Value of f(1/2) using estimation for the remainder

[mathmari]

Hey! :giggle:

Let $f(x)=e^{-x}\sin (x)$, $x\in \mathbb{R}$.

a) Calculate the Taylor polynomial of order $4$ at $0$.

b) Calculate the value of $f \left (\frac{1}{2}\right )$ using estimation for the remainder with an error not more than $\frac{1}{400}$.I have done question a) :
\begin{align*}T_{0,4}(x)&=\sum_{k=0}^4\frac{f^{(k)}(0)(x-0)^k}{k!}=\frac{f^{(0)}(0)x^0}{0!}+\frac{f^{(1)}(0)x^1}{1!}+\frac{f^{(2)}(0)x^2}{2!}+\frac{f^{(3)}(0)x^3}{3!}+\frac{f^{(4)}(0)x^4}{4!} \\ & =\frac{0\cdot x^0}{0!}+\frac{1\cdot x^1}{1!}+\frac{(-2)\cdot x^2}{2!}+\frac{2\cdot x^3}{3!}+\frac{0\cdot x^4}{4!}=x+\frac{(-2)\cdot x^2}{2}+\frac{2\cdot x^3}{6}\\ & =x- x^2+\frac{ x^3}{3}\end{align*} As for question b) :

Do we have to check for which order the error is less or equal to $\frac{1}{400}$ ?
After having found the order we have that $\left |R_n \left (\frac{1}{2}\right )\right |\leq \frac{1}{400}$ and that $R_n\left (\frac{1}{2}\right )=f\left (\frac{1}{2}\right )-T_n\left (\frac{1}{2}\right )$ and from that inequality we get value of $f \left (\frac{1}{2}\right )$ ?

Or could we use part a) ? But there we have the point $0$ and now the point $\frac{1}{2}$

:unsure:

 

[I like Serena]

Hey mathmari!

The estimation of the remainder is each subsequent term.
If we approximate $f(\frac 12)$ with $f(x)\approx x$ then the estimation of the remainder is $R_1(x)\approx|-x^2|=|-(\frac 12)^2|=\frac 14$.
That is not small enough so we need at least one more term.

 

[mathmari]

The estimation of the remainder is each subsequent term.
If we approximate $f(\frac 12)$ with $f(x)\approx x$ then the estimation of the remainder is $R_1(x)\approx|-x^2|=|-(\frac 12)^2|=\frac 14$.
That is not small enough so we need at least one more term.

So do we have to calculate $R_1\left (\frac{1}{2}\right )$, $R_2\left (\frac{1}{2}\right )$, $R_3\left (\frac{1}{2}\right )$, ... till it is smaller or equal to $\frac{1}{400}$ ? :unsure:

 

[I like Serena]

I think so yes. (Nod)

 

[mathmari]

I think so yes. (Nod)

So question b) is independent from question a), isn't it? I mean we cannot use a) for b), right? :unsure:

 

[I like Serena]

We can use (a).
Each subsequent term in (a) is an approximation of the remainder.

 

[mathmari]

We can use (a).
Each subsequent term in (a) is an approximation of the remainder.

Does this hold although at (a) we had the point $0$ and at (b) we have the point $\frac{1}{2}$ ? :unsure:

 

[I like Serena]

Does this hold although at (a) we had the point $0$ and at (b) we have the point $\frac{1}{2}$ ?

Yes.
Strictly speaking we have $R_n(x) = \frac{f^{(n+1)}(\xi)\,x^{n+1}}{(n+1)!}$ for some $\xi$ between $0$ and $x$.
We can approximate it with $R_n(x) \approx \frac{f^{(n+1)}(0)\,x^{n+1}}{(n+1)!}$.

 

[mathmari]

Yes.
Strictly speaking we have $R_n(x) = \frac{f^{(n+1)}(\xi)\,x^{n+1}}{(n+1)!}$ for some $\xi$ between $0$ and $x$.
We can approximate it with $R_n(x) \approx \frac{f^{(n+1)}(0)\,x^{n+1}}{(n+1)!}$.

We have that $$R_1\left (\frac{1}{2}\right ) \approx \frac{f^{(1+1)}(0)\,\left (\frac{1}{2}\right )^{1+1}}{(1+1)!}=\frac{1}{4} \\ R_2\left (\frac{1}{2}\right ) \approx \frac{f^{(2+1)}(0)\,\left (\frac{1}{2}\right )^{2+1}}{(2+1)!} =\frac{1}{1536}<\frac{1}{400} $$

So we have that $
R_2\left (\frac{1}{2}\right )=f\left (\frac{1}{2}\right )-T_2\left (\frac{1}{2}\right )
$, right?

But is $T_2$ the polynomial we calculated at (a) or do we have to calculate the polynomial again around $\frac{1}{2}$ ?

:unsure:

 

[I like Serena]

We had in (a) that $f(x)=x-x^2+\frac {x^3}3 +\ldots$.
It follows that $R_0(x)\approx x,\, R_1(x)\approx -x^2,\, R_2(x)\approx\frac {x^3}3$.
Since $|R_2(\frac 12)| <\frac{1}{400}$, we get that the corresponding approximation for $x=\frac 12$ is $f(x)\approx x-x^2$ with an error of less than $\frac 1{400}$.
Since we have already calculated the third term, we can throw that one in as well and make it $f(x)\approx x-x^2+\frac {x^3}3$, which will have an even smaller error.

 

[mathmari]

We had in (a) that $f(x)=x-x^2+\frac {x^3}3 +\ldots$.
It follows that $R_0(x)\approx x,\, R_1(x)\approx -x^2,\, R_2(x)\approx\frac {x^3}3$.
Since $|R_2(\frac 12)| <\frac{1}{400}$, we get that the corresponding approximation for $x=\frac 12$ is $f(x)\approx x-x^2$ with an error of less than $\frac 1{400}$.
Since we have already calculated the third term, we can throw that one in as well and make it $f(x)\approx x-x^2+\frac {x^3}3$, which will have an even smaller error.

Why do we need also the third term? Isn't it enough that $f(x)\approx x-x^2$ ? :unsure:

 

[I like Serena]

Why do we need also the third term? Isn't it enough that $f(x)\approx x-x^2$ ?

Yes, that is enough. (Nod)

 

[mathmari]

Yes, that is enough. (Nod)

So we get that $f\left (\frac{1}{2}\right )\approx \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$, right?

But why can we use here the Taylor polynomial at point $0$ here with $x=\frac{1}{2}$ ? Or are the point around the one we calculate the polynomial and the value of $x$ independent?

:unsure:

 

[Theia]

But why can we use here the Taylor polynomial at point $0$ here with $x=\frac{1}{2}$ ? Or are the point around the one we calculate the polynomial and the value of $x$ independent

The reason to use of a Taylor expansion is usually, that one can approximate the value of function at point $x_0 + h$ based on information of the function at point $x_0$.

Reasoning is that if the remainder is small enough (as it's in this case), or in other words: if the taylor expansion will converge at point $x_0+h$, then one can evaluate the $f(x_0+h)$ using the expansion around the point $x_0$. In this example, the Taylor series of the function in question will converge regardless of the point where it is calculated, and for what value one wants to use the series. But this is not always true. For example for $\arctan x$ around $x_0=0$ one can't use the Taylor expansion to evaluate $\arctan 1.1$.

 

[I like Serena]

But why can we use here the Taylor polynomial at point $0$ here with $x=\frac{1}{2}$ ? Or are the point around the one we calculate the polynomial and the value of $x$ independent?

Yep.
What do you mean?
We use the Taylor polynomial approximation around 0 at a distance of $\frac 12$.
Some Taylor formulas write $f(a+h)=f(a)+f'(a)h+\ldots$. We then have $a=0$ and $h=x$.