Real Analysis: Expanding a Function at Different Points

[Silviu]

Hello! Can someone explain to me, in real analysis, what is the difference in expanding a function as a Taylor series around 2 different point. So we have $f(x)=\sum c_k (z-z_1)^k = \sum d_k (z-z_2)^k$ and as $k \to \infty$ the series equals f in both cases, but why would one choose a point in the favor of the other?

 

[PeroK]

Hello! Can someone explain to me, in real analysis, what is the difference in expanding a function as a Taylor series around 2 different point. So we have $f(x)=\sum c_k (z-z_1)^k = \sum d_k (z-z_2)^k$ and as $k \to \infty$ the series equals f in both cases, but why would one choose a point in the favor of the other?

If you are interested in the function at $z_1$ you choose $z_1$ and if you are interested in the function at $z_2$ you choose $z_2$.

 

[FactChecker]

There are a few reasons to expand around different points. One is as @PeroK says, that you are interested in a region around a different point. Another is if you have information about all the derivatives around a different point. A third is that the rate of convergence of the Taylor series may be better at a different point. A fourth is that the expansion around another point may have a circle of convergence that is not included in the region of convergence of the first point. (The fourth reason may be considered related to the third reason.)

 

[mathwonk]

to illustrate one of the previous answers, suppose you are interested in a taylor expansion of the real logarithm function ln(x). it is defined for all positive x, but if you expand around the center a>0, it will only converge on the interval of radius a about that center, i.e. on (0,2a). so if you are interested in the values at some larger number than 2a you need to re expand at a larger center.

 

[zwierz]

So we have f(x)=∑ck(z−z1)k=∑dk(z−z2)kf(x)=\sum c_k (z-z_1)^k = \sum d_k (z-z_2)^k and as k→∞k \to \infty the

it can happen that this formula does not make sense. For example
$$f(z)=\frac{1}{z^2+1}+\frac{1}{(z-1)^2+1}$$ and $z_1=-10,\quad z_2=10$

 

[FactChecker]

it can happen that this formula does not make sense. For example
$$f(z)=\frac{1}{z^2+1}+\frac{1}{(z-1)^2+1}$$ and $z_1=-10,\quad z_2=10$

Good point. A simpler example is 1/z at z₁=-1 and z₂ = +1 since there is no value of z where both series converge.

 

[zwierz]

I wanted the function f to be defined on $\mathbb{R}$ :)

 

[FactChecker]

I wanted the function f to be defined on $\mathbb{R}$ :)

Oh. I missed that point. Good example.

 

[Silviu]

If you are interested in the function at $z_1$ you choose $z_1$ and if you are interested in the function at $z_2$ you choose $z_2$.

Thank you for your reply! However I am still a bit confused. If I have let's say $f(x)=e^x$ and I want to calculate $e^7$, I am interested in the function at $z=7$, not at $z_1$ or $z_2$. My question is, what information do $z_1$ and $z_2$ give me?

 

[FactChecker]

The Taylor series expansion at each point z₁ and z₂ has a radius of convergence which may or may not include the point z=7. Even if both converge at z=7, they may have different rates of convergence there. In the case of the exponential function, all Taylor series converge at z=7, so it is not the best example. Still, I think it would be obvious that an expansion at z₁ = 6 will converge faster at z=7 than an expansion at z₂=0.
Consider the example that @zwierz gave. The expansion at z₁=-10 will not converge at z=7 but the expansion at z₂=+10 will.

 

[mathwonk]

e^x of course will converge everywhere, no matter what center of expansion is chosen. But the best center to choose for computation is one for which you actually know the coefficients, namely x=0. Expanding e^x about the point x=6 would require first making an infinite number of coefficient calculations, each comparable to the one computation you want.

 

[FactChecker]

e^x of course will converge everywhere, no matter what center of expansion is chosen. But the best center to choose for computation is one for which you actually know the coefficients, namely x=0. Expanding e^x about the point x=6 would require first making an infinite number of coefficient calculations, each comparable to the one computation you want.

Right. But I was trying to make a general point that applied to more than the exponential function. Also, you can expand the series once and, even if you need to calculate a lot of coefficients, it could give you a tool to use for general values of z.