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Maclaurin Series

Page history last edited by Jason 15 years 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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