3rd order, multi variable taylor polynomial

In summary: My apologies, you are correct. I have made a mistake when multiplying the series. Here is the corrected formula:f(x,y)\sim\sum_{0\leq n+k\leq 3}\frac{\partial ^{n+k}f (0,0)}{\partial x^{n}\partial y^{k}}\cdot\frac{x^{n}}{n!}\cdot\frac{y^{k}}{k!}=f(0,0)+\left( f_{x}(0,0)x+f_{y}(0,0)y\right) +\left(\frac1{2}f_{xx}(0,0)x^2+f_{xy}(0,0)xy +
  • #1
chy1013m1
15
0
any insight to this question? .. i mean.. usually people just do up to order 2..

find the taylor polynomial of order 3 based at (x, y) = (0, 0) for the function f(x, y) = (e^(x-2y)) / (1 + x^2 - y)

how large do you have to take k so that the kth order taylor polynomial f about (0, 0) approximates f within 0.45 for
|x| < sqrt(x^2 + y^2) <= 1/10

my guess is...3rd order.. otherwise they won't be explicitly asking us to for the 3rd order?
 
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  • #2
chy1013m1 said:
find the taylor polynomial of order 3 based at (x, y) = (0, 0) for the function f(x, y) = (e^(x-2y)) / (1 + x^2 - y)

The taylor series expansion of a function [tex]f(x,y)[/tex] about the point [tex](x_0,y_0)[/tex] is given by

[tex]f(x,y)=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\partial ^{n}f (x_0,y_0)}{\partial x^{n}}\cdot\frac{\partial ^{k}f (x_0,y_0)}{\partial y^{k}}\cdot\frac{(x-x_0)^{n}}{n!}\cdot\frac{(y-y_0)^{k}}{k!}[/tex]​

notice that to obtain the third order Taylor polynomial, we need only sum over [tex]n+k\leq 3[/tex]

We have need of some calculations...

[tex]f(x,y)=\frac{e^{x-2y}}{(1+x^2-y)^2}[/tex]
[tex]\Rightarrow f(0,0)=1[/tex]

[tex]f_{x}(x,y)=e^{x-2y}\frac{1-2x+x^2-y}{(1+x^2-y)^2}[/tex]
[tex]\Rightarrow f_{x}(0,0)=1[/tex]

[tex]f_{xx}=e^{x-2y}\frac{-1-4x+8x^2-4x^3+x^4+4xy-2x^2y+y^2}{(1+x^2-y)^3} [/tex]
[tex]\Rightarrow f_{xx}(0,0)=-1[/tex]

[tex]f_{xxx}(x,y)=e^{x-2y}\frac{-5+18x+15x^2-36x^3+21x^4-6x^5+x^6+9y-12xy-18x^2y+12x^3y-3x^4y-3y^2-6xy^2+3x^2y^2-y^3}{(1+x^2-y)^4}[/tex]
[tex] \Rightarrow f_{xxx}(0,0)=-5[/tex]

[tex]f_{y}(x,y)=-e^{x-2y}\frac{1+2x^2-2y}{(1+x^2-y)^2}[/tex]
[tex]\Rightarrow f_{y}(0,0)=-1[/tex]

[tex]f_{yy}(x,y)=2e^{x-2y}\frac{1+2x^2+2x^4-2y-4x^2y+2y^2}{(1+x^2-y)^3}[/tex]
[tex]\Rightarrow f_{yy}(0,0)=2[/tex]

[tex]f_{yyy}(x,y)=-2e^{x-2y}\frac{1+6x^2+6x^4+4x^6-6y-12x^2y-12x^4y+6y^2+12x^2y^2-4y^3}{(1+x^2-y)^4}[/tex]
[tex]\Rightarrow f_{yyy}(0,0)=-2[/tex]

Our Taylor polynomial of order three is then

[tex]f(x,y)\sim\sum_{0\leq n+k\leq 3}\frac{\partial ^{n}f (0,0)}{\partial x^{n}}\cdot\frac{\partial ^{k}f (0,0)}{\partial y^{k}}\cdot\frac{x^{n}}{n!}\cdot\frac{y^{k}}{k!}[/tex]
[tex]=f(0,0)+f_{x}(0,0)x+f_{y}(0,0)y +\frac1{2}f_{xx}(0,0)x^2+f_{x}(0,0)f_{y}(0,0)xy+\frac1{2}f_{yy}(0,0)y^2[/tex]​
[tex]+\qquad\qquad\frac1{6}f_{xxx}(0,0)x^3+\frac1{2}f_{xx}(0,0)f_{y}(0,0)x^2y+\frac1{2}f_{x}(0,0)f_{yy}(0,0)xy^2+\frac1{6}f_{yyy}(0,0)y^3[/tex]
[tex]\boxed{=1+x-y-\frac1{2}x^2-xy+y^2-\frac5{6}x^3+\frac1{2}x^2y+xy^2-\frac1{3}y^3}[/tex]​
 
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  • #3
IGNORE LAST POST: I had the wrong formula for a multivariate Taylor series. Here's the fix:

The taylor series expansion of a function [tex]f(x,y)[/tex] about the point [tex](x_0,y_0)[/tex] is given by

[tex]f(x,y)=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty} \frac{\partial ^{n+k}f (x_0,y_0)}{\partial x^{n}\partial y^{k}} \cdot\frac{(x-x_0)^{n}}{n!} \cdot\frac{(y-y_0)^{k}}{k!}[/tex]​

notice that to obtain the third order Taylor polynomial, we need only sum over [tex]n+k\leq 3[/tex]

We have need of some calculations...

[tex]f(x,y)=\frac{e^{x-2y}}{(1+x^2-y)^2}[/tex]
[tex]\Rightarrow f(0,0)=1[/tex]

[tex]f_{x}(x,y)=e^{x-2y}\frac{1-2x+x^2-y}{(1+x^2-y)^2}[/tex]
[tex]\Rightarrow f_{x}(0,0)=1[/tex]

[tex]f_{xx}=e^{x-2y}\frac{-1-4x+8x^2-4x^3+x^4+4xy-2x^2y+y^2}{(1+x^2-y)^3} [/tex]
[tex]\Rightarrow f_{xx}(0,0)=-1[/tex]

[tex]f_{xxx}(x,y)=e^{x-2y}\frac{-5+18x+15x^2-36x^3+21x^4-6x^5+x^6+9y-12xy-18x^2y+12x^3y-3x^4y-3y^2-6xy^2+3x^2y^2-y^3}{(1+x^2-y)^4}[/tex]
[tex] \Rightarrow f_{xxx}(0,0)=-5[/tex]

[tex]f_{y}(x,y)=-e^{x-2y}\frac{1+2x^2-2y}{(1+x^2-y)^2}[/tex]
[tex]\Rightarrow f_{y}(0,0)=-1[/tex]

[tex]f_{yy}(x,y)=2e^{x-2y}\frac{1+2x^2+2x^4-2y-4x^2y+2y^2}{(1+x^2-y)^3}[/tex]
[tex]\Rightarrow f_{yy}(0,0)=2[/tex]

[tex]f_{yyy}(x,y)=-2e^{x-2y}\frac{1+6x^2+6x^4+4x^6-6y-12x^2y-12x^4y+6y^2+12x^2y^2-4y^3}{(1+x^2-y)^4}[/tex]
[tex]\Rightarrow f_{yyy}(0,0)=-2[/tex]

[tex]f_{xy}(x,y)=-e^{x-2y}\frac{1+3x^2-4x^3+2x^4-3y+4xy-4x^2y+2y^2}{(1+x^2-y)^3}[/tex]
[tex]\Rightarrow f_{xy}(0,0)=-1[/tex]

[tex]f_{xyy}(x,y)=2e^{x-2y}\frac{1-2x+3x^2+4x^4-4x^5+2x^6-3y-8x^2y+8x^3y-6x^4y+4y^2-4xy^2+6x^2y^2-2y^3}{(1+x^2-y)^4}[/tex]
[tex]\Rightarrow f_{xyy}(0,0)=2[/tex]

[tex]f_{xxy}(x,y)=-e^{x-2y}\frac{1-8x^2-8x^3+17x^4-8x^5+2x^6+8xy-18x^2y+16x^3y-6x^4y+y^2-8xy^2+6x^2y^2-2y^3}{(1+x^2-y)^4}[/tex]
[tex]\Rightarrow f_{xxy}(0,0)=-1[/tex]

Our Taylor polynomial of order three is then

[tex]f(x,y)\sim\sum_{0\leq n+k\leq 3}\frac{\partial ^{n+k}f (0,0)}{\partial x^{n}\partial y^{k}}\cdot\frac{x^{n}}{n!}\cdot\frac{y^{k}}{k!}[/tex]
[tex]=f(0,0)
+\left( f_{x}(0,0)x+f_{y}(0,0)y\right) [/tex]
[tex]+\left(\frac1{2}f_{xx}(0,0)x^2+f_{xy}(0,0)xy +\frac1{2}f_{yy}(0,0)y^2\right)[/tex]​
[tex]+\left(\frac1{6}f_{xxx}(0,0)x^3+\frac1{2}f_{xxy}(0,0)x^2y +\frac1{2}f_{xyy}(0,0)xy^2+\frac1{6}f_{yyy}(0,0)y^3\right)[/tex]

[tex]\boxed{=1+(x-y)+\left( -\frac1{2}x^2-xy+y^2\right) +\left( -\frac5{6}x^3-\frac1{2}x^2y+xy^2-\frac1{3}y^3\right)}[/tex]​
 
  • #4
for the numerator:

[tex]\exp^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...[/tex]

so we can get:
Expression N:

[tex]\exp^{x-2y}=1+(x-2y)+\frac{(x-2y)^{2}}{2!}+\frac{(x-2y)^{3}}{3!}+...[/tex]

for the denominator:

[tex]\frac{1}{1-x} = 1+x+x^{2}+x^{3}+...[/tex]

so we can get:
Expression D:

[tex]\frac{1}{1-(y-x^{2})} = 1+(y-x^{2})+(y-x^{2})^{2}+(y-x^{2})^{3}+...[/tex]

Get the product of N*D we can get:

[tex]\frac{\exp^{x-2y}}{1+x^{2}-y)} =1+(x-y)+\left( -\frac1{2}x^2-xy+y^2\right) +\left( -\frac5{6}x^3-\frac1{2}x^2y+xy^2-\frac1{3}y^3\right)+...[/tex]
 
  • #5
I don't understand how you multyplied in that formula , you have 10 terms , shouldn'd you have 16? 4 by 4 since there are 0-3 for x and 0-3 for y ??
 

Related to 3rd order, multi variable taylor polynomial

What is a 3rd order, multi variable Taylor polynomial?

A 3rd order, multi variable Taylor polynomial is a mathematical expression that approximates a function using a series of polynomial terms up to the third degree. It takes into account multiple variables and their respective derivatives to create a more accurate approximation of the function.

What is the purpose of using a 3rd order, multi variable Taylor polynomial?

The purpose of using a 3rd order, multi variable Taylor polynomial is to approximate a function with higher accuracy than a simple linear or quadratic approximation. It takes into account multiple variables and their higher-order derivatives, making it a more precise representation of the function.

How is a 3rd order, multi variable Taylor polynomial calculated?

A 3rd order, multi variable Taylor polynomial is calculated by taking the function's value and its derivatives at a specific point, and plugging them into the general form of a Taylor polynomial. The polynomial is then expanded to include higher-order terms up to the third degree, resulting in a more accurate approximation of the function.

What are the advantages of using a 3rd order, multi variable Taylor polynomial?

One advantage of using a 3rd order, multi variable Taylor polynomial is that it provides a more precise approximation of a function compared to simpler methods. It also allows for the evaluation of the function at different points, providing insight into its behavior and properties. Additionally, the use of multiple variables allows for a more comprehensive analysis of the function.

What are some limitations of using a 3rd order, multi variable Taylor polynomial?

One limitation of using a 3rd order, multi variable Taylor polynomial is that it can only approximate a function within a limited range of values. Outside of this range, the approximation may become increasingly inaccurate. Additionally, calculating higher-order derivatives can be time-consuming and computationally intensive, making it impractical for certain applications.

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