Question

Problem with Taylor series?

Answer 1

Write a second degree polynomial equation of McLaurin for the function \[F(x)=(f(x))^{g(x)}\] knowing that f(x) ang g(x) taylor series are respectively \[T _{f}^{}(x)=2-3x+2x ^{2} \]and\[T _{g}^{}(x)=x-3x ^{2}\]

Answer 2

hmm

what are the derivatives of f^g ?

Answer 3

spose y = f^g

ln(y) = g ln(f)

y'/y = g' ln(f) + g f'/f

y' = yg' ln(f) + yg f'/f

Answer 4

a maclaurin and a taylor are basically the same thing, so arent we just creating a series of coefficients?

Answer 5

Mac are centered about x=0, which prolly reduces the work by a ton

Answer 6

f(0) ^g(0) = 2^0 = 1 for the first coefficient, and also tells us y=1

y' = yg' ln(f) + yg f'/f

y' = g' ln(2) + 0 f'/2

y' = g' ln(2), g' = 1 - <x>, at x=0 is 1

y' = ln(2)

so far we have 1 + ln(2)x + y'' x^2/2!

what is y''(0) ?

Answer 7

wait a sec

Answer 8

can you please explain to me the first part?

Answer 9

I did not understand this:
spose y = f^g

ln(y) = g ln(f)

y'/y = g' ln(f) + g f'/f

y' = yg' ln(f) + yg f'/f

Answer 10

what did you do here?

Answer 11

i took the derivative of f^g

Answer 12

its best to use logs to play with exponential functions like this

Answer 13

But the derivative of lny is 1/y and not y'/y

Answer 14

for example: what is the derivative of 2^x?

y = 2^x

ln(y) = x ln(2)

y'/y = x' ln(2) + x (0)

y' = y ln(2)

y' = 2^x ln(2)

Answer 15

the derivative of ln(y) .. with respect to x, is a chain rule

Answer 16

SO you mean that everytime I have to do the derivative of lny I should write y'/y?

Answer 17

the derivative of ln(y) *with respect to y* is 1/y ... but that is not what we are working out here. think implicit

Answer 18

yes

Answer 19

whnever you have the derivative with respect to x, you should also write x' .... but it just so happens that x' = dx/dx = 1

Answer 20

y' = dy/dx so we cant automatically assume it is equal to 1

Answer 21

Thank you so much

Answer 22

youre welcome

solving for y'', at x=0 should give us the x^2 coefficient of our Mac

Answer 23

y' = yg' ln(f) + yg f'/f

y'' = y' g' ln(f) + y g'' ln(f) + yg' f'/f
+ y' g f'/f + y g' f'/f + yg f''/f - yg f'/f^2

if i kept it in sight lol

Answer 24

Omg you're so fast, slow down a little bit xD

Answer 25

f(0)=2−3x+2x2 = 2
f'(0) = -3+4x = -3
f''(0) = 4 = 4

g(0)=x−3x2 = 0
g'(0)=1−6x = 1
g''(0)=−6 = -6

y(0) = 2^0 = 1
y'(0) = ln(2)

y''(0) = ln(2) ln(2) -6 ln(2) - 3/2
- 3/2

y''(0) = ln(2)( ln(2) -6) - 3

with any luck lol

Answer 26

http://www.wolframalpha.com/input/?i=y%3D%282-3x%2B2x%5E2%29%5E%28x-3x%5E2%29%2C+y%3D1%2Bx*ln%282%29%2Bx%5E2*%28ln%282%29%28ln%282%29-6%29-3%29%2F2

looks good to me :)

Answer 27

sorry, i was just having too much fun with this :)

Answer 28

Ok there are some passages that I don't get '.'

Answer 29

product rule ... a lot of work with product rules for y'' is all

Answer 30

y' = yg' ln(f) + yg f'/f
3 functions 4 functions

[abc]' = a'bc + ab'c + abc'

[abcd]' = a'bcd + ab'cd + abc'd + abcd'

the rest is filling in the 'at 0' values

Answer 31

Ah thanks. But in the original passage you wrote a minus. Is is actually a plus or a mistake?

Answer 32

*it

Answer 33

and thank again for helping me

Answer 34

1/f = f^-1, derivaticve is (-1)f^(-1-1) = - f^(-2)

Answer 35

its just a consequence of the derivative of what the 4th function in the set is all.

Answer 36

slight error, which didnt make a difference, 1/f derives to -f'/f^2 silly me forgot the chain rule ... but that term zeroed out in the end so it was an insignificant error

Answer 37

Ah ok, now I understood how it works. Thanks

Answer 38

You're good at this xD

Answer 39

thnx ;) good luck

Answer 40

Um... can i ask for your help for other problems next time?

Answer 41

if im not busy, sure. this is just what i do when there is a lull in my real job :/

Answer 42

What kind of job do you have?

Answer 43

Sorry for my curiosity xD