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 May 7th, 2016, 02:55 PM #1 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 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. Thanks from Joppy May 8th, 2016, 10:49 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics Bump Nobody likes me  May 8th, 2016, 02:58 PM #3 Global Moderator   Joined: May 2007 Posts: 6,854 Thanks: 744 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. Thanks from 123qwerty May 10th, 2016, 07:38 AM   #4
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Quote:
 Originally Posted by 123qwerty 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. May 10th, 2016, 09:00 AM   #5
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Quote:
 Originally Posted by Country Boy You mean $\displaystyle \left|f(x)- P_n(x)\right|$, surely?
Yeah, I noticed that when mathman pointed it out.  Tags correct, proof, taylor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post irunktm Algebra 3 September 6th, 2012 06:03 PM confused Algebra 6 May 31st, 2010 03:51 PM starwok Algebra 3 December 16th, 2009 12:21 AM balste Advanced Statistics 2 September 3rd, 2009 12:20 PM Gunuu Applied Math 5 September 21st, 2008 08:17 AM

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