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

jenyasd209 [6]

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) =

3 0

3 years ago

A car weighs 1323 N. what is its mass?

Shalnov [3]

If the car is on the moon, its mass is about 817 kg.

If it's on the Earth, its mass is about 135 kg.

6 0

3 years ago

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A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-

Alisiya [41]

First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
$1 rev= 2 \pi rad$
$1 min = 60 s$
the angular speed is
$\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s$

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
$v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s$

3 0

3 years ago

HEY CAN ANYONE PLS HELP ME OUT IN DIS PLS

Zanzabum

Answer:

Its the second choice

Explanation:

4 0

3 years ago

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Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]