Maclaurin series expansion 4th order I'm trying to grasp how Maclaurin series with two variables develop into the forth order. I would appreciate if you correct my guess below.
$$f(x,y)=f(0,0)+f_{1}(0,0)x+f_{2}(0,0)y\\
+\frac{1}{2!}(f_{11}(0,0)x^2+f_{12}(0,0)xy+f_{22}(0,0)y^2)\\
+\frac{1}{3!}(f_{111}(0,0)x^3+f_{112}(0,0)x^2y+f_{122}(0,0)xy^2+f_{222}(0,0)y^3)\\
+\frac{1}{4!}(f_{1111}(0,0)x^4+f_{1112}(0,0)x^3y+f_{1122}(0,0)x^2y^2+f_{1222}(0,0)xy^3+f_{2222}(0,0)y^4)\\$$
EDIT: Updated the coefficients. Tried to use
https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Multivariable_Calculus/3%3A_Topics_in_Partial_Derivatives/Taylor__Polynomials_of_Functions_of_Two_Variables#mjx-eqn-tpn
 A: There is actually an easy way to remember this series, consider:
Consider the following expression:
$$ \frac{1}{n!} \bigg[ \frac{ \partial}{\partial x} + \frac{ \partial }{ \partial y} \bigg]^n = \sum_{i+j=n}^n \frac{1}{k! j!} \frac{\partial^i}{\partial x^i} \frac{\partial^{j} }{\partial y^{j} }$$
So, we can write $(i,j)$ term as:
$$ C_{ij} =  \frac{1}{k! j!} \frac{\partial^i}{\partial x^i} \frac{\partial^{j} }{\partial y^{j} }$$
Then, you'd notice that, the taylor expansion has the form:
$$ f(x,y) = \sum_{i,j=0}^{\infty} \bigg[ C_{ij} f(x,y) \bigg]_{x=a,y=b} (x-a)^i (y-b)^j$$
So, suppose for the 'nth' degree expansion, you want the $i+j=n$ .. because we want homogeneity of the polynomial. Hence, we can write as:
$$ f(x,y) = \sum_{n=0}^{\infty} \bigg[\sum_{i=0}^{n} \bigg[ C_{i,(n-i)} f(x,y) \bigg]_{x=a,y=b} (x-a)^i (y-b)^{n-i}\bigg]$$
The inner sum gives you the component introduced by increasing the degree of Taylor expansion from $n-1 $ to $n$. Let's take it out:
$$ Q= \bigg[\sum_{i=0}^{n} \bigg[ C_{i,(n-i)} f(x,y)\bigg]_{x=a,y=b} (x-a)^i (y-b)^{n-i}\bigg]$$
For $n=2$,
$$ Q= \bigg[ \bigg[C_{0,2} f(x,y)\bigg]_{a,b} (y-b)^2 + \bigg[C_{1,1} f(x,y) \bigg]_{a,b} (x-a)(y-b)+ \bigg[C_{2,0} f(x,y)\bigg]_{a,b}  (x-a)^2 \bigg]$$
computing coefficients:
$$ Q= \bigg[  \bigg[\frac{\partial^2}{\partial y^2}f(x,y)\bigg]_{a,b} (y-b)^2 + \bigg[\frac{\partial^2}{\partial y \partial x}f(x,y) \bigg]_{a,b} (x-a)(y-b)+ \bigg[\frac{\partial^2}{\partial x^2}ff(x,y)\bigg]_{a,b}  (x-a)^2 \bigg]$$
For the additional terms gained by changing order from three to four,
$$ Q= \bigg[\sum_{i=0}^{4} \bigg[ C_{i,(4-i)} f(x,y)\bigg]_{x=a,y=b} (x-a)^i (y-b)^{4-i}\bigg]$$
Now I'll leave it up to you to evaluate it 

Explanation on the notation:
$$ \frac{ \partial^i }{\partial x^i} \frac{ \partial^j}{\partial y^j} = \frac{ \partial^{i+j} }{ \partial x^i \partial y^j}$$
$$ \frac{ \partial^i }{\partial x^i} f(x,y)= \frac{ \partial^i f(x,y)}{ \partial^i x}$$
$$ \bigg( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} \bigg)^2= \frac{\partial^2}{\partial x^2} + 2 \frac{\partial }{\partial x} \frac{\partial }{\partial y} + \frac{ \partial^2 }{\partial y^2}$$
$$ \frac{ \partial}{\partial x} \frac{ \partial }{\partial x} = \frac{\partial^2}{\partial x^2}$$
Note: $C_{0,0} = 1$
For the maclaurain, set $ a=b=0$
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