could someone check my answers for these two power series problems?

[dim frigate]

Your formula for antiderivative seems to be right when k=2, 3, ... but not when k = 0 or k = 1

If you want you could just start your summation at k = 2 and just add the first 2 terms

[soft heart]

what formula would I use then?

also, #27 looks good?

[dim frigate]

Your formula is good, I would keep it as it is but start the summation at k=2

And then add the cases when k = 0 and k = 1

That way you don't need to find a formula that has ALL the terms, but all but 2

[soft heart]

so becaues its

I need to add the first two terms

?

[dim frigate]

Well currently your formula doesn't seem to work when k = 0 and k = 1

[soft heart]

why wouldn't it?

i'm sorry, my professor gave us the formula, but he didn't really explain it

so idk what i'm doing lol

[dim frigate]

Well when k = 0 here you get 2^0(x-3)^0 which is just 1

And when you integrate 1 you get x

So when k = 0 your formula should spit out x

But instead your formula gives (x-3)

[soft heart]

i get this bit now

[soft heart]

how do i fix that?

[dim frigate]

So you could try to make your formula work for k = 0 and k = 1, or you can simply add those cases to the summation starting from k = 2

You can add the integral of 1 and the integral of 2(x-3)

It's the same thing as a summation from k = 0 to infinity

You're just adding them because your formula doesn't include them starting from k = 2

[soft heart]

Like this? Z

Or as in adding an addendum that it’s 1 and x at k=0 and 1 respectively?

[dim frigate]

When you write out the sum you get 1+2(x-3)+ ... and so on

To get antiderivative you're integrating each term

[soft heart]

i get that bit for x then, but i still dont get it for (x^2-6)

[dim frigate]

Oh that's just the integral of 2(x-3)

[soft heart]

i get it now

[soft heart]

ty, do you mind if I ask you some more questions?

[paper raven]

@soft heart has given 1 rep to @dim frigate

[soft heart]

All of my stuff is due midnight and I was out sick for a few days so I'm way behind on concepts

[dim frigate]

Yep I think that should be okay now

[soft heart]

I owe you one my guy

[dim frigate]

Thank you. I like these problems

[soft heart]

could you explain to me what it means when it asks me to find its domain? I was able to get the first few terms of the tatlor series, but IDK how to write a whole taylor serries and I dpn't get what it means by domain

[dim frigate]

I think thats another way to say radius of convergence

Like if you move away from a=2 how long is your approximation valid

Or what values of x it works for

[soft heart]

How would I do that for a basic Taylor series? I thought you could only do that with power series

[dim frigate]

Not sure actually

Yea that's the formula

You just keep taking derivatives of e^x

[soft heart]

I don't understand how that would help get me closer to the domain

[dim frigate]

Let me read about it a bit

[soft heart]

k. i'll read too

I could send you my professor's power point if you'd like. I don't really understand it, but all of this stuff seems to be under unit 3.4

starts on page 224

[dim frigate]

Aha, I see

So once you write your Taylor Series for e^x, you just apply the ratio test to it and examine the limit at infinity

[soft heart]

IDK how to do that since my tylor series has infintie terms in it

[dim frigate]

When you look at the limit you're only looking at the sequence attached to the series

[soft heart]

my brain cells do not understand this sentence.

I understnad using ratio test on stuff like (2+x)^k/k!

I don't understand how to apply it to a bunch of taylor polynomial terms

[dim frigate]

You mean how do you apply it to 1 + x + x^2 + x^3 + ... ?

Taylor Series of e^x

[soft heart]

I recognize that it's a geometric sieries, so I go back to x^n

[dim frigate]

Yea x^n/n! is your sequence and you apply the ratio test to that

[soft heart]

that's the sequence for e^x?

[dim frigate]

Yea, that's the sequence corresponding with the series

[soft heart]

how'd you find that for e^x? or is it just the same for any taylor polynomial?

[dim frigate]

You can find it from the Taylor Series

[soft heart]

how would I go about doing that?

[dim frigate]

Well for the Taylor Series about a = 0 for e^x its pretty simple, since e^0 = 1 and all the derivatives f'(a) are 1, and a = 0 so (x-a)^n is just x^n/n!

Just erase all derivatives from Taylor's Theorem basically

[soft heart]

you don't need to care about it actually being about a = 2?

is that cause it goes to infinity?

[dim frigate]

I think it's because e^x has an infinite domain of convergence, by the ratio test

[soft heart]

as in e^x+1/e^x?

I don't get that

1, right?

[dim frigate]

No the ratio of the sequences a_n+1 / a_n

The sequences associated with e^x

So some x^(n+1)/(n+1)! all divided by x^n/n!

[soft heart]

Where do I go from here?

[soft heart]

I've gotten myself in the unfortunate situation where I have x/(n+1) as somehow less than 1 for my domain

IDK what to do

[dim frigate]

It's 0

[soft heart]

how can that be my domain?

I thought you needed an interval for a domain?

[dim frigate]

The limit of the ratio is 0, so you can conclude the Taylor Series convergences absolutely for every value of x

[soft heart]

so domain is all real numbers?

[dim frigate]

Yeah, exactly

[soft heart]

thanks

[paper raven]

@soft heart has given 1 rep to @dim frigate

[dim frigate]

What you wrote is correct

[soft heart]

Do you understand what part C is asking?

[dim frigate]

I'm a bit tired, so I'll let someone else answer that question

[soft heart]

ok, ty anywats

[paper raven]

@soft heart has given 1 rep to @dim frigate