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Limits and Taylor Series

1. 6. If we have y = 0 in the xy- plane, then in polar coordinates, does this refer to or . I assume both if we're not given any bounds for x. If we are told additionally that x > 0, then I assume and if x < 0, then .
2. I was just wondering that:
If we had f(x) <= g(x) for x >= a and f(x) is decreasing, then if g(x) --> 0 as x --> ∞, is it that f(x) also --> 0 as x --> ∞?

3. If I wanted to find the Taylor series of f(x) as x --> ∞, is it correct to find the Taylor series of f(1/y) at y = 0 and then use x = 1/y substitution to find what we want.
Original post by Takeover Season
1. 6. If we have y = 0 in the xy- plane, then in polar coordinates, does this refer to or . I assume both if we're not given any bounds for x. If we are told additionally that x > 0, then I assume and if x < 0, then .


Yep.

2. I was just wondering that:
If we had f(x) <= g(x) for x >= a and f(x) is decreasing, then if g(x) --> 0 as x --> ∞, is it that f(x) also --> 0 as x --> ∞?


Yep, this is a simple application of the sandwich theorem.
(edited 3 years ago)
Original post by Takeover Season

3. If I wanted to find the Taylor series of f(x) as x --> ∞, is it correct to find the Taylor series of f(1/y) at y = 0 and then use x = 1/y substitution to find what we want.


Do you mean Taylor expansion at infinity? Otherwise any expansion is valid as x tends to infinity as long as you dont omit any terms.

At infinity, you want to subsitute in y=1/x.
As x tends to infinity, y tends to 0. So expand your function about y=0 and sub back in for x.
(edited 3 years ago)
Original post by RDKGames
Yep.



Yep, this is a simple application of the sandwich theorem.

Thank you! Don't know why but for some reason I thought to use the Sandwich theorem I needed to have a lower bound for f tending to 0 too.

Original post by RDKGames
Do you mean Taylor expansion at infinity? Otherwise any expansion is valid as x tends to infinity as long as you dont omit any terms.

At infinity, you want to subsitute in y=1/x.
As x tends to infinity, y tends to 0. So expand your function about y=0 and sub back in for x.

Yes, I meant Taylor expansion at infinity. Thank you, I understand now!
Original post by RDKGames
Yep, this is a simple application of the sandwich theorem.

For 2, don't we need another inequality to use the sandwich theorem? As stated, that isn't true, e.g. f(x)=-x and g(x)=e^-x. It could be made true by saying that f(x) is increasing, I think?
Original post by I hate maths!
For 2, don't we need another inequality to use the sandwich theorem? As stated, that isn't true, e.g. f(x)=-x and g(x)=e^-x. It could be made true by saying that f(x) is increasing, I think?

I think this is what I meant in my previous post, I was unclear on how to say it.
Original post by I hate maths!
For 2, don't we need another inequality to use the sandwich theorem? As stated, that isn't true, e.g. f(x)=-x and g(x)=e^-x. It could be made true by saying that f(x) is increasing, I think?


Original post by Takeover Season
I think this is what I meant in my previous post, I was unclear on how to say it.


It is also true if f(x) > 0, which gives it its lower bound. I had misread the original post in such a way that I took f(x) as positive.

But yes, original statement doesnt hold as it stands and needs an extra tweak or two to make it correct.
Original post by RDKGames
It is also true if f(x) > 0, which gives it its lower bound. I had misread the original post in such a way that I took f(x) as positive.

But yes, original statement doesnt hold as it stands and needs an extra tweak or two to make it correct.


Thanks for the response. What if f(x) is non-negative? Does it still hold?
Original post by Takeover Season
Thanks for the response. What if f(x) is non-negative? Does it still hold?


Im sure you dont need to ask this.

0 is still a lower bound for f(x) when its non-negative. :smile:
Original post by RDKGames
Im sure you dont need to ask this.

0 is still a lower bound for f(x) when its non-negative. :smile:


Oh yes lol, sorry. Now I get where you're coming from with the Sandwich Theorem as 0 tends to 0 as x tends to infinity.
Original post by Takeover Season
I think this is what I meant in my previous post, I was unclear on how to say it.

It's actually still not true even if we say f(x) is increasing, e.g. f(x)=-1/x - 1 and g(x)=e^-x. I think your statement ought to include a lower bound as RDK says.
Original post by I hate maths!
It's actually still not true even if we say f(x) is increasing, e.g. f(x)=-1/x - 1 and g(x)=e^-x. I think your statement ought to include a lower bound as RDK says.


Thanks buddy, I understand now. Great example!

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