I saw the following  that provides which should actually be a false proof since $0^0$ is not defined to be $1$.     $(a+b)^{n}=\sum_{k=0}^{n}\binom n{ k}a^kb^{n-k}$.   Put $a=0, b=1, n=0$. This gives $1^0=\binom0{0}1^00^0\implies 0^0=1$.   Also the person claiming this to be true says that:  "Also, $0^0$ is the number of mappings from the empty set to the empty set. Surely there’s no other answer to this number of mappings than $1$. Any arguments against setting $0^0=1$ are based on limiting forms $f(x)^{g(x)}$ where, as $x$ approaches some number, $f(x)$ and $g(x)$ approach zero but $f(x)^{g(x)}$ does not approach $1$. That doesn’t matter one bit: the limit of $⌊x⌋$ as $x$ tends to $0$ does not exist, but everyone is happy with defining $⌊0⌋=0$ all the same."  Why is this claim true if it is and why otherwise? What is the possible discrepancy in this proof and defining $0^0$ as $1$? Why is it beneficial to define it so?
1330
Could someone provide me with a good explanation of why $0^0=1$?  My train of thought:  $x&gt;0$  $0^x=0^{x-0}=0^x/0^0$, so  $0^0=0^x/0^x=\,?$  Possible answers:   $0^0\cdot0^x=1\cdot0^0$, so $0^0=1$ $0^0=0^x/0^x=0/0$, which is undefined   PS. I've read the  on mathforum.org, but it isn't clear to me.
19041
I just found this identity but without any proof, could you just give me an hint how I could prove it? $$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$  I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
103787
I am thinking of $\lim_{x \to 0} \ {f(x)} = 1$ but I am confused as $x=0$ is not in the domain and also I want to write an $\epsilon$-$\delta$ proof. So any help is much appreciated!  Thanks!
45167
My question is whether $x/x$ is always equal to $1$. I am mostly intersted in real numbers and particularly wonder whether $x/x$ is defined at $x=0$.  On one hand the division should simplify to $1$, on the other hand you should not be allowed to divide by zero.   I have been trying to find whether the simplification 'goes first' or whether the division causes trouble first, but it has proven impossible for me to find usefull search terms.     Note that this question arose after reading  and its first comment.
16105
How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$? I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me. (This series arises in studying the first passage time of a simple random walk.)
4721
So I answered this question:  I believe my answer is correct, however one thing I couldn't explain fully, and which is bugging me, is why the rationals $\mathbb Q$, integers $\mathbb Z$ and naturals $\mathbb N$ all have cardinality $| \mathbb Q |=| \mathbb Z | = |\mathbb N| = \aleph_0$, when $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$.  The basic proof of the other question, from my answer, is that for any $r$ in $\mathbb Q$ there is a single function $f(r)$ such that $f(r) \in \mathbb Z$, and the same holds between $\mathbb Z$ and $\mathbb N$. Because there is this 1:1 transformation possible, there must be the same number of numbers in the three sets, because otherwise there would be a number in one set for which the bijection could not produce a number of the other set, and this is not so.  Now, Belgi explained why $1&lt;2$ in the other question by defining the value 0 as the cardinality of the empty set $\emptyset$, 1 as the cardinality of a set of sets containing only the empty set, and 2 as the set of sets containing the empty set and a set containing the empty set, then proceeding as follows:     Now that we have defined the natural numbers we can define when one number   is smaller than another. The definition is $x&lt;y$ is $x\subset y$   and $x\neq y$ .      Clearly $\{\emptyset\}\neq\{\emptyset,\{\emptyset\}\}$ and $\{\emptyset\}\subset\neq\{\emptyset,\{\emptyset\}\}$   So this a proof, by definition, of why $1&lt;2$.   ... however, by the same definition, because $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$, then $| \mathbb N | &lt; | \mathbb Z | &lt; |\mathbb Q|$ and thus a maximum of one of these quantities can be the ordinal quantity $\aleph_0$.
9682
My question is:  show $$\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$$ $$n\geq m\geq 1$$ I tried to do this via induction and failed. there has to be another way of doing this.  We could either explain combinatorically that we are counting the same thing in 2 different ways, in this sense, we are counting the number of ways to choose $m$ elements from a set with $n+1$ elements.  Or we could try like me to prove it algebraically, both ways are valid...I'd like some help, or a push in the right direction.
52031
I'm studying on my own over the summer and I'm having a bit of trouble with this question. Also, I don't have any background in this, I haven't taken any classes in this yet so, if you choose to help, please go slow. Thanks. Anyway, here is the problem:   For any set A, finite or infinite, let B be the set of all functions mapping A into the set {0,1}. Show that the cardinality of B is the same as the cardinality of P(A) (power set of A).   Here's what progress I've made. Let f be a mapping from B to P(A). Now I have to show it's injective. Let g,h be elements of B and let x be an element of A:   f(g(x)) = f(h(x))  g, h are elements of B and B maps x to {0,1}. If g(x) = h(x) = 1 or g(x) = h(x) = 0 then we are done. But what about the possibility of g(x) = 1 and h(x) = 0 or vice-versa. How do I deal with that possibility?
88840
Usually, when someone says something like $\left(\frac12\right)!$, they are probably referring to the Gamma function, which extends the factorial to any value of $x$.  The usual definition of the factorial is $x!=1\times2\times3\times\dots x$, but for $x\notin\mathbb{N}$, the Gamma function results in $x!=\int_0^\infty t^xe^{-t}dt$.  However, back a while ago, someone mentioned that there may be more than one way to define the factorial for non-integer arguments, and so, I wished to disprove that statement with some assumptions about the factorial function.    the factorial   is a  for $x\in\mathbb{C}$ except at $\mathbb{Z}_{&lt;0}$ because of singularities, which we will see later. is a monotone increasing function that is concave up for $x&gt;1$. satisfies the relation $x!=x(x-1)!$ and lastly $1!=1$     From $3$ and $4$, one can define $x!$ for $x\in\mathbb{N}$, and we can see that for negative integer arguments, the factorial is undefined.  We can also see that $0!=1$.  Since we assumed $2$, we should be able to sketch the factorial for $x&gt;1$, using our points found from $3,4$ as guidelines.  At the same time, when sketching the graph, we remember $1$, so there can be no jumps or gaps from one value of $x$ to the next.  Then we reapply $3$, correcting values for $x\in\mathbb{R}$, since all values of $x$ must satisfy this relationship.  Again, because of $1$, we must re-correct our graph, since having $3$ makes the derivative of $x!$ for $x\in\mathbb N$ undefined.  So, because of $1$ and $3$, I realized that there can only be one way to define the factorial for $x\in\mathbb R$.  Is my reasoning correct?  And can there be only one extension to the factorial?    Oh, and here is a  to how I almost differentiated the factorial only with a few assumptions, like that it is even possible to differentiate.  Putting that in mind, it could be possible to define the factorial with Taylor's theorem?
81480
Infinity, is clearly not . But there is some dissent on whether or not zero is finite. I have seen authors use "finite" to indicate the value of $0$ is excluded as well as infinity.  Is there any rule about this?
14558
We all know that $|\mathbb{N}| = \aleph_0$. Since $|\{-1\} \cup \mathbb{N}| = \aleph_0$ as well, I guess you could say that $\aleph_0 + 1 = \aleph_0$.  You can go on to derive that $\aleph_0 + \aleph_0 = \aleph_0$, and by induction $n \times \aleph_0 = \aleph_0$ (assuming that $n$ is finite, anyway).  Now, here's a thing: What is $\aleph_0 - \aleph_0$?  Well, $|\mathbb{P}| = \aleph_0$, and $|\mathbb{P} \backslash \mathbb{N}| = \aleph_0$, so perhaps $\aleph_0 - \aleph_0 = \aleph_0$?  No, wait. Consider the set $S = \{n \in \mathbb{N}: n &gt; 5\}$. Now we have $|S| = \aleph_0$, and yet $|\mathbb{N} \backslash S| = 5$. So maybe $\aleph_0 - \aleph_0 = 5$?  But that is absurd, since we can redefine $S$ to make the result any finite number we wish. Or we can define $S$ such that the result is countably infinite.  Does this mean that the notion of $\aleph_0 - \aleph_0$ simply has no definite answer? Or am I just being too simplistic here?
66723
The question: Let $f_n$ be fibonacci numbers. Prove: $\sum_{k=0}^{n-1} \binom{n+k}{2k+1} = f_{2n-1}$ and $\sum_{k=0}^n \binom{n+k}{2k} = f_{2n}$  For every $n\in N$.  $f_0=f_1=1$, $f_n=f_{n-1}+f_{n-2}$ for $n\ge 2$  I'm supposed to do this through induction.  What I did, first, is that I've seen that for $1$ both statements are true, then assumed for $n$.  For $n+1$ I got that the first sum is equal to the first sum for $n$ and the second sum for $n$, which under the definition of fibonacci numbers, makes the assumption true.  Similarly I did the process for the second sum.  Is this all that I had to do to solve this problem? What's bugging me is that I need to use both of the sums for a certain $n$ in proving that either one of those sums is valid for $n+1$, which I haven't encountered yet in exercises containing induction.
7049
As the title says, I'm wondering if there is a continuous function such that $f$ is nonzero on $[0, 1]$, and for which $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$. I am trying to solve a problem proving that if (on $C([0, 1])$) $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 0$, then $f$ must be identically zero. I presume then we do require the $n=0$ case to hold too, otherwise it wouldn't be part of the statement. Is there ay function which is not identically zero which satisfies $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$?  The statement I am attempting to prove is homework, but this is just idle curiosity (though I will tag it as homework anyway since it is related). Thank you!
104791
I would like your help with proving that for every  $0 \leq k \leq n$,      $$\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx  . $$   I tried to integration by parts and to get a pattern or to use the binomial formula somehow, but it didn't go well.  Thanks a lot!
42494
Why does:   $$ \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np $$ ? Taking the derivative of:   $$ \sum_{k=0}^{n}  \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 $$   does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that:   $$ \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p  $$   assuming that:   $$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$
36022
the background I'm allowed to deal with to solve this problem is as follows:  Definition of +: \begin{equation} m+0=m\quad \text{for all}\quad m \in \mathbb{N} \\ m+(k+1) = (m+k)+1  \end{equation} in conjunction with the induction property of $\mathbb{N}$. In a previous task I showed that $ 0+m=m \quad \forall \, m \in \mathbb{N}$.  Task: Prove $m+n=n+m$ by induction on $n$.  Now, it follows directly that $0+m=m=m+0$. With the assumption that $k+m=m+k$ for a $k\in\mathbb{N}$, it can be deduced: \begin{eqnarray} m+(k+1) &amp; = &amp; (m+k)+1\quad \text{by def. of}+\\ &amp; = &amp; (k+m) + 1\quad \text{with assumption}\\ &amp; = &amp; k + (m+1)\quad \text{by def. of}+. \end{eqnarray} Here, it looks to me that the true base step would be $m+1 = 1+m$, because then it would be possible to bring the proof straight to an end. But I tried hard to hack this last nut without any results. Is there anybody who could help me with this? Or maybe there is a completely different way to come up with the proof requested?  Thanks in advance for your help
65114
Let $ Y = \{ (1,1),(0,0\} \subset \mathbb A_k^2$. Find the $ \mathbb I(Y)$.   As if $f(x,y) \in \mathbb I(Y)$ then as $f(0,0)=0$ hence constant term of $f$ is zero and as $f(1,1)=0$ therefore sum of the coefficients of $f(x,y)$ is zero. But how can we describe this set properly?