Maths Uni Chat Watch

Dadeyemi
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#3221
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#3221
(Original post by My Alt)
Conclusion: Anime is for kids.


QED
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.matt
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#3222
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#3222
I've never thought about this before:

\displaystyle\sum_{r=0}^n \frac{x^r}{r!} has n roots in \mathbb{C}, so the number of roots tends to \infty as n \rightarrow \infty, but \displaystyle\sum_{r=0}^{\infty} \frac{x^r}{r!} has no roots.

Maffs is weird.
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My Alt
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#3223
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#3223
(Original post by .matt)
I've never thought about this before:

\displaystyle\sum_{r=0}^n \frac{x^r}{r!} has n roots in \mathbb{C}, so the number of roots tends to \infty as n \rightarrow \infty, but \displaystyle\sum_{r=0}^{\infty} \frac{x^r}{r!} has no roots.

Maffs is weird.
Do we know what happens to the modulae of the roots. (plural fail? perhaps) It would be neat if something like this happens:

root=\alpha_i+\beta_i i

with

\alpha_i \to -\infty, \beta_i \to 0

Obviously this is a very wishy washy idea of what I want to happen, as the number of roots grows as n increases, but perhaps the means or something...
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SimonM
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#3224
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#3224
I can prove that for all R, there exists N st for n>N \displaystyle \sum_0^n \frac{z^r}{r!} has no zeros in |z|<R.

(But my proof relies on the fact that exp has no roots)
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.matt
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#3225
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#3225
(Original post by My Alt)
Do we know what happens to the modulae of the roots. (plural fail? perhaps) It would be neat if something like this happens:

root=\alpha_i+\beta_i i

with

\alpha_i \to -\infty, \beta_i \to 0

Obviously this is a very wishy washy idea of what I want to happen, as the number of roots grows as n increases, but perhaps the means or something...
Just had a play around with the means of the roots in Mathematica, and realised that their mean for all n is -1, which is quite a nice result. Follows from some Further Maths stuff that I haven't encountered once whilst at university: the sum of the roots of \displaystyle\sum_{r=0}^{n}a_n x^n is \displaystyle -\frac{a_{n-1}}{a_n}, so for the nth exponential partial sum the sum is -\frac{n!}{(n-1)!} = -n, thus their mean is -1.

Another thing that I find quite weird is that if you have two intersecting lines in \mathbb{R}^n with a positive angle \varepsilon between them, then the intersection will just be a single point for all \varepsilon &gt; 0, yet when \varepsilon = 0 the intersection is an uncountable set of points.
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majikthise
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#3226
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#3226
That doesn't seem particularly weird to me!
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.matt
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#3227
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#3227
(Original post by majikthise)
That doesn't seem particularly weird to me!
Haha, I think I find it weird just because the number of points of intersection jumps to infinity rather than tending to it - a bazillionth of a degree in one direction can be the difference between one point and infinite points of intersection, yet in the other direction 100 degrees won't make a difference at all. That said, I guess the number of points could be given by 1+\delta(\varepsilon), extended to be periodic, and that isn't particularly obscure...
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Totally Tom
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#3228
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#3228
Anyone ever noticed that the best fraction is \displaystyle \frac{14}{6}?
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around
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#3229
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#3229
no, it's blates \dfrac{16}{64}
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Simplicity
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#3230
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#3230
1/3 is

since 1/3=0.333333... then 1=0.99999....

However, clearly 0.9999.. doesn't equal 1. So there is something magical about 1/3.
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majikthise
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#3231
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#3231
Around is definitely correct.
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sonofdot
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#3232
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#3232
I know they got their mathematical legends a bit confused and all, but I would love to see Doctor Who and Fermat in a duel
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My Alt
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#3233
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#3233
(Original post by Simplicity)
1/3 is

since 1/3=0.333333... then 1=0.99999....

However, clearly 0.9999.. doesn't equal 1. So there is something magical about 1/3.
I made that joke ages ago ;_;
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Glutamic Acid
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#3234
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#3234
(Original post by My Alt)
I made that joke ages ago ;_;
That's nothing to be proud of. exp(x).
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majikthise
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#3235
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#3235
(Original post by sonofdot)
I know they got their mathematical legends a bit confused and all, but I would love to see Doctor Who and Fermat in a duel
Haha, think I caught that bit just as I was walking past the living room- did he confuse Fermat and Galois?
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Hathlan
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#3236
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#3236
(Original post by My Alt)
I made that joke ages ago ;_;
I don't understand.
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Simplicity
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#3237
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#3237
(Original post by sonofdot)
I know they got their mathematical legends a bit confused and all, but I would love to see Doctor Who and Fermat in a duel
Yeah, that was shameful. How can they confuse Fermat with Galois?

Even then Galois didn't really die in a duel, but more of a game of russian roulette.

P.S. Even, then there is no evidence that Fermat had a proof and he had ages to actually prove he did.

Besides that it was a good episode.
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baldwin7121
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#3238
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#3238
bum
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Hedgeman49
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#3239
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#3239
(Original post by Simplicity)
1/3 is

since 1/3=0.333333... then 1=0.99999....

However, clearly 0.9999.. doesn't equal 1. So there is something magical about 1/3.
Please, please tell me that you're trying to be funny?
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Totally Tom
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#3240
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#3240
(Original post by Hedgeman49)
Please, please tell me that you're trying to be funny?
Asif you actually think that little of poor simpy.

:suith: Simpster.
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