Understanding Taylor Expansion near a Point

[Blanchdog]

Homework Statement

Taylor's theorem states that for any reasonable function $f(x)$, the value of f at a point $(x + \delta)$ can be expressed as an infinite series involving f and its derivatives at the point x:
$$f(x+\delta) = f(x) + f'(x)\delta +\frac{1}{2!}f''(x)\delta^2+\frac{1}{3!}f'''(x)\delta^3$$ Find the Taylor series for $ln(1+\delta)$

Relevant Equations

See above

I'm just trying to understand how this works, because what I've been looking at online seems to indicate that I evaluate at $\delta =0$ for some reason, but that would make the given equation for the Taylor series wrong since every derivative term is multiplied by some power of $\delta$. Could someone walk me through at least the first derivative term?

 

[phyzguy]

I think what is confusing you is the value of 1 in ln(1+δ). Trying setting f(x) = ln(x), and then evaluate the series for ln(x+δ). Then, after you have done the derivatives, plug in 1 for x. Does this help?

 

[FactChecker]

The first derivative is the slope going away from x so multiplying it by $\delta$ makes sense. It approximates the function with a straight line. The next term is the curvature going away from x, so it corrects the linear estimate around x. It is a second-order polynomial approximation of the function. It continues like that.

 

[Blanchdog]

I think what is confusing you is the value of 1 in ln(1+δ). Trying setting f(x) = ln(x), and then evaluate the series for ln(x+δ). Then, after you have done the derivatives, plug in 1 for x. Does this help?

Not really to be honest, that's already done in the set up for the problem. Right now I've got for the expansion up to the second derivative
$$ln(1) + \frac{1}{1+\delta} \cdot \delta + \frac{1}{2!} \frac{-1}{(1+\delta)^2}\cdot \delta^2$$
Which simplifies slightly to $$\frac{\delta}{1+\delta} - \frac{\delta^2}{2(1+\delta)^2}$$

Problem is, that doesn't match what a later problem says the Taylor series to two derivatives is for $ln(1+\delta)$, which is $\delta -\frac{\delta^2}{2}$. Where am I going wrong?

 

[WWGD]

Not really to be honest, that's already done in the set up for the problem. Right now I've got for the expansion up to the second derivative
$$ln(1) + \frac{1}{1+\delta} \cdot \delta + \frac{1}{2!} \frac{-1}{(1+\delta)^2}\cdot \delta^2$$
Which simplifies slightly to $$\frac{\delta}{1+\delta} - \frac{\delta^2}{2(1+\delta)^2}$$

Problem is, that doesn't match what a later problem says the Taylor series to two derivatives is for $ln(1+\delta)$, which is $\delta -\frac{\delta^2}{2}$. Where am I going wrong?

Have you tried to simplify the above expression? It looks. similar to your objective expression.

 

[hutchphd]

Have you tried to simplify the above expression? It looks. similar to your objective expression.

? I don't understand this.?

Anyhow

Where am I going wrong?

You must evaluate the $n^{th}$ derivative at $\delta=0$ and then multiply that value by the $\delta^n$.

 

[Blanchdog]

Have you tried to simplify the above expression? It looks. similar to your objective expression.

It looks similar, but simplifying only gives me $\frac{\delta(2+\delta)}{2(\delta+1)^2}$, not the desired expression.

 

[Blanchdog]

You must evaluate the $n^{th}$ derivative at $\delta=0$ and then multiply that value by the $\delta^n$.

I mean that looks like it will give the right answer, but I have no idea why I would evaluate the derivative at $\delta = 0$ only to then treat $\delta$ as not being zero later. Can you help me understand what's going on there?

 

[SammyS]

Not really to be honest, that's already done in the set up for the problem. Right now I've got for the expansion up to the second derivative
$$ln(1) + \frac{1}{1+\delta} \cdot \delta + \frac{1}{2!} \frac{-1}{(1+\delta)^2}\cdot \delta^2$$
Which simplifies slightly to $$\frac{\delta}{1+\delta} - \frac{\delta^2}{2(1+\delta)^2}$$

Problem is, that doesn't match what a later problem says the Taylor series to two derivatives is for $ln(1+\delta)$, which is $\delta -\frac{\delta^2}{2}$. Where am I going wrong?

You're incorrectly substituting into the expression for the Taylor Expansion.

You are to find an approximation for $\ln (1+\delta)$. Basically, you are finding an approximation for $\ln (x)$ for values of $x$ near $x=1$.

Think of this as using the following:
$f(x+\delta) = f(x) + f'(x)\delta +\frac{1}{2!}f''(x)\delta^2+\frac{1}{3!}f'''(x)\delta^3$

for the case where $x=1$, i.e.
$f(1+\delta) = f(1) + f'(1)\delta +\frac{1}{2!}f''(1)\delta^2+\frac{1}{3!}f'''(1)\delta^3$

Now apply the to the case where the function $f$ is replaced with the function $\ln$..

 

[hutchphd]

You are using what you know about the function at x to extrapolate (continue) it to $x+\delta$. This gives a power series that is arbitrarily correct over some range of values.

 

[Blanchdog]

You're incorrectly substituting into the expression for the Taylor Expansion.

You are to find an approximation for $\ln (1+\delta)$. Basically, you are finding an approximation for $\ln (x)$ for values of $x$ near $x=1$.

Think of this as using the following:
$f(x+\delta) = f(x) + f'(x)\delta +\frac{1}{2!}f''(x)\delta^2+\frac{1}{3!}f'''(x)\delta^3$

for the case where $x=1$, i.e.
$f(1+\delta) = f(1) + f'(1)\delta +\frac{1}{2!}f''(1)\delta^2+\frac{1}{3!}f'''(1)\delta^3$

Now apply the to the case where the function $f$ is replaced with the function $\ln$..

This helped a lot, and I was able to get the answer to the question. I am curious however about whether $\delta$ was was the independent variable in this case, and why this formulation of the Taylor series seems different than the one on the wikipedia page.

 

[vela]

This helped a lot, and I was able to get the answer to the question. I am curious however about whether $\delta$ was was the independent variable in this case, and why this formulation of the Taylor series seems different than the one on the wikipedia page.

The expansion you have in the first post is probably confusing you. Usually, the expansion is written like
$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots.$$ Essentially, if we know the value of $f$ and its derivatives at the point $x=a$, we can calculate the value of $f$ at other points.

Now let $\delta = x-a$. Then $x = a+\delta$ and you get
$$f(x) = f(a+\delta) = f(a) + f'(a)\delta + \frac{f''(a)}{2!}\delta^2+\cdots.$$ The coefficients don't depend on $\delta$ at all. You're not evaluating the function when $\delta=0$. You're evaluating the function at $x=a$. $\delta$ is just how far away $x$ is from $a$.

 

[WWGD]

The expansion you have in the first post is probably confusing you. Usually, the expansion is written like
$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots.$$ Essentially, if we know the value of $f$ and its derivatives at the point $x=a$, we can calculate the value of $f$ at other points.

Now let $\delta = x-a$. Then $x = a+\delta$ and you get
$$f(x) = f(a+\delta) = f(a) + f'(a)\delta + \frac{f''(a)}{2!}\delta^2+\cdots.$$ The coefficients don't depend on $\delta$ at all. You're not evaluating the function when $\delta=0$. You're evaluating the function at $x=a$. $\delta$ is just how far away $x$ is from $a$.

@Blanchdog : Like Vela wrote, If we were using $\delta=0$ it would just be computing $ln(1)$ which is 0. We want to approximate the value near 1; not _at_ 1.