HP 17B and HP 27S: Derivatives to the Nth Order (and TI-84 Plus CE Python 5.7 update)

 HP 17B and HP 27S: Derivatives to the Nth Order

With the Solver of the HP 17B family, HP 27S, and the HP 19B calculator, we can calculate derivatives of any order.  Four derivatives are presented here.  You can use any name you want other than those presented here.  Each derivative is to the kth order. 

Derivative 1:   d^k/dx^k a × x^n 

N and K must be positive integers, D is the value of the derivative  

DER1: D=A×X^(N-K)×PERM(N:K)

Example:

Input:  N = 2, K = 1, A = 3, X = 1.5

Result:  D = 9

Derivative 2:   d^k/dx^k e^(a × x) 

K must be a positive integer, D is the value of the derivative

DER2: D=A^K×EXP(A×X)

Example:

Input:  A = 1.8, K =3, X = 3

Result:  D = 213.4409

Derivatives 3 and 4 will require trigonometric functions, which are not available on the HP 17B family.   It is recommended you set the calculator to Radian angle mode.

Derivative 3:  d^k/dx^k sin(a × x)

DER3: D=IF(MOD(K:2)=0:(-1)^(K÷2)×A^K×SIN(A×X):(-1)^((K+3)÷2)×A^K×COS(A×X))

Examples:

Input:  A = 0.75, X = 0.66, K = 2

Result: D = -0.2672

Input:  A = 0.75, X = 0.66, K = 3

Result: D = -0.3712

Derivative 4:  d^k/dx^k cos(a × x)

DER4: D=IF(MOD(K:2)=0:(-1)^(K÷2)×A^K×COS(A×X):(-1)^(K÷2+1÷2)×A^K×SIN(A×X))

Examples:

Input:  A = 0.75, X = 0.66, K = 2

Result: D = -0.4950

Input:  A = 0.75, X = 0.66, K = 3

Result: D = 0.2004

Eddie

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Σ(1 / (a^n)) from n=1 to m

 Σ(1 / (a^n)) from n=1 to m

This blog entry covers the sum of the series:

Σ[1 / (a^n), n=1 to m] with n and m positive integers

Specific Cases:  a = 2 and a = 3

When a = 2:

m = 1:   1/2

m = 2:   1/2 + 1/4  = (2 + 1)/4 = 3/4

m = 3:   1/2 + 1/4 + 1/8 = (4 + 2 + 1)/8 = 7/8

m = 4:   1/2 + 1/4 + 1/8 + 1/16 = (8 + 4 + 2 + 1)/16 = 15/16

Going from the pattern,

Σ[1 / (2^n), n=1 to m] = 1/(2^m) * Σ[(2^n), n=0 to m-1] = (2^m - 1) / 2^m 

When a = 3:

m = 1:  1/3

m = 2:  1/3 + 1/9  = (3 + 1)/9 = 4/9

m = 3:  1/3 + 1/9 + 1/27 = (9 + 3 + 1)/27 = 13/27

m = 4:  1/3 + 1/9 + 1/27 + 1/81 = (27 + 9 + 3 + 1)/81 = 40/81

Going from the pattern,

Σ[1 / (3^n), n=1 to m] = 1/(3^m) * Σ[(3^n), n=0 to m-1]

Finding the General Formula and Proof

Let's presume that, for any a:

Σ[1 / (a^n), n=1 to m-1] = 1/(a^(m-1) * Σ[(a^n), n=0 to m-2]

Let's add 1/(a^m) to the series:  

Σ[1 / (a^n), n=1 to m-1]  + 1/(a^m)

= (1 + a + a^2 + ... + a^(m-3) + a^(m-2)) / (a^(m-1)) + 1 / (a^m)

= (a * (1 + a + a^2 + ... + a^(m-3) + a^(m-2)) + 1) / (a^m)

= (a + a^2 + a^3 + ... + a^(m-2) + a^(m-1) + 1) / (a^m)

= 1/(a^(m)) * Σ[(a^n), n=0 to m-1]

The general formula is now: 

Σ[1 / (a^n), n=1 to m] = 1/(a^(m)) * Σ[(a^n), n=0 to m-1]

Until next time,

Eddie

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Sum of Increasing Series

 Sum of Increasing Series

The Sum From 1 to N - Increasing by 1

It is well known that sum from 1 to N, which each term increasing by 1 is:

1 + 2 + 3 + 4 + ... + N = ∑ x from x = 1 to N = (N + 1) * N / 2

The derivation is fairly easy.  Let S be the sum:

S = 1 + 2 + 3 + ... + N-1 + N

Add S to both sides:

2 * S = 1 + 2 + 3 + ... + N-1 + N + 1 + 2 + 3 ... + N-1 + N

Thanks to the commutative property of addition, we can arrange terms, and with a clever and creative way of arranging terms:

2 * S = ( 1 + N ) + ( 2 + N - 1 ) + ( 3 + N - 2 ) + ... + ( N - 1 + 2 ) + ( N + 1 )

Note that:

2 * S = 

1   +    2   +   3   + ... +  N-1  + N +

N  +  N-1 + N-2  + ..  +   2   +  1

Written this way, there are N "pairs".  Hence:

2 * S = ( 1 + N ) + ( N + 1 ) + ( N +1 ) + ... + ( N + 1 ) + ( N + 1 )

2 * S = ( N + 1 ) * N

Solving for S:

S =  ( N + 1 ) * N / 2

The Sum From 1 to N - Increasing by 2

Now lets consider the sum:

1 + 3 + 5 + 7 + ...

Fun fact: Adding odd numbers in this fashion will always total perfect squares.

1 + 3 = 4 = 2^2

1 + 3 + 5 = 9 = 3^2

1 + 3 + 5 + 7  = 16 = 4^2

1 + 3 + 5 + 7 + 9  = 25 = 5^2

... and so on

Let's define S as the sum:

S = 1 + 3 + 5 + ... + N-4 + N-2 + N

Note that

1 = 2 * 0 + 1

3 = 2 * 1 + 1

5 = 2 * 2 + 1

and so on...

For integer q,

N = 2 * q + 1

Then:

S = 1 + 3 + 5 + ... + ( 2*(q-2) + 1 ) + ( 2*(q-1) + 1) + ( 2*q + 1 )

Using the same strategy as last time:

2 * S = 1 + 3 + 5 + ... + ( 2*(q-2) + 1 ) + ( 2*(q-1) + 1) + ( 2*q + 1 ) 

           + 1 + 3 + 5 + ... + ( 2*(q-2) + 1 ) + ( 2*(q-1) + 1) + ( 2*q + 1 )

2 * S =

     1         +          3            +         5            + ... + (2*(q-2) + 1) + (2*(q-1) + 1) +   (2*q + 1) +

(2*q + 1) + (2*(q-1) + 1)  + (2*(q-2) + 1) +  ... +             5        +            3        +     1

Combine each pair as such:

2  * S = (2*q + 2) + (2*q + 2) + (2*q + 2) + ... + (2*q + 2) + (2*q + 2) + (2*q + 2)

Since q starts at q = 0, there are q+1 "pairs".  

2 * S = (q + 1) * (2*q + 2)

S = (q + 1) * (2*q + 2) / 2

To show that the sum is a perfect square:

S = (q + 1) * (2*q + 2) / 2

= (2*q^2 + 2*q  + 2*q + 2) / 2

= (2*q^2 + 4*q + 2) / 2

= (q^2 + 2*q + 1)

= (q + 1)^2

Example:

Calculate 7^2 (albeit the "longer" way):

7 = q+1;  q = 6:

S = (6 + 1) * (2*6 + 2) / 2 = 7 * 14 / 2 = 7 * 7 = 49

The Sum From 1 to N - Increasing by 3

Now add the series:

S = 1 + 4 + 7 + 10 + ... + N - 3 + N

where each term increases by 3.  Letting q be a positive integer and noting that

1 = 3*0 + 1

4 = 3*1 + 1

7 = 3*2 + 1

10 = 3*3 + 1

and so on...

Then:

S = (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + .... + (3(q-1) + 1) + (3q + 1)

2* S 

= (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + ... + (3(q-1) + 1) + (3q + 1) 

+ (3*0 + 1) + (3*1 + 1) + (3*2 + 1) + ... + (3(q-1) + 1) + (3q + 1)

2 * S 

= (3*0 + 1) + (3*1 + 1)     + (3*2 + 1)     + ... + (3(q-1) + 1) + (3q + 1) 

+ (3*q + 1) + (3(q-1) + 1) +(3(q-2) + 1) + ...  +  (3*1 + 1)    + (3*0 + 1)

There are q+1 pairs.

2 * S 

= (3*q + 2)  + (3*q + 2) + (3*q + 2) + ... + (3*q + 2) + (3*q + 2)

2 *S = (q + 1) * (3*q + 2)

S = (q + 1) * (3*q + 2) / 2

Example:

q = 5  (terms from 0 to 5)

S = (5 + 1) * (3*5 + 2) / 2 = 6 * 17 / 2 = 51

1 + 4 + 7 + 10 + 13 + 16 = 51 

The Sum From A for q terms, increasing by D

Let A ≥ 0, D > 0 and q be a positive integer, and A is a starting term, let the sum S be:

S = A + (A + D) + (A + 2*D) + (A + 3*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D)

Since we start at q = 0, there are q+1 terms.

2*S =

A + (A + D) + (A + 2*D) + (A + 3*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D) +

A + (A + D) + (A + 2*D) + (A + 3*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D)

2*S = 

A + (A + D) + (A + 2*D) + ... + (A + (q-2)*D) + (A + (q-1)*D) + (A + q*D) +

(A + q*D) + (A + (q-1)*D) + (A + (q-2)*D) + ... + (A + 2*D) + (A + D) + A

2*S = (2*A + q*D) + (2*A + q*D) + (2*A + q*D) + ... + (2*A + q*D) + (2*A + q*D) + (2*A + q*D) 

2*S = (q + 1) * (2*A + q*D)

S = (q + 1) * (2*A + q*D) / 2

Example:

A = 10, D = 7; q = 4  (5 terms, increase by 7, initial term is 10)

(4 + 1) * (2*10 + 4*7) / 2 = 5 * (20 + 28) / 2 = 120

Note:  10 + 17 + 24 + 31 + 38  = 120

Source:

Knott, Dr. Ron "Proving that 1+2+3+...+n is n(n+1)/2"  February 12, 2003.  Accessed December 6, 2020.  http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html

Eddie

All original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Graph Gallery Series with GeoGebra

Graph Gallery Series with GeoGebra

The following graphs are from one to five terms of an infinite series of the form:

Σ f(n,x) from n = 0 to ∞

I use the GeoGebra online graphing tool, it is a great online, and free, graphing, mathematics, and geometry app.  Check GeoGebra out at https://www.geogebra.org. 

Enjoy!

Series 1: 

Σ( n * x ) from n = 0 to ∞

Series 2:

Σ( x^n * e^(-n*x) ) from n = 0 to ∞

Series 3:

Σ( x^(-n) ) from n = 0 to ∞

Series 4:

Σ( n * erf((n * x) / (n + 1)) ) from n = 0 to ∞

Series 5:

Σ( n * sin x + sin( x * n ) ) from n = 0 to ∞

Series 6:

Σ( x^(-n) * cos( x * n ) ) from n = 0 to ∞

One of my favorite pictures.  I used this as a wallpaper on my PC.  Eddie 

Eddie

All original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

The Series ( ((((0 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + … + 1)^-1 and Fibonacci Numbers

The Series ( ((((0 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + … + 1)^-1 and Fibonacci Numbers

  

Add One Then Reciprocate

Define the series t as:

t = ( ((((0 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + … + 1)^-1    (an infinite amount of terms)

This is a sum that can’t be easily stated in summation statement (Σ f(x)). 

On the HP Prime, I programmed this as:

EXPORT TEST1112(n)

BEGIN

LOCAL k, t:=0;

FOR k FROM 1 TO n DO

t:=(t+1)^-1;

END;

RETURN t;

END;

The result seems to converge at 0.6180339785 when n ≥ 27.  Note that 0.6180339785 = ϕ – 1, where ϕ is the Golden Ratio ( ϕ = (√5 + 1)/2)

Fibonacci Gets Involved

Note that:

k =	t =
1	1
2	(1 + 1)^-1 = 1/2
3	(1 + 1/2)^-1 = (3/2)^-1 = 2/3
4	(1 + 2/3)^-1 = (5/3)^-1 = 3/5
5	(1 + 3/5)^-1 = (8/5)^-1 = 5/8
6	(1 + 5/8)^-1 = (13/8)^-1 = 8/13
7	(1 + 8/13)^-1 = (21/13)^-1 = 13/21

We get a sequence of terms {1, 1/2, 2/3, 3/5, 5/8, 8/13, 13/21, 21/34, 34/55, 55/89, 89/144, …} where each term takes the fraction a/b, a is the kth Fibonacci number and b is the (k+1)th Fibonacci number.  Can we show that this sequence of partial sums is convergent?

Each partial sums of the series takes the form F_k / F_k+1 where F is the Fibonacci number.

The closed formula for the Fibonacci number is:

F_k = ( ϕ^k – α^k )/√5 , where ϕ  = (1 + √5)/2 and α = (1 - √5)/2.

Then:

F_k / F_k+1

=( ϕ^k – α^k )/√5 * √5/( ϕ^k+1 – α^k+1)

=( ϕ^k – α^k) / ( ϕ^k+1 – α^k+1 )

= ( ϕ^k / ϕ^k+1) * ( (1 – (α/ϕ)^k) / (1 – (α/ϕ)^k+1) )

= 1/ϕ * ( (1 – (α/ϕ)^k) / (1 – (α/ϕ)^k+1) )

Note that α/ϕ = (1 - √5)/(1 + √5 ) ≈ -0.38197 < 1

As k → ∞,  α/ϕ → 0. 

Hence,

lim k → ∞ (F_k / F_k+1)

= lim k → ∞ (1/ϕ * ( (1 – (α/ϕ)^k) / (1 – (α/ϕ)^k+1) ) )

= 1/ϕ

Simplifying:

1/ϕ

= 2/(1 + √5)

= 2*(1 - √5) / ((1 + √5)*(1 - √5))

= 2*(1 - √5)/-4

= (√5 – 1)/2

= √5/2 – 1/2

Adding and subtracting 1/2:

√5/2 – 1/2 + (1/2 – 1/2)

= (√5 + 1)/2 – 1

= ϕ – 1

Since the sequence of partial sums converge to ϕ – 1, the series

t = ( ((((0 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + 1)^-1 + … + 1)^-1   

converges to ϕ – 1.

This blog is property of Edward Shore.  2015.

HP Prime: Plotting Sums of Series

Introduction

To explore the sum of the following series:

S(x) = Σ( f(N,x), N, 1, x)

As x varies from 1 to 250.

Goals

To explore different series to determine if the converge and at what approximate value. Each sum will be determined using 250 terms. I used the HP Prime to determine:

* The sum of each series after 250 terms: Σ(f(N,x), N, 1, 250), and,

* The change between the sum using 249 terms to the sum using 250 terms:
Σ(f(N,x), N, 1, 250) - Σ(f(N,x), N, 1, 249)

Decimal Accuracy for the HP Prime: 13 digits

* I also make a Scatterplot for each series using the Statistics 2Var app and turning the Fit option turned off. Plot setup: X Range = [-5, 255], Y Range = [-2, 2], X Tick = 10, Y Tick = 0.25

Each scatter plot is presented before the results.

Σ( 1/(1 - 2^N), N, 1, X)

250th Value: -1.60669515242
Change from 249th to 250th value: 0 (internal calculator accuracy has been met)

Σ( 1/(2^N - 1), N, 1, X)

250th Value: 1.60669515242
Change from 249th to 250th value: 0 (internal calculator accuracy has been met)

Σ( 1/(N^2), N, 1, X)

250th Value: 1.6409205624
Change from 249th to 250th value: 0.000016

If we allow this series to go to infinity, we get zeta(2) = π^2/6

Σ( 1/(2^N), N, 1, X)

250th Value: 1
Change from 249th to 250th value: 0 (internal calculator accuracy has been met)

Σ( 1/(2^N + N^2), N, 1, 250)

250th Value: 5.52714787527 x 10^-76
Change from 249th to 250th value: -5.527147875323 x 10^-76

Could this sum eventually converge to 0 as x approaches infinity?


This blog is property of Edward Shore. 2014

Calculus Revisited #18: Series

Welcome to Part 18 of our wonderful 21 part of our Calculus Revisited Series. I hope you are enjoying this series. An announcement: Part 21 will be a "catch all" section - covering some of the topics we don't get to in detail in this series.

The next three blog entries will be about series. Today, the basics.

Series:

A series is a sum of terms in a sequence. The series can be finite or infinite. An infinite series is convergent (has a value) if the following is true:

∞
∑ a_n = S
n=i

and S < ∞

Here are some famous series:

Arithmetic Series:

a + (a + d) + (a + 2d) + (a + 3d) + ...

Sum for a finite arithmetic series:

n-1
∑ a + k * d
k=0

= a * n + (d * n * (n -1))/2

Geometric Series:

a + a * r + a * r^2 + a * r^3 + ...

Sum for an infinite geometric series, provided that |r| < 1:

∞
∑ a * r^k = a / (1 - r)
k = 0

Harmonic Series:

∞
∑ 1/n
n=1

= 1 + 1/2 + 1/3 + 1/4 + 1/5 + ....

There is no closed formula for a finite harmonic series. In addition the infinite harmonic series diverges (has no sum).

Series Properties:

∑ c * a(n) = c * ∑ a(n) (c is a constant)

∑ a(n) + b(n) = ∑ a(n) + ∑ b(n)

n
∑ k = n * (n + 1)/2
k=1

n
∑ k^2 = n * (n + 1) * (2n + 1)/6
k = 1

n
∑ k^3 = n^2 *(n + 1)^2 / 4
k = 1


Subtracting from a whole:
If t > 1:

k
∑ a(n) =
n = t

∑( a(n) for n = 1 to k) - ∑( a(n) for n = 1 to t - 1)

Problems

1. Calculate

6
∑ 50 - 2k
k = 0

This is an arithmetic series with n = 6 + 1 = 7, a = 50, and d = -2.

Then the sum is:
(50)(7) + (-2 * 7 * 6)/2 = 308


2. Show why the (infinite) harmonic series is divergent.

Here is one way:

Let S be the sum of the harmonic series, that is:

S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

Note that:
1 + 1/2 > 2/2
1/3 + 1/4 > 2/4
1/5 + 1/6 > 2/6
1/7 + 1/8 > 2/8
and so on...

Then we have 2/2 + 2/4 + 2/6 + 2/8 + ....

Which simplifies to 1 + 1/2 + 1/3 + 1/4 + .... = S

Which implies S > S, which is impossible.

The harmonic series is divergent.

3. Find the sum:

∞
∑ .5^n + .3^n
n=0

We can first break the sum up:

∑(.5^n for n=0 to ∞) + ∑(.3^n for n=0 to ∞)

Both terms are geometric series. r = .5 for the first term and r = .3 for the second term.

Then:

∑(.5^n for n=0 to ∞) + ∑(.3^n for n=0 to ∞)
= 1 / (1 - .5) + 1 / (1 - .3)
= 1 / .5 - 1 / .7
= 24/7 ≈ 3.42857

4. Find the sum:

∞
∑ 2 / (5^n)
n = 1

This looks like a geometric series with r = 1/5. If we can put the sum into it's proper form, perhaps by adding and subtracting 2 / 5^0:

∑( 2 / (5^n) for n = 1 to ∞)
= ∑ ( 2 / (5^n) for n = 0 to ∞) - 2 / 5^0
= ∑ ( 2 / (5^n) for n = 0 to ∞) - 2
= 2 / ( 1 - 1/5) - 2
= 2 / (4/5) - 2
= 5/2 - 2 = 1/2

5. Find the sum:

20
∑ k^2
k=10

Again, looks like a k^2 series, but use the subtraction from the whole technique and:

∑( k^2 for k = 10 to 20)
= ∑(k^2 for k = 1 to 20) - ∑(k^2 for k = 1 to 10 - 1)
= (20)(20 + 1)(2 * 20 + 1)/6 - (9)(9 + 1)(2 * 9 + 1)/6
= 17220 / 6 + 1710 / 6
= 15510 / 6 = 2585

Thank you once again for joining us. Next time we will work some series convergence tests.

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited # 20: Taylor Series and Maclaurin Series

Welcome to Part 20 of 21: Taylor and Maclaurin Series.

The Taylor and Maclaurin series are representation of the function f(x) by using an infinite series. If we use a finite number of terms, the series can (I stress can), but a good approximation f(x).

A Taylor series of f(x) is centered on a focus point x = a. Generally, approximations are best when x is around a, and gets worse the further x gets from a.

Taylor Series:

About the point x = a:

f(x) = f(a) + f'(a) * (x - 1) + f''(a) * (x - a)^2 / 2! + f'''(a) * (x - a)^3 / 3! + ....

If the series is cut off at n terms, the final term of the Taylor series is:

f^(n+1)(t) / (n+1)! * (x - a)^(n+1). This is known as the error term.

Maclaurin Series:

The Taylor series with a = 0.

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

With error term f^(n+1)(0) / (n+1)! * x^(n+1)

Some famous series:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + .... + x^n/n! + ....

sin x = x - x^3/3! + x^5/5! - x^7/7! + .... + ( (-1)^n * x^(2n+1) ) / (2n+1)! + ....

cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ... + ( (-1)^n * x^(2n) ) / (2n)! + ....

Problems

1. Find a Macluarin series for

f(x) = ln(x + 1) to four terms

f(x) = ln(x + 1)
f(0) = ln(0 + 1) = 0

f'(x) = 1/(x+1)
f'(0) = 1/1 = 1

f''(x) = -1/(x+1)^2
f''(0) = -1/(1^2) = -1

f'''(x) = 2/(x+1)^3
f'''(0) = 2/(1^3) = 2

Then:

ln(x + 1) = 0 + 1 * x + (-1) * x^2/2! + 2 * x^3/3! + ....
= x - x^2/2! + 2x^3/3! + ....

2. Find the Macluarin series for

f(x) = e^(-x^2) to three nonzero terms. Approximate its integral.

f(x) = e^(-x^2)
f(0) = 1

f'(x) = e^(-x^2) * (-2x)
f'(0) = 0

f''(x) = e^(-x^2) * (4x^2 - 2)
f''(0) = -2

f'''(x) = e^(-x^2) * (-8x^3 + 12x)
f'''(0) = 0

f''''(x) = e^(-x^2) * (16x^4 - 48x^2 + 12)
f''''(0) = 12

Then:

e^(-x^2) = 1 - 2 * x^2/2! + 12 * x^4/4! + ....
= 1 - x^2 + x^4/2 + ...

∫ e^(-x^2) dx = x -x^3/3 + x^5/10 + .... + C

3. Find the Taylor series for cos x at a = π/4 to three nonzero terms.

f(x) = cos x
f(π/4) = √2/2

f'(x) = - sin x
f'(π/4) = -√2/2

f''(x) = -cos x
f''(π/4) = -√2/2

Then:

cos x = √2/2 - √2/2 * (x - π/4) + √2/2 * (x - π/4)^2/2! + ....

That concludes our section on Taylor and Maclaurin series. The next and last entry will be our "catch all" entry.

Thank you as always,

Eddie

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