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May 7th, 2016, 02:55 PM   #1
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Is my proof correct?

This is actually a part of a multiple-choice question. I think I get the intuition behind it, but I'd like to make sure my proof is right as well. Thanks!

For an integer $\displaystyle n \geq 0$, let $\displaystyle P_n(x)$ be the order n Taylor polynomial of a differentiable function $\displaystyle f(x)$ on $\displaystyle \mathbb{R}$ at $\displaystyle a \in \mathbb{R}$. Which of the following statements is/are true for every differentiable function $\displaystyle f$?

I. For any given $\displaystyle x \in \mathbb{R}$, $\displaystyle |f(x) = P_n(x)|$ gets smaller as $\displaystyle n$ gets bigger.
II. $\displaystyle f(a) = P_0(a)$ and $\displaystyle f^{(k)}(a) = P_n^{(k)}(a)$ for every $\displaystyle 1 \leq k \leq n$
III. $\displaystyle f(a) = P_0(a)$ and $\displaystyle f^{(k)}(a) = P_n^{(k)}(a)$ for every $\displaystyle k \geq 1$

For II, here's my work:

Define $\displaystyle P_n(x) = \sum_{i = 0}^n \frac{f^{(i)}(a)}{i!} (x-a)^i$

Then

$\displaystyle
\begin{align*}
P_n'(x) &= \sum_{i = 1}^n \frac{f^{(i)}(a)}{i!} i(x-a)^{i-1}(1-0) + \frac{d}{dx}\frac{f^{(i)}(a)}{i!} (1-0)\\
&= \sum_{i = 1}^n \frac{f^{(i)}(a)}{i!} i(x-a)^{i-1}\\
P_n''(x) &= \sum_{i = 2}^n \frac{f^{(i)}(a)}{i!} i(i-1)(x-a)^{i-2} \\
P_n^{(k)}(x) &= \sum_{i = k}^n \frac{f^{(i)}(a)}{i!} (\Pi_{j = i - k + 1}^i j) (x-a)^{i-k} \\
P_n^{(k)}(a) &= \sum_{i = k+1}^n \frac{f^{(i)}(a)}{i!} (\Pi_{j = i - k + 1}^i j) (a-a)^{i-k} + \frac{f^{(k)}(a)}{k!} (\Pi_{j = k - k + 1}^i j) (x-a)^{k-k} \\
&= \frac{f^{(k)}(a)}{k!} (k!)\\
&= f^{(k)}(a)
\end{align*}$

For III, I only need to show it's false for k > n. Since $\displaystyle P_n(a)$ is a polynomial, $\displaystyle P_n^{(k)}(a) = 0$. But $\displaystyle f(x)$ may not be a polynomial, which implies $\displaystyle f^{(k)}(a) = 0$ isn't always true. Thus III is false.
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May 8th, 2016, 10:49 AM   #2
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May 8th, 2016, 02:58 PM   #3
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Looks good (I didn't check the details, but you have the right idea).
There is a typo in I: should have a - sign, not an = sign.
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May 10th, 2016, 07:38 AM   #4
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Quote:
Originally Posted by 123qwerty View Post
This is actually a part of a multiple-choice question. I think I get the intuition behind it, but I'd like to make sure my proof is right as well. Thanks!

For an integer $\displaystyle n \geq 0$, let $\displaystyle P_n(x)$ be the order n Taylor polynomial of a differentiable function $\displaystyle f(x)$ on $\displaystyle \mathbb{R}$ at $\displaystyle a \in \mathbb{R}$. Which of the following statements is/are true for every differentiable function $\displaystyle f$?

I. For any given $\displaystyle x \in \mathbb{R}$, $\displaystyle |f(x) = P_n(x)|$ gets smaller as $\displaystyle n$ gets bigger.
You mean $\displaystyle \left|f(x)- P_n(x)\right|$, surely?

Quote:
II. $\displaystyle f(a) = P_0(a)$ and $\displaystyle f^{(k)}(a) = P_n^{(k)}(a)$ for every $\displaystyle 1 \leq k \leq n$
III. $\displaystyle f(a) = P_0(a)$ and $\displaystyle f^{(k)}(a) = P_n^{(k)}(a)$ for every $\displaystyle k \geq 1$

For II, here's my work:

Define $\displaystyle P_n(x) = \sum_{i = 0}^n \frac{f^{(i)}(a)}{i!} (x-a)^i$

Then

$\displaystyle
\begin{align*}
P_n'(x) &= \sum_{i = 1}^n \frac{f^{(i)}(a)}{i!} i(x-a)^{i-1}(1-0) + \frac{d}{dx}\frac{f^{(i)}(a)}{i!} (1-0)\\
&= \sum_{i = 1}^n \frac{f^{(i)}(a)}{i!} i(x-a)^{i-1}\\
P_n''(x) &= \sum_{i = 2}^n \frac{f^{(i)}(a)}{i!} i(i-1)(x-a)^{i-2} \\
P_n^{(k)}(x) &= \sum_{i = k}^n \frac{f^{(i)}(a)}{i!} (\Pi_{j = i - k + 1}^i j) (x-a)^{i-k} \\
P_n^{(k)}(a) &= \sum_{i = k+1}^n \frac{f^{(i)}(a)}{i!} (\Pi_{j = i - k + 1}^i j) (a-a)^{i-k} + \frac{f^{(k)}(a)}{k!} (\Pi_{j = k - k + 1}^i j) (x-a)^{k-k} \\
&= \frac{f^{(k)}(a)}{k!} (k!)\\
&= f^{(k)}(a)
\end{align*}$

For III, I only need to show it's false for k > n. Since $\displaystyle P_n(a)$ is a polynomial, $\displaystyle P_n^{(k)}(a) = 0$. But $\displaystyle f(x)$ may not be a polynomial, which implies $\displaystyle f^{(k)}(a) = 0$ isn't always true. Thus III is false.
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May 10th, 2016, 09:00 AM   #5
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Quote:
Originally Posted by Country Boy View Post
You mean $\displaystyle \left|f(x)- P_n(x)\right|$, surely?
Yeah, I noticed that when mathman pointed it out.
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