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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)


Sep
30
revised Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
added 515 characters in body
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
awarded  Peer Pressure
Sep
29
revised Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
added 241 characters in body; edited title
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.) :)
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
@wnoise: by using the derived formula (partial sum formula) I then claim that if |r| < 1, the expression tends to a limiting values (i.e. 'sum' S) as successive differences tend to zero as n tends to infinity. I need to prove that successive differences (|$S_{n+1} - S_n$|) tend to zero (for n $\rightarrow \infty$). .. I'm currently trying to find the exact epsilon-delta argument to demonstrate this.
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
@wnoise: I need to demonstrate that the series converges; if it does, the sum has meaning. I can't use the concept of sum if I cannot first demonstrate that the series converges.
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: Can you provide a full epsilon-delta proof for this? This is where I'm stuck [for first part].
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
@J.M.: I need a sort of epsilon-delta proof if I'm not mistaken.
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
[Derivation for $S_n = a \frac{(r^n-1)}{r-1}$ was by simple algebraic manipulation. I don't need any assistance with that.] ... The specific issue I have with first part of first proof is, how to prove above claim.
Sep
29
comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
Basically, 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$_{n \rightarrow \infty}$ |$ar^n$| $\rightarrow$ 0. But I'm wondering how to prove this. ...
Sep
29
asked Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$
Sep
17
comment Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors
Ok. Thanks. - regards
Sep
17
comment Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors
Ref: math.stackexchange.com/questions/tagged/education ... Plus, the description field is empty. ...
Sep
17
comment Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors
I've looked through the posts with that tag. I found mostly questions similar to mine. ... I think the posts with that tag are different for the two sites. ... (Need clarification here.)
Sep
17
revised Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors
edited tags
Sep
17
comment Show that $\int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$
Yup! :) ... I'm using it to appetize my tummy. :) ... Once I finish the chapter I plan on using the standard textbooks I have.