Mathematics: Jake has proved that a function, f(x), is a geometric sequence. How did he prove that? He showed that an explicit formula could be created. He showed that a recursive formula could be created. He showed that f(n) ÷ f(n − 1) was a constant ratio. He showed that f(n) − f(n − 1) was a constant difference.

You re looking for a solution in the formDifferentiating, we getSubstitute these for y and y in the differential equation:Then the coefficients of y are given by the recurrenceorBut we cannot assume that and depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so thatSo in the power series solution, we split off the constant term and we re left withso that the fundamental solutions areand...

Jake has proved that a function, f(x), is a geometric

Ответ:

royalty67

20.10.2020

Mathematics

The characteristic of a geometric progression is that the ratio of a term and its consecutive term remains the same. Thus, if he can show that f(n)/f(n-1) remains constant, he can prove f(x) to be a geometric progression.

The third option is correct.

Hope this helps :)

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Ответ:

averycipher

17.08.2020

Mathematics

You're looking for a solution in the form

$y(x) = \displaystyle \sum_{n=0}^\infty a_nx^n$

Differentiating, we get

$y'(x) = \displaystyle \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1}x^n$

$y''(x) = \displaystyle \sum_{n=0}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=1}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n$

Substitute these for y' and y'' in the differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^\infty (n+1)a_{n+1}x^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1)a_{n+2}-(n+1)a_{n+1}\bigg)x^n = 0$

Then the coefficients of y are given by the recurrence

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\frac{a_{n+1}}{n+2}&\text{for }n\ge0\end{cases}$

$a_n = \dfrac{a_{n-1}}n$

But we cannot assume that a_0 and a_1 depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so that

$a_2=\dfrac{a_1}2 \\\\ a_3=\dfrac{a_2}3=\dfrac{a_1}{3\times2} \\\\ a_4=\dfrac{a_3}4=\dfrac{a_1}{4\times3\times2} \\\\ \vdots \\\\ a_n=\dfrac{a_1}{n!}$

So in the power series solution, we split off the constant term and we're left with

$y(x) = a_0 + a_1 \displaystyle \sum_{n=1}^\infty \frac{x^n}{n!}$

so that the fundamental solutions are

y_1=1

and

$y_2=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots$

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