MHB Taylor Expansion: Wondering Which is Right?

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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
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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)
 
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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)
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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