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Maclaurin Expansions

Given a "nice" function f(x) (e.g. y = sin(x)) and x = a, we want to find a numerical method in calculating f(x)for x around a. Maclaurin Expansions

Given a "nice" function f(x) (e.g. y = sin(x)) and x = a, we want to find a numerical method in calculating f(x)for x around a. Maclaurin Expansions It turns out that being "nice" in this case means that the derivative exists.

Given a "nice" function f(x) (e.g. y = sin(x)) and x = a, we want to find a numerical method in calculating f(x)for x around a. Maclaurin Expansions It turns out that being "nice" in this case means that the derivative exists. The higher the derivative we can find, the nicer the function is, the better the approximation we may achieve.

Given a "nice" function f(x) (e.g. y = sin(x)) and x = a, we want to find a numerical method in calculating f(x)for x around a. Maclaurin Expansions It turns out that being "nice" in this case means that the derivative exists. The higher the derivative we can find, the nicer the function is, the better the approximation we may achieve. We will assume our functions are of the best kind, i.e. they're infinitely differentiable and we will approximate the points around x = 0.

Given a "nice" function f(x) (e.g. y = sin(x)) and x = a, we want to find a numerical method in calculating f(x)for x around a. Maclaurin Expansions It turns out that being "nice" in this case means that the derivative exists. The higher the derivative we can find, the nicer the function is, the better the approximation we may achieve. We will assume our functions are of the best kind, i.e. they're infinitely differentiable and we will approximate the points around x = 0.Given such a nice function, we'll try to find polynomials of higher and higher degree that give better and better approximation.

Given a "nice" function f(x) (e.g. y = sin(x)) and x = a, we want to find a numerical method in calculating f(x)for x around a. Maclaurin Expansions It turns out that being "nice" in this case means that the derivative exists. The higher the derivative we can find, the nicer the function is, the better the approximation we may achieve. We will assume our functions are of the best kind, i.e. they're infinitely differentiable and we will approximate the points around x = 0.Given such a nice function, we'll try to find polynomials of higher and higher degree that give better and better approximation. We'll find the power series that is the same as the function in some cases.

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial =pn(x)Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial =f(0)pn(x)f(0)(0)0!

Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)x=f(0)+pn(x)f(0)(0)0!

f(1)(0)1!

Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(1)(0)1!

Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(3)(0)+3!x3..

f(1)(0)1!

Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(3)(0)+3!x3..

f(n)(0)+n!xn

f(1)(0)1!

Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(3)(0)+3!x3..

f(n)(0)+n!xn

f(1)(0)1!

or pn(x) = k=0nxkk!

f(k)(0)Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(3)(0)+3!x3..

f(n)(0)+n!xn

f(1)(0)1!

or pn(x) = k=0nxkk!

f(k)(0)This is called the n'th (degree) Maclaurin polynomial (Mac-poly) for f(x). Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(3)(0)+3!x3..

f(n)(0)+n!xn

f(1)(0)1!

or pn(x) = k=0nxkk!

f(k)(0)This is called the n'th (degree) Maclaurin polynomial (Mac-poly) for f(x). If we set n to be , we get the Maclaurin series (Mac-series):P(x) = k=0xk.k!

f(k)(0)Maclaurine Expansions

Given a function f(x) that is infinitely differentiable at x = 0, we define the polynomial f '(0)xf(2)(0)+2!=f(0)+x2pn(x)f(0)(0)0!

f(3)(0)+3!x3..

f(n)(0)+n!xn

f(1)(0)1!

or pn(x) = k=0nxkk!

f(k)(0)This is called the n'th (degree) Maclaurin polynomial (Mac-poly) for f(x). If we set n to be , we get the Maclaurin series (Mac-series):P(x) = k=0xk.k!

f(k)(0)Maclaurine Expansions We call them the Mac-expansions of f(x).

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). =f(0)pn(0)In other words, Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 pn(0)In other words, =pn(0)(1)(1)Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 = f (0)pn(0)In other words, =pn(0)(1)(1)(1)Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 = f (0)pn(0)In other words, =pn(0)(1)f (0)+ #x + #x2 + ..#xn-2|x=0 =pn(0)(2)(2)(1)(1)Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 = f (0)pn(0)In other words, =pn(0)(1)f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)(2)(2)(2)(1)(1)Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 = f (0)pn(0)In other words, =pn(0)(1)f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)(2)(2)(2)(1)(1)and so on, up to pn(0) = f (0). (n)(n)Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 = f (0)pn(0)In other words, =pn(0)(1)f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)(2)(2)(2)(1)(1)and so on, up to pn(0) = f (0). (n)(n)In fact, pn(x) is the only degree n (or less) polynomial whose values of the first n'th derivatives agree with those of f(x) at x = 0. Maclaurine Expansions

pn(x) is a polynomial whose values of the first n'th derivatives at x=0 are the same as f(x). f (0)=f(0)+ #x + #x2 + ..#xn-1|x=0 = f (0)pn(0)In other words, =pn(0)(1)f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)(2)(2)(2)(1)(1)and so on, up to pn(0) = f (0). (n)(n)In fact, pn(x) is the only degree n (or less) polynomial whose values of the first n'th derivatives agree with those of f(x) at x = 0. In similar arguments, the Mac-series is the only power series that has all derivatives agree with those of f(x) at x = 0.Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x): Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 (1)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 (2)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 f (0) = 2! (2)(2)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 f (0) = 2! (2)(2)f (x) = 3*2 + 4*3*2x (3)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 f (0) = 2! (2)(2)f (x) = 3*2 + 4*3*2x f (0) = 3! (3)(3)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 f (0) = 2! (2)(2)f (x) = 3*2 + 4*3*2x f (0) = 3! (3)(3)f (x) = 4*3*2 (4)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 f (0) = 2! (2)(2)f (x) = 3*2 + 4*3*2x f (0) = 3! (3)(3)f (x) = 4*3*2 f (0) = 4! (4)(4)Maclaurine Expansions

The following are some examples of the Mac-polys and Mac-series of basic functions calculated via the definition. Later, we will use these expansions to calculate the expansions of other functions (in stead of using the definition).Example: A. Find the Mac-expansion of f(x) = 1 + x + x2 + x3 + x4 around x = 0.We need the derivatives of f(x):f (x) = 1 + 2x + 3x2 + 4x3 f (0) = 1. (1)(1)f (x) = 2 + 3*2x + 4*3x2 f (0) = 2! (2)(2)f (x) = 3*2 + 4*3*2x f (0) = 3! (3)(3)f (x) = 4*3*2 f (0) = 4! (4)(4)f (x) = 0 for n > 5

(n)Maclaurine Expansions

Hence p0(x) = f(0) = 1Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x=f(0)+p1(x)Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2f(2)(0)+2!x2

f(3)(0)+3!x3

f '(0)x =f(0)+p3(x)Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2f(2)(0)+2!x2

f(3)(0)+3!x3

f '(0)x =f(0)+p3(x)+x2+x31x =1+2!2!

3!3!

Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2f(2)(0)+2!x2

f(3)(0)+3!x3

f '(0)x =f(0)+p3(x)+x2+x31x =1+2!2!

3!3!

= 1 + x + x2 + x3Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)=p4(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2f(2)(0)+2!x2

f(3)(0)+3!x3

f '(0)x =f(0)+p3(x)+x2+x31x =1+2!2!

3!3!

= 1 + x + x2 + x3+x2+x31x 1+2!2!

3!3!

+x44!4!

Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)=p4(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2f(2)(0)+2!x2

f(3)(0)+3!x3

f '(0)x =f(0)+p3(x)+x2+x31x =1+2!2!

3!3!

= 1 + x + x2 + x3+x2+x31x 1+2!2!

3!3!

+x44!4!

= 1 + x + x2 + x3 + x4Maclaurine Expansions

Hence p0(x) = f(0) = 1f '(0)x = 1 + 1x=f(0)+p1(x)=p4(x)f(2)(0)+2!x2

f '(0)x =f(0)+p2(x)= 1 + 1x + 2!2!x2

= 1 + x + x2f(2)(0)+2!x2

f(3)(0)+3!x3

f '(0)x =f(0)+p3(x)+x2+x31x =1+2!2!

3!3!

= 1 + x + x2 + x3+x2+x31x 1+2!2!

3!3!

+x44!4!

= 1 + x + x2 + x3 + x4For n > 5, pn(x)

= 1 + x + x2 + x3 + x4 = f(x)Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenp0(x) = a0Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenp0(x) = a0p1(x) = a0 + a1xMaclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenp0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenp0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 ..Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenp0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 ..pk(x) = a0 + a1x + a2x2.. + akxk = f(x)Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenp0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 ..pk(x) = a0 + a1x + a2x2.. + akxk = f(x)and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = f(x). Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenExample: B. Find the Mac-expansion of f(x) = ex around x = 0.p0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 ..pk(x) = a0 + a1x + a2x2.. + akxk = f(x)and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = f(x). Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenExample: B. Find the Mac-expansion of f(x) = ex around x = 0.We need the derivatives of f(x): p0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 ..pk(x) = a0 + a1x + a2x2.. + akxk = f(x)and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = f(x). Maclaurine Expansions

In general, if f(x) = a0 + ax + a2x2 + .. +akxk is a polynomial, thenExample: B. Find the Mac-expansion of f(x) = ex around x = 0.We need the derivatives of f(x):f (x) = ex f (0) = 1 for all n. (n)(n)p0(x) = a0p1(x) = a0 + a1xp2(x) = a0 + a1x + a2x2 ..pk(x) = a0 + a1x + a2x2.. + akxk = f(x)and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = f(x). Maclaurine Expansions

f(1)(0)xf(2)(0)+2!=f(0)+x2pn(x)

f(3)(0)+3!x3..

f(n)(0)+n!xn

Therefore the n'th Mac-polynomial of ex isMaclaurine Expansions

f(1)(0)xf(2)(0)+2!=f(0)+x2pn(x)

f(3)(0)+3!x3..

f(n)(0)+n!xn

Therefore the n'th Mac-polynomial of ex isx+2!=1+x2+ ..+

+3!x3

n!xn

Maclaurine Expansions

f(1)(0)xf(2)(0)+2!=f(0)+x2pn(x)

f(3)(0)+3!x3..

f(n)(0)+n!xn

P(x) = k=0k! .

xkTherefore the n'th Mac-polynomial of ex isx+2!=1+x2+ ..+

+3!x3

n!xn

The Mac-series of ex isx+2!1+x2+ ..+

+3!x3

n! ..xn

=Maclaurine Expansions

y = exy=x+1The graphs of Mac-polys for y = ex Maclaurine Expansions

y = exy=x+1y=x2/2+x+1

Maclaurine Expansions The graphs of Mac-polys for y = ex

y = exy=x+1y=x2/2+x+1y=x3/6+x2/2+x+1Maclaurine Expansions The graphs of Mac-polys for y = ex

f(1)(0)xf(2)(0)+2!=f(0)+x2pn(x)

f(3)(0)+3!x3..

f(n)(0)+n!xn

P(x) = k=0k! .

xkTherefore the n'th Mac-polynomial of ex isx+2!=1+x2+ ..+

+3!x3

n!xn

The Mac-series of ex isExample: C. Find the Mac-expansion of f(x) = sin(x), around x = 0.x+2!1+x2+ ..+

+3!x3

n! ..xn

=Maclaurine Expansions

f(1)(0)xf(2)(0)+2!=f(0)+x2pn(x)

f(3)(0)+3!x3..

f(n)(0)+n!xn

P(x) = k=0k! .

xkTherefore the n'th Mac-polynomial of ex isx+2!=1+x2+ ..+

+3!x3

n!xn

The Mac-series of ex isExample: C. Find the Mac-expansion of f(x) = sin(x), around x = 0.We need the derivatives of sin(x). x+2!1+x2+ ..+

+3!x3

n! ..xn

=Maclaurine Expansions

f(1)(0)xf(2)(0)+2!=f(0)+x2pn(x)

f(3)(0)+3!x3..

f(n)(0)+n!xn

P(x) = k=0k! .

xkTherefore the n'th Mac-polynomial of ex isx+2!=1+x2+ ..+

+3!x3

n!xn

The Mac-series of ex isExample: C. Find the Mac-expansion of f(x) = sin(x), around x = 0.We need the derivatives of sin(x). We can arrange them in a circle. x+2!1+x2+ ..+

+3!x3

n! ..xn

=Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

derivative: 0th, 4th, 8th, .. Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

derivative: 0th, 4th, 8th, .. derivative: 1st, 5th, 9th, .. Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

derivative: 0th, 4th, 8th, .. derivative: 1st, 5th, 9th, .. derivative: 2nd, 6th, 10th, .. Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

derivative: 0th, 4th, 8th, .. derivative: 1st, 5th, 9th, .. derivative: 2nd, 6th, 10th, .. derivative: 3rd, 7th, 11th, .. Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

derivative: 0th, 4th, 8th, .. derivative: 1st, 5th, 9th, .. derivative: 2nd, 6th, 10th, .. derivative: 3rd, 7th, 11th, .. At x = 0: Maclaurine Expansions

sin(x) cos(x) -sin(x) -cos(x)

derivative: 0th, 4th, 8th, .. derivative: 1st, 5th, 9th, .. derivative: 2nd, 6th, 10th, .. derivative: 3rd, 7th, 11th, .. At x = 0: 0 1 0 -1

derivative: 0th, 4th, 8th, .. derivative: 1st, 5th, 9th, .. derivative: 2nd, 6th, 10th, .. derivative: 3rd, 7th, 11th, .. Maclaurine Expansions

Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:Maclaurine Expansions

1x0+2!=0+x2P(x)

-1+3!x3

Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:0+4!x4

1+5!x5

+6!x6

0+7!x7..

-1Maclaurine Expansions

1x0+2!=0+x2P(x)

-1+3!x3

Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:0+4!x4

1+5!x5

+6!x6

0+7!x7..

-11x=P(x) 3!x3

+5!x5

+..7!x7

Maclaurine Expansions

1x0+2!=0+x2P(x)

-1+3!x3

Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:0+4!x4

1+5!x5

+6!x6

0+7!x7..

-11x=P(x) 3!x3

+5!x5

+..7!x7

The alternating odd numbers {1, -3, 5, -7, ..} may be written as {(-1)k(2k + 1)} for k = 0, 1, 2, 3, .. Maclaurine Expansions

1x0+2!=0+x2P(x)

-1+3!x3

k=0Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:0+4!x4

1+5!x5

+6!x6

0+7!x7..

-11x=P(x) 3!x3

+5!x5

+..7!x7

The alternating odd numbers {1, -3, 5, -7, ..} may be written as {(-1)k(2k + 1)} for k = 0, 1, 2, 3, .. 1x=P(x) 3!x3

+5!x5

+ .. =7!x7

Maclaurine Expansions

1x0+2!=0+x2P(x)

-1+3!x3

k=0(2k+1)!

(-1)kx2k+1is the Mac-series of sin(x).Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:0+4!x4

1+5!x5

+6!x6

0+7!x7..

-11x=P(x) 3!x3

+5!x5

+..7!x7

The alternating odd numbers {1, -3, 5, -7, ..} may be written as {(-1)k(2k + 1)} for k = 0, 1, 2, 3, .. 1x=P(x) 3!x3

+5!x5

+ .. =7!x7

Maclaurine Expansions

1x0+2!=0+x2P(x)

-1+3!x3

k=0(2k+1)!

(-1)kx2k+1is the Mac-series of sin(x).Set 0, 1, 0, -1, 0, 1, 0, -1, .. for f(n)(0) in the expansion:0+4!x4

1+5!x5

+6!x6

0+7!x7..

-11x=P(x) 3!x3

+5!x5

+..7!x7

The alternating odd numbers {1, -3, 5, -7, ..} may be written as {(-1)k(2k + 1)} for k = 0, 1, 2, 3, .. 1x=P(x) 3!x3

+5!x5

+ .. =7!x7

Verify that the Mac-series of f(x) = cos(x) isk=0(2k)!

(-1)kx2k=P(x)+ 4!x4

6!x6

8!x8

+ 1 .. =2!x2

Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 (1)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 (2)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 so at x = 0, f (x) = 2! (2)(2)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 so at x = 0, f (x) = 2! (2)(2)f (x) = 3*2(1 x)-4 (3)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 so at x = 0, f (x) = 2! (2)(2)f (x) = 3*2(1 x)-4 so at x = 0, f (x) = 3! (3)(3)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 so at x = 0, f (x) = 2! (2)(2)f (x) = 3*2(1 x)-4 so at x = 0, f (x) = 3! (3)(3)In general, f (x) = n! (n)Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 so at x = 0, f (x) = 2! (2)(2)f (x) = 3*2(1 x)-4 so at x = 0, f (x) = 3! (3)(3)In general, f (x) = n! (n)Therefore, P(x) = 1x+2!1+x2

+3!x3

+4!x4

+ 2!3!4!Maclaurine Expansions

Example: D. Find the Mac-expansion of f(x) = (1 x)-1 around x = 0.Again, we compute and obtain a pattern of the derivatives first. f(x) = (1 x)-1 so at x = 0, f(0) = 1f (x) = (1 x)-2 so at x = 0, f (x) = 1 (1)(1)f (x) = 2(1 x)-3 so at x = 0, f (x) = 2! (2)(2)f (x) = 3*2(1 x)-4 so at x = 0, f (x) = 3! (3)(3)In general, f (x) = n! (n)Therefore, P(x) = 1x+2!1+x2

+3!x3

+4!x4

+ or2!3!4!P(x) = 1 + x + x2 + x3 + x4 .. is the Mac-series for (1- x) . 1

Maclaurine Expansions

Summary of the Mac-seriesI. For polynomials P, Mac-poly of degree k consists the first k-terms of the polynomial P. Mac-series of polynomials are themselves.II. For ex, its k=0k! .

xkx+2!1+x2+ ..+

+3!x3

n! ..xn

=k=0(2k+1)!

(-1)kx2k+1x 3!x3

+5!x5

+ .. =7!x7

III. For sin(x), itsIV. For cos(x), itsk=0(2k)!

(-1)kx2k+ 4!x4

6!x6

8!x8

+ 1 .. =2!x2

V. For , its(1 x ) 1

1 + x + x2 + x3 + x4 .. =k=0xkComputation Techniques for Maclaurin Expansions VI. For Ln(1 + x), its+ 3x3

4x4

5x5

+ x 2x2

..k=1k .(-1)k+1xk=