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Two linear homogeneous diﬀerence equations of third order, and an homogeneous diﬀerence equation of fourth order.. Find the roots of the characteristic polynomials and then apply a solut[r]

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Examples of Sequences

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Examples of Sequences Calculus 3c-1

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Calculus 3c-1 – Examples of Sequences

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Calculus 3c-1 Preface

Preface

Here follows a collection of sequences, including sequences, which satisfy some simple diﬀerence

equa-tions The reader is also referred to Calculus 3b Since my aim also has been to demonstrate some

solution strategy I have as far as possible structured the examples according to the following form

A Awareness, i.e a short description of what is the problem

D Decision, i.e a reﬂection over what should be done with the problem

I Implementation, i.e where all the calculations are made

C Control, i.e a test of the result

This is an ideal form of a general procedure of solution It can be used in any situation and it is not

linked to Mathematics alone I learned it many years ago in the Theory of Telecommunication in a

situation which did not contain Mathematics at all The student is recommended to use it also in

other disciplines

One is used to from high school immediately to proceed to I Implementation However, examples

and problems at university level are often so complicated that it in general will be a good investment

also to spend some time on the ﬁrst two points above in order to be absolutely certain of what to do

in a particular case Note that the ﬁrst three points, ADI, can always be performed.

This is unfortunately not the case with C Control, because it from now on may be diﬃcult, if possible,

to check one’s solution It is only an extra securing whenever it is possible, but we cannot include it

always in our solution form above

I shall on purpose not use the logical signs These should in general be avoided in Calculus as a

shorthand, because they are often (too often, I would say) misused Instead of ∧ I shall either write

“and”, or a comma, and instead of ∨ I shall write “or” The arrows ⇒ and ⇔ are in particular

misunderstood by the students, so they should be totally avoided Instead, write in a plain language

what you mean or want to do

It is my hope that these examples, of which many are treated in more ways to show that the solutions

procedures are not unique, may be of some inspiration for the students who have just started their

studies at the universities

Finally, even if I have tried to write as careful as possible, I doubt that all errors have been removed

I hope that the reader will forgive me the unavoidable errors

Leif Mejlbro13th May 2008

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is convergent or divergent Find its limit, if it is convergent.

Here we have several possibilities:

the rules of calculations that

−

1 + 1n

(n + 1)n =−2 +

1n

n + 1 → 0 for n → ∞

It is seen in all three variants that the sequence is convergent and its limit is 0 ♦

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is convergent or divergent In case of convergence, ﬁnd its limit.

the denominator in the two fractions, we get

− n − 1

n → ∞ − ∞ − 0

This is an illegal type of convergence and nothing can be concluded in this way

The latter calculation can of course be performed more or less elegant ♦

a1= 0, a2=−1, a3= 0, a4= 1,

are repeated cyclically, i.e they all occur inﬁnitely often Thus we have four candidates of the limit,

but since any possible limit is unique, it does not exist in this case, and the sequence is divergent ♦

Sequences in General

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8

an= n(−1)n

is convergent or divergent Find the limit in case of convergence

Since the subsequence

a2n= (2n)(−1)2n = 2n

is divergent, the “bigger sequence” (an) (it contains more elements) must also be divergent ♦

an= an

n, a∈ R,

This sequence contains a parameter, and the question of convergence depends on the the size of the

parameter

1) If|a| > 1, it follows from the magnitudes that

|an| = 1

n|a|n → ∞ for n→ ∞

(The exponential function “dominates” the power function in n) In this case we have divergence

2) If|a| ≤ 1, we get the estimate

|an− 0| = |an| = 1

n|a|n ≤ 1

n → 0 for n→ ∞

It follows immediately from the deﬁnition that (an) is convergent and that its limit is 0 ♦

an= ln(n2+ 1)− 2 ln n

The type of convergence is “∞ − ∞, so we ﬁrst apply the functional equation of the logarithm Thus

n2 → 1 for n → ∞, we can interchange lnand the limit,

lim

n→∞an = ln

limn→∞

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Calculus 3c-1

and it follows that the sequence is convergent towards the limit 0

→ 0 + 0 · 0 = 0 for n → ∞,hence the sequence is convergent and its limit is 0 ♦

an= (−1)n

n +

1 + (−1)n2

Due to the change of sign (−1)na good strategy would be to consider odd and even indices separately.

Thus we shall consider the two subsequences,

It follows that we have two diﬀerent candidates of the limit, and since a limit is always unique, we

conclude that it does not exist and the sequence is divergent ♦

an= 1

nsin

5n

is convergent or divergent Find the limit if the sequence is convergent

This example is trying to pull the reader’s leg, because one is persuaded to concentrate on the

mys-terious term sin5n, which apparently cannot be controlled

Notice that we always have | sin x| ≤ 1, so

|an− 0| = |an| = 1

n| sin5n| ≤ 1

n → 0 for n→ ∞,and we conclude that

|an− 0| → 0 for n→ ∞,

The sequence is convergent according to the deﬁnition and its limit is 0 ♦

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10

an=√

n + 1−√n

is convergent or divergent Find its limit if it is convergent

This example is of the type “∞ − ∞ It follows from

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is convergent or divergent Find its limit if it is convergent.

The function f (x) = x4 is continuous and independent of n, and the “inner part” converges,

2n− 33n + 7

= f

23

=

23

3 + 7n

⎞

⎟

⎠ →

23

4

= 16

81 for n→ ∞,which is correct, as long as the exponent is a constant, i.e it does not depend on n ♦

an= n

4/3cos

n!π/(√2)n

n + 1

We ﬁrst rewrite an in the following way,

an= n4/3

n + 1 cos

n!π/(√2)n

.The ﬁrst factor tends to ∞,

which, however, in general is not suﬃcient, because cos x during the limit might lie close to 0, so we

get the type of convercence “∞ · 0”

Notice that if we only consider even indices, then we get rid of the square root By the chance of

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2 π

= (2n)4/32n + 1 → ∞ for n→ ∞,and the sequence is divergent ♦

an= cot1

n− n

This is a tricky example, in which one must

1) replace 1/n by x = 1/n, i.e x→ 0+ for n → ∞,

2) apply that cot x = cos x/ sin x, followed by putting everything on the same fraction line,

3) apply Taylor’s formula in both the numerator and the denominator, followed by some reduction,

4) ﬁnally take the limit x→ 0+

The details of this program look like the following:

an = cos1

n− n = cot x −1

x=

cos xsin x −1

We conclude that the sequence is convergent with the limit 0 ♦

convergence

an= 3 + (−2)n

3n+1+ (−2)n+1, n∈ N

When we estimate expressions consisting of two terms the trick is to put the numerically larger and

dominating term outside the expression as a factor We get by using this principle in both the

numerator and the denominator that

n

1 +

−23

n

1 +

−23

n+1

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→ 0 for n → ∞ (standard sequences), hence according

to the rules of calculations,

n

1 +

−23

The type of convergence is “∞ − ∞” We note that both terms behave approximately as n, so we

subtract n from both terms:

an = n(n + 2)

n + 1 − n3

n2+ 1 =

n(n + 2)

→ 1 − 0 + 0 = 1 for n→ ∞

We see that the sequence is convergent with the limit 1

A simpler variant is obtained if we immediately see that

an=3

n3+ 1− n

The type is “∞ − ∞” In this case the trick a − b = (a2− b2)/(a + b) does not work However, we

succeed by a small modiﬁcation First notice that the cubic is removed by taking the third power, i.e

we start by considering a3− b3 where a =√3

n3+ 1)2+ n√3

n3+ 1 + n2 → 0

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14

for n→ ∞, and we see that the sequence is convergent with the limit 0

1 + x, i.e

(1 + x)1/3= 1 +

1/31

x + xε(x) = 1 +1

3x + xε(x),where ε(x)→ 0 for x → 0 By a small rearrangement, in which we put x = 1/n3 → 0 for n → ∞ we

→ 0 for n → ∞,and the sequence is convergent with the limit 0 ♦

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Calculus 3c-1

an= (2n+ 3n)1/n

The trick in case of expressions with several terms is always to put the dominating term as a factor

n1/n

We get from 1 < 1 +

23

n1/n

< 3√n

2→ 3 for n→ ∞

Then all terms an lie between 3 and a sequence which converges towards 3, hence (an) is convergent

with the limit 3

In a variant we can instead take the logarithm of (1),

n

→ ln 3 + 0 · 0 = ln 3 for n→ ∞,which shows that (an) is convergent with the limit 3 ♦

Prove that (bn) is convergent and ﬁnd its limit

(Hint: One may apply Taylor’s formula for ln(1 + x).)

Taylor’s formula for ln(1 + x) gives

a2n

n2ε

ann

a2n

n2ε

ann

→ a − 0 + 0 = a for n→ ∞

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16

Since exp is continuous, we ﬁnally conclude that

bn= exp(ln bn)→ exp a = ea for n→ ∞,

and (bn) is convergent with the limit ea ♦

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Prove that if (an) is convergent with the limit a, then (bn) is also convergent with the limit a.

Give an example of a divergent sequence (an), for which the corresponding sequence (bn) is convergent

We say that a sequence (an) is summable, if its corresponding sequence (bn) deﬁned as above is

convergent We shall prove that if (an) is convergent, then (an) is also summable Then we shall

construct an example of a summable sequence (an), which is not convergent Hence, there are more

summable sequences that convergent ones

1) Assume that an → a for n → ∞ This means that one to every ε > 0 can ﬁnd some N = N(ε) ∈ N,

|a − ak| + 1

n

k=N(ε)+1

|a − ak| ≤ 1

n

N(ε)k=1

|a − ak| +n− N(ε)

n ·ε2

≤ 1n

N(ε)k=1

|a − ak| + ε

2,since the n− N(ε) terms of the latter sum are all < ε/2 by (2)

Since N (ε) is ﬁxed (corresponding to the given ε > 0), the sum is

Summable sequences

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Prove that if an→ ∞ for n → ∞, then bn→ ∞ for n → ∞.

Give an example of a sequence (an) which does not tend towards ∞ for n tending towards ∞, for

which the corresponding sequence (bn) fulﬁls bn → ∞ for n → ∞

First note that if|an| ≤ c, then also |bn| ≤ c It follows that if bn→ ∞, then (an) must be unbounded

Assume that an → ∞ for n → ∞ This means that we to every c > 0 can ﬁnd an N = N(c) ∈ N,

such that (e.g.)

n

k=N(c)+1

ak > 1n

N(c)k=1

n < 1

for n≥ N1

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we conclude that bn→ ∞ for n → ∞.

By the introducing remark, bn → ∞ implies that (an) in unbounded We note that an unbounded

sequence does not necessarily tend towards ∞ Choose e.g

b2n+1 = 1

2n + 1(2 + 4 +· · · + 2n) = n(n + 1)

2n + 1 → ∞,thus bn→ ∞

We have here used that

(3) 1 + 2 +· · · + n = 1

2n(n + 1), n∈ N

For completeness, we see that this is true for n = 1, 2, 3 If (3) holds for some n∈ N, then we get for

the following term by using (3) that

1 + 2 +· · · + n + (n + 1) = 1

2n(n + 1) + (n + 1) =

1

2(n + 1)(n + 2),which we recognize as (3) where n has been replaced by n + 1 Then (3) follows in general by induction

(the boot strap principle), because if (3) holds for some n, then it also holds for the following term,

etc Since (3) is true for n = 1 (in the beginning), we see that (3) is true for all n∈ N ♦

Summable sequences

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20

3 Recursively given sequences

Prove that (an) is convergent and ﬁnd the limit

We shall ﬁrst ﬁnd the possible limit

Assume that the sequence is convergent, an → a for n → ∞ Since taking the square root of

nonnegative numbers is a continuous function, we get by taking the limit in

an+1=

an+1

2 > 0,that

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Calculus 3c-1

It does not yet follow that the sequence actually is convergent We continue in the following way

A sequence (an) is convergent, if it is (weakly) increasing and bounded from above

1) The sequence is bounded from above.

2 =

1 +√3

2 ,where we have used that

2 .Then it follows by induction that (an) is bounded from above

2) The sequence is increasing.

2 =

1

√

2 = a1, i.e a2> a1.Then assume that an> an−1for some n≥ 2 (This is at least true for n = 2) Then

and it follows by induction that (an) is increasing

3) We have now proved that (an) is convergent

We showed in the beginning that a = (1 +√

3)/2 is the only possible limitSince the limit exists, we must have the limit a = (1 +√

3)/2 ♦

Recursively given sequences

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22

a1= 1, an+1=√

3an, n∈ N

Prove that (an) is convergent, and ﬁnd its limit

If an→ a for n → ∞, it follows by taking the limit in the recursion formula,

a =√

3a, or a2= 3a by a squaring

We use here that taking the square root is a continuous operation, so the limit and the square root

can be interchanged Thus we conclude that a = 0 and a = 3 are the only possible limits We shall

prove that the sequence indeed is convergent (Our assumption above)

1) The sequence is bounded.

If a≥ 1, then an+1≥√3≥ 1, i.e (an) is bounded from below It follows in particular that if the

limit a exists, then we must have a

If an< 3, then an+1=√

3an <√

3· 3 = 3, and (an) is also bounded from above

2) The sequence is increasing.

An increasing bounded sequence (an) is convergent, so the given sequence is convergent The only

possible limits were a = 0 and a = 3 But since all an ≥ 1, we can exclude a = 0, and

lim

n→∞an= 3. ♦

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where k > 0 and p > 0 are any positive numbers Prove that a2n ≥ p for every n ≥ 2 and that the

sequence a2, a3, a4,· · · , is weakly decreasing Then prove that (an) is convergent with the limit √

, t > 0, then

n≥ p for all n ≥ 2

2) Proof of the claim that (an) is weakly decreasing for n ≥ 2.

We ﬁrst prove that a2≥ a3 This follows from

a1 − 1

a2

= a1− a22a2

k−pk

12

k−pk

= 18a2

k−pk

2

≥ 0

Then assume that an−1≥ an, i.e an−1− an ≥ 0 for n ≥ 3 This is true for n = 3, according to

our ﬁrst result then

By 1 we have an≥ √p and √an−1≥ √p, so anan−1− p ≥ 0

By the inductions assumption we get an−1− an≥ 0

We see that we also have an− an+1≥ 0, hence an≥ an+1

Then by induction, at an−1≥ an for all n≥ 2, and the sequence is weakly decreasing

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is continuous, hence we can ﬁnd the limit value by taking the

limit in the recursion formula, i.e replace an+1and an by the limit value a We get the equation

Since every an ≥ √p, we have a ≥ √p, it is in particular positive It therefore follows that a = √p

5) When k = 2 and p = 3, we get

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Calculus 3c-1

6) Comparison with the Newton-Raphson iteration method.

Let F (x) = x2− p, then F(x) = 2x, hence

g(x) = x− F (x)

F(x) = x−x2− p

2x =

12

x + px

.The iteration formula becomes

the arithmetical mean of the two preceding terms:

(−1)n

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26

f (x) = x + x2 The function f has the ﬁx point x0 = 0 Show graphically that the ﬁx point is

attractive for some values of c < 0, and repelling for every c > 0

It follows from the equation f (x0) = x0, i.e x2= 0, that x0= 0, hence x0= 0 is a ﬁx point Since

f (x) = x2+ x =

x +12

–1.5 –1 –0.5 0.5 1 1.5

x

It is diﬃcult to sketch on the ﬁgure in MAPLE the lines which shows the convergence, so this is left

to the reader We see that x0is attractive for c∈ [−1, 0] and repelling for c ∈ R \ [−1, 0]

We shall now prove these claims

1) If c > 0, then f (c) = c2+ c > c, and f (c) moves away from x0= 0, thus the point is repelling

2) If r c <−1, then f(c) = c2+ c =|c|(|c| − 1) > 0, and we are back in case 1

3) If either c = 0 or c =−1, then f(c) = 0 Since trivially f(0) = 0 in all the following iterations, it

follows that x0= 0 is attractive for these values of c

4) Finally, if−1 < c < 0, then f(c) = c + c2= c(1 + c) < 0, and f (c) > c, hence

c < f (c) < 0,

and we conclude that the ﬁx point is attractive for c∈ ] − 1, 0[ ♦

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Calculus 3c-1

a1= c, an+1=−a3

n.1) Compute for c =−1 the terms a2, , a5, and give a graphical discription

2) The same for c =−1/2

Again, it is difficult to give all necessary details on a figure in MAPLE-figure, so these additions are

left to the reader

–1 –0.5

0.5 1

an= (−1)n −(3 n−1 ), n∈ N ♦

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28

a1= 100, an+1= 2√

an− 1

1) Prove by induction that an> 1 for every n∈ N

2) Prove by induction that the sequence (an) is decreasing

3) Finally, prove that the sequence (an) is convergent, and ﬁnd its limit value

1) Assume that an> 1 Then an+1= 2√

an− 1 > 2 · 1 − 1 = 1, and we see that an > 1 implies thatalso the successor satisﬁes an+1 > 1 Since a1 = 100 > 1, the claim follows by induction, hence

(an) is bounded from below

2) We get by insertion, a2= 2√

100− 1 = 19 < 100 = a1, thus a2< a1.Assume that an−1> an (this is true for n = 2) When n is replaced by n + 1, we get

an− an+1= (2√

an−1− 1) − (2√an− 1) = 2(√an−1−√an) > 0,hence an> an+1 We conclude by induction that

a1> a2> a3>· · · > an−1> an> an+1>· · · ≥ 1

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Calculus 3c-1

3) Since (an) is decreasing and bounded from below, it follows that (an) is convergent Denote the

limit value by a By taking the limit of the recursion formula we get

a = 1, and thus limn→∞an = a = 1

x−1, where f(x) = √1

x, then f

(x0) = f(1) = 1,corresponding to a limiting case, in which one usually can say nothing about the convergence This is

also demonstrated by an iteration on a pocket calculator, because the approximation becomes slower

the closer one is to a = 1,

a5= 3, 26904, a10= 1, 69909, a15= 1, 39299, a20= 1, 26997 ♦

cos x = 1

1 + 25· 10−6/ sin x

with four decimals by using a convenient iteration method Then comment on the convergence

The diﬃculty of this example apparently stems from the denominator on the right hand side First

we rewrite in the following way

into (4) and compare,

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We are just inside the range of the accuracy of the pocket calculator, because the factor 40000 gives

an error of rounding oﬀ which is 40000 times bigger than usual (i.e 40000 times 10−12)

It is quite ironical that our ﬁrst approximation by using Taylor’s formula in fact gives the best

approximation with four decimals,

Even if we choose the value x0= 0, 037 as our start [where we have an eye to the value 0,03684] the

iteration is extremely slow,

F(x) = 40000 cos 2x− 40000 cos x − sin x,

and we get by the Newton-Raphson iteration

xn+1= xn−20000 sin 2xn− 40000 sin xn+ cos xn

40000 cos 2xn− 40000 cos xn− sin xnwith starting value x0= 0, 037, the ﬁrst values

x1= 0, 036837, x2= 0, 036836, x3= 0, 036836

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Calculus 3c-1

F (x) = cos x + 1

cosh x.1) Write down the Newton-Raphson iteration formula for the solution of the equation F (x) = 0

2) Apply a programmable pocket calculator to the Newton-Raphson iteration and ﬁnd the ﬁrst four

positive zeros of F (x) = 0 by choosing the starting values

1) Since

F(x) =− sin x − sinh x

cosh2x =−sin x· cosh2x + sinh x

cosh2x ,the Newton-Raphson iteration is written

xn+1= xn− F (xn)

F(xn) = xn+

cos xn· cosh2xn+ cosh xnsin xn· cosh2xn+ sinh xn.2) The requested iterations give with 5 decimals,

Prove that the equation F (x) = 0 has precisely one solution α, and ﬁnd this by an iteration in two

steps

Since both ex and sin x are increasing in

0,π2

, and F (x) is continuous with F (0) = −1 < 0 andF

F(x) = ex(sin x + cos x),

Trang 32

is fairly robust.

If we choose α0= 1, then we get successively

α1= 0, 657, α2= 0, 591, α3= 0, 5885

If we instead choose α0= 0, 5 then we also get α3= 0, 5885

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Trang 33

Hint Apply the General principle of convergence.

Apply a programmable pocket calculator to sketch the graph of fn(x) for some large n

It can be proved that fn converges uniformly on the interval [−1, 1] towards the function

for x

1(n + j)2.

It can be proved from the Theory of Fourier Series that !∞

1(n + j)2 =

∞

j=n+1

1

j < ε for alle n≥ N(ε)

By insertion we get that|fn+m(x)−fn(x)| < ε for n ≥ N(ε), not just pointwisely, but even uniformly

Since every fn, n∈ N is an odd function, we shall only sketch the graph of fn for x∈ [0, 1]

That the limit function f (x) is precisely the given function can either be shown by a formula from

f (x) = 1 + x

4(1 + x2)(n + x4 , n∈ N,

converges uniformly towards the function 0

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34

0 0.2 0.4 0.6 0.8 1

uniformly towards 0 all overR

Trang 35

Calculus 3c-1

towards a C1-function f : [0, 1] → R, and for which the sequence of derivatives (f

n) is pointwiseconvergent, but not converging towards the function f

The best strategy must be ﬁrst to choose some convenient pointwisely convergent sequence of functions

(fn), where the limit function is not continuous, and then integrate the terms of this sequence from

n(x) = g(x).

Sequences of functions

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Trang 36

36

By using the example above (one introduces a singularity at x = 1) it is possible to construct a

sequence of functions (fn), which converges uniformly towards 0 in [0, 1] where

lim

n→∞f

n(x) for ethvert x∈ [0, 1] ∩ Q

The construction, however, uses series which formally have not yet been introduced

limn→∞f (x) dx.

1) Pointwise convergence What is here obvious? When x = 0 is ﬁxed, then

Trang 37

f (x) dx =

" 10

1− e−n =1

2

" 10

limn→∞f (x) dx.

interchanged, the convergence of (fn) can never be uniform ♦

f (x) = x

2n

1 + x2n, n∈ N

1) Prove that the sequence (fn) is pointwise convergent, but not uniformly convergent

2) Prove that the sequence (fn) is uniformly convergent in the interval

−1

2,

12

.1) This example is tricky, because one must split it up into three cases,

2 for|x| = 1,

1 for|x| > 1

Since every fn is continuous and the limit function f is not continuous [cf the ﬁgure], it follows

that the sequence is not uniformly convergent

Sequences of functions

Trang 38

38

0 0.2 0.4 0.6 0.8 1

x

2) Get rid of x! When|x| ≤1

2, we get the estimate

|fn(x)− 0| = fn(x) = x2n

1 + x2n ≤ x2n≤

12

2n

→ 0 for n → ∞

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Trang 13

→ for n → ∞ (standard sequences) , hence according... diﬃculty of this example apparently stems from the denominator on the right hand side First

we rewrite in the following way

into (4) and compare,

Recursively given sequences. .. 35

Calculus 3c-1

towards a C1-function f : [0, 1] → R, and for which the sequence of derivatives (f

n)

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