errang
asked on
How to simplify a polynomial divided by a natural number?
Hey,
I had a question about simplifying a polynomial divided by a natural number... I start off with
((9^n) + 24n + 15)/48
And I have to show that this works for n+1 too... So I have...
((9^(n+1)) + 24n + 24 + 15)/48
Its the 9^(n+1) that's killing me... I've tried everything I could think of, but I couldn't get rid of it...
Appreciate any tips/pointers on this.
I had a question about simplifying a polynomial divided by a natural number... I start off with
((9^n) + 24n + 15)/48
And I have to show that this works for n+1 too... So I have...
((9^(n+1)) + 24n + 24 + 15)/48
Its the 9^(n+1) that's killing me... I've tried everything I could think of, but I couldn't get rid of it...
Appreciate any tips/pointers on this.
9^(n+1) = 9*9^n
ASKER
Yea, got that far...
I'm trying to reduce the n+ 1 polynomial to the n polynomial + some stuff.
I'm trying to reduce the n+ 1 polynomial to the n polynomial + some stuff.
do you need an n+1 after the 24 too?
BTW - ((9^n) + 24n + 15) is not a polynomial http://en.wikipedia.org/wiki/PolynomialThe 9^n term is an exponential. http://www.purplemath.com/modules/expofcns.htm
And I have to show that this works for n+1 too (my emphasis)
what is the 'this' that you are refering to?
AW
what is the 'this' that you are refering to?
AW
ASKER
I am basically trying to reduce ((9^(n+1)) + 24n + 24 + 15)/48 to ((9^n) + 24n + 15)/48 + some numbers.
"I am basically trying to reduce ((9^(n+1)) + 24n + 24 + 15)/48 to ((9^n) + 24n + 15)/48 + some numbers"
can't be done, since the term 9^(n+1) = 9*9^n and that can't be shown as 9^n plus some numbers.
((9^(n+1)) + 24(n+1) + 15)/48 = (9^n + 24n + 15)/48 + (8*9^n + 24)/48
AW
can't be done, since the term 9^(n+1) = 9*9^n and that can't be shown as 9^n plus some numbers.
((9^(n+1)) + 24(n+1) + 15)/48 = (9^n + 24n + 15)/48 + (8*9^n + 24)/48
AW
You start off with: ((9^n) + 24n + 15)/48
Where did you get this, what does it mean, and what is it for?
This isn't a very complicated expression. You could evaluate using Excel for n = 1 ...
I've done that. It is interesting that all the answers are integers. Is that what you are talking about?
Also, once n is larger than 4 or so, the exponential term dominates. So each term is approx 9x its predecessor.
Where did you get this, what does it mean, and what is it for?
This isn't a very complicated expression. You could evaluate using Excel for n = 1 ...
I've done that. It is interesting that all the answers are integers. Is that what you are talking about?
Also, once n is larger than 4 or so, the exponential term dominates. So each term is approx 9x its predecessor.
1 1.00
2 3.00
3 17.00
4 139.00
5 1233.00
6 11075.00
7 99649.00
8 896811.00
9 8071265.00
10 72641347.00
11 653772081.00
12 5883948683.00
13 52955538097.00
14 476599842819.00
15 4289398585313.00
16 38604587267755.00
17 347441285409729.00
ASKER
I'm trying to show that ((9^n) + 24n + 15) divides by 48 for all n, I know that I could just go dividing it in excel, but I need to come up with a general expression.
I assume you wish to try things out rather than getting a complete solution from us.
First define your function
f(n) = 9^n + 24n + 15 (1)
and assume that there exists an integer k such that for some particular integer n, then
f(n) = 48 k (2)
Then, as you did in your OP, compute
f(n+1) = 9^(n+1) + 24(n+1) + 15
You want to show that if (1) is true, then
f(n+1) = 48 k2 for some integer k2 (3)
What you are trying to show is that if (2) is true, then (3) is also true. This is proof by induction.
http://en.wikipedia.org/wiki/Mathematical_induction
The base case is f(1) = 48
You have already computed
f(n+1) = 9^(n+1) + 24(n+1) + 15
Tip #1:
Now, using d-glitch formula in first post: 9^(n+1) = 9*9^n
f(n+1) = 9*9^n + 24(n+1) + 15 (4)
Tip #2:
Looking at (1) and (4), you can see a common expression, 9^n
Solve for 9^n in (1)
Tip #3:
When you do a few more steps, you will find a number of terms each having a factor of 48 in them. You can factor this 48 out and come up with the final desired result.
First define your function
f(n) = 9^n + 24n + 15 (1)
and assume that there exists an integer k such that for some particular integer n, then
f(n) = 48 k (2)
Then, as you did in your OP, compute
f(n+1) = 9^(n+1) + 24(n+1) + 15
You want to show that if (1) is true, then
f(n+1) = 48 k2 for some integer k2 (3)
What you are trying to show is that if (2) is true, then (3) is also true. This is proof by induction.
http://en.wikipedia.org/wiki/Mathematical_induction
The base case is f(1) = 48
You have already computed
f(n+1) = 9^(n+1) + 24(n+1) + 15
Tip #1:
Now, using d-glitch formula in first post: 9^(n+1) = 9*9^n
f(n+1) = 9*9^n + 24(n+1) + 15 (4)
Tip #2:
Looking at (1) and (4), you can see a common expression, 9^n
Solve for 9^n in (1)
Tip #3:
When you do a few more steps, you will find a number of terms each having a factor of 48 in them. You can factor this 48 out and come up with the final desired result.
"I'm trying to show that ((9^n) + 24n + 15) divides by 48 for all n,
Of course it does.
But do you want the result to be an integer (or something else definite)?
Of course it does.
But do you want the result to be an integer (or something else definite)?
"((9^n) + 24n + 15)/48
And I have to show that this works for n+1 too... So I have...
((9^(n+1)) + 24n + 24 + 15)/48"
--
If you replace the first n by n+1 you must replace the second n by n+1 too.
And I have to show that this works for n+1 too... So I have...
((9^(n+1)) + 24n + 24 + 15)/48"
--
If you replace the first n by n+1 you must replace the second n by n+1 too.
ASKER
>>But do you want the result to be an integer (or something else definite)?
yea, I need the answer to be an Integer, I'm trying out the tips phoffric mentioned right now...
one thing though, I do realize 9^(n+1) = 9^n * 9.
I haven't tried f(n+1) = 48 k2 for some integer k2 yet tho.
yea, I need the answer to be an Integer, I'm trying out the tips phoffric mentioned right now...
one thing though, I do realize 9^(n+1) = 9^n * 9.
I haven't tried f(n+1) = 48 k2 for some integer k2 yet tho.
To emphasize agreement with your OP:
f(n+1) = 9^(n+1) + 24(n+1) + 15
= 9^(n+1) + 24n + 24 +15
You want to show that there exists an Integer, k2, such that
f(n+1) = 48 k2
f(n+1) = 9^(n+1) + 24(n+1) + 15
= 9^(n+1) + 24n + 24 +15
You want to show that there exists an Integer, k2, such that
f(n+1) = 48 k2
ASKER
>>((9^(n+1)) + 24n + 24 + 15)/48"
--
If you replace the first n by n+1 you must replace the second n by n+1 too.
yes, I know that, as I wrote in my original question:
>>And I have to show that this works for n+1 too... So I have...
((9^(n+1)) + 24n + 24 + 15)/48
--
If you replace the first n by n+1 you must replace the second n by n+1 too.
yes, I know that, as I wrote in my original question:
>>And I have to show that this works for n+1 too... So I have...
((9^(n+1)) + 24n + 24 + 15)/48
You know that if an integer A is divisible by 48, then you can write
A/48 = m, an integer (i.e., no remainder)
So, obviously A = 48 m
If you work with the f(n) = 9^n + 24n + 15, and not divide by 48, you may find it easier (although either way should work).
A/48 = m, an integer (i.e., no remainder)
So, obviously A = 48 m
If you work with the f(n) = 9^n + 24n + 15, and not divide by 48, you may find it easier (although either way should work).
ultimately, it comes down to demonstrating that 8*9^n + 24 is always divisible by 48
8*9^n + 24 = 8*(9^n +3) so, is 9^n+3 always divisble by 6
that reduces to 8*9^n being divisible by 6 which is true (8*9^n = 4*2*(3^(2n))= 4*6*3^(2n-1) which IS divible by 6)
That proves the original hypothesis: (9^n+24n+15) is always divisible by 48
AW
8*9^n + 24 = 8*(9^n +3) so, is 9^n+3 always divisble by 6
that reduces to 8*9^n being divisible by 6 which is true (8*9^n = 4*2*(3^(2n))= 4*6*3^(2n-1) which IS divible by 6)
That proves the original hypothesis: (9^n+24n+15) is always divisible by 48
AW
ASKER
>>ultimately, it comes down to demonstrating that 8*9^n + 24 is always divisible by 48
Sorry... bit confused how you got 8*9^n + 24
Sorry... bit confused how you got 8*9^n + 24
SOLUTION
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from my earlier post:
((9^(n+1)) + 24(n+1) + 15)/48 = (9^n + 24n + 15)/48 + (8*9^n + 24)/48
so if (9^n + 24n + 15) is divisible by 48, and (8*9^n + 24) is also divisible by 48, then it follows that ((9^(n+1)) + 24(n+1) + 15) will be divisible by 48, which proves the original assertion.
AW
((9^(n+1)) + 24(n+1) + 15)/48 = (9^n + 24n + 15)/48 + (8*9^n + 24)/48
so if (9^n + 24n + 15) is divisible by 48, and (8*9^n + 24) is also divisible by 48, then it follows that ((9^(n+1)) + 24(n+1) + 15) will be divisible by 48, which proves the original assertion.
AW
Multiple ways to prove your problem.
To see the 8 * 9^n, you can compute f(n+1) - f(n) and watch most of terms cancel out.
>> (8*9^n + 24) is also divisible by 48
This shows you that it would be useful for you to work out both approaches if you want to obtain more in-depth learning. Each approach has nice things to learn.
See if you can prove (without induction) that (8*9^n + 24) is also divisible by 48.
To see the 8 * 9^n, you can compute f(n+1) - f(n) and watch most of terms cancel out.
>> (8*9^n + 24) is also divisible by 48
This shows you that it would be useful for you to work out both approaches if you want to obtain more in-depth learning. Each approach has nice things to learn.
See if you can prove (without induction) that (8*9^n + 24) is also divisible by 48.
SOLUTION
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Another way to approach the problem is to separately show that the polynomial is always divisible by both 3 and 16.
ASKER CERTIFIED SOLUTION
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ASKER
Thanks
>> phoffric>> why do you say "See if you can prove (without induction) that (8*9^n + 24) is also divisible by 48." (my emphasis) what is the problem with proving that WITH induction (which I demonstrated above is quite straight forward) ?
We have been showing multiple families of solutions. I offered another one since it is not obvious after looking at:
(9^n + 3) for a second or two that this expression is divisible by 2.
If the author goes through every approach, then there will be a nice bag of tricks for the future.
We have been showing multiple families of solutions. I offered another one since it is not obvious after looking at:
(9^n + 3) for a second or two that this expression is divisible by 2.
If the author goes through every approach, then there will be a nice bag of tricks for the future.