#epsilon delta continuity relation between δ1 and δ2

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worldly storm
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so far i have found d = e/3 and my assumption would be that d1 and d2 are equal or less than e/6

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nocturne rampartBOT
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gilded lotus
worldly storm
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i wish to mathematically prove the correct answer, it is not simply a pick the correct answer question

gilded lotus
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well iii is a sufficient condition for |f(x, y)| <= epsilon

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i don't know if it's a necessary condition

worldly storm
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fair enough , how do i get d2 and d1 involved though

gilded lotus
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well you just need to check that they work

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here's one useful trick though

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$\lvert x \rvert \leq (x^2 + y^2)^{1/2}$

slate lakeBOT
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ARIMA(1,0,1)(2,1,1)_[12]

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gilded lotus
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likewise y

worldly storm
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i did use that to get the value of d and e respectively, what confuses me is the separation of deltas

gilded lotus
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now that I think about it

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it seems like answer (i) will actually work as well

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it's a looser bound

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but still does the job

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you can try to check out why

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$f(x, y) = 3 \times \frac{y^2}{x^2 + y^2} \times x$

slate lakeBOT
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ARIMA(1,0,1)(2,1,1)_[12]

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gilded lotus
worldly storm
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lol

gilded lotus
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so from there I think you can show that (i) is sufficient as well

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(note that i is implied by iii, so i is a weaker sufficient condition)

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but i'm still not convinced that it's sufficient

worldly storm
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i am lost , i do not wish to be spoonfed an answer but i do not understand how to check the answer
my train of thought currently is that given d = e/3
and (x^2 + y^2)^1/2 is less or equal than |x| + |y| it will be not be greater nor equal to d1+ d2 thus we conclude that d1+ d2 < e
given that in the preveous mathematical steps i have calculated d i can assume that they will be less than or equal to e/6 at the same time

gilded lotus
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I don't intend to spoonfeed you the answer

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my point is that what they probably want from you is the weakest sufficient condition

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basically, the weakest A such that A implies |f(x, y)| < epsilon

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you can see that some of the answers entail other answers, here

worldly storm
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i also had a thought , would it be possible to form the function given the nature of the problem as in -d1 < x-x- < d1 and form the function within it for both d1 and d2?

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gilded lotus
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for example, if iii is true, you can easily see that i is true

worldly storm
gilded lotus
# worldly storm o

so for example, iii implies |f(x, y)| < epsilon, is a weaker statement than i implies |f(x, y)| < epsilon

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because condition iii is harder to meet

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do you get the logic?

worldly storm
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i do yes , let me retry it one sec

worldly storm
# gilded lotus do you get the logic?

so at the end of the day , which one am i to choose , i am trying to find a correlation between d1 and d2 , do i simply explain my option using logic ?

gilded lotus
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because there are multiple sufficient ones

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if you assume (i), you can prove |f(x, y)| < epsilon too

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same if you assume (iii) and (iv)

worldly storm
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thanks for the help

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might be a question on my upcoming exam , will look into it , have a good day

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+close

upbeat perchBOT
# worldly storm +close
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upbeat perchBOT
# upbeat perch

Thank you for your feedback! ARIMA(1,0,1)(2,1,1)_[12] has been awarded 1 helper_points. They now have 62 helper_points. They have 2 helper_points daily left for today.

gilded lotus
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as in, |f(x, y)| < epsilon implies A, is the same as (not A) implies |f(x, y)| >= epsilon

upbeat perchBOT
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@worldly storm

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