B Can We Simplify the Taylor Series Expansion for e^(f(x,y))?

AI Thread Summary
The discussion centers on the Taylor series expansion for the exponential function e^(f(x,y)) in multiple variables. It is clarified that while the formula e^(f(x,y)) = Σ (f(x,y)^n/n!) is valid, it does not represent the Taylor series for the function of two variables. The conversation highlights that the expansion may not yield a power series in x or y, and specific cases can lead to the Taylor series, such as when f(x) is a simple function like x^2. A suggestion is made to factor out e^(f(0,0)) to simplify the expression, although complications arise when dealing with higher-order terms. Ultimately, the challenge of simplifying the series for two variables remains unresolved.
thatboi
Messages
130
Reaction score
20
I know that for 1 variable, one can write ##e^{f(x)} = \sum_{n = 0}^{\infty}\frac{(f(x))^n}{n!}##. In the case of 2-variables ##f(x,y)##, I assume we cannot write ##e^{f(x,y)} = \sum_{n = 0}^{\infty}\frac{(f(x,y))^n}{n!}## right (because of how the Taylor series is defined for multiple variables)? Is there still a compact way of writing this expansion?
 
Mathematics news on Phys.org
For any variable ##z## it is the case that
$$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$
Substituting any expression for ##z## it remains valid, as long as the expression delivers a real number.
Substituting ##f(x)## for ##z## gives your first formula.
Substituting ##f(x,y)## for ##z## gives your second formula.
Both are valid.
The definition of a Taylor series for two variables is not relevant. In neither case is the formula a Taylor series for the function of ##x## or ##x,y##.
 
  • Like
Likes PeroK and thatboi
thatboi said:
I know that for 1 variable, one can write ##e^{f(x)} = \sum_{n = 0}^{\infty}\frac{(f(x))^n}{n!}##. In the case of 2-variables ##f(x,y)##, I assume we cannot write ##e^{f(x,y)} = \sum_{n = 0}^{\infty}\frac{(f(x,y))^n}{n!}## right (because of how the Taylor series is defined for multiple variables)?
Those equations are perfectly valid. But, they may not represent the Taylor series for the given function. In general, it won't even be a power series in ##x##:
$$e^{f(x)} = \sum_{n=0}^\infty \frac{f(x)^n}{n!} = 1 + f(x) + \frac{f(x)^2}{2!} + \dots$$Which is fine, but it's not necessarily the Taylor series for ##e^{f(x)}##

In some cases, you do get the Taylor series. For example, if we let ##z = x^2##, then (as above) we get the power series:
$$e^{x^2} = \sum_{n=0}^\infty \frac{x^{2n}}{n!}$$Which is the Taylor series for ##e^{x^2}##.
 
Suggestion is that if ## f(0,0) ## is not approximately zero, that you factor out ## e^{f(0,0)} ## from the expression for the Taylor type series of ## f(x,y) ##.

Edit: It leaves you with ## e^{f(0,0)}e^{\Delta}=e^{f(0,0)}(1+ \Delta+ \Delta^2/2+...)##, = maybe it will work...Edit 2=even ## \Delta ## is complicated to second order in ## \Delta x ## and ## \Delta y ##=I don't see an easy way to simplify it.

In any case, ignore=my calculus was rusty today...:confused:
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
1
Views
1K
Replies
5
Views
2K
Replies
33
Views
3K
Replies
2
Views
2K
Replies
3
Views
3K
Replies
10
Views
2K
Back
Top