# 3.1.1: Exercises 3.1

## Terms and Concepts

Exercise (PageIndex{1})

Is it possible to solve a cubic statement?

Yes, but it can be quite difficult, especially if it has many parameters.

Exercise (PageIndex{2})

What are the possible types of solutions when solving a quadratic statement?

2 distinct real roots; 1 repeated real root; a complex conjugate pair of roots.

Exercise (PageIndex{3})

What is the maximum number of different solutions that a seventh degree statement could have?

7

Exercise (PageIndex{4})

T/F: A cubic statement can have only complex solutions. Explain.

F; it must have at least one real solution since complex solutions come in pairs

## Problems

In exercises (PageIndex{5}) - (PageIndex{10}), determine the type of statement in terms of the given variable.

Exercise (PageIndex{5})

(x^3y+2x^2yz-6xz^2 = yz^2 -10) in terms of (x)

cubic

Exercise (PageIndex{6})

(x^3y+2x^2yz-6xz^2 = yz^2 -10) in terms of (y)

linear

Exercise (PageIndex{7})

(x^3y+2x^2yz-6xz^2 = yz^2 -10) in terms of (z)

Exercise (PageIndex{8})

(xt + cos{( heta)}=x^4t^3-6t) in terms of ( heta)

trigonometric

Exercise (PageIndex{9})

(xt + cos{( heta)}=x^4t^3-6t) in terms of (x)

quartic, or a statement of degree 4

Exercise (PageIndex{10})

(xt + cos{( heta)}=x^4t^3-6t) in terms of (t)

cubic

In exercises (PageIndex{11}) - (PageIndex{19}), determine if it is possible to solve the statement for the given variable. If it is possible, solve but do not simplify your answer(s). If it is not possible, explain why.

Exercise (PageIndex{11})

(xy^2-xy=5y-3x) for (x)

It is possible to solve; (x=displaystyle frac{5y}{y^2-y+3})

Exercise (PageIndex{12})

(xy^2-xy=5y-3x) for (y)

(displaystyle frac{x+5 pmsqrt{-11x^2+10x+25}}{2x})

Exercise (PageIndex{13})

(3t^2-5mq=8qt+2m^3) for (q)

It is possible to solve; (q=displaystyle frac{3t^2-2m^3}{8t+5m})

Exercise (PageIndex{14})

(2a^2bc^3+3abc^2+4a^2c^2-3b=4c) for (a)

It is possible to solve; (a = displaystyle frac{-(3bc^2) pm sqrt{(3bc^2)^2 - 4 (2bc^3+4c^2)(-3b-4c)}}{2(2bc^3+4c^2)})

Exercise (PageIndex{15})

(2a^2bc^3+3abc^2+4a^2c^2-3b=4c) for (b)

It is possible to solve; (b = displaystyle frac{4c-4a^2c^2}{2a^2c^3+3ac^2-3})

Exercise (PageIndex{16})

(2a^2bc^3+3abc^2+4a^2c^2-3b=4c) for (c)

It is possible to solve; but it would require using the cubic root formula

Exercise (PageIndex{17})

(log_2{(xy)=x+e^z}) for x

Not possible to solve for x; it is inside of a logarithm and has a linear term

Exercise (PageIndex{18})

(log_2{(xy)=x+e^z}) for y

It is possible to solve; (displaystyle y= 2^{x+e^z-log_2{(x)}}) or (displaystyle y= frac{2^{x+e^z}}{x})

Exercise (PageIndex{19})

(log_2{(xy)=x+e^z}) for z

It is possible to solve; (displaystyle z= ln{[log_2{(xy)} -x]})

In exercises (PageIndex{20}) - (PageIndex{28}), solve for (x). Be sure to list all possible values of (x).

Exercise (PageIndex{20})

(x^2-16=0)

(x=-4,4)

Exercise (PageIndex{21})

(x^2+16=0)

(x=-4i,4i)

Exercise (PageIndex{22})

(x^2-4x-7=2)

(x=2+sqrt{13}, 2- sqrt{13})

Exercise (PageIndex{23})

(x^2-2x+7=2)

(x=1+2i, 1-2i)

Exercise (PageIndex{24})

(5x^2+2x=-1)

(displaystyle x=frac{-1+2i}{5}, frac{-1-2i}{5})

Exercise (PageIndex{25})

(x^3=8)

(x=2)

Exercise (PageIndex{26})

(x^3+x^2=4x+4)

(x=-2, -1, 2)

Exercise (PageIndex{27})

(2(x-3)^2-7 = -4x+9)

(x=2-sqrt{3}, 2+ sqrt{3})

Exercise (PageIndex{28})

((x+2)^3 = 2x^2+8x+7)

(displaystyle x=-1, frac{-3+sqrt{5}}{2}, frac{-3-sqrt{5}}{2})

In exercises (PageIndex{29}) - (PageIndex{33}), classify the type(s) of solution(s) from the given exercise.

Exercise (PageIndex{29})

Exercise 3.1.1.20

Two real solutions

Exercise (PageIndex{30})

Exercise 3.1.1.21

A complex conjugate pair

Exercise (PageIndex{31})

Exercise 3.1.1.22

Two real solutions

Exercise (PageIndex{32})

Exercise 3.1.1.25

One repeated solution

Exercise (PageIndex{33})

Exercise 3.1.1.26

Three real solutions

## 3.1.1: Exercises 3.1

We will follow in part the proof of the Lindeberg CLT given in class. We first derive the characteristic function of a Bernoulli r.v. X with success probability :

Now let , be independent Bernoulli random variables with success probability , and put = . The characteristic function of is

We wish to show that this converges to the characteristic function of the Poisson distribution with mean . Let us derive what that characteristic function is. If T is Poisson with mean , then

Since = and exponentiation is a continuous function, it follows that

Now we would like to apply the Lemma from class that was used in the proof of the Lindeberg CLT, namely:

See Lemma 1 in section 27 of Billingsley. Now

since it is the characteristic function of a Bernoulli r.v. (characteristic functions are always bounded by 1 in modulus). Similarly,

since it is the characteristic function of a Poisson r.v. with mean . Thus, we obtain from (10) and Lemma 1.1 without further difficulty that

we see that the summands in (14) are simply the remainders in the first order Taylor expansion of at . It is not necessarily so easy to estimate what this remainder is since is a general complex number, i.e. has nonzero real and imaginary parts. We know what the remainder is like for Taylor series expansions of a real variable, but not a general complex variable.

However, let us assume for now that the remainder has the same order, that is

Thus, there exists an M such that and a such that

We wish to apply this result to = . We are given in condition (ii) that 0 as . Since and , condition (ii) is the same as simply 0. Thus,

So there exists such that for all ,

and hence for all and all j , ,

Using this back in (14) we obtain

This proves (8), so to complete the problem we need to verify (16).

Now we turn to proving (16). This is a well known result in complex variables, but one can show it with a real variable argument applied to the real and imaginary parts. Denote a complex number z by z = where and are both real. Now = = = . Further, we know

Since , of course can be replaced with . Thus,

Here, equation (29) follows by just multiplying out the terms. Equation (30) follows since = as , and

## 3.1.1: Exercises 3.1

To see why this expression works, the first part consists of all strings in which every 1 is followed by a 0. To that, we have only to add the possibility that there is a 1 at the end, which will not be followed by a 0. That is the job of (ε+1).

Now, we can rethink the question as asking for strings that have a prefix with no adjacent 1's followed by a suffix with no adjacent 0's. The former is the expression we developed, and the latter is the same expression, with 0 and 1 interchanged. Thus, a solution to this problem is (10+0)*( ε +1)(01+1)*( ε +0) . Note that the ε+1 term in the middle is actually unnecessary, as a 1 matching that factor can be obtained from the (01+1)* factor instead.

## Solutions for Section 3.2

### Exercise 3.2.1

Part (b): Here all expression names are R (1) we again list only the subscripts. R11 = 1 * R12 = 1*0 R13 = phi R21 = 11* R22 = ε+ 11*0 R23 = 0 R31 = phi R32 = 1 R33 = ε+ 0 .

Part (e): Here is the transition diagram:

If we eliminate state q2 we get:

Applying the formula in the text, the expression for the ways to get from q1 to q3 is: [1 + 01 + 00(0+10)*11]*00(0+10)*

### Exercise 3.2.8

Basis: R 0 ij1 is the number of arcs (or more precisely, arc labels) from state i to state j . R 0 ii0 = 1, and all other R 0 ijm 's are 0.

Induction: R (k) ijm is the sum of R (k-1) ijm and the sum over all lists (p1,p2. pr) of positive integers that sum to m , of R (k-1) ikp1 * R (k-1) kkp2 *R (k-1) kkp3 *. * R (k-1) kkp(r-1) * R (k-1) kjpr . Note r must be at least 2.

The answer is the sum of R (k) 1jn , where k is the number of states, 1 is the start state, and j is any accepting state.

## NCERT Solutions for Class 6 Maths Exercise 3.1

NCERT Solutions Class 6 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 6 Maths includes text book solutions from Class 6 Maths Book . NCERT Solutions for CBSE Class 6 Maths have total 14 chapters. 6 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 6 solutions PDF and Maths ncert class 6 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

## 3.1.1: Exercises 3.1

Let (f(n)) and (g(n)) be asymptotically nonnegative functions. Using the basic definition of (Theta)-notation, prove that (max(f(n),g(n)) = Theta(f(n) + g(n))).

(f(n)) and (g(n)) being asympototically nonnegative functions means that (0 leq f(n)) and (0 leq g(n)) which implies:

Taking the max of two integers results in one of those two values which gives us (f(n) leq max(f(n), g(n))) and (g(n) leq max(f(n), g(n))) and happens to imply (frac <2>leq max(f(n), g(n))). We can combine this with our above inequality by first multiplying both sides by (frac<1><2>):

Lastly, (max(f(n), g(n))) is always less than or equal to the sum of its two terms (since we know they are not negative). This tells us (max(f(n), g(n)) leq f(n) + g(n)). This gives us the following inequality:

[0 leq frac <2>leq max(f(n), g(n)) leq f(n) + g(n)]

Which is the definition of (Theta)-notation with (c_1 = frac<1><2>) and (c_2 = 1).

## 3.1.1: Practice Problems- Nuclear Chemistry and Radioactive Decay (Optional)

Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:

1. 25 protons, 20 neutrons, 24 electrons
2. 45 protons, 24 neutrons, 43 electrons
3. 53 protons, 89 neutrons, 54 electrons
4. 97 protons, 146 neutrons, 97 electrons Which of the following nuclei lie within the band of stability?

1. chlorine-37
2. calcium-40
3. 204 Bi
4. 56 Fe
5. 206 Pb
6. 211 Pb
7. 222 Rn
8. carbon-14 Which of the following nuclei lie within the band of stability?

1. argon-40
2. oxygen-16
3. 122 Ba
4. 58 Ni
5. 205 Tl
6. 210 Tl
7. 226 Ra
8. magnesium-24

Write a brief description or definition of each of the following:

1. nucleon
2. &alpha particle
3. &beta particle
4. positron
5. &gamma ray
6. nuclide
7. mass number
8. atomic number

collective term for protons and neutrons in a nucleus

(&alpha or (ce<^4_2He>) or (ce<^4_2&alpha>)) high-energy helium nucleus a helium atom that has lost two electrons and contains two protons and two neutrons

antiparticle to the electron it has identical properties to an electron, except for having the opposite (positive) charge

(&gamma or (ce<^0_0&gamma>)) short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality

nucleus of a particular isotope

sum of the numbers of neutrons and protons in the nucleus of an atom

number of protons in the nucleus of an atom

Complete each of the following equations by adding the missing species:

1. (ce<^<27>_<13>Al + ^4_2He⟶:? + ^1_0n>)
2. (ce<^<239>_<94>Pu +, ? ⟶ ^<242>_<96>Cm + ^1_0n>)
3. (ce<^<14>_7N + ^4_2He⟶:? + ^1_1H>)
4. (ce<^<235>_<92>U⟶:? + ^<135>_<55>Cs + 4^1_0n>)

Complete each of the following equations:

Write a balanced equation for each of the following nuclear reactions:

1. the production of 17 O from 14 N by &alpha particle bombardment
2. the production of 14 C from 14 N by neutron bombardment
3. the production of 233 Th from 232 Th by neutron bombardment
4. the production of 239 U from 238 U by (ce<^2_1H>) bombardment

Technetium-99 is prepared from 98 Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a &beta particle to yield an excited form of technetium-99, represented as 99 Tc * . This excited nucleus relaxes to the ground state, represented as 99 Tc, by emitting a &gamma ray. The ground state of 99 Tc then emits a &beta particle. Write the equations for each of these nuclear reactions.

What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?

1. an &alpha particle is emitted
2. a &beta particle is emitted
4. a positron is emitted
5. an electron is captured

Since an &alpha particle is the same as a (ce<^4_2He>) nucleus, the mass number will decrease by 4 and the atomic number will decrease by 2.

Since a &beta particle is the same as (ce<^0_<-1>e>), the mass number will not change but the atomic number will increase by 1.

Since a &gamma ray has no mass (it is energy) the mass number and atomic number do not change.

A positron is the opposite of a &beta particle, it is (ce<^0_<+1>e>), the mass number will not change but the atomic number will decrease by 1

Electron capture has the same effect on the nucleus as positron emission: The atomic number is decreased by one and the mass number does not change.

What is the change in the nucleus that results from the following decay scenarios?

conversion of a neutron to a proton: (ce<^1_0n ⟶ ^1_1p + ^0_<+1>e>)

conversion of a proton to a neutron the positron has the same mass as an electron and the same magnitude of positive charge as the electron has negative charge

when the n:p ratio of a nucleus is too low, a proton is converted into a neutron with the emission of a positron: (ce<^1_1p ⟶ ^1_0n + ^0_<+1>e>)

In a proton-rich nucleus, an inner atomic electron can be absorbed. In simplest form, this changes a proton into a neutron: (ce<^1_1p + ^0_<-1>e ⟶ ^1_0p>) Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability if

(a) they are below the band of stability and

(b) they are above the band of stability

Nuclei below the band of stability will undergo positron decay, while those above the band of stability will undergo beta decay. Heavy nuclei past the band of stability will undergo alpha decay

Which of the following nuclei is most likely to decay by positron emission? Explain your choice.

Manganese-51 is most likely to decay by positron emission. The n:p ratio for Cr-53 is (dfrac<29><24>) = 1.21 for Mn-51, it is (dfrac<26><25>) = 1.04 for Fe-59, it is (dfrac<33><26>) = 1.27. Positron decay occurs when the n:p ratio is low. Mn-51 has the lowest n:p ratio and therefore is most likely to decay by positron emission. Besides, (ce<^<53>_<24>Cr>) is a stable isotope, and (ce<^<59>_<26>Fe>) decays by beta emission. The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.

Above the band of stability, beta decay is expected

Beyond the band of stability, heavy nuclei undergo alpha decay

Below the band of stability, positron decay is expected

Above the band of stability, beta decay is expected

Beyond the band of stability, heavy nuclei undergo alpha decay

Write a nuclear reaction for each step in the formation of (ce<^<218>_<84>Po>) from (ce<^<238>_<92>U>) , which proceeds by a series of decay reactions involving the step-wise emission of &alpha, &beta, &beta, &alpha, &alpha, &alpha, &alpha particles, in that order.

Write a nuclear reaction for each step in the formation of (ce<^<208>_<82>Pb>) from (ce<^<228>_<90>Th>), which proceeds by a series of decay reactions involving the step-wise emission of &alpha, &alpha, &alpha, &alpha, &beta, &beta, &alpha particles, in that order.

## Bump dns-packet from 1.3.1 to 1.3.4 in /coding-exercise #118

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Q: If |4= 4 and 4, B are 3x 3 matrices, det 33)* A>-. What is B?

A: Click to see the answer

Q: (3) If a projectile is launched into the air at 29.4 feet per second from a height of 60 feet, its h.

Q: (a) State the value of [I-1 (b) Let f, g:Zx Z → Z × Z be two functions defined as follows: f (a,b) =.

A: Since you have asked multiple question, we will solve the first question for you. If you want any sp.

Q: Only solve part (iii). Please show the full steps.

A: in part (iii) we have to find the KKT point and the minimum value of function For this, we impose t.

Q: Find the power series representation for the given function: 1 f(x): (1 + x²)²

A: Power series of function is represented by:f(x)=∑n=0∞anxn Where an's are coefficients. Also we know .

Q: Q1// Use Tow Phase method to solve the Mathematical model (Stop at Table 1 in Stage Two) MinZ= x,-2 .

A: Question: Min Z=x1-2x2 Subject to x1+x2≥2-x1+x2≥10x1+x2≤3x1,x2≥0 Use two phase method(stop at Table .

Q: y(n+2)+2y(n+1)+ y(n) = n where y(1)=y(0)=0

A: This is the difference equation of order two. we solve this by finding solution of homogeneous apart.

Q: If G is an open set in complex plan and f : G → C is differentiable, then on G, f is: إختر أحد الخيا.

A: Click to see the answer

Q: Show that the solution of the equation YUx – xUy = 0 containing the curve x² + y² = a², u= y, does n.

A: Given that y ux – x uy = 0, The Lagrange’s auxiliary equation for the given equation is given as: Pp.

## NCERT Solutions class 12 Maths Exercise 3.1

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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