Question

Find Maclaurin Series for function y:

lny = xy

Answer 1

hi!

Answer 2

Ikram :D

Answer 3

Maclaurin series is just a special type of Taylor series, which is the null case case of Taylor about x=0.
Do you know how to construct Taylor series of the given function ?

Answer 4

I know how to construct Taylor for any function but this.

I have the function in both sides.

Answer 5

oh, you don't know how to make Taylor for f(x,y) function ?

Answer 6

Yeah.

Answer 7

can u make sense of this rule ?
http://prntscr.com/9krawx

Answer 8

know that
ln y=xy
f(x,y)=xy-ln y

Answer 9

@ikram002p , Isn't the question asking for the expansion for function y ?

and lny = xy ?

lny - xy = 0 ?

if we got that , we got the expansion for the 2 variables.

Answer 10

yes my bad!
but the thing is its not separable hmmm
lets think of another way

Answer 11

@TrojanPoem know it does not important... id we separate it or not
lets do it on the casual way

Answer 12

ikram002p
know that
ln y=xy
f(x,y)=xy-ln y

Trojo: The above makes sense to me, but for two reasons I can't be of any significant help right now: 1) Never tired finding a Mac series for a fun. of two variables and (2) have just finished a marathon session with another student and my head is swimming. Glad to hear from you, tho'.

Answer 13

@ikram002p , Do you know about something like this ?


y = e^(xy)

y' = e^(xy) * [ xy' + y]

e^xy * xy' + e^(xy) y = y'

y' ( 1- e^(xy)) = e^(xy)*y

y' = e^(xy)*y/(1-e^(xy))

?

Then x = 0

lny = 0
lny = ln 1

y = 1

y' = e^(0) * 1 / (1-e^0) = 1 * 1/(1-1) = 1/0

Check my calculations.

Answer 14

@ikram002p , I mean do you think of

Answer 15

so as it asked for y only i think we can do this

y'=y^2/1-xy
y''=..... ect

Answer 16

I agree to ur solution, there is no harm for doing that!

Answer 17

BUT if it asked u to write a series with both of x,y then that solution is not valid.

Answer 18

It asks for the expansion of function y.

I didn't take the 2 variables expansion yet.

Answer 19

We had better remember that y > 0, before doing that.

Answer 20

@ikram002p , By the way, How did you get this nice derivative ?

mine is nasty.

Answer 21

would that make a problem @tkhunny ? x_0 of the condition have no strict AND it can be 0.

Answer 22

...and we haven't really lost that when writing y = e^(xy). I just want to keep track of it explicitly. We sometimes wander off and forget that the solution is limited to the open half-plane.

Answer 23

@TrojanPoem i'm lazy http://www.wolframalpha.com/input/?i=%28ln+y%3Dxy%29+first++derivative+

but y>0 has nothing with it , ln y=x+y when x=0 then ln y=y solve that we would get a complex constant which is not important as u haven't studies 2 variables yet.

Answer 24

studied**

Answer 25

i suggest that you write the formula first ... and then lets see

Answer 26

f(x) = f(0) f'(0)x + [ f''(0)x^2 ]/2! +.....

Answer 27

Still can't find how it got that cool derivative.

Answer 28

@ikram002p , what's f(x) here ?

It used to be simple f(x) = e^x

so e^x = ... the expansion

Answer 29

hint
ln y= x+y

(ln y)' =(x+y)'
1/y y' =x'+y'

Answer 30

well y itself ...

Answer 31

@ikram002p Sorry for dumbness
how did you get it ?

lny = x+ y?

Answer 32

why they teasing u kids with such problems xD

Answer 33

wait pardon
lny=xy
1/y y'=x'y+xy'

Answer 34

@ikram002p , My prof doesn't like seeing grade A^+ , A ... till K

Answer 35

Ok , y'/y = xy' + y

y' = xyy' + y^2

Answer 36

@TrojanPoem focus

(ln y)'=(xy)'
(1/y) y' =xy'+y
y'({1/y}-x)=y
y'(1-x/y)=y
y'=y^2/1-xy

Answer 37

@ikram002p , I got it first :P

Answer 38

well i just broke my phone cause of this question :P

Answer 39

Trololo.

Answer 40

What's f(x) here ? xD

Answer 41

y only :P (what ever it is)

Answer 42

Ok another stupid question ?

f(0) = ?

Answer 43

As a eighth grader, I have no clue what this is

Answer 44

I don't know what's the problem with the cute

sinx expansion.

Answer 45

XD

Answer 46

:D

Answer 47

@ikram002p , what is f(0) ? OK you don't have to say it : focus.

Answer 48

ln y= xy
ln y=0y
ln y=0 so y=1

Answer 49

at first i said ln y=x+y AS i made a typo. sorry!

Answer 50

@ikram002p, Never mind.

Thanks a lot. I think that way the question is done.

Answer 51

yw =)

Answer 52

If it has nothing to do with "y > 0", you have created an extension of the problem. That's fine, you just need to know that.

Answer 53

=) everyone knows that

Answer 54

pretty trivial

Answer 55

Disagree. Domain is never trivial. Of course, over the years, I have come to understand that I seem to care more than most others about Domain issues.

Answer 56

i'm just trolling you @tkhunny i understand ur point.
but being sarcasm is one of my way :P *Pardon*