qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,728,707 | <p>I would like to solve this linear diophantine equation:
<span class="math-container">$$
40x_1+296x_2+945x_3+2048x_4+4500x_5+8640x_6=616103
$$</span>
All the answers have to be an integer number in the interval <span class="math-container">$\{[10] \cup [29,95]\}$</span>.</p>
<p>As a first step, I started by finding a particular solution of the equation without taking the constraints into account. I used the following procedure:</p>
<ul>
<li>Find a particular solution for <span class="math-container">$x_6$</span>:
<span class="math-container">$$
gcd(40,296,945,2048,4500)w_6+8640x_6=616103
$$</span>
I found <span class="math-container">$x_6=71$</span> and <span class="math-container">$w_6=2663$</span>. I can also determine a general solution: <span class="math-container">$X_6=71-n_6$</span> and <span class="math-container">$W_6=2663+8640n_6$</span>.</li>
<li>Use <span class="math-container">$w_6$</span> to find the other solutions by repeating the same procedure:
<span class="math-container">$$
40x_1+296x_2+945x_3+2048x_4+4500x_5=gcd(40,296,945,2048,4500)w_6 = 2663
$$</span>
<span class="math-container">$$
gcd(40,296,945,2048)w_5+4500x_5=2663
$$</span>
<span class="math-container">$$
...
$$</span></li>
</ul>
<p>That way, I could determine one particular solution of this equation:
<span class="math-container">$$
x_1=6876450,
x_2=-916860,
x_3=-3885,
x_4=1,
x_5=1,
x_6=71
$$</span>
I also introduce some intermediate variables to compute a general solution:
<span class="math-container">$$X_6=71-n_6, W_6=2663+8640n_6$$</span>
<span class="math-container">$$X_5=1+n_5,W_5=-1837-4500n_5$$</span>
<span class="math-container">$$X_4=1+n_4,W_4=-3885-2018n_4$$</span>
<span class="math-container">$$X_3=-3885+8n_3,W_3=458430-945n_3$$</span>
<span class="math-container">$$X_2=-916860-5n_2$$</span>
<span class="math-container">$$X_1=6876450+37n_2$$</span></p>
<p>Now, I'm stuck with my general solutions and I don't know what I can do to fit my particular solutions with the constraints. Here are the thoughts I got to solve this problem and the associated issue:</p>
<ul>
<li>I wanted to find what values of <span class="math-container">$n_2$</span> make <span class="math-container">$x_1$</span> in the interval defined above. I found <span class="math-container">$n_2=\{-185849,-185848\}$</span> that gives <span class="math-container">$x_1=\{37,74\}$</span> but <span class="math-container">$x_2$</span> is out of the interval with these values.</li>
<li>I wanted to rewrite the equation that way:
<span class="math-container">$$40\cdot(6876450+37n_2)+296\cdot(-916860-5n_2)+945\cdot(-3885+8n_3)+2048\cdot(1+n_4)+4500\cdot(1+n_5)+8640\cdot(71-n_6)=616103$$</span>
However, I can't because <span class="math-container">$n_2$</span>, <span class="math-container">$n_3$</span>, <span class="math-container">$n_4$</span>, <span class="math-container">$n_5$</span> and <span class="math-container">$n_6$</span> are not independent and the choice of the value of <span class="math-container">$n_6$</span>, for instance, has an impact on <span class="math-container">$w_6$</span> that has an impact on all general solutions.</li>
</ul>
<p>What can I do to find a solution that corresponds to the constraints?</p>
| Jan Eerland | 226,665 | <blockquote>
<p>Not a 'real' answer, but it was too big for a comment.</p>
</blockquote>
<p>I wrote and ran some Mathematica code:</p>
<pre><code>In[1]:=FullSimplify[
Solve[{40*x1 + 296*x2 + 945*x3 + 2048*x4 + 4500*x5 + 8640*x6 ==
616103, 1 <= x1 <= x2 <= x3 <= x4 <= x5 <= x6 <= 1000}, {x1, x2,
x3, x4, x5, x6}, PositiveIntegers]]
</code></pre>
<p>Running the code gives:</p>
<pre><code>Out[1]={{x1 -> 1, x2 -> 2, x3 -> 3, x4 -> 12, x5 -> 27, x6 -> 54}, {x1 -> 1,
x2 -> 2, x3 -> 7, x4 -> 12, x5 -> 30, x6 -> 52}, {x1 -> 1, x2 -> 2,
x3 -> 11, x4 -> 12, x5 -> 33, x6 -> 50}, {x1 -> 1, x2 -> 3, x3 -> 3,
x4 -> 5, x5 -> 9, x6 -> 65}, {x1 -> 1, x2 -> 5, x3 -> 7, x4 -> 36,
x5 -> 40, x6 -> 41}, {x1 -> 1, x2 -> 6, x3 -> 19, x4 -> 29,
x5 -> 31, x6 -> 46}, {x1 -> 1, x2 -> 6, x3 -> 23, x4 -> 29,
x5 -> 34, x6 -> 44}, {x1 -> 1, x2 -> 6, x3 -> 27, x4 -> 29,
x5 -> 37, x6 -> 42}, {x1 -> 1, x2 -> 8, x3 -> 11, x4 -> 15,
x5 -> 37, x6 -> 47}, {x1 -> 1, x2 -> 8, x3 -> 15, x4 -> 15,
x5 -> 40, x6 -> 45}, {x1 -> 1, x2 -> 12, x3 -> 15, x4 -> 32,
x5 -> 32, x6 -> 45}, {x1 -> 1, x2 -> 12, x3 -> 19, x4 -> 32,
x5 -> 35, x6 -> 43}, {x1 -> 1, x2 -> 12, x3 -> 23, x4 -> 32,
x5 -> 38, x6 -> 41}, {x1 -> 1, x2 -> 18, x3 -> 19, x4 -> 35,
x5 -> 39, x6 -> 40}, {x1 -> 1, x2 -> 25, x3 -> 27, x4 -> 31,
x5 -> 31, x6 -> 44}, {x1 -> 1, x2 -> 25, x3 -> 31, x4 -> 31,
x5 -> 34, x6 -> 42}, {x1 -> 1, x2 -> 31, x3 -> 31, x4 -> 34,
x5 -> 38, x6 -> 39}, {x1 -> 2, x2 -> 3, x3 -> 3, x4 -> 15, x5 -> 39,
x6 -> 47}, {x1 -> 2, x2 -> 3, x3 -> 7, x4 -> 15, x5 -> 42,
x6 -> 45}, {x1 -> 2, x2 -> 4, x3 -> 7, x4 -> 8, x5 -> 24,
x6 -> 56}, {x1 -> 2, x2 -> 7, x3 -> 7, x4 -> 32, x5 -> 34,
x6 -> 45}, {x1 -> 2, x2 -> 7, x3 -> 11, x4 -> 32, x5 -> 37,
x6 -> 43}, {x1 -> 2, x2 -> 7, x3 -> 15, x4 -> 32, x5 -> 40,
x6 -> 41}, {x1 -> 2, x2 -> 8, x3 -> 19, x4 -> 25, x5 -> 25,
x6 -> 50}, {x1 -> 2, x2 -> 8, x3 -> 23, x4 -> 25, x5 -> 28,
x6 -> 48}, {x1 -> 2, x2 -> 10, x3 -> 11, x4 -> 11, x5 -> 31,
x6 -> 51}, {x1 -> 2, x2 -> 14, x3 -> 19, x4 -> 28, x5 -> 29,
x6 -> 47}, {x1 -> 2, x2 -> 14, x3 -> 23, x4 -> 28, x5 -> 32,
x6 -> 45}, {x1 -> 2, x2 -> 14, x3 -> 27, x4 -> 28, x5 -> 35,
x6 -> 43}, {x1 -> 2, x2 -> 20, x3 -> 23, x4 -> 31, x5 -> 36,
x6 -> 42}, {x1 -> 2, x2 -> 20, x3 -> 27, x4 -> 31, x5 -> 39,
x6 -> 40}, {x1 -> 3, x2 -> 3, x3 -> 11, x4 -> 25, x5 -> 27,
x6 -> 50}, {x1 -> 3, x2 -> 3, x3 -> 15, x4 -> 25, x5 -> 30,
x6 -> 48}, {x1 -> 3, x2 -> 3, x3 -> 19, x4 -> 25, x5 -> 33,
x6 -> 46}, {x1 -> 3, x2 -> 3, x3 -> 23, x4 -> 25, x5 -> 36,
x6 -> 44}, {x1 -> 3, x2 -> 5, x3 -> 7, x4 -> 11, x5 -> 36,
x6 -> 49}, {x1 -> 3, x2 -> 5, x3 -> 11, x4 -> 11, x5 -> 39,
x6 -> 47}, {x1 -> 3, x2 -> 9, x3 -> 11, x4 -> 28, x5 -> 31,
x6 -> 47}, {x1 -> 3, x2 -> 9, x3 -> 15, x4 -> 28, x5 -> 34,
x6 -> 45}, {x1 -> 3, x2 -> 9, x3 -> 19, x4 -> 28, x5 -> 37,
x6 -> 43}, {x1 -> 3, x2 -> 9, x3 -> 23, x4 -> 28, x5 -> 40,
x6 -> 41}, {x1 -> 3, x2 -> 11, x3 -> 11, x4 -> 14, x5 -> 43,
x6 -> 44}, {x1 -> 3, x2 -> 15, x3 -> 15, x4 -> 31, x5 -> 38,
x6 -> 42}, {x1 -> 3, x2 -> 16, x3 -> 23, x4 -> 24, x5 -> 26,
x6 -> 49}, {x1 -> 3, x2 -> 22, x3 -> 23, x4 -> 27, x5 -> 30,
x6 -> 46}, {x1 -> 3, x2 -> 22, x3 -> 27, x4 -> 27, x5 -> 33,
x6 -> 44}, {x1 -> 4, x2 -> 4, x3 -> 7, x4 -> 28, x5 -> 36,
x6 -> 45}, {x1 -> 4, x2 -> 4, x3 -> 11, x4 -> 28, x5 -> 39,
x6 -> 43}, {x1 -> 4, x2 -> 5, x3 -> 11, x4 -> 21, x5 -> 21,
x6 -> 54}, {x1 -> 4, x2 -> 5, x3 -> 15, x4 -> 21, x5 -> 24,
x6 -> 52}, {x1 -> 4, x2 -> 5, x3 -> 19, x4 -> 21, x5 -> 27,
x6 -> 50}, {x1 -> 4, x2 -> 7, x3 -> 7, x4 -> 7, x5 -> 30,
x6 -> 53}, {x1 -> 4, x2 -> 11, x3 -> 11, x4 -> 24, x5 -> 25,
x6 -> 51}, {x1 -> 4, x2 -> 11, x3 -> 15, x4 -> 24, x5 -> 28,
x6 -> 49}, {x1 -> 4, x2 -> 11, x3 -> 19, x4 -> 24, x5 -> 31,
x6 -> 47}, {x1 -> 4, x2 -> 11, x3 -> 23, x4 -> 24, x5 -> 34,
x6 -> 45}, {x1 -> 4, x2 -> 17, x3 -> 19, x4 -> 27, x5 -> 35,
x6 -> 44}, {x1 -> 4, x2 -> 17, x3 -> 23, x4 -> 27, x5 -> 38,
x6 -> 42}, {x1 -> 5, x2 -> 6, x3 -> 7, x4 -> 24, x5 -> 30,
x6 -> 49}, {x1 -> 5, x2 -> 6, x3 -> 11, x4 -> 24, x5 -> 33,
x6 -> 47}, {x1 -> 5, x2 -> 6, x3 -> 15, x4 -> 24, x5 -> 36,
x6 -> 45}, {x1 -> 5, x2 -> 6, x3 -> 19, x4 -> 24, x5 -> 39,
x6 -> 43}, {x1 -> 5, x2 -> 7, x3 -> 15, x4 -> 17, x5 -> 18,
x6 -> 56}, {x1 -> 5, x2 -> 12, x3 -> 15, x4 -> 27, x5 -> 40,
x6 -> 42}, {x1 -> 5, x2 -> 13, x3 -> 15, x4 -> 20, x5 -> 22,
x6 -> 53}, {x1 -> 5, x2 -> 13, x3 -> 19, x4 -> 20, x5 -> 25,
x6 -> 51}, {x1 -> 5, x2 -> 19, x3 -> 19, x4 -> 23, x5 -> 29,
x6 -> 48}, {x1 -> 5, x2 -> 19, x3 -> 23, x4 -> 23, x5 -> 32,
x6 -> 46}, {x1 -> 6, x2 -> 7, x3 -> 7, x4 -> 27, x5 -> 42,
x6 -> 42}, {x1 -> 6, x2 -> 8, x3 -> 11, x4 -> 20, x5 -> 27,
x6 -> 51}, {x1 -> 6, x2 -> 8, x3 -> 15, x4 -> 20, x5 -> 30,
x6 -> 49}, {x1 -> 6, x2 -> 8, x3 -> 19, x4 -> 20, x5 -> 33,
x6 -> 47}, {x1 -> 6, x2 -> 12, x3 -> 35, x4 -> 37, x5 -> 37,
x6 -> 39}, {x1 -> 6, x2 -> 14, x3 -> 15, x4 -> 23, x5 -> 34,
x6 -> 46}, {x1 -> 6, x2 -> 14, x3 -> 19, x4 -> 23, x5 -> 37,
x6 -> 44}, {x1 -> 6, x2 -> 14, x3 -> 23, x4 -> 23, x5 -> 40,
x6 -> 42}, {x1 -> 6, x2 -> 15, x3 -> 15, x4 -> 16, x5 -> 16,
x6 -> 57}, {x1 -> 7, x2 -> 7, x3 -> 27, x4 -> 37, x5 -> 39,
x6 -> 39}, {x1 -> 7, x2 -> 9, x3 -> 11, x4 -> 23, x5 -> 39,
x6 -> 44}, {x1 -> 7, x2 -> 9, x3 -> 15, x4 -> 23, x5 -> 42,
x6 -> 42}, {x1 -> 7, x2 -> 10, x3 -> 11, x4 -> 16, x5 -> 21,
x6 -> 55}, {x1 -> 7, x2 -> 10, x3 -> 15, x4 -> 16, x5 -> 24,
x6 -> 53}, {x1 -> 7, x2 -> 16, x3 -> 19, x4 -> 19, x5 -> 31,
x6 -> 48}, {x1 -> 8, x2 -> 9, x3 -> 27, x4 -> 33, x5 -> 33,
x6 -> 43}, {x1 -> 8, x2 -> 9, x3 -> 31, x4 -> 33, x5 -> 36,
x6 -> 41}, {x1 -> 8, x2 -> 11, x3 -> 11, x4 -> 19, x5 -> 33,
x6 -> 48}, {x1 -> 8, x2 -> 11, x3 -> 15, x4 -> 19, x5 -> 36,
x6 -> 46}, {x1 -> 8, x2 -> 11, x3 -> 19, x4 -> 19, x5 -> 39,
x6 -> 44}, {x1 -> 8, x2 -> 15, x3 -> 27, x4 -> 36, x5 -> 37,
x6 -> 40}, {x1 -> 9, x2 -> 10, x3 -> 15, x4 -> 36, x5 -> 36,
x6 -> 42}, {x1 -> 9, x2 -> 10, x3 -> 19, x4 -> 36, x5 -> 39,
x6 -> 40}, {x1 -> 9, x2 -> 13, x3 -> 15, x4 -> 15, x5 -> 30,
x6 -> 50}, {x1 -> 9, x2 -> 17, x3 -> 31, x4 -> 32, x5 -> 34,
x6 -> 42}, {x1 -> 9, x2 -> 23, x3 -> 27, x4 -> 35, x5 -> 35,
x6 -> 41}, {x1 -> 9, x2 -> 23, x3 -> 31, x4 -> 35, x5 -> 38,
x6 -> 39}, {x1 -> 10, x2 -> 12, x3 -> 19, x4 -> 32, x5 -> 33,
x6 -> 44}, {x1 -> 10, x2 -> 12, x3 -> 23, x4 -> 32, x5 -> 36,
x6 -> 42}, {x1 -> 10, x2 -> 12, x3 -> 27, x4 -> 32, x5 -> 39,
x6 -> 40}, {x1 -> 10, x2 -> 14, x3 -> 15, x4 -> 18, x5 -> 42,
x6 -> 43}, {x1 -> 10, x2 -> 18, x3 -> 19, x4 -> 35, x5 -> 37,
x6 -> 41}, {x1 -> 10, x2 -> 25, x3 -> 31, x4 -> 31, x5 -> 32,
x6 -> 43}, {x1 -> 10, x2 -> 31, x3 -> 31, x4 -> 34, x5 -> 36,
x6 -> 40}, {x1 -> 11, x2 -> 14, x3 -> 23, x4 -> 28, x5 -> 30,
x6 -> 46}, {x1 -> 11, x2 -> 14, x3 -> 27, x4 -> 28, x5 -> 33,
x6 -> 44}, {x1 -> 11, x2 -> 20, x3 -> 23, x4 -> 31, x5 -> 34,
x6 -> 43}, {x1 -> 11, x2 -> 20, x3 -> 27, x4 -> 31, x5 -> 37,
x6 -> 41}, {x1 -> 12, x2 -> 15, x3 -> 15, x4 -> 31, x5 -> 36,
x6 -> 43}, {x1 -> 12, x2 -> 15, x3 -> 19, x4 -> 31, x5 -> 39,
x6 -> 41}, {x1 -> 12, x2 -> 16, x3 -> 23, x4 -> 24, x5 -> 24,
x6 -> 50}, {x1 -> 12, x2 -> 22, x3 -> 23, x4 -> 27, x5 -> 28,
x6 -> 47}, {x1 -> 12, x2 -> 22, x3 -> 27, x4 -> 27, x5 -> 31,
x6 -> 45}, {x1 -> 13, x2 -> 17, x3 -> 19, x4 -> 27, x5 -> 33,
x6 -> 45}, {x1 -> 13, x2 -> 17, x3 -> 23, x4 -> 27, x5 -> 36,
x6 -> 43}, {x1 -> 13, x2 -> 17, x3 -> 27, x4 -> 27, x5 -> 39,
x6 -> 41}, {x1 -> 13, x2 -> 23, x3 -> 23, x4 -> 30, x5 -> 40,
x6 -> 40}, {x1 -> 14, x2 -> 19, x3 -> 19, x4 -> 23, x5 -> 27,
x6 -> 49}, {x1 -> 14, x2 -> 19, x3 -> 23, x4 -> 23, x5 -> 30,
x6 -> 47}, {x1 -> 16, x2 -> 16, x3 -> 19, x4 -> 19, x5 -> 29,
x6 -> 49}, {x1 -> 17, x2 -> 17, x3 -> 19, x4 -> 22, x5 -> 41,
x6 -> 42}, {x1 -> 18, x2 -> 23, x3 -> 31, x4 -> 35, x5 -> 36,
x6 -> 40}, {x1 -> 19, x2 -> 31, x3 -> 31, x4 -> 34, x5 -> 34,
x6 -> 41}, {x1 -> 20, x2 -> 20, x3 -> 23, x4 -> 31, x5 -> 32,
x6 -> 44}, {x1 -> 20, x2 -> 20, x3 -> 27, x4 -> 31, x5 -> 35,
x6 -> 42}, {x1 -> 20, x2 -> 20, x3 -> 31, x4 -> 31, x5 -> 38,
x6 -> 40}, {x1 -> 20, x2 -> 26, x3 -> 27, x4 -> 34, x5 -> 39,
x6 -> 39}, {x1 -> 21, x2 -> 22, x3 -> 27, x4 -> 27, x5 -> 29,
x6 -> 46}, {x1 -> 22, x2 -> 23, x3 -> 23, x4 -> 30, x5 -> 38,
x6 -> 41}, {x1 -> 27, x2 -> 29, x3 -> 31, x4 -> 38, x5 -> 38,
x6 -> 38}}
</code></pre>
<hr />
<p>If we, for example, want to extend the search to <span class="math-container">$10^6$</span> the number of solutions is given by:</p>
<pre><code>In[2]:=Length[FullSimplify[
Solve[{40*x1 + 296*x2 + 945*x3 + 2048*x4 + 4500*x5 + 8640*x6 ==
616103, 1 <= x1 <= x2 <= x3 <= x4 <= x5 <= x6 <= 10^6}, {x1, x2,
x3, x4, x5, x6}, PositiveIntegers]]]
Out[2]=128
</code></pre>
|
192,391 | <p>I know that this is sometimes the case, but that some matrices are not tensors.
So what is the intuitive and specific demands of a matrix to also be a tensor?
Does it need to be quadratic, singular or something else?</p>
<p>Some sources I read seem to suggest that all rank 2 matrices are tensors while other just claims that "some" matrices are rank 2 tensors.</p>
<p>What's the connection between tensors and matrices?</p>
| Jim Belk | 1,726 | <p>This question doesn't have a single good answer, because there isn't a universally agreed upon definition of "tensor" in mathematics. In particular:</p>
<ol>
<li><p>Tensors are sometimes defined as multidimensional arrays, in the same way that a matrix is a two-dimensional array. From this point of view, a matrix is certainly a special case of a tensor.</p></li>
<li><p>In differential geometry and physics, "tensor" refers to a certain kind of object that can be described at a point on a manifold (though the word "tensor" is often used to refer to a <em>tensor field</em>, in which one tensor is chosen for every point). From this point of view, a matrix can be used to describe a rank-two tensor in local coordinates, but a rank-two tensor is not itself a matrix.</p></li>
<li><p>In linear algebra, "tensor" sometimes refers to an element of a <a href="http://en.wikipedia.org/wiki/Tensor_product">tensor product</a>, and sometimes refers to a certain kind of multilinear map. Again, neither of these is a generalization of "matrix", though you can get a matrix from a rank-two tensor if you choose a basis for your vector space.</p></li>
</ol>
<p>You run into the same problem if you ask a question like "Is a vector just a tuple of numbers?" Sometimes a vector is defined as a tuple of numbers, in which case the answer is yes. However, in differential geometry and physics, the word "vector" refers to an element of the tangent space to a manifold, while in linear algebra, a "vector" may be any element of a vector space.</p>
<p>On a basic level, the statement "a vector is a rank 1 tensor, and a matrix is a rank 2 tensor" is roughly correct. This is certainly the simplest way of thinking about tensors, and is reflected in the <a href="http://en.wikipedia.org/wiki/Einstein_notation">Einstein notation</a>. However, it is important to appreciate the subtleties of this identification, and to realize that "tensor" often means something slightly different and more abstract than a multidimensional array.</p>
|
3,232,998 | <p>In Velleman's How to Prove It, the strategy given for proving goal of the form <span class="math-container">$P \lor Q$</span> goes like this:</p>
<blockquote>
<p>If <span class="math-container">$P$</span> is true, then clearly <span class="math-container">$P \lor Q$</span> is true. Now suppose <span class="math-container">$P$</span> is false.</p>
<p><code>[Proof of Q goes here]</code></p>
<p>Thus, <span class="math-container">$P \lor Q$</span> is true.</p>
</blockquote>
<p>I feel like the first sentence is redundant. After all, when proving <span class="math-container">$P \lor Q$</span>, it's equivalent to proving <span class="math-container">$\lnot P \to Q$</span>, thus we only need to suppose <span class="math-container">$\lnot P$</span> to begin with. </p>
| 雨が好きな人 | 438,600 | <p><em>Any</em> statement in <em>any</em> proof is redundant if you consider it ‘too obvious’.</p>
<p>In this case, maybe you consider it obvious that <span class="math-container">$P \implies P \lor Q$</span> (and I wouldn’t blame you). It does follow directly from the definition of <span class="math-container">$\lor$</span>, but it’s nice to be especially clear when writing a proof and so it doesn’t hurt to add it in. It is certainly true, though, that proving <span class="math-container">$\neg Q \implies P$</span> will suffice for most readers.</p>
|
3,232,998 | <p>In Velleman's How to Prove It, the strategy given for proving goal of the form <span class="math-container">$P \lor Q$</span> goes like this:</p>
<blockquote>
<p>If <span class="math-container">$P$</span> is true, then clearly <span class="math-container">$P \lor Q$</span> is true. Now suppose <span class="math-container">$P$</span> is false.</p>
<p><code>[Proof of Q goes here]</code></p>
<p>Thus, <span class="math-container">$P \lor Q$</span> is true.</p>
</blockquote>
<p>I feel like the first sentence is redundant. After all, when proving <span class="math-container">$P \lor Q$</span>, it's equivalent to proving <span class="math-container">$\lnot P \to Q$</span>, thus we only need to suppose <span class="math-container">$\lnot P$</span> to begin with. </p>
| matt | 401,452 | <p>I think that those are two different ways of proving the same thing. If we want to prove that P or Q is true, we can prove that P is true, or we can prove that Q is true. We could also assume P is false and show that it implies Q is true. Then we would have proven the disjunction as it is logically equivalent to the implication.</p>
|
1,294,792 | <p>Can a 2-dimensional sphere be decomposed into two disjoint homeomorphic subspaces? If yes, can these subspaces be non-discrete / connected / have some other good properties?</p>
| Chappers | 221,811 | <p>You can actually prove this using Fourier series: consider the Fourier series of $\cos{\alpha x}$ on $[-\pi,\pi]$. The function is continuous, and differentiable except at the endpoints, so the Fourier series converges to the function everywhere. Some easy integration will show you that the Fourier coefficients are such that the series is
$$ \cos{\alpha x} = \frac{2\alpha \sin{\pi \alpha}}{\pi} \left( \frac{1}{2\alpha^2} + \sum_{n=1}^{\infty} \frac{(-1)^n}{\alpha^2-n^2} \cos{nx} \right). $$
Now setting $x=\pi$ and $\alpha=z$ gives
$$ \frac{\pi \cot{\pi z}}{z} = \sum_{n=-\infty}^{\infty} \frac{1}{z^2-n^2}. $$</p>
<hr>
<p>What the question might actually be looking for, on the other hand, is the method explained on <a href="https://archive.org/stream/courseofmodernan00whit#page/134/mode/2up" rel="nofollow">p.134 of Whittaker and Watson</a>, which is a valuable resource for these sorts of slightly off-the-beaten-track results.</p>
|
1,342,951 | <p>I was thinking what function I should compare it to. If I say whether a function is smaller or bigger than this one, then I must prove that. I was thinking of (x+1)^2 but I realized that this converges so that won't prove the integral diverges. So, what function should I use?</p>
| Michael Hardy | 11,667 | <p>It would be circular only if you were to take the equality $P(A,B) = P(A \cap B) = P(A \mid B)P(B)$ to be a definition. So don't do that.</p>
|
47,728 | <p>Function <code>CreateTemporary</code> can generate a unique file name like</p>
<pre><code>file = CreateTemporary[]
(* "C:\Documents and Settings\kkouptsov\Local \Settings\Temp\m-263a0380-3a03-49ac-95cb-d21390b2d3fd" *)
</code></pre>
<p>How do we generate a unique file name such as above example? The reason I don't use <code>CreateTemporary</code> because I don't need to actually create the file, but only need the file path.</p>
<p>I plan to generate many of these names in parallel, and they should be unique to each other.</p>
| halirutan | 187 | <p>If it is merely about creating something unique, why don't you use the current time to build a readable file-name which is unique?</p>
<pre><code>DateString[{"file", "Year", "-", "MonthNameShort",
"Day", "-", "Hour24", "Minute", "Second", ".txt"}]
(* "file2014-May13-013029.txt" *)
</code></pre>
<p>Since it contains the time it will be unique (in this case up to the second). When you create many files and additionally, do it in parallel, you can add a counter and the kernel id. This lets you identify the kernel which created the name and it makes the files unique in one run. When you start it again, the time will be different and the counter can start over. </p>
<pre><code>ParallelEvaluate[i = 0];
ParallelTable[
DateString[{ToString@$KernelID, "Year", "-", "MonthNameShort", "Day",
"-", "Hour24", "Minute", "Second", "-", IntegerString[i++, 16, 5],
".txt"}], {10}]
(* {"82014-May13-015507-00000.txt", \
"82014-May13-015507-00001.txt", "72014-May13-015507-00000.txt", \
"72014-May13-015507-00001.txt", "62014-May13-015507-00000.txt", \
"52014-May13-015507-00000.txt", "42014-May13-015507-00000.txt", \
"32014-May13-015507-00000.txt", "22014-May13-015507-00000.txt", \
"12014-May13-015507-00000.txt"} *)
</code></pre>
<p>I tested this with 10^5 file-names and all were unique.</p>
|
786,187 | <p>I can't seem to figure this one out:</p>
<p>$\int \frac{t^5}{\sqrt{t^2+2}}dt$</p>
<p>$t^5 / \sqrt{t^2+2}$</p>
<p>I know I need to substitute t for a trigonometric function, $\tan$ I think</p>
<p>Any hints are greatly appreciated!</p>
| David H | 55,051 | <p>Alternatively, you don't necessarily need trigonometric substitution to solve this problem.</p>
<p>Substituting $u=t^2+2$,</p>
<p>$$\int \frac{t^5}{\sqrt{t^2+2}} dt = \int \frac{(u-2)^2}{2\sqrt{u}}du\\
=\int\frac12u^{3/2}du-\int 2\sqrt{u}\,du+\int\frac{2}{\sqrt{u}}du\\
=\frac15u^{5/2}-\frac43 u^{3/2}+4\sqrt{u}+constant\\
=\frac15(t^2+2)^{5/2}-\frac43 (t^2+2)^{3/2}+4\sqrt{t^2+2}+constant$$</p>
<hr>
<p>Notes:</p>
<p>Let $u=t^2+2$. Then,</p>
<p>$$t^2=u-2\implies t^4=(t^2)^2=(u-2)^2.$$</p>
<p>From $u=t^2+2$, we also have $du=2t\,dt\implies \frac12du=t\,dt$. Hence,</p>
<p>$$t^5dt=(t^4)\cdot(t\,dt)=\frac{(u-2)^2}{2}du$$</p>
|
193,355 | <p>I am taking a calculus course, and in one of the exercises in the book, I am asked to find the limits for both sides of $\sqrt{x}$ where $x \to 0$.</p>
<p>Graph for sqrt(x) from <a href="http://www.wolframalpha.com/input/?i=sqrt%28x%29" rel="nofollow noreferrer">WolframAlpha</a>:<br>
<img src="https://i.stack.imgur.com/ahbjy.gif" alt="Graph for sqrt(x) from WolframAlpha"></p>
<p>This is how I solved the exercise:</p>
<blockquote>
<p>For simplicity, I choose to disregard the negative result of
$\pm\sqrt{x}$. Since we are looking at limits for $x \to 0$, both
results will converge at the same point, and will thus have the same
limits.</p>
<p>$\sqrt{x}$ = $0$ for $x = 0$.<br></p>
<p>$\sqrt{x}$ is a positive real number for all $x > 0$.<br> $\displaystyle \lim_{x \to 0^+} \sqrt{x} = \sqrt{+0} = 0$</p>
<p>$\sqrt{x}$ is a complex number for all $x < 0$.<br > $\displaystyle \lim_{x \to 0^-} \sqrt{x} = \sqrt{-0} = 0 \times \sqrt{-1} = 0i = 0$</p>
</blockquote>
<p>The solution in the book, however, does not agree that there exists a limit for $x \to 0-$.</p>
<p>I guess there are three questions in this post, although some of them probably overlaps:</p>
<ul>
<li>Does $\sqrt{x}$ have a limit for $x \to 0$?</li>
<li>Are square root functions defined to have a range of only real numbers, unless specified otherwise?</li>
<li>Is $\sqrt{x}$ continuous for $-\infty < x < \infty$?</li>
</ul>
<p>WolframAlpha says the limit for x=0 is 0: <a href="http://www.wolframalpha.com/input/?i=limit+x+to+0%2C+sqrt%28x%29" rel="nofollow noreferrer">limit (x to 0) sqrt(x)</a></p>
<p>And also that both the positive and negative limits are 0: <a href="http://www.wolframalpha.com/input/?i=limit+%28x+to+0-%29++sqrt%28x%29" rel="nofollow noreferrer">limit (x to 0-) sqrt(x)</a></p>
<p>If my logic is flawed, please correct me.</p>
| gerardlouw | 29,729 | <p>The answers to your questions depend on whether you are working with $\mathbb{R_{\geq 0}}$ or $\mathbb{R}$ as your domain. $\sqrt{x}$ is really not the same function in these two cases.</p>
<p>With a domain of $\mathbb{R_{\geq 0}}$:</p>
<ul>
<li>Clearly $\lim_{x \to 0^+} \sqrt{x} = 0$. However $\lim_{x \to 0^-} \sqrt{x}$ is not defined for negative numbers in this case, so $\lim_{x \to 0^-} \sqrt{x}$ is undefined.</li>
<li>If your course is on real analysis, they most likely assume that the domain is $\mathbb{R_{\geq 0}}$. However, this could depend very much on the context. If you are unsure what your course's convention is, you should consult a teaching assistant.</li>
<li>In this case, $\sqrt{x}$ is not continuous on $\mathbb{R}$, since it is not even defined everywhere.</li>
</ul>
<p>With a domain of $\mathbb{R}$:</p>
<ul>
<li>As your plot shows you, both the real and the imaginary parts tend to 0, and so $\sqrt{x}$ tends to 0.</li>
<li>See above.</li>
<li>Once again, your plot shows you that both the real and imaginary parts are continuous functions over $\mathbb{R}$, and so $\sqrt{x}$ is continuous over $\mathbb{R}$.</li>
</ul>
<p>Note that there is an even more general case, where the domain is $\mathbb{C}$. This is where things get very strange.</p>
|
2,612,538 | <p>I want to find the Eigenvalues and Eigenvectors of the following matrix $$A=\begin{pmatrix}0&1\\-3&0\end{pmatrix}.$$</p>
<p>I'm able to calculate the Eigenvalues, which are simply the zero points of A's characteristic polynom $\lambda^2 + 3$:<br>
$$\lambda_{1,2} = \pm i\sqrt{3}\\$$ </p>
<p>But how do I calculate the Eigenvectors? I know that in theory the process is the same as in $\Re$ (clearly I lack some understanding even in $\Re$), but no matter what I do I cannot reach the same solution as is suggested by <a href="https://www.wolframalpha.com/input/?i=%7B%20%7B0,%201%7D,%20%7B-3,%200%7D%20%7D" rel="nofollow noreferrer">Wolframlpha</a>:
$$v_1=\begin{pmatrix}-i\sqrt{3}\\1\end{pmatrix}$$
$$v_2=\begin{pmatrix}+i\sqrt{3}\\1\end{pmatrix}$$</p>
<p>For example, when I try to solve the two equations
\begin{equation}
(i\sqrt{3})c_1+c_2 = 0 \\
-3c_1 + (i\sqrt{3})c_2 = 0
\end{equation}</p>
<p>I first multiply the upper row by $(-i\sqrt{3})$ which gives me $3c_1 - (-i\sqrt{3})c_2$. Adding this to the lower equation gives me $0=0$.</p>
<p>I've tried applying row operations as well (like <a href="https://math.stackexchange.com/questions/1867002/how-to-find-eigenvectors-given-complex-eigenvalues">here</a>), but ultimately I end up at</p>
<p>$$\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$</p>
<p>I would appreciate it if you could point my mistake out, or give me some advice on how to solve this problem.</p>
| amd | 265,466 | <p>I suggest checking your input to Wolfram|Alpha since neither vector that you’ve listed in your question is an eigenvector of $A$. For instance, $A[i\sqrt3,1]^T=[1,-3i\sqrt3]^T$, which is not a scalar multiple of $[i\sqrt3,1]^T$.</p>
<p>That aside, you should expect to end up with $0=0$ when you combine the two equations the way you did. Since $i\sqrt3$ is an eigenvalue of $A$, $A-i\sqrt3 I$ must be singular: its rows are linearly dependent, and so are the two equations that they represent—one is a multiple of the other. The system has an infinite number of solutions, and any non-zero solution will be an eigenvector. </p>
<p>The first equation says that $c_2=-i\sqrt3 c_1$. So, if you choose $c_1$ as the free variable, then any nonzero vector of the form $(c_1,-i\sqrt3 c_1)^T$ will be a corresponding eigenvector.</p>
|
11,267 | <p>John Von Neumann once said to Felix Smith, "Young man, in mathematics you don't understand things. You just get used to them." This was a response to Smith's fear about the method of characteristics. </p>
<p>Did he mean that with experience and practice, one obtains understanding? </p>
| S. P. Forman | 236,021 | <p>He meant that when you think you understand math, you have in fact fallen under the sway of an <em>illusion</em>. This illusion of understanding is brought on by familiarity ("getting used to it") coupled with intellectual laziness. People get used to doing math and having it work and they forget that no one knows <em>why</em> it works, they forget that it is fundamentally mysterious.</p>
<p>If you read this and object, let me ask you,</p>
<p>What is number?</p>
|
11,267 | <p>John Von Neumann once said to Felix Smith, "Young man, in mathematics you don't understand things. You just get used to them." This was a response to Smith's fear about the method of characteristics. </p>
<p>Did he mean that with experience and practice, one obtains understanding? </p>
| Len Margolin | 869,981 | <p>Poincare once wrote that all of mathematics is one plus one equals two, and a lot of definitions. Speaking as a physicist, I believe the most important thing to understand about mathematics is whether those definitions are useful in describing reality (i.e., observations). I don't believe mathematicians are similarly concerned, though certainly von Neumann himself is a most prominent counter-example.</p>
<p>Just to be provocative, I have found Newton's calculus to be sufficient for my needs and have never understood \delta -- \epsilon proofs. There seems to be no place for "infinity" in physics. On the other hand, nonstandard analysis has proven quite useful to me.</p>
|
522,179 | <p>Everyone says that the dot product is interpreted as the projection of A onto B (if you are dot producting A and B), but isn't that length just equal to |A|$\cos \left( \theta \right)$? Why does the dot product have an extra |B|?</p>
| Aaron Golden | 37,535 | <p>If we didn't account for the length of $B$ then the order of the operands to the dot product would matter. $A$ dot $B$ would not be equal to $B$ dot $A$, and that would be irritating.</p>
<p>I suppose you could think of taking the dot product of $A$ and $B$ as going in three steps. First scaling $B$ to be unit length, then measuring the projection of $A$ onto (unit length) $B$, then <em>undoing</em> that first scaling. I'm not sure if that's the most intuitive way to go. The main thing to remember is that you've got to do <em>something</em> to account for the length of $B$, or the dot product couldn't be commutative.</p>
|
2,153,580 | <p>I'm trying to prove that $K((x))/K(x)$ is not an algebraic extension, where $K$ is a field of characteristic $p>0$, and $K((x))$ is the field of fractions of $K[[x]]:=\{\sum_{i=0}^{\infty}a_ix^i\mid a_i\in K\}$.</p>
<p>To make it more concrete, I'm considering the case $K=\mathbb{Z}/p\mathbb{Z}$ and $x=p$, so that we need to find $\tau\in\mathbb{Q}_p$ transcendental in respect to the field extension $\mathbb{Q}_p|\mathbb{Q}$. </p>
<p>I thought about $\tau:=\sum_{i=0}^{\infty}\frac{p^n}{n!}$, which I've checked converges in the $p$-adic metric, but I'm not sure is transcendental or not. It's very tempting to say that $e^p=\sum_{i=0}^{\infty}\frac{p^n}{n!}$, and since $e$ is transcendental, we're done, but I know this probably doesn't make any formal sense. How do I resolve this?</p>
| Sungjin Kim | 67,070 | <p>Maybe showing that a certain element is transcendental is not easy. I think that about the same or more work is required as to prove that $e^p\in\mathbb{R}$ is transcendental. </p>
<p>One way to get around is to use the cardinality comparison. This way is not constructive, but we can prove that there is $\tau\in\mathbb{Q}_p$ which is transcendental over $\mathbb{Q}$. </p>
<p>Since $\mathbb{Q}$ is countable, the set of algebraic elements inside $\mathbb{Q}_p$ over $\mathbb{Q}$ is countable. However, the set $\mathbb{Q}_p$ is uncountable. Thus, there must be a transcendental element $\tau\in\mathbb{Q}_p$. </p>
<p>This argument is possible for any countably infinite field $K$. Set-theoretically stating, the algebraic closure $\overline{K}$ over $K$ satisfies
$$
|\overline{K}|=|K| = \aleph_0 < 2^{\aleph_0} = |K^{\mathbb{N}}|=|K((x))|.
$$</p>
|
65,865 | <p>There are so many different optimization algorithms out there, and lots of research going on. However, I have difficulties to find good comparison between them, and all articles / books / papers seem to evaluate them in different ways.</p>
<p>Isn't there some set of standard functions to run "standardized" benchmarks in order to test and compare all such algorithms? Is there some reference website showing how each algorithm performs?</p>
<p>(I'm especially interested in numerical derivative-free optimization techniques)</p>
| H. Vabri | 410,305 | <p>Due to optimisation problems varying considerably you might want to creating a library of test problems that somewhat represent the problems you are trying to solve, optimising these using multiple optimisation algorithms and comparing their performance. For that an increasingly accepted method to compare optimisation software is <em>performance profiles</em>, as stated by @Dominique. A method to generate these in Python can be found <a href="https://github.com/Jvanschoubroeck/performance-profiles-in-python" rel="nofollow noreferrer">here</a>. For MATLAB see the post of @Dominique.</p>
|
779,328 | <p>I am almost a beginner in the topics of Fourier transform. So, I am asking this question here.</p>
<p>Let <span class="math-container">$n=3$</span> and let <span class="math-container">$\mu_t$</span> denote surface measure on the sphere <span class="math-container">$|x|=t$</span>. Then how do we show that
<span class="math-container">$$
\frac{\text{sin} 2\pi t|\xi|}{2\pi|\xi|}=\frac{1}{4\pi t}\hat{\mu_{t}}(\xi)
$$</span>
where <span class="math-container">$\hat{\mu_{t}}$</span> is the Fourier transform of <span class="math-container">$\mu_t$</span>.</p>
| Giuseppe Negro | 8,157 | <p>It is easy if we use Kirchhoff's formula for the solution to the following wave equation:
\begin{equation}
\begin{cases}
(\partial^2_t-\Delta)u=0, & x\in \mathbb{R}^3,\ t\in \mathbb{R}\\
u(0, x)=0 \\
u_t(0, x)=f
\end{cases}
\end{equation}
Kirchhoff's formula says that
\begin{equation}
u(t,x)=\frac{1}{4\pi t} \int_{\lvert y-x\rvert=t} f(y)\, dS(y) =\frac{1}{4\pi t}d\sigma_t\ast f,
\end{equation}
so taking the spatial Fourier transform one gets
\begin{equation}
\frac{1}{4\pi t} \widehat{d\sigma}_t \hat{f}= \hat{u}(t, \xi).
\end{equation}
On the other hand, for fixed $\xi$ the function $\hat{u}(\cdot, \xi)$ solves the following Cauchy problem, which is a harmonic oscillator with respect to the time variable $t$:
\begin{equation}
\begin{cases}
(\partial^2_t +4\pi^2\lvert \xi\rvert^2)\hat{u}(t,\xi)= 0,& t\in \mathbb{R} \\
\hat{u}(0, \xi)=0 \\
\hat{u}_t(0, \xi)=\hat{f}(\xi).
\end{cases}
\end{equation}
Therefore
\begin{equation}
\hat{u}(t, \xi)=\frac{\sin 2\pi t\lvert \xi\rvert}{2\pi \lvert \xi\rvert}\hat{f}(\xi).
\end{equation}
Equating the two expressions for $\hat{u}$ yields the desired formula. </p>
<hr>
<p>One might think that this answer is a cheat, since one usually <em>derives</em> Kirchhoff's formula by first computing that Fourier transform. However, that's not the only way to derive Kirchhoff's formula. Another popular approach uses spherical means and is completely devoid of any Fourier transform methods. So the computation presented in this answer is not circular. </p>
<p>I'll leave a reference for a more direct computation just in case: you can find it in exercises to the chapter "Topics in Fourier Analysis" of the book "Real Analysis" by G. Folland (both editions).</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| guest | 13,465 | <p>The simple answer I would give is that there is a good chance they go into the business world at SOME TIME in the future of their life. Perhaps they will get an MBA at some point to help them move up. And you'd laugh at how deer in the headlights the English major types were with finance math. And how the STEM majors ate it up like Valentines Day heart candy,</p>
<p>As such, no they won't be doing algebra, per se. But some basic training in algebra is helpful to be more numerate when looking at spreadsheets or presentation slides where percentages, ratios, compounding, etc. occur.</p>
<p>I do think you could cut the cord before trig/calc, though. Of course that means they can't handle high school physics. So they won't know that either. But high school chem is fine with algebra 2. </p>
<p>P.s. I disagree on the counterpoint that STEM people can skip humanities. You absolutely do need a knowledge of writing (more and more). Also a knowledge of your culture from history. I would draw the line on lib'rul college foolishness like Rigoberto Menchu. But classical Western Civ (or Chinese for a Chinese) is absolutely something an educated...person should know (to read the newspaper, discuss politics, etc.)</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| Lawnmower Man | 13,468 | <h1>Citizenship</h1>
<p>The health of a democracy depends on the electorate being thoughtful and <em>informed</em>. And said citizens can only be informed if they understand the information around them (e.g., the news). Very few policies boil down to: "A is morally right and B is morally wrong." Those issues were mostly settled long ago or established in the Constitution. The issues facing most legislators are questions like: "Is A better than B? By how much? What are the costs involved? How much uncertainty do we have?" And while we will always take qualitative values into consideration, the difference between A and B often comes down to <em>quantities</em>.</p>
<p>Only a few people debate whether we should <em>have</em> a minimum wage. Most folks instead debate: <em>how large</em>? To do this reasonably, we need studies with results, which means <em>statistics</em>. The very process of conducting an election is rife with statistics. Which means, at the very least, every citizen should have a solid understanding of basic statistics, and the stronger that understanding, the better. But it's hard to get a foundational knowledge of statistics without some algebra, and even calculus is helpful in understanding continuous distributions.</p>
<h1>Economics</h1>
<p>Related to the citizenship issues above, but worth mentioning on its own, is the centrality of economics to every citizen. We all need to know how to spend money wisely (except the few of us born into extreme wealth). It's hard to answer a basic question like: "Which is cheaper: the 16 oz jar for \$3.88 or the 24 oz jar for $4.96?" without knowing some algebra. And even if you can do this math without having passed an algebra course, you are doing algebra whether you call it such or not. So students would probably be more motivated if algebra were taught in the context of personal finance.</p>
<p>And that's not even getting into more difficult questions like: "Should I pay an extra $10,000 on my mortgage to take off a point?" Or: "How much should I invest in my 401(k) vs. directly buying ETFs?" Anyone who lives a life without math lives a life without money. Capitalism is rife with value judgments of a very precise numerical nature. Even computing the tip at a restaurant or coffee shop requires math skills. Formulating a retirement strategy at the start of your career requires more than simple arithmetic.</p>
<h1>Computer Science</h1>
<p>As a professional software engineer, I obviously think that programming is valuable (but the salary : education ratio of CS also speaks quite loudly). I think most of us recognize that the future is automation. Virtually every scientific field is inundated with software and automation. Every startup wants to call itself a "tech company" (including obviously not-tech companies like mattress-maker Casper). But every corporation benefits from tech-savvy employees, even if that just means knowing how to put together a few formulas in Excel to aid planning and scheduling. The ubiquity of personal electronics, from laptops to smart phones, means that the youngest generation is practically born as "power users", especially relative to older generations.</p>
<p>At the very least, computer science needs combinatorics. And for combinatorics, you need algebra. Eventually, some very basic scripting capability will be required of virtually all white collar jobs. So I would double down and say that not only <em>should</em> math be required education for all high schoolers, but <em>also</em> computer science. Anyone who says we don't need math and CS is living in the past.</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| Reese Johnston | 10,081 | <p>Math, taught well, teaches modes of thought, not facts. A person who is exposed to a wider variety of modes of thought will be more prepared to critically consider new ideas. It's the same reason we teach history and literature - I can't honestly say that I've had reason to use my middle-school knowledge of Macbeth for anything, but the tools of critical reading and strategies for considering interpretations of a text have been useful for spotting articles that are meant to provoke a certain feeling and bias the reader. I've never needed to know when my hometown was founded, but a basic understanding of what kind of historical facts we know and what evidence we have of them helps me distinguish between historical fact and conspiracy theory. And I've never needed to be able to do algebra (except, of course, in my capacity as a math instructor) but the ability to mix systematic thought with creative strategizing -- the skills that come out of long algebra problems, especially word problems -- allows me to manage my finances, cook a decent meal, organize my closet, or schedule a vacation. The ability to pay attention to key details without losing the big picture allows me to follow detailed instructions well (have you ever assembled a piece of Ikea furniture?). The ability to reason precisely, as in a geometry proof, helps me spot when other people's arguments are appealing to emotion rather than reason.</p>
<p>To be honest, of all the skills I learned at or below the high-school level, math seems like the <em>most</em> useful.</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| A. I. Breveleri | 13,476 | <p>Many people think that the clientele of our public school system is <em>the students</em>. Others act as if the clientele were the <em>parents</em> of the students. </p>
<p>Those people are wrong. The client is the <em>citizenry of his state and nation</em>, who need the electorate to be educated and informed, in order to secure the benefits of a well-ordered government. </p>
<p>The question, "How does the study of mathematics<sup>1</sup> benefit the student at any point in his life?" is irrelevant. The student has not contracted for the education he is receiving, and has no right to dictate its form. </p>
<p>The proper question to ask of publicly funded universal compulsory education is, "How does the study of mathematics benefit the state, nation, and society that the student will eventually be called upon to supervise and control?" </p>
<p>When the studies, that contribute to a citizen's ability to keep his republic, also augment his ability successfully to pursue happiness, that is a fortunate serendipity. But the educational purposes of the student do not trump the educational purposes of the state. </p>
<p>So the answer to "Why is learning mathematics compulsory?" is "Because the people have decided that you need to know mathematics in order to be a worthy citizen." </p>
<hr>
<ol>
<li>For "mathematics" substitute any subject that the <em>people</em> consider necessary to an educated voter. </li>
</ol>
|
473,972 | <p>Let $X$ be an infinite set. Prove that there is a bijective function $f: X \rightarrow X$ with the property that for every $x \in X$ and all $n > 0$: $f^n(x) \neq x$.</p>
<p>I've tried to proved this by considering a bijective function $g: \mathbb{Z} \times X \rightarrow X$ in a certain way (by the composition of a function $f: \mathbb{Z} \times X \rightarrow \mathbb{Z} \times X$), but that's all i've got at the moment.</p>
| Community | -1 | <p><strong>Hint</strong>: By changing the order of integration:</p>
<p>$$\int_0^1 \int_x^1 e^{x/y} dy dx = \int_0^1 \int_0^y e^{x/y} dx dy$$</p>
<p>Now the inside integral can be evaluated by</p>
<p>$$\int_0^y e^{x/y} dx = y e^{x/y} |_0^y = y e^1 - y e^0$$</p>
|
72,994 | <p>I am not able to solve this question from chapter "Similar & Congruent Triangles" in my book.</p>
<p>Can some one help to calculate <strong>AC</strong>?</p>
<p><img src="https://i.stack.imgur.com/5ZTn9.jpg" alt="HERE">.</p>
| Ajay Kumar | 375,350 | <p><a href="https://i.stack.imgur.com/XCTvV.png" rel="nofollow noreferrer">enter image description here</a> using triangle similarity
here the triangle 1 is similar to the triangle 2, also the triangle 2 is similar to the triangle 3</p>
|
2,932,510 | <p><a href="https://i.stack.imgur.com/9kTCs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9kTCs.png" alt="enter image description here"></a>
Apostol tells the reader to verify the last sentence.
I've verified the backward implication (<span class="math-container">$a_kx+b_k=0\implies \text{equality}$</span>).
I'm having trouble verifying the forward implication. Could someone show me a proof of the forward implication?</p>
| Siong Thye Goh | 306,553 | <p>The input to <span class="math-container">$f$</span> is <span class="math-container">$2x+6$</span>,</p>
<p>hence <span class="math-container">$$y = -2(2x+6)^4 - 4$$</span></p>
<p>We can also write it as </p>
<p><span class="math-container">$$y = -2(2(x+\color{red}3))^4 - 4$$</span></p>
<p>since <span class="math-container">$2x+6=2(x+3)$</span>.</p>
|
123,675 | <p>What are some "applications to" / "connections with" topology that one could hope to reasonably cover in a first course on topos theory (for master students)? I have an idea of what parts of the theory I would like to cover, however, I would love some more nice examples and applications. Of course, I have some ideas of my own, but am open to suggestions. Thank you!</p>
<p>P.S.</p>
<p>I am also interested in perhaps learning about some new connections myself, which would be out of reach for such a course, so feel free to leave these as well, qualified as such.</p>
| Andrej Bauer | 1,176 | <p>Here is one that is not too grand.</p>
<p>Suppose you want to embed some (essentially small) category of spaces into a cartesian-closed category, so as to extend it to a <a href="http://ncatlab.org/nlab/show/convenient+category+of+topological+spaces" rel="nofollow">convenient category</a>. Then you can use a gros topos to do it. The site is your category of spaces, a family $\lbrace e_i : X_i \to Y\rbrace_i$ is covering when the $e_i$ are open embeddings and they cover $Y$ (I hope I got that right). The Yoneda lemma embeds the original category.</p>
<p>It is not hard to come up with interesting exercises that are not too demanding.</p>
|
4,090,775 | <p>Let <span class="math-container">$\mu$</span> be the Lebesgue measure and, <span class="math-container">$E$</span> and <span class="math-container">$F$</span> compact subsets of <span class="math-container">$\mathbb{R}$</span> such that <span class="math-container">$E\subset F$</span> and <span class="math-container">$\mu(E)=1$</span> and <span class="math-container">$\mu(F)=3$</span>. Prove there is a compact subset <span class="math-container">$K$</span> of <span class="math-container">$\mathbb{R}$</span> such that <span class="math-container">$E \subseteq K \subseteq F$</span> and <span class="math-container">$\mu(K)=2$</span>. First and foremost, as <span class="math-container">$\mu(E)=1$</span> and <span class="math-container">$\mu(F)=3$</span> are compact subsets of <span class="math-container">$\mathbb{R}$</span> there are open subsets of <span class="math-container">$\mathbb{R}$</span>, <span class="math-container">$E'$</span> and <span class="math-container">$F'$</span>, such as <span class="math-container">$E \subset E'$</span> and <span class="math-container">$F' \subset F$</span> and <span class="math-container">$\mu^{\ast}(E'- E) < \epsilon$</span> and <span class="math-container">$\mu^{\ast}(F'- F) < \epsilon$</span>. Where <span class="math-container">$\mu(E')=\mu(E)=1$</span> and <span class="math-container">$\mu(F')=\mu(F)=3$</span>. Obviously, <span class="math-container">$\mu^{\ast}$</span> stands for Lebesgue outer measure. So far, I´ve been run out of ideas and aproaches to solve this. I have googled properties about Lebesgue measure of compact sets and there is nothing about them that I have found helpful, for instance, all compact sets are Lebesgue measurable and <span class="math-container">$\mu(E)< \infty$</span> for every compact set in <span class="math-container">$\mathbb{R}$</span>. But I dont know how I´m supposed to construct this compact set <span class="math-container">$K$</span> such as <span class="math-container">$E \subseteq K \subseteq F$</span> and <span class="math-container">$\mu(K)=2$</span>. Thanks!</p>
| DanielWainfleet | 254,665 | <p>There seems to be a new style of teaching measure theory without mentioning <span class="math-container">$inner$</span> measure. Let <span class="math-container">$m^o$</span> denote outer measure.</p>
<p>For <span class="math-container">$S\subseteq \Bbb R,$</span> let <span class="math-container">$C(S)$</span> be the set of compact subsets of S.</p>
<p>The inner measure of <span class="math-container">$S$</span> is <span class="math-container">$m^i(S)=\sup \{m^o(T): T\in C(S)\}=\sup \{m(T): T\in C(S)\}.$</span></p>
<p>If <span class="math-container">$m^0(S)<\infty$</span> then <span class="math-container">$S$</span> is measurable iff <span class="math-container">$m^i(S)=m^o(S)$</span> iff <span class="math-container">$m(S)=m^i(S)=m^o(S)$</span>.</p>
<p>If <span class="math-container">$E,F$</span> are compact with <span class="math-container">$E\subset F$</span> and <span class="math-container">$m(E)=1$</span> and <span class="math-container">$m(F)=3$</span> then <span class="math-container">$m(F\setminus E)=m^i(F\setminus E)=2,$</span> so take <span class="math-container">$T\in C( F$</span> \ <span class="math-container">$E)$</span> with <span class="math-container">$1<m(T)\le 2.$</span></p>
<p>For <span class="math-container">$r\in [0,\infty )$</span> let <span class="math-container">$f(r)=m(T$</span> \ <span class="math-container">$(-r,r)\,).$</span> Now <span class="math-container">$f$</span> is monotonic decreasing with <span class="math-container">$f(0)=m(T)>2,$</span> and <span class="math-container">$f(r)=0$</span> for some <span class="math-container">$r>0$</span> because <span class="math-container">$F$</span> \ <span class="math-container">$E$</span> is a bounded set.</p>
<p>So let <span class="math-container">$r_1=\sup \{r>0:f(r)\ge 1\}.$</span></p>
<p><span class="math-container">$(\bullet).\;$</span> We can now show that <span class="math-container">$f(r_1)=1.$</span></p>
<p>So we can let <span class="math-container">$K=E\cup (T$</span> \ <span class="math-container">$(-r_1,r_1)\,).$</span></p>
<p>Remark:<span class="math-container">$f$</span> is Lipschitz-continuous with Lipschitz constant <span class="math-container">$2$</span> because if <span class="math-container">$0\le r<r'$</span> then <span class="math-container">$0\le f(r)-f(r')\le m(\,(-r',-r]\cup [r,r')\,)= 2(r'-r).$</span></p>
|
2,315,599 | <p>I am trying to evaluate the integral<br>
$\int (\frac{1}{z(\exp(z)-1)}dz$ $\mathbb{C}:|z|=1\}$</p>
<p>using Laurent theorem. $0$ is a singularity point and both denominators becomes $0$ when $z=0$. So it's hard to find a method to evaluate this using Residue theorem. Could someone please show me how to solve this.
Thank you.</p>
| Deepak | 151,732 | <p>On that "special day", the girl walks for an extra $12 - 3 = 9$ minutes to get to the new meeting point. The tram has travelled for $3$ minutes less than usual to get to that new meeting point. The distance of the new meeting point from the usual one is the same for both. Therefore the tram is travelling $\frac 93 = 3$ times as fast as the girl, so its speed is $(5.5)(3) = 16.5 \mathrm{km/h}$.</p>
<p>Addendum: as I have been asked for a solution "by equations", I shall oblige. Referencing the equations you have written as comments to my answer (please do not delete or alter them),</p>
<p>The first equation you have written is correct: $d = (r_0 + r_1)t$.</p>
<p>The second equation is incorrect. It should be $d = r_0(t - \frac 1{20} + \frac 15) + r_1(t - \frac 1{20})$. The girl starts $12 \mathrm{
min}$ (one fifth of an hour) sooner, but meets the tram $3\mathrm{
min}$ (one twentieth of an hour) sooner also. The effects of these two are opposing - starting sooner gives her more of a head start and increases the travelled distance so you add that time, but meeting the tram sooner lessens the time and decreases the travelled distance, so you must subtract that time.</p>
<p>Equating those two (to eliminate $d$) immediately allows one to find the ratio between $r_0$ and $r_1$.</p>
<p>It is unclear why you have written the term $t$ twice in the "girl component" of your second equation causing it to cancel out.</p>
<p>Sometimes it's important to take a step back and look at the problem in a simple fashion. Equations may cloud the issue. If it can be solved without algebra, all the better.</p>
|
2,482,083 | <p>I want to know if the following is a true statement:</p>
<blockquote>
<p>If $A$ is real a symmetric and $V$ is orthogonal then $V^TAV=D$ where $D$ is a diagonal matrix. </p>
</blockquote>
<p>To prove it I can say </p>
<p>$$(V^TAV)^T=V^TA^TV = V^TAV.$$
So $V^TAV$ is itself symmetric. So if I know $V^TAV$ is triangular then I'm done. How do I know $V^TAV$ is triangular (if it is)?</p>
| user334639 | 221,027 | <p>It is false: take $V$ to be the identity and $A$ to be symmetric but not diagonal.</p>
<p>Then $D=V^TAV=A$ so $D$ is not diagonal.</p>
|
2,810,755 | <p>I'm trying to set up a double integral on $e^{-(x+y)}$ and the range listed is: $0 < x < y < \infty.$</p>
<p>I'm interpreting this as $0 < x < y$ for $x$'s range and $x < y < \infty$ as $y$'s range. I've put those bounds on my two integrals and proceeded with the mission. However, $y$ is still lingering in my result. (The homework is already turned in and I'm sure I've done it wrong; now I just want to know how to do it properly.)</p>
<p>My result of the double integral was: $-((e^{-2y})/2) + (1/2)$.
I suspect that how I set up the double integral was the source of the problem. How should it have looked?</p>
<p>The integral I calculated was
$$\int_0^y\int_x^\infty e^{-(x+y)}dydx$$</p>
| tien lee | 557,074 | <p>Inductive Hypothesis:-</p>
<p>Assume $P(k)$ is true</p>
<p>$1\cdot2+2\cdot3+3\cdot4+.....+\sum _{ n=1 }^{ k }{ n(n+1) } =\frac{k(k+1)(k+2)}{3}$</p>
<p>Now show that Inductive Hypothesis is true. Show that $P(k+1)$ is true
$$\sum _{ n=1 }^{ k+1 }{ n(n+1) } =\frac{k+1(k+1+1)(k+1+2)}{3}=\frac{k+1(k+2)(k+3)}{3}$$
Now,
$$\sum _{ n=1 }^{ k+1 }{ n(n+1) }=\sum _{ n=1 }^{ k }{ n(n+1) }++(k+1)(k+1+1)$$
$$=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$
$$=\frac{k\color{red}{(k+1)(k+2)}+3\color{red}{(k+1)(k+2)}}{3}$$
$$=\frac{(k+1)(k+2)}{3}[k+3]$$
$$=\frac{(k+1)(k+2)(k+3)}{3}$$</p>
<p>By the principle of mathematical induction $1\cdot2+2\cdot3+3\cdot4+.....+n(n+1)=\frac{n(n+1)(n+2)}{3}$</p>
|
118,658 | <p>I have set of files in the format 20160615-1-0.asc, 20160615-2-0.asc,... 20160615-5-0.asc etc. The following commands can List the file names at once which can be used to import later. </p>
<pre><code> files = Range[1, 5];
c = "D:\\20160615\\20160615-" <> ToString[#] <> "-0.asc" & /@files
</code></pre>
<p>I want to use "0" part of the "-0.asc" string as a variable so that I can make different lists of multiple files like {-1-0.asc, -2-0.asc, ...-5-0.asc}, {-1-1.asc, -2-1.asc, ...-5-1.asc},{-1-3.asc, -2-3.asc, ...-5-3.asc}..and so on.</p>
| m_goldberg | 3,066 | <p>It seems that a simple modification of your code will work. </p>
<pre><code>fileList[prefix_String, m_Integer, n_Integer] :=
With[{dash = "-"},
prefix <> dash <> ToString[#] <> dash <> ToString[n] <> ".asc"] & /@ Range[m]
</code></pre>
<p>Then</p>
<pre><code>fileList["D:\\20160615\\20160615", 5, 1]
</code></pre>
<p>gives</p>
<blockquote>
<p><code>{"D:\\20160615\\20160615-1-1.asc",
"D:\\20160615\\20160615-2-1.asc",
"D:\\20160615\\20160615-3-1.asc",
"D:\\20160615\\20160615-4-1.asc",
"D:\\20160615\\20160615-5-1.asc"}</code></p>
</blockquote>
|
90,965 | <p>Im wondering if there's an existing literature on this binary operation involving graphs wherein you identity $n$ vertices from one graph with $n$ vertices from the other such that the resulting structure is still a graph (no loops and multiple edges). For instance, given two paths $[a,b,c]$ and $[d,e,f]$, letting $a=f$ and $c=d$ produces $C_4$.</p>
| Marc Olschok | 22,141 | <p>Since I cannot add comments, I put this as an answer.
I do not know about specific graph-theoretical literature about the
operation you describe, but it seems that it can be exhibited as
a pushout in a suitable category.</p>
<p>For simplicity, I suppose you consider only undirected graphs.</p>
<p>Take the category where objects are sets equipped with a reflexive
and symmetric relation, and where a morphism $h: (A,\alpha)\to(B,\beta)$
is an (ordinary) map of sets $h:A\to B$ that satisfies
$\forall a,a'\in A: (a,a')\in\alpha\Rightarrow (ha,ha')\in\beta$.</p>
<p>Then the objects of that category correspond to (undirected simple) graphs
and therefore also give a notion of graph homomorphism. Let $1$ be the
graph with exactly one vertex and write $n$ for the discrete graph with
$n$ vertices. Then, given two graphs $G$ and $H$, choosing $n$ vertices
in $G$ and $H$ is the same as giving monomorphisms $n\to G$ and $n\to H$,
and your gluing operation is then given as a pushout of
$G\leftarrow $n$ \rightarrow H$</p>
|
90,965 | <p>Im wondering if there's an existing literature on this binary operation involving graphs wherein you identity $n$ vertices from one graph with $n$ vertices from the other such that the resulting structure is still a graph (no loops and multiple edges). For instance, given two paths $[a,b,c]$ and $[d,e,f]$, letting $a=f$ and $c=d$ produces $C_4$.</p>
| Felix Goldberg | 22,051 | <p>Have a look at this:</p>
<p><a href="http://www.win.tue.nl/~hholst/Research%20slides/minimum%20rank.pdf" rel="nofollow">http://www.win.tue.nl/~hholst/Research%20slides/minimum%20rank.pdf</a></p>
<p>I think that what van der Holst calls <em>Separations</em> is what you need.</p>
|
3,556,109 | <p>Suppose that <span class="math-container">$f : (0, 2] \to \mathbb{R}$</span> is a continuous function and that <span class="math-container">$\lim_{x \to 0}f(x) = L$</span> for some <span class="math-container">$L \in \mathbb{R}$</span>.
Define <span class="math-container">$g : [0, 2] \to \mathbb{R}$</span> via <span class="math-container">$g(0)=L$</span> and <span class="math-container">$g(x)= f(x)$</span> for <span class="math-container">$x \in (0,2]$</span>.</p>
<p>(i) Show that <span class="math-container">$g$</span> is continuous on <span class="math-container">$[0, 2]$</span>.</p>
<p>(ii) Show that <span class="math-container">$f$</span> is uniformly continuous on <span class="math-container">$(0, 2]$</span>.</p>
<p>I'm struggling to know where to start! I understand that <span class="math-container">$|g(y)-g(x)|< \varepsilon$</span> for all <span class="math-container">$y \in [0,2]$</span> with <span class="math-container">$|y-x|<\delta$</span>.</p>
<p>I just can't get the formal proof started.</p>
| Leander Tilsted Kristensen | 631,468 | <p>"Clearly the former implies the latter via the triangle inequality." No, it is actually the other way around. The triangle inequality states
<span class="math-container">$$|\sum_i x_iy_i| \leq \sum_i |x_i y_i|,$$</span></p>
<p>so clearly Hölders inequality implies Cauchy-Schwarz and it would seem that Hölders is the stronger statement, although they are equivalent. To see that Cauchy-Schwarz inequality implies this special case of Hölders inequality, apply Cauchy-Schwarz inequality to the vectors <span class="math-container">$(|x_1|,...,|x_n|)$</span> and <span class="math-container">$(|y_1|,...,|y_n|)$</span>.</p>
|
2,559,639 | <p>In the book of <em>General Topology</em> by Munkres, on page 104, it is given that</p>
<p><a href="https://i.stack.imgur.com/Pv9MU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pv9MU.png" alt="enter image description here"></a></p>
<p>However, as far as I know, the lower limit topology <span class="math-container">$\tau_l$</span> corresponds to the intervals of the form <span class="math-container">$[a,b)$</span> where <span class="math-container">$a < b$</span>. So how can <span class="math-container">$(a,b)$</span> be open in this topology? </p>
| Angina Seng | 436,618 | <p>$(a,b)=\bigcup_{n=1}^\infty[a+1/n,b)$</p>
|
40,669 | <p><a href="http://en.wikipedia.org/wiki/Wang_tile">Wang tiles</a> are (by Wikipedia): "equal-sized squares with a color on each edge which can be arranged side by side (on a regular square grid) so that abutting edges of adjacent tiles have the same color; the tiles cannot be rotated or reflected."</p>
<p>The usual decision problem associated with them is: given a set of tiles, can they tile the plane?</p>
<p>This problem is known to be undecidable and has a nice history (Wang originally gave a decision procedure, but it works only if the tiles form a periodic tiling, while there exists sets which give rise only to nonperiodic tilings). It has fascinated me personally for a very long time</p>
<p>The state-of-the-art proof of undecidability I'm familiar with is that of Raphael Robinson ("Undecidability and Nonperiodicity for Tilings of the Plane," Inventiones Mathematicae, 12(3), 1971 pp. 177–209). It is quite a difficult proof, and I have never seen it in textbooks.</p>
<p>I'm familiar with a much easier version - this time the set only needs to tile a quadrant of the plane, and the corner tile is already given. This problem is much easier to handle - the given corner tile enables us to "run a Turing machine" in the tiling (each row is a configuration) and undecidability follows. However, this result is primarily a "folklore" version - I don't recall it in textbooks at all.</p>
<p>This leads to my question: has Robinson's method been improved but remains folklore, and so not shown in textbooks? Is there a relatively simple proof of the undecidability of the general tiling problem I'm missing?</p>
| blatmalt | 562,720 | <p>There is also the newer work of Kari "On the Undecidability of the Tiling Problem", which shows the undecidability of the domino problem by a reduction to the immortality problem for affine functions.</p>
<p>It shows the undecidability of the domino problem in the hyperbolic plane which was not shown by Robinson.
I am not sure if it is easier than the proof by Robinson.</p>
<p>It is available here, page 74:
<a href="https://link.springer.com/content/pdf/10.1007%2F978-3-540-77566-9.pdf" rel="nofollow noreferrer">https://link.springer.com/content/pdf/10.1007%2F978-3-540-77566-9.pdf</a></p>
|
2,204,834 | <p>I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is
$$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$.</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>If $a_n=(2b+1)^n$</p>
<p>$$a_{m+2}-a_m=8(2b+1)^m\cdot\dfrac{b(b+1)}2$$ which is multiple of $8$ as $b(b+1)$ is even.</p>
|
1,057,429 | <p>In mathematics statistics. I'm a bit confused by the terminology <em>normal approximation</em>. What is it? Is it just something you say when you approximate, for example the normal distribution? </p>
| j4GGy | 186,141 | <p>By the <a href="http://en.wikipedia.org/wiki/Spectral_theorem" rel="nofollow">finite-dimensional spectral theorem</a> there exists an orthogonal matrix $C$ (i.e. $C^-1 = C^T$ such that $C^TAC = D = diag(d_1,\ldots, d_n)$ is a diagonal matrix. Let $e_k \in \mathbb{R}^n$ denote the vector, which $k$-th coordinate is $1$ and all others are $0$. Then
$$d_k = e_k^T D e_k = e_k^T C^T A Ce_k = (Ce_k)^T A (Ce_k) \geq 0.$$
Hence, all entries of $D$ are $\geq 0$. This means we can consider the square root $\sqrt{d_k}$ of $d_k$ and define $E := diag(\sqrt{d_1},\ldots, \sqrt{d_k})$. It follows that
$$E^2 = diag(\sqrt{d_1},\ldots, \sqrt{d_k}) \cdot diag(\sqrt{d_1},\ldots, \sqrt{d_k}) = diag(d_1,\ldots, d_k) = D = C^T A C.$$
Using that $C^T = C^{-1}$, we see that
$$A = CE \cdot EC^T = CEC^T \cdot CEC^T.$$
Now, if we define $B = CEC^T$, then
$$B^T = (CEC^T)^T = (C^T)^T E^T C^T = CEC^T = B$$
and $A = B^2$.</p>
|
3,048,340 | <blockquote>
<p>Suppose <span class="math-container">$f$</span> is a convex function and differentiable everywhere on <span class="math-container">$(0, \infty)$</span> and satisfies
<span class="math-container">$$\lim_{x \rightarrow\infty}{f(x)}=A$$</span>
where <span class="math-container">$A$</span> is an arbitrary real number. Show that <span class="math-container">$$\lim_{x\rightarrow\infty}{f'(x)}=0$$</span></p>
</blockquote>
<p><strong>My attempt:</strong> Suppose <span class="math-container">$f'(x_0)>0$</span> for <span class="math-container">$x_0 \in (0, \infty)$</span> such that
<span class="math-container">$$f(x) \geq f(x_0)+f'(x_0)(x-x_0)$$</span>
Taking limit that tends to infinity on both sides obtain
<span class="math-container">$$A=\lim_{x \rightarrow \infty}{f(x)} \geq \lim_{x \rightarrow \infty}{[f(x_0)+f'(x_0)(x-x_0)]}=\infty$$</span>
which contradicts the fact that <span class="math-container">$A$</span> is finite. Similar to the case when <span class="math-container">$f'(x_0)<0$</span>, Hence, <span class="math-container">$f'(x_0)=0$</span> for <span class="math-container">$x_0 \in (0, \infty)$</span>. Take <span class="math-container">$x_0 \rightarrow \infty$</span> and it follows the conclusion.</p>
<p><strong>Note:</strong> Any constant function <span class="math-container">$f$</span> is a trivial case on this question. </p>
<p>In this question, I illustrated an example <span class="math-container">$f(x)=e^{-x}$</span> and referring to its geometry interpretation of derivative, but I hope to know that if my attempt is wrong as I make <span class="math-container">$x_0\rightarrow \infty$</span> in my last progess, which is weird and correct it.</p>
| JimmyK4542 | 155,509 | <p><strong>Hint</strong>: You've successfully come up with a contradiction if <span class="math-container">$f'(x_0) > 0$</span> for any <span class="math-container">$x_0 \in (0,\infty)$</span>. Therefore, <span class="math-container">$f'(x) \le 0$</span> for all <span class="math-container">$x \in (0,\infty)$</span>. </p>
<p>Now, since <span class="math-container">$f$</span> is convex, i.e. <span class="math-container">$f'(x)$</span> is increasing. Since <span class="math-container">$f'(x)$</span> is increasing and bounded above by <span class="math-container">$0$</span>, <span class="math-container">$\displaystyle\lim_{x \to \infty}f'(x)$</span> exists. Can you find a contradiction if this limit is less than zero?</p>
|
1,887,315 | <p>$e$ is one of the most important numbers in our universe, it is everywhere. When I try to find out why the most common explanation is some reverse-engineered physics or finance problem. But these are just one off examples of why $e$ <em>is</em> important, they fall short of illuminating the origin of its significance. What I'm looking for is some fundamental definition of $e$ that explains it's significance and omnipresence, something akin to $\pi$ relating circumference and diameter. </p>
<p>Thanks, I hope I'm being clear.</p>
<p>EDIT*: A common saying is "$e$ is the most natural base." What does that mean?</p>
| Wavelet | 336,350 | <p>$e^{\lambda x}$ is the solution to the equation $Df=\lambda f$. This is equivalent to saying that $e^{\lambda x}$ is an eigenfunction of the differentiation operator D for any value $\lambda \in \mathbb{R}$. Now how ubiquitous is differentiation to physics?</p>
|
7,900 | <p>This semesters I am teaching a course to 12-13 year old kids in which they are supposed to learn the basics of spreadsheet usage.</p>
<p>I am having difficult in coming up with fun / interesting exercises to teach them the basic functions, such as average, max, minimum and the like. One of the biggest problems is having only a 50 minutes long lesson every two weeks.</p>
<p>I've tried so far:</p>
<ul>
<li>Teaching them to plot relations using a dispersion graphic: they would produce the axes and plot relations such as linear and quadratic formulas, exponential, etc.</li>
<li>Introduced them to simple recurrences such as the Fibonacci sequence and the discrete logistic map.</li>
<li>Using stock data to teach them some descriptive statistics.</li>
</ul>
<p>What else would be proper, and preferably fun, to teach them? </p>
| Sue VanHattum | 60 | <p>I used Excel to solve many of the problems I encountered in <a href="https://projecteuler.net/" rel="nofollow">Project Euler</a>. I bet they'd find some of those problems interesting and accessible.</p>
|
3,562,324 | <blockquote>
<p>Find all positive integers x such that <span class="math-container">$x^4-8x+16$</span> is a full square.</p>
</blockquote>
<p>I presented it as <span class="math-container">$(x-4)^2+x^4-x^2$</span> and found that <span class="math-container">$x = +1$</span> satisfy the condition. Then I equalized it <span class="math-container">$(x-4)^2+x^4-x^4=a^2+2ab+b^2$</span> then <span class="math-container">$a=x-4$</span> and nothing else happened. Help prove that there are no more solutions or find the rest of the solutions</p>
| J. W. Tanner | 615,567 | <p><strong>Hint:</strong></p>
<p>For <span class="math-container">$x>2$</span>, <span class="math-container">$(x^2-1)^2<x^4-8x+16<(x^2)^2$</span>.</p>
|
2,626,454 | <blockquote>
<p>I used Wolfram Alpha to try and compute the integral $$\int_0^{\pi/2}(1+\cos2x)^{-0.5}\,dx$$ and I'm told that the integral does not converge. However, if I try
$$\lim_{a\to0.5^-}\int_0^{\pi/2}\frac{dx}{(1+\cos2x)^a}$$
by substituting in values of $a$ close to $0.5$ e.g $a\in\{0.49,0.499,0.4999...\}$ the integral appears to converge to $26.8877$.</p>
</blockquote>
<p>Is it possible to determine analytically if the integral in the title converges? If so, what technique should I use to deduce this?</p>
| mickep | 97,236 | <p>In the interval you are looking at
$$
\frac{1}{(1+\cos 2x)^a}=\frac{1}{2^a\cos^{2a}x}.
$$
Hence, the integral you are looking at is
$$
\frac{1}{2^a}\int_0^{\pi/2}\frac{1}{\cos^{2a}x}\,dx=\frac{1}{2^{a+1}}B(1/2,1/2-a)
=\frac{1}{2^{a+1}}\sqrt{\pi}\frac{\Gamma(1/2-a)}{\Gamma(1-a)}.
$$
Here $B$ denotes the Beta function. The expression in the right-hand side blows up to $+\infty$ as $a\to 1/2^-$. This is goind hand-in-hand with other answers, showing that the integral with $a$ replaced by $1/2$ is divergent.</p>
|
2,328 | <p>There is an interesting discussion taking place at MO over the suitability of this question: <a href="https://mathoverflow.net/questions/210089/complex-structure-on-s6-gets-published-in-journ-math-phys">Complex structure on $S^6$ gets published in Journ. Math. Phys</a>. The question is about the correctness of a paper that has recently appeared in a peer-reviewed journal, purporting to settle a long-standing open problem regarding existence of a complex structure on the 6-dimensional sphere <span class="math-container">$S^6$</span>.</p>
<p>The basic question here is whether the community thinks this <em>type</em> of question could be suitable for MO, but a subsidiary consideration is how to edit this particular question to remove at least some of the features that some commenters found objectionable. Namely, they found the question brusque and somewhat confrontational or less than respectful to the paper's author. Let me then take a crack at rewording the question:</p>
<blockquote>
<p>A paper purporting to solve a major open problem has recently appeared:</p>
<ul>
<li>Gabor Etesi, Complex structure on the six dimensional sphere from a spontaneous symmetry breaking, Journ. Math. Phys. 56, 043508-1-043508-21 (2015). Link: <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a></li>
</ul>
<p>An earlier preprint by this author on this topic, <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a> (now 10 years old), had some problems [<em>insert specific criticisms here</em>]. My question is whether these difficulties have been satisfactorily addressed in the current paper, and if so, whether this paper has indeed answered this outstanding question (whether a complex structure on <span class="math-container">$S^6$</span> exists).</p>
</blockquote>
<p>This rewording could perhaps be improved, but the idea would be to try and pinpoint the specific mathematical considerations within the body of the question, and do so in a more or less polite and decorous way.</p>
<p>Onto the more general question: personally, I'm somewhat torn. We've generally declared to would-be solvers of the Riemann Hypothesis (etc.) that MathOverflow is not to be used as a vetting service, and that certainly seems like wise policy to me. Some users (I'll quote Joseph O'Rourke in his comment) take it a little further and say "It is inappropriate to discuss the flaws in a paper in a public forum." Other commenters seem perplexed by this stance.</p>
<p>I'd like to hear more discussion on this. I can definitely see that we do not want to be in the business of raking people over the coals, whether they be cranks or serious researchers. But if we manage to take the personal element out of it and get straight down to the business of pinpointing really specific mathematical difficulties encountered in peer-reviewed published papers, it's hard for me to think of really convincing counterarguments. Anyway, what do you think about this?</p>
<p>Last I checked there are three votes to reopen, so maybe it's good to discuss this openly here.</p>
<p>Oh, by the way: what about Community Wiki? There's another closely related question <a href="https://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere">Is there a complex structure on the 6-sphere?</a> which mentions the same paper that is CW. (I think one could argue that a suitably reworded question with specific mathematical content would not be a duplicate.)</p>
| The Masked Avenger | 35,626 | <p>I think Joseph O'Rourke meant to say something else. I will guess that something like
the following is intended.</p>
<p>MathOverflow is not a forum for judging, critiquing, or refereeing preprints or published
articles. MathOverflow is for answering specific questions in mathematics, including
but not limited to judging, critiquing, or refereeing a part of a proof or statement,
possibly a whole proof or statement if concise enough for this forum. To keep
the quality high and the helpful and welcoming atmosphere of the forum,
subjective opinions and similar statements which detract from such atmosphere
and quality are discouraged.</p>
<p>With the appropriate specific details included, Todd's suggested edit seems fine to me.
It is about the math, not about fame, infamy, or if a paper is wrong. I hope
Misha Verbitsky agrees to the change in tone, for the sake of the forum.</p>
|
607,353 | <p>So I've recently been looking at the <a href="http://www.photographyblogger.net/stereographic-projection-photography-100-little-planets/" rel="nofollow">Tiny Planet</a> images.</p>
<p>I've been googling a few things to try and find out how images are converted from normal to a tiny planet. Some phone apps, as well as photoshop do this.</p>
<p>I think Photoshop does it by converting Cartesian coordinates to Polar coordinates.
I found a good explanation of converting them <a href="http://www.teacherschoice.com.au/maths_library/coordinates/polar_-_rectangular_conversion.htm" rel="nofollow">here</a>.</p>
<p>However, I am yet to find somewhere that describes an algorithm of the process.</p>
<p>Is there a set formula? or is it more complex than that?</p>
<p>I'm not sure if it's as easy as moving pixels and converting coordinates because I think some form of stretching must be included... hence, looking for an algorithm and this exchange as a resource. Thanks.</p>
| Logan Keefe | 115,380 | <p>The article you have linked has "Stereographic Projection" in the title. <a href="https://en.wikipedia.org/wiki/Stereographic_projection#Definition" rel="nofollow">Here</a> is a link to the relevant part of the Wikipedia article. It is difficult to tell solely from looking at the images, but it seems as if the original photograph (which is a effectively a portion of a plane) is mapped to a hemisphere using a similar transformation. Then, in a second transformation, each point $(x,y,z)$ of the hemisphere is mapped to $(x,y,0)$ so that all of the points lie in a plane. It seems to me that this would create the desired effect.</p>
|
2,760,113 | <p>Suppose $f'$ is continuous on $[a,b]$ and $\epsilon >0$ is given.
Prove that $\exists\; \delta>0$ such that $$\left\lvert \frac{f(t)-f(x)}{t-x}-f'(x)\right\rvert<\epsilon$$</p>
<p>$\forall \;0<|t-x|<\delta, a\le x \le b ,a\le t\le b$.</p>
<p>How to solve this problem since using compactness argument ?</p>
| Mark Pineau | 432,691 | <p>Following from my hint I gave in the comments:</p>
<p>$f'$ is continuous on the compact interval $[a,b]$, and thus is uniformly continuous on $[a,b]$. </p>
<p>$\therefore \ \forall \epsilon>0$, $\exists \ \delta>0$, such that for all $x,t\in [a,b]$, we have </p>
<p>$$0<|x-t|<\delta\implies|f'(x)-f'(t)|<\epsilon$$</p>
<p>Because $f'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, by the mean value theorem, there exists some $c\in[x,t]$, such that</p>
<p>$$\frac {f(t)-f(x)}{t-x}=f'(c)$$</p>
<p>The claim then follows</p>
<p>$$\bigg| \frac {f(t)-f(x)}{t-x}-f'(t) \bigg|=|f'(c)-f'(t)|<\epsilon$$</p>
<p>For $0<|x-t|=|t-x|<\delta$</p>
|
1,820,853 | <p>How to derive the Gregory series for the inverse tangent function?
Why is Gregory series applicable only to the set $ [-\pi/4,\pi/4] $ ?</p>
| user353587 | 353,587 | <p>I think the most straightforward way is to yes the expansions of <strong>log(1+x) and log(1-x)</strong>
<strong>log(1+x)=x - x^2/2 + x^3/3....and so on.</strong></p>
<p><strong>log(1-x)=-x-x^2/2-x^3/3....and so on.</strong></p>
<p>Now subtract the two series to get
<strong>log[(1+x)/(1-x)]=
2[x+x^3/3+x^5/5+x^7/7......inf ]</strong>
Now <strong>replace x</strong> with <strong>[¡.arctanx]</strong>
After a little bit of manipulation [using <strong>Euler's form</strong>]
You will get the famous
Madhav-Gregory Series .</p>
|
2,338,797 | <p>A bag contains $3$ red balls and $4$ blue balls. Find the probability of picking, at random, two balls of </p>
<p>a) the same color </p>
<p>b) different colors</p>
| Zack Miodownik | 450,075 | <p>The probability of picking 2 blues balls (lol) is 4/7 times 3/6 which is 2/7. The probability of picking 2 red balls is 3/7 times 2/6 which is 1/7. Thus the probability of picking two same colored balls is 3/7 (The sum of the two). Thus the probability of picking different colored balls is 1-3/7 which is 4/7.</p>
|
2,338,797 | <p>A bag contains $3$ red balls and $4$ blue balls. Find the probability of picking, at random, two balls of </p>
<p>a) the same color </p>
<p>b) different colors</p>
| BruceET | 221,800 | <p>Answers by @Mandar, @Zack, and @David are in agreement, clear, and correct (+1's).</p>
<p>I just wanted to show you the role of the 'hypergeometric distribution' in solving
problems such as this. Let $X$ be the number of red balls in two draws without replacement. Then
$$P(X = k) = \frac{{3 \choose k}{4 \choose {2-k}}}{{7 \choose 2}},$$
for $k = 0, 1, 2.$
The two balls are of the same color if $X = 0$ or $2$, and of different
colors if $X = 1$.</p>
<p>You can get the numerical values of the probabilities by evaluating the
'binomial coefficients', or by using software. In R statistical software
the computation is as follows:</p>
<pre><code> k = 0:2; pdf=dhyper(k, 3, 4, 2)
cbind(k, pdf, 7*pdf)
# k pdf
#[1,] 0 0.2857143 2
#[2,] 1 0.5714286 4
#[3,] 2 0.1428571 1
</code></pre>
<p>Ignore the row numbers in brackets. The PDF column shows the three
probabilities as decimal fractions. The last column indicates that these
probabilities are $2/7, 4/7,$ and $1/7.$ So $P(\mathrm{Same\; Color}) = 3/7$
and $P(\mathrm{Different\; Color}) = 4/7.$</p>
<p>Perhaps you have seen the hypergeometric distribution already, and
perhaps you will encounter it later in your course. Either way, the
most difficult thing about it may be the name 'hypergeometric'.</p>
|
1,544,789 | <blockquote>
<p>Calculate $\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx$ where $a,c > 0$ and $b>1$.</p>
</blockquote>
<p>The best I could do about this integral is to find an upperbound for it:</p>
<p>$\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx \leq \int_0^{\infty} \frac{x^{2N+1}}{x^{-b}} e^{-c x^2} dx= \int_0^{\infty} x^{2N+b+1} e^{-c x^2} dx$</p>
<p>Then I can use $\int_{0}^{\infty} x^{n} e^{-ax^2}\,\mathrm{d}x =
\begin{cases}
\frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\
\frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{integer}, a>0) \\
\frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{integer}, a>0)
\end{cases} $</p>
<p>Any idea that how I can calculate the integral, not jut an upper bound?</p>
| mrprottolo | 84,266 | <p>The reason is that the integral you evaluated doesn't give the area of the quadrilateral $R$. Indeed, called $A$ its area, you have
$$\int\int_R dx dy=A$$
and
$$\int\int_R x dx dy\neq \int\int_R dx dy.$$</p>
|
267,161 | <p>I would like to write <code>x^5 + 1/x^5</code> in terms of <code>x^2 + 1/x^2</code>, <code>x^3 + 1/x^3</code> and <code>x + 1/x</code> or express <code>x^6 + 1/x^6</code> in terms of <code>x^2 + 1/x^2</code> and <code>x^4 + 1/x^4</code> etc.</p>
<p>How can I do that in <em>Mathematica</em>?</p>
| Daniel Lichtblau | 51 | <p>@Artes lready showed how one might use <code>Solve</code> or <code>Eliminate</code> for this task. Since it involves algebraic rewriting of a rational expression, a standard method from computational algebra involves Groebner bases. Technical-sounding, yes, but also quite powerful.</p>
<p>The idea is to write the inputs as polynomials. I use <code>y</code> as the reciprocal of <code>x</code> and add a polynomial relation <code>x*y-1</code> to enforce this. I equate each target polynomial to a new "marker" (or "tag") variable. The code for this is written in a way that is intended to be readily automated, that is, to systems with more relations, different variables, etc.</p>
<pre><code>basepolys = {x^3 + y^3, x^2 + y^2, x + y};
avars = Array[a, Length[basepolys]];
allpolys = Join[avars - basepolys, {x*y - 1}];
allvars = Join[Variables[basepolys], avars];
</code></pre>
<p>Next compute a Groebner basis using a term order that places the original variables "larger" than the marker variables. This will make rewrites favor the marker variables over the polynomial variables.</p>
<pre><code>gb = GroebnerBasis[polys, allvars];
</code></pre>
<p>Now we can rewrite that input `x^5+1/x^5 like so:</p>
<pre><code>PolynomialReduce[x^5 + y^5, gb, allvars][[2]]
(* Out[1322]= 5 a[3] - 5 a[3]^3 + a[3]^5 *)
</code></pre>
<p>I left out a couple of niceties. One can programatically (as in, exercise for the reader) replace negative powers of <code>x</code>, for example. Or replace the result with explicit uses of the original expressions <code>x^3+1/x^3,...</code>.</p>
|
625,395 | <blockquote>
<p>Let $f:\mathbb{R}^3 \to \mathbb{R}$ be a smooth function rapidly
decreasing to zero as $|(x,y,z)| \to \infty$, and let $D(t)$ denote
$$D(t)=\left\{(x,y,z)\in \mathbb{R}^3 \mid ax+by+cz \leq t\right\},$$
where $a,b,c\in \mathbb{R}$. Compute $$\frac{d}{dt}
\int_{D(t)}f(x,y,z)\,dx\,dy\,dz.$$</p>
</blockquote>
<p>I'd like to change variables so that the region no longer depends on $t$. Let $$R:=\{(x,y,z)\mid x+y+z\leq 1\}.$$ I note that $$\varphi_t:(x,y,z)\mapsto \left( \frac tax, \frac tb y, \frac tc z \right):R\to D_t$$</p>
<p>Then following the rule $$\int_S \varphi^*(\omega) = \int_{\varphi(S)}\omega$$ I get $$\int_R \frac {t^3}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right)dx dy dz=\int_{D(t)}f(x,y,z)dxdydz.$$</p>
<p>Now $$\frac{d}{dt}\int_{D(t)}f(x,y,z)\,dx\,dy\,dz = \frac{d}{dt} \int_R \frac {t^3}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right)dx dy dz$$$$= \int_R \frac{3t^2}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right) + \frac{t^3}{abc} \left\langle \nabla f\left( \frac ta x, \frac tb y, \frac tc z \right), \left( \begin{matrix}t/a \\t/b\\t/c\end{matrix} \right) \right\rangle \tag{1}$$ where we still need to justify switching the integral with the derivative.</p>
<p>Can I simplify that any more, do you think?</p>
<p>To justify the swap of integral and derivative, correct me if I'm wrong, but we would just need the right hand side of (1) uniformly convergent in $t$, for some compact interval of any $t$, right? Can we deduce that from "rapid decay"?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#0000ff}{\large%
\totald{}{t}\int_{D\pars{t}}\fermi\pars{x,y,z}\,\dd x\,\dd y\,\dd z}
=
\totald{}{t}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,z}\Theta\pars{t - ax - by - cz}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,z}\delta\pars{t - ax - by - cz}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,z}\,{\delta\pars{z - \bracks{t - ax - by}/c} \over \verts{c}}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\color{#0000ff}{\large{1 \over \verts{c}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,{t - ax - by \over c}}\,\dd x\,\dd y}
\end{align}</p>
|
3,575,867 | <p>I need to prove that <span class="math-container">${n\choose 0} < {n\choose 1} < \ldots < {n\choose (n+1)/2}$</span> (for every <span class="math-container">$n$</span>)
I tried to do it with induction but I got stuck.</p>
| Paras Khosla | 478,779 | <p><strong>Hint</strong>:</p>
<p>By definition <span class="math-container">${n\choose k}=\frac{n!}{k!(n-k)!}$</span>, where <span class="math-container">$n, k \in \mathbb{N}$</span> and <span class="math-container">$n\ge k $</span>. Use this definition assuming the proposition to be true and show that it satisfies the constraints on the definition. Also note that the inequality can go on till <span class="math-container">${n\choose \frac{1}{2}(n-1)}$</span>. </p>
|
3,575,867 | <p>I need to prove that <span class="math-container">${n\choose 0} < {n\choose 1} < \ldots < {n\choose (n+1)/2}$</span> (for every <span class="math-container">$n$</span>)
I tried to do it with induction but I got stuck.</p>
| SomeStrangeUser | 68,387 | <p>Note that <span class="math-container">$\binom{n}{k}=\frac{n-k+1}{k}\cdot\binom{n}{k-1}$</span>. Hence, the inequality is true as long as <span class="math-container">$n-k+1>k$</span>, or rearranging: <span class="math-container">$\frac{n+1}{2}>k$</span>.</p>
|
4,181,235 | <p>The topology on <span class="math-container">$\mathcal D$</span> is the locally convex topology induced by seminorms like <span class="math-container">$\|\partial_\alpha f\|_\infty$</span>, where <span class="math-container">$\alpha=(d_1,\ldots,d_n) $</span> and <span class="math-container">$\partial_\alpha=\partial_{x_1}^{d_1}\ldots\partial_{x_n}^{d_n}$</span>.</p>
<p><span class="math-container">$\mathcal D'$</span> is its continuous dual space.</p>
<p>The weak^<span class="math-container">$^*$</span> topology on <span class="math-container">$\mathcal D'$</span> is induced by seminorms like <span class="math-container">$\|\phi\|_f=|\phi(f)|$</span>, where <span class="math-container">$f\in \mathcal D$</span>.</p>
<p>It is easy to define the limit of a Cauchy net on <span class="math-container">$\mathcal D'$</span>: <span class="math-container">$\Phi(f)=\lim_{\lambda}\phi(f)$</span>.</p>
<p>However, I don't know how to show <span class="math-container">$\Phi$</span> is continuous, that is, <span class="math-container">$\Phi(f_\eta)\to 0$</span> whenever <span class="math-container">$\|\partial_\alpha f_\eta\|\to 0$</span> for all <span class="math-container">$\alpha\in \mathbb N^n$</span>.</p>
<p>Could someone please give me some hints?</p>
| Aaron Hendrickson | 271,783 | <p>This is not the most direct approach but it does show where hypergeometric functions appear as mentioned by @DrSonnhardGraubner.</p>
<hr />
<p>We write
<span class="math-container">$$
\begin{align}
\sum_{j=0}^n\binom{a+j}{b}
&=\sum_{j=0}^n\frac{\Gamma(a+1+j)}{\Gamma(b+1)\Gamma(a-b+1+j)}\\
&=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\sum_{j=0}^n\frac{(a+1)_j}{(a-b+1)_j}\\
&=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\sum_{j=0}^n\frac{(a+1)_j(1)_j}{(a-b+1)_j\,j!},
\end{align}
$$</span>
where <span class="math-container">$(s)_n=\Gamma(s+n)/\Gamma(s)$</span> is the Pochhammer symbol and <span class="math-container">$(1)_n=n!$</span>. The result is a partial hypergeometric series, which may be evaluated using <a href="http://dlmf.nist.gov/16.2.E4" rel="nofollow noreferrer">DLMF 16.2.4</a> to give
<span class="math-container">$$
\sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\frac{(a+1)_n(1)_n}{(a-b+1)_n\,n!}{_3F_2}\left({-n,1,b-a-n\atop -n,-a-n};1\right).
$$</span>
After some simplification
<span class="math-container">$$
\sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1+n)}{\Gamma(b+1)\Gamma(a-b+1+n)}{_2F_1}\left({1,b-a-n\atop -a-n};1\right).
$$</span>
<a href="https://dlmf.nist.gov/15.4.E20" rel="nofollow noreferrer">Gauss's hypergeometric theorem</a> then permits us to write
<span class="math-container">$$
\sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1+n)}{\Gamma(b+1)\Gamma(a-b+1+n)}\frac{\Gamma(-a-n)\Gamma(-1-b)}{\Gamma(-a-n-1)\Gamma(b)},
$$</span>
which can be further reduced to give a sum of two binomial coefficients.</p>
|
2,880,377 | <p>As the title states, I need to find the limit for $x\left(x + 1 - \frac{1}{\sin(\frac{1}{1+x})}\right)$ as $x \rightarrow \infty$, as part of a larger proof I am working on.</p>
<p>I believe the answer is 0. I think that to start, I can show that $\frac{1}{\sin(\frac{1}{1+x})} \rightarrow x + 1$ for large x. By looking at the series expansion for Sin, it's clear that Sin approximates to $\frac{1}{1+x}$ for large x, as the higher-power entries in the series $\frac{1}{1+x}^3 + \frac{1}{1+x}^5 + ...$ would disappear faster, but would it be sufficient to state this? Is there not a more rigorous way of showing this to be true?</p>
<p>If my approach is entirely wrong, or there is a more elegant way of reaching the answer, please share.</p>
| user1337 | 62,839 | <p>According to Taylor's theorem with remainder, we have
$$\sin t=t-\frac{t^3}{6}+R_3(t) $$
where $|R_3(t)| \leq \frac{t^4}{24}$. Taking $t=\frac{1}{x+1}$ and plugging in your expression gives
$$\frac{x (x+1) \left(6 R_3(t) (x+1)^3-1\right)}{6 R_3(t) (x+1)^3+6 x (x+2)+5}=\frac{-x^2+o(x^2)}{6x^2+o(x^2)} $$
which approaches $-1/6$ as $x \to \infty$.</p>
|
2,226,307 | <p>Let <span class="math-container">$c\in \mathbb{R}$</span> and <span class="math-container">$f:\mathbb{R}\to \mathbb{R}$</span> is continuous at <span class="math-container">$c$</span>. If for every positive <span class="math-container">$\delta$</span>, there is a point <span class="math-container">$y\in (c-\delta, c+\delta)$</span> such that <span class="math-container">$f(y)=0$</span>, show that <span class="math-container">$f(c)=0.$</span></p>
<p>I am unable to solve the problem. I don't know how to start the problem. I need a help.</p>
<p>Edit:</p>
<blockquote>
<p>From the definition of continuity of <span class="math-container">$f(x)$</span> at <span class="math-container">$x=c$</span>. For any <span class="math-container">$\epsilon > 0$</span>, <span class="math-container">$\exists \delta > 0$</span>:</p>
<p><span class="math-container">$|f(x) - f(c)| < \epsilon \, \, \, \, \mbox{whenever} \, \, \, \, |x-c| < \delta$</span>.</p>
<p>Let <span class="math-container">$x=y\in (c-\delta, c+\delta)$</span> st <span class="math-container">$f(y)=0$</span> then <span class="math-container">$|f(c)|<\epsilon$</span> whenever <span class="math-container">$ |x-c| < \delta$</span></p>
<p>then <span class="math-container">$f(c)=0$</span> (proved)</p>
<p><strong>Whether the proof by <span class="math-container">$\epsilon-\delta$</span> method is correct</strong>?</p>
</blockquote>
| Tsemo Aristide | 280,301 | <p>Take $\delta=1/n$, there exists $y_n\in (c-1/n,c+1/n)$ with $f(y_n)=0$, $lim_ny_n=c$ implies $f(c)=lim_nf(y_n)=0$ since $f$ is continue.</p>
|
849,510 | <p>Let $f:\mathbb{R}\to\mathbb{R}$ be a $T$-periodic function, that is $f(t+T)=f(t)$ for all $t\in \mathbb{R}$. Assume that
$$\int_0^{+\infty}|f(s)|ds<+\infty.$$
Now if we assume in addition that $f$ is continuous, my intuition tells me that we must have necessarily $f=0$, is this correct ? </p>
| Community | -1 | <p><strong>Hint</strong>: </p>
<p>$$\int_0^{\infty} |f(s)| ds = \sum_{k = 0}^{\infty} \int_{kT}^{(k + 1)T} |f(s)| ds$$</p>
|
25,412 | <p>This is hopefully a simpler version of this previous unanswered <a href="https://mathematica.stackexchange.com/questions/23496/solving-recursion-relations-using-mathematica">question</a> of mine. </p>
<p>Let me just focus on the two expressions $F_2^{(s)}$ and $F_3^{(s)}$ given in A.3 and A.4 of page 19 of <a href="http://arxiv.org/pdf/1301.7182v2.pdf" rel="nofollow noreferrer">this paper</a>. </p>
<ul>
<li>How do I get Mathematica to just even manipulate such vector expressions? Like if I want to calculate $(F_2^{(s)})^2$ or $F_2^{(s)} F_3^{(s)}$ etc? </li>
</ul>
<hr>
<p>To make the question clear let me add in some more details about what I exactly want, </p>
<p>I define the function F2s as,</p>
<pre><code>F2s[q_, k1_] := (5/
14) + (3 (Norm[k1])^2)/(28 (Norm[q])^2) + (3 Norm[
k1]^2)/(28 (Norm[q - k1])^2) - (5)/(28 (Norm[q])^2 (Norm[
q - k1])^(-2)) - (5)/(28 (Norm[q])^(-2) (Norm[
q - k1])^(2)) + ( (Norm[
k1])^4)/(14 (Norm[q])^2 (Norm[q - k1])^2 )
</code></pre>
<p>But when I ask it to be squared all I get is this!
(basically nothing has been done and the situation doesn't change with taking a FullSimplify either) </p>
<pre><code>(2 Norm[k1]^4 - 5 (Norm[q]^2 - Norm[-k1 + q]^2)^2 + 3 Norm[k1]^2 (Norm[q]^2 + Norm[-k1 + q]^2))^2/(784 Norm[q]^4 Norm[-k1 + q]^4)
</code></pre>
<p>I would have wanted the answer to be given in the way I gave the functions $F2s$ - as a sum of fractions each of which is a product of powers of $q$, $k1$ and $\vert \vec{q} - \vec{k1}\vert$. How do I get that? </p>
| TransferOrbit | 3,236 | <p>Based on the assumption that those F functions produce scalar results…</p>
<p>If you just want to manipulate the vectors as entities without considering their components, then you can enter the function for F2s</p>
<pre><code>f2s[q_, k1_] := (5/
14) + (3 (Norm[k1])^2)/(28 (Norm[q])^2) + (3 Norm[
k1]^2)/(28 (Norm[q - k1])^2) - (5)/(28 (Norm[q])^2 (Norm[
q - k1])^(-2)) - (5)/(28 (Norm[q])^(-2) (Norm[
q - k1])^(2)) + ((Norm[
k1])^4)/(14 (Norm[q])^2 (Norm[q - k1])^2)
</code></pre>
<p>Then fill in two symbols which represent vectors (and optionally use TraditionalForm):</p>
<pre><code>f2s[q, k1] // TraditionalForm
</code></pre>
<p>which produces</p>
<p><img src="https://i.stack.imgur.com/EwD4X.jpg" alt="enter image description here"></p>
<p>You can then take the square and to review it apply <a href="http://reference.wolfram.com/mathematica/ref/Expand.html?q=Expand&lang=en" rel="nofollow noreferrer">Expand</a> and TraditionalForm:</p>
<pre><code>f2s[q, k1]^2 // Expand // TraditionalForm
</code></pre>
<p>which produces</p>
<p><img src="https://i.stack.imgur.com/u21fu.jpg" alt="enter image description here"></p>
<p>Alternatively, if you wish to work with the components of the vectors (using numbers here as example)</p>
<pre><code>q1 = {1, 2, 3}
q2 = {4, 5, 6}
</code></pre>
<p>write a function such as F2 (simplified here relative to the paper you mention)</p>
<pre><code>f2[v1_, v2_] := (5/7) + (1/2) (v1.v2)/(Norm[v1] Norm[v2])
</code></pre>
<p>use it to evaluate for any two vectors</p>
<pre><code>f2[q1, q2]
(* 5/7 + (8 Sqrt[2/11])/7 *)
</code></pre>
<p>then square it if you wish</p>
<pre><code>f2[q1, q2]^2
</code></pre>
<p>If you want to work with generalized vector components and know the dimensions of each vector, use for example</p>
<pre><code>q1 = {x1, y1, z1}
q2 = {x2, y2, z2}
f2[q1, q2]
</code></pre>
|
2,509,583 | <p>How is intermediate value theorem valid for $\sin x$ in $[0,\pi]$?</p>
<p>It has max value $1$ in the interval $[0,\pi]$ which doesn't lie between values given by $\sin0$ and $\sin\pi$.</p>
| José Carlos Santos | 446,262 | <p>Let $f\colon[a,b]\longrightarrow\mathbb R$ be a continuous function and let $y\in\mathbb R$. The intermediate value theorem says that <em>if</em> $f(a)\geqslant y\geqslant f(b)$ or if $f(a)\leqslant y\leqslant f(b)$, then there is a $c\in[a,b]$ such that $f(c)=y$. But it says <em>nothing</em> if $y$ lies outside the interval bounded by $f(a)$ and $f(b)$. So, there is no contradiction here.</p>
|
2,662,007 | <p>I'd like to show that the group given by presentation $G=\langle a,b \ | \ a^2b^2 \rangle$ is not isomorphic to any of $\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, or $\mathbb{Z} \times \mathbb{Z}$.</p>
<p>For $G \ncong \mathbb{Z}$, I note that $G$ is not free because it has a relator in its presentation.</p>
<p>For $G \ncong \mathbb{Z}/2\mathbb{Z}$, I claim that $G$ cannot be of order 2, since the subgroup $\langle a \rangle$ in $G$ is infinite: if I write $a^n=1$ for some $n \in \mathbb{Z}\ \backslash\{0\}$ then $a^n$ has to be obtained by a finite number of composition of self-inversions, or self-multiplications, or conjugations on the relator $a^2b^2$. All of these operations preserve the presence of $b$, so we can never get to $a^n$.</p>
<p>For $G \ncong \mathbb{Z} \times \mathbb{Z}$ I argue that $G$ is not abelian. The abelianization of $G$ is $G_{ab}=\langle a,b \ | \ a^2b^2, aba^{-1}b^{-1} \rangle$. How do I show $G_{ab}$ is not trivial?</p>
<p>I'm new to presentations and am completely unsure if the above reasonings are allowed or not. Any help appreciated.</p>
| Jeremy Rickard | 88,262 | <p>The empty $G$-set is the initial object.</p>
<p>The coproduct of $G$-sets is the disjoint union.</p>
<p>Call a $G$-set “indecomposable” if it is not the coproduct of two non-empty $G$-sets, so an indecomposable $G$-set is just a transitive one.</p>
<p>An epimorphism of $G$-sets is just a surjective map of $G$-sets.</p>
<p>Up to isomorphism, there is a unique transitive $G$-set with an epimorphism to every other transitive $G$-set, namely the regular $G$-set, whose automorphism group is $G$.</p>
<p>So $G$ can be recovered from the category of $G$-sets.</p>
|
247,622 | <p>Find the dimension of the vector subspace in $\mathbb{R}^{4}$ spanned by the columns </p>
<p>$$
\begin{pmatrix}
1 & 1 & 2 & 4 \\
t & 3 & 2 & 8 \\
0 & 0 & 2 & 2 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$</p>
<p>(discussion over real $t$).</p>
<p>Thanks :) </p>
| kahen | 1,269 | <p>Basic facts about <a href="https://en.wikipedia.org/wiki/Net_%28mathematics%29" rel="nofollow">nets</a>: A function $f: X\to Y$ is continuous at $x \in X$ if and only if for every net $(x_\alpha)$ in $X$ converging to $x$ we have that $f(x_\alpha) \to f(x)$; and a set $A \subseteq X$ is closed if and only if for every convergent net $(x_\alpha)$ taking values in $A$ the limit lies in $A$.</p>
<p><strong>Lemma 1</strong>. The graph of a continuous function $f: X\to Y$ is homeomorphic to $X$ as a subspace of $X\times Y$ via the projection map $\pi_1((x,y)) = x$.</p>
<p><em>Proof</em>. We only need to verify that the inverse $x \mapsto (x,(f(x))$ is continuous. Let $(x_\alpha)$ be a net converging to $x$. Then $f(x_\alpha) \to f(x)$ by continuity of $f$ and by the properties of the product topology $(x_\alpha,f(x_\alpha)) \to (x,f(x))$ as desired.</p>
<p><strong>Lemma 2</strong>. The graph of a continuous function into a Hausdorff space is closed.</p>
<p><em>Proof</em>. Let $f: X\to Y$ be as in Lemma 1 with $Y$ Hausdorff. Suppose we have a net $(x_\alpha,f(x_\alpha))$ converging to $(x,y) \in X \times Y$. We are to show that $(x,y)$ is in the graph of $f$, i.e. $y = f(x)$. Certainly we have $x_\alpha \to x$, so $f(x_\alpha) \to f(x)$ and since limits of nets in Hausdorff spaces are unique, we have $y=f(x)$ as desired.</p>
<p>Now we're ready for the proof that your set is closed. </p>
<p><strong>Proposition 3</strong>. Let $X$ be Hausdorff and $f,g: X \to X$ continuous. Then $E = \{x \in X\mathrel| f(x)=g(x)\}$ is closed in $X$.</p>
<p><em>Proof</em>. By the lemmas above $F$ and $G$, the graphs of $f$ and $g$ respectively, are closed in $X\times X$ and $E = \pi_1(F \cap G)$ is the image of a closed set under a homeomorphism, so $E$ is closed in $X$.</p>
<p>That your set is closed follows from using the above with the identity map as $g$.</p>
|
247,622 | <p>Find the dimension of the vector subspace in $\mathbb{R}^{4}$ spanned by the columns </p>
<p>$$
\begin{pmatrix}
1 & 1 & 2 & 4 \\
t & 3 & 2 & 8 \\
0 & 0 & 2 & 2 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$</p>
<p>(discussion over real $t$).</p>
<p>Thanks :) </p>
| user46128 | 46,128 | <p>Let $G=\{x \in X : f(x)=x\}$.
We want to prove that $G$ is closed, hence that $G^c$ is open.</p>
<p>So take $y\in G^c$.
Since X is Hausdorff we can find a two open sets U and V so that:
$y\in U$, $f(y)\in V$ and $U\cap V= \emptyset$.
Since f is continuous we can find a Set $V'$ s.t. $f(V')\subset V$.
The set $W=V' \cap U$ is open and non empty, since $y\in W$.
Furthermore, we see: $$f(W)\cap W=\emptyset$$
Hence $G^c$ is open. </p>
|
247,622 | <p>Find the dimension of the vector subspace in $\mathbb{R}^{4}$ spanned by the columns </p>
<p>$$
\begin{pmatrix}
1 & 1 & 2 & 4 \\
t & 3 & 2 & 8 \\
0 & 0 & 2 & 2 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$</p>
<p>(discussion over real $t$).</p>
<p>Thanks :) </p>
| Julian Kuelshammer | 15,416 | <p>Since there are several answers addressing the first question, let me just give an example for the second one. A good example for a T1-space that is not T2 is $\mathbb{N}$ with the cofinite topology, i.e. the closed sets are the finite ones. Then take the map $f:\mathbb{N}\to \mathbb{N}$, given as $f(x)=\begin{cases}x&\text{$x$ odd}\\\frac{x}{2}&\text{$x$ even.}\end{cases}$
I'll leave it as an exercise for you to show that this map is continous (you have to show that preimages of finite sets are finite). But $\{f(x)=x\}$ is the set of odd numbers, which is certainly not finite.</p>
|
59,105 | <p>Let $n$ be a positive integer. $\; $ Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere <a href="http://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative">Frechet differentiable</a>.
<br>
Let $S$ be a subset of $\mathbb{R}^n$ with <a href="http://en.wikipedia.org/wiki/Lebesgue_measure">Lebesgue measure</a> zero.</p>
<p>Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?</p>
<p>(I know it would if $D(f)$ is continuous.)</p>
| George Lowther | 1,321 | <p>Yes. In fact, if $f$ has a Frechet derivative $Df$ on a measurable set $S\subseteq{R}^n$ then
$$
\begin{align}
\mu(f(S))\le\int_S\left\vert{\rm det} Df\right\vert\,d\mu&&(1)
\end{align}
$$
where $\mu$ is the Lebesgue measure on $\mathbb{R}^n$. In particular, if $\mu(S)=0$ then the right hand side is zero regardless of the value of $Df$. Actually, (1) is an immediate consequence of the <a href="http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables">change of variables</a> formula for multidimensional integration, which can be shown to apply in the generality required here. Note that a general form of <a href="http://en.wikipedia.org/wiki/Sard%27s_theorem">Sard's theorem</a> also follows from (1), and says that the image of the set of points where $Df$ is singular has zero measure.</p>
<p>The Vitali covering lemma gives a quick proof that $f(S)$ has zero Lebesgue measure. If $f$ is Frechet differentiable at a point $x$ then, for any $K > \Vert Df\Vert$ we have $f(B_r(x))\subseteq B_{Kr}(f(x))$ for all small enough $r > 0$. So, $\mu(f(B_r(x))\le K^n\mu(B_r(x))$. Now, if $\Vert Df\Vert < K$ on a measurable set $S$ then you can choose open $U\supseteq S$ with $\mu(U)$ as close to $\mu(S)$ as you wish. For each $x\in S$ choose $r_x\in(0,1)$ such that $B_{r_x}(x)\subseteq U$ and $f(B_{r}(x))\subseteq B_{Kr}(f(x))$ for all $r\le 5r_x$. The <a href="http://en.wikipedia.org/wiki/Vitali_covering_lemma">Vitali covering lemma</a> implies that there is a countable subset $S^\prime\subseteq S$ such that $\{B_{r_x}(x)\colon x\in S^\prime\}$ are disjoint and
$$
\bigcup_{x\in S^\prime}B_{5r_x}(x)\supseteq\bigcup_{x\in S}B_{r_x}(x)\supseteq S.
$$
So,
$$
\begin{align}
\mu(f(S))&\le\sum_{x\in S^\prime}\mu(f(B_{5r_x}(x)))\\
&\le 5^nK^n\sum_{x\in S^{\prime}}\mu(B_{r_x}(x))\\
&=5^nK^n\mu\left(\bigcup_{x\in S^\prime}B_{r_x}(S)\right)\\
&\le5^nK^n\mu(U).
\end{align}
$$
Letting $\mu(U)$ decrease to $\mu(S)$ gives $\mu(f(S))\le5^nK^n\mu(S)$. In particular, if $\mu(S)=0$ then $\mu(f(S))=0$. Even if the Frechet derivative just exists on $S$, but is not bounded, we can let $S_n$ be the subset of $S$ on which $\Vert Df\Vert\le n$. Then, using monotone convergence, $\mu(f(S))=\lim_{n\to\infty}\mu(f(S_n))=0$.</p>
<p>This gives a positive answer to the question asked, where $\mu(S)=0$. We can go further and use the Vitali covering theorem to prove (1). In fact, the full change of variables formula
$$
\int_S\vert{\rm det} Df\vert\,d\mu=\int_{f(S)}\#\left(f^{-1}(\{y\})\cap S\right)\,d\mu(y)
$$
can be shown to hold in the generality asked here. This can be proven with a similar argument as above involving the Vitali covering theorem, and also using the limits $\lim_{r\to0}\mu(f(B_r(x))/\mu(B_r(x))=\vert{\rm det}Df(x)\vert$ on $S$.</p>
|
1,598,255 | <p>Consider the system
$$
\dot{x}=y,\qquad\dot{y}=x+x^2-y.
$$
It has two equilibria, namely $(0,0)$ and $(-1,0)$.</p>
<p>I would like to linearize the system in both equilibria.</p>
<p>My start is to set $\Delta x=x-x_0,\qquad \Delta y=y-y_0$. Then
$$
\dot{\Delta x}=\dot{x}-\dot{x_0}=y-y_0=\Delta y
$$
and
$$
\dot{\Delta y}=\dot{y}-\dot{y_0}=\Delta x-\Delta y+x^2-x_0^2.
$$</p>
<p>How can I get rid of the summand $x^2-x_0^2$?</p>
<p>Can I approximate the function $f(x)=x^2$ by Taylor, getting $x^2-x_0^2\approx 2x_0\Delta x$?</p>
<p>I then would get the linearization matrices
$$
\begin{pmatrix}0 & 1\\ 1 & -1\end{pmatrix}\text{ for }(0,0)
$$
and, similarly,
$$
\begin{pmatrix}0 & 1\\-1 & -1\end{pmatrix}\text{ for }(-1,0).
$$</p>
| Lutz Lehmann | 115,115 | <p>No, that is wrong. $x_0$ and $y_0$ are constants, only $Δx$ and $Δy$ are functions of $t$. Thus
\begin{align}
(Δx)˙=\frac d{dt}(x_0+Δx(t))&=y_0\quad+\quad Δy
\\ \\
(Δy)˙=\frac d{dt}(y_0+Δy(t))
&=(x_0+Δx)+(x_0+Δx)^2−(y_0+Δy)
\\
&=x_0+x_0^2-y_0\quad+\quad (1+2x_0)Δx-Δy\quad+\quad Δx^2
\end{align}</p>
|
221,137 | <p>What's the difference between Fourier transformations and Fourier Series? </p>
<p>Are they the same, where a transformation is just used when its applied (i.e. not used in pure mathematics)?</p>
| Vitali Pom | 692,929 | <p>I see you’re asking probably about the most “meaningful two ones” that are seemingly differ, but yet comparable if you compare them one to one and these are discrete time signal’s fourier transform and discrete time signal’s fourier series, while both signals are periodic. </p>
<p>I’ll explain it in the simplest way possible. </p>
<p>Imagine you have a drawing in a map (don’t imagine an nth dimensional map, but it might be). Both of them represent a point on this drawing on this map. (Map is a globe map, don’t go too far.) Now it’s simple, each representation solves a different problem, but both are representing the same point. It’s just one representation represent a point as a point as we know it (alas series, with e and i or j and w0) and one representation (the one with the deltas) represents the same point as + <0,delta(y)>. We know the first representation. I won’t explain. But in the second representation if we sum up all e-s with i-s or j-s we’ll get what? A superposition. Meaning that we’ll know the final superposition, but won’t know what the projection is. We won’t know what x,y is because in signal x,y are not trivial to see as a point on the map while they are in superposition, so we need a coordinate system instead. The Fourier Transform shows us the same point, but like in the map - with coordinates. Alas we see the coefficients. And we see the coefficients thanks to the deltas. Otherwise we’d had a sum of e’s which is sort of “summable” sometimes and in order to find each coefficient we’d had to project and calculate the projection for each coefficient. Here we SEE the coefficients WITHOUT calculating each of them thanks to the deltas. </p>
<p>All this is because the inner product in signal’s dimension is “less straight forward” than the inner product in a globe map. But yet it is straight forward for a computer if you do it repeatedly and ineffectively. :)</p>
<p>That’s it, hope it helps. </p>
|
1,148,624 | <p>This is for teaching math. I'm wondering if someone knows some striking near-equalities between simple arithmetic expressions. I vaguely remember that such things exist (e.g., numbers that look alike out 10 digits, but that are really different), but I don't remember where I saw them or what they are. I think there are even some famous historical instances where it was debated whether certain expressions were equal or not.</p>
<p>One possibility I am particularly interested in is integer combinations of (integer) square roots that turn out to be very very close to zero (but that are nonzero). Does anyone know how to construct these? I am assuming they exist because I read somewhere (and forgotten again!) that there is currently no efficient algorithm---or maybe even no algorithm at all, can't remember---for determining the sign of such a sum.</p>
<p>Thank you!</p>
<p>PS: I'm also interested in sources (e.g., book of number puzzles or relevant number theory, etc).</p>
| Asinomás | 33,907 | <p>$e^{\sqrt{163}\pi}=262537412640768743.99999999999925$ is almost an integer. It was conjectured for a long while this number could in fact be an integer.</p>
<p>You can find more on this here:</p>
<p><a href="https://www.youtube.com/watch?v=DRxAVA6gYMM" rel="nofollow noreferrer">https://www.youtube.com/watch?v=DRxAVA6gYMM</a></p>
<p><a href="https://math.stackexchange.com/questions/4544/why-is-e-pi-sqrt163-almost-an-integer">Why is $e^{\pi \sqrt{163}}$ almost an integer?</a></p>
|
2,948,673 | <p>I was thinking about the squeeze theorem here. We can denote the <span class="math-container">$\left[\frac{1}{|x|}\right] =n$</span>, and then try something like:</p>
<p><span class="math-container">$$x^2(1+2+3+...+(n-1)+(n-1)) \leq x^2 \frac{n(n+1)}{2} \leq x^2(1+2+3+...+(n-1)+(n+1))$$</span></p>
<p>But I don't know what to do with <span class="math-container">$x^2$</span>. Given that <span class="math-container">$\left[ \frac{1}{|x|} \right]=n$</span>, how do we proceed to find <span class="math-container">$x^2$</span>?</p>
| ajotatxe | 132,456 | <p>Given that
<span class="math-container">$$\left[\frac1{|x|}\right]=n$$</span>
we have
<span class="math-container">$$n\le\frac1{|x|}<n+1$$</span>
Therefore
<span class="math-container">$$\frac1n\ge|x|>\frac1{n+1}$$</span>
and thus,
<span class="math-container">$$\frac1{(n+1)^2}<x^2\le\frac1{n^2}$$</span></p>
<p>Because you seem worry only about an upper bound of the function, I feel that you expect that the limit is <span class="math-container">$0$</span>. I think it isn't...</p>
|
313,597 | <p>Let $V$ be a finite-dimensional vector space with an ordered base $\beta$ over a field $F$.</p>
<p>Let $x\in V$ an $[x]_{\beta}$ be a coordinate vector of $x$ relative to $\beta$.</p>
<p>Is this an element of $M_{n\times 1}(F)$ or $F^n$?</p>
<p>Since we usually treat them as a matrix, i think it should be defined as a matrix so that $[x]_{\beta} \in M_{n\times 1}(F)$. However, the text i'm studying states that it is an element of $F^n$. (Note that $F^n\neq M_{n\times 1}(F)=F^{n\times 1}$)</p>
<p>The reason why I'm distinguishing these two even though they have the same meaning, is to formally define a matrix multiplication. For example, an element in $F^n$ is not only sometimes treated as a column vector, but also treated as a row vector. If these two are not distinguished then multiplication cannot be well defined.</p>
<p>Am i correct? Or if i am wrong or if what i'm worrying is not a matter, please explain me why. Thank you in advance.</p>
| Berci | 41,488 | <p>You are absolutely correct.</p>
<p>However, -- and if you read some specific text, then it's probably written somewhere explicitly -- it is usual to <em>identify</em> $F^n$ with the <em>culomn vectors</em>, i.e. $n\times 1$ matrices. As you say, $F^n\ne M_{n\times 1}(F)$, but they are isomorphic. It is also usual that an $n\times m$ matrix $A$ represents a linear map $F^m\to F^n$, namely
$$v\mapsto A\cdot v$$
where the product is given by matrix multiplication. Here $v$ was considered as a culomn vector (this is the default, if composition of (linear) functions is written to the left: $f\circ g=x\mapsto f(g(x))$.)</p>
|
1,221,368 | <p>An inheritance is divided between three people in the ratio $4:7:2$. If the least amount received is $\$\ 2300$ calculate how much the other two people received?</p>
| Mufasa | 49,003 | <p>If the amount is distributed in the ratio $4:7:2$ then you know the person who got the least got:$$\frac{2}{4+7+2}=\frac{2}{13}$$of the total amount.</p>
<p>Use this fact to arrive at the answer.</p>
|
103,358 | <p>I am trying to illustrate some simple ideas with exponents. I can manually express something like $5^4$ as $5 \cdot 5 \cdot 5 \cdot 5$, but wondered how to get <em>Mathematica</em> to do that for me.</p>
<p>I found the example below in the documentation, but can't figure out how to "massage" it to work with a number that only has one factor, for example I would like $625$ to be represented as $5 \cdot 5 \cdot 5 \cdot 5$:</p>
<pre><code>CenterDot @@ (Superscript @@@ FactorInteger[20!])
</code></pre>
<p>Any ideas would be appreciated.</p>
| Zhi-Bang Yao | 84,879 | <p>I have just solved this problem by turning off the automatic updates in the preferences. Hope it will work for you too!</p>
|
650,381 | <p>the definition of variance is $V(X) = E((X-E(X))^2 )$</p>
<p>For a discrete random variable: if we have put $Y = g(X)$ , where $g$ is a real function</p>
<p>$E(Y) = E(g(X)) = \sum\limits_{k} g(k)p_X(k)$ , where $p_X(k) = P(X = k) $ </p>
<p>how can i use this formula to show that:</p>
<p>$V(X) = \sum\limits_{k} (k-E(X))^2 p_X(k)$</p>
| jaranna | 119,935 | <p>@Amzoti: Could you tell me where is my mistake: $A= \frac{i}{3} \begin{bmatrix} 1&-2&1\\-2&1&1\\1&1&-2\end{bmatrix} = \begin{bmatrix} \frac{i}{3}&\frac{-2i}{3}&\frac{i}{3}\\\frac{-2i}{3}&\frac{i}{3}&\frac{i}{3}\\\frac{i}{3}&\frac{i}{3}&\frac{-2i}{3}\end{bmatrix} \\ A_{\lambda} = \begin{bmatrix} \frac{i}{3}-\lambda&\frac{-2i}{3}&\frac{i}{3}\\ \frac{-2i}{3}&\frac{i}{3}-\lambda&\frac{i}{3}\\ \frac{i}{3}&\frac{i}{3}&\frac{-2i}{3}-\lambda\end{bmatrix}$</p>
<p>After transformations($C_{1}-C_{2}$ and $W_{2}+W_{1}$), I have:
$A_{\lambda} = \begin{bmatrix} i-\frac{\lambda}{3}&\frac{-2i}{3}&\frac{i}{3}\\ 0&\frac{-i-\lambda}{3}&\frac{2i}{3}\\ 0&\frac{i}{3}&\frac{-2i-\lambda}{3}\end{bmatrix}$</p>
<p>So then I wanted to develop it with the first column:
$(i-\frac{\lambda}{3})(-1)^{2} \begin{bmatrix} \frac{-1-\lambda}{3}&\frac{2i}{3}\\\frac{i}{3}&\frac{-2i-\lambda}{3} \end{bmatrix}$</p>
<p>And the CP: $\lambda(i-\frac{\lambda}{3})(i+\frac{\lambda}{3})=0$</p>
|
359,816 | <p>Let $(f_n)$ and $(g_n)$ be sequences of nonnegative function in $L^1(\mathbb R)$, for which
$$
f_n \to 0, \\
g_n \to 0,
$$
almost everywhere.
Show
$$
\int_A \frac{2f_n g_n}{1+f_n^2+g_n^2} \to 0,
$$
when $A \subset \mathbb R$ is a set of finite measure.</p>
<p>Define
$$
h_n = \frac{2f_n g_n}{1+f_n^2+g_n^2}.
$$
I know
$$
h_n \le 2f_n g_n,
$$
but I can't use a convergence theorem, because the product of integrable function may fail to be integrable.<br>
I considered uniform integrability
$$
\text{Since }
f_n \in L^1: \forall \epsilon, \exists \delta : m(B) < \delta \to \int_B f_n <\epsilon,
$$
where $m(\cdot)$ is Lebesgue measure.<br>
To use that, I write $A=\cup_m B_{n,m}$ where the index $n$ is associated to $f_n$.<br>
I can't go further.</p>
| Davide Giraudo | 9,849 | <p><strong>Hint:</strong> Egorov's theorem. Where there is uniform convergence, no problem, and on "the small set", the integrand is bounded by $1$.</p>
|
2,850,435 | <p>The equation is
$$
\int f(x) = \frac{1}{f'(x)}
$$
I saw it on a shirt so I thought it is probably a joke, but I couldn't get it. So I tried to solve it. On the first look I thought this is never true. Then I thought of differentiating the whole thing.</p>
<p>$$
f(x) = - \frac{f''(x)}{f'(x)^2}
$$ </p>
<p>This is what I got. I don't see any easy solution. This looks like a complicated differential equation. The only diff. eq. that I know how to solve are the most basic ones that appear in newtonian mechanics.</p>
<p>So, where is the joke if there is one. Or is this equation special in any other way.</p>
| Exit path | 161,569 | <p>Since pointed maps $X \wedge Y \to Z$ are precisely pointed maps $X \times Y \to Z$ that send the wedge sum of $X$ and $Y$ to the base point in $Z$, by the adjunction we know that such maps correspond to certain maps $X \to Maps(Y,Z)$. Can you figure out which these are?</p>
<p>These adjunctions work for all reasonable spaces. </p>
|
1,425,961 | <p>I am about to take a course in undergraduate Algebra which is using Artin's Algebra. I have been studying from it and haven't really appreciated the combination of text and problems (granted, I haven't gone too deep into the text). </p>
<p>Now, this is the text for the class, so I would expect a nice compliment from the lecture to go along with Artin. I was thinking about getting Dummit and Foote, which seems to have high praise across the board. What are the strengths and weaknesses of Dummit and Foote, compared to Artin? Is there a better text for a class at this level?</p>
<p>EDIT: I really don't want a recommendation for another text, but how does Artin compare to other texts. </p>
| JeffW89 | 200,928 | <p>Dummit and Foote has a more standard format, and covers a wider variety of topics, but it's extremely dry (and feels very "text-booky" to me). I didn't like it much. Take a look at Vinberg instead. Takes a very similar approach to Artin, but it's more readable in my opinion and has a more manageable number of problems (which are embedded in the section, so you're forced to solve them as you go along). I find the massive number of exercises in Artin a bit discouraging. </p>
|
1,808,019 | <p>I'm trying to better understand how the following group action looks like on a torus. The group $\mathbb{R}$ acts on the torus $C\times C=\{(e^{ix},e^{iy}):x,y\in \mathbb{R}\}$ ($C$ being the unit circle in $\mathbb{C}$) via $$t\cdot (e^{ix},e^{iy})\mapsto (e^{i(x+t)},e^{iy})$$</p>
<p>I am able (I think) to visualize the torus based on the definition of $C\times C$: the first coordinate gives us the ring of unit length in the donut and, as $x$ sweeps through $\mathbb{R}$, the second coordinate gives us the "hollow body" of the donut, again of unit length, as $y$ sweeps through $\mathbb{R}$. So here is where the problems begin for me.</p>
<ol>
<li><p>What am I doing wrong to show that this a homomorphism? $$st\cdot (e^{ix},e^{iy})\mapsto (e^{i(x+st)},e^{iy})$$ but $$[s\cdot (e^{ix},e^{iy})][t\cdot (e^{ix},e^{iy})] = (e^{i(x+s)},e^{iy})(e^{i(x+t)},e^{iy}) = (e^{i(2x+s+t)},e^{i2y})$$</p></li>
<li><p>For $t\in \mathbb{R}$, the orbit $\mathbb{R}(t)$ is $\{(e^{i(x+t)},e^{iy}):t\in \mathbb{R}\}$. Is this any different from $C\times C$?</p></li>
</ol>
| Behrouz Maleki | 343,616 | <p>for $n=1$ we have
$${{(x+y)}^{\,1}}=\ 1\ =\sum\limits_{i=0}^{1}{\left( \begin{matrix}
1 \\
0 \\
\end{matrix} \right)}\,{{x}^{1-i}}{{y}^{i}}=\left( \begin{matrix}
1 \\
0 \\
\end{matrix} \right)\,{{x}^{1}}+\left( \begin{matrix}
1 \\
1 \\
\end{matrix} \right)\,{{y}^{1}}=x+y$$</p>
<p>for $n=k$ let
$$(x+y){{\,}^{k}}=\ \sum\limits_{i=0}^{k}{\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)}\ {{x}^{k-i}}{{y}^{i}}$$</p>
<p>for $n=k+1$ we show
$$\left( x+y \right){{\,}^{k+1}}\,=\ \sum\limits_{i=0}^{k+1}{\left( \begin{matrix}
k+1 \\
i \\
\end{matrix} \right)}\ {{x}^{k-i+1}}{{y}^{i}}$$
<strong>proof</strong></p>
<p>$$\left( x+y \right){{\,}^{k}}\left( x+y \right)\ =\ \left( x+y \right)\ \sum\limits_{i=0}^{k}{\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)\ {{x}^{k-i}}}\ {{y}^{i}}\quad $$
as a result
$$\left( x+y \right){{\,}^{k+1}}=\sum\limits_{i=0}^{k}{\,\,\,\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)}\ {{x}^{k\,-\,i\,\,+\,1}}{{y}^{i}}\ +\,\sum\limits_{i=0}^{k}{\,\,\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)}\ {{x}^{k\,-\,i}}{{y}^{i\,+\,1}}$$</p>
<p>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =x\sum\limits_{i=0}^{k}{\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)}\ {{x}^{k-i}}{{y}^{i}}+y\ \sum\limits_{i=0}^{k}{\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)}\ {{x}^{k-i}}{{y}^{i}}
$$
By expanding the last equation
\begin{align}
& {{\left( x+y \right)}^{\,k+1}}\ =\,\left( \begin{matrix}
k \\
0 \\
\end{matrix} \right)\,\,{{x}^{\,k+1}}\,\,+\,\,\left( \begin{matrix}
k \\
1 \\
\end{matrix} \right)\,\,{{x}^{k}}y+\cdots +\,\left( \begin{matrix}
k \\
k-1 \\
\end{matrix} \right)\,\,{{x}^{2}}{{y}^{k-1}}+\left( \begin{matrix}
k \\
k \\
\end{matrix} \right)\,\,x{{y}^{k}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{align}
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( \begin{matrix}
k \\
0 \\
\end{matrix} \right)\,\,{{x}^{k}}y+\cdots +\,\left( \begin{matrix}
k \\
k-2 \\
\end{matrix} \right)\,{{x}^{2}}{{y}^{k-1}}+\,\left( \begin{matrix}
k \\
k-1 \\
\end{matrix} \right)\,\,x{{y}^{k}}+\,\left( \begin{matrix}
k \\
k \\
\end{matrix} \right)\,\,{{y}^{k+1}}$$
we know$$\left( \begin{matrix}
k \\
i \\
\end{matrix} \right)=\left( \begin{matrix}
k-1 \\
i \\
\end{matrix} \right)+\left( \begin{matrix}
k-1 \\
i-1 \\
\end{matrix} \right)$$
then
$$\,\left( x+y \right){{\,}^{k+1}}\,=\ \sum\limits_{i=0}^{k+1}{\left( \begin{matrix}
k+1 \\
i \\
\end{matrix} \right)}\ {{x}^{k-i+1}}{{y}^{i}}$$</p>
|
1,320,659 | <p>Is the ratio test applicable for testing convergence of infinite products?</p>
<p>In other words, consider the sequence $(a_i)_{i=1}^\infty$ of non-zero real numbers.</p>
<p>Also, consider the product $\displaystyle\mathcal P=\prod_{k=1}^\infty a_k$</p>
<p>Are the following statements true?</p>
<blockquote>
<p>$$\lim_{n\to\infty}\left|\frac{a_{k+1}}{a_k}\right|\lt 1\implies \mathcal P\textrm{ converges}\\ \lim_{n\to\infty}\left|\frac{a_{k+1}}{a_k}\right|\gt 1\implies \mathcal P\textrm{ diverges}$$</p>
</blockquote>
| Zach Stone | 38,565 | <p>With products, $a_k\to 1$ is the right assumption, otherwise stupid things happen. Just as in series, we usually just assume $a_k \to 0$. So in this context,
$$
\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_{k}}\right| =1
$$
should happen every time.</p>
<p>It's worth noting that
$$\prod_{k=1}^{\infty}a_k$$
converges iff
$$\sum_{i=1}^k\log(a_k)$$ </p>
<p>Think about the exponential function to see why this is. So really, what you want to test is </p>
<p>$$
\lim_{k\to\infty}\left|\frac{\log(a_{k+1})}{\log(a_{k})}\right|
$$</p>
|
2,666,390 | <p>How do I show that for $x>0$:
$$\frac{x-(x^2+1)\arctan(x)}{x^2(x^2+1)} < 0$$</p>
<p>I tried to do it somehow using the fact that $$\frac{\arctan(x)}{x} < 1$$ but still didn't figure it out...</p>
| Jack D'Aurizio | 44,121 | <p>The question is equivalent to proving $\arctan(x)>\frac{x}{x^2+1}$ for any $x>0$, but this is trivial since
$$ \arctan(x)=\int_{0}^{x}\frac{dt}{t^2+1}> \int_{0}^{x}\frac{dt}{x^2+1}=\frac{x}{x^2+1}.$$</p>
|
389,912 | <p>I'm writing a paper and want to cite some references to efficiently prove that over any field <span class="math-container">$k$</span> of characteristic zero, every irreducible representation of a product of symmetric groups, say</p>
<p><span class="math-container">$$ S_{n_1} \times \cdots \times S_{n_p} $$</span></p>
<p>is isomorphic to a tensor product <span class="math-container">$\rho_1 \otimes \cdots \otimes \rho_p$</span> where <span class="math-container">$\rho_i$</span> is an irreducible representation of <span class="math-container">$S_{n_i}$</span>.</p>
<p>I have an open mind about this, but I'm imagining doing it by finding references for these two claims:</p>
<ol>
<li><p>If <span class="math-container">$k$</span> is an algebraically closed field of characteristic zero, every irreducible representation of a product <span class="math-container">$G_1 \times G_2$</span> of finite groups is of the form <span class="math-container">$\rho_1 \otimes \rho_2$</span> where <span class="math-container">$\rho_i$</span> is an irreducible representation of <span class="math-container">$G_i$</span>.</p>
</li>
<li><p>If <span class="math-container">$k$</span> has characteristic zero and <span class="math-container">$\overline{k}$</span> is its algebraic closure, every finite-dimensional representation of <span class="math-container">$S_n$</span> over <span class="math-container">$\overline{k}$</span> is isomorphic to one of the form <span class="math-container">$\overline{k} \otimes_k \rho$</span> where <span class="math-container">$\rho$</span> is a representation of <span class="math-container">$S_n$</span> over <span class="math-container">$k$</span>.</p>
</li>
</ol>
<p>Serre's book <em>Linear Representations of Finite Groups</em> states the first fact for <span class="math-container">$k = \mathbb{C}$</span> but apparently not for a general algebraically closed field of characteristic zero. (It's Theorem 10.) It could be true already for any field of characteristic zero, which would simplify my life.</p>
<p>The second fact should be equivalent to saying that <span class="math-container">$\mathbb{Q}$</span> is a splitting field for any symmetric group, which seems to be something everyone knows - yet I haven't found a good reference.</p>
| Geoff Robinson | 14,450 | <p>A positive answer to the first question for <span class="math-container">$S_{n}$</span> follows from the work of Alfred Young. He proved that every irreducible representation of <span class="math-container">$S_{n}$</span> over <span class="math-container">$\mathbb{C}$</span> is afforded by an explicit <span class="math-container">$\mathbb{Q}G$</span>-module. Another way of saying this is that any irreducible <span class="math-container">$\mathbb{Q}S_{n}$</span>-module <span class="math-container">$V$</span> is absolutely irreducible, and we have <span class="math-container">${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$</span> This last property is already enough to answer the question (positively) for symmetric groups for the field <span class="math-container">$k = \mathbb{Q}$</span>, and then for every characteristic zero field <span class="math-container">$k$</span>, since the prime subfield of <span class="math-container">$k$</span> is isomorphic to <span class="math-container">$\mathbb{Q}.$</span> This answers 2 positively.</p>
<p>The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible <span class="math-container">$kG_{i}$</span>-module is isomorphic to <span class="math-container">$k$</span>. This certainly holds for any algebraically closed field <span class="math-container">$k$</span> of characteristic zero.</p>
<p>I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of <span class="math-container">$S_{n}$</span> (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).</p>
<p>Later edit: Perhaps a more succinct way to say all this is that <span class="math-container">$\mathbb{Q}$</span> is a splitting field for <span class="math-container">$S_{n}$</span>, and consequently, any field <span class="math-container">$k$</span> of characteristic
zero is a splitting field for <span class="math-container">$S_{n}$</span>. On the other hand, if field <span class="math-container">$k$</span> of characteristic zero is a splitting field for each finite group <span class="math-container">$G_{i}: 1 \leq i \leq n$</span>, then <span class="math-container">$k$</span> is a splitting field for <span class="math-container">$G_{1} \times G_{2} \times \ldots \times G_{n}$</span>, and any irreducible <span class="math-container">$kG$</span>-module is the tensor product of <span class="math-container">$n$</span> irreducible modules, one for each <span class="math-container">$G_{i}$</span>. This is a consequence of Schur's Lemma and Clifford's Theorem.</p>
|
3,720,856 | <p><strong>Question:</strong>
Prove that <span class="math-container">$\sin(nx) \cos((n+1)x)-\sin((n-1)x)\cos(nx) = \sin(x) \cos(2nx)$</span> for <span class="math-container">$n \in \mathbb{R}$</span>.</p>
<p><strong>My attempts:</strong></p>
<blockquote>
<p>I initially began messing around with the product to sum identities, but I couldn't find any way to actually use them.<br />
I also tried compound angles to expand the expression, but it became too difficult to work with.</p>
</blockquote>
<p>Any help or guidance would be greatly appreciated</p>
| J.G. | 56,861 | <p>The left-hand side is<span class="math-container">$$\begin{align}&\sin nx(\cos nx\cos x-\sin nx\sin x)-(\sin nx\cos x-\cos nx\sin x)\cos nx\\&=(\cos^2nx-\sin^2nx)\sin x\\&=\cos 2nx\sin x.\end{align}$$</span></p>
|
2,744,172 | <p>(Note: This question is tangentially related to this <a href="https://math.stackexchange.com/q/3609304">later one</a>.)</p>
<p>Let <span class="math-container">$$\sigma(x) = \sum_{d \mid x}{d}$$</span> denote the <em>sum of divisors</em> of <span class="math-container">$x \in \mathbb{N}$</span>, where <span class="math-container">$\mathbb{N}$</span> is the set of natural numbers or positive integers.</p>
<p>Recall that a <em>Descartes number</em> is an odd number <span class="math-container">$n = km$</span>, with <span class="math-container">$1 < k$</span>, <span class="math-container">$1 < m$</span>, satisfying <span class="math-container">$$\sigma(k)(m+1)=2km.$$</span> (<span class="math-container">$m$</span> is called the quasi-Euler prime of <span class="math-container">$n$</span>.) Note that we define <span class="math-container">$\sigma(m) := m + 1$</span> even when <span class="math-container">$m$</span> is composite (that is, we <em>pretend</em> that <span class="math-container">$m$</span> is prime).</p>
<p>Notice that the lone Descartes number that is known is
<span class="math-container">$$\mathscr{D} = k'm' = {{3003}^2}\cdot{22021}.$$</span></p>
<p>In particular, note that:</p>
<p><strong>(1)</strong> <span class="math-container">$k$</span> is a square.</p>
<p><strong>(2)</strong> <span class="math-container">$\sigma(k)/m = 2k - \sigma(k)$</span></p>
<p><strong>(3)</strong> <span class="math-container">$m \equiv 1 \pmod 4$</span></p>
<p>I want to prove that it must necessarily be the case that <span class="math-container">$m < k$</span>, for a Descartes number <span class="math-container">$n = km$</span>.</p>
<p><strong>Lemma 1</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span>, then <span class="math-container">$k \neq m$</span>.</p>
<p>To this end, suppose that <span class="math-container">$m = k$</span>. Then we have
<span class="math-container">$$\frac{\sigma(k)}{k} = \frac{2m}{m + 1} = \frac{2k}{k + 1},$$</span>
from which it follows that
<span class="math-container">$$\sigma(k) = \frac{2k^2}{k + 1} = \frac{2k^2 - 2}{k + 1} + \frac{2}{k + 1} = \frac{2(k - 1)(k + 1)}{k + 1} + \frac{2}{k + 1} = 2(k - 1) + \frac{2}{k + 1}.$$</span></p>
<p>Since <span class="math-container">$\sigma(k)$</span> and <span class="math-container">$2(k - 1)$</span> are integers, it follows that <span class="math-container">$2/(k+1)$</span> is also an integer, which means that <span class="math-container">$(k + 1) \mid 2$</span>. This implies that <span class="math-container">$k + 1 \leq 2$</span>, from which wet get <span class="math-container">$m = k \leq 1$</span>. This last inequality contradicts the condition <span class="math-container">$1 < k$</span>, <span class="math-container">$1 < m$</span>.</p>
<p><strong>Lemma 2</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m,k)=1$</span>, then
<span class="math-container">$$\frac{\sigma(m)}{k} \neq \frac{\sigma(k)}{m}.$$</span></p>
<p>Suppose to the contrary that <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m,k)=1$</span>, and that <span class="math-container">$\sigma(m)/k = \sigma(k)/m$</span>. Then it follows that
<span class="math-container">$$\frac{\sigma(m)}{k} = \frac{\sigma(k)}{m} = r \in \mathbb{N},$$</span>
from which we obtain
<span class="math-container">$$\frac{\sigma(m)}{k}\cdot\frac{\sigma(k)}{m} = r^2 \in \mathbb{N},$$</span>
contradicting
<span class="math-container">$$\sigma(k)\sigma(m) = \sigma(k)(m+1) = 2km,$$</span>
since the last two equations imply that <span class="math-container">$r^2 = 2$</span>.</p>
<p><strong>Lemma 3</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k) = 1$</span>, then we have</p>
<p><strong>(a)</strong> <span class="math-container">$\sigma(m) \neq \sigma(k)$</span></p>
<p><strong>(b)</strong> <span class="math-container">$\sigma(m) \neq k$</span></p>
<p><strong>(c)</strong> <span class="math-container">$\sigma(k) \neq m$</span></p>
<p>Proof of <strong>(a)</strong>: Suppose that <span class="math-container">$\sigma(k) = \sigma(m) = m + 1 \equiv 2 \pmod 4$</span>. This contradicts the fact that <span class="math-container">$k$</span> is a square, since then <span class="math-container">$\sigma(k) \equiv 1 \pmod 2$</span>.</p>
<p>Proof of <strong>(b)</strong>: Suppose that <span class="math-container">$\sigma(m) = k$</span>. Then the even number <span class="math-container">$m + 1 = \sigma(m)$</span> is equal to the odd number <span class="math-container">$k$</span>, which is a clear contradiction.</p>
<p>Proof of <strong>(c)</strong>: Suppose to the contrary that <span class="math-container">$\sigma(k) = m$</span>. Then we obtain the estimate
<span class="math-container">$$\frac{\sigma(k)}{m} + \frac{\sigma(m)}{k} = 1 + 2 = 3,$$</span>
which contradicts the known quantity
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$</span></p>
<p><strong>Lemma 4</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k)=1$</span>, then the following biconditionals hold:
<span class="math-container">$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}$$</span></p>
<p>We consider three different cases:</p>
<p><strong>Case (1)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} = \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$</span></p>
<p>Case (1) is equivalent to <span class="math-container">$\sigma(m) = \sigma(k)$</span> (which is ruled out by Lemma 3 (a)) or <span class="math-container">$k = m$</span> (which is ruled out by Lemma 1).</p>
<p><strong>Case (2)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$</span></p>
<p>Case (2) implies the estimate
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{9 + 1}{9} + 2 = \frac{28}{9},$$</span>
which again contradicts the known quantity
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$</span></p>
<p><strong>Case (3)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{\sigma(m)}{k} + \frac{\sigma(k)}{m}$$</span></p>
<p>Case (3) is equivalent to the truth of the biconditional <span class="math-container">$m < k \iff \sigma(m) < \sigma(k)$</span> (by virtue of Lemma 1, Lemma 2, and Lemma 3), which in turn is equivalent to the truth of the biconditional
<span class="math-container">$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}.$$</span></p>
<p>By Lemma 4, we have the following possibilities:</p>
<p><strong>(A)</strong> <span class="math-container">$k < \sigma(k) < m < \sigma(m)$</span></p>
<p><strong>(B)</strong> <span class="math-container">$m < \sigma(m) < k < \sigma(k)$</span></p>
<p>Note that Case <strong>(A)</strong> implies that
<span class="math-container">$$\frac{\sigma(k)}{m} = 2k - \sigma(k) < 1$$</span>
forcing <span class="math-container">$2k - \sigma(k) = 0$</span> (i.e. <span class="math-container">$k$</span> must be perfect). This contradicts the fact that <span class="math-container">$k$</span> is a square.</p>
<p>Hence we necessarily have Case <strong>(B)</strong>, and a proof for the following theorem:</p>
<p><strong>THEOREM</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k) = 1$</span>, then <span class="math-container">$k$</span> is <em>not</em> an odd almost perfect number.</p>
<p><strong>QUESTIONS</strong></p>
<p><em>(I)</em> Can we remove the reliance of the proof on the condition
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002?$$</span></p>
<p><em>(II)</em> To what extent can we relax the condition <span class="math-container">$\gcd(m, k)=1$</span> in the <strong>THEOREM</strong>?</p>
| Servaes | 30,382 | <p>Let <span class="math-container">$k$</span> and <span class="math-container">$m$</span> be odd integers greater than <span class="math-container">$1$</span> such that
<span class="math-container">$$(m+1)\sigma(k)=2mk.$$</span>
Then clearly <span class="math-container">$\sigma(k)<2k$</span>, and hence the above is equivalent to
<span class="math-container">$$m=\frac{\sigma(k)}{2k-\sigma(k)}.$$</span>
Now the desired result that <span class="math-container">$m<k$</span> is equivalent to
<span class="math-container">$$\frac{\sigma(k)}{2k-\sigma(k)}<k,$$</span>
which in turn, by clearing denominators and rearranging, is equivalent to
<span class="math-container">$$\sigma(k)<\frac{2k^2}{k+1}=2k-2+\frac{2}{k+1}.$$</span>
In particular, as <span class="math-container">$k>1$</span> this is equivalent to
<span class="math-container">$$\sigma(k)\leq 2k-2.$$</span>
As we already saw that <span class="math-container">$\sigma(k)<2k$</span>, this is equivalent to <span class="math-container">$\sigma(k)\neq2k-1$</span>.</p>
<p>So your desired result that <span class="math-container">$m<k$</span> is equivalent to <span class="math-container">$k$</span> <em>not</em> being an odd almost perfect number. Whether any odd almost perfect number greater than <span class="math-container">$1$</span> exists is an open problem, to my knowledge.</p>
<p>Put differently; the (integral) solutions to
<span class="math-container">$$(m+1)\sigma(k)=2mk,$$</span>
with <span class="math-container">$k$</span> and <span class="math-container">$m$</span> both odd that do <em>not</em> satisfy <span class="math-container">$m<k$</span>, are precisely the pairs
<span class="math-container">$$(k,m)=\left(k,\sigma(k)\right),$$</span>
where <span class="math-container">$k$</span> is an almost perfect number. In particular <span class="math-container">$k$</span> is a perfect square.</p>
|
2,563,779 | <p>Let $(X,d)$ be a metric space. If every subset of $X$ is bounded, does it mean that the space itself is bounded?</p>
| José Carlos Santos | 446,262 | <p>Of course. After all, $X$ is a subset of itself. On the other hand, if $X$ is bounded, every subset of $X$ is bounded too.</p>
|
4,552,988 | <p>I want to prove by diagonalization that the set of surjective total computable functions from N to N is not recursively enumerable. I know that the result is trivial using Rice's theorem, but I am trying to prove it only by a direct diagonalization argument. However, supposing that we can enumerate the functions of the set, I am unable to construct a proper surjective and total function that cannot belong to the set.</p>
| insipidintegrator | 1,062,486 | <p>Now you have (after the edit(s)) that <span class="math-container">$\displaystyle f(0)=\int_0^0\frac{dt}{\sqrt{1+t^4}}=0$</span> so <span class="math-container">$f^{-1}(0)=g(0)=0$</span> so <span class="math-container">$g’(0)=\sqrt{1+(g(0))^4}=1.$</span></p>
|
4,552,988 | <p>I want to prove by diagonalization that the set of surjective total computable functions from N to N is not recursively enumerable. I know that the result is trivial using Rice's theorem, but I am trying to prove it only by a direct diagonalization argument. However, supposing that we can enumerate the functions of the set, I am unable to construct a proper surjective and total function that cannot belong to the set.</p>
| xpaul | 66,420 | <p>This is what you have to do next. Let <span class="math-container">$x=0$</span> in
<span class="math-container">$$f'(g(x))g'(x)=1.$$</span>
Then you have
<span class="math-container">$$ f'(g(0))g'(0)=1. $$</span>
Noting <span class="math-container">$g(0)=0$</span> and <span class="math-container">$f'(0)=\frac1{\sqrt{1+x^4}}\bigg|_{x=0}=1$</span>,
you have
<span class="math-container">$$ g'(0)=1. $$</span></p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| JTP - Apologise to Monica | 64 | <p>I only became aware of τ a few weeks ago. It's clever, and in theory, might simplify the teaching a bit. If one could wave a wand and rewrite every text book, and change every pi key on a calculator to tau, I'm in. Even though it would be tough for me to rewire my brain away from the pi related equations I've come to memorize. Not to mention the <a href="http://en.wikipedia.org/wiki/Piphilology" rel="nofollow">mnemonic poem</a> that helped me memorize the first 20 digits of pi. </p>
<p>Short of that, the question is one of transition. There will be a period of time one needs to learn both numbers as a way to bridge the gap. I am still getting over the trauma of losing an entire planet. I wonder when the last textbook calling Pluto a planet will be retired. How do you propose to replace every math text? </p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| Eddie | 307 | <p>I feel as though a lot of this misses the point. Teaching the unit circle and trig functions in terms of "one turn" makes a lot more intuitive sense to young students than do arbitrary variables tau or pi. It just so happens that Tau is equivalent to one turn. Once students grasp the concept of the unit circle, then the pi conventions can be explored.</p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| Benoît Kloeckner | 187 | <p>After coming back to this question, it strikes me that in fact the best thing might be to explicitly teach both. Not to accommodate the advantages of $\tau$ with the entrenchment of $\pi$, but simply because it makes clearer that pi is a convention, has a definition, and is not just some magical formula given to us. It is a choice. The real thing that is given as a fact is that a circle twice as large is twice as long, after that we name one ratio or the other (perimeter over diameter or over radius), but it does not matter that much which one.</p>
<p>I once had to teach some trigonometry using complex numbers, and asked myself for one quarter of an hour how damn to prove that $e^{i\theta}$ equals $1$ precisely at $\theta=2\pi$, until I realized that in that setup it was nothing else than the definition of $\pi$. I guess many students also need to be reminded that once in a while, and teaching $\tau$ is probably a good opportunity.</p>
|
1,833,495 | <p>I tried to solve this like this.</p>
<p>$x=1,y=1$ is solution.</p>
<p>And Let $x=a y=b\, (a\geq 1,b\geq2)$</p>
<p>Then, $11$ can divide $11^b = 10^a+1$</p>
<p>so $10^a = 10 \pmod{11}$ but order of $11(10) = 2$.</p>
<p>Then there is contradiction.</p>
<p>Can I solve this problem while I make $x=a y=b\, (a\geq2 , b\geq1)$?</p>
| Joss | 338,303 | <p>If you take logarithms in both sides you get $x=log(11^y-1)$ so you have a solution for each $y$ greater than $0$. This works in case you aren´t looking for integer solutions. </p>
|
2,922,290 | <p>Lemma 2.2</p>
<p>Let $(X,d)$ be a compact metric space and suppose that $U$ is a clopen subset of $X$. </p>
<p>There exists $\delta >0 $ s.t $d(x,y) \geq \delta $ $\forall x \in U $ and $y\in X-U $</p>
<p>Proof attempt</p>
<p>Suppose that it is not true then for all $\delta >0 $ $\exists x \in U $ and $\exists y\in X-U $s.t $d(x,y) < \delta $</p>
<p>Consider $d(x_i,y_i) <\frac{1}{i}$ pick a sub-sequence converging to $x$ then from the sequence $d(x,y_i)< \frac {1}{i} $ pick a sub-sequence converging to $y$. Then this pair $x,y$ has the property that for $\delta >0 $ we have $d(x,y) <\delta$ hence $x=y$ but this is a contradiction as if $ x\in U $ then $x\notin X-U$</p>
<p>Can anyone give a constructive proof? (ie not with negation)</p>
| mathworker21 | 366,088 | <p>Take $y \in U^c$. For each $n \ge 1$, let $E_n = \{x \in U : d(x,y) \ge \frac{1}{n}\}$. Each $E_n$ is closed and $\cap_{n \ge 1} E_n = \emptyset$. By compactness (and the fact that $E_n \subseteq E_m$ for $n \ge m$), some $E_N \not = \emptyset$.</p>
<p>For $n \ge 1$, let $F_n = \{y \not \in U : \exists x \in U, d(x,y) \le \frac{1}{n}\}$. Each $F_n$ is closed. The above shows that $\cap_n F_n \not = \emptyset$. Therefore, some $F_N \not = \emptyset$. We're done.</p>
|
3,511,565 | <p>I am in the middle of an exercise and I am stuck in a final step in which I want to show that <span class="math-container">$y-x^2 \notin \langle x(x-1), (x-1)y\rangle$</span> in <span class="math-container">$\mathbb{C}[x,y]$</span>. My first thought was that <span class="math-container">$x-1$</span> does not divide <span class="math-container">$y-x^2$</span>, but I have no idea how to prove this properly.</p>
| g.kov | 122,782 | <p><a href="https://i.stack.imgur.com/YY0aH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YY0aH.png" alt="enter image description here"></a></p>
<p>The property checks out
with
a straightforward proof using
complex numbers representation, with
the circumcenter at the origin,</p>
<p><span class="math-container">\begin{align}
O&=0
,\\
A&=R
,\\
B&=R\,\exp(\mathbf i\,\phi);
,\\
C&=R\,\exp(\mathbf i\,\psi);
,\\
P&=|PO|\,\exp(\mathbf i\,\theta)
.
\end{align}</span> </p>
<p><span class="math-container">\begin{align}
a^2&=(B-C)\overline{(B-C)}
,\\
b^2&=(A-C)\overline{(A-C)}
,\\
c^2&=(A-C)\overline{(B-A)}
,\\
x^2=|PA|^2&=(P-A)\overline{(P-A)}
,\\
y^2=|PB|^2&=(P-B)\overline{(P-B)}
,\\
z^2=|PC|^2&=(P-C)\overline{(P-C)}
.
\end{align}</span><br>
where <span class="math-container">$\overline{z}$</span> is the complex conjugate of <span class="math-container">$z$</span>.</p>
<p>The square of the areas is calculated using a variant of the Heron’s formula:</p>
<p><span class="math-container">\begin{align}
S^2&=
\tfrac1{16}\,(4a^2b^2-(a^2+b^2-c^2)^2)
,\\
S_0^2&=
\tfrac1{16}\,(4x^2a^2y^2b^2-(x^2a^2+y^2b^2-z^2c^2)^2)
,\\
\frac{S_0^2}{S^2}
&=(R^2-|PO|^2)^2
.
\end{align}</span></p>
<p>Corresponding <code>Maxima</code> session:</p>
<pre><code>Maxima 5.38.1 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.12
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) _(z):=conjugate(z)<span class="math-container">$
(%i2) fS2(a2,b2,c2):=1/16*(4*a2*b2-(a2+b2-c2)^2)$</span>
(%i3) declare([A,B,C,P,O],complex)<span class="math-container">$
(%i4) declare([R,PO,phi,psi,theta,a2,b2,c2,PA2,PB2,PC2,S2,S02],real)$</span>
(%i5) A:R<span class="math-container">$
(%i6) B:R*exp(%i*phi)$</span>
(%i7) C:R*exp(%i*psi)<span class="math-container">$
(%i8) P:PO*exp(%i*theta)$</span>
(%i9) a2:(B-C)*_(B-C)<span class="math-container">$
(%i10) b2:(A-C)*_(A-C)$</span>
(%i11) c2:(B-A)*_(B-A)<span class="math-container">$
(%i12) PA2:(P-A)*_(P-A)$</span>
(%i13) PB2:(P-B)*_(P-B)<span class="math-container">$
(%i14) PC2:(P-C)*_(P-C)$</span>
(%i15) S2:fS2(a2,b2,c2)<span class="math-container">$
(%i16) S02:fS2(a2*PA2,b2*PB2,c2*PC2)$</span>
(%i17) factor(S02/S2);
2 2
(%o17) (R - PO) (R + PO)
</code></pre>
|
204,003 | <p>I have a sequence of numbers
like 1,7,22,45,12,96,21,45,65,36,85,14,51,16,18,17,16....65...</p>
<blockquote>
<p>IS there any formula to check whether the sequence is random or not ?</p>
</blockquote>
<p>In my case</p>
<ol>
<li>odd numbers are not random since previous+2 </li>
<li>even numbers are not random </li>
<li>the numbers can be repeated in a sequence but must be far away,</li>
<li>we cannot generate the sequence using any formula like (x+2)^2 -2x like...</li>
</ol>
| Godot | 38,875 | <p>First of all you have to tell what "random" means...</p>
<p>But in the algorithmic context there is a nice definition.</p>
<p>To be more specific: a number $\alpha\in(0,1)$ is algorithmically random if for every $n$ the size of the shortest program giving you the first $n$ bits of $\alpha$ is at least $n$ bits long. </p>
<p>You can encode your sequence as a binary representation of some number from the interval $[0,1]$ and try to check if it is algorithmically random.</p>
<p>But do not count on a formula that can tell you if a particular number (sequence) is random since such a formula cannot exist.</p>
<p>See "Algorithmically random sequence" "Chaitin constant" and "halting problem" in Wikipedia for more details.</p>
|
2,588,271 | <blockquote>
<p>Show that $\mathcal{L}\{t^{1/2}\}=\sqrt{\pi}/(2s^{3/2}), \: s>0$</p>
</blockquote>
<p>By the definition of Laplace transform we get:</p>
<p>$$\mathcal{L}\{t^{1/2}\} = \int_0^\infty t^{1/2}e^{-st} \, dt = \{x = \sqrt{st} \} = \dfrac{2}{s^{3/2}} \int_0^\infty e^{-x^2} x^2 \, dx. $$</p>
<p>A known and easily proved result is $\int_0^\infty e^{-x^2} \, dx = \dfrac{\sqrt{\pi}}{2}$. How (if possible) can I use this result to determine the integral above? I was thinking either integration by parts or perhaps a suitable coordinate transformation (e.g. polar coordinates) but my attempts failed. </p>
| Fimpellizzeri | 173,410 | <p><strong>Hint:</strong> Consider integration by parts on</p>
<p>$$
\left(
x\,e^{-x^2}
\right)
\cdot x
$$</p>
|
2,588,271 | <blockquote>
<p>Show that $\mathcal{L}\{t^{1/2}\}=\sqrt{\pi}/(2s^{3/2}), \: s>0$</p>
</blockquote>
<p>By the definition of Laplace transform we get:</p>
<p>$$\mathcal{L}\{t^{1/2}\} = \int_0^\infty t^{1/2}e^{-st} \, dt = \{x = \sqrt{st} \} = \dfrac{2}{s^{3/2}} \int_0^\infty e^{-x^2} x^2 \, dx. $$</p>
<p>A known and easily proved result is $\int_0^\infty e^{-x^2} \, dx = \dfrac{\sqrt{\pi}}{2}$. How (if possible) can I use this result to determine the integral above? I was thinking either integration by parts or perhaps a suitable coordinate transformation (e.g. polar coordinates) but my attempts failed. </p>
| Tob Ernack | 275,602 | <p>Rewrite the integral as $$\int\limits_0^\infty-\frac{x}{2} \left(-2xe^{-x^2}\right)dx$$ and use integration by parts with $f(x) = -\frac{x}{2}$ and $g(x) = e^{-x^2}$ to obtain $$\left. -\frac x 2 e^{-x^2}\right|_0^\infty + \frac{1}{2} \int\limits_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{4}$$</p>
|
474,260 | <p>We know that irrational number has not periodic digits of finite number as rational number.<br>
All this means that we can find out which digit exist in any position of rational number.<br>
But what about non-rational or irrational numbers?<br>
For example:<br>
How to find out which digit exists in Fortieth position of $\sqrt[2]{2}$ which equals 1,414213.......<br>
Is it possible to solve such kind of problem for any irrational number?</p>
| Foo Barrigno | 73,411 | <p>In some cases, yes. See <a href="http://en.wikipedia.org/wiki/Spigot_algorithm" rel="nofollow">spigot algorithms</a> and digit extraction algorithms. In most cases however, no. Most irrational numbers have no such algorithm.</p>
|
1,970,082 | <p>How would you go about proving that:</p>
<p>$\lim_{x\to2^-}\frac{1}{x-2}= -\infty$
(i.e. x approaching 2 from left hand side)</p>
<p>Been struggling to make much headway on this. Not sure how I'm supposed to use Epsilon in this case. So far, I have: <br/>
Given any negative number M there exists $\delta$ s.t. $f(x) < M$ if $0<|x-2|<\delta$ <br/>
-> $\frac{1}{x-2} < M$ if "..." <br/>
-> $\frac{1}{M} > x-2$ if "..." <br/></p>
<p>Really not sure how to proceed from here. Can't find any online examples like it and the one in my book is for $\frac{1}{x^2}$ (I understand it, but not sure how to apply it here).</p>
<p>Any help/feedback/advice would be greatly appreciated.</p>
| Alexis Olson | 11,246 | <p>For a given $\epsilon$, having $\frac{1}{q} \ge \epsilon$ means that $q$ must be bounded above by $q \le \frac{1}{\epsilon}$ so you only have finitely many $q$. Now since all the $\frac{p}{q}$ must be in a bounded interval, you can only have finitely many $p$ as well (or else the ratio would grow arbitrarily large).</p>
|
190,905 | <p>I want to make two general graphics something like the following.</p>
<p><a href="https://i.stack.imgur.com/1SNYt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1SNYt.png" alt="enter image description here"></a></p>
<p>I want to illustrate the approach to the function by means of a polynomial. I have tried the following.</p>
<pre><code>Plot[{Sin[x] + 2, Sin[2 x] + 2}, {x, 0, 8}, AxesLabel -> {x, y},
Ticks -> {{0.8, 2.4, 4, 5.5, 7.1}, {0, 1, 2, 3}}]
</code></pre>
<p>I also want to put <span class="math-container">$ x_{0}, x_{1}, x_{2},...,x_{n} $</span> in the points where the graphs differ the most. I appreciate any help.</p>
| kglr | 125 | <pre><code>f[x_] := 1 + x + Sin[10 Sqrt[1 + x]];
g[x_] := 1 + x + Cos[10 Sqrt[2 + x]];
xx = x /. NSolve[{f'[x] == g'[x], 0 <= x <= 8}, x];
ticks = MapIndexed[{#, Style[ Subscript["x", ToString@ #2[[1]]], 16]} &, xx];
Plot[{f[x], g[x]}, {x, 0, 3 Pi}, AxesLabel -> {x, y},
MeshFunctions -> {f'[#] - g'[#] &}, Mesh -> {{0}},
MeshStyle -> Directive[Red, PointSize[Large]],
GridLines -> {xx, None}, Ticks -> {ticks, Automatic},
AxesOrigin -> {0, 0}]
</code></pre>
<p><a href="https://i.stack.imgur.com/FDVlM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FDVlM.png" alt="enter image description here"></a></p>
|
14,167 | <p>I randomly pick a natural number <em>n</em>. Assuming that I would have picked each number with the same probability, what was the probability for me to pick <em>n</em> before I did it?</p>
| Anne van Rossum | 63,110 | <p>It would require a definition of a probability distribution with an infinitesimal probability mass at every natural number. Dirac delta functions of infinitesimal length. </p>
<p>In (Bayesian) statistics, functions that relax the axiom of "summation to one" are known as <a href="https://en.wikipedia.org/wiki/Prior_probability" rel="nofollow">improper priors</a> and have an infinite total mass. This loss of the <strong>axiom on tautology</strong> ($p(T)=1$ for any tautology $T$) comes with severe repercussions. You'll see Dutch book arguments that make this clear. </p>
<p>Note that there are many infinite sequences that people try to tackle with tools from probability theory. Most famous are the attempts to postulate hypotheses about the <a href="https://www.dartmouth.edu/~chance/chance_news/for_chance_news/Riemann/cramer.pdf" rel="nofollow">distribution of subsequent primes</a>. So, this means that questions like the one you have are very good to ask ourselves.</p>
<p>There is also the concept of <a href="https://en.wikipedia.org/wiki/Natural_density" rel="nofollow">natural density</a>. The natural density of picking an even natural number is $1/2$.</p>
|
1,073,730 | <p>Is it possible to prove the proposition</p>
<p>if $P \land Q \Rightarrow R$</p>
<p>then</p>
<p>$P \land R \Rightarrow \lnot Q $</p>
<p>or in other words</p>
<p>($P \land Q \Rightarrow R) \Rightarrow (P \land R \Rightarrow \lnot Q )$</p>
| parsiad | 64,601 | <p>Consider $P,Q,R=1$. The first statement is true and the second statement is...</p>
|
1,496,676 | <p>let's $F(x)= x'Ax + a'x $</p>
<p>where $x'Ax$ is a quadratic form and $a'$ is defined as a vector. </p>
<p>$$A:= \left[ \begin{matrix} 6 &1&1 \\ 1&2&0 \\ 1&0&4\end{matrix}\right] $$</p>
<p>Does there exist a global minimum point (absolute min) for this function $F(x)$?</p>
<p>note: I know that if the function is given as a linear structure, whether this is convex or concave is not important. we only exemine $x'Ax$. </p>
<p>but i cannot perfectly prove this question. thank you for helping. </p>
| John Alexiou | 3,301 | <p>Isn't minimum or maximum were the gradient is zero (vector)?</p>
<p>$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 A x + a' =0 $$</p>
<p>So the unique solution </p>
<p>$$ x = - \frac{1}{2} A^{-1} a' $$</p>
<p>if it exists must be a minimum or maximum. </p>
<h2>Edit 1</h2>
<p>So for your example suppose $a'=(a_1,a_2,a_3)$ and $x=(x_1,x_2,x_3)$ then</p>
<p>$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 \begin{pmatrix} 6 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix}+ \begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} = \begin{pmatrix} 12 x_1+2 x_2+2 x_3 + a_1 \\ 2 x_1+4 x_2 + a_2 \\ 2 x_1 + 8 x_4 + a_3\end{pmatrix}$$</p>
<p>To make this the zero vector you need</p>
<p>$$ \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix} =-\frac{1}{2} \begin{pmatrix} 6 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 4 \end{pmatrix}^{-1} \begin{pmatrix} a_1 \\a_2 \\a_3 \end{pmatrix} = \begin{pmatrix} -\frac{4 a_1-2 a_2-a_3}{42} \\ \frac{4 a_1-23 a_3-a_3}{84} \\ \frac{2 a_1-a_2 -11 a_3}{84} \end{pmatrix} $$</p>
<p>This makes $${\rm extrema}(F(x)) = \frac{-8 a_1^2+8 a_1 a_2+4 a_1 a_3-23 a_2^2-2 a_2 a_3 -11 a_3}{168} $$</p>
|
708,138 | <p>Show that the maximum of a convex function on a convex polytope occurs at one of the extreme points of the polytope. I am really stuck here and don't know where to begin.</p>
| Jimmy R. | 128,037 | <p>Hint: For the following reasoning use</p>
<ol>
<li>The definition of convexity of a function</li>
<li>The fact that the convex polytope over which the function is maximized contains all convex combinations of points in it. In fact every point in the convex polytope can be expressed as a convex combination of it's extreme points.</li>
<li>The definition of an exterme point (that it cannot be written as a linear combination of two distinct points).</li>
</ol>
<p>General idea of proof (you need to write it in more detail): Consider the point $$x_0=λx_1+(1-λ)x_2 \quad \in C$$ where $λ \in [0,1]$ and $x_1, x_2 \in C$. Since $f$ is convex (I named your convex function $f$) you know that $$f(λx_1+(1-λ)x_2)\le λf(x_1)+(1-λ)f(x_2)$$ for every $x_1, x_2 \in C$ (I named the convex polytope $C$), $λ \in [0,1]$. But this implies that the inner point $x_0:=λx_1+(1-λ)x_2$, for $λ \in (0,1)$ (which is in $C$, since $C$ is convex) cannot have a maximum since it's value under $f$ is dominated by the RHS of the above expression (where $x_1$ and $x_2$ can be chosen to be extreme points). </p>
<p>(Note of course, than when the inequality holds with a equality, then the maximum is attained in the inner point $x_0$ as well as in the extreme points $x_1, x_2$ so the theorem is again valid.) </p>
|
2,027,888 | <h2>Exercise</h2>
<p>If $H$ is the Heaviside function, prove, using the definition below, that $\lim \limits_{t \to 0}{H(t)}$ does not exist.</p>
<hr>
<h2>Definition</h2>
<blockquote>
<p>Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write
$$\lim \limits_{x \to a}{f(x)} = L$$
if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that
$$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$</p>
</blockquote>
<hr>
<h2>Hint</h2>
<blockquote class="spoiler">
<p> Use an indirect proof as follows. Suppose that the limit is $L$. Take $\epsilon = \frac{1}{2}$ in the definition of a limit and try to arrive at a contradiction.</p>
</blockquote>
<hr>
<h2>Attempt</h2>
<p>Let $\delta$ be any (preferably small) positive number.</p>
<p>$H(0 - \delta) = H(-\delta) = 0$</p>
<p>$H(0 + \delta) = H(\delta) = 1$</p>
<p>$H(0 - \delta) =^? H(0 + \delta) \implies 0 =^? 1 \implies 0 \neq 1 \implies H(0 - \delta) \neq H(0 + \delta)$</p>
<p>$\lim \limits_{t \to 0^-}{H(t)} \neq \lim \limits_{t \to 0^+}{H(t)} \implies \lim \limits_{t \to 0}{H(t)}$ does not exist</p>
<hr>
<h2>Request</h2>
<p>I don't even know where to begin, even with the hint.</p>
<p><strong>Can someone kickstart the proof for me?</strong>$^1$</p>
<p><strong>$^1$ Update:</strong> I've come up with an attempt. Is it valid? It seems that I don't use the hint to my advantage; so if indeed my attempt is correct, what is the alternative proof using the hint?</p>
| Doug M | 317,162 | <p>$\forall$ means "for all" or "for any"</p>
<p>$\exists$ means "there exists"</p>
<p>For any $\delta > 0$ there exist a $t$ such that $|t|<\delta$ and $|H(t) - L|>\epsilon$</p>
<p>What are we doing.</p>
<p>If the limit exist. Then:</p>
<p>For any $\epsilon > 0$ there exist a $\delta>0$ such that when $|t|<\delta,|H(t) - L|<\epsilon$</p>
<p>And what does that mean. When $t$ is in the neighborhood of $0, H(t)$ is in a neighborhood of $L.$ The $\epsilon-\delta$ tell us the size of the neighborhoods. </p>
<p>But we need to show that the limit does not exist. There is a $t$ in the neighborhood 0, such that $H(t)$ is not in the neighborhood of $L$</p>
<p>How to do that? Fix the size of the target neighborhood of $H(t)$ small enough such that no matter how small of a neighborhood we find around $t$ we are sure to miss the target neighborhood around $H(t)$</p>
|
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