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Sucesiones y seriesSucesión es un conjunto de elementos (o términos o miembros) en un cierto orden.

Finita {a1, a2 a3,a4} Infinita {a1, a2 a3, …}

Ejemplos:

Finitas: {1,3,5,7}, {0.99, 0.99, 0.99}, {1,4,7,10}, { 1,3,3,1}

Infinitas. {1,-1, 2, -2, 3, -3…}; {1,2,3,4…}; {six, sin2x, sin3x, sin4, …}

Término enésimo

Ejemplos:

an = 2n + 1, n=0, 1, 2… Genera: {1, 3, 5, 7, 9…}

ao=1, a1= xn

n ! Genera: {1, x

1!, x2

2!, x3

3! …}

Donde Factorial: 3!=1x2x3=6; 2!=2x1=2; 1!=1; 0!=1

an=n2−10n+1

, n=4,5,6,.. Genera: {1.2, 2.5, 3.7142…, 4.875, 6, 7.1…}

Ejercicio: Genere las series.

an=n2

2n ,−2≤ n ≤5

an=1−1❑

n , n ≥ 1

bn=1+(−1)n 1n

, n ≥ 1

cn=(−1)n+ 1n

, n ≥ 1

Encontrar (o determinar) el término enésimo de las sucesiónes

U= {3,8,13…} an=5 n−2

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V={0, 14

, 49 , 9

16 , 1625 , 25

36… } an=

n2

(n+1)n

U ={1,2,4,8,16, 32…} an=¿ 2n

{4, 2, 1, 12, 1

4 , 18 , 1

16 , … } an=4 /2−n

Algunas sucesiones especiales

Números triangulares: {1,3,6,10,15…}

Números cuadrados: {1,4,9,16…}

Números cúbicos: {1,8, 27, 64..}

Fibonacci: an=an−1+an−2

Una sucesión es aritmética cuando an−1−an−2 = constante

Una sucesión es geométrica an+1/an = constante

Ejercicios.

Determine de las series anteriores cuales son aritméticas y cuales geométricas.

Dada an=

n2−10n+1

sume a4+a9=

Convergencia y divergencia de las sucesiones

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UNA SERIE ES LA SUMA DE LOS TÉRMINOS DE UNA SUCESIÓN.

Ejemplos. Sucesión {1,3,5,7} Serie: S=1+3+5+7=16

Serie de potencias: ex=1+ x1 !

+ x2

2 !+ x3

3 !+…,−∞<x<∞

Series de Fourier: f ( x )=a0+∑n=1

∞

(an cos nπxL

+bn sin nπxL )

Teorema binomial ( x+a )n=∑k=0

n

(nk )xk an−k

For example, for the sum: S = 1492 + 1499 + 1506 +… 2003 of consecutive terms of a succession of reason 7, we found\ frac {2003 - 1492} {7} + 1 = 74 terms, and the 74 sum is \ frac {\ times (1492 + 2003)}{2} = 129 315.

S=( 2003−14927 + 1)(2003+1492

2 ) = (74)(1747.5) = 129 315

Ejemplos.

∑n=1

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( 2n+1¿¿2 (1 )+1 )+¿2 (2 )+1+2 (3 )+1+2 ( 4 )+1=3+5 9=24 ¿

∑n=0

∞

( 12n ) = 1 + 1

2+ 14 + 1

8 + 116

+…Sumar hasta el infinitode términos de la serie .r

∑n=1

∞

( 310n ) = 3

10+ 3100 + 3

1000 +… Sumar hasta elinfinito de términos de la serie .

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Sumar hasta el infinito (∞ ¿ parece no tener significado aún.

Serie geométrica:

∑k =1

∞

( a rkn−1 ) =a+ar+ar2+ar3+… donde a ≠ 0

La serie geométrica a la suma S=a

1−r si |r |<1, pero diverge si |r |≥1

Ejercicio

Determine si converge la serie y obtenga el límite.

∑n=1

∞

( 43n ) Aquí a=4

3 r=13 Como |1

3 | < 1 converge

La suma será S= lim ¿n ∞ Sn=a

1−r¿=

43

1−13

r = 4

3−1=2

∑n=1

∞

( 43n ) = 2

Le toca a usted.

∑n=1

∞

( 310n ) = 3

10+ 3100 + 3

1000 +…=

Serie armónica.

∑n=1

∞

(1n )=1 + 1

2+13+… + 1

n + … Esta serie DIVERGE

Demostrarlo tomando grupos de términos que sumen 12

Así aunque lim ¿n→ ∞ ¿ (1/n) = 0 el término enésimo tiende a cero

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La serie ∑n=1

∞

( n3

3 n3+2n2 ) pruebe que diverge.

lim ¿n→∞( n3

3 n3+2n2 )¿ = 13 Por la prueba del enésimo término la serie

diverge.

La serie ∑k =1

∞

( 1(k+2 )(k+3)❑ ) Demuestre que converge.

Uso de fracciones parciales más

lim ¿n→ ∞ Sn=13¿ La serie converge y tiene esa suma ∑

k =1

∞

( 1(k+2 )(k+3)❑ ) = 1

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Series de Taylor y de Maclaurin

Taylor Series

 

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If we use  , so we are talking about the Taylor Series about  , we call the series a Maclaurin Series for   or, Maclaurin Series

Example 1  Find the Taylor Series for   about  . SolutionThis is actually one of the easier Taylor Series that we’ll be asked to compute.  To find the Taylor Series for a function we will need to determine a general

formula for  .  This is one of the few functions where this is easy to do right from the start. 

To get a formula for   all we need to do is recognize that,

                                           

and so,                                     

 

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Therefore, the Taylor series for   about x=0 is,

                                                       

Example 2  Find the Taylor Series for   about  . SolutionThere are two ways to do this problem.  Both are fairly simple, however one of them requires significantly less work.  We’ll work both solutions since the longer one has some nice ideas that we’ll see in other examples. Solution 1

As with the first example we’ll need to get a formula for  .  However, unlike the first one we’ve got a little more work to do.  Let’s first take some derivatives and evaluate them at x=0.                           

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After a couple of computations we were able to get general formulas for

both   and  .  We often won’t be able

to get a general formula for   so don’t get too excited about getting that formula.  Also, as we will see it won’t always be easy to get a general

formula for  . So, in this case we’ve got general formulas so all we need to do is plug these into the Taylor Series formula and be done with the problem.

                                                          

 Solution 2The previous solution wasn’t too bad and we often have to do things in that manner.  However, in this case there is a much shorter solution method.  In the previous section we used series that we’ve already found to help us find a new series.  Let’s do the same thing with this one.  We already know a Taylor Series for   about   and in this case the only difference is we’ve got a “-x” in the exponent instead of just an x.  

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So, all we need to do is replace the x in the Taylor Series that we found in the first example with “-x”.

                                                 

 This is a much shorter method of arriving at the same answer so don’t forget about using previously computed series where possible (and allowed of course).

Example 3  Find the Taylor Series for   about  . SolutionFor this example we will take advantage of the fact that we already have a Taylor Series for   about  .  In this example, unlike the previous example, doing this directly would be significantly longer and more difficult. 

                                                     

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 To this point we’ve only looked at Taylor Series about   (also known as Maclaurin Series) so let’s take a look at a Taylor Series that isn’t about  .  Also, we’ll pick on the exponential function one more time since it makes some of the work easier.  This will be the final Taylor Series for exponentials in this section.

Example 4  Find the Taylor Series for   about  . Solution

Finding a general formula for   is fairly simple.                                 

 The Taylor Series is then,

                                                    

 Okay, we now need to work some examples that don’t involve the exponential function since these will tend to require a little more work.

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Example 5  Find the Taylor Series for 

 about  . SolutionFirst we’ll need to take some derivatives of the function and evaluate them at x=0.

                                      

 In this example, unlike the previous ones, there is not an easy formula for either the general derivative or the evaluation of the derivative.  However, there is a clear pattern to the evaluations.  So, let’s plug what we’ve got into the Taylor series and see what we get,

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 So, we only pick up terms with even powers on the x’s.  This doesn’t really help us to get a general formula for the Taylor Series.  However, let’s drop the zeroes and “renumber” the terms as follows to see what we can get.

                                           

 By renumbering the terms as we did we can actually come up with a general formula for the Taylor Series and here it is,

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This idea of renumbering the series terms as we did in the previous example isn’t used all that often, but occasionally is very useful.  There is one more series where we need to do it so let’s take a look at that so we can get one more example down of renumbering series terms.

Example 6  Find the Taylor Series for   about  . SolutionAs with the last example we’ll start off in the same manner.

                                      

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 So, we get a similar pattern for this one.  Let’s plug the numbers into the Taylor Series.

                                           

 In this case we only get terms that have an odd exponent on x and as with the last problem once we ignore the zero terms there is a clear pattern and formula.  So renumbering the terms as we did in the previous example we get the following Taylor Series.

                                                       

 We really need to work another example or two in which f(x) isn’t about  .

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Example 7  Find the Taylor Series for   about  . SolutionHere are the first few derivatives and the evaluations. 

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 Note that while we got a general formula here it doesn’t work for .  This will happen on occasion so don’t worry about it when it does. In order to plug this into the Taylor Series formula we’ll need to strip out the   term first.

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 Notice that we simplified the factorials in this case.  You should always simplify them if there are more than one and it’s possible to simplify them. Also, do not get excited about the term sitting in front of the series.  Sometimes we need to do that when we can’t get a general formula that will hold for all values of n.

Example 8  Find the Taylor Series for   about  .SolutionAgain, here are the derivatives and evaluations.                     

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Notice that all the negative signs will cancel out in the evaluation.  Also, this formula will work for all n, unlike the previous example. Here is the Taylor Series for this function.

                                                    

 Now, let’s work one of the easier examples in this section.  The problem for most students is that it may not appear to be that easy (or maybe it will appear to be too easy) at first glance.

Example 9  Find the Taylor Series for 

 about  .SolutionHere are the derivatives for this problem.                                

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 This Taylor series will terminate after  .  This will always happen when we are finding the Taylor Series of a polynomial.  Here is the Taylor Series for this one.               

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 When finding the Taylor Series of a polynomial we don’t do any simplification of the right hand side.  We leave it like it is.  In fact, if we were to multiply everything out we just get back to the original polynomial! While it’s not apparent that writing the Taylor Series for a polynomial is useful there are times where this needs to be done.  The problem is that they are beyond the scope of this course and so aren’t covered here.  For example, there is one application to series in the field of Differential Equations where this needs to be done on occasion.

 So, we’ve seen quite a few examples of Taylor Series to this point and in all of them we were able to find general formulas for the series.  This won’t always be the case.  To see an example of one that doesn’t have a general formula check out the last example in the next section. Before leaving this section there are three important Taylor Series that we’ve derived in this section that we should summarize up in one place.  In my class I will assume that you know these formulas from this point on. 

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http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx

  

Example 2  For the function   plot

the function as well as  ,  , and   on the same graph for the interval [-4,4].  SolutionHere is the general formula for the Taylor polynomials for cosine.

                                                        

 The three Taylor polynomials that we’ve got are then,

                                             

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 Here is the graph of these three Taylor polynomials as well as the graph of cosine.

As we can see from this graph as we increase the degree of the Taylor polynomial it starts to look more and more like the function itself.  In

fact by the time we get to   the only difference is right at the ends.  The higher the degree of the Taylor polynomial the better it approximates the function. Also the larger the interval the higher degree Taylor polynomial we need to get a good approximation for the whole interval.Binomial TheoremIf n is any positive integer then,                            

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where,                                

 

This is useful for expanding   for large n when straight forward multiplication wouldn’t be easy to do.  Let’s take a quick look at an example. 

Example 1  Use the Binomial Theorem to expand  SolutionThere really isn’t much to do other than plugging into the theorem.      

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 Now, the Binomial Theorem required that n be a positive integer.  There is an extension to this however that allows for any number at all. 

Mathematical successionOne mathematical succession is one application defined on natural numbers. It is custom to use the letters or, v, w… to designate them, instead of f, g, h… that are used for the functions. In the same way, the variable note usually n (by natural ) instead of x, habitual for the real variables .

By convention, un is written time of u (n), the image of n by the succession or, that is the term number n+1 of succession u (the first term is habitually u0). 

Two ways exist essentially to define a succession:The definition is explicit when a formula occurs that allows to find un by means of a unique calculation where another variable does not take part that n . In other words, un is a function of n : un = f (n) .

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The definition is implicit when… she is not explicit. This means that un not only depends on n but also on other terms of the succession, that will have to calculate antes.

For example > 0 can be fixed uo = 1 and be decided that for any natural n, un = n·un-1. In order to find u3 we say, is necessary to calculate u2 what the knowledge needs u1 which calculates with uo. We obtain: u1 = 1×u0 = 1, soon u2 = 2×u1 = 2 and finally u3 = 3×u2 = 6. They are Factorial

Another example very known is Succession of Fibonacci defined by un+2  =  un+1  +  un.The formula that defines a term in relation to the previous ones calls relation of induction .

The geometric successions

One geometric succession can be defined as function of n : 

u_n = b \ cdot r^n \ qquad (r \ in \ mathbb {R})

Also it can be defined by induction of the following form: 

\begin{matrix} u_0 & = & b \ qquad & (b \ in \ mathbb {R}) \ \ u_ {n+1} & = & r \ cdot u_n & (r \ in \ mathbb {R}) \ end {matrix}

To the real number r also denominates to him reason of the succession.

Example:

The behavior of the geometric succession depends on the sign of the first term and the value of its reason.

If the reason is positive, then the succession is monotonous, and has a very regular aspect, that can be prolonged by a function of Exponential type basic r : u_n = b \ cdot r^n extends in f ( x ) = b · r x .

Four cases, as are distinguished it is seen in the following figure; the ordinates of the black points are the values of the succession, and the curve represents the function:

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If the reason is negative, then the succession is oscillating . Two cases based on if are distinguished r is minor who -1 or no. The sign of the first term does not modify general aspect of the succession (to change of sign is equivalentto a symmetry around the horizontal axis, and here not note much). The powers rn with negative r do not become general to the real ones, except for particular convention, and therefore does not exist a natural function that prolongs the succession. In the following figure the function has multiplied | r | x by the factor cos π x to simulate the periodic change of sign.

If the initial term is null, or if the reason is worth -1, 0 or 1, the succession does not enter the previous classification, but it does not matter then in such case lacks interest.

Discarding these particular cases, it is possible to be said that the convergence of the succession depends on theabsolute value of the reason: 

if | r | > 1, does not converge, and if | r | < 1, converges towards Zero .

We notice q the reason, and suppose q ≠ 1. Then the sum of numbers in geometric progression is given by the following formula, under three equivalent forms:

If -1 < q < 1, the sum of all the terms of the succession is: S = \ frac {u_0} {1 - q} .

Formulas

Supposing that A_n is the term anyone, A_k the term that occupies the position " k", and A_1 the first term of the succession:

In order to find a term anyone in a geometric succession, is due to use: A_n = A_k \ cdot r^ {n-k}

In order to add " n" first terms of a geometric succession: S_n = \ frac {A_n \ cdot r - A_1} {r - 1}

In order to add all the numbers of a succession (Extreme infinite): S_ \ infty = \ frac {A_1} {1-r} . This formula is only applicable when \ 0< r <1

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In order to calculate the product of nº first terms of a succession: P_n = \ sqrt {(A_n \ cdot A_1) ^n}

The aritmeticogeométricas successions

It is, as it indicates his name, a mixture of the two previous definitions. They are possible to be defined by induction of the following form: 

\begin{matrix} w_0 & = & c \ qquad \ quad \ & (c \ in \ mathbb {R}) \ quad \ \ w_ {n+1} & = & q \ cdot w_n + r & (q, r \ in \ mathbb {R}) \ end {matrix}

The induction formula makes take part the sum of the arithmetical succession, and the product of the geometric succession.

We discard the cases q = 1 (arithmetical succession) and r = 0 (geometric succession). Then it is possible to be affirmed that the behavior of the succession is of geometric type, and determined by q, and that  arithmeticcharacter only appears like a translation.

More indeed, he is l the unique number that verifies l = ql + r .

If w 0 = l (what is equivalent to w 1 = w 0) then w is a constant succession. If it is not easy to see that v 1 = wn - l is a geometric succession (non-zero) of reason q, and that therefore: 

if | q | > 1, w does not converge (because does not do v) if | q | < 1, w converges towards l (because v tends towards 0).

Logically, the classification of the previous paragraph according to the values of q continues being valid if we transferred the curves of l vertically units.