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visits member for 2 years, 7 months
seen Apr 24 '13 at 10:58

Undergraduate theoretical physics student.

Interests:

Physics: all areas of fundamental physics.

Math: all areas related to the above.

(specifically, the interface of theoretical physics and mathematics)


Oct
17
comment Resource request: history of and interconnections between math and physics
@WillieWong: I want to see if - and how - it is possible, since on some <underline>very rare</underline> situations it is the most effectual use of the tool. (Pun very much intended! :) )
Oct
17
comment Resource request: history of and interconnections between math and physics
@Willie Wong: [Note on formatting]: I would prefer to underline the word "seriously" in the question- (both because it more precisely & accurately conveys the 'sense' of my question, and because I want to know how to do it on SE, (I perused Meta.Stackoverflow to find out how - but couldn't make out head or tails.)) -cheers
Oct
16
comment Resource request: history of and interconnections between math and physics
I have this book. - Looking specifically for a book / paper / online video / (and/or other) educational resource on this specific topic (math-physics link) and their historical origin, and interrelations.
Nov
21
comment Show that the function $h(x) = \int^x_0 g(t) \, dt$ is $C^2$ but not $C^3$ at $x = 0$
Yes; now I see. Thank you.
Nov
21
comment Show that the function $h(x) = \int^x_0 g(t) \, dt$ is $C^2$ but not $C^3$ at $x = 0$
Function $f$ is $C^1$ but not $C^0$ at $x = 0$ as $f'(x) = \frac{1}{3}x^{-2/3}$ for $x \neq 0$, and 'undefined' for $x = 0$.
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
So, -to clarify this whole debate-, in the space of real numbers I cannot write an expression like $S$, and claim that I can recast it [as $1 + x^2\,S$] unless the expression is convergent, (or defined within a specified radius of convergence)?
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
... of convergence.]
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
@t.b.: Within the radius of convergence the series is identified with / [or possibly mathematically equivalent to(?)] $\frac{1}{1-x^2}$. My question is: [preliminarily, and from receiving various responses: can we recast it the way we have if $S$ is not convergent?, and] (if we can recast it regardless or not the series is convergent or divergent [which I now believe the answer is a 'no' to]) how is it mathematically justified? .. [Part of my question I think has already been answered since I don't believe you can recast it as above if the series isn't convergent [or outside its domain...
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
If I understand correctly, the series in the question can be recast only if it is convergent; otherwise it can't. [via @Jyrki | ... so, how then are "formal power series" related to the "usual" series above? ...
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
@Jyrki: Ok. Thanks for the response. -- May I courteously request a formal answer to the above question? (Use whatever mathematics you like, as long as your answer is completely rigorous.) - cheers
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
Hi @Mark, thanks for the response. | May I know the formal power series version of your response? (I think that's what I'm searching for.)
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
@Jyrki: This would only apply if the series converged. | However, for this series, the recasting is valid regardless of convergence or divergence of the series, (i.e. the issue revolves around the specific infinite number of terms for the series - which allows recasting). ... My question is: how (in a rigorous mathematical sense) is this mathematically meaningful / (possible)?
Oct
5
comment Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$
Well, I'm basically trying to understand why we can do this recasting. Normally - (I'm assuming) - we cannot recast a finite series. But the above recasting is allowed since the terms go to infinity. ... What is the exact mathematical (i.e. rigorous) basis which allows this manipulation?
Oct
5
comment General question on relation between infinite series and complex numbers
Hi @anon, your answer seems well-thought-out and valuable to someone with complex analysis under their belt, but I'm afraid currently it's somewhat over my head. (I'm still giving it +1 since I will review it once I gain sufficient knowledge / background in the subject.) - cheers
Sep
30
comment General question on relation between infinite series and complex numbers
Can you provide a summary which is comprehensible to a 2nd-3rd year undergrad with modest knowledge of real analysis, and almost none of complex analysis (apart from complex numbers and some basic identities)?
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
[Plus, as per FredrikMeyer's comment ["You want to show that $r^n \rightarrow 0$ as $n \rightarrow \infty$ provided |r|<1. So let ϵ > 0 be given. Then $r^n \lt ϵ$ is equivalent to n ≥ ln ϵ/ ln r. Such an n exists."] I'm not sure why you would claim a "formal" $\epsilon$-$\delta$ proof of the statement would not be applicable.]
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
@Arturo: The issue of the epsilon-delta proof comes from the following: [comment # 5] "first I derive the formula for the finite series as $S_n = a\frac{(r^n−1)}{r−1}$; then, I claim that, if |r| < 1, convergence of the series is equivalent to claiming that: $\lim_{n \rightarrow \infty} |S_{n+1}−S_n| = \lim_{n \rightarrow \infty} |ar^n| = 0$. But I'm wondering how to prove this." ... I'm wondering if this is a valid approach to the initial problem [the first part of]?
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
Srivatsan, +1. Thanks. Appreciate greatly.
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
@Arturo: Can you provide your perspective on how best to prove the above [in question body]? I think my general approach is in the right direction, but maybe you can suggest a better method / methods. -cheers
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
@Fredrik: I've never done a formal $\epsilon - \delta$ proof before. Can you provide a detailed version for a complete newbie to fully understand. ... (Would appreciate greatly.) :)