New to Mathematica - comp.soft-sys.math.mathematica

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.

On May 3, 9:40 am, "J.Jack.J." <jack.j.jep...@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.

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...@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@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

On 05/05/2012 09:16, J.Jack.J. wrote: > On May 3, 9:40 am, "J.Jack.J."<jack.j.jep...@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. > If you press F1, you will enter the help system. Once there, you could try looking up log and product. The examples will give you the basics of Mathematica syntax, and you can cut and paste them into your notebook, and alter them as desired. David Bailey http://www.dbaileyconsultancy.co.uk

Thanks for replying. Responses embedded: On May 6, 8:24 am, Murray Eisenberg <mur...@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.

On May 6, 8:25 am, David Bailey <d...@removedbailey.co.uk> wrote: > On 05/05/2012 09:16, J.Jack.J. wrote: > > > > > > > > > > > On May 3, 9:40 am, "J.Jack.J."<jack.j.jep...@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 whethe= r > >> 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. > > If you press F1, you will enter the help system. Once there, you could > try looking up log and product. The examples will give you the basics of > Mathematica syntax, and you can cut and paste them into your notebook, > and alter them as desired. > > David Baileyhttp://www.dbaileyconsultancy.co.uk Have tried and failed, tried and failed, with this method! Can somebody give me the requisite inputs? Note that I need to add the condition that x be the highest integer for which <definition of r(k)>. With thanks.

On Mon, 07 May 2012 01:30:43 +0100, J.Jack.J. <jack.j.jepper@googlemail.com> wrote: > On May 6, 8:25 am, David Bailey <d...@removedbailey.co.uk> wrote: >> On 05/05/2012 09:16, J.Jack.J. wrote: >> >> > On May 3, 9:40 am, "J.Jack.J."<jack.j.jep...@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. >> >> If you press F1, you will enter the help system. Once there, you could >> try looking up log and product. The examples will give you the basics of >> Mathematica syntax, and you can cut and paste them into your notebook, >> and alter them as desired. >> >> David Baileyhttp://www.dbaileyconsultancy.co.uk > > > Have tried and failed, tried and failed, with this method! Can > somebody give me the requisite inputs? Note that I need to add the > condition that x be the highest integer for which <definition of > r(k)>. > > With thanks. > Just give up. There is no such integer x for any integer k between 0 and 3000 inclusive, and Mathematica can prove it: Exists[ x, x \[Element] Integers, (x/Log[x]) (1 + 1/Log[x]) == 541/5 + k && k \[Element] Integers && 0 <= k <= 3000 ] // Resolve False However, your problem was not well stated from the beginning, so I doubt if you really meant what you wrote above. Here is one of the 3^3001 possible solutions (type this into Mathematica; it is an absurdly large number, too big to post here) to your problem consistent with the definition of r(k) originally given, and with the additional assumption that r(k) is the largest *real* x such that (x/Log[x]) (1 + 1/Log[x]) == 541/5 + k: r[k_?NumericQ] := Block[{x}, x /. FindRoot[ (x/Log[x]) (1 + 1/Log[x]) == 541/5 + k, {x, 5}, WorkingPrecision -> $MachinePrecision ] ]; Product[1 - 1/r[k], {k, 0, 3000}] 0.6308341356354897 I don't know if you will be able to find a closed form for this, but if an approximation good to 100 places will do, you could use the second root of: 0 == -70 - 13047 x + 9183 x^2 - 5019 x^3 + 35491 x^4 - 3680 x^5 - 14978 x^6 + 6707 x^7 + 28462 x^8 + 9675 x^9 + 30267 x^10 + 19638 x^11 + 3179 x^12 + 15750 x^13 - 21197 x^14 - 5200 x^15 + 37078 x^16 - 4243 x^17 + 10128 x^18 - 15288 x^19 + 27082 x^20

Please start over and carefully rewrite your question, stating exactly what you are trying to calculate. Take the trouble to use clear language. When you do so, I and others here will be glad to help if we can. If you have more general questions about how to tell Mathematica to produce what you want, please also state them clearly, so that some of us can help. For better or worse, learning how to use Mathematica is a significant job. Do as much studying as possible on your own, then if you are really stumped, ask for help with a specific difficulty. Ralph On Sun, May 6, 2012 at 8:29 PM, J.Jack.J. <jack.j.jepper@googlemail.com> wrote: > On May 6, 8:25 am, David Bailey <d...@removedbailey.co.uk> wrote: >> On 05/05/2012 09:16, J.Jack.J. wrote: >> >> >> >> >> >> >> >> >> >> > On May 3, 9:40 am, "J.Jack.J."<jack.j.jep...@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. >> >> If you press F1, you will enter the help system. Once there, you could >> try looking up log and product. The examples will give you the basics of >> Mathematica syntax, and you can cut and paste them into your notebook, >> and alter them as desired. >> >> David Baileyhttp://www.dbaileyconsultancy.co.uk > > > Have tried and failed, tried and failed, with this method! Can > somebody give me the requisite inputs? Note that I need to add the > condition that x be the highest integer for which <definition of > r(k)>. > > With thanks. >

Build one step at a time. For a given value of k (say k=2) use exact values (i.e., 1082/10 rather than 108.2) and restrict the domain of Solve to Reals With[{k = 2}, Solve[(x/Log[x]) (1 + 1/Log[x]) == 1082/10 + k, x, Reals]] {{x -> Root[{-551 Log[#1]^2 + 5 #1 + 5 Log[#1] #1 &, 0.91655356819195758220}]}, {x -> Root[{-551 Log[#1]^2 + 5 #1 + 5 Log[#1] #1 &, 1.11136974029598568569}]}, {x -> Root[{-551 Log[#1]^2 + 5 #1 + 5 Log[#1] #1 &, 611.73540466624378639}]}} Use N to convert Root objects to their values (last entry in Root object) With[{k = 2}, Solve[ (x/Log[x]) (1 + 1/Log[x]) == 1082/10 + k, x, Reals] // N] {{x -> 0.916554}, {x -> 1.11137}, {x -> 611.735}} Use ReplaceAll to extract values from rules With[{k = 2}, x /. Solve[ (x/Log[x]) (1 + 1/Log[x]) == 1082/10 + k, x, Reals] // N] {0.916554, 1.11137, 611.735} Use Part, Last, or Max to extract largest value With[{k = 2}, x /. Solve[ (x/Log[x]) (1 + 1/Log[x]) == 1082/10 + k, x, Reals][[-1]] // N] 611.735 Convert to a function of k with k restricted to Integers r[k_Integer] := x /. Solve[ (x/Log[x]) (1 + 1/Log[x]) == 1082/10 + k, x, Reals][[-1]] // N r[2] 611.735 Plot of r ListPlot[{#, r[#]} & /@ Range[-20, 20, 4], Frame -> True, Axes -> False] Define your Product as a function of kmax p[kmax_Integer] := Product[1 - 1/r[k], {k, 0, kmax}] p[10] 0.982714 This is quite slow so calculation for large values of kmax will take considerable time. Plot of p ListPlot[{#, p[#]} & /@ Range[0, 20, 2], Frame -> True, Axes -> False] Bob Hanlon On Sun, May 6, 2012 at 8:28 PM, J.Jack.J. <jack.j.jepper@googlemail.com> wr= ote: > Thanks for replying. Responses embedded: > > > On May 6, 8:24 am, Murray Eisenberg <mur...@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. >