PLEASE HELP ME? 10 POINTS??? HELP ME ASAP

Mathematics

1 answer:

tankabanditka [31]3 years ago

8 0

<h3>Third answer</h3>

$p\%=\dfrac{p}{100}\\\\4\%=\dfrac{4}{100}=0.04\\\\\text{Therefore the equation is:}\\\\\boxed{y=3000(1+0.04)^n}$

Why?

4% per year.

100% + 4% = 1 + 0.04

after the first year

$y=3000(1+0.04)$

after the second year

$y=3000(1+0.04)(1+0.04)=3000(1+0.04)^2$

after the third year

$y=3000(1+0.04)^2(1+0.04)=3000(1+0.04)^3$

. . .

after the nth year

$y=3000(1+0.04)^n$

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Find the Maclaurin series for the f(x)=xe^2x and find<br> itsradius of convergence.

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Answer:

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Maclaurin series is a special form of Taylor series where a = 0,

$f(x)=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)(x-a)^n}{n!}=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)x^n}{n!}$

Hence,

$f\left(x\right)=f\left(0\right)+\frac{f^{'}\left(0\right)}{1!}\left(x\right)+\frac{f^{''}\left(0\right)}{2!}\left(x\right)^2+\frac{f^{'''}\left(0\right)}{3!}\left(x\right)^3+\ldots$

For $f(x)=xe^{2x}$ ,

$f\left(x\right)=0\cdot \:e^{2\cdot \:0}+\frac{\frac{d}{dx}\left(xe^{2x}\right)\left(0\right)}{1!}x+\frac{\frac{d^2}{dx^2}\left(xe^{2x}\right)\left(0\right)}{2!}x^2+\frac{\frac{d^3}{dx^3}\left(xe^{2x}\right)\left(0\right)}{3!}x^3+\ldots$

Where,

$0\cdot \:e^{2\cdot \:0}\quad :\quad 0\\\\\frac{d}{dx}\left(xe^{2x}\right)\left(0\right)\quad :\quad 1\\\\\frac{d^2}{dx^2}\left(xe^{2x}\right)\left(0\right)\quad :\quad 4\\\\\frac{d^3}{dx^3}\left(xe^{2x}\right)\left(0\right)\quad :\quad 12\\\\\frac{d^4}{dx^4}\left(xe^{2x}\right)\left(0\right)\quad :\quad 32\\\\\frac{d^5}{dx^5}\left(xe^{2x}\right)\left(0\right)\quad :\quad 80$

Thus,

$f\left(x\right)=0+\frac{1}{1!}x+\frac{4}{2!}x^2+\frac{12}{3!}x^3+\frac{32}{4!}x^4+\frac{80}{5!}x^5+\ldots=\\\\=x+2x^2+2x^3+\frac{4}{3}x^4+\frac{2}{3}x^5+\ldots$