Mathematics Thread

[vcuber13]

how do you prove

\(
\frac{Tan \theta + Cot \alpha}{Cot \theta + Tan \alpha} = \frac{Tan \theta}{Tan \alpha}
\)

 

[Christopher Mowla]

how do you prove

\(
\frac{Tan \theta + Cot \alpha}{Cot \theta + Tan \alpha} = \frac{Tan \theta}{Tan \alpha}
\)

I can verify that it's true using basic identities and algebra.

Spoiler

\( \text{Prove (verify) that }\frac{\tan \left( \theta \right)+\cot \left( a \right)}{\cot \left( \theta \right)+\tan \left( a \right)}=\frac{\tan \left( \theta \right)}{\tan \left( a \right)}. \)

\( =\frac{\frac{\sin \left( \theta \right)}{\cos \left( \theta \right)}+\frac{\cos \left( a \right)}{\sin \left( a \right)}}{\frac{\cos \left( \theta \right)}{\sin \left( \theta \right)}+\frac{\sin \left( a \right)}{\cos \left( a \right)}} \)

\( =\frac{\frac{\sin \left( \theta \right)\sin \left( a \right)+\cos \left( a \right)\cos \left( \theta \right)}{\cos \left( \theta \right)\sin \left( a \right)}}{\frac{\cos \left( \theta \right)\cos \left( a \right)+\sin \left( a \right)\sin \left( \theta \right)}{\sin \left( \theta \right)\cos \left( a \right)}} \)

\( =\frac{\sin \left( \theta \right)\sin \left( a \right)+\cos \left( a \right)\cos \left( \theta \right)}{\cos \left( \theta \right)\sin \left( a \right)}\frac{\sin \left( \theta \right)\cos \left( a \right)}{\cos \left( \theta \right)\cos \left( a \right)+\sin \left( a \right)\sin \left( \theta \right)} \)

\( =\left( \frac{\sin \left( \theta \right)\sin \left( a \right)+\cos \left( a \right)\cos \left( \theta \right)}{\cos \left( \theta \right)\cos \left( a \right)+\sin \left( a \right)\sin \left( \theta \right)} \right)\left( \frac{\sin \left( \theta \right)\cos \left( a \right)}{\cos \left( \theta \right)\sin \left( a \right)} \right) \)

\( =1\times \frac{\sin \left( \theta \right)\cos \left( a \right)}{\cos \left( \theta \right)\sin \left( a \right)} \)

\( =\frac{\frac{\sin \left( \theta \right)}{\cos \left( \theta \right)}}{\frac{\sin \left( a \right)}{\cos \left( a \right)}}=\frac{\tan \left( \theta \right)}{\tan \left( a \right)} \)

 

[ben1996123]

Finished every past exam paper for C2 in 3 days. Started C3 on wednesday last week. Finished C3 today. Starting C4 tomorrow. In college (I start in September), I'm hopefully doing C3, C4, M1, M2, S2, FP1, FP2, FP3, FP4, D1, D2 and maybe S3, S4 or M3, M4.

My teacher told me today that I should be able to do C3, C4, M1, S2, FP1, FP2, FP3, FP4 and D1 all in the first year, but I'm not quite sure about that

Did C1 exam in January this year, got 100% in that. Hopefully I will get 100% in C2 in a few weeks.

maths

 

[Christopher Mowla]

Sum of Powers Image (Revisited)

I have found an even simpler image which shows how to calculate the sum of powers (using Pascal's Triangle).

The image should be self-explanatory.

​

 

[MattMcConaha]

The hardest class I've taken (and still am taking) is differential equations. The math itself isn't really that hard, but I'm basically learning out of the book because the teacher doesn't really help that much and I don't fully understand everything that's going on.

 

[ben1996123]

\( i^{i}\sqrt{e^{\pi}}+e^{i\pi}=0 \)

 

[cube fan]

I'm bad at algebra,but I'm good at the rest of them(I think calculus is interesting)

 

[Georgeanderre]

I'm hopefully doing C3, C4, M1, M2, S2, FP1, FP2, FP3, FP4, D1, D2 and maybe S3, S4 or M3, M4.

Did C1 exam in January this year, got 100% in that. Hopefully I will get 100% in C2 in a few weeks.

maths

No Numerical Methods =O

and as a side note... that looks like an insane amount of maths but given the overlap in maths these days it probably isn't. Little bit like Mechanics and Physics, buy one get one free A levels

Gratz on 100% on C1! I only got 93% =(

Hope you get results like that for C2 and whatever else you took in May, I know I messed up Q7 & Q10 in C2 so not expecting much more than 80%

 

[Christopher Mowla]

Here's another argument that 0^0 could be 0 or 1 (this example shows that we actually have a choice in the matter, as opposed to limit examples from Wikipedia which demand it to be one or the other. In other words, we can give it both values in the same example).

Spoiler: Argument

From the sum of positive integer powers generalized formula,

\( \sum\limits_{i=1}^{n}{\left[ i^{R} \right]}=\sum\limits_{i=1}^{n}{\left[ \left( i^{R}-\left( i-1 \right)^{R} \right)\left( n-\left( i-1 \right) \right) \right]} \) (well, they are equal for positive integers \( \ge \) 1).

\( =\sum\limits_{i=1}^{n}{\left[ \left( i^{R}-\left( i-1 \right)^{R}-\left( i-1 \right)^{R-1}+\left( i-1 \right)^{R-1} \right)\left( n-\left( i-1 \right) \right) \right]} \)

\( =\sum\limits_{i=1}^{n}{\left[ \left( i^{R}-\left( i-1 \right)^{R}-\left( i-1 \right)^{R-1} \right)\left( n-\left( i-1 \right) \right)+\left( i-1 \right)^{R-1}\left( n-\left( i-1 \right) \right) \right]} \)

\( \text{For }R=1,\text{ that is, for }\sum\limits_{i=1}^{n}{i^{1}}=\sum\limits_{i=1}^{n}{i}, \)

\( =\sum\limits_{i=1}^{n}{\left[ \left( i^{1}-\left( i-1 \right)^{1}-\left( i-1 \right)^{1-1} \right)\left( n-\left( i-1 \right) \right)+\left( i-1 \right)^{1-1}\left( n-\left( i-1 \right) \right) \right]} \)

\( =\sum\limits_{i=1}^{n}{\left[ \left( 1-\left( i-1 \right)^{0} \right)\left( n-i+1 \right)+\left( i-1 \right)^{0}\left( n-i+1 \right) \right]} \)

=
\( \left[ \left( 1-\left( 1-1 \right)^{0} \right)\left( n-1+1 \right)+\left( 1-1 \right)^{0}\left( n-1+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 2-1 \right)^{0} \right)\left( n-2+1 \right)+\left( 2-1 \right)^{0}\left( n-2+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 3-1 \right)^{0} \right)\left( n-3+1 \right)+\left( 3-1 \right)^{0}\left( n-3+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 4-1 \right)^{0} \right)\left( n-4+1 \right)+\left( 4-1 \right)^{0}\left( n-4+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 5-1 \right)^{0} \right)\left( n-5+1 \right)+\left( 5-1 \right)^{0}\left( n-5+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 6-1 \right)^{0} \right)\left( n-6+1 \right)+\left( 6-1 \right)^{0}\left( n-6+1 \right) \right]+\cdot \cdot \cdot \)

\( =\left[ \left( 1-0^{0} \right)n+0^{0}n \right]+\left[ \left( 1-1^{0} \right)\left( n-1 \right)+1^{0}\left( n-1 \right) \right] \)

\( +\left[ \left( 1-2^{0} \right)\left( n-2 \right)+2^{0}\left( n-2 \right) \right]+\left[ \left( 1-3^{0} \right)\left( n-3 \right)+3^{0}\left( n-3 \right) \right] \)

\( +\left[ \left( 1-4^{0} \right)\left( n-4 \right)+4^{0}\left( n-4 \right) \right]+\left[ \left( 1-5^{0} \right)\left( n-5 \right)+5^{0}\left( n-5 \right) \right]+\cdot \cdot \cdot \)


\( =\left[ \left( 1-0^{0} \right)n+0^{0}n \right]+\left[ \left( 1-1 \right)\left( n-1 \right)+1\left( n-1 \right) \right] \)

\( +\left[ \left( 1-1 \right)\left( n-2 \right)+1\left( n-2 \right) \right]+\left[ \left( 1-1 \right)\left( n-3 \right)+1\left( n-3 \right) \right] \)

\( +\left[ \left( 1-1 \right)\left( n-4 \right)+1\left( n-4 \right) \right]+\left[ \left( 1-1 \right)\left( n-5 \right)+1\left( n-5 \right) \right]+\cdot \cdot \cdot \)


\( =\left[ \left( 1-0^{0} \right)n+0^{0}n \right]+\left[ 0\left( n-1 \right)+1\left( n-1 \right) \right] \)

\( +\left[ 0\left( n-2 \right)+1\left( n-2 \right) \right]+\left[ 0\left( n-3 \right)+1\left( n-3 \right) \right] \)

\( +\left[ 0\left( n-4 \right)+1\left( n-4 \right) \right]+\left[ 0\left( n-5 \right)+1\left( n-5 \right) \right]+\cdot \cdot \cdot \)

Just looking at \( \left[ \left( 1-0^{0} \right)n+0^{0}n \right] \), we need to choose a value for \( \text{0}^{0} \) so that the result in that bracket is \( \left( n-0 \right)=n. \)

Since the second value in the second bracket is \( \left( n-1 \right), \) the value in the third bracket is \( \left( n-2 \right), \) etc.,

we can either choose \( \text{}0^{0}=1\text{ or }0^{0}=0,\ \) since

\( \left[ \left( 1-\left[ 1 \right] \right)n+\left[ 1 \right]n \right]=\left[ \left( 1-\left[ 0 \right] \right)n+\left[ 0 \right]n \right]=\left[ 0n+n \right]=\left[ n+0n \right]=n=\left( n-0 \right) \).

 

[ben1996123]

Here's another argument that 0^0 could be 0 or 1 (this example shows that we actually have a choice in the matter, as opposed to limit examples from Wikipedia which demand it to be one or the other. In other words, we can give it both values in the same example).

Spoiler: Argument

From the sum of positive integer powers generalized formula,

\( \sum\limits_{i=1}^{n}{\left[ i^{R} \right]}=\sum\limits_{i=1}^{n}{\left[ \left( i^{R}-\left( i-1 \right)^{R} \right)\left( n-\left( i-1 \right) \right) \right]} \) (well, they are equal for positive integers \( \ge \) 1).

\( =\sum\limits_{i=1}^{n}{\left[ \left( i^{R}-\left( i-1 \right)^{R}-\left( i-1 \right)^{R-1}+\left( i-1 \right)^{R-1} \right)\left( n-\left( i-1 \right) \right) \right]} \)

\( =\sum\limits_{i=1}^{n}{\left[ \left( i^{R}-\left( i-1 \right)^{R}-\left( i-1 \right)^{R-1} \right)\left( n-\left( i-1 \right) \right)+\left( i-1 \right)^{R-1}\left( n-\left( i-1 \right) \right) \right]} \)

\( \text{For }R=1,\text{ that is, for }\sum\limits_{i=1}^{n}{i^{1}}=\sum\limits_{i=1}^{n}{i}, \)

\( =\sum\limits_{i=1}^{n}{\left[ \left( i^{1}-\left( i-1 \right)^{1}-\left( i-1 \right)^{1-1} \right)\left( n-\left( i-1 \right) \right)+\left( i-1 \right)^{1-1}\left( n-\left( i-1 \right) \right) \right]} \)

\( =\sum\limits_{i=1}^{n}{\left[ \left( 1-\left( i-1 \right)^{0} \right)\left( n-i+1 \right)+\left( i-1 \right)^{0}\left( n-i+1 \right) \right]} \)

=
\( \left[ \left( 1-\left( 1-1 \right)^{0} \right)\left( n-1+1 \right)+\left( 1-1 \right)^{0}\left( n-1+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 2-1 \right)^{0} \right)\left( n-2+1 \right)+\left( 2-1 \right)^{0}\left( n-2+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 3-1 \right)^{0} \right)\left( n-3+1 \right)+\left( 3-1 \right)^{0}\left( n-3+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 4-1 \right)^{0} \right)\left( n-4+1 \right)+\left( 4-1 \right)^{0}\left( n-4+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 5-1 \right)^{0} \right)\left( n-5+1 \right)+\left( 5-1 \right)^{0}\left( n-5+1 \right) \right]+ \)

\( \left[ \left( 1-\left( 6-1 \right)^{0} \right)\left( n-6+1 \right)+\left( 6-1 \right)^{0}\left( n-6+1 \right) \right]+\cdot \cdot \cdot \)

\( =\left[ \left( 1-0^{0} \right)n+0^{0}n \right]+\left[ \left( 1-1^{0} \right)\left( n-1 \right)+1^{0}\left( n-1 \right) \right] \)

\( +\left[ \left( 1-2^{0} \right)\left( n-2 \right)+2^{0}\left( n-2 \right) \right]+\left[ \left( 1-3^{0} \right)\left( n-3 \right)+3^{0}\left( n-3 \right) \right] \)

\( +\left[ \left( 1-4^{0} \right)\left( n-4 \right)+4^{0}\left( n-4 \right) \right]+\left[ \left( 1-5^{0} \right)\left( n-5 \right)+5^{0}\left( n-5 \right) \right]+\cdot \cdot \cdot \)


\( =\left[ \left( 1-0^{0} \right)n+0^{0}n \right]+\left[ \left( 1-1 \right)\left( n-1 \right)+1\left( n-1 \right) \right] \)

\( +\left[ \left( 1-1 \right)\left( n-2 \right)+1\left( n-2 \right) \right]+\left[ \left( 1-1 \right)\left( n-3 \right)+1\left( n-3 \right) \right] \)

\( +\left[ \left( 1-1 \right)\left( n-4 \right)+1\left( n-4 \right) \right]+\left[ \left( 1-1 \right)\left( n-5 \right)+1\left( n-5 \right) \right]+\cdot \cdot \cdot \)


\( =\left[ \left( 1-0^{0} \right)n+0^{0}n \right]+\left[ 0\left( n-1 \right)+1\left( n-1 \right) \right] \)

\( +\left[ 0\left( n-2 \right)+1\left( n-2 \right) \right]+\left[ 0\left( n-3 \right)+1\left( n-3 \right) \right] \)

\( +\left[ 0\left( n-4 \right)+1\left( n-4 \right) \right]+\left[ 0\left( n-5 \right)+1\left( n-5 \right) \right]+\cdot \cdot \cdot \)

Just looking at \( \left[ \left( 1-0^{0} \right)n+0^{0}n \right] \), we need to choose a value for \( \text{0}^{0} \) so that the result in that bracket is \( \left( n-0 \right)=n. \)

Since the second value in the second bracket is \( \left( n-1 \right), \) the value in the third bracket is \( \left( n-2 \right), \) etc.,

we can either choose \( \text{}0^{0}=1\text{ or }0^{0}=0,\ \) since

\( \left[ \left( 1-\left[ 1 \right] \right)n+\left[ 1 \right]n \right]=\left[ \left( 1-\left[ 0 \right] \right)n+\left[ 0 \right]n \right]=\left[ 0n+n \right]=\left[ n+0n \right]=n=\left( n-0 \right) \).

and, since \( \frac{0}{0}=\frac{0^{1}}{0^{1}}=\left( 0^{1} \right)\left( 0^{-1} \right)=0^{1-1}=0^{0}, \) this is indirectly an argument why 0/0 can be 1 or zero too.

Can't \( 0^0 \) have any value, not just 0 or 1?

\( 0^{0}=0^{n}\times 0^{-n}=\frac{0^{n}}{0^{n}}=\frac{0}{0} \), which is indeterminate and can have any value, eg.

\( \lim_{x\rightarrow 1} \frac{x}{x-1}-\frac{1}{ln(x)}=\lim_{x\rightarrow 1} \frac{xln(x)-x+1}{xln(x)-ln(x)}\rightarrow \frac{0}{0} \)

Using L'hopital's rule, this limit must be the same as \( \lim_{x\rightarrow 1} \frac{\frac{d}{dx}(xln(x)-x+1)}{\frac{d}{dx}(xln(x)-ln(x))} \)

\( \frac{d}{dx}(xln(x)-x+1)=\frac{d}{dx}(x)ln(x)+\frac{d}{dx}(ln(x))x-1=ln(x)+\frac{x}{x}-1=ln(x) \)

\( \frac{d}{dx}(xln(x)-ln(x))=\frac{d}{dx}(x)ln(x)+\frac{d}{dx}(ln(x))x-\frac{d}{dx}(ln(x)) \)\( =ln(x)+\frac{x}{x}-\frac{1}{x}=ln(x)+1-\frac{1}{x} \)

\( \lim_{x\rightarrow 1} \frac{\frac{d}{dx}(xln(x)-x+1)}{\frac{d}{dx}(xln(x)-ln(x))}=\lim_{x\rightarrow 1} \frac{ln(x)}{ln(x)+1-\frac{1}{x}}\rightarrow \frac{0}{0} \)

Using L'hopital's rule again:

\( \lim_{x\rightarrow 1} \frac{ln(x)}{ln(x)+1-\frac{1}{x}}=\lim_{x\rightarrow 1} \frac{\frac{d}{dx}(ln(x))}{\frac{d}{dx}ln(x)+1-\frac{1}{x}} \)

\( \frac{d}{dx}(ln(x))=\frac{1}{x} \)

\( \frac{d}{dx}(ln(x)+1-\frac{1}{x})=\frac{1}{x}+\frac{1}{x^{2}} \)

\( \lim_{x\rightarrow 1} \frac{\frac{d}{dx}(ln(x))}{\frac{d}{dx}ln(x)+1-\frac{1}{x}}=\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^{2}}}=\frac{1}{2} \)

So in this case, \( \frac{0}{0}=\frac{1}{2} \), which also means that (in this case) \( 0^{0}=\frac{1}{2} \). Maybe I'm not noticing something, but I think this is correct.

 

[Christopher Mowla]

Can't \( 0^0 \) have any value, not just 0 or 1?

Oops! I edited my post just as you quoted me! Well, the reason I edited my post is because I realized that I made the mistake of dividing by zero (so 0^0 doesn't equal 0/0 because you would have to accept that 0/0 is defined when it is not). Besides, the limit as x->a of a function (or in the case of L'hopital's rule) doesn't actually equal the function evaluated at a unless the function is defined at x=a. So I don't think we can say that \( \underset{x\to 1}{\mathop{\lim }}\,\left( \frac{x}{x-1}-\frac{1}{\ln \left( x \right)} \right) \) approaches 0/0 because that doesn't make sense. We say that "..." 0/0 in order to say that we are currently not looking (analyzing) the given function in the correct manner.

So sorry for the confusion, but I do think that my argument is correct for 0^0.

 

[CarlBrannen]

I voted for Basic Algebra. The toughest class I ever had used a text titled "Basic Algebra I" and "Basic Algebra II". It's also known as "Jacobson".

It has something to do with Rubik's cubes. Section 1.7 is about orbits and cosets:

[/IMG]

 

[DYGH.Tjen]

Went to a small competition yesterday, will probably be cake to some of you so here are the questions hope I can get steps/workings plus solutions within the hour Thanks!

1. Find the area of a convex quadrilateral which has perpendicular diagonals of lengths 15 and 18.

2. A sequence a₀, a₁, a₂ ... is defined by a₀ = 2012 and a_k+1 = a_k^2 +1 for all k ≥ 0. Find the last digit of a₂₀₁₂. whaaat I can't type subscripts. Hope you guys know what I mean. On a side note, what program/online tool do you use to type math symbols and the like?

3. Given three distinct positive integers a,b,c such that a!b!c! = 10!, find a+b+c.

4. An obtuse triangle has side lengths d,30,40 where d is an integer. How many possible values of d are there?

5. The product of two four-digit positive integers is 4^8 + 6^8 + 9^8. What is the smaller integer?

6. A positive integer ending with 8888 has 16 odd positive factors. How many even positive factors does this number have?

Part B

1. Given a triangle ABC and a point M on the side BC. Let [1 and [2 be the circumcircles of triangles ABM and ACM respectively. The perpendicular bisector of BC intersects AM at P. Line BP intersects [1 at D (different from B), and line CP intersects [2 at E (different from C). Prove that ED is parallel to BC.

Note: the circumcircle of a triangle XYZ is the unique circle that passes through X, Y and Z.

2. Let T_n = 1 + 1/n - 1/(n^2) - 1/(n^3)

Find the smallest integer k such that T₂T₃T₄...T_k > 2012

3. There are 15 keys on a keyboard, arranged in a straight line. Two keys are one semitone apart if they are adjacent, and are one tone apart if they are separated by one key. A general scale is a sequence of keys from left to right such that the first key is the leftmost key, and two consecutive keys are either one tone or semitone apart.

Find the number of general scales with 8 keys.

 

[Stefan]

3. Given three distinct positive integers a,b,c such that a!b!c! = 10!, find a+b+c.
I will. As soon as they're actually given.

6. A positive integer ending with 8888 has 16 odd positive factors. How many even positive factors does this number have?
48

 

[DYGH.Tjen]

Reasons/workings/steps please Stefan? Especially for Q6?

 

[Stefan]

You can divide the number by 2 three times, so its prime factorization includes 2^3. For every odd factor x, you get even factors x*2, x*4, x*8. Thus 16*3=48 even factors.

On a side note, what program/online tool do you use to type math symbols and the like?
http://www.speedsolving.com/forum/showthread.php?18755-LaTeX-on-Forum-and-Wiki

 

[mr. giggums]

3. Given three distinct positive integers a,b,c such that a!b!c! = 10!, find a+b+c.
I will. As soon as they're actually given.

3, 5, 7

So the answer is 15.

 

[Stefan]

Bah, you should have only given me the three numbers.

 

[DYGH.Tjen]

Went to a small competition yesterday, will probably be cake to some of you so here are the questions hope I can get steps/workings plus solutions within the hour Thanks!

1. Find the area of a convex quadrilateral which has perpendicular diagonals of lengths 15 and 18.

2. A sequence a₀, a₁, a₂ ... is defined by a₀ = 2012 and a_k+1 = a_k^2 +1 for all k ≥ 0. Find the last digit of a₂₀₁₂. whaaat I can't type subscripts. Hope you guys know what I mean. On a side note, what program/online tool do you use to type math symbols and the like?

3. Given three distinct positive integers a,b,c such that a!b!c! = 10!, find a+b+c.

4. An obtuse triangle has side lengths d,30,40 where d is an integer. How many possible values of d are there?

5. The product of two four-digit positive integers is 4^8 + 6^8 + 9^8. What is the smaller integer?

6. A positive integer ending with 8888 has 16 odd positive factors. How many even positive factors does this number have?

Part B

1. Given a triangle ABC and a point M on the side BC. Let [1 and [2 be the circumcircles of triangles ABM and ACM respectively. The perpendicular bisector of BC intersects AM at P. Line BP intersects [1 at D (different from B), and line CP intersects [2 at E (different from C). Prove that ED is parallel to BC.

Note: the circumcircle of a triangle XYZ is the unique circle that passes through X, Y and Z.

2. Let T_n = 1 + 1/n - 1/(n^2) - 1/(n^3)

Find the smallest integer k such that T₂T₃T₄...T_k > 2012

3. There are 15 keys on a keyboard, arranged in a straight line. Two keys are one semitone apart if they are adjacent, and are one tone apart if they are separated by one key. A general scale is a sequence of keys from left to right such that the first key is the leftmost key, and two consecutive keys are either one tone or semitone apart.

Find the number of general scales with 8 keys.

=O bump, can anyone help to complete those questions? :/ I thought they would be easy considering we have people like qq and Ben here

@qqwref @ben1996123

 

[Hyprul 9-ty2]

Question 3 is just logic/brute force if you will.

10! is just 10x9x8..x2. Leave the biggest prime, 7, as a factorial because that way you no longer need to think about replicating primes.
Now you're left with 10x9x8x7!
Prime factorize 10, 9 and 8.. and then have fun?

10 = 5 2
9 = 3 3
8 = 2 2 2

5x2x2x3 x 3x2
5x4x3 x 3x2

---
I think you're trying to say \( "a_{k+1}=a_{k}^2+1" \)

a_0 = 2012
a_1 = a_0 ^2 + 1 = 2012 ^ 2 + 1 = .....4 + 1 = ...5
a_2 = a_1^2 + 1 = ....5^2 + 1 = ....25+1 = .....6
So on and so forth. It loops back at a_6 = .....2

Which leads me to believe that the last digit of a_2012 is 6. I'm sure I've done something/everything wrong, I'm so tired it's not even funny >_>