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Need to practice these for next year, so will get on to them soon  thanks
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Your answers for #1 and #2 are a bit off! I haven't looked at #3!
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On December 09 2011 21:12 Day[9] wrote:
Your answers for #1 and #2 are a bit off! I haven't looked at #3!
A bit off how ? Incorrect numbers ? o.O
+ Show Spoiler +OMFG Day9 replied to my post! Bronze to diamond thanks to you, and my mother thinks you're funny<3
(Please do a daily on how to spin pens pleaaaaase :D)
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is the first question y=(3x+2)/(x-4)?
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On December 09 2011 21:23 plutonius wrote:
is the first question y=(3x+2)/(x-4)?
Yep, it is like this...
3x+2
(divided by)
x-4
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3# is not a problem, all you need to do is to apply the law of cosines. Result is a bit off by the way
+ Show Spoiler +cos-1(4.2/6.4) = 48.9855°
edit : if you like maths problems, you should read evanthebouncy!'s blog entries !
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On December 09 2011 21:22 achristes wrote:Show nested quote +On December 09 2011 21:12 Day[9] wrote:
Your answers for #1 and #2 are a bit off! I haven't looked at #3! A bit off how ? Incorrect numbers ? o.O + Show Spoiler +OMFG Day9 replied to my post! Bronze to diamond thanks to you, and my mother thinks you're funny<3
(Please do a daily on how to spin pens pleaaaaase :D)
In #1 take a look at the denominator.. Suffice it to say, strange things happen when the denominator approaches zero..
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On December 09 2011 21:25 endy wrote:3# is not a problem, all you need to do is to apply the law of cosines. Result is a bit off by the way + Show Spoiler +cos-1(4.2/6.4) = 48.9855°
Maybe not a problem, but we just finished a chapter about trigonometry and I thought it was cool
I did (after calculating BC)
+ Show Spoiler +4.8^2=6.4^2+4.2^2-2*6.4*4.2*cosA
A=cos-1 (6.4^2+4.2^2-4.8^2/2*6.4*4.2)
A=48.5888
If it's off I blame my calculator! (casio fx-9860GII)
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For the second question, I'm assuming you are finding the roots or values of x that yield y = 0.
+ Show Spoiler +Applying the quadratic formula:
a = 2
b = 7
c = -4
x = [- b + root(b^2-4ac)]/2a
x = [- b - root(b^2-4ac)]/2a
x = 1/2,-4
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Oh, shit, big blunder by me. Editing OP.
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Unfortunately, all of your answers are incorrect.
1.
y = 3x + 2 / x - 4. I'll assume that you didn't post the function with missing brackets. What's written is y = (3 x) + (2 / x) - 4
+ Show Spoiler +
Asymptotes are line that the function creeps closer and closer to but never reaches. To determine a horizontal asymptote, simply look at what the function does as x tends to positive and negative infinity. An easy test is to compute f(100), f(1000), f(10000), etc... and similarly for the negative numbers, to find the value that the function tends to. In this case, the function is ever increasing (for large positive x) or decreasing (large negative x), so there is no horizontal asymptote.
A vertical asymptote can be found by finding a value of x where the function is not defined. In this case, at x = 0, the function is not defined (division by zero in the second term). Fill in small value of x and see if the function goes to infinity to test if it is indeed an asymptote. As you fill in f(0.01), f(0.001), etc.. you'll notice that the function values explode. So the vertical asymptote is x = 0.
Note that there is also an oblique asymptote: y = 3 x.
edit: You edited the first question. Here the updated solution:
1. edited
(3x + 2) / (x - 4)
+ Show Spoiler +
For large positive or negative values of x, the constant terms in the numerator and denominator no longer matter much, so the function will tend to 3x / x = 3. Consequently, the horizontal asymptote is y = 3.
The vertical asymptote will most likely lie where the function is undefined: At x = 4. Testing by filling in values very close to x = 4, f(4.001), f(4.0001), etc... we see the function-value blow up as we approach x = 4. So the vertical asymptote is x = 4.
2.
2x^2 + 7x - 4 = 0
+ Show Spoiler +
The solutions are given by the "abc" formula:
x = -b / (2 a) + sqrt(b^2 - 4 a c) / (2 a)
If you fill in the relevant values (a = 2, b = 7, c = -4) and work it out, you get that x = 0.5 or x = -4.
3.
+ Show Spoiler +
You don't need the cosine law or anything like that. What you need to do is to recognize that you have a triangle with a right angle whose side have a ratio of 1 : sqrt(3) : 2. AC is the hypothenuse (not sure about the spelling of that word in english) and is twice as long as AB. A property of 1 : sqrt(3) : 2 triangles is that the angles are 30, 60 and 90 degrees. In this case, angle A is 60 degrees.
Draw a picture to make it more clear.
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Here's how to find horizontal asymptotes for when you have a polynomial divided by another polynomial:
Look at the biggest exponent in both the numerator and denominator. Then:
1. If the numerator's biggest exponent is larger than the denominator's biggest exponent, then there's no asymptote.
2. If the numerator's biggest exponent is smaller than the denominator's biggest exponent, then there's a horizontal asymptote at zero.
3. If the biggest exponents match, then take the leading coefficients (the coefficients of the x-values that have the highest exponents) and find the numerator's coefficient divided by the denominator's coefficient. There's an asymptote at that number.
(You can literally ignore all other smaller terms when evaluating the horizontal asymptote of a polynomial divided by another polynomial.)
Vertical asymptote of a polynomial divided by another polynomial is wherever you're dividing by zero.
Good luck!
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On December 09 2011 20:41 achristes wrote:
Mathematical english is really different (IMO) than mathematical norwegian, sorry in advance for any confusion.
When i did my thesis, I had to learn mathematical french because the area i was working in only had one good elementary textbook and of course it was only in french at the time. One good thing about maths is that there is definitely truth in the cliché that mathematics is a universal language
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On December 09 2011 22:44 sigma_x wrote:Show nested quote +On December 09 2011 20:41 achristes wrote:
Mathematical english is really different (IMO) than mathematical norwegian, sorry in advance for any confusion.
When i did my thesis, I had to learn mathematical french because the area i was working in only had one good elementary textbook and of course it was only in french at the time. One good thing about maths is that there is definitely truth in the cliché that mathematics is a universal language
Wow, I would be really mad if I had to learn french for any subject ^^
It's universal enough, but the expressions are a little bit different, for example
ligning = eqaution
andregradsligning = quadratic eqation
potens = power
standardform = scientific notation
pytagorassetningen = pythagorean theoren
etc
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What is this like 9th grade?
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On December 10 2011 00:02 jamesr12 wrote:
What is this like 9th grade?
Almost, + 2 years and then you're right. Anything wrong with that ?
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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....?
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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....?
By separating the problems
+ Show Spoiler +a^1/2*a^2*(a^1/3)^-2
----------------------------
a^2/3
(a^1/3)^-2 = a^-2/3
a^1/2*a^2
------------------
a^2/3*a^2/3
a^2/2
--------
a^4/3
It is what I think is the answer. Although i'm not completely sure.
And FFS, did excactly what I did just a few hours ago, posting answers without double checking >.<
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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
