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[Axelato]

Hey! I got a problem for my homework that i cant solve.
I want to find an for the problem below. The word "där" means "where an is" in my language.


I am happy for all the help i can get. Thx.

 

[Steven G]

I am lost by your question. You say that you want to find a_n, but a_n is given. Can you please try again explaining what you need help with?

 

[Dr.Peterson]

Hey! I got a problem for my homework that i cant solve.
I want to find a_n for the problem below. The word "där" means "where a_n is" in my language.
View attachment 32044

I am happy for all the help i can get. Thx.

I presume what you want to find is the sum, presumably as an explicitly defined function of x.

Can you tell us what you have been learning, including what particular methods you have seen?

 

[JeffM]

First of all, the definition of $a_n$ is slightly ambiguous.

$$a_n = \dfrac{\dfrac{1}{2} * \displaystyle \prod_{j=1}^n \left ( \dfrac{1}{2} - j \right )}{n! * \left ( \dfrac{1}{2} - n \right )}$$
That does not make any sense for $a_0$ because j = 1 starts j off above n. So I suspect what is meant is

$$a_n = \dfrac{\displaystyle \left (\prod_{j=0}^n \dfrac{1}{2} - j \right )}{n! * \left ( \dfrac{1}{2} - n \right )}.$$
Let's compute some values.

$$a_0 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )}{0! * \left ( \dfrac{1}{2} - 0 \right )} = 1.$$
$$a_1 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )}{1! * \left ( \dfrac{1}{2} - 1 \right )} = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}$$
$$a_2 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )\left ( \dfrac{1}{2} - 2 \right )}{2! * \left ( \dfrac{1}{2} - 2 \right )} =\dfrac{\left ( \dfrac{1}{2}\right )\left ( - \dfrac{1}{2} \right)}{2} = - \dfrac{1}{8}$$
$$a_3 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )\left ( \dfrac{1}{2} - 2 \right )\left ( \dfrac{1}{2} - 3 \right ) }{3! * \left ( \dfrac{1}{2} - 3 \right )} =\\ \dfrac{\left ( \dfrac{1}{2}\right )\left ( - \dfrac{1}{2} \right)\left ( - \dfrac{3}{2} \right)}{3 * 2} = \dfrac{1}{16}$$
This looks like a pattern to me. If n is an integer $\ge$ 0, it appears that $a_{n+1} = \text {minus } \dfrac{(2n - 1)a_n}{2n + 2}.$ Can we prove it?

$$\text {We calculated } a_0 = 1 \text { and } a_1 = \dfrac{1}{2} \text { above}.\\ \therefore a_1 = \dfrac{1}{2} = \dfrac{1}{2} * 1 = \dfrac{1 }{2} * a_0 = \dfrac{(1 - 2 * 0)a_0}{2 * 0 + 2} \implies\\ a_1 = \text {minus } \dfrac{(2 * 0 - 1)a_0}{2 * 0 + 2}.$$
Pattern if n = 1. Now consider any integer k $\ge$ 0 such that

$$a_{k+1} = \text {minus } * \dfrac{(2k - 1)a_k}{2k + 2}.\\ \text {By definition, } a_{(k+1)+1} = \dfrac{\displaystyle \left ( \prod_{j=0}^{(k+1)+1} \dfrac{1}{2} - j \right ) }{\{(k+1) + 1\}! * \left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )} =\\ \dfrac{\displaystyle \left ( \prod_{j=0}^{k+1} \dfrac{1}{2} - j \right ) * \cancel{\left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )} }{\{(k+1) + 1\}! * \cancel{\left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )}} =\\ \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right ) * \left \{ \dfrac{1}{2} - (k + 1) \right \}}{\{(k+1) + 1\}(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}} =\\ \dfrac{\dfrac{1}{2} - (k + 1)}{(k+1) + 1} * \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right )}{(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}}.\\ \text {But, by definition, } \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right )}{(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}} = a_{k+1}.\\ \therefore a_{(k+1)+1} = \dfrac{\dfrac{1}{2} - (k + 1)}{(k+1) + 1} * a_{k+1} = \dfrac{\{1 - 2(k + 1)\}a_{k+1}}{2(k+1) + 2} \implies\\ a_{(k+1)+1} = \text {minus } \dfrac{\{2(k + 1) - 1\}a_{k+1}}{2(k+1) + 2}.$$
So we can eliminate the product elements and just focus on the sum. Moreover, we have telescoping possibilities in the sum.