Gouilleul Sequence.

Mathematics

1 answer:

harkovskaia [24]3 years ago

6 0

Answer:

Step-by-step explanation:

No idea what a gouilleul Sequence is, but hopefully it will help to show how I would find the answer.

A geometric sequence means a term is multiplied by something over and over So say we start with 100 and the common ratio is .5 then the sequence would go 100,50,25,12.5,6.25 and so on. Another way of writing it is:

100*.5^0 = 100

100*5^1 = 50

100*.5^2 = 100*.25 = 25

and so on.

ANyway, here we have the first and fourth term. so that's 3.5*x^0 and 3.5*x^3. Keep in mind the exponent will be n-1 when the term you want to find is the nth term. Anyway, we know 3.5*x^3 = 28, so we just use algebra to find x, which is the common ratio.

3.5*x^3 = 28

x^3 = 8

x = 2

So the common ratio is 2, or in other words it keeps doubling. You could try it real quick, keep doubling the starting term 3.5

3.5, 7, 14, 28 and 28 is the fourth term.

So we know the common ratio. you could just add the four terms listed then for part b, but there is an equation, for in case they ask you to find the sum of say the first 100 terms. that wouldn't be fun. Anyway, the formula is a((1-r^n)/(1-r)) where a is the starting term, r is the common ratio and n is the number term. In your case that's 3.5((1-2^4)/(1-2)) = 52.5. You can actually check since it's only 4 terms. 3.5 + 7 + 14 + 28 = 52.5.

If you want to know how that formula was gotten let em know and I can explain or give a link to a video or something to show.

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Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

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In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$