Can someone please let me know how to find x in π/4=arctan(x+arctan(x+...)) ?
Nov 7, 2011 #1 XYZ313 21 0 Can someone please let me know how to find x in π/4=arctan(x+arctan(x+...)) ?
Nov 7, 2011 #2 micromass Staff Emeritus Science Advisor Homework Helper Insights Author 22,129 3,300 Let me do a similar one: 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??
Let me do a similar one: 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??
Nov 7, 2011 #3 XYZ313 21 0 micromass: isn't it the use of inverse operator (tan in this case)? like z=arctan(x+arctan(x+...)) and make the use of tanz? and then use tan addition rule??? I am really confused... Can you explain further please?
micromass: isn't it the use of inverse operator (tan in this case)? like z=arctan(x+arctan(x+...)) and make the use of tanz? and then use tan addition rule??? I am really confused... Can you explain further please?
Nov 7, 2011 #4 lurflurf Homework Helper 2,445 140 π/4=arctan(x+arctan(x+...)) take tan of both sides tan(π/4)=tan(arctan(x+arctan(x+...)) )
Nov 7, 2011 #6 XYZ313 21 0 what is tan(tanx)??? Is above equation some kind of sum of geometric progress???
Nov 7, 2011 #8 lurflurf Homework Helper 2,445 140 tan(π/4)=tan(arctan(x+arctan(x+...)) ) 1=x+arctan(x+arctan(x+...))
Nov 7, 2011 #9 XYZ313 21 0 lurflurf: thank you, this is obvious and not the solution then tan(1-x) = x + arctan(x+arctan(x+...)) and where it is leading???
lurflurf: thank you, this is obvious and not the solution then tan(1-x) = x + arctan(x+arctan(x+...)) and where it is leading???
Nov 8, 2011 #11 XYZ313 21 0 if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...
Nov 8, 2011 #12 spamiam 360 0 XYZ313 said: if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread... What? You said what this equals in your first post. It's x that is the unknown.
XYZ313 said: if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread... What? You said what this equals in your first post. It's x that is the unknown.
Nov 8, 2011 #13 XYZ313 21 0 spamiam: right, sorry, I'm really dumb x = 1 - π/4 Thank you all for bearing with me...
Nov 8, 2011 #14 XYZ313 21 0 This is dumbest question I have ever asked. Hope this is never going to happen again...
Nov 8, 2011 #15 spamiam 360 0 You're not dumb--this question is difficult until you see the trick to use.