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On December 10 2011 01:06 ymir233 wrote:
How would you even do 4 with digital help unless you have something abusive like a TI-89/NSpire or some overkill stuff like WAlpha/Mathematica/MATLAB....?
What do you mean, how would you do 4 without digital help? It's just multiplication... Why would you even need digital help? The answer in the OP is wrong, BTW.
+ Show Spoiler + Assuming you mean the parentheses to look like ![[image loading]](http://i.imgur.com/G2DZO.jpg) , that reduces to a^(1/2 + 2 + -2/3 - 2/3) = a^(7/6). Remember that multiplying things with the same base adds the exponents, dividing subtracts the exponents, and raising a power to a power multiplies the exponents.
It would be REALLY nice if TL supported LaTeX. Is that possible? I know some forums can, but I don't know much about what it would take to implement that functionality.
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Hey achristes, I was just working on these problems for fun and I think I noticed a mistake on your most recent question (#4):
+ Show Spoiler +a^1/2*a^2
------------------
a^2/3*a^2/3
a^2/2
--------
a^4/3
Shouldn't the numerator be a^5/2, because you have to add the exponents of 1/2 and 2? In your case you seem to multiply the top two, yet for the bottom two exponents you add them. If you add the top ones you would need to first get a common denominator - that is 1/2 + 4/2, leaving you with 5/2. Overall I got a^7/6 upon simplification.
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On December 10 2011 01:37 radscorpion9 wrote:Hey achristes, I was just working on these problems for fun and I think I noticed a mistake on your most recent question (#4): + Show Spoiler +a^1/2*a^2
------------------
a^2/3*a^2/3
a^2/2
--------
a^4/3
Shouldn't the numerator be a^5/2, because you have to add the exponents of 1/2 and 2? In your case you seem to multiply the top two, yet for the bottom two exponents you add them. If you add the top ones you would need to first get a common denominator - that is 1/2 + 4/2, leaving you with 5/2. Overall I got a^7/6 upon simplification.
There could very well be a mistake, but if you multiply two fractions aren't you spupposed to just do numerator*numerator and denominator*denominator, which would be like saying
+ Show Spoiler +
Because a^2 == a^2/1 or am I just being wrong here ?
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On December 10 2011 01:46 achristes wrote:Show nested quote +On December 10 2011 01:37 radscorpion9 wrote:Hey achristes, I was just working on these problems for fun and I think I noticed a mistake on your most recent question (#4): + Show Spoiler +a^1/2*a^2
------------------
a^2/3*a^2/3
a^2/2
--------
a^4/3
Shouldn't the numerator be a^5/2, because you have to add the exponents of 1/2 and 2? In your case you seem to multiply the top two, yet for the bottom two exponents you add them. If you add the top ones you would need to first get a common denominator - that is 1/2 + 4/2, leaving you with 5/2. Overall I got a^7/6 upon simplification.
There could very well be a mistake, but if you multiply two fractions aren't you spupposed to just do numerator*numerator and denominator*denominator, which would be like saying + Show Spoiler +Because a^2 == a^2/1 or am I just being wrong here ?
No, no, the problem is that you shouldn't be multiplying 1/2 and 2, you should be adding them. See my post above for rules of exponentiation.
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I think there was probably a misunderstanding - let me try to be more clear:
+ Show Spoiler +
You're right if you are multiplying two fractions, then it is indeed just numerator*numerator and denominator*denominator. But in this case you shouldn't be multiplying the fractions but adding them, because that's the special rule for multiplying the same bases with exponents (where the base is "a" and the exponent is the superscripted number).
In fact that's what you (in my view correctly) did for the lower case - you had a^2/3*a^2/3 and the result was a^4/3 - I was just surprised because you didn't do the same for the numerator of the general fraction.
Anyways its probably better to look at Iranon's overview, it covers it pretty succinctly. Don't mean to repeat the same information, Iranon just happened to post while I was writing my post.
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really clever homework blog
props
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On December 10 2011 01:56 radscorpion9 wrote:I think there was probably a misunderstanding - let me try to be more clear: + Show Spoiler +
You're right if you are multiplying two fractions, then it is indeed just numerator*numerator and denominator*denominator. But in this case you shouldn't be multiplying the fractions but adding them, because that's the special rule for multiplying the same bases with exponents (where the base is "a" and the exponent is the superscripted number).
In fact that's what you (in my view correctly) did for the lower case - you had a^2/3*a^2/3 and the result was a^4/3 - I was just surprised because you didn't do the same for the numerator of the general fraction.
Anyways its probably better to look at Iranon's overview, it covers it pretty succinctly. Don't mean to repeat the same information, Iranon just happened to post while I was writing my post.
Yeah, I see that now TY for pointing it out
On December 10 2011 03:06 Divinek wrote:
really clever homework blog
props
Are you suggesting I made this to help my homework ?
All the answers to our questions are in the back of our books so that would be pointless 
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I really like math. Too bad I have to go through Geometry before trig and calc.
I honestly hate that class. I'm also way better than others.
+ Show Spoiler +Day[9] should do a math daily! ^.^
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On December 10 2011 01:33 Iranon wrote:Show nested quote +On December 10 2011 01:06 ymir233 wrote:
How would you even do 4 with digital help unless you have something abusive like a TI-89/NSpire or some overkill stuff like WAlpha/Mathematica/MATLAB....?
What do you mean, how would you do 4 without digital help? It's just multiplication... Why would you even need digital help? The answer in the OP is wrong, BTW. + Show Spoiler + Assuming you mean the parentheses to look like , that reduces to a^(1/2 + 2 + -2/3 - 2/3) = a^(7/6). Remember that multiplying things with the same base adds the exponents, dividing subtracts the exponents, and raising a power to a power multiplies the exponents. It would be REALLY nice if TL supported LaTeX. Is that possible? I know some forums can, but I don't know much about what it would take to implement that functionality.
- ______ -;; as in, you might as well just fuckin add the numbers/exponents, but you can't just input variable expressions (as in, the original expression above) and press simplify with digital help unless you have 89 or above. If the OP was talking about needing digital help to add -2/3 to -2/3, well I'm not sure what to say to that other than trololols.
EDIT: Doing anything except for the most complicated-looking stuff (and things that require a lot of symbols) on Latex is either obnoxious or dumb. Trust me.
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Now do this:
A fibonacci number n (as given by the recursion A0 = 0, A1 = 1, An+2 = An+1 + An) can be expressed in closed form as
An = (Θ^n - ω^n)/sqrt(5) where ω = (1-sqrt(5))/2 and Θ = (1+sqrt(5))/2 (that is, Θ^2 = Θ+1 and ω^2 = ω+1.)
Prove that the statement holds for all positive integers n (by induction).
(totally not for my upcoming exam, just testing your abilities people.)
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On December 10 2011 06:23 hifriend wrote:
Now do this:
A fibonacci number n (as given by the recursion A0 = 0, A1 = 1, An+2 = An+1 + An) can be expressed in closed form as
An = (Θ^n - ω^n)/sqrt(5) where ω = (1-sqrt(5))/2 and Θ = (1+sqrt(5))/2 (that is, Θ^2 = Θ+1 and ω^2 = ω+1.)
Prove that the statement holds for all positive integers n (by induction).
(totally not for my upcoming exam, just testing your abilities people.)
Proof:
+ Show Spoiler +People love golden ratios 
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Really? Now even the TL math blogs are becoming crappier? Who would've guessed 
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On December 10 2011 06:23 hifriend wrote:
Now do this:
A fibonacci number n (as given by the recursion A0 = 0, A1 = 1, An+2 = An+1 + An) can be expressed in closed form as
An = (Θ^n - ω^n)/sqrt(5) where ω = (1-sqrt(5))/2 and Θ = (1+sqrt(5))/2 (that is, Θ^2 = Θ+1 and ω^2 = ω+1.)
Prove that the statement holds for all positive integers n (by induction).
(totally not for my upcoming exam, just testing your abilities people.)
Bonus: Derive the formula itself using power series. Don't worry, I already had my exam and there were much harder derivations than the fibonacci series.
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did #5 for math class last year =) by expressing as a dynamical system and solving by classical diffeq solving methods. very fun proof
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A really satisfying problem (b/c I like analysis :3):
Evaluate the integral from 0 to 2pi of 1/(3+2cos(theta))*dtheta (hopefully the easy way...)
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On December 10 2011 06:23 hifriend wrote:
Now do this:
A fibonacci number n (as given by the recursion A0 = 0, A1 = 1, An+2 = An+1 + An) can be expressed in closed form as
An = (Θ^n - ω^n)/sqrt(5) where ω = (1-sqrt(5))/2 and Θ = (1+sqrt(5))/2 (that is, Θ^2 = Θ+1 and ω^2 = ω+1.)
Prove that the statement holds for all positive integers n (by induction).
(totally not for my upcoming exam, just testing your abilities people.)
hm kinda just assumed it was difficult at first glance but not really
+ Show Spoiler +![[image loading]](http://img15.imageshack.us/img15/1180/87193276.gif) ![[image loading]](http://img714.imageshack.us/img714/1512/49324732.gif) assume statement holds for some n = p & n = p+1 then ![[image loading]](http://img208.imageshack.us/img208/7062/18098541.gif) statement holds for all positive n by the principle of induction q.e.d bro
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On December 10 2011 11:28 ymir233 wrote:
A really satisfying problem (b/c I like analysis :3):
Evaluate the integral from 0 to 2pi of 1/(3+2cos(theta))*dtheta (hopefully the easy way...)
The easy way is to forget every formula and write a Gaussian quadrature algorithm.
I may have only gotten a B in Numerical Analysis but I learned a far more important lesson. :D
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EPT
