MHB Taylor Expansion: Wondering Which is Right?

Click For Summary
The discussion centers on the Taylor expansion of the function f(x, y) = x^2(3y - 2x^2) - y^2(1 - y)^2 at the point (0, 1). One participant presents their expansion, while a friend offers a different result. A third participant references Wolfram Alpha, confirming that the first expansion is correct and the second is not. The original poster expresses gratitude for the clarification. The conversation concludes with a consensus on the accuracy of the first expansion.
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

I want to find the taylor expansion of $f(x, y)=x^2 (3y-2x^2)-y^2 (1-y)^2$ at the point $(0, 1)$ and I got the following:

$$f(x, y)=3x^2-(y-1)^2+3x^2 (y-1)-2 (y-1)^3-2x^4-(y-1)^4$$

but a friend of mine got the following result:

$$f(x, y)=3x^2-(y-1)^2+3x^2 (y-1)+3x (y-1)^2+2(y-1)^3-2x^4-(y-1)^4$$

which of them is the right one?? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

I want to find the taylor expansion of $f(x, y)=x^2 (3y-2x^2)-y^2 (1-y)^2$ at the point $(0, 1)$ and I got the following:

$$f(x, y)=3x^2-(y-1)^2+3x^2 (y-1)-2 (y-1)^3-2x^4-(y-1)^4$$

but a friend of mine got the following result:

$$f(x, y)=3x^2-(y-1)^2+3x^2 (y-1)+3x (y-1)^2+2(y-1)^3-2x^4-(y-1)^4$$

which of them is the right one?? (Wondering)

Hi! (Wave)

According to Wolfram f(x,y) is equal to your expansion, but it is not equal to the second expansion.

So I believe your expansion is the right one. (Nod)
 
I like Serena said:
Hi! (Wave)

According to Wolfram f(x,y) is equal to your expansion, but it is not equal to the second expansion.

So I believe your expansion is the right one. (Nod)

Great! Thank you! (Smile)
 
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
9
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K