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The equation `1 + (x/(1!))^2 + (x^2/(2!))^2 + (x^3/(3!))^2 + ... = 2.3`  has a...

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user8346721 | eNotes Newbie

Posted May 28, 2013 at 12:36 PM via web

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The equation

`1 + (x/(1!))^2 + (x^2/(2!))^2 + (x^3/(3!))^2 + ... = 2.3` 

has a positive solution. Calculate it with an error of less than 10^(-4)

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mathsworkmusic | (Level 1) Educator

Posted May 28, 2013 at 5:54 PM (Answer #1)

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Take quite a large number of terms and use trial and error (bifurcation, say):

eg, solve `1 + (x/(1!))^2 + (x^2/(2!))^2 + (x^3/(3!))^2 + (x^4/(4!))^2 -2.3 = 0`

Crudely, from the first two terms, the solution is 1 to 1 sf.

So, try `x=1` . The is residual = -0.0205

Try `x = 1.01`, residual = 0.0116

Try `x = 1.005`, residual = -0.0045

Try `x = 1.0075`, residual = 0.0035

Try `x = 1.0063`, residual = -0.0003

Try `x = 1.007`, residual =  0.0019

Try `x = 1.0067`, residual =  0.00096

Try `x = 1.0065`, residual =  0.00031

Try `x = 1.0064`, residual = -8.66 x 10^-6

Therefore a suitable solution is x = 1.0064 to 4dp (so that the error is less than 10^-4)

 

 

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mathsworkmusic | (Level 1) Educator

Posted May 28, 2013 at 6:01 PM (Answer #2)

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Check by adding more of the terms of the series that this solution meets the criterion. You will find that the terms diminish in size quickly because of the squared factorial term in the denominator, so this solution does meet the criterion of having an error less than 10^-4.

 

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