1 21p+31p 1 4p

Chapter 11 Problems Plus

1. We already know that sin x = x ^x_3!³ +^x_5!⁵ ::: + ( 1)²ⁿ⁺¹ _(2n+1)!^x2n+1 + :::.

Substitute x by x³we have sin x³ = x³ x⁹

3! +x¹⁵

5! ::: + ( 1)²ⁿ⁺¹ x⁶ⁿ⁺³ (2n + 1)!+ :::

and f⁽¹⁵⁾(0) is the coe¢ cient of x¹⁵ above, which is _5!¹. 6.

1 +¹₂+¹₃+₂¹2 +_{2 3}¹ +₂¹3 +₃¹2 + :::

= P₁

n;m=0 1 2ⁿ 3^m

= P₁

n=0 1 2ⁿ

P₁

m=0 1 3^m

= P1

n=0 1 2ⁿ

1 1 ¹₃

= ₁¹1 2

1 1 ¹₃

= 3

10. Substitute a_k= a₀ a₁ :::: a_{k 1} so that we have

nlim!1

Pk 1 p=0ap p

n + p p n + k

= lim

n!1

Pk 1

p=0ap p k 2n+p+k

= Pk 1 p=0a_p lim

n!1 p k 2n+p+k

= 0 14.

1 ₂¹p+₃¹p 1 4^p+ :::

= 1 + ₂¹p 2₂¹p +₃¹p+ ₄¹p 2₄¹p + :::

= 1 + ₂¹p+₃¹p+₄¹p+ ::::

2₂¹p 1 +₂¹p+₃¹p+₄¹p + :::

Hence de…ne A = 1 +₂¹p+₃¹p+₄¹p+ :::. Then 1 +₂¹p +₃¹p +₄¹p + :::

1 ₂¹p +₃¹p 1

4^p + ::: = A

A ₂²pA = 2^{p 1} 2^{p 1} 1:

21. As x 0, all the term are positive, hence there is no root on the right side of 0:

For x < 0, the series is

1 jxj 2! +jxj²

4! ::: = X1 n=0

( 1)ⁿ p jxj ²ⁿ

(2n)! = cosp jxj

Hence the root occur whilep

jxj = 2 + m for m = 0; 1; 2; ::: as x < 0.

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25. One check that the derivative of f (x) := u³+ v³+ w³ 3uvw is zero and f (0) = 1: One will use the following relation

u⁰(x) = w (x) ; w⁰(x) = v (x) ;

v⁰(x) = u (x) : 26.Follow the hint.

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