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Re: New to Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg126400] Re: New to Mathematica
  • From: Jaebum Jung <jaebum at wolfram.com>
  • Date: Tue, 8 May 2012 04:07:35 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

You could try like the following:

r[k_?IntegerQ] :=
 Block[{x}, 
  Max[x /. NSolve[(x/Log[x])*(1 + 1/Log[x]) == 108.2 + k, x, Reals]]]

NProduct[1 - 1/r[k], {k, 0, 3000}]


----- Original Message -----
From: "J.Jack.J." <jack.j.jepper at googlemail.com>
To: mathgroup at smc.vnet.net
Sent: Sunday, May 6, 2012 7:28:55 PM
Subject: [mg126400] Re: New to Mathematica

Thanks for replying. Responses embedded:


On May 6, 8:24 am, Murray Eisenberg <mur... at math.umass.edu> wrote:
> First, perhaps folks were reluctant to respond because this looked like
> it could be a homework exercise.
>
> Second, you don't even have proper Mathematica syntax in your equation
> relating x and k. Did you even try to read the documentation to learn
> the very basics?

I tried and tried for hours but couldn't so much as find any section
that would even tell me how to write the condition that k be an
integer.

 For example, proper syntax for the equation would be:
>
>    (x/Log[x]) (1 + 1/Log[x]) == 108.2 + k
>
> Function arguments must be enclosed in brackets, not parentheses, and
> the equality is a doubled "=" sign. Moreover, your original expression
> had an unbalanced terminal parenthesis.
>

Thanks.

> Third, the equation itself looks really nasty. Aside from the fact that
> it mixes exact formulas with an approximate real (108.2), the left-hand
> side is transcendental.
>
> Fourth, the equation does not seem to uniquely define x as a function of
> k! For example, form the difference between the two sides...
>
>    f[x_] := (x/Log[x]) (1 + 1/Log[x]) - 108.2 - k
>
> ... and plot f for, say, k = 2:
>
>    Plot[Evaluate[f[x] /. k -> 2], {x, 0.5, 2}, Exclusions -> {x ==
= 1},
>   AxesOrigin -> {0, 0}, PlotRange -> {-5, 5}]
>
> The graph crosses the x-axis twice. And indeed, if you use FindRoot with
> initial guesses above and below 1, you'll see that this is so:
>
>     FindRoot[Evaluate[f[x] /. k -> 2], {x, 0.9}]
> {x -> 0.916554}
>
>     FindRoot[Evaluate[f[x] /. k -> 2], {x, 1.1}]
> {x -> 1.11137}
>

Am sorry, I should have said "let r(k) be the highest x such that..."

Can you give me the required inputs? That would help me immensely.




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