#math - Fri 9 Mar 2007 between 03:15 and 03:29

NY Lost Funds



excelbluelet y = x + Sqrt[x^2 + 2x], find the limit on ln y
and then after doing that, get the proper thing by doing e^(limit of ln y) again
JabberWalkieprobably multiply by x-Sqrt[x^2+2x]
and divide..
godlingthat gives me -2x in the numerator
hang on
yeah, -2x/x-Sqrt[x^2+2x]
excelblueyay, L'Hospital :)
kmhzbrown : actually i kinda told you nonsense go with excelblue's substitution
excelbluethe place you need to go when you have a math problem :P
zbrown;
lol ok
excelbluewell, you could use the Euler formula definitions of sin and cos, work from there using partial fractions
kmhzbrown : because partial fraction of that will yield you the expression you already have
excelbluejust don't expect to be done within 5 pages
(of tiny work, that is)
kmhzbrown : if you think about it for a minute you'll see why
zbrownya just did
godlingexcelblue: was the bit about e^... towards me?
excelblueyeah
godlingi see
excelblueit's one of the hardest parts about first year calculus IMO
unless you make lots of simple arithmetic errors - then it becomes the riemann sum w/o a calculator
zbrownexcelblue: ok so in the case of the int((x+1)/(x^2+4)^2), one side is x, one side is 2, and one side is (x^2+4) right?
godlingexcelblue: how is e involved in computing the limit of this function?
zbrownfor trig substitution that is
kmhzbrown : not quite
zbrownkmh: trying to figure it out, not sure where im going wrong?
kmhint((x+1)/(x^2+4)^2)= int((x)/(x^2+4)^2)+ int(1/(x^2+4)^2)
first part is handled with u=x^2+4, second part with x=tan(u)
zbrownkmh: well ok... so for the sides of the triangle, one side is x, one is 2, and one is x^2+4?
kmhexcelblue*s suggestion was concerning the 2nd part of the sum
zbrown : oh you mean a triangle to visualize the trig substitution ?
zbrownkmh: ya
kmhok
zbrown : i don't think you need that but yes it looks ok
zbrown : basically it is pythagoras
zbrownkmh: ok thanks
kmh: just helps me visualize w/ the triangle :)
actionzbrown likes pictures
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