Taylor Expansion: Wondering Which is Right?

In summary, the conversation is about finding the Taylor expansion of a function at a specific point. One person got a result and another person got a different result. After checking with Wolfram, it is determined that the first person's expansion is correct.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

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
  • #2
mathmari said:
Hey! :eek:

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)
 
  • #3
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)
 

Related to Taylor Expansion: Wondering Which is Right?

1. What is a Taylor Expansion?

A Taylor Expansion is a mathematical technique used to approximate a function with a polynomial. It is named after mathematician Brook Taylor and is also known as a Taylor series.

2. How is a Taylor Expansion calculated?

A Taylor Expansion is calculated by taking the derivative of a function at a specific point and evaluating it at that point, then taking the second derivative and evaluating it at the point, and so on. The resulting terms are then combined to form a polynomial that represents the original function.

3. What is the purpose of a Taylor Expansion?

The purpose of a Taylor Expansion is to approximate a function with a simpler polynomial that is easier to work with. This allows for easier calculations and can also help in understanding the behavior of a function near a specific point.

4. How accurate is a Taylor Expansion?

The accuracy of a Taylor Expansion depends on the number of terms used in the polynomial. The more terms included, the more accurate the approximation will be. However, in some cases, the Taylor Expansion may not accurately represent the original function, especially for functions with complex behavior.

5. Can a Taylor Expansion be used for all functions?

No, a Taylor Expansion can only be used for functions that are infinitely differentiable. This means that the function must have derivatives of all orders at the point of expansion. If a function is not infinitely differentiable, the Taylor Expansion cannot be used.

Similar threads

I Taylor expansion of f(x)=arctan(x) at infinity
    • Calculus
Replies
2
Views
2K
B Mean-Value Theorem, Taylor's formula, and error estimation
    • Calculus
Replies
3
Views
1K
B Calculating shaded area: I'm getting discrepancies between methods
    • Calculus
Replies
3
Views
552
MHB Evaluate the surface integral $\iint\limits_{\sum}f\cdot d\sigma$
    • Calculus
Replies
1
Views
2K
B Why is this definite integral a single number?
    • Calculus
Replies
20
Views
2K
I Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?
    • Calculus
Replies
9
Views
1K
MHB Definite integral involving a lot of exponentials.
    • Calculus
Replies
2
Views
981
MHB -s8.3.6.46 Find Slant asymptote
    • Calculus
Replies
3
Views
1K
I Lipschitz continuity of vector-valued function
    • Calculus
Replies
3
Views
1K
I The Euler-Lagrange equation and the Beltrami identity
    • Calculus
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top