View Poll Results: What is the hardest math you learned?

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  • Calculus (specify in comments)

    38 36.19%
  • Trigonometry

    11 10.48%
  • (Enriched) Geometry

    11 10.48%
  • Equations (specify)

    4 3.81%
  • Graphing Equations (specify)

    2 1.90%
  • Basic Algebra (specify)

    7 6.67%
  • I can do better!

    24 22.86%
  • What kind of math is that?!

    8 7.62%
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Thread: Calculator/Math Thread

  1. #221
    Member ben1996123's Avatar
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    Quote Originally Posted by cmowla View Post
    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).

    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.
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  2. #222

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    Quote Originally Posted by ben1996123 View Post
    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.

  3. #223
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    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][/IMG]

  4. #224
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    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 :P Thanks!

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

    2. A sequence a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub> ... is defined by a<sub>0</sub> = 2012 and a<sub>k+1</sub> = a<sub>k</sub>^2 +1 for all k ≥ 0. Find the last digit of a<sub>2012</sub>. 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<sub>n</sub> = 1 + 1/n - 1/(n^2) - 1/(n^3)

    Find the smallest integer k such that T<sub>2</sub>T<sub>3</sub>T<sub>4</sub>...T<sub>k</sub> > 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.
    Last edited by DYGH.Tjen; 07-10-2012 at 02:25 AM.

  5. #225

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    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
    Last edited by Stefan; 07-07-2012 at 02:08 PM.

  6. #226
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    Reasons/workings/steps please Stefan? :P Especially for Q6?

  7. #227

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    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/sh...Forum-and-Wiki

  8. #228
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    Quote Originally Posted by Stefan View Post
    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.
    There are 10 kinds of people in the world. Those who get it, those who don't, and those who didn't see a ternary joke coming. ~Petrus Solver~

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

  10. #230
    Member DYGH.Tjen's Avatar
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    Quote Originally Posted by DYGH.Tjen View Post
    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 :P Thanks!

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

    2. A sequence a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub> ... is defined by a<sub>0</sub> = 2012 and a<sub>k+1</sub> = a<sub>k</sub>^2 +1 for all k ≥ 0. Find the last digit of a<sub>2012</sub>. 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<sub>n</sub> = 1 + 1/n - 1/(n^2) - 1/(n^3)

    Find the smallest integer k such that T<sub>2</sub>T<sub>3</sub>T<sub>4</sub>...T<sub>k</sub> > 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 :P

    @qqwref @ben1996123

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