Question

help! finding a taylor polynomial of f(x,y)= sin(x+y)/x+y at point (0,0)

Answer 1

hmm, i ve never tried to do a multivariable taylor before; im sure its based on teh same principles tho

Answer 2

gonna have to download the chrome to be of any use on this site ... i need to read up on this a bit as well

Answer 3

in the mean time this migh tbe useful

http://crab.rutgers.edu/~maslen/Courses/PChemII/1_Maths_Review/multi_variable_taylor_series_homework_solutions_totex.pdf

Answer 4

interesting thats
awesum enough

Answer 5

f
fx
fxx
fy
fyy
fxy

are all the taylor stuff that correspond to their variable parts too

Answer 6

\[ f= sin(x+y)\ (x+y)^{-1}\]
\[ fx= cos(x+y)\ (x+y)^{-1}+sin(x+y)\ -(x+y)^{-2}\]
i have got to write this down on paper; cant keep track of it in this accursed latex coding

Answer 7

\[f= \frac{sin(x+y)}{(x+y)}\]
\[fx= \frac{cos(x+y)}{(x+y)}-\frac{sin(x+y)}{(x+y)^2}\]
\[fxx= -\frac{sin(x+y)}{(x+y)}-2\frac{cos(x+y)}{(x+y)^2}+2\frac{sin(x+y)}{(x+y)^3}\]

now for the y parts

Answer 8

actually, the y parts i beleive are the exact same things as the x parts since there is no variation among the variables that would consititute a major change up.

so on to the xy part :)

Answer 9

yep, since all we do is pop out +1s all over the place the xy parts the same as an xx part as well

Answer 10

so we double up the fx: fx fy
we get to more versions of fxx: fxx fyy fxy

and we should be good to go

Answer 11

the issue of course now is that there is no place for 0,0 to be in the denominator since /0 is bad mojo

Answer 12

ow..thats interesting

Answer 13

i checked with the wolf and it likes my deriving :) but the issue I still have is in determining values using the point 0,0

Answer 14

if its a third degree what happens?

Answer 15

we still get the x+y on the bottom so its really not gonna matter how many iterations we do; they will all fail once you put in the 0,0

Answer 16

http://web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/CalcPlot3D.htm

we might be able to take the limit as this apporaches 0,0

Answer 17

in that sense it approaches 1

Answer 18

lim (x,y) -(pi/2;pi/2) of cosxsiny+ytanx

Answer 19

cos(pi/2) sin(pi/2) = 0 so thats not an issue

Answer 20

y going to pi/2 isnt an issue either but might play into the picture if tanx dont go bad

Answer 21

sin(x)
----- as x goes to pi/2 hmmm
cos(x)

Answer 22

i beleive that blows up into infinity but id have to chk with the wolf to be certain

Answer 23

http://www.wolframalpha.com/input/?i=limit+tan%28x%29+x+to+pi%2F2

yeah, so the limit does not exist

Answer 24

thank you that helped