Question

Taylor's Formula
Write taylor's formula for F(x,y)= sin(x)sin(y) using a=0, b=0, and n=2

Answer 1

umadbro?

Answer 2

He mad son.

Answer 3

:P

Answer 4

what's this from? university maths?

Answer 5

Yup

Answer 6

which year?

Answer 7

which year am i?

Answer 8

yes are you an undergrad?

Answer 9

yes. i'm 2nd year.

Answer 10

n=2, so we only have go up to the second degree correct?

Answer 11

correct.

Answer 12

So then the first term would merely be f(0, 0)=0

Answer 13

Now we need to make a list of 5 partial derivatives\[\large f_{x}\]\[\large f_{y}\]\[\large f_{xx}\]\[\large f_{yy}\]\[\large f_{xy}=f_{yx}\]

Answer 14

\[\sin(h)\sin(k) = hk - (1/6)h(h^2 + 3k^2)\cos \theta h \sin \theta k - (1/6)k (3h^2 + k^2) \sin \theta h \cos \theta k\] that's the answer.

Answer 15

From now on, when I type one of the partial derivatives, assume I'm evaluating it at the point (0, 0).

The formula you now need to calculate is given by\[\large f(0, 0) +f_x+f_y+f_{xy}\cdot ( xy)+{f_{xx} \over 2} \cdot(x^2)+{f_{yy} \over 2}\cdot(x^2)\]

Answer 16

Since your function is originally, \(F(x)=\sin(x)\sin(y)\), the only nonzero term will be the \(f_{xy}\) term.

All other terms will contain a \(\sin(0)=0\) term.

Answer 17

Hence, the taylor polynomial of degree two approximating \(F(x, y)=\sin(x)\sin(y)\) is equal to \(G(x, y)=xy\)

Answer 18

I honestly have no idea where the answer you were given comes from

Answer 19

Comes from the back of the book.

Answer 20

What you were doing is the taylor series i believe. We gotta use the taylor's formula.

Answer 21

Apparently, there's a difference... I have no idea.

Answer 22

As far as I can tell, there's no difference. But if there is, I've never heard of the Taylor Formula.

btw, in the answer you gave me, is there some stuff cut off at the edge? Or is that all of it?

Answer 23

there should be a cos(theta)k at the end of it.

Answer 24

If that is the correct answer, I'm really confused where the theta's came from. Also, is the question given in x and y and the answer in h and k?

Answer 25

Hahaha, then you're in the same boat as me. Eliassaab is an ex math professor, hopefully he'll be able to help out :)

Answer 26

I've got to sleep now, so good luck with this problem.

Answer 27

Thanks KG! I'll let you know how it goes :)

Answer 28

I was going to do this, but then I realized that this is probably harder math than I know. Is this after calculus, or calculus itself?

Answer 29

This is calculus in its most theoretical form.

Answer 30

Oh,I see why I don't get it. THere are partial derivatives in here.

Answer 31

(x,y) lol

Answer 32

If you really need help, go to r/cheatatmathhomework, or math stack exchange

Answer 33

Noooo idea what you are talking about.

Answer 34

Reddit. Or mathstackexchange.

Answer 35

www.reddit.com/r/cheatatmathhomework/
math.stackexchange.com/

Answer 36

good luck ;)

Answer 37

Awww you suck reb.

Answer 38

here is tour ans.. @kevo

Answer 39

I know..but I'm so good at it ;) hehe

Answer 40

so the formula you want to apply reads
F(a+h,b+k)=∑ i,j=0 n−1 1 i!j! ⋅∂ i+j F ∂x i ∂y j (a,b)h i k j +∑ i+j=n 1 i!j! ⋅∂ 2 F ∂x i ∂y j (a+θh,b+θk)h i k j
and tells you how F behaves in a neighbourhood of (a,b) . The corresponding theorem says the for each (h,k) there is some θ∈[0,1] such that this formula holds. Now plug in your F , your n , and your (a,b) .

Answer 41

then you should get your answer

Answer 42

\[F(0,0)
+x
F^{(1,0)}(0,0)+y
F^{(0,1)}(0,0)+\\
\frac{1}{2} \left(x^2
F^{(2,0)}(0,0)+2x y
F^{(1,1)}(0,0)+y^2
F^{(0,2)}(0,0)\right)= x y
\]
All the terms are zero at (0,0) because of sin(x) except the coeeficient of x y. That is why the answer is x y