Can somene tell if I have started this right please.

  • Thread starter ptex
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  • #1
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I need to find the explicit formuala for this recursive sequence.
Code:
a[sub]1[/sub] = 3
a[sub]2[/sub] = 7
a[sub]n[/sub] = 7a[sub]n-1[/sub] - 10a[sub]n-2[/sub]
a[sub]n[/sub] - 7a[sub]n-1[/sub] + 10a[sub]n-2[/sub] = 0
t[sup]n[/sup] - 7k[sup]n-1[/sup] + 10t[sup]n-2[/sup] = 0
t[sup]2[/sup] - 7t + 10 = 0
(t - 2)(t - 5)
(t = 2)(t = 5) :confused: 
if that is right I can move on.
 
Last edited:

Answers and Replies

  • #2
1,036
1
Move on. :smile:
 
  • #3
42
0
Thank you I will go ahead and stop if I have more questions I will stop. Ok?
 
  • #4
42
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Is this the formula?
Code:
a[sub]n[/sub] = 8/5(2)[sup]n[/sup] - 1/15(5)[sup]n[/sup]
 
  • #5
1,036
1
No. Probably just a mistake someplace in your algebra.

You know, you can check your solution by using the recurrence to determine the value of a3 and comparing that to the value generated by your formula.
 
  • #6
42
0
To find B;
Code:
2(4A + 25B = 7) = 8A + 50B = 14
4(2A + 5B =  3) = 8A + 20B = 12
30B = 2
B = 1/15
To find A;
Code:
2A + 5B = 3
2A + 1/5B = 3
10A + 1 = 15
10A = 16
A = 16/10
A = 8/5
A = 8/5 B= 1/15
?
 
  • #7
1,036
1
> 2A + 5B = 3
> 2A + 1/5B = 3 ......what is this? How did 5B become 1/5 B? You already know B. Plug it in to the first eqn & solve for A.
 
  • #8
42
0
B = 15 and A = 16?
 
  • #9
1,036
1
No.

This was correct:
2(4A + 25B = 7) = 8A + 50B = 14
4(2A + 5B = 3) = 8A + 20B = 12
30B = 2
B = 1/15
Now use this equation to find A :

2A + 5B = 3
 
Last edited:
  • #10
42
0
I think
Code:
a[sub]3[/sub] = 19
 
  • #11
1,036
1
You think?

You're supposed to know. :wink:
 
  • #12
42
0
Is A=1/2 if so
Code:
the formula would be a[sub]n[/sub] = 1/2(2[sup]n[/sup]) - 1/15(5[sup]n[/sup])
 
  • #13
1,036
1
I don't understand what is confusing you.

You have already solved the simultaneous equations:
2(4A + 25B = 7) = 8A + 50B = 14
4(2A + 5B = 3) = 8A + 20B = 12
30B = 2
B = 1/15
so you know that B=1/15.

So now take one of your original equations (either one) and solve for A. Show me what you're doing, step by step.
 
  • #14
42
0
woops
Code:
2A + 5B = 3
2A + 5(1/15) = 3
2A + 1 /3 = 3
2A = 3 1/3
A = 3 1/3 / 2
A = 5/3
 
Last edited:
  • #15
1,036
1
[tex]B =\frac{1}{15}[/tex]
 
  • #16
42
0
2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
2a = 3 1/3
A = 3 1/3 / 2
A = 5/3
 
  • #17
1,036
1
2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
2a = 3 1/3 <<<<check this again
A = 3 1/3 / 2
A = 5/3
 
  • #18
42
0
2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
2a = 10/3
A = 10/3 / 2
A = 5/3
 
  • #19
1,036
1
2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
.....- 1/3 = -1/3
2a = 8/3
A = 4/3
 
  • #20
42
0
Thank you
the 1/3 turns negitive? then ?
 
  • #21
1,036
1
The 1/3 doesn't turn negative. You subtract 1/3 from both sides of the equation. Then you divide both sides of the equation by 2. Then the equation is solved. A = 4/3.

So, you've never studied algebra and yet you are interested in solving linear homogeneous recurrence relations with constant coefficients. Very interesting.
 
  • #22
42
0
The last time I took algebra was 10 years ago. This class Discrete Mathematics is a requirerment for my degree (all I want to do is write code). Yeah I know its a long time to get a degree but I have a bit of a skiing career so I can't go to school during the winter. Anyway thank you very much now I must move on to the second problem and try to recall algebra which by the seems more defficult then some of this stuff.
 

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