I'm pretty sure I got 1 right. That's not the right answer.
I'm pretty sure I got 1 right. That's not the right answer.
BREAKTHROUGH!
Back to problem number one. The sigma can be further simplified. The sigma that I am specifying is the one in my old final answer. Starting at i=1 until n of 2i.
Write out the first terms of the series
a1=2
a2=4
a3=6
a4=8
a5=10
Using those find the first sums of the series.
s1=2
s2=6
s3=12
s4=20
s5=30
Now here is where it gets interesting. Check out this neat pattern I noticed.
2=1*2
6=2*3
12=3*4
20=4*5
30=5*6
So you can tell that the nth sum (which is what the sigma is looking for) is equal to n(n+1).
Now lets substitute that in for the sigma in my old answer.
n+(1/n)sigma(2i)
n+(1/n)[n(n+1)]
n+(n+1)
2n+1
There you go. That entire mess reduces to 2n+1
Dude it's an easy problem using integration. I get 2n-1/n-1, I double checked it.
EDIT: Wait, nvm I forgot we're adding integers only. My bad.
Last edited by PooFoo; May 17, 2007 at 09:45 PM.
It sure doesn't look like a calculus class to me so integration isn't an option. And if you checked yours definitly does not work anyways. Mine is definitely correct.
Lets pick an arbitrary value for n, say 5.
If n is 5 then the original equation goes to:
(1+2/5)+(1+4/5)+(1+6/5)+(1+8/5)+(1+10/5)
5+1/5(2+4+6+8+10)
5+1/5(30)
5+6=11
Using my equation
2n+1
2(5)+1=11
Using your equation
(2n-1)/(n-1)
(2*5-1)/(5-1)
=9/4
Lets pick another arbitrary n value, 8.
I'll skip showing all my work but you can check it. It comes out to 17.
Using my equation.
2(8)+1=17
Your equation.
(2*8-1)/(8-1)=15/7
So you must have done the integration incorrectly.
#4 was definitely log (base something) (x^2 - 3x)
that base, judging by my poor handwriting, could have been either x or 2.
The equation seems straightforward if 2 is the base, but what if x is it?
Or is that impossible due to being too high level for basic math?
If it were log base x don't let that throw you off. Follow the same rules. When you convert to exponential you would get x^2 = x^2 - 3x which then of course reduces to x=-1/3. That works out nicely, but it just seems like it is too nicely. If it is another base other than 10 I would guess it would be log base 2. It works out nicely.
If its log base 2.
2^2 = x^2 - 3x
x^2 - 3x - 4 = 0
(x-4)(x+1)=0
x=4,-1
Wait, wouldn't x have to be 0 in that case?When you convert to exponential you would get x^2 = x^2 - 3x which then of course reduces to x=-1/3.
As for number 1, do you just pick arbitrary values and try to notice a pattern?
As for 6, is it me, or is there a wide range of numbers it can be?
Seems to me that approximately anything lower than 2/3 and higher than 2 can be it (the problem has a > sign in case I put that wrong in the equation).
Last edited by RusskiSoldat; May 17, 2007 at 10:54 PM.
My mistake, I lost the inequality. In that case find all of the "special" points. This includes the solution which I showed as zero, plus any points where the expression is undefined. When x is 2/3 the numerator is zero. The log of zero is always undefined. When x is 2 the denominator is zero so it is undefined. Then make a sign chart and test the values in between to see when it is positive. As it appears you have already figured out that it is positive when x is less than 2/3 but greater than 2.
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You have bad handwriting.
Incidentally, in my elementary real analysis class we were given a way to do this using Taylor's Theorem, but I always forget the exact statement of the theorem.
Different sigmas. Capital Σ is used to indicate summation, lowercase σ indicates standard deviation. Of course, summation is important in probability.
You were probably expected to know that Σi=1n i = n(n+1)/2. A simple proof of that is attributed to Gauss, as follows. Take the sum
1 + 2 + 3 + . . . + n-2 + n-1 + n
and the identical sum
n + n-1 + n-2 + . . . + 3 + 2 + 1.
It's easy to see that by adding the two you'll get
(n + 1) + (n + 1) + . . . + (n + 1),
which is twice the value of the original sum. But that's just n times n + 1. Thus the sum is n(n+1)/2.
But yes, going through the values and looking for patterns can work.
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umm a bio question, with the GFP gene thats been added to the genes of a pig. If i was to breed 2 GFP pigs together would they always produce GFP offspring? or would they breed only half GFP and the other half normal? Sorry for vagueness i dont really know how to ask this question.
Well, GFP is a tracking gene, tacked onto other genes you want to follow. Thus, its inheritance is tied to that of the gene it's attached to. So let's say, as an example, I wanted to track a gene that codes for a structural protein (let's say actin). I would add the GFP gene to the end of the actin sequence (before the part that codes for the stop codon), and the cells that are produced to be fluorescent. When it comes to passing this on, though, it depends on whether I have added it to both copies of the DNA of both pigs (in which case it will always be passed on), or only to one copy of their DNA ( in which case the expression of the GFP gene will depend on the expression of the gene it is attached to).
For example, let's say I tag the actin on one copy of each pig. That makes each pig Gg (where G=actin+attached GFP gene). In that case, the pigs will have roughly 50 % of the actin in their cells fluoresce. If we then put them together, their offspring (if it follows strict Mendellian inheritance) will be: GG, Gg, Gg, gg. This means that, on average, 25% of their offspring will have ALL of their actin lighting up, 50% will have half of it lighting up (like their parents), and 25% will not light up at all.
What this means, though, is that the original gene in question follows Mendellian inheritance patterns, and doesn't have any weird inheritance properties.
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:EDIT:
Never mind, I figured it out after all, and I feel silly forgetting the procedure
Last edited by LSJ; June 10, 2007 at 04:10 PM. Reason: D'OH!
I have one other problem i need to do a dibybrid cross and im not sure on what to use. So far ive done 2 monohybrid crosses of Gg x Gg and GG x gg but im not sure what other trait to compare it to. also i cant give you rep for some reason.