My Math Forum Derivative problem

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 October 16th, 2010, 06:30 AM #1 Newbie   Joined: Oct 2010 Posts: 14 Thanks: 0 Derivative problem I've never been good with math . Here is the question: There is a polynomial with many(infinite) terms which corresponds exactly this function y=e^x. Guided by this particular function is its own derivative, try to solve the five first terms in this polynomial. Guidence: the first term is 1. I've no idea , thank you.
October 16th, 2010, 07:29 AM   #2
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Re: Derivative problem

Hello, azelio!

Quote:
 There is a polynomial with infinite terms which corresponds exactly to:[color=beige] .[/color]$y\:=\:e^x$ This particular function is its own derivative. Guided by this fact, find the five first terms of this polynomial. Hint: the first term is 1.

$\text{The function is: }\;\;\;y \;=\;1\,+\,a_1x\,+\,a_2x^2\,+\,a_3x^3\,+\,a_4x^4\, +\,\cdots$

$\text{Its derivative is: }\;y' \;=\;a_1\,+\,2a_2x\,+\,3a_3x^2\,+\,4a_x^3\,+\,\cdo ts$

$\text{Since the function equals its derivative, equate coefficients:}$

[color=beige]. . [/color]$\begin{array}{ccc} a_1=&1 \\ 2a_2=&a_1 \\ 3a_3=&a_2 \\ 4a_4=&a_3 \\ \vdots=&\vdots \end{array}=$

$\begin{array}{cccccccc}\text{Then we have: }=&a_1=&1 \\ \\ \\=&a_2=&\frac{1}{2}a_1=&\frac{1}{1\cdot2}=&\frac{1}{2!} \\ \\ \\=&a_3=&\frac{1}{3}a_2=&\frac{1}{1\cdot2\cdot3}=&\frac{1}{3!} \\ \\ \\=&a_4=$

$\text{Therefore: }\;y \;=\;1\,+\,x\,+\,\frac{x^2}{2!}\,+\,\frac{x^3}{3!} \,+\,\frac{x^4}{4!}\,+\,\cdots$

 October 17th, 2010, 05:24 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Derivative problem Quite interesting... the Maclaurin expansions for sin(x) and cos(x) can be derived similarly by using: y = -y'' and for sin(x) begin the polynomial with x, and for cos(x) begin with 1.

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