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[✓] Solve the following inequality?


Here's the code:

NSolve[(10!)/(10^n)*(1/(10 - n)!) < 0.5, n]

but when I use the NSolve Command (as I want a numerical solution), I just get the following:

NSolve[(567 2^(8 - n) 5^(2 - n))/(10 - n)! < 0.5, n]

Please help me with this. Thanks in advance :)

POSTED BY: Raivat Shah
10 months ago

Dear Raivat Shah,

are you looking for a solution over the Reals or Integers? I guess that it is a good idea to start with the equalities such as:

FindInstance[(10!)/(10^n)*(1/(10 - n)!) == 0.5 && n > 0, n, Reals]
{{n -> 4.01848}}

or else

FindInstance[(10!)/(10^n)*(1/(10 - n)!) == 0.5 && n > 0, n, Integers]

You might want to look at:

Plot[{(10!)/(10^n)*(1/(10 - n)!), 0.5}, {n, 0, 20}, PlotRange -> All]


Plot[{(10!)/(10^n)*(1/(10 - n)!), 0.5}, {n, 0, 80}, PlotRange -> All]

Also look at this

Table[(10!)/(10^n)*(1/(10 - n)!), {n, 1, 1000}]

and this (you might want to play with the PlotPoints option)

LogPlot[Abs@{(10!)/(10^n)*(1/(10 - n)!), 1/2}, {n, 0, 60}, PlotRange -> All, PlotPoints -> 10000]

enter image description here

Can you guess what is going on? Once you see that, it might help you to solve the original inequality.

Best wishes,


PS: This here might also be useful:

Plot[n!, {n, -10, 0}]
POSTED BY: Marco Thiel
10 months ago

Dear Marco, Thank you so much. I was not only able to solve the inequality, but you have given me an excellent idea/way to present it graphically. I don't have enough words to thank you. I also realized that Mathematica is so powerful that it can do the computations for very large numbers. For example,

 FindInstance[(10000!)/(10000^n)*(1/(10000 - n)!) == 0.5 && n > 0, n, Reals]
 {{n -> 118.011}}

Thank you once again,

Raivat Shah

POSTED BY: Raivat Shah
10 months ago

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