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Makovka662 [10]
3 years ago
8

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

ot show that Rn(x) → 0.] Find the associated radius of convergence R. f(x) = cos(3x)/5
Physics
1 answer:
jenyasd209 [6]3 years ago
3 0

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

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man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
Explain why dogs pant during hot summer days using the evaporation concept?
damaskus [11]
All, or almost all, warm-blooded creatures get rid of excess heat by evaporating moisture from their bodies. It's a great system, because evaporation takes a lot of heat. That's the reason people perspire when we're active and build up a lot of heat inside. The evaporation of sweat from our skin carries away heat with it. Dogs do not sweat on their skin. The only place they can evaporate moisture is through their mouth. Panting speeds up the evaporation by blowing air across the moisture.
5 0
3 years ago
⚠️ PLEASE HELP ⚠️
Svetllana [295]

Answer:

-79.6

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-81.2

Explanation:

5 0
3 years ago
Read 2 more answers
An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec
Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0
3 years ago
A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.
victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A)  = N₀(B)

N = N₀ e^{-\lambda t}

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ e^{-\lambda_1 t}

For B

N(B) = N(B)₀ e^{-\lambda_2 t}

N(A) = N(B) , for t = 2 h

N(A)₀ e^{-\lambda_1 t} = N(B)₀ e^{-\lambda_2 t}

N(A)₀ e^{-\lambda_1 t} = 5 x N₀(A)  e^{-\lambda_2 t}

e^{-\lambda_1 t} = 5  e^{-\lambda_2 t}

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

half life = .693 / λ

For A

.77 =  .693 / λ₁

λ₁ = .9 h⁻¹

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

Putting t = 2 h , λ₁ = .9 h⁻¹

e^{\lambda_2\times  2} = 5  e^{.9\times  2}

e^{\lambda_2\times  2} = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0
3 years ago
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