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Re: New to Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg126386] Re: New to Mathematica
  • From: Murray Eisenberg <murray at math.umass.edu>
  • Date: Sun, 6 May 2012 03:23:32 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <jntga4$2c$1@smc.vnet.net> <201205050814.EAA22939@smc.vnet.net>
  • Reply-to: murray at math.umass.edu

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? 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.

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}



On 5/5/12 4:14 AM, J.Jack.J. wrote:
> On May 3, 9:40 am, "J.Jack.J."<jack.j.jep... at googlemail.com>  wrote:
>> Hi,
>>
>>   I have just downloaded my free trial version of Mathematica.
>>   I only need it for one thing (as yet, anyway), and I wonder whether
>> someone can help me in step-by-step fashion to get what I want. I need
>> the solution to the following (I might make amendments to the values):
>>
>> "For any integer k, let r(k) be x such that
>>
>> (x/ln(x))*(1 + 1/ln(x)) = 108.2 + k)
>>
>> product (k = 0 to 3000) (1-1/r(k))"
>>
>> With many thanks in advance.
>
> Can nobody help me with this? Just for a Newbie?
> As most will realise, the first lines define my function and the
> product (k = 0 to 3000) (1-1/r(k))
> is my desired calculation. I need to know what inputs to use.
>
> With thanks in advance.
>

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305



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