Convergence: |x - a| < R. Divergence: |x - a| > R idek???: |x - a| = R Now, for part b, 2 = 7-5
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Replying to @EmilyGorcenski
But if R needs to be greater than 2 for x = 3 to be convergent, wouldn’t x = 7 be convergent too?
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Replying to @EmilyGorcenski @emilymhorsman
Ah, right. Both points are on what we can best guess to be the boundary. So idek?? holds there, but we're given x=3 is good.
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Replying to @EmilyGorcenski @emilymhorsman
I can write that up more formally if you want :)
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Replying to @EmilyGorcenski
If you have time that’d be rad! Thanks so much
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Replying to @EmilyGorcenski
Aha! That makes sense, by saying x=3 converges we’re making an assurance on that boundary, doesn’t help the other one.
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Replying to @emilymhorsman @EmilyGorcenski
and thus R doesn’t _need_ to be _greater than_ 2. thanks so much!!
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Correct. This might seem silly in the real-valued case, but in complex analysis the radius of convergence is a circle.
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Replying to @EmilyGorcenski
ohh, I realized another nuance while typing out my own answer to cement my understanding http://math.stackexchange.com/a/2124819/412218 …
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