Current Events > x/2 + (x^2)/4! + (x^3)/6! + ... + (x^n)/(2n)! = 1

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FLOUR
05/07/20 2:43:46 PM
#1:


Similar to the previous problem, the left side of the equation represents an infinite series. What are the real value(s) of x that satisfy the equation?

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MC_BatCommander
05/07/20 2:44:06 PM
#2:


Do your own fucking homework

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UnholyMudcrab
05/07/20 2:44:07 PM
#3:


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Tired-Insomniac
05/07/20 2:44:20 PM
#4:


Elon Musk had another kid?

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R1masher
05/07/20 2:44:33 PM
#5:


Im no botanist, but yikes

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MC_BatCommander
05/07/20 2:45:11 PM
#6:


Tired-Insomniac posted...
Elon Musk had another kid?

LOL, legit made me laugh. Good one

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Austin_Era_II
05/07/20 2:46:38 PM
#7:


n3rds

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Anteaterking
05/07/20 2:50:35 PM
#8:


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philsov
05/07/20 2:52:34 PM
#9:


1....

1/2 + 1/4 + 1/6 + 1/8.............. = 0.99999999999999999999999~ = 1
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EmbraceOfDeath
05/07/20 2:56:43 PM
#10:


philsov posted...
1....

1/2 + 1/4 + 1/6 + 1/8.............. = 0.99999999999999999999999~ = 1
That's not the series being used here. If x=1, it would be

1/2 + 1/24 + 1/720 + ... + 1/(2n)! , which is going to be less than 1.

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FLOUR
05/08/20 6:03:43 PM
#11:


anyone?

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EmbraceOfDeath
05/08/20 8:02:37 PM
#12:


FLOUR posted...
anyone?
Don't know the exact value, but it appears to be somewhere between 1.84 and 1.85.

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DirkDiggles
05/08/20 8:13:12 PM
#13:


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