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Difference between revisions of "User:Temperal/The Problem Solver's Resource9"

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(Added basic Quadratic Recipricoty, will add more theorems later)
 
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{{User:Temperal/testtemplate|page 9}}
 
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==<span style="font-size:20px; color: blue;">Derivatives</span>==
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==<span style="font-size:20px; color: blue;">Advanced Number Theory</span>==
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This page will cover derivatives and their applications, as well as some advanced limits. The Fundamental Theorem of Calculus is covered on the [[User:Temperal/The Problem Solver's Resource10|integral page]].
 
  
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===Definition===
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[[User:Temperal/The Problem Solver's Resource8|Back to page 8]] | [[User:Temperal/The Problem Solver's Resource10|Continue to page 10]]
  
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*<math>\frac{df(x)}{dx}=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}</math>, where <math>f(x)</math> is a function continuous in <math>L</math>, and <math>x_0</math> is an arbitrary constant such that <math>x_0\subset L</math>.
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===Quadratic Reciprocity===
  
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*Multiple derivatives are taken by evaluating the innermost first, and can be notated as follows: <math>\frac{d^2f(x)}{dx^2}</math>.
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A number m is a quadratic residue mod n if and only there exists a number a such that <math>a^2 \equiv m \pmod{n}</math> and <math>0 \le m < n.</math>. For example, since <math>5^2 \equiv 4 \pmod{7},</math> 4 is a quadratic residue mod 7.  
  
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*The derivative of <math>f(x)</math> can also be expressed as <math>f'(x)</math>, or the <math>n</math>th derivative of <math>f(x)</math> can be expressed as <math>f^{(n)}(x)</math>.
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Let p be an odd prime. There are some theorems regarding quadratic residues mod p. Usually, 0 is a "special case." So 0 is not a residue, but it is also not a non-residue.
  
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===Basic Facts===
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Adopting these definitions, we prove some theorems.
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*<math>\frac{df(x)\pm g(x)}{dx}=f'(x)\pm g'(x)</math>
 
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*<math>\frac{df(x)\cdot g(x)}{dx}=f'(x)\cdot g(x)+ g'(x)\cdot f(x)</math>
 
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*<math>\frac{d\frac{f(x)}{g(x)}}{dx}=\frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}</math>
 
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*<math>\frac{d\sin x}{dx}=\cos x</math>
 
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*<math>\frac{d\cos x}{dx}=-\sin x</math>
 
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*<math>\frac{d\tan x}{dx}=\sec^2 x</math>
 
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*<math>\frac{d\csc x}{dx}=-\csc x\cot x</math>
 
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*<math>\frac{d\cot x}{dx}=-\csc^2 x</math>
 
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*<math>\frac{d\ln |x|}{dx}=\frac{1}{x}</math>
 
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*<math>\frac{dc}{dx}=0</math> for a constant <math>c</math>.
 
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====The Power Rule====
 
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*<math>\frac{dx^n}{dx}=nx^{n-1}</math>
 
  
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===Rolle's Theorem===
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Theorem 9.1. There are an equal number of non-residues and residues mod p.  
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If <math>f(x)</math> is differentiable in the open interval <math>(a,b)</math>, continuous in the closed interval <math>[a,b]</math>, and if <math>f(a)=f(b)</math>, then there is a point <math>c</math> between <math>a</math> and <math>b</math> such that <math>f'(c)=0</math>
 
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====Extension: Mean Value Theorem====
 
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If <math>f(x)</math> is differentiable in the open interval <math>(a,b)</math> and continuous in the closed interval <math>[a,b]</math>, then there is a point <math>c</math> between <math>a</math> and <math>b</math> such that <math>f(b)-f(a)=f'(c)\cdot(b-a)</math>.
 
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===L'Hopital's Rule===
 
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<math>\lim \frac{f(x)}{g(x)}=\lim \frac{f'(x)}{g'(x)}</math>
 
  
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Note that this inplies that <math>\lim \frac{f(x)}{g(x)}=\lim \frac{f^{(n)}(x)}{g^{(n)}(x)}</math> for any <math>n</math>.
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'''Proof: '''First, we prove that <math>1^2, 2^2, ... (\frac{p-1}{2})^2</math> are distinct modulo p. Assume <math>a^2 \equiv b^2 \pmod{p},</math> where a and b are distinct integers from <math>1</math> to <math>\frac{p-1}{2}.</math> Then, <math>a^2-b^2</math> would have to be a multiple of <math>p</math>. <math>a^2-b^2</math> can be factored as <math>(a-b)(a+b).</math> However, since a and b are distinct integers from <math>1</math> to <math>\frac{p-1}{2},</math> <math>a-b</math> can't be a multiple of p. Therefore, <math>a+b</math> is a multiple of p. We know that <math>\frac{p-1}{2} \ge a,b \ge 1,</math> so <math>p-1 \ge a+b \ge 2,</math> so <math>a+b</math> also can't be a multiple of p, a contradiction.
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===Taylor's Formula===
 
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Let <math>a</math> be a point in the domain of the function <math>f(x)</math>, and suppose that <math>f^{(n+1)}(x)</math> (that is, the <math>n+1</math>th derivative of <math>f(x)</math>) exists in the neighborhood of <math>a</math> (where <math>n</math> is a nonnegative integer). For each <math>x</math> in the neighborhood,
 
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<cmath>f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+...+\frac{f^{(n)}(a)}{n!}(x-a)^n+\frac{f^{(n+1)}(a)}{n!}(x-a)^{n+1}</cmath>
 
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where <math>c</math> is in between <math>x</math> and <math>a</math>.  
 
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===Chain Rule===
 
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If <math>h(x) = f(g(x))</math>, then <math>h'(x)=f'(g(x))\cdot g'(x)</math>
 
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===Applications===
 
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*The slope of <math>f(x)</math> at any given point is the derivative of <math>f(x)</math>. (The obvious one.)
 
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*Acceleration is the derivative of velocity in relation to time; velocity is the derivative of position in relation to time.
 
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*The derivative of work (in Joules) in relation to time is power (in watts).
 
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[[User:Temperal/The Problem Solver's Resource8|Back to page 8]] | [[User:Temperal/The Problem Solver's Resource10|Continue to page 10]]
 

Latest revision as of 15:08, 11 January 2009

Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 9.

Advanced Number Theory

Back to page 8 | Continue to page 10

Quadratic Reciprocity

A number m is a quadratic residue mod n if and only there exists a number a such that $a^2 \equiv m \pmod{n}$ and $0 \le m < n.$. For example, since $5^2 \equiv 4 \pmod{7},$ 4 is a quadratic residue mod 7.

Let p be an odd prime. There are some theorems regarding quadratic residues mod p. Usually, 0 is a "special case." So 0 is not a residue, but it is also not a non-residue.

Adopting these definitions, we prove some theorems.

Theorem 9.1. There are an equal number of non-residues and residues mod p.

Proof: First, we prove that $1^2, 2^2, ... (\frac{p-1}{2})^2$ are distinct modulo p. Assume $a^2 \equiv b^2 \pmod{p},$ where a and b are distinct integers from $1$ to $\frac{p-1}{2}.$ Then, $a^2-b^2$ would have to be a multiple of $p$. $a^2-b^2$ can be factored as $(a-b)(a+b).$ However, since a and b are distinct integers from $1$ to $\frac{p-1}{2},$ $a-b$ can't be a multiple of p. Therefore, $a+b$ is a multiple of p. We know that $\frac{p-1}{2} \ge a,b \ge 1,$ so $p-1 \ge a+b \ge 2,$ so $a+b$ also can't be a multiple of p, a contradiction.

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