Mathematics, 27.08.2019 23:30 kaylaw332

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the maclaurin series for f.

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$f(x)=\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Note that, by the ratio test, this converges for

$\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|$

Integrating once, you have

$\displaystyle\int f(x)\,\mathrm dx=\int\sum_{n=0}^\infty(n+1)x^n\,\mathrm dx=\sum_{n=0}^\infty(n+1)\int x^n\,\mathrm dx$
$\displaystyle\int f(x)\,\mathrm dx=\sum_{n=0}^\infty\frac{(n+1)x^{n+1}}{n+1}+C=\sum_{n=0}^\infty x^{n+1}+C$

Given that f(0)=0

, it's clear that C=0

, and so

$\displaystyle\int f(x)\,\mathrm dx=\displaystyle\sum_{n=0}^\infty x^{n+1}=x\sum_{n=0}^\infty x^n=\frac x{1-x}$

Differentiating both sides recovers f(x)

:

$f(x)=\dfrac{(1-x)(1)-x(-1)}{(1-x)^2}=\dfrac1{(1-x)^2}$

Answer from: Quest

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

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