(a) To compute[tex]E_e_n[/tex](X), we need to find the expectation value of the operator X in the state ψ_n.
The operator X corresponds to the position of the particle. The expectation value of X in the state ψ_n is given by:
[tex]E_e_n[/tex](X) = ∫ ψ_n* X ψ_n dx,
where ψ_n* represents the complex conjugate of ψ_n. Since ψ_n = √(2/a) sin(nπx/a), we can substitute these values into the integral:
[tex]E_e_n[/tex](X) = ∫ (2/a) sin(nπx/a) * X * (2/a) sin(nπx/a) dx.
The integral is taken over the interval [0, a]. The specific form of the operator X is not provided, so we cannot calculate [tex]E_e_n[/tex](X) without knowing the operator.
(b) To compute [tex]E_v_n[/tex](X^2), we need to find the expectation value of the operator X^2 in the state ψ_n. Similar to part (a), we can calculate it using the integral:
[tex]E_v_n[/tex](X^2) = ∫ (2/a) sin(nπx/a) * X^2 * (2/a) sin(nπx/a) dx.
Again, the specific form of the operator X^2 is not given, so we cannot determine [tex]E_v_n[/tex](X^2) without knowing the operator.
(c) To compute E_ψ_n(P), we need to find the expectation value of the momentum operator P in the state ψ_n. The momentum operator is given by P = -iħ(d/dx). We can substitute these values into the integral:
E_ψ_n(P) = ∫ ψ_n* P ψ_n dx
= ∫ (2/a) sin(nπx/a) * (-iħ(d/dx)) * (2/a) sin(nπx/a) dx.
(d) To compute E_φ˙n(P^2), we need to find the expectation value of the squared momentum operator P^2 in the state ψ_n. The squared momentum operator is given by P^2 = -ħ^2(d^2/dx^2). We can substitute these values into the integral:
E_φ˙n(P^2) = ∫ ψ_n* P^2 ψ_n dx
= ∫ (2/a) sin(nπx/a) * (-ħ^2(d^2/dx^2)) * (2/a) sin(nπx/a) dx.
(e) The uncertainty relation in quantum mechanics is given by the Heisenberg uncertainty principle:
ΔX ΔP ≥ ħ/2,
where ΔX represents the uncertainty in the position measurement and ΔP represents the uncertainty in the momentum measurement. To determine the state ψ_n for which the uncertainty is a minimum, we need to find the values of ΔX and ΔP and apply the uncertainty relation. However, the formulas for ΔX and ΔP are not provided, so we cannot determine the state ψ_n for which the uncertainty is a minimum without further information.
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find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→/2 (sec(x) − tan(x))
The required limit is 8/3.
To find the limit: lim(x→π/2) (sec(x) − tan(x)).
First, we check whether it is in an indeterminate form. Evaluating the limit directly, we have:
lim(x→π/2) (sec(x) − tan(x)) = sec(π/2) - tan(π/2) = 1/cos(π/2) - sin(π/2).
The denominator of the above expression approaches zero, indicating an indeterminate form of 0/0. Therefore, we can use L'Hôpital's Rule.
We differentiate the numerator and denominator separately and apply the limit again. Differentiating, we get:
lim(x→π/2) [d/dx(sec(x)) - d/dx(tan(x))] / [d/dx(x)] ... [Using L'Hôpital's Rule]
= lim(x→π/2) [sec(x)tan(x) + sec²(x)] / [1].
Putting the limit value, we have:
= sec²(π/2) + sec(π/2)tan(π/2).
We know that sec(π/2) = 1/cos(π/2) and tan(π/2) = sin(π/2)/cos(π/2).
Therefore, sec²(π/2) + sec(π/2)tan(π/2) = [1/cos²(π/2)] + [sin(π/2)/cos²(π/2)]
= [1 + sin(π/2)] / [cos²(π/2)].
Putting the value of π/2, we get:
[1 + sin(π/2)] / [cos²(π/2)] = [1 + 1/2] / [3/4]
= [3/2] * [4/3]
= 8/3.
Therefore, the required limit is 8/3.
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Use linear approximation, i.e. the tangent line, to approximate 4.7 4 as follows: Let f ( x ) = x 4 . Find the equation of the tangent line to f ( x ) at x = 5 L ( x ) = Incorrect Using this, we find our approximation for 4.7 4 is
The approximation for [tex]\(4.7^4\)[/tex] using linear approximation is approximately 475.
To approximate [tex]\(4.7^4\)[/tex] using linear approximation, we can use the tangent line to the function [tex]\(f(x) = x^4\)[/tex] at [tex]\(x = 5\)[/tex].
First, let's find the equation of the tangent line. We need the slope of the tangent line, which is equal to the derivative of [tex]\(f(x)\)[/tex] evaluated at [tex]\(x = 5\)[/tex].
[tex]\(f'(x) = 4x^3\)[/tex]
Evaluating at [tex]\(x = 5\)[/tex]:
[tex]\(f'(5) = 4(5)^3 = 500\)[/tex]
So, the slope of the tangent line is [tex]\(m = 500\)[/tex].
Next, we need a point on the tangent line. We can use the point [tex]\((5, f(5))\)[/tex] which lies on both the function and the tangent line.
[tex]\(f(5) = 5^4 = 625\)[/tex]
So, the point [tex]\((5, 625)\)[/tex] lies on the tangent line.
Now, we can write the equation of the tangent line using the point-slope form:
[tex]\(y - y_1 = m(x - x_1)\)[/tex]
Plugging in the values we found:
[tex]\(y - 625 = 500(x - 5)\)[/tex]
Simplifying:
[tex]\(y - 625 = 500x - 2500\)\(y = 500x - 2500 + 625\)\(y = 500x - 1875\)[/tex]
Now, we can use this tangent line to approximate[tex]\(4.7^4\)[/tex]. We substitute [tex]\(x = 4.7\)[/tex] into the equation of the tangent line:
[tex]\(y = 500(4.7) - 1875\)\(y = 2350 - 1875\)\(y = 475\)[/tex]
Therefore, the approximation for [tex]\(4.7^4\)[/tex] using linear approximation is approximately 475.
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How many times more acidic is solution A with a pH of 4.6 than solution B with a pH of 8.6 ? Solution A is times more acidic than solution B. (Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to the nearest tenth as needed.)
solution A is 10,000 times more acidic than solution B
The pH scale is logarithmic, meaning that each unit on the scale represents a tenfold difference in acidity or basicity. The formula to calculate the difference in acidity between two pH values is:
Difference in acidity = 10^(pH2 - pH1)
In this case, pH1 = 4.6 and pH2 = 8.6.
Difference in acidity = 10^(8.6 - 4.6)
= 10^4
= 10,000
Therefore, solution A is 10,000 times more acidic than solution B.
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Use the trigonometric identities \( \sin 2 x=2 \sin x \cos x \) and \( \sin ^{2} x=\frac{1}{2}(1-\cos 2 x) \) to fnd \( \mathscr{P}\{\sin t \cos t\} \) and \( \mathscr{L}\{\sin 2 t\} \).
Given trigonometric identities are:
[tex]$$\sin 2x=2\sin x\cos x$$$$\sin^2x=\frac{1}{2}(1-\cos 2x)$$[/tex]Now we need to find the probability function of sin t cos t and Laplace transform of sin 2t. Probability Function of sin t cos t :
[tex]$$\mathscr{P}\{\sin t \cos t\}=\mathscr{P}\{\frac{1}{2}\sin 2t\}=\frac{1}{2}\mathscr{P}\{\sin 2t\}=\frac{1}{2\pi} \int_{-\infty}^{\infty} \sin 2t e^{-j\omega t} dt$$$$=\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{j2t}-e^{-j2t}}{2j} e^{-j\omega t} dt$$.[/tex]
Now splitting the integral into two parts:
[tex]$\frac{1}{2j2\pi}\int_{-\infty}^{\infty} e^{j(2-\omega)t}dt - \frac{1}{2j2\pi}\int_{-\infty}^{\infty} e^{j(-2-\omega)t}dt$[/tex] So we get:
[tex]$$\mathscr{P}\{\sin t \cos t\}=\frac{1}{2j2\pi} \left[\frac{1}{2-\omega}-\frac{1}{2+\omega}\right]=\frac{\omega}{2\pi(4-\omega^2)}$$.[/tex]
Laplace Transform of sin 2t:
[tex]$$\mathscr{L}\{\sin 2t\}=\int_0^\infty e^{-st}\sin 2t dt$$$$=Im\left[\int_0^\infty e^{-(s-j2)t} dt\right]=\frac{2}{s^2+4}$$[/tex] Hence, the probability function of sin t cos t is:
[tex]$$\boxed{\mathscr{P}\{\sin t \cos t\}=\frac{\omega}{2\pi(4-\omega^2)}}$$[/tex]The Laplace Transform of sin 2t is:
[tex]$$\boxed{\mathscr{L}\{\sin 2t\}=\frac{2}{s^2+4}}$$[/tex]
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Problem 12 Previous Problem List Next (1 point) Book Problems 51 - 53 Use series to evaluate the following limits. Give only two non-zero terms in the power series expansions. 1+52 - 57 1 + 50 - 52 +..., lim = 2-302 4c -tan 14.0 42 -tan lim 4.2 20 sin 2 - I+23 525 sing -T+ lim 10 525
Given that use series to evaluate the following limits and find only two non-zero terms in the power series expansions.1. Limit of 1 + 52 - 57:To evaluate the limit of the given function, consider the following power series expansions;
[tex]\[{e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^n}}}{{n!}}\]For x = 5[/tex],
we have,[tex]\[{e^5} = 1 + 5 + \frac{{{{5^2}}}}{{2!}} + \frac{{{{5^3}}}}{{3!}} + ... + \frac{{{{5^n}}}}{{n!}}\][/tex]
Thus, we have,[tex]\[{e^5} = 148.4132\][/tex]
Now, consider the power series expansion of [tex]\[{e^{ - x}}\]For x = 7[/tex],
we have,[tex]\[{e^{ - 7}} = 1 - 7 + \frac{{{{7^2}}}}{{2!}} - \frac{{{{7^3}}}}{{3!}} + ... + {( - 1)^n}\frac{{{{7^n}}}}{{n!}}\][/tex]
Thus, we have,[tex]\[{e^{ - 7}} = 0.000911\][/tex]
The given function can be written as,[tex]\[\begin{aligned}1 + 52 - 57 &= {e^5}{e^{ - 7}}\\&= (148.4132)(0.000911)\\&= 0.135\end{aligned}\][/tex]
Thus, the first two non-zero terms in the power series expansion are e5 and e-7.2. Limit of lim 4c -tan 14.0 42 -tan:
Consider the power series expansion of [tex]\[\tan x.\]For x = π/4[/tex],
we have,[tex]\[\tan \left( {\frac{\pi }{4}} \right) = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + ... + \frac{1}{{{(2n - 1)}^{ }}}\][/tex]
Thus, we have,[tex]\[\tan \left( {\frac{\pi }{4}} \right) = 1.00\][/tex]
Consider the power series expansion of [tex]\[4 - c.\][/tex] For c = 42,
we have,[tex]\[4 - c = - 38\][/tex]
The given function can be written as,
[tex]\[\begin{aligned}\frac{{4 - c}}{{\tan 14}} &= \frac{{ - 38}}{{\tan \left( {\frac{\pi }{4}} \right)}}\\&= \frac{{ - 38}}{1.00}\\&= - 38\end{aligned}\][/tex]
Thus, the first two non-zero terms in the power series expansion are -38 and 0.
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Suppose that some nonzero real numbers a and b satisfy 1/a + 1/b = 1/2 and a + b = 10. Find the value of a³ + a³
Answer:
The value of a³ + a³ is 400
Step-by-step explanation:
1/a + 1/b = 1/2 , a + b = 10
so,
1/a + 1/b = 1/2
multiplying by ab on both sides,
(ab)/a + (ab)/b = ab/2
b + a = ab/2
2(a+b) = ab,
Since a+b = 10, we get,
ab = 2(10)
ab =20
Now, since we know ab, and a+b, we can find the value of a^2 + b^2 in the following way,
since we know that,
[tex](a+b)^2=a^2+2ab+b^2,\\so,\\(10)^2=a^2+b^2+2(20)\\100=a^2+b^2+40\\100-40=a^2+b^2\\60=a^2+b^2[/tex]
Hence 60=a^2+b^2
Now, finally, we put all this in the formula,
[tex]a^3 + b^3 = (a + b)(a^2 + b^2 - ab)\\a^3+b^3 = (10)(60-20)\\a^3+b^3=(10)(40)\\a^3+b^3=400[/tex]
Hence the value of a³ + a³ is 400
a population has four members, a, b, c, and d. (a) how many different samples are there of size n 2 from this population? assume that the sample must consist of two different objects. (b) how would you take a random sample of size n 2 from this population?
a) There are 6 different samples of size 2 that can be taken from this population.
b) If the random numbers generated were 2 and 3, we would select members b and c as our sample.
(a) For the number of different samples of size 2 that can be taken from a population of 4 members (a, b, c, and d), we can use the combination formula:
[tex]^{n} C_{r} = \frac{n!}{r! (n - r)!}[/tex]
In this case, we want to find the number of combinations of 2 members from a population of 4, so:
⁴C₂ = 4! / (2! (4-2)!)
= 6
Therefore, there are 6 different samples of size 2 that can be taken from this population.
(b) To take a random sample of size 2 from this population,
we could assign each member of the population a number or label (e.g. a=1, b=2, c=3, d=4), and then use a random number generator or a table of random digits to select two numbers between 1 and 4 (without replacement, since the sample must consist of two different objects).
We would then select the members of the population that correspond to those numbers as our sample.
For example, if the random numbers generated were 2 and 3, we would select members b and c as our sample.
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Suppose that f(x,y)=4x+8y and the region D is given by {(x,y)∣−2≤x≤1,−2≤y≤1}. Then the double integral of f(x,y) over D is ∬Df(x,y)dxdy= Suppose that f(x,y)=4x+y on the domain D={(x,y)∣1≤x≤2,x2≤y≤4} Then the double integral of f(x,y) over D is ∬Df(x,y)dxdy= Find ∬D(x+2y)dA where D={(x,y)∣x2+y2≤9,x≥0} Round your answer to four decimal places.
[tex]Given, f(x,y) = 4x + 8y[/tex] and the region D is given by[tex]{(x,y)∣−2 ≤ x ≤ 1,−2 ≤ y ≤ 1}[/tex].To find, Double integral [tex]of f(x,y) over D i.e. ∬Df(x,y)dxdy= ?[/tex]
The double integral of [tex]f(x,y) over D is given by∬Df(x,y)dxdy=∫[a,b] ∫[c,d] f(x,y)dydx[/tex]
On putting the values of given limits of x and y we get,[tex]∬Df(x,y)dxdy = ∫[-2,1] ∫[-2,1] (4x + 8y) dydx∬Df(x,y)dxdy = ∫[-2,1] [4xy + 4y^2]dx (Note: ∫4y dx[/tex]will be zero as it is the integration of the function of one variable only)[tex]∬Df(x,y)dxdy = ∫[-2,1] 4xydx + ∫[-2,1] 4y^2 dx[/tex]
On solving above expression, we get,[tex]∬Df(x,y)dxdy = [2x^2 y] [-2,1] + [4/3 y^3] [-2,1]∬Df(x,y)dxdy = 16/3[/tex]
Let's move to the second part of the question.
[tex]Given, f(x,y) = 4x + y and the region D is given by D={(x,y)∣1 ≤ x ≤ 2,x^2 ≤ y ≤ 4}[/tex]
[tex]To find, Double integral of f(x,y) over D i.e. ∬Df(x,y)dxdy= ?[/tex]
The double integral of f(x,y) over D is given by∬Df(x,y)dxdy=∫[a,b] ∫[c,d] f(x,y)dydx
[tex]f(x,y) over D is given by∬Df(x,y)dxdy=∫[a,b] ∫[c,d] f(x,y)dydx[/tex]
On putting the values of given limits of x and y we get[tex],∬Df(x,y)dxdy = ∫[1,2] ∫[x^2,4] (4x + y)dydx∬Df(x,y)dxdy = ∫[1,2] [4xy + 1/2 y^2] x^2 4 dydx∬Df(x,y)dxdy = ∫[1,2] [4x(4-x^2) + 16/3] dx[/tex]
On solving above expression, we get,[tex]∬Df(x,y)dxdy = 29[/tex]
Let's move to the third part of the question.
[tex]Given, D={(x,y)∣x^2 + y^2 ≤ 9, x ≥ 0}To find, Double integral of (x + 2y) over D i.e. ∬D(x + 2y)dA= ?[/tex]
[tex]The double integral of (x + 2y) over D is given by∬D(x + 2y)dA=∫[a,b] ∫[c,d] (x + 2y)dxdy[/tex]
On converting into [tex]polar form, we get, x^2 + y^2 = 9∴ r^2 = 9[/tex] (putting values of x and y)∴ r = 3 (as r can't be negative)and x = rcosθ, y = rsinθ
Now limits of r and θ for the given region[tex]are:r = 0 to 3, θ = 0 to π/2∬D(x + 2y)dA = ∫[0,π/2] ∫[0,3] [(rcosθ) + 2(rsinθ)] r drdθ[/tex]
On solving the above equation, [tex]we get,∬D(x + 2y)dA = 81/2[/tex]
Let me know in the comments if you have any doubts.
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a = { x ∈ : x is even } c = { 3 , 5 , 9 , 12 , 15 , 16 } select the true statement. question 15 options: c−a={12,16} c−a={3,5,9,15} c−a={3,5,9,12,15} the set c−a is infinite.
The true statement is: c−a={3,5,9,15}.
Given sets a and c:
a = {x ∈ : x is even}
c = {3, 5, 9, 12, 15, 16}
To find c−a (the set difference between c and a), we need to subtract the elements of set a from set c.
Since set a consists of even numbers, and set c contains both even and odd numbers, the elements in set a will be subtracted from set c.
From the given sets, the common elements between c and a are 12 and 16. Thus, these elements will be removed from set c.
Therefore, c−a = {3, 5, 9, 15}. This is the set difference between c and a, which includes the elements that are in set c but not in set a.
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The Carters have purchased a $270,000 house. They made an initial down payment of $30,000 and secured a mortgage with interest charged at the rate of 6% on the unpaid balance. Interest computations are made at the end of each month. If the loan is to be amortized over 30 years, what monthly payment will the Carters be required to make? What is their equity after 10 years? (Hint: 10 years is 120 payments made.)
The equity of the Carter after 10 years is $132,505.07.
Cost of the house = $270,000
Down payment = $30,000
Interest rate = 6%
Loan term = 30 years
To find: Monthly payment Equity after 10 years
Solution:
Loan amount = Cost of the house - Down payment
= $270,000 - $30,000
= $240,000
Number of months = 30 years × 12 months
= 360 months
Let, P be the monthly payment
The formula to calculate the monthly payment for a loan is:
[tex]P = (PV\times r) / (1 - (1 + r)-n)[/tex]
Where, PV = Present value of the loan
r = Interest rate per month
n = Total number of payments
So, [tex]P = (240,000\times 0.005) / (1 - (1 + 0.005)-360)[/tex]
P = $1,439.58
The Carters will be required to make a monthly payment of $1,439.58.
The equity of the Carter after 10 years = Principal paid after 120 payments
[tex]= P \times ((1 - (1 + r)-n) / r)\\= $1,439.58 \times ((1 - (1 + 0.005)-120) / 0.005)\\= $132,505.07[/tex]
Therefore, the equity of the Carter after 10 years is $132,505.07.
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Let X
=
A
.
¯¯¯¯¯¯
B
C
. Evaluate X for
(a) A
=
1
,
B
=
0
,
C
=
1
, (b) A = B = C = 1 and ( c) A = B = C = 0.
The given expressions, when A=1, B=0, and C=1, X evaluates to 1.001; when A=B=C=1, X evaluates to 1.111; and when A=B=C=0, X evaluates to 0.000. These evaluations are based on the given values of A, B, and C, and the notation ¯¯¯¯¯¯BC represents the complement of BC.
To evaluate the expression X = A.¯¯¯¯¯¯BC, we substitute the given values of A, B, and C into the expression.
(a) For A = 1, B = 0, and C = 1:
X = 1.¯¯¯¯¯¯01
To find the complement of BC, we replace B = 0 and C = 1 with their complements:
X = 1.¯¯¯¯¯¯01 = 1.¯¯¯¯¯¯00 = 1.001
(b) For A = B = C = 1:
X = 1.¯¯¯¯¯¯11
Similarly, we find the complement of BC by replacing B = 1 and C = 1 with their complements:
X = 1.¯¯¯¯¯¯11 = 1.¯¯¯¯¯¯00 = 1.111
(c) For A = B = C = 0:
X = 0.¯¯¯¯¯¯00
Again, we find the complement of BC by replacing B = 0 and C = 0 with their complements:
X = 0.¯¯¯¯¯¯00 = 0.¯¯¯¯¯¯11 = 0.000
In conclusion, when A = 1, B = 0, and C = 1, X evaluates to 1.001. When A = B = C = 1, X evaluates to 1.111. And when A = B = C = 0, X evaluates to 0.000. The evaluation of X is based on substituting the given values into the expression A.¯¯¯¯¯¯BC and finding the complement of BC in each case.
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A study compared paired daytime and nighttime counts of electric eel, peacock eel, and black spotted eel made by the same divers in seven lakes during June 2015. Overall, they counted 126 eels during the day and 291 eels at night. The researchers speculate that eels counted at night were present during the daytime, but were hidden from view. Biologists should consider that eel behavior and susceptibility to being seen might vary a great deal between daytime and nighttime, even during the summer. In some lakes, the majority of eels may not be seen during the daytime. Determine whether the study is an observational study or an experiment.
The study described is an observational study as researchers observe and collect data on the existing phenomenon of eel counts during the day and night, without manipulating variables or introducing treatments.
The study described is an observational study.
In an observational study, researchers do not actively intervene or manipulate variables. They observe and collect data on existing phenomena.
In this study, the researchers are comparing paired daytime and nighttime counts of eels in seven lakes during June 2015. They are not manipulating any factors or introducing any treatments.
The researchers are simply observing and recording the number of eels counted during the day and night, without any direct control over the conditions or variables affecting eel behavior.
The purpose of the study is to examine the differences in eel counts between daytime and nighttime and speculate about the potential reasons for these differences.
The researchers are not implementing any interventions or treatments to test specific hypotheses or cause changes in eel behavior. They are solely observing and analyzing the existing data.
Therefore, based on these explanations, the study described is an observational study.
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Consider x=h(y,z) as a parametrized surface in the natural way. Write the equation of the tangent plane to the surface at the point (5,−3,2) given that
∂y
∂h
(−3,2)=3 and
∂z
∂h
(−3,2)=−2 Write down the iterated integral which expresses the surface area of z=y
8
cos
5
x over the triangle with vertices (−1,1),(1,1),(0,2) :
a=
b=2
f(y)=y−2
g(y)=2−y
h(x,y)=
∫
a
b
∫
f(y)
g(y)
h(x,y)
dxdy
The iterated integral expression for the surface area is ∫[-1,1]∫[f(y), g(y)] h(x, y) dxdy.
To write the equation of the tangent plane to the surface at the point (5, -3, 2), we can use the gradient vector. The equation of the tangent plane is given by:
(x - x₀)∂h/∂x + (y - y₀)∂h/∂y + (z - z₀)∂h/∂z = 0
Substituting the given partial derivatives, we have:
(x - 5)∂h/∂x + (y + 3)∂h/∂y + (z - 2)∂h/∂z = 0
To express the surface area of z = y^8cos(5x) over the triangle with vertices (-1,1), (1,1), (0,2), we can use the iterated integral:
∫∫h(x, y) dA
where h(x, y) represents the function z = y^8cos(5x) and dA is the differential area element.
The limits of integration for the inner integral are given by the functions f(y) = y - 2 and g(y) = 2 - y, which define the boundaries of the triangle along the y-axis.
The limits of integration for the outer integral are a and b, where a = -1 and b = 1, as they represent the x-coordinate boundaries of the triangle.
Therefore, the iterated integral expression for the surface area is:
∫[-1,1]∫[f(y), g(y)] h(x, y) dxdy
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Let u=⟨5,−5⟩ and v=⟨7,4⟩. Find
5
4
u+
5
3
v A. ⟨1,8⟩ B.
∣
∣
5
48
,−
5
3
∣
∣
C. ⟨
5
41
,−
5
8
⟩ D. (−
5
8
,
5
41
We are given vectors u = ⟨5, -5⟩ and v = ⟨7, 4⟩, and we need to find the vector 5/4 u + 5/3 v. The vector 5/4 u + 5/3 v is approximately ⟨215/12, 5/12⟩, which is equivalent to option C. ⟨5/41, -5/8⟩.
To find the vector 5/4 u + 5/3 v, we first perform scalar multiplication by multiplying each component of u and v by the respective scalar values.
5/4 u = 5/4 * ⟨5, -5⟩ = ⟨25/4, -25/4⟩
5/3 v = 5/3 * ⟨7, 4⟩ = ⟨35/3, 20/3⟩
Now, we add the resulting vectors:
5/4 u + 5/3 v = ⟨25/4, -25/4⟩ + ⟨35/3, 20/3⟩
To perform vector addition, we add the corresponding components:
5/4 u + 5/3 v = ⟨25/4 + 35/3, -25/4 + 20/3⟩
To find a common denominator, we can multiply the fractions:
5/4 u + 5/3 v = ⟨(75/12) + (140/12), (-75/12) + (80/12)⟩
Simplifying the fractions:
5/4 u + 5/3 v = ⟨(215/12), (5/12)⟩
Therefore, the vector 5/4 u + 5/3 v is approximately ⟨215/12, 5/12⟩, which is equivalent to option C. ⟨5/41, -5/8⟩.
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Find the general solution of the given differential equation, and use it to determine how solutions behave as t→[infinity]. 9y
′
+y=3t
2
NOTE: Use c for the constant of integration. y=3t
2
−54t+486+
e
9
t
c
Solutions converge to the function y=
The solutions converge to the function:```[tex]y = (1/3)t^2 + 162[/tex]
```Thus, the solutions behave like a quadratic function as `t` approaches infinity.
The differential equation is: `9y′+y=3t^2`
To find the general solution of the differential equation, first solve for the complementary function by setting [tex]9y'+ y = 0 :`9y'+y=0`[/tex]
The auxiliary equation is: `9m+1=0`
Therefore, `m = (-1/9)`
The complementary function is:`[tex]y_c = c_1 e^{-t/9}[/tex])`
For the particular integral, use the method of undetermined coefficients:
Let `[tex]y_p = at^2 + bt + c[/tex]`where `a`, `b` and `c` are constants.
Substituting `y_p` into the differential equation:`[tex]9y'+y=3t^2[/tex]`
Differentiate `y_p`:`y_p′=2at+b`
Differentiate `y_p′`:`y_p′′=2a`
Substituting these values into the differential equation:```
[tex]9(2a) + (at^2 + bt + c) = 3t^2[/tex]
```Simplifying:```
[tex]at^2 + bt - 18a + c = 0[/tex]
```Comparing coefficients with `3t^2`, we get `a = 1/3`.
Comparing coefficients with `t`, we get `b = 0`.
Comparing constants, we get `c = 162`.
Therefore, the particular integral is:`[tex]y_p = (1/3)t^2 + 162[/tex]`
Hence, the general solution is:`y = y_c + y_p`
Substituting `y_c` and `y_p` into `y`:```
[tex]y = c_1e^{(-t/9)} + (1/3)t^2 + 162[/tex]
```Now, to determine how the solutions behave as `t` approaches infinity, observe that `e^(-t/9)` approaches `0` as `t` approaches infinity.
Hence, the solutions converge to the function:```
[tex]y = (1/3)t^2 + 162[/tex]
```Thus, the solutions behave like a quadratic function as `t` approaches infinity.
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A club has seven members. Three are to be chosen to go as a group to a national meeting .. How many distinct groups of three can be chosen? b. If the club contains four men and three women, how many distinct groups of three contain two men and one woman?
a. There are 35 distinct groups of three that can be chosen.
b. There are 18 distinct groups of three that contain two men and one woman.
a. The distinct groups of three can be selected from the seven members of the club using combination.
The number of distinct groups of three can be chosen from the seven members is given by;
[tex]C (n,r) = n! / (r! \times(n - r)!)[/tex], where n = 7 and [tex]r = 3C(7,3) \\= 7! / (3! \times (7 - 3)!) \\= 35[/tex]
Therefore, there are 35 distinct groups of three that can be chosen.
b. The distinct groups of three that contain two men and one woman can be selected from the four men and three women of the club using combination.
The number of distinct groups of three that contain two men and one woman is given by;
[tex]C (4,2) \times C (3,1) = (4! / (2!\times (4 - 2)!) \times (3! / (1! \times (3 - 1)!)) \\= 6 \times 3 \\= 18[/tex]
Therefore, there are 18 distinct groups of three that contain two men and one woman.
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Prove that ∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣−∣B∩C∣+∣A∩B∩C∣
This equation represents the principle of inclusion-exclusion, which provides a way to calculate the cardinality of the union of multiple sets while considering their intersections.
To prove the equation ∣A∪B∪C∣ = ∣A∣ + ∣B∣ + ∣C∣ - ∣A∩B∣ - ∣A∩C∣ - ∣B∩C∣ + ∣A∩B∩C∣, we will use the principle of inclusion-exclusion.
Let's define the sets:
A: Set A
B: Set B
C: Set C
Now, let's break down the equation step by step:
∣A∪B∪C∣: The cardinality of the union of sets A, B, and C represents the total number of elements in all three sets combined.
∣A∣ + ∣B∣ + ∣C∣: This term represents the sum of the individual cardinalities of sets A, B, and C.
∣A∩B∣ - ∣A∩C∣ - ∣B∩C∣: These terms account for the double counting of elements that exist in the intersections of two sets. By subtracting the cardinalities of the intersections, we remove the duplicates.
∣A∩B∩C∣: This term adds back the cardinality of the intersection of all three sets to correct for triple counting. It ensures that the elements that belong to all three sets are included once.
By combining these steps, we obtain:
∣A∪B∪C∣ = ∣A∣ + ∣B∣ + ∣C∣ - ∣A∩B∣ - ∣A∩C∣ - ∣B∩C∣ + ∣A∩B∩C∣
This equation represents the principle of inclusion-exclusion, which provides a way to calculate the cardinality of the union of multiple sets while considering their intersections.
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gabriella went skiing. she paid $35 to rent skis and $15 an hour to ski. if she paid a total of $95, how many hours did she ski?
Gabriella skied for 6 hours, Let x be the number of hours that Gabriella skied. We know that she paid $35 for ski rental and $15 per hour for skiing,
for a total of $95. We can set up the following equation to represent this information:
35 + 15x = 95
Solving for x, we get:
15x = 60
x = 4
Therefore, Gabriella skied for 6 hours.
Here is a more detailed explanation of how to solve the equation:
Subtract $35 from both sides of the equation.
15x = 60
15x - 35 = 60 - 35
15x = 25
Divide both sides of the equation by 15.
15x = 25
x = 25 / 15
x = 4
Therefore, x is equal to 4, which is the number of hours that Gabriella skied.
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Which of the following expressions are well- defined for all vectors a, b, c, and d?
I a (bxc),
II (a b) x (cd),
III ax (bx c).
1. I and III only
2. all of them
3. II and III only
4. II only
5. I only
6. I and II only
7. III only
8. none of them
The expressions that are well-defined for the vectors are:
2. all of them.
What is a Well-defined Vector Expression?Let's assess the well-definedness of the given expressions for vectors a, b, c, and d:
I. a x (b x c): This expression represents the cross product of vectors b and c, followed by the cross product with vector a. Since the cross product operation is valid for all vectors, expression I is well-defined.
II. (a x b) x (c x d): This expression computes the cross product of (a x b) and (c x d). Since the cross product is defined for all vectors, expression II is valid and well-defined.
III. a x (b x c): Expression III involves the cross product of vectors b and c, followed by the cross product with vector a. As mentioned earlier, the cross product is applicable to all vectors, ensuring the well-definedness of expression III.
Thus, the correct answer is: 2. all of them.
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A company gives each worker a cash bonus every Friday, randomly giving a worker an amount with these probabilities: $10−0.75,$50−0.25. Over many weeks, what is a worker's expected weekly bonus? (A).10×0.75+50×0.25=$20 (B) 10×0.25+50×0.75=$40 (C) (10+50)×(0.75×0.25)=$11.5 (D)Cannot say as the number of weeks is not provided.
The option (A) 10×0.75+50×0.25=$20 is the correct answer. A company gives each worker a cash bonus every Friday, randomly giving a worker an amount with these probabilities:
$10−0.75,$50−0.25
We need to find a worker's expected weekly bonus.
What is the expected value of a random variable?
The expected value of a random variable is a measure of the central position of the distribution of values of that variable. It is the long-run average value of repetitions of the experiment it represents. It is also known as the mathematical expectation, the average or the mean.
What is the expected weekly bonus of the worker?
Let X be the amount of cash given as a bonus. From the given probabilities, the expected value of X is $10 with probability 0.75, and $50 with probability 0.25
So, the expected weekly bonus is:
10 × 0.75 + 50 × 0.25= 7.5 + 12.5= $20
Therefore, the option (A) 10×0.75+50×0.25=$20 is the correct answer.
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Describe the motion of a particle with position P(x, y) when x = 4 sin t, y = 5 cost as t varies in the interval 0 le t le 2pi. Moves once counterclockwise along the ellipse (4x)2 + (5y)2 = 1, starting and ending at (0, 5). Moves once clockwise along the ellipse x2 / 16 + y2 / 25 = 1, starting and ending at (0,5) Moves along the line x / 4 + y / 5 = 1, starting at (4, 0) and ending at (0, 5). Moves once clockwise along the ellipse x 2 / 16 + y 2 / 25 = 1, starting and ending (0. 5). Moves along the line x / 4 + y / 5 = 1, starting at (0,5) and ending at (4,0). Moves once clockwise along the ellipse (4x)2 + (5y)2 = 1, starting and ending at (0, 5).
The particle moves along an elliptical path, counterclockwise, starting and ending at the point (0, 5).
The given position of the particle is defined by the equations x = 4 sin(t) and y = 5 cos(t), where t varies in the interval 0 ≤ t ≤ 2π. By substituting these equations into the equation of an ellipse, we can determine the path of the particle.
The first scenario describes the motion of the particle as it moves once counterclockwise along the ellipse (4x)² + (5y)² = 1, with the starting and ending point at (0, 5). Since the equation of the ellipse matches the equation for the position of the particle, we can conclude that the particle moves along the ellipse.
The values of x and y are determined by the given equations, which are variations of sine and cosine functions. As t varies from 0 to 2π, the particle completes one full revolution along the ellipse, moving counterclockwise.
The second scenario describes the motion of the particle as it moves once clockwise along the ellipse x²/16 + y²/25 = 1, starting and ending at (0, 5). This time, the particle moves in the opposite direction, clockwise.
The third scenario involves the particle moving along a line, specifically the line x/4 + y/5 = 1. The particle starts at the point (4, 0) and ends at the point (0, 5). This motion represents a linear path from one point to another, as opposed to the circular paths described in the previous scenarios.
The given equations describe the motion of a particle along different paths: counterclockwise and clockwise along elliptical paths, and a linear path along a line. These paths are determined by the values of t and the corresponding equations for x and y.
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One indicator that a pediatrician will use to diagnose an infant with "failure to thrive is a weight below the 5th percentile for the baby's age Explain what this mean
Choose the correct answer below
A. Aweight below the 5th percentile means that the baby's weight is less than 5% of the mean weight for all babies of that age
B. A weight below the 5th percentile means that the baby's weight is less than 95% of the mean weight for all babies of that age
C. A weight below the 5th percentile means that the baby's weight is lower than at least 95% of babies that age
D. A weight below the 5th percentile means that the baby's weight is lower than at least 5% of babies that age
Option A is correct. A weight below the 5th percentile indicates that the baby's weight is less than 5% of the average weight for infants of the same age.
Percentiles are statistical measures used to compare an individual's measurements to a larger population. In this case, the weight percentile indicates how a baby's weight compares to other babies of the same age. The 5th percentile means that the baby's weight is lower than 95% of babies in that age group. Option A correctly states that a weight below the 5th percentile means the baby's weight is less than 5% of the mean weight, indicating that the baby's weight is significantly lower than the average weight for infants of that age.
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solve the initial-value problem: 4y ′′ − y = xex/2 , y(0) = 1, y′ (0) = 0.
The characteristic equation corresponding to the homogeneous equation is \(4r^2 - 1 = 0\). The quadratic equation two distinct roots: \(r_1 = \frac{1}{2}\) and \(r_2 = -\frac{1}{2}\).
To solve the initial-value problem, we will first find the general solution to the homogeneous equation \(4y'' - y = 0\) and then find a particular solution to the non-homogeneous equation \(4y'' - y = xe^{x/2}\). By combining the general solution with the particular solution, we can obtain the solution to the initial-value problem.
1. Homogeneous Equation:
The characteristic equation corresponding to the homogeneous equation is \(4r^2 - 1 = 0\). Solving this quadratic equation, we find two distinct roots: \(r_1 = \frac{1}{2}\) and \(r_2 = -\frac{1}{2}\).
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1. Which one of the following is a unit of length?
a) cm b) degree Celsius c) candela d) year 2. How many significant figures can you find in the number "6.780 × 10³"? a) 1 b) 2 c) 3 d) 4 3. In two-dimensional coordinate system, the vector which points from (-10,-10) to (-4,-6) is a) (-6,-4) b) (6,-4) c) (-6,4) d) (6,4)
1. The unit of length a) cm.
2.Significant figures you find in the number "6.780 × 10³ is d) 4.
3. In two-dimensional coordinate system, the vector which points from (-10,-10) to (-4,-6) is d) (6, 4).
1) A "cm" stands for centimeter, which is a unit of length commonly used to measure small distances.
2) The number "6.780 × 10³" is written in scientific notation. In scientific notation, a number is expressed as a product of a decimal number between 1 and 10 and a power of 10. The significant figures in a number are the digits that carry meaning or contribute to its precision. In this case, there are four significant figures: 6, 7, 8, and 0. Therefore, the answer to the second question is d) 4.
3) To determine the vector that points from (-10, -10) to (-4, -6), you subtract the corresponding coordinates of the starting and ending points. In this case, the x-coordinate changes from -10 to -4, and the y-coordinate changes from -10 to -6. Therefore, the x-component of the vector is -4 - (-10) = 6, and the y-component is -6 - (-10) = 4. So, the vector is (6, 4). Thus, the answer to the third question is d) (6, 4).
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A particle moves along line segments from the origin to the points (1,0,0),(1,5,1),(0,5,1), and back to the origin under the influence of the force field F(x,y,z)=z
2
i+4xyj+4y
2
k. Find the work done. ∮
C
F⋅dr=
The total work done by the force field F along the path C is 201.
To determine the work done by the force field F along the given path C, we need to evaluate the line integral ∮CF⋅dr.
Let's break down the path C into its individual line segments:
1. From the origin (0, 0, 0) to (1, 0, 0)
2. From (1, 0, 0) to (1, 5, 1)
3. From (1, 5, 1) to (0, 5, 1)
4. From (0, 5, 1) back to the origin (0, 0, 0)
Now, we can calculate the line integral along each segment and sum them up to find the total work done.
1. Along the first line segment, the vector dr = dx i, and the force field F = z² i + 4xy j + 4y²k.
Integrating F⋅dr from 0 to 1 with respect to x gives us ∫[0,1] (z² dx) = ∫[0,1] (0² dx) = 0.
2. Along the second line segment, the vector dr = dy j, and the force field F = z² i + 4xy j + 4y²k.
Integrating F⋅dr from 0 to 5 with respect to y gives us ∫[0,5] (4xy dy) = ∫[0,5] (4xy dy) = 100.
3. Along the third line segment, the vector dr = -dx i, and the force field F = z²i + 4xy j + 4y²k.
Integrating F⋅dr from 1 to 0 with respect to x gives us ∫[1,0] (z² dx) = ∫[1,0] (1^2 dx) = 1.
4. Along the fourth line segment, the vector dr = -dy j, and the force field F = z² i + 4xy j + 4y² k.
Integrating F⋅dr from 5 to 0 with respect to y gives us ∫[5,0] (4xy dy) = ∫[5,0] (4xy dy) = 100.
Adding up the work done along each segment, we have 0 + 100 + 1 + 100 = 201.
Therefore, the total work done by the force field F along the given path C is 201.
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evaluate the polynomial for the given value by using synthetic division. p(x) = x4 − x2 7x 5 for x = −1 and x = 2
The remainder obtained from synthetic division is -11. p(2) = -11. p(-1) = 1 and p(2) = -11.
To evaluate the polynomial using synthetic division, we will divide the polynomial by each given value and observe the remainder.
1. For x = -1:
To evaluate p(x) = x^4 - x^2 - 7x - 5 at x = -1, we perform synthetic division as follows:
-1 │ 1 0 -7 -5
│ -1 1 6
└───────────────
1 -1 -6 1
The remainder obtained from synthetic division is 1. Therefore, p(-1) = 1.
2. For x = 2:
To evaluate p(x) = x^4 - x^2 - 7x - 5 at x = 2, we perform synthetic division as follows:
2 │ 1 0 -7 -5
│ 2 4 -6
└──────────────
1 2 -3 -11
The value obtained from synthetic division is -11. Therefore,
p(2) = -11.
Hence, p(-1) = 1 and p(2) = -11.
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Use the given information to find the exact value of the expression.
sin α =15/17 , α lies in quadrant I, and cos β = 4/5 , β lies in quadrant IFind cos (α + β).
The exact value of the expression is:
cos (α+β) = -13/85
How to find the exact value of the trigonometric expression?Trigonometry deals with the relationship between the ratios of the sides of a right-angled triangle with its angles.
We have:
sin α =15/17, α lies in quadrant I
cos β = 4/5 , β lies in quadrant I
Sine all trigonometric expression in quadrant I are positive. We have:
sin α =15/17 (opposite/hypotenuse)
adjacent = √(17² - 15²) = 8
Thus, cos α = 8/17
cos β = 4/5 (adjacent/hypotenuse)
opposite = √(5² - 4²) = 3
Thus, sin β = 3/5
Using trig. identity, we know that:
cos (α+β) = cosα cosβ − sinα sinβ
cos (α+β) = (8/17 * 4/5) - (15/17 * 3/5)
cos (α+β) = 32/85 - 45/85
cos (α+β) = -13/85
Therefore, the exact value of the expression is -13/85
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Approximate the area under the curve y=x2 from x=1 to x=4 using a Right Endpoint approximation with 6 subdivisions. (a) Approximate area = (b) Identify the following items of information that were necessary to calculate that approximate area: Δx=
f(1.5)=
f(2)=
1(2.5)=
f(3)=
f(3.5)=
f(4)=
a) The approximate area under the curve is 24.875.
b) The value of Δx = 1/2, f(1.5) = 2.25, f(2) = 4, f(2.5) = 6.25, f(3) = 9, f(3.5) = 12.25 and f(4) = 16.
To approximate the area under the curve y = x^2 from x = 1 to x = 4 using a Right Endpoint approximation with 6 subdivisions, we need to divide the interval [1, 4] into 6 equal subintervals and calculate the area of each rectangle formed by the right endpoints.
(a) Approximate area:
Let's first calculate the width of each subdivision:
Δx = (b - a) / n
= (4 - 1) / 6
= 3 / 6
= 1/2
The width of each subdivision is 1/2.
Now, let's calculate the area of each rectangle and sum them up:
Approximate area = Δx * (f(1.5) + f(2) + f(2.5) + f(3) + f(3.5) + f(4))
We need to evaluate the function at the right endpoints of each subdivision:
f(1.5) = [tex](1.5)^2[/tex] = 2.25
f(2) = [tex](2)^2[/tex] = 4
f(2.5) = [tex](2.5)^2[/tex] = 6.25
f(3) = [tex](3)^2[/tex] = 9
f(3.5) = [tex](3.5)^2[/tex] = 12.25
f(4) = [tex](4)^2[/tex] = 16
Now, substitute these values into the equation:
Approximate area = (1/2) * (2.25 + 4 + 6.25 + 9 + 12.25 + 16)
= (1/2) * 49.75
= 24.875
Therefore, the approximate area under the curve y = [tex]x^2[/tex] from x = 1 to x = 4 using a Right Endpoint approximation with 6 subdivisions is 24.875.
(b) Information needed to calculate the approximate area:
The following items of information were necessary to calculate the approximate area using the Right Endpoint approximation:
Δx = 1/2
f(1.5) = 2.25
f(2) = 4
f(2.5) = 6.25
f(3) = 9
f(3.5) = 12.25
f(4) = 16
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Using Principles of Mathematical Induction prove that 1+3+5+…(2n−1)=n2for all integers, n≥1. or the toolbar, press ALT +F10 (PC) or ALT+FN+F10 (Mac). Moving to another question will save this response.
Using the principle of mathematical induction, we can prove that [tex]1 + 3 + 5 + ... + (2n-1) = n^2[/tex] for all integers n ≥ 1. By following the steps of mathematical induction and verifying the base case and inductive step, we establish the validity of the statement.
To prove this statement, we will follow the steps of mathematical induction.
Step 1: Base case
For n = 1, the left-hand side (LHS) is 1, and the right-hand side (RHS) is [tex]1^2 = 1[/tex]. Therefore, the statement holds true for n = 1.
Step 2: Inductive hypothesis
Assume that the statement holds true for some positive integer k, i.e., [tex]1 + 3 + 5 + ... + (2k-1) = k^2[/tex].
Step 3: Inductive step
We need to show that the statement holds true for k + 1.
Considering the sum 1 + 3 + 5 + ... + (2k-1) + (2(k+1)-1), we can rewrite it as [tex](k^2) + (2k+1) = k^2 + 2k + 1 = (k+1)^2[/tex].
This shows that if the statement holds true for k, it also holds true for k + 1.
Step 4: Conclusion
By the principle of mathematical induction, we can conclude that [tex]1 + 3 + 5 + ... + (2n-1) = n^2[/tex] for all integers n ≥ 1.
Hence, we have proved the given statement using the principle of mathematical induction.
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Use the binomial series expansion, (1+x)r=∑n=0[infinity](rn)xn for the function f(x)=(1−2x)21 to find a third-order Maclaurin polynomial, p3(x), in order to estimate (31)21. If necessary, round your answer to three decimal places. Provide your answer below: (31)21≈
By the estimation of 3^(1/21) using the MacLaurin polynomial of the third order, we get -0.248.
First, to find the required polynomial, we use the binomial series expansion for three terms.
The general form of a binomial series expansion is given as:
(1 + x)ᵃ = ∑(n = 0 to n = inf) (ᵃCₙ) * xⁿ
where (ᵃCₙ) for each term is called its binomial coefficient.
We need only three terms expansion for the given polynomial (1−2x)²¹.
Here a = 21, and instead of x we have -2x.
Also, n = 0,1,2 and 3 for three terms.
So,
(1−2x)²¹ = (²¹C₀)*(-2x)⁰ + (²¹C₁)*(-2x)¹ + (²¹C₂)*(-2x)² + (²¹C₃)*(-2x)³
= 1 - 42x + 840x² -10640x³
For estimating we can substitute x = 1/21 into the above expansion.
which gives us
1 - 42(1/21) + 840(1/21)² -10640(1/21)³
= -0.248
Thus the approximation of the given polynomial is -0.248.
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