MHB Can't understand/solve Taylor exercises.

[Velo]

So, the information they give me is the following:
$(1) f \in {C}^{3}({\rm I\!R})$
$(2) f(x) = 5 -2(x+2) - (x+2)^2 + (x+2)^3 + R3(x+2)$
$(3) \lim_{{x}\to{-2}} \frac{R3(x+2)}{(x+2)^3}=0$

And they ask me for the equation of the tangent line... Which would be simple if that R3 wasn't there, I'd just derive the function, use it on the tangent line equation and that'd be that... I have no idea how the derive that function with the R3 there tho.

Edit: Typo in the exercise.

 

[I like Serena]

So, the information they give me is the following:
$(1) f \in {C}^{3}({\rm I\!R})$
$(2) f(x) = 5 -2(x+2) - x(x+2)^2 + (x+2)^3 + R3(x+2)$
$(3) \lim_{{x}\to{-2}} \frac{R3(x+2)}{(x+2)^3}=0$

And they ask me for the equation of the tangent line... Which would be simple if that R3 wasn't there, I'd just derive the function, use it on the tangent line equation and that'd be that... I have no idea how the derive that function with the R3 there tho.

Hi Velo!

We can write the derivative of R3(x+2) at x=-2 as:
$$R3'(x+2) = R3'(0)= \lim_{h\to 0} \frac{R3(h) - R3(0)}{h}$$
Can we find what it is using equation (3)?
How do the expressions compare?
Which is bigger for $|h|<1$?

 

[I like Serena]

Oh, I'm just realizing that this is a Taylor series exercise and that R3 is the remainder term for the 3rd order expansion.

What is the general formula for the Taylor expansion of f(x) at x=-2?
Can we compare it to the given f(x)?
We should be able to read off f(-2) directly.

 

[Velo]

Oh, I'm just realizing that this is a Taylor series exercise and that R3 is the remainder term for the 3rd order expansion.

What is the general formula for the Taylor expansion of f(x) at x=-2?
Can we compare it to the given f(x)?
We should be able to read off f(-2) directly.

I'm not sure I'm following... I know a Taylor expansion should look something like this when we're trying to get a function similar to another function on $x = -2$:
$f(-2) + f'(-2)(x+2) + \frac{f''(-2)}{2}(x+2)^2 + o(x+2)^2$
I also get that this new function will be very close to the original function next to $x = 2$, though I don't get what part that $o(x+2)^2$ plays on it exactly...

The f(x) they give us looks very similar to that formula... It's missing the derivatives though, I can't really tell how to relate the two :/

 

[I like Serena]

I'm not sure I'm following... I know a Taylor expansion should look something like this when we're trying to get a function similar to another function on $x = -2$:
$f(-2) + f'(-2)(x+2) + \frac{f''(-2)}{2}(x+2)^2 + o(x+2)^2$
I also get that this new function will be very close to the original function next to $x = 2$, though I don't get what part that $o(x+2)^2$ plays on it exactly...

The f(x) they give us looks very similar to that formula... It's missing the derivatives though, I can't really tell how to relate the two :/

Indeed.
Normally we expand by evaluating f(-2), f'(-2), and so on giving constants to multiply $(x+2)^k$ with.
But this has already been done.
We can directly match them.
In other words f'(-2) is the constant that gets multiplied by (x+2).
And that's the slope we need for the tangent line.

The notation $O((x+2)^2)$ means that the remainder is less than a constant times $(x+2)^2$.
Note that $(x+2)^k$ is an increasingly small number for higher powers when x is close to -2.

 

[Velo]

Indeed.
Normally we expand by evaluating f(-2), f'(-2), and so on giving constants to multiply $(x+2)^k$ with.
But this has already been done.
We can directly match them.
In other words, f'(-2) is the constant that gets multiplied by (x+2).
And that's the slope we need for the tangent line.

The notation $O((x+2)^2)$ means that the remainder is less than constant times $(x+2)^2$.
Note that $(x+2)^k$ is an increasingly small number for higher powers when x is close to -2.

OOOhhhhh! I see! So,
$f'(-2) = -2 \implies y = -2x + b$,

From the original f(x) we also know that $f(-2) = 5$ so we can just replace that int he equations and get b!
$y = -2x + b \implies 5 = (-2) * (-2) + b \Leftrightarrow b = 1$

And so the final equation is:
$y = -2x + 1$

So if we wanted to find out for example f'''(-2) we'd just have to do:
$\frac{f'''(-2)}{!3} = 1 \Leftrightarrow f'''(-2) = \frac{1}{6}$

Thank you so much, can't believe I finally understood how this works! :')

 

[I like Serena]

OOOhhhhh! I see! So,
$f'(-2) = -2 \implies y = -2x + b$,

From the original f(x) we also know that $f(-2) = 5$ so we can just replace that int he equations and get b!
$y = -2x + b \implies 5 = (-2) * (-2) + b \Leftrightarrow b = 1$

And so the final equation is:
$y = -2x + 1$

So if we wanted to find out for example f'''(-2) we'd just have to do:
$\frac{f'''(-2)}{!3} = 1 \Leftrightarrow f'''(-2) = \frac{1}{6}$

Thank you so much, can't believe I finally understood how this works! :')

Yep. (Nod)

But let's make it $\frac{f'''(-2)}{!3} = 1 \Leftrightarrow f'''(-2) = 6$. (Wink)

One more thing, we can read the tangent equation also directly from the expansion: it's the expansion up to the first order.
That is:
$$y=f(-2)+f'(-2)(x+2)=5-2(x+2)=-2x+1$$

 

[Velo]

Classic :')