

what's that supposed to proove

that others are constant?

[vale abyss]

I don't understand why you're talking about n

n is constant

[gloomy prism]

wait i can explain

so earlier when you gave me that sum

it also has a maximum number n yk

which was 5

so like

[vale abyss]

Yeah

That's fine

I mean all that matters is that it's a finite sum

[gloomy prism]

yeah its finite

n is unknown

so it could be infinite

[vale abyss]

no, you said it's a natural number

therefore the sum is finite

[gloomy prism]

yeah

huh

but

we cant know..

ok ok

[vale abyss]

Differentiating infinite sums is indeed more complicated

but we are not in such a dilemma

so just go for it

[gloomy prism]

..

kay ig

thanks..

[vale abyss]

you don't sound all too convinced by my explanation

I mean, that's fine, I'm not asking you to trust me blindly

[gloomy prism]

i'm just a bit lost haha

thank you in every case

[vale abyss]

Well here, n is a FIXED natural number

certainly, it can be anything, but it's not a function variable

[gloomy prism]

yeah

[vale abyss]

You study the function, which name is $f_n$, given by:
$$f_n : x \mapsto \exp(-x) + \sum_{k=0}^{n} \frac{(-1)^k}{k!}x^k$$

[undone leaf]

Rion

[gloomy prism]

that's indeed the function

the thing is

from the example you gave earlier

you cant just do the same with this one

[vale abyss]

You can, really

the example with the polynomial had 6 terms

here, instead of 6, we have n+1

which is also fixed

at least with respect to the function variable, which is x

[gloomy prism]

but

n isnt rlly a

known number..

[vale abyss]

That doesn't mean you can't work with it

I think your confusion stems from the fact that differentiation only means with respect to the function variable

which in our case will be x

so just because you don't know n, doesn't change anything to that, it's still a natural number like 5

just that instead of 5, you put n, to generalize

[gloomy prism]

generalization is both a curse and a blessing at this rate..

think i got enough

i'll keep trying

[vale abyss]

Sure

[gloomy prism]

sorry for the bother and thank you

[vale abyss]

👍

[cloud gull]

Does that even matter? I've seen infinite series differentiated the same way.

[vale abyss]

It matters because it's nontrivial yes

Especially for power series exactly on the circle of convergence

where the behavior can be pretty erratic

[cloud gull]

Wouldn't that just be because the series doesn't converge at the endpoints of the interval of convergence, in which case a lack of convergence of the derivative doesn't tell us anything we didn't already know or give us problems we didn't already have?

[vale abyss]

No, it can converge

but there is no guarantee that the power series will yield the expected result on the circle of convergence

or if it does (continuity on the circle), differentiation is not guaranteed

[cloud gull]

Of course, this is all moot in this instance anyway.