What's the Taylor series for $f(x+h,y+k)$?

Question

My course notes give the Taylor series up to degree 2 as $$f(x+h,y+k)\approx f(x,y)+\frac{\partial f}{\partial x}h+\frac{\partial f}{\partial y}k+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}h^2+\frac{\partial^2 f}{\partial x \partial y}hk+\frac{1}{2}\frac{\partial^2 f}{\partial y^2}k^2.$$

It looks like the formula could be something like $$f(x+h,y+k)=\sum_{i=0}^\infty \left(\sum_{j=0}^i \left(\frac{\partial f^i}{\partial (x^{i-j}y^j)}\cdot\frac{h^{i-j}k^j}{i!} \right)\right),$$

except, in degree 2, the factorial doesn't seem to apply equally to the mixed term, and I can't see what the pattern would be without more terms.

You are missing a factor of $\binom{i}j$ (and using a peculiar notation for mixed second partial derivatives). — Angina Seng, Apr 23 '20 at 21:35
@AnginaSeng Fixed the notation. Sounds good; care to make that an answer? — mjc, Apr 23 '20 at 21:45
This is arguably a duplicate of https://math.stackexchange.com/questions/221669/derivation-of-multivariable-taylor-series — Travis Willse, Apr 23 '20 at 22:15

Bernard · Accepted Answer · 2020-04-23T22:07:19.577

It is indeed applied to the miwed term, but to see it plainly, you have to use a symbolic binomial expansion: $$f(x+h,y+k)=\sum_{n=0}^\infty \frac 1{n!}\left(\frac{\partial f} {\partial x}\,h+\frac{\partial f}{\partial y}\,k\right)^{\mkern -5mu(n)},$$ where we set $$\left(\frac{\partial f} {\partial x}\,h+\frac{\partial f}{\partial y}\,k\right)^{\mkern -5mu(n)}=\sum_{i=0}^n\binom ni\frac{\partial^n\mkern-2mu f}{\partial x^i\,\partial y^{n-i}}\,h^i k^{n-i}.$$

For instance, the terms of degree $2$ are

$$\frac1{2!}\biggl(\frac{\partial f}{\partial x}\,h+\frac{\partial f}{\partial y}\,k\biggr)^{\mkern -5mu(2)}= \frac12\biggl(\frac{\partial^2\mkern-2mu f}{\partial x^i}\,h^2 + 2\frac{\partial^2\mkern-2mu f}{\partial x\,\partial y } \,h \, k + \frac{\partial^2\mkern-2mu f}{\partial y^2}\,k^{2}\biggr),$$ which are exactly what you have, after a simplification.

Should $i$ start at $0$ in the second sum? Otherwise you get no unmixed terms. — mjc, Apr 23 '20 at 22:03
Yes, sorry for the typo, and thank you for pointing it! . It's fixed now. — Bernard, Apr 23 '20 at 22:09

score 1 · Answer 2 · answered Apr 23 '20 at 22:01

If we denote $x^1 := x, x^2 := y$ and $h^1 := h, h^2 := k$, then the terms of total degree $i$ are those of the sum $$\frac{1}{i!} \sum_{k_i = 1}^2 \cdots \sum_{k_1 = 1}^2 \frac{\partial^i f}{\partial x^{k_1} \cdots \partial x^{k_i}}(x^1_0, x^2_0) h^{k_1} \cdots h^{k_i} .$$ Expanding this in the case $i = 2$ indeed gives the quadratic terms from the formula in your notes.

If $f$ has continuous $k$th derivatives, then we may reorganize this sum by grouping terms according to the number of indices $k_j$ that are $1$ and the number that are $2$. For any $j \in \{0, \ldots, i\}$ among the $2^i$ strings $(h^{k_1}, \ldots, h^{k_i})$ there are $i \choose j$ with $i - j$ of the $k_1$ equal to $1$ and $j$ equal to $2$. Thus, indexing the sum by $j$ gives $$\frac{1}{i!} \sum_{j=0}^i {i \choose j} \frac{\partial^i f}{\partial (x^1)^{i - j} \,\partial (x^2)^j}(x^1_0, x^2_0) (h^1)^{i - j} (h^2)^j .$$ Rewriting using our original variable names gives the easier-to-read expression $$\frac{1}{i!} \sum_{j = 0}^i {i \choose j} \frac{\partial^i f}{\partial x^{i - j} \,\partial y^j}(x_0, y_0) h^{i - j} k^j ,$$ and so the multivariable Taylor series is $$\sum_{i = 0}^\infty \frac{1}{i!} \sum_{j = 0}^i {i \choose j} \frac{\partial^i f}{\partial x^{i - j} \,\partial y^j}(x_0, y_0) h^{i - j} k^j .$$ One can make this proof and the formulae a little cleaner using multi-index notation, in which case the final formula is $$\sum_{i = 0}^\infty \sum_{|\alpha| = 1}^i \frac{1}{|\alpha|!} \frac{\partial^\alpha f}{\partial x^\alpha}(x) \,h^\alpha$$

What's the Taylor series for $f(x+h,y+k)$?

2 Answers2