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*hoards the cake for herself* :D

oooooh...... okay. i understand!.... :D thanyouuuuuuu! *gives you cake*

OK- for the integral of (x^2 + 1)^(1/2), You said to let x = tan@, correct? My tutor told me to try the problem by substituting: u - x = (x^2 + 1)^(1/2), then squaring both sides to make x^2 disappear. Which I've tried to do- but it doesn't seem to work in actually getting the integral.

oooh- okay- thanks. :) One thing- with the (x^2 + 1)^(1/2) problem, my tutor said something about substituting (x^2 + 1)^(1/2) = u - x which i dont really understand; he said it'd work if I managed to make the x^2's cancel out... 0.o

Hey, i'm stuck on a few integration questions, any help would be very nice... :) Find the indefinite integrals of: 1. (x^2 + 1)^(1/2) 2. cotx / [ln(esinx)] 3. [1/ (2x - x^2)^1/2] Thanks in advance!

e ^ (-i) is not in the form x^x... nor is 10^100 and the only reason the thing says error calc can't handle that because it's too large a number... try doing it on your comp...

haha, i'll say... but can you use the limit as x ---> -0 ? because when x < 0 you dont get a smooth curve... i think... would it matter if it wasnt a smooth curve?

and lim(x --->0) x^x = 1 but... Isnt that... as x approaches... ? not when x = 0? So it wouldnt be the actual answer to x^x, would it? 0.o

we had to draw x^x once... perhaps that has something to do with it... but then again... the graph was undefined at x = 0... so maybe not. just a thought.

lol... yeah, i could've... -.-' oops, oh well. i got it in the end... :) yay!

Thanks for the help, guys. CM's method works! (very smart!) yayyyyyy! =D thanks!

ooh, i see how you have to do it now.. yeah... that makes much more logical sense than mine. for (e), i'm also not quite sure about my answer but here goes: 0 =cos@ + cos2@ + cos3@ + cos4@ = [sin(n@/2)/sin(@/2)][cis(n+1)@/2] ......... where n = 4. 0 = [ sin2@ / sin(@/2) ] *...

Okay guys, it's a pretty stupid question, because of the not being allowed to use the double angle rule, but I just cannot see any other way of doing it! If you can come up with a solution, it would be absolutely wonderful! Thanks in advance. Prove that: tan[2@] + cot@ = sec[2@] *...

oooh, okay- thanks! Ive got those tucked away some where. i'll go take a look. =)

ok.. i think i have the solution for (d)- |cos@ + cos2@ + cos3@ +...+ cosn@| = |[sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]| .... (from (c)) The Inequality becomes: |[sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]| <= |cosec(@/2)| multiplying both sides by |sin(@/2)|...

.. 0.o ... oops. lol, yeah, i meant that. i'm stuck on d... :( havent even loked at e yet... geez... i should be memorizing my chem assessment... =P this is too distracting -.-'

(c) cos@ + cos(2@) + cos(3@) + ... + cos(n@) = Re(z + z^2 + z^3 + ... + z^n) = [sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]

for (b) z + z^2 + z^3 + z^4 + .... + z^n = [z(z^n - 1)]/(z-1) ... sum of n terms in a geometric sequence. [z(z^n - 1)]/(z-1) = (cis@{2i*sin(n@ / 2)*[cos(n@ /2) + isin(n@ / 2)]})/(cis@-1) = [2i(sin(n@/2)(cis(2+n)@/2)]/{2i[sin(@/2)][cos(@/2) + isin(@/2)] =...

1. draw y = sin2x 2. draw y = x/2 on the same set of axes- (Important- it MUST be to scale). 3. You will observe 2 points of intersection if you drew it correctly- one at (0,0), and one between x = 0 and x = pi/2. 4. At the 2nd point of intersection, draw a line down from the point to...

Well, in the HSC syllabus they ask us to approximate values graphically. Do you have the 2u fitzpatrick book?