- #1

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Can someone please let me know how to find x in

*π/4=arctan(x+arctan(x+...))*?- Thread starter XYZ313
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- #1

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Can someone please let me know how to find x in

*π/4=arctan(x+arctan(x+...))* ?

- #2

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Let [itex]x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}[/itex]. Then [itex]x^2-2=x[/itex].

Solving the quadratic equation gives us the answer.

Can you do your problem now along similar lines??

- #3

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like

and then use tan addition rule??? I am really confused...

Can you explain further please?

- #4

lurflurf

Homework Helper

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π/4=arctan(x+arctan(x+...))

take tan of both sides

tan(π/4)=tan(arctan(x+arctan(x+...)) )

take tan of both sides

tan(π/4)=tan(arctan(x+arctan(x+...)) )

- #5

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lurflurf: and then what?

tan(π/4) = x + arctan(x+arctan(x+...)) ???

tan(π/4) = x + arctan(x+arctan(x+...)) ???

- #6

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what is tan(tanx)??? Is above equation some kind of sum of geometric progress???

- #7

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is it so easy that no one is going even to comment?

- #8

lurflurf

Homework Helper

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tan(π/4)=tan(arctan(x+arctan(x+...)) )

1=x+arctan(x+arctan(x+...))

1=x+arctan(x+arctan(x+...))

- #9

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then

tan(1-x) = x + arctan(x+arctan(x+...))

and where it is leading???

- #10

lurflurf

Homework Helper

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arctan(x+arctan(x+...)) ?

- #11

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if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...

- #12

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What? You said what this equals in yourif I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...

- #13

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spamiam: right, sorry, I'm really dumb

x = 1 - π/4

Thank you all for bearing with me...

x = 1 - π/4

Thank you all for bearing with me...

- #14

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This is dumbest question I have ever asked. Hope this is never going to happen again...

- #15

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You're not dumb--this question is difficult until you see the trick to use.

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