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Maclaurin Series

Page history last edited by Jason 15 years, 9 months ago

 

Maclaurin Series

 

SOUND COMPLICATED???

 

ITS PROBABLY THE EASIEST THING IN FP3 IF YOU KNOW YOUR WAY AROUND DIFFERENTIATION!!!!

 

BASICALLY, MACLAURIN SAID THAT:

 

 f(x)=f(0)+f^'(0)x+(f^('')(0))/(2!)x^2+(f^((3))(0))/(3!)x^3+...+(f^((n))(0))/(n!)x^n+....

 

Where f(x) is a function we want to find an expansion for.

 

EASY EXAMPLE:

 

Q: Find the first 3 non-zero terms in the Maclaurin Expansion of f(x) = Sin x

 

Well f(x) = Sin x

So f(0) = Sin 0 = 0

 

f'(x) = Cos x

So f'(0) = Cos 0 = 1

 

f''(x) = -Sin x

So f''(0) = - Sin 0 = 0

 

f'''(x) = - Cos x

So f'''(0) = - Cos 0 = -1

 

f''''(x) = Sin x

So f''''(0) = Sin 0 = 0

 

f'''''(x) = Cos x

So f'''''(0) = Cos 0 = 1

 

WE HAVE NOW FOUND OUR 3 NON-ZERO TERMS!!!!!

 

AS A SUMMARY:

 

f(0) = 0

f'(0) = 1

f''(0) = 0

f'''(0) = -1

f''''(0) = 0

f'''''(0) = 1

 

MACLAURIN SERIES:

 

 f(x)=f(0)+f^'(0)x+(f^('')(0))/(2!)x^2+(f^((3))(0))/(3!)x^3+...+(f^((n))(0))/(n!)x^n+....

 

 So:

 

 

 

 

THEREFORE THE ANSWER IS :

 

 

 

 

NOW LETS CHECK OUR ANSWER.

 

USING A CALCULATOR, SIN 0.5 = 0.479425538

 

PLUGGING 0.5 INTO THE EXPANSION, WE GET  0.479427083

 

EASY, ISN'T IT?????????

 

ALL YOU NEED TO KNOW IS DIFFERENTIATION

 

PLEASE FEEL FREE TO LEAVE HARDER EXAMPLES AND/OR CORRECT ANY MISTAKES ON THIS PAGE

 

Comments (2)

Jason said

at 1:25 pm on May 11, 2009

PLEASE LEAVE COMMENTS IF I MADE A MISTAKE :)

Steph Richards said

at 7:44 pm on May 11, 2009

Thank you for your contribution Jason. It is REALLY appreciated!

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