Subsection 1.23.2

Problems

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Problem 1.23.1

Solve the equation graphically, indicating for each root an isolation interval of length not exceeding 0.5:

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(1)

x3−6x+2=0x^{3}-6 x+2 = 0;

(2)

x4−4x−1=0x^{4}-4 x-1 = 0;

(3)

x3+2x+7,8=0x^{3}+2 x+7,8 = 0;

(4)

x3−1,75x+0,75=0x^{3}-1,75 x+0,75 = 0;

(5)

2x3−x2−x−3=02 x^{3}-x^{2}-x-3 = 0;

(6)

x3+x2−5x−12=0x^{3}+x^{2}-5 x-12 = 0;

(7)

x3−0,2x2−0,2x−1,2=0x^{3}-0,2 x^{2}-0,2 x-1,2 = 0;

(8)

0,3x4−0,7x3−0,3x2−2=00,3 x^{4}-0,7 x^{3}-0,3 x^{2}-2 = 0.

Problem 1.23.2

Solve the equation graphically, indicating for each root an isolation interval of length not exceeding 0.1:

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(1)

2x3+x+1=02 x^{3}+x+1 = 0;

(2)

x4−x−1=0x^{4}-x-1 = 0;

(3)

x+ex=0x+e^{x} = 0;

(4)

x−sin⁡x−1=0x-\sin x-1 = 0;

(5)

x=ctg⁡x,x∈(0;π)x = \operatorname {ctg} x, x \in (0 ; \pi );

(6)

x2−cos⁡x=0x^{2}-\cos x = 0;

(7)

4x−5ln⁡x=54 x-5 \ln x = 5;

(8)

xx=10x^{x} = 10.

Problem 1.23.3

Prove that, by means of a shift along the axis OxOx or a contraction (dilation) along the axis OxOx, the equation

a0xn+a1xn−1+…+an−1x+an=0,a0≠0 a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n} = 0, \quad a_{0} \neq 0

can be reduced to the form

b0tn+b2tn−2+…+bn−1t+bn=0 b_{0} t^{n}+b_{2} t^{n-2}+\ldots +b_{n-1} t+b_{n} = 0

or to the form

c0zn+c1zn−1+…+cn=0 c_{0} z^{n}+c_{1} z^{n-1}+\ldots +c_{n} = 0

where ∣c0∣=∣cn∣\left|c_{0}\right| = \left|c_{n}\right|.

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Problem 1.23.4

Let the coefficients of the polynomial

Pn(x)=a0xn+a1xn−1+…+an−1x+an,a0≠0 P_{n}(x) = a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n}, \quad a_{0} \neq 0

satisfy the equalities ai=an−i(i=0,1,…,n)a_{i} = a_{n-i}(i = 0,1, \ldots , n). Prove that:

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(1)

if nn is odd, then x=−1x = -1 is a root of the polynomial, and the coefficients of the quotient

Qn−1=b0xn−1+b1xn−2+…+bn−2x+bn−1 Q_{n-1} = b_{0} x^{n-1}+b_{1} x^{n-2}+\ldots +b_{n-2} x+b_{n-1}

from dividing Pn(x)P_{n}(x) by x+1x+1 satisfy the equalities bi=bn−1−i(i=0,1,…,n−1)b_{i} = b_{n-1-i} (i = 0,1, \ldots , n-1);

(2)

if nn is even, then the substitution z=x+1/xz = x+1 / x reduces the equation Pn(x)=0P_{n}(x) = 0 to an equation of degree n/2n / 2 and to n/2n / 2 quadratic equations;

(3)

if nn is even, then the substitution

z=(x+1x−1)2 z = \left(\frac{x+1}{x-1}\right)^{2}

also reduces the equation Pn(x)=0P_{n}(x) = 0 to an equation of degree n/2n / 2 and to n/2n / 2 quadratic equations.

Problem 1.23.5

Let the function ff be defined and twice continuously differentiable on R\mathbb {R}, and let f′′f^{\prime \prime } not change sign on R\mathbb {R}. Prove that:

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(1)

the equation f(x)=0f(x) = 0 cannot have more than two real roots;

(2)

if f(x0)f′(x0)<0,f(x0)f′′(x0)<0f\left(x_{0}\right) f^{\prime }\left(x_{0}\right) < 0, f\left(x_{0}\right) f^{\prime \prime }\left(x_{0}\right) < 0, then the equation f(x)=0f(x) = 0 has a unique root in the interval (x0;x0−f(x0)/f′(x0))(x_{0} ; x_{0}-f\left(x_{0}\right) / f^{\prime }\left(x_{0}\right));

(3)

if f′(x0)=0,f(x0)f′′(x0)<0f^{\prime }\left(x_{0}\right) = 0, f\left(x_{0}\right) f^{\prime \prime }\left(x_{0}\right) < 0, then the equation f(x)=0f(x) = 0 has one root each in the intervals (−∞;x0),(x0;+∞)\left(-\infty ; x_{0}\right),\left(x_{0} ;+\infty \right).

Problem 1.23.6
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(1)

Let the function ff be continuous on [a;+∞)[a ;+\infty ), differentiable on (a;+∞)(a ;+\infty ), f(a)<0f(a) < 0, f′(x)⩾m>0f^{\prime }(x) \geqslant m > 0 on (a;+∞)(a ;+\infty ). Prove that the equation f(x)=0f(x) = 0 has exactly one real root on (a;+∞)(a ;+\infty );

(2)

give an example showing that the condition " f(x)⩾m>0f(x) \geqslant m > 0 on (a;+∞)(a ;+\infty ) " cannot be replaced by the condition " f′(x)>0f^{\prime }(x) > 0 ";

(3)

prove that under the conditions of 1) the root of the equation belongs to the interval (a;a−f(a)/m](a ; a-f(a) / m].

Problem 1.23.7

A root ξ\xi of the equation f(x)=0f(x) = 0 is called pp-fold (p∈N)(p \in \mathbb {N}) if f(x)=(x−ξ)pφ(x)f(x) = (x-\xi )^{p} \varphi (x), where φ(ξ)≠0\varphi (\xi ) \neq 0. A single root is called simple. Prove that:

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(1)

if ff is differentiable in a neighborhood of a pp-fold (p⩾2)(p \geqslant 2) root ξ\xi of the equation f(x)=0f(x) = 0, then ξ\xi is a (p−1)(p-1)-fold root of the equation f(x)=0f(x) = 0;

(2)

if a2−3b<0a^{2}-3 b < 0, then the equation x3+ax2+bx+c=0x^{3}+a x^{2}+b x+c = 0 has one simple real root.

Problem 1.23.8

Prove that the polynomial

xn+a1xn−1+a2xn−2+…+an−1x−an x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots +a_{n-1} x-a_{n}

where ai⩾0(i=1,2,…,n−1),an>0a_{i} \geqslant 0(i = 1,2, \ldots , n-1), a_{n} > 0, has only one positive root, and this root is simple and does not exceed ann\sqrt[n]{a_{n}}.

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Problem 1.23.9

Prove that all the roots of the derivative of the polynomial

P4(x)=(x+1)(x−1)(x−2)(x−3) P_{4}(x) = (x+1)(x-1)(x-2)(x-3)

are real, and find their isolation intervals of length at most 0.5.

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Problem 1.23.10

Prove that the polynomial

a0xn+a1xn−1+…+an−1x+an,a0≠0 a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n}, \quad a_{0} \neq 0

has at least one root on the interval (0;1)(0 ; 1) if

a0/(n+1)+a1/n+…+an−1/2+an=0 a_{0} /(n+1)+a_{1} / n+\ldots +a_{n-1} / 2+a_{n} = 0
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Problem 1.23.11

Arrange in increasing order of index the coefficients a0,a1,…,ana_{0}, a_{1}, \ldots , a_{n} of the polynomial

Pn(x)=a0xn+a1xn−1+…+an−1x+an P_{n}(x) = a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n}

omitting the coefficients equal to zero. The number of sign changes of the resulting ordered set is called the number of pairs of adjacent elements having different signs.

Prove that the number of positive roots of the polynomial is not greater than the number of sign changes in the set of its nonzero coefficients.

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Problem 1.23.12

Prove that the polynomial:

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(1)

x5−2x4−x2−5=0x^{5}-2 x^{4}-x^{2}-5 = 0 has exactly one real root; indicate for this root an isolation interval of length at most 0.1;

(2)

x6+2x4+x−3=0x^{6}+2 x^{4}+x-3 = 0 has exactly two real roots; indicate for these roots isolation intervals of length at most 0.1.

Problem 1.23.13

Prove that all the roots of the equation

a0xn+a1xn−1+…+an−1x+an=0,a0≠0: a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n} = 0, \quad a_{0} \neq 0:
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(1)

satisfy the inequality ∣ξ∣<1+M1/∣a0∣\left|\xi \right| < 1+M_{1} /\left|a_{0}\right|, where

M1=max⁡{∣a1∣,∣a2∣,…,∣an∣} M_{1} = \max \left\{ \left|a_{1}\right|,\left|a_{2}\right|, \ldots ,\left|a_{n}\right|\right\}
(2)

for an≠0a_{n} \neq 0 satisfy the inequality ∣ξ∣>(1+M2/∣an∣)−1\left|\xi \right| > \left(1+M_{2} /\left|a_{n}\right|\right)^{-1}, where

M2=max⁡{∣a0∣,∣a1∣,…,∣an−1∣} M_{2} = \max \left\{ \left|a_{0}\right|,\left|a_{1}\right|, \ldots ,\left|a_{n-1}\right|\right\}
Problem 1.23.14

Let, in the equation

a0xn+a1xn−1+…+an−1x+an=0,a0>0 a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n} = 0, \quad a_{0} > 0

ak(k⩾1)a_{k}(k \geqslant 1) be the negative coefficient with the smallest index, and AA the greatest of the absolute values of the negative coefficients. Prove that all real roots of this equation satisfy the inequality

ξ<1+A/a0k \xi < 1+\sqrt[k]{A / a_{0}}

(Lagrange's theorem).

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Problem 1.23.15

Let, for the polynomial

Pn(x)=a0xn+a1xn−1+…+an−1x+an P_{n}(x) = a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n-1} x+a_{n}

there exist a number c>0c > 0 such that

Pn(k)(c)⩾0(k=0,1,…,n) P_{n}^{(k)}(c) \geqslant 0 \quad (k = 0,1, \ldots , n)

and let a0>0a_{0} > 0. Prove that all real roots of this polynomial satisfy the inequality ξ⩽c\xi \leqslant c (Newton's theorem).

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Problem 1.23.16

Let the polynomial Pn(x)P_{n}(x) be represented in the form

Pn(x)=Q(x)+R(x) P_{n}(x) = Q(x)+R(x)

where Q(x)Q(x) contains the term of Pn(x)P_{n}(x) of highest degree with a positive coefficient and all terms with negative coefficients, and R(x)R(x) contains all the remaining terms of Pn(x)P_{n}(x). Prove that if Q(c)>0Q(c) > 0 for some c>0c > 0, then all real roots of Pn(x)P_{n}(x) are less than cc.

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Problem 1.23.17

Prove that the equation P(x)=0P(x) = 0 has no negative roots, and, using the results of problems 13-16, indicate an interval [a;b][a ; b], 0<a<b0 < a < b, containing all the positive roots, if:

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(1)

P(x)=2x5−100x2+2x−1P(x) = 2 x^{5}-100 x^{2}+2 x-1;

(2)

P(x)=x4−35x3+380x2−1350x+1000P(x) = x^{4}-35 x^{3}+380 x^{2}-1350 x+1000.

Problem 1.23.18

Indicate intervals [a1;b1]\left[a_{1} ; b_{1}\right], a1<b1<0a_{1} < b_{1} < 0, and [a2;b2]\left[a_{2} ; b_{2}\right], 0<a2<b20 < a_{2} < b_{2}, containing respectively all the negative and all the positive roots of the equation 3x5+7x4−8x3+5x2−2x−1=03 x^{5}+7 x^{4}-8 x^{3}+5 x^{2}-2 x-1 = 0. For the positive roots, indicate isolation intervals of length not exceeding 0.5.

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Problem 1.23.19

Prove that the equation x5+x4+x2+10x−5=0x^{5}+x^{4}+x^{2}+10 x-5 = 0 has a unique root, and indicate its isolation interval of length not exceeding 0.1.

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Problem 1.23.20
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(1)

Prove that if 4p3+27q2>04 p^{3}+27 q^{2} > 0, then the equation x3+px+q=0,p,q∈Rx^{3}+p x + q = 0, p, q \in \mathbb {R}, has one real root, and if 4p3+27q2<04 p^{3} + 27 q^{2} < 0, then this equation has three real roots;

(2)

under what condition on a,b,ca,b,c are all the roots of the equation x3+ax2+bx+c=0x^{3}+a x^{2} + b x+c = 0 real?

Problem 1.23.21

Under what condition on pp and qq are all the roots of the equation x5−5px2+5p2x+2q=0x^{5}-5 p x^{2} + 5 p^{2} x+2 q = 0 real?

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Problem 1.23.22

Under what condition on pp and qq are all the roots of the equation xm+pxn+q=0x^{m}+p x^{n} + q = 0, where m>n>0,m,n∈N,q≠0,p≠0m > n > 0, m, n \in \mathbb {N}, q \neq 0, p \neq 0, real?

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Problem 1.23.23

Prove that the equation

x2n+1+a1x2n−1+…+anx=a0 x^{2 n+1}+a_{1} x^{2 n-1}+\ldots +a_{n} x = a_{0}

where ai∈R(i=0,1,…,n)a_{i} \in \mathbb {R}(i = 0,1, \ldots , n):

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(1)

has at least one root in the interval (−2;2)(-2 ; 2) if ∣a0∣<2\left|a_{0}\right| < 2;

(2)

has a root in the interval (−2l;2l)(-2 l ; 2 l), where l=(∣a0∣/2)1/(2n+1)l = \left(\left|a_{0}\right| / 2\right)^{1 /(2 n+1)} (Chebyshev's theorem).

Problem 1.23.24

Indicate, depending on the values of a∈Ra \in \mathbb {R}, the number of real roots of the equation:

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(1)

x4−4ax3−2=0x^{4}-4 a x^{3}-2 = 0;

(2)

2x3−3ax2+1=02 x^{3}-3 a x^{2}+1 = 0.

Problem 1.23.25

Find all values a∈Ra \in \mathbb {R} for which the equation has the indicated number of real roots:

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(1)

2x3+13x2−20x+a=02 x^{3}+13 x^{2}-20 x+a = 0, one root;

(2)

3x3+1,5x2−12x+a=03 x^{3}+1,5 x^{2}-12 x+a = 0, one double root and one simple root;

(3)

3x4+8x3+a=03 x^{4}+8 x^{3}+a = 0, two simple roots;

(4)

9x4+14x3−15x2+a=09 x^{4}+14 x^{3}-15 x^{2}+a = 0, four distinct roots.

Problem 1.23.26

Prove that the equation 2ex+x2+18x−6=02 e^{x}+x^{2}+18 x-6 = 0 has a unique positive root; indicate its isolation interval of length not exceeding 0.2.

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Problem 1.23.27

Indicate, depending on the values of aa, the number of roots of the equation:

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(1)

ln⁡x+ax=0\ln x+a x = 0

(2)

xln⁡x=ax \ln x = a;

(3)

ex=ax2e^{x} = a x^{2}

(4)

ch⁡x=ax\operatorname {ch} x = a x;

(5)

cos⁡3xsin⁡x=a;x∈[0;π]\cos^{3} x \sin x = a ; x \in [0 ; \pi ].

Problem 1.23.28

Indicate all values a∈Ra \in \mathbb {R} for which the equation has the indicated number of real roots:

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(1)

ex=a+x−x2e^{x} = a+x-x^{2}, two distinct roots;

(2)

x2+x−ln⁡x+a=0x^{2}+x-\ln x+a = 0, one double root;

(3)

x2=aln⁡xx^{2} = a \ln x, two distinct roots;

(4)

6arctg⁡x−x3+a=06 \operatorname {arctg} x-x^{3}+a = 0, three distinct roots.

Problem 1.23.29

Given the equation ln⁡x−a−bx=0\ln x-a-b x = 0:

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(1)

prove that this equation cannot have more than two roots;

(2)

indicate on the coordinate plane the sets of points (a;b)(a ; b) for which the number of roots of the given equation equals: 0,1,20,1,2.

Problem 1.23.30

Given the equation ex−a−bx3=0e^{x}-a-b x^{3} = 0:

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(1)

prove that this equation cannot have more than four roots;

(2)

indicate on the coordinate plane the sets of points (a;b)(a ; b) for which the number of roots of the equation equals: 0,1,2,3,40,1,2,3,4.

Problem 1.23.31

Let a>1a > 1. Prove that the equation ax=bxa^{x} = b x:

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(1)

has two real roots for b>eln⁡ab > e \ln a;

(2)

has one double real root for b=eln⁡ab = e \ln a;

(3)

has no real roots for 0⩽b<eln⁡a0 \leqslant b < e \ln a;

(4)

has one simple root for b<0b < 0.

Problem 1.23.32

Indicate the number of roots of the equation xn=aex,n∈Nx^{n} = a e^{x}, \quad n \in \mathbb {N}, depending on a∈Ra \in \mathbb {R}.

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Problem 1.23.33

Separate the real roots of the equation, indicating isolation intervals of length not exceeding unity:

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(1)

x3+2x−7=0x^{3}+2 x-7 = 0;

(2)

x3−27x−17=0x^{3}-27 x-17 = 0

(3)

x3−6x2+9x−10=0x^{3}-6 x^{2}+9 x-10 = 0;

(4)

x4−2x3−5x2+2x+2=0x^{4}-2 x^{3}-5 x^{2}+2 x+2 = 0;

(5)

x4−4x3−3x+23=0x^{4}-4 x^{3}-3 x+23 = 0

(6)

x4+6x3−2x2+1=0x^{4}+6 x^{3}-2 x^{2}+1 = 0

(7)

x5−10x3+6x+1=0x^{5}-10 x^{3}+6 x+1 = 0;

(8)

x5+5x3−7x+2=0x^{5}+5 x^{3}-7 x+2 = 0

(9)

x5+7x3−5x+11=0x^{5}+7 x^{3}-5 x+11 = 0

(10)

x5−2x4−5x3+19x2−17x+1=0x^{5}-2 x^{4}-5 x^{3}+19 x^{2}-17 x+1 = 0.

Problem 1.23.34

Separate the roots of the equation p2a2+x+q2b2+x+r2c2+x=1\frac{p^{2}}{a^{2}+x}+\frac{q^{2}}{b^{2}+x}+\frac{r^{2}}{c^{2}+x} = 1, where p,q,r,a,b,c∈R,a2>b2>c2,pqr≠0p, q, r, a, b, c \in \mathbb {R}, a^{2} > b^{2} > c^{2}, \quad p q r \neq 0.

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Problem 1.23.35

Separate the real roots of the equation, indicating for each root an isolation interval of length not exceeding 0.1:

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(1)

x−sin⁡2x=0x-\sin 2 x = 0

(2)

x=4cos⁡xx = 4 \cos x

(3)

ex−x−5/4=0e^{x}-x-5 / 4 = 0

(4)

2x2−ex+1=02 x^{2}-e^{x+1} = 0

(5)

x+ln⁡(x+2)=0x+\ln (x+2) = 0.

Problem 1.23.36

Using the bisection method, solve the equation with the indicated error Δ\Delta:

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(1)

x4+x−10=0,Δ=10−2x^{4}+x-10 = 0, \Delta = 10^{-2};

(2)

x3−12x−8=0,Δ=10−2x^{3}-12 x-8 = 0, \Delta = 10^{-2}

(3)

x3+x2−5x−12=0,Δ=10−2x^{3}+x^{2}-5 x-12 = 0, \Delta = 10^{-2}

(4)

x4+2x3+6x2−1=0,Δ=10−3x^{4}+2 x^{3}+6 x^{2}-1 = 0, \Delta = 10^{-3}

(5)

x4−2x3+x−1=0,Δ=10−3x^{4}-2 x^{3}+x-1 = 0, \Delta = 10^{-3};

(6)

(x−1)2−2sin⁡x=0,Δ=10−2(x-1)^{2}-2 \sin x = 0, \Delta = 10^{-2};

(7)

ex=2(1−x)2,Δ=10−2e^{x} = 2(1-x)^{2}, \Delta = 10^{-2};

(8)

10(x−1)=sin⁡x,Δ=10−310(x-1) = \sin x, \Delta = 10^{-3};

(9)

10ln⁡x=x3−3,Δ=10−310 \ln x = x^{3}-3, \Delta = 10^{-3};

(10)

xex=1,Δ=10−3x e^{x} = 1, \Delta = 10^{-3}.

Problem 1.23.37

A ball of radius 1 m with specific density 0.75 floats in water. Compute the height of the part of the ball protruding from the water with an error of 0,05mm0,05 \mathrm{mm}.

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Problem 1.23.38

Using the graphs of the hyperbolas xy=y+1xy = y+1 and x2−y2=1x^{2}-y^{2} = 1 and the bisection method, compute with an error of 0.01 the coordinates of the points of intersection of these hyperbolas.

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Problem 1.23.39
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(1)

Give an example of an equation f(x)=0f(x) = 0 with a root ξ\xi and its approximate value x∗x^{*} such that the inequalities

f′(x)>0 and ∣x∗−ξ∣⩾103,∣f(x∗)∣⩽10−3, f^{\prime }(x) > 0 \text{ and }\left|x^{*}-\xi \right| \geqslant 10^{3}, \quad \left|f\left(x^{*}\right)\right| \leqslant 10^{-3},

i.e., verify that from the "closeness" of f(x∗)f\left(x^{*}\right) to zero it does not follow that x∗x^{*} is "close" to the root ξ\xi;

(2)

give an example of an equation f(x)=0f(x) = 0 with a root ξ\xi and its approximate value x∗x^{*} such that the inequalities

f′(x)>0 and ∣x∗−ξ∣⩽10−3,∣f(x∗)∣⩾103, f^{\prime }(x) > 0 \text{ and }\left|x^{*}-\xi \right| \leqslant 10^{-3}, \quad \left|f\left(x^{*}\right)\right| \geqslant 10^{3},

i.e., verify that from the "closeness" of the approximate value x∗x^{*} to the root ξ\xi it does not follow that f(x∗)f\left(x^{*}\right) is "close" to zero.

Problem 1.23.40

The equation

3x4−16x3+192=0 3 x^{4}-16 x^{3}+192 = 0

for x>0x > 0 is equivalent to each of the following equations: а) x=2x3−1234x = 2 \sqrt[4]{\frac{x^{3}-12}{3}}; б) x=1232(x4+64)3x = \frac{1}{2} \sqrt[3]{\frac{3}{2}\left(x^{4}+64\right)}; в) x=163(1−12x3)x = \frac{16}{3}\left(1-\frac{12}{x^{3}}\right); г) x=32(x28+8x2)x = \frac{3}{2}\left(\frac{x^{2}}{8}+\frac{8}{x^{2}}\right).

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(1)

Prove that equation (29) has two positive roots, and separate them;

(2)

indicate which of the equations a)-d) has iterations converging fastest to the smaller positive root of equation (29); compute this root to an accuracy of 10−310^{-3};

(3)

carry out, for the larger root of equation (29), a task analogous to 2).

Problem 1.23.41

Using the method of iterations, find the real roots of the equation with the indicated error Δ\Delta:

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(1)

x3+x=1000,Δ=10−4x^{3}+x = 1000, \Delta = 10^{-4}

(2)

x3−3x2+8x+10=0,Δ=10−5x^{3}-3 x^{2}+8 x+10 = 0, \Delta = 10^{-5};

(3)

x5+5x+1=0,Δ=10−5x^{5}+5 x+1 = 0, \Delta = 10^{-5}

(4)

10x=e−x,Δ=5⋅10−410 x = e^{-x}, \Delta = 5 \cdot 10^{-4};

(5)

4x=8x,Δ=10−54^{x} = 8 x, \Delta = 10^{-5}

(6)

4ex=5(x+1),Δ=10−44 e^{x} = 5(x+1), \Delta = 10^{-4};

(7)

x3−2x−5=0,Δ=10−10x^{3}-2 x-5 = 0, \Delta = 10^{-10};

(8)

sin⁡x=2x−0,5,Δ=5⋅10−5\sin x = 2 x-0,5, \Delta = 5 \cdot 10^{-5};

(9)

x−sin⁡x=0,25,Δ=10−3x-\sin x = 0,25, \Delta = 10^{-3}.

Problem 1.23.42

Determine for which values:

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(1)

a>0a > 0 the equation x=axx = a^{x} has a solution;

(2)

of aa the iterative sequence x0=a,xn=axn−1,n∈Nx_{0} = a, x_{n} = a^{x_{n-1}}, n \in \mathbb {N}, converges.

Problem 1.23.43

Find, with four correct decimal digits:

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(1)

the smallest positive root of the equation xsin⁡x+1=0x \sin x+1 = 0;

(2)

two positive roots — the smallest and the one nearest to it — of the equation cos⁡xch⁡x=1\cos x \operatorname {ch} x = 1.

Problem 1.23.44

Solve, with an error not exceeding 10−410^{-4}, the equation 4x−5ln⁡x=54 x -5 \ln x = 5, choosing for each root a convergent iterative process.

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Problem 1.23.45

Using the method of chords, solve the equation with the indicated error Δ\Delta:

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(1)

x3−4x+2=0,Δ=10−3x^{3}-4 x+2 = 0, \Delta = 10^{-3};

(2)

x4+x−1=0,Δ=10−3x^{4}+x-1 = 0, \Delta = 10^{-3};

(3)

x3−3,2x2+3,1x−2,2=0,Δ=10−3x^{3}-3,2 x^{2}+3,1 x-2,2 = 0, \Delta = 10^{-3};

(4)

0,1sin⁡x=x+2,Δ=10−3;0,1 \sin x = x+2, \Delta = 10^{-3} ;

(5)

cos⁡x=x2,Δ=10−3\cos x = x^{2}, \Delta = 10^{-3};

(6)

x3−5x+1=0,Δ=10−5x^{3}-5 x+1 = 0, \Delta = 10^{-5}.

Problem 1.23.46

Using the method of chords and the method of tangents, solve the equation with the indicated error Δ\Delta:

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(1)

x2=13,Δ=10−8x^{2} = 13, \Delta = 10^{-8};

(2)

2x3+2x2−11x+3=0,Δ=10−62 x^{3}+2 x^{2}-11 x+3 = 0, \Delta = 10^{-6};

(3)

x3+x2+2x−3=0,Δ=5⋅10−5x^{3}+x^{2}+2 x-3 = 0, \Delta = 5 \cdot 10^{-5}.

Problem 1.23.47

Using the method of tangents, solve the equation with the indicated error Δ\Delta:

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(1)

x3−2x−2=0,Δ=5⋅10−5x^{3}-2 x-2 = 0, \Delta = 5 \cdot 10^{-5};

(2)

x3+x−3=0,Δ=5⋅10−5x^{3}+x-3 = 0, \Delta = 5 \cdot 10^{-5};

(3)

2x3−7x2+x+9=0,Δ=10−32 x^{3}-7 x^{2}+x+9 = 0, \Delta = 10^{-3}

(4)

x2+1/x2=10x,Δ=10−3;x^{2}+1 / x^{2} = 10 x, \Delta = 10^{-3} ;

(5)

x=cos⁡x,Δ=10−6x = \cos x, \Delta = 10^{-6};

(6)

xlg⁡x=1,Δ=10−4x \lg x = 1, \Delta = 10^{-4};

(7)

x+ex=1,Δ=10−5x+e^{x} = 1, \Delta = 10^{-5};

(8)

xth⁡x=1,Δ=10−6x \operatorname {th} x = 1, \Delta = 10^{-6};

(9)

tg⁡x=x,Δ=5⋅10−5\operatorname {tg} x = x, \Delta = 5 \cdot 10^{-5} (the smallest positive root);

(10)

ctg⁡x=1/x−x/2,Δ=10−3\operatorname {ctg} x = 1 / x-x / 2, \Delta = 10^{-3} (find only two positive roots — the smallest and the one nearest to it).

Problem 1.23.48

Using the method of tangents, compute the negative root of the equation x4−3x2+75x−10000=0x^{4}-3 x^{2}+75 x-10000 = 0 with three correct decimal digits.

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Problem 1.23.49

Find, with four correct digits, the purely imaginary roots of the equation zsin⁡z+1=0z \sin z+1 = 0.

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Problem 1.23.50

Solve the equation x3+1,1x2+0,9x−1,4=0x^{3}+1,1 x^{2}+0,9 x-1,4 = 0:

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(1)

using the bisection method to an accuracy of 5⋅10−35 \cdot 10^{-3};

(2)

using the method of chords to an accuracy of 5⋅10−45 \cdot 10^{-4};

(3)

using the method of tangents to an accuracy of 5⋅10−45 \cdot 10^{-4}.

Problem 1.23.51

To which of the roots of the equation x3−x=0x^{3}-x = 0 does the sequence constructed by the method of tangents converge, depending on the choice of the initial approximation x0∈Rx_{0} \in \mathbb {R}?

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Problem 1.23.52

Prove that for the equation ∣x∣sign⁡x=0\sqrt{\left|x\right|} \operatorname {sign} x = 0 the method of tangents does not converge.

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Problem 1.23.53

Verify that for the function f(x)=ln⁡2πsin⁡(2πln⁡xln⁡2)+1f(x) = \frac{\ln 2}{\pi } \sin \left(2 \pi \frac{\ln x}{\ln 2}\right)+1, the sequence constructed by the method of tangents, starting from x0=1x_{0} = 1, converges to a number that is not a root of the equation f(x)=0f(x) = 0.

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Problem 1.23.54

Using the method of tangents, construct a second-order iterative process for computing a\sqrt{a}.

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Problem 1.23.55

Prove that if x0>0,xn+1=12(xn+axn)x_{0} > 0, x_{n+1} = \frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right), then

xn+1−18m(xn−axn)2⩽a⩽xn+1 x_{n+1}-\frac{1}{8 m}\left(x_{n}-\frac{a}{x_{n}}\right)^{2} \leqslant \sqrt{a} \leqslant x_{n+1}

where m=min⁡{xn,a/xn}m = \min \left\{ x_{n}, a / x_{n}\right\}.

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Problem 1.23.56

Prove that if ξ\xi is a double root of the equation f(x)=0f(x) = 0, then for finding it one can use the iterative process

xn+1=xn−2f(xn)/f′(xn) x_{n+1} = x_{n}-2 f\left(x_{n}\right) / f^{\prime }\left(x_{n}\right)

(the modified Newton method). Indicate sufficient conditions for the convergence of the sequence {xn}\left\{ x_{n}\right\} to the root ξ\xi.

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Problem 1.23.57

What accuracy does a twofold application of the combined method give when computing the root of the equation x4−x−1=0x^{4}-x-1 = 0 starting from the interval (1,22;1,23)(1,22 ; 1,23)?

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Problem 1.23.58

Solve the equation x3−4,1x2+6,1x−1,6=0x^{3}-4,1 x^{2}+6,1 x-1,6 = 0:

?
(1)

using the bisection method to an accuracy of 5⋅10−35 \cdot 10^{-3};

(2)

using the method of chords to an accuracy of 5⋅10−45 \cdot 10^{-4};

(3)

using the method of tangents to an accuracy of 5⋅10−45 \cdot 10^{-4};

(4)

using the combined method to an accuracy of 5⋅10−45 \cdot 10^{-4}.

Problem 1.23.59

Using the combined method, compute the greatest root of the equation x5−x−0,2=0x^{5}-x-0,2 = 0 to an accuracy of 10−410^{-4}.

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Problem 1.23.60

Using the combined method, compute the roots of the equation

x4−3x2+75x−10000=0 x^{4}-3 x^{2}+75 x-10000 = 0

from the intervals (−11;−10)(-11 ;-10) and (9;10)(9 ; 10) to an accuracy of 10−410^{-4}.

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Problem 1.23.61

Compute the smallest positive root of the equation to an accuracy of 10−610^{-6}.

xsin⁡x=0,5 x \sin x = 0,5
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Problem 1.23.62

Using the combined method, solve, with the indicated error Δ\Delta, the equation:

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(1)

2x2−e1−x=0,Δ=10−52 x^{2}-e^{1-x} = 0, \Delta = 10^{-5};

(2)

2−x−lg⁡x=0,Δ=10−62-x-\lg x = 0, \Delta = 10^{-6}.

Problem 1.23.63

Prove that Newton's method (the method of tangents) has second order.

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Problem 1.23.64

Prove that the iterative formula xn+1=xn3+2axn3xn2+ax_{n+1} = \frac{x_{n}^{3}+2 a x_{n}}{3 x_{n}^{2}+a} for solving the equation x2=a,a>0x^{2} = a, a > 0, has third order.

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Problem 1.23.65

Using the expansion of the function by Taylor's formula up to second order, obtain an iterative formula for solving the equation f(x)=0f(x) = 0 and prove that it has third order:

?
(1)

xn+1=xn−f(xn)f′(xn)(f′(xn))2−f(xn)f′′(xn)/2x_{n+1} = x_{n}-\frac{f\left(x_{n}\right) f^{\prime }\left(x_{n}\right)}{\left(f^{\prime }\left(x_{n}\right)\right)^{2}-f\left(x_{n}\right) f^{\prime \prime }\left(x_{n}\right) / 2};

(2)

xn+1=xn−f(xn)f′(xn)−f2(xn)f′′(xn)2(f′(xn))3x_{n+1} = x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime }\left(x_{n}\right)}-\frac{f^{2}\left(x_{n}\right) f^{\prime \prime }\left(x_{n}\right)}{2\left(f^{\prime }\left(x_{n}\right)\right)^{3}} (Chebyshev's formula).

Problem 1.23.66

Prove that the smallest positive root ξ\xi of the equation ctg⁡x=λ−x,λ>2\operatorname {ctg} x = \lambda -x, \quad \lambda > 2, satisfies the inequality

ξ<2λ+λ2−4 \xi < \frac{2}{\lambda +\sqrt{\lambda ^{2}-4}}
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Problem 1.23.67

Let {xn}\left\{ x_{n}\right\} be the increasing sequence of positive roots of the equation xsin⁡x=1x \sin x = 1. Prove that as n→∞n \rightarrow \infty:

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(1)

xn=πn+o(1)x_{n} = \pi n+o(1);

(2)

xn=πn+(−1)n/(πn)+o(1/n)x_{n} = \pi n+(-1)^{n} /(\pi n)+o(1 / n);

(3)

xn=πn+(−1)n/(πn)−(1/(π3n3))(1−(−1)n/6)+o(1/n3)x_{n} = \pi n+(-1)^{n} /(\pi n)-\left(1 /\left(\pi^{3} n^{3}\right)\right)\left(1-(-1)^{n} / 6\right)+o\left(1 / n^{3}\right).

Problem 1.23.68

Let {xn}\left\{ x_{n}\right\} be the increasing sequence of positive roots of the equation tg⁡x=1/(1+x2)\operatorname {tg} x = 1 /\left(1+x^{2}\right). Prove that as n→∞n \rightarrow \infty

xn=πn+1/(πn)2+o(1/n2). x_{n} = \pi n+1 /(\pi n)^{2}+o\left(1 / n^{2}\right).
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Problem 1.23.69

Let {xn}\left\{ x_{n}\right\} be the increasing sequence of positive roots of the equation tg⁡x=x\operatorname {tg} x = x. Prove that as n→∞n \rightarrow \infty:

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(1)

xn=π2+πn+o(1)x_{n} = \frac{\pi }{2}+\pi n+o(1);

(2)

xn=π2+πn−1π/2+πn+o(1n)x_{n} = \frac{\pi }{2}+\pi n-\frac{1}{\pi / 2+\pi n}+o\left(\frac{1}{n}\right);

(3)

xn=π2+πn−1π/2+πn−23(π/2+πn)3+o(1n3)x_{n} = \frac{\pi }{2}+\pi n-\frac{1}{\pi / 2+\pi n}-\frac{2}{3(\pi / 2+\pi n)^{3}}+o\left(\frac{1}{n^{3}}\right).